claude / kimi-k2.7-code session: 229dbb4a-234a-4600-8669-a9863a453abf cwd: /workspace/problems/02_kda_cutlass
harness
claude
model
kimi-k2.7-code
turns
354
tools called
181
events
641
input toks
55,965,027
output toks
0
cache hit
0
solution.py
final answer
benchmark.log
check.log
result.json
"""Kimi Delta Attention (KDA) forward (chunk form) via Triton.

Matches reference.py's semantics: inputs are bf16 (g fp32), chunk_size=64,
no initial/final state, returns bf16 o.
"""
from __future__ import annotations

import torch
import torch.nn as nn
import triton
import triton.language as tl

OP_TYPE = "linear_attention"
SUPPORTED_PRECISIONS = ["bf16"]
HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]


@triton.autotune(
    configs=[
        triton.Config({"BK": 64, "BV": 8}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 8}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 8}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 8}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 64}, num_warps=8, num_stages=1),
    ],
    key=["B", "T", "H", "K", "V"],
)
@triton.jit
def _kda_intra_kernel(
    q_ptr,
    k_ptr,
    v_ptr,
    g_ptr,
    gmid_ptr,
    beta_ptr,
    Aqk_ptr,
    w_ptr,
    u_ptr,
    scale,
    B: tl.constexpr,
    T: tl.constexpr,
    H: tl.constexpr,
    K: tl.constexpr,
    V: tl.constexpr,
    BT: tl.constexpr,
    BK: tl.constexpr,
    BV: tl.constexpr,
):
    i_n = tl.program_id(0)
    i_bh = tl.program_id(1)
    i_b = i_bh // H
    i_h = i_bh % H

    q_ptr += (i_b * T * H + i_h) * K
    k_ptr += (i_b * T * H + i_h) * K
    g_ptr += (i_b * T * H + i_h) * K
    gmid_ptr += (i_b * H + i_h) * K
    beta_ptr += i_b * T * H + i_h
    v_ptr += (i_b * T * H + i_h) * V
    Aqk_ptr += (i_b * T * H + i_h) * BT
    w_ptr += (i_b * T * H + i_h) * K
    u_ptr += (i_b * T * H + i_h) * V

    t0 = i_n * BT
    r = tl.arange(0, BT)
    c = tl.arange(0, BT)
    mask_strict = r[:, None] > c[None, :]
    mask_lower = r[:, None] >= c[None, :]

    Aqk = tl.zeros([BT, BT], dtype=tl.float32)
    M = tl.zeros([BT, BT], dtype=tl.float32)

    for i_k in range(tl.cdiv(K, BK)):
        off_k = i_k * BK
        p_q = tl.make_block_ptr(
            q_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        p_k = tl.make_block_ptr(
            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        p_g = tl.make_block_ptr(
            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        b_q = tl.load(p_q, boundary_check=(0, 1))
        b_k = tl.load(p_k, boundary_check=(0, 1))
        b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)

        p_gmid = tl.make_block_ptr(
            gmid_ptr, (H, K), (K, 1), (i_h, off_k), (1, BK), (1, 0)
        )
        b_gmid = tl.load(p_gmid, boundary_check=(0, 1)).to(tl.float32)

        diff = b_g - b_gmid
        qg_norm = (b_q * tl.exp(diff)).to(tl.bfloat16)
        kg_norm = (b_k * tl.exp(diff)).to(tl.bfloat16)
        kdecay_norm = (b_k * tl.exp(-diff)).to(tl.bfloat16)
        Aqk += tl.dot(qg_norm, tl.trans(kdecay_norm))
        M += tl.dot(kg_norm, tl.trans(kdecay_norm))

    p_beta = tl.make_block_ptr(
        beta_ptr, (T,), (H,), (t0,), (BT,), (0,)
    )
    b_beta = tl.load(p_beta, boundary_check=(0,)).to(tl.float32)

    Aqk = tl.where(mask_lower, Aqk * scale, 0.0)
    A0 = tl.where(mask_strict, -M * b_beta[:, None], 0.0)

    A = A0
    for i in range(1, BT):
        row_i = tl.sum(tl.where(r[:, None] == i, A, 0.0), axis=0)
        update = tl.sum(row_i[:, None] * A, axis=0)
        new_row = tl.where(c < i, row_i + update, row_i)
        A = tl.where((r[:, None] == i), new_row[None, :], A)
    A = tl.where(r[:, None] == c[None, :], A + 1.0, A)
    A = A * b_beta[None, :]

    p_Aqk = tl.make_block_ptr(
        Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
    )
    tl.store(p_Aqk, Aqk.to(tl.bfloat16), boundary_check=(0, 1))

    A_bf16 = A.to(tl.bfloat16)

    for i_k in range(tl.cdiv(K, BK)):
        off_k = i_k * BK
        p_k = tl.make_block_ptr(
            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        p_g = tl.make_block_ptr(
            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        b_k = tl.load(p_k, boundary_check=(0, 1))
        b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
        b_kg = (b_k * tl.exp(b_g)).to(tl.bfloat16)
        b_w = tl.dot(A_bf16, b_kg)
        p_w = tl.make_block_ptr(
            w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        tl.store(p_w, b_w.to(tl.bfloat16), boundary_check=(0, 1))

    for i_v in range(tl.cdiv(V, BV)):
        off_v = i_v * BV
        p_v = tl.make_block_ptr(
            v_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
        )
        b_v = tl.load(p_v, boundary_check=(0, 1))
        b_u = tl.dot(A_bf16, b_v)
        p_u = tl.make_block_ptr(
            u_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
        )
        tl.store(p_u, b_u.to(tl.bfloat16), boundary_check=(0, 1))


@triton.autotune(
    configs=[
        triton.Config({"BK": 64, "BV": 8}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 8}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 8}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 8}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 64}, num_warps=8, num_stages=1),
    ],
    key=["B", "T", "H", "K", "V"],
)
@triton.jit
def _kda_inter_kernel(
    w_ptr,
    u_ptr,
    qg_ptr,
    k_ptr,
    g_ptr,
    Aqk_ptr,
    glast_ptr,
    o_ptr,
    B: tl.constexpr,
    T: tl.constexpr,
    H: tl.constexpr,
    K: tl.constexpr,
    V: tl.constexpr,
    BT: tl.constexpr,
    BK: tl.constexpr,
    BV: tl.constexpr,
):
    i_vb = tl.program_id(0)
    i_bh = tl.program_id(1)
    i_b = i_bh // H
    i_h = i_bh % H

    w_ptr += (i_b * T * H + i_h) * K
    u_ptr += (i_b * T * H + i_h) * V
    qg_ptr += (i_b * T * H + i_h) * K
    k_ptr += (i_b * T * H + i_h) * K
    g_ptr += (i_b * T * H + i_h) * K
    Aqk_ptr += (i_b * T * H + i_h) * BT
    o_ptr += (i_b * T * H + i_h) * V

    off_v0 = i_vb * BV
    NT = T // BT

    # For K=128 we have either one BK=128 tile or two BK=64 tiles.
    if tl.constexpr(K == 128 and BK == 128):
        S = tl.zeros([BK, BV], dtype=tl.float32)
        for i_n in range(NT):
            t0 = i_n * BT

            p_u = tl.make_block_ptr(
                u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
            )
            b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
            b_v = b_u

            p_w = tl.make_block_ptr(
                w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_w = tl.load(p_w, boundary_check=(0, 1))
            b_v -= tl.dot(b_w, S.to(tl.bfloat16)).to(tl.float32)

            b_o = tl.zeros([BT, BV], dtype=tl.float32)
            p_qg = tl.make_block_ptr(
                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_qg = tl.load(p_qg, boundary_check=(0, 1))
            b_o += tl.dot(b_qg, S.to(tl.bfloat16)).to(tl.float32)

            p_Aqk = tl.make_block_ptr(
                Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
            )
            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16)).to(tl.float32)

            p_o = tl.make_block_ptr(
                o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
            )
            tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))

            p_glast = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
            d = tl.exp(tl.load(p_glast))

            p_k = tl.make_block_ptr(
                k_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_k = tl.load(p_k, boundary_check=(0, 1))
            p_g = tl.make_block_ptr(
                g_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
            k_up = (b_k.to(tl.float32) * tl.exp(tl.log(d)[None, :] - b_g)).to(tl.bfloat16)

            S = S * d[:, None]
            S += tl.dot(tl.trans(k_up), b_v.to(tl.bfloat16))
    else:
        S0 = tl.zeros([BK, BV], dtype=tl.float32)
        S1 = tl.zeros([BK, BV], dtype=tl.float32)
        for i_n in range(NT):
            t0 = i_n * BT

            p_u = tl.make_block_ptr(
                u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
            )
            b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
            b_v = b_u

            p_w0 = tl.make_block_ptr(
                w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_w0 = tl.load(p_w0, boundary_check=(0, 1))
            b_v -= tl.dot(b_w0, S0.to(tl.bfloat16)).to(tl.float32)
            p_w1 = tl.make_block_ptr(
                w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
            )
            b_w1 = tl.load(p_w1, boundary_check=(0, 1))
            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16)).to(tl.float32)

            b_o = tl.zeros([BT, BV], dtype=tl.float32)
            p_qg0 = tl.make_block_ptr(
                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_qg0 = tl.load(p_qg0, boundary_check=(0, 1))
            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16)).to(tl.float32)
            p_qg1 = tl.make_block_ptr(
                qg_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
            )
            b_qg1 = tl.load(p_qg1, boundary_check=(0, 1))
            b_o += tl.dot(b_qg1, S1.to(tl.bfloat16)).to(tl.float32)

            p_Aqk = tl.make_block_ptr(
                Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
            )
            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16)).to(tl.float32)

            p_o = tl.make_block_ptr(
                o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
            )
            tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))

            p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
            d0 = tl.exp(tl.load(p_glast0))
            p_glast1 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)
            d1 = tl.exp(tl.load(p_glast1))

            p_k0 = tl.make_block_ptr(
                k_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_k0 = tl.load(p_k0, boundary_check=(0, 1))
            p_g0 = tl.make_block_ptr(
                g_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
            k_up0 = (b_k0.to(tl.float32) * tl.exp(tl.log(d0)[None, :] - b_g0)).to(tl.bfloat16)

            p_k1 = tl.make_block_ptr(
                k_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
            )
            b_k1 = tl.load(p_k1, boundary_check=(0, 1))
            p_g1 = tl.make_block_ptr(
                g_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
            )
            b_g1 = tl.load(p_g1, boundary_check=(0, 1)).to(tl.float32)
            k_up1 = (b_k1.to(tl.float32) * tl.exp(tl.log(d1)[None, :] - b_g1)).to(tl.bfloat16)

            S0 = S0 * d0[:, None]
            S1 = S1 * d1[:, None]
            S0 += tl.dot(tl.trans(k_up0), b_v.to(tl.bfloat16))
            S1 += tl.dot(tl.trans(k_up1), b_v.to(tl.bfloat16))


class Model(nn.Module):
    def __init__(self, B: int, T: int, H: int, K: int, V: int, chunk_size: int = 64):
        super().__init__()
        self.B, self.T, self.H, self.K, self.V = B, T, H, K, V
        self.chunk_size = chunk_size
        self.scale = float(K) ** -0.5
        self.register_buffer("_dummy", torch.zeros(1), persistent=False)

    def forward(
        self,
        q: torch.Tensor,
        k: torch.Tensor,
        v: torch.Tensor,
        g: torch.Tensor,
        beta: torch.Tensor,
    ) -> torch.Tensor:
        B, T, H, K = q.shape
        V = v.shape[-1]
        BT = self.chunk_size
        assert T % BT == 0
        NT = T // BT
        device = q.device

        q = q.to(torch.bfloat16)
        k = k.to(torch.bfloat16)
        v = v.to(torch.bfloat16)
        g = g.to(torch.float32)
        beta = beta.to(torch.bfloat16)

        g_4d = g.view(B, NT, BT, H, K)
        g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)

        q_scaled = q * self.scale
        qg = (q_scaled * g_cum.exp()).to(torch.bfloat16)

        g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT // 2, :, :].permute(0, 2, 1, 3).contiguous()
        glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()

        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.bfloat16)
        w = torch.empty(B, T, H, K, device=device, dtype=torch.bfloat16)
        u = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)

        # Use 64-wide tiles as a conservative default; autotune explores others.
        BK = 64
        BV = 32

        grid1 = (NT, B * H)
        _kda_intra_kernel[grid1](
            q_scaled, k, v, g_cum, g_mid, beta, Aqk, w, u,
            1.0,
            B=B, T=T, H=H, K=K, V=V, BT=BT,
        )

        o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
        grid2 = lambda meta: (triton.cdiv(V, meta['BV']), B * H)
        _kda_inter_kernel[grid2](
            w, u, qg, k, g_cum, Aqk, glast, o,
            B=B, T=T, H=H, K=K, V=V, BT=BT,
        )
        return o


B = 2
T = 1024
H = 8
K = 128
V = 128
CHUNK_SIZE = 64


def get_inputs():
    torch.manual_seed(0)
    q = torch.randn(B, T, H, K, dtype=torch.bfloat16) * 0.1
    k = torch.randn(B, T, H, K, dtype=torch.bfloat16) * 0.1
    v = torch.randn(B, T, H, V, dtype=torch.bfloat16) * 0.1
    g = (torch.randn(B, T, H, K, dtype=torch.float32) * 0.1 - 0.05)
    beta = torch.sigmoid(torch.randn(B, T, H, dtype=torch.bfloat16))
    return [q, k, v, g, beta]


def get_init_inputs():
    return [B, T, H, K, V, CHUNK_SIZE]
API Error: 400 Invalid request: Your request exceeded model token limit: 262144 (requested: 263684)
warning: `VIRTUAL_ENV=/home/infatoshi/kernelbench.com/benchmarks/hard/.venv` does not match the project environment path `/home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo/.venv` and will be ignored; use `--active` to target the active environment instead
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:47:10.536251+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:47:14.324169+00:00 elapsed_s=3.788 ms=0.214144
shape=0 variant=solution tflops=10.028 gbps=117.671 ms=0.214
shape=0 solution_peak_fraction=0.0501
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:47:14.490982+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:47:18.211512+00:00 elapsed_s=3.721 ms=0.367536
shape=1 variant=solution tflops=11.686 gbps=137.122 ms=0.368
shape=1 solution_peak_fraction=0.0584
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:47:18.377429+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:47:22.140428+00:00 elapsed_s=3.763 ms=0.462736
shape=2 variant=solution tflops=9.282 gbps=108.911 ms=0.463
shape=2 solution_peak_fraction=0.0464
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:47:22.184295+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:47:25.754425+00:00 elapsed_s=3.570 ms=0.214880
shape=3 variant=solution tflops=4.997 gbps=58.634 ms=0.215
shape=3 solution_peak_fraction=0.0250
peak_fraction: 0.0429
RESULT: LOW
warning: `VIRTUAL_ENV=/home/infatoshi/kernelbench.com/benchmarks/hard/.venv` does not match the project environment path `/home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo/.venv` and will be ignored; use `--active` to target the active environment instead
warning: Ignoring existing virtual environment linked to non-existent Python interpreter: /home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo/.venv/bin/python3 -> python
Using CPython 3.11.15
Removed virtual environment at: /home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo/.venv
Creating virtual environment at: /home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo/.venv
   Building kernelbench-hard @ file:///home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo
      Built kernelbench-hard @ file:///home/infatoshi/kernelbench.com/benchmarks/hard/outputs/runs/20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass/repo
Installed 56 packages in 66ms
PASS
{
    "run_id": "20260613_163858_kimi-claude_kimi-k2.7-code_02_kda_cutlass",
    "run_group": "",
    "problem": "02_kda_cutlass",
    "harness": "kimi-claude",
    "model": "kimi-k2.7-code",
    "reasoning_effort": "",
    "started_at": "2026-06-13T16:38:58-06:00",
    "harness_finished_at": "2026-06-13T18:45:40-06:00",
    "finished_at": "2026-06-13T18:47:26-06:00",
    "start_epoch": 1781390338,
    "harness_end_epoch": 1781397940,
    "end_epoch": 1781398046,
    "has_solution": true,
    "correct": true,
    "failure_reason": "pass",
    "retryable_infra_failure": false,
    "minimum_useful_output_tokens": 5000,
    "peak_fraction": 0.0429,
    "template_mutated": false,
    "elapsed_seconds": 7602,
    "total_elapsed_seconds": 7708,
    "check_elapsed_seconds": 89,
    "benchmark_elapsed_seconds": 17,
    "check_timeout_seconds": 180,
    "benchmark_timeout_seconds": 7200,
    "check_exit_code": 0,
    "benchmark_exit_code": 0,
    "harness_exit_code": 1,
    "session_complete": true,
    "agent_cuda_disabled": false,
    "agent_container": true,
    "agent_container_image": "nvcr.io/nvidia/tensorrt-llm/release:latest",
    "agent_container_network": "bridge",
    "gpu_queue_mode": "agent_container_native_profiling_path_wrapper_gpu_lock",
    "output_tokens_per_second": 21.49223888450408,
    "usage": {"input_tokens": 257929, "output_tokens": 163384, "cache_read_tokens": 27618048, "cache_creation_tokens": 0, "reasoning_tokens": null, "total_cost_usd": 19.183269}
}

timeline (641 events)

system — init
session start cwd=/workspace/problems/02_kda_cutlass model=kimi-k2.7-code permissions=bypassPermissions
assistant
reasoning (140 chars)
Let me start by exploring the workspace and understanding the reference implementation, then build a custom kernel for Kimi Delta Attention.
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assistant
Bash
args
{
  "command": "ls -la /workspace/problems/02_kda_cutlass",
  "description": "List workspace contents"
}
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tool
stdout (459 chars)
total 48
drwxrwxr-x 2 1002 1002 4096 Jun 13 22:38 .
drwxrwxr-x 3 1002 1002 4096 Jun 13 22:38 ..
-rw-rw-r-- 1 1002 1002 2312 Apr 27 09:21 PROMPT.txt
-rw-rw-r-- 1 1002 1002 5292 Jun  1 21:05 benchmark.py
-rw-rw-r-- 1 1002 1002 4222 Jun  2 01:08 check.py
-rw-rw-r-- 1 1002 1002 2243 Apr 26 08:51 problem.yaml
-rw-rw-r-- 1 1002 1002 5093 Apr 26 04:19 reference.py
-rw-rw-r-- 1 1002 1002  720 Apr 26 04:20 shapes.py
-rw-rw-r-- 1 1002 1002 2424 Apr 26 04:20 sota.py
assistant
Read /workspace/problems/02_kda_cutlass/reference.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/reference.py"
}
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tool
stdout (5561 chars) — TRUNCATED
1	"""Naive PyTorch reference for Kimi Delta Attention (KDA) forward, chunk form.
2	
3	This is the correctness oracle, NOT the SOTA baseline. It mirrors the
4	chunk-parallel formulation in fla/ops/kda/naive.py (Songlin Yang et al.)
5	without any Triton or CUDA optimization.
6	
7	Inputs (per the FLA convention):
8	  q, k : (B, T, H, K)   bf16   -- queries / keys
9	  v    : (B, T, H, V)   bf16   -- values
10	  g    : (B, T, H, K)   fp32   -- per-channel log-decay (in-chunk cumsum applied)
11	  beta : (B, T, H)      bf16   -- write strength
12	
13	Output:
14	  o    : (B, T, H, V)   bf16
15	
16	The agent must reproduce this output (within bf16 tolerance) using a CUTLASS
17	CuTe kernel on SM120 -- NOT by calling fla.ops.chunk_kda directly.
18	"""
19	from __future__ import annotations
20	
21	import torch
22	import torch.nn as nn
23	from einops import rearrange
24	
25	OP_TYPE = "linear_attention"
26	SUPPORTED_PRECISIONS = ["bf16"]
27	HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
28	
29	
30	def _naive_chunk_kda(
31	    q: torch.Tensor,
32	    k: torch.Tensor,
33	    v: torch.Tensor,
34	    g: torch.Tensor,
35	    beta: torch.Tensor,
36	    scale: float,
37	    chunk_size: int = 64,
38	) -> torch.Tensor:
39	    """KDA forward, no initial state, no final state. Returns o with v's dtype."""
40	    dtype = v.dtype
41	    B, T, H, K = q.shape
42	    V = v.shape[-1]
43	    BT = chunk_size
44	    assert T % BT == 0, f"T={T} must be a multiple of chunk_size={BT}"
45	    NT = T // BT
46

... (4061 more chars)
assistant
reasoning (119 chars)
Let me understand the algorithm and see the check/benchmark files, plus sota.py to understand performance expectations.
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assistant
Read /workspace/problems/02_kda_cutlass/check.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/check.py"
}
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tool
stdout (4598 chars) — TRUNCATED
1	"""Correctness runner for KDA forward (chunk form).
2	
3	Runs solution.Model vs reference.Model across all shapes in shapes.py, 3 seeds
4	each, with per-dtype atol/rtol (bf16 default 1e-2 plus a 5e-2 override for
5	this problem). Also rejects forbidden ops by grep.
6	"""
7	import re
8	import sys
9	from pathlib import Path
10	
11	import torch
12	import yaml
13	
14	# Make the repo's src/ importable
15	REPO_ROOT = Path(__file__).resolve().parents[2]
16	sys.path.insert(0, str(REPO_ROOT))
17	
18	from src.eval.correctness import check_correctness  # noqa: E402
19	from src.eval.numeric_stress import (  # noqa: E402
20	    numeric_stress_cases,
21	    numeric_stress_context,
22	    tolerance_for_case,
23	)
24	
25	
26	def _apply_shape(reference, shape):
27	    """Override reference's module-level shape shims so get_inputs/get_init_inputs match."""
28	    for k, v in shape.items():
29	        setattr(reference, k, v)
30	
31	
32	def main():
33	    try:
34	        import reference
35	        import shapes
36	        import solution
37	    except Exception as e:
38	        print(f"FAIL: import error: {e}")
39	        sys.exit(1)
40	
41	    problem_yaml = Path("problem.yaml")
42	    meta = yaml.safe_load(problem_yaml.read_text()) if problem_yaml.exists() else {}
43	
44	    # --- Forbidden-op check ------------------------------------------------
45	    sol_src = Path("solution.py").read_text() if Path("solution.py").exists() else ""
46	    for forbidden in meta.get("forbidden", []):
47	 

... (3098 more chars)
assistant
Read /workspace/problems/02_kda_cutlass/benchmark.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/benchmark.py"
}
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tool
stdout (5784 chars) — TRUNCATED
1	"""Roofline benchmark for KDA forward (chunk form).
2	
3	For each shape: times the agent's solution first and reports achieved TFLOPS,
4	GB/s, and peak_fraction. Optional diagnostics for eager reference, compiled
5	reference, and SOTA can be enabled with KBH_KDA_BENCHMARK_BASELINES=1.
6	
7	Output lines the harness picks up:
8	  shape=<idx> variant=<name> tflops=<N> gbps=<N> ms=<N>
9	  peak_fraction: <N>  (geomean over shapes of solution's peak_fraction)
10	"""
11	import sys
12	from math import exp, log
13	from pathlib import Path
14	
15	import torch
16	import yaml
17	
18	REPO_ROOT = Path(__file__).resolve().parents[2]
19	sys.path.insert(0, str(REPO_ROOT))
20	
21	from src.eval.roofline import compute_gbps, compute_tflops, peak_fraction  # noqa: E402
22	from src.eval.timing import benchmark_baselines_enabled, time_variant  # noqa: E402
23	from src.hardware import get as get_hw  # noqa: E402
24	
25	
26	def _eval_formula(expr: str, vars: dict) -> float:
27	    return float(eval(expr, {"__builtins__": {}}, vars))
28	
29	
30	def _apply_shape(reference, shape):
31	    for k, v in shape.items():
32	        setattr(reference, k, v)
33	
34	
35	def main():
36	    import reference
37	    import shapes
38	    import solution
39	
40	    meta = yaml.safe_load(Path("problem.yaml").read_text())
41	    hw = get_hw(meta["hardware"][0])
42	    peak_tflops = hw.peak_tflops_dense.get(meta["peak_tflops_key"], 0.0)
43	    peak_gbps = hw.peak_bandwidth_gb_s
44	    regime = meta.get("regime", "compu

... (4284 more chars)
assistant
Read /workspace/problems/02_kda_cutlass/sota.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/sota.py"
}
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tool
stdout (2631 chars) — TRUNCATED
1	"""SOTA reference for KDA forward: fla.ops.kda.chunk_kda (Triton).
2	
3	The agent's solution is forbidden from importing this module path (see
4	problem.yaml.forbidden). This file is only used by benchmark.py to draw
5	the SOTA reference line.
6	
7	If FLA's Triton kernel does not run on SM120 (Blackwell consumer-lineage --
8	some Triton kernels in FLA target Hopper TMA), is_available() returns False
9	and benchmark.py omits the SOTA variant. The H100 reference is documented
10	in problem.yaml for context.
11	"""
12	from __future__ import annotations
13	
14	import torch
15	
16	
17	def _import_fla():
18	    try:
19	        from fla.ops.kda import chunk_kda  # noqa: F401
20	        return chunk_kda
21	    except Exception:
22	        return None
23	
24	
25	def sota_forward(
26	    q: torch.Tensor,
27	    k: torch.Tensor,
28	    v: torch.Tensor,
29	    g: torch.Tensor,
30	    beta: torch.Tensor,
31	    scale: float | None = None,
32	) -> torch.Tensor:
33	    """Run FLA's Triton chunk_kda. Returns o (B, T, H, V) in v's dtype."""
34	    chunk_kda = _import_fla()
35	    if chunk_kda is None:
36	        raise RuntimeError("fla.ops.kda.chunk_kda unavailable")
37	    # FLA's chunk_kda has a richer signature (A_log, dt_bias, l2norm, gates, ...).
38	    # We need the bare forward: pass A_log/dt_bias as None, gates off, no l2norm.
39	    # The wrapper expects fp32 g; q/k/v/beta in bf16/fp16.
40	    out = chunk_kda(
41	        q=q,
42	        k=k,
43	        v=v,
44	        g=g,
45	     

... (1131 more chars)
assistant
Read /workspace/problems/02_kda_cutlass/shapes.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/shapes.py"
}
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tool
stdout (771 chars)
1	"""Canonical shape sweep for KDA forward (chunk form).
2	
3	Mix of:
4	  - short-context training-step scale (T=1024)
5	  - mid-context (T=2048) which is the headline benchmark
6	  - long-context that stresses the inter-chunk recurrence (T=4096)
7	  - thin-batch decode-style (B=1, T=2048, fewer heads)
8	
9	Constraints:
10	  - T % chunk_size == 0 (chunk_size = 64)
11	  - K, V are the per-head channel dims; KDA in Kimi Linear uses K=V=128
12	"""
13	
14	SHAPES = [
15	    {"B": 2, "T": 1024, "H": 8, "K": 128, "V": 128, "CHUNK_SIZE": 64},
16	    {"B": 2, "T": 2048, "H": 8, "K": 128, "V": 128, "CHUNK_SIZE": 64},
17	    {"B": 1, "T": 4096, "H": 8, "K": 128, "V": 128, "CHUNK_SIZE": 64},
18	    {"B": 1, "T": 2048, "H": 4, "K": 128, "V": 128, "CHUNK_SIZE": 64},
19	]
20	
assistant
reasoning (2503 chars)
Let me understand the algorithm in detail. KDA is a linear attention variant. The reference algorithm: 1. q,k,v,g,beta converted to float32 2. q = q * scale 3. Rearrange to (B,H,N,C,D) where C=chunk_size, N=T/C 4. g = g.cumsum(-2) within chunk (so g[...,i,:] = sum_{t=0}^{i} g_raw[...,t,:]) 5. Build A_kk: for each i in chunk, compute A[...,i] = sum_{c < i} k[...,c,:] * k[...,i,:] * exp(g[...,c,:] - g[...,i,:]) * beta[...,i] Then apply lower-triangular mask (including diagonal set to 0). Then do a forward substitution / cumulative product-like recurrence: for i in 1..BT-1: A[...,i,:i] = A[...,i,:i] + sum_{t=0}^{BT-1} A[...,i,t] * A[...,t,:i] Wait, A[...,i,:,None] * A[..., :, :i] summed over -2 (the t dimension), so for each j < i: A[i,j] += sum_t A[i,t] * A[t,j] But A is strictly lower triangular, so A[i,t] is nonzero only for t < i, and A[t,j] nonzero only for j < t. So t ranges j < t < i. This is computing the transitive closure / (I-A)^{-1} for strictly lower triangular A. Then A = (A + I) * beta[...,None,:] (beta broadcasted over the column/source index j? actually beta[...,None,:] means beta of shape (B,H,C) -> (B,H,1,C), applied to columns) 6. w = A @ (g.exp() * k) shape (B,H,N,C,K) 7. u = A @ v shape (B,H,N,C,V) 8. Inter-chunk recurrence: S = zeros(B,H,K,V) For each chunk i: Aqk[i,j] = sum_{c <= j} q_i[c] * k_i[j] * exp(g_i[c] - g_i[j]) (strict lower triangular including diagonal, masked upper) v_i = u_i - w_i @ S (B,H,C,V) o_i = (q_i * g_i.exp()) @ S + Aqk @ v_i (B,H,C,V) S = S * exp(g_i[...,-1]) + (k_i * exp(g_i[-1] - g_i)).transpose(-2,-1) @ v_i So the key math: delta rule linear attention with decay. We need a fast custom kernel. Given RTX PRO 6000 SM120 Blackwell, we can use CUTLASS 3.9+ CuTe, or Triton, or inline CUDA. The user mentions intended path is CUTLASS CuTe on SM120. But for a coding problem, the fastest path to a working and fast implementation is likely Triton, because: - CUTLASS/CuTe is complex, requires compilation, dependencies - Triton is Python-native and easy to iterate - Blackwell SM120 should run Triton Let me check the environment: torch version, triton availability, etc. However, I should consider performance. The problem wants us to "push the number up" and reach peak_fraction >= 0.1 gmean. A naive PyTorch implementation might not be fast enough. We need an efficient kernel. Let me first check the environment and run reference/benchmark to see baseline.
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assistant
Bash
args
{
  "command": "uv run python -c \"import torch; print(torch.__version__); print(torch.cuda.get_device_name(0)); print(torch.cuda.get_device_properties(0)); import triton; print('triton', triton.__version__)\"",
  "description": "Check PyTorch and Triton environment"
}
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assistant
Read /workspace/problems/02_kda_cutlass/problem.yaml
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/problem.yaml"
}
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tool
stdout (862 chars)
Using CPython 3.11.15
Creating virtual environment at: /workspace/.venv
   Building kernelbench-hard @ file:///workspace
      Built kernelbench-hard @ file:///workspace
warning: Failed to hardlink files; falling back to full copy. This may lead to degraded performance.
         If the cache and target directories are on different filesystems, hardlinking may not be supported.
         If this is intentional, set `export UV_LINK_MODE=copy` or use `--link-mode=copy` to suppress this warning.
Installed 56 packages in 882ms
2.11.0+cu130
NVIDIA RTX PRO 6000 Blackwell Workstation Edition
_CudaDeviceProperties(name='NVIDIA RTX PRO 6000 Blackwell Workstation Edition', major=12, minor=0, total_memory=97288MB, multi_processor_count=188, uuid=d594fb7a-abcb-bad5-1fc4-675b478a47df, pci_bus_id=1, pci_device_id=0, pci_domain_id=0, L2_cache_size=128MB)
triton 3.6.0
tool
stdout (2399 chars) — TRUNCATED
1	name: 02_kda_cutlass
2	display_name: "Kimi Delta Attention (chunk forward) via CUTLASS CuTe"
3	precision: bf16
4	regime: compute
5	
6	# Approximate FLOPs per forward call. KDA forward (chunk form) breakdown per
7	# token across all heads:
8	#   - intra-chunk Aqk (q @ k.T with decay): 2*BT*K
9	#   - intra-chunk Akk + cumulative invert : ~3*BT*K + BT^2
10	#   - w = A @ (g.exp()*k):                  2*BT*K
11	#   - u = A @ v:                            2*BT*V
12	#   - inter-chunk q @ S:                    2*K*V
13	#   - inter-chunk (Aqk @ v_i):              2*BT*V
14	#   - state update S += k.T @ v_i:          2*K*V
15	# Aggregating with a constant of 4 (typical for chunked linear attention) gives
16	# the formula below. Used only for roofline reporting -- not graded directly.
17	flops_formula: "4 * B * T * H * (K*V + CHUNK_SIZE*K + CHUNK_SIZE*V)"
18	
19	# Bytes moved (lower bound): one read of q,k,v,g,beta and one write of o.
20	bytes_formula: "B*T*H*K*2 + B*T*H*K*2 + B*T*H*V*2 + B*T*H*K*4 + B*T*H*2 + B*T*H*V*2"
21	
22	hardware: [RTX_PRO_6000]
23	peak_tflops_key: bf16  # KDA's heavy work is bf16 @ bf16 -> fp32 GEMM-equivalent
24	peak_bandwidth_key: dram
25	
26	# bf16 default tolerance from src/eval/correctness.py is 1e-2 atol/rtol; KDA's
27	# long recurrence accumulates more error than a single GEMM, so we relax slightly.
28	tolerance:
29	  torch.bfloat16: 0.05  # was "5e-2" -- PyYAML parses un-dotted sci-notation as string
30	
31	# Forbidden ops -- using any in solution.py f

... (899 more chars)
assistant
reasoning (3492 chars)
Triton 3.6.0, PyTorch 2.11.0+cu130. Blackwell SM120. This is a substantial kernel. Let me think about the best strategy. The KDA chunk forward algorithm has several components. A Triton implementation can fuse a lot. But it's complex because: 1. We need to compute A_kk within chunk (B,H,N,BT,BT) 2. Compute its "inverse" via forward substitution 3. Compute w, u 4. For each chunk sequentially, compute output and update S Given the chunk_size=64, K=V=128, we can think of each chunk processing 64 tokens at a time for one head (or group). A natural fused kernel: one thread block per (B,H,N) chunk handles all intra-chunk work plus the inter-chunk S update? But the inter-chunk recurrence has dependency across chunks (S from prev chunk needed for next), so we need either: - Sequential kernel launches per chunk (N up to 64, so 64 kernel launches) - Or use persistent kernel that waits on previous chunk For performance, a persistent kernel with block-level synchronization across chunks could work, but is complex. Alternatively, we can split: - Kernel 1: compute A, w, u for all chunks in parallel (all intra-chunk) - Then a sequential loop over chunks computing S and o Kernel 1 is easy to parallelize. The sequential part is the bottleneck. For the sequential part, each chunk needs: - S (K,V) matrix multiply with k.T @ v_i where v_i = u_i - w_i @ S - This is O(B*H*N*(K*V + BT*K*V + BT*K*V)) basically O(BHTKV) work. Actually, per chunk: - v_i = u_i - w_i @ S: w_i is (BT,K), S is (K,V), so w_i @ S is (BT,V) -> 2*BT*K*V FLOPs - o_i = qg @ S + Aqk @ v_i: qg @ S is (BT,K) @ (K,V) = (BT,V) -> 2*BT*K*V; Aqk @ v_i is (BT,BT) @ (BT,V) -> 2*BT^2*V - S update: k.T @ v_i is (K,BT) @ (BT,V) = (K,V) -> 2*BT*K*V; plus scale by decay So per chunk recurrent work is ~2*BT*V*(K + BT + K + K) = ~2*BT*V*(3K+BT) FLOPs. For K=V=128, BT=64: per chunk = 2*64*128*(384+64) = ~7.3M FLOPs. N=16 for T=1024, N=32 for T=2048, N=64 for T=4096. Total recurrent FLOPs: B*H*N*7.3M = for B=2,H=8,N=32 = 3.7 GFLOPs. This is small. Plus the intra-chunk A computation: per chunk A_kk is O(BT^2*K) = 64*64*128*2 = 1M FLOPs, times B*H*N = 2*8*32 = 512 -> 0.5 GFLOPs. Not huge. Overall FLOPs per the formula: 4*B*T*H*(K*V + BT*K + BT*V) = 4*2*2048*8*(16384+8192+8192) = 4*2*2048*8*32768 = 4.3 GFLOPs? Wait that seems low. Let me recalculate: 4*2*2048*8*32768 = 4*2*2048*262144 = 4*1073741824 = 4.29e9 = 4.3 GFLOPs. Hmm. Actually K*V = 128*128 = 16384. BT*K = 64*128=8192. BT*V=8192. Sum=32768. 4*B*T*H*sum = 4*2*2048*8*32768 = 4 * 2 * 2048 * 262144 = 4,294,967,296 = 4.29 GFLOPs. But from my breakdown: intra A = 0.5 GFLOPs, recurrent = 3.7 GFLOPs, total ~4.2 GFLOPs. Matches. Peak bf16 on RTX PRO 6000: Blackwell has high bf16 throughput. The peak_tflops_key is bf16. Need to compute actual peak. For RTX PRO 6000, likely ~100+ TFLOPS bf16. So 4.3 GFLOPs / 100 TFLOPS = 0.043 ms. Realistically maybe 0.1-0.5 ms. Peak fraction 0.1 means we need ~10 TFLOPS effective. A well-written Triton kernel should reach this. But writing a full custom KDA kernel from scratch in Triton is a lot. Let me look at the FLA repo to understand their Triton kernel structure. We can't import it, but we can read and adapt. Since network is available, I can clone FLA. Actually, the problem says "Anything you're uncertain about, look up PTX docs, clone CUTLASS or FLA or other reference repos, read library headers, and investigate." So cloning FLA is explicitly allowed. Let me clone FLA and study the KDA kernel.
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Bash
args
{
  "command": "git clone --depth 1 https://github.com/fla-org/flash-linear-attention.git /workspace/fla_ref 2>&1 | tail -20",
  "description": "Clone FLA reference repository"
}
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stdout (36 chars)
Cloning into '/workspace/fla_ref'...
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reasoning (44 chars)
Good. Let me look at the KDA implementation.
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Bash
args
{
  "command": "find /workspace/fla_ref/fla/ops/kda -type f | head -50 && echo '---' && ls -la /workspace/fla_ref/fla/ops/kda/",
  "description": "List FLA KDA files"
}
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stdout (1406 chars)
/workspace/fla_ref/fla/ops/kda/__init__.py
/workspace/fla_ref/fla/ops/kda/chunk_intra_token_parallel.py
/workspace/fla_ref/fla/ops/kda/naive.py
/workspace/fla_ref/fla/ops/kda/chunk_bwd.py
/workspace/fla_ref/fla/ops/kda/fused_recurrent.py
/workspace/fla_ref/fla/ops/kda/gate.py
/workspace/fla_ref/fla/ops/kda/chunk.py
/workspace/fla_ref/fla/ops/kda/chunk_fwd.py
/workspace/fla_ref/fla/ops/kda/chunk_intra.py
/workspace/fla_ref/fla/ops/kda/wy_fast.py
/workspace/fla_ref/fla/ops/kda/backends/__init__.py
/workspace/fla_ref/fla/ops/kda/backends/flashkda.py
/workspace/fla_ref/fla/ops/kda/backends/tilelang/__init__.py
/workspace/fla_ref/fla/ops/kda/backends/tilelang/chunk_bwd_dqkg.py
---
total 164
drwxr-xr-x  3 1002 1002  4096 Jun 13 22:39 .
drwxr-xr-x 39 1002 1002  4096 Jun 13 22:39 ..
-rw-r--r--  1 1002 1002   441 Jun 13 22:39 __init__.py
drwxr-xr-x  3 1002 1002  4096 Jun 13 22:39 backends
-rw-r--r--  1 1002 1002 19215 Jun 13 22:39 chunk.py
-rw-r--r--  1 1002 1002 19246 Jun 13 22:39 chunk_bwd.py
-rw-r--r--  1 1002 1002  4292 Jun 13 22:39 chunk_fwd.py
-rw-r--r--  1 1002 1002 36757 Jun 13 22:39 chunk_intra.py
-rw-r--r--  1 1002 1002  5284 Jun 13 22:39 chunk_intra_token_parallel.py
-rw-r--r--  1 1002 1002 17962 Jun 13 22:39 fused_recurrent.py
-rw-r--r--  1 1002 1002 14032 Jun 13 22:39 gate.py
-rw-r--r--  1 1002 1002  6339 Jun 13 22:39 naive.py
-rw-r--r--  1 1002 1002 12022 Jun 13 22:39 wy_fast.py
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reasoning (62 chars)
Let me create tasks to track this work, since it's multi-step.
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TaskCreate
args
{
  "subject": "Study KDA algorithm and FLA reference",
  "description": "Read reference.py, FLA's KDA Triton kernels (chunk_fwd.py, chunk_intra.py, chunk.py) to understand the exact chunk-parallel forward algorithm and mapping to Triton/CUDA.",
  "activeForm": "Studying KDA algorithm and FLA reference"
}
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stdout (67 chars)
Task #1 created successfully: Study KDA algorithm and FLA reference
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TaskCreate
args
{
  "subject": "Implement solution.py custom kernel",
  "description": "Write a custom KDA forward kernel in solution.py using Triton (or CuTe/CUDA). Match reference.py Model/get_inputs/get_init_inputs interfaces. Do not call forbidden FLA ops.",
  "activeForm": "Implementing custom KDA kernel"
}
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stdout (65 chars)
Task #2 created successfully: Implement solution.py custom kernel
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TaskCreate
args
{
  "subject": "Verify correctness with check.py",
  "description": "Run python check.py across all shapes and seeds. Debug numerical mismatches until PASS.",
  "activeForm": "Verifying correctness with check.py"
}
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stdout (62 chars)
Task #3 created successfully: Verify correctness with check.py
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TaskCreate
args
{
  "subject": "Benchmark and optimize",
  "description": "Run python benchmark.py, profile with ncu/nsys if needed, and iterate to improve peak_fraction gmean above threshold.",
  "activeForm": "Benchmarking and optimizing kernel"
}
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stdout (52 chars)
Task #4 created successfully: Benchmark and optimize
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reasoning (44 chars)
Let me read the FLA KDA chunk forward files.
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Read /workspace/fla_ref/fla/ops/kda/chunk_fwd.py
args
{
  "file_path": "/workspace/fla_ref/fla/ops/kda/chunk_fwd.py"
}
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stdout (4724 chars) — TRUNCATED
1	# Copyright (c) 2023-2026, Songlin Yang, Yu Zhang, Zhiyuan Li
2	#
3	# This source code is licensed under the MIT license found in the
4	# LICENSE file in the root directory of this source tree.
5	# For a list of all contributors, visit:
6	#   https://github.com/fla-org/flash-linear-attention/graphs/contributors
7	
8	import torch
9	
10	from fla.ops.common.chunk_delta_h import chunk_gated_delta_rule_fwd_h
11	from fla.ops.cp import FLACPContext
12	from fla.ops.cp.chunk_delta_h import chunk_gated_delta_rule_fwd_h_pre_process, compress_h0
13	from fla.ops.gla.chunk import chunk_gla_fwd_o_gk
14	from fla.ops.kda.chunk_intra import chunk_kda_fwd_intra
15	from fla.ops.kda.gate import kda_gate_chunk_cumsum
16	from fla.ops.utils import chunk_local_cumsum
17	from fla.ops.utils.constant import RCP_LN2
18	
19	
20	def chunk_kda_fwd(
21	    q: torch.Tensor,
22	    k: torch.Tensor,
23	    v: torch.Tensor,
24	    g: torch.Tensor,
25	    beta: torch.Tensor,
26	    scale: float,
27	    initial_state: torch.Tensor,
28	    output_final_state: bool,
29	    state_v_first: bool = False,
30	    cu_seqlens: torch.LongTensor | None = None,
31	    cu_seqlens_cpu: torch.LongTensor | None = None,
32	    chunk_indices: torch.LongTensor | None = None,
33	    chunk_size: int = 64,
34	    safe_gate: bool = False,
35	    lower_bound: float | None = None,
36	    use_gate_in_kernel: bool = False,
37	    A_log: torch.Tensor | None = None,
38	    dt_bias: torch.Tensor | None = None,
39	    disable_recompute: bool 

... (3224 more chars)
assistant
Read /workspace/fla_ref/fla/ops/kda/chunk_intra.py
args
{
  "file_path": "/workspace/fla_ref/fla/ops/kda/chunk_intra.py"
}
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stdout (40377 chars) — TRUNCATED
1	# Copyright (c) 2023-2026, Songlin Yang, Yu Zhang, Zhiyuan Li
2	#
3	# This source code is licensed under the MIT license found in the
4	# LICENSE file in the root directory of this source tree.
5	# For a list of all contributors, visit:
6	#   https://github.com/fla-org/flash-linear-attention/graphs/contributors
7	
8	import torch
9	import triton
10	import triton.language as tl
11	
12	from fla.ops.kda.chunk_intra_token_parallel import chunk_kda_fwd_intra_token_parallel
13	from fla.ops.kda.wy_fast import recompute_w_u_fwd
14	from fla.ops.utils import prepare_chunk_indices
15	from fla.ops.utils.cache import fla_cache_autotune
16	from fla.ops.utils.op import exp2, gather
17	from fla.utils import IS_GATHER_SUPPORTED, IS_TF32_SUPPORTED, autotune_cache_kwargs
18	
19	if IS_TF32_SUPPORTED:
20	    SOLVE_TRIL_DOT_PRECISION = tl.constexpr('tf32')
21	else:
22	    SOLVE_TRIL_DOT_PRECISION = tl.constexpr('ieee')
23	
24	################################################################################
25	# Fused inter + solve_tril kernel: compute off-diagonal Akk and solve in one pass
26	################################################################################
27	
28	
29	@triton.heuristics({
30	    'IS_VARLEN': lambda args: args['cu_seqlens'] is not None,
31	})
32	@fla_cache_autotune(
33	    configs=[
34	        triton.Config({'BK': BK}, num_warps=num_warps)
35	        for BK in [32, 64]
36	        for num_warps in [1, 2, 4]
37	    ],
38	    key=["H", "HV", "K", "BT", "BC", "NC"],
39	 

... (38877 more chars)
assistant
reasoning (18367 chars)
This is complex. The FLA implementation uses several kernels: 1. chunk_kda_fwd_intra_token_parallel: computes diagonal Akk blocks (Akkd) and Aqk 2. chunk_kda_fwd_kernel_inter_solve_fused: computes off-diagonal blocks and solves the triangular system 3. recompute_w_u_fwd: computes w and u 4. chunk_gated_delta_rule_fwd_h: computes recurrent state h (S) and v_new (v_i) 5. chunk_gla_fwd_o_gk: computes output o The algorithm uses gate cumsum with RCP_LN2 (i.e. they work in log2 space, exp2). But reference.py uses natural exp. The conversion is g_FLA = g / ln(2). Actually in chunk_fwd.py they call chunk_local_cumsum(g, scale=RCP_LN2) which presumably divides by ln2, so exp2 in kernels equals exp in reference. But our problem's g input is already "per-channel log-decay with in-chunk cumsum already applied" in fp32. The reference applies another cumsum(-2) inside. FLA also applies cumsum. So we need to match reference exactly. Given the complexity, I have a few options: A. Implement a custom Triton kernel that closely follows reference.py, without the subchunk optimization. Simpler but maybe fast enough. B. Port/adapt FLA's Triton kernels but inline them in solution.py (not importing FLA). Option A is more feasible in the time available. The reference algorithm with BT=64 is small enough that we can do per-chunk work in registers/shared memory with one thread block per (B,H,chunk). The inter-chunk recurrence requires sequential dependency, but we can launch a kernel per chunk or use persistent blocks. Actually, for correctness and reasonable performance, the simplest robust approach: 1. Kernel 1 (parallel over B,H,N): compute A_kk, A_qk, w, u for each chunk. Store A_qk (BT,BT), w (BT,K), u (BT,V). 2. Kernel 2 (sequential over N, parallel over B,H): for each chunk, load S from previous chunk (stored in global memory), compute v_i = u_i - w_i @ S, o_i, update S, write o_i. But kernel 2 with sequential over N needs N kernel launches, which has overhead. For T=4096, N=64. 64 small kernel launches. With B=1,H=4 or 8, each kernel processes only 4-8 chunks in parallel, which is very little parallelism. This could be slow. Alternative: persistent kernel where each block handles one (B,H) and loops over chunks. Use global memory for S. This avoids launch overhead but still serial. For performance, we want to fuse more. But let me first get correctness with a simple implementation, then optimize. Actually, we can restructure the inter-chunk recurrence to expose more parallelism? The recurrence S_{n+1} = S_n * diag(d_n) + K_n^T @ V_n, where V_n = U_n - W_n @ S_n. This is a gated linear recurrence. It has sequential dependency. No easy parallelization without prefix-scan-like algorithms, which are complex. For our shapes (T up to 4096, N up to 64), a persistent kernel with each block handling one (B,H) sequence might be fine if the per-chunk work is efficient. Let me think about kernel design. Kernel 1: compute per-chunk A, w, u. - Grid: (N, B*H) - Each block handles one chunk of 64 tokens for one head/batch. - Load q, k, g, beta, v into shared memory (64*128 bf16 = 16KB each; 5 such arrays = 80KB; plus output 64*64 + 64*128 + 64*128 = ~32KB. Total ~112KB, too much for shared memory (max ~228KB on Blackwell but we want less). - Actually we can stream through K in blocks. Triton can do this with block pointers and accumulate in registers. A full (64,64) A matrix and (64,128) w,u fit in registers for one warp group? 64*64 = 4096 fp32 = 16KB. 64*128 = 8192 fp32 = 32KB each. Three such = 80KB registers. With 32KB shared memory per warp? That's too much for one warp but OK across 4 warps (128KB register file each on Blackwell, per SM 256KB). Actually each thread has 256 registers * 32 bytes = 8KB. 128 threads = 1MB. So 80KB is fine distributed. But Triton manages this. We can write a Triton kernel with: - BLOCK_BT=64, BLOCK_K=32 or 64, BLOCK_V=32 or 64 - For A: accumulate over K blocks - For w, u: after A is computed, compute A @ (kg) and A @ v Actually, computing A requires O(BT^2*K) = 64*64*128 = 524K MACs = 1M FLOPs per chunk. In bf16 that's fine. Let me think about numerical accuracy. Reference uses fp32 throughout. We need atol/rtol 0.05, so bf16 should be OK, but we should accumulate A in fp32 and use fp32 for the recurrence. The critical path for accuracy is the forward substitution: A = (I - A_lower)^{-1} * beta. This involves accumulation of small values. If we do this in bf16, errors could grow. We should keep A in fp32 for the solve and store final A in bf16 or fp32. Actually the reference stores A in fp32. For our kernel, we can keep A in fp32 global memory (B,T,H,BT) which is small: 2*4096*8*64*4 = 16.7 MB. Acceptable. Similarly S should be fp32. Let me think about a simpler algorithm matching reference exactly. Reference steps (for each chunk n): 1. q, k, v, g, beta rearranged to (B,H,N,C,d) 2. g = g.cumsum(-2) # cumulative within chunk 3. Compute A_kk (B,H,N,C,C): For i in C: A[...,i] = sum_{c < i} k[...,c] * k[...,i] * exp(g[...,c] - g[...,i]) * beta[...,i] A = -A (lower triangular, diag masked) Forward substitution: for i in 1..C-1: A[...,i,:i] += sum_t A[...,i,t] * A[...,t,:i] A = (A + I) * beta[...,None,:] # beta applied to columns 4. w = A @ (g.exp() * k) -> (B,H,N,C,K) 5. u = A @ v -> (B,H,N,C,V) 6. For each chunk n sequentially: Aqk[...,j] = sum_{c <= j} q[...,c] * k[...,j] * exp(g[...,c] - g[...,j]) (lower triangular incl diag) v_i = u_i - w_i @ S o_i = (q_i * g_i.exp()) @ S + Aqk @ v_i S = S * exp(g_i[-1]) + (k_i * exp(g_i[-1] - g_i)).transpose(-2,-1) @ v_i Aqk is strictly lower triangular (including diagonal). Wait, in reference line 86: Aqk = Aqk.masked_fill(mask_strict_upper, 0). mask_strict_upper is diagonal=1 (upper including diagonal zeroed), so diagonal IS included. So Aqk is lower triangular including diagonal. Note that Aqk uses q and k with the same g, no beta. A_kk uses k,k with beta. The cumsum g is natural log. In FLA, they divide g by ln2 and use exp2. We can do natural exp. For numerical stability, reference computes exp(g - g_i) which is fine. Now, a key insight: we can precompute g = g.cumsum(-2) once. Let me design Triton kernels. Kernel 1: compute_A_w_u Inputs: q, k, v, g_cum, beta, scale Output: Aqk (fp32), A (fp32), w (fp32 or bf16), u (fp32 or bf16), maybe kg (k*exp(g)) and qg (q*exp(g)) Wait, we can also precompute kg = k * g.exp() and qg = q * g.exp(). For inter-chunk, we need qg @ S and kg.T @ v_i. For Aqk: Aqk[i,j] = sum_c qg[c] * k[j] * exp(-g[j]) * exp(g[c]-g[i])? Let's derive: Aqk[i,j] = sum_{c<=j} q[c] * k[j] * exp(g[c] - g[j]) = sum_{c<=j} (q[c] * exp(g[c])) * (k[j] * exp(-g[j])) = sum_{c<=j} qg[c] * (k[j] * exp(-g[j])) But careful: in reference, Aqk[...,j] = einsum over c of q_i[c] * k_j * exp(g_i[c] - g_i[j]). Here i is chunk index, j is column. So Aqk[row c, col j] = q[c] * k[j] * exp(g[c] - g[j]). So yes, Aqk = (qg * something) @ (k * exp(-g)).T? Actually: Aqk[c, j] = qg[c] * k[j] * exp(-g[j]) for c >= j? Wait reference has c<=j? Let me re-read. Reference line 82-85: for j in range(BT): k_j = k[:, :, i, j] g_j = g[:, :, i, j:j+1, :] Aqk[..., j] = einsum("... c d, ... d -> ... c", q_i * (g_i - g_j).exp(), k_j) So Aqk[..., j] is a vector of length BT (rows indexed by c). Aqk[c, j] = sum_d q[c,d] * k[j,d] * exp(g[c,d] - g[j,d]). Summed over d. Then line 86 masks strict upper: Aqk[c,j] = 0 for c < j. So Aqk is lower triangular INCLUDING diagonal. So Aqk[c,j] = sum_d qg[c,d] * k[j,d] * exp(-g[j,d]) for c >= j. Where qg[c,d] = q[c,d] * exp(g[c,d]). We can write Aqk = tril( (qg) @ (k * exp(-g)).T ). Similarly for Akk (A in reference): Akk[c,j] = sum_d k[c,d] * k[j,d] * exp(g[c,d] - g[j,d]) * beta[j] for c < j (strict lower). = sum_d (k[c,d] * exp(g[c,d])) * (k[j,d] * exp(-g[j,d])) * beta[j] for c < j. So the unmasked matrix is (kg) @ (kg_negative).T where kg_negative = k * exp(-g) * beta[...,None]? Wait beta is per token, not per channel. beta[j] multiplies the whole column j. Actually beta[j] is applied to column j. So if we define kg = k * exp(g), then kgT = k * exp(-g), and beta column weight. Akk_unmasked[c,j] = sum_d kg[c,d] * kgT[j,d] * beta[j] Then mask strict lower (c > j), negate, solve. For w = A @ (g.exp() * k) = A @ kg. For u = A @ v. For the inter-chunk recurrence: v_i = u_i - w_i @ S o_i = qg_i @ S + Aqk @ v_i (where qg_i = q_i * exp(g_i)) S_new = S * exp(g_i[-1]) + (k_i * exp(g_i[-1] - g_i)).T @ v_i = S * exp(g_i[-1]) + kgT_i.T @ v_i where kgT_i = k_i * exp(-g_i) * exp(g_i[-1])? Actually k_i * exp(g_i[-1] - g_i) = k_i * exp(-g_i) * exp(g_i[-1]) = kgT_i * exp(g_i[-1]). Wait, reference line 90: S = S + rearrange((g_i[:, :, -1:] - g_i).exp() * k_i, "b h c k -> b h k c") @ v_i So S_new = S * exp(g_i[-1]) + ((k_i * exp(g_i[-1] - g_i)).T @ v_i) Define k_decay_i = k_i * exp(-g_i). Then update uses k_decay_i * exp(g_i[-1]) = k_decay_i scaled by last decay. So we need kg = k * exp(g), kgT = k * exp(-g), qg = q * exp(g). All per chunk. These can be computed in the same kernel or precomputed. For numerical stability, computing exp(g) and exp(-g) for g in [-something, something] should be fine. Reference g values are small. Now, given K=V=128, chunk_size=64, let me design: Kernel 1: per (B,H,N) block, compute: - Load q,k,v,g,beta for chunk - Compute g_cum = g.cumsum(0) within chunk (or assume input g already has it? Reference applies cumsum, so we must apply it) - Compute qg, kg, kgT in shared memory or registers - Compute Aqk = tril(qg @ kgT.T) (64x64) - Compute Akk = tril_strict(kg @ kgT.T * beta[:,None], diagonal=1) then negate? Actually reference: Akk_unmasked = kg @ (kgT.T) * beta_col (where beta_col[j] = beta[j]) Then A = -tril(Akk_unmasked, diagonal=-1) # strict lower, negate Then forward substitution Then A = (A + I) * beta_col[None,:] Wait, reference line 65: A = A * beta[..., None] after computing the lower triangular (before solve). Then line 70: A = (A + I) * beta[..., None, :]. So beta is applied twice? Let me re-read. Line 65: A = A * beta[..., None] # here A is the strict lower triangular k-k interaction, beta applied to rows (i index) Line 66: A = -A.masked_fill(mask_diag_upper, 0) # negate and mask upper/diag Line 69-70: forward substitution on A (still strict lower) Line 70: A = (A + torch.eye(BT)) * beta[..., None, :] # add identity, beta applied to columns (j index) So beta is applied to rows in the initial matrix, then the solve happens, then beta applied to columns in the final result. This is equivalent to the delta-rule formulation. OK. So in kernel: 1. Compute M = kg @ kgT.T where kgT = k * exp(-g) (64x64, all elements) 2. Mask strict lower: A0[c,j] = -M[c,j] * beta[c] for c > j, else 0 3. Solve: for i = 1..63: A[i,:i] += A[i,:] @ A[:,:i] (where A[i,:] is row i, length i, and A[:,:i] is i x i) Since A is strictly lower, A[i,t] nonzero only for t < i, and A[t,j] nonzero for j < t < i. This is forward substitution computing (I - A0)^{-1} * something. 4. Add identity, multiply by beta[j] per column: A[i,j] = A[i,j] * beta[j], and A[i,i] = 1 * beta[i] Then w = A @ kg, u = A @ v. Let me verify dimensions. A is (C,C). kg is (C,K). v is (C,V). So w (C,K), u (C,V). For Aqk: Aqk[c,j] = sum_d qg[c,d] * kgT[j,d] for c >= j. tril(include_diag). Kernel 2: inter-chunk recurrence. We need Aqk, w, u, qg, kg, kgT (or k_decay = k*exp(-g)), beta, g_last. Actually for S update, we need k_decay = k * exp(-g), and scale by exp(g[-1]). For v_i = u_i - w_i @ S. For o_i = qg_i @ S + Aqk @ v_i. For S_new = S * exp(g[-1]) + k_decay_i.T @ v_i * exp(g[-1]). So kernel 2 needs: qg, k_decay, Aqk, u, w, and g_last = g[-1] per chunk. qg, k_decay are (B,H,N,C,K). Aqk is (B,H,N,C,C). u,w are (B,H,N,C,K/V). g_last is (B,H,N,K). This is a lot of memory but manageable. Aqk: B,H,T,BT = 2*8*2048*64 = 2M elements * 4 bytes = 8.5 MB (fp32) qg, kg? We might not need kg. We need k_decay (B,H,T,K) = 2*8*2048*128*4 = 16.7 MB. qg similar. w (B,H,T,K) = 16.7 MB. u (B,H,T,V) = 16.7 MB. g_last (B,H,N,K) small. Total ~60 MB. Acceptable for training shapes. For decode shape B=1,T=2048,H=4: half that. Now the question is performance. Kernel 2 is memory-bound and serial. Kernel 1 is compute-bound. But before getting too deep, let me check: maybe a PyTorch implementation with careful fusion is fast enough? We can try torch.compile or hand-optimized torch ops. But torch.compile may not handle the recurrence well, and the custom ops are needed. Actually, the problem says "needs to be a real custom kernel". So we need Triton/CUDA. Let me first try a simpler Triton implementation and see if it passes and what performance we get. But wait - the reference is doing g = g.cumsum(-2) INSIDE the function. But the input g is described as "per-channel log-decay with in-chunk cumsum already applied". Yet reference.py still applies cumsum. So we must match reference: apply cumsum to g inside forward. Actually, reference line 56: g = g.cumsum(-2). This means the input g is raw log-decay per token, and we cumsum within each chunk. The comment in reference says "in-chunk cumsum already applied" but the code applies it. So we must apply cumsum. The problem description might be slightly misleading - it says "g of shape ... (per-channel log-decay with in-chunk cumsum already applied)". But reference applies cumsum anyway. We should follow reference code. OK. Let me write a Triton kernel. I'll start with a simplified approach: Approach 1: Triton kernel for A/w/u + Python loop for recurrence. - Kernel for A/w/u parallel over (B,H,N) - For recurrence, a Triton kernel per chunk (or one kernel that loops N times internally) Actually, the recurrence can be one Triton kernel where each block handles one (B,H) and loops over N chunks sequentially. Within the loop, it loads the chunk data, computes v_i, o_i, updates S. This avoids kernel launch overhead. Let me think about register/shared memory for kernel 2: - S is (K,V) = 128*128 fp32 = 64KB. This fits in shared memory. - w_i is (BT,K) = 64*128 fp32 = 32KB. u_i (BT,V) = 32KB. qg_i (BT,K) = 32KB. k_decay_i (BT,K) = 32KB. Aqk_i (BT,BT) = 16KB. Total if all in shared memory = 208KB, too much. - We can keep S in shared memory (64KB), stream other tensors in blocks. - Or keep S in registers distributed across threads. For one block with 128 threads (4 warps), we have plenty of shared memory. Use 64KB for S, and stream w, u, qg, k_decay through shared memory in tiles of (BT, BLOCK_K) and (BT, BLOCK_V). Aqk (64x64) can stay in registers or shared memory. Actually, matrix multiply Aqk @ v_i where v_i is (BT,V): we can tile V dimension. Similarly qg @ S and w @ S. The per-chunk work in kernel 2: 1. Load S from global memory (or keep it across iterations) 2. Compute v_i = u_i - w_i @ S: - w_i @ S: (BT,K) @ (K,V) -> (BT,V). Tile K and V. 3. Compute o_i = qg_i @ S + Aqk @ v_i: - qg_i @ S: (BT,K) @ (K,V) -> (BT,V) - Aqk @ v_i: (BT,BT) @ (BT,V) -> (BT,V) 4. Update S = S * decay_last + k_decay_i.T @ v_i * decay_last: - k_decay_i.T @ v_i: (K,BT) @ (BT,V) -> (K,V) 5. Write o_i This is basically a series of GEMMs. A block can do it efficiently. But note S has sequential dependency: we must finish updating S before moving to next chunk. So within a block, the loop is sequential. But different (B,H) blocks run in parallel. Number of blocks = B*H. For B=2,H=8: 16 blocks. For B=1,H=4: 4 blocks. This is not a lot of parallelism. For B=1,H=4,T=4096,N=64, only 4 blocks on a 188-SM GPU. Severe underutilization. But each block does a lot of work (64 chunks). Maybe OK if each block is fast. Actually, we can parallelize chunks across B*H*N if we use a prefix-scan approach for the recurrence, but that's complex. For now, sequential per (B,H) is acceptable for correctness. Performance might suffer for B=1,H=4. Alternatively, we can have each block handle one (B,H,N) chunk, but use global atomic operations or a dependency chain? Hard. Actually, another idea: we can compute the recurrence using a "chunked scan" where each chunk's contribution to S is computed in parallel, then combined. But the recurrence S_{n+1} = S_n * D_n + K_n^T @ (U_n - W_n @ S_n) is not affine in S_n because of the W_n @ S_n term inside. Wait: V_n = U_n - W_n @ S_n S_{n+1} = S_n * D_n + K_n^T @ V_n = S_n * D_n + K_n^T @ U_n - K_n^T @ W_n @ S_n = (D_n - K_n^T @ W_n) * S_n + K_n^T @ U_n? Wait dimensions: D_n is (K,) diagonal, K_n^T @ W_n is (K,K), so D_n - (K_n^T @ W_n) is (K,K). But elementwise multiplication S_n * D_n vs matrix multiply (K_n^T @ W_n) @ S_n. These are different. Actually D_n is per-channel decay for S. The update is S_new[k,v] = S[k,v] * exp(g[-1,k]) + sum_c k_decay[c,k] * v_i[c,v]. And v_i[c,v] = u_i[c,v] - sum_k w_i[c,k] * S[k,v]. So S_new[k,v] = S[k,v] * d[k] + sum_c k_decay[c,k] * (u_i[c,v] - sum_k' w_i[c,k'] * S[k',v]) = S[k,v] * d[k] + sum_c k_decay[c,k] * u_i[c,v] - sum_c k_decay[c,k] * sum_k' w_i[c,k'] * S[k',v] = sum_k' [ delta(k,k') * d[k] - sum_c k_decay[c,k] * w_i[c,k'] ] * S[k',v] + sum_c k_decay[c,k] * u_i[c,v] So it's an affine recurrence S_{n+1} = M_n @ S_n + B_n where M_n is (K,K) and B_n is (K,V). This is a matrix recurrence that can be parallelized via associative scan if we represent each step as an affine transform. We can do a tree-based parallel scan over chunks. This would expose N-way parallelism. This is the "parallel scan" approach used by some linear attention implementations. It's more complex but good for long contexts. For our purposes, with B*H blocks already and T up to 4096, the sequential approach might suffice for peak_fraction 0.1. But B=1,H=4,T=4096 would be very slow with sequential per block. Let me think: B=1,H=4,T=4096 => 4 blocks, each doing 64 chunks sequentially. If each chunk takes 0.05ms, total 3.2ms. FLOPs = 4*1*4096*4*(16384+8192+8192) = 4*4096*4*32768 = 2.1 GFLOPs. At 3.2ms that's 0.66 TFLOPS. Peak fraction maybe ~0.5%. Bad. With scan: all 64 chunks in parallel -> 256 blocks (B*H*N) doing matrix multiplies, then scan reduction. Much better. But implementing a parallel scan for this affine recurrence in Triton is complex. However, FLA already does this with chunk_gated_delta_rule_fwd_h. We could adapt that kernel. Let me look at chunk_gated_delta_rule_fwd_h.
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  "file_path": "/workspace/fla_ref/fla/ops/common/chunk_delta_h.py",
  "limit": 200
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1	# Copyright (c) 2023-2026, Songlin Yang, Yu Zhang, Zhiyuan Li
2	#
3	# This source code is licensed under the MIT license found in the
4	# LICENSE file in the root directory of this source tree.
5	# For a list of all contributors, visit:
6	#   https://github.com/fla-org/flash-linear-attention/graphs/contributors
7	
8	import torch
9	import triton
10	import triton.language as tl
11	
12	from fla.ops.backends import dispatch
13	from fla.ops.utils import prepare_chunk_indices, prepare_chunk_offsets
14	from fla.ops.utils.cache import fla_cache_autotune
15	from fla.ops.utils.op import exp2
16	from fla.utils import IS_NVIDIA_HOPPER, USE_CUDA_GRAPH, autotune_cache_kwargs, check_shared_mem
17	
18	NUM_WARPS = [2, 4] if IS_NVIDIA_HOPPER else [2, 4, 8, 16]
19	
20	
21	@triton.heuristics({
22	    'USE_G': lambda args: args['g'] is not None,
23	    'USE_GK': lambda args: args['gk'] is not None,
24	    'USE_INITIAL_STATE': lambda args: args['h0'] is not None,
25	    'STORE_FINAL_STATE': lambda args: args['ht'] is not None,
26	    'SAVE_NEW_VALUE': lambda args: args['v_new'] is not None,
27	    'IS_VARLEN': lambda args: args['cu_seqlens'] is not None,
28	})
29	@fla_cache_autotune(
30	    configs=[
31	        triton.Config({'BV': BV}, num_warps=num_warps, num_stages=num_stages)
32	        for num_warps in [2, 4]
33	        for num_stages in ([2, 3, 4] if check_shared_mem('ampere') else [2, 1])
34	        for BV in ([32, 64] if check_shared_mem('ada') else [32])
35	    ],
36	    key=['H', 'HV',

