Fix review issues #17-#22: 恢复 #11/#12/#14 (StreamK fixpipe 单次计账/K=1 AIV单缓冲/advice) + 占位方案不可评估 + 转置 dValue 判据三处同源(form c 双缓冲适配修复) + 恢复 #13/#15 回归测试 + 清理临时 csv/.gitignore + 文档同步
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@@ -43,22 +43,33 @@ class IterBatchBranch(Branch):
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return "b", k, f"双batch乒乓: 2*(MK+KN)*dtype={2*single/1024:.0f}KB <= L1"
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# c) 一侧驻留 + 对侧切 K, 预算按 b_core 分档
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# dValue 守卫与条件 4 / constraints 同源 (issue#19): 转置感知连续维判据
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# (dv_a=M*dt 当 A 转置, 否则 k_l1*dt; dv_b=N*dt 当 B 不转置, 否则 k_l1*dt).
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from ..models import dvalue_contig_dims
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l1_budget = s.l1_bytes / min(b_core, 2)
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for resident, side in ((m * k * dt, "A"), (k * n * dt, "B")):
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other = n * dt if side == "A" else m * dt
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if resident <= l1_budget:
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k_l1 = min(int((l1_budget - resident) / other / 2), k)
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k_l1 = align_down(max(k_l1, s.fractal), s.fractal)
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if k_l1 >= s.fractal and k_l1 * dt >= s.dvalue_min:
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dv_a, dv_b = dvalue_contig_dims(case, k_l1)
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# 双缓冲适配: 驻留侧 + 对侧 k_l1 段 x2 必须 <= 预算
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# (resident 恰占满预算时 k_l1 被抬到 16 会造成超预算的假方案)
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if (k_l1 >= s.fractal and
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resident + 2 * k_l1 * other <= l1_budget and
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dv_a >= s.dvalue_min and dv_b >= s.dvalue_min):
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return "c", k_l1, (
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f"一侧驻留({side})+对侧切K: {side}驻留{resident/1024:.0f}KB, "
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f"预算L1/{min(b_core,2)}, k_L1={k_l1}")
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f"预算L1/{min(b_core,2)}, k_L1={k_l1}, "
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f"dValueA={dv_a:.0f}B/dValueB={dv_b:.0f}B")
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# b_core>=2 时另一半 L1 预取下一 batch 驻留侧, 边界无气泡
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# d) 两侧都切 K (兜底)
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k_l1 = align_down(int(s.l1_bytes / (2 * (m + n) * dt)), s.fractal)
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if k_l1 >= s.fractal and k_l1 * dt >= s.dvalue_min:
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return "d", k_l1, f"两侧都切K: k_L1={k_l1}, K段成对流水, batch边界天然无缝"
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dv_a, dv_b = dvalue_contig_dims(case, k_l1)
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if (k_l1 >= s.fractal and dv_a >= s.dvalue_min and dv_b >= s.dvalue_min):
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return "d", k_l1, (f"两侧都切K: k_L1={k_l1}, K段成对流水, batch边界天然无缝; "
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f"dValueA={dv_a:.0f}B/dValueB={dv_b:.0f}B")
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return None, 0, "L1 四形态均不满足 (M/N 相对 L1 过大)"
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@@ -90,24 +101,18 @@ class IterBatchBranch(Branch):
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form is not None, form_desc))
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# 条件 4: 搬移效率下限 (c/d 切分后)
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# 转置影响 (参考 bmmv3): A 不转置时 K 向连续, dValue 判 K*dt;
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# A 转置时 M 向连续, dValue 判 M*dt;
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# B 不转置时 N 向连续, dValue 判 N*dt;
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# B 转置时 K 向连续, dValue 判 K*dt.
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# dValue 判定按转置调整连续维 (与 l1_form 守卫/constraints 同源, issue#19):
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# A 不转置: K 向连续 -> k_l1*dt; A 转置: M 向连续 -> M*dt;
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# B 不转置: N 向连续 -> N*dt; B 转置: K 向连续 -> k_l1*dt.
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from ..models import dvalue_contig_dims
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m, n, k = case.m, case.n, case.k
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dt = case.dtype_in_bytes
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if form in ("c", "d"):
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tile_ok = (k_l1 * m * dt >= s.min_tile_size) or (k_l1 * n * dt >= s.min_tile_size)
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# dValue 判定按转置调整连续维
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if case.trans_a:
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dv_a = m * dt # A 转置: M 向连续
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else:
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dv_a = k_l1 * dt # A 不转置: K 向连续 (切分后)
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if case.trans_b:
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dv_b = k_l1 * dt # B 转置: K 向连续 (切分后)
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else:
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dv_b = n * dt # B 不转置: N 向连续
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dv_ok = dv_a >= s.dvalue_min or dv_b >= s.dvalue_min # 至少一侧满足
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dv_a, dv_b = dvalue_contig_dims(case, k_l1)
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# 切 K 时 A/B 两侧分段各自搬移, 两侧连续维 dValue 均须 >= 下限 (与生成守卫/约束同源,
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# issue#19; 非转置时两侧同为 k_l1*dt, 等价于历史口径)
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dv_ok = dv_a >= s.dvalue_min and dv_b >= s.dvalue_min
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c4 = tile_ok and dv_ok
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checks.append(ConditionCheck(
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"4_搬移效率: 搬移分块>=min_TileSize 且 dValue>=128B (转置调整连续维)",
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@@ -24,11 +24,14 @@ class SpecialBranch(Branch):
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c1 = case.k <= 1
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checks = [ConditionCheck("1_K<=1 (Cube 无用)", c1, f"K={case.k}")]
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if case.k == 1:
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# K=1 触发 AIV 通路需 B >= 2*AIV核数 且单 batch 输入输出能驻留 UB
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c2 = case.batch_c >= 2 * self.spec.aiv_num
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# K=1 的 AIV 通路恒可用 (issue#12/#17): B>=2*AIV 开 UB 乒乓; B<128 退化为
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# AIV 单缓冲 (无乒乓, 逐 batch 串行搬入), 不再是无方案空洞.
