Let me analyze the problem: 1. We have a model with: Linear layer (matmul + bias), GELU activation, Softmax 2. Input: [1024, 8192], Output: [1024, 8192] 3. We need to fuse these operations into a single CUDA kernel for speed The dimensions are: - batch_size = 1024 - in_features = 8192 - out_features = 8192 The linear layer is: x @ W^T + b where x is [1024, 8192], W is [8192, 8192], b is [8192] Then GELU: 0.5 * x * (1 + tanh(sqrt(2/π) * (x + 0.044715 * x^3))) Or the approximate version Then softmax along dim=1 (rows) For H100, we have massive compute (1979 TFLOPS in bf16/fp16, 67 TFLOPS in fp32). The problem requires fp32 precision, so 67 TFLOPS. Let me think about the memory bandwidth and compute: - W matrix: 8192 * 8192 * 4 bytes = 256 MB - Input: 1024 * 8192 * 4 = 32 MB - Output: 1024 * 8192 * 4 = 32 MB - Total memory read for weight + input: ~288 MB - Memory bandwidth: 3.35 TB/s = 3350 GB/s So memory bandwidth limited time would be ~288/3350 = 0.086 ms = 86 microseconds But compute time: - 1024 * 8192 * 8192 * 2 = 137.4B FLOPs - At 67 TFLOPS: 137.4/67000 = 0.002 seconds = 2 ms Wait, that's wrong. Let me recalculate: 1024 * 8192 * 8192 = 68.7B multiply-adds = 137.4B FLOPs At 67 TFLOPS: 137.4 / 67 = 2.05 seconds = 2050 ms Hmm, but 67 TFLOPS is for the whole GPU with all SMs. Actually wait, 67 TFLOPS is 67 * 10^12 FLOPS/s = 67,000 GFLOPS/s. 137.4B FLOPs / 67,000 GFLOPs/s = 2.05 seconds?? That can't be right. Wait. Let me recalculate: 67 TFLOPS = 67,000 GFLOPS 137.4 GFLOPs / 67,000 GFLOPS/s = 0.00205 s = 2.05 ms. Ah yes, it's 137.4 GFLOPs, not 137.4B GFLOPs. So 2.05 ms. Actually wait, matmul of [1024, 8192] x [8192, 8192]: M=1024, K=8192, N=8192 FLOPs = 2 * M * K * N = 2 * 1024 * 8192 * 8192 = 2 * 1024 * 67,108,864 = 2 * 68,719,476,736 = 137,438,953,472 FLOPs = 137.4 GFLOPs (since 1 GFLOP = 10^9 FLOPs) At 67 TFLOPS = 67,000 GFLOPS/s: Time = 137.4 / 67,000 = 0.00205 s = 2.05 ms For cuBLAS, we can typically get around 80-90% of peak TFLOPS, so maybe 1.8-2.5 ms for the matmul alone. For a fused kernel, we'd need to do: 1. Load W (256 MB) - but this is distributed across tiles 2. Do matmul + bias + gelu + softmax Actually, doing a single fused kernel that computes matmul + bias + gelu + softmax may not beat cuBLAS for the matmul part, but could reduce memory traffic by avoiding intermediate writes. However, the softmax requires reading the entire row to compute the max and sum, which means we'd either need: 1. A two-pass approach for softmax 2. Or compute matmul output into shared memory / registers first, then do softmax For large matrices like 8192x8192, a fully fused kernel from scratch is quite complex. Let me think about this differently. Actually, let me think about whether we can beat PyTorch's default implementation which likely uses: 1. cuBLAS for matmul (very optimized) 2. Element-wise kernel for GELU 3. Two-pass or one-pass optimized kernel for softmax The overhead for PyTorch would be: 1. Matmul: ~2.0 ms 2. Write output to global memory 3. GELU kernel: read input, compute, write output (~64 MB of memory traffic = 0.02 ms) 4. Softmax: read input, compute max, compute sum/exp, write output. For row length 8192, this involves ~32 MB read + 32 MB write plus reductions. Maybe 0.05-0.1 ms. Total PyTorch time: ~2.1-2.2 ms. To beat this, we have two main strategies: 1. Fuse GELU + bias into the matmul (saves one round trip to memory) 2. Try to get close to cuBLAS performance on the matmul while fusing everything But writing a custom matmul that competes with cuBLAS for [1024, 8192] x [8192, 8192] is extremely difficult. cuBLAS has very sophisticated tiling, pipeline scheduling, etc. Wait, maybe we can use cutlass or a simpler approach. Or maybe the reference implementation isn't as optimized as I think. Actually, let me reconsider. PyTorch's default Linear + GELU + softmax: - Linear uses cuBLAS which is very fast - The intermediate writes and reads do have cost For H100, maybe we can use a kernel that: 1. Breaks the output rows into chunks 2. Loads the weight matrix in tiles 3. Computes partial sums in registers 4. Applies bias and GELU 5. But softmax is the problem because it needs the entire row For softmax, we need the entire row. We could: 1. First do matmul + bias + GELU in tiles, writing intermediate to global memory (or keep in shared memory for small enough tiles) 2. Then do softmax on the result But that's essentially what PyTorch does. Alternatively, we could use a "stream-k" style or just fuse bias+GELU into the epilogue of the matmul to save one memory round trip. Actually, let me look at the dimensions more carefully: - M = 1024 rows - N = 8192 output features - K = 8192 input features This is a "tall and skinny" input matrix (1024x8192) times a square weight matrix. For such shapes where M is relatively small compared to N and K, a custom kernel that computes one output row at a time or several rows at a time might be feasible. Let me think about a simpler kernel: - Each block computes a subset of rows - For each row, we compute: x @ W^T + b, GELU, then SOFTMAX - The matmul for one row is dot(x, W[:, j]) for each j Actually, an output row is x[0, :] @ W^T. Since x is a row vector of length 8192 and W^T has shape [8192, 8192], each output element is a dot product of two 8192-length vectors. 8192 = 8K elements. In fp32, that's 32 KB per column of W. If we have each thread compute one output element: - Read 32 KB from x (but x is shared across all threads in a block) - Read 32 KB from W column - Do 8192 multiply-adds With shared memory, we can tile this. But 8192 is quite large. Let's think about a block that computes a tile of outputs. For example, a block of 128 threads computing a 128-element chunk of an output row. For a 128-element chunk: 1. Load x into shared memory (or have it in registers with k-tiling) 2. For k=0 to 8192 in steps of K_Tile: - Load W[k:k+K_Tile, j_start:j_start+128] into shared memory - Compute partial dot products Wait, W is stored as [out_features, in_features] in PyTorch Linear (W^T is [in_features, out_features] for the matmul). Actually: - self.linear.weight has shape [out_features, in_features] = [8192, 8192] - In forward, it does x @ W^T, so W^T has shape [8192, 8192] - In memory, W is row-major: W[i, j] where i is out_feature index, j is in_feature index - W^T would have W^T[j, i] = W[i, j] So when computing the matmul x @ W^T, we're doing: output[m, n] = sum_k x[m, k] * W^T[k, n] = sum_k x[m, k] * W[n, k] So for output element (m, n), we multiply row m of x with row n of W. This is great! Because each output row n just needs row n of W, and W is stored row-major. But wait, in PyTorch nn.Linear, the weight is [out_features, in_features], and forward does x @ weight.t(). So yes, each output column n corresponds to weight row n. Wait no, output is [batch_size, out_features]. Let me clarify: - x: [M, K] = [1024, 8192] - weight: [N, K] = [8192, 8192] (out_features, in_features) - x @ weight.t(): [M, K] @ [K, N] = [M, N] = [1024, 8192] So output[m, n] = sum_k x[m, k] * weight[n, k] Each thread computing output[m, n] needs to read weight[n, :] which is a contiguous row of 8192 fp32 values = 32KB. This is actually interesting: each output element requires reading a different 32KB row of weights. With 8192 output elements per row, that's 256MB of weights total. If a block of 256 threads each computes one output element for a given row: - They all need to read x[m, :] (32KB, shared) - Thread n reads weight[n, :] (its own 32KB row) - Total weight read per block = 256 * 32KB = 8MB This is an enormous amount of register/shared memory pressure. Let me think about tiling differently. If we process K in tiles: - All threads in a block process the same batch of output elements for different n - For k_tile of say 64 elements: - Load x[m, k:k+64] into shared memory (256B) - Each thread loads weight[n, k:k+64] for its n (256B per thread) - Compute partial dot products - Move to next k tile For 256 threads and K=8192 with k_tile=64: - Number of tiles = 128 - Each thread does 128 iterations - Each iteration: load 64 fp32 weights (256B) into registers, load 64 fp32 x values into shared mem For H100: - L2 cache size is quite large (50MB) - 256MB weights won't fit in L2, but maybe 8MB for a block will partially fit Actually, with K=8192, and if we have 256 threads, and K_tile=64: - X shared mem: 64 * 4 = 256 bytes - Each thread keeps an accumulator - Weight access: each thread reads 64 contiguous floats per tile - For 8192 output elements per row, with 256 threads, we'd need 32 blocks per row Total weight reads = 8192 rows * 8192 cols * 4 bytes = 256MB This has to be read once no matter what. Actually, let me think about what PyTorch does: 1. x @ W^T: cuBLAS is highly optimized, probably gets near peak compute/bandwidth 2. The weight is 256MB, input is 32MB, output is 32MB Wait, 1024 * 8192 * 8192 * 2 = 137 GFLOPs. If we can get 60 TFLOPS, that's 2.3 ms. Memory bandwidth of 3.35 TB/s. 256MB weights + 32MB input + 32MB output = 320MB. At 3.35 TB/s pure bandwidth = 0.095 ms. But matmul is compute bound. For H100 with tensor cores: 989 TFLOPS in TF32, but the problem says fp32 precision is required, so we can't use TF32/TF16 tensor cores with reduced precision. We need actual fp32 computes. Actually wait, the problem says "Produce numerically equivalent outputs at fp32 precision." Does that mean the kernel must use fp32 internally? Or can we use tensor cores and still get fp32 accuracy? TF32 has 10-bit mantissa vs fp32's 23-bit, so numerical equivalence to fp32 would be difficult. So we should probably stick to actual fp32 computation, getting up to 67 TFLOPS. Actually, let me reconsider the kernel approach. For a 1024x8192 x 8192x8192 matmul: - M=1024, N=8192, K=8192 - cuBLAS should handle this very well This is compute bound. For compute bound problems, fusion doesn't help much because the bottleneck is the multiply-adds, not memory bandwidth. A fused kernel would save at most the memory bandwidth of two intermediate tensors (the linear output before GELU and the GELU output before softmax): - 1024 * 8192 * 4 bytes * 2 reads + 2 writes = relatively small compared to compute Wait, for a compute bound matmul, writing the output and then reading it again adds very little overhead because the computation is the bottleneck. So fusing the elementwise ops might not help much. However, there might be another angle. The reference PyTorch code has three separate kernel launches: 1. Linear forward (cuBLAS) 2. GELU (elementwise) 3. Softmax The overhead of kernel launches plus memory traffic for the elementwise operations might be non-trivial. But for H100, a kernel launch is ~5 microseconds, and the memory bandwidth is 3.35 TB/s. Let me compute more precisely: - Elementwise intermediate: 1024 * 8192 * 4 bytes = 32MB - GELU: read 32MB, write 32MB = 64MB total memory - Softmax: this is trickier. For a row of 8192 elements: - First pass: read all elements, compute max - Second pass: read all elements, compute exp(x-max), sum - Third pass: read/sum, write normalized values - Or fused two-pass. Either way, roughly 2x read + 1x write of the row. - For 1024 rows: ~96MB of memory traffic So total extra memory traffic beyond the matmul: ~160MB. At 3.35 TB/s = 0.048 ms. This is tiny compared to the 2ms matmul. So the matmul dominates, and we can't beat cuBLAS with a custom matmul easily. But wait! There might be something I'm missing. For M=1024, N=8192, K=8192, using cublas is probably the way to go. What if our fused kernel does: 1. Call cublas for the matmul (or use torch.matmul) 2. Fuse GELU + softmax in a single custom kernel? That would save one round trip to memory. But is that enough to beat reference? Actually, let me think about a different approach. What if we do a custom kernel that: - First computes the linear projection x @ W^T + b - Then does GELU - Then does softmax But compute x @ W^T on the GPU using a custom kernel that tiles appropriately and fuses everything. For the H100, let's think about what CUTLASS or a simple custom kernel can do. Actually, wait - maybe the reference implementation in PyTorch doesn't use an optimized softmax. Let me think about what PyTorch does for softmax. PyTorch softmax on a [1024, 8192] tensor along dim=1: - Usually launched as one block per row or multiple blocks per row depending on size - For 8192 elements per row, probably multiple blocks per row or one warp per row, etc. It's quite optimized. Hmm, what about an alternative: use FlashAttention-style online softmax during the matmul computation? Online softmax computes the softmax while streaming through the data, maintaining running max and sum. This is typically used in attention where you have a softmax(QK^T). The idea is: - For each tile of output, compute partial max and partial sum - Update running max and sum - At the end, normalize But wait, for our case, we can't do online softmax with the matmul because we need all the terms of the dot product to be complete before we can compute GELU, and then we need the entire row of GELU outputs to compute softmax. So we must complete the matmul for the entire row before starting softmax. What about a two-stage kernel? 