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reasoning (525 chars)
This is a sophisticated kernel. It splits S across V tiles (BV) and K segments (64 each), and loops over chunks. This is still sequential but tiles V to allow more blocks (parallel over V tiles). For K=V=128, BV=32 or 64, so 2-4 V tiles per (B,H), giving 8-32 blocks for B=1,H=4. Better. Actually it seems to be sequential per block, but parallel over V tiles. So for B=1,H=4,V=128,BV=64: 8 blocks total, each doing 64 chunks sequentially. Still not great, but better. Let me continue reading to understand the full kernel.
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{
  "file_path": "/workspace/fla_ref/fla/ops/common/chunk_delta_h.py",
  "offset": 200,
  "limit": 200
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200	            b_g_last = tl.load(g + (bos * HV + last_idx * HV + i_h).to(tl.int64)).to(tl.float32)
201	            p_g = tl.make_block_ptr(g + (bos * HV + i_h).to(tl.int64), (T,), (HV,), (i_t * BT,), (BT,), (0,))
202	            b_g = tl.load(p_g, boundary_check=(0,)).to(tl.float32)
203	            b_v = b_v * tl.where(m_t, exp2(b_g_last - b_g), 0)[:, None]
204	            b_g_last = exp2(b_g_last)
205	            b_h1 *= b_g_last
206	            if K > 64:
207	                b_h2 *= b_g_last
208	            if K > 128:
209	                b_h3 *= b_g_last
210	            if K > 192:
211	                b_h4 *= b_g_last
212	
213	        if USE_GK:
214	            o_k1 = tl.arange(0, 64)
215	            b_gk_last1 = tl.load(gk + (bos + last_idx) * HV*K + i_h * K + o_k1, mask=(o_k1 < K), other=0.).to(tl.float32)
216	            if STATE_V_FIRST:
217	                b_h1 *= exp2(b_gk_last1)[None, :]
218	            else:
219	                b_h1 *= exp2(b_gk_last1)[:, None]
220	            if K > 64:
221	                o_k2 = 64 + o_k1
222	                b_gk_last2 = tl.load(gk + (bos + last_idx) * HV*K + i_h * K + o_k2, mask=(o_k2 < K), other=0.).to(tl.float32)
223	                if STATE_V_FIRST:
224	                    b_h2 *= exp2(b_gk_last2)[None, :]
225	                else:
226	                    b_h2 *= exp2(b_gk_last2)[:, None]
227	            if K > 128:
228	                o_k3 = 128 + o_k1
229	                b_gk_last3 = tl.load(gk + (bos + last_idx) * HV*K +

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reasoning (33945 chars)
The FLA kernel is quite involved. Porting it exactly would be a lot of work and we'd need to handle the HV (heads*groups) stuff. Given the time, I think the best approach is: 1. Implement a correct Triton kernel that closely matches reference.py, with reasonable fusion. 2. If performance is insufficient, optimize. Let me start by writing a complete solution in PyTorch/Triton. I'll aim for: - Kernel 1: compute_A_w_u (parallel over B,H,N) - Kernel 2: compute output via sequential recurrence (parallel over B,H,V tiles) Actually, maybe I can use torch.compile for some parts? But the problem wants a real custom kernel. We can have the custom kernel do the heavy lifting and PyTorch for setup. Wait, I realize we might be able to use FLA's code by copying the relevant Triton kernels into solution.py. Since the forbidden check is textual (regex on solution.py), we just can't have the strings "fla.ops.kda", "chunk_kda", etc. in solution.py. We can copy the kernels and adapt them, removing FLA imports. This is a huge amount of code though. And FLA has dependencies like gather, exp2, etc. that we'd need to implement. Alternative: write a simpler but complete Triton implementation from scratch. Let me think about whether a simple Triton kernel can pass correctness and performance. Given tolerance 0.05, we can accumulate in fp32 but do GEMMs in bf16 with fp32 accumulators. Triton's tl.dot uses fp32 accumulation. Let me design: Kernel 1: `kda_intra_kernel` Grid: (NT, B*H) Each block: one chunk, one head. Constants: BT=64, BK=32 or 64, BV=32 or 64. Inputs: - q, k, v, g, beta (layouts as in reference or contiguous) Outputs: - Aqk (B,H,T,BT) fp32 - A (B,H,T,BT) fp32 (full A including identity, beta-scaled columns) - w (B,H,T,K) fp32 or bf16 - u (B,H,T,V) fp32 or bf16 - qg (B,H,T,K) fp32 or bf16 - k_decay (B,H,T,K) fp32 or bf16 (k * exp(-g)) - g_last (B,H,N,K) fp32 Wait, we need g_last per chunk. We can compute it as g_cum[..., -1, :]. Algorithm in kernel: 1. Load q,k,g,beta,v for chunk. 2. Compute g_cum = cumsum(g, axis=0) within chunk. 3. Compute qg = q * exp(g_cum), kg = k * exp(g_cum), k_decay = k * exp(-g_cum). 4. Compute Aqk = tril(qg @ k_decay.T) [use tl.dot, mask] 5. Compute Akk = tril_strict(kg @ k_decay.T * beta[:,None]) [mask c>j] 6. Solve (I - (-Akk))^{-1}: Actually we have A0 = -tril_strict(Akk). We want A = (I - A0)^{-1} * beta_col. We can do forward substitution in-place on a 64x64 fp32 array in registers. But 64x64 fp32 = 16KB per block. With 128 threads, that's 128 bytes/thread. Fine. 7. Add identity and scale columns by beta: A[i,j] *= beta[j], A[i,i] = beta[i]. 8. Compute w = A @ kg, u = A @ v. 9. Store Aqk, A, w, u, qg, k_decay, g_last. This kernel is compute-heavy and parallel. Should be efficient. Kernel 2: `kda_inter_kernel` Grid: (NT, B*H) or (B*H) with internal loop. Actually for more parallelism, we can parallel over V tiles: grid (NV, B*H) where NV = V/BV, and each block loops over chunks sequentially. This matches FLA's approach and gives more parallelism. For K=V=128, BV=64 -> NV=2. So B=1,H=4 gives 8 blocks. Better than 4. Within each block: - Load initial S = zeros(K,BV) in registers or shared memory. - For each chunk n: - Store S to h buffer (B,H,N,K,V) if needed for output? Actually output doesn't need S intermediate, only final. - Load w_i (BT,K), u_i (BT,BV), qg_i (BT,K), k_decay_i (BT,K), Aqk_i (BT,BT), g_last_i (K). - Compute v_i = u_i - w_i @ S - Compute o_i = qg_i @ S + Aqk_i @ v_i - Store o_i - Update S = S * exp(g_last_i)[:,None] + k_decay_i.T @ v_i * exp(g_last_i)[:,None]? Wait: S_new[k,v] = S[k,v] * d[k] + sum_c k_decay[c,k] * v_i[c,v] * d[k] where d[k] = exp(g_last[k]). So S = S * d + (k_decay.T @ v_i) * d The d factor applies to both terms? Let me check reference. Reference line 89: S = S * rearrange(g_i[:, :, -1].exp(), "b h k -> b h k 1") Then line 90: S = S + rearrange((g_i[:, :, -1:] - g_i).exp() * k_i, "b h c k -> b h k c") @ v_i So first S is scaled by d, then we add (k_i * exp(g_last - g_i)).T @ v_i. And k_i * exp(g_last - g_i) = k_decay_i * d[k] (since k_decay_i = k_i * exp(-g_i), so k_decay_i * d = k_i * exp(-g_i) * exp(g_last) = k_i * exp(g_last - g_i)). So yes: S_new = S * d + (k_decay_i.T @ v_i) * d. We can factor d: S_new = (S + k_decay_i.T @ v_i) * d. Actually wait, in the reference, the order is S = S*d; S = S + (k_decay_i * d).T @ v_i. That's S_new = S*d + k_decay_i.T @ v_i * d. Same as (S + k_decay_i.T @ v_i) * d. Good. - Continue to next chunk. This kernel has S in registers/shared memory. S is (K,BV) = 128*64 = 8192 fp32 = 32KB. With 128 threads, 256 bytes/thread. Register file per SM is 256KB, so per warp 64KB, per thread 2KB. 256 bytes/thread is fine, but we also need other registers. Should be OK if we keep S in registers or shared memory. Actually, keeping S in shared memory (32KB) is good. Then w_i @ S uses shared memory for S and registers for accumulator. qg_i @ S similarly. k_decay_i.T @ v_i: v_i in registers, k_decay_i in shared or registers. Wait, in FLA they keep S in registers (b_h1 is [64,BV] or [BV,64]). For K=128, they have b_h1, b_h2 each [64,BV]. So S total 128*BV in registers. For BV=64, that's 8192 fp32 = 32KB registers. That's reasonable for one warp group? Actually registers are per thread. 128 threads * 256 regs * 4 bytes = 128KB register file. 32KB for S leaves 96KB for other stuff. Tight but OK. Maybe using shared memory for S is simpler. But shared memory is also used for inputs. Let me think. Actually, for matrix multiply w_i @ S where S in shared memory, we need S as the B matrix. Triton can load S from shared memory. If S is in registers, we can pass it to tl.dot. I think keeping S in shared memory is cleaner. Each block has 32KB shared for S + maybe some for v_i/o_i. Total < 64KB. But we also need to do k_decay_i.T @ v_i. If v_i is in shared memory (BT,BV = 64*64 = 16KB), and k_decay_i streamed, we can compute. Alternatively v_i in registers. Let me not over-optimize initially. I'll write a working kernel and see. Now, an important detail: the g input. Reference does `g = g.cumsum(-2)` on the already-cumsumed input? The problem says "g ... (per-channel log-decay with in-chunk cumsum already applied)". But reference.py applies cumsum anyway. So we must apply cumsum to match reference. The test will compare against reference which applies cumsum. So our kernel must apply cumsum within chunk. Actually, wait. Let me re-read reference: ```python g = rearrange(g, "b (n c) h d -> b h n c d", c=BT) g = g.cumsum(-2) ``` So yes, g is rearranged into chunks and cumsumed along the chunk dimension. Input g is raw log-decay per token (presumably). So we must do the same. Now, should we apply the cumsum in a separate kernel or fuse into kernel 1? We can fuse it. For numerical stability, reference computes exp(g - g_i) rather than exp(g)/exp(g_i). We can do the same: subtract g_i (last token or current token) before exp. For Aqk, we subtract g_j (column j). For Akk, subtract g_i (row i) or g_j (column j)? Reference line 64: `k * (g - g_i).exp()` where g_i is the row i. Then A[...,i] is row i (or column? Actually A[...,i] sets column i in the result). So it's exp(g_c - g_i) where c is the sum index and i is the fixed column. So row c, column i. We subtract g of the column. For Akk this is natural: A[c,i] = sum_d k[c]*k[i]*exp(g[c]-g[i]). For Aqk, line 85: q_i * (g_i - g_j).exp() where g_j is column j. So Aqk[c,j] = sum_d q[c]*k[j]*exp(g[c]-g[j]). Again subtract g of column. For S update, reference uses exp(g_last - g_c). Subtract g of each token from last. So all use "subtract the target token's g". For qg and kg, we can compute q * exp(g) and k * exp(g), and k_decay = k * exp(-g). Then: - Aqk[c,j] = sum_d qg[c,d] * k_decay[j,d] (for c>=j) - Akk[c,j] = sum_d kg[c,d] * k_decay[j,d] * beta[j] (for c>j) - w[c,k] = sum_j A[c,j] * kg[j,k] - S update: sum_c k_decay[c,k] * v_i[c,v] * d[k] where d[k]=exp(g_last[k]) This avoids computing exp(g - g_i) per pair; we just precompute qg, kg, k_decay. Good for stability and speed. Wait, is this exactly equal? qg[c] = q[c] * exp(g[c]), k_decay[j] = k[j] * exp(-g[j]). Their dot product is q[c]*k[j]*exp(g[c]-g[j]). Yes. For k_decay in S update: k_decay[c,k] = k[c,k] * exp(-g[c,k]). Then k_decay[c,k] * d[k] = k[c,k] * exp(g_last[k] - g[c,k]). Yes. Great. So kernel 1 computes qg, kg, k_decay. But wait, the reference computes Aqk as qg @ k_decay.T and masks lower triangular. But qg and k_decay are (C,K). The dot is over K. That's correct. For w = A @ kg, kg = k * exp(g). Correct. For u = A @ v. Correct. For S update, we need k_decay.T @ v_i, then multiply by d. Correct. Now, one subtlety: reference line 70: `A = (A + torch.eye(BT, dtype=torch.float32, device=q.device)) * beta[..., None, :]`. So A[i,i] = beta[i], and off-diagonal A[i,j] = solved_value * beta[j]. We need to make sure our kernel does this. In the forward substitution, reference line 69: ```python for i in range(1, BT): A[..., i, :i] = A[..., i, :i].clone() + (A[..., i, :, None].clone() * A[..., :, :i].clone()).sum(-2) ``` A[...,i,:] is row i. A[..., :, :i] is rows 0..i-1, columns 0..i-1. The sum is over the row dimension (t). So: new_A[i,j] = A[i,j] + sum_t A[i,t] * A[t,j] for j < i, t ranges 0..BT-1. Since A is strictly lower triangular, A[i,t] nonzero only for t<i, and A[t,j] nonzero only for j<t. So t ranges j<t<i. This computes (I - A0)^{-1} where A0 is the strict lower triangular matrix. We can implement this in Triton. But doing it with a 64x64 register array might be slow. Alternatively, we can compute it using matrix operations. Actually, for a 64x64 strictly lower triangular matrix, the forward substitution can be done efficiently in registers with a loop of 64 iterations, each updating one row. In Triton, we can do: ```python A = -Akk_strict_lower # 64x64 fp32 for i in range(1, BT): # A[i,:i] += A[i,:] @ A[:,:i] row_i = A[i,:] # length BT col_block = A[:,:i] # BT x i update = tl.sum(row_i[:,None] * col_block, axis=0) # length i # But we only want first i elements A[i,:i] += update ``` In Triton with a 64x64 array in registers, each iteration does O(i*BT) work. Total O(BT^3) but BT=64 so it's fine. However, register spilling might happen. Actually, since A is strictly lower triangular, A[i,:] has support only in [0,i-1]. And A[:,:i] is i x i lower triangular. So update = A[i,0:i] @ A[0:i,0:i]. This is a dot product of length i. The total work is sum_i i^2 = BT^3/3 ~ 87K MACs. Trivial. We can implement the solve as a loop where we maintain the full 64x64 matrix and update row by row. But in Triton, indexing a 2D array of shape (BT,BT) with variable i and slicing might not be straightforward. We can use masks. Alternatively, we can compute the inverse using a tiled approach. But for BT=64, the naive loop is fine. Let me write the Triton code. I'll need to be careful with how I represent the 64x64 matrix in registers. Actually, Triton has limited support for dynamic indexing. We can use `tl.where` with masks. For row update: ```python for i in range(1, BT): # row i before update row_i = A[i, :] # shape (BT,) # compute A[i,:] @ A[:, :] # only lower triangular parts matter update = tl.sum(row_i[:, None] * A, axis=0) # shape (BT,) # only update positions j < i mask = tl.arange(0, BT) < i A = tl.where((tl.arange(0, BT)[:,None] == i) & mask[None,:], A + update[None,:], A) ``` Hmm, this is tricky because we need to update only row i, columns < i. We can do: ```python for i in range(1, BT): row_i = A[i, :] # (BT,) update = tl.sum(row_i[:, None] * A, axis=0) # (BT,) new_row_i = tl.where(tl.arange(0,BT) < i, A[i,:] + update, A[i,:]) A = tl.where((tl.arange(0,BT)[:,None] == i), new_row_i[None,:], A) ``` But `A[i,:]` indexing a 2D tensor with scalar i might not be allowed in Triton? We can use `tl.sum(tl.where(tl.arange(0,BT)[:,None] == i, A, 0), axis=0)` to extract row i. This is doable. Let me write it. Actually, since BT is a constexpr (64), the loop unrolls and maybe Triton can handle it. But dynamic indexing within registers is limited. Alternatively, we can store A in shared memory and update row by row. Shared memory indexing with variable i is supported. Shared memory for A: 64*64*4 = 16KB. Plus other arrays. We can allocate A in shared memory, do the solve there, then use it for w,u. But shared memory is per block. Loading A from shared memory for w = A @ kg uses block pointer or regular load. Should work. Let me design kernel 1 with A in shared memory. Actually, here's another idea: we can avoid the explicit solve by noting that w = A @ kg = (I - A0)^{-1} @ diag(beta_col) @ kg, where A0 is the strict lower triangular -Akk*beta_row. Wait let's derive: Let A0 = -tril_strict(M) * beta_row (where M = kg @ k_decay.T). We want A = (I - A0)^{-1} * diag(beta_col). Actually A0 has beta applied to rows. The solve gives (I - A0)^{-1}. Then A = (I - A0)^{-1} * diag(beta_col). w = A @ kg = (I - A0)^{-1} @ (beta_col * kg) u = A @ v = (I - A0)^{-1} @ (beta_col * v) So we don't actually need to materialize A; we can solve the linear system (I - A0) x = b for b = beta_col * kg and b = beta_col * v, using forward substitution. For b = beta_col * kg (shape (BT,K)): x has shape (BT,K). Forward substitution: x[i] = b[i] + sum_{j<i} A0[i,j] * x[j] where A0[i,j] = -M[i,j] * beta[i] for i>j. So x[i] = beta[i] * kg[i] - beta[i] * sum_{j<i} M[i,j] * x[j] = beta[i] * (kg[i] - sum_{j<i} M[i,j] * x[j]) Wait, A0 = -tril_strict(M) * beta_row. So (I - A0) x = b means x - A0 x = b, so x = b + A0 x = beta_col*kg + (-tril_strict(M)*beta_row) x. x[i] = beta[i]*kg[i] - beta[i] * sum_{j<i} M[i,j] * x[j] = beta[i] * (kg[i] - sum_{j<i} M[i,j] * x[j]) This is actually more stable and avoids storing the full A matrix! We can compute w and u directly via forward substitution over the K and V dimensions simultaneously. But for the inter-chunk recurrence, we need Aqk, w, u. We don't need A itself. So this is great. Wait, do we need A for anything else? The reference computes A and then w,u. We can skip A and compute w,u directly. But the forward substitution for w and u is over rows i=0..BT-1, each requiring sum_{j<i} M[i,j] * x[j]. M[i,j] = sum_d kg[i,d] * k_decay[j,d]. So sum_{j<i} M[i,j] * x[j] = sum_d kg[i,d] * sum_{j<i} k_decay[j,d] * x[j]. This looks like we can maintain an accumulator acc[d] = sum_{j<i} k_decay[j,d] * x[j] (for each channel d). Then x[i] = beta[i] * (kg[i] - kg[i] * acc? Wait no. Let me recompute: sum_{j<i} M[i,j] * x[j] = sum_{j<i} sum_d kg[i,d] * k_decay[j,d] * x[j] = sum_d kg[i,d] * (sum_{j<i} k_decay[j,d] * x[j]) = sum_d kg[i,d] * acc[d] where acc[d] = sum_{j<i} k_decay[j,d] * x[j]. Then x[i] = beta[i] * (kg[i] - sum_d kg[i,d] * acc[d]). And acc[d] += k_decay[i,d] * x[i] for next iteration. This is very efficient! For each row i, we compute a dot product kg[i] @ acc, update x[i], then update acc with k_decay[i] * x[i]. This is exactly the "delta rule" forward substitution. We can do it in vector form for all K/V channels simultaneously. So kernel 1 can: 1. Load kg, k_decay, beta, v for chunk. 2. For w (x_w): initialize acc_w = zeros(K). For i=0..BT-1: x_w[i] = beta[i] * (kg[i] - kg[i] * acc_w) where the product is elementwise? Wait no, kg[i] is vector (K), acc_w is vector (K), dot product is scalar. We need: dot = sum_d kg[i,d] * acc_w[d] x_w[i,d] = beta[i] * (kg[i,d] - dot) # same dot for all d acc_w[d] += k_decay[i,d] * x_w[i,d] Wait, this doesn't look right dimensionally. Let me re-derive. Actually, w has shape (BT,K). The forward substitution solves (I - A0) w = beta_col * kg. A0 is (BT,BT), kg is (BT,K). A0 w is (BT,K). Row i: (A0 w)[i,d] = sum_j A0[i,j] * w[j,d] = -beta[i] * sum_{j<i} M[i,j] * w[j,d] = -beta[i] * sum_{j<i} sum_{d'} kg[i,d'] * k_decay[j,d'] * w[j,d] Note the inner sum is over d' (the K dimension), and w[j,d] has its own d. So: = -beta[i] * sum_{d'} kg[i,d'] * (sum_{j<i} k_decay[j,d'] * w[j,d]) Let acc_w[d', d] = sum_{j<i} k_decay[j,d'] * w[j,d]. This is a (K,K) matrix! Not a vector. Then (A0 w)[i,d] = -beta[i] * sum_{d'} kg[i,d'] * acc_w[d',d] So x_w[i,d] = beta[i] * kg[i,d] - (A0 w)[i,d] = beta[i] * (kg[i,d] + sum_{d'} kg[i,d'] * acc_w[d',d]) Hmm, this requires maintaining a KxK accumulator. For K=128, that's 16K fp32 = 64KB. Per block. Plus for u, another KxV accumulator = 64KB. Total 128KB just for accumulators. Too much. Wait, maybe I'm overcomplicating. Let's go back to storing A explicitly. With A in shared memory (16KB), computing w = A @ kg is straightforward and doesn't need KxK accumulator. Alternatively, we can solve for w column by column: for each k in K, maintain acc[k] = sum_{j<i} k_decay[j,k] * w[j,k]. Then x_w[i,k] = beta[i] * (kg[i,k] - kg[i,k]*acc[k])? No, the dot is still over d' for each output channel d. Hmm. Wait, if we solve column-by-column (fixed output channel d), then: x_w[i,d] = beta[i] * (kg[i,d] - sum_{d'} kg[i,d'] * acc_w[d']) where acc_w[d'] = sum_{j<i} k_decay[j,d'] * w[j,d]. But this acc depends on the output channel d (because w[j,d] depends on d). So we still need K values per output channel, i.e., KxK matrix. So column-by-column doesn't reduce storage. OK, storing A is simpler. Let's stick with A in shared memory. For A (64x64 fp32) in shared memory, we need 16KB. For w = A @ kg, we tile over K. For u = A @ v, tile over V. Let me now think about the Triton implementation details. Kernel 1 pseudo-code: ``` def kda_intra_kernel(q, k, v, g, beta, Aqk, w, u, qg, k_decay, g_last, scale, ...): i_n, i_bh = tl.program_id(0), tl.program_id(1) i_b, i_h = i_bh // H, i_bh % H # compute base pointers for chunk q_ptr += (i_b*T*H + i_h)*K + i_n*BT*H*K ... # load g and compute cumsum # g shape (B,T,H,K), stride (T*H*K, H*K, K, 1) # Load g as (BT,K) in tiles of BK g_cum = zeros(BT,K) in shared or registers for i_k in range(K//BK): p_g = tl.make_block_ptr(g, (T,K), (H*K,1), (i_n*BT, i_k*BK), (BT,BK), (1,0)) b_g = tl.load(p_g) # cumsum along BT b_g_cum = tl.cumsum(b_g, axis=0) # Triton has tl.cumsum? Yes. store to g_cum_shared or use directly ``` Triton has `tl.cumsum`. Good. Then compute qg, kg, k_decay: ``` for i_k in range(K//BK): load q,k tiles qg = q * exp(g_cum) kg = k * exp(g_cum) k_decay = k * exp(-g_cum) store qg, kg, k_decay ``` Then compute Aqk and Akk: We need dot products over K. We can accumulate in registers. ``` Aqk_acc = zeros(BT,BT) Akk_acc = zeros(BT,BT) for i_k in range(K//BK): load qg_tile (BT,BK), k_decay_tile (BT,BK), kg_tile (BT,BK) Aqk_acc += tl.dot(qg_tile, tl.trans(k_decay_tile)) Akk_acc += tl.dot(kg_tile, tl.trans(k_decay_tile)) # mask mask_lower = tl.arange(0,BT)[:,None] >= tl.arange(0,BT)[None,:] mask_strict = tl.arange(0,BT)[:,None] > tl.arange(0,BT)[None,:] Aqk = tl.where(mask_lower, Aqk_acc, 0) * scale Akk = tl.where(mask_strict, Akk_acc, 0) ``` Wait, Aqk should be masked lower triangular INCLUDING diagonal. So >=. Akk should be strict lower (c > j). So >. Then solve for A: ``` # Akk[c,j] = M[c,j] * beta[j] for c>j, where M = kg @ k_decay.T # A0 = -Akk (strict lower) # Forward substitution: (I - A0) A = I * beta_col? Wait we want A = (I - A0)^{-1} * diag(beta_col) # So solve (I - A0) X = diag(beta_col). X is (BT,BT). Each column j of X is the solution with RHS beta[j] * e_j. # But then w = A @ kg. We can instead solve (I - A0) x = beta_col * kg for w, and (I - A0) x = beta_col * v for u. ``` OK so we don't need full A. We need to solve two systems with the same matrix (I - A0) but different RHS. We can do forward substitution on both simultaneously if we keep A0 in shared memory. Let's store A0 in shared memory: A0[c,j] = -M[c,j] * beta[c] for c>j, else 0. (Note beta applied to rows.) Actually wait, A0[i,j] = -M[i,j] * beta[i] for i>j. Reference line 65 applies beta[..., None] which is beta[i] for row i (since None adds a dimension at the end, making beta shape (B,H,C,1), and A has shape (B,H,C,C), so beta is broadcast over columns). Yes, beta applied to rows. So A0 = -tril_strict(M) * beta[:,None]. Then solve (I - A0) X = B where B = beta[:,None] * kg for w, and B = beta[:,None] * v for u. Forward substitution: For i = 0..BT-1: X[i] = B[i] + sum_{j<i} A0[i,j] * X[j] = beta[i] * kg[i] + sum_{j<i} (-M[i,j] * beta[i]) * X[j] = beta[i] * (kg[i] - sum_{j<i} M[i,j] * X[j]) And sum_{j<i} M[i,j] * X[j] = sum_d kg[i,d] * sum_{j<i} k_decay[j,d] * X[j]. If X is (BT,K) for w, then sum_{j<i} k_decay[j,d] * X[j,k] is acc[d,k]. We need KxK accumulator. 64KB. Too much. If we process K in tiles of BK, then for each tile k_out of output, we need acc[d, k_out] for all d (K) and k_out in tile (BK). That's K*BK = 128*32 = 4K fp32 = 16KB. Manageable. Alternatively, we can just keep A0 in shared memory (16KB) and compute w = A @ kg via GEMM. That's simpler and uses less register pressure. The solve is O(BT^3) = 262K FLOPs, trivial. So I'll store A0 in shared memory (actually the full A = (I - A0)^{-1} * diag(beta_col)). Let's just compute A explicitly. Algorithm: 1. Compute M = kg @ k_decay.T (BT,BT) in registers. 2. Apply masks and beta to get A0 = -tril_strict(M) * beta_row. Store in shared memory. 3. Solve forward substitution in shared memory to get A_inv = (I - A0)^{-1}. Wait, reference solves for (I - A0)^{-1} and then multiplies by diag(beta_col). Let's store the full A = (I - A0)^{-1} * diag(beta_col) in shared memory. 4. Compute w = A @ kg, u = A @ v via GEMM. Forward substitution in shared memory with variable indexing: ```python # A0 in shared memory as (BT,BT) A = A0 # start with A0 (BT,BT) # Actually we want to compute X = (I - A0)^{-1}. We can do: # Initialize A as A0. # for i in range(1, BT): # for j in range(i): # A[i,j] += sum_{t=0}^{i-1} A[i,t] * A[t,j] # But this is the same as reference. We can implement row by row. ``` In Triton, shared memory 2D array with dynamic row indexing. We can use a 1D shared memory pointer and index manually. Let me allocate `A_s = tl.zeros([BT, BT], dtype=tl.float32)` in shared memory? Actually Triton allows `tl.zeros([BT,BT])` but we need to store/load. Or we can allocate a tensor in shared memory using `make_block_ptr` with shape (BT,BT). Hmm, `make_block_ptr` requires block dimensions that are constexpr. We can use `(BT,BT)` block. But dynamic row indexing with `tl.load(A_ptr + i*BT + tl.arange(0,BT))` works with a 1D pointer. Let me allocate a shared memory pointer: ```python A_s = tl.zeros([BT*BT], dtype=tl.float32) # this creates a local tensor, not shared? In Triton, zeros is in registers? ``` Actually `tl.zeros` creates a tensor in registers (or shared if used with block ptr?). We want shared memory. In Triton, we can use `tl.zeros` and it will be placed in registers or shared memory by the compiler based on use. For a 64x64 fp32 array, shared memory is likely. But for dynamic indexing, we can treat it as a 1D array and use `tl.load(A_s + i*BT + tl.arange(0,BT))`. This should work if A_s is in shared memory. Let me write the solve: ```python # M is (BT,BT) in registers mask_strict = row_idx[:,None] > col_idx[None,:] A0 = tl.where(mask_strict, -M * beta[:,None], 0.0) # Store A0 to shared memory A_s (BT,BT) tl.store(A_s + row_idx[:,None]*BT + col_idx[None,:], A0) # Forward substitution for i in range(1, BT): # Load row i of A row_i = tl.load(A_s + i*BT + col_idx) # shape (BT,) # Load block A[0:i, 0:i] from shared memory? We need sum_t A[i,t] * A[t,j] for j<i. # For each j<i, sum_t A[i,t]*A[t,j]. We can compute this as dot(row_i, A[:,j]) for each j, but only t<i matters. # Simpler: for each j, load column j of A (full BT), dot with row_i, mask t<i. # We can load A[:, :i] as a block? But i is dynamic. # Alternative: compute update = sum_t row_i[t] * A[t, :], then mask j<i. # Since A[t,j]=0 for j>=t, the sum over t effectively only contributes for t>j. For j<i and t up to BT-1, the relevant t are j<t<BT. But row_i[t] is nonzero only for t<i. So t ranges j<t<i. update = tl.zeros(BT) for t in range(i): # unrolled? i is dynamic, loop up to BT row_t = tl.load(A_s + t*BT + col_idx) update += row_i[t] * row_t # Update row i, columns < i new_row_i = tl.where(col_idx < i, row_i + update, row_i) tl.store(A_s + i*BT + col_idx, new_row_i) ``` This has an inner loop over t up to i, total O(BT^3). With BT=64, 64^3/6 ~ 43K iterations. Each iteration loads a row from shared memory. That's 43K * 64 loads = 2.7M loads per block. That seems like a lot. But shared memory is fast. Actually we can do better. The update for row i is row_i @ A[:i, :i] (matrix multiply of 1xi and ixi). We can compute this with a single matrix multiply if we tile. But i is dynamic. Since BT=64 is small, we can just do the naive loop. But 43K shared memory loads per block might be slow. Alternatively, we can compute the entire solve using a matrix multiply approach: A = I + A0 + A0^2 + A0^3 + ... (since A0 is nilpotent, A0^BT = 0) So (I - A0)^{-1} = I + A0 + A0^2 + ... + A0^{BT-1} Then A_final = (I + A0 + A0^2 + ...) * diag(beta_col) We can compute powers of A0 iteratively: P = I Sum = I for p in 1..BT-1: P = A0 @ P (or P @ A0) Sum += P A_final = Sum * diag(beta_col) Each iteration is a 64x64 @ 64x64 matrix multiply = 64^3*2 = 524K FLOPs. 63 iterations = 33M FLOPs per chunk. For 512 chunks (B=2,H=8,N=32), total 17 GFLOPs. This is more FLOPs than the whole problem! Not good. So the power series is too expensive. The forward substitution O(BT^3) = 262K FLOPs is much cheaper than O(BT^4) = 16M. Wait, forward substitution for (I-A0)^{-1} is O(BT^3). The naive loop I wrote is O(BT^3) because inner loop up to i and each update is a vector of length i. Total ~BT^3/3 operations. But in terms of memory access, it's O(BT^3) loads from shared memory. 262K loads of 64 elements? Actually each inner operation is a vector multiply-add of length i. Total memory is sum_i i^2 ~ BT^3/3 = 87K elements read. Wait no, sum_i i^2 = ~87K scalar operations. Each operation reads one element of A[i,t] and one element of A[t,j]. So 174K scalar reads. That's tiny. Actually the loop I wrote: for i in 1..BT, for t in 0..i-1, load row t (BT elements) and multiply by scalar row_i[t]. That's sum_i i * BT = BT * BT*(BT-1)/2 = ~BT^3/2 = 128K loads of BT elements = 8M elements loaded. Hmm, that's more. But we can optimize: update = sum_t row_i[t] * A[t, :]. This is a dot product of a length-i vector with an i x BT matrix. We can compute it by loading A block [0:i, 0:BT] and doing row_i[0:i] @ A[0:i, 0:BT]. In Triton, if A is in shared memory, we can load the block and use tl.dot? But i is dynamic. Actually we can always load the full block A[0:BT, 0:BT] and mask. Since A is strictly lower triangular, the upper parts are zero. So update = row_i @ A where row_i has zeros for t>=i and A has zeros for upper triangle. So we can do: update = tl.sum(row_i[:,None] * A, axis=0) This is a 64x64 elementwise multiply and sum over axis 0. That's 64*64 = 4K ops. Doing this for each i gives 64*4K = 256K ops. Very cheap! And no inner loop. But we need to read the full A matrix each iteration. 64 * 64*64 = 262K elements read from shared memory. Shared memory bandwidth is huge, so this is fine. So: ```python for i in range(1, BT): row_i = tl.load(A_s + i*BT + col_idx) # (BT,) # Compute update = row_i @ A (full A) update = tl.sum(row_i[:,None] * A_full, axis=0) # (BT,) new_row_i = tl.where(col_idx < i, row_i + update, row_i) # Store back A_full = tl.where((row_idx == i)[:,None], new_row_i[None,:], A_full) # Or store to shared memory and reload ``` Wait, A_full is in registers or shared? If A_full is in registers (4096 elements), then each iteration we update it. That's 64 iterations, each reading/writing 4096 elements in registers. Register operations are fast. So we can keep A_full in registers! Then we don't need shared memory for the solve. We just compute A0 in registers, then update in registers. ```python # A0 is (BT,BT) in registers, strictly lower triangular A = A0 for i in range(1, BT): row_i = A[i, :] # extract row i update = tl.sum(row_i[:,None] * A, axis=0) new_row_i = tl.where(col_idx < i, row_i + update, row_i) A = tl.where((row_idx == i)[:,None], new_row_i[None,:], A) # Add identity and scale columns by beta A = tl.where(row_idx[:,None] == col_idx[None,:], A + 1.0, A) A = A * beta[None,:] ``` But extracting row i from a register tensor A in Triton: `A[i, :]` might not support dynamic i. We can use `tl.sum(tl.where(row_idx[:,None] == i, A, 0.0), axis=0)` to extract row i. And updating row i: `A = tl.where((row_idx[:,None] == i), new_row_i[None,:], A)`. This should work. The loop has 64 iterations, each doing 64x64 elementwise ops. Total ~262K ops. Very fast in registers. Great. So kernel 1 keeps A in registers, does solve in registers, then computes w = A @ kg and u = A @ v via tl.dot. But A is 64x64 fp32 in registers. With 128 threads, each thread holds 32 elements = 128 bytes. Fine. Now, w = A @ kg. A is fp32, kg is bf16. tl.dot requires both inputs same dtype? In Triton, we can cast A to bf16. But A might have values that overflow bf16. However, for KDA with small inputs, A values should be moderate. We can accumulate in fp32. Actually, to maintain accuracy, we might want to keep A in fp32 and do the GEMM in fp32. Triton tl.dot with fp32 inputs and fp32 accumulator should work. But fp32 GEMM is slower than bf16 GEMM on Tensor Cores. For accuracy tolerance 0.05, bf16 GEMM for w,u might be OK. But the solve involves subtractions that could lose precision in bf16. We should keep A in fp32. For w = A @ kg, we can cast A to bf16 for tl.dot. Or use fp32 dot (if supported). Actually Triton on Blackwell might not support fp32 dot. Let me check. The standard is bf16 dot with fp32 accumulation. We can cast A to bf16. But A can have values up to ~1/beta? With beta ~ sigmoid(normal) ~ 0.5. A diagonal is beta ~ 0.5. Off-diagonal could be smaller. So A fits in bf16 OK. I'll cast A to bf16 for w,u GEMMs to use tensor cores. Alternatively, we can compute w directly without storing A by using the forward substitution with A in registers and streaming kg/v. But A @ kg is a 64x64 @ 64x128 GEMM, easy. OK. Kernel 1 design: - Grid: (NT, B*H) - Each block loads chunk q,k,v,g,beta. - Computes g_cum. - Computes qg, kg, k_decay. - Computes Aqk (64x64) and M (64x64). - Computes A0, solves in registers to get A. - Computes w = A @ kg, u = A @ v. - Stores Aqk, w, u, qg, k_decay, g_last. Wait, we also need g_last for the inter-chunk recurrence. g_last is g_cum[-1,:] per chunk. Kernel 2 design: - Grid: (NV, B*H) where NV = V/BV - Each block loops over chunks n=0..NT-1. - Maintains S (K,BV) in registers/shared memory. - For each chunk: - Load w (BT,K), u (BT,BV), qg (BT,K), k_decay (BT,K), Aqk (BT,BT), g_last (K). - Compute v_i = u - w @ S - Compute o = qg @ S + Aqk @ v_i - Store o - Update S = (S + k_decay.T @ v_i) * d[:,None] This is sequential per block but parallel over V tiles and B,H. Number of blocks: NV*B*H. For B=2,H=8,V=128,BV=64: NV=2, blocks=32. Good. For B=1,H=4: blocks=8. Acceptable. For K=128, we can split K into tiles of 64 or 32 for the GEMMs. Since S is (128,BV), and we have w @ S (BT,128) @ (128,BV) -> (BT,BV), we can tile K into 64 segments. Actually, FLA's kernel keeps K in 64-segment registers (b_h1, b_h2). We can do similar. In kernel 2, S is (K,BV). We can store it in shared memory as (K,BV) fp32 = 128*64*4 = 32KB. Good. Then w @ S: w is (BT,K). We can load w in tiles of (BT,64), do tl.dot with S[64:,:] in shared memory. Accumulate to get (BT,BV). qg @ S similarly. For k_decay.T @ v_i: k_decay is (BT,K), v_i is (BT,BV). We want (K,BV). We can tile K into 64, load k_decay[:,64:128] and v_i, do tl.dot(k_decay_tile.T, v_i) -> (64,BV), add to S. This is straightforward. Let me now write the code. But I need to be careful about memory layouts and strides. Input layout: reference uses (B,T,H,K) with default contiguous layout (row-major). Strides: - q: (T*H*K, H*K, K, 1) - k: same - v: (T*H*V, H*V, V, 1) - g: (T*H*K, H*K, K, 1) - beta: (T*H, H, 1) actually beta shape (B,T,H), so strides (T*H, H, 1) I'll use these strides in the kernels. Output o shape (B,T,H,V), strides (T*H*V, H*V, V, 1). Intermediate tensors: - Aqk: (B,T,H,BT) fp32, strides (T*H*BT, H*BT, BT, 1) - w: (B,T,H,K) fp32 or bf16 - u: (B,T,H,V) fp32 or bf16 - qg: (B,T,H,K) bf16 - k_decay: (B,T,H,K) bf16 - g_last: (B,H,N,K) fp32 Actually, we can store w,u in bf16 to save memory and bandwidth. But they participate in the recurrence, and errors can accumulate. For tolerance 0.05, bf16 should be OK. But let's use fp32 for safety, or at least for u (v_new) which is critical. We can try bf16 and see if it passes. Actually, qg and k_decay can be bf16. w and u might need fp32. Let me use fp32 for w,u,Aqk,g_last. Memory for intermediates (B=2,T=2048,H=8): - Aqk: 2*2048*8*64*4 = 8.4 MB - w fp32: 2*2048*8*128*4 = 16.8 MB - u fp32: same 16.8 MB - qg bf16: 2*2048*8*128*2 = 8.4 MB - k_decay bf16: 8.4 MB - g_last: 2*8*32*128*4 = 0.26 MB Total: ~59 MB. Fine. OK let me write the kernel. This is going to be long. I'll write it incrementally and test. Actually, before writing everything, let me first write a minimal solution that passes correctness, even if slow, then optimize. The minimal solution could be a pure PyTorch implementation that's reasonably optimized. But PyTorch might not handle the recurrence well. Alternatively, I can write a Triton kernel that's simpler but might be slower. Let me start coding solution.py with the two-kernel Triton approach.
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TaskUpdate
args
{
  "taskId": "1",
  "status": "in_progress"
}
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stdout (22 chars)
Updated task #1 status
assistant
reasoning (682 chars)
Actually, I realize there's a smarter and simpler way to verify correctness first: write a Python/PyTorch reference-like implementation but more efficient, then write the Triton kernel. But the problem wants a custom kernel in solution.py. Let me start writing the Triton kernel. I'll write it in a scratch file first, then move to solution.py. Actually, I'll write directly to solution.py and iterate. Let me create the initial version. Key decisions: 1. Use natural exp (not log2) to match reference. The reference uses torch.exp on fp32. 2. Apply g.cumsum(-2) inside kernel. 3. Compute Aqk, w, u, qg, k_decay in kernel 1. 4. Compute output in kernel 2. Let me write the code.
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Write /workspace/problems/02_kda_cutlass/solution.py
diff (401 lines — TRUNCATED)
--- /dev/null
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -0,0 +1,398 @@
+"""Kimi Delta Attention (KDA) forward (chunk form) via Triton.
+
+Matches reference.py's semantics: inputs are bf16 (g fp32), chunk_size=64,
+no initial/final state, returns bf16 o.
+"""
+from __future__ import annotations
+
+import torch
+import torch.nn as nn
+import triton
+import triton.language as tl
+
+OP_TYPE = "linear_attention"
+SUPPORTED_PRECISIONS = ["bf16"]
+HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
+
+# ---------------------------------------------------------------------------
+# Kernel 1: intra-chunk work.
+#   - apply in-chunk cumsum to g
+#   - compute qg = q * exp(g), kg = k * exp(g), k_decay = k * exp(-g)
+#   - compute Aqk = tril(qg @ k_decay.T) * scale
+#   - compute A = (I - A0)^{-1} * diag(beta)  where A0 = -tril_strict(kg @ k_decay.T) * beta_row
+#   - compute w = A @ kg, u = A @ v
+# Grid: (NT, B*H)
+# ---------------------------------------------------------------------------
+
+
+@triton.jit
+def _kda_intra_kernel(
+    q_ptr,
+    k_ptr,
+    v_ptr,
+    g_ptr,
+    beta_ptr,
+    Aqk_ptr,
+    w_ptr,
+    u_ptr,
+    qg_ptr,
+    kdecay_ptr,
+    glast_ptr,
+    scale,
+    B: tl.constexpr,
+    T: tl.constexpr,
+    H: tl.constexpr,
+    K: tl.constexpr,
+    V: tl.constexpr,
+    BT: tl.constexpr,
+    BK: tl.constexpr,
+    BV: tl.constexpr,
+):
+    i_n = tl.program_id(0)
+    i_bh = tl.program_id(1)
+    i_b = i_bh // H
+    i_h = i_bh % H
+
+    # base offsets
+    bos = i_b * T
+    off_qk = (bos * H + i_h) * K
+    off_v = (bos * H + i_h) * V
+    off_g = (bos * H + i_h) * K
+    off_beta = bos * H + i_h
+    t0 = i_n * BT
+
+    # row/col index helpers
+    r = tl.arange(0, BT)
+    c = tl.arange(0, BT)
+    mask_strict = r[:, None] > c[None, :]
+    mask_lower = r[:, None] >= c[None, :]
+
+    # --- load g and cumsum ---
+    # use shared-like local accumulator; g_cum kept in registers as fp32
+    g_cum = tl.zeros([BT, K], dtype=tl.float32)
+    for i_k in range(tl.cdiv(K, BK)):
+        off_k = i_k * BK
+        o_k = off_k + tl.arange(0, BK)
+        m_k = o_k < K
+        p_g = tl.make_block_ptr(
+            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        b_g = tl.load(p_g, boundary_check=(0, 1), mask=m_k[None, :]).to(tl.float32)
+        b_g_cum = tl.cumsum(b_g, axis=0)
+        # scatter into g_cum
+        for jj in range(BT):
+            g_cum = tl.where(
+                (tl.arange(0, BT)[:, None] == jj) & m_k[None, :],
+                b_g_cum[jj, :][None, :],
+                g_cum,
+            )
+
+    # --- compute qg, kg, k_decay and store them; also accumulate Aqk/Akk ---
+    Aqk_acc = tl.zeros([BT, BT], dtype=tl.float32)
+    M_acc = tl.zeros([BT, BT], dtype=tl.float32)  # kg @ k_decay.T
+
+    for i_k in range(tl.cdiv(K, BK)):
+        off_k = i_k * BK
+        o_k = off_k + tl.arange(0, BK)
+        m_k = o_k < K
+
+        p_q = tl.make_block_ptr(
+            q_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        p_k = tl.make_block_ptr(
+            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32)
+        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+
+        b_g = tl.zeros([BT, BK], dtype=tl.float32)
+        # load the corresponding g_cum tile
+        for jj in range(BT):
+            row_g = g_cum[jj, off_k + tl.arange(0, BK)]
+            b_g = tl.where(
+                (tl.arange(0, BT)[:, None] == jj) & m_k[None, :],
+                row_g[None, :],
+                b_g,
+            )
+
+        b_qg = b_q * tl.exp(b_g)
+        b_kg = b_k * tl.exp(b_g)
+        b_kdecay = b_k * tl.exp(-b_g)
+
+        # store qg / kdecay (bf16)
+        p_qg = tl.make_block_ptr(
+            qg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        p_kd = tl.make_block_ptr(
+            kdecay_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        tl.store(p_qg, b_qg.to(tl.bfloat16), boundary_check=(0, 1))
+        tl.store(p_kd, b_kdecay.to(tl.bfloat16), boundary_check=(0, 1))
+
+        Aqk_acc += tl.dot(b_qg, tl.trans(b_kdecay))
+        M_acc += tl.dot(b_kg, tl.trans(b_kdecay))
+
+    # --- beta row vector ---
+    p_beta = tl.make_block_ptr(
+        beta_ptr, (T,), (H,), (t0,), (BT,), (0,)
+    )
+    b_beta = tl.load(p_beta, boundary_check=(0,)).to(tl.float32)
+
+    Aqk = tl.where(mask_lower, Aqk_acc * scale, 0.0)
+    A0 = tl.where(mask_strict, -M_acc * b_beta[:, None], 0.0)
+
+    # --- solve (I - A0) in registers via forward substitution ---
+    A = A0
+    for i in range(1, BT):
+        row_i = tl.sum(tl.where(r[:, None] == i, A, 0.0), axis=0)
+        update = tl.sum(row_i[:, None] * A, axis=0)
+        new_row = tl.where(c < i, row_i + update, row_i)
+        A = tl.where((r[:, None] == i), new_row[None, :], A)
+    # add identity and scale columns by beta
+    A = tl.where(r[:, None] == c[None, :], A + 1.0, A)
+    A = A * b_beta[None, :]
+
+    # store Aqk
+    p_Aqk = tl.make_block_ptr(
+        Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
+    )
+    tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
+
+    # --- w = A @ kg, u = A @ v ---
+    # Recompute kg on the fly while streaming K/V.
+    for i_k in range(tl.cdiv(K, BK)):
+        off_k = i_k * BK
+        o_k = off_k + tl.arange(0, BK)
+        m_k = o_k < K
+        p_k = tl.make_block_ptr(
+            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+        b_g = tl.zeros([BT, BK], dtype=tl.float32)
+        for jj in range(BT):
+            row_g = g_cum[jj, off_k + tl.arange(0, BK)]
+            b_g = tl.where(
+                (tl.arange(0, BT)[:, None] == jj) & m_k[None, :],
+                row_g[None, :],
+                b_g,
+            )
+        b_kg = b_k

... (7449 more chars)
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "content": "<12966 chars \u2014 see diff>"
}
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stdout (142 chars)
File created successfully at: /workspace/problems/02_kda_cutlass/solution.py (file state is current in your context — no need to Read it back)
assistant
reasoning (815 chars)
I wrote a draft solution. There are likely bugs, especially with the scatter/gather patterns using tl.where. Let me run check.py to see what happens. Actually, before running check.py, let me think about potential compile errors: 1. `g_cum[jj, off_k + tl.arange(0, BK)]` - indexing a 2D tensor with scalar + range might not work. 2. The scatter patterns with `tl.where` and broadcasting are suspicious. Triton doesn't support arbitrary indexing of register tensors very well. We should probably use shared memory for g_cum, qg, kg, kdecay. Let me rewrite kernel 1 to use shared memory for intermediate tensors. But first, let's see if it compiles. If not, I'll refactor. Also, I need to make sure q,k,v,g,beta are contiguous or use proper strides. They should be contiguous by default. Let me run a quick test.
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Bash
args
{
  "command": "uv run python -c \"import solution; import torch; m = solution.Model(2,1024,8,128,128).cuda(); q,k,v,g,beta = [x.cuda() for x in solution.get_inputs()]; o=m(q,k,v,g,beta); print(o.shape, o.dtype, o.abs().max())\" 2>&1 | tail -40",
  "description": "Quick compile/run test of solution"
}
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stdout (2541 chars) — TRUNCATED
Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "/workspace/.venv/lib/python3.11/site-packages/torch/nn/modules/module.py", line 1779, in _wrapped_call_impl
    return self._call_impl(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/torch/nn/modules/module.py", line 1790, in _call_impl
    return forward_call(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/problems/02_kda_cutlass/solution.py", line 359, in forward
    _kda_intra_kernel[grid1](
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 370, in <lambda>
    return lambda *args, **kwargs: self.run(grid=grid, warmup=False, *args, **kwargs)
                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 720, in run
    kernel = self._do_compile(key, signature, device, constexprs, options, attrs, warmup)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 849, in _do_compile
    kernel = self.compile(src, target=target, options=options.__dict__)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 304, in compile
    module = src.make_ir(

... (1041 more chars)
assistant
reasoning (1984 chars)
Block pointer load doesn't take mask argument. The boundary_check handles the mask. Also, K is a multiple of BK so we don't need mask. Let me fix. But the bigger issue is the register indexing. I should refactor to use shared memory for g_cum, qg, kg, kdecay. Let me redesign kernel 1 with shared memory: - Allocate shared memory for g_cum, qg, kg, kdecay? Actually we can use block pointers to load/store these from global memory, no need for shared memory. Wait, the issue is computing g_cum across K tiles. We can compute and store g_cum directly to global memory in a separate kernel, or compute it on the fly. Simpler approach: 1. Precompute g_cum in PyTorch: `g_cum = g.view(B, T//BT, BT, H, K).cumsum(dim=2).view(B, T, H, K)`. This is a simple op. Then pass g_cum to kernels. 2. Precompute qg, kg, kdecay in PyTorch: `qg = q * g_cum.exp()`, etc. Then kernel 1 only needs to: - Compute Aqk = tril(qg @ kdecay.T) * scale - Compute M = kg @ kdecay.T - Solve for A - Compute w = A @ kg, u = A @ v - Store glast = g_cum[-1] This avoids the g_cum computation and indexing issues in Triton. But wait, the problem says we need a real custom kernel. Preprocessing in PyTorch is fine if the heavy work is in the kernel. Actually, reference does all this in Python; we can do the same but fused into kernels. But to get Triton working, let me first do the preprocessing in PyTorch and have a simpler Triton kernel. Then if performance is good, keep it. If not, optimize by fusing. Actually, computing g_cum, qg, kg, kdecay in PyTorch is cheap and vectorized. The heavy work is the GEMMs and solve. So this is fine. But the reference semantics: g = g.cumsum(-2). In our PyTorch preprocessing, g_cum = rearrange(g).cumsum(2).rearrange back. Same. Let me rewrite solution.py to do preprocessing in PyTorch and simpler Triton kernels. Actually, to keep it a "real custom kernel", I should have the bulk of the work in Triton. The preprocessing is a tiny fraction. Let me rewrite.
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assistant
Write /workspace/problems/02_kda_cutlass/solution.py
diff (316 lines — TRUNCATED)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -14,29 +14,17 @@
 SUPPORTED_PRECISIONS = ["bf16"]
 HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
 
-# ---------------------------------------------------------------------------
-# Kernel 1: intra-chunk work.
-#   - apply in-chunk cumsum to g
-#   - compute qg = q * exp(g), kg = k * exp(g), k_decay = k * exp(-g)
-#   - compute Aqk = tril(qg @ k_decay.T) * scale
-#   - compute A = (I - A0)^{-1} * diag(beta)  where A0 = -tril_strict(kg @ k_decay.T) * beta_row
-#   - compute w = A @ kg, u = A @ v
-# Grid: (NT, B*H)
-# ---------------------------------------------------------------------------
-
 
 @triton.jit
 def _kda_intra_kernel(
-    q_ptr,
-    k_ptr,
+    qg_ptr,
+    kg_ptr,
+    kdecay_ptr,
     v_ptr,
-    g_ptr,
     beta_ptr,
     Aqk_ptr,
     w_ptr,
     u_ptr,
-    qg_ptr,
-    kdecay_ptr,
     glast_ptr,
     scale,
     B: tl.constexpr,
@@ -53,86 +41,35 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
-    # base offsets
     bos = i_b * T
-    off_qk = (bos * H + i_h) * K
-    off_v = (bos * H + i_h) * V
-    off_g = (bos * H + i_h) * K
-    off_beta = bos * H + i_h
     t0 = i_n * BT
 
-    # row/col index helpers
     r = tl.arange(0, BT)
     c = tl.arange(0, BT)
     mask_strict = r[:, None] > c[None, :]
     mask_lower = r[:, None] >= c[None, :]
 
-    # --- load g and cumsum ---
-    # use shared-like local accumulator; g_cum kept in registers as fp32
-    g_cum = tl.zeros([BT, K], dtype=tl.float32)
+    # accumulate Aqk and M = kg @ kdecay.T over K tiles
+    Aqk_acc = tl.zeros([BT, BT], dtype=tl.float32)
+    M_acc = tl.zeros([BT, BT], dtype=tl.float32)
+
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
-        o_k = off_k + tl.arange(0, BK)
-        m_k = o_k < K
-        p_g = tl.make_block_ptr(
-            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        b_g = tl.load(p_g, boundary_check=(0, 1), mask=m_k[None, :]).to(tl.float32)
-        b_g_cum = tl.cumsum(b_g, axis=0)
-        # scatter into g_cum
-        for jj in range(BT):
-            g_cum = tl.where(
-                (tl.arange(0, BT)[:, None] == jj) & m_k[None, :],
-                b_g_cum[jj, :][None, :],
-                g_cum,
-            )
-
-    # --- compute qg, kg, k_decay and store them; also accumulate Aqk/Akk ---
-    Aqk_acc = tl.zeros([BT, BT], dtype=tl.float32)
-    M_acc = tl.zeros([BT, BT], dtype=tl.float32)  # kg @ k_decay.T
-
-    for i_k in range(tl.cdiv(K, BK)):
-        off_k = i_k * BK
-        o_k = off_k + tl.arange(0, BK)
-        m_k = o_k < K
-
-        p_q = tl.make_block_ptr(
-            q_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        p_k = tl.make_block_ptr(
-            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32)
-        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
-
-        b_g = tl.zeros([BT, BK], dtype=tl.float32)
-        # load the corresponding g_cum tile
-        for jj in range(BT):
-            row_g = g_cum[jj, off_k + tl.arange(0, BK)]
-            b_g = tl.where(
-                (tl.arange(0, BT)[:, None] == jj) & m_k[None, :],
-                row_g[None, :],
-                b_g,
-            )
-
-        b_qg = b_q * tl.exp(b_g)
-        b_kg = b_k * tl.exp(b_g)
-        b_kdecay = b_k * tl.exp(-b_g)
-
-        # store qg / kdecay (bf16)
         p_qg = tl.make_block_ptr(
             qg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
+        p_kg = tl.make_block_ptr(
+            kg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
         p_kd = tl.make_block_ptr(
             kdecay_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        tl.store(p_qg, b_qg.to(tl.bfloat16), boundary_check=(0, 1))
-        tl.store(p_kd, b_kdecay.to(tl.bfloat16), boundary_check=(0, 1))
-
-        Aqk_acc += tl.dot(b_qg, tl.trans(b_kdecay))
-        M_acc += tl.dot(b_kg, tl.trans(b_kdecay))
-
-    # --- beta row vector ---
+        b_qg = tl.load(p_qg, boundary_check=(0, 1)).to(tl.float32)
+        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.float32)
+        b_kd = tl.load(p_kd, boundary_check=(0, 1)).to(tl.float32)
+        Aqk_acc += tl.dot(b_qg, tl.trans(b_kd))
+        M_acc += tl.dot(b_kg, tl.trans(b_kd))
+
     p_beta = tl.make_block_ptr(
         beta_ptr, (T,), (H,), (t0,), (BT,), (0,)
     )
@@ -141,48 +78,35 @@
     Aqk = tl.where(mask_lower, Aqk_acc * scale, 0.0)
     A0 = tl.where(mask_strict, -M_acc * b_beta[:, None], 0.0)
 
-    # --- solve (I - A0) in registers via forward substitution ---
+    # solve (I - A0) in registers
     A = A0
     for i in range(1, BT):
         row_i = tl.sum(tl.where(r[:, None] == i, A, 0.0), axis=0)
         update = tl.sum(row_i[:, None] * A, axis=0)
         new_row = tl.where(c < i, row_i + update, row_i)
         A = tl.where((r[:, None] == i), new_row[None, :], A)
-    # add identity and scale columns by beta
     A = tl.where(r[:, None] == c[None, :], A + 1.0, A)
     A = A * b_beta[None, :]
 
-    # store Aqk
     p_Aqk = tl.make_block_ptr(
         Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
     )
     tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
 
-    # --- w = A @ kg, u = A @ v ---
-    # Recompute kg on the fly while streaming K/V.
+    # w = A @ kg
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
-        o_k = off_k + tl.arange(0, BK)
-        m_k = o_k < K
-        p_k = tl.make_block_ptr(
-            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
-        b_g = tl.zeros([BT, BK], dtype=tl.float32)
-        for jj in range(BT):
-            row_g = g_cum[jj, off_k + tl.arange(0, BK)]
-            b_g = tl.where(
-                (tl.ara

... (5103 more chars)
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "content": "<9625 chars \u2014 see diff>"
}
in 73449out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (1962 chars)
There are still issues with `S[off_k + kk, :]` indexing a 2D register tensor with scalar. This likely won't compile. Also the scatter patterns are inefficient. Let me think of a better way to handle S. We can store S in shared memory instead of registers, and use block pointers for slices. But Triton block pointers require constexpr offsets. Alternative: keep S as separate segments for each K tile. Since K=128 and BK=64, we have 2 K tiles. We can have S1, S2 as separate [BK,BV] tensors in registers. Then we don't need dynamic indexing. For K=128, BK=64: S1 = S[0:64,:], S2 = S[64:128,:]. In the loop over chunks, we load w[:,0:64] and dot with S1, load w[:,64:128] and dot with S2. Similarly for qg. For kdecay.T @ v_i, compute dS1 and dS2 separately. This is much cleaner and avoids dynamic indexing. K is always 128 in our problem, so we can specialize. But to be general, we can have a loop over K tiles and use an array of S tiles. Triton doesn't support arrays of tensors easily, but we can unroll. Since K=128, BK=64, we have exactly 2 tiles. We can write the kernel assuming K=128, BK=64, BV=64, V=128. But the problem wants K=V=128. Let me check shapes.py - yes all shapes have K=V=128. So we can specialize to K=V=128, BT=64, BK=64, BV=64. But the code might be more robust if we handle general K,V via loops. Let me try the segment approach with an unrolled loop. Actually for K=128, we can just have b_S1 and b_S2. But we also need to update S after kdecay.T @ v_i. We update S1 += dS1, S2 += dS2. Then scale by d: S1 *= d[0:64][:,None], S2 *= d[64:128][:,None]. This is clean. Let me also handle the solve. A is 64x64 in registers. The solve loop uses dynamic row extraction with tl.where. That should work. Let me rewrite kernel 2 with S as two segments. And kernel 1 with simpler layout. Also, for `tl.dot(A.to(tl.bfloat16), b_kg.to(tl.bfloat16))`, if K tile is 64 and BT=64, the dot is 64x64 @ 64x64, which works. Let me rewrite.
in 77722out 0cache_r 0cache_w 0
assistant
Write /workspace/problems/02_kda_cutlass/solution.py
diff (219 lines — TRUNCATED)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -25,7 +25,6 @@
     Aqk_ptr,
     w_ptr,
     u_ptr,
-    glast_ptr,
     scale,
     B: tl.constexpr,
     T: tl.constexpr,
@@ -41,7 +40,6 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
-    bos = i_b * T
     t0 = i_n * BT
 
     r = tl.arange(0, BT)
@@ -49,7 +47,6 @@
     mask_strict = r[:, None] > c[None, :]
     mask_lower = r[:, None] >= c[None, :]
 
-    # accumulate Aqk and M = kg @ kdecay.T over K tiles
     Aqk_acc = tl.zeros([BT, BT], dtype=tl.float32)
     M_acc = tl.zeros([BT, BT], dtype=tl.float32)
 
@@ -93,34 +90,31 @@
     )
     tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
 
-    # w = A @ kg
+    A_bf16 = A.to(tl.bfloat16)
+
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
         p_kg = tl.make_block_ptr(
             kg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.float32)
-        b_w = tl.dot(A.to(tl.bfloat16), b_kg.to(tl.bfloat16))
+        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_w = tl.dot(A_bf16, b_kg)
         p_w = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
         tl.store(p_w, b_w.to(tl.float32), boundary_check=(0, 1))
 
-    # u = A @ v
     for i_v in range(tl.cdiv(V, BV)):
         off_v = i_v * BV
         p_v = tl.make_block_ptr(
             v_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        b_v = tl.load(p_v, boundary_check=(0, 1)).to(tl.float32)
-        b_u = tl.dot(A.to(tl.bfloat16), b_v.to(tl.bfloat16))
+        b_v = tl.load(p_v, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_u = tl.dot(A_bf16, b_v)
         p_u = tl.make_block_ptr(
             u_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
         tl.store(p_u, b_u.to(tl.float32), boundary_check=(0, 1))
-
-    # g_last: last row of kg/kdecay not stored; read from global g_exp/cum instead
-    # We pass glast precomputed from Python.
 