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b = case.batch_c
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pingpong = b >= 2 * self.spec.aiv_num
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mode = "UB乒乓" if pingpong else "AIV单缓冲(逐batch串行, B<2*AIV)"
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checks.append(ConditionCheck(
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"2_K=1的AIV触发: B >= 2*AIV核数 (开UB乒乓)",
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c2, f"B={case.batch_c} vs {2*self.spec.aiv_num}"))
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"2_K=1的AIV通路: 恒可用 (B>=128 开UB乒乓, 否则单缓冲)",
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True, f"B={b}, 模式={mode}"))
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return checks
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# ------------------------------------------------------------------
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@@ -36,11 +39,15 @@ class SpecialBranch(Branch):
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s = self.spec
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if case.k == 0:
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sub = "K=0纯写值"
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mode = ""
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note = "无任何计算, C=bias 或 0, 纯 AIV 写值; 按行均分到 AIV 核"
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else:
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sub = "K=1逐元素乘"
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note = ("退化为 C=A⊙B 无累加深度, Cube 16x16x16 粒度浪费 15/16; "
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"走 AIV 通路 GM->UB->Mul->GM, UB 乒乓")
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pingpong = case.batch_c >= 2 * s.aiv_num
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mode = "UB乒乓" if pingpong else "AIV单缓冲"
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note = (f"退化为 C=A⊙B 无累加深度, Cube 16x16x16 粒度浪费 15/16; "
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f"走 AIV 通路 GM->UB->Mul->GM, {mode} "
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f"({'B>=2*AIV 双batch乒乓流水' if pingpong else 'B<2*AIV 逐batch单缓冲串行'})")
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return ImplPlan(
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case_id=case.case_id, branch=self.name,
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used_core_num=s.aiv_num, # 用 AIV 核
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@@ -48,7 +55,8 @@ class SpecialBranch(Branch):
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core_map="AIV 核间按行均分 (无 Cube tile 概念)",
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b_core=0, merge_b0=1,
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single_core_m=0, single_core_n=0, single_core_k=case.k,
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k_l1=0, b_l1=1, l1_form="UB驻留(AIV)",
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k_l1=0, b_l1=1,
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l1_form="UB驻留(AIV)" if case.k == 0 else "UB驻留(AIV) " + mode,
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base_m=0, base_n=0, base_k=0,
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l2_policy_in="allocate", l2_policy_out="direct_gm",
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swizzle_w=0, workspace_bytes=0,
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@@ -145,13 +145,15 @@ class StreamKBranch(Branch):
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t_mmad = t_mmad_tile / grid_k
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t_mte2 = t_mte2_tile / grid_k
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# 归约: 部分和 4B 驻留 L2, AIV 归约 (含最终按 C dtype 写回)
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# 归约: 部分和 4B 驻留 L2, AIV 归约 (含部分和写/读回/求和/最终按 C dtype 写回)
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# 口径 (issue#11/#17): 归约整体为串行追加 (t_drain=t_reduce, reduce_serial=True),
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# 部分和写出已计入 eval_streamk_reduce 的 t_write_partial —— 稳态 Fixpipe 不再
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# 重复计账. 若再按 grid_k*tile*4B/单核带宽份额另计一次, 既重复计账又把整组
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# 部分和串行压到单核写口, 高估 grid_k 倍 (streamk_demo 曾虚高到 55us/FIXPIPE;
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# 第三轮曾回退该修复, issue#17 恢复).
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t_reduce = eval_streamk_reduce(tile_elems, grid_k, out_b, s)
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# Fixpipe: 部分和写出按 4B (L0C dtype, 防精度丢失), 驻留 L2.
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# 最终归约结果的 C dtype 写回已在 t_reduce 内计, 此处不重复 (issue#9 口径对齐).
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fix_bytes = grid_k * tile_elems * 4
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t_fix = fix_bytes / s.bw_l2_pc
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fix_bytes = 0.0
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t_fix = 0.0
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flops_pc = 2.0 * tile_elems * k / grid_k
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gm_bytes = k * (2 * math.sqrt(tile_elems)) * dt / grid_k
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