1. First kernel: matmul + bias + GELU, output to intermediate buffer 2. Second kernel: softmax on intermediate buffer This is what PyTorch does, but we can fuse bias+GELU into the matmul epilogue. Or wait, can we do everything in shared memory? For a row of 8192 fp32 elements, that's 32KB. That fits in shared memory (227KB max per block). Ah! Here's an idea: - Each block processes one or more rows - For each row, compute the linear output in tiles, accumulating in registers/shared mem - Once a subset of output elements is computed, apply GELU and store in shared memory - But softmax needs the entire row... Actually, here's a better idea: process one row per block, multiple blocks per row for the matmul reduction, or one block per row if K dimension can fit/work well. For one block per row of output (M=1024, so 1024 blocks): - Each block has 256 threads (8 warps) - K = 8192 - We can tile K. Each thread processes N/256 = 8192/256 = 32 output elements. Wait, 8192/256 = 32. So each thread would compute 32 output elements. That's a lot. Alternatively, each block computes a tile of output elements for one row: - Block computes N_TILE output elements, say N_TILE = 256 - Number of blocks per row = 8192/256 = 32 - Total blocks = 1024 * 32 = 32768 blocks For K tiling: - K is 8192, maybe tile size 64 or 128 - Load x[k:k+K_TILE] to shared memory (all threads in block share it) - Each thread computes dot product for its assigned output elements by loading weight rows Actually, each output element n corresponds to weight row n. If a block handles output elements n_start to n_end, then for each k, we need to load weight[n_start:n_end, k]. If block handles 256 output elements and K_TILE=64, then we need weight[256, 64] = 16K floats = 64KB per tile. This could fit in shared memory. But weight is stored as [8192, 8192] row-major. So weight[n_start:n_end, k:k+K_TILE] is NOT contiguous for K_TILE. It's contiguous within a row but strided across rows. If we load weight[n, k:k+K_TILE] for n from n_start to n_end, each row has K_TILE contiguous elements. So we can load them coalesced. Actually, for matmul x @ W^T, output element (m, n) = dot(x[m, :], weight[n, :]). So for a block handling n in [n_start, n_end): - For each k_tile: - Load x[m, k:k+K_TILE] into shared memory (K_TILE floats) - Thread j loads weight[n_start + j, k:k+K_TILE] (but this is K_TILE floats per thread, which is a lot) Wait, if we have 256 threads and each handles one output element (one n), then: - Thread j computes output element n = n_start + j - For each k in K_TILE, thread j loads weight[n, k]... but that's not coalesced if each thread loads a different row. Actually, if threads in a warp read weight[n0, k], weight[n1, k], weight[n2, k], etc., where n differs by 1, and the weight is stored row-major, then weight[n, k] and weight[n+1, k] are separated by 8192 floats = 32KB. That's a huge stride. Not coalesced at all! For better coalescing, we'd want threads in a warp to access weight[n, k], weight[n, k+1], weight[n, k+2], etc., but that doesn't match our problem structure. Wait, actually for computing output[m, n] = dot(x[m, :], weight[n, :]), the reads for the inner dimension k are contiguous within weight[n, :]. So if thread j reads weight[n_j, k] for fixed k and varying n, that's strided. But if thread j reads weight[n_j, k0+k_tid] or something, we need to think differently. Actually, coalescing is about threads in a warp accessing contiguous memory. Let's say warp has threads 0..31. - If thread tid computes output for n = n_start + tid, then for a given k, thread tid reads weight[n_start + tid, k]. - Memory layout: weight[n, k] is at base + n * K + k. - Thread 0 reads: base + n_start * K + k - Thread 1 reads: base + (n_start+1) * K + k - ... - These are spaced K floats apart = 8192*4 = 32768 bytes apart. - This is terrible for coalescing! Each read would be a separate cache line. So to get coalescing, we should have threads in a warp work on the same n but different k, or adjacent k for the same n. Let me rethink. If a warp works on a single output element n, and each thread in the warp computes a partial sum over different k ranges: - Thread 0: k in [0, 256) - Thread 1: k in [256, 512) - ... - Then each thread reads contiguous weight[n, k_start:k_end] - This is perfectly coalesced in the sense that each thread reads contiguous, but different threads read different ranges, which is also fine. Then we do a warp reduction. But with K=8192 and 32 threads, each thread does 256 multiply-adds. Then we reduce. Hmm, but then the shared memory for x is minimal. And each warp handles one output element. With 32 warps per block (1024 threads), we can compute 32 output elements per block. For 8192 output elements per row, we need 256 blocks per row. For 1024 rows, 262,144 blocks. That's a lot but manageable. Actually, for H100, let's think about what occupancy we need. Let me consider a simpler approach that might work better: 1. For each output row, compute all linear outputs using simple loops + shared memory 2. Then do softmax on the row But a custom matmul kernel with poor coalescing will be much slower than cuBLAS. OK, let me try a completely different approach. What if we: 1. Use PyTorch's matmul for the linear layer (which calls cuBLAS) 2. Write a custom fused kernel for bias + GELU + softmax Wait, the linear layer in PyTorch already includes bias. So: 1. cuBLAS does matmul + bias 2. Then GELU kernel 3. Then softmax kernel So our custom kernel can fuse GELU + softmax. But actually, the output of linear has 32MB. Fusing GELU + softmax means reading 32MB and writing 32MB instead of: read 32MB write 32MB for GELU, then read 32MB write 32MB for softmax. That saves 32MB of writes + 32MB of reads = 64MB total memory traffic, which at 3.35 TB/s is about 0.02 ms. Not much. What if we can fuse bias + GELU into the matmul using a CUTLASS-like custom kernel? Again, the bias is already fused in cuBLAS. Wait, I think I'm overthinking this. For a compute-bound matmul, the speedup from fusing elementwise ops is minimal because the matmul dominates the time. Unless... is there a way to optimize softmax specifically? For softmax, PyTorch does something like: 1. Find max per row 2. Compute exp(x - max) and sum per row 3. Divide by sum For a row of length 8192, this touches 32KB per row. With 1024 rows, that's 32MB. The time is memory bandwidth bound. Can we make it faster? Actually, what if the issue is that PyTorch's softmax is not well optimized for this particular shape? Or what if we use warp-level primitives more efficiently? Hmm, let me think about this from a different angle. The problem says "Run faster than the PyTorch reference." The reference is: ```python x = self.linear(x) x = torch.nn.functional.gelu(x) x = torch.nn.functional.softmax(x, dim=1) ``` Maybe the speedup comes from: 1. Using Tensor Cores with FP32 accumulation for the matmul (if allowed) 2. Or using a custom kernel that parallelizes differently Wait, let me re-read: "Produce numerically equivalent outputs at fp32 precision." This likely means the OUTPUT should be fp32 precision equivalent to the reference. For H100, we could potentially use FP16/BF16 Tensor Cores with FP32 accumulation and still get fp32-equivalent results for typical random inputs. The tolerance is atol=rtol=1e-4 for fp32. Actually no, the problem states "at fp32 precision" which suggests we should use fp32 arithmetic, not fp16/bf16. But wait, can we use the code from the reference to compute W and b, and then do a custom fused kernel? The linear layer weights are fixed after initialization. Let me think about what we can actually implement quickly and correctly. A full custom matmul that beats cuBLAS is extremely hard. But maybe we can do: 1. Use torch.matmul for the linear part (cuBLAS is very fast) 2. For GELU and softmax, write a fused custom kernel This would save one kernel launch and one memory round trip. Would that be enough to beat the reference? Actually wait, can we do even better? What if we can write a kernel that does softmax more efficiently using vectorized loads/stores? Alternatively, what if we use CUTLASS or something similar? But with load_inline, we can write custom CUDA. Let me look at this from the perspective of what actually might make a difference: - The matmul is ~2ms - The GELU is ~0.02ms - The softmax is ~0.05ms - Total reference: ~2.07ms To get a meaningful speedup (say 1.1x or 1.2x), we'd need to significantly optimize the matmul or cut the overhead dramatically. A 10% speedup would require saving ~0.2ms. Actually, I realize I might be miscalculating the softmax time. For [1024, 8192], softmax along dim=1: - Each row: find max of 8192 elements, then compute exp and sum, then normalize. - Typical optimized softmax kernel on H100 might process multiple rows per block. - Time might actually be closer to 0.02-0.03ms or even less because of cache effects. Hmm, let me think about whether we can actually get a speedup at all with just fusing elementwise ops into a custom kernel after a standard matmul. It seems challenging to beat cuBLAS + PyTorch's native ops by any significant margin. Wait, but the evaluation only says "Run faster than the PyTorch reference." Any speedup counts. Even 1% might be enough. But let me think about whether the evaluation does multiple runs and what variance is like. Actually, I wonder if the reference itself is running on CPU or something? No, it says PyTorch reference. Let me think about another approach: can we compute the linear layer more efficiently using block-sparse or some other approximation? No, we need exact results. Wait, I have another idea. What if instead of doing x @ W^T, we can precompute something? No, W is learnable and initialized randomly. What about using CUDA streams or overlapping? Probably not helpful for this sequential computation. OK, let me consider writing a kernel that: 1. Does the matmul using tiling in shared memory 2. Fuses bias + GELU + softmax Even if it's slower than cuBLAS for the matmul, maybe the fusion of softmax (avoiding two intermediate memory round trips) makes up for it? For a fully fused kernel: - Read x: 32MB - Read W: 256MB - Read b: 32KB - Write output: 32MB - Total memory: ~320MB - At 3.35 TB/s: 0.095ms But compute: 137 GFLOPs at 67 TFLOPS = 2.05ms. So compute bound. If we can get 80% of peak on our custom kernel (1.6ms matmul equivalent), it would be faster overall because we avoid kernel launch overhead and intermediate memory traffic. But getting 80% of cuBLAS performance with a handwritten kernel is very hard for large square-ish matmuls. For H100, maybe we can use texture memory or something? No, that doesn't help for compute-bound code. Actually, wait. Let me reconsider the dimensions. M=1024, K=8192, N=8192. This is a case where M is much smaller than K and N. This is often called a "batched small-M" scenario, and depending on the exact dimensions, custom kernels can sometimes be competitive or even better than cuBLAS because cuBLAS is optimized for large M and N. For M=1024, K=8192, N=8192, the effective matrix multiplication is: - A: [1024, 8192] - B: [8192, 8192] (W^T) - C: [1024, 8192] This is a GEMM with M relatively "small" compared to N and K. There exist papers and implementations showing that for certain M-N-K ratios, custom kernels can outperform cuBLAS (e.g., Cutlass, FlashAttention-style kernels). Actually, 1024 is not that small. cuBLAS should handle this very well. But here's another thought: what if we use a kernel where each block computes a tile of output, and we process multiple rows per block? Or we process the K dimension in a particular way? Let me design a kernel architecture: - Grid: (num_row_tiles, num_col_tiles) - Each block computes a tile of output rows and columns - Tile size: say 32 rows x 32 cols, or 64x64, or 128x128 With M=1024, N=8192: - If tile is 64x64: grid = (16, 128) = 2048 blocks - Each block: 64 threads computing 64x64 output elements? No, usually more threads. Actually, for output tile of size BM x BN with BK as the K-tile: - Block has (BM * BN) / (elements per thread) threads - Or block has BM warps, each warp handles a sub-tile This is getting quite complex. Let me think if there's a simpler path. What about using CUTLASS through PyTorch? Not easily possible with load_inline. Hmm, what about a purely element-wise fusion? We can definitely write a fused bias+GELU+softmax kernel that operates on the output of a standard matmul. But the linear layer already includes bias in PyTorch's implementation (if we use nn.Linear). So we'd do: 1. torch.matmul(x, weight.t()) -- no bias to let us fuse it 2. Custom kernel: add bias, GELU, softmax Wait, actually if we use self.linear in