 
 @triton.jit
@@ -146,12 +140,16 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
-    bos = i_b * T
     off_v0 = i_vb * BV
     NT = T // BT
 
-    # S as contiguous fp32 local tensor (K,BV)
-    S = tl.zeros([K, BV], dtype=tl.float32)
+    # S split into K tiles of shape (BK, BV)
+    # We keep them as separate register tensors to avoid dynamic indexing.
+    num_k_tiles = tl.cdiv(K, BK)
+    # Triton does not support arrays of tensors; unroll manually via repeated code.
+    # For K=128, BK=64 we have two tiles.
+    S0 = tl.zeros([BK, BV], dtype=tl.float32)
+    S1 = tl.zeros([BK, BV], dtype=tl.float32)
 
     for i_n in range(NT):
         t0 = i_n * BT
@@ -162,43 +160,36 @@
         b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
 
         b_v = b_u
-        for i_k in range(tl.cdiv(K, BK)):
-            off_k = i_k * BK
-            p_w = tl.make_block_ptr(
-                w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-            )
-            b_w = tl.load(p_w, boundary_check=(0, 1)).to(tl.float32)
-            # slice S[off_k:off_k+BK, :]
-            b_S = tl.zeros([BK, BV], dtype=tl.float32)
-            for kk in range(BK):
-                b_S = tl.where(
-                    tl.arange(0, BK)[:, None] == kk,
-                    S[off_k + kk, :][None, :],
-                    b_S,
-                )
-            b_v -= tl.dot(b_w, b_S)
+        # w @ S, K tile 0
+        p_w0 = tl.make_block_ptr(
+            w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
+        )
+        b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))
+        # K tile 1
+        p_w1 = tl.make_block_ptr(
+            w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
+        )
+        b_w1 = tl.load(p_w1, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))
 
         b_o = tl.zeros([BT, BV], dtype=tl.float32)
-        for i_k in range(tl.cdiv(K, BK)):
-            off_k = i_k * BK
-            p_qg = tl.make_block_ptr(
-                qg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-            )
-            b_qg = tl.load(p_qg, boundary_check=(0, 1)).to(tl.float32)
-            b_S = tl.zeros([BK, BV], dtype=tl.float32)
-            for kk in range(BK):
-                b_S = tl.where(
-                    tl.arange(0, BK)[:, None] == kk,
-                    S[off_k + kk, :][None, :],
-                    b_S,
-                )
-            b_o += tl.dot(b_qg, b_S)
+        p_qg0 = tl.make_block_ptr(
+            qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
+        )
+        b_qg0 = tl.load(p_qg0, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))
+        p_qg1 = tl.make_block_ptr(
+            qg_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
+        )
+        b_qg1 = tl.load(p_qg1, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_o += tl.dot(b_qg1, S1.to(tl.bfloat16))
 
         p_Aqk = tl.make_block_ptr(
             Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
         )
-        b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.float32)
-        b_o += tl.dot(b_Aqk, b_v)
+        b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.bfloat16)
+        b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))
 
         p_o = tl.make_block_ptr(
             o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
@@ -206,30 +197,25 @@
         tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))
 
         # S += kdecay.T @ v_i
-        for i_k in range(tl.cdiv(K, BK)):
-            off_k = i_k * BK
-            p_kd = tl.make_block_ptr(
-                kdecay_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-            )
-            b_kd = tl.load(p_kd, boundary_check=(0, 1)).to(tl.float32)
-            dS = tl.d

... (1838 more chars)
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "content": "<9217 chars \u2014 see diff>"
}
in 77722out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (143 chars)
The kernel still has `d[:BK]` and `d[BK:]` which might not work in Triton. Also `S0.to(tl.bfloat16)` for dot might be OK. Let me test compile.
in 82055out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python -c \"import solution; import torch; m = solution.Model(2,1024,8,128,128).cuda(); q,k,v,g,beta = [x.cuda() for x in solution.get_inputs()]; o=m(q,k,v,g,beta); print(o.shape, o.dtype, o.abs().max())\" 2>&1 | tail -60",
  "description": "Test compile and run"
}
in 82055out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (3088 chars) — TRUNCATED
Traceback (most recent call last):
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/core.py", line 43, in wrapper
    return fn(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/core.py", line 1101, in __getitem__
    raise ValueError(f"unsupported tensor index: {sl}")
ValueError: unsupported tensor index: <triton.language.core.slice object at 0x783edc0737d0>

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "/workspace/.venv/lib/python3.11/site-packages/torch/nn/modules/module.py", line 1779, in _wrapped_call_impl
    return self._call_impl(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/torch/nn/modules/module.py", line 1790, in _call_impl
    return forward_call(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/problems/02_kda_cutlass/solution.py", line 269, in forward
    _kda_inter_kernel[grid2](
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 370, in <lambda>
    return lambda *args, **kwargs: self.run(grid=grid, warmup=False, *args, **kwargs)
                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 720, in run
    kernel = self._do_compile(key, 

... (1588 more chars)
assistant
reasoning (1901 chars)
Slicing `d[:BK]` doesn't work. Need to use `tl.arange(0, BK)` to index. Also, the dot `tl.dot(tl.trans(b_kd0), b_v.to(tl.bfloat16))` requires both inputs same dtype. b_kd0 is bf16, b_v is bf16. OK. For d0 and d1, use: ```python d0 = tl.load(glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)) d1 = tl.load(glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)) ``` Also `S0.to(tl.bfloat16)` for dot with b_w0: b_w0 is bf16, S0 is fp32. Casting S0 to bf16 for dot is OK but loses precision. Maybe cast b_w0 to fp32 and use tl.dot with fp32? Triton might support fp32 dot for 64x64? Actually tensor cores prefer bf16/fp16. But for S in fp32, we want to keep accumulation in fp32. We can cast b_w0 to fp32, dot in fp32, then result is fp32. But tl.dot with fp32 inputs may not use tensor cores and could be slow. Alternatively, keep S in fp32, cast to bf16 for dot, accumulate in fp32? tl.dot returns fp32 accumulator if inputs are bf16. So `tl.dot(b_w0, S0.to(tl.bfloat16))` returns fp32. Good. The accumulation in fp32 is preserved, only S values are rounded to bf16 during the multiply. That might be OK for tolerance 0.05. But to be safe, we can try fp32 dot by casting both to fp32. Let's see if Triton supports it. If not, fall back to bf16. Actually, Triton tl.dot supports fp32 on some arch. On Blackwell, it should support bf16/fp16 well. fp32 dot might fall back to SIMD and be slow. For accuracy, let's try fp32 first and see if it passes. But the heavy GEMM in w,u is A @ kg where A is bf16 and kg is bf16. That uses tensor cores. The recurrence GEMMs (w @ S, qg @ S, kdecay.T @ v) are 64x64 @ 64x64, small but many. Using bf16 tensor cores is fine. For S scaling by d, S0 *= d0[:,None] works. Let me fix the d0/d1 indexing and try again. Also, I realized the `b_v -= tl.dot(...)` pattern: b_v is fp32, dot returns fp32. Good. Let me fix and re-run.
in 83024out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (16 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -209,11 +209,8 @@
         S1 += tl.dot(tl.trans(b_kd1), b_v.to(tl.bfloat16))
 
         # S *= exp(g_last) per k
-        o_k = tl.arange(0, K)
-        p_glast = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + o_k
-        d = tl.exp(tl.load(p_glast))
-        d0 = d[:BK]
-        d1 = d[BK:]
+        d0 = tl.exp(tl.load(glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)))
+        d1 = tl.exp(tl.load(glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)))
         S0 *= d0[:, None]
         S1 *= d1[:, None]
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<262 chars \u2014 see diff>",
  "new_string": "<281 chars \u2014 see diff>"
}
in 83024out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (18 chars)
Let me test again.
in 83939out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python -c \"import solution; import torch; m = solution.Model(2,1024,8,128,128).cuda(); q,k,v,g,beta = [x.cuda() for x in solution.get_inputs()]; o=m(q,k,v,g,beta); print(o.shape, o.dtype, o.abs().max())\" 2>&1 | tail -60",
  "description": "Test compile and run"
}
in 83939out 0cache_r 0cache_w 0
tool
stdout (1708 chars) — TRUNCATED
Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "/workspace/.venv/lib/python3.11/site-packages/torch/nn/modules/module.py", line 1779, in _wrapped_call_impl
    return self._call_impl(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/torch/nn/modules/module.py", line 1790, in _call_impl
    return forward_call(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/problems/02_kda_cutlass/solution.py", line 266, in forward
    _kda_inter_kernel[grid2](
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 370, in <lambda>
    return lambda *args, **kwargs: self.run(grid=grid, warmup=False, *args, **kwargs)
                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 743, in run
    launch_metadata = kernel.launch_metadata(grid, stream, *bound_args.values())
                      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 482, in launch_metadata
    self._init_handles()
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 456, in _init_handles
    raise_(OutOfResources(self.metadata.shared, max_shared, "shared memory"))
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compile

... (208 more chars)
assistant
reasoning (1192 chars)
Shared memory overflow. We're using too much shared memory. The main culprit is likely the register tensors being spilled to shared memory. In particular, A (64x64 fp32 = 16KB), plus b_qg, b_kg, b_kd, b_Aqk, etc. Triton might allocate a lot of shared memory. To reduce shared memory, I can add `num_stages=1` and reduce block size. But the issue is the register arrays. Actually, I think the problem is the way Triton is allocating shared memory for block pointers and intermediate tensors. With BT=64, BK=64, BV=64, the block tiles are large. Let me reduce BK and BV to 32 to see if it compiles. But for K=V=128, that means 4 K tiles and 4 V tiles. More loops but less shared memory per tile. Also, in kernel 2, S0 and S1 are 64x64 fp32 = 16KB each. Plus other tiles. With BK=32, S tiles would be 32x64 = 8KB each, 4 of them = 32KB. Plus other buffers. Let me try BK=BV=32. But first, let me understand the shared memory calculation. With block size 1 (one block per program), and block dimensions (BT,BK) = (64,64), the shared memory for block loads might be 64*64*2 bytes * 3 = 24KB. Plus A 16KB, plus S tiles. Triton might allocate more than needed. Let me reduce BK=BV=32 and see.
in 84477out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (25 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -257,17 +257,17 @@
             qg, kg, kdecay, v, beta, Aqk, w, u,
             1.0,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=min(64, triton.next_power_of_2(K)),
-            BV=min(64, triton.next_power_of_2(V)),
+            BK=32,
+            BV=32,
         )
 
         o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
-        grid2 = (triton.cdiv(V, 64), B * H)
+        grid2 = (triton.cdiv(V, 32), B * H)
         _kda_inter_kernel[grid2](
             w, u, qg, kdecay, Aqk, glast, o,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=min(64, triton.next_power_of_2(K)),
-            BV=min(64, triton.next_power_of_2(V)),
+            BK=32,
+            BV=32,
         )
         return o
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<635 chars \u2014 see diff>",
  "new_string": "<507 chars \u2014 see diff>"
}
in 84477out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (3788 chars)
Wait, if BK=32, then in kernel 2 we need 4 S tiles (K=128/32=4). Currently kernel 2 hardcodes S0, S1 for K=128/BK=64. With BK=32, we need S0,S1,S2,S3. I need to update kernel 2. But before doing that, let me check if kernel 1 compiles with BK=32. Actually the kernel is generic for num_k_tiles, but I hardcoded only 2 in kernel 2. Let me rewrite kernel 2 to handle multiple K tiles properly. Since Triton doesn't support arrays of tensors, I need to unroll. For K=128 and BK=32, 4 tiles. For BK=64, 2 tiles. I can write the kernel with BK=64 and V=128,BV=64 fixed, which gives 2 K tiles and 2 V tiles. The shared memory issue might be from other places. Actually, the shared memory overflow with BK=BV=64 might be because Triton allocated too much for block loads. Let me try num_stages=1 explicitly, or use smaller tiles but handle the K tiles. Let me rewrite kernel 2 with generic K tile handling using a Python loop to generate code? No, Triton kernel is compiled; we can't use Python loops inside. Alternative: use shared memory for S and access slices via block pointers. In Triton, we can allocate a shared memory buffer and use make_block_ptr with dynamic offsets? Offsets can be variables, so we can loop over K tiles and load S slices from shared memory. Actually, we can store S in shared memory as a (K,BV) fp32 tensor. Then in the loop over K tiles, we load slices from shared memory using block_ptr. But Triton doesn't have a direct way to allocate shared memory with a name. We can use `tl.zeros([K,BV])` and it will be placed in shared memory if used across the kernel. Or we can use a 1D tensor. Actually, we can use a local tensor `S = tl.zeros([K, BV], dtype=tl.float32)`. If we index it with `S[off_k + tl.arange(0,BK), :]` it might not work because of dynamic first index. But we can access rows with `tl.sum(tl.where(tl.arange(0,K)[:,None] == (off_k+kk), S, 0), axis=0)` to extract one row. This is expensive. Alternatively, keep S in shared memory and use `tl.load(S_ptr + off_k*BV + row*BV + col)` but S_ptr would need to be a shared memory pointer. Hmm. Let me think differently. We can store S as a 1D tensor `S = tl.zeros([K*BV], dtype=tl.float32)` in shared memory. Then for each K tile, load a (BK,BV) block: ```python p_S = tl.make_block_ptr(S, (K, BV), (BV, 1), (off_k, 0), (BK, BV), (1, 0)) b_S = tl.load(p_S) ``` Does `make_block_ptr` work on a local tensor? I think it only works on global memory pointers. Let me check. Actually, Triton 3.x supports `make_block_ptr` on tensors in shared memory? I'm not sure. Usually block pointers are for global memory. Alternative: don't use block ptr for S, use manual indexing: ```python b_S = tl.zeros([BK, BV], dtype=tl.float32) for kk in range(BK): for vv in range(BV): b_S = tl.where((tl.arange(0,BK)[:,None] == kk) & (tl.arange(0,BV)[None,:] == vv), S[(off_k + kk) * BV + vv], b_S) ``` This is O(BK*BV) operations per load, too expensive. Another approach: use `tl.reshape` and `tl.trans` with masks to slice. But dynamic slicing is hard. Maybe the cleanest is to unroll for the known K=128 and choose BK=64 (2 tiles). Then we only need S0, S1. The shared memory overflow with BK=64 might be solved by reducing num_stages or other parameters. Let me try keeping BK=64,BV=64 but add `num_stages=1` to the kernel launch to reduce shared memory. Also maybe reduce warps. Actually, Triton kernels without specifying num_warps default to 4. With block size (BT,BK)=(64,64), 4 warps = 128 threads. Shared memory for pipeline stages might be the issue. Let me try `_kda_intra_kernel[grid1](..., num_stages=1, num_warps=4)` and `_kda_inter_kernel[grid2](..., num_stages=1, num_warps=4)`. But I already changed to BK=32. Let me revert to BK=64 and add num_stages=1.
in 85362out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (25 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -3,8 +3,10 @@
             qg, kg, kdecay, v, beta, Aqk, w, u,
             1.0,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=32,
-            BV=32,
+            BK=64,
+            BV=64,
+            num_stages=1,
+            num_warps=4,
         )
 
         o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
@@ -12,6 +14,8 @@
         _kda_inter_kernel[grid2](
             w, u, qg, kdecay, Aqk, glast, o,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=32,
-            BV=32,
+            BK=64,
+            BV=64,
+            num_stages=1,
+            num_warps=4,
         )
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<507 chars \u2014 see diff>",
  "new_string": "<609 chars \u2014 see diff>"
}
in 85362out 0cache_r 0cache_w 0
tool
stderr (585 chars)
<tool_use_error>String to replace not found in file.
String:         grid1 = (NT, B * H)
        _kda_intra_kernel[grid1](
            qg, kg, kdecay, v, beta, Aqk, w, u,
            1.0,
            B=B, T=T, H=H, K=K, V=V, BT=BT,
            BK=32,
            BV=32,
        )

        o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
        grid2 = (triton.cdiv(V, 64), B * H)
        _kda_inter_kernel[grid2](
            w, u, qg, kdecay, Aqk, glast, o,
            B=B, T=T, H=H, K=K, V=V, BT=BT,
            BK=32,
            BV=32,
        )</tool_use_error>
assistant
reasoning (53 chars)
Let me read the current file to see what was changed.
in 87144out 0cache_r 0cache_w 0
assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 100,
  "offset": 220
}
in 87144out 0cache_r 0cache_w 0
tool
stdout (2460 chars) — TRUNCATED
220	        super().__init__()
221	        self.B, self.T, self.H, self.K, self.V = B, T, H, K, V
222	        self.chunk_size = chunk_size
223	        self.scale = float(K) ** -0.5
224	        self.register_buffer("_dummy", torch.zeros(1), persistent=False)
225	
226	    def forward(
227	        self,
228	        q: torch.Tensor,
229	        k: torch.Tensor,
230	        v: torch.Tensor,
231	        g: torch.Tensor,
232	        beta: torch.Tensor,
233	    ) -> torch.Tensor:
234	        B, T, H, K = q.shape
235	        V = v.shape[-1]
236	        BT = self.chunk_size
237	        assert T % BT == 0
238	        NT = T // BT
239	        device = q.device
240	
241	        g_4d = g.view(B, NT, BT, H, K)
242	        g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
243	
244	        q_scaled = q * self.scale
245	        qg = q_scaled * g_cum.exp()
246	        kg = k * g_cum.exp()
247	        kdecay = k * (-g_cum).exp()
248	
249	        glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()
250	
251	        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.float32)
252	        w = torch.empty(B, T, H, K, device=device, dtype=torch.float32)
253	        u = torch.empty(B, T, H, V, device=device, dtype=torch.float32)
254	
255	        grid1 = (NT, B * H)
256	        _kda_intra_kernel[grid1](
257	            qg, kg, kdecay, v, beta, Aqk, w, u,
258	            1.0,
259	            B=B, T=T, H=H, K=K, V=V, BT=BT,
260	            BK=32,
261	            BV=32,
262

... (960 more chars)
assistant
reasoning (1950 chars)
I need to update kernel 2 to handle BK=32 (4 K tiles). But that's a lot of hardcoding. Let me think of a better approach. Actually, I realize I can use a different strategy for kernel 2: instead of keeping S in registers, I can keep it in global memory and load/store each chunk. But then the recurrence would require each block to read/write S per chunk, which is slow. Or, I can keep S in registers but use `tl.load` with a 1D pointer and gather the needed rows. For a GEMM w @ S where w is (BT,BK) and S is (BK,BV), we need S as a block. If S is a 1D tensor of length K*BV in registers, we can load a (BK,BV) slice by using `tl.reshape` + `tl.trans` + masking? Not straightforward. Another idea: process only one K tile at a time. For w @ S, we iterate over K tiles, load w tile and S tile, accumulate. The S tile is BKxBV = 32x64 = 2K fp32 = 8KB. With 4 tiles, we need 4 such tensors. Hardcoded. Actually, we can avoid storing all S tiles by recomputing? No, S is updated each chunk. Let me just hardcode for K=128 and BK=32: S0,S1,S2,S3. And for V=128 and BV=32: 4 V tiles. The block would process one V tile and all K tiles. This is manageable. But the code becomes long. Maybe I should use BK=64 and hardcode 2 K tiles, and accept that we need num_stages=1 or smaller V tile. Let me try BK=64,BV=32. Then S has 2 tiles (S0,S1), and V is processed in 4 tiles (i_vb 0..3). The shared memory for S tiles is 2*64*32*4 = 16KB. Plus other buffers. Might fit. Actually, the shared memory overflow earlier was with BV=64. With BV=32, the b_u, b_v, b_o tiles are 64x32 = 8KB each. Much less. Let me try BK=64, BV=32. Grid2 = V/BV = 128/32 = 4. But I also need to check the inter kernel with S0,S1 (BK=64 each). The code currently has S0,S1. Good. Also, kernel 1 with BK=64,BV=32: w uses K tiles of 64 (2 tiles), u uses V tiles of 32 (4 tiles). A is 64x64. Should be OK. Let me update the launch with BK=64, BV=32, num_stages=1, num_warps=4.
in 88190out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (26 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -257,8 +257,10 @@
             qg, kg, kdecay, v, beta, Aqk, w, u,
             1.0,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=32,
+            BK=64,
             BV=32,
+            num_stages=1,
+            num_warps=4,
         )
 
         o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
@@ -266,8 +268,10 @@
         _kda_inter_kernel[grid2](
             w, u, qg, kdecay, Aqk, glast, o,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=32,
+            BK=64,
             BV=32,
+            num_stages=1,
+            num_warps=4,
         )
         return o
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<507 chars \u2014 see diff>",
  "new_string": "<609 chars \u2014 see diff>"
}
in 88190out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (17 chars)
Now test compile.
in 89333out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python -c \"import solution; import torch; m = solution.Model(2,1024,8,128,128).cuda(); q,k,v,g,beta = [x.cuda() for x in solution.get_inputs()]; o=m(q,k,v,g,beta); print(o.shape, o.dtype, o.abs().max())\" 2>&1 | tail -60",
  "description": "Test compile and run"
}
in 89333out 0cache_r 0cache_w 0
tool
stdout (98 chars)
torch.Size([2, 1024, 8, 128]) torch.bfloat16 tensor(0.0376, device='cuda:0', dtype=torch.bfloat16)
assistant
reasoning (50 chars)
It compiled and ran. Now let me check correctness.
in 89498out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run correctness check"
}
in 89498out 0cache_r 0cache_w 0
tool
stdout (125 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case small_qkv: solution contains NaN
assistant
reasoning (451 chars)
NaN. This could be due to numerical instability in exp(g) where g is large, or the solve producing invalid values, or division by zero. The reference g values are small (after cumsum, could grow). For T=1024, cumsum of values ~ N(0, 0.1) over 64 steps -> std = 0.1*sqrt(64) = 0.8. exp(0.8) ~ 2.2, exp(-0.8) ~ 0.45. Reasonable. But the test uses `numeric_stress_cases` which might include `small_qkv` scaling inputs. Let me check what cases are used.
in 89621out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "find /workspace -path '*src/eval/numeric_stress.py' -type f -exec cat {} \\; | head -120",
  "description": "Check numeric stress cases"
}
in 89621out 0cache_r 0cache_w 0
tool
stdout (4021 chars) — TRUNCATED
"""Numeric distribution stress cases for correctness checks.

These cases are not hidden shapes. They rerun the same shape/seed validation
under a few scale regimes that catch kernels that only work for the nominal
N(0, 1)-ish inputs.
"""
from __future__ import annotations

import os
from collections.abc import Iterator, Mapping, Sequence
from contextlib import contextmanager
from dataclasses import dataclass, field

import torch

ToleranceOverride = Mapping[str, float | Mapping[str, float]]


@dataclass(frozen=True)
class NumericStressCase:
    name: str
    input_scales: Mapping[int, float] = field(default_factory=dict)
    state_scales: Mapping[str, float] = field(default_factory=dict)
    tolerance: ToleranceOverride | None = None


NOMINAL_CASE = NumericStressCase("nominal")

_SMALL_BF16 = {"bfloat16": {"atol": 1e-4, "rtol": 5e-2}}
_MED_BF16 = {"bfloat16": {"atol": 5e-4, "rtol": 5e-2}}
_LARGE_BF16 = {"bfloat16": {"atol": 2e-1, "rtol": 5e-2}}
_TINY_FP32 = {"float32": {"atol": 1e-7, "rtol": 1e-4}}
_LARGE_FP32 = {"float32": {"atol": 1e-1, "rtol": 1e-4}}

_CASES: dict[str, tuple[NumericStressCase, ...]] = {
    "01_fp8_gemm": (
        NumericStressCase("small_input", input_scales={0: 1e-3}, tolerance=_SMALL_BF16),
        NumericStressCase("large_input", input_scales={0: 64.0}, tolerance=_LARGE_BF16),
        NumericStressCase(
            "small_weight",
            state_scales={"weight": 1e-2},
            tolerance=_SMALL_BF16,
        ),
    ),
    "02_kda_cutlass": (


... (2521 more chars)
assistant
reasoning (3253 chars)
For small_qkv, inputs are scaled by 1e-2. With small q,k,v, g values after cumsum could be ~0.8, exp(-g) ~ 0.45, exp(g) ~ 2.2. The delta rule might produce A values that cause NaN if the matrix (I - A0) is singular or if beta is very small. Wait, beta is sigmoid(randn), so beta in (0,1). With small q,k, M = kg @ kdecay.T is small. A0 is small. I - A0 is well-conditioned. Shouldn't produce NaN. Maybe the issue is in `A = A * b_beta[None, :]` where if beta is 0, diagonal becomes 0 and I - A0 might be singular? But beta is sigmoid, never 0. Let me check the NaN location. Maybe it's in the solve: the update `new_row = row_i + update` could amplify if A0 has eigenvalues near 1. With small inputs, A0 is tiny. Actually, the issue might be `tl.dot(A.to(tl.bfloat16), b_kg.to(tl.bfloat16))`. If A has values that overflow bf16, it becomes inf. With small inputs, A should be small. Let me debug by printing max values of intermediates. But also, the small_qkv case scales q,k,v by 1e-2. g is not scaled (it's index 3, not in input_scales). So g stays the same. Then exp(g) is normal, but q*k is tiny. Aqk and A are tiny. No overflow. Wait, maybe NaN comes from `0 * inf` or similar in the exp. Let me check. Actually, I think the issue might be the solve loop. When row_i is extracted using `tl.where(r[:,None] == i, A, 0.0)`, for the first iteration i=1, row_i is row 1 of A0. Then `update = tl.sum(row_i[:,None] * A, axis=0)`. A includes row 0. This computes row_1 @ A which is a length-BT vector. We update row 1 columns < 1. That's correct. But the issue: A0 has zeros on and above diagonal. For i=1, row_1 has nonzero only at column 0. update = A[1,0] * A[0,:]. A[0,:] has only A[0,0] nonzero (since A0 is strict lower). So update[0] = A[1,0]*A[0,0]. We update A[1,0] += update[0]. Correct. For i=2, row_2 has nonzero at columns 0,1. update = A[2,0]*A[0,:] + A[2,1]*A[1,:]. Since A[0,:] has A[0,0] updated (from add identity? Wait we add identity AFTER solve). During solve, A is A0 (no identity). A[0,:] has only A[0,0]=0 (since strict lower). So update for j=0 is 0. A[1,:] has A[1,0] updated. So update[1] = A[2,1]*A[1,0]. We update A[2,1] += update[1]. A[2,0] stays A[2,0]. Correct. Then after solve, we add identity: A[i,i] = 1. Then scale columns by beta. This matches reference. So solve is correct. Where could NaN come from? Let me add debug prints in forward to check intermediates for NaN. Actually, maybe the issue is in the Python preprocessing: `qg = q_scaled * g_cum.exp()`. With g_cum up to ~0.8, exp fine. With small qkv, q_scaled is small, qg is small. No NaN. `kg = k * g_cum.exp()`: k small, exp fine. `kdecay = k * (-g_cum).exp()`: fine. In kernel 1, Aqk and M are computed from qg, kg, kdecay. Small values. A0 small. Solve produces A ~ I * beta + small off-diagonal. Then w = A @ kg, u = A @ v. Small values. In kernel 2, S starts at 0. v_i = u_i - w_i @ S = u_i (small). o_i = qg @ S + Aqk @ v_i = Aqk @ v_i (small). S update = (S + kdecay.T @ v_i) * d = small. So everything should be small and finite. NaN is strange. Maybe the NaN is from `tl.dot` with bf16 inputs when values are subnormal/underflow? No, that gives 0. Let me run with nominal case first (disable numeric stress) to see if it passes.
in 91054out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "KBH_NUMERIC_STRESS=0 uv run python check.py 2>&1 | tail -60",
  "description": "Run check without numeric stress"
}
in 91054out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (124 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 123 case nominal: solution contains NaN
assistant
reasoning (129 chars)
NaN even in nominal case. Let me debug by comparing with reference and finding NaN locations. Let me write a quick debug script.
in 92247out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<1100 chars \u2014 see below>",
  "description": "Debug NaN issue"
}
command (1100 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(123)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]

ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
with torch.no_grad():
    ro = ref(q,k,v,g,beta)
    so = sol(q,k,v,g,beta)

print('ref has nan', torch.isnan(ro).any().item(), 'max', ro.abs().max().item())
print('sol has nan', torch.isnan(so).any().item(), 'max', so.abs().max().item())
print('sol inf', torch.isinf(so).any().item())

# Check intermediates
sol_g_4d = g.view(B, T//CS, CS, H, K)
sol_g_cum = sol_g_4d.cumsum(dim=2).view(B, T, H, K)
print('g_cum max', sol_g_cum.max().item(), 'min', sol_g_cum.min().item())
print('exp(g) max', sol_g_cum.exp().max().item())
print('exp(-g) max', (-sol_g_cum).exp().max().item())
PY
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stdout (194 chars)
ref has nan False max 0.01171875
sol has nan False max 0.03759765625
sol inf False
g_cum max 0.9883767366409302 min -6.218692302703857
exp(g) max 2.6868696212768555
exp(-g) max 502.0462646484375
assistant
reasoning (1400 chars)
Wait, no NaN in this run! But check.py reported NaN. Maybe check.py uses a different seed or scales? It said seed 123 case nominal. Let me check what's different. Oh, maybe the issue is with the state_dict load. In check.py, it loads ref_model.state_dict() into sol_model. Both have only `_dummy`. That should be fine. Wait, but my debug script manually set module-level variables and didn't load state_dict. Let me mimic check.py exactly. Also, the output max is 0.037 vs ref 0.0117. That's a big difference. So either my kernel is wrong or reference is different. Actually, reference applies q = q * scale inside. I did q_scaled = q * self.scale in Python. That should match. But the max difference suggests my kernel is numerically wrong, not just tolerance. Let me check the diff. Actually, wait - maybe check.py sets solution's module-level variables too? No, check.py only sets reference's variables. solution.py uses the variables from its own module. But the Model init receives B,T,H,K,V from reference.get_init_inputs(). So it's fine. Let me run check.py with more output to see exactly where NaN is. Actually, maybe my debug run didn't trigger NaN because of warmup/cache? Let me run check.py again and also print first few outputs. Also, the output max 0.037 vs 0.011 suggests my kernel might be producing wrong values (maybe due to solve error). Let me compute max absolute diff.
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args
{
  "command": "<994 chars \u2014 see below>",
  "description": "Debug NaN and diff"
}
command (994 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(123)
torch.cuda.manual_seed_all(123)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]

ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sd = ref.state_dict()
sol.load_state_dict(sd, strict=True)
with torch.no_grad():
    ro = ref(q,k,v,g,beta)
    so = sol(q,k,v,g,beta)

print('ref nan', torch.isnan(ro).any().item(), 'max', ro.abs().max().item())
print('sol nan', torch.isnan(so).any().item(), 'max', so.abs().max().item())
print('sol inf', torch.isinf(so).any().item())
print('max abs diff', (ro-so).abs().max().item())
print('mean abs diff', (ro-so).abs().mean().item())
PY
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stdout (128 chars)
ref nan False max 0.01171875
sol nan False max 0.03125
sol inf False
max abs diff 0.0277099609375
mean abs diff 0.00140380859375
assistant
reasoning (709 chars)
No NaN and diff is within 0.05. So why does check.py report NaN? Let me run check.py and see the exact failure. Maybe it's a different seed or numeric stress case. Wait, check.py uses `reference.get_inputs()` which uses module-level B,T,etc. I set solution's module-level variables but maybe check.py doesn't set solution's. Actually, Model receives init args, so module-level doesn't matter for Model. But get_inputs uses module-level. check.py only overrides reference's module-level variables. It calls reference.get_inputs(). Then passes to both ref_model and sol_model. So both use reference's inputs. Fine. Maybe the NaN is from the first run (seed 42) and my debug used seed 123. Let me test seed 42.
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args
{
  "command": "<919 chars \u2014 see below>",
  "description": "Check all seeds"
}
command (919 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

def check(seed):
    torch.manual_seed(seed)
    torch.cuda.manual_seed_all(seed)
    q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
    ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
    sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
    sol.load_state_dict(ref.state_dict(), strict=True)
    with torch.no_grad():
        ro = ref(q,k,v,g,beta)
        so = sol(q,k,v,g,beta)
    print(f'seed {seed}: ref nan', torch.isnan(ro).any().item(), 'sol nan', torch.isnan(so).any().item(), 'max diff', (ro-so).abs().max().item())

for s in [42,123,456]:
    check(s)
PY
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stdout (160 chars)
seed 42: ref nan False sol nan False max diff 0.032958984375
seed 123: ref nan False sol nan True max diff nan
seed 456: ref nan False sol nan True max diff nan
assistant
reasoning (14369 chars)
Interesting, seed 42 no NaN, seeds 123 and 456 have NaN. This is non-deterministic? Or depends on the actual values. The issue is likely numerical instability in the solve or exp. Let me find where NaN first appears. Let me modify the kernel or add debug prints to see intermediates. Since Triton kernels are hard to debug, I can compute intermediates in Python/PyTorch and compare. Actually, let me think about what could cause NaN in my kernel but not reference. Reference uses fp32 throughout. My kernel uses bf16 for some GEMMs. Maybe bf16 underflow/overflow or non-associative accumulation causes NaN? No, bf16 dot can't produce NaN from finite inputs unless there's inf intermediate. Wait, `A.to(tl.bfloat16)`: if A has inf or NaN, it propagates. Where could A have inf? In the solve, if (I - A0) is close to singular, the inverse could have huge values. But with these small inputs, A0 is small. Actually, maybe the issue is that I apply scale in Python (`q_scaled = q * self.scale`) and pass scale=1.0 to kernel. Reference applies scale inside and computes Aqk with scale. In my kernel, Aqk uses qg = q * scale * exp(g). That's correct. But wait, the reference computes q = q * scale in fp32, then does everything in fp32. I do q * scale in bf16 (since q is bf16 and scale is float, result bf16), then cast to fp32 inside kernel. This is slightly different but shouldn't cause NaN. Let me check if the NaN is from `(-g_cum).exp()`. g_cum min can be negative (since input g has mean -0.05). If g_cum is -6.2, exp(-(-6.2)) = exp(6.2) ~ 500. That's fine for bf16 (max ~57300). But k is small, so kdecay is finite. Wait, in my earlier debug, g_cum min was -6.2. That means some token has very negative cumsum. exp(g) for that token is exp(-6.2) ~ 0.002, exp(-g) ~ 500. kdecay = k * 500. If k ~ 0.1, kdecay ~ 50. Then kg = k * 0.002 ~ 0.0002. Then M = kg @ kdecay.T ~ 0.0002 * 50 * 128 = 1.28. A0 ~ -1.28 * beta. Still fine. But if g has larger negative cumsum, kdecay could be huge. The test seed 123/456 might produce g_cum min < -10, leading to exp(10) ~ 22000, kdecay ~ 2200, M ~ huge. Then A0 has large values, (I-A0) might be singular, solve produces inf. Reference uses fp32 and handles this without NaN? Actually reference would also produce huge values if g_cum is very negative. But reference computes exp(g - g_i) which cancels. Wait, reference computes A using `k * (g - g_i).exp()`. This is k[c] * k[i] * exp(g[c] - g[i]). If both g[c] and g[i] are very negative, exp(g[c]-g[i]) might be moderate. But my approach computes kg = k * exp(g) and kdecay = k * exp(-g). If g is very negative, exp(-g) is huge and exp(g) is tiny. kg and kdecay have extreme values, and their dot product kg @ kdecay.T = k[c]*k[i]*exp(g[c]-g[i]) is moderate, but computing it as product of extreme values can overflow/underflow in bf16. This is the numerical stability issue! The reference's formulation is more stable because it computes exp(g - g_i) directly. My factorization into exp(g) * exp(-g_i) loses precision when g values are far from 0. I need to compute Aqk and M more stably, similar to reference. For each row c and column i, compute exp(g[c] - g[i]) directly, not as exp(g[c]) * exp(-g[i]). But that requires computing pairwise differences. For Aqk[c,i] = sum_d q[c,d] * k[i,d] * exp(g[c,d] - g[i,d]). We can compute this by, for each column i, subtract g[i] from all rows. This is what reference does in the loop. In a Triton kernel, we can compute Aqk and M by iterating over columns (or rows) and using the difference. For each column i: g_i = g[i, :] # (K,) diff = g - g_i[None, :] # (BT,K) q_part = q * exp(diff) # (BT,K) Aqk[:,i] = q_part @ k[i,:] # dot over K This matches reference's loop: ```python for i in range(BT): k_i = k[..., i, :] g_i = g[..., i:i+1, :] A[..., i] = einsum("... c d, ... d -> ... c", k * (g - g_i).exp(), k_i) ``` So I should compute Aqk and M this way, using g differences directly. This is more stable. But it requires an outer loop over columns and loading k_i, g_i per column. That's BT iterations. For each iteration, we load a (BT,K) block of g (or use the full g_cum in shared memory). We can keep g_cum in shared memory (64*128*4 = 32KB) and stream q,k through. Actually, we can compute Aqk and M in a single pass over K tiles, but for each K tile, we need to compute pairwise exp(g[c,k] - g[i,k]) for all c,i. That's a (BT,BT,BK) tensor. We can compute it as: diff = g_cum[:, None, :] - g_cum[None, :, :] # (BT,BT,K) But this requires g_cum in registers or shared memory. If we load g_cum into shared memory as (BT,K) fp32 = 32KB, then for each column i, we can broadcast g_i to all rows and compute diff. Then q * exp(diff) and k * exp(diff). Wait, for Aqk: Aqk[c,i] = sum_k q[c,k] * k[i,k] * exp(g[c,k] - g[i,k]). For each i, we can compute: q_exp = q * exp(g - g_i[None,:]) # (BT,K) Aqk[:,i] = q_exp @ k[i,:] # (BT,) This needs q (BT,K) and k_i (K,). We can load q and k in K tiles. Similarly for M (Akk): M[c,i] = sum_k k[c,k] * k[i,k] * exp(g[c,k] - g[i,k]). For each i: k_exp = k * exp(g - g_i[None,:]) # (BT,K) M[:,i] = k_exp @ k[i,:] # (BT,) So both Aqk and M can be computed in the same loop over i, with one dot product each (over K tiles). Total FLOPs: BT * 2 * BT * K = 2*BT^2*K = 2M per chunk, same as before. In terms of memory, we need g_cum in shared memory (32KB). For each i, we load k_i and g_i (K elements each). And we stream q,k in K tiles. Wait, we can compute q_exp for the full (BT,K) block if it fits. q is bf16 (BT,K) = 64*128*2 = 16KB. g_cum is fp32 32KB. Total in shared memory ~48KB. Fine. We can load q,k,g_cum into shared memory and compute Aqk/M without tiling K. Actually, if we load full q,k,g into shared memory, we can compute everything in registers. But shared memory might be tight: q 16KB + k 16KB + g 32KB + maybe k_i 1KB = 65KB. Plus output Aqk 16KB, M 16KB. Total ~97KB. Might still fit in 101KB? Close. Better to tile K. For each i (column), and each K tile: load q_tile, k_tile, g_tile compute diff = g_tile - g_i_tile[None,:] compute q_exp = q_tile * exp(diff), k_exp = k_tile * exp(diff) Aqk[:,i] += q_exp @ k_i_tile M[:,i] += k_exp @ k_i_tile g_i_tile needs to be loaded per i. We can load it once per i and reuse across K tiles. But loading q_tile, k_tile for each i means we read q,k BT times. That's a lot: q is read BT times = 64 * 16KB = 1MB per block. Not great. Alternative: load q,k,g into shared memory once, then loop over i and K tiles reading from shared memory. This is better. Shared memory budget: - q: BT*K*2 = 16KB - k: 16KB - g: BT*K*4 = 32KB - Aqk: BT*BT*4 = 16KB - M: 16KB - workspace for k_i, g_i: small Total: 96KB. This is close to the 101KB limit. Might fail. If we use bf16 for g (or exp(g) precomputed), we can reduce. But we need g differences in fp32 for stability. Actually, we can precompute exp(g) and exp(-g) in shared memory? No, that's the unstable approach. Alternatively, keep q,k in shared memory as bf16, g in shared memory as fp32. Aqk and M in registers (32KB total). Then we don't need shared memory for Aqk/M. Total shared: 16+16+32 = 64KB. Good. But we need to compute Aqk/M in registers. For each i, we compute column i of Aqk and M. We can accumulate a column vector of length BT in registers, and do this for each i, then assemble into the 64x64 matrix. Algorithm: 1. Load q,k (bf16) and g (fp32) into shared memory. 2. For i in 0..BT-1: a. Load k_i (K) and g_i (K) from shared memory. b. acc_Aqk = zeros(BT), acc_M = zeros(BT) c. For each K tile (or full K): - load q_tile, k_tile, g_tile from shared memory - diff = g_tile - g_i[None,:] - q_exp = q_tile * exp(diff) - k_exp = k_tile * exp(diff) - acc_Aqk += q_exp @ k_i_tile (dot over K tile) - acc_M += k_exp @ k_i_tile d. Store acc_Aqk to Aqk[:,i], acc_M to M[:,i] in registers. If K=128 fits in one tile (BK=128), then each i does one dot. We need q_tile, k_tile, g_tile for full K. But we can read directly from shared memory using block pointers. Wait, block pointers into shared memory? Not sure. We can use manual indexing with a shared memory pointer. Actually, in Triton, we can allocate shared memory by declaring a local tensor and using it. But indexing with variable i is tricky. Alternative: use a separate kernel to compute Aqk and M with the stable formulation, then another kernel for solve and w,u. Or do Aqk/M computation in PyTorch with loops (slow but stable). Actually, we can compute Aqk and M in PyTorch efficiently using broadcasting: ```python # g_cum: (B,H,N,C,K) # q,k: (B,H,N,C,K) # For each i, we want q * exp(g - g[:,:,:,i:i+1,:]) @ k[:,:,:,i,:].T # Compute diff = g_cum.unsqueeze(3) - g_cum.unsqueeze(2) # (B,H,N,C,C,K) # Aqk = torch.einsum('b h n c k, b h n i k -> b h n c i', q * diff.exp(), k) * scale # M = torch.einsum('b h n c k, b h n i k -> b h n c i', k * diff.exp(), k) ``` This computes the full (C,C,K) diff tensor: 64*64*128 = 524K elements per chunk, times B*H*N = 512 -> 268M elements = 1GB memory. Too much. But we can do it in chunks along K: for each K tile, compute diff_tile = g_tile[:,:,:,:,None,:] - g_tile[:,:,:,None,:,:], then accumulate q_exp @ k.T. Actually: ```python Aqk = torch.zeros(B,H,N,C,C) for i_k in range(K//BK): q_tile = q[..., i_k*BK:(i_k+1)*BK] k_tile = k[..., i_k*BK:(i_k+1)*BK] g_tile = g_cum[..., i_k*BK:(i_k+1)*BK] diff = g_tile.unsqueeze(3) - g_tile.unsqueeze(2) # (B,H,N,C,C,BK) Aqk += torch.einsum('b h n c k, b h n i k -> b h n c i', q_tile * diff.exp(), k_tile) ``` diff shape per K tile: B*H*N*C*C*BK = 2*8*16*64*64*64 = 67M elements = 268MB fp32. Still too much for K tile of 64. We need to loop over columns i to avoid the C*C*K materialization: ```python Aqk = torch.zeros(B,H,N,C,C) M = torch.zeros(B,H,N,C,C) for i in range(C): k_i = k[..., i, :] g_i = g_cum[..., i:i+1, :] diff = g_cum - g_i # (B,H,N,C,K) Aqk[..., i] = torch.einsum('... c k, ... k -> ... c', q * diff.exp(), k_i) * scale M[..., i] = torch.einsum('... c k, ... k -> ... c', k * diff.exp(), k_i) ``` This loops over 64 columns, each doing einsum of (B,H,N,C,K) @ (B,H,N,K) -> (B,H,N,C). This is 64 matrix-vector products per chunk. With B*H*N = 512 chunks, total 32768 such products. In PyTorch, each is a batched vector dot. Might be slow but correct. But we want a custom kernel. We can do this column loop in Triton, with q,k,g in shared memory. Actually, I think we can compute Aqk and M in Triton with the column loop and keep q,k,g in global memory, using block pointers. For each column i, we load g_i and k_i from global memory (K elements each), and for each K tile, we load q_tile, k_tile, g_tile and compute. But this reads q,k repeatedly for each i. For K tile size 32, q tile is 64*32*2 = 4KB. Read 64 times = 256KB per block. With shared memory, we can cache q,k,g. Let me try to write a kernel that loads q,k,g into shared memory. But Triton shared memory with manual indexing is tricky. Alternatively, we can keep the qg/kg/kdecay factorization but compute Aqk and M in a more stable way by normalizing. For example, compute relative to the column: for each column i, compute exp(g - g_i) directly, not via exp(g)*exp(-g_i). This is what we want. Hmm, let me think about another approach. We can compute `qg = q * exp(g - g_last)` and `kdecay = k * exp(g_last - g)` for some anchor g_last (last token in chunk). Then Aqk[c,i] = sum_d qg[c,d] * kdecay[i,d] * exp(g_last[i] - g_last[c])? Not simpler. Actually, the reference uses per-column subtraction. We should too. Let me write kernel 1 with the stable formulation, using global memory loads and an outer loop over columns. It might be slower but correct. Then optimize later. Kernel 1 stable: - Load g_cum (or compute from g), keep in registers? For K=128, g_cum (64,128) = 32KB. Too big for registers per thread but OK across block. - Actually we can compute g_cum in Python and pass it. - For each column i in 0..BT-1: - Load g_i and k_i from global memory. - For each K tile: - Load q_tile, k_tile, g_tile. - diff = g_tile - g_i[None,:] - q_exp = q_tile * exp(diff) - k_exp = k_tile * exp(diff) - dot_q = q_exp @ k_i_tile (BT length) - dot_k = k_exp @ k_i_tile - accumulate into Aqk_col and M_col - Store Aqk_col and M_col to global memory? But we need Aqk and M matrices for later. We can store column by column. Actually, we can store Aqk/M column by column to global memory. Then in a second phase, load Aqk/M and solve. But solving needs the full A0 matrix. We can have kernel 1 compute Aqk/M and store them, then kernel 1b solve and compute w,u. Or combine. For simplicity, let me have kernel 1 do: 1. Load q,k,v,g,beta for chunk. 2. Compute g_cum (or receive precomputed). 3. Loop over columns i: - Compute Aqk[:,i] and M[:,i] using stable diff. 4. After full matrix, solve for A in registers. 5. Compute w,u. But storing Aqk/M column by column to global memory and then reading back is inefficient. We can keep them in registers (64x64 each = 32KB total). With 128 threads, 256 bytes/thread. OK. So kernel 1: - Inputs: q,k,v,g,beta (global) - Load q,k,g,beta,v into shared memory or registers. For q,k,g, we need full (BT,K) per column computation. We can load them into shared memory to avoid repeated global loads. Shared memory: - q_s: BT*K*2 = 16KB - k_s: 16KB - g_s: BT*K*4 = 32KB - v_s: BT*V*2 = 16KB - beta_s: BT*2 = 128B Total: 80KB. OK, under 101KB. Then for each column i, we read k_s[i,:] and g_s[i,:] from shared memory. We accumulate Aqk_col and M_col in registers. For K=128, we can process full K in one go: load q_s[:,:] (64x128), k_s[i,:] (128), diff = g_s[:,:] - g_s[i,:][None,:], q_exp = q_s * exp(diff), dot over K. In Triton, reading from shared memory with dynamic row index i is the challenge. But if q_s, k_s, g_s are local tensors in shared memory, can we index them with variable i? Triton supports indexing local tensors with scalars: `q_s[i, :]` might work? Actually earlier we saw `d[:BK]` doesn't work (slice), but scalar indexing might work. Let me test. Actually, Triton supports `tensor[scalar, :]` for loading a row. But the scalar must be a constexpr or a runtime int? I think runtime scalar indexing of local tensors is supported. Let me try a small test kernel.
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args
{
  "command": "<507 chars \u2014 see below>",
  "description": "Test scalar indexing"
}
command (507 chars)
uv run python - <<'PY' 2>&1 | tail -40
import torch
import triton, triton.language as tl

@triton.jit
def test_index(in_ptr, out_ptr, K: tl.constexpr):
    i = tl.program_id(0)
    # load full tensor into local memory
    x = tl.load(in_ptr + tl.arange(0, K))
    # extract element i
    val = tl.sum(tl.where(tl.arange(0, K) == i, x, 0.0))
    tl.store(out_ptr, val)

a = torch.randn(10, device='cuda')
out = torch.empty(1, device='cuda')
test_index[(1,)](a, out, K=10)
print(out.item(), a.sum().item())
PY
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tool
stdout (1468 chars)
Traceback (most recent call last):
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 461, in __init__
    self.raw_src, self.starting_line_number = inspect.getsourcelines(fn)
                                              ^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/uv-cache/python/cpython-3.11.15-linux-x86_64-gnu/lib/python3.11/inspect.py", line 1240, in getsourcelines
    lines, lnum = findsource(object)
                  ^^^^^^^^^^^^^^^^^^
  File "/uv-cache/python/cpython-3.11.15-linux-x86_64-gnu/lib/python3.11/inspect.py", line 1077, in findsource
    raise OSError('could not get source code')
OSError: could not get source code

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 935, in jit
    return decorator(fn)
           ^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 923, in decorator
    return JITFunction(
           ^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 756, in __init__
    super().__init__(fn)
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 463, in __init__
    raise ValueError("@jit functions should be defined in a Python file") from e
ValueError: @jit functions should be defined in a Python file
assistant
reasoning (58 chars)
Triton can't compile from stdin. Let me write a test file.
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args
{
  "command": "<972 chars \u2014 see below>",
  "description": "Test Triton indexing"
}
command (972 chars)
cat > /workspace/problems/02_kda_cutlass/test_triton.py <<'PY'
import torch
import triton, triton.language as tl

@triton.jit
def test_index(in_ptr, out_ptr, K: tl.constexpr):
    i = tl.program_id(0)
    x = tl.load(in_ptr + tl.arange(0, K))
    val = tl.sum(tl.where(tl.arange(0, K) == i, x, 0.0))
    tl.store(out_ptr, val)

a = torch.randn(10, device='cuda')
out = torch.empty(1, device='cuda')
test_index[(1,)](a, out, K=10)
print('sum extract', out.item(), 'expected', a.sum().item())

@triton.jit
def test_row_mat(in_ptr, out_ptr, N: tl.constexpr, K: tl.constexpr):
    i = tl.program_id(0)
    x = tl.load(in_ptr + tl.arange(0, N*K))
    x = tl.reshape(x, (N, K))
    row = x[i, :]
    tl.store(out_ptr + tl.arange(0, K), row)

a = torch.randn(4, 8, device='cuda')
out = torch.empty(8, device='cuda')
test_row_mat[(1,)](a, out, N=4, K=8)
print('row extract', out, 'expected', a[0])
PY
uv run python /workspace/problems/02_kda_cutlass/test_triton.py 2>&1 | tail -40
in 99188out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (2390 chars) — TRUNCATED
Traceback (most recent call last):
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/core.py", line 43, in wrapper
    return fn(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/core.py", line 1644, in arange
    return _semantic.arange(start, end)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/semantic.py", line 583, in arange
    raise ValueError("arange's range must be a power of 2")
ValueError: arange's range must be a power of 2

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/test_triton.py", line 13, in <module>
    test_index[(1,)](a, out, K=10)
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 370, in <lambda>
    return lambda *args, **kwargs: self.run(grid=grid, warmup=False, *args, **kwargs)
                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 720, in run
    kernel = self._do_compile(key, signature, device, constexprs, options, attrs, warmup)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 849, in _do_compile
    kernel = self.compile(src, target=

... (890 more chars)
assistant
reasoning (46 chars)
K must be power of 2. Let me use K=8 for test.
in 100323out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<970 chars \u2014 see below>",
  "description": "Test Triton indexing with power of 2"
}
command (970 chars)
cat > /workspace/problems/02_kda_cutlass/test_triton.py <<'PY'
import torch
import triton, triton.language as tl

@triton.jit
def test_index(in_ptr, out_ptr, K: tl.constexpr):
    i = tl.program_id(0)
    x = tl.load(in_ptr + tl.arange(0, K))
    val = tl.sum(tl.where(tl.arange(0, K) == i, x, 0.0))
    tl.store(out_ptr, val)

a = torch.randn(8, device='cuda')
out = torch.empty(1, device='cuda')
test_index[(1,)](a, out, K=8)
print('sum extract', out.item(), 'expected', a.sum().item())

@triton.jit
def test_row_mat(in_ptr, out_ptr, N: tl.constexpr, K: tl.constexpr):
    i = tl.program_id(0)
    x = tl.load(in_ptr + tl.arange(0, N*K))
    x = tl.reshape(x, (N, K))
    row = x[i, :]
    tl.store(out_ptr + tl.arange(0, K), row)

a = torch.randn(4, 8, device='cuda')
out = torch.empty(8, device='cuda')
test_row_mat[(1,)](a, out, N=4, K=8)
print('row extract', out, 'expected', a[0])
PY
uv run python /workspace/problems/02_kda_cutlass/test_triton.py 2>&1 | tail -40
in 100323out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (2323 chars) — TRUNCATED
sum extract 0.1483558565378189 expected -1.0120272636413574
Traceback (most recent call last):
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/core.py", line 43, in wrapper
    return fn(*args, **kwargs)
           ^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/language/core.py", line 1101, in __getitem__
    raise ValueError(f"unsupported tensor index: {sl}")
ValueError: unsupported tensor index: int32[]

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/test_triton.py", line 26, in <module>
    test_row_mat[(1,)](a, out, N=4, K=8)
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 370, in <lambda>
    return lambda *args, **kwargs: self.run(grid=grid, warmup=False, *args, **kwargs)
                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 720, in run
    kernel = self._do_compile(key, signature, device, constexprs, options, attrs, warmup)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 849, in _do_compile
    kernel = self.compile(src, target=target, options=options.__dict__)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/work

... (823 more chars)
assistant
reasoning (19241 chars)
So `x[i, :]` with runtime scalar i doesn't work. And my `test_index` didn't work because `tl.where(arange == i)` with i as program_id (0) extracted element 0, but I summed expecting sum. Actually it extracted element i=0 correctly, but I printed `a.sum()` as expected. My bad. It extracted `a[0]`. Let me verify. Actually the issue: `tl.where(arange == i, x, 0)` and sum gives x[i]. Since i=0, it gives a[0]. Output 0.148 matches a[0]. Good. So the `tl.where` trick works. So to extract row i from a matrix, we can do: ```python row = tl.sum(tl.where((tl.arange(0,N)[:,None] == i) & mask, x, 0.0), axis=0) ``` This is O(N*K) operations per extraction. For N=64,K=128, that's 8K ops per row. Extracted 64 rows -> 512K ops. Acceptable. But for computing Aqk/M column i, we need k_i (length K) and g_i (length K). We can extract them this way. Alternatively, we can avoid extracting rows by transposing the problem. If we process row by row instead of column by column, we need column vectors. Same issue. Or, we can load q,k,g from global memory using block pointers for each K tile, and for each column i, load k_i and g_i from global memory directly using pointer arithmetic: ```python g_i = tl.load(g_ptr + (t0 + i) * H * K + tl.arange(0, K)) k_i = tl.load(k_ptr + (t0 + i) * H * K + tl.arange(0, K)) ``` This is straightforward and avoids shared memory row extraction. For each column i, we load k_i and g_i once. And for each K tile, we load q_tile, k_tile, g_tile from global memory. But this means we read q,k,g from global memory BT times (once per column i). For q: BT * (BT*K) = 64 * 16KB = 1MB per block. K tile size doesn't matter for total q reads; we always read full q per column. Actually if we process full K in one tile, q is read once per column i (64 times). If we tile K into 2 tiles, q is read 2*64 = 128 times. So larger K tile is better. For K=128, we can use BK=128 (power of 2). Then q_tile is full q. Read 64 times per block = 1MB. For B=2,H=8,N=16: 256 blocks. Total q reads = 256MB. Plus k reads similar. Global memory bandwidth 1.8TB/s, so 512MB takes ~0.3ms. Maybe OK. But we can do better with shared memory caching. Let's see if we can load q,k,g into shared memory once and extract rows with `tl.where`. Actually, we can store q,k,g in global memory but access them via block pointers in the outer loop. For each i, load q_block, k_block, g_block for full K? No, block pointers require constexpr offsets. We can load the full (BT,K) block once into shared memory, then extract rows. Let me try the shared memory approach with `tl.where` row extraction. But Triton shared memory allocation is implicit. If I declare `q_s = tl.zeros([BT*K], dtype=tl.bfloat16)`, will it go to shared memory? Probably yes if it's large. But indexing with `tl.where` is in registers? Actually `tl.where` operates on the tensor and returns a tensor. If q_s is in shared memory, `tl.where(mask, q_s, 0)` creates a copy in registers? That would be expensive. Alternative: use a 1D shared memory pointer and load/store individual rows. But we can't easily load a row with dynamic index. Hmm. Let me try yet another approach: compute Aqk and M in PyTorch with loops (batched over B,H,N), using the stable formulation. This will be correct but might be slow. Then the Triton kernel only does solve, w, u, and the inter recurrence. Actually, if PyTorch loop is too slow, we can use Triton for that too. But first, let me check if the PyTorch loop is fast enough for correctness. If it takes < 10ms, it's fine. PyTorch computation: ```python g_cum = g.view(B, NT, BT, H, K).cumsum(dim=2).view(B, T, H, K) Aqk = torch.zeros(B, T, H, BT, device=device, dtype=torch.float32) M = torch.zeros(B, T, H, BT, device=device, dtype=torch.float32) # Rearrange to (B,H,N,C,K) for easier looping qg = (q * scale).view(B, NT, BT, H, K).permute(0,3,1,2,4) # (B,H,N,C,K) kg = k.view(B, NT, BT, H, K).permute(0,3,1,2,4) g_cum = g_cum.view(B, NT, BT, H, K).permute(0,3,1,2,4) for i in range(BT): k_i = kg[..., i, :] g_i = g_cum[..., i:i+1, :] diff = g_cum - g_i Aqk[..., i] = torch.einsum('... c k, ... k -> ... c', qg * diff.exp(), k_i) M[..., i] = torch.einsum('... c k, ... k -> ... c', kg * diff.exp(), k_i) ``` This loops 64 times, each doing a batched einsum over B*H*N=256 chunks. Each einsum is (256,64,128) * (256,128) -> (256,64). This is a batched matrix-vector product. PyTorch can do this efficiently with bmm. Actually, `torch.einsum('b h n c k, b h n k -> b h n c', ...)` for B*H*N chunks. We can reshape to (B*H*N, C, K) and (B*H*N, K) and use bmm: `(BHN, C, K) @ (BHN, K, 1) -> (BHN, C, 1)`. This uses GEMV. For 64 iterations, each GEMV is (256,64,128). FLOPs per iteration = 2*256*64*128 = 4.2M. 64 iterations = 268M FLOPs. At 100 TFLOPS, ~2.7ms. Plus the same for M. Total ~5.4ms for Aqk+M. Plus solve, w, u, inter recurrence. Might be OK but not great. But this is in PyTorch fp32, using tensor cores for bf16 if we cast. Actually q,k are bf16, g is fp32. diff.exp() is fp32. qg * diff.exp() is fp32. The GEMV is fp32. On Blackwell, fp32 GEMV is not tensor-core accelerated and might be slow. Alternatively, cast to bf16: `qg * diff.exp().to(bf16)` etc. But stability is the issue. Hmm. Let me think about a Triton implementation that's stable and doesn't need dynamic row extraction. We can compute Aqk and M using the factorization but with care: For each column i, we want exp(g[c,k] - g[i,k]). Instead of exp(g) * exp(-g_i), we can compute: max_g = max(g[c,k], g[i,k]) or use g[i,k] as anchor. exp(g - g_i) = exp(g - g_i) directly. We can compute this by, for each K tile, loading g_tile (BT,BK) and g_i_tile (BK), then diff = g_tile - g_i_tile[None,:]. This is stable. The issue is accessing g_i_tile for variable i. We can load g_i_tile directly from global memory: `g_ptr + (t0+i)*H*K + off_k + tl.arange(0,BK)`. Similarly for k_i_tile. This avoids row extraction from shared memory. Then for each K tile, load q_tile, k_tile, g_tile from global memory. Compute diff, q_exp, k_exp. Dot with k_i_tile. Accumulate Aqk_col, M_col. Total global reads per block: - For each i (64), for each K tile (1 if BK=128, 2 if BK=64): - q_tile: BT*BK*2 bytes - k_tile: BT*BK*2 bytes - g_tile: BT*BK*4 bytes - k_i_tile: BK*2 bytes - g_i_tile: BK*4 bytes With BK=128, 1 K tile: per i: 64*128*2 + 64*128*2 + 64*128*4 + 128*2 + 128*4 = 16KB+16KB+32KB+0.25KB+0.5KB = 65KB total 64*65KB = 4MB per block. With BHN=256 blocks: 1GB. At 1.8TB/s: 0.56ms. OK. Actually, q and k are read for each i, but we can cache them in shared memory. g is also read for each i. If we put q,k,g in shared memory (64KB), we avoid most global reads. Then only k_i, g_i are read from global memory per i (small). So shared memory approach: - Load q_s, k_s (bf16) and g_s (fp32) into shared memory once. - For each i, extract g_i, k_i from shared memory using `tl.where` trick. - For each K tile (or full K), load q_tile, k_tile, g_tile from shared memory (but if they're already in shared memory, do we need to tile? We can operate on full K). If q_s, k_s, g_s are in shared memory, we can compute for each i: g_i = extract_row(g_s, i) # using tl.where, O(BT*K) ops k_i = extract_row(k_s, i) diff = g_s - g_i[None,:] # (BT,K) q_exp = q_s * exp(diff) k_exp = k_s * exp(diff) Aqk[:,i] = q_exp @ k_i M[:,i] = k_exp @ k_i But q_s * exp(diff) requires q_s in registers? Actually elementwise ops on shared memory tensors might load to registers. The `tl.where` extraction of g_i from shared memory g_s creates a register tensor of length K. Then diff = g_s - g_i[None,:] might load g_s to registers. If g_s is 32KB, loading it to registers per column i (64 times) means 2MB of register traffic. Registers are fast. OK. But the `tl.where` extraction itself is O(BT*K) per column = 8K ops, 64 columns = 512K ops. Acceptable. Actually, we can avoid extracting g_i and k_i by loading them from global memory directly. That might be simpler and faster than the `tl.where` extraction from shared memory. Plan: 1. Load q_s, k_s, g_s into shared memory (or keep in global and use block pointers). 2. For each i, load g_i, k_i from global memory. 3. For each K tile, load q_tile, k_tile, g_tile from shared memory using block pointers? Or directly as local tensor slices? Wait, if q_s is a local tensor in shared memory, can we use `make_block_ptr` on it? Probably not. We can reshape q_s to (BT,K) and use `q_s[:, off_k:off_k+BK]` but slicing doesn't work. Alternatively, keep q,k,g in global memory and use block pointers for tiles. For each column i, loop over K tiles, load q_tile, k_tile, g_tile via block pointers. Also load g_i_tile, k_i_tile. This is what I described earlier with ~4MB per block. That's OK. Let me implement this simpler approach: no shared memory for q,k,g in kernel 1. Just global loads per column. Actually, we can at least keep q,k in shared memory and only load g_i,k_i per column. But the slicing issue remains. You know what, let me just try the global-memory-per-column approach and see performance. It might be fast enough. Also, for the solve and w,u, we can keep in registers as before. Let me rewrite kernel 1 with stable column-loop formulation. Wait, but we also need kg for w = A @ kg. kg = k * exp(g). With stable approach, we can compute kg similarly by loading k and g. But we already have k_s? No, we're not using shared memory. We can compute kg in Python and pass it, or compute it in the kernel. Actually, we can compute kg and kdecay in Python using stable exp? But kg = k * exp(g) is the unstable factorization. For w = A @ kg, if kg is computed unstably, it could be wrong. But w is used in the recurrence v_i = u_i - w_i @ S. If kg has extreme values, w might have extreme values, causing numerical issues. Reference computes w = A @ (g.exp() * k) directly. A itself is computed stably. So kg in w is indeed k * exp(g), but then A cancels some of the instability? Not necessarily. Actually, reference line 72: `w = A @ (g.exp() * k)`. Here g is the cumsum. For tokens with very negative g, exp(g) is tiny, so kg is tiny. For tokens with very positive g, kg is large. A @ kg mixes them. If we compute kg unstably and then w = A @ kg, the result should match reference's w = A @ kg. The issue is only numerical precision. For tolerance 0.05, bf16 might be OK if values don't overflow. But we saw NaN with the unstable factorization. The NaN likely came from A being computed unstably (via kg @ kdecay.T) leading to singular-like behavior. If we compute A stably, then even if kg is computed unstably, A might be well-conditioned and w finite. But for better accuracy, we can compute kg and kdecay in Python using fp32: `kg = k.float() * g_cum.exp(); kdecay = k.float() * (-g_cum).exp()`. Then cast to bf16 for the kernel. This avoids bf16 overflow in exp. The values themselves might be extreme, but finite as fp32, then cast to bf16 (could become inf if too large). For g_cum min -6.2, exp(6.2) ~ 500, k~0.1 -> kdecay~50, fits bf16. For other seeds, g_cum might be more extreme. Let me check the range for seeds 123,456. Actually, the NaN in my current kernel might come from the unstable A computation. Let me first fix A computation to be stable and see if NaN disappears. We can keep kg/kdecay for w and S update computed unstably for now. Wait, for S update we use kdecay = k * exp(-g). If g is very negative, exp(-g) huge, kdecay huge, S update huge. Then S grows, next chunk w @ S huge, v_i = u - w@S could be unstable. So we should compute kdecay for S update stably too. Reference computes k_i * exp(g_last - g_i) directly. We can do that: kdecay_for_update = k * exp(g_last - g) per chunk. But g_last depends on chunk. Hmm. Actually for the S update, we can avoid precomputing kdecay. In kernel 2, for each chunk, we have g_last (K). We can compute `k_for_update = k * exp(g_last[None,:,None] - g)` on the fly. But g varies per token, so we need g for each token. Actually, reference line 90: `(g_i[:, :, -1:] - g_i).exp() * k_i`. So for chunk i, token c: k[c] * exp(g_last - g[c]). We can compute this in kernel 2 by loading k and g for the chunk, computing exp(g_last - g), and doing the GEMM. This is stable. So in kernel 2, instead of passing kdecay, we pass k and g_cum. Then: - For v_i = u_i - w_i @ S: w is precomputed (A @ kg). - For o_i = qg @ S + Aqk @ v_i: qg is precomputed (q * exp(g)). - For S update: compute k_update = k_i * exp(g_last[None,:] - g_i) (BT,K), then k_update.T @ v_i. This avoids extreme kdecay in S update. For w = A @ kg, kg = k * exp(g). If g has very negative values, kg is tiny but finite. If g has very positive values, kg can be large. But A's diagonal is beta ~0.5, so w values are bounded by beta * max(kg). If kg overflows bf16, we have a problem. But g_cum for these inputs probably doesn't exceed ~2 (exp~7.4), and k~0.1, so kg~0.74. With large_qkv scale 2.0, k~0.2, kg~1.5. Fine. For kdecay in S update, we use exp(g_last - g) which is bounded because g_last - g is the cumsum from token c to end of chunk, so it's a sum of ~BT random values, std ~0.1*sqrt(64)=0.8. exp(0.8)~2.2. So k_update ~ 0.2*2.2 = 0.44. Stable. So the main instability is in computing A via kg @ kdecay.T. If we compute A stably, everything else should be fine. Let me rewrite kernel 1 to compute Aqk and M stably with column loop, but still compute w from kg precomputed in Python. Actually, we can compute w in the same kernel after A is known, using kg. kg we can compute in Python and pass. Plan: - Python: compute g_cum, qg = q*scale*exp(g_cum), kg = k*exp(g_cum). - Kernel 1: compute Aqk and M stably (column loop with exp(g-g_i)), solve for A, compute w = A @ kg, u = A @ v. - Kernel 2: use qg, w, u, Aqk. Also pass k and g_cum for S update. Compute k_update = k * exp(g_last - g_cum) per chunk. This avoids kdecay tensor entirely. Let me implement this. For kernel 1 stable A computation: ```python for i in range(BT): # load g_i, k_i for column i p_gi = tl.make_block_ptr(g_ptr, (T, K), (H*K, 1), (t0+i, 0), (1, K), (1, 0)) p_ki = tl.make_block_ptr(k_ptr, (T, K), (H*K, 1), (t0+i, 0), (1, K), (1, 0)) g_i = tl.load(p_gi, boundary_check=(0,1)).to(tl.float32) # shape (1,K) k_i = tl.load(p_ki, boundary_check=(0,1)).to(tl.float32) # loop over K tiles (or full K) acc_Aqk = tl.zeros([BT], dtype=tl.float32) acc_M = tl.zeros([BT], dtype=tl.float32) for i_k in range(tl.cdiv(K, BK)): off_k = i_k * BK p_q = tl.make_block_ptr(q_ptr, (T,K), (H*K,1), (t0, off_k), (BT,BK), (1,0)) p_k = tl.make_block_ptr(k_ptr, (T,K), (H*K,1), (t0, off_k), (BT,BK), (1,0)) p_g = tl.make_block_ptr(g_ptr, (T,K), (H*K,1), (t0, off_k), (BT,BK), (1,0)) b_q = tl.load(p_q, boundary_check=(0,1)).to(tl.float32) b_k = tl.load(p_k, boundary_check=(0,1)).to(tl.float32) b_g = tl.load(p_g, boundary_check=(0,1)).to(tl.float32) g_i_tile = g_i[:, off_k:off_k+BK] # slicing (1,BK) - does this work? diff = b_g - g_i_tile q_exp = b_q * tl.exp(diff) k_exp = b_k * tl.exp(diff) k_i_tile = k_i[:, off_k:off_k+BK] acc_Aqk += tl.sum(q_exp * k_i_tile, axis=1) acc_M += tl.sum(k_exp * k_i_tile, axis=1) # store acc_Aqk to Aqk[:,i], acc_M to M[:,i] # Aqk is register tensor (BT,BT), set column i Aqk = tl.where(c[None,:] == i, acc_Aqk[:,None], Aqk) M = tl.where(c[None,:] == i, acc_M[:,None], M) ``` Slicing `g_i[:, off_k:off_k+BK]` might not work. We can instead load g_i and k_i in K tiles from the start: ```python for i_k in range(tl.cdiv(K, BK)): off_k = i_k * BK p_gi = tl.make_block_ptr(g_ptr, (T, K), (H*K, 1), (t0+i, off_k), (1, BK), (1, 0)) p_ki = tl.make_block_ptr(k_ptr, (T, K), (H*K, 1), (t0+i, off_k), (1, BK), (1, 0)) g_i_tile = tl.load(p_gi, boundary_check=(0,1)).to(tl.float32) k_i_tile = tl.load(p_ki, boundary_check=(0,1)).to(tl.float32) # load q,k,g tile ... diff = b_g - g_i_tile ... ``` This avoids slicing. Good. But `tl.sum(q_exp * k_i_tile, axis=1)` is elementwise multiply then sum, not a dot product using tensor cores. For BT=64, BK=64, it's 64*64=4K muls and 64 sums of 64 elements. Per column i, per K tile. 64 columns * 2 K tiles = 128 such operations. Total 512K muls. Not too bad, but not using tensor cores. We can use tl.dot: `tl.dot(q_exp, tl.trans(k_i_tile))` gives (BT,1). Then squeeze. But k_i_tile is (1,BK). tl.trans gives (BK,1). q_exp is (BT,BK). Dot gives (BT,1). Good. Similarly for k_exp. So: ```python acc_Aqk += tl.dot(q_exp, tl.trans(k_i_tile))[:,0] # or reshape acc_M += tl.dot(k_exp, tl.trans(k_i_tile))[:,0] ``` Actually tl.dot returns (BT,1). We can store as column. Let me accumulate a (BT,1) tensor and then set Aqk[:,i]. ```python col_Aqk = tl.zeros([BT, 1], dtype=tl.float32) col_M = tl.zeros([BT, 1], dtype=tl.float32) for i_k in ...: ... col_Aqk += tl.dot(q_exp, tl.trans(k_i_tile)) col_M += tl.dot(k_exp, tl.trans(k_i_tile)) # set column i Aqk = tl.where(c[None,:] == i, col_Aqk, Aqk) M = tl.where(c[None,:] == i, col_M, M) ``` Wait, Aqk shape is (BT,BT). col_Aqk is (BT,1). `tl.where(c[None,:] == i, col_Aqk, Aqk)` broadcasts col_Aqk across columns? Actually c[None,:] is (1,BT), and we want to set column i. `c[None,:] == i` gives (1,BT) mask. For column i, it's True. col_Aqk is (BT,1), broadcast to (BT,BT) across columns. So it sets all columns to col_Aqk, not just column i. Wrong. We need mask of shape (BT,BT) where column i is True. `mask_col_i = (c[None,:] == i) & (r[:,None] < BT)` or just `c[None,:] == i` which is (1,BT), broadcast to (BT,BT) by repeating rows. That sets entire column i to col_Aqk. Wait, `tl.where(mask, col_Aqk, Aqk)` with mask (1,BT) and col_Aqk (BT,1): broadcasting rules? In Triton, tl.where broadcasts all operands to the same shape. mask (1,BT), col_Aqk (BT,1), Aqk (BT,BT) -> broadcast to (BT,BT). mask expanded along rows, so column i is True for all rows. col_Aqk expanded along columns, so each column gets the same col_Aqk values. So column i of result is col_Aqk, other columns from Aqk. That's exactly what we want! Actually yes: `tl.where(c[None,:] == i, col_Aqk, Aqk)` sets column i to col_Aqk and keeps other columns from Aqk. Good. Let me verify broadcasting in Triton: tl.where with (1,BT) mask, (BT,1) x, (BT,BT) y -> result (BT,BT) where result[r,c] = x[r,0] if mask[0,c] else y[r,c]. For column i, mask[0,i]=True, so result[r,i] = x[r,0]. Correct. OK, this works. Now, loading q,k,g tiles for each column i means we read q,k,g BT times. We can use BK=128 to minimize K tile loop. With BK=128, one K tile. q tile = 64*128*2 = 16KB. Read 64 times per block = 1MB. For 256 blocks = 256MB. Fine. But we also need to load q,k,g for kg computation? No, kg is computed in Python. Wait, for w = A @ kg, we need kg. We compute kg in Python and pass. Good. For u = A @ v, need v. Pass v. For kernel 2, we need qg (computed in Python), w, u, Aqk, k, g_cum. Let me rewrite solution.py with this stable approach. Actually, I realize we can also compute Aqk in Python with the stable formulation using bmm. But let's do it in Triton to keep custom kernel. Let me write the new solution.
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assistant
Write /workspace/problems/02_kda_cutlass/solution.py
diff (215 lines — TRUNCATED)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -17,10 +17,11 @@
 
 @triton.jit
 def _kda_intra_kernel(
-    qg_ptr,
+    q_ptr,
+    k_ptr,
+    v_ptr,
+    g_ptr,
     kg_ptr,
-    kdecay_ptr,
-    v_ptr,
     beta_ptr,
     Aqk_ptr,
     w_ptr,
@@ -41,39 +42,62 @@
     i_h = i_bh % H
 
     t0 = i_n * BT
-
     r = tl.arange(0, BT)
     c = tl.arange(0, BT)
     mask_strict = r[:, None] > c[None, :]
     mask_lower = r[:, None] >= c[None, :]
 
-    Aqk_acc = tl.zeros([BT, BT], dtype=tl.float32)
-    M_acc = tl.zeros([BT, BT], dtype=tl.float32)
-
-    for i_k in range(tl.cdiv(K, BK)):
-        off_k = i_k * BK
-        p_qg = tl.make_block_ptr(
-            qg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        p_kg = tl.make_block_ptr(
-            kg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        p_kd = tl.make_block_ptr(
-            kdecay_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        b_qg = tl.load(p_qg, boundary_check=(0, 1)).to(tl.float32)
-        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.float32)
-        b_kd = tl.load(p_kd, boundary_check=(0, 1)).to(tl.float32)
-        Aqk_acc += tl.dot(b_qg, tl.trans(b_kd))
-        M_acc += tl.dot(b_kg, tl.trans(b_kd))
+    # Compute Aqk and M = kg @ kdecay.T using the stable formulation
+    # Aqk[c,i] = sum_k q[c,k]*k[i,k]*exp(g[c,k]-g[i,k])
+    # M[c,i]  = sum_k k[c,k]*k[i,k]*exp(g[c,k]-g[i,k])
+    Aqk = tl.zeros([BT, BT], dtype=tl.float32)
+    M = tl.zeros([BT, BT], dtype=tl.float32)
+
+    for i in range(BT):
+        col_Aqk = tl.zeros([BT, 1], dtype=tl.float32)
+        col_M = tl.zeros([BT, 1], dtype=tl.float32)
+        for i_k in range(tl.cdiv(K, BK)):
+            off_k = i_k * BK
+            # column i's g and k
+            p_gi = tl.make_block_ptr(
+                g_ptr, (T, K), (H * K, 1), (t0 + i, off_k), (1, BK), (1, 0)
+            )
+            p_ki = tl.make_block_ptr(
+                k_ptr, (T, K), (H * K, 1), (t0 + i, off_k), (1, BK), (1, 0)
+            )
+            g_i = tl.load(p_gi, boundary_check=(0, 1)).to(tl.float32)
+            k_i = tl.load(p_ki, boundary_check=(0, 1)).to(tl.float32)
+
+            # full chunk tiles
+            p_q = tl.make_block_ptr(
+                q_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+            )
+            p_k = tl.make_block_ptr(
+                k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+            )
+            p_g = tl.make_block_ptr(
+                g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+            )
+            b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32)
+            b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+            b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
+
+            diff = b_g - g_i
+            q_exp = b_q * tl.exp(diff)
+            k_exp = b_k * tl.exp(diff)
+            col_Aqk += tl.dot(q_exp, tl.trans(k_i))
+            col_M += tl.dot(k_exp, tl.trans(k_i))
+
+        Aqk = tl.where(c[None, :] == i, col_Aqk, Aqk)
+        M = tl.where(c[None, :] == i, col_M, M)
 
     p_beta = tl.make_block_ptr(
         beta_ptr, (T,), (H,), (t0,), (BT,), (0,)
     )
     b_beta = tl.load(p_beta, boundary_check=(0,)).to(tl.float32)
 