ModelNew, it will do matmul+bias. Then we apply our custom kernel for GELU+softmax. That saves one round trip. Alternatively, we can do x @ W^T in PyTorch, then our custom fused kernel does bias+GELU+softmax. This saves fusing bias into GELU+softmax. But again, the matmul is ~2ms, and the elementwise savings might be 0.02-0.05ms total. That's only 1-2% speedup. Is that enough? Let me think about whether PyTorch's matmul+GELU+softmax might actually be slower than I estimated. The matmul calls cublas GEMM. GELU is an elementwise kernel. Softmax might be more expensive than I estimated. Actually, PyTorch softmax for large inner dimension generally involves: 1. A reduction to find the max 2. A second pass to compute exponentials and sum 3. A normalization pass For 8192 elements per row, the reduction and the computation can be done quite efficiently. But if PyTorch's implementation is generic and handles arbitrary dimensions, there might be some overhead. Let me try a different strategy: what if I write a custom kernel that does everything, but specifically optimized for this exact shape (M=1024, N=8192, K=8192)? A shape-specific kernel can often outperform general-purpose libraries because it doesn't need dimension dispatch overhead, can use optimal tile sizes, etc. For such a kernel: - Each block handles a fixed number of output rows, say 1 or 2 or 4 - Within a block, we use shared memory to cache x rows and weight tiles - Process K in tiles of say 64 or 128 Actually, here's a potentially effective design: 1. Use 256 threads per block 2. Each block computes a complete output row (all N=8192 elements) 3. But 8192 elements is a lot for one block. Instead: - Each block computes 1 row x 256 cols (or similar) - Grid: 1024 rows x 32 col_blocks = 32768 blocks For K tiling with BK=64: - Load x[m, k:k+64] into shared memory (256B, all threads in block share) - Thread tid computes output[m, n_start + tid] - To do this, thread tid needs weight[n_start + tid, k:k+64] - As discussed before, this has terrible memory coalescing because each thread reads a different row To fix coalescing, we can transpose our access pattern. Instead of assigning each thread a unique n, we can have warps work on sub-tiles where threads within a warp access adjacent elements of weight. Wait, for weight[n, k] with fixed n, threads accessing adjacent k are coalesced. But in our dot product, for a given output n and step k, we need weight[n, k]. What if we have the threads in a warp cooperate to load a tile of weight into shared memory, then compute? For example: - Block computes output for rows in [m_start, m_end) and n in [n_start, n_end) - Let's say block computes 4 rows x 64 cols (256 outputs total) - BK = 64 - Each iteration: - Load x[m_start:m_end, k:k+64] into shared memory: 4 x 64 = 256 floats = 1KB - Load weight[n_start:n_end, k:k+64] into shared memory: 64 x 64 = 4096 floats = 16KB - Total shared memory: 17KB, fits easily - Then compute dot products using shared memory For loading weight[n_start:n_end, k:k+64]: - This is a 64x64 tile of weight - weight is stored row-major as [8192, 8192] - So weight[n_start:n_end, k:k+64] has 64 rows, each row has 64 contiguous floats - Threads can load this coalesced: thread 0 loads weight[n_start, k:k+4], thread 1 loads weight[n_start, k+4:k+8], etc. - Or better, have each thread load a contiguous chunk This is much better! Because we're loading a tile of weight into shared memory first (coalesced), then all threads in the block can read from shared memory without worrying about coalescing. So the kernel structure is: ``` for m_block in grid: for n_block in grid: for k_tile: load x_tile into shared mem (coalesced) load w_tile into shared mem (coalesced) accumulate in registers apply bias, GELU ``` Then after computing the full linear output, we need softmax. But wait - softmax requires the full row. If a block only computes 64 columns of a row, it doesn't have the full row for softmax. So we'd either need: 1. A separate kernel for softmax after writing to global memory 2. Have blocks cooperate via shared memory / global atomics for softmax 3. Have each block compute an entire row (but N=8192 might be too large) If each block computes an entire row of 8192 elements: - We need 8192 accumulators per thread, or we need many threads - With 1024 threads per block, each thread computes 8 output elements - But we'd need to store all 8192 elements in registers / shared memory for softmax - 8192 floats = 32KB, which fits in shared memory (227KB) So one block could compute one full row: - Block size: 1024 threads (or 512, or 256) - Grid: 1024 blocks (one per row) - K tiles of 64: need to load 64 elements of x into shared memory each iteration - For output, if we have 1024 threads, each computes 8 outputs. Wait, 8192/1024 = 8. - So thread tid computes outputs for n in [tid*8, tid*8 + 8) Wait, but then for loading weight, we'd need to load weight for all n in [0, 8192) for each k tile. That's 8192 floats per k tile = 32KB. With BK=64, each k tile requires loading 32KB of weight. At 8192/64 = 128 tiles, total weight read = 128 * 32KB = 4MB per row. But the actual weight is 32KB * 8192 = 256MB. Hmm, no. Weight for the entire row is 8192 * 8192 floats = 256MB. For one row of output and BK=64, we load 8192 * 64 floats = 524,288 floats = 2MB per k_tile. Over 128 tiles: 256MB total. Yes. Actually wait, for one output row, we need all of W (256MB total). If we process in tiles of BK=64, for each tile we load a [8192, 64] chunk of W. That's 8192*64*4 = 2MB per tile. Over 128 tiles: 256MB total, correct. So per block: - Shared memory for W tile: 8192 * 64 * 4 bytes = 2MB. TOO BIG! Max shared memory is 227KB. Ah, so we can't load the entire N dimension into shared memory at once. We'd need to tile N as well. Let me think about this differently. We can tile both N and K: - N_TILE = 256 - K_TILE = 64 - For each N tile, we iterate over K tiles - Shared memory: K_TILE * N_TILE = 64 * 256 = 16K floats = 64KB. Fits! - Also need K_TILE elements of x: 64 * 4 = 256 bytes. Wait, but if we tile N, then we process the output in N_TILE chunks. For each row, we'd need to: 1. Compute the first N_TILE linear outputs (by iterating over K) 2. Compute the second N_TILE linear outputs 3. etc. 4. Then we have all N outputs and can do softmax But step 4 requires having all outputs available. If a single block computes an entire row with N-tiling, it would need to keep all N=8192 outputs in registers/shared mem, or write to global memory and read back. Actually, a block could: - Process the row in N-tiles, accumulating partial results in registers - But it can't finalize softmax until all N outputs are computed - So it would need to store 8192 intermediate values 8192 * 4 bytes = 32KB. A block can have 32KB in shared memory, or threads can hold values in registers. - With 1024 threads, each holds 8 floats in registers = doable - But we'd need to tile K, so for each N-tile and K-tile, we do a matrix multiply micro-kernel Actually, for a single block doing one row with N-tiles: - N_TILE = 256, so 32 N-tiles total - For each N_TILE: - Initialize accumulators for the 256 outputs - For each K-tile (BK=64, 128 tiles): - Load x[k:k+64] into shared memory (256B) - Load weight[n:n+256, k:k+64] into shared memory (64KB) - Have 256 threads, each computing one output element - Wait, 256 threads * 1 output each = 256 outputs, perfect! - Thread tid accumulates for n = n_start + tid - After all K tiles, we have 256 linear outputs - Apply bias and GELU - Now, we can't do softmax yet because we need all 8192 outputs for the row So we'd either: 1. Store the GELU outputs in global memory as a temporary buffer 2. Or accumulate all 8192 outputs in the block, then do softmax Option 2: accumulate all 8192 outputs in the block. - We process N-tiles sequentially - For each N-tile of 256 elements, we compute and store in shared memory - After all 32 N-tiles, shared memory has 8192 floats = 32KB - Then do softmax on the 32KB - Then write to global memory Shared memory budget: - x tile: 64 * 4 = 256B - W tile: 256 * 64 * 4 = 64KB - Output buffer: 8192 * 4 = 32KB - Total: 96.25KB, well under 227KB! Thread count: - We have 256 threads computing the N-tile. - Each thread computes one output element for the current N-tile. - Over 32 N-tiles, each thread computes 32 output elements total. - Grid: 1024 blocks (one per row) Wait, 1024 blocks for 1024 rows. For an H100 with 132 SMs (I think H100 has ~132 SMs for the SXM version, or 114 for PCIe). Either way, 1024 blocks / ~132 SMs = ~8 blocks per SM. With good occupancy, this should work. Let me check H100 SMs: - H100 has 132 SMs for SXM5, 114 for PCIe - Each SM can have up to 32 blocks - But limited by registers and shared memory With 96KB shared memory per block and 228KB per SM: - At most 2 blocks per SM (96 * 2 = 192KB < 228KB, 96 * 3 = 288KB > 228KB) - So 2 blocks per SM concurrently Total concurrent blocks = 132 * 2 = 264 blocks concurrently. Total blocks = 1024. So it takes about 1024/264 ≈ 3.9 waves. That's fine. But wait, we're launching only 256 threads per block. On H100, each SM can have up to 2048 threads. With 2 blocks per SM, that's 512 threads per SM, which is 25% occupancy. Not great, but might be OK since we're compute bound. Can we increase to more threads? If we use 512 threads per block: - Each thread computes 256/512 = 0.5 outputs for an N-tile. Hmm, awkward. - Or we can make N_TILE = 512 - Then W tile = 512 * 64 * 4 = 128KB - Output buffer = still 32KB - Total shared = 160KB - Blocks per SM = 1 (since 160KB > 114KB, actually might only fit 1) - Threads per block = 512 - Concurrent blocks = 132 - Waves = 1024/132 ≈ 7.8 - Occupancy per SM = 512/2048 = 25% Or N_TILE = 128, threads = 128: - W tile = 128 * 64 * 4 = 32KB - Output buffer = 32KB - Total = 64.25KB - Blocks per SM = 3 (64 * 3 = 192KB, close to limit) - Actually, let's say 3 blocks per SM - Threads per block = 128 - Total threads per SM = 384 - Occupancy = 384/2048 = 18.75%. Worse. Let me try N_TILE = 256, threads = 256, shared = 96KB, 2 blocks per SM = 512 threads = 25%. Actually, maybe we can use registers more aggressively and reduce shared memory for the output buffer. 8192 floats across 256 threads = 32 floats per thread in registers. That's very manageable! Each thread holds 32 floats. Then the kernel becomes: - Thread tid in block handles output elements: tid, tid+256, tid+512, ..., tid+(31)*256 - Wait, for 256 threads and 8192 outputs per row, each thread handles 32 outputs. - If we process K-tiles in the inner loop, we can compute all 32 outputs simultaneously. Actually, for K-tile approach: - For each N-tile of 256 elements: - But if each thread handles 32 scattered outputs, we can't process N-tiles sequentially that way easily. Let me go back to the sequential N-tile approach but keep the 8192 outputs in registers: - Block handles 1 row, 256 threads - Each thread computes 32 output elements (spaced by 256) - Thread 0: n=0, 256, 512, ..., 7936 - Thread 1: n=1, 257, 513, ..., 7937 - ... - For each K-tile (BK=64): - Load x[k:k+64] into shared memory - For each of the 32 output positions per thread: - Load weight[n, k:k+64] - Compute dot product partial Wait, thread 0 needs weight[0, :], weight[256, :], weight[512, :], etc. These are rows 0, 256, 512, ... of W. In memory, they're separated by 8192*4 = 32KB. So for a given k, reading weight[0, k], weight[256, k], weight[512, k] across a warp would be very strided and uncoalesced. Hmm, this is a fundamental problem with our access pattern. In PyTorch, W is [out_features, in_features] = [8192, 8192] row-major. When computing x @ W^T, output[m, n] = dot(x[m, :], W[n, :]). So each output element n needs row n of W. For coalescing, if we have thread tid computing output n = base + tid, then for a given k, the threads read W[base + tid, k]. Since adjacent threads read adjacent rows, and rows are separated by K floats, the accesses are strided by K*4 bytes. For K=8192, stride = 32768 bytes = 32KB. This means each thread in a warp reads from a different cache line -> terrible coalescing. The solution is to have the THREADS in a warp work on the SAME output but different K values. Or have them load weight into shared memory in a coalesced way. For the N-tile approach: - We want to load W[n_start:n_start+N_TILE, k:k+BK] into shared memory - This is an N_TILE x BK tile - W[n, k] is at address base + n*K + k - For fixed k, W[n, k] across n is strided by K - But for fixed n, W[n, k:k+BK] is contiguous So to load this tile coalesced: - Have thread tid load a contiguous chunk of a row - Or transpose: have threads load columns of the tile Actually, W[n_start:n_start+N_TILE, k:k+BK] is a contiguous sub-matrix. If we view it as N_TILE rows of length BK: - Row 0: W[n_start, k:k+BK] - contiguous BK floats - Row 1: W[n_start+1, k:k+BK] - contiguous BK floats, starts K floats after row 0 To load the tile into shared memory: - Option A: Load row by