-    Aqk = tl.where(mask_lower, Aqk_acc * scale, 0.0)
-    A0 = tl.where(mask_strict, -M_acc * b_beta[:, None], 0.0)
+    Aqk = tl.where(mask_lower, Aqk * scale, 0.0)
+    A0 = tl.where(mask_strict, -M * b_beta[:, None], 0.0)
 
     # solve (I - A0) in registers
     A = A0
@@ -122,7 +146,8 @@
     w_ptr,
     u_ptr,
     qg_ptr,
-    kdecay_ptr,
+    k_ptr,
+    g_ptr,
     Aqk_ptr,
     glast_ptr,
     o_ptr,
@@ -143,11 +168,7 @@
     off_v0 = i_vb * BV
     NT = T // BT
 
-    # S split into K tiles of shape (BK, BV)
-    # We keep them as separate register tensors to avoid dynamic indexing.
-    num_k_tiles = tl.cdiv(K, BK)
-    # Triton does not support arrays of tensors; unroll manually via repeated code.
-    # For K=128, BK=64 we have two tiles.
+    # S split into K tiles of (BK, BV)
     S0 = tl.zeros([BK, BV], dtype=tl.float32)
     S1 = tl.zeros([BK, BV], dtype=tl.float32)
 
@@ -160,13 +181,11 @@
         b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
 
         b_v = b_u
-        # w @ S, K tile 0
         p_w0 = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
         )
         b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.bfloat16)
         b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))
-        # K tile 1
         p_w1 = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
         )
@@ -196,21 +215,34 @@
         )
         tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))
 
-        # S += kdecay.T @ v_i
-        p_kd0 = tl.make_block_ptr(
-            kdecay_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
-        )
-        b_kd0 = tl.load(p_kd0, boundary_check=(0, 1)).to(tl.bfloat16)
-        S0 += tl.dot(tl.trans(b_kd0), b_v.to(tl.bfloat16))
-        p_kd1 = tl.make_block_ptr(
-            kdecay_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_kd1 = tl.load(p_kd1, boundary_check=(0, 1)).to(tl.bfloat16)
-        S1 += tl.dot(tl.trans(b_kd1), b_v.to(tl.bfloat16))
-
-        # S *= exp(g_last) per k
-        d0 = tl.exp(tl.load(glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)))
-        d1 = tl.exp(tl.load(glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)))
+        # S += k_update.T @ v_i where k_update = k * exp(g_last - g)
+        p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
+        d0 = tl.exp(tl.load(p_glast0))
+        p_glast1 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)
+        d1 = tl.exp(tl.load(p_glast1))
+
+        p_k0 = tl.make_block_ptr(
+   

... (1898 more chars)
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "content": "<10470 chars \u2014 see diff>"
}
in 101325out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (464 chars)
Wait, in kernel 1 I'm passing q_scaled (bf16) and scale=1.0, but reference uses q * scale inside kernel in fp32. I compute q_scaled in bf16. The difference is small. Let me keep it for now. Also, in kernel 2, `tl.exp(tl.log(d0)[None, :] - b_g0)` computes exp(g_last - g) stably. log(d0) = g_last. Good. But `tl.log(d0)` for d0 = exp(g_last). If g_last is very negative, d0 small, log fine. If g_last positive, d0 up to ~2.7, log fine. Good. Let me test compile.
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Bash
args
{
  "command": "uv run python -c \"import solution; import torch; m = solution.Model(2,1024,8,128,128).cuda(); q,k,v,g,beta = [x.cuda() for x in solution.get_inputs()]; o=m(q,k,v,g,beta); print(o.shape, o.dtype, o.abs().max())\" 2>&1 | tail -60",
  "description": "Test compile and run"
}
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stdout (98 chars)
torch.Size([2, 1024, 8, 128]) torch.bfloat16 tensor(0.0159, device='cuda:0', dtype=torch.bfloat16)
assistant
reasoning (36 chars)
Good, it ran. Now check correctness.
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Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run correctness check"
}
in 111951out 0cache_r 0cache_w 0
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tool
stdout (230 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case nominal: tolerance exceeded: max_abs_diff=4.75199e+37 max_rel_diff=inf bad=508073/2097152 worst_idx=(0, 0, 3, 118) (atol=0.05, rtol=0.05)
assistant
reasoning (100 chars)
Huge diff. The kernel is producing wrong values. Let me debug with seed 42 and compare to reference.
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Bash
args
{
  "command": "<887 chars \u2014 see below>",
  "description": "Debug large diff"
}
command (887 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol.load_state_dict(ref.state_dict(), strict=True)
with torch.no_grad():
    ro = ref(q,k,v,g,beta)
    so = sol(q,k,v,g,beta)
print('ref max', ro.abs().max().item(), 'sol max', so.abs().max().item())
print('sol nan', torch.isnan(so).any().item(), 'sol inf', torch.isinf(so).any().item())
print('max diff', (ro-so).abs().max().item())
PY
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tool
stdout (107 chars)
ref max 0.01171875 sol max 4.751990084931074e+37
sol nan False sol inf False
max diff 4.751990084931074e+37
assistant
reasoning (5856 chars)
Sol max is huge. This suggests A is huge or w is huge. Let me check intermediates in Python to see what's wrong. Actually, maybe the issue is the solve. With the stable A computation, A0 might still be large if M is large. Let me compute A in Python with reference-like code and see. Wait, I think the issue might be that I compute Aqk with `mask_lower` but reference masks strict upper and includes diagonal for Aqk. That's correct (>=). For M (Akk), reference uses strict lower (c > j). I use `mask_strict`. Correct. But the solve: I compute A = (I - A0)^{-1} * diag(beta_col). Let me verify with a small example in Python. Actually, let me just compute A in Python using reference's algorithm and compare to what my kernel produces. But I can't easily extract A from kernel. Let me write a small Python script to compute A, w, u with the reference algorithm and the stable formulation, and compare. Wait, the stable formulation for A should be identical to reference. Let me check if my Python preprocessing matches reference. Reference: - q = q * scale (fp32) - g = g.cumsum(-2) - For i: A[...,i] = einsum(k * exp(g - g_i), k_i) * beta? Wait let me re-read. Reference line 58-65: ```python A = torch.zeros(*q.shape[:-1], BT, dtype=torch.float32, device=q.device) for i in range(BT): k_i = k[..., i, :] g_i = g[..., i:i+1, :] A[..., i] = torch.einsum("... c d, ... d -> ... c", k * (g - g_i).exp(), k_i) A = A * beta[..., None] A = -A.masked_fill(mask_diag_upper, 0) ``` So A[c,i] = sum_d k[c,d] * k[i,d] * exp(g[c,d] - g[i,d]) * beta[c] for c,i (all), then mask upper/diag to 0 and negate. So A0[c,i] = -sum_d k[c]*k[i]*exp(g[c]-g[i]) * beta[c] for c > i. This matches my M * beta_row, negated, strict lower. Then solve line 69: ```python for i in range(1, BT): A[..., i, :i] = A[..., i, :i].clone() + (A[..., i, :, None].clone() * A[..., :, :i].clone()).sum(-2) ``` This is the forward substitution. Then line 70: A = (A + I) * beta[..., None, :] So A_final[c,i] = solved_A[c,i] * beta[i], and diagonal = beta[i]. My kernel: after solve, A = (A + I) * beta[None,:]. Correct. Hmm, but my kernel produces huge values. Let me think... Oh wait! In kernel 1, `Aqk = tl.where(c[None,:] == i, col_Aqk, Aqk)`. But `col_Aqk` is shape (BT,1). When broadcasting with Aqk (BT,BT), `tl.where` with mask (1,BT) and col_Aqk (BT,1): result[r,c] = col_Aqk[r,0] if mask[0,c] else Aqk[r,c]. So column i gets col_Aqk. Good. But I also have `Aqk = tl.where(mask_lower, Aqk * scale, 0.0)`. mask_lower is (BT,BT) with c>=r. This masks Aqk to lower triangular including diagonal. Good. Wait, I think the issue is in the solve loop. Let me re-examine: ```python for i in range(1, BT): row_i = tl.sum(tl.where(r[:,None] == i, A, 0.0), axis=0) ``` `r[:,None]` is (BT,1). `r[:,None] == i` is (BT,1). A is (BT,BT). tl.where broadcasts mask (BT,1) to (BT,BT). For row i, mask True. So it extracts row i. Then sum over axis=0? Wait, axis=0 sums over rows, but we want to keep columns. Let me think. `tl.where(r[:,None] == i, A, 0.0)` gives a tensor where only row i is non-zero. Then `sum(axis=0)` sums over the row dimension, leaving a vector of length BT (columns). So row_i[c] = A[i,c]. Correct. Then `update = tl.sum(row_i[:,None] * A, axis=0)`. row_i[:,None] is (BT,1). A is (BT,BT). Product is (BT,BT). Sum over axis=0 gives vector of length BT: update[c] = sum_r row_i[r] * A[r,c] = sum_r A[i,r] * A[r,c]. Correct. Then `new_row = tl.where(c < i, row_i + update, row_i)`. For columns < i, update. Correct. Then `A = tl.where((r[:,None] == i), new_row[None,:], A)`. mask (BT,1), new_row[None,:] is (1,BT). Broadcast: result[r,c] = new_row[c] if mask[r,c] (i.e., r==i) else A[r,c]. So row i updated. Correct. So solve is correct. Hmm, where is the huge value coming from? Maybe Aqk is huge? But Aqk is not used in solve; M is. M might be huge. Actually, M is computed stably as sum_k k[c]*k[i]*exp(g[c]-g[i]). For c < i, g[c]-g[i] could be negative or positive. If g grows within chunk (positive drift), then for c < i, g[c]-g[i] < 0, exp small. If g decreases, exp large. With input g mean -0.05, cumsum tends to decrease. So g[c]-g[i] for c < i could be positive, exp > 1. But k ~ 0.1, so M ~ 0.01 * 128 * exp(diff). If diff ~ 1, M ~ 1.6. Not huge. Wait, but the huge value is 1e37. That's like exp(85). Could come from exp overflow if diff is large. But diff = g[c] - g[i], and g values are within chunk, max diff bounded by sum of 64 g values ~ 64 * 0.3 = 19 (with std 0.1*sqrt(64)=0.8, but outliers could be larger). exp(19) ~ 1.8e8. Times k^2 ~ 0.01 -> 1.8e6. Sum over K ~ 2e8. Still not 1e37. But if g_cum has huge range due to outliers... Actually g input is N(-0.05, 0.1). Over 64 tokens, max cumsum could be ~0.8 + a few std, maybe 2. Min ~ -2. diff max ~ 4. exp(4) ~ 55. M ~ 55 * 0.01 * 128 = 70. Still moderate. So why 1e37? Maybe exp is overflowing because of bug. Let me check: `diff = b_g - g_i`. b_g is (BT,BK), g_i is (1,BK). Broadcast OK. exp(diff). If diff somehow is ~85, overflow. Actually, maybe g_i is not loaded correctly. Block pointer with shape (1,BK) and strides (H*K,1), offset (t0+i, off_k). Should load row i, columns off_k:off_k+BK. OK. But maybe `boundary_check=(0,1)` with shape (1,BK) is weird? It might load garbage. Actually, I think the issue might be that I use `q_scaled` (bf16) in kernel, and `b_q = tl.load(...).to(tl.float32)`. But q_scaled is small. Fine. Wait, I just realized: in kernel 1, I pass `q_scaled` but I also compute `qg` in Python. But kernel doesn't use qg. Kernel uses q_scaled to compute Aqk. Python computes qg for kernel 2. OK. But q_scaled in Python is `q * self.scale` in bf16. In reference, q is converted to fp32 then scaled. Difference tiny. Hmm. Let me add some debug by computing Aqk and M in Python with the same formula and see if they're reasonable.
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assistant
Bash
args
{
  "command": "<869 chars \u2014 see below>",
  "description": "Check intermediate ranges"
}
command (869 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
import reference

B,T,H,K,V,CS = 2,1024,8,128,128,64
torch.manual_seed(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]

# mimic reference for first chunk
ref_model = reference.Model(B,T,H,K,V,CS).cuda()
scale = ref_model.scale

qg = (q * scale).float()
kg = k.float()
g_cum = g.view(B, CS, T//CS, H, K).permute(0,2,3,1,4).cumsum(dim=3).permute(0,2,3,1,4).contiguous().view(B,T,H,K)
kg = kg * g_cum.exp()

# first chunk, head 0
i=0
A_col = torch.einsum('b h c d, b h d -> b h c', kg[:,:CS] * (g_cum[:,:CS] - g_cum[:,i:i+1]).exp(), kg[:,i])
print('A_col max', A_col.abs().max().item())
print('g_cum range', g_cum.min().item(), g_cum.max().item())
print('exp diff range', (g_cum[:,:CS] - g_cum[:,i:i+1]).exp().min().item(), (g_cum[:,:CS] - g_cum[:,i:i+1]).exp().max().item())
PY
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stdout (398 chars)
Traceback (most recent call last):
  File "<stdin>", line 19, in <module>
  File "/workspace/.venv/lib/python3.11/site-packages/torch/functional.py", line 373, in einsum
    return _VF.einsum(equation, operands)  # type: ignore[attr-defined]
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
RuntimeError: einsum(): subscript h has size 8 for operand 1 which does not broadcast with previously seen size 64
assistant
reasoning (79 chars)
My reshaping was wrong. Let me just use reference's rearrange logic or simpler.
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Bash
args
{
  "command": "<784 chars \u2014 see below>",
  "description": "Check intermediate ranges correctly"
}
command (784 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
from einops import rearrange
import reference

B,T,H,K,V,CS = 2,1024,8,128,128,64
torch.manual_seed(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]

scale = float(K)**-0.5
q = (q * scale).float()
k = k.float()
g = g.float()
g = rearrange(g, 'b (n c) h d -> b h n c d', c=CS).cumsum(dim=3)
k = rearrange(k, 'b (n c) h d -> b h n c d', c=CS)
kg = k * g.exp()

n=0
i=0
A_col = torch.einsum('b h c d, b h d -> b h c', kg[:,:,n] * (g[:,:,n] - g[:,:,n,i:i+1]).exp(), kg[:,:,n,i])
print('A_col max', A_col.abs().max().item())
print('g range', g.min().item(), g.max().item())
print('exp diff range', (g[:,:,n] - g[:,:,n,i:i+1]).exp().min().item(), (g[:,:,n] - g[:,:,n,i:i+1]).exp().max().item())
PY
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stdout (128 chars)
A_col max 1.4974877834320068
g range -6.218692302703857 0.9883767366409302
exp diff range 0.00359784415923059 2.0869579315185547
assistant
reasoning (2648 chars)
Ranges are reasonable. So why does my kernel produce 1e37? Let me check if the issue is in the solve. Actually, wait. In my kernel, I compute M = sum_k k[c]*k[i]*exp(g[c]-g[i]) without applying scale or beta. Then A0 = -tril_strict(M) * beta_row. With M up to ~1.5 and beta up to 1, A0 up to ~1.5. I - A0 has eigenvalues near 0 if A0 has eigenvalue 1. A0 being 1.5 on some entry could make (I-A0) singular? No, a single entry of 1.5 doesn't make eigenvalue 1. But if A0 is large enough, the inverse could have large entries. But reference computes A0 from k*k*exp(g-g_i) * beta. Same as mine. Then solves. Reference doesn't produce 1e37. Wait, maybe the issue is that in my kernel, I compute Aqk and M with q and k where q is already scaled, but I also multiply by scale again? Let me check: `Aqk = tl.where(mask_lower, Aqk * scale, 0.0)`. And I pass scale=1.0 from Python because q is already scaled. So Aqk is correct. M is not scaled. Good. But wait, in Python I compute `qg = q_scaled * g_cum.exp()` and pass it to kernel 2. But kernel 1 computes Aqk using q_scaled and stable diff. That's equivalent to qg * kdecay, but stable. Good. Hmm. Let me add debug prints inside the kernel? Not possible. But I can write a smaller test kernel that outputs M or A. Actually, let me check if the issue is the `tl.where(c[None,:] == i, col_Aqk, Aqk)` broadcasting. Maybe Triton's broadcasting behavior is different. Let me test with a simple kernel. Actually, I think the issue might be that `col_Aqk` is shape (BT,1) and when broadcasting with `Aqk` (BT,BT), Triton might broadcast col_Aqk across rows, not columns. Let me think about Triton's broadcasting semantics. In numpy/torch, (BT,1) broadcast with (BT,BT) -> (BT,BT) by repeating columns. So col_Aqk[r,c] = col_Aqk[r,0]. That means all columns get the same values from col_Aqk. With mask (1,BT), only column i is selected. So column i gets col_Aqk[:,0], other columns unchanged. Correct. But in my code, mask is `c[None,:] == i` which is (1,BT). Broadcast to (BT,BT) repeats along rows. col_Aqk (BT,1) broadcast repeats along columns. So for mask True (column i), result[r,i] = col_Aqk[r,0]. Correct. If Triton broadcasts mask (1,BT) and x (BT,1) to (BT,BT), it should work. But what if `tl.where` requires x and y to have the same shape as mask? In Triton, broadcasting should happen. Actually, maybe the issue is that `Aqk` starts as zeros, and `tl.where(c[None,:] == i, col_Aqk, Aqk)` works. But after the loop, Aqk has columns. Good. Let me test a minimal kernel with this pattern. Actually, let me first check if the huge value is in w (output of kernel 1) by checking w's max.
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Bash
args
{
  "command": "<1619 chars \u2014 see below>",
  "description": "Debug kernel 1 outputs"
}
command (1619 chars)
cat > /workspace/problems/02_kda_cutlass/debug_sol.py <<'PY'
import torch
import torch.nn as nn
import triton, triton.language as tl

class DebugModel(nn.Module):
    def __init__(self, B,T,H,K,V,CS=64):
        super().__init__()
    def forward(self, q,k,v,g,beta):
        B,T,H,K = q.shape
        V=v.shape[-1]; CS=64; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled = q * scale
        kg = k * g_cum.exp()
        Aqk = torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w = torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u = torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        # call kernel 1
        import solution
        solution._kda_intra_kernel[(NT,B*H)](
            q_scaled, k, v, g_cum, kg, beta, Aqk, w, u,
            1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,
            num_stages=1,num_warps=4,
        )
        return Aqk, w, u

import reference
B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T = B, T
reference.H, reference.K, reference.V, reference.CHUNK_SIZE = H,K,V,CS
torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
dm = DebugModel(B,T,H,K,V).cuda()
Aqk,w,u = dm(q,k,v,g,beta)
print('Aqk max', Aqk.abs().max().item(), 'nan', Aqk.isnan().any().item())
print('w max', w.abs().max().item(), 'nan', w.isnan().any().item())
print('u max', u.abs().max().item(), 'nan', u.isnan().any().item())
PY
uv run python /workspace/problems/02_kda_cutlass/debug_sol.py 2>&1 | tail -40
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tool
stdout (107 chars)
Aqk max 2.6868696212768555 nan False
w max 0.3508794605731964 nan False
u max 0.32112348079681396 nan False
assistant
reasoning (1274 chars)
Kernel 1 outputs are reasonable. So the huge value comes from kernel 2. Let me debug kernel 2. Maybe the issue is in the S update: `k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)`. If d0 is exp(g_last), log(d0) = g_last. Then exp(g_last - g). For tokens with g much larger than g_last, this exp is large. But g_last is the max in chunk, so g_last - g >= 0. exp range [1, exp(range)]. g range within chunk ~ 7, so exp up to ~1000. k ~0.1, k_up ~100. Then k_up.T @ v_i: v_i ~0.3, sum over BT ~ 64*100*0.3 = 1920. S grows. Then next chunk S scaled by d ~ exp(g_last). g_last could be negative, so d < 1, S shrinks. But if g_last is positive, S grows. Over many chunks, S could accumulate. But reference handles it without 1e37. Wait, the issue might be that in kernel 2, I use `b_k0` and `b_g0` loaded as bf16? No, I load as fp32. Good. But then `k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)`. If d0 = 0 (underflow of exp(g_last) if g_last very negative), log(0) = -inf, then exp(-inf - g) = 0, k_up = 0. OK. But if d0 is small but not zero, log is finite. Actually, d0 = exp(g_last). If g_last is positive, d0 > 1. If g_last > ~88, exp overflows. But g_last is cumsum of 64 values ~ mean -3.2, std 0.8. Unlikely > 88. Hmm. Let me check kernel 2 outputs directly.
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Bash
args
{
  "command": "<1894 chars \u2014 see below>",
  "description": "Debug kernel 2 output"
}
command (1894 chars)
cat > /workspace/problems/02_kda_cutlass/debug_sol2.py <<'PY'
import torch
import torch.nn as nn
import triton, triton.language as tl
import solution

class DebugModel2(nn.Module):
    def __init__(self, B,T,H,K,V,CS=64):
        super().__init__()
    def forward(self, q,k,v,g,beta):
        B,T,H,K = q.shape
        V=v.shape[-1]; CS=64; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled = q * scale
        qg = q_scaled * g_cum.exp()
        kg = k * g_cum.exp()
        glast = g_cum.view(B,NT,CS,H,K)[:,:,-1,:,:].permute(0,2,1,3).contiguous()
        Aqk = torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w = torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u = torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](
            q_scaled, k, v, g_cum, kg, beta, Aqk, w, u,
            1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,
            num_stages=1,num_warps=4,
        )
        o = torch.empty(B,T,H,V,device='cuda',dtype=torch.bfloat16)
        solution._kda_inter_kernel[(triton.cdiv(V,32),B*H)](
            w, u, qg, k, g_cum, Aqk, glast, o,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,
            num_stages=1,num_warps=4,
        )
        return o

import reference
B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T = B, T
reference.H, reference.K, reference.V, reference.CHUNK_SIZE = H,K,V,CS
torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
dm = DebugModel2(B,T,H,K,V).cuda()
o = dm(q,k,v,g,beta)
print('o max', o.abs().max().item(), 'nan', o.isnan().any().item(), 'inf', o.isinf().any().item())
print('first values', o[0,0,0,:5])
PY
uv run python /workspace/problems/02_kda_cutlass/debug_sol2.py 2>&1 | tail -40
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tool
stdout (173 chars)
o max 0.015869140625 nan False inf False
first values tensor([ 2.2984e-04, -1.0347e-04, -3.3140e-05,  1.6499e-04, -6.4261e-08],
       device='cuda:0', dtype=torch.bfloat16)
assistant
reasoning (1353 chars)
Wait, now kernel 2 output is reasonable! But check.py reported huge diff. Why? Oh, in check.py, it uses `sol.load_state_dict(sd, strict=True)`. My DebugModel2 doesn't load state_dict. Maybe loading state_dict changes something? But both models have only `_dummy`. Shouldn't matter. Wait, but check.py calls `reference.get_inputs()` which uses reference's module-level B,T,etc. I set reference's variables in my debug. But check.py might set solution's variables too? No. Actually, the issue might be that check.py uses the actual `solution.Model` and my DebugModel2 has a different forward (but same code). Let me run check.py again to see if it still fails. Actually wait, my debug_sol2 gave o max 0.0159, but earlier the full solution Model gave 1e37. Why the difference? Let me compare DebugModel2 forward to solution.Model forward. They should be identical. Oh! I see: in DebugModel2, I compute `qg = q_scaled * g_cum.exp()` and `kg = k * g_cum.exp()`. Then in kernel 1, I pass `q_scaled` (not qg) and compute Aqk stably. In solution.Model, I also pass `q_scaled` to kernel 1 and compute qg for kernel 2. So same. Wait, but solution.Model also computes `kg = k * g_cum.exp()` and passes to kernel 1. Same. Why different results? Maybe because in debug_sol2 I didn't move inputs to cuda? No, I did. Let me check solution.Model directly again.
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Bash
args
{
  "command": "<755 chars \u2014 see below>",
  "description": "Check solution Model directly"
}
command (755 chars)
uv run python - <<'PY' 2>&1 | tail -80
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol.load_state_dict(ref.state_dict(), strict=True)
with torch.no_grad():
    so = sol(q,k,v,g,beta)
print('sol max', so.abs().max().item(), 'nan', so.isnan().any().item(), 'inf', o.isinf().any().item())
PY
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stdout (129 chars)
Traceback (most recent call last):
  File "<stdin>", line 16, in <module>
NameError: name 'o' is not defined. Did you mean: 'so'?
assistant
reasoning (17 chars)
Typo, let me fix.
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Bash
args
{
  "command": "<756 chars \u2014 see below>",
  "description": "Check solution Model directly"
}
command (756 chars)
uv run python - <<'PY' 2>&1 | tail -20
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol.load_state_dict(ref.state_dict(), strict=True)
with torch.no_grad():
    so = sol(q,k,v,g,beta)
print('sol max', so.abs().max().item(), 'nan', so.isnan().any().item(), 'inf', so.isinf().any().item())
PY
in 118908out 0cache_r 0cache_w 0
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stdout (42 chars)
sol max 0.015869140625 nan False inf False
assistant
reasoning (577 chars)
Now solution.Model gives 0.0159, same as debug. But earlier check.py reported 1e37. Why? Maybe the first run cached a bad kernel? Or check.py uses different code path? Wait, earlier when I ran check.py, the solution.py was different (the unstable version). After I rewrote, maybe check.py is importing a cached bad version? No, Python imports fresh. Actually, check.py reported `max_abs_diff=4.75199e+37` for seed 42 case nominal. But now solution.Model gives 0.0159. Maybe the difference is that check.py sets solution module-level variables? No. Let me run check.py again.
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Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run check again"
}
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stdout (230 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case nominal: tolerance exceeded: max_abs_diff=4.75199e+37 max_rel_diff=inf bad=508073/2097152 worst_idx=(0, 0, 3, 118) (atol=0.05, rtol=0.05)
assistant
reasoning (1144 chars)
Still fails in check.py but my direct test passes. There must be a difference. Let me look at check.py more carefully. check.py: ```python init_args = reference.get_init_inputs() ref_model = reference.Model(*init_args).to(device).eval() sol_model = solution.Model(*init_args).to(device).eval() sd = ref_model.state_dict() sol_model.load_state_dict(sd, strict=True) for seed in (42, 123, 456): torch.manual_seed(seed) torch.cuda.manual_seed_all(seed) base_inputs = [t.to(device) if hasattr(t, 'to') else t for t in reference.get_inputs()] for case in numeric_stress_cases(...): with numeric_stress_context(ref_model, sol_model, base_inputs, case) as inputs: with torch.no_grad(): ref_out = ref_model(*inputs) sol_out = sol_model(*inputs) ``` So it uses `reference.get_inputs()` and passes to both. Model receives init args from reference. That matches my direct test. But wait, `numeric_stress_context` might modify inputs. For nominal case, it doesn't. For small_qkv, it scales q,k,v. But the failure is case nominal seed 42. Hmm. Let me look at `numeric_stress_context`.
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Bash
args
{
  "command": "find /workspace -path '*src/eval/numeric_stress.py' -type f -exec sed -n '80,160p' {} \\;",
  "description": "Check numeric stress context"
}
in 119530out 0cache_r 0cache_w 0
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stdout (2795 chars) — TRUNCATED
),
    "07_w4a16_gemm": (
        NumericStressCase("small_activation", input_scales={0: 1e-3}, tolerance=_SMALL_BF16),
        NumericStressCase(
            "large_activation",
            input_scales={0: 64.0},
            tolerance={"bfloat16": {"atol": 1.0, "rtol": 5e-2}},
        ),
    ),
    "09_fmha_preattn_mrope": (
        NumericStressCase("small_qk", input_scales={0: 1e-2, 1: 1e-2}, tolerance=_SMALL_BF16),
        NumericStressCase("large_qk", input_scales={0: 16.0, 1: 16.0}, tolerance=_LARGE_BF16),
    ),
    "10_patch_embed_conv3d_gemm": (
        NumericStressCase("small_input", input_scales={0: 1e-2}, tolerance=_SMALL_BF16),
        NumericStressCase("large_input", input_scales={0: 16.0}, tolerance=_LARGE_BF16),
    ),
}


def numeric_stress_cases(problem_name: str) -> tuple[NumericStressCase, ...]:
    if os.environ.get("KBH_NUMERIC_STRESS", "1").lower() in {"0", "false", "no"}:
        return (NOMINAL_CASE,)
    return (NOMINAL_CASE, *_CASES.get(problem_name, ()))


def tolerance_for_case(
    base: dict | None,
    case: NumericStressCase,
) -> dict | None:
    if case.tolerance is None:
        return base
    merged = dict(base or {})
    merged.update(case.tolerance)
    return merged


@contextmanager
def numeric_stress_context(
    ref_model: torch.nn.Module,
    sol_model: torch.nn.Module,
    inputs: Sequence[object],
    case: NumericStressCase,
) -> Iterator[list[object]]:
    backups = _scale_states([ref_model, sol_model], case.state_scales)
   

... (1295 more chars)
assistant
reasoning (570 chars)
For nominal case, scales empty, so inputs unchanged. So check.py should give same as my direct test. But it doesn't. Why? Wait, maybe check.py imports solution and reference, and the module-level variables in solution are used by get_inputs? But check.py uses reference.get_inputs(). Model receives init_args from reference.get_init_inputs(). So solution's module-level variables don't matter for Model init. But maybe my direct test set solution's variables, and check.py doesn't. But Model init overrides them. Hmm. Let me run a script that exactly mimics check.py.
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Bash
args
{
  "command": "<961 chars \u2014 see below>",
  "description": "Mimic check.py exactly"
}
command (961 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check.py <<'PY'
import torch
import reference, solution

shape = {"B":2,"T":1024,"H":8,"K":128,"V":128,"CHUNK_SIZE":64}
for k,v in shape.items():
    setattr(reference,k,v)

init_args = reference.get_init_inputs()
ref_model = reference.Model(*init_args).cuda().eval()
sol_model = solution.Model(*init_args).cuda().eval()
sd = ref_model.state_dict()
sol_model.load_state_dict(sd, strict=True)

seed = 42
k = 'torch.bfloat16'
torch.manual_seed(seed)
torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
with torch.no_grad():
    ref_out = ref_model(*inputs)
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('nan', sol_out.isnan().any().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check.py 2>&1 | tail -20
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stdout (85 chars)
ref max 0.01171875
sol max 4.751990084931074e+37
nan False
diff 4.751990084931074e+37
assistant
reasoning (2718 chars)
Now I reproduce it. The difference from my direct test must be that I set `solution.B, solution.T, ...` in my direct test, but mimic_check doesn't. Wait, but Model init receives args. Why would module-level variables matter? Oh! In solution.py, the `Model.forward` calls `self.scale = float(K) ** -0.5` where K is the init arg. That's fine. But maybe the module-level `B,T,H,K,V` are used somewhere? No, forward uses q.shape. Wait, in mimic_check.py, I didn't import solution's get_inputs, only reference's. That's same as check.py. Hmm, but my direct test set solution's module-level variables. Maybe that affects the Triton kernel compilation? The kernel is compiled with constexprs from the launch args, not from module-level variables. So no. Wait, let me compare my direct test code vs mimic_check.py. In direct test, I set: ```python solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS ``` And I used `solution.Model(B,T,H,K,V,CS)`. In mimic_check, I use `solution.Model(*init_args)` where init_args = reference.get_init_inputs(). That's [B,T,H,K,V,CS]. Same. But wait, in my direct test, I set solution's module-level variables BEFORE creating the model. Maybe that changes something? No. Actually, in my direct test I used `sol = solution.Model(B,T,H,K,V,CS).cuda().eval()` and then `sol.load_state_dict(ref.state_dict(), strict=True)`. In mimic_check, same. Let me check if there's a difference in the inputs. In my direct test, I called `reference.get_inputs()` which uses reference's module-level variables. In mimic_check, I set reference's variables. Same. Wait, in my direct test, I also set `reference.B,...`. So same. Let me check the actual solution.Model in both cases. Maybe there's caching issue? No. Actually, wait. Let me look at my direct test output again. It printed `sol max 0.015869140625`. That was for seed 42. And mimic_check gives 1e37. The only difference is that in direct test I set solution's module-level variables. Let me try not setting them. Actually, in my direct test I had: ```python reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS ``` Maybe setting solution's module-level B/T/H/K/V changes the Triton kernel's behavior? No, kernel gets launch args. Wait, could it be that the Triton kernel caches based on the source code, and the first compilation (when solution.B etc were default 2,1024,8,128,128) somehow compiled with different constants? No, kernel launch uses runtime args. Let me run mimic_check but set solution's variables too, to see if that's the difference.
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args
{
  "command": "<989 chars \u2014 see below>",
  "description": "Mimic check with solution vars set"
}
command (989 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check2.py <<'PY'
import torch
import reference, solution

shape = {"B":2,"T":1024,"H":8,"K":128,"V":128,"CHUNK_SIZE":64}
for k,v in shape.items():
    setattr(reference,k,v)
    setattr(solution,k,v)

init_args = reference.get_init_inputs()
ref_model = reference.Model(*init_args).cuda().eval()
sol_model = solution.Model(*init_args).cuda().eval()
sd = ref_model.state_dict()
sol_model.load_state_dict(sd, strict=True)

seed = 42
k = 'torch.bfloat16'
torch.manual_seed(seed)
torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
with torch.no_grad():
    ref_out = ref_model(*inputs)
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('nan', sol_out.isnan().any().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check2.py 2>&1 | tail -20
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stdout (85 chars)
ref max 0.01171875
sol max 4.751990084931074e+37
nan False
diff 4.751990084931074e+37
assistant
reasoning (142 chars)
Still 1e37. So setting solution vars doesn't matter. Then why did my direct test give 0.0159? Let me re-run my direct test exactly as before.
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args
{
  "command": "<756 chars \u2014 see below>",
  "description": "Re-run direct test"
}
command (756 chars)
uv run python - <<'PY' 2>&1 | tail -20
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol.load_state_dict(ref.state_dict(), strict=True)
with torch.no_grad():
    so = sol(q,k,v,g,beta)
print('sol max', so.abs().max().item(), 'nan', so.isnan().any().item(), 'inf', so.isinf().any().item())
PY
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stdout (42 chars)
sol max 0.015869140625 nan False inf False
assistant
reasoning (537 chars)
Direct test gives 0.0159. mimic_check gives 1e37. What's different? Let me look carefully. In direct test: ```python ref = reference.Model(B,T,H,K,V,CS).cuda().eval() sol = solution.Model(B,T,H,K,V,CS).cuda().eval() ``` In mimic_check: ```python init_args = reference.get_init_inputs() ref_model = reference.Model(*init_args).cuda().eval() sol_model = solution.Model(*init_args).cuda().eval() ``` init_args = [B,T,H,K,V,CS]. Same. Wait, maybe `reference.get_init_inputs()` in mimic_check returns something different? Let me print it.
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args
{
  "command": "<957 chars \u2014 see below>",
  "description": "Debug mimic_check init_args"
}
command (957 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check3.py <<'PY'
import torch
import reference, solution

shape = {"B":2,"T":1024,"H":8,"K":128,"V":128,"CHUNK_SIZE":64}
for k,v in shape.items():
    setattr(reference,k,v)

init_args = reference.get_init_inputs()
print('init_args', init_args)

ref_model = reference.Model(*init_args).cuda().eval()
sol_model = solution.Model(*init_args).cuda().eval()
sd = ref_model.state_dict()
sol_model.load_state_dict(sd, strict=True)

seed = 42
torch.manual_seed(seed)
torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
print('inputs shapes', [x.shape if hasattr(x,'shape') else x for x in inputs])
with torch.no_grad():
    ref_out = ref_model(*inputs)
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check3.py 2>&1 | tail -20
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stdout (250 chars)
init_args [2, 1024, 8, 128, 128, 64]
inputs shapes [torch.Size([2, 1024, 8, 128]), torch.Size([2, 1024, 8, 128]), torch.Size([2, 1024, 8, 128]), torch.Size([2, 1024, 8, 128]), torch.Size([2, 1024, 8])]
ref max 0.01171875
sol max 4.751990084931074e+37
assistant
reasoning (1212 chars)
Same init_args, same inputs. But different results. This is very strange. Wait, could it be that in direct test, the Triton kernel was already compiled/cached from debug_sol2.py, and that cached version is different? No, kernel source code is the same. Actually, maybe the issue is that check.py (and mimic_check) run BEFORE my direct test in the same process? No, they're separate processes. Wait, in mimic_check, I don't set solution's module-level variables. But Model init receives args. However, `solution.get_init_inputs` isn't called. The module-level variables in solution remain default (2,1024,8,128,128). But they're not used. Hmm. Unless... the Triton kernel uses `B: tl.constexpr` from launch args, but maybe it also somehow uses global `B`? No. Let me try: in mimic_check, after setting reference variables, also import solution and print solution.B etc. They are default. But Model uses init args. Wait, could there be a name collision? In solution.py, module-level B,T,H,K,V are defined. The Triton kernels have `B: tl.constexpr` as a parameter. No collision. I'm stumped. Let me try to make mimic_check identical to direct test by also creating ref_model and sol_model with explicit args.
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args
{
  "command": "<978 chars \u2014 see below>",
  "description": "Mimic direct test exactly"
}
command (978 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check4.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
sd = ref_model.state_dict()
sol_model.load_state_dict(sd, strict=True)

seed = 42
torch.manual_seed(seed)
torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for x in reference.get_inputs()]
with torch.no_grad():
    ref_out = ref_model(*inputs)
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check4.py 2>&1 | tail -20
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stdout (537 chars)
Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/mimic_check4.py", line 16, in <module>
    inputs = [t.cuda() if hasattr(t,'to') else t for x in reference.get_inputs()]
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/problems/02_kda_cutlass/mimic_check4.py", line 16, in <listcomp>
    inputs = [t.cuda() if hasattr(t,'to') else t for x in reference.get_inputs()]
                                  ^
NameError: name 't' is not defined. Did you mean: 'T'?
assistant
reasoning (17 chars)
Typo. Let me fix.
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args
{
  "command": "<978 chars \u2014 see below>",
  "description": "Mimic direct test exactly"
}
command (978 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check4.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
sd = ref_model.state_dict()
sol_model.load_state_dict(sd, strict=True)

seed = 42
torch.manual_seed(seed)
torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
with torch.no_grad():
    ref_out = ref_model(*inputs)
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check4.py 2>&1 | tail -20
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tool
stdout (75 chars)
ref max 0.01171875
sol max 4.751990084931074e+37
diff 4.751990084931074e+37
assistant
reasoning (874 chars)
Even with explicit args, mimic_check4 gives 1e37, but direct test gives 0.0159. The only difference is the script file. That doesn't make sense. Wait, in direct test, I used `reference.get_inputs()` after setting reference variables. In mimic_check4, same. Actually, in mimic_check4, I set solution variables. In direct test, I also set solution variables. Same. Let me look at the direct test output again: it was `sol max 0.015869140625`. The mimic_check4 gives 1e37. Could it be that the direct test ran in a different working directory or imported a different solution.py? No. Wait, in the direct test, I have: ```python with torch.no_grad(): so = sol(q,k,v,g,beta) ``` In mimic_check4: ```python with torch.no_grad(): sol_out = sol_model(*inputs) ``` Same. Let me run the direct test again in the same command as mimic_check4 to see if it's reproducible.
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args
{
  "command": "<773 chars \u2014 see below>",
  "description": "Run both direct and mimic"
}
command (773 chars)
uv run python - <<'PY' 2>&1 | tail -20
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol.load_state_dict(ref.state_dict(), strict=True)
with torch.no_grad():
    so = sol(q,k,v,g,beta)
print('sol max', so.abs().max().item())
PY

uv run python /workspace/problems/02_kda_cutlass/mimic_check4.py 2>&1 | tail -5
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stdout (98 chars)
sol max 0.015869140625
ref max 0.01171875
sol max 4.751990084931074e+37
diff 4.751990084931074e+37
assistant
reasoning (1262 chars)
Direct test: 0.0159. mimic_check4: 1e37. The code is essentially identical. This is bizarre. Wait, in direct test, I create `ref = reference.Model(...)` and `sol = solution.Model(...)`. In mimic_check4, I create `ref_model` and `sol_model`. Same. Oh! Maybe the issue is that in mimic_check4, I import `reference` and `solution`, and then in the same process, the module-level variables of `solution` are set AFTER the Triton kernels are defined (they're defined at import time, not affected). No. Wait, maybe the Triton kernel caches are keyed by the source file path, and mimic_check4.py imports solution from /workspace/problems/02_kda_cutlass/solution.py, same as direct test. Same source. Actually, I wonder if the issue is that in mimic_check4, the model variable is named `sol_model` which shadows something? No. Let me print the actual model object and its forward source in both cases. Actually, let me add a print in solution.py forward to see inputs. Wait, another idea: in mimic_check4, I call `reference.get_inputs()` after creating the models. In direct test, I call `reference.get_inputs()` after creating the models too (in the line `q,k,v,g,beta = ...`). Same. Hmm. Let me literally copy the exact direct test code into a file and run it.
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args
{
  "command": "<795 chars \u2014 see below>",
  "description": "Run direct test as file"
}
command (795 chars)
cat > /workspace/problems/02_kda_cutlass/direct_file.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()]
ref = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol.load_state_dict(ref.state_dict(), strict=True)
with torch.no_grad():
    so = sol(q,k,v,g,beta)
print('sol max', so.abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/direct_file.py 2>&1 | tail -5
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tool
stdout (19 chars)
sol max 0.052734375
assistant
reasoning (1029 chars)
Now as a file, direct test gives 0.0527, not 0.0159! So running from stdin gave a different result. That's very suspicious. Actually, when running from stdin, maybe the Triton kernel source is from stdin and compiles differently? No, Triton reads the function source from the module file. Wait, from stdin, maybe `inspect.getsourcelines` fails for Triton kernels, but it didn't error. Actually, when I ran test kernels from stdin earlier, it errored. But the direct test from stdin worked? Maybe because solution.py kernels are defined in a file, so source is available. But why different results between stdin and file? Maybe non-determinism in Triton? Or the Triton cache is different? Actually, I think the issue might be that from stdin, the Python process has different `sys.argv[0]` or working directory, affecting Triton cache location. But the cache should return same compiled kernel. Wait, 0.0527 is still reasonable. mimic_check4 gave 1e37. So even file vs file differs. Let me run direct_file.py multiple times.
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args
{
  "command": "for i in 1 2 3; do uv run python /workspace/problems/02_kda_cutlass/direct_file.py 2>&1 | tail -1; done",
  "description": "Run direct file multiple times"
}
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tool
stdout (68 chars)
sol max 0.015869140625
sol max 0.015869140625
sol max 0.015869140625
assistant
reasoning (174 chars)
Now it gives 0.0159 consistently. The first run gave 0.0527 due to cold cache or something. But 0.0159 is stable. mimic_check4 gives 1e37. Why? Let me run it multiple times.
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args
{
  "command": "for i in 1 2 3; do uv run python /workspace/problems/02_kda_cutlass/mimic_check4.py 2>&1 | tail -1; done",
  "description": "Run mimic check multiple times"
}
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tool
stdout (80 chars)
diff 4.751990084931074e+37
diff 4.751990084931074e+37
diff 4.751990084931074e+37
assistant
reasoning (2890 chars)
mimic_check4 consistently gives 1e37. The difference must be something in the code. Let me diff the two files carefully. direct_file.py: ```python import torch import reference, solution B,T,H,K,V,CS = 2,1024,8,128,128,64 reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS torch.manual_seed(42) torch.cuda.manual_seed_all(42) q,k,v,g,beta = [x.cuda() if hasattr(x,'cuda') else x for x in reference.get_inputs()] ref = reference.Model(B,T,H,K,V,CS).cuda().eval() sol = solution.Model(B,T,H,K,V,CS).cuda().eval() sol.load_state_dict(ref.state_dict(), strict=True) with torch.no_grad(): so = sol(q,k,v,g,beta) print('sol max', so.abs().max().item()) ``` mimic_check4.py: ```python import torch import reference, solution B,T,H,K,V,CS = 2,1024,8,128,128,64 reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval() sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval() sd = ref_model.state_dict() sol_model.load_state_dict(sd, strict=True) seed = 42 torch.manual_seed(seed) torch.cuda.manual_seed_all(seed) inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()] with torch.no_grad(): ref_out = ref_model(*inputs) sol_out = sol_model(*inputs) print('ref max', ref_out.abs().max().item()) print('sol max', sol_out.abs().max().item()) print('diff', (ref_out-sol_out).abs().max().item()) ``` The key difference: in mimic_check4, models are created BEFORE setting seeds and generating inputs. In direct_file.py, seeds are set and inputs generated BEFORE creating models. Does creating the model before setting seeds affect the Triton kernel randomness? Triton compilation might use random numbers? That would be very surprising. Actually, Triton autotune might use randomness? But we don't have autotune. Or maybe creating the model triggers Triton kernel compilation, and compilation uses CUDA context that affects subsequent random number generation? No, torch.cuda.manual_seed sets RNG state. Wait, actually: in mimic_check4, `ref_model = reference.Model(...).cuda().eval()` triggers reference's forward? No, eval() doesn't call forward. But `.cuda()` moves the dummy buffer. Then `sol_model = solution.Model(...).cuda().eval()` moves dummy buffer. Doesn't call forward. So no forward calls before input generation. Triton kernels not compiled yet. But wait, maybe the issue is the order of randomness? If model creation consumes random numbers? No. Hmm, let me try reordering in mimic_check4: set seeds and generate inputs first, then create models. Actually, that would match direct_file. Let me test.
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args
{
  "command": "<964 chars \u2014 see below>",
  "description": "Mimic with seeds before model creation"
}
command (964 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check5.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
sd = ref_model.state_dict()
sol_model.load_state_dict(sd, strict=True)

with torch.no_grad():
    ref_out = ref_model(*inputs)
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check5.py 2>&1 | tail -5
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stdout (75 chars)
ref max 0.01171875
sol max 4.751990084931074e+37
diff 4.751990084931074e+37
assistant
reasoning (746 chars)
Still 1e37. So the difference is not about seed ordering. Wait, in direct_file.py, I create `ref` and `sol` and call `sol.load_state_dict(ref.state_dict())`. In mimic_check5, same. But in direct_file.py, I don't call `ref` forward, only `sol` forward. In mimic_check5, I call both ref and sol forward. Could calling ref forward before sol forward affect sol's output? No, they are independent. Actually, wait! In mimic_check5, I call `ref_out = ref_model(*inputs)` BEFORE `sol_out = sol_model(*inputs)`. Reference uses a lot of memory and might leave CUDA memory in a state that affects sol's Triton kernel? No. But maybe reference's forward modifies the inputs in-place? No, it doesn't. Let me try calling sol forward first in mimic_check5.
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args
{
  "command": "<956 chars \u2014 see below>",
  "description": "Call sol before ref"
}
command (956 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check6.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol_model.load_state_dict(ref_model.state_dict(), strict=True)

with torch.no_grad():
    sol_out = sol_model(*inputs)
    ref_out = ref_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check6.py 2>&1 | tail -5
in 127332out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (62 chars)
ref max 0.01171875
sol max 0.015869140625
diff 0.0142822265625
assistant
reasoning (2045 chars)
Aha! Calling sol before ref gives correct result. Calling ref before sol gives 1e37. So reference's forward somehow corrupts the state for my solution's forward. This is very weird. Reference uses Python loops and PyTorch ops. It shouldn't affect CUDA global state. Wait, maybe reference's forward allocates huge intermediate tensors that cause OOM, and CUDA silently corrupts? But no OOM reported. Or maybe reference's forward sets some CUDA device property? No. Actually, maybe reference's forward uses more memory, and my solution's Triton kernel has a bug that manifests when there's memory pressure or when the memory pool is in a certain state? Unlikely. Wait, more likely: reference's forward modifies the input tensors? Let me check. Reference line 47: `q, k, v, g, beta = (x.to(torch.float32) for x in (q, k, v, g, beta))`. This creates new tensors, doesn't modify inputs. Then `q = q * scale`. New tensor. Then rearrange. New tensors. No in-place ops on inputs. Hmm. Actually, maybe the issue is that reference's forward and solution's forward share some intermediate tensor due to module-level variables? No. Wait, could it be that reference's forward triggers a different CUDA context or stream? No. Let me think about what could cause a kernel to produce different results depending on prior GPU work. Triton kernels are deterministic given inputs. Unless there's a race condition or shared memory not initialized properly. Actually, I wonder if the issue is that my kernel 2 uses uninitialized shared memory or has a race condition. When run after reference (which uses many kernels and synchronizations), maybe the timing is different and a race manifests? But Triton kernels within a block are synchronous; no races within a block. Across blocks, no communication. Wait, maybe the issue is that reference's forward leaves the GPU in a state where my kernel reads uninitialized memory due to L2 cache? No, Triton loads from global memory. Hmm. Let me try running reference forward, then a small sync, then sol forward.
in 127902out 0cache_r 0cache_w 0
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Bash
args
{
  "command": "<985 chars \u2014 see below>",
  "description": "Sync between ref and sol"
}
command (985 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check7.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol_model.load_state_dict(ref_model.state_dict(), strict=True)

with torch.no_grad():
    ref_out = ref_model(*inputs)
    torch.cuda.synchronize()
    sol_out = sol_model(*inputs)
print('ref max', ref_out.abs().max().item())
print('sol max', sol_out.abs().max().item())
print('diff', (ref_out-sol_out).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check7.py 2>&1 | tail -5
in 127902out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (75 chars)
ref max 0.01171875
sol max 4.751990084931074e+37
diff 4.751990084931074e+37
assistant
reasoning (144 chars)
Sync doesn't help. So it's not a race. What if reference's forward changes the input tensors? Let me check inputs before and after ref forward.
in 128739out 0cache_r 0cache_w 0
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Bash
args
{
  "command": "<763 chars \u2014 see below>",
  "description": "Check if ref modifies inputs"
}
command (763 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check8.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()

q0 = q.clone()
with torch.no_grad():
    ref_out = ref_model(q,k,v,g,beta)
print('q changed', (q != q0).any().item())
print('q0 max', q0.abs().max().item(), 'q max', q.abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check8.py 2>&1 | tail -10
in 128739out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (50 chars)
q changed False
q0 max 0.33203125 q max 0.33203125
assistant
reasoning (588 chars)
Inputs unchanged. So why does running ref before sol break sol? Let me check if the issue is deterministic: maybe sol's Triton kernel result depends on the order it's launched relative to other kernels due to some persistent state. Actually, wait. I have a hypothesis: reference's forward uses `torch.triu` with `diagonal=0` and `diagonal=1` which might set some CUDA global state? No. Or, reference's forward allocates a tensor of size (BT,BT) with `torch.ones` which triggers some PyTorch initialization? No. Let me try running a dummy PyTorch op before sol to see if it breaks sol.
in 129108out 0cache_r 0cache_w 0
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Bash
args
{
  "command": "<729 chars \u2014 see below>",
  "description": "Dummy op before sol"
}
command (729 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check9.py <<'PY'
import torch
import solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in solution.get_inputs()]

# dummy op
dummy = torch.randn(1000, 1000, device='cuda') @ torch.randn(1000, 1000, device='cuda')
torch.cuda.synchronize()

sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
with torch.no_grad():
    sol_out = sol_model(q,k,v,g,beta)
print('sol max', sol_out.abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check9.py 2>&1 | tail -5
in 129108out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (22 chars)
sol max 0.015869140625
assistant
reasoning (6067 chars)
Dummy op before sol is fine. So it's something specific about reference's forward. Let me try running only part of reference's forward to isolate. But reference is monolithic. I can copy it and comment parts. Actually, let me think about what reference does that dummy doesn't. Reference uses `rearrange` which calls `einops`, creates many views. It uses `torch.triu`, `torch.einsum`, `cumsum`. It might use more memory and trigger memory allocation patterns. Wait, maybe reference's forward causes my Triton kernel to use a different memory allocation that exposes a bug. For example, if my kernel reads from a tensor that was freed and reallocated. But my kernel reads from its own intermediate tensors (Aqk, w, u, etc.) allocated inside forward. They should be valid. Unless... my kernel has a bug where it reads out of bounds, and after reference's forward, the memory at those out-of-bounds locations has different values (maybe inf or huge). In direct test (sol only), the memory might be zero or small. That would explain why sol only works but after ref it breaks: out-of-bounds reads get garbage. So I likely have an out-of-bounds access in my kernel. Let me check the block pointers and indices. Kernel 1: - Grid (NT, B*H). For i_n = NT-1, t0 = (NT-1)*BT. Block ptr `p_g = make_block_ptr(g, (T,K), ..., (t0, off_k), (BT,BK), ...)`. With boundary_check, should handle T not multiple of BT? But T is multiple of BT. So no OOB. Kernel 2: - Grid (V/BV, B*H). Loop i_n in range(NT). t0 = i_n*BT. Block ptrs for w,u,qg,k,g,Aqk with (BT,BK) or (BT,BV) and boundary_check. T multiple of BT. OK. - `p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)`. glast shape (B,H,NT,K). Strides: contiguous, so (H*NT*K, NT*K, K, 1). Offset formula: i_b*H*NT*K + i_h*NT*K + i_n*K + arange. I wrote `(i_b * H + i_h) * NT * K + i_n * K + arange`. That's correct if glast is contiguous with shape (B,H,NT,K). Wait, glast in Python: `glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()`. So shape (B, H, NT, K). Strides: after permute(0,2,1,3) and contiguous, stride is (H*NT*K, NT*K, K, 1). Yes. So p_glast offset is correct. Hmm, where else? The `A = tl.where((r[:,None] == i), new_row[None,:], A)` - all within registers. OK. Maybe the issue is in kernel 2: `b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))`. b_w0 is (BT,BK)=(64,64), S0 is (BK,BV)=(64,32). Dot returns (64,32). b_v is (64,32). OK. Wait, `b_v` is loaded from u (BT,BV). b_v is fp32. `tl.dot(b_w0, S0.to(tl.bfloat16))` returns fp32. Subtract. OK. `b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))`. b_Aqk (64,64), b_v (64,32). OK. `S0 += tl.dot(tl.trans(b_kd0), b_v.to(tl.bfloat16))`. b_kd0 (64,64) transposed to (64,64), b_v (64,32). Dot returns (64,32). OK. `S0 *= d0[:,None]`. d0 (64,), S0 (64,32). OK. All dimensions seem correct. Maybe the issue is that in kernel 1, I compute A with `A = A * b_beta[None, :]` but b_beta shape is (BT,), and A is (BT,BT). OK. Wait, in kernel 1, `Aqk = tl.where(mask_lower, Aqk * scale, 0.0)`. mask_lower is r[:,None] >= c[None,:], i.e., rows >= columns. But Aqk[c,i] where c is row, i is column. We want lower triangular including diagonal, i.e., c >= i. mask_lower with r as rows and c as columns gives True for r>=c, which is c<=r. Wait, r is rows (c in my notation), c is columns (i). So mask_lower is row >= column, which is c >= i. Correct (lower triangular including diagonal). OK. Actually, wait. I just realized: in kernel 1, I load `k_i` from `k_ptr` which is the original k (bf16). But I also load `b_k` from `k_ptr` for the full chunk. Then `k_exp = b_k * exp(diff)`. Then `col_M += tl.dot(k_exp, tl.trans(k_i))`. This computes M[c,i] = sum_k k[c,k] * k[i,k] * exp(g[c,k]-g[i,k]). Correct. But for w = A @ kg, I load `kg_ptr` which is kg = k * exp(g). And A is computed from M (which uses original k and g). Correct. For u = A @ v. Correct. For kernel 2 S update, I load `k_ptr` (original k) and `g_ptr` (g_cum) and compute k_update = k * exp(g_last - g). Correct. Hmm. Let me try to find the exact OOB. Maybe the issue is that `tl.dot(A_bf16, b_kg)` with A shape (64,64) and b_kg shape (64,64) returns (64,64). But A_bf16 might contain values that are huge or inf after the solve. Wait, in debug_sol.py, w max was 0.35, so A is reasonable. Maybe the issue is in kernel 2 reading Aqk. After reference's forward, Aqk buffer might be reused and contain stale data? No, Aqk is allocated fresh in solution's forward. Wait, but when reference runs first, it allocates many tensors. Then solution runs and allocates Aqk, w, u, o, etc. The memory pool might give solution buffers that were previously used by reference. If solution's kernel writes to them correctly, they should be fine. But if there's an OOB write in solution's kernel, it might corrupt memory. Actually, the huge value 1e37 could be from reading uninitialized memory. If a block pointer boundary_check is wrong or an index is wrong, we might read garbage. Let me carefully check all strides and offsets. In kernel 1: - q_ptr, k_ptr, g_ptr, kg_ptr: layout (B,T,H,K), stride (T*H*K, H*K, K, 1). Offsets use i_b, i_h, t0. Correct. - v_ptr: layout (B,T,H,V), stride (T*H*V, H*V, V, 1). Offsets use i_b, i_h, t0. Correct. - beta_ptr: layout (B,T,H), stride (T*H, H, 1). Offsets use i_b, i_h, t0. Correct. - Aqk_ptr, w_ptr, u_ptr: outputs, same layouts. Correct. In kernel 2: - w_ptr, u_ptr, qg_ptr, k_ptr, g_ptr, Aqk_ptr: same as above. - glast_ptr: shape (B,H,NT,K), stride (H*NT*K, NT*K, K, 1). Offset `(i_b * H + i_h) * NT * K + i_n * K + arange`. Correct. - o_ptr: (B,T,H,V), correct. Hmm. All seem correct. Wait, in kernel 2, `i_vb = tl.program_id(0)`, `i_bh = tl.program_id(1)`. Grid is `(triton.cdiv(V, 32), B * H)`. So program_id(0) is V tile, program_id(1) is B*H. Correct. But in kernel 1, `i_n = tl.program_id(0)`, `i_bh = tl.program_id(1)`. Grid `(NT, B*H)`. Correct. Maybe the issue is that the Triton kernel name collides or cache is corrupted? Unlikely. Let me try clearing Triton cache and running mimic_check4.
in 129555out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "rm -rf ~/.triton/cache /tmp/triton_* 2>/dev/null; uv run python /workspace/problems/02_kda_cutlass/mimic_check4.py 2>&1 | tail -5",
  "description": "Clear Triton cache and rerun"
}
in 129555out 0cache_r 0cache_w 0
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tool
stdout (75 chars)
ref max 0.01171875
sol max 4.751990084931074e+37
diff 4.751990084931074e+37
assistant
reasoning (685 chars)
Still 1e37. So it's not cache corruption. Wait, I just realized something. In mimic_check4, I call `reference.get_inputs()` which uses reference's module-level B,T,etc. I set those. But `solution.get_inputs()` is never called. The default solution module-level variables remain 2,1024,8,128,128. Not used. But wait, in solution.py, the module-level `B,T,H,K,V,CHUNK_SIZE` are used by `get_inputs` and `get_init_inputs`. But I don't call them. Hmm. Actually, I think I need to check if reference's forward and solution's forward are using the same random seed or something. No. Let me try a different approach: run sol forward twice in mimic_check4, once before ref and once after.
in 131631out 0cache_r 0cache_w 0
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Bash
args
{
  "command": "<937 chars \u2014 see below>",
  "description": "Sol before and after ref"
}
command (937 chars)
cat > /workspace/problems/02_kda_cutlass/mimic_check10.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42)
torch.cuda.manual_seed_all(42)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
sol_model = solution.Model(B,T,H,K,V,CS).cuda().eval()
sol_model.load_state_dict(ref_model.state_dict(), strict=True)

with torch.no_grad():
    sol1 = sol_model(*inputs)
    print('sol1 max', sol1.abs().max().item())
    ref_out = ref_model(*inputs)
    sol2 = sol_model(*inputs)
    print('sol2 max', sol2.abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/mimic_check10.py 2>&1 | tail -5
in 131631out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (36 chars)
sol1 max 0.015869140625
sol2 max nan
assistant
reasoning (3885 chars)
sol1 is fine, sol2 after ref is NaN. So reference's forward corrupts something that makes sol's second forward produce NaN. This strongly suggests that my kernel has an out-of-bounds write that corrupts memory used by reference, or reference has an OOB write that corrupts memory used by sol's second run. Actually, reference's forward uses a lot of memory. If my kernel has a subtle OOB write into reference's buffers, it would corrupt them. But sol2 reads from inputs and its own intermediates. If inputs are unchanged (we checked), then maybe sol2's intermediates are allocated in memory previously used by reference and my kernel writes OOB into them. But sol1 works fine. After sol1, the memory contains sol1's results. Then ref runs and overwrites much memory. Then sol2 runs and allocates new buffers. If my kernel writes OOB into sol2's buffers (e.g., writing past the end of Aqk), it might write into adjacent memory that happens to contain ref's leftover data. But that would only affect values beyond Aqk's bounds, not Aqk itself. Wait, but if the OOB write corrupts the input tensors or other intermediates that sol2 reads, then sol2 output is wrong. We checked inputs unchanged. But maybe my kernel reads from a tensor using a wrong pointer and gets ref's leftover data. Actually, maybe the issue is simpler: my kernel has an OOB read that, after ref, reads from a memory location containing NaN or huge value, and that propagates. Let me check for OOB in my kernels. The most suspicious is the `tl.where` row extraction in kernel 1 solve, and the `d0,d1` indexing in kernel 2. Actually, I just noticed: in kernel 2, I have: ```python p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK) d0 = tl.exp(tl.load(p_glast0)) ``` But glast has shape (B,H,NT,K). The offset `(i_b * H + i_h) * NT * K` assumes the stride between (i_b, i_h) and (i_b, i_h+1) is NT*K. For contiguous (B,H,NT,K), stride for H dim is NT*K. For B dim is H*NT*K. So offset = i_b * (H*NT*K) + i_h * (NT*K). I wrote `(i_b * H + i_h) * NT * K` which equals `i_b * H * NT * K + i_h * NT * K`. That's correct. OK. What about kernel 1 `p_gi` and `p_ki` with shape (1,BK)? `make_block_ptr(g, (T,K), (H*K,1), (t0+i, off_k), (1,BK), (1,0))`. This loads 1 row, BK cols. boundary_check=(0,1). Should be fine. Hmm. Actually, maybe the issue is that reference's forward leaves the CUDA device in a bad state because it uses too much memory and triggers some memory allocation that my kernel doesn't handle. But sol1 works, so memory allocation works. Wait, sol1 works, ref runs, sol2 fails. If my kernel had an OOB write, it would corrupt memory during sol1 too. Unless the corrupted memory isn't reused until after ref. Actually, here's a thought: my kernel might write to output `o` beyond its bounds. If `o` has shape (B,T,H,V) and I write with block ptr (BT,BV) where BV=32 and V=128, grid is V/BV=4. For i_vb=0..3, off_v0=0,32,64,96. Each writes (BT,32). Last writes off_v0=96, covers 96..127. OK. No OOB. What about Aqk output? Shape (B,T,H,BT). Block ptr (BT,BT). For i_n=NT-1, t0=(NT-1)*BT. Block covers t0..t0+BT-1 = T-BT..T-1. OK. w output: shape (B,T,H,K). Block ptr (BT,BK) with BK=64, K=128. For i_k=0, covers cols 0..63. For i_k=1, covers 64..127. OK. u output: shape (B,T,H,V). Block ptr (BT,BV) with BV=32, V=128. i_v=0..3. OK. What about reads? Same. Hmm. Actually, maybe the issue is in kernel 1: `Aqk = tl.where(c[None,:] == i, col_Aqk, Aqk)` with col_Aqk shape (BT,1). If Triton's broadcasting is wrong, it might write to wrong columns. But that would just make Aqk wrong, not cause huge values necessarily. Wait, but if col_Aqk is broadcast across rows when it should be across columns, then Aqk could end up with values from a single column repeated, leading to wrong (but not necessarily huge) values. Let me test by printing Aqk from sol1 and sol2.
in 132312out 0cache_r 0cache_w 0
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Bash
args
{
  "command": "<1930 chars \u2014 see below>",
  "description": "Check kernel 1 outputs before and after ref"
}
command (1930 chars)
cat > /workspace/problems/02_kda_cutlass/debug_kernel2.py <<'PY'
import torch
import torch.nn as nn
import triton, triton.language as tl
import solution, reference

class DebugK2(nn.Module):
    def __init__(self,B,T,H,K,V,CS=64):
        super().__init__()
    def forward(self, q,k,v,g,beta):
        B,T,H,K = q.shape; V=v.shape[-1]; CS=64; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled = q * scale
        qg = q_scaled * g_cum.exp()
        kg = k * g_cum.exp()
        glast = g_cum.view(B,NT,CS,H,K)[:,:,-1,:,:].permute(0,2,1,3).contiguous()
        Aqk = torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w = torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u = torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk, w, u, qg, kg, glast, g_cum

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [x.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
dk = DebugK2(B,T,H,K,V).cuda()

with torch.no_grad():
    Aqk1,w1,u1,qg,kg,glast,g_cum = dk(q,k,v,g,beta)
    print('before ref: Aqk max', Aqk1.abs().max().item(), 'w max', w1.abs().max().item(), 'u max', u1.abs().max().item())
    _ = ref_model(q,k,v,g,beta)
    Aqk2,w2,u2,_,_,_,_ = dk(q,k,v,g,beta)
    print('after ref: Aqk max', Aqk2.abs().max().item(), 'w max', w2.abs().max().item(), 'u max', u2.abs().max().item())
    print('Aqk equal', torch.allclose(Aqk1,Aqk2,atol=0,rtol=0))
PY
uv run python /workspace/problems/02_kda_cutlass/debug_kernel2.py 2>&1 | tail -10
in 132312out 0cache_r 0cache_w 0
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tool
stdout (523 chars)
Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/debug_kernel2.py", line 27, in <module>
    q,k,v,g,beta = [x.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/problems/02_kda_cutlass/debug_kernel2.py", line 27, in <listcomp>
    q,k,v,g,beta = [x.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
                    ^
NameError: name 'x' is not defined
assistant
reasoning (5 chars)
Typo.
in 134353out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<1930 chars \u2014 see below>",
  "description": "Check kernel 1 outputs before and after ref"
}
command (1930 chars)
cat > /workspace/problems/02_kda_cutlass/debug_kernel2.py <<'PY'
import torch
import torch.nn as nn
import triton, triton.language as tl
import solution, reference

class DebugK2(nn.Module):
    def __init__(self,B,T,H,K,V,CS=64):
        super().__init__()
    def forward(self, q,k,v,g,beta):
        B,T,H,K = q.shape; V=v.shape[-1]; CS=64; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled = q * scale
        qg = q_scaled * g_cum.exp()
        kg = k * g_cum.exp()
        glast = g_cum.view(B,NT,CS,H,K)[:,:,-1,:,:].permute(0,2,1,3).contiguous()
        Aqk = torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w = torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u = torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk, w, u, qg, kg, glast, g_cum

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
dk = DebugK2(B,T,H,K,V).cuda()

with torch.no_grad():
    Aqk1,w1,u1,qg,kg,glast,g_cum = dk(q,k,v,g,beta)
    print('before ref: Aqk max', Aqk1.abs().max().item(), 'w max', w1.abs().max().item(), 'u max', u1.abs().max().item())
    _ = ref_model(q,k,v,g,beta)
    Aqk2,w2,u2,_,_,_,_ = dk(q,k,v,g,beta)
    print('after ref: Aqk max', Aqk2.abs().max().item(), 'w max', w2.abs().max().item(), 'u max', u2.abs().max().item())
    print('Aqk equal', torch.allclose(Aqk1,Aqk2,atol=0,rtol=0))
PY
uv run python /workspace/problems/02_kda_cutlass/debug_kernel2.py 2>&1 | tail -10
in 134353out 0cache_r 0cache_w 0
tool
stdout (188 chars)
before ref: Aqk max 0.036124639213085175 w max 2.6868696212768555 u max 0.32112348079681396
after ref: Aqk max 0.96484375 w max 2.6868696212768555 u max 0.32112348079681396
Aqk equal False
assistant
reasoning (1548 chars)
Aqk changes after ref runs! w and u stay same. So kernel 1's Aqk output is corrupted by reference's forward. This means my kernel 1 has an out-of-bounds write that writes into memory later used by reference, OR reference has an OOB write that overwrites my Aqk buffer. But Aqk is allocated by my kernel (DebugK2) after ref runs. So ref can't write to it unless it writes to freed memory and my Aqk happens to be allocated there. Actually, Aqk2 is allocated after ref. If ref writes to freed memory (OOB in ref), it could corrupt Aqk2. But reference uses safe PyTorch ops, unlikely OOB. More likely: my kernel 1 has an OOB write. When run before ref, the OOB write goes into memory that's not yet allocated or is zero, so Aqk1 looks fine. When run after ref, the memory pool has ref's data, and my OOB write corrupts something? But Aqk2 itself is wrong, not some other tensor. Wait, Aqk2 is the output of my kernel. If my kernel writes OOB into Aqk's memory, it could write past the end of Aqk. But Aqk2's values within bounds are different from Aqk1. So it's not just OOB; the in-bounds values are wrong. Hmm. Unless the OOB write corrupts the input q,k,g,beta or kg that my kernel reads, causing different Aqk output. Actually, in debug_kernel2, I allocate Aqk, w, u. My kernel reads q,k,v,g_cum,kg,beta (inputs). If my kernel writes OOB, it could corrupt these input tensors. Then Aqk2 would use corrupted inputs. But we checked earlier that ref doesn't modify inputs. But my kernel might. Let me check if inputs are modified by my kernel.
in 135208out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<1684 chars \u2014 see below>",
  "description": "Check if kernel modifies inputs"
}
command (1684 chars)
cat > /workspace/problems/02_kda_cutlass/debug_kernel3.py <<'PY'
import torch
import torch.nn as nn
import triton, triton.language as tl
import solution, reference

class DebugK3(nn.Module):
    def __init__(self,B,T,H,K,V,CS=64):
        super().__init__()
    def forward(self, q,k,v,g,beta):
        B,T,H,K = q.shape; V=v.shape[-1]; CS=64; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled = q * scale
        qg = q_scaled * g_cum.exp()
        kg = k * g_cum.exp()
        Aqk = torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w = torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u = torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk, w, u

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]
ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
dk = DebugK3(B,T,H,K,V).cuda()

q0,k0,v0,g0,beta0 = q.clone(),k.clone(),v.clone(),g.clone(),beta.clone()
with torch.no_grad():
    Aqk1,w1,u1 = dk(q,k,v,g,beta)
    print('input changed by dk?', (q!=q0).any().item(), (k!=k0).any().item(), (v!=v0).any().item(), (g!=g0).any().item(), (beta!=beta0).any().item())
    print('kg changed?', 'no kg')
PY
uv run python /workspace/problems/02_kda_cutlass/debug_kernel3.py 2>&1 | tail -10
in 135208out 0cache_r 0cache_w 0
tool
stdout (68 chars)
input changed by dk? False False False False False
kg changed? no kg
assistant
reasoning (1459 chars)
Inputs not modified. But Aqk2 differs after ref. So either: 1. My kernel reads from uninitialized memory (not inputs) and the content changes after ref. 2. My kernel's Aqk output is corrupted by some other write. Wait, in debug_kernel2, Aqk2 is computed after ref. The inputs to kernel 1 are the same. The kernel code is the same. The only thing different is the memory pool state. If my kernel reads from a tensor using a pointer that's wrong, it could read garbage. But it reads from q,k,v,g,beta,kg which are all valid inputs. Actually, wait. In debug_kernel2, I compute `kg = k * g_cum.exp()` in Python. After ref runs, is kg still the same? Yes, ref doesn't modify kg. Hmm. Actually, maybe the issue is that my kernel uses `scale=1.0` but I pass q_scaled. Wait, in debug_kernel2 I pass `q_scaled` and `scale=1.0`. In my solution.Model, I also pass `q_scaled` and `scale=1.0`. Same. But wait, in my kernel signature, `scale` is the first argument after pointers. In debug_kernel2, I pass `1.0`. In solution.Model, I pass `1.0`. Same. Actually, looking at kernel 1 signature: `_kda_intra_kernel(qg_ptr, kg_ptr, kdecay_ptr, v_ptr, beta_ptr, Aqk_ptr, w_ptr, u_ptr, glast_ptr, scale, ...)`. But in my current solution.py, kernel 1 signature is `_kda_intra_kernel(q_ptr, k_ptr, v_ptr, g_ptr, kg_ptr, beta_ptr, Aqk_ptr, w_ptr, u_ptr, scale, ...)`. Let me verify. Actually, I rewrote solution.py. Let me read it to confirm the kernel signature and launch.
in 136276out 0cache_r 0cache_w 0
assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 120
}
in 136276out 0cache_r 0cache_w 0
tool
stdout (4100 chars) — TRUNCATED
1	"""Kimi Delta Attention (KDA) forward (chunk form) via Triton.
2	
3	Matches reference.py's semantics: inputs are bf16 (g fp32), chunk_size=64,
4	no initial/final state, returns bf16 o.
5	"""
6	from __future__ import annotations
7	
8	import torch
9	import torch.nn as nn
10	import triton
11	import triton.language as tl
12	
13	OP_TYPE = "linear_attention"
14	SUPPORTED_PRECISIONS = ["bf16"]
15	HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
16	
17	
18	@triton.jit
19	def _kda_intra_kernel(
20	    q_ptr,
21	    k_ptr,
22	    v_ptr,
23	    g_ptr,
24	    kg_ptr,
25	    beta_ptr,
26	    Aqk_ptr,
27	    w_ptr,
28	    u_ptr,
29	    scale,
30	    B: tl.constexpr,
31	    T: tl.constexpr,
32	    H: tl.constexpr,
33	    K: tl.constexpr,
34	    V: tl.constexpr,
35	    BT: tl.constexpr,
36	    BK: tl.constexpr,
37	    BV: tl.constexpr,
38	):
39	    i_n = tl.program_id(0)
40	    i_bh = tl.program_id(1)
41	    i_b = i_bh // H
42	    i_h = i_bh % H
43	
44	    t0 = i_n * BT
45	    r = tl.arange(0, BT)
46	    c = tl.arange(0, BT)
47	    mask_strict = r[:, None] > c[None, :]
48	    mask_lower = r[:, None] >= c[None, :]
49	
50	    # Compute Aqk and M = kg @ kdecay.T using the stable formulation
51	    # Aqk[c,i] = sum_k q[c,k]*k[i,k]*exp(g[c,k]-g[i,k])
52	    # M[c,i]  = sum_k k[c,k]*k[i,k]*exp(g[c,k]-g[i,k])
53	    Aqk = tl.zeros([BT, BT], dtype=tl.float32)
54	    M = tl.zeros([BT, BT], dtype=tl.float32)
55	
56	    for i in range(BT):
57	        col_Aqk = tl.zeros([BT, 1], dtype=tl.float32)


... (2600 more chars)
assistant
reasoning (483 chars)
OK signature matches. Wait, I see `Aqk = tl.where(c[None, :] == i, col_Aqk, Aqk)`. `c[None,:]` is (1,BT). `col_Aqk` is (BT,1). `Aqk` is (BT,BT). The broadcasting might be wrong in Triton. Let me test this exact pattern in a small kernel. Actually, I think the issue might be that `tl.where` with operands of different shapes might not broadcast as expected, and could produce undefined behavior or wrong results that depend on memory layout. Let me test the column update pattern.
in 138178out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<697 chars \u2014 see below>",
  "description": "Test tl.where column update"
}
command (697 chars)
cat > /workspace/problems/02_kda_cutlass/test_where.py <<'PY'
import torch
import triton, triton.language as tl

@triton.jit
def test_where_col(out_ptr, N: tl.constexpr):
    r = tl.arange(0, N)
    c = tl.arange(0, N)
    x = tl.zeros([N, N], dtype=tl.float32)
    for i in range(N):
        col = tl.full([N, 1], float(i+1), dtype=tl.float32)
        x = tl.where(c[None, :] == i, col, x)
    tl.store(out_ptr + tl.arange(0, N*N), tl.reshape(x, (N*N,)))

N = 8
out = torch.empty(N*N, device='cuda', dtype=torch.float32)
test_where_col[(1,)](out, N=N)
out = out.view(N,N)
print(out)
print('expected diagonal-ish?')
PY
uv run python /workspace/problems/02_kda_cutlass/test_where.py 2>&1 | tail -20
in 138178out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (1357 chars)
kernel = self._do_compile(key, signature, device, constexprs, options, attrs, warmup)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 849, in _do_compile
    kernel = self.compile(src, target=target, options=options.__dict__)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 304, in compile
    module = src.make_ir(target, options, codegen_fns, module_map, context)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 80, in make_ir
    return ast_to_ttir(self.fn, self, context=context, options=options, codegen_fns=codegen_fns,
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
triton.compiler.errors.CompilationError: at 6:30:
def test_where_col(out_ptr, N: tl.constexpr):
    r = tl.arange(0, N)
    c = tl.arange(0, N)
    x = tl.zeros([N, N], dtype=tl.float32)
    for i in range(N):
        col = tl.full([N, 1], float(i+1), dtype=tl.float32)
                              ^
TypeError("float() argument must be a string or a real number, not 'tensor'")
assistant
reasoning (50 chars)
`i+1` is a tensor in Triton. Use `i + 1` directly.
in 138924out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<660 chars \u2014 see below>",
  "description": "Test tl.where column update"
}
command (660 chars)
cat > /workspace/problems/02_kda_cutlass/test_where.py <<'PY'
import torch
import triton, triton.language as tl

@triton.jit
def test_where_col(out_ptr, N: tl.constexpr):
    r = tl.arange(0, N)
    c = tl.arange(0, N)
    x = tl.zeros([N, N], dtype=tl.float32)
    for i in range(N):
        col = tl.full([N, 1], i + 1, dtype=tl.float32)
        x = tl.where(c[None, :] == i, col, x)
    tl.store(out_ptr + tl.arange(0, N*N), tl.reshape(x, (N*N,)))

N = 8
out = torch.empty(N*N, device='cuda', dtype=torch.float32)
test_where_col[(1,)](out, N=N)
out = out.view(N,N)
print(out)
PY
uv run python /workspace/problems/02_kda_cutlass/test_where.py 2>&1 | tail -20
in 138924out 0cache_r 0cache_w 0
tool
stdout (353 chars)
tensor([[1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.],
        [1., 2., 3., 4., 5., 6., 7., 8.]], device='cuda:0')
assistant
reasoning (4598 chars)
So `tl.where(c[None,:] == i, col, x)` with col shape (N,1) sets each row to [1,2,3,...,N], i.e., it broadcasts col across columns. Wait, the output has row 0 = [1,2,3,...]. That means column j got value j+1. So col (N,1) was broadcast across columns, and column j got col[:,0]. For column 0, value 1. Column 1, value 2. Correct. So the column update pattern works. Good. Then why does Aqk change after ref? Let me check if it's specifically Aqk or all outputs. w and u stayed same. Aqk changed. Wait, in debug_kernel2, I computed Aqk before and after ref. Aqk1 max 0.036, Aqk2 max 0.965. That's a huge change. Could it be that the inputs q,k,g are read differently due to memory layout changes after ref? No, inputs are unchanged. Wait, maybe my kernel 1 has a race condition or uses uninitialized shared memory, and the behavior depends on prior GPU state. When run first, it happens to produce one result. When run after ref, it produces another. But the kernel has no shared memory declared. It uses registers and global memory. Triton manages shared memory for block loads. Could there be uninitialized shared memory that affects results? Only if my kernel uses values from shared memory without initializing them. But all shared memory use is through block pointer loads with boundary_check, which initialize. Hmm. Actually, wait. In kernel 1, I have a loop `for i in range(BT):`. Inside, for each i, I load p_gi, p_ki, p_q, p_k, p_g. These are block pointer loads. Triton might use shared memory as a staging buffer. If a previous kernel (reference) left shared memory with values, and my kernel reuses that shared memory but doesn't fully overwrite it due to boundary_check or partial loads, it could read stale data. But boundary_check should fill with 0 or other. And partial loads (shape (1,BK) at end of K) are fully within K since K is multiple of BK. So no partial. Wait, K=128, BK=64, so 2 K tiles. Each tile is fully loaded. OK. Actually, maybe the issue is that `p_gi` with shape (1,BK) and boundary_check loads only 1 row and BK cols. But Triton might allocate shared memory for a (BT,BK) tile and only initialize part of it, reading garbage for the rest. Then when it later loads the full (BT,BK) tile p_g, it might reuse the same shared memory buffer that has stale data in some rows. This is getting speculative. Let me try a different approach: add `tl.debug_barrier()` or avoid block pointers for the column loads. Actually, maybe the issue is the `tl.dot` with shapes (BT,BK) @ (BK,1). In Triton, dot with one operand of shape (BK,1) might not be optimized and could behave unexpectedly. Let me try computing col_Aqk with elementwise multiply and sum instead of tl.dot: ```python col_Aqk += tl.sum(q_exp * k_i[None,:], axis=1)[:,None] ``` This avoids tl.dot with (BK,1). But it uses broadcasting and sum. Actually, `q_exp * k_i[None,:]` is (BT,BK) * (1,BK) -> (BT,BK). Sum over axis=1 -> (BT,). Then [:,None] -> (BT,1). Good. Let me try this instead of tl.dot. But first, let me see if the issue is reproducible with a minimal kernel that just computes Aqk. Actually, debug_kernel2 already does that. Let me try changing the column update to a different pattern, like using a loop over rows. Actually, maybe the issue is that `Aqk = tl.where(c[None,:] == i, col_Aqk, Aqk)` uses a mask that is recomputed each iteration, and Triton has a bug with this pattern on SM120. Let me try precomputing the mask or using a different update method. Alternative: store col_Aqk to a shared/global buffer column by column, then load back as full matrix. But that's extra memory. Or, compute Aqk by rows instead of columns: Aqk[c,i] = sum_k q[c,k]*k[i,k]*exp(g[c,k]-g[i,k]). For each row c: q_c, g_c loaded. For each i: diff = g_c - g_i, dot with k_i. This requires loading k_i, g_i for each i (same as before). And we can accumulate Aqk row by row. But we can also compute Aqk as a matrix multiply: Aqk = (q * exp(g)) @ (k * exp(-g)).T with per-token normalization? No, that's the unstable approach. Hmm. Actually, we can compute Aqk stably by normalizing each row/col. For each column i, subtract g_i. That's what we do. Let me try the elementwise sum instead of tl.dot for col_Aqk and col_M. Actually, I realize the issue might be the `tl.dot(q_exp, tl.trans(k_i))` where `k_i` has shape (1,BK). `tl.trans(k_i)` has shape (BK,1). tl.dot expects 2D tensors. The result has shape (BT,1). This should work. But maybe on Blackwell/SM120, tl.dot with one dimension being 1 has a bug? Let me test. Let me change to elementwise sum and see if Aqk becomes stable.
in 139452out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (15 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -85,8 +85,10 @@
             diff = b_g - g_i
             q_exp = b_q * tl.exp(diff)
             k_exp = b_k * tl.exp(diff)
-            col_Aqk += tl.dot(q_exp, tl.trans(k_i))
-            col_M += tl.dot(k_exp, tl.trans(k_i))
+            # col_Aqk += tl.dot(q_exp, tl.trans(k_i))
+            # col_M += tl.dot(k_exp, tl.trans(k_i))
+            col_Aqk += tl.sum(q_exp * k_i, axis=1)[:, None]
+            col_M += tl.sum(k_exp * k_i, axis=1)[:, None]
 
         Aqk = tl.where(c[None, :] == i, col_Aqk, Aqk)
         M = tl.where(c[None, :] == i, col_M, M)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<208 chars \u2014 see diff>",
  "new_string": "<330 chars \u2014 see diff>"
}
in 139452out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (9 chars)
Now test.
in 140936out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python /workspace/problems/02_kda_cutlass/debug_kernel2.py 2>&1 | tail -10",
  "description": "Test elementwise sum change"
}
in 140936out 0cache_r 0cache_w 0
tool
stdout (186 chars)
before ref: Aqk max 0.036135099828243256 w max 2.6868696212768555 u max 0.3211267590522766
after ref: Aqk max 0.96484375 w max 2.6868696212768555 u max 0.3211267590522766
Aqk equal False
assistant
reasoning (236 chars)
Still Aqk differs. So not tl.dot issue. Let me try a completely different approach for computing Aqk/M: use shared memory or compute by rows. Actually, let me first verify that Aqk1 (before ref) is correct. Compare to reference's Aqk.
in 141085out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<2050 chars \u2014 see below>",
  "description": "Compare Aqk to reference"
}
command (2050 chars)
cat > /workspace/problems/02_kda_cutlass/check_Aqk.py <<'PY'
import torch
from einops import rearrange
import solution, reference

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

# reference Aqk
ref_model = reference.Model(B,T,H,K,V,CS).cuda().eval()
with torch.no_grad():
    _ = ref_model(q,k,v,g,beta)

# We can't extract Aqk from reference. Compute manually.
scale = float(K)**-0.5
qf = (q * scale).float()
kf = k.float()
gf = g.float()
gf = rearrange(gf, 'b (n c) h d -> b h n c d', c=CS).cumsum(dim=3)
kf = rearrange(kf, 'b (n c) h d -> b h n c d', c=CS)
qf = rearrange(qf, 'b (n c) h d -> b h n c d', c=CS)
Aqk_ref = torch.zeros(B,H,T//CS,CS,CS,device='cuda')
for i in range(CS):
    Aqk_ref[...,i] = torch.einsum('b h n c d, b h n d -> b h n c', qf * (gf - gf[...,i:i+1,:]).exp(), kf[...,i,:])
Aqk_ref = Aqk_ref.view(B,T,H,CS)

# my Aqk
class DK(torch.nn.Module):
    def forward(self,q,k,v,g,beta):
        B,T,H,K=q.shape; V=v.shape[-1]; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled=q*scale; kg=k*g_cum.exp()
        Aqk=torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w=torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u=torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk

dk = DK().cuda()
Aqk_mine = dk(q,k,v,g,beta)
print('Aqk_ref max', Aqk_ref.abs().max().item())
print('Aqk_mine max', Aqk_mine.abs().max().item())
print('diff', (Aqk_ref - Aqk_mine).abs().max().item())
print('mine==ref?', torch.allclose(Aqk_ref, Aqk_mine, atol=1e-3, rtol=1e-3))
PY
uv run python /workspace/problems/02_kda_cutlass/check_Aqk.py 2>&1 | tail -10
in 141085out 0cache_r 0cache_w 0
tool
stdout (103 chars)
Aqk_ref max 1.8088372945785522
Aqk_mine max 6.7505998611450195
diff 6.7210493087768555
mine==ref? False
assistant
reasoning (2526 chars)
Aqk_mine is wrong even before ref! It has max 6.75 vs ref 1.81. So my kernel 1 Aqk computation is buggy, not just affected by ref. But earlier when running sol only, the final output matched reasonably. How can Aqk be wrong but output be OK? Maybe Aqk errors cancel or are small relative to other terms. Actually, max diff 6.72 is huge. But sol output max was 0.0159. That's because Aqk @ v_i might be small if v_i is small. Aqk errors could be large but multiplied by small v_i. But for check.py, after ref, Aqk becomes even more wrong, leading to huge output. So I need to fix kernel 1 Aqk computation. Let me debug the Aqk computation. The formula: Aqk[c,i] = sum_k q[c,k] * k[i,k] * exp(g[c,k] - g[i,k]) In my kernel: - For each column i: - load g_i (1,BK), k_i (1,BK) - for each K tile: - load q (BT,BK), k (BT,BK), g (BT,BK) - diff = g - g_i - q_exp = q * exp(diff) - col_Aqk += sum(q_exp * k_i, axis=1)[:,None] This computes col_Aqk[c] = sum_k q[c,k] * exp(g[c,k]-g[i,k]) * k_i[k]. Correct. But Aqk_ref uses q * scale. In my kernel, I pass q_scaled = q * scale. And multiply by scale=1.0. So Aqk should match Aqk_ref. But it doesn't. Wait, Aqk_ref was computed with qf = (q * scale).float(). In my kernel, q_scaled = q * scale in bf16, then loaded as fp32. That slight difference shouldn't cause 6x error. Also, Aqk_ref computes exp(g - g_i) directly in fp32. My kernel computes exp(diff) in fp32 (since b_g and g_i are fp32). Same. Hmm. Let me check if the issue is the column update pattern. Maybe `Aqk = tl.where(c[None,:] == i, col_Aqk, Aqk)` doesn't set the column correctly in some cases. Actually, from my test, the column update pattern sets column j to j+1 when col is full of j+1. Wait, in my test, col = full([N,1], i+1). Output column j is j+1. So column i is set to i+1. Correct. But in my kernel, col_Aqk is a computed column vector. So column i should be set to col_Aqk. Correct. Maybe the issue is that `col_Aqk` is accumulated across K tiles and iterations, and somewhere it gets wrong values. Wait, in my kernel, `col_Aqk = tl.zeros([BT, 1], dtype=tl.float32)` is reset for each i. Good. Let me try computing Aqk using a different method: directly as matrix multiply with per-column normalization, but in Python or with a simpler kernel. Actually, let me first check if the issue is that I'm reading k_i from k_ptr but should read from kg_ptr or something. No, Aqk uses q and k (not kg). Let me print a few values of Aqk_mine vs Aqk_ref for a specific chunk/head.
in 142030out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<297 chars \u2014 see below>",
  "description": "Print Aqk values"
}
command (297 chars)
cat >> /workspace/problems/02_kda_cutlass/check_Aqk.py <<'PY'
# add to end
print('ref first chunk head0 first few:')
print(Aqk_ref[0,0,0,:5,:5])
print('mine first chunk head0 first few:')
print(Aqk_mine[0,0,0,:5,:5])
PY
uv run python /workspace/problems/02_kda_cutlass/check_Aqk.py 2>&1 | tail -30
in 142030out 0cache_r 0cache_w 0
tool
stdout (368 chars)
Aqk_ref max 1.8088372945785522
Aqk_mine max 6.7505998611450195
diff 6.7210493087768555
mine==ref? False
ref first chunk head0 first few:
Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/check_Aqk.py", line 50, in <module>
    print(Aqk_ref[0,0,0,:5,:5])
          ~~~~~~~^^^^^^^^^^^^^
IndexError: too many indices for tensor of dimension 4
assistant
reasoning (100 chars)
Aqk_ref is (B,T,H,BT). So index [0,0,0,:5] gives first 5 columns of token 0. Let me print correctly.
in 143197out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<1818 chars \u2014 see below>",
  "description": "Print Aqk row values"
}
command (1818 chars)
cat > /workspace/problems/02_kda_cutlass/check_Aqk2.py <<'PY'
import torch
from einops import rearrange
import solution, reference

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

scale = float(K)**-0.5
qf = (q * scale).float()
kf = k.float()
gf = g.float()
gf = rearrange(gf, 'b (n c) h d -> b h n c d', c=CS).cumsum(dim=3)
kf = rearrange(kf, 'b (n c) h d -> b h n c d', c=CS)
qf = rearrange(qf, 'b (n c) h d -> b h n c d', c=CS)
Aqk_ref = torch.zeros(B,H,T//CS,CS,CS,device='cuda')
for i in range(CS):
    Aqk_ref[...,i] = torch.einsum('b h n c d, b h n d -> b h n c', qf * (gf - gf[...,i:i+1,:]).exp(), kf[...,i,:])
Aqk_ref = Aqk_ref.view(B,T,H,CS)

class DK(torch.nn.Module):
    def forward(self,q,k,v,g,beta):
        B,T,H,K=q.shape; V=v.shape[-1]; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled=q*scale; kg=k*g_cum.exp()
        Aqk=torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w=torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u=torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk

dk = DK().cuda()
Aqk_mine = dk(q,k,v,g,beta)

b,h,t = 0,0,0
print('ref[0,0,0,:5]', Aqk_ref[b,t,h,:5])
print('mine[0,0,0,:5]', Aqk_mine[b,t,h,:5])
print('ref[0,0,1,:5]', Aqk_ref[b,t+1,h,:5])
print('mine[0,0,1,:5]', Aqk_mine[b,t+1,h,:5])
PY
uv run python /workspace/problems/02_kda_cutlass/check_Aqk2.py 2>&1 | tail -20
in 143197out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (341 chars)
ref[0,0,0,:5] tensor([-0.0089, -0.0020, -0.0021, -0.0175,  0.0126], device='cuda:0')
mine[0,0,0,:5] tensor([-0.0089,  0.0000,  0.0000,  0.0000,  0.0000], device='cuda:0')
ref[0,0,1,:5] tensor([-0.0112, -0.0155, -0.0040, -0.0035,  0.0011], device='cuda:0')
mine[0,0,1,:5] tensor([-0.0034,  0.0102,  0.0000,  0.0000,  0.0000], device='cuda:0')
assistant
reasoning (3376 chars)
My Aqk has many zeros in the upper triangular part. But Aqk should be lower triangular including diagonal. For row 0 (token 0), columns 0.. should be nonzero (since 0 <= all columns). Mine has only column 0 nonzero, rest zero. Wait, Aqk_ref[0,0,0,:5] has all 5 values nonzero. Mine has only first nonzero. This means my column update is not setting columns correctly. Only column 0 got set. Why? Because `tl.where(c[None,:] == i, col_Aqk, Aqk)` sets column i. But for i=1,2,3,4, it seems col_Aqk is zero? Or the update isn't happening. Wait, for row 0, column 1: Aqk[0,1] = q[0] @ k[1] * exp(g[0]-g[1]) * scale. Should be nonzero. But mine is zero. So either col_Aqk for i=1 is zero, or the column update failed. Actually, looking at mine[0,0,1,:5]: row 1 has columns 0,1 nonzero, others zero. So columns 2,3,4 are zero. Same pattern: only columns <= row are nonzero, but for row 0, only column 0. Wait, for row 0, column 0 is nonzero. For row 1, columns 0,1 are nonzero. For row 0, column 1 should be nonzero (0 >= 1? No, 0 < 1, so it's in upper triangle). Aqk is lower triangular INCLUDING diagonal, so Aqk[0,1] should be 0 (row 0, column 1 is upper triangle). Oh wait, Aqk[c,i] with c=row, i=column. Lower triangular including diagonal means c >= i. So row 0 has only column 0. Row 1 has columns 0,1. That's exactly what mine shows! But Aqk_ref[0,0,0,:5] has row 0 columns 0..4 all nonzero. That means Aqk_ref is NOT lower triangular? Let me check reference's Aqk masking. Reference line 86: `Aqk = Aqk.masked_fill(mask_strict_upper, 0)`. mask_strict_upper is `torch.triu(..., diagonal=1)`, i.e., upper triangle excluding diagonal. So Aqk is lower triangular INCLUDING diagonal. Row 0 should have only column 0 nonzero. But Aqk_ref row 0 has all 5 nonzero. That means my Aqk_ref computation is wrong! Let me re-check. In reference, Aqk[...,i] = einsum over c of q[c]*k[i]*exp(g[c]-g[i]). So Aqk[c,i] = q[c] @ k[i] * exp(g[c]-g[i]). Then masked_fill upper (c < i) to 0. So for c=0, i=1: Aqk[0,1] = q[0] @ k[1] * exp(g[0]-g[1]). This is not masked because c=0 < i=1 is upper triangle, so it IS masked to 0. Wait, upper triangle is c < i. Yes, masked. So Aqk_ref[0,1] should be 0. But my Aqk_ref[0,0,0,:5] (row 0) has columns 0..4 all nonzero. That's wrong. So my Aqk_ref computation is wrong. Oh! I computed Aqk_ref with: ```python Aqk_ref[...,i] = torch.einsum('... c d, ... d -> ... c', qf * (gf - gf[...,i:i+1,:]).exp(), kf[...,i,:]) ``` Here `... c d` with c being the token dimension, and the result has shape (B,H,N,C). Then `Aqk_ref[...,i]` sets the i-th index of the last dimension. So Aqk_ref[b,h,n,c,i] = result[b,h,n,c]. Then Aqk_ref shape (B,H,N,C,C). View to (B,T,H,C). So Aqk_ref[b,t,h,i] = Aqk_ref[b,h,n,c,i] = result[b,h,n,c] = sum_d q[c,d]*k[i,d]*exp(g[c,d]-g[i,d]). This is correct. Then I should mask upper triangle. I didn't mask! So my Aqk_ref is the unmasked full matrix. That's why row 0 has all columns nonzero. So my kernel is correct in masking. Aqk_mine matches the masked version of Aqk_ref. Let me verify. Actually, mine row 0 col 0 = -0.0089 matches ref row 0 col 0. Mine row 1 col 0 = -0.0034 vs ref row 1 col 0 = -0.0112. Wait, that doesn't match. Hmm. Actually, ref Aqk_ref row 1 col 0 = -0.0112, mine row 1 col 0 = -0.0034. Different. But both are in lower triangle. So mine is still wrong. Let me mask Aqk_ref and compare.
in 144229out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<595 chars \u2014 see below>",
  "description": "Compare masked Aqk"
}
command (595 chars)
cat >> /workspace/problems/02_kda_cutlass/check_Aqk2.py <<'PY'
# mask upper triangle
mask = torch.arange(CS)[:,None] >= torch.arange(CS)[None,:]
Aqk_ref_m = Aqk_ref * mask.to(Aqk_ref.device).view(1,1,1,CS,CS).permute(0,2,3,1,4).contiguous().view(B,T,H,CS)
print('masked ref[0,0,0,:5]', Aqk_ref_m[b,t,h,:5])
print('masked ref[0,0,1,:5]', Aqk_ref_m[b,t+1,h,:5])
print('mine[0,0,0,:5]', Aqk_mine[b,t,h,:5])
print('mine[0,0,1,:5]', Aqk_mine[b,t+1,h,:5])
print('masked diff', (Aqk_ref_m - Aqk_mine).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/check_Aqk2.py 2>&1 | tail -20
in 144229out 0cache_r 0cache_w 0
tool
stdout (759 chars)
ref[0,0,0,:5] tensor([-0.0089, -0.0020, -0.0021, -0.0175,  0.0126], device='cuda:0')
mine[0,0,0,:5] tensor([-0.0089,  0.0000,  0.0000,  0.0000,  0.0000], device='cuda:0')
ref[0,0,1,:5] tensor([-0.0112, -0.0155, -0.0040, -0.0035,  0.0011], device='cuda:0')
mine[0,0,1,:5] tensor([-0.0034,  0.0102,  0.0000,  0.0000,  0.0000], device='cuda:0')
Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/check_Aqk2.py", line 45, in <module>
    Aqk_ref_m = Aqk_ref * mask.to(Aqk_ref.device).view(1,1,1,CS,CS).permute(0,2,3,1,4).contiguous().view(B,T,H,CS)
                          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
RuntimeError: shape '[2, 1024, 8, 64]' is invalid for input of size 4096
assistant
reasoning (2452 chars)
My mask reshape is wrong. Aqk_ref is already (B,T,H,CS) but logically contains the full matrix flattened. Actually I computed Aqk_ref = Aqk_ref.view(B,T,H,CS) where the last dim is columns, and the rows are along T dim. So Aqk_ref[b, t, h, i] is row (t % CS), column i, of chunk (t // CS). To mask, I need mask of shape (CS,CS) applied per chunk: for token t in chunk n, row r = n*CS + r? Wait, T dim is n*CS + r. So Aqk_ref[b, n*CS + r, h, c] is Aqk[r,c]. Mask[r,c] = r >= c. So: ```python rows = torch.arange(T) % CS cols = torch.arange(CS) mask = rows[:,None] >= cols[None,:] # (T,CS) Aqk_ref_m = Aqk_ref * mask[None,:,None,:].to(device) ``` Let me just compare specific elements. Mine row 0 col 0 = -0.0089 matches ref. Row 1 col 0 = -0.0034 vs ref -0.0112. Row 1 col 1 = 0.0102 vs ref -0.0155. So mine is wrong. Why? Let me think. For row 1 col 0: Aqk[1,0] = q[1] @ k[0] * exp(g[1]-g[0]) * scale. Mine = -0.0034, ref = -0.0112. Different by factor ~3. Hmm. Maybe my q_scaled is bf16, ref q is fp32. But difference of 3x is too large for bf16 rounding. Wait, maybe I'm loading the wrong column for k_i or g_i. For Aqk column i=0, I load k_i = k[t0+0], g_i = g[t0+0]. Then compute q_exp = q[t0:t0+BT] * exp(g - g_i). Then col_Aqk[c] = sum_k q_exp[c,k] * k_i[k]. This should give Aqk[c,0]. For c=1, mine is -0.0034, ref is -0.0112. Different. Maybe the issue is that I'm loading `q_scaled` from `q_ptr` but `q_scaled` is in bf16, and reference uses fp32 q. But bf16 q has ~3 decimal digits, so relative error ~1%. Not 3x. Wait, maybe the issue is that I compute `q_scaled = q * scale` in bf16, but reference does `q = q.to(fp32) * scale`. If q is small and scale is ~0.088, the bf16 product might underflow or lose precision. But q is ~0.1, scale ~0.088, product ~0.009. bf16 can represent ~0.009 with ~1% relative error. Hmm. Let me check q_scaled values. Actually, wait. In my kernel, I pass `q_scaled` which is `q * self.scale` in bf16. But `self.scale = float(K)**-0.5 = 1/sqrt(128) ≈ 0.0884`. Reference also uses this scale. But in kernel, I multiply Aqk by `scale` parameter which is 1.0. So Aqk = sum(q_scaled * exp(diff) * k_i) * 1.0. Reference Aqk = sum((q_fp32 * scale) * exp(diff) * k_i). Same. But q_scaled_bf16 vs q_fp32*scale might differ. Let me compute the expected difference. Actually, the huge difference in row 1 col 0 suggests something else. Let me compute Aqk manually in Python using bf16 q_scaled and compare.
in 145967out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<1962 chars \u2014 see below>",
  "description": "Compare Aqk with bf16 inputs"
}
command (1962 chars)
cat > /workspace/problems/02_kda_cutlass/check_Aqk3.py <<'PY'
import torch
from einops import rearrange
import solution, reference

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

scale = float(K)**-0.5
q_scaled = (q * scale).bfloat16().float()  # bf16 then fp32
qf = q_scaled
kf = k.bfloat16().float()
gf = g.float()
gf = rearrange(gf, 'b (n c) h d -> b h n c d', c=CS).cumsum(dim=3)
kf = rearrange(kf, 'b (n c) h d -> b h n c d', c=CS)
qf = rearrange(qf, 'b (n c) h d -> b h n c d', c=CS)
Aqk_bf16 = torch.zeros(B,H,T//CS,CS,CS,device='cuda')
for i in range(CS):
    Aqk_bf16[...,i] = torch.einsum('b h n c d, b h n d -> b h n c', qf * (gf - gf[...,i:i+1,:]).exp(), kf[...,i,:])
Aqk_bf16 = Aqk_bf16.view(B,T,H,CS)

# my kernel
class DK(torch.nn.Module):
    def forward(self,q,k,v,g,beta):
        B,T,H,K=q.shape; V=v.shape[-1]; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled=q*scale; kg=k*g_cum.exp()
        Aqk=torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w=torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u=torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk

dk = DK().cuda()
Aqk_mine = dk(q,k,v,g,beta)

b,h,t = 0,0,0
print('bf16 ref[0,0,0,:5]', Aqk_bf16[b,t,h,:5])
print('bf16 ref[0,0,1,:5]', Aqk_bf16[b,t+1,h,:5])
print('mine[0,0,0,:5]', Aqk_mine[b,t,h,:5])
print('mine[0,0,1,:5]', Aqk_mine[b,t+1,h,:5])
print('diff', (Aqk_bf16 - Aqk_mine).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/check_Aqk3.py 2>&1 | tail -20
in 145967out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (374 chars)
bf16 ref[0,0,0,:5] tensor([-0.0089, -0.0020, -0.0021, -0.0175,  0.0126], device='cuda:0')
bf16 ref[0,0,1,:5] tensor([-0.0112, -0.0155, -0.0040, -0.0035,  0.0011], device='cuda:0')
mine[0,0,0,:5] tensor([-0.0089,  0.0000,  0.0000,  0.0000,  0.0000], device='cuda:0')
mine[0,0,1,:5] tensor([-0.0034,  0.0102,  0.0000,  0.0000,  0.0000], device='cuda:0')
diff 2.687044858932495
assistant
reasoning (3989 chars)
Even with bf16 inputs, the bf16 reference Aqk matches the fp32 reference Aqk (same values), but mine is wrong. So my kernel computation is incorrect. Let me think why. Wait, in my kernel, I compute `col_Aqk += tl.sum(q_exp * k_i, axis=1)[:, None]`. `q_exp` is (BT,BK), `k_i` is (1,BK). `q_exp * k_i` broadcasts k_i to (BT,BK), then sum over axis=1 gives (BT,). This computes sum_k q_exp[c,k] * k_i[k]. Correct. But wait, `k_i` has shape (1,BK). In Triton, `q_exp * k_i` with (BT,BK) and (1,BK) -> (BT,BK). Sum axis=1 -> (BT,). Correct. Hmm. But the values are wrong. Let me check if maybe `b_q` is not q_scaled but something else. Wait, in kernel 1 launch, I pass `q_scaled`. But `q_scaled` is `q * self.scale` computed in bf16. Then in kernel, I load it as fp32. Correct. But in the bf16 reference, I also used `q_scaled.bfloat16().float()`. Same. Maybe the issue is `b_g - g_i`. `b_g` is loaded from `g_ptr` which is `g_cum`. `g_i` loaded from `g_ptr` at row t0+i. Correct. Wait, could the block pointer `p_gi` with shape (1,BK) load from the wrong location? Let me verify the pointer arithmetic. `tl.make_block_ptr(g_ptr, (T,K), (H*K,1), (t0+i, off_k), (1,BK), (1,0))`. Base: g_ptr + (t0+i)*H*K + off_k*1. Shape (1,BK), strides (H*K,1). Order (1,0) means row-major? Actually order parameter: (1,0) means first dimension is fastest changing? Let me check Triton docs. `make_block_ptr(base, shape, strides, offsets, block_shape, order)`: - shape: logical shape of the tensor - strides: strides in elements - offsets: starting offsets - block_shape: shape of the block to load - order: the order of the dimensions in memory (0 = innermost?) Actually, I think `order=(1,0)` means column-major? No, for a 2D tensor, order=(1,0) means the second dimension is contiguous (innermost), i.e., row-major because within a row, columns are contiguous. Wait, order list gives the order from fastest-changing to slowest? Or slowest to fastest? In Triton, `order` in make_block_ptr specifies the order of dimensions from innermost (fastest) to outermost (slowest). For row-major, the last dimension (columns) is innermost, so order should be (1,0). Yes, that's what I have. So p_gi loads row t0+i, cols off_k:off_k+BK. Correct. p_q loads rows t0:t0+BT, cols off_k:off_k+BK. Correct. Hmm. Wait, I think I see the issue! In my kernel, `for i in range(BT):` is a runtime loop with `i` as a tensor (tl.tensor). But `tl.make_block_ptr` offsets can be runtime ints? Yes, offsets are tensors. Actually, Triton might not handle dynamic offsets in block pointers well in some cases. Let me check by printing the loaded g_i values. Actually, let me test a simpler kernel that computes Aqk using the same pattern but for one chunk, and compare. But maybe I can avoid the issue by computing Aqk/M with a different approach: use matrix multiply with per-token normalization. Actually, let me try using the unstable factorization but with the stable reference formula for Aqk/M. Wait, that's what I'm doing. Hmm, let me check if `tl.exp(diff)` is correct. diff = b_g - g_i. b_g shape (BT,BK), g_i shape (1,BK). Correct. Actually, wait. `g_i` is loaded with shape (1,BK). When I do `b_g - g_i`, Triton broadcasts g_i to (BT,BK). Good. But when I do `tl.sum(q_exp * k_i, axis=1)[:,None]`, `k_i` is (1,BK). `q_exp` is (BT,BK). `q_exp * k_i` broadcasts k_i. Good. Let me add a `tl.debug_barrier()` after each column? No. Actually, I wonder if the issue is that `Aqk = tl.where(c[None,:] == i, col_Aqk, Aqk)` with runtime `i` doesn't work as expected because `i` is a tensor and `c[None,:] == i` compares a constexpr vector with a tensor. My earlier test used `tl.full` not computed column. Let me test with actual computed col. Actually, my earlier test with `col = tl.full([N,1], i+1, ...)` showed that columns are set correctly. So the comparison works. Wait, but in my earlier test, the output was all rows equal. It didn't test if col values vary per row. Let me test with row-varying col.
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Bash
args
{
  "command": "<713 chars \u2014 see below>",
  "description": "Test column update with row-varying values"
}
command (713 chars)
cat > /workspace/problems/02_kda_cutlass/test_where2.py <<'PY'
import torch
import triton, triton.language as tl

@triton.jit
def test_where_col(out_ptr, N: tl.constexpr):
    c = tl.arange(0, N)
    x = tl.zeros([N, N], dtype=tl.float32)
    for i in range(N):
        col = (tl.arange(0, N) + i * N)[:, None].to(tl.float32)
        x = tl.where(c[None, :] == i, col, x)
    tl.store(out_ptr + tl.arange(0, N*N), tl.reshape(x, (N*N,)))

N = 8
out = torch.empty(N*N, device='cuda', dtype=torch.float32)
test_where_col[(1,)](out, N=N)
out = out.view(N,N)
print(out)
print('expected: column i has values [i*N, i*N+1, ..., i*N+N-1]')
PY
uv run python /workspace/problems/02_kda_cutlass/test_where2.py 2>&1 | tail -20
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stdout (474 chars)
tensor([[ 0.,  8., 16., 24., 32., 40., 48., 56.],
        [ 1.,  9., 17., 25., 33., 41., 49., 57.],
        [ 2., 10., 18., 26., 34., 42., 50., 58.],
        [ 3., 11., 19., 27., 35., 43., 51., 59.],
        [ 4., 12., 20., 28., 36., 44., 52., 60.],
        [ 5., 13., 21., 29., 37., 45., 53., 61.],
        [ 6., 14., 22., 30., 38., 46., 54., 62.],
        [ 7., 15., 23., 31., 39., 47., 55., 63.]], device='cuda:0')
expected: column i has values [i*N, i*N+1, ..., i*N+N-1]
assistant
reasoning (1889 chars)
Column update works correctly with row-varying values. So why is my Aqk wrong? Let me check if maybe `k_i` is being loaded incorrectly. In my kernel, I load `k_i` from `k_ptr`. But `k` is the original bf16 k. In the bf16 reference, k is bf16. Same. Wait, maybe the issue is that I load `g_i` and `k_i` with shape (1,BK) but Triton interprets it as a column vector instead of row? In my test, `tl.trans(k_i)` for k_i shape (1,BK) gives (BK,1). But in `tl.sum(q_exp * k_i, axis=1)`, k_i is (1,BK) and broadcasts correctly. Actually, wait. `k_i` shape is (1,BK). When I do `q_exp * k_i`, it broadcasts to (BT,BK). Sum axis=1 -> (BT,). But the sum is over K dimension. Good. But maybe `k_i` is actually stored as (BK,1) due to block pointer order? No, I specified shape (1,BK) and order (1,0). It should be row vector. Let me verify by checking `k_i` values. Actually, I can write a small kernel that extracts k_i and compares to CPU. Actually, I think I should just compute Aqk/M in Python and pass them to the kernel, to avoid this debugging. The Python computation is correct and reasonably fast. Then kernel 1 only does solve, w, u. We can also move Aqk computation to Python. But we want a custom kernel. Still, the solve and w/u are the heavy/complex parts; Aqk is just a batched GEMV. We can do it in PyTorch for correctness and speed. Actually, let me first try to fix the kernel. Let me check if the issue is the block pointer `p_ki` with shape (1,BK). Maybe I should use a scalar pointer load instead: ```python k_i = tl.load(k_ptr + (t0+i)*H*K + off_k + tl.arange(0,BK)).to(tl.float32) g_i = tl.load(g_ptr + (t0+i)*H*K + off_k + tl.arange(0,BK)).to(tl.float32) ``` This avoids block pointer for a single row. Let me try. Also, I should use `tl.full` or ensure k_i is shape (1,BK). With scalar load, k_i is shape (BK,). I need to add a dimension: `k_i[None,:]`. Let me try.
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Edit /workspace/problems/02_kda_cutlass/solution.py
diff (25 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -58,15 +58,13 @@
         col_M = tl.zeros([BT, 1], dtype=tl.float32)
         for i_k in range(tl.cdiv(K, BK)):
             off_k = i_k * BK
-            # column i's g and k
-            p_gi = tl.make_block_ptr(
-                g_ptr, (T, K), (H * K, 1), (t0 + i, off_k), (1, BK), (1, 0)
-            )
-            p_ki = tl.make_block_ptr(
-                k_ptr, (T, K), (H * K, 1), (t0 + i, off_k), (1, BK), (1, 0)
-            )
-            g_i = tl.load(p_gi, boundary_check=(0, 1)).to(tl.float32)
-            k_i = tl.load(p_ki, boundary_check=(0, 1)).to(tl.float32)
+            # column i's g and k (scalar row load)
+            g_i = tl.load(
+                g_ptr + (t0 + i) * H * K + off_k + tl.arange(0, BK)
+            ).to(tl.float32)[None, :]
+            k_i = tl.load(
+                k_ptr + (t0 + i) * H * K + off_k + tl.arange(0, BK)
+            ).to(tl.float32)[None, :]
 
             # full chunk tiles
             p_q = tl.make_block_ptr(
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<460 chars \u2014 see diff>",
  "new_string": "<348 chars \u2014 see diff>"
}
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stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (15 chars)
Test Aqk again.
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Bash
args
{
  "command": "uv run python /workspace/problems/02_kda_cutlass/check_Aqk3.py 2>&1 | tail -10",
  "description": "Test scalar row load"
}
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stdout (374 chars)
bf16 ref[0,0,0,:5] tensor([-0.0089, -0.0020, -0.0021, -0.0175,  0.0126], device='cuda:0')
bf16 ref[0,0,1,:5] tensor([-0.0112, -0.0155, -0.0040, -0.0035,  0.0011], device='cuda:0')
mine[0,0,0,:5] tensor([-0.0089,  0.0000,  0.0000,  0.0000,  0.0000], device='cuda:0')
mine[0,0,1,:5] tensor([-0.0034,  0.0102,  0.0000,  0.0000,  0.0000], device='cuda:0')
diff 2.687044858932495
assistant
reasoning (9366 chars)
Still wrong. So not the block pointer issue. Let me try a completely different approach: compute Aqk/M using matrix multiply with broadcasting in Python, and only do solve/w/u in Triton. This will be correct and might be fast enough. Actually, wait. The diff 2.687 is exactly `Aqk_ref.abs().max() * scale`? No, Aqk_ref max is 1.81. 2.687 is close to exp(1) ~ 2.718. Hmm. Actually, 2.687 = exp(0.988) which is exp(g_cum max). Interesting. Maybe my kernel is multiplying by exp(g) or something extra. Wait, let me look at mine row 1 col 1 = 0.0102. bf16 ref row 1 col 1 = -0.0155. Ratio ~ -0.66. Not exp. Row 1 col 0: mine -0.0034 vs ref -0.0112. Ratio 0.30. Hmm. Wait, maybe my kernel is computing Aqk for the wrong chunk? No, grid is per chunk. Actually, I just realized: in my kernel, I load `b_q` from `q_ptr` which is `q_scaled`. But `q_scaled` is bf16. Then I cast to fp32. In the bf16 reference, I also cast q to bf16 then fp32. Same. But wait, in the kernel, `q_exp = b_q * tl.exp(diff)`. b_q is q_scaled (already scaled). In reference, qf is also scaled. Same. Maybe the issue is that I pass `scale=1.0` to kernel, but I also compute q_scaled in Python. So Aqk gets scale once. In reference, q is scaled once. Same. Hmm. Let me check if maybe the issue is in `tl.sum(q_exp * k_i, axis=1)`. The axis numbering: q_exp is (BT,BK) = (64,64). axis=1 is the K dimension (64). Sum over K. Correct. But wait, when I do `q_exp * k_i`, if k_i is (1,BK) and q_exp is (BT,BK), the result is (BT,BK). Sum axis=1 gives (BT,). Correct. Unless Triton's axis numbering is different? No, axis=1 is second dimension. Let me print the actual `col_Aqk` values for a specific i to debug. I can modify the kernel to store col_Aqk to a debug buffer. But that's annoying. Alternative: compute Aqk with a simpler kernel that uses block pointers for matrix multiply, not column loop. Actually, I just realized I might have a fundamental issue: in the kernel, `for i in range(BT):` where `i` is a runtime tensor. Inside the loop, `tl.where(c[None,:] == i, col_Aqk, Aqk)` uses `i` as a scalar. But in Triton, loop variable `i` might not be a simple scalar; it could be a tensor representing the loop index. The comparison `c[None,:] == i` should work elementwise. But maybe the issue is that `i` is being treated as a tensor of shape () and the comparison broadcasts. My test confirmed it works. OK, let me try yet another approach: compute Aqk by rows instead of columns. Aqk[c,i] = sum_k q[c,k] * k[i,k] * exp(g[c,k] - g[i,k]). For each row c: load q_c, g_c For each column i: load k_i, g_i diff = g_c - g_i # (1,K) - (1,K)? Actually g_c is (1,K), g_i is (1,K). diff (1,K). q_exp = q_c * exp(diff) # (1,K) Aqk[c,i] = sum_k q_exp[0,k] * k_i[k] We can compute this in a loop over c and i. But that's BT*BT iterations, each a dot over K. 4096 dots per chunk. Too slow. Better: for each row c, compute a vector q_c * exp(g_c) and k * exp(-g) dot? That's unstable. Hmm. Actually, we can compute Aqk stably using the column loop but avoid the `tl.where` update by storing columns to a global memory buffer and reading back. But that's extra memory. Or, use a 3D tensor in registers: `Aqk = tl.zeros([BT, BT], dtype=tl.float32)` and directly assign `Aqk = tl.where((r[:,None] == c) & (c[None,:] == i), col_Aqk[r], Aqk)`? No. Actually, we can use a loop over c inside the loop over i: ```python for i in range(BT): for c in range(BT): # but c can't be runtime loop variable for block loads ``` No. Let me try using `tl.store` to write col_Aqk to the i-th column of a global Aqk buffer directly, instead of accumulating in registers and using tl.where. In kernel 1, I can allocate Aqk output as global memory. For each column i, store col_Aqk to Aqk[:,i] using a scalar pointer store: ```python for c in range(BT): tl.store(Aqk_ptr + (t0 + c) * H * BT + i, col_Aqk[c]) ``` But this requires runtime `i` and `c` in pointer arithmetic. Triton supports scalar index in pointer arithmetic. Then I can load the full Aqk matrix back after the loop using a block pointer. But that's extra global memory traffic. Actually, I can just keep writing columns to Aqk global memory, then use it directly for the rest. But the solve needs M, not Aqk. I still need M in registers or memory. Let me try storing col_Aqk and col_M to global memory per column, then load back the full matrix for solve. This avoids the `tl.where` column assembly. But wait, my test showed `tl.where` column assembly works. So that's not the issue. Actually, maybe the issue is that `col_Aqk` is wrong, not the assembly. Let me check col_Aqk values. I can add a debug output to store col_Aqk for a specific i. But I don't want to modify the kernel too much. Let me instead compute one column in Python and compare to what the kernel would produce. For chunk 0, head 0, column 0: Aqk[c,0] = sum_k q_scaled[c,k] * k[0,k] * exp(g[c,k] - g[0,k]). In my kernel, col_Aqk[c] = sum_k q_exp[c,k] * k_i[k] where q_exp = q_scaled[c,k] * exp(g[c,k] - g_i[k]), k_i = k[0,k]. Correct. Mine col_Aqk[0] = -0.0089 matches ref. col_Aqk[1] = -0.0034 vs ref -0.0112. So for c=1, i=0, my kernel computed wrong value. Why? Let me compute manually: q_scaled[1], k[0], g[1], g[0]. Actually, maybe the issue is that in my kernel, `b_q` is loaded for the full chunk but `b_g` is loaded for the full chunk, and `g_i` is loaded for column i. All correct. Wait, could it be that the block pointer `p_q` loads rows t0..t0+BT-1 but the data layout is (B,T,H,K), so row t0+c is at offset (t0+c)*H*K + i_h*K + off_k. Block pointer handles this. Correct. Hmm. Oh! I think I see it. In my kernel, I have: ```python b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32) ``` But `p_q` is created with `q_ptr` and shape (T,K). The base pointer `q_ptr` is the start of the q tensor. Offsets (t0, off_k). Strides (H*K, 1). But wait, q layout is (B,T,H,K) with strides (T*H*K, H*K, K, 1). The block pointer shape is (T,K) with strides (H*K, 1), but T dimension actually spans across B and H too. For batch i_b and head i_h, the actual offset is i_b*T*H*K + i_h*K + t*H*K + k. But in my kernel, I don't add i_b and i_h offsets to q_ptr! Wait, let me check. In kernel 1: ```python t0 = i_n * BT ``` No adjustment for i_b and i_h! The q_ptr passed is the base of the q tensor. The block pointer uses (T,K) shape and (H*K,1) strides. But to get to batch i_b and head i_h, we need to offset q_ptr by i_b*T*H*K + i_h*K. Wait, in my previous kernels, I had `q += (bos * H + i_h) * K` to adjust the pointer. In the current kernel, I removed that! That's the bug! Same for k_ptr, v_ptr, g_ptr, kg_ptr, beta_ptr, Aqk_ptr, w_ptr, u_ptr. None are adjusted for i_b and i_h. So all blocks process batch 0, head 0, chunk i_n. That's why Aqk is wrong (always from b=0,h=0) and why after ref (which fills memory with other data) it changes. Wait, but my debug_kernel2 showed Aqk1 and Aqk2 differ. If the kernel always reads b=0,h=0, then Aqk1 should be the same as Aqk2 because inputs are same. But Aqk2 differs because the memory at b=0,h=0 might be different? No, inputs are unchanged. Actually, if the kernel reads b=0,h=0, it writes to Aqk output at b=0,h=0 for all blocks. So Aqk for b>0,h>0 remains uninitialized (whatever memory was there). Aqk1 happens to have zeros or small values for those. After ref, Aqk2 has ref's leftover data for b>0,h>0. But for b=0,h=0, Aqk1 and Aqk2 should be same. Let me check specific element b=0,h=0,t=0. In debug_kernel2, Aqk1[0,0,0,0] = -0.0089 and Aqk2[0,0,0,0] = ? I didn't print. But Aqk2 max was 0.964, so it changed. Wait, if all blocks write to b=0,h=0, there's a race condition! Multiple blocks (for different b,h) write to the same output locations. The result depends on order. That explains non-determinism and corruption. Yes! The missing pointer offset for batch/head is the bug. I need to add `q += (i_b * T * H + i_h) * K` etc. back. Actually, using block pointers, the base pointer should be adjusted by `(i_b * T * H + i_h) * K` for q/k/g/kg. For v/w/u, by `(i_b * T * H + i_h) * V` or `*K`. For beta, by `i_b * T * H + i_h`. For Aqk/w outputs, same. Let me fix kernel 1 by adjusting base pointers. Actually, the clean way is to adjust the pointer at the start: ```python q_ptr += (i_b * T * H + i_h) * K k_ptr += (i_b * T * H + i_h) * K g_ptr += (i_b * T * H + i_h) * K kg_ptr += (i_b * T * H + i_h) * K beta_ptr += i_b * T * H + i_h v_ptr += (i_b * T * H + i_h) * V Aqk_ptr += (i_b * T * H + i_h) * BT w_ptr += (i_b * T * H + i_h) * K u_ptr += (i_b * T * H + i_h) * V ``` Wait, Aqk shape is (B,T,H,BT). Strides (T*H*BT, H*BT, BT, 1). So offset for (i_b, i_h, t0) = i_b*T*H*BT + t0*H*BT + i_h*BT. But with block ptr (T,BT) strides (H*BT,1), offset (t0,0) gives t0*H*BT. So we need base offset i_b*T*H*BT + i_h*BT. Similarly for w: shape (B,T,H,K), strides (T*H*K, H*K, K, 1). Base offset for block ptr (T,K): i_b*T*H*K + i_h*K. For u: base offset i_b*T*H*V + i_h*V. For v: same. For beta: shape (B,T,H), strides (T*H,H,1). Base offset i_b*T*H + i_h. For g: same as q. OK. Let me fix kernel 1 and kernel 2. Actually, in kernel 2, I also need to adjust base pointers for w,u,qg,k,g,Aqk,glast,o. glast shape (B,H,NT,K), strides (H*NT*K, NT*K, K, 1). Base offset i_b*H*NT*K + i_h*NT*K. o base offset i_b*T*H*V + i_h*V. Let me fix.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -41,6 +41,17 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
+    # adjust base pointers for batch/head
+    q_ptr += (i_b * T * H + i_h) * K
+    k_ptr += (i_b * T * H + i_h) * K
+    g_ptr += (i_b * T * H + i_h) * K
+    kg_ptr += (i_b * T * H + i_h) * K
+    beta_ptr += i_b * T * H + i_h
+    v_ptr += (i_b * T * H + i_h) * V
+    Aqk_ptr += (i_b * T * H + i_h) * BT
+    w_ptr += (i_b * T * H + i_h) * K
+    u_ptr += (i_b * T * H + i_h) * V
+
     t0 = i_n * BT
     r = tl.arange(0, BT)
     c = tl.arange(0, BT)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<112 chars \u2014 see diff>",
  "new_string": "<489 chars \u2014 see diff>"
}
in 150819out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (27 chars)
Now fix kernel 2 similarly.
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assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 100,
  "offset": 155
}
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stdout (3819 chars) — TRUNCATED
155	@triton.jit
156	def _kda_inter_kernel(
157	    w_ptr,
158	    u_ptr,
159	    qg_ptr,
160	    k_ptr,
161	    g_ptr,
162	    Aqk_ptr,
163	    glast_ptr,
164	    o_ptr,
165	    B: tl.constexpr,
166	    T: tl.constexpr,
167	    H: tl.constexpr,
168	    K: tl.constexpr,
169	    V: tl.constexpr,
170	    BT: tl.constexpr,
171	    BK: tl.constexpr,
172	    BV: tl.constexpr,
173	):
174	    i_vb = tl.program_id(0)
175	    i_bh = tl.program_id(1)
176	    i_b = i_bh // H
177	    i_h = i_bh % H
178	
179	    off_v0 = i_vb * BV
180	    NT = T // BT
181	
182	    # S split into K tiles of (BK, BV)
183	    S0 = tl.zeros([BK, BV], dtype=tl.float32)
184	    S1 = tl.zeros([BK, BV], dtype=tl.float32)
185	
186	    for i_n in range(NT):
187	        t0 = i_n * BT
188	
189	        p_u = tl.make_block_ptr(
190	            u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
191	        )
192	        b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
193	
194	        b_v = b_u
195	        p_w0 = tl.make_block_ptr(
196	            w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
197	        )
198	        b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.bfloat16)
199	        b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))
200	        p_w1 = tl.make_block_ptr(
201	            w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
202	        )
203	        b_w1 = tl.load(p_w1, boundary_check=(0, 1)).to(tl.bfloat16)
204	        b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))
205	
206	        b_o = tl.zero

... (2319 more chars)
assistant
reasoning (306 chars)
In kernel 2, I already compute glast pointer with `(i_b * H + i_h) * NT * K`. But I don't adjust other base pointers. Need to add. For glast, the base offset `(i_b * H + i_h) * NT * K` is correct if glast is contiguous (B,H,NT,K). Yes. For other pointers: w,u,qg,k,g,Aqk,o need base offsets. Let me add.
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Edit /workspace/problems/02_kda_cutlass/solution.py
diff (18 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -176,6 +176,15 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
+    # adjust base pointers for batch/head
+    w_ptr += (i_b * T * H + i_h) * K
+    u_ptr += (i_b * T * H + i_h) * V
+    qg_ptr += (i_b * T * H + i_h) * K
+    k_ptr += (i_b * T * H + i_h) * K
+    g_ptr += (i_b * T * H + i_h) * K
+    Aqk_ptr += (i_b * T * H + i_h) * BT
+    o_ptr += (i_b * T * H + i_h) * V
+
     off_v0 = i_vb * BV
     NT = T // BT
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<135 chars \u2014 see diff>",
  "new_string": "<441 chars \u2014 see diff>"
}
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stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (39 chars)
Now test Aqk again and then full check.
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Bash
args
{
  "command": "uv run python /workspace/problems/02_kda_cutlass/check_Aqk3.py 2>&1 | tail -10",
  "description": "Test pointer offset fix"
}
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tool
stdout (375 chars)
bf16 ref[0,0,0,:5] tensor([-0.0089, -0.0020, -0.0021, -0.0175,  0.0126], device='cuda:0')
bf16 ref[0,0,1,:5] tensor([-0.0112, -0.0155, -0.0040, -0.0035,  0.0011], device='cuda:0')
mine[0,0,0,:5] tensor([-0.0089,  0.0000,  0.0000,  0.0000,  0.0000], device='cuda:0')
mine[0,0,1,:5] tensor([-0.0034,  0.0102,  0.0000,  0.0000,  0.0000], device='cuda:0')
diff 1.8088372945785522
assistant
reasoning (1077 chars)
Still wrong. Wait, diff is now 1.808 which is Aqk_ref max. So maybe mine is now all zeros except some values? No, mine still has values. Hmm, the pointer offsets didn't fix Aqk. Wait, but the values are exactly the same as before. So maybe the check_Aqk3.py is using a cached kernel? Let me clear cache. Actually, maybe my offset calculation is wrong. Let me double-check. q shape (B,T,H,K) = (2,1024,8,128). Strides (T*H*K, H*K, K, 1) = (1024*8*128, 8*128, 128, 1) = (1048576, 1024, 128, 1). For i_b=0, i_h=0: offset = 0. For i_b=0, i_h=1: offset = 1*128 = 128. For i_b=1, i_h=0: offset = 1*1048576 = 1048576. My formula `(i_b * T * H + i_h) * K` = (i_b * 1024*8 + i_h) * 128 = i_b * 1048576 + i_h * 128. Correct. But wait, I added the offset to q_ptr. In the block pointer, the shape is (T,K) = (1024,128) with strides (H*K,1) = (1024,1). For t0=0, off_k=0, the block pointer computes base + t0*1024 + off_k*1 = base. With base offset i_b*1048576 + i_h*128, we get correct position. Good. So why Aqk still wrong? Let me run mimic_check4 again to see if output changed.
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Bash
args
{
  "command": "uv run python /workspace/problems/02_kda_cutlass/mimic_check4.py 2>&1 | tail -5",
  "description": "Run mimic check after offset fix"
}
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tool
stdout (63 chars)
ref max 0.01171875
sol max 0.0111083984375
diff 0.0103759765625
assistant
reasoning (1357 chars)
mimic_check4 now passes (diff 0.01 < 0.05). So the offset fix worked for the full model. But check_Aqk3 still shows wrong Aqk? Wait, check_Aqk3 uses a custom DK class that calls `_kda_intra_kernel` directly. It passes `q_scaled, k, v, g_cum, kg, beta, Aqk, w, u`. But after my fix, the kernel expects base pointers to the start of tensors and adjusts internally. DK passes the tensors directly, so base pointers are correct. It should work. But check_Aqk3 shows wrong values. Maybe check_Aqk3 is using an old cached kernel? Let me clear cache and rerun. Actually, wait. In check_Aqk3.py, it imports solution and calls `solution._kda_intra_kernel`. But the kernel was recompiled when I changed solution.py. So it should use the new version. Hmm, but the output values are exactly the same as before the offset fix. That suggests the cached kernel is still being used. But I edited solution.py, so Triton should recompile. Unless the kernel cache key doesn't include the offset code? It includes source code hash. Source changed. Actually, maybe the issue is that check_Aqk3.py is in a separate file and imports solution, but solution's kernel was already cached from a previous run with the old source, and Triton doesn't detect the change because the file modification time isn't checked? No, Triton hashes the source. Let me explicitly delete cache.
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Bash
args
{
  "command": "rm -rf ~/.triton/cache /tmp/triton_* 2>/dev/null; uv run python /workspace/problems/02_kda_cutlass/check_Aqk3.py 2>&1 | tail -10",
  "description": "Clear cache and check Aqk"
}
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tool
stdout (375 chars)
bf16 ref[0,0,0,:5] tensor([-0.0089, -0.0020, -0.0021, -0.0175,  0.0126], device='cuda:0')
bf16 ref[0,0,1,:5] tensor([-0.0112, -0.0155, -0.0040, -0.0035,  0.0011], device='cuda:0')
mine[0,0,0,:5] tensor([-0.0089,  0.0000,  0.0000,  0.0000,  0.0000], device='cuda:0')
mine[0,0,1,:5] tensor([-0.0034,  0.0102,  0.0000,  0.0000,  0.0000], device='cuda:0')
diff 1.8088372945785522
assistant
reasoning (3276 chars)
Still wrong. But mimic_check4 with full Model works. Why? Wait, in mimic_check4, I call solution.Model which does Python preprocessing and launches both kernels. The final output is correct. But check_Aqk3 only calls kernel 1 and checks Aqk. Aqk is wrong. This means Aqk might be wrong, but the final output is still correct because the errors cancel or are small in the final result? That seems unlikely given diff 1.8 in Aqk. Wait, maybe the Aqk output in check_Aqk3 is reading from the wrong location due to layout? Let me check Aqk_mine shape and strides. Actually, Aqk_mine is (B,T,H,CS) = (2,1024,8,64). Index [b,t,h,:] gives row t%CS, all columns. Correct. But wait, in my kernel, I store Aqk using block ptr: ```python p_Aqk = tl.make_block_ptr(Aqk_ptr, (T, BT), (H*BT, 1), (t0, 0), (BT, BT), (1,0)) tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0,1)) ``` With Aqk_ptr adjusted by `(i_b*T*H + i_h)*BT`. Block ptr shape (T,BT) strides (H*BT,1). For t0, offset = t0*H*BT + 0*1. With base offset, total = i_b*T*H*BT + i_h*BT + t0*H*BT. Correct. So Aqk_mine[b,t,h,:] should be correct for b=0,h=0,t=0. But it shows only column 0 nonzero, while ref row 0 should have only column 0 nonzero (after masking). Wait, ref row 0 col 0 = -0.0089 matches. Row 1 col 0 = -0.0034 vs ref -0.0112. These are lower triangular, should match. So Aqk is wrong for row 1. Why? Wait, maybe my kernel 1 Aqk computation is wrong, but kernel 2 somehow produces correct output because Aqk errors are in upper triangle or something? No, row 1 col 0 is lower triangle and wrong. But mimic_check4 output is correct (diff 0.01). How? Unless my check_Aqk3 reference computation is wrong. Wait, in check_Aqk3, I compute Aqk_ref using bf16 q and k. But maybe the reference Aqk is not what I think. Let me recompute carefully. Actually, in reference, Aqk is computed from q (scaled) and k (not scaled), with g cumsum. I do the same. But reference applies scale to q in fp32. I apply in bf16. That shouldn't cause 3x diff. Wait, maybe in my kernel, I compute Aqk but then overwrite it? No. Actually, let me check if check_Aqk3 is using the same q,k as reference. Yes. Hmm. Let me check Aqk_mine for b=0,h=0,t=0 row 1 col 0 manually. Actually, wait. mimic_check4 gave sol max 0.0111 and ref max 0.0117, diff 0.01. That's good. But if Aqk is wrong by 0.008 in lower triangle, the final output should be wrong more than 0.01. Unless Aqk errors are mostly in upper triangle (which are masked to 0 in kernel). But row 1 col 0 is lower triangle and wrong. Wait, actually maybe my Aqk_mine values are correct and the reference computation in check_Aqk3 is wrong. Let me verify reference Aqk manually with the reference code. Actually, I can extract Aqk from reference by adding prints, but reference doesn't expose it. Let me compute it with reference's exact logic. Reference Aqk computation: ```python Aqk = torch.zeros(B, H, N, C, C) for i in range(C): Aqk[...,i] = einsum('... c d, ... d -> ... c', q_i * (g_i - g_i[:,:,:,i:i+1,:]).exp(), k_i[:,:,:,i,:]) Aqk = Aqk.masked_fill(mask_strict_upper, 0) ``` where mask_strict_upper is triu(diagonal=1). For row 1 (c=1), col 0 (i=0): Aqk[1,0] = sum_d q[1,d]*k[0,d]*exp(g[1,d]-g[0,d]) * scale. Let me compute this manually in Python.
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Bash
args
{
  "command": "<1503 chars \u2014 see below>",
  "description": "Manual Aqk value"
}
command (1503 chars)
cat > /workspace/problems/02_kda_cutlass/manual_Aqk.py <<'PY'
import torch
import reference

B,T,H,K,V,CS = 2,1024,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS

torch.manual_seed(42); torch.cuda.manual_seed_all(42)
q,k,v,g,beta = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

scale = float(K)**-0.5
q = (q * scale).float()
k = k.float()
g = g.float().cumsum(dim=1)  # cumsum along T (since no chunking)

# chunk 0, head 0
c, i = 1, 0
val = (q[0, c, 0] * k[0, i, 0] * (g[0, c, 0] - g[0, i, 0]).exp()).sum()
print(f'manual Aqk[{c},{i}] =', val.item())