row. Each row is contiguous. We can have 1 thread load a whole row, or multiple threads load parts of a row. - If BK=64 and N_TILE=256: 256 rows of 64 floats each. - With 256 threads: each thread loads 1 row of 64 floats (coalesced within each thread). - Thread 0 loads W[n_start, k:k+64] - Thread 1 loads W[n_start+1, k:k+64] - These are separated by K*4 = 32KB in memory. Not coalesced across threads in a warp! Option B: Transpose the load. Have threads in a warp load a column of the tile. - Column j consists of W[n_start, k+j], W[n_start+1, k+j], ..., W[n_start+31, k+j] - In memory: addresses are base + n_start*K + k+j, base + (n_start+1)*K + k+j, etc. - Stride between consecutive elements is K*4 = 32KB. - Reading with a stride of 32KB within a warp is also terrible. So loading weight tiles from global memory into shared memory is problematic because W is not stored in a way that's friendly to reading a subset of rows in a tiled manner. We'd get poor cache utilization. Wait, but GPUs have L2 cache. If threads in a block read different rows, they might hit different cache lines, but the cache can handle some amount of strided access as long as it's not too bad. With 32KB between rows, each float read is a separate 128-byte cache line for a warp of 32 threads. That's 32 cache lines per warp per read. That uses 4096 bytes of cache for 128 bytes of useful data - a 1/32 cache efficiency. This is terrible. There's really no way around this fundamental issue: for x @ W^T with W stored as [N, K], each output column corresponds to a different row of W, and accessing different rows from different threads in a warp has strided access. The only way to get coalesced access to W is if all threads in a warp access the same row n, but different k values. That means all threads in a warp should compute the SAME output element n, but different partial sums over k. Then each thread reads W[n, k_start:k_end] which is contiguous. This is the "reductive GEMM" approach: - Warp computes one output element (one n) - Threads in the warp split the K dimension - Each thread reads a contiguous chunk of W[n, :] and x[m, :] - Do a warp shuffle reduction to accumulate - Apply bias, GELU - Then for softmax, we have a problem because each warp only produces one output If we use this approach, to compute a full row of 8192 elements, we need 8192 warps per row. With 32 warps per block, that's 256 blocks per row. For 1024 rows, 262,144 blocks total. Each block has very few threads (1024 threads but each warp works independently). For H100 with 132 SMs, that's 1987 blocks per SM. With good scheduling, this could work. Let's analyze the memory access pattern: - For a block computing 32 output elements (one per warp), and K=8192: - Each warp reads x[m, k_start:k_end] where k_end-k_start = 8192/32 = 256 elements - Each warp reads W[n, k_start:k_end] for its n - x is read from shared memory or all threads load it cooperatively - Actually, x can be loaded into shared memory once per block: 8192 floats = 32KB - 32KB shared memory per block, which allows multiple blocks per SM Wait, but each warp in the block handles a DIFFERENT n. So the block handles n in [n_start, n_start+31). The x values are shared across all warps in the block. For a block of 32 warps (1024 threads): - Load x[m, :] into shared memory = 32KB (all warps share this) - Warp w computes output[m, n_start + w] - Thread t in warp w computes partial sum of W[n_start + w, t*256:(t+1)*256] dotted with corresponding x elements - After loading x into shared mem, threads read from shared mem (fast, no coalescing issues) - Threads also read W[n_start + w, t*256:(t+1)*256] from global memory. Since each thread reads its own contiguous chunk, and threads in a warp read the same row w but different k ranges, these accesses ARE coalesced! - Wait, all threads in warp w access W[n, :], but thread t accesses offset t*256. Are these coalesced? - Thread 0: W[n, 0:1] (first float) - Thread 1: W[n, 256:257] - In memory, these are 256 floats apart = 1024 bytes apart. - That's a stride of 1024 bytes per thread. Not coalesced at all! For coalesced access within a warp, adjacent threads should access adjacent memory. So: - Thread t in warp w should read W[n, t] or W[n, t*VECT:] where VEC is the vector size and threads read contiguous vectors. For K=8192, if we have 32 threads per warp, each thread could compute a partial sum over K/32 = 256 elements. But to make reads coalesced: - Thread t reads W[n, t], W[n, t+32], W[n, t+64], ..., W[n, t+256-32] - Memory access: W[n, t] for t=0..31 is contiguous (32 floats = 128 bytes = one cache line) - Perfect coalescing! Yes! So each thread in a warp reads elements with a stride of 32, but the initial offset t makes the first read of all threads contiguous. Let's detail this: - Warp computes one output: n fixed - K = 8192, K / 32 = 256 elements per thread - But for coalescing, thread t reads W[n, t], W[n, t+32], W[n, t+64], ..., W[n, t+32*(L-1)] where L = 256 - First access: thread t reads W[n, t]. In a warp, t=0..31. Addresses: base + n*K + 0, base + n*K + 1, ..., base + n*K + 31. Contiguous! Coalesced! - Second access: thread t reads W[n, t+32]. Addresses: base + n*K + 32, ..., base + n*K + 63. Contiguous! Coalesced! - And so on. This is perfect coalescing for W. For x: - Thread t reads x[m, t], x[m, t+32], ..., x[m, t+32*(L-1)] - Same pattern, perfect coalescing. So each warp computes ONE output element using perfect coalescing. For a block with 32 warps: - Computes 32 output elements (n_start to n_start+31) - All warps share x[m, :], which is loaded into shared memory - Wait, if each thread reads x with coalescing, we can just have all threads read x from global memory directly with coalescing. - But x is shared across all 32 warps. If each warp independently reads x, we'd read x 32 times per block. Since x is 32KB, and blocks per row might be many, this is a lot of redundant reads. - Better: load x[m, :] into shared memory once per block. 1024 threads cooperate to load 8192 floats = each thread loads 8 floats. Coalesced. - Then all warps read x from shared memory. For W: each warp reads a different row W[n, :]. Since each row is 32KB, and a block has 32 warps, the block reads 32 * 32KB = 1MB of W per row. - For 8192 output elements and 32 elements per block, we need 256 blocks per row. - Total blocks: 1024 rows * 256 = 262144 blocks. Actually, W is different for each warp - each warp reads its own row. So there's no sharing of W across warps. Each block reads 1MB of W. Total W reads across all blocks: 1024 * 256 * 1MB? No. - Each W row is read exactly once (by the block computing that output element). - Total W reads: 8192 rows * 32KB/row = 256MB. Good, no redundant reads. - Wait, each block computes 32 outputs. There are 256 blocks per row. Each block reads 32 rows of W. Total: 256 blocks * 32 rows/block = 8192 rows. Correct. For x: each block reads one row of x (32KB). There are 256 blocks per row * 1024 rows = 262144 blocks. But x only has 1024 rows. So each x row is read 256 times (once per block working on that row). Total x reads: 1024 * 256 * 32KB = 8GB. That's 256x redundant reads! That's terrible. So we need to avoid redundant x reads. One option: have a single block per row compute all 8192 outputs. But then the block needs 8192 warps? No, we can have fewer warps but each warp compute multiple outputs. Option: one block per row, with 256 warps? Impossible, max 32 warps per block. Option: one block per row, with 32 warps, each warp computes 256 outputs sequentially. - Warp w handles n in {w, w+32, w+64, ..., w+256*31}... no wait, 32 warps * 256 outputs each = 8192. - So warp w computes outputs for n = w, w+32, w+64, ...? Actually, each warp needs to compute 8192/32 = 256 outputs. - But if a warp computes 256 outputs, and each output requires reading a different row of W, and the warp processes them one at a time: - For output n=w: warp reads W[n, :] and dot with x[:] - For output n=w+32: warp reads W[n, :] and dot with x[:] - But x is in shared memory, so no redundant x reads! - W reads: 256 rows * 32KB = 8MB per block - Total W reads: 8192 per row * 1024 rows * 32KB = 256MB. Correct, no redundancy. - x reads: 1024 rows * 32KB = 32MB. Correct, no redundancy. This is much better! One block per row, 32 warps per block, each warp computes 256 outputs. For each warp: - Handles 256 different n values - For each n: reads W[n, :] from global memory (32KB) with coalesced pattern - x is already in shared memory (32KB) - Computes dot product - Apply bias, GELU - Store result somewhere Wait, but 256 outputs per warp means 256 iterations. Each iteration reads 32KB of W. Total W reads per warp = 8MB. At memory bandwidth of 3.35 TB/s = 3350 GB/s, one warp needs 8MB / 3350GB/s = 0.0024 ms = 2.4 microseconds. But with 32 warps in parallel (across SMs, not just within a block), and the GPU processing blocks... Actually, we have 1024 blocks total (one per row). On 132 SMs, about 7.8 blocks per SM. Each block has 32 warps. With each block reading 8192*32KB = 256MB of W and 32KB of x. Wait, a block computes all outputs for one row. So it reads the entire W for that row? No! The block computes ALL n for a single row m. That means it needs ALL rows of W (since output n needs W[n, :]). So each block reads the ENTIRE W matrix: 256MB! That can't be right. Let me re-examine. For row m of output: output[m, n] = dot(x[m, :], W[n, :]) for n=0..8191. To compute ALL outputs for row m, we need ALL rows of W. So each block reads the entire W matrix: 256MB. For 1024 rows, total W reads = 1024 * 256MB = 256GB. That's 1024x redundancy! Oh no, that's much worse. The W matrix is shared across all rows. For computing different rows of output, the same W rows are reused, but if each block computes one row independently, each block must read all of W. That's 1024 redundant reads of W. This is terrible unless W is cached. The L2 cache on H100 is about 50MB. W is 256MB. So W doesn't fit in L2 cache. Each block would read W from HBM. 256GB of W reads at 3.35 TB/s = 76ms. That's absurdly slow. So the "one block per row" approach is terrible for W reuse. We need multiple rows to share the same W reads. That means a block should compute multiple rows simultaneously, or we need to tile by N such that a block computes a subset of n for multiple m. Back to tiling. Let me think about what tiling makes sense. In standard GEMM, we tile by M, N, and K to maximize reuse. For our problem, the standard approach is: - Assign a block to compute a tile of outputs: e.g., BM rows x BN cols - The block loads a tile of A (BM x BK) and a tile of B (BK x BN) into shared memory - B is W^T. But W^T is [K, N]. Wait, W is [N, K], so W^T is [K, N]. - If we work with W^T, then B = W^T has shape [K, N]. - B[k, n] = W[n, k] - A tile of B with k in [k_start, k_start+BK) and n in [n_start, n_start+BN): - B[k_start, n_start] = W[n_start, k_start] - B[k_start, n_start+1] = W[n_start+1, k_start] - This is strided by 1 in the n dimension in B, but in W it's strided by K. Wait, if B is W^T in row-major: - B[k, n] is at address base_B + k*N + n - W[n, k] is at address base_W + n*K + k - B[k_start:k_start+BK, n_start:n_start+BN] is a BK x BN tile in B - In W, this corresponds to W[n_start:n_start+BN, k_start:k_start+BK] - The mapping is: B[row r, col c] = W[n_start + c, k_start + r] So in B's layout, a tile is contiguous in BN dimension but strided in BK dimension (stride N). Actually, B[k, n] with n varying is contiguous (since it's row-major). Wait, B[k, n] where k is fixed and n varies from n_start to n_start+BN: these are B[k, n_start], B[k, n_start+1], etc. These are contiguous in memory! So for loading B[k_start:k_start+BK, n_start:n_start+BN]: - For a fixed k, W[n_start:n_start+BN, k] accessed as B[k, n_start:n_start+BN] is contiguous in B! - But B is W^T. Is PyTorch's nn.Linear stored with W or W^T? - In PyTorch, nn.Linear weight is [out_features, in_features]. So W[n, k]. - W^T would be [in_features, out_features], but W^T is not explicitly stored - it's just a concept for the matmul. When we do x @ W^T in PyTorch, it calls cuBLAS with matrices A=x [M, K], B=W.t() [K, N]. PyTorch passes W.t() as the B matrix. In cuBLAS, the B matrix is accessed as B[k, n] where k is the contracting index and n is the output index. But W.t() is the transpose of W. In memory, the elements of W.t() are at positions corresponding to W[n, k]. So W.t()[k, n] = W[n, k]. The memory layout of W.t() as a [K, N] matrix: element (k, n) is at the same memory location as W[n, k]. But W is stored row-major. So W[n, k] is at base_W + n*K + k. So W.t()[k, n] is at base_W + n*K + k. If we view this as a [K, N] matrix in row-major order: - Element (k, n) should be at base + k*N + n for row-major. - But W.t()[k, n] is at base_W + n*K + k. - This is NOT row-major in k! It's transposed. So when we pass W.t() to cuBLAS, cuBLAS handles the non-standard striding. But for our custom kernel, if we want to read a tile of W corresponding to W[n_start:n_start+BN, k_start:k_start+BK], we need to think about the memory layout. For our custom kernel computing output[m, n] = dot(x[m, :], W[n, :]): - We need W[n, k] for n in tile and k in tile - W[n, k] in memory: base + n*K + k If we want a block to compute a BN-wide tile of outputs for BM rows: - We need to load W[n_start:n_start+BN, k_start:k_start+BK] - This is BN rows, each of length BK, starting at k_start - In memory: row n is at base + n*K + k_start, length BK - To load this into shared memory, for a given k column, W[n_start, k], W[n_start+1, k], ... are separated by K floats The standard way to handle this in a custom GEMM kernel is: 1. Use K-tiling in the outer loop 2. For each K-tile, load a tile of A into shared memory and a tile of B into shared memory 3. But for the B tile (W[n_start:n_start+BN, k_start:k_start+BK]), we need to read strided memory. Actually, in shared memory, we can store it in any layout we want (e.g., transpose it during load). The load from global memory: - Have threads load W[n, k_start + local_k] for their assigned n - Even though adjacent threads in a warp might read different n (stride K), this is not coalesced. But we can transpose the assignment! - Have thread (local_k, local_n) load W[n_start + local_n, k_start + local_k] - If local_k is the threadIdx.x (fast varying dimension), then for fixed local_n, threads with different local_k read W[n, k_start + local_k] - These are contiguous in memory! Coalesced! - And each thread loads one element. Wait yes! If we assign threads as: - Thread tid corresponds to (local_n = tid / BK, local_k = tid % BK) in some sense... no. Let's say BN = 64, BK = 64. - We have a block of 256 threads. - We want each thread to load one element of the W tile. - For coalescing, adjacent threads in a warp should access adjacent memory. - Adjacent memory means adjacent k for the same n. - So we can map threadIdx.x to k, and threadIdx.y to n. - E.g., blockDim.x = 64, blockDim.y = 4. - Thread (tx, ty) loads W[n_start + ty, k_start + tx] - Adjacent threads in x-dimension read W[n, k_start], W[n, k_start+1], etc. -> contiguous, coalesced! - Different y values read different rows, but that's OK because threads in a warp have the same y (if warp size 32 > blockDim.x? Actually, if blockDim.x=64, a warp contains threads with x=0..31, y=0). Wait, in CUDA, threads in a warp are ordered by threadIdx.x first, then y, etc.: - Thread 0: (0, 0) - Thread 1: (1, 0) - ... - Thread 63: (63, 0) - Thread 64: (0, 1) - etc. So if blockDim.x = 64: - Warp 0: threads (0..31, 0) - These threads read W[n_start, k_start + 0..31] -> contiguous, coalesced! - Warp 1: threads (32..63, 0) - These threads read W[n_start, k_start + 32..63] -> contiguous, coalesced! - Warp 2: threads (0..31, 1) - These threads read W[n_start+1, k_start + 0..31] -> contiguous, coalesced! This is great! Each warp reads a contiguous row chunk. And across warps, we read different rows. The pattern is very coalesced because each warp handles a contiguous chunk. For the x tile [m_start:m_start+BM, k_start:k_start+BK]: - Similar logic. x[m, k_start + tx] for fixed m is contiguous. - If blockDim.x = 64, threads read x[m_start, k_start + tx], x[m_start, k_start + tx], etc. - Wait, if BM=4 and blockDim.y=4, thread (tx, ty) reads x[m_start + ty, k_start + tx] - Warp 0: threads (0..31, 0) read x[m_start, k_start + 0..31] - Warp 1: threads (32..63, 0) read x[m_start, k_start + 32..63] - Warp 2: threads (0..31, 1) read x[m_start+1, k_start + 0..31] - Also coalesced! Now in shared memory, we store the W tile. But how we store it matters for computation. In the compute phase, we want to access W[n, k] and x[m, k] where k is the contracting dimension. We'd iterate k inside shared memory. If we store W_tile[BN][BK] in row-major (n varying, k varying): - To compute output[m, n], we want to access W[n, k] for k=0..BK-1 - So for fixed n, we access W[n, 0], W[n, 1], etc. - contiguous in shared memory. Good. Similarly, x_tile[BM][BK] in row-major: - For fixed m, x[m, 0], x[m, 1], etc. - contiguous. Good. So the kernel design: - Block computes BM rows x BN cols of output - Grid: (ceil(N/BN), ceil(M/BM)) - For each k_tile (BK elements): - Load x_tile[BM][BK] into shared memory (coalesced) - Load w_tile[BN][BK] into shared memory (coalesced, using the thread mapping described) - Synchronize - Each thread computes partial dot products for its assigned (m, n) - We need to synchronize before loading next tile - After all k_tiles, apply bias, GELU - Then we need softmax. But again, a block only computes BN columns, not the full row. For softmax, we'd need a second pass or a separate kernel. But we can avoid writing to global memory between the two by having a cleverer approach. Actually, for a fixed row m, if we want to compute softmax over all 8192 elements, and our blocks compute only BN=64 elements at a time, we'd need to: 1. Compute all linear outputs for the row 2. Find max over the row 3. Compute exp and sum 4. Normalize This requires coordination across blocks. We could do an atomic reduction. But maybe there's a simpler approach. Since M=1024 and N=8192, what if we process the row in a single block? But BN=8192 is a lot. Wait, let's revisit. For M=1024, N=8192, K=8192: - We could set BM=1 (one row per block) and have multiple blocks per row for different N-ranges. - Then each row has 8192/BN blocks. - To do softmax, we'd need to: a) First compute all linear outputs for the row into a temporary global memory buffer b) Then do softmax on the buffer This is what PyTorch does! Alternatively, can we reduce the matmul time enough to make up for the separate softmax? Probably not, since cuBLAS dominates. But our custom kernel can fuse bias+GELU into the matmul epilogue. Actually, I think the most practical approach is: 1. Use cuBLAS for the matmul: x @ W^T (fastest matmul) 2. Write a custom fused kernel for bias + GELU + softmax But the issue is that nn.Linear already includes bias. If we use nn.Linear, we get matmul+bias. Then we can only fuse GELU+softmax. That's one memory round trip saved. Alternatively: 1. Use F.linear or manual matmul without bias 2. Custom kernel for bias + GELU + softmax This way we still get cuBLAS for matmul, and fuse everything else. Wait, can we use torch.mm and fuse bias+GELU+softmax? Yes: ```python out = torch.mm(x, self.weight.t()) # no bias out = fused_kernel(out, self.bias) # adds bias, applies gelu, applies softmax ``` This avoids the bias in cuBLAS, but cuBLAS+fused bias in a custom kernel is about the same as cuBLAS with bias. Actually, maybe cuBLAS can fuse bias more efficiently. So maybe: ```python out = F.linear(x, self.weight, self.bias) # uses cuBLAS with bias out = fused_gelu_softmax(out) ``` The second approach gives us the best matmul and still allows fusing GELU+softmax. The question is: is fusing GELU+softmax enough to beat the reference? Actually, I just realized something important. PyTorch's nn.Linear forward with bias typically writes to an output tensor, then GELU reads and writes, then softmax reads and writes. That's: - Linear: reads x (32MB), W (256MB), writes out (32MB), reads b (32KB) - GELU: reads out (32MB), writes out2 (32MB) - Softmax: reads out2 (32MB), writes final (32MB) Total memory traffic: 32 + 256 + 32*4 + 32*3 = 32 + 256 + 128 + 96 = 512MB? Wait, that's approximate. Actually: - Linear: read x (32MB), read W (256MB), read b (32KB), write temp1 (32MB) - GELU: read temp1 (32MB), write temp2 (32MB) - Softmax: read temp2 (32MB), write output (32MB) Total HBM traffic: ~416MB. For our fused GELU+softmax after linear: - Linear: same as above, ~320MB - Fused kernel: read temp1 (32MB), write output (32MB) - Total: ~384MB Wait, that saves 32MB of memory traffic. At 3.35 TB/s, 32MB / 3350GB/s = 0.0095ms = 9.5 microseconds. That's negligible compared to the 2ms matmul. But what if the GELU kernel has launch overhead and the softmax kernel has launch overhead? Launch overhead is ~5 microseconds each. So we save ~10 microseconds of launch overhead plus 10 microseconds of memory traffic. Total ~20 microseconds. On a 2ms operation, that's 1% speedup. Hmm, but what if we can do a custom kernel that is faster than cuBLAS for this particular shape by a more significant amount? Or what if we can use the GPU more effectively? Actually, wait. For small M relative to N and K, I wonder if a custom kernel that fuses everything might be competitive because: 1. cuBLAS has some overhead for small batch sizes 2. Kernel fusion eliminates temporary allocations And for H100, memory bandwidth is so high that a bandwidth-efficient but compute-inefficient kernel might still be fast. But no, this is compute-bound. Let me think about another angle: can we use multiple streams or some clever scheduling? Probably not in a single forward pass. Wait - what if we use a TILE-based custom kernel where each block computes a tile of outputs but we DO tile the matmul well? For example: - Use BM=4, BN=64, BK=64 - Block size: 256 threads - Grid for M=1024, N=8192: (128, 256) = 32768 blocks - Each block computes a 4x64 tile - Block reads x[4][64] into shared memory and W[64][64] into shared memory - Accumulates in registers - After all K tiles (128 iterations), apply bias and GELU, write to temporary global memory - Then a separate softmax kernel For this kernel, the shared memory per block: - x_tile: 4 * 64 * 4 = 1KB - W_tile: 64 * 64 * 4 = 16KB - Total: 17KB - Blocks per SM: 228KB / 17KB ≈ 13 blocks (but limited by 32 blocks/SM and thread count) - Threads per block: 256. Threads per SM: 2048. So 8 blocks per SM max by threads. - 8 * 17KB = 136KB < 228KB, so thread-limited to 8 blocks per SM. - Occupancy: 8 * 256 = 2048 threads = 100% of threads per SM. Good! But the compute per thread is small. Each thread computes 4*64/256 = 1 output element. Actually, if block computes 256 outputs with 256 threads, each thread computes 1 output. That seems low. With BK=64 and 128 K-tiles, each thread does 128 multiply-adds. That's very little work. The kernel would be dominated by memory loads and synchronization overhead. If we increase the tile size: - BM=8, BN=128, BK=128 - 8*128 = 1024 outputs per block - Threads: 256 (each computes 4 outputs) - Shared mem: 8*128*4 + 128*128*4 = 4KB + 64KB = 68KB - Blocks per SM by shared mem: 3 (204KB) - Blocks per SM by threads: 8 - So limited to 3 blocks per SM - Occupancy: 3 * 256 = 768 / 2048 = 37.5% - Each thread does 128 K-tiles * 4 outputs * 128 K-elements... wait. - K=8192, BK=128, so 64 K-tiles. - Each thread computes 4 output elements. - For each K-tile, each thread loops over BK=128 elements in shared memory. - Total MACs per thread: 4 * 64 * 128 = 32768 MACs = 65536 FLOPs. - At 67 TFLOPS for the whole GPU with 132 SMs: per SM is 508 GFLOPS/s. - At 37.5% occupancy, effective FLOPS per SM is lower. But H100 can sustain high IPC even with lower occupancy due to memory latency hiding. Actually, I realize that for a compute-bound kernel, we need to ensure each thread does enough work to amortize the memory loads. 32768 MACs per thread at ~2 FLOPs/cycle at high frequency... H100 runs at ~2GHz or so? Actually 67 TFLOPS / (132 SMs * something)... 67 TFLOPS = 67,000 GFLOPS. Per SM: 67,000/132 ≈ 507 GFLOPS. At say 1.5 GHz, that's about 338 FLOPs/cycle per SM. With 2048 threads per SM, that's about 0.165 FLOPs/cycle/thread. So 32768 FLOPs = 32768 / 0.165 ≈ 198,000 cycles per thread. That's actually a decent amount of work. Wait, I need to think about this differently. With 3 blocks per SM and 256 threads per block, we have 768 active threads per SM. If the scheduler can keep them all busy (which requires enough warps to hide latency), then we get good utilization. 768 threads = 24 warps. That's a lot of warps for latency hiding. Actually, on H100, 24 warps per SM is plenty for hiding L1/shared memory latency. For global memory, it might be a bit less than ideal (32-64 warps is better), but with shared memory tiling, most accesses are to shared memory after the load. So this tiling approach might work! Let me design the kernel: Grid: (N/BN, M/BM) = (8192/128 = 64, 1024/8 = 128) = 8192 blocks Each SM can run 3 blocks concurrently. Total waves = 8192 / (132 * 3) ≈ 20.7 waves. That's a lot of waves, but each wave processes 512 rows... wait, each block does 8 rows. 396 blocks per wave do 3168 rows. With 1024 rows total, that's more than enough. Wait, I made a mistake. Grid is (N/BN, M/BM) = (64, 128) = 8192 blocks. Each block processes 8 rows * 128 cols = 1024 outputs. Total outputs = 1024 * 8192 = 8,388,608. 8192 blocks * 1024 outputs/block = 8,388,608. Correct. On 132 SMs with 3 blocks each = 396 concurrent blocks. 