# via kernel
class DK(torch.nn.Module):
    def forward(self,q,k,v,g,beta):
        B,T,H,K=q.shape; V=v.shape[-1]; NT=T//CS; scale=float(K)**-0.5
        g_cum = g.view(B,NT,CS,H,K).cumsum(dim=2).view(B,T,H,K)
        q_scaled=q*scale; kg=k*g_cum.exp()
        Aqk=torch.empty(B,T,H,CS,device='cuda',dtype=torch.float32)
        w=torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u=torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        import solution
        solution._kda_intra_kernel[(NT,B*H)](q_scaled,k,v,g_cum,kg,beta,Aqk,w,u,1.0,
            B=B,T=T,H=H,K=K,V=V,BT=CS,BK=64,BV=32,num_stages=1,num_warps=4)
        return Aqk

dk = DK().cuda()
Aqk = dk(q.bfloat16(),k.bfloat16(),v.bfloat16(),g.unsqueeze(-2).expand(-1,-1,H,-1),beta)
# wait g shape
PY
uv run python /workspace/problems/02_kda_cutlass/manual_Aqk.py 2>&1 | tail -10
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stdout (528 chars)
manual Aqk[1,0] = -0.003372868988662958
Traceback (most recent call last):
  File "/workspace/problems/02_kda_cutlass/manual_Aqk.py", line 35, in <module>
    Aqk = dk(q.bfloat16(),k.bfloat16(),v.bfloat16(),g.unsqueeze(-2).expand(-1,-1,H,-1),beta)
                                                    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
RuntimeError: expand(torch.cuda.FloatTensor{[2, 1024, 8, 1, 128]}, size=[-1, -1, 8, -1]): the number of sizes provided (4) must be greater or equal to the number of dimensions in the tensor (5)
assistant
reasoning (1724 chars)
Manual Aqk[1,0] = -0.00337, which matches my kernel's mine[0,0,1,0] = -0.0034! So my kernel is correct, and my reference computation in check_Aqk3 was wrong. Wait, manual uses q scaled in fp32 and k fp32. My kernel uses q scaled in bf16 and k bf16. They match -0.0034. So kernel is correct. But check_Aqk3's reference gave -0.0112 for Aqk[1,0]. So check_Aqk3 reference is wrong. Why? In check_Aqk3, I used: ```python qf = (q * scale).bfloat16().float() kf = k.bfloat16().float() gf = g.float() gf = rearrange(gf, 'b (n c) h d -> b h n c d', c=CS).cumsum(dim=3) ``` Wait, `g` is already the raw input. I rearrange and cumsum. In reference, g is also rearranged and cumsumed. Same. But then I compute `qf * (gf - gf[...,i:i+1,:]).exp()` and einsum with `kf[...,i,:]`. For c=1,i=0, this is q[1]*k[0]*exp(g[1]-g[0]). Should be -0.00337. But check_Aqk3 gave -0.0112. So there must be a bug in check_Aqk3's reshaping or indexing. Actually, I see: in check_Aqk3, I do: ```python Aqk_bf16 = torch.zeros(B,H,T//CS,CS,CS,device='cuda') for i in range(CS): Aqk_bf16[...,i] = torch.einsum('b h n c d, b h n d -> b h n c', qf * (gf - gf[...,i:i+1,:]).exp(), kf[...,i,:]) Aqk_bf16 = Aqk_bf16.view(B,T,H,CS) ``` Aqk_bf16 shape (B,H,N,C,C). After view to (B,T,H,CS), the indexing is Aqk_bf16[b, n*C + r, h, c]. But `view` might not preserve the intended layout. The original Aqk_bf16 has last two dims (C,C) where first C is rows and second C is columns. After view (B, N*C, H, C), the rows are n*C + r and columns are c. Correct. Hmm. But manual gave -0.00337 and check_Aqk3 gave -0.0112. So check_Aqk3 has a bug. It doesn't matter; my kernel is correct. So the offset fix worked. mimic_check4 now passes. Let me run check.py.
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Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run check.py"
}
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stdout (230 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case large_qkv: tolerance exceeded: max_abs_diff=0.101654 max_rel_diff=342803 bad=621/2097152 worst_idx=(0, 581, 5, 30) (atol=0.05, rtol=0.05)
assistant
reasoning (6037 chars)
Good progress! Nominal and small_qkv pass, but large_qkv fails (max diff 0.10 > 0.05). This is likely a numerical precision issue with bf16 in the solve or GEMMs. For large_qkv, inputs scaled by 2.0. q,k,v are larger. The delta rule solve might amplify errors. I should use fp32 accumulation and maybe fp32 for the solve and some GEMMs. Currently A is bf16 for w,u GEMMs. For large inputs, A values could be larger and bf16 rounding hurts. Let me try keeping A in fp32 for the GEMMs (w,u and kernel 2). Triton tl.dot with fp32 inputs might be supported. Or I can cast inputs to bf16 but accumulate in fp32 (which is default). Actually, the issue might be in the solve itself: I do forward substitution in fp32, which is good. But then I cast A to bf16 for GEMMs. For large inputs, A's off-diagonal values might be more significant, and bf16 rounding changes them. Let me try keeping A in fp32 and doing GEMMs in fp32. In Triton, `tl.dot(A, b_kg)` where both are fp32 should work. But it might not use tensor cores. Alternatively, I can compute w and u in Python using the solved A (computed in kernel) and fp32 GEMMs. But A is only in the kernel. Actually, I can have kernel 1 store A in fp32 global memory, then compute w = A @ kg and u = A @ v in Python using fp32. This avoids bf16 GEMM errors. But A is (B,T,H,BT) = 2*1024*8*64 = 1M elements * 4 bytes = 4MB. Storing it is fine. Then kernel 2 can also use A in fp32 for Aqk @ v_i. But Aqk is already stored in fp32. Wait, Aqk is separate from A. I was computing Aqk in kernel 1 and storing it. The A matrix used for w,u is internal to kernel 1. Actually, I can restructure: - Kernel 1: compute Aqk and M (A0), solve for A. Store Aqk (fp32), A (fp32), and maybe not w,u. - Python: compute w = A @ kg, u = A @ v using fp32 GEMM. - Kernel 2: use w,u,Aqk as before, but Aqk in fp32. This moves w,u computation to Python (fast with cuBLAS) and avoids bf16 errors. But we still need A for the recurrence. Actually w and u are A @ kg and A @ v. If we compute them in Python fp32, they're accurate. For kernel 2, Aqk @ v_i is currently bf16 dot. We can keep Aqk in fp32 and do dot in fp32. But Aqk is (BT,BT) small; bf16 might be OK. For large inputs, Aqk values are larger (scale * 2 * 2 = 4x), still moderate. Actually, the tolerance for large_qkv is 0.05. Let me see if using fp32 for A and w,u helps. Alternatively, the issue might be in kernel 2's S update or recurrence, where errors accumulate. Using fp32 for w,u and S might help. Let me modify kernel 1 to store A, and compute w,u in Python. Wait, but kernel 2 also uses w and u. If I compute them in Python, I pass them to kernel 2. Same as before, just computed in Python. Actually, I can keep kernel 1 computing w,u but in fp32 by casting A to bf16? No, that loses precision. I can do `tl.dot(A, b_kg)` with A and b_kg as fp32. Let me check if Triton supports fp32 dot. Actually, I think Triton tl.dot supports fp32 inputs but uses FMA, not tensor cores. For 64x64 @ 64x128, it's small, so FMA is fine. Let me try. Change: ```python b_w = tl.dot(A, b_kg) # both fp32 b_u = tl.dot(A, b_v) # both fp32 ``` And load b_kg, b_v as fp32 instead of bf16. Similarly in kernel 2, use fp32 for all dots. But that might be slower. Actually, the large_qkv error is small (0.10 vs tolerance 0.05). Maybe just using fp32 for A->w,u is enough. Let me also consider: reference uses fp32 throughout. My kernel uses bf16 for qg, k, g loads in kernel 2. For large inputs, bf16 might not have enough precision. But tolerance is 0.05, so bf16 should mostly be OK. The main issue is likely the solve: A is computed in fp32, then cast to bf16 for GEMM. The bf16 rounding of A propagates to w,u and then to the recurrence. For large inputs, the relative error might exceed tolerance. Let me try storing A and computing w,u in Python fp32. I'll also keep Aqk in fp32. Actually, storing A requires adding an output buffer. Then in Python: ```python A = torch.empty(B,T,H,BT, device=device, dtype=torch.float32) _kda_intra_kernel(..., Aqk, A, ...) w = torch.einsum('b t h i, b t h i k -> b t h k', A, kg) u = torch.einsum('b t h i, b t h i v -> b t h v', A, v) ``` But wait, A is (B,T,H,BT). kg is (B,T,H,K). The einsum is over the chunk token index i. For each output token t, A[t,i] is the i-th column? Wait, A[t,:] is the row corresponding to token t (as the target), and columns are source tokens. w[t,k] = sum_i A[t,i] * kg[i,k] where i ranges over the chunk containing t. But A is stored per token t, with columns within the chunk. So for token t = n*C + r, A[t, c] is A[r,c] for chunk n. Then w[t,k] = sum_{c=0}^{C-1} A[t,c] * kg[n*C + c, k]. In einsum, I need to handle the chunking. A shape (B,T,H,C), kg shape (B,T,H,K). For token t, only tokens in same chunk contribute. So: ```python A_5d = A.view(B, NT, C, H, C) # A[b,n,r,h,c] kg_5d = kg.view(B, NT, C, H, K) # kg[b,n,c,h,k] w = torch.einsum('b n r h c, b n c h k -> b n r h k', A_5d, kg_5d).view(B,T,H,K) u = torch.einsum('b n r h c, b n c h v -> b n r h v', A_5d.view(B,NT,C,H,C), v.view(B,NT,C,H,V)).view(B,T,H,V) ``` This is a batched GEMM over chunks. PyTorch can do it efficiently. Let me implement this. It might be faster than Triton anyway because cuBLAS is optimized. Also for kernel 2, I can keep using bf16 or switch to fp32 for Aqk @ v_i. Let me first try with Aqk in bf16 (store from kernel as bf16 to save memory). Actually, I can store Aqk as fp32 and cast to bf16 in kernel 2. Let me modify kernel 1 to output A (fp32) instead of w,u. Then compute w,u in Python. Actually, I can have kernel 1 output both A and Aqk, and skip w,u. Then Python computes w,u. Wait, but kernel 1 already computes w,u. I can just change them to use fp32 dot. Let me try that first (simpler), before restructuring. Change in kernel 1: ```python b_kg = tl.load(p_kg, boundary_check=(0,1)).to(tl.float32) # keep fp32 b_w = tl.dot(A, b_kg) # fp32 dot ``` Similarly for u. In kernel 2, keep bf16 for now but maybe use fp32. Let me try fp32 for kernel 1 w,u.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (41 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -125,31 +125,30 @@
     )
     tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
 
-    A_bf16 = A.to(tl.bfloat16)
-
+    # compute w = A @ kg and u = A @ v in fp32 for better accuracy
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
         p_kg = tl.make_block_ptr(
             kg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_w = tl.dot(A_bf16, b_kg)
+        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.float32)
+        b_w = tl.dot(A, b_kg)
         p_w = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        tl.store(p_w, b_w.to(tl.float32), boundary_check=(0, 1))
+        tl.store(p_w, b_w, boundary_check=(0, 1))
 
     for i_v in range(tl.cdiv(V, BV)):
         off_v = i_v * BV
         p_v = tl.make_block_ptr(
             v_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        b_v = tl.load(p_v, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_u = tl.dot(A_bf16, b_v)
+        b_v = tl.load(p_v, boundary_check=(0, 1)).to(tl.float32)
+        b_u = tl.dot(A, b_v)
         p_u = tl.make_block_ptr(
             u_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        tl.store(p_u, b_u.to(tl.float32), boundary_check=(0, 1))
+        tl.store(p_u, b_u, boundary_check=(0, 1))
 
 
 @triton.jit
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<941 chars \u2014 see diff>",
  "new_string": "<934 chars \u2014 see diff>"
}
in 160259out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (148 chars)
Let me also use fp32 for kernel 2 dots to reduce recurrence error. But first check if fp32 dot is supported. If not, fallback. Let me run check.py.
in 162971out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run check.py with fp32 w/u"
}
in 162971out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (230 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case large_qkv: tolerance exceeded: max_abs_diff=0.101669 max_rel_diff=345043 bad=622/2097152 worst_idx=(0, 581, 5, 30) (atol=0.05, rtol=0.05)
assistant
reasoning (804 chars)
Same error. So fp32 w,u in kernel 1 didn't fix it. The error is likely in kernel 2 recurrence. Let me use fp32 for kernel 2 as well. Specifically, load w,u,qg,k,g,Aqk as fp32 and do dots in fp32. S in fp32. Actually, the issue might be that Aqk is stored as fp32 but loaded as bf16 in kernel 2. Let me load as fp32. Also, w,u are fp32 already. qg,k are bf16 (from Python). g is fp32. Aqk is fp32. In kernel 2, dots: - w @ S: w (BT,BK) fp32, S (BK,BV) fp32. Use fp32 dot. - qg @ S: qg (BT,BK) bf16, S fp32. Cast qg to fp32. - Aqk @ v_i: Aqk (BT,BT) fp32, v_i (BT,BV) fp32. Use fp32 dot. - k_update.T @ v_i: k_update (BT,BK) fp32, v_i (BT,BV) fp32. Use fp32 dot. This should be more accurate. Let me modify kernel 2. Also, S scaling: d0 = exp(g_last). Use fp32. Let me rewrite kernel 2 dots in fp32.
in 163172out 0cache_r 0cache_w 0
assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "offset": 155,
  "limit": 130
}
in 163172out 0cache_r 0cache_w 0
tool
stdout (4894 chars) — TRUNCATED
155	def _kda_inter_kernel(
156	    w_ptr,
157	    u_ptr,
158	    qg_ptr,
159	    k_ptr,
160	    g_ptr,
161	    Aqk_ptr,
162	    glast_ptr,
163	    o_ptr,
164	    B: tl.constexpr,
165	    T: tl.constexpr,
166	    H: tl.constexpr,
167	    K: tl.constexpr,
168	    V: tl.constexpr,
169	    BT: tl.constexpr,
170	    BK: tl.constexpr,
171	    BV: tl.constexpr,
172	):
173	    i_vb = tl.program_id(0)
174	    i_bh = tl.program_id(1)
175	    i_b = i_bh // H
176	    i_h = i_bh % H
177	
178	    # adjust base pointers for batch/head
179	    w_ptr += (i_b * T * H + i_h) * K
180	    u_ptr += (i_b * T * H + i_h) * V
181	    qg_ptr += (i_b * T * H + i_h) * K
182	    k_ptr += (i_b * T * H + i_h) * K
183	    g_ptr += (i_b * T * H + i_h) * K
184	    Aqk_ptr += (i_b * T * H + i_h) * BT
185	    o_ptr += (i_b * T * H + i_h) * V
186	
187	    off_v0 = i_vb * BV
188	    NT = T // BT
189	
190	    # S split into K tiles of (BK, BV)
191	    S0 = tl.zeros([BK, BV], dtype=tl.float32)
192	    S1 = tl.zeros([BK, BV], dtype=tl.float32)
193	
194	    for i_n in range(NT):
195	        t0 = i_n * BT
196	
197	        p_u = tl.make_block_ptr(
198	            u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
199	        )
200	        b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
201	
202	        b_v = b_u
203	        p_w0 = tl.make_block_ptr(
204	            w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
205	        )
206	        b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.bfloat16)
20

... (3394 more chars)
assistant
reasoning (538 chars)
Let me change kernel 2 to use fp32 for all computations. I'll load qg as fp32, do fp32 dots, store o as bf16. Actually, keeping qg,k as bf16 and casting to fp32 for dot gives fp32 accumulation. That's fine. Change: - b_w0, b_w1 loaded as fp32 (w is fp32), dot in fp32 with S0/S1 fp32. - b_qg0, b_qg1 loaded as bf16, cast to fp32, dot in fp32 with S fp32. - b_Aqk loaded as fp32 (Aqk is fp32), dot in fp32 with b_v fp32. - k_up0, k_up1 computed in fp32, dot in fp32 with b_v fp32. This avoids bf16 rounding in all recurrence operations.
in 165338out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (44 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -203,31 +203,31 @@
         p_w0 = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
         )
-        b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))
+        b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.float32)
+        b_v -= tl.dot(b_w0, S0)
         p_w1 = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
         )
-        b_w1 = tl.load(p_w1, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))
+        b_w1 = tl.load(p_w1, boundary_check=(0, 1)).to(tl.float32)
+        b_v -= tl.dot(b_w1, S1)
 
         b_o = tl.zeros([BT, BV], dtype=tl.float32)
         p_qg0 = tl.make_block_ptr(
             qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
         )
-        b_qg0 = tl.load(p_qg0, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))
+        b_qg0 = tl.load(p_qg0, boundary_check=(0, 1)).to(tl.float32)
+        b_o += tl.dot(b_qg0, S0)
         p_qg1 = tl.make_block_ptr(
             qg_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
         )
-        b_qg1 = tl.load(p_qg1, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_o += tl.dot(b_qg1, S1.to(tl.bfloat16))
+        b_qg1 = tl.load(p_qg1, boundary_check=(0, 1)).to(tl.float32)
+        b_o += tl.dot(b_qg1, S1)
 
         p_Aqk = tl.make_block_ptr(
             Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
         )
-        b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.bfloat16)
-        b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))
+        b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.float32)
+        b_o += tl.dot(b_Aqk, b_v)
 
         p_o = tl.make_block_ptr(
             o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1216 chars \u2014 see diff>",
  "new_string": "<1131 chars \u2014 see diff>"
}
in 165338out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (38 chars)
Also update the S update dots to fp32.
in 166609out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -249,7 +249,7 @@
         )
         b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
         k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)
-        S0 += tl.dot(tl.trans(k_up0.to(tl.bfloat16)), b_v.to(tl.bfloat16))
+        S0 += tl.dot(tl.trans(k_up0), b_v)
 
         p_k1 = tl.make_block_ptr(
             k_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
@@ -260,7 +260,7 @@
         )
         b_g1 = tl.load(p_g1, boundary_check=(0, 1)).to(tl.float32)
         k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
-        S1 += tl.dot(tl.trans(k_up1.to(tl.bfloat16)), b_v.to(tl.bfloat16))
+        S1 += tl.dot(tl.trans(k_up1), b_v)
 
         S0 *= d0[:, None]
         S1 *= d1[:, None]
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<620 chars \u2014 see diff>",
  "new_string": "<556 chars \u2014 see diff>"
}
in 166609out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (17 chars)
Now run check.py.
in 167205out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run check with fp32 kernel 2"
}
in 167205out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (230 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case large_qkv: tolerance exceeded: max_abs_diff=0.101654 max_rel_diff=342803 bad=621/2097152 worst_idx=(0, 581, 5, 30) (atol=0.05, rtol=0.05)
assistant
reasoning (4075 chars)
Same error. So fp32 in kernel 2 didn't help. The issue might be in kernel 1 Aqk or A computation, or in Python preprocessing. For large_qkv, q,k,v scaled by 2.0. g unchanged. q_scaled ~ 0.2 * 0.088 = 0.0177. k ~ 0.2. Aqk values ~ 0.0177*0.2*128*exp(diff) ~ 0.45*exp(diff). A0 ~ -0.45*exp(diff)*beta. With diff up to ~1, A0 ~ -1.2. (I - A0) might have large condition number, and the forward substitution could amplify errors. But reference uses fp32 and gets correct. My kernel uses fp32 for solve. Should match. Wait, but in kernel 1, I compute Aqk and M with q_scaled in bf16. For large_qkv, q_scaled values ~0.018. bf16 has ~3 digits, so relative error ~0.5%. That could propagate. But the error is 0.10 in output, which is large. Maybe the bf16 q in Aqk computation is the issue. Actually, in kernel 1, I load q as bf16 then cast to fp32. The bf16 q has rounding error. For Aqk, this is q_scaled_bf16 vs q_scaled_fp32. The relative error in q is ~0.4%. In Aqk, ~0.4%. In A (after solve), could be amplified by condition number. In w = A @ kg, errors in A and kg (bf16) multiply. In recurrence, errors accumulate. For large inputs, the condition number of (I - A0) might be larger, amplifying bf16 errors in q and k beyond tolerance. Solution: load q and k as fp32 in kernel 1. But they're bf16 tensors. We can cast in Python to fp32 and pass fp32 q,k to kernel 1. Then all Aqk/M/solve/w/u are fp32. Only qg for kernel 2 might be bf16. Actually, for accuracy, we can pass q,k,v,g as fp32 to the kernels. The problem says inputs are bf16, but we can upcast inside forward. Reference upcasts to fp32. In solution.Model.forward, I can convert q,k,v,beta to fp32, compute qg, kg, etc. in fp32, pass fp32 tensors to kernels. The kernels then operate in fp32. Store outputs as bf16. This will be slower (2x memory traffic) but more accurate. Actually, reference already does everything in fp32. So matching reference accuracy requires fp32. But the problem says performance matters. We need a balance. For large_qkv, tolerance is 0.05. With bf16 q,k in Aqk, error might exceed. Let me try upcasting q,k to fp32 for kernel 1 and see. Actually, I can upcast all inputs in Python: ```python q = q.to(torch.float32) k = k.to(torch.float32) v = v.to(torch.float32) g = g.to(torch.float32) beta = beta.to(torch.float32) ``` Then compute g_cum, qg, kg, etc. in fp32. Pass fp32 tensors to kernels. But qg and k need to be stored as fp32, which uses more memory. We can cast qg to bf16 for kernel 2 if needed, or keep fp32. Let me first try upcasting q,k for kernel 1 only (Aqk computation) and see if large_qkv passes. Keep qg,k for kernel 2 as bf16. Actually, the error might also come from kernel 2 using bf16 qg. For large_qkv, qg = q*scale*exp(g) ~ 0.018 * 2.7 ~ 0.048. bf16 can represent this with ~0.5% error. Recurrence accumulates over T/64=16 chunks, error could grow to ~8% relative, which at output max ~0.2 is 0.016. Plus other errors. Might still pass if condition number not too bad. But the actual error is 0.10. So likely kernel 1 Aqk/A error dominates. Let me try passing fp32 q,k to kernel 1. I'll modify Python to compute q_scaled_fp32 and k_fp32, g_cum_fp32, kg_fp32, and pass those. Kernel 1 loads as fp32. w,u computed in fp32. Then cast w,u to bf16? Or keep fp32. For kernel 2, keep w,u as fp32. qg for kernel 2 can be computed from q_scaled_fp32 * exp(g_cum) and cast to bf16, or keep fp32. Let me keep qg,k as bf16 for memory, and see. Actually, if w,u are fp32 and Aqk is fp32, and kernel 2 uses fp32 for dots, the only bf16 inputs are qg and k. qg error might matter. Let me try keeping qg fp32 too. So full fp32 path: - Python: cast q,k,v,g,beta to fp32. - Compute g_cum, qg, kg in fp32. - Kernel 1: all fp32. - Kernel 2: w,u,Aqk,g,k,glast in fp32, compute in fp32. Store o as bf16. Memory: qg fp32 = 33MB for B=2,T=2048,H=8,K=128. k fp32 same. w,u fp32 33MB each. Aqk fp32 8MB. Total ~140MB. Acceptable. Performance: 2x memory vs bf16. Might be slower. But let's first get correctness. Let me modify forward to upcast and pass fp32.
in 167515out 0cache_r 0cache_w 0
assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "offset": 270,
  "limit": 100
}
in 167515out 0cache_r 0cache_w 0
tool
stdout (2638 chars) — TRUNCATED
270	    def __init__(self, B: int, T: int, H: int, K: int, V: int, chunk_size: int = 64):
271	        super().__init__()
272	        self.B, self.T, self.H, self.K, self.V = B, T, H, K, V
273	        self.chunk_size = chunk_size
274	        self.scale = float(K) ** -0.5
275	        self.register_buffer("_dummy", torch.zeros(1), persistent=False)
276	
277	    def forward(
278	        self,
279	        q: torch.Tensor,
280	        k: torch.Tensor,
281	        v: torch.Tensor,
282	        g: torch.Tensor,
283	        beta: torch.Tensor,
284	    ) -> torch.Tensor:
285	        B, T, H, K = q.shape
286	        V = v.shape[-1]
287	        BT = self.chunk_size
288	        assert T % BT == 0
289	        NT = T // BT
290	        device = q.device
291	
292	        g_4d = g.view(B, NT, BT, H, K)
293	        g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
294	
295	        q_scaled = q * self.scale
296	        qg = q_scaled * g_cum.exp()
297	        kg = k * g_cum.exp()
298	
299	        glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()
300	
301	        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.float32)
302	        w = torch.empty(B, T, H, K, device=device, dtype=torch.float32)
303	        u = torch.empty(B, T, H, V, device=device, dtype=torch.float32)
304	
305	        grid1 = (NT, B * H)
306	        _kda_intra_kernel[grid1](
307	            q_scaled, k, v, g_cum, kg, beta, Aqk, w, u,
308	            1.0,
309	            B=B, T=T, H=H, K=K, V=V

... (1138 more chars)
assistant
reasoning (474 chars)
Let me modify forward to upcast inputs to fp32 for the kernels. I'll keep the input tensors bf16 but upcast internally. Actually, the simplest is to upcast at the start: ```python q = q.to(torch.float32) k = k.to(torch.float32) v = v.to(torch.float32) g = g.to(torch.float32) beta = beta.to(torch.float32) ``` Then all computations are fp32. Output o is bf16. But this means the Model receives bf16 and returns bf16, but internally uses fp32. That's fine. Let me do this.
in 169710out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (16 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -289,6 +289,13 @@
         NT = T // BT
         device = q.device
 
+        # Upcast to fp32 for numerical accuracy (reference works in fp32).
+        q = q.to(torch.float32)
+        k = k.to(torch.float32)
+        v = v.to(torch.float32)
+        g = g.to(torch.float32)
+        beta = beta.to(torch.float32)
+
         g_4d = g.view(B, NT, BT, H, K)
         g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1481 chars \u2014 see diff>",
  "new_string": "<1723 chars \u2014 see diff>"
}
in 169710out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (17 chars)
Now run check.py.
in 171084out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run check with fp32 inputs"
}
in 171084out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (230 chars)
FAIL: shape 0 {'B': 2, 'T': 1024, 'H': 8, 'K': 128, 'V': 128, 'CHUNK_SIZE': 64} seed 42 case large_qkv: tolerance exceeded: max_abs_diff=0.101669 max_rel_diff=342803 bad=620/2097152 worst_idx=(0, 581, 5, 30) (atol=0.05, rtol=0.05)
assistant
reasoning (401 chars)
Still same error. So upcasting to fp32 didn't fix large_qkv. The issue is not bf16 rounding. This means my algorithm has a bug that's more fundamental. The same element (0,581,5,30) has error 0.10. Let me debug for large_qkv. The error is at batch 0, token 581, head 5, channel 30. Token 581 is in chunk 9 (581 // 64 = 9). Head 5. Let me compare sol and ref outputs for large_qkv around that token.
in 171251out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "<1318 chars \u2014 see below>",
  "description": "Debug large_qkv error"
}
command (1318 chars)
cat > /workspace/problems/02_kda_cutlass/debug_large.py <<'PY'
import torch
import reference, solution
from src.eval.numeric_stress import numeric_stress_cases, numeric_stress_context

shape = {"B":2,"T":1024,"H":8,"K":128,"V":128,"CHUNK_SIZE":64}
for k,v in shape.items():
    setattr(reference,k,v)
    setattr(solution,k,v)

init_args = reference.get_init_inputs()
ref_model = reference.Model(*init_args).cuda().eval()
sol_model = solution.Model(*init_args).cuda().eval()
sol_model.load_state_dict(ref_model.state_dict(), strict=True)

seed = 42
torch.manual_seed(seed); torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

# large_qkv case
for case in numeric_stress_cases("02_kda_cutlass"):
    print('case', case.name)
    with numeric_stress_context(ref_model, sol_model, inputs, case) as inps:
        with torch.no_grad():
            ro = ref_model(*inps)
            so = sol_model(*inps)
        diff = (ro-so).abs()
        print('max diff', diff.max().item(), 'mean diff', diff.mean().item())
        idx = (diff == diff.max()).nonzero()[0]
        print('worst idx', idx.tolist())
        print('ref at idx', ro[tuple(idx)].item(), 'sol', so[tuple(idx)].item())
PY
uv run python /workspace/problems/02_kda_cutlass/debug_large.py 2>&1 | tail -30
in 171251out 0cache_r 0cache_w 0
tool
stdout (450 chars)
case nominal
max diff 0.0103759765625 mean diff 0.000415802001953125
worst idx [1, 196, 6, 100]
ref at idx 0.01068115234375 sol 0.0003147125244140625
case small_qkv
max diff 1.0360963642597198e-08 mean diff 4.5656634029001e-10
worst idx [1, 196, 6, 100]
ref at idx 1.0710209608078003e-08 sol 3.4378899727016687e-10
case large_qkv
max diff 0.1015625 mean diff 0.00506591796875
worst idx [0, 581, 5, 30]
ref at idx 0.09814453125 sol -0.0035247802734375
assistant
reasoning (2118 chars)
For large_qkv, ref at (0,581,5,30) = 0.098, sol = -0.0035. Huge difference. This suggests the recurrence is way off for that token. For nominal, worst diff is at (1,196,6,100), ref=0.0107, sol=0.0003. Also a factor of 30 difference, but absolute is 0.01. For large_qkv, same factor gives 0.10. So there's a systematic error in the recurrence for some tokens/heads. The output is much smaller than reference. This might be because the delta rule solve or recurrence is wrong. Let me verify by computing the output with a PyTorch implementation that matches reference exactly but uses my Aqk,w,u. Actually, let me first check if my Aqk,w,u match reference's. If they do, then the bug is in kernel 2 recurrence. If not, bug is in kernel 1. Let me extract Aqk, w, u from my solution and compare to reference. I can modify solution to return intermediates. Actually, I can compute reference's Aqk,w,u in Python. But easier: I can check if using reference's recurrence with my w,u,Aqk gives the right output. Wait, reference computes Aqk,w,u internally. I can write a PyTorch function that does the inter-chunk recurrence using my w,u,Aqk and reference's formula, and compare to my kernel 2 output. Actually, let me just implement the reference recurrence in Python with my w,u,Aqk,qg,k,g_cum and see if it matches reference output. If it does, then kernel 2 is wrong. If it doesn't, then my w,u,Aqk are wrong. But the recurrence needs Aqk (which I have), w, u, qg, k, g_cum. Let me write it. Reference recurrence: For each chunk n: q_i = q[:,:,n], k_i = k[:,:,n], u_i = u[:,:,n], g_i = g[:,:,n] Aqk[...,j] = einsum(q_i * exp(g_i - g_j), k_j) (lower triangular incl diag) v_i = u_i - w_i @ S o_i = (q_i * exp(g_i)) @ S + Aqk @ v_i S = S * exp(g_i[-1]) + (k_i * exp(g_i[-1] - g_i)).T @ v_i I can compute this in Python with my intermediates. But I need Aqk. I have Aqk from kernel 1. Let me compare. Actually, the Aqk in reference is also computed from q,k,g. If my Aqk matches reference's, then the recurrence using my Aqk,w,u should match reference output. Let me write a Python recurrence function.
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assistant
Bash
args
{
  "command": "<3637 chars \u2014 see below>",
  "description": "Check if recurrence with my intermediates matches ref"
}
command (3637 chars)
cat > /workspace/problems/02_kda_cutlass/debug_inter.py <<'PY'
import torch
from einops import rearrange
import reference, solution
from src.eval.numeric_stress import numeric_stress_cases, numeric_stress_context

shape = {"B":2,"T":1024,"H":8,"K":128,"V":128,"CHUNK_SIZE":64}
for k,v in shape.items():
    setattr(reference,k,v)
    setattr(solution,k,v)

init_args = reference.get_init_inputs()
ref_model = reference.Model(*init_args).cuda().eval()
sol_model = solution.Model(*init_args).cuda().eval()
sol_model.load_state_dict(ref_model.state_dict(), strict=True)

seed = 42
torch.manual_seed(seed); torch.cuda.manual_seed_all(seed)
inputs = [t.cuda() if hasattr(t,'to') else t for t in reference.get_inputs()]

# large_qkv
for case in numeric_stress_cases("02_kda_cutlass"):
    if case.name != 'large_qkv':
        continue
    with numeric_stress_context(ref_model, sol_model, inputs, case) as inps:
        q,k,v,g,beta = inps
        B,T,H,K = q.shape; V=v.shape[-1]; BT=64; NT=T//BT
        
        # my intermediates
        q_f = q.to(torch.float32)
        k_f = k.to(torch.float32)
        v_f = v.to(torch.float32)
        g_f = g.to(torch.float32)
        g_cum = g_f.view(B,NT,BT,H,K).cumsum(dim=2).view(B,T,H,K)
        qg = (q_f * float(K)**-0.5) * g_cum.exp()
        kg = k_f * g_cum.exp()
        
        # run my kernel 1 to get Aqk,w,u
        Aqk = torch.empty(B,T,H,BT,device='cuda',dtype=torch.float32)
        w = torch.empty(B,T,H,K,device='cuda',dtype=torch.float32)
        u = torch.empty(B,T,H,V,device='cuda',dtype=torch.float32)
        solution._kda_intra_kernel[(NT,B*H)](
            q_f*float(K)**-0.5, k_f, v_f, g_cum, kg, beta.to(torch.float32), Aqk, w, u, 1.0,
            B=B,T=T,H=H,K=K,V=V,BT=BT,BK=64,BV=32,num_stages=1,num_warps=4)
        
        # reference recurrence in Python with my Aqk,w,u
        qg_5d = rearrange(qg, 'b (n c) h d -> b h n c d', c=BT)
        k_5d = rearrange(k_f, 'b (n c) h d -> b h n c d', c=BT)
        Aqk_5d = rearrange(Aqk, 'b (n c) h d -> b h n c d', c=BT)
        w_5d = rearrange(w, 'b (n c) h d -> b h n c d', c=BT)
        u_5d = rearrange(u, 'b (n c) h v -> b h n c v', c=BT)
        g_5d = rearrange(g_cum, 'b (n c) h d -> b h n c d', c=BT)
        S = torch.zeros(B,H,K,V,device='cuda',dtype=torch.float32)
        o_py = torch.empty(B,H,NT,BT,V,device='cuda',dtype=torch.float32)
        for n in range(NT):
            q_i = qg_5d[:,:,n]
            k_i = k_5d[:,:,n]
            Aqk_i = Aqk_5d[:,:,n]
            w_i = w_5d[:,:,n]
            u_i = u_5d[:,:,n]
            g_i = g_5d[:,:,n]
            v_i = u_i - torch.einsum('b h c k, b h k v -> b h c v', w_i, S)
            o_i = torch.einsum('b h c k, b h k v -> b h c v', q_i, S) + torch.einsum('b h c j, b h j v -> b h c v', Aqk_i, v_i)
            o_py[:,:,n] = o_i
            d = g_i[:,:,-1,:].exp()
            S = S * d.unsqueeze(-1) + torch.einsum('b h c k, b h c v -> b h k v', k_i * (g_i[:,:,-1:,:] - g_i).exp(), v_i)
        o_py = rearrange(o_py, 'b h n c v -> b (n c) h v').to(torch.bfloat16)
        
        with torch.no_grad():
            ro = ref_model(q.to(torch.bfloat16),k.to(torch.bfloat16),v.to(torch.bfloat16),g,beta.to(torch.bfloat16))
            so = sol_model(q.to(torch.bfloat16),k.to(torch.bfloat16),v.to(torch.bfloat16),g,beta.to(torch.bfloat16))
        
        print('ref vs py_recurrence max diff', (ro - o_py).abs().max().item())
        print('ref vs sol max diff', (ro - so).abs().max().item())
        print('py_rec vs sol max diff', (o_py - so).abs().max().item())
PY
uv run python /workspace/problems/02_kda_cutlass/debug_inter.py 2>&1 | tail -20
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subagent · 16 events · 5 tools
tool
stdout (283 chars)
Command running in background with ID: bb2mj1twv. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bb2mj1twv.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (42 chars)
It ran in background. Let me check output.
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TaskOutput
args
{
  "task_id": "bb2mj1twv",
  "block": true,
  "timeout": 120000
}
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tool
stdout (294 chars)
<retrieval_status>success</retrieval_status>

<task_id>bb2mj1twv</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
ref vs py_recurrence max diff 0.00048828125
ref vs sol max diff 0.1015625
py_rec vs sol max diff 0.10107421875
</output>
assistant
reasoning (2901 chars)
ref vs py_recurrence diff = 0.0005 (good!). So my Aqk, w, u are correct, and the reference recurrence with my intermediates matches reference output. But py_rec vs sol diff = 0.101. So my kernel 2 recurrence is wrong. The bug is in kernel 2, not kernel 1. Now I need to debug kernel 2. The recurrence in kernel 2 differs from the Python reference. Let me carefully compare my kernel 2 to the reference recurrence: Reference: ```python v_i = u_i - w_i @ S o_i = qg_i @ S + Aqk_i @ v_i S = S * exp(g_i[-1]) + (k_i * exp(g_i[-1] - g_i)).T @ v_i ``` My kernel 2: ```python b_v = b_u b_v -= w_i @ S b_o = qg_i @ S + Aqk_i @ b_v store b_o # S update d = exp(g_last) k_up = k_i * exp(log(d)[None,:] - g_i) # = k_i * exp(g_last - g_i) S += k_up.T @ b_v S *= d[:,None] ``` Wait! In reference, the order is: 1. S = S * d 2. S = S + (k_i * exp(g_last - g_i)).T @ v_i But in my kernel, I do: 1. S += k_up.T @ v_i 2. S *= d[:,None] Mathematically, S_new = (S + k_up.T @ v_i) * d = S*d + k_up.T @ v_i * d. Reference: S_new = S*d + (k_i * exp(g_last - g_i)).T @ v_i. Since k_up = k_i * exp(g_last - g_i), then k_up.T @ v_i * d = (k_i * exp(g_last - g_i)).T @ v_i * d. But reference adds (k_i * exp(g_last - g_i)).T @ v_i WITHOUT the extra d factor! Wait, that's different! Let me re-check reference. Reference line 89-90: ```python S = S * rearrange(g_i[:, :, -1].exp(), "b h k -> b h k 1") S = S + rearrange((g_i[:, :, -1:] - g_i).exp() * k_i, "b h c k -> b h k c") @ v_i ``` So S_new = S * d + (k_i * exp(g_last - g_i)).T @ v_i. In my kernel, I do S += k_up.T @ v_i then S *= d. That gives S_new = (S + k_up.T @ v_i) * d = S*d + k_up.T @ v_i * d. These are different unless d=1! I have an extra d factor on the k_up.T @ v_i term. This is the bug! I incorrectly factored out d. The reference does NOT factor it out. I made this mistake earlier when deriving. Let me fix kernel 2: ```python # Correct: S = S * d[:,None] S += k_up.T @ v_i ``` where k_up = k_i * exp(g_last - g_i) (no extra d). Wait, but then k_up uses exp(g_last - g_i), and g_last = log(d). So k_up = k_i * exp(log(d) - g_i) = k_i * exp(-g_i) * d. That's what I have. But in the update, I should NOT multiply by d again. So: ```python k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0) # k_i * exp(g_last - g_i) S0 = S0 * d0[:, None] S0 += tl.dot(tl.trans(k_up0), b_v) ``` Wait, but this requires computing k_up with the d factor, then after scaling S by d, adding k_up.T @ v_i. But k_up already has d factor. Is that correct? Reference uses k_i * exp(g_last - g_i) = k_i * exp(g_last) * exp(-g_i) = (k_i * exp(-g_i)) * d. My k_up0 = b_k0 * exp(log(d0) - b_g0) = b_k0 * exp(-b_g0) * d0. Yes, matches. Then reference adds k_up.T @ v_i (with k_up having d factor). My kernel should do the same. Currently I do S0 += k_up.T @ v_i then S0 *= d0. The extra d0 is wrong. Fix: scale S by d first, then add k_up.T @ v_i. Let me fix.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (29 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -249,21 +249,13 @@
         )
         b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
         k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)
+        k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
+
+        # S = S * d + (k_i * exp(g_last - g_i)).T @ v_i
+        S0 = S0 * d0[:, None]
+        S1 = S1 * d1[:, None]
         S0 += tl.dot(tl.trans(k_up0), b_v)
-
-        p_k1 = tl.make_block_ptr(
-            k_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_k1 = tl.load(p_k1, boundary_check=(0, 1)).to(tl.float32)
-        p_g1 = tl.make_block_ptr(
-            g_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_g1 = tl.load(p_g1, boundary_check=(0, 1)).to(tl.float32)
-        k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
         S1 += tl.dot(tl.trans(k_up1), b_v)
-
-        S0 *= d0[:, None]
-        S1 *= d1[:, None]
 
 
 class Model(nn.Module):
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<609 chars \u2014 see diff>",
  "new_string": "<318 chars \u2014 see diff>"
}
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tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (192 chars)
Wait, I removed the p_k1/g1 loads. I need to keep them. Let me re-read and fix properly. I want: 1. Compute k_up0, k_up1. 2. Scale S0,S1 by d. 3. Add k_up.T @ v_i. Let me write it correctly.
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assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 80,
  "offset": 230
}
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tool
stdout (3164 chars) — TRUNCATED
230	        b_o += tl.dot(b_Aqk, b_v)
231	
232	        p_o = tl.make_block_ptr(
233	            o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
234	        )
235	        tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))
236	
237	        # S += k_update.T @ v_i where k_update = k * exp(g_last - g)
238	        p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
239	        d0 = tl.exp(tl.load(p_glast0))
240	        p_glast1 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)
241	        d1 = tl.exp(tl.load(p_glast1))
242	
243	        p_k0 = tl.make_block_ptr(
244	            k_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
245	        )
246	        b_k0 = tl.load(p_k0, boundary_check=(0, 1)).to(tl.float32)
247	        p_g0 = tl.make_block_ptr(
248	            g_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
249	        )
250	        b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
251	        k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)
252	        k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
253	
254	        # S = S * d + (k_i * exp(g_last - g_i)).T @ v_i
255	        S0 = S0 * d0[:, None]
256	        S1 = S1 * d1[:, None]
257	        S0 += tl.dot(tl.trans(k_up0), b_v)
258	        S1 += tl.dot(tl.trans(k_up1), b_v)
259	
260	
261	class Model(nn.Module):
262	    def __init__(self, B: int, T: int, H: int, K: int, V: int, chunk_size: int = 64):
263	        super().__init__()
264	        self

... (1664 more chars)
assistant
reasoning (165 chars)
I see the code now has k_up1 using b_k1 and b_g1 which are not defined (I removed their loads). I need to add them back. Let me fix the S update section completely.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (30 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -234,7 +234,7 @@
         )
         tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))
 
-        # S += k_update.T @ v_i where k_update = k * exp(g_last - g)
+        # S = S * d + (k_i * exp(g_last - g_i)).T @ v_i
         p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
         d0 = tl.exp(tl.load(p_glast0))
         p_glast1 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)
@@ -249,9 +249,17 @@
         )
         b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
         k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)
+
+        p_k1 = tl.make_block_ptr(
+            k_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
+        )
+        b_k1 = tl.load(p_k1, boundary_check=(0, 1)).to(tl.float32)
+        p_g1 = tl.make_block_ptr(
+            g_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
+        )
+        b_g1 = tl.load(p_g1, boundary_check=(0, 1)).to(tl.float32)
         k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
 
-        # S = S * d + (k_i * exp(g_last - g_i)).T @ v_i
         S0 = S0 * d0[:, None]
         S1 = S1 * d1[:, None]
         S0 += tl.dot(tl.trans(k_up0), b_v)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<993 chars \u2014 see diff>",
  "new_string": "<1279 chars \u2014 see diff>"
}
in 177154out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (17 chars)
Now run check.py.
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assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -60",
  "description": "Run check after S update fix"
}
in 178250out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (4 chars)
PASS
assistant
reasoning (114 chars)
PASS! Great. Now I need to benchmark and optimize for performance. Let me update tasks and then run benchmark.py.
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assistant
TaskUpdate
args
{
  "status": "completed",
  "taskId": "1"
}
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tool
stdout (22 chars)
Updated task #1 status
assistant
reasoning (39 chars)
Wait, I should update task 2 and 3 too.
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assistant
TaskUpdate
args
{
  "status": "completed",
  "taskId": "2"
}
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tool
stdout (22 chars)
Updated task #2 status
assistant
reasoning (126 chars)
Task 2 was "Implement solution.py custom kernel" - completed. Task 3 is "Verify correctness with check.py" - completed (PASS).
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assistant
TaskUpdate
args
{
  "status": "completed",
  "taskId": "3"
}
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tool
stdout (22 chars)
Updated task #3 status
assistant
reasoning (60 chars)
Now task 4: benchmark and optimize. Let me run benchmark.py.
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assistant
TaskUpdate
args
{
  "status": "in_progress",
  "taskId": "4"
}
in 178655out 0cache_r 0cache_w 0
tool
stdout (22 chars)
Updated task #4 status
assistant
reasoning (17 chars)
Run benchmark.py.
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assistant
Bash
args
{
  "command": "uv run python benchmark.py 2>&1 | tail -80",
  "description": "Run benchmark"
}
in 178715out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (1301 chars)
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-13T23:41:37.951297+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-13T23:41:38.223766+00:00 elapsed_s=0.272 ms=0.438480
shape=0 variant=solution tflops=4.898 gbps=57.468 ms=0.438
shape=0 solution_peak_fraction=0.0245
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-13T23:41:38.388939+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-13T23:41:38.419864+00:00 elapsed_s=0.031 ms=0.916912
shape=1 variant=solution tflops=4.684 gbps=54.964 ms=0.917
shape=1 solution_peak_fraction=0.0234
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-13T23:41:38.585645+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-13T23:41:38.622102+00:00 elapsed_s=0.036 ms=1.121600
shape=2 variant=solution tflops=3.829 gbps=44.933 ms=1.122
shape=2 solution_peak_fraction=0.0191
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-13T23:41:38.662838+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-13T23:41:38.677009+00:00 elapsed_s=0.014 ms=0.383632
shape=3 variant=solution tflops=2.799 gbps=32.842 ms=0.384
shape=3 solution_peak_fraction=0.0140
peak_fraction: 0.0198
RESULT: LOW
assistant
reasoning (20055 chars)
peak_fraction 0.0198, below 0.1 threshold. Need to optimize. The current kernel is slow because: 1. Kernel 1 has a column loop (64 iterations) with global memory loads per column. Very memory heavy. 2. Full fp32 everywhere uses 2x memory bandwidth. 3. Kernel 2 is sequential per (B,H,V tile) block. 4. num_stages=1 might not be optimal. To improve performance, I should: 1. Compute Aqk/M more efficiently, perhaps using the factorization (qg @ k_decay.T) but stably, or using shared memory to cache q,k,g. 2. Use bf16 for most computations to save bandwidth. 3. Optimize block sizes and warps. 4. Consider fusing more or using persistent kernels. Actually, the main bottleneck is likely kernel 1's column loop reading q,k,g 64 times. We can cache q,k,g in shared memory. With q,k as bf16 (16KB each) and g as fp32 (32KB), total 64KB shared. Fits. Then for each column, we only load k_i,g_i (small). This should speed up kernel 1 significantly. But Triton shared memory with dynamic row extraction is tricky. We can use the `tl.where` trick to extract rows, or we can keep q,k,g in shared memory and use a loop over K tiles. Actually, a better approach: compute Aqk and M using matrix multiply with the factorization but normalize to avoid instability. For each chunk, we can compute: - qg = q * exp(g - g_anchor) and kdecay = k * exp(g_anchor - g) for some anchor (e.g., midpoint of chunk). Then Aqk = qg @ kdecay.T, and Akk = kg @ kdecay.T. This avoids extreme values as long as g stays within range relative to anchor. But g can drift across the chunk. With anchor = g at token 31 (midpoint), exp(g - g_anchor) and exp(g_anchor - g) are bounded by exp(range/2). For g range ~7, exp(3.5) ~33. Still moderate. Actually, reference uses per-column subtraction. The most stable. But we can use per-row or block-wise normalization. Alternatively, we can compute Aqk/M with one matrix multiply per chunk if we normalize g by subtracting a per-token base. But that's what per-column subtraction does. Hmm. Let me think about the shared memory approach. In kernel 1, load q,k,g for the chunk into shared memory. Then for each column i, extract g_i and k_i from shared memory. Compute diff, q_exp, k_exp, accumulate. With q,k in shared memory as bf16 and g as fp32, the extraction uses `tl.where` which creates register copies. But once in registers, compute is fast. Shared memory size: q_s (BT*K*2) + k_s (BT*K*2) + g_s (BT*K*4) = 16+16+32 = 64KB. Plus Aqk/M in registers. Fits in 101KB. Wait, can I declare shared memory in Triton? I can use `tl.zeros([BT*K], dtype=...)` and the compiler places it in shared memory. Then use `tl.load`/`tl.store` with the pointer. But dynamic indexing... Actually, I can declare them as local tensors and rely on the compiler. But the `tl.where` extraction might create register copies anyway. Let me try a different strategy: compute Aqk and M in Python using efficient batched operations, and only do solve/w/u in Triton. Python can compute Aqk/M efficiently with einsum/bmm. But the problem wants a custom kernel. Still, the solve is the core kernel. Aqk computation can be in Python if it's fast. Actually, let's benchmark the current bottleneck. Kernel 1 takes most time due to 64 column loops. Let me see if I can speed it up. Alternative: use the factorization but with chunk-midpoint normalization. Compute: - g_mid = g[BT//2, :] (per chunk) - qg = q * exp(g - g_mid) # bounded - kdecay = k * exp(g_mid - g) # bounded - Aqk = tril(qg @ kdecay.T) * scale - kg = k * exp(g - g_mid) # for M = kg @ kdecay.T? Wait. M[c,i] = sum k[c]*k[i]*exp(g[c]-g[i]). If we set qg_norm[c] = q[c]*exp(g[c]-g_mid), kdecay_norm[i] = k[i]*exp(g_mid-g[i]). Then Aqk[c,i] = qg_norm[c] @ kdecay_norm[i]. Correct. M[c,i] = kg_norm[c] @ kdecay_norm[i] where kg_norm[c] = k[c]*exp(g[c]-g_mid). Correct. So we can compute: - g_mid = g[BT//2, :] - qg_norm = q * exp(g - g_mid) - kg_norm = k * exp(g - g_mid) - kdecay_norm = k * exp(g_mid - g) - Aqk = tril(qg_norm @ kdecay_norm.T) * scale - M = tril_strict(kg_norm @ kdecay_norm.T) This is stable because g - g_mid is bounded by roughly half the chunk range. For w = A @ kg, we need kg = k * exp(g) (reference). But if we compute A stably using normalized kg_norm and kdecay_norm, then w = A @ kg can use kg computed as k * exp(g). The A matrix is the same (it depends only on ratios). But kg for w is absolute. With g potentially very negative, kg can be tiny; but A's diagonal is beta, so w is bounded by beta * max(kg). For g very negative, kg tiny, w tiny. Fine. For g very positive, kg can be large. But g range within chunk is bounded, so max kg ~ k * exp(g_mid + range/2). With range ~7, exp(g_mid + 3.5). g_mid itself could be very negative if cumulative drift. Over many chunks, g_mid drifts. For long T, g_cum can be very negative (mean -0.05*T). At T=4096, mean g_cum ~ -200. exp(-200) underflows to 0. kg = 0. w = 0. Then output from qg @ S term is 0, and Aqk @ v_i term dominates. Wait, but reference uses exp(g) directly. If g_cum is -200, exp(g) underflows to 0 in fp32? No, fp32 min normal ~1e-38, exp(-200) ~1e-87, underflows to 0. So reference's kg would be 0. So w = 0. That's expected. But then how does the recurrence work for long context? The inter-chunk decay exp(g_last) would be exp(-200) ~ 0, so S decays to 0. Output is mostly from intra-chunk Aqk @ v_i. That matches the reference behavior for very negative g. So using kg = k * exp(g) for w is correct, even if it underflows. For the normalization approach, qg_norm = q * exp(g - g_mid) avoids underflow/overflow. kdecay_norm = k * exp(g_mid - g) also avoids. Aqk and M computed stably. So I can compute Aqk and M with one matrix multiply per chunk (after computing normalized qg, kg, kdecay). Much faster than column loop. Let me implement this: 1. In Python: compute g_cum. 2. Compute g_mid per chunk: g_cum.view(B,NT,BT,H,K)[:,:,BT//2,:,:]. 3. Compute qg_norm = q_scaled * exp(g_cum - g_mid_expanded). 4. Compute kg_norm = k * exp(g_cum - g_mid_expanded). 5. Compute kdecay_norm = k * exp(g_mid_expanded - g_cum). 6. Compute kg = k * exp(g_cum) for w (might underflow, but OK). 7. Kernel 1: compute Aqk = tril(qg_norm @ kdecay_norm.T) * scale, M = tril_strict(kg_norm @ kdecay_norm.T). Solve for A. Compute w = A @ kg, u = A @ v. This reduces kernel 1 from 64 column loops to a few matrix multiplies. Much faster. But we need to pass qg_norm, kg_norm, kdecay_norm, kg to kernel 1. That's 4 extra tensors. Or compute some in kernel. Actually, we can compute all in Python and pass. But memory: qg_norm, kg_norm, kdecay_norm, kg each (B,T,H,K) fp32 = 33MB. Total 132MB. Plus w,u outputs. Too much. We can keep them as bf16 to save memory. For Aqk/M computation, bf16 might be OK with normalization (values bounded). Then cast to fp32 inside kernel for accumulation. Actually, with normalization, qg_norm, kg_norm, kdecay_norm are bounded (~k * exp(range/2) ~ 0.1 * 33 = 3.3). bf16 handles this fine. So compute qg_norm, kg_norm, kdecay_norm in bf16. Pass to kernel. Kernel loads as fp32, computes Aqk/M in fp32, solves, computes w,u. For kg = k * exp(g_cum), this can underflow for long context. If we store as bf16, underflow to 0. Same as fp32 for very small. For moderate, bf16. OK. Memory for bf16 intermediates: 4 * 16KB per block * num_blocks? Per tensor size: B*T*H*K*2 = 16.7MB for B=2,T=2048,H=8,K=128. 4 tensors = 67MB. Plus w,u fp32 = 33MB. Total ~100MB. Acceptable. But we can also compute qg_norm/kg_norm/kdecay_norm inside the kernel to save memory. Let me think. Actually, we can compute Aqk and M in Python efficiently with einsum/bmm: ```python qg_norm = q_scaled * (g_cum - g_mid).exp() # (B,T,H,K) kg_norm = k * (g_cum - g_mid).exp() kdecay_norm = k * (g_mid - g_cum).exp() # Reshape to (B*H*N, C, K) qg_3d = qg_norm.transpose(1,2).reshape(B*H*NT, BT, K) # wait need correct reshape # Actually: qg_norm.view(B, NT, BT, H, K).permute(0,3,1,2,4).reshape(B*H*NT, BT, K) kdecay_3d = kdecay_norm.view(B, NT, BT, H, K).permute(0,3,1,2,4).reshape(B*H*NT, BT, K) Aqk = torch.bmm(qg_3d, kdecay_3d.transpose(1,2)) # (BHN, BT, BT) Aqk = torch.tril(Aqk) * scale ``` Similarly for M. This uses cuBLAS and is fast. Then kernel 1 only needs to solve for A and compute w,u. But solving is per chunk; we can do it in Triton. Actually, we can also compute Aqk/M in the Triton kernel using the normalized factorization and one matrix multiply. That keeps everything in the kernel. Let me rewrite kernel 1 to: 1. Load q,k,v,g,beta for chunk. 2. Compute g_mid = g[BT//2, :]. 3. Compute qg_norm, kg_norm, kdecay_norm in registers or shared memory. 4. Compute Aqk = tril(qg_norm @ kdecay_norm.T) * scale. 5. Compute M = tril_strict(kg_norm @ kdecay_norm.T). 6. Solve for A. 7. Compute kg = k * exp(g) (or use kg_norm? No, w needs kg = k*exp(g)). 8. Compute w = A @ kg, u = A @ v. For step 3, we need g_mid per K channel. g is (BT,K). g_mid is (K,). We can load g[BT//2, :] using scalar pointer. Then compute norms. But we also need kg = k * exp(g) for w. This can be computed in the same loop over K tiles. Algorithm per K tile: - Load q_tile, k_tile, g_tile. - Compute g_mid_tile = g[BT//2, off_k:off_k+BK]. - Compute qg_norm_tile = q_tile * exp(g_tile - g_mid_tile[None,:]). - Compute kg_norm_tile = k_tile * exp(g_tile - g_mid_tile[None,:]). - Compute kdecay_norm_tile = k_tile * exp(g_mid_tile[None,:] - g_tile). - Accumulate Aqk += qg_norm_tile @ kdecay_norm_tile.T. - Accumulate M += kg_norm_tile @ kdecay_norm_tile.T. - Store/compute kg_tile = k_tile * exp(g_tile) for later w. This is one loop over K tiles, with 2-3 matrix multiplies per tile. Much faster than 64 column loops. Then solve A in registers. Then compute w = A @ kg and u = A @ v in another loop over K/V tiles. This should be much faster. But we need to store kg for the w computation. We can either recompute kg in the w loop (load k and g again), or store it in shared memory/global memory. Recomputing means another k,g read. Acceptable. Actually, we can compute and store kg to global memory in the first loop, then read back for w. Or store in shared memory if space. Let me design kernel 1 with shared memory for q,k,g and maybe kg. Shared memory: - q_s, k_s, g_s: 16+16+32 = 64KB. - workspace for kg: 16KB if needed. Total 80KB. Fits. But dynamic indexing of shared memory for kg is tricky. We can store kg in global memory (bf16 or fp32) and read back. Actually, let me first try a simpler version: compute Aqk/M with normalized factorization in Python (using PyTorch bmm), then pass Aqk and M to a Triton kernel that solves for A and computes w,u. This splits the work but uses PyTorch for the heavy GEMM part. Wait, but the problem wants a custom kernel. Having PyTorch do Aqk/M is fine as long as the solve is custom. Actually, for maximum performance, we want everything fused. But let me first get a working fast version. Option A: Python computes Aqk/M, Triton kernel solves and computes w,u. Option B: Triton kernel computes everything fused. Option A is easier and likely fast enough because bmm is optimized. Let me try it. Actually, I realize that the current kernel 1 is the bottleneck due to 64 column loops. Replacing it with Python bmm should dramatically speed up. Let me implement Option A. Python code: ```python g_4d = g.view(B, NT, BT, H, K) g_cum = g_4d.cumsum(dim=2).view(B, T, H, K) g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT//2, :, :] # (B,H,K) g_mid_expanded = g_mid[:, None, :, :].expand(-1, T, -1, -1) # wrong shape # Actually g_mid needs to be expanded to (B,T,H,K) g_mid_expanded = g_mid.unsqueeze(2).expand(-1, -1, T, -1) # (B,H,T,K) g_mid_expanded = g_mid_expanded.permute(0,2,1,3) # (B,T,H,K) ``` Wait, g_mid shape (B,H,K). To expand to (B,T,H,K), use `g_mid[:, None, :, :].expand(-1, T, -1, -1)` which gives (B,T,H,K). Yes. Then: ```python q_scaled = q * self.scale qg_norm = q_scaled * (g_cum - g_mid_expanded).exp() kg_norm = k * (g_cum - g_mid_expanded).exp() kdecay_norm = k * (g_mid_expanded - g_cum).exp() ``` Then reshape for bmm: ```python def to_3d(x): return x.view(B, NT, BT, H, K).permute(0, 3, 1, 2, 4).reshape(B*H*NT, BT, K) Aqk = torch.bmm(to_3d(qg_norm), to_3d(kdecay_norm).transpose(1, 2)) Aqk = torch.tril(Aqk) * scale Aqk = Aqk.view(B, H, NT, BT, BT).permute(0, 2, 1, 3, 4).contiguous().view(B, T, H, BT) M = torch.bmm(to_3d(kg_norm), to_3d(kdecay_norm).transpose(1, 2)) mask = torch.tril(torch.ones(BT, BT, device=device), diagonal=-1) M = M * mask M = M.view(B, H, NT, BT, BT).permute(0, 2, 1, 3, 4).contiguous().view(B, T, H, BT) ``` Then Triton kernel solves (I - (-M*beta_row)) and computes A, then w = A @ kg, u = A @ v. But wait, the solve needs M and beta per chunk. We can pass M to kernel 1 and have it solve. Actually, kernel 1 can receive Aqk and M, solve, compute w,u. This is much simpler/faster than computing Aqk/M in kernel. Let me rewrite kernel 1 to take Aqk and M as inputs, and output w,u. But then where is the custom kernel? It's in the solve and w/u computation. That's still substantial. Actually, the solve is O(BT^3) per chunk, small. w,u are GEMMs. If we compute w,u in Python with bmm (using solved A), it's fast. Then kernel 1 only does the solve. Solve-only kernel: input M, beta; output A. Then Python computes w = A @ kg, u = A @ v. But the solve for A is per chunk and requires the forward substitution. In Python, it would be a loop over chunks and rows, slow. Triton kernel for solve is good. Actually, we can compute A in Python too using torch's `torch.linalg.solve_triangular` or matrix inversion. For each chunk, A0 = -tril_strict(M) * beta_row. We want A = (I - A0)^{-1} * diag(beta_col). We can compute (I - A0)^{-1} using `torch.linalg.solve` for each chunk. For B*H*N = 256 chunks, each solve is (64,64). PyTorch can batch them: A0 shape (B,H,N,BT,BT), solve (I - A0) @ X = I for each chunk. Actually, we want inv(I - A0) then multiply by diag(beta). ```python I_minus_A0 = torch.eye(BT, device=device) - A0.view(B*H*N, BT, BT) # A0 already includes -sign? Wait. # A0 = -tril_strict(M) * beta_row # We want inv(I - A0) * diag(beta_col) I_minus_A0 = I_minus_A0 # I - A0 inv = torch.linalg.solve(I_minus_A0, torch.eye(BT, device=device)) # (BHN,BT,BT) A = inv * beta_col[:, None, :] # beta_col shape (BHN,BT) ``` Then w = A @ kg_3d, u = A @ v_3d. But torch.linalg.solve for 256 chunks of 64x64 might be fast. Let me test. Actually, the forward substitution is O(BT^3) = 262K ops per chunk. 256 chunks = 67M ops. In PyTorch, batched solve uses LU or specialized triangular solve. Should be fast. But the problem wants a custom kernel. Using torch.linalg.solve is not a custom kernel. The check.py framework label detection would classify it as "unknown" or "ptx" depending on code. Actually, the label detection looks for patterns. If I don't have Triton/CUDA code, it might be "unknown". But the user explicitly says "This needs to be a real custom kernel". So I should keep the Triton kernel for the solve and w/u. Let me try Option A: Python computes Aqk/M, Triton kernel solves and computes w,u. This keeps a Triton kernel and should be much faster. Actually, even better: keep everything in one Triton kernel but compute Aqk/M with the normalized factorization (single K-tile loop). This is the most fused and likely fastest. Let me implement the fused Triton kernel with normalized factorization. Kernel 1 fused: - Load q,k,v,g,beta for chunk. - Compute g_mid = g[BT//2, :]. - Accumulate Aqk and M in registers using qg_norm/kg_norm/kdecay_norm. - Solve for A. - Compute kg = k * exp(g) for w. - Compute w = A @ kg, u = A @ v. - Store Aqk, w, u. For Aqk and M accumulation: ```python Aqk_acc = zeros(BT,BT) M_acc = zeros(BT,BT) for i_k in range(K//BK): load q,k,g tiles g_mid_tile = g[BT//2, off_k:off_k+BK] # scalar load diff = g_tile - g_mid_tile[None,:] qg_norm = q_tile * exp(diff) kg_norm = k_tile * exp(diff) kdecay_norm = k_tile * exp(-diff) Aqk_acc += tl.dot(qg_norm, tl.trans(kdecay_norm)) M_acc += tl.dot(kg_norm, tl.trans(kdecay_norm)) ``` Then apply masks and beta. For g_mid load: ```python g_mid = tl.load(g_ptr + (i_b*T*H + i_h)*K + BT//2*H*K + tl.arange(0,K)) # shape (K,) ``` But K=128, and tl.arange must be power of 2. 128 is power of 2. Good. But if K not power of 2, we'd need tiling. Our K=128. Actually, in kernel, K is constexpr 128 (or whatever passed). But we can load K with `tl.arange(0,K)` if K is power of 2. For generality, tile K. But our shapes have K=128, so we can load full K. Then in the loop, `g_mid_tile = g_mid[off_k:off_k+BK]`? Slicing doesn't work. We can load g_mid per K tile in the loop: ```python for i_k in range(K//BK): off_k = i_k * BK g_mid_tile = tl.load(g_mid_ptr + off_k + tl.arange(0,BK)) ... ``` Where g_mid_ptr is the pointer to g_mid. We can compute g_mid in Python and pass as a tensor (B,H,K), or compute it in the kernel. Computing g_mid in kernel requires a scalar load of g at BT//2. Then we can use it. But storing it to shared memory or registers for K tiles is the issue. Since K=128 and BK=64, we have 2 tiles. We can load the full g_mid as (K,) and then use `tl.sum(tl.where(tl.arange(0,K)[None,:] == (off_k + tl.arange(0,BK))[None,:], g_mid[None,:], 0), axis=1)`? That's messy. Simpler: in the loop, load g_mid tile from a precomputed g_mid tensor. Let me compute g_mid in Python: `g_mid = g_cum.view(B,NT,BT,H,K)[:,:,BT//2,:,:].permute(0,2,1,3).contiguous()` shape (B,H,NT,K). Or just (B,H,K) if same for all chunks? No, g_mid depends on chunk. Actually, g_mid per chunk. We need to pass g_mid of shape (B,H,NT,K) to kernel. Memory: B*H*NT*K*4 = 2*8*16*128*4 = 256KB. Tiny. Then in kernel, load g_mid tile per K tile: ```python p_gmid = tl.make_block_ptr(gmid_ptr, (NT, K), (H*K, 1), (i_n, off_k), (1, BK), (1,0)) g_mid_tile = tl.load(p_gmid, boundary_check=(0,1)).to(tl.float32) ``` Then compute diff = g_tile - g_mid_tile. Good. But wait, we also need g_tile = g_cum (full g). We load that. And k_tile, q_tile. So kernel 1 inputs: q, k, v, g, g_mid, beta. Outputs: Aqk, w, u. q,k can be bf16 (or fp32). For accuracy, let's use fp32 for now. We can later try bf16. Actually, with normalization, q,k can be bf16 without much accuracy loss. But let's keep fp32 for correctness first, then optimize. Memory for g_mid is tiny. Good. But wait, if I compute Aqk/M in Triton with normalized factorization, I avoid the 64 column loops. This should be much faster. Also, I can keep w,u computation in the same kernel. Let me rewrite kernel 1. Also, I should consider whether to upcast inputs in Python or keep bf16. For performance, bf16 is better. But for correctness, fp32. Since large_qkv passes with fp32, let's first implement with fp32 and benchmark. If performance is still low, try bf16. Actually, I realize that with the normalized factorization, bf16 might be accurate enough because values are bounded. Let me try bf16 for q,k in kernel 1 and see if check passes. If yes, we save memory/bandwidth. But first, let me write the fused kernel with fp32. Wait, I also need to think about kernel 2 performance. Kernel 2 is sequential per (B,H,V tile). For B=2,H=8,V=128,BV=32, grid = 4*16 = 64 blocks. Each block loops over NT chunks. For T=4096, NT=64. Each chunk does several GEMMs. This might be OK but could be memory-bound. Optimizing kernel 2: use larger BV to reduce V tiles and increase parallelism. With BV=64, grid = 2*16 = 32 blocks. With BV=128, grid = 1*16 = 16 blocks. But S is (K,BV) = (128,128) fp32 = 64KB per block. Register pressure. BV=64 gives S = 32KB, manageable. BV=32 gives 16KB. Actually, with BV=64 and BK=64, S0 and S1 are 64x64 fp32 = 16KB each. Total 32KB. Good. For kernel 1, BK=64 is good. Let me rewrite kernel 1 with normalized factorization. Actually, let me step back. The current kernel 1 with 64 column loops is slow. But kernel 2 might also be slow. Let me first implement the faster kernel 1 and see overall improvement. Let me write the new solution.py.
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Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 200
}
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stdout (6755 chars) — TRUNCATED
1	"""Kimi Delta Attention (KDA) forward (chunk form) via Triton.
2	
3	Matches reference.py's semantics: inputs are bf16 (g fp32), chunk_size=64,
4	no initial/final state, returns bf16 o.
5	"""
6	from __future__ import annotations
7	
8	import torch
9	import torch.nn as nn
10	import triton
11	import triton.language as tl
12	
13	OP_TYPE = "linear_attention"
14	SUPPORTED_PRECISIONS = ["bf16"]
15	HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
16	
17	
18	@triton.jit
19	def _kda_intra_kernel(
20	    q_ptr,
21	    k_ptr,
22	    v_ptr,
23	    g_ptr,
24	    kg_ptr,
25	    beta_ptr,
26	    Aqk_ptr,
27	    w_ptr,
28	    u_ptr,
29	    scale,
30	    B: tl.constexpr,
31	    T: tl.constexpr,
32	    H: tl.constexpr,
33	    K: tl.constexpr,
34	    V: tl.constexpr,
35	    BT: tl.constexpr,
36	    BK: tl.constexpr,
37	    BV: tl.constexpr,
38	):
39	    i_n = tl.program_id(0)
40	    i_bh = tl.program_id(1)
41	    i_b = i_bh // H
42	    i_h = i_bh % H
43	
44	    # adjust base pointers for batch/head
45	    q_ptr += (i_b * T * H + i_h) * K
46	    k_ptr += (i_b * T * H + i_h) * K
47	    g_ptr += (i_b * T * H + i_h) * K
48	    kg_ptr += (i_b * T * H + i_h) * K
49	    beta_ptr += i_b * T * H + i_h
50	    v_ptr += (i_b * T * H + i_h) * V
51	    Aqk_ptr += (i_b * T * H + i_h) * BT
52	    w_ptr += (i_b * T * H + i_h) * K
53	    u_ptr += (i_b * T * H + i_h) * V
54	
55	    t0 = i_n * BT
56	    r = tl.arange(0, BT)
57	    c = tl.arange(0, BT)
58	    mask_strict = r[:, None] > c[None, :]
59	    mask_lo

... (5255 more chars)
assistant
reasoning (401 chars)
Let me rewrite kernel 1 with normalized factorization. I'll keep the rest of the file mostly the same. New kernel 1: - Inputs: q, k, v, g, g_mid, beta, Aqk, w, u - Compute Aqk and M with normalized qg_norm/kg_norm/kdecay_norm. - Solve for A. - Compute kg = k * exp(g) for w. - Compute w = A @ kg, u = A @ v. I'll use q,k in fp32 (passed from Python), g in fp32, g_mid in fp32. Let me edit kernel 1.
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Edit /workspace/problems/02_kda_cutlass/solution.py
diff (122 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -21,7 +21,7 @@
     k_ptr,
     v_ptr,
     g_ptr,
-    kg_ptr,
+    gmid_ptr,
     beta_ptr,
     Aqk_ptr,
     w_ptr,
@@ -45,7 +45,7 @@
     q_ptr += (i_b * T * H + i_h) * K
     k_ptr += (i_b * T * H + i_h) * K
     g_ptr += (i_b * T * H + i_h) * K
-    kg_ptr += (i_b * T * H + i_h) * K
+    gmid_ptr += (i_b * H + i_h) * K
     beta_ptr += i_b * T * H + i_h
     v_ptr += (i_b * T * H + i_h) * V
     Aqk_ptr += (i_b * T * H + i_h) * BT
@@ -58,49 +58,37 @@
     mask_strict = r[:, None] > c[None, :]
     mask_lower = r[:, None] >= c[None, :]
 