8192 / 396 ≈ 20.7 waves. Since M=1024 rows, and each block does 8 rows, the row dimension (M/BM=128) means we have 128 blocks along the M dimension. These blocks for the same N-tile all read the same W-tile. With 3 blocks per SM, we can have 396/64 ≈ 6 blocks for each N-tile. These 6 blocks could be from different M-positions but same N-range. They all read the same W[n_start:n_start+128, k:k+128] tile. In L2 cache, the W tile is 128*128*4 = 64KB. Multiple blocks can share this via L2 cache. And since we have 128 M-positions, each W tile is read by 128 blocks. If we process waves such that the same W tile is reused, L2 can help significantly. But with 20 waves, L2 might not be big enough. Wait, the issue is that blocks across different M-values read the SAME W tile for a given N-tile and K-tile. So for each K-tile, the W tile needs to be read 128 times (once per M-block). With K=8192/BK=128 = 64 K-tiles, each W tile is read 128*64 = 8192 times. But W is 256MB, so most of it won't stay in L2. Actually, if we process M-blocks in some order: for a fixed N-tile, all 128 M-blocks read the same W tile. With 396 concurrent blocks, we can process about 3 M-blocks per N-tile at a time. So we'd still need many passes over W. This is the classic GEMM issue. cuBLAS handles this with very sophisticated tiling, prefetching, and pipeline scheduling. For our custom kernel to be competitive, we need to be smart about this. But honestly, for M=1024, K=8192, N=8192, cuBLAS is going to be very hard to beat. Let me reconsider whether fusion alone is sufficient. What if we: 1. Use cuBLAS torch.matmul for the linear layer (without bias? or with bias?) 2. Fused kernel for GELU + softmax Actually, I realize I should just try a custom kernel and measure. Let me design a kernel that uses torch.matmul for the heavy compute and then fuses the rest. But first, let me think about whether there's something else we can optimize. In PyTorch: ```python x = self.linear(x) # cuBLAS GEMM + bias x = F.gelu(x) # elementwise x = F.softmax(x, dim=1) # softmax ``` For the softmax, PyTorch might do: 1. max per row 2. exp(x - max) and sum per row 3. divide per row If we write a custom kernel that does GELU + online softmax, maybe we can be faster? Online softmax would compute the max and sum in a single pass, but actually the standard online softmax also needs two passes over the data. Or we can do: 1. First pass: read row, apply GELU, track max 2. Second pass: read row again, apply GELU again, compute exp(x-max) and sum 3. Third pass: normalize and write Wait, that's worse. A better approach: 1. Read the row into shared memory (or registers) 2. Find max 3. Compute exp and sum 4. Normalize and write Since each row is 8192 floats = 32KB, we can't fit many rows in shared memory, but one row fits. So a block per row with 256 threads, processing the row in stages. For 8192 elements and 256 threads, each thread handles 32 elements. We can load the entire row into registers? 32 floats per thread = manageable. 1. All threads read 32 elements each (coalesced) 2. Do a block-wide reduction to find max 3. Compute exp(x - max) for each element, do block-wide reduction for sum 4. Divide each element by sum, write back For step 2 and 3, we can use warp shuffles and shared memory for the reductions. This approach is quite efficient because: - The row is loaded once and kept in registers - Reductions can be fast with warp shuffles - We minimize memory traffic Actually, if the linear output is in global memory, we read it once to do GELU + softmax. But wait, we need GELU before softmax. So in the kernel: 1. Read 32 floats per thread from linear output 2. Apply GELU 3. Find max via reduction 4. Compute exp(GELU - max), find sum via reduction 5. Divide, write back We read the linear output once and write the final output once. No intermediate writes! For PyTorch reference: 1. Linear writes temp1 2. GELU reads temp1, writes temp2 3. Softmax reads temp2, writes output Memory traffic after linear: temp1 read (32MB), temp2 write (32MB), temp2 read (32MB), output write (32MB) = 128MB. Our fused kernel: read linear output (32MB), write output (32MB). We save 64MB of memory traffic! At 3.35 TB/s = 0.019ms = 19 microseconds. Plus we save one kernel launch (maybe 5 microseconds). Total ~24 microseconds. On a 2ms+ operation, this is about 1% speedup. Hmm, but what if we can make the matmul portion itself faster? I keep coming back to this. A 1% speedup might barely pass. Let me see if there's any other advantage. Wait! What if PyTorch's softmax is slower than our custom implementation? For a [1024, 8192] tensor, PyTorch might use a generic kernel that handles arbitrary dimensions. A custom kernel optimized for this exact size could be faster. Also, what if the linear layer is not as fast as I think? For M=1024, N=8192, K=8192, the effective throughput of cuBLAS might not be at peak because M is only 1024. Let me look up some approximate cuBLAS performance for H100: - For large square matrices, H100 gets maybe 50-60 TFLOPS in FP32. - For M=1024, K=8192, N=8192, let me think about the tiling efficiency. cuBLAS likely selects an algorithm. For M much smaller than K and N, there are different algorithms. But 1024 is large enough that cuBLAS should do well. What about using a custom kernel for the entire thing but leveraging the fact that M=1024? If we have blocks where each block computes multiple rows and handles softmax on those rows, we might reduce some overhead. Actually, let me try a completely different approach. What if we use a "stream_k" style or just do persistent kernels? Probably overkill. What if we implement the matmul using tensor cores in FP32? H100 supports FP32 tensor cores through the Tensor Memory Accelerator or similar. Actually, H100 has TF32 tensor cores, but TF32 is lower precision than FP32. What about using wmma or mma PTX instructions for FP32 accumulation on H100? The H100 has FP64, FP32, TF32, FP16, etc. For FP32, the standard path is CUDA cores. OK, let me try a hybrid approach in code: 1. Use PyTorch matmul + biasadd for linear layer 2. Custom fused GELU + softmax kernel I'll also try a full custom kernel to see if it helps. Let me start coding. Wait, actually, I should also consider whether we can use the TMA (Tensor Memory Accelerator) on H100 for faster global/shared transfers, but that's only available in Hopper-specific features and might require PTX inline assembly. That could be complex. Let me try a simpler full custom matmul that uses shared memory tiling. I'll need to be careful about the layout. Actually, since W is accessed by row (W[n, k] for varying k), and x is accessed by row (x[m, k] for varying k), the contracting dimension k is contiguous for both. This is actually great for tiling! If we tile by N and K, a block computes a subset of outputs for one or more rows: - For a block computing BN outputs for BM rows: - For each K-tile: - Load x[m_start:m_start+BM, k:k+BK] into shared memory - This is BM rows of BK contiguous elements each. In memory, these are strided by K. - If BM=1, it's just one row of BK contiguous elements. - If BM>1, the second row starts K elements after the first. - Load W[n_start:n_start+BN, k:k+BK] into shared memory - This is BN rows of BK contiguous elements each. In memory, strided by K. The load pattern for x: - Memory layout: x[m, k] at base + m*K + k - To load x[m_start..m_start+BM, k_start..k_start+BK]: - We want adjacent threads to access adjacent memory. - A common pattern: thread (ty, tx) loads x[m_start+ty, k_start+tx] - For coalescing, tx should be the fast dimension and we want it to match warp width. - If BK >= 32, then threads with tx=t and tx=t+1 (same ty) are contiguous in memory. Similarly for W: - thread (ty, tx) loads W[n_start+ty, k_start+tx] - This is coalesced when tx varies. BlockDim could be (BK, some_y) or (some_x, some_y). Let's say BN=64, BK=64. - We need to load 64x64 elements of W and BMx64 elements of x. - If BM=4, then 4x64 elements of x. - Total elements to load per tile: 64*64 + 4*64 = 4096 + 256 = 4352. - With 256 threads: each thread loads 4352/256 = 17 elements. Not a whole number. Better: 256 threads, each loads a fixed number. - For W: 64*64 = 4096 elements. 4096/256 = 16 elements per thread. - For x: if BM=4, 4*64=256 elements. 256/256 = 1 element per thread. - Total: 17 elements per thread. Actually, threads can load a different number of elements. Let's say half the threads load x and half load W? That gets complicated. A simpler approach: use blockDim = 256 threads. Each thread is responsible for: - Loading some W elements - Loading some x elements - Computing some output elements For W tile [BN][BK] = [64][64]: - Thread tid loads W[n_start + tid/64, k_start + tid%64] if tid < 4096 - But we have 256 threads, not 4096. - So each thread loads 4096/256 = 16 W elements. - Thread tid loads 16 contiguous W elements? Or strided? - For coalescing, we want adjacent threads to load adjacent elements at each step. - We can have a loop: ``` for i = 0 to 15: idx = i * 256 + tid ty = idx / 64 tx = idx % 64 sW[ty][tx] = W[n_start + ty, k_start + tx] ``` Here, for i=0, threads 0..255 load elements 0..255 of the tile. Elements are stored as row-major: index = ty*64 + tx. So elements 0..255 correspond to ty=0, tx=0..63 (first 4 rows?). Wait, 64*64 = 4096. Elements 0..63 are row 0, 64..127 are row 1, etc. So elements 0..255 are rows 0..3, all 64 columns. Thread tid loads element tid for the first 256 elements. - Thread 0: element 0 = (ty=0, tx=0) -> W[n_start, k_start] - Thread 1: element 1 = (ty=0, tx=1) -> W[n_start, k_start+1] - ... - Thread 63: element 63 = (ty=0, tx=63) -> W[n_start, k_start+63] - Thread 64: element 64 = (ty=1, tx=0) -> W[n_start+1, k_start] Are these accesses coalesced? Thread 0..31 access W[n_start, k_start + 0..31] -> contiguous, coalesced! Thread 32..63 access W[n_start, k_start + 32..63] -> contiguous, coalesced! Thread 64..95 access W[n_start+1, k_start + 0..31] -> contiguous, coalesced! So yes, this works well for loading W. Similarly for x: - x tile: [BM][BK] = [4][64], 256 elements total. - Each thread loads 1 x element. - We can use the same pattern: thread tid loads x[m_start + tid/64, k_start + tid%64] for tid < 256. - Since BM=4 and BK=64, tid/64 gives m offset (0..3), tid%64 gives k offset. - Thread 0: x[m_start, k_start] - Thread 1: x[m_start, k_start+1] - Coalesced. Now for computation: - Each thread computes some output elements. - Block outputs: BM x BN = 4 x 64 = 256 outputs. - Threads: 256. So each thread computes 1 output element! - Thread tid computes output for (m = m_start + tid/64, n = n_start + tid%64). - Wait, that means: - Threads 0..63: m=m_start, n=n_start + 0..63 - Threads 64..127: m=m_start+1, n=n_start + 0..63 - etc. - For each K-tile in shared memory: - Thread tid reads s_x[m_offset][k] and s_W[n_offset][k] for k=0..63 - Computes dot product - Since each thread is reading from shared memory independently, there's no issue with coalescing. This seems like a reasonable kernel. Let's analyze performance: - 256 threads per block, 8192 blocks. - Blocks per SM: shared mem = 68KB, so SM can hold 3 blocks (204KB). - Thread-limited: 2048/256 = 8 blocks per SM. - So shared memory limits us to 3 blocks per SM. - Total concurrent blocks: 132 * 3 = 396. - Waves: 8192 / 396 = 20.7 waves. - Each thread computes 1 output with 64 K-tiles * 64 MACs = 4096 MACs = 8192 FLOPs per thread. - Total FLOPs per block: 256 * 8192 = 2,097,152 FLOPs. At 67 TFLOPS for the whole GPU: - Time for compute: 137GF / 67TF = 2.05ms. - But our kernel might not hit peak. Wait, the issue is that each block only computes 4 rows x 64 cols. There are 128 N-tiles and 256 M-tiles. The work per block is quite small. The overhead of threadblock scheduling across 20 waves might be significant. Can we make blocks larger to reduce the number of waves? - BM=8, BN=128, BK=128 - Block outputs: 1024 outputs - Shared mem: 8*128*4 + 128*128*4 = 4KB + 64KB = 68KB - Threads: 1024 (1 thread per output) - Blocks: M/BM * N/BN = 128 * 64 = 8192 blocks - Shared mem per block = 68KB -> 3 blocks per SM - Threads per block = 1024 -> 2 blocks per SM (2048 threads max) - So limited to 2 blocks per SM - Concurrent blocks: 264 - Waves: 8192/264 = 31 waves! Even worse. Wait, 1024 threads per block means fewer blocks per SM, but each block does more work. Total blocks is still 8192 because M/BM*N/BN = 128*64 = 8192. With 2 blocks per SM, we have 264 concurrent, 31 waves. Actually, this is worse for occupancy. What if we make the block larger in terms of outputs per thread but keep thread count lower? - BM=16, BN=128, BK=64 - Outputs per block: 2048 - Threads: 512 (each computes 4 outputs) - Shared mem: 16*64*4 + 128*64*4 = 4KB + 32KB = 36KB - Blocks per SM by shared mem: 6 - By threads: 4 - So 4 blocks per SM - Blocks total: M/BM * N/BN = 64 * 64 = 4096 - Concurrent: 132 * 4 = 528 - Waves: 4096 / 528 = 7.8 waves. Much better! - Work per thread: 4 outputs * 128 K-tiles * 64 MACs = 32768 MACs. Good amount of work. Hmm, but each thread computes 4 outputs. How do we assign threads? - Thread tid computes outputs for some (m, n) combinations. - If we have 512 threads and 2048 outputs, each thread does 4. - A simple mapping: thread tid handles n_offset = tid % 128, m_offset = tid/128 + some multiple. - But m ranges from 0..15 and n from 0..127. - With 512 threads: tid=0..511. - For the first set: m = tid/128, n = tid%128. This gives threads for m=0..3, n=0..127 (512 threads). - But we need threads for m=0..15. So each thread computes 4 outputs: for j=0..3, output at (m=tid%4 + j*4, n=tid/4)? - This is getting messy. A cleaner approach: have each thread compute a sub-tile. For instance, with BM=16, BN=128: - We can divide into 4 sub-tiles in M: each sub-tile is 4 rows x 128 cols. - 512 threads handle one sub-tile (4*128=512 outputs, 1 per thread). - Each block has 4 such sub-tiles... no wait, 16*128 = 2048 outputs. If 512 threads each do 1 in the first sub-tile, we'd need 4 iterations over sub-tiles. That's complex. Actually, a simpler assignment: - Block computes BM=16 rows, BN=128 cols. - Threads = 512. - We can view the output tile as 16 * 128 = 2048 elements. - Each thread computes 2048/512 = 4 elements. - The 4 elements are assigned with some stride to improve shared memory access patterns. Actually, in the K-loop, we access s_x[m][k] and s_w[n][k]. If a thread computes output for multiple (m, n), it might have good locality. For simplicity, let's say thread tid computes 4 outputs: - (m0, n0), (m1, n1), (m2, n2), (m3, n3) - A simple layout: tid maps to (m, n) as: - sub_tid = tid % 128 -> n = sub_tid - group = tid / 128 -> m = group, group + 4, group + 8, group + 12... no. Actually, with 16 rows and 128 cols, and 512 threads computing 4 outputs each, we can think of it as: - 128 groups of 4 threads each, where group g handles column n=g. - In group g, 4 threads handle rows m=0, 4, 8, 12 and also m=1, 5, 9, 13? No. A standard approach with thread-level tiling: - Each thread computes a TM x TN tile of outputs. - For BM=16, BN=128, BK=64. - 512 threads. - We could have TM=2, TN=2, so each thread computes 4 outputs. - Number of output tiles = (BM/TM) * (BN/TN) = 8 * 64 = 512 tiles. Exactly 1 per thread! - Thread tid corresponds to tile (tile_m = tid / 64, tile_n = tid % 64) - Thread computes outputs for m in [tile_m*TM, tile_m*TM+TM) and n in [tile_n*TN, tile_n*TN+TN) In the compute loop over k in shared memory: ``` for k = 0 to BK-1: for i = 0 to TM-1: x_val[i] = s_x[tile_m*TM + i][k] for j = 0 to TN-1: w_val[j] = s_w[tile_n*TN + j][k] for i = 0 to TM-1: for j = 0 to TN-1: acc[i][j] += x_val[i] * w_val[j] ``` This gives good instruction-level parallelism and register usage. The shared memory access for s_x is linear (same k, different m), and for s_w is also linear. For loading shared memory: - s_x[16][64] = 1024 floats - s_w[128][64] = 8192 floats - Total = 9216 floats = 36.8KB... wait, earlier I calculated 4KB + 32KB = 36KB. Let me recheck. - s_x: 16 rows * 64 cols = 1024 elements. But we need to store it as [16][64] or [64][16]. For the compute pattern above, we want s_x[m][k], so [16][64] row-major is fine. 1024 * 4 = 4KB. - s_w: 128 rows * 64 cols = 8192 elements. s_w[n][k] in row-major. 8192 * 4 = 32KB. - Total = 36KB. Yes. Loading s_w[128][64] with 256 threads (half the threads) in a coalesced manner: - Total elements = 8192. Each thread loads 32 elements. - Loop over i=0..31: - idx = i * 256 + tid - ty = idx / 64 - tx = idx % 64 - s_w[ty][tx] = W[n_start + ty][k_start + tx] Wait, but we have 512 threads, and half of them (256) load W. The other half load x. For W: 8192 elements / 256 threads = 32 elements per thread. For x: 1024 elements / 256 threads = 4 elements per thread. Let's say threads 0..255 load W, threads 256..511 load x. For W load: ``` for i in range(32): idx = i * 256 + tid ty = idx >> 6 // divide by 64 tx = idx & 63 // mod 64 s_w[ty][tx] = W[n_start + ty][k_start + tx] ``` For x load: ``` for i in range(4): idx = i * 256 + (tid - 256) ty = idx >> 6 // divide by 64 tx = idx & 63 s_x[ty][tx] = x[m_start + ty][k_start + tx] ``` Wait, x has 16 rows and 64 cols. Elements 0..63 are row 0, 64..127 are row 1, etc. 0..1023 total. With 256 threads, 4 elements each = 1024. idx ranges 0..1023. ty = idx / 64 (0..15), tx = idx % 64 (0..63). Thread (tid-256)=0 loads element 0 = x[m_start+0][k_start+0] Thread (tid-256)=1 loads element 1 = x[m_start+0][k_start+1] ... This is coalesced because adjacent threads load adjacent k for the same m. For W, same pattern: Thread 0 loads element 0 = W[n_start+0][k_start+0] Thread 1 loads element 1 = W[n_start+0][k_start+1] ... Thread 63 loads element 63 = W[n_start+0][k_start+63] Thread 64 loads element 64 = W[n_start+1][k_start+0] ... Also coalesced. This looks like a solid kernel design. Let me also make sure I handle the boundary conditions (M=1024 is divisible by 16? 1024/16=64, yes! N=8192/128=64, yes! K=8192/64=128, yes!). All dimensions are powers of 2 and divisible by our tile sizes. Perfect. Now, does this custom matmul kernel stand a chance against cuBLAS? - cuBLAS is highly optimized with sophisticated pipelines, tensor core usage (if applicable), and prefetching. - But for FP32, H100 uses CUDA cores, and the theoretical peak is 67 TFLOPS. - Our kernel might achieve maybe 40-50% of peak due to shared memory bandwidth limits, bank conflicts, and other inefficiencies. - If cuBLAS gets 60 TFLOPS and our kernel gets 30 TFLOPS, the matmul alone would take 4.5ms vs cuBLAS's 2.3ms. That's much worse. There's a risk our custom matmul is slower than cuBLAS by a lot. But if we fuse bias+GELU+softmax into it, we avoid 2 intermediate global memory round trips. Still, 2x slower on the matmul would be hard to make up. What if we use cuBLAS for matmul, and our custom kernel just fuses GELU+softmax? It's safer and can still provide a small but meaningful speedup. Let me think about whether there are any other optimizations for the reference. In PyTorch: - `self.linear(x)` computes `x @ W^T + b` - If W is [8192, 8192] and x is [1024, 8192], this is a single GEMM call. - GELU is elementwise - Softmax is a reduction + elementwise What if we use `torch.compile` or something? No, we need a custom kernel. Wait! I just realized that for the Linear layer, if we use a custom kernel that computes x @ W^T but doesn't use cuBLAS, it might be slower. But what if we can exploit some property of the kernel? Another thought: what if we compute multiple output rows in parallel within a block and do warp-level reductions for the dot product? Since W is shared, this can reduce the total W reads. Actually, I think the hybrid approach is best: 1. Use `torch.matmul(x, self.weight.t())` for the linear projection 2. Add bias manually (or let linear include it) 3. Custom fused kernel for the rest But wait, the evaluation setup creates `Model` with `self.linear = nn.Linear(...)`. For `ModelNew`, we're free to use this or avoid it. We can store `self.weight` and `self.bias` and compute manually. Actually, using `self.linear` in ModelNew will call cuBLAS which is optimal. Then a custom fused kernel for GELU+softmax. Let me code this up first and see if it compiles and passes correctness. For the fused GELU+softmax kernel: ```cuda __global__ void fused_gelu_softmax(const float* input, const float* bias, float* output, int M, int N) { int m = blockIdx.x; int tid = threadIdx.x; int num_threads = blockDim.x; if (m >= M) return; // Load data into registers/shared memory, apply GELU+bias, find max // Each thread handles a portion of the row int elements_per_thread = (N + num_threads - 1) / num_threads; int start = tid * elements_per_thread; int end = min(start + elements_per_thread, N); // But this might not be nicely divisible. Since N=8192, let's fix blockDim.x // For N=8192, we can use 1024 threads, each handling 8 elements. // Or 512 threads, each handling 16 elements. // Or 256 threads, each handling 32 elements. extern __shared__ float sdata[]; // We can use shared memory for reductions. } ``` For a row of 8192 elements, we want to load, add bias, apply GELU, find max, compute exp and sum, normalize. A typical efficient softmax kernel for large N: - Use multiple warps per block. - Each warp computes a local max and local sum for its portion of the row. - Block-level reduction for global max. - Another block-level reduction for global sum. - Final normalization write. With N=8192 and block size = 256 threads (8 warps): - Each warp handles 8192/8 = 1024 elements. - Wait, 256 threads = 8 warps. 8192/8 = 1024 elements per warp. - Each thread in a warp handles 1024/32 = 32 elements. - This is a good amount of work. For GELU, we can apply it as we load the elements. Kernel design: ```cuda __global__ void fused_bias_gelu_softmax(const float* input, const float* bias, float* output, int M, int N) { int row = blockIdx.x; int tid = threadIdx.x; int lane = tid % 32; int warp_id = tid / 32; int warps_per_block = blockDim.x / 32; if (row >= M) return; const float* row_input = input + row * N; float* row_output = output + row * N; int elements_per_warp = N / warps_per_block; int start = warp_id * elements_per_warp + lane * (elements_per_warp / 32); int stride = elements_per_warp / 32; int end = start + stride; // Each thread processes [start, end) with stride // Actually, better: each warp handles a contiguous chunk int warp_start = warp_id * elements_per_warp; int thread_start = warp_start + lane; int thread_stride = 32; int thread_count = elements_per_warp / 32; float local_max = -INFINITY; float vals[32]; // Each thread stores up to 32 values in registers // Load, add bias, apply GELU, track local max for (int i = 0; i < thread_count; i++) { int idx = thread_start + i * thread_stride; float val = row_input[idx] + bias[idx]; val = gelu(val); vals[i] = val; local_max = fmaxf(local_max, val); } // Warp shuffle reduce for local_max local_max = warp_reduce_max(local_max); // Block reduce for global max using shared memory __shared__ float shared_max[8]; // one per warp if (lane == 0) shared_max[warp_id] = local_max; __syncthreads(); float global_max = -INFINITY; if (tid < warps_per_block) { global_max = shared_max[tid]; } global_max = warp_reduce_max(global_max); // Only first warp needs to do this, but we can have all participate or just use shfl // Broadcast global_max global_max = __shfl_sync(0xFFFFFFFF, global_max, 0); // Compute exp and local sum float local_sum = 0.0f; for (int i = 0; i < thread_count; i++) { float exp_val = expf(vals[i] - global_max); vals[i] = exp_val; local_sum += exp_val; } // Warp shuffle reduce for local_sum local_sum = warp_reduce_sum(local_sum); // Block reduce for global sum __shared__ float shared_sum[8]; if (lane == 0) shared_sum[warp_id] = local_sum; __syncthreads(); float global_sum = 0.0f; if (tid < warps_per_block) { global_sum = shared_sum[tid]; } global_sum = warp_reduce_sum(global_sum); global_sum = __shfl_sync(0xFFFFFFFF, global_sum, 0); // Normalize and write for (int i = 0; i < thread_count; i++) { int idx = thread_start + i * thread_stride; row_output[idx] = vals[i] / global_sum; } } ``` Wait, `warp_reduce_max` and `warp_reduce_sum` can be implemented with warp shuffles. But there might be issues with the block-level reduction if not all threads participate. Actually, for the block-level reduction: 1. After warp reduce, store in shared memory. 2. One warp loads and reduces the shared values. 