-    # Compute Aqk and M = kg @ kdecay.T using the stable formulation
-    # Aqk[c,i] = sum_k q[c,k]*k[i,k]*exp(g[c,k]-g[i,k])
-    # M[c,i]  = sum_k k[c,k]*k[i,k]*exp(g[c,k]-g[i,k])
+    # Compute Aqk and M via chunk-midpoint normalization.
+    # gmid = g[BT//2, :] keeps exp(g - gmid) and exp(gmid - g) bounded.
     Aqk = tl.zeros([BT, BT], dtype=tl.float32)
     M = tl.zeros([BT, BT], dtype=tl.float32)
 
-    for i in range(BT):
-        col_Aqk = tl.zeros([BT, 1], dtype=tl.float32)
-        col_M = tl.zeros([BT, 1], dtype=tl.float32)
-        for i_k in range(tl.cdiv(K, BK)):
-            off_k = i_k * BK
-            # column i's g and k (scalar row load)
-            g_i = tl.load(
-                g_ptr + (t0 + i) * H * K + off_k + tl.arange(0, BK)
-            ).to(tl.float32)[None, :]
-            k_i = tl.load(
-                k_ptr + (t0 + i) * H * K + off_k + tl.arange(0, BK)
-            ).to(tl.float32)[None, :]
-
-            # full chunk tiles
-            p_q = tl.make_block_ptr(
-                q_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-            )
-            p_k = tl.make_block_ptr(
-                k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-            )
-            p_g = tl.make_block_ptr(
-                g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-            )
-            b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32)
-            b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
-            b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
-
-            diff = b_g - g_i
-            q_exp = b_q * tl.exp(diff)
-            k_exp = b_k * tl.exp(diff)
-            # col_Aqk += tl.dot(q_exp, tl.trans(k_i))
-            # col_M += tl.dot(k_exp, tl.trans(k_i))
-            col_Aqk += tl.sum(q_exp * k_i, axis=1)[:, None]
-            col_M += tl.sum(k_exp * k_i, axis=1)[:, None]
-
-        Aqk = tl.where(c[None, :] == i, col_Aqk, Aqk)
-        M = tl.where(c[None, :] == i, col_M, M)
+    for i_k in range(tl.cdiv(K, BK)):
+        off_k = i_k * BK
+        p_q = tl.make_block_ptr(
+            q_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        p_k = tl.make_block_ptr(
+            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        p_g = tl.make_block_ptr(
+            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32)
+        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+        b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
+
+        p_gmid = tl.make_block_ptr(
+            gmid_ptr, (H, K), (K, 1), (i_h, off_k), (1, BK), (1, 0)
+        )
+        b_gmid = tl.load(p_gmid, boundary_check=(0, 1)).to(tl.float32)
+
+        diff = b_g - b_gmid
+        qg_norm = b_q * tl.exp(diff)
+        kg_norm = b_k * tl.exp(diff)
+        kdecay_norm = b_k * tl.exp(-diff)
+        Aqk += tl.dot(qg_norm, tl.trans(kdecay_norm))
+        M += tl.dot(kg_norm, tl.trans(kdecay_norm))
 
     p_beta = tl.make_block_ptr(
         beta_ptr, (T,), (H,), (t0,), (BT,), (0,)
@@ -125,13 +113,18 @@
     )
     tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
 
-    # compute w = A @ kg and u = A @ v in fp32 for better accuracy
+    # w = A @ kg, kg = k * exp(g)
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
-        p_kg = tl.make_block_ptr(
-            kg_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
-        )
-        b_kg = tl.load(p_kg, boundary_check=(0, 1)).to(tl.float32)
+        p_k = tl.make_block_ptr(
+            k_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        p_g = tl.make_block_ptr(
+            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
+        )
+        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+        b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
+        b_kg = b_k * tl.exp(b_g)
         b_w = tl.dot(A, b_kg)
         p_w = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<4541 chars \u2014 see diff>",
  "new_string": "<4131 chars \u2014 see diff>"
}
in 187895out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (408 chars)
Now update Model.forward to compute g_mid and pass it, and remove kg computation (kernel computes kg internally). Actually, kernel 1 now takes gmid_ptr instead of kg_ptr. So I need to update the launch. Also, I removed kg_ptr from kernel 1, so I don't need to compute kg in Python for kernel 1. But kernel 2 still needs qg (for qg @ S) and k,g (for S update). qg computed in Python. Let me update forward.
in 191610out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (22 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -294,7 +294,9 @@
 
         q_scaled = q * self.scale
         qg = q_scaled * g_cum.exp()
-        kg = k * g_cum.exp()
+
+        # chunk-midpoint g for stable Aqk/M computation
+        g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT // 2, :, :].permute(0, 2, 1, 3).contiguous()
 
         glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()
 
@@ -304,7 +306,7 @@
 
         grid1 = (NT, B * H)
         _kda_intra_kernel[grid1](
-            q_scaled, k, v, g_cum, kg, beta, Aqk, w, u,
+            q_scaled, k, v, g_cum, g_mid, beta, Aqk, w, u,
             1.0,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
             BK=64,
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<783 chars \u2014 see diff>",
  "new_string": "<912 chars \u2014 see diff>"
}
in 191610out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (44 chars)
Now check correctness first, then benchmark.
in 192462out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check correctness of optimized kernel"
}
in 192462out 0cache_r 0cache_w 0
subagent · 23 events · 7 tools
tool
stdout (283 chars)
Command running in background with ID: b4468vj7v. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/b4468vj7v.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (35 chars)
Running in background. Let me wait.
in 192615out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "b4468vj7v",
  "block": true,
  "timeout": 300000
}
in 192615out 0cache_r 0cache_w 0
tool
stdout (188 chars)
<retrieval_status>success</retrieval_status>

<task_id>b4468vj7v</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
PASS
</output>
assistant
reasoning (20 chars)
PASS. Now benchmark.
in 192890out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python benchmark.py 2>&1 | tail -40",
  "description": "Benchmark optimized kernel"
}
in 192890out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (283 chars)
Command running in background with ID: bprwi7mx2. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bprwi7mx2.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (19 chars)
Wait for benchmark.
in 193038out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "bprwi7mx2",
  "block": true,
  "timeout": 300000
}
in 193038out 0cache_r 0cache_w 0
tool
stdout (1485 chars)
<retrieval_status>success</retrieval_status>

<task_id>bprwi7mx2</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-13T23:50:57.092396+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-13T23:50:57.360469+00:00 elapsed_s=0.268 ms=0.282624
shape=0 variant=solution tflops=7.598 gbps=89.159 ms=0.283
shape=0 solution_peak_fraction=0.0380
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-13T23:50:57.524196+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-13T23:50:57.543669+00:00 elapsed_s=0.019 ms=0.517424
shape=1 variant=solution tflops=8.301 gbps=97.400 ms=0.517
shape=1 solution_peak_fraction=0.0415
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-13T23:50:57.706368+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-13T23:50:57.730669+00:00 elapsed_s=0.024 ms=0.698960
shape=2 variant=solution tflops=6.145 gbps=72.103 ms=0.699
shape=2 solution_peak_fraction=0.0307
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-13T23:50:57.770321+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-13T23:50:57.782234+00:00 elapsed_s=0.012 ms=0.313776
shape=3 variant=solution tflops=3.422 gbps=40.154 ms=0.314
shape=3 solution_peak_fraction=0.0171
peak_fraction: 0.0302
RESULT: LOW
</output>
assistant
reasoning (1805 chars)
Improved from 0.0198 to 0.0302, but still below 0.1. I need to optimize further. The current kernel uses fp32 everywhere. Switching to bf16 would roughly double performance. Also, the block sizes/warps might not be optimal. Let me try using bf16 for q,k,v inputs to kernels (but keep g, g_mid fp32). This saves bandwidth for kernel 1 and 2. Also, I can fuse the preprocessing (qg, g_mid, glast) into the kernels to avoid extra memory, but preprocessing is small. Let me first try bf16 q,k,v. In Model.forward, don't upcast q,k,v. Keep them bf16. Pass bf16 q,k,v to kernel 1. Kernel 1 loads and casts to fp32 for computation. Same for kernel 2. But qg for kernel 2: currently computed in Python as fp32 (because q is fp32). If q is bf16, qg is bf16. Kernel 2 loads as bf16 and casts to fp32. Same accuracy as before? Actually, before we upcast q to fp32, computed qg in fp32, passed to kernel 2 as fp32, loaded as fp32. If we keep q bf16, qg bf16, kernel 2 loads bf16 -> fp32. The qg values have bf16 precision. For tolerance 0.05, should be OK. Similarly for w,u: computed in kernel 1 in fp32, stored as fp32. Kernel 2 loads as fp32. OK. Let me try not upcasting q,k,v. But wait, for Aqk/M computation in kernel 1, q,k are bf16, g,g_mid fp32. Kernel casts q,k to fp32 and computes. Aqk/M accuracy depends on q,k precision. With bf16 q,k, relative error ~0.5%. For tolerance 0.05, should be OK. Let me try. Also, I can try larger BV in kernel 2 (e.g., BV=64) to reduce V tiles and increase work per block. But with BV=64, S is 32KB per block. Fine. And increase num_warps or num_stages. Let me experiment with autotune configs. Actually, Triton has @triton.autotune. I can use it to try different configs. But for simplicity, let me manually try a few. First, remove upcast and use bf16 q,k,v.
in 193786out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (29 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -13,12 +13,12 @@
         NT = T // BT
         device = q.device
 
-        # Upcast to fp32 for numerical accuracy (reference works in fp32).
-        q = q.to(torch.float32)
-        k = k.to(torch.float32)
-        v = v.to(torch.float32)
+        # Keep activations in bf16 for bandwidth; only g stays fp32.
+        q = q.to(torch.bfloat16)
+        k = k.to(torch.bfloat16)
+        v = v.to(torch.bfloat16)
         g = g.to(torch.float32)
-        beta = beta.to(torch.float32)
+        beta = beta.to(torch.bfloat16)
 
         g_4d = g.view(B, NT, BT, H, K)
         g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
@@ -37,7 +37,7 @@
 
         grid1 = (NT, B * H)
         _kda_intra_kernel[grid1](
-            q_scaled, k, v, g_cum, kg, beta, Aqk, w, u,
+            q_scaled, k, v, g_cum, g_mid, beta, Aqk, w, u,
             1.0,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
             BK=64,
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1490 chars \u2014 see diff>",
  "new_string": "<1491 chars \u2014 see diff>"
}
in 193786out 0cache_r 0cache_w 0
tool
stderr (1568 chars) — TRUNCATED
<tool_use_error>String to replace not found in file.
String:     def forward(
        self,
        q: torch.Tensor,
        k: torch.Tensor,
        v: torch.Tensor,
        g: torch.Tensor,
        beta: torch.Tensor,
    ) -> torch.Tensor:
        B, T, H, K = q.shape
        V = v.shape[-1]
        BT = self.chunk_size
        assert T % BT == 0
        NT = T // BT
        device = q.device

        # Upcast to fp32 for numerical accuracy (reference works in fp32).
        q = q.to(torch.float32)
        k = k.to(torch.float32)
        v = v.to(torch.float32)
        g = g.to(torch.float32)
        beta = beta.to(torch.float32)

        g_4d = g.view(B, NT, BT, H, K)
        g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)

        q_scaled = q * self.scale
        qg = q_scaled * g_cum.exp()

        # chunk-midpoint g for stable Aqk/M computation
        g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT // 2, :, :].permute(0, 2, 1, 3).contiguous()

        glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()

        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.float32)
        w = torch.empty(B, T, H, K, device=device, dtype=torch.float32)
        u = torch.empty(B, T, H, V, device=device, dtype=torch.float32)

        grid1 = (NT, B * H)
        _kda_intra_kernel[grid1](
            q_scaled, k, v, g_cum, kg, beta, Aqk, w, u,
            1.0,
            B=B, T=T, H=H, K=K, V=V, BT=BT,
            BK=64,
            BV=32,
         

... (68 more chars)
assistant
reasoning (84 chars)
The string has `->` which might be encoded. Let me read the file and edit carefully.
in 195934out 0cache_r 0cache_w 0
assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 80,
  "offset": 265
}
in 195934out 0cache_r 0cache_w 0
tool
stdout (2732 chars) — TRUNCATED
265	        self.B, self.T, self.H, self.K, self.V = B, T, H, K, V
266	        self.chunk_size = chunk_size
267	        self.scale = float(K) ** -0.5
268	        self.register_buffer("_dummy", torch.zeros(1), persistent=False)
269	
270	    def forward(
271	        self,
272	        q: torch.Tensor,
273	        k: torch.Tensor,
274	        v: torch.Tensor,
275	        g: torch.Tensor,
276	        beta: torch.Tensor,
277	    ) -> torch.Tensor:
278	        B, T, H, K = q.shape
279	        V = v.shape[-1]
280	        BT = self.chunk_size
281	        assert T % BT == 0
282	        NT = T // BT
283	        device = q.device
284	
285	        # Upcast to fp32 for numerical accuracy (reference works in fp32).
286	        q = q.to(torch.float32)
287	        k = k.to(torch.float32)
288	        v = v.to(torch.float32)
289	        g = g.to(torch.float32)
290	        beta = beta.to(torch.float32)
291	
292	        g_4d = g.view(B, NT, BT, H, K)
293	        g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
294	
295	        q_scaled = q * self.scale
296	        qg = q_scaled * g_cum.exp()
297	
298	        # chunk-midpoint g for stable Aqk/M computation
299	        g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT // 2, :, :].permute(0, 2, 1, 3).contiguous()
300	
301	        glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()
302	
303	        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.float32)
304	        w = torch.empty(B, T, H, K, device=device, dtype=tor

... (1232 more chars)
assistant
reasoning (52 chars)
Let me edit the forward to keep bf16 for q,k,v,beta.
in 197064out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -282,12 +282,12 @@
         NT = T // BT
         device = q.device
 
-        # Upcast to fp32 for numerical accuracy (reference works in fp32).
-        q = q.to(torch.float32)
-        k = k.to(torch.float32)
-        v = v.to(torch.float32)
+        # Keep activations in bf16 for bandwidth; only g stays fp32.
+        q = q.to(torch.bfloat16)
+        k = k.to(torch.bfloat16)
+        v = v.to(torch.bfloat16)
         g = g.to(torch.float32)
-        beta = beta.to(torch.float32)
+        beta = beta.to(torch.bfloat16)
 
         g_4d = g.view(B, NT, BT, H, K)
         g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<240 chars \u2014 see diff>",
  "new_string": "<238 chars \u2014 see diff>"
}
in 197064out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (210 chars)
Now kernel 1 receives q,k,v,beta as bf16. It casts to fp32 for computation. Good. Kernel 2 receives qg (bf16), k (bf16), g_cum (fp32), w,u,Aqk (fp32). It casts as needed. Good. Let me run check and benchmark.
in 197308out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10 && uv run python benchmark.py 2>&1 | tail -20",
  "description": "Check and benchmark bf16 inputs"
}
in 197308out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (1307 chars)
PASS
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-13T23:51:55.252414+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-13T23:51:55.522941+00:00 elapsed_s=0.271 ms=0.259376
shape=0 variant=solution tflops=8.279 gbps=97.151 ms=0.259
shape=0 solution_peak_fraction=0.0414
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-13T23:51:55.688609+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-13T23:51:55.706122+00:00 elapsed_s=0.018 ms=0.437952
shape=1 variant=solution tflops=9.807 gbps=115.075 ms=0.438
shape=1 solution_peak_fraction=0.0490
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-13T23:51:55.872893+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-13T23:51:55.894617+00:00 elapsed_s=0.022 ms=0.612784
shape=2 variant=solution tflops=7.009 gbps=82.243 ms=0.613
shape=2 solution_peak_fraction=0.0350
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-13T23:51:55.934895+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-13T23:51:55.946717+00:00 elapsed_s=0.012 ms=0.301184
shape=3 variant=solution tflops=3.565 gbps=41.833 ms=0.301
shape=3 solution_peak_fraction=0.0178
peak_fraction: 0.0336
RESULT: LOW
assistant
reasoning (1856 chars)
peak_fraction 0.0336, still low. bf16 helped a bit (0.0302 -> 0.0336) but not enough. I need more aggressive optimization. The peak fraction is computed against bf16 peak TFLOPS of RTX PRO 6000. Let me find the actual peak. The hardware has 188 SMs. Blackwell SM120. Each SM has high throughput for bf16 tensor cores. RTX PRO 6000 might have ~100-150 TFLOPS bf16 dense. The benchmark reports tflops ~8-10, which is low. For reference, RTX 4090 has ~80 TFLOPS bf16. RTX PRO 6000 likely higher. So 8-10 TFLOPS is ~10% of peak. To reach 0.1 peak_fraction, we need ~10-15 TFLOPS effective. Currently at 3-10. The main bottlenecks: 1. Kernel 1 still does multiple passes over data (Aqk/M accumulation, then w, then u). 2. Kernel 2 is sequential and memory-bound. 3. Python preprocessing adds overhead. 4. Small block count for some shapes. Let me profile with ncu to see where time is spent. Actually, before profiling, let me think about algorithmic improvements. A key optimization: fuse kernel 1 and kernel 2? Hard due to recurrence dependency. Another: use a persistent kernel for the recurrence to reduce launch overhead and improve occupancy. Another: expose more parallelism in kernel 2 by processing multiple chunks in parallel with a prefix-scan-like approach. But the recurrence S_{n+1} = f(S_n) is sequential. Wait, we can process the recurrence with more V tiles to increase parallelism. Currently BV=32, grid V/BV=4. For shape 3 (B=1,H=4,V=128), grid=4*4=16 blocks. For shape 0 (B=2,H=8,V=128), grid=4*16=64 blocks. If we reduce BV to 16, grid=8, giving 128 blocks for shape 0. More parallelism, but more memory traffic. Let me try BV=16. Also BK can be 32, giving 4 K tiles and more loop overhead, but maybe better occupancy? Probably not. Actually, the main issue might be kernel 1. Let me profile. Let me run ncu on a single shape.
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assistant
Bash
args
{
  "command": "<448 chars \u2014 see below>",
  "description": "Warmup for profiling"
}
command (448 chars)
uv run python -c "
import torch, solution, reference
reference.B,reference.T,reference.H,reference.K,reference.V,reference.CHUNK_SIZE = 2,2048,8,128,128,64
solution.B,solution.T,solution.H,solution.K,solution.V,solution.CHUNK_SIZE = 2,2048,8,128,128,64
m=solution.Model(2,2048,8,128,128).cuda()
q,k,v,g,beta=[x.cuda() for x in reference.get_inputs()]
for _ in range(5): m(q,k,v,g,beta)
torch.cuda.synchronize()
print('warmup done')
" 2>&1 | tail -5
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subagent · 0 events · 0 tools
tool
stdout (11 chars)
warmup done
assistant
reasoning (15 chars)
Let me run ncu.
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assistant
Bash
args
{
  "command": "which ncu && ncu --version 2>&1 | head -5 || echo 'ncu not found'",
  "description": "Check ncu availability"
}
in 198655out 0cache_r 0cache_w 0
subagent · 7 events · 2 tools
tool
stdout (283 chars)
Command running in background with ID: bldd2r54m. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bldd2r54m.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (19 chars)
Wait for ncu check.
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assistant
TaskOutput
args
{
  "task_id": "bldd2r54m",
  "block": true,
  "timeout": 60000
}
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tool
stdout (135 chars)
<retrieval_status>timeout</retrieval_status>

<task_id>bldd2r54m</task_id>

<task_type>local_bash</task_type>

<status>running</status>
assistant
reasoning (93 chars)
ncu might be slow. Let me try a different approach: use torch.profiler to see time breakdown.
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assistant
Bash
args
{
  "command": "<940 chars \u2014 see below>",
  "description": "Profile with torch.profiler"
}
command (940 chars)
cat > /workspace/problems/02_kda_cutlass/profile.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 2,2048,8,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

m = solution.Model(B,T,H,K,V,CS).cuda().eval()
torch.manual_seed(0)
q,k,v,g,beta = [x.cuda() for x in reference.get_inputs()]

for _ in range(3):
    _ = m(q,k,v,g,beta)

torch.cuda.synchronize()
with torch.profiler.profile(
    activities=[torch.profiler.ProfilerActivity.CUDA, torch.profiler.ProfilerActivity.CPU],
    record_shapes=False,
    with_stack=False,
) as prof:
    for _ in range(5):
        _ = m(q,k,v,g,beta)
    torch.cuda.synchronize()

print(prof.key_averages().table(sort_by="cuda_time_total", row_limit=30))
PY
uv run python /workspace/problems/02_kda_cutlass/profile.py 2>&1 | tail -40
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system — task_started
↳ subagent started
tool
stdout (6812 chars) — TRUNCATED
aten::as_strided         0.23%       5.680us         0.23%       5.680us       0.284us       0.000us         0.00%       0.000us       0.000us            20  
                                          aten::permute         0.36%       8.970us         0.44%      10.990us       1.099us       0.000us         0.00%       0.000us       0.000us            10  
                                       aten::empty_like         0.25%       6.270us         0.80%      19.990us       1.999us       0.000us         0.00%       0.000us       0.000us            10  
                                            aten::empty         1.50%      37.581us         1.50%      37.581us       1.253us       0.000us         0.00%       0.000us       0.000us            30  
                                  cudaDeviceSynchronize        61.84%       1.548ms        61.84%       1.548ms     773.947us       0.000us         0.00%       0.000us       0.000us             2  
-------------------------------------------------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  
Self CPU time total: 2.503ms
Self CUDA time total: 2.011ms

-------------------------------------------------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  
                                                   Name    Self CPU %      Self CPU   CPU tot

... (5312 more chars)
assistant
reasoning (6930 chars)
Profile shows: - _kda_inter_kernel: 897us (5 calls) -> 179us per call - _kda_intra_kernel: 859us (5 calls) -> 172us per call - cumsum: 144us - mul/exp: ~100us For shape 1 (T=2048), benchmark reported ms=0.438. The profile per call is ~0.35ms for kernels + overhead. Close. Both kernels take similar time. To reach 0.1 peak fraction (~10 TFLOPS for this shape), we need ~0.2ms total. Currently 0.35ms. Need ~1.75x speedup. Optimizations to try: 1. Use bf16 for Aqk, w, u storage and computation where possible. 2. Increase num_warps or use autotune. 3. Fuse preprocessing into kernels. 4. Optimize kernel 2 with larger BV or persistent approach. Actually, a major optimization: avoid storing Aqk, w, u in fp32. Store as bf16. This halves memory traffic in kernel 2. But Aqk is used in kernel 2; bf16 Aqk might reduce accuracy. With tolerance 0.05, it might be OK. But large_qkv already borderline. Alternatively, keep Aqk/w/u in fp32 but use larger tiles to improve compute efficiency. Let me try autotuning block sizes and warps. Actually, let me first check if num_warps=8 helps. And try num_stages=2. Also, I can try BK=128 (full K) in kernel 1, which eliminates the K tile loop and allows one big matrix multiply. But shared memory for (BT,BK)=(64,128) block is 64*128*2 = 16KB per tensor, times q,k,g = 48KB. Plus Aqk/M registers. Might fit. With BK=128, kernel 1 does one tl.dot per matrix (Aqk, M, w). Much faster. Similarly for kernel 2, BK=128 means S is one tile (128,BV). We can have a single S tensor instead of S0,S1. This simplifies and might be faster. But S size = 128*BV fp32. With BV=32, 16KB. With BV=64, 32KB. Manageable. Let me try BK=128, BV=64 in both kernels. But with BV=64, grid2 = V/64 = 2. For shape 0 (B=2,H=8), blocks = 2*16 = 32. Less parallelism but more work per block. For shape 3 (B=1,H=4), blocks = 2*4 = 8. Might be OK. Actually, kernel 2 is sequential per block, so fewer blocks means less SM utilization. But each block does more work per chunk. Need balance. Let me try BK=128, BV=32 first in kernel 1, and keep kernel 2 as is. Then try BV=64. Actually, let me add Triton autotune to both kernels. This will try configs and pick best. Triton autotune: ```python @triton.autotune( configs=[ triton.Config({'BK': 64, 'BV': 32}, num_warps=4, num_stages=1), triton.Config({'BK': 64, 'BV': 32}, num_warps=8, num_stages=1), triton.Config({'BK': 128, 'BV': 32}, num_warps=4, num_stages=1), triton.Config({'BK': 128, 'BV': 32}, num_warps=8, num_stages=1), triton.Config({'BK': 128, 'BV': 64}, num_warps=4, num_stages=1), triton.Config({'BK': 128, 'BV': 64}, num_warps=8, num_stages=1), ], key=['B','T','H','K','V'], ) @triton.jit def _kda_intra_kernel(..., BK: tl.constexpr, BV: tl.constexpr, ...): ``` But autotune adds compile time. And the key should include relevant dims. Actually, for kernel 2, BV matters more. For kernel 1, BK matters. Let me add autotune to both kernels with a few configs. Also, I should consider using a different strategy for kernel 2: instead of S0,S1 hardcoded, handle general K tiles with a loop. But with BK=128 and K=128, only 1 tile. So S is single tensor. Let me rewrite kernel 2 for BK=128, BV=64 with single S. But I want to keep generality. Actually, since K=V=128 always, I can specialize. Wait, the problem says shapes have K=V=128. But Model init can receive arbitrary K,V. The check uses the given shapes. I can specialize to K=V=128 for performance, but the Model should still work for other dims if K,V multiples of BK,BV. Let me first try BK=128, BV=32/64 with autotune. Actually, with BK=128, kernel 2 needs S shape (128,BV). I can handle it as a single tensor or split into two. Let me use a single S tensor and loop over K tiles if needed. But Triton doesn't support dynamic number of S tiles. I can hardcode for K=128: if BK=128, one tile; if BK=64, two tiles. Let me write a generic version with a loop over K tiles, storing S tiles in separate variables via unrolling. But that's complex. Simpler: use BK=128 always for K=128. Then S is one tensor. For V, use BV=64 (grid=2) or BV=32 (grid=4). Let me try both. Actually, for kernel 1, BK=128 means we load q,k,g as (64,128) blocks. That's 64*128*2 = 16KB for q,k each, 32KB for g. Total 64KB in shared memory at once. Plus Aqk/M 16KB in registers. Should fit. Let me rewrite kernel 1 for BK=128. But I want to keep it generic, so I'll use `tl.cdiv(K,BK)` loops. With BK=128, one iteration. With BK=64, two. For kernel 2, if BK=128, S is (128,BV). If BK=64, S0,S1. To handle both, I can write separate code paths or use a single S with slicing? Slicing doesn't work well. Actually, I can keep S as a list of tensors S_tiles = [S0, S1, ...] but Triton doesn't support lists. I can unroll for the expected number of tiles. Since K=128, BK can be 64 or 128. I'll handle both with if/else or separate kernels. But autotune chooses BK. Let me write kernel 2 with a loop over K tiles using a pre-allocated array of S tiles. Since Triton doesn't support lists, I'll use a tuple? No. Alternative: use shared memory for S. Allocate S_shared = tl.zeros([K,BV]) in shared memory? Actually `tl.zeros([K,BV])` creates a register tensor. To put in shared memory, we need to use it in a way the compiler decides. But we can load/store slices using block pointers if S_shared is a global pointer? No. Hmm. Let me just hardcode two versions: BK=64 uses S0,S1; BK=128 uses S_single. I can use a compile-time check: ```python if tl.constexpr(BK == 128): S = tl.zeros([BK, BV], dtype=tl.float32) # use S else: S0 = tl.zeros([BK, BV], ...) S1 = tl.zeros([BK, BV], ...) ``` But `tl.constexpr` comparisons in if statements work for compile-time branches. Actually, Triton supports `if tl.constexpr(BK == 128):` for compile-time branching. Let me use that. But it makes the kernel code messy. Alternatively, I can just set BK=128 and BV=64 fixed, since K=V=128. Let me try fixed BK=128, BV=64 and see performance. Rewrite kernel 1 with BK=128: - One K tile loop iteration. - Aqk/M computed with single tl.dot each. - w = A @ kg, u = A @ v with single tile each. Rewrite kernel 2 with BK=128, BV=64: - S shape (128,64). - One w @ S, one qg @ S. - One k_update.T @ v_i. - Grid = (V/64, B*H) = (2, B*H). This should be much faster due to fewer loops and larger tiles. But register pressure: S is 128*64 = 8192 fp32 = 32KB. Plus b_v (64*64=16KB), b_o (64*64=16KB), etc. Total ~80KB registers per block. For 128 threads, register file 128*256*4 = 128KB. 80KB is tight but might fit with num_warps=8 or 16. Actually, register file per SM is 256KB. With num_warps=8 (256 threads), 256KB. So 80KB for S plus other is OK. But occupancy might be low. Let me try. Also, I can reduce register pressure by computing o in bf16 or smaller tiles. But let's try first. Let me write a new version with BK=128, BV=64.
in 201521out 0cache_r 0cache_w 0
assistant
Write /workspace/problems/02_kda_cutlass/solution.py
diff (308 lines — TRUNCATED)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -41,7 +41,6 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
-    # adjust base pointers for batch/head
     q_ptr += (i_b * T * H + i_h) * K
     k_ptr += (i_b * T * H + i_h) * K
     g_ptr += (i_b * T * H + i_h) * K
@@ -58,8 +57,6 @@
     mask_strict = r[:, None] > c[None, :]
     mask_lower = r[:, None] >= c[None, :]
 
-    # Compute Aqk and M via chunk-midpoint normalization.
-    # gmid = g[BT//2, :] keeps exp(g - gmid) and exp(gmid - g) bounded.
     Aqk = tl.zeros([BT, BT], dtype=tl.float32)
     M = tl.zeros([BT, BT], dtype=tl.float32)
 
@@ -98,7 +95,6 @@
     Aqk = tl.where(mask_lower, Aqk * scale, 0.0)
     A0 = tl.where(mask_strict, -M * b_beta[:, None], 0.0)
 
-    # solve (I - A0) in registers
     A = A0
     for i in range(1, BT):
         row_i = tl.sum(tl.where(r[:, None] == i, A, 0.0), axis=0)
@@ -113,7 +109,6 @@
     )
     tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
 
-    # w = A @ kg, kg = k * exp(g)
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
         p_k = tl.make_block_ptr(
@@ -168,7 +163,6 @@
     i_b = i_bh // H
     i_h = i_bh % H
 
-    # adjust base pointers for batch/head
     w_ptr += (i_b * T * H + i_h) * K
     u_ptr += (i_b * T * H + i_h) * V
     qg_ptr += (i_b * T * H + i_h) * K
@@ -180,83 +174,132 @@
     off_v0 = i_vb * BV
     NT = T // BT
 
-    # S split into K tiles of (BK, BV)
-    S0 = tl.zeros([BK, BV], dtype=tl.float32)
-    S1 = tl.zeros([BK, BV], dtype=tl.float32)
-
-    for i_n in range(NT):
-        t0 = i_n * BT
-
-        p_u = tl.make_block_ptr(
-            u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
-        )
-        b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
-
-        b_v = b_u
-        p_w0 = tl.make_block_ptr(
-            w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
-        )
-        b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.float32)
-        b_v -= tl.dot(b_w0, S0)
-        p_w1 = tl.make_block_ptr(
-            w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_w1 = tl.load(p_w1, boundary_check=(0, 1)).to(tl.float32)
-        b_v -= tl.dot(b_w1, S1)
-
-        b_o = tl.zeros([BT, BV], dtype=tl.float32)
-        p_qg0 = tl.make_block_ptr(
-            qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
-        )
-        b_qg0 = tl.load(p_qg0, boundary_check=(0, 1)).to(tl.float32)
-        b_o += tl.dot(b_qg0, S0)
-        p_qg1 = tl.make_block_ptr(
-            qg_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_qg1 = tl.load(p_qg1, boundary_check=(0, 1)).to(tl.float32)
-        b_o += tl.dot(b_qg1, S1)
-
-        p_Aqk = tl.make_block_ptr(
-            Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
-        )
-        b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.float32)
-        b_o += tl.dot(b_Aqk, b_v)
-
-        p_o = tl.make_block_ptr(
-            o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
-        )
-        tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))
-
-        # S = S * d + (k_i * exp(g_last - g_i)).T @ v_i
-        p_glast0 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
-        d0 = tl.exp(tl.load(p_glast0))
-        p_glast1 = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + BK + tl.arange(0, BK)
-        d1 = tl.exp(tl.load(p_glast1))
-
-        p_k0 = tl.make_block_ptr(
-            k_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
-        )
-        b_k0 = tl.load(p_k0, boundary_check=(0, 1)).to(tl.float32)
-        p_g0 = tl.make_block_ptr(
-            g_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
-        )
-        b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
-        k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)
-
-        p_k1 = tl.make_block_ptr(
-            k_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_k1 = tl.load(p_k1, boundary_check=(0, 1)).to(tl.float32)
-        p_g1 = tl.make_block_ptr(
-            g_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
-        )
-        b_g1 = tl.load(p_g1, boundary_check=(0, 1)).to(tl.float32)
-        k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
-
-        S0 = S0 * d0[:, None]
-        S1 = S1 * d1[:, None]
-        S0 += tl.dot(tl.trans(k_up0), b_v)
-        S1 += tl.dot(tl.trans(k_up1), b_v)
+    # For K=128 we have either one BK=128 tile or two BK=64 tiles.
+    if tl.constexpr(K == 128 and BK == 128):
+        S = tl.zeros([BK, BV], dtype=tl.float32)
+        for i_n in range(NT):
+            t0 = i_n * BT
+
+            p_u = tl.make_block_ptr(
+                u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
+            )
+            b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
+            b_v = b_u
+
+            p_w = tl.make_block_ptr(
+                w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
+            )
+            b_w = tl.load(p_w, boundary_check=(0, 1)).to(tl.float32)
+            b_v -= tl.dot(b_w, S)
+
+            b_o = tl.zeros([BT, BV], dtype=tl.float32)
+            p_qg = tl.make_block_ptr(
+                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
+            )
+            b_qg = tl.load(p_qg, boundary_check=(0, 1)).to(tl.float32)
+            b_o += tl.dot(b_qg, S)
+
+            p_Aqk = tl.make_block_ptr(
+                Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
+            )
+            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.float32)
+            b_o += tl.dot(b_Aqk, b_v)
+
+            p_o = tl.make_block_ptr(
+                o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
+            )
+            tl.store(p_o, b_o.to(tl.bfloat16), boundary_check=(0, 1))
+
+            p_glast = glast_ptr + (i_b * H + i_h) * NT * K + i_n * K + tl.arange(0, BK)
+            d = tl.exp(

... (5559 more chars)
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "content": "<12868 chars \u2014 see diff>"
}
in 201521out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (39 chars)
Let me check correctness and benchmark.
in 208795out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10 && uv run python benchmark.py 2>&1 | tail -20",
  "description": "Check and benchmark BK=128 BV=64"
}
in 208795out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (1306 chars)
PASS
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:01:57.325076+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:01:57.599600+00:00 elapsed_s=0.275 ms=0.474160
shape=0 variant=solution tflops=4.529 gbps=53.144 ms=0.474
shape=0 solution_peak_fraction=0.0226
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:01:57.771661+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:01:57.804076+00:00 elapsed_s=0.032 ms=0.965248
shape=1 variant=solution tflops=4.450 gbps=52.212 ms=0.965
shape=1 solution_peak_fraction=0.0222
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:01:57.973811+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:01:58.021786+00:00 elapsed_s=0.048 ms=1.492192
shape=2 variant=solution tflops=2.878 gbps=33.774 ms=1.492
shape=2 solution_peak_fraction=0.0144
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:01:58.064771+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:01:58.086584+00:00 elapsed_s=0.022 ms=0.632528
shape=3 variant=solution tflops=1.698 gbps=19.919 ms=0.633
shape=3 solution_peak_fraction=0.0085
peak_fraction: 0.0158
RESULT: LOW
assistant
reasoning (505 chars)
Worse! BK=128, BV=64 is slower. Probably due to register pressure or lower occupancy. Let me go back to BK=64, BV=32 which gave 0.0336. Actually, the previous best was 0.0336 with bf16 inputs. Let me try autotune to find best config. Add @triton.autotune with various BK,BV,num_warps. But autotune increases compile time. Let me add a few configs. Actually, before autotune, let me try BK=64, BV=64 with num_warps=4/8. And BK=64, BV=32 with num_warps=8. Let me revert to BK=64,BV=32 and add autotune.
in 209370out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (33 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -344,9 +344,9 @@
         w = torch.empty(B, T, H, K, device=device, dtype=torch.float32)
         u = torch.empty(B, T, H, V, device=device, dtype=torch.float32)
 
-        # Use full K tile for K=128 to minimize loops.
-        BK = min(128, triton.next_power_of_2(K))
-        BV = 64
+        # Use 64-wide tiles as a conservative default; autotune explores others.
+        BK = 64
+        BV = 32
 
         grid1 = (NT, B * H)
         _kda_intra_kernel[grid1](
@@ -355,8 +355,6 @@
             B=B, T=T, H=H, K=K, V=V, BT=BT,
             BK=BK,
             BV=BV,
-            num_stages=1,
-            num_warps=8,
         )
 
         o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
@@ -366,8 +364,6 @@
             B=B, T=T, H=H, K=K, V=V, BT=BT,
             BK=BK,
             BV=BV,
-            num_stages=1,
-            num_warps=8,
         )
         return o
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<743 chars \u2014 see diff>",
  "new_string": "<634 chars \u2014 see diff>"
}
in 209370out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (3595 chars)
Now add autotune. Let me add to both kernels. Configs to try: - BK=64, BV=32, num_warps=4, num_stages=1 - BK=64, BV=32, num_warps=8, num_stages=1 - BK=64, BV=64, num_warps=4, num_stages=1 - BK=64, BV=64, num_warps=8, num_stages=1 - BK=32, BV=32, num_warps=4, num_stages=1 - BK=32, BV=64, num_warps=4, num_stages=1 But too many configs slow compile. Let me try 4-6. Actually, I realize the issue might be that I'm passing q_scaled,k etc. as bf16 and the kernel casts to fp32. The cast itself has overhead. If I pass fp32 tensors, the kernel doesn't cast. But that doubles memory bandwidth. Hmm. For kernel 1, q,k are read once. Casting bf16->fp32 happens on load. The hardware can do this efficiently. Memory bandwidth is halved vs fp32. So bf16 is better. For kernel 2, qg and k are bf16, w,u,Aqk are fp32. Casting bf16->fp32 on load is fine. I think the bottleneck is the sequential recurrence in kernel 2 and the multiple passes in kernel 1. Actually, let me think about kernel 1 again. It reads q,k,g (and g_mid) once, computes Aqk/M with 2 dots, solves, then reads k,g again for w, and v for u. That's 3 passes over chunk data. We can fuse: compute Aqk/M and kg simultaneously, then after solve, use stored kg for w. But storing kg needs memory. Or recompute k,g for w (current approach) which is another read. Actually, we can store kg in shared memory during the first loop and reuse for w. With kg (BT,K) bf16 = 16KB, it fits. Then w loop just reads kg from shared memory. Similarly, we could store kdecay_norm or other. But kg for w is the main extra read. Let me try storing kg in shared memory in kernel 1. But Triton shared memory allocation is implicit. I can declare `kg_s = tl.zeros([BT*K], dtype=tl.bfloat16)` and store kg tiles. Then in w loop, load from kg_s. But dynamic indexing for kg_s tiles is tricky. Actually, since K=128 and BK=64, we have 2 K tiles. We can declare `kg0 = tl.zeros([BT,BK], dtype=tl.bfloat16)` and `kg1 = ...` as register tensors. Store kg tiles there during first loop. Then use for w. Register usage: kg0 + kg1 = 2 * 64*64*2 = 16KB. Plus Aqk/M 32KB. Plus A 16KB. Total 64KB registers. With 128 threads, 512 bytes/thread. Register file 256 regs * 4 bytes = 1KB/thread. So 512 bytes is OK. Let me try storing kg in registers. Actually, the w loop with BK=64 reads k and g again. That's 2 reads of 64*64*2 = 8KB each per block. For 256 blocks, 4MB read each. Total 8MB extra. Not huge. The time saved might be small. The bigger issue is kernel 2. Let me focus there. Kernel 2 with BK=64,BV=32: each block processes one V tile. For each chunk, it reads w (BT,K=128) fp32 = 32KB, u (BT,BV=32) fp32 = 8KB, qg (BT,K=128) bf16 = 16KB, k (BT,K=128) bf16 = 16KB, g (BT,K=128) fp32 = 32KB, Aqk (BT,BT) fp32 = 16KB. Total per chunk ~120KB. For NT=32, ~3.8MB per block. With 32 blocks (shape 1), ~122MB total memory reads. At 1.8TB/s, 0.07ms. But benchmark kernel time is 0.18ms. So memory not the only bottleneck; compute and serialization matter. The sequential loop has synchronization between chunks (S update must finish before next chunk). This limits instruction-level parallelism. One optimization: unroll the loop? NT is runtime (T/BT). Can't unroll fully. Another: use larger BV to amortize loop overhead. BV=64 doubles work per iteration and halves iterations. But fewer blocks. Let me try BV=64 with BK=64. Actually, the previous BV=64 test was with BK=128. Let me try BK=64,BV=64. Also, we can try num_stages=2 or 3 for kernel 2 to overlap memory and compute. But sequential dependency limits this. Let me add autotune and test.
in 210145out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -15,6 +15,17 @@
 HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
 
 
+@triton.autotune(
+    configs=[
+        triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 32, "BV": 32}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 32, "BV": 64}, num_warps=4, num_stages=1),
+    ],
+    key=["B", "T", "H", "K", "V"],
+)
 @triton.jit
 def _kda_intra_kernel(
     q_ptr,
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<34 chars \u2014 see diff>",
  "new_string": "<542 chars \u2014 see diff>"
}
in 210145out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (29 chars)
Add autotune to kernel 2 too.
in 211724out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -150,6 +150,17 @@
         tl.store(p_u, b_u, boundary_check=(0, 1))
 
 
+@triton.autotune(
+    configs=[
+        triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 32, "BV": 32}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 32, "BV": 64}, num_warps=4, num_stages=1),
+    ],
+    key=["B", "T", "H", "K", "V"],
+)
 @triton.jit
 def _kda_inter_kernel(
     w_ptr,
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<34 chars \u2014 see diff>",
  "new_string": "<542 chars \u2014 see diff>"
}
in 211724out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (82 chars)
Now run benchmark. Autotune will try configs and pick best. It might take a while.
in 212074out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check with autotune"
}
in 212074out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (869 chars)
benchmark()
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/autotuner.py", line 229, in benchmark
    timings = {config: self._bench(*args, config=config, **kwargs) for config in pruned_configs}
              ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/autotuner.py", line 229, in <dictcomp>
    timings = {config: self._bench(*args, config=config, **kwargs) for config in pruned_configs}
                       ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/autotuner.py", line 139, in _bench
    raise ValueError(f"Conflicting meta-parameters: {', '.join(conflicts)}."
ValueError: Conflicting meta-parameters: BK, BV. Make sure that you don't re-define auto-tuned symbols.
assistant
reasoning (148 chars)
I passed BK=64,BV=32 in the launch but also autotune them. Conflict. I should not pass them in launch if autotuning. Remove BK,BV from launch args.
in 212351out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (21 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -375,17 +375,13 @@
             q_scaled, k, v, g_cum, g_mid, beta, Aqk, w, u,
             1.0,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=BK,
-            BV=BV,
         )
 
         o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
-        grid2 = (triton.cdiv(V, BV), B * H)
+        grid2 = (triton.cdiv(V, 32), B * H)
         _kda_inter_kernel[grid2](
             w, u, qg, k, g_cum, Aqk, glast, o,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
-            BK=BK,
-            BV=BV,
         )
         return o
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<520 chars \u2014 see diff>",
  "new_string": "<444 chars \u2014 see diff>"
}
in 212351out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (1896 chars)
Also, grid2 uses triton.cdiv(V, 32) but if BV=64 is chosen, grid should be V/64. I need to pass a conservative grid that works for all configs. If I use grid=(V/32, B*H), then for BV=64 config, the block will process 64 columns but grid has too many blocks (some will be out of bounds). Need to handle in kernel. Actually, the grid should be based on the chosen BV. With autotune, I can't know BV at Python level. I can use grid=(triton.cdiv(V, 32), B*H) and in kernel check if off_v0 >= V and return. For BV=64, blocks with i_vb >= V/64 will return. But there will be extra blocks. Actually, for BV=64, grid has 2x too many blocks. They return immediately. Not ideal but works. Alternatively, use a grid function. Triton autotune supports `grid` as a function of kwargs. But simpler to add boundary check in kernel. In kernel 2, at start: ```python off_v0 = i_vb * BV if off_v0 >= V: return ``` But Triton doesn't support early return? Actually, it does with `if` and no return? Let me check. In Triton, you can use `tl.device_assert` but early return might not be supported in JIT. Actually, I think you can use a guard like: ```python if off_v0 >= V: pass else: # main code ``` But that indents everything. Not nice. Alternatively, use `tl.constexpr` to unroll? No. Actually, I can just use grid=(triton.cdiv(V, 32), B*H) and for BV=64 configs, the extra blocks will process out-of-bounds V and write to invalid memory. That's bad. Better: use a grid function. Triton launch supports `grid=lambda meta: (triton.cdiv(V, meta['BV']), B*H)`. But with autotune, the meta comes from the chosen config. Let me use that. Actually, Triton's autotune expects grid to be passed as a function if it depends on meta. Let me change grid2 to: ```python grid2 = lambda meta: (triton.cdiv(V, meta['BV']), B * H) ``` Same for grid1? Grid1 doesn't depend on BK/BV. So fine. Let me update.
in 212878out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (11 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -378,7 +378,7 @@
         )
 
         o = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
-        grid2 = (triton.cdiv(V, 32), B * H)
+        grid2 = lambda meta: (triton.cdiv(V, meta['BV']), B * H)
         _kda_inter_kernel[grid2](
             w, u, qg, k, g_cum, Aqk, glast, o,
             B=B, T=T, H=H, K=K, V=V, BT=BT,
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<150 chars \u2014 see diff>",
  "new_string": "<171 chars \u2014 see diff>"
}
in 212878out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (1126 chars)
Also, in kernel 2, the branch `if tl.constexpr(K == 128 and BK == 128):` uses BK. With autotune, BK can be 64, 32, or 128. For K=128, BK=128 branch only if config chooses BK=128. For other BK, else branch. Good. But for K != 128, we need to handle. Currently if K != 128 and BK=64, else branch uses S0,S1 assuming 2 tiles. But if K=256, BK=64, we'd need 4 tiles. The code only handles 2. So it only works for K=128. That's fine for our shapes. But autotune might choose BK=32 for K=128. Then else branch with S0,S1 assumes BK=64 and covers 128. Good. Wait, BK=32 means 4 tiles, but else branch only has S0,S1 (2 tiles). So BK=32 config will be wrong for K=128! I should remove BK=32 configs, or add more S tiles. Let me remove BK=32 configs. K=128, BK can be 64 or 128. Also BV=32/64. For V=128, BV=32 -> 4 tiles, BV=64 -> 2 tiles. The else branch doesn't use BV-specific code (it processes BV columns in b_v,b_o). BV can be 32 or 64. Good. But the K==128 and BK==128 branch uses S single. For K=128,BK=64, else branch uses S0,S1. Good. So configs should be BK in {64,128}, BV in {32,64}. Let me update autotune configs.
in 213607out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (15 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -21,8 +21,10 @@
         triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
-        triton.Config({"BK": 32, "BV": 32}, num_warps=4, num_stages=1),
-        triton.Config({"BK": 32, "BV": 64}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 64}, num_warps=8, num_stages=1),
     ],
     key=["B", "T", "H", "K", "V"],
 )
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<542 chars \u2014 see diff>",
  "new_string": "<690 chars \u2014 see diff>"
}
in 213607out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (27 chars)
Same for kernel 2 autotune.
in 214602out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (15 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -158,8 +158,10 @@
         triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
-        triton.Config({"BK": 32, "BV": 32}, num_warps=4, num_stages=1),
-        triton.Config({"BK": 32, "BV": 64}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 64}, num_warps=8, num_stages=1),
     ],
     key=["B", "T", "H", "K", "V"],
 )
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<542 chars \u2014 see diff>",
  "new_string": "<690 chars \u2014 see diff>"
}
in 214602out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (41 chars)
Now run check. Autotune will try configs.
in 215238out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check with autotune v2"
}
in 215238out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (4 chars)
PASS
assistant
reasoning (20 chars)
PASS. Now benchmark.
in 215310out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python benchmark.py 2>&1 | tail -25",
  "description": "Benchmark with autotune v2"
}
in 215310out 0cache_r 0cache_w 0
subagent · 20 events · 6 tools
tool
stdout (283 chars)
Command running in background with ID: b6065eljf. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/b6065eljf.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (19 chars)
Wait for benchmark.
in 215460out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "b6065eljf",
  "block": true,
  "timeout": 600000
}
in 215460out 0cache_r 0cache_w 0
tool
stdout (1486 chars)
<retrieval_status>success</retrieval_status>

<task_id>b6065eljf</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:07:58.108971+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:08:00.115675+00:00 elapsed_s=2.007 ms=0.256832
shape=0 variant=solution tflops=8.361 gbps=98.113 ms=0.257
shape=0 solution_peak_fraction=0.0418
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:08:00.284566+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:08:02.176489+00:00 elapsed_s=1.892 ms=0.445664
shape=1 variant=solution tflops=9.637 gbps=113.083 ms=0.446
shape=1 solution_peak_fraction=0.0482
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:08:02.347351+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:08:04.277261+00:00 elapsed_s=1.930 ms=0.621936
shape=2 variant=solution tflops=6.906 gbps=81.033 ms=0.622
shape=2 solution_peak_fraction=0.0345
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:08:04.318039+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:08:06.145222+00:00 elapsed_s=1.827 ms=0.303824
shape=3 variant=solution tflops=3.534 gbps=41.469 ms=0.304
shape=3 solution_peak_fraction=0.0177
peak_fraction: 0.0333
RESULT: LOW
</output>
assistant
reasoning (8812 chars)
peak_fraction 0.0333, still low. Autotune didn't help much. I need a more aggressive optimization. The current approach has two separate kernels with global memory intermediates. The recurrence in kernel 2 is sequential. A better approach might be to fuse the entire per-chunk computation into one kernel that processes chunks sequentially but with better parallelism, or use a persistent kernel. Actually, let me reconsider the algorithm. The inter-chunk recurrence is the bottleneck. We have N chunks, and each chunk's S depends on the previous. But we can compute the recurrence with a parallel scan if we formulate it as an affine transformation. Recall: S_{n+1} = M_n @ S_n + B_n, where M_n is (K,K) and B_n is (K,V). Actually, from earlier: S_new[k,v] = S[k,v] * d[k] + sum_c k_decay[c,k] * v_i[c,v] where v_i[c,v] = u[c,v] - sum_k' w[c,k'] * S[k',v]. So S_new = diag(d) @ S + k_decay.T @ (u - w @ S) = diag(d) @ S + k_decay.T @ u - k_decay.T @ w @ S = (diag(d) - k_decay.T @ w) @ S + k_decay.T @ u Let P_n = diag(d_n) - k_decay_n.T @ w_n (K,K) Let Q_n = k_decay_n.T @ u_n (K,V) Then S_{n+1} = P_n @ S_n + Q_n. This is an affine recurrence. We can parallelize it with a parallel scan (prefix scan) over chunks, where each chunk's "operator" is (P_n, Q_n) and composition is: (P2, Q2) ∘ (P1, Q1) = (P2 @ P1, P2 @ Q1 + Q2) Then S_n = (composite of chunks 0..n-1) applied to S_0 = 0. This exposes N-way parallelism: compute all (P_n, Q_n) in parallel, then do a parallel scan to combine them, then compute outputs. The work is: 1. Compute P_n, Q_n for all chunks in parallel. This requires w_n, u_n, k_decay_n, d_n for each chunk. We already compute w,u in kernel 1. k_decay_n = k_n * exp(g_last_n - g_n). d_n = exp(g_last_n). 2. Parallel scan over N chunks to get S_n for all n. 3. Compute o_n = qg_n @ S_n + Aqk_n @ (u_n - w_n @ S_n) in parallel. This is more complex but can be much faster for long contexts because it exposes parallelism across chunks. However, implementing a parallel scan for matrix operators in Triton is complex. But it could give significant speedup, especially for shape 2 (T=4096, N=64) and shape 3 (thin batch). For our shapes, N up to 64. The sequential kernel 2 loops 64 times. A scan would reduce this to log2(64)=6 steps, but with more work per step. Actually, the scan approach has overhead and might not be faster for N=64. But for shape 3 (B=1,H=4,N=32), it would give 32x more parallelism. Let me think if there's a simpler optimization. Alternative: fuse kernel 1 and kernel 2 into a single persistent kernel per (B,H,V tile). The kernel loops over chunks, computing intra-chunk work and inter-chunk recurrence on the fly. This avoids storing/reading Aqk,w,u intermediates, saving memory bandwidth. In this fused kernel: - For each chunk n: - Load q,k,v,g,beta for chunk. - Compute Aqk, M, solve A. - Compute w,u. - Compute v_i = u - w @ S, o_i = qg @ S + Aqk @ v_i, store o_i. - Update S. - Keep S in registers/shared memory across chunks. This eliminates global memory for Aqk,w,u. The only global reads are q,k,v,g,beta and write o. This halves memory traffic. But the kernel is more complex and has more registers. Also, it's still sequential per block. Let me try this fused approach. It might give 1.5-2x speedup from reduced memory bandwidth. Actually, the issue is that kernel 1 already computes Aqk,w,u and writes them, then kernel 2 reads them. Fusing avoids this round-trip. Let me design a fused kernel. Fused kernel grid: (B*H, V/BV). Each block handles one (B,H) and a V tile. Loop over chunks. Per chunk: 1. Load q,k,v,g,beta for chunk into shared memory or registers. 2. Compute Aqk, M, solve A. 3. Compute w = A @ kg, u = A @ v_tile. 4. Compute v_i = u - w @ S, o = qg @ S + Aqk @ v_i, store o. 5. Update S. S is (K,BV) in registers/shared memory. The challenge is fitting everything. Aqk/M/A are 64x64 = 16KB each in registers. S is 128*BV fp32 = 16KB (BV=32) or 32KB (BV=64). q,k,g,v tiles need shared memory or registers. If we load q,k,g,v into shared memory: q 16KB, k 16KB, g 32KB, v_tile 8KB (BV=32). Total 72KB. Plus Aqk/M/A registers. Shared memory total 72KB < 101KB. OK. But dynamic indexing of shared memory for Aqk/M computation with normalized factorization: we can use block pointers since q,k,g are in global memory, not shared. Actually, we can use global memory block pointers directly and not use shared memory. The hardware caches in L1/L2. With the fused kernel, we read q,k,v,g,beta once per chunk from global memory, compute everything, write o. No intermediate global memory. This is optimal memory-wise. Let me try implementing the fused kernel. Actually, this is getting complex. Let me first see if I can improve with simpler optimizations. Current peak fraction 0.033. Need 0.1. That's 3x. Fused kernel might give 1.5-2x. Still need more. Other optimizations: - Use bf16 for Aqk,w,u in kernel 2 (currently fp32). This halves memory traffic in kernel 2. But accuracy might suffer. - Use bf16 for S (currently fp32). Reduces register pressure but accuracy suffers. - Optimize the solve with matrix operations instead of scalar loop. The solve loop is O(BT^3) = 262K ops per chunk. With 256 chunks, 67M ops. In fp32 FMA, at 100 TFLOPS, 0.67ms. But our total time is ~0.35ms for shape 1. So solve is not the bottleneck (it's in registers, fast). The main bottleneck is memory bandwidth and sequential recurrence. Let me try the fused kernel. It should reduce memory bandwidth significantly. Actually, I realize that the reference implementation in PyTorch is also sequential. But PyTorch is slow. Our Triton kernel is faster. Let me also consider: maybe I should use CUTLASS CuTe as the problem suggests. But Triton is easier. If Triton can't reach 0.1, maybe I need CuTe. But let me push Triton further. Let me implement a fused persistent kernel. Design: Grid: (B*H, V//BV) where BV=32 or 64. Each block: - S = zeros(K, BV) in registers or shared memory. - For n in 0..NT-1: - t0 = n*BT - Load q,k,v,g,beta for chunk. - Compute Aqk and M with normalized factorization (using g_mid). - Solve for A. - Compute kg = k*exp(g), w = A @ kg, u = A @ v_tile. - Compute v_i = u - w @ S. - Compute o = qg @ S + Aqk @ v_i, store o. - Compute k_update = k * exp(g_last - g), update S = S*d + k_update.T @ v_i. Wait, but g_mid for Aqk/M normalization needs to be computed. We can compute g_mid = g[BT//2, :] within the kernel. Also, we need g_last for S update. We can compute it as g[BT-1,:]. All in one kernel. This avoids storing Aqk,w,u. But it recomputes Aqk/M/solve/A for each V tile! Since grid has V/BV tiles per (B,H), each tile recomputes the same intra-chunk work. That's redundant. For V=128,BV=32, 4 tiles, so 4x redundant compute for kernel 1 work. That's bad. The current split avoids this: kernel 1 computes once, kernel 2 uses for all V tiles. So fused kernel with V-tile grid has redundant compute. Not good. Alternative: grid (B*H) only, each block processes all V. Then S is (K,V) = 128*128 = 64KB. Plus other registers. Very high register pressure. But no redundant compute. For shape 0 (B=2,H=8), blocks = 16. Each block processes all V and all chunks. Sequential. Low occupancy but no redundancy. Actually, current kernel 2 has grid (V/BV, B*H) = (4,16)=64 blocks, each processing one V tile. This has good parallelism. Fusing would reduce blocks to 16 and increase work per block. Hmm. Tradeoff. A better fused approach: grid (NT, B*H) for kernel 1 style, but each block computes intra-chunk AND stores partial outputs? No, recurrence needs S. Actually, we can split differently: - Kernel 1: compute intra-chunk (Aqk, w, u) per chunk. Grid (NT, B*H). No redundancy. - Kernel 2: process V tiles. Grid (V/BV, B*H). Sequential over chunks. This is what we have. The issue is kernel 2 reads Aqk,w,u from global memory. We can't avoid that without redundancy. But we can make Aqk,w,u bf16 to halve bandwidth. Let me try that. Also, we can avoid storing Aqk if we don't need it? No, kernel 2 needs Aqk. Actually, w and u are A @ kg and A @ v. If we store A (instead of w,u), kernel 2 can compute w_i @ S and u_i on the fly: w_i @ S = (A_i @ kg_i) @ S = A_i @ (kg_i @ S). But kg_i @ S is (K,V) GEMM. And u_i = A_i @ v_i. This might be more compute but less memory. Not obviously better. Let me try bf16 for Aqk,w,u and see if correctness still passes. If yes, bandwidth halves. Change Aqk,w,u storage to bf16. In kernel 1, store as bf16. In kernel 2, load as bf16 and cast to fp32 for compute. Actually, large_qkv was borderline with fp32 intermediates. With bf16, it might fail. Let me try. Also, I can try bf16 for S to reduce register pressure and allow larger BV. But accuracy... Let me first try bf16 Aqk,w,u.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (15 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -366,9 +366,9 @@
         g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT // 2, :, :].permute(0, 2, 1, 3).contiguous()
         glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()
 
-        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.float32)
-        w = torch.empty(B, T, H, K, device=device, dtype=torch.float32)
-        u = torch.empty(B, T, H, V, device=device, dtype=torch.float32)
+        Aqk = torch.empty(B, T, H, BT, device=device, dtype=torch.bfloat16)
+        w = torch.empty(B, T, H, K, device=device, dtype=torch.bfloat16)
+        u = torch.empty(B, T, H, V, device=device, dtype=torch.bfloat16)
 
         # Use 64-wide tiles as a conservative default; autotune explores others.
         BK = 64
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<218 chars \u2014 see diff>",
  "new_string": "<221 chars \u2014 see diff>"
}
in 216208out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (677 chars)
In kernel 1, Aqk,w,u are stored as bf16 now. The kernel currently stores Aqk as fp32 (`.to(tl.float32)`). Need to change to bf16. And w,u are stored as fp32. Need to change. In kernel 2, Aqk,w,u are loaded. Currently loaded as fp32. Need to load as bf16 and cast to fp32? Actually for bf16 storage, load as bf16 then cast to fp32 for compute. But my kernel currently does `.to(tl.float32)` on load. If the tensor is bf16, tl.load returns bf16, then .to(fp32). Good. But in kernel 1, I need to store as bf16: ```python tl.store(p_Aqk, Aqk.to(tl.bfloat16), ...) tl.store(p_w, b_w.to(tl.bfloat16), ...) tl.store(p_u, b_u.to(tl.bfloat16), ...) ``` Let me update kernel 1 stores.
in 219089out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (29 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -120,7 +120,7 @@
     p_Aqk = tl.make_block_ptr(
         Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
     )
-    tl.store(p_Aqk, Aqk.to(tl.float32), boundary_check=(0, 1))
+    tl.store(p_Aqk, Aqk.to(tl.bfloat16), boundary_check=(0, 1))
 
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
@@ -137,7 +137,7 @@
         p_w = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        tl.store(p_w, b_w, boundary_check=(0, 1))
+        tl.store(p_w, b_w.to(tl.bfloat16), boundary_check=(0, 1))
 
     for i_v in range(tl.cdiv(V, BV)):
         off_v = i_v * BV
@@ -149,7 +149,7 @@
         p_u = tl.make_block_ptr(
             u_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        tl.store(p_u, b_u, boundary_check=(0, 1))
+        tl.store(p_u, b_u.to(tl.bfloat16), boundary_check=(0, 1))
 
 
 @triton.autotune(
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1239 chars \u2014 see diff>",
  "new_string": "<1272 chars \u2014 see diff>"
}
in 219089out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (142 chars)
Now kernel 2 already loads as fp32 via `.to(tl.float32)`. Since tensors are bf16, load returns bf16 then cast. Good. Run check and benchmark.
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assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10 && uv run python benchmark.py 2>&1 | tail -20",
  "description": "Check and benchmark bf16 intermediates"
}
in 220439out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (283 chars)
Command running in background with ID: bh8qj61fd. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bh8qj61fd.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (5 chars)
Wait.
in 220646out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "bh8qj61fd",
  "block": true,
  "timeout": 600000
}
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tool
stdout (1493 chars)
<retrieval_status>success</retrieval_status>

<task_id>bh8qj61fd</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
PASS
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:12:14.810870+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:12:16.803796+00:00 elapsed_s=1.993 ms=0.229072
shape=0 variant=solution tflops=9.375 gbps=110.003 ms=0.229
shape=0 solution_peak_fraction=0.0469
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:12:16.968145+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:12:18.841520+00:00 elapsed_s=1.873 ms=0.397936
shape=1 variant=solution tflops=10.793 gbps=126.646 ms=0.398
shape=1 solution_peak_fraction=0.0540
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:12:19.006591+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:12:20.916009+00:00 elapsed_s=1.909 ms=0.535120
shape=2 variant=solution tflops=8.026 gbps=94.179 ms=0.535
shape=2 solution_peak_fraction=0.0401
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:12:20.956805+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:12:22.767170+00:00 elapsed_s=1.810 ms=0.256080
shape=3 variant=solution tflops=4.193 gbps=49.201 ms=0.256
shape=3 solution_peak_fraction=0.0210
peak_fraction: 0.0382
RESULT: LOW
</output>
assistant
reasoning (439 chars)
Improved to 0.0382. Shape 1 now reaches 0.054. But still below 0.1 overall. I need more. Let me think about the peak TFLOPS assumption. The benchmark computes peak_fraction based on hardware peak bf16. For RTX PRO 6000, what's the actual peak? Let me check the hardware config used by benchmark. Actually, the problem is compute regime. peak_tflops_key: bf16. The hardware file likely has a peak_tflops_dense['bf16'] value. Let me check.
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assistant
Bash
args
{
  "command": "uv run python -c \"from src.hardware import get; hw=get('RTX_PRO_6000'); print(hw); print('peak_tflops_dense', hw.peak_tflops_dense); print('peak_bandwidth', hw.peak_bandwidth_gb_s)\" 2>&1 | tail -10",
  "description": "Check hardware peak specs"
}
in 221397out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (498 chars)
HardwareTarget(name='RTX PRO 6000 Blackwell Workstation', sm='sm_120a', vram_gb=96, peak_bandwidth_gb_s=1800.0, peak_tflops_dense={'fp4': 800.0, 'nvfp4': 800.0, 'mxfp4': 800.0, 'fp6': 800.0, 'fp8': 400.0, 'bf16': 200.0, 'fp16': 200.0, 'tf32': 100.0, 'fp32': 12.0, 'int8': 400.0, 'int4': 800.0})
peak_tflops_dense {'fp4': 800.0, 'nvfp4': 800.0, 'mxfp4': 800.0, 'fp6': 800.0, 'fp8': 400.0, 'bf16': 200.0, 'fp16': 200.0, 'tf32': 100.0, 'fp32': 12.0, 'int8': 400.0, 'int4': 800.0}
peak_bandwidth 1800.0
assistant
reasoning (10790 chars)
Peak bf16 is 200 TFLOPS. We need 20 TFLOPS effective for 0.1 peak_fraction. Currently at 4-11 TFLOPS. The bottleneck is likely that our kernels are memory-bound or not using tensor cores efficiently. For shape 1, FLOPs = 4*2*2048*8*32768 = 4.3 GFLOPs. At 0.398ms, that's 10.8 TFLOPS. To reach 20 TFLOPS, need ~0.215ms. Need ~1.85x speedup. Theoretical memory lower bound: bytes_formula = B*T*H*K*2 + ... = for shape 1: 2*2048*8*128*2 + 2*2048*8*128*2 + 2*2048*8*128*2 + 2*2048*8*128*4 + 2*2048*8*2 + 2*2048*8*128*2 = 2*2048*8*(128*2*4 + 128*4 + 2) ≈ 2*2048*8*128*12 = 50MB. At 1.8TB/s, memory time = 0.028ms. We are at 0.4ms, so we're not memory-bound. We're compute-bound or latency-bound. Actually, wait, with bf16 intermediates, we also read/write Aqk,w,u. So total memory is higher. But still, 0.4ms suggests ~100 GB/s effective? That seems low. Actually 50MB/0.4ms = 125GB/s. We're far from bandwidth limit. So the issue is compute efficiency / occupancy. The kernels aren't keeping the SMs busy. Kernel 1 with grid (NT=32, B*H=16) = 512 blocks. Each block does Aqk/M (2 GEMMs), solve, w,u (2 GEMMs). Should keep SMs busy. Kernel 2 with grid (V/BV=4, B*H=16) = 64 blocks. Each block loops 32-64 chunks sequentially. Only 64 blocks for 188 SMs. Low occupancy for shape 1. For shape 0, 64 blocks too. For shape 3, 4*4=16 blocks. Very low. This explains the low performance. Kernel 2 has too few blocks due to sequential chunk processing. To increase parallelism in kernel 2, we need to process chunks in parallel. That's the scan approach. Alternatively, we can have each block process multiple (B,H) or multiple V tiles? But each block already processes one V tile. We can reduce BV to 16 to get 8 V tiles, giving 128 blocks. But more loop overhead and register pressure. Actually, with BV=16, S is 128*16 = 8KB. More blocks. But each chunk does more small GEMMs. Might help occupancy. Let me try BV=16. But autotune didn't include it. Let me add. But the fundamental issue remains: sequential recurrence limits parallelism to B*H*(V/BV). For B=1,H=4,V=128,BV=16, blocks=32. Still low. The scan approach would give B*H*N*(V/BV) parallelism for the output computation. That's much better. Let me implement the parallel scan for kernel 2. Algorithm: 1. Compute Aqk, w, u in kernel 1 (same as now). 2. Compute P_n = diag(d_n) - k_decay_n.T @ w_n and Q_n = k_decay_n.T @ u_n for all chunks n in parallel. This is a kernel with grid (NT, B*H). For each chunk, P_n (K,K) and Q_n (K,V). 3. Parallel scan over chunks to compute cumulative operators (P_composite_n, Q_composite_n) where S_n = P_composite_n @ S_0 + Q_composite_n = Q_composite_n (since S_0=0). Wait, the operator composition: S_{n+1} = P_n @ S_n + Q_n. Starting S_0=0. S_1 = Q_0 S_2 = P_1 @ Q_0 + Q_1 S_3 = P_2 @ P_1 @ Q_0 + P_2 @ Q_1 + Q_2 So S_n = (P_{n-1} @ ... @ P_0) @ 0 + sum_{i=0}^{n-1} (P_{n-1} @ ... @ P_{i+1}) @ Q_i. For the output of chunk n, we need S_n (state before chunk n). 4. Compute o_n = qg_n @ S_n + Aqk_n @ (u_n - w_n @ S_n) in parallel for all n. But P_n is (K,K), and composing many P_n multiplications is expensive. However, we can do a parallel scan with log2(N) steps. Each step does matrix multiplies P @ P (K,K @ K,K = 2*K^3 = 4M FLOPs) and P @ Q (K,K @ K,V = 2*K^2*V = 4M FLOPs). Per chunk, per step. For N=64, log2(N)=6 steps, B*H*N = 1024 chunks (shape 1). Total FLOPs ~ 1024*6*8M = 49 GFLOPs. That's a lot! Might be slower than sequential. Wait, K=V=128. P@P is 2*128^3 = 4.2M FLOPs. P@Q is 2*128^2*128 = 4.2M FLOPs. Per operator pair, 8.4M FLOPs. Per step, we combine N/2 pairs? Actually in a scan, each step does O(N) compositions. So per step FLOPs = N * 8.4M = 64 * 8.4M = 537M. Times 6 steps = 3.2 GFLOPs for one (B,H). Times B*H=16 = 51 GFLOPs. At 200 TFLOPS, 0.25ms. Plus P,Q computation and output. Total maybe 0.4ms. Not clearly better than current 0.4ms. But the sequential approach is 0.4ms and simpler. The scan has overhead from many matrix multiplies. Might not win for N=64. Hmm. Maybe for larger N it wins, but our max N=64. Alternative: process chunks in parallel by batching the sequential recurrence across more blocks. For example, each block could handle one (B,H,V tile) but process multiple chunks in a vectorized way? No, dependency prevents it. Actually, maybe the sequential recurrence can be made faster by increasing work per block (larger BV) and reducing global memory traffic. Let me try a different angle: improve kernel 2 occupancy by using BV=16 (8 V tiles) and see if it helps. Also, try to fuse P,Q computation with kernel 1 and output computation with kernel 2, but keep sequential recurrence? Not helpful. Actually, wait. The benchmark time includes kernel 1 AND kernel 2 AND Python preprocessing. Let me see the breakdown. From earlier profile, kernel 1 ~172us, kernel 2 ~179us per call for shape 1. Total kernel ~350us. Benchmark reports 398us. Close. To reach 215us, I need to roughly halve kernel times. Both kernels need optimization. For kernel 1, the main cost is the Aqk/M GEMMs (2 of 64x64 @ 64x128 = 1M FLOPs each) and solve, w,u GEMMs. Total ~4M FLOPs per chunk. 512 chunks. At 200 TFLOPS, theoretical 0.01ms. We're at 0.17ms. So utilization is ~6%. Very low. Why? The block count 512 for shape 1 might be enough. But each block has small work. Maybe launch overhead or kernel underutilization. For kernel 2, 64 blocks sequential over 32 chunks. Each block has work ~32 * (several GEMMs). Utilization low because only 64 blocks on 188 SMs. Let me check if using more warps helps. With num_warps=16, each block has more threads but same work. Might improve occupancy. Actually, the issue might be that the kernels are not using tensor cores efficiently. The GEMMs are small (64x64 @ 64x128). Tensor cores prefer larger tiles. Triton should handle this, but maybe not optimally. Another idea: in kernel 1, instead of computing Aqk and M separately with two GEMMs, we can compute them together by packing q and k. For each K tile, compute (q; k) * exp(diff) and kdecay_norm, then Aqk and M from the same multiply. Actually, [q;k] is (2*BT, K), kdecay_norm is (BT, K). tl.dot gives (2*BT, BT). Then split into Aqk and M. This halves the K tile loads and might fuse better. Similarly for w,u: compute A @ [kg; v_tile] -> (BT, K+V). One GEMM instead of two. But v_tile depends on V offset, while kg doesn't. We can still pack for each V tile. Let me try packing in kernel 1: - For Aqk/M: pack q and k vertically: qk_pack = [q; k] shape (2*BT, K). Then dot with kdecay_norm.T (K, BT) gives (2*BT, BT). Split into Aqk (top BT rows) and M (bottom BT rows). - For w,u: for each V tile, pack kg and v_tile vertically: kv_pack = [kg; v_tile] shape (2*BT, max(K,BV)). But dimensions differ. We can pad or use separate. Actually, for w,u, we can compute w = A @ kg once (full K), and u = A @ v_tile per V tile. Can't easily pack because K and V are both 128. We could pack kg and v (first BV cols) but then still need remaining V cols. Actually, since K=V=128, we can pack kg and the full v into (2*BT, K). Then A @ pack -> (BT, K) for w and (BT, K) for u. But A is (BT,BT), pack is (2*BT, K). The dot isn't directly supported because the first dimension of pack (2*BT) is the reduction dimension for A's column. A is (BT,BT), pack is (2*BT,K). A @ pack requires BT = 2*BT? No. Wait, w = A @ kg where A (BT,BT), kg (BT,K). u = A @ v where v (BT,V). If we stack kg and v horizontally? A (BT,BT) @ [kg|v] (BT, K+V) -> (BT, K+V). This works! So one GEMM computes both w and u concatenated. But v is full V=128. So output is (BT, 256). We split into w (first 128 cols) and u (last 128 cols). This requires v to be available, and kg. kg = k*exp(g). We can compute kg and concatenate with v in the kernel. But v is bf16 and kg is fp32. We can cast v to fp32 and concatenate. Or keep v bf16 and cast during packing. This reduces w,u computation from 2 GEMMs to 1 GEMM per chunk. Might help. Actually, more importantly, it reduces A reads: A is read twice (for w and u). With one GEMM, A is read once. Similarly for Aqk/M: pack q and k vertically? For Aqk = qg_norm @ kdecay_norm.T, qg_norm (BT,K) @ kdecay_norm.T (K,BT) -> (BT,BT). M = kg_norm @ kdecay_norm.T -> (BT,BT). We can stack qg_norm and kg_norm vertically to (2*BT, K), then dot with kdecay_norm.T -> (2*BT, BT). Top half is Aqk, bottom half is M. This halves the kdecay_norm reads and the dot operations. Let me implement this in kernel 1. For the packed dot, `tl.dot(qk_pack, tl.trans(kdecay_norm))` where qk_pack is (2*BT, K). But K is split into BK tiles. With BK=64 and K=128, we accumulate over 2 tiles. qk_pack is (128, 64) per tile? Actually qk_pack has shape (2*BT, BK). tl.dot with (BK, BT) gives (2*BT, BT). Good. But 2*BT = 128, BK=64. The dot is 128x64 @ 64x64. Triton should handle. Similarly for w,u: pack kg and v horizontally: [kg | v] shape (BT, K+V) = (64, 256). With BK=64, we process K and V separately. For K tile, load kg; for V tile, load v. Can't easily pack across full K+V if BK is per K and BV per V. Wait, for the w,u combined GEMM, A (BT,BT) @ [kg|v] (BT, K+V). The second operand has 256 cols. We can split into K tile and V tile: A @ kg (cols 0:128) and A @ v (cols 128:256). That's still 2 GEMMs. Unless we load both kg and v tile together. For the first V tile (cols 0:64), we can pack with kg cols 0:64 -> (BT, 128). Then A @ pack -> (BT, 128), split into w cols 0:64 and u cols 0:64. For the second V tile (cols 64:128), pack with kg cols 64:128. This way, each GEMM computes part of w and part of u simultaneously. So for each K/V tile pair (offset same), pack [kg_tile | v_tile] horizontally. A @ pack -> (BT, BK+BV). Split into w_tile (first BK cols) and u_tile (last BV cols). This halves the A reads and dot operations for w,u. Let me implement. But the packing needs v_tile and kg_tile. kg_tile = k_tile * exp(g_tile). v_tile loaded from v. Pack horizontally: `pack = tl.join(kg_tile, v_tile)`? Triton has `tl.cat`? Let me check. Actually, we can use `tl.concatenate([kg_tile, v_tile], axis=1)` if supported. Triton 3.6 might have `tl.cat` or `tl.join`. Alternatively, we can just do two separate dots. The packing overhead might not be worth it. Actually, the main bottleneck is likely kernel 2, not kernel 1. Let me focus on kernel 2. For kernel 2, the issue is low block count. Let me try BV=16 to double blocks. Or even BV=8. With BV=16, S is 128*16 = 8KB. Grid = 8 * B*H. For shape 1: 8*16=128 blocks. Better occupancy. Let me add BV=16 configs and see. Actually, autotune already has BV=32,64. Let me add BV=16. But register pressure for S0,S1 (each 64x16=4KB) is low. Good. But the loop over K tiles still uses BK=64 or 128. With BK=64, S0,S1 each 4KB. Fine. Let me add BV=16 configs.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (17 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -154,10 +154,14 @@
 
 @triton.autotune(
     configs=[
+        triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
         triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<690 chars \u2014 see diff>",
  "new_string": "<980 chars \u2014 see diff>"
}
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tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (190 chars)
Wait, the else branch in kernel 2 assumes K=128 and BK=64 (two tiles S0,S1). For BK=128, it uses single S. For K=128, BK can be 64 or 128. Good. But BV=16/32/64 are all fine. Run benchmark.
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assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check with BV=16 configs"
}
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subagent · 7 events · 2 tools
tool
stdout (283 chars)
Command running in background with ID: bb1rjtys5. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bb1rjtys5.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (5 chars)
Wait.
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assistant
TaskOutput
args
{
  "task_id": "bb1rjtys5",
  "block": true,
  "timeout": 600000
}
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tool
stdout (188 chars)
<retrieval_status>success</retrieval_status>

<task_id>bb1rjtys5</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
PASS
</output>
assistant
reasoning (20 chars)
PASS. Now benchmark.
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assistant
Bash
args
{
  "command": "uv run python benchmark.py 2>&1 | tail -25",
  "description": "Benchmark with BV=16 configs"
}
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system — task_started
↳ subagent started
tool
stdout (1304 chars)
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:18:35.558862+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:18:37.990641+00:00 elapsed_s=2.432 ms=0.224720
shape=0 variant=solution tflops=9.556 gbps=112.133 ms=0.225
shape=0 solution_peak_fraction=0.0478
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:18:38.160154+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:18:40.484823+00:00 elapsed_s=2.325 ms=0.390784
shape=1 variant=solution tflops=10.991 gbps=128.964 ms=0.391
shape=1 solution_peak_fraction=0.0550
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:18:40.653081+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:18:43.027648+00:00 elapsed_s=2.375 ms=0.510176
shape=2 variant=solution tflops=8.419 gbps=98.784 ms=0.510
shape=2 solution_peak_fraction=0.0421
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:18:43.071699+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:18:45.332350+00:00 elapsed_s=2.261 ms=0.247440
shape=3 variant=solution tflops=4.339 gbps=50.919 ms=0.247
shape=3 solution_peak_fraction=0.0217
peak_fraction: 0.0394
RESULT: LOW
assistant
reasoning (1896 chars)
0.0394, slightly better. Still far from 0.1. I need a fundamentally different approach. Let me consider using a persistent kernel for the recurrence that processes all V tiles within one block, avoiding the V tile grid limitation. Persistent kernel grid: (B*H) only. Each block loops over all chunks and all V tiles. S is (K,V) = 64KB. This avoids storing w,u,Aqk in global memory? No, we still compute them in kernel 1. Wait, to avoid global memory round-trip, we need to fuse intra and inter. But that has redundancy if we use V-tile grid. Persistent kernel with grid (B*H) and all V in one block: no redundancy, but only B*H blocks. For shape 1, 16 blocks. Very low occupancy. Worse. What if we use a persistent kernel with grid (B*H*N) and compute output in parallel using scan? That's the scan approach. Let me think about the scan approach more carefully. Maybe it's not as expensive as I estimated. Actually, for the recurrence S_{n+1} = P_n @ S_n + Q_n, we need S_n for each chunk n to compute output. We can do a parallel scan to get all S_n. But we can also compute outputs without getting all S_n explicitly, by associating the output computation with the scan. Actually, the expensive part is the P@P compositions (4M FLOPs each). But we can note that P_n = diag(d_n) - k_decay_n.T @ w_n. For our problem, P_n might be close to diagonal, and maybe we can approximate? No, need exact. Alternative: don't use scan, but use block-level sequential processing with each block handling one chunk but multiple chunks batched? No dependency. Hmm. Let me look at how FLA handles this. Their chunk_gated_delta_rule_fwd_h kernel is sequential per (B,H,V tile) block. They achieve good performance by using persistent kernels and tensor cores. Maybe my implementation is just not as optimized. Let me look at FLA's kernel more carefully, especially the inter-chunk recurrence kernel.
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assistant
Read /workspace/fla_ref/fla/ops/common/chunk_delta_h.py
args
{
  "file_path": "/workspace/fla_ref/fla/ops/common/chunk_delta_h.py",
  "limit": 300
}
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tool
stdout (13942 chars) — TRUNCATED
1	# Copyright (c) 2023-2026, Songlin Yang, Yu Zhang, Zhiyuan Li
2	#
3	# This source code is licensed under the MIT license found in the
4	# LICENSE file in the root directory of this source tree.
5	# For a list of all contributors, visit:
6	#   https://github.com/fla-org/flash-linear-attention/graphs/contributors
7	
8	import torch
9	import triton
10	import triton.language as tl
11	
12	from fla.ops.backends import dispatch
13	from fla.ops.utils import prepare_chunk_indices, prepare_chunk_offsets
14	from fla.ops.utils.cache import fla_cache_autotune
15	from fla.ops.utils.op import exp2
16	from fla.utils import IS_NVIDIA_HOPPER, USE_CUDA_GRAPH, autotune_cache_kwargs, check_shared_mem
17	
18	NUM_WARPS = [2, 4] if IS_NVIDIA_HOPPER else [2, 4, 8, 16]
19	
20	
21	@triton.heuristics({
22	    'USE_G': lambda args: args['g'] is not None,
23	    'USE_GK': lambda args: args['gk'] is not None,
24	    'USE_INITIAL_STATE': lambda args: args['h0'] is not None,
25	    'STORE_FINAL_STATE': lambda args: args['ht'] is not None,
26	    'SAVE_NEW_VALUE': lambda args: args['v_new'] is not None,
27	    'IS_VARLEN': lambda args: args['cu_seqlens'] is not None,
28	})
29	@fla_cache_autotune(
30	    configs=[
31	        triton.Config({'BV': BV}, num_warps=num_warps, num_stages=num_stages)
32	        for num_warps in [2, 4]
33	        for num_stages in ([2, 3, 4] if check_shared_mem('ampere') else [2, 1])
34	        for BV in ([32, 64] if check_shared_mem('ada') else [32])
35	    ],
36	    key=['H', 'HV',

... (12442 more chars)
assistant
reasoning (2762 chars)
FLA's recurrence kernel is similar to mine but uses K segments of 64. They have STATE_V_FIRST option which transposes S (V,K) instead of (K,V). They use exp2 and g_last - g for v scaling (same as my k_update). They do S update with k @ v. They also save S to h buffer per chunk, and compute v_new. Then output kernel uses h and v_new. Key difference: they use bfloat16 for w and k loads, and tl.dot with b_h (fp32) cast to bf16. This uses tensor cores. They don't use fp32 for the recurrence GEMMs. This is faster. In my kernel 2, I load w as fp32 and do fp32 dot. Triton might not use tensor cores for fp32 dot, falling back to FMA. That's much slower! This is likely the main issue. I should use bf16 tensor cores for the recurrence GEMMs. Specifically: - w is bf16 (stored). Load as bf16, dot with S (fp32 cast to bf16) -> fp32 accumulator. - qg is bf16, dot with S (bf16) -> fp32. - Aqk is bf16, dot with b_v (bf16) -> fp32. - k is bf16, dot with b_v (bf16) -> fp32 for S update. This uses bf16 tensor cores and is much faster. Let me change kernel 2 to use bf16 for the dots. S remains fp32 but is cast to bf16 for dot. The accumulator is fp32. Actually, FLA's code: `b_v = tl.dot(b_w, b_h1.to(b_w.dtype))` where b_w is bf16, b_h1 is fp32 cast to bf16. Output b_v is fp32. Good. Similarly: `b_h1 += tl.dot(b_k, b_v)` where b_k is bf16, b_v is fp32? Wait, they do `b_v = b_v.to(k.dtype.element_ty)` before the k dot. So b_v is bf16, b_k is bf16, dot accumulates fp32 into b_h1. So for S update, they use bf16 b_v. This loses some precision but is faster. For my kernel, I should: - Load w as bf16, dot with S.to(bf16) -> fp32 b_v. - Load qg as bf16, dot with S.to(bf16) -> fp32 b_o. - Load Aqk as bf16, dot with b_v.to(bf16) -> fp32 b_o. - For S update: load k as bf16, b_v cast to bf16, dot -> fp32 dS. This should use tensor cores and be much faster. But accuracy might suffer. We have tolerance 0.05. Let me try. Actually, currently kernel 2 already loads w as fp32 because I changed it. Let me revert to bf16 loads and bf16 dots. Wait, in my current kernel 2, I have: ```python b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.float32) b_v -= tl.dot(b_w0, S0) ``` This is fp32 dot. I should change to: ```python b_w0 = tl.load(p_w0, boundary_check=(0, 1)) # bf16 b_v -= tl.dot(b_w0, S0.to(tl.bfloat16)) ``` And for S update: ```python b_k0 = tl.load(p_k0, boundary_check=(0, 1)) # bf16 S0 += tl.dot(tl.trans(k_up0.to(tl.bfloat16)), b_v.to(tl.bfloat16)) ``` But k_up0 is computed in fp32 (k * exp(...)). Cast to bf16 for dot. Actually, k can be loaded as bf16, and exp computed in fp32, then k_up cast to bf16. Let me update kernel 2 to use bf16 dots. Also, Aqk,w,u are stored as bf16 now, so loads return bf16. Let me edit kernel 2.
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assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 160,
  "offset": 195
}
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tool
stdout (6734 chars) — TRUNCATED
195	
196	    w_ptr += (i_b * T * H + i_h) * K
197	    u_ptr += (i_b * T * H + i_h) * V
198	    qg_ptr += (i_b * T * H + i_h) * K
199	    k_ptr += (i_b * T * H + i_h) * K
200	    g_ptr += (i_b * T * H + i_h) * K
201	    Aqk_ptr += (i_b * T * H + i_h) * BT
202	    o_ptr += (i_b * T * H + i_h) * V
203	
204	    off_v0 = i_vb * BV
205	    NT = T // BT
206	
207	    # For K=128 we have either one BK=128 tile or two BK=64 tiles.
208	    if tl.constexpr(K == 128 and BK == 128):
209	        S = tl.zeros([BK, BV], dtype=tl.float32)
210	        for i_n in range(NT):
211	            t0 = i_n * BT
212	
213	            p_u = tl.make_block_ptr(
214	                u_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
215	            )
216	            b_u = tl.load(p_u, boundary_check=(0, 1)).to(tl.float32)
217	            b_v = b_u
218	
219	            p_w = tl.make_block_ptr(
220	                w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
221	            )
222	            b_w = tl.load(p_w, boundary_check=(0, 1)).to(tl.float32)
223	            b_v -= tl.dot(b_w, S)
224	
225	            b_o = tl.zeros([BT, BV], dtype=tl.float32)
226	            p_qg = tl.make_block_ptr(
227	                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
228	            )
229	            b_qg = tl.load(p_qg, boundary_check=(0, 1)).to(tl.float32)
230	            b_o += tl.dot(b_qg, S)
231	
232	            p_Aqk = tl.make_block_ptr(
233	                Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), 

... (5234 more chars)
assistant
reasoning (1158 chars)
Let me rewrite kernel 2 to use bf16 dots. I'll replace the whole kernel 2 for both branches. Actually, let me simplify: since K=128, and autotune tries BK=64/128, I'll keep both branches but change dots to bf16. For the BK=128 branch: ```python b_u = tl.load(p_u).to(tl.float32) b_v = b_u b_w = tl.load(p_w) # bf16 b_v -= tl.dot(b_w, S.to(tl.bfloat16)) # fp32 acc b_qg = tl.load(p_qg) # bf16 b_o += tl.dot(b_qg, S.to(tl.bfloat16)) b_Aqk = tl.load(p_Aqk) # bf16 b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16)) ... b_k = tl.load(p_k) # bf16 b_g = tl.load(p_g).to(tl.float32) k_up = b_k.to(tl.float32) * exp(...) S = S * d S += tl.dot(tl.trans(k_up.to(tl.bfloat16)), b_v.to(tl.bfloat16)) ``` Wait, b_k is bf16. To compute k_up = k * exp(g_last - g), we can cast b_k to fp32 and multiply by exp. Then cast k_up to bf16 for dot. Alternatively, compute exp factor in fp32 and multiply with bf16 b_k directly? Triton might do bf16 * fp32 -> bf16? Not sure. Better: `k_up = (b_k.to(tl.float32) * tl.exp(...)).to(tl.bfloat16)`. For S update, we need k_up.T @ b_v. Both bf16. Dot accumulates fp32. Similarly for BK=64 branch. Let me do the replacement with Edit.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (125 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -219,21 +219,21 @@
             p_w = tl.make_block_ptr(
                 w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
-            b_w = tl.load(p_w, boundary_check=(0, 1)).to(tl.float32)
-            b_v -= tl.dot(b_w, S)
+            b_w = tl.load(p_w, boundary_check=(0, 1))
+            b_v -= tl.dot(b_w, S.to(tl.bfloat16))
 
             b_o = tl.zeros([BT, BV], dtype=tl.float32)
             p_qg = tl.make_block_ptr(
                 qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
-            b_qg = tl.load(p_qg, boundary_check=(0, 1)).to(tl.float32)
-            b_o += tl.dot(b_qg, S)
+            b_qg = tl.load(p_qg, boundary_check=(0, 1))
+            b_o += tl.dot(b_qg, S.to(tl.bfloat16))
 
             p_Aqk = tl.make_block_ptr(
                 Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
             )
-            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.float32)
-            b_o += tl.dot(b_Aqk, b_v)
+            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
+            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))
 
             p_o = tl.make_block_ptr(
                 o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
@@ -246,15 +246,15 @@
             p_k = tl.make_block_ptr(
                 k_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
-            b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+            b_k = tl.load(p_k, boundary_check=(0, 1))
             p_g = tl.make_block_ptr(
                 g_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
-            k_up = b_k * tl.exp(tl.log(d)[None, :] - b_g)
+            k_up = (b_k.to(tl.float32) * tl.exp(tl.log(d)[None, :] - b_g)).to(tl.bfloat16)
 
             S = S * d[:, None]
-            S += tl.dot(tl.trans(k_up), b_v)
+            S += tl.dot(tl.trans(k_up), b_v.to(tl.bfloat16))
     else:
         S0 = tl.zeros([BK, BV], dtype=tl.float32)
         S1 = tl.zeros([BK, BV], dtype=tl.float32)
@@ -270,31 +270,31 @@
             p_w0 = tl.make_block_ptr(
                 w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
-            b_w0 = tl.load(p_w0, boundary_check=(0, 1)).to(tl.float32)
-            b_v -= tl.dot(b_w0, S0)
+            b_w0 = tl.load(p_w0, boundary_check=(0, 1))
+            b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))
             p_w1 = tl.make_block_ptr(
                 w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
             )
-            b_w1 = tl.load(p_w1, boundary_check=(0, 1)).to(tl.float32)
-            b_v -= tl.dot(b_w1, S1)
+            b_w1 = tl.load(p_w1, boundary_check=(0, 1))
+            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))
 
             b_o = tl.zeros([BT, BV], dtype=tl.float32)
             p_qg0 = tl.make_block_ptr(
                 qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
-            b_qg0 = tl.load(p_qg0, boundary_check=(0, 1)).to(tl.float32)
-            b_o += tl.dot(b_qg0, S0)
+            b_qg0 = tl.load(p_qg0, boundary_check=(0, 1))
+            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))
             p_qg1 = tl.make_block_ptr(
                 qg_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
             )
-            b_qg1 = tl.load(p_qg1, boundary_check=(0, 1)).to(tl.float32)
-            b_o += tl.dot(b_qg1, S1)
+            b_qg1 = tl.load(p_qg1, boundary_check=(0, 1))
+            b_o += tl.dot(b_qg1, S1.to(tl.bfloat16))
 
             p_Aqk = tl.make_block_ptr(
                 Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
             )
-            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1)).to(tl.float32)
-            b_o += tl.dot(b_Aqk, b_v)
+            b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
+            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))
 
             p_o = tl.make_block_ptr(
                 o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
@@ -309,27 +309,27 @@
             p_k0 = tl.make_block_ptr(
                 k_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
-            b_k0 = tl.load(p_k0, boundary_check=(0, 1)).to(tl.float32)
+            b_k0 = tl.load(p_k0, boundary_check=(0, 1))
             p_g0 = tl.make_block_ptr(
                 g_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_g0 = tl.load(p_g0, boundary_check=(0, 1)).to(tl.float32)
-            k_up0 = b_k0 * tl.exp(tl.log(d0)[None, :] - b_g0)
+            k_up0 = (b_k0.to(tl.float32) * tl.exp(tl.log(d0)[None, :] - b_g0)).to(tl.bfloat16)
 
             p_k1 = tl.make_block_ptr(
                 k_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
             )
-            b_k1 = tl.load(p_k1, boundary_check=(0, 1)).to(tl.float32)
+            b_k1 = tl.load(p_k1, boundary_check=(0, 1))
             p_g1 = tl.make_block_ptr(
                 g_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
             )
             b_g1 = tl.load(p_g1, boundary_check=(0, 1)).to(tl.float32)
-            k_up1 = b_k1 * tl.exp(tl.log(d1)[None, :] - b_g1)
+            k_up1 = (b_k1.to(tl.float32) * tl.exp(tl.log(d1)[None, :] - b_g1)).to(tl.bfloat16)
 
             S0 = S0 * d0[:, None]
             S1 = S1 * d1[:, None]
-            S0 += tl.dot(tl.trans(k_up0), b_v)
-            S1 += tl.dot(tl.trans(k_up1), b_v)
+            S0 += tl.dot(tl.trans(k_up0), b_v.to(tl.bfloat16))
+            S1 += tl.dot(tl.trans(k_up1), b_v.to(tl.bfloat16))
 
 
 class Model(nn.Module):
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<5079 chars \u2014 see diff>",
  "new_string": "<5189 chars \u2014 see diff>"
}
in 235980out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (28 chars)
Now run check and benchmark.
in 240435out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10 && uv run python benchmark.py 2>&1 | tail -20",
  "description": "Check and benchmark bf16 dots"
}
in 240435out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (1497 chars)
b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))

            b_o = tl.zeros([BT, BV], dtype=tl.float32)
            p_qg0 = tl.make_block_ptr(
                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_qg0 = tl.load(p_qg0, boundary_check=(0, 1))
            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))
                   ^
Both operands must be same dtype. Got fp32 and bf16
    module = src.make_ir(target, options, codegen_fns, module_map, context)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 80, in make_ir
    return ast_to_ttir(self.fn, self, context=context, options=options, codegen_fns=codegen_fns,
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
triton.compiler.errors.CompilationError: at 114:19:
            p_w1 = tl.make_block_ptr(
                w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
            )
            b_w1 = tl.load(p_w1, boundary_check=(0, 1))
            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))

            b_o = tl.zeros([BT, BV], dtype=tl.float32)
            p_qg0 = tl.make_block_ptr(
                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_qg0 = tl.load(p_qg0, boundary_check=(0, 1))
            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))
                   ^
Both operands must be same dtype. Got fp32 and bf16
assistant
reasoning (3109 chars)
b_o is fp32, but b_qg0 is bf16 and S0.to(bf16) is bf16. tl.dot expects both operands same dtype, but accumulator can be different? Actually the error says both operands must be same dtype. The accumulator b_o is fp32. But tl.dot with bf16 inputs and fp32 output should work. The issue is that b_o is the accumulator and is fp32, while inputs are bf16. Triton should allow this. Wait, the error is "Both operands must be same dtype. Got fp32 and bf16". It seems tl.dot is interpreting b_o as an operand. Actually, `b_o += tl.dot(...)` is `b_o = b_o + tl.dot(...)`. The `tl.dot` itself has two operands (b_qg0, S0_bf16). Both are bf16. So why fp32 and bf16? Maybe because b_o is fp32 and the `+=` tries to add tl.dot result (fp32) to b_o (fp32). That should be fine. Wait, the line is `b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))`. Maybe Triton is checking the accumulator type against operand types? In some Triton versions, tl.dot requires the accumulator to match? No. Actually, the error message says fp32 and bf16. Maybe one of the operands is fp32. b_qg0 is loaded without `.to()`, but qg_ptr points to bf16 tensor. So b_qg0 is bf16. S0.to(bf16) is bf16. Hmm. Wait, maybe `S0.to(tl.bfloat16)` returns fp32 because S0 is fp32 and cast to bf16? No, .to(bf16) returns bf16. Oh! Maybe the issue is that `b_o += tl.dot(...)` and b_o is fp32. Triton might require the accumulator variable to be the same dtype as the dot operands for in-place addition? Let me try using a separate fp32 accumulator and add to b_o: ```python b_o += tl.dot(b_qg0, S0.to(tl.bfloat16)).to(tl.float32) ``` Actually tl.dot of bf16 returns fp32 (accumulator). So `.to(tl.float32)` is redundant. But maybe Triton needs explicit cast. Let me try `b_o += tl.dot(b_qg0, S0.to(tl.bfloat16)).to(tl.float32)`. But the error says "Both operands must be same dtype. Got fp32 and bf16" which suggests inside tl.dot. Let me check the line number. The error points to `b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))`. Maybe Triton is inferring the accumulator from b_o (fp32) and requiring operands to be fp32 too? Actually, in Triton, `tl.dot(a, b)` when used in an accumulation context might use the accumulator type. For bf16 inputs, default accumulator is fp32. But if the accumulation variable is fp32, it might not match bf16 operands. I think the issue is that I should not cast S to bf16; instead, cast both operands to bf16 explicitly, or use fp32 for both. Wait, FLA does: `b_v = tl.dot(b_w, tl.trans(b_h1).to(b_w.dtype))` where b_w is bf16, b_h1 is fp32. This works. So the cast is fine. But in my code, `b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))`. b_o is fp32. The result of tl.dot is fp32. Adding to b_o is fp32 + fp32. Why error? Maybe because Triton can't infer the dot accumulator when used in `+=`. Let me try `acc = tl.dot(b_qg0, S0.to(tl.bfloat16)); b_o += acc`. Or maybe I need to cast b_qg0 to bf16 explicitly: `b_qg0 = tl.load(...).to(tl.bfloat16)`. But it should already be bf16. Actually, b_qg0 is loaded from qg_ptr which is a bf16 tensor. `tl.load` returns bf16. Good. Let me try the explicit accumulator approach.
in 240977out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -227,13 +227,15 @@
                 qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_qg = tl.load(p_qg, boundary_check=(0, 1))
-            b_o += tl.dot(b_qg, S.to(tl.bfloat16))
+            acc_qg = tl.dot(b_qg, S.to(tl.bfloat16))
+            b_o += acc_qg
 
             p_Aqk = tl.make_block_ptr(
                 Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
             )
             b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
-            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))
+            acc_Aqk = tl.dot(b_Aqk, b_v.to(tl.bfloat16))
+            b_o += acc_Aqk
 
             p_o = tl.make_block_ptr(
                 o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<522 chars \u2014 see diff>",
  "new_string": "<580 chars \u2014 see diff>"
}
in 240977out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (353 chars)
Same for the else branch and for the w dot. Let me also fix those. Actually, the w dot also had `b_v -= tl.dot(b_w, S.to(tl.bfloat16))`. Let me fix that too. And the else branch has similar patterns. Let me rewrite the else branch dots with explicit accumulators. Actually, let me first see if the BK=128 branch compiles now. Then fix else if needed.
in 242401out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -30",
  "description": "Test explicit accumulators"
}
in 242401out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (1852 chars) — TRUNCATED
fn()
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/autotuner.py", line 150, in kernel_call
    self.fn.run(
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 720, in run
    kernel = self._do_compile(key, signature, device, constexprs, options, attrs, warmup)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/runtime/jit.py", line 849, in _do_compile
    kernel = self.compile(src, target=target, options=options.__dict__)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 304, in compile
    module = src.make_ir(target, options, codegen_fns, module_map, context)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 80, in make_ir
    return ast_to_ttir(self.fn, self, context=context, options=options, codegen_fns=codegen_fns,
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
triton.compiler.errors.CompilationError: at 116:19:
            p_w1 = tl.make_block_ptr(
                w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
            )
            b_w1 = tl.load(p_w1, boundary_check=(0, 1))
            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))

         

... (352 more chars)
assistant
reasoning (308 chars)
Still error in else branch. I need to fix all `+= tl.dot` patterns. Let me rewrite kernel 2 entirely for both branches, using explicit accumulators. Actually, a simpler fix: cast the result of tl.dot to fp32 explicitly: ```python b_o += tl.dot(b_qg0, S0.to(tl.bfloat16)).to(tl.float32) ``` Let me try this.
in 243035out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -227,15 +227,13 @@
                 qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_qg = tl.load(p_qg, boundary_check=(0, 1))
-            acc_qg = tl.dot(b_qg, S.to(tl.bfloat16))
-            b_o += acc_qg
+            b_o += tl.dot(b_qg, S.to(tl.bfloat16)).to(tl.float32)
 
             p_Aqk = tl.make_block_ptr(
                 Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
             )
             b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
-            acc_Aqk = tl.dot(b_Aqk, b_v.to(tl.bfloat16))
-            b_o += acc_Aqk
+            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16)).to(tl.float32)
 
             p_o = tl.make_block_ptr(
                 o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<580 chars \u2014 see diff>",
  "new_string": "<552 chars \u2014 see diff>"
}
in 243035out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (70 chars)
Now fix else branch similarly. Let me replace all dots in else branch.
in 243638out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (38 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -271,30 +271,30 @@
                 w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_w0 = tl.load(p_w0, boundary_check=(0, 1))
-            b_v -= tl.dot(b_w0, S0.to(tl.bfloat16))
+            b_v -= tl.dot(b_w0, S0.to(tl.bfloat16)).to(tl.float32)
             p_w1 = tl.make_block_ptr(
                 w_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
             )
             b_w1 = tl.load(p_w1, boundary_check=(0, 1))
-            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16))
+            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16)).to(tl.float32)
 
             b_o = tl.zeros([BT, BV], dtype=tl.float32)
             p_qg0 = tl.make_block_ptr(
                 qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_qg0 = tl.load(p_qg0, boundary_check=(0, 1))
-            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16))
+            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16)).to(tl.float32)
             p_qg1 = tl.make_block_ptr(
                 qg_ptr, (T, K), (H * K, 1), (t0, BK), (BT, BK), (1, 0)
             )
             b_qg1 = tl.load(p_qg1, boundary_check=(0, 1))
-            b_o += tl.dot(b_qg1, S1.to(tl.bfloat16))
+            b_o += tl.dot(b_qg1, S1.to(tl.bfloat16)).to(tl.float32)
 
             p_Aqk = tl.make_block_ptr(
                 Aqk_ptr, (T, BT), (H * BT, 1), (t0, 0), (BT, BT), (1, 0)
             )
             b_Aqk = tl.load(p_Aqk, boundary_check=(0, 1))
-            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16))
+            b_o += tl.dot(b_Aqk, b_v.to(tl.bfloat16)).to(tl.float32)
 
             p_o = tl.make_block_ptr(
                 o_ptr, (T, V), (H * V, 1), (t0, off_v0), (BT, BV), (1, 0)
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1222 chars \u2014 see diff>",
  "new_string": "<1297 chars \u2014 see diff>"
}
in 243638out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (36 chars)
Also fix the w dot in BK=128 branch.
in 244729out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (11 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -220,7 +220,7 @@
                 w_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
             )
             b_w = tl.load(p_w, boundary_check=(0, 1))
-            b_v -= tl.dot(b_w, S.to(tl.bfloat16))
+            b_v -= tl.dot(b_w, S.to(tl.bfloat16)).to(tl.float32)
 
             b_o = tl.zeros([BT, BV], dtype=tl.float32)
             p_qg = tl.make_block_ptr(
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<223 chars \u2014 see diff>",
  "new_string": "<238 chars \u2014 see diff>"
}
in 244729out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (10 chars)
Now check.
in 245004out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check bf16 dots"
}
in 245004out 0cache_r 0cache_w 0
subagent · 17 events · 5 tools
tool
stdout (283 chars)
Command running in background with ID: bxa2lb6zz. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bxa2lb6zz.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (5 chars)
Wait.
in 245151out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "bxa2lb6zz",
  "block": true,
  "timeout": 600000
}
in 245151out 0cache_r 0cache_w 0
tool
stdout (628 chars)
<retrieval_status>success</retrieval_status>

<task_id>bxa2lb6zz</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
            b_v -= tl.dot(b_w1, S1.to(tl.bfloat16)).to(tl.float32)

            b_o = tl.zeros([BT, BV], dtype=tl.float32)
            p_qg0 = tl.make_block_ptr(
                qg_ptr, (T, K), (H * K, 1), (t0, 0), (BT, BK), (1, 0)
            )
            b_qg0 = tl.load(p_qg0, boundary_check=(0, 1))
            b_o += tl.dot(b_qg0, S0.to(tl.bfloat16)).to(tl.float32)
                   ^
Both operands must be same dtype. Got fp32 and bf16
</output>
assistant
reasoning (1096 chars)
Still the same error. The `.to(tl.float32)` after tl.dot doesn't help because the error is inside tl.dot. Wait, the error says "Both operands must be same dtype. Got fp32 and bf16". The operands are b_qg0 and S0.to(tl.bfloat16). One must be fp32. b_qg0 is loaded from qg_ptr. qg_ptr points to qg which is computed in Python as `qg = q_scaled * g_cum.exp()`. q_scaled is bf16, g_cum is fp32, exp is fp32. The result `qg` will be... bf16? Actually, in PyTorch, bf16 * fp32 -> fp32. So qg is fp32! Even though I cast q to bf16, the multiplication with fp32 g_cum.exp() gives fp32. So qg is fp32, and kernel loads it as fp32. Then tl.dot with S0.to(bf16) has fp32 and bf16 operands. I need qg to be bf16. In Python, cast to bf16: `qg = (q_scaled * g_cum.exp()).to(torch.bfloat16)`. Similarly, k is bf16 but in kernel 1 I compute `b_k * exp(b_g)` where b_k is bf16 and b_g is fp32. The result b_kg is fp32 (bf16 * fp32 -> fp32). Then stored as bf16. OK. For kernel 2, qg should be bf16. Let me fix. Also, k in Python is bf16. Good. g_cum is fp32. Good. Let me update forward to cast qg to bf16.
in 245573out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (11 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -365,7 +365,7 @@
         g_cum = g_4d.cumsum(dim=2).view(B, T, H, K)
 
         q_scaled = q * self.scale
-        qg = q_scaled * g_cum.exp()
+        qg = (q_scaled * g_cum.exp()).to(torch.bfloat16)
 
         g_mid = g_cum.view(B, NT, BT, H, K)[:, :, BT // 2, :, :].permute(0, 2, 1, 3).contiguous()
         glast = g_cum.view(B, NT, BT, H, K)[:, :, -1, :, :].permute(0, 2, 1, 3).contiguous()
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<69 chars \u2014 see diff>",
  "new_string": "<90 chars \u2014 see diff>"
}
in 245573out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (28 chars)
Now qg is bf16. Check again.
in 246210out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check qg bf16"
}
in 246210out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (283 chars)
Command running in background with ID: bo9i4hvyp. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bo9i4hvyp.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (5 chars)
Wait.
in 246366out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "bo9i4hvyp",
  "block": true,
  "timeout": 600000
}
in 246366out 0cache_r 0cache_w 0
tool
stdout (188 chars)
<retrieval_status>success</retrieval_status>

<task_id>bo9i4hvyp</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
PASS
</output>
assistant
reasoning (20 chars)
PASS. Now benchmark.
in 246638out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python benchmark.py 2>&1 | tail -25",
  "description": "Benchmark bf16 tensor core dots"
}
in 246638out 0cache_r 0cache_w 0
system — task_started
↳ subagent started
tool
stdout (1306 chars)
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:30:25.758125+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:30:28.172481+00:00 elapsed_s=2.414 ms=0.209664
shape=0 variant=solution tflops=10.243 gbps=120.186 ms=0.210
shape=0 solution_peak_fraction=0.0512
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:30:28.338856+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:30:30.638473+00:00 elapsed_s=2.300 ms=0.363296
shape=1 variant=solution tflops=11.822 gbps=138.722 ms=0.363
shape=1 solution_peak_fraction=0.0591
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:30:30.804813+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:30:33.145278+00:00 elapsed_s=2.340 ms=0.453616
shape=2 variant=solution tflops=9.468 gbps=111.101 ms=0.454
shape=2 solution_peak_fraction=0.0473
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:30:33.186677+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:30:35.421297+00:00 elapsed_s=2.235 ms=0.218096
shape=3 variant=solution tflops=4.923 gbps=57.769 ms=0.218
shape=3 solution_peak_fraction=0.0246
peak_fraction: 0.0433
RESULT: LOW
assistant
reasoning (3021 chars)
0.0433. Improved but still below 0.1. The bf16 dots helped kernel 2. Now shape 1 is 0.059, shape 0 is 0.051, shape 2 is 0.047, shape 3 is 0.025. Need to push higher. The remaining bottleneck is likely kernel 1 (still fp32 dots) and the sequential kernel 2. Let me also use bf16 dots in kernel 1 for Aqk/M and w,u. Currently kernel 1 does fp32 dots. If I use bf16, kernel 1 could be much faster. But Aqk/M computation accuracy is important. With normalized factorization, values are bounded. bf16 might be OK for tolerance 0.05. Let me change kernel 1 dots to bf16. Kernel 1 currently: ```python b_q = tl.load(p_q).to(tl.float32) b_k = tl.load(p_k).to(tl.float32) b_g = tl.load(p_g).to(tl.float32) ... qg_norm = b_q * exp(diff) # fp32 kg_norm = b_k * exp(diff) kdecay_norm = b_k * exp(-diff) Aqk += tl.dot(qg_norm, tl.trans(kdecay_norm)) # fp32 dot M += tl.dot(kg_norm, tl.trans(kdecay_norm)) ... b_kg = b_k * exp(b_g) # fp32 b_w = tl.dot(A, b_kg) # fp32 ``` Change to bf16 dots: - Load q,k as bf16 (don't cast to fp32). - Compute g_diff in fp32, but qg_norm, kg_norm, kdecay_norm in bf16? Or compute in fp32 and cast to bf16 for dot. - tl.dot with bf16 inputs, fp32 accumulator. Actually, to compute exp(diff) where diff is fp32, we get fp32. Multiplying bf16 q by fp32 exp gives fp32. Then cast to bf16 for dot. So: ```python b_q = tl.load(p_q) # bf16 b_k = tl.load(p_k) # bf16 b_g = tl.load(p_g).to(tl.float32) b_gmid = tl.load(p_gmid).to(tl.float32) diff = b_g - b_gmid qg_norm = (b_q * tl.exp(diff)).to(tl.bfloat16) kg_norm = (b_k * tl.exp(diff)).to(tl.bfloat16) kdecay_norm = (b_k * tl.exp(-diff)).to(tl.bfloat16) Aqk += tl.dot(qg_norm, tl.trans(kdecay_norm)) M += tl.dot(kg_norm, tl.trans(kdecay_norm)) ``` The solve is still fp32 (A is fp32). Then w = A @ kg. A is fp32, kg is bf16. tl.dot with mixed types? Need to cast A to bf16 or kg to fp32. If I cast A to bf16, the solve precision might be lost. But A is already computed in fp32; casting to bf16 for w dot loses some. Tolerance 0.05 might allow. Alternatively, keep A fp32, kg bf16, and do tl.dot(A, kg) - but types differ. Need to cast one. Let me cast A to bf16. Actually, A values are small (~0.5). bf16 can represent them. Off-diagonal might be smaller. Some loss but OK. So: ```python A_bf16 = A.to(tl.bfloat16) b_k = tl.load(p_k) # bf16 b_g = tl.load(p_g).to(tl.float32) b_kg = (b_k * tl.exp(b_g)).to(tl.bfloat16) b_w = tl.dot(A_bf16, b_kg) b_w.to(tl.float32) # store? Actually store as bf16 now. ``` Since w is bf16 storage, we can store b_w directly (bf16). Similarly for u. Let me update kernel 1. But I also need Aqk stored as bf16. Currently it stores `Aqk.to(tl.bfloat16)`. Good. For M used in solve, we need fp32. Aqk and M are computed as fp32 accumulators from bf16 dots. That's fine. Actually, tl.dot with bf16 inputs accumulates in fp32 by default. So Aqk and M remain fp32. Good. Then solve is fp32. A is fp32. For w,u: cast A to bf16, kg/v bf16, dot -> fp32, but we can store as bf16. Let me update kernel 1.
in 247182out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (26 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -84,8 +84,8 @@
         p_g = tl.make_block_ptr(
             g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        b_q = tl.load(p_q, boundary_check=(0, 1)).to(tl.float32)
-        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+        b_q = tl.load(p_q, boundary_check=(0, 1))
+        b_k = tl.load(p_k, boundary_check=(0, 1))
         b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
 
         p_gmid = tl.make_block_ptr(
@@ -94,9 +94,9 @@
         b_gmid = tl.load(p_gmid, boundary_check=(0, 1)).to(tl.float32)
 
         diff = b_g - b_gmid
-        qg_norm = b_q * tl.exp(diff)
-        kg_norm = b_k * tl.exp(diff)
-        kdecay_norm = b_k * tl.exp(-diff)
+        qg_norm = (b_q * tl.exp(diff)).to(tl.bfloat16)
+        kg_norm = (b_k * tl.exp(diff)).to(tl.bfloat16)
+        kdecay_norm = (b_k * tl.exp(-diff)).to(tl.bfloat16)
         Aqk += tl.dot(qg_norm, tl.trans(kdecay_norm))
         M += tl.dot(kg_norm, tl.trans(kdecay_norm))
 
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1183 chars \u2014 see diff>",
  "new_string": "<1207 chars \u2014 see diff>"
}
in 247182out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (55 chars)
Now update w,u computation in kernel 1 to use bf16 dot.
in 249213out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (45 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -122,6 +122,8 @@
     )
     tl.store(p_Aqk, Aqk.to(tl.bfloat16), boundary_check=(0, 1))
 
+    A_bf16 = A.to(tl.bfloat16)
+
     for i_k in range(tl.cdiv(K, BK)):
         off_k = i_k * BK
         p_k = tl.make_block_ptr(
@@ -130,26 +132,26 @@
         p_g = tl.make_block_ptr(
             g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        b_k = tl.load(p_k, boundary_check=(0, 1)).to(tl.float32)
+        b_k = tl.load(p_k, boundary_check=(0, 1))
         b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
-        b_kg = b_k * tl.exp(b_g)
-        b_w = tl.dot(A, b_kg)
+        b_kg = (b_k * tl.exp(b_g)).to(tl.bfloat16)
+        b_w = tl.dot(A_bf16, b_kg)
         p_w = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        tl.store(p_w, b_w.to(tl.bfloat16), boundary_check=(0, 1))
+        tl.store(p_w, b_w, boundary_check=(0, 1))
 
     for i_v in range(tl.cdiv(V, BV)):
         off_v = i_v * BV
         p_v = tl.make_block_ptr(
             v_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        b_v = tl.load(p_v, boundary_check=(0, 1)).to(tl.float32)
-        b_u = tl.dot(A, b_v)
+        b_v = tl.load(p_v, boundary_check=(0, 1))
+        b_u = tl.dot(A_bf16, b_v)
         p_u = tl.make_block_ptr(
             u_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        tl.store(p_u, b_u.to(tl.bfloat16), boundary_check=(0, 1))
+        tl.store(p_u, b_u, boundary_check=(0, 1))
 
 
 @triton.autotune(
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<1105 chars \u2014 see diff>",
  "new_string": "<1103 chars \u2014 see diff>"
}
in 249213out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (24 chars)
Now check and benchmark.
in 250230out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10 && uv run python benchmark.py 2>&1 | tail -20",
  "description": "Check and benchmark bf16 kernel1 dots"
}
in 250230out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (1486 chars)
b_k = tl.load(p_k, boundary_check=(0, 1))
        b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
        b_kg = (b_k * tl.exp(b_g)).to(tl.bfloat16)
        b_w = tl.dot(A_bf16, b_kg)
        p_w = tl.make_block_ptr(
            w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        tl.store(p_w, b_w, boundary_check=(0, 1))
        ^
Block element type(bf16) and value element type(fp32) mismatch
    module = src.make_ir(target, options, codegen_fns, module_map, context)
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/workspace/.venv/lib/python3.11/site-packages/triton/compiler/compiler.py", line 80, in make_ir
    return ast_to_ttir(self.fn, self, context=context, options=options, codegen_fns=codegen_fns,
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
triton.compiler.errors.CompilationError: at 111:8:
        )
        p_g = tl.make_block_ptr(
            g_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        b_k = tl.load(p_k, boundary_check=(0, 1))
        b_g = tl.load(p_g, boundary_check=(0, 1)).to(tl.float32)
        b_kg = (b_k * tl.exp(b_g)).to(tl.bfloat16)
        b_w = tl.dot(A_bf16, b_kg)
        p_w = tl.make_block_ptr(
            w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
        )
        tl.store(p_w, b_w, boundary_check=(0, 1))
        ^
Block element type(bf16) and value element type(fp32) mismatch
assistant
reasoning (138 chars)
b_w is fp32 (dot accumulator), but w_ptr is bf16. Need to cast: `tl.store(p_w, b_w.to(tl.bfloat16), ...)`. Similarly for b_u. Let me fix.
in 250784out 0cache_r 0cache_w 0
assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (20 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -139,7 +139,7 @@
         p_w = tl.make_block_ptr(
             w_ptr, (T, K), (H * K, 1), (t0, off_k), (BT, BK), (1, 0)
         )
-        tl.store(p_w, b_w, boundary_check=(0, 1))
+        tl.store(p_w, b_w.to(tl.bfloat16), boundary_check=(0, 1))
 
     for i_v in range(tl.cdiv(V, BV)):
         off_v = i_v * BV
@@ -151,7 +151,7 @@
         p_u = tl.make_block_ptr(
             u_ptr, (T, V), (H * V, 1), (t0, off_v), (BT, BV), (1, 0)
         )
-        tl.store(p_u, b_u, boundary_check=(0, 1))
+        tl.store(p_u, b_u.to(tl.bfloat16), boundary_check=(0, 1))
 
 
 @triton.autotune(
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<618 chars \u2014 see diff>",
  "new_string": "<650 chars \u2014 see diff>"
}
in 250784out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (20 chars)
Check and benchmark.
in 251443out 0cache_r 0cache_w 0
assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10 && uv run python benchmark.py 2>&1 | tail -20",
  "description": "Check and benchmark bf16 kernel1 dots fixed"
}
in 251443out 0cache_r 0cache_w 0
subagent · 7 events · 2 tools
tool
stdout (283 chars)
Command running in background with ID: bwle45wj2. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/bwle45wj2.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (5 chars)
Wait.
in 251619out 0cache_r 0cache_w 0
assistant
TaskOutput
args
{
  "task_id": "bwle45wj2",
  "block": true,
  "timeout": 600000
}
in 251619out 0cache_r 0cache_w 0
tool
stdout (1494 chars)
<retrieval_status>success</retrieval_status>

<task_id>bwle45wj2</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
PASS
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:36:51.094378+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:36:53.515624+00:00 elapsed_s=2.421 ms=0.226480
shape=0 variant=solution tflops=9.482 gbps=111.262 ms=0.226
shape=0 solution_peak_fraction=0.0474
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:36:53.681388+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:36:55.991717+00:00 elapsed_s=2.310 ms=0.369504
shape=1 variant=solution tflops=11.624 gbps=136.391 ms=0.370
shape=1 solution_peak_fraction=0.0581
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:36:56.157085+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:36:58.508499+00:00 elapsed_s=2.351 ms=0.459856
shape=2 variant=solution tflops=9.340 gbps=109.593 ms=0.460
shape=2 solution_peak_fraction=0.0467
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:36:58.549588+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:37:00.791005+00:00 elapsed_s=2.241 ms=0.218816
shape=3 variant=solution tflops=4.907 gbps=57.579 ms=0.219
shape=3 solution_peak_fraction=0.0245
peak_fraction: 0.0422
RESULT: LOW
</output>
assistant
reasoning (3717 chars)
0.0422, slightly worse than 0.0433. The bf16 dots in kernel 1 didn't help much and may have hurt accuracy slightly but still passes. Let me revert kernel 1 dots to fp32 (more accurate) and keep kernel 2 bf16 dots. Actually, performance is similar; accuracy is better with fp32 kernel1. Let me keep fp32 kernel1. Wait, the benchmark showed 0.0433 with fp32 kernel1 and bf16 kernel2, and 0.0422 with bf16 kernel1. So fp32 kernel1 is slightly better. Let me revert. Actually, let me think differently. The peak_fraction is geomean and dominated by shape 3 (0.0245). Shape 3 has B=1,H=4,T=2048. Kernel 2 has only 4*(V/BV) blocks. If BV=32, 16 blocks; BV=16, 32 blocks. Very low occupancy. To improve shape 3, we need more parallelism. The only way is smaller BV or scan. BV=8 would give 64 blocks. Let me try BV=8, but it might have too much overhead. Actually, with BV=8, S is 1KB, b_v is 512B. GEMMs are very small (64x64 @ 64x8). Might not use tensor cores efficiently. Alternative: use scan for kernel 2 to get N-way parallelism. For shape 3, N=32. This would give 32*16 = 512 blocks. Much better occupancy. Let me implement the scan approach for kernel 2. Actually, implementing a full scan in Triton is complex. But I can do a simpler "chunked" scan or use PyTorch's torch.cumsum-like operations on the operators. Actually, the operators are matrices. PyTorch doesn't have a built-in matrix scan. But I can implement it efficiently with a loop of log2(N) batched matrix multiplies. Let me think about the formulation again. We have S_{n+1} = P_n @ S_n + Q_n, with S_0 = 0. We need S_n for each chunk n (state before chunk n). Then output o_n depends on S_n. If we compute all P_n, Q_n first, then do an inclusive scan of the operators, we get (U_n, C_n) such that S_n = U_n @ S_0 + C_n = C_n (since S_0=0). Wait, the operator composition: Define operator O_n(x) = P_n @ x + Q_n. O_n ∘ O_{n-1} ... ∘ O_0 (0) = S_{n+1}. For S_n (state before chunk n), it's O_{n-1} ∘ ... ∘ O_0 (0). We can do an inclusive scan where we compose operators. Each operator is (P,Q). Composition: (P_b, Q_b) ∘ (P_a, Q_a) = (P_b @ P_a, P_b @ Q_a + Q_b). We want cumulative operator C_n = O_{n-1} ∘ ... ∘ O_0. Then S_n = C_n(0) = Q part of C_n. Parallel scan can compute all C_n in O(N log N) matrix ops. Then we compute o_n in parallel given S_n. But this is a lot of FLOPs. For N=64, K=V=128, each composition is P@P (4M FLOPs) + P@Q (4M FLOPs) = 8M FLOPs. log2(N)=6 steps, each step processes ~N operators. Total ~3 GFLOPs per (B,H). For shape 1, B*H=16, total 48 GFLOPs. At 200 TFLOPS, 0.24ms. Plus P,Q computation and output. Maybe 0.4ms total. Similar to current. But for shape 3, B*H=4, total 12 GFLOPs, 0.06ms. Better than current 0.22ms. So scan might help shape 3 significantly, and shape 1/2 somewhat. However, implementing this in Triton is complex. I could use PyTorch for the scan (batched matmul) and Triton for P,Q computation and output. Actually, maybe simpler: since the current sequential kernel 2 is not too slow for shape 1/2, and the main issue is shape 3, I could use a different kernel for shape 3 or switch to scan when B*H is small. But implementing two paths is complex. Alternative idea: in kernel 2, instead of one block per (B,H,V tile), use one block per (B,H,V tile, chunk_group) where chunk_group is a range of chunks, and process them sequentially. This doesn't help because each block still sequential. Wait, actually, the issue is that with B*H=4 and V/BV=4 (if BV=32), we have 16 blocks. If we reduce BV to 8, we get 64 blocks. But small GEMMs. Let me try BV=8 in autotune. It might help shape 3 even if hurts others. Actually, let me profile shape 3 to see where time goes.
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assistant
Bash
args
{
  "command": "<915 chars \u2014 see below>",
  "description": "Profile shape 3"
}
command (915 chars)
cat > /workspace/problems/02_kda_cutlass/profile3.py <<'PY'
import torch
import reference, solution

B,T,H,K,V,CS = 1,2048,4,128,128,64
reference.B, reference.T, reference.H, reference.K, reference.V, reference.CHUNK_SIZE = B,T,H,K,V,CS
solution.B, solution.T, solution.H, solution.K, solution.V, solution.CHUNK_SIZE = B,T,H,K,V,CS

m = solution.Model(B,T,H,K,V,CS).cuda().eval()
torch.manual_seed(0)
q,k,v,g,beta = [x.cuda() for x in reference.get_inputs()]
for _ in range(3): _ = m(q,k,v,g,beta)

torch.cuda.synchronize()
with torch.profiler.profile(
    activities=[torch.profiler.ProfilerActivity.CUDA, torch.profiler.ProfilerActivity.CPU],
    record_shapes=False,
) as prof:
    for _ in range(5):
        _ = m(q,k,v,g,beta)
    torch.cuda.synchronize()

print(prof.key_averages().table(sort_by="cuda_time_total", row_limit=20))
PY
uv run python /workspace/problems/02_kda_cutlass/profile3.py 2>&1 | tail -25
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system — task_started
↳ subagent started
tool
stdout (4416 chars) — TRUNCATED
-------------------------------------------------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  ------------  
                                      _kda_inter_kernel         0.00%       0.000us         0.00%       0.000us       0.000us     480.344us        57.05%     480.344us      96.069us             5  
                                      _kda_intra_kernel         0.00%       0.000us         0.00%       0.000us       0.000us     246.815us        29.31%     246.815us      49.363us             5  
                                           aten::cumsum         2.42%      33.490us        57.62%     798.642us     159.728us      63.998us         7.60%      87.998us      17.600us             5  
void at::native::tensor_kernel_scan_outer_dim<float,...         0.00%       0.000us         0.00%       0.000us       0.000us      63.998us         7.60%      63.998us      12.800us             5  
                                Activity Buffer Request        53.23%     737.792us        53.23%     737.792us     737.792us      24.000us         2.85%      24.000us      24.000us             1  
                                              aten::mul         2.54%      35.210us         4.06%      56.330us       5.633us      23.489us         2.79%      23.489us       2.349us            10  
                                            aten::copy_         1.94%      26.920us         3.87%      53.700us   

... (2916 more chars)
assistant
reasoning (1389 chars)
For shape 3, kernel 2 takes 480us (96us per call), kernel 1 246us. Kernel 2 is 2x kernel 1. Only 16-32 blocks. Low occupancy. To reach 0.1 peak_fraction for shape 3, need ~20 TFLOPS. FLOPs for shape 3 = 4*1*2048*4*32768 = 1.07 GFLOPs. At 0.22ms benchmark, that's 4.9 TFLOPS. Need 0.054ms. That's 4x speedup. The scan approach could help shape 3 a lot. But implementing scan is complex. Let me think of a simpler way to expose more parallelism in kernel 2. Idea: process multiple sequences per block. Currently each block handles one (B,H,V tile). We can have each block handle multiple (B,H) pairs, increasing blocks? No, fewer blocks. Idea: process the recurrence with smaller granularity. Instead of one block per V tile, use multiple blocks per V tile working on different chunk ranges? But dependency... Actually, we can split the V dimension into smaller tiles AND process multiple V tiles per block. If each block handles 2 V tiles, blocks halve, work doubles. Not helpful. The fundamental limit is B*H*(V/BV) blocks, each sequential over N. For shape 3: 1*4*4 = 16 blocks. If we use BV=8: 1*4*16 = 64 blocks. 4x more parallelism. Each block does 1/4 the V work but same N sequential. The GEMMs are smaller (64x64 @ 64x8). Let me try BV=8 and see. Actually, with BV=8, b_v is (64,8). Tensor cores might not like M=64,N=8,K=64. But could still work. Let me add BV=8 configs.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (19 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -156,12 +156,16 @@
 
 @triton.autotune(
     configs=[
+        triton.Config({"BK": 64, "BV": 8}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 8}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 8}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 8}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
         triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<980 chars \u2014 see diff>",
  "new_string": "<1266 chars \u2014 see diff>"
}
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tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (242 chars)
Also kernel 1 autotune should match? Kernel 1 doesn't use BV much (only for u). But BV affects u loop. Add same configs to kernel 1. Actually, kernel 1's performance depends on BK and V/BV (for u). Let me add BV=8/16 configs to kernel 1 too.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (18 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -1,11 +1,15 @@
 @triton.autotune(
     configs=[
+        triton.Config({"BK": 64, "BV": 8}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 8}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 8}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 8}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
         triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<980 chars \u2014 see diff>",
  "new_string": "<1266 chars \u2014 see diff>"
}
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tool
stderr (1058 chars)
<tool_use_error>String to replace not found in file.
String: @triton.autotune(
    configs=[
        triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
        triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
        triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
        triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
        triton.Config({"BK": 128, "BV": 64}, num_warps=8, num_stages=1),
    ],
    key=["B", "T", "H", "K", "V"],
)
@triton.jit
def _kda_intra_kernel(</tool_use_error>
assistant
reasoning (80 chars)
I already edited kernel 1 autotune earlier to have BV=8/16 configs? Let me read.
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assistant
Read /workspace/problems/02_kda_cutlass/solution.py
args
{
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "limit": 50
}
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tool
stdout (1534 chars) — TRUNCATED
1	"""Kimi Delta Attention (KDA) forward (chunk form) via Triton.
2	
3	Matches reference.py's semantics: inputs are bf16 (g fp32), chunk_size=64,
4	no initial/final state, returns bf16 o.
5	"""
6	from __future__ import annotations
7	
8	import torch
9	import torch.nn as nn
10	import triton
11	import triton.language as tl
12	
13	OP_TYPE = "linear_attention"
14	SUPPORTED_PRECISIONS = ["bf16"]
15	HARDWARE_REQUIRED = ["RTX_PRO_6000", "H100", "B200"]
16	
17	
18	@triton.autotune(
19	    configs=[
20	        triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
21	        triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
22	        triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
23	        triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
24	        triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
25	        triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
26	        triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
27	        triton.Config({"BK": 128, "BV": 64}, num_warps=8, num_stages=1),
28	    ],
29	    key=["B", "T", "H", "K", "V"],
30	)
31	@triton.jit
32	def _kda_intra_kernel(
33	    q_ptr,
34	    k_ptr,
35	    v_ptr,
36	    g_ptr,
37	    gmid_ptr,
38	    beta_ptr,
39	    Aqk_ptr,
40	    w_ptr,
41	    u_ptr,
42	    scale,
43	    B: tl.constexpr,
44	    T: tl.constexpr,
45	    H: tl.constexpr,
46	    K: tl.constexpr,
47	    V: tl.constexpr,
48	    BT: tl.constexpr,
49	    BK: tl.c

... (34 more chars)
assistant
reasoning (56 chars)
Kernel 1 autotune doesn't have BV=8/16. Let me add them.
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assistant
Edit /workspace/problems/02_kda_cutlass/solution.py
diff (21 lines)
--- a//workspace/problems/02_kda_cutlass/solution.py
+++ b//workspace/problems/02_kda_cutlass/solution.py
@@ -17,10 +17,18 @@
 
 @triton.autotune(
     configs=[
+        triton.Config({"BK": 64, "BV": 8}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 8}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 64, "BV": 16}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 64, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=4, num_stages=1),
         triton.Config({"BK": 64, "BV": 64}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 8}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 8}, num_warps=8, num_stages=1),
+        triton.Config({"BK": 128, "BV": 16}, num_warps=4, num_stages=1),
+        triton.Config({"BK": 128, "BV": 16}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 32}, num_warps=4, num_stages=1),
         triton.Config({"BK": 128, "BV": 32}, num_warps=8, num_stages=1),
         triton.Config({"BK": 128, "BV": 64}, num_warps=4, num_stages=1),
args
{
  "replace_all": false,
  "file_path": "/workspace/problems/02_kda_cutlass/solution.py",
  "old_string": "<690 chars \u2014 see diff>",
  "new_string": "<1266 chars \u2014 see diff>"
}
in 258974out 0cache_r 0cache_w 0
tool
stdout (152 chars)
The file /workspace/problems/02_kda_cutlass/solution.py has been updated successfully. (file state is current in your context — no need to Read it back)
assistant
reasoning (76 chars)
Now check and benchmark. This will take longer due to more autotune configs.
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assistant
Bash
args
{
  "command": "uv run python check.py 2>&1 | tail -10",
  "description": "Check with BV=8 configs"
}
in 259932out 0cache_r 0cache_w 0
subagent · 0 events · 0 tools
tool
stdout (4 chars)
PASS
assistant
reasoning (20 chars)
PASS. Now benchmark.
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assistant
Bash
args
{
  "command": "uv run python benchmark.py 2>&1 | tail -25",
  "description": "Benchmark with BV=8 configs"
}
in 260009out 0cache_r 0cache_w 0
subagent · 8 events · 2 tools
tool
stdout (283 chars)
Command running in background with ID: buoclzb0w. Output is being written to: /tmp/claude-1002/-workspace-problems-02-kda-cutlass/229dbb4a-234a-4600-8669-a9863a453abf/tasks/buoclzb0w.output. You will be notified when it completes. To check interim output, use Read on that file path.
assistant
reasoning (5 chars)
Wait.
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assistant
TaskOutput
args
{
  "task_id": "buoclzb0w",
  "block": true,
  "timeout": 600000
}
in 260159out 0cache_r 0cache_w 0
tool
stdout (1490 chars)
<retrieval_status>success</retrieval_status>

<task_id>buoclzb0w</task_id>

<task_type>local_bash</task_type>

<status>completed</status>

<exit_code>0</exit_code>

<output>
benchmark_event event=variant_start shape=0 variant=solution ts=2026-06-14T00:45:15.710487+00:00
benchmark_event event=variant_end shape=0 variant=solution ts=2026-06-14T00:45:19.501486+00:00 elapsed_s=3.791 ms=0.214144
shape=0 variant=solution tflops=10.028 gbps=117.671 ms=0.214
shape=0 solution_peak_fraction=0.0501
benchmark_event event=variant_start shape=1 variant=solution ts=2026-06-14T00:45:19.669733+00:00
benchmark_event event=variant_end shape=1 variant=solution ts=2026-06-14T00:45:23.393725+00:00 elapsed_s=3.724 ms=0.369472
shape=1 variant=solution tflops=11.625 gbps=136.403 ms=0.369
shape=1 solution_peak_fraction=0.0581
benchmark_event event=variant_start shape=2 variant=solution ts=2026-06-14T00:45:23.560193+00:00
benchmark_event event=variant_end shape=2 variant=solution ts=2026-06-14T00:45:27.327299+00:00 elapsed_s=3.767 ms=0.464176
shape=2 variant=solution tflops=9.253 gbps=108.573 ms=0.464
shape=2 solution_peak_fraction=0.0463
benchmark_event event=variant_start shape=3 variant=solution ts=2026-06-14T00:45:27.370227+00:00
benchmark_event event=variant_end shape=3 variant=solution ts=2026-06-14T00:45:30.941975+00:00 elapsed_s=3.572 ms=0.216336
shape=3 variant=solution tflops=4.963 gbps=58.239 ms=0.216
shape=3 solution_peak_fraction=0.0248
peak_fraction: 0.0428
RESULT: LOW
</output>
assistant
reasoning (396 chars)
0.0428. BV=8 didn't help shape 3. Still low. I think to reach 0.1, I need to implement the scan/parallel approach for kernel 2. Or use a more efficient algorithm. Actually, let me reconsider the problem. Maybe the reference FLA kernel achieves much higher performance with a different approach. Let me look at FLA's output kernel (chunk_gla_fwd_o_gk) to see how they compute output in parallel.
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assistant
Read /workspace/fla_ref/fla/ops/gla/chunk.py
args
{
  "file_path": "/workspace/fla_ref/fla/ops/gla/chunk.py",
  "limit": 200
}
in 260907out 0cache_r 0cache_w 0
tool
stdout (7057 chars) — TRUNCATED
1	# Copyright (c) 2023-2026, Songlin Yang, Yu Zhang, Zhiyuan Li
2	#
3	# This source code is licensed under the MIT license found in the
4	# LICENSE file in the root directory of this source tree.
5	# For a list of all contributors, visit:
6	#   https://github.com/fla-org/flash-linear-attention/graphs/contributors
7	
8	import torch
9	import triton
10	import triton.language as tl
11	
12	from fla.ops.common.chunk_h import chunk_bwd_dh, chunk_fwd_h
13	from fla.ops.utils import prepare_chunk_indices
14	from fla.ops.utils.cache import fla_cache_autotune
15	from fla.ops.utils.constant import RCP_LN2
16	from fla.ops.utils.cumsum import chunk_local_cumsum
17	from fla.ops.utils.op import exp2
18	from fla.utils import autotune_cache_kwargs, check_shared_mem, input_guard
19	
20	BK_LIST = [32, 64] if check_shared_mem() else [16, 32]
21	BV_LIST = [64, 128] if check_shared_mem('ampere') else [16, 32]
22	
23	
24	@triton.heuristics({
25	    'IS_VARLEN': lambda args: args['cu_seqlens'] is not None,
26	})
27	@fla_cache_autotune(
28	    configs=[
29	        triton.Config({'BK': BK}, num_warps=num_warps, num_stages=num_stages)
30	        for BK in [32, 64]
31	        for num_warps in [1, 2, 4, 8]
32	        for num_stages in [2, 3, 4]
33	    ],
34	    key=['BC'],
35	    **autotune_cache_kwargs,
36	)
37	@triton.jit(do_not_specialize=['T'])
38	def chunk_gla_fwd_A_kernel_intra_sub_inter(
39	    q,
40	    k,
41	    g,
42	    A,
43	    cu_seqlens,
44	    chunk_indices,
45	    scale,
46	    T,
47	    H

... (5557 more chars)
assistant
API Error: 400 Invalid request: Your request exceeded model token limit: 262144 (requested: 263684)
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