3. Broadcast result. Let me clean this up. Also, I need to be careful about the softmax numerical stability. For GELU, PyTorch uses the approximate form: `gelu(x) = 0.5 * x * (1.0 + tanh(sqrt(2/π) * (x + 0.044715 * x^3)))` Or the exact form with erf. PyTorch's default GELU uses the approximate form with tanh. Wait, let me check. In PyTorch, `torch.nn.functional.gelu(x, approximate='none')` or `approximate='tanh'`. The default since some version is 'none' (exact). Let me look at the reference code - it just calls `torch.nn.functional.gelu(x)` with no approximate argument. In recent PyTorch, `F.gelu` uses the exact formulation by default: `gelu(x) = 0.5 * x * (1 + erf(x / sqrt(2)))` Hmm, implementing erf in CUDA is tricky. There's `erf()` in the standard math library, so we can use `erff()` for float. But wait, using `erff()` in CUDA is fine but might be slower than the tanh approximation. For correctness, we need to match PyTorch's exact output. Actually, let me check: does PyTorch use exact or approximate GELU by default in the version being tested? The problem says "numerically equivalent outputs at fp32 precision" with tolerance 1e-4. For speed, if PyTorch uses exact GELU, we should too. But exact GELU with erf is quite expensive. There might not be a good approximation that stays within 1e-4. Let me think... Actually, for fp32 tolerance 1e-4, the tanh approximation might be close enough? Let me test. PyTorch's `gelu(x, approximate='tanh')` uses: `0.5 * x * (1.0 + torch.tanh(math.sqrt(2.0 / math.pi) * (x + 0.044715 * torch.pow(x, 3))))` While exact is: `0.5 * x * (1.0 + torch.erf(x / math.sqrt(2.0)))` The max difference between these two depends on x, but for typical random values, it might be less than 1e-4. Let me think... The tanh approximation is known to be quite accurate, with errors typically around 1e-5 to 1e-6 for most inputs. Since the tolerance is 1e-4, and assuming PyTorch uses exact GELU by default, if our kernel uses the tanh approximation, it might still pass if the difference is < 1e-4. Actually, wait. In PyTorch 2.0+, `torch.nn.functional.gelu` without specifying `approximate` uses the exact formulation. Let me verify. Yes, in recent PyTorch, the default is exact. But using `erff()` in CUDA might be slow. However, since GELU is elementwise and our bottleneck is the matmul, the GELU computation time is small either way. For the fused kernel, we just need to match the reference. Wait, but the reference calls `torch.nn.functional.gelu(x)`. If we use the tanh approximation in our kernel, and the reference uses exact, will the difference be within 1e-4? Let me think about the worst case. The tanh approximation for GELU is: `x * 0.5 * (1 + tanh(0.7978845608 * (x + 0.044715 * x^3)))` The exact is: `x * 0.5 * (1 + erf(x / sqrt(2)))` The max difference occurs somewhere in the tails or near the inflection points. According to some sources, the approximation error is around 1e-7 relative or about 1e-6 absolute for typical ranges. But near x=0 or for large x, it's very accurate. For x in [-5, 5], the absolute difference should be well under 1e-5. For x in [-10, 10], maybe 1e-4? I'm not entirely sure. To be safe, I should implement exact GELU using `erff()`. But `erff()` in CUDA is a device function from `<cmath>` or `<math.h>` that's available. Wait, does CUDA have `erff`? Yes, `erff` is the float version of the error function and is available in CUDA's math library. But `erff` might be slower. For just a GELU+softmax kernel after a matmul, the slowdown from `erff` is minimal because we're memory bandwidth bound, not compute bound, for that part. OK, so for a safe correctness approach, I'll use exact GELU with `erff()`. For speed, I can try the tanh approximation later. Wait, actually, I just realized: if the matmul time dominates, maybe I should focus more on whether we can make a custom matmul that's competitive. Let me calculate: if reference takes ~2.1ms and we save 0.02ms from fusion, that's ~1% speedup. If reference is used for competition and others submit better kernels, 1% might still pass if it's the simplest approach, but the task says "faster", not "significantly faster". Any measurable speedup should pass. But let me consider whether there's a bigger win. Maybe PyTorch's cuBLAS call isn't as fast as I think for this shape? For M=1024, N=8192, K=8192, maybe cuBLAS takes 3-4ms? Custom kernels might be more competitive? Also, the weight matrix is 256MB. If we can somehow stream through it more efficiently... Actually, I just thought of another optimization: for the custom kernel, if we process multiple rows in a block, we can reuse the W tile across those rows via shared memory. In standard GEMM, a block loads a W tile once and uses it for multiple rows of x. This is exactly what the tiling approach does! In our custom kernel, when BM=16, we process 16 rows with the same W tile. This means W is read from global memory once per K-tile and N-tile, and reused for 16 rows. That's the standard GEMM optimization. The main question is whether our implementation is efficient enough to beat cuBLAS. Since cuBLAS does exactly this but with much more optimization, our implementation is likely to be slower. However, the total fused kernel time = matmul time + saved memory time. If our matmul is 2.5ms and we save 0.1ms from fusion, total is 2.4ms. If reference is 2.1ms, we lose. If our matmul is 2.0ms and we save 0.1ms, total is 1.9ms vs reference 2.1ms, we win slightly. I think the safe approach might be to write a full custom fused kernel for everything and measure. If it's slower than reference by >5%, we can fall back to the hybrid approach. For the full custom kernel, I'll design a tile-based matmul with fusing. Let me write the full custom kernel. It will be long. Actually, here's another idea: since we need to call compile to check, maybe I should start with a simpler fused approach (cuBLAS + custom GELU/softmax) and measure. Then if it's not enough, try a full custom kernel. Let me write the cuBLAS + fused kernel approach first. Wait, but what if we can use torch.addmm for the linear layer, which might be well optimized, and then do a custom kernel? Or just use self.linear. Actually, let me look at whether we can save time by fusing something else. What if PyTorch's softmax implementation is not optimal for our shape? For [1024, 8192], softmax along dim=1: - PyTorch dispatches to `_softmax` kernel which handles the reduction. - For large inner dimensions, it likely uses a block per row with many threads. Our custom kernel for softmax alone might not be faster than PyTorch's, but combined with GELU fusion, it saves memory traffic and a kernel launch. OK, I'll code the hybrid approach first. Let me write the kernel. Wait, before I write code, let me think about whether we should even use self.linear. The constructor needs to accept the same arguments. We can do: ```python class ModelNew(nn.Module): def __init__(self, in_features, out_features): super().__init__() self.in_features = in_features self.out_features = out_features self.weight = nn.Parameter(torch.empty(out_features, in_features)) self.bias = nn.Parameter(torch.empty(out_features)) nn.init.kaiming_uniform_(self.weight, a=math.sqrt(5)) fan_in, _ = nn.init._calculate_fan_in_and_fan_out(self.weight) bound = 1 / math.sqrt(fan_in) nn.init.uniform_(self.bias, -bound, bound) ``` Or we can just create our own linear and extract weight/bias from it. Actually, a simpler approach: in __init__, create a reference linear to get initialized weights, then optionally use them. ```python class ModelNew(nn.Module): def __init__(self, in_features, out_features): super().__init__() linear = nn.Linear(in_features, out_features) self.weight = nn.Parameter(linear.weight.detach().clone()) self.bias = nn.Parameter(linear.bias.detach().clone()) ``` This ensures the same initialization as the reference. For forward: ```python def forward(self, x): out = torch.matmul(x, self.weight.t()) return fused_kernel(out, self.bias) ``` Wait, but the problem says "Accept the same constructor arguments as Model." So we need in_features, out_features. It doesn't say we need to use nn.Linear inside. Now, for the CUDA kernel in the hybrid approach: ```cpp __global__ void fused_bias_gelu_softmax(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N) { int row = blockIdx.x; int tid = threadIdx.x; int lane = tid & 31; int warp_id = tid >> 5; const int WARPS_PER_BLOCK = 8; // 256 threads / 32 const int THREADS_PER_BLOCK = 256; if (row >= M) return; const float* row_in = input + row * N; float* row_out = output + row * N; // Each warp handles a chunk of 8192 / 8 = 1024 elements // Each thread handles 1024 / 32 = 32 elements const int chunk_size = N / WARPS_PER_BLOCK; // 1024 const int thread_stride = 32; const int thread_num_iters = chunk_size / 32; // 32 int warp_start = warp_id * chunk_size; int thread_start = warp_start + lane; float thread_max = -1e20f; // -INFINITY might cause issues float vals[32]; #pragma unroll for (int i = 0; i < thread_num_iters; i++) { int idx = thread_start + i * thread_stride; float val = row_in[idx] + bias[idx]; // Exact GELU float cdf = 0.5f * (1.0f + erff(val * 0.7071067811865475f)); val = val * cdf; vals[i] = val; thread_max = fmaxf(thread_max, val); } // Warp max reduction #pragma unroll for (int offset = 16; offset > 0; offset >>= 1) { float other = __shfl_down_sync(0xFFFFFFFF, thread_max, offset); thread_max = fmaxf(thread_max, other); } __shared__ float shared_max[WARPS_PER_BLOCK]; __shared__ float shared_sum[WARPS_PER_BLOCK]; if (lane == 0) { shared_max[warp_id] = thread_max; } __syncthreads(); // Block-wide max using first warp float block_max = -1e20f; if (tid < WARPS_PER_BLOCK) { block_max = shared_max[tid]; } #pragma unroll for (int offset = WARPS_PER_BLOCK / 2; offset > 0; offset >>= 1) { float other = __shfl_down_sync(0xFFFFFFFF, block_max, offset); block_max = fmaxf(block_max, other); } block_max = __shfl_sync(0xFFFFFFFF, block_max, 0); // Compute exp and sum float thread_sum = 0.0f; #pragma unroll for (int i = 0; i < thread_num_iters; i++) { float exp_val = expf(vals[i] - block_max); vals[i] = exp_val; thread_sum += exp_val; } // Warp sum reduction #pragma unroll for (int offset = 16; offset > 0; offset >>= 1) { float other = __shfl_down_sync(0xFFFFFFFF, thread_sum, offset); thread_sum += other; } if (lane == 0) { shared_sum[warp_id] = thread_sum; } __syncthreads(); // Block-wide sum using first warp float block_sum = 0.0f; if (tid < WARPS_PER_BLOCK) { block_sum = shared_sum[tid]; } #pragma unroll for (int offset = WARPS_PER_BLOCK / 2; offset > 0; offset >>= 1) { float other = __shfl_down_sync(0xFFFFFFFF, block_sum, offset); block_sum += other; } block_sum = __shfl_sync(0xFFFFFFFF, block_sum, 0); // Normalize and write #pragma unroll for (int i = 0; i < thread_num_iters; i++) { int idx = thread_start + i * thread_stride; row_out[idx] = vals[i] / block_sum; } } ``` Actually, a few potential issues: 1. `-1e20f` might not be small enough if values can be very negative. Better to use `-(1e20f)` or `-CUDART_INF_F`. Actually, for float, we can use `-1e30f` or `-CUDART_INF_F` from `<cuda_runtime.h>`. Or just `-CUDART_INF_F` which is better. But including `<cuda_runtime.h>` is usually fine. 2. The exact GELU: `val * 0.5f * (1.0f + erff(val * 0.7071067811865475f))`. This matches PyTorch exactly. 3. For N=8192 and WARPS_PER_BLOCK=8, chunk_size = 1024. thread_num_iters = 32. `vals[32]` is fine. Wait, we need to make sure the block-level reduction for max and sum works correctly. With WARPS_PER_BLOCK=8: - `__shfl_down_sync(0xFFFFFFFF, block_max, offset)` with offset=4: threads 0..3 get values from 4..7. offset=2: threads 0..1 get from 2..3 and 4..5 get from 6..7 (but threads 2+ don't care about the result since block_max is only read from thread 0). Actually, for a correct reduction in the first warp, we need all 8 threads in the warp to participate. The warp has 32 threads but only 8 have valid data. So we need to make sure the other threads don't corrupt the reduction. In the code above: ```cpp if (tid < WARPS_PER_BLOCK) { block_max = shared_max[tid]; } ``` For threads 8..31, `block_max` is uninitialized or -1e20f. Then: ```cpp for (int offset = 16; offset > 0; offset >>= 1) { float other = __shfl_down_sync(0xFFFFFFFF, block_max, offset); block_max = fmaxf(block_max, other); } ``` When tid=0, offset=16: gets block_max from tid=16. What's in tid=16's block_max? It's uninitialized! Unless we initialize it. So we should do: ```cpp float block_max = -CUDART_INF_F; // or -1e30f if (tid < WARPS_PER_BLOCK) { block_max = shared_max[tid]; } ``` This ensures threads 8..31 have block_max = -inf. Then when tid=0 does shfl_down to tid=16, it gets -inf. fmaxf(current, -inf) = current. This works! For the sum, same logic but initialize to 0.0f. Actually, `fmaxf` with `-inf` works. But what if all values are `-inf`? That shouldn't happen with GELU. For the sum reduction: ```cpp float block_sum = 0.0f; if (tid < WARPS_PER_BLOCK) { block_sum = shared_sum[tid]; } for (int offset = 16; offset > 0; offset >>= 1) { float other = __shfl_down_sync(0xFFFFFFFF, block_sum, offset); block_sum += other; } ``` For thread 0 getting from thread 16: 0.0f + 0.0f = 0.0f. So threads 8..31 having 0 doesn't affect the sum of threads 0..7. Correct! Let me double check with smaller example: - WARPS_PER_BLOCK = 8 - tid = 0..7 have block_sum = [1, 2, 3, 4, 5, 6, 7, 8], tid = 8..31 have 0. - After offset=16: tid=0 gets from tid=16 (0). sum=1+0=1. tid=1 gets from 17 (0). sum=2. etc. No change. - After offset=8: tid=0 gets from tid=8 (0). sum=1+0=1. tid=1 gets from 9 (0). etc. No change for tid<8. - After offset=4: tid=0 gets from tid=4 (5). sum=1+5=6. tid=1 gets from 5 (6). sum=8. tid=2 gets from 6 (7). sum=10. tid=3 gets from 7 (8). sum=12. - After offset=2: tid=0 gets from 2 (10). sum=16. tid=1 gets from 3 (12). sum=20. - After offset=1: tid=0 gets from 1 (20). sum=36. But wait, we're doing this in all threads with `__shfl_down_sync` using full warp mask. Threads 0..31 all run the loop. For threads >= 8, block_sum is 0 initially. After the reductions, thread 0 has 36. But threads 8..31 will have some garbage values. That's fine as long as we broadcast from thread 0. Actually, `__shfl_sync(0xFFFFFFFF, block_sum, 0)` broadcasts thread 0's value to everyone. Let me also consider numerical precision. For softmax, we subtract max for stability. The reference PyTorch also does this. One more thing: the block-level max and sum reductions use `for (int offset = 16; offset > 0; offset >>= 1)`, but with only 8 valid elements, `offset=16` and `offset=8` are no-ops effectively (adding 0 or taking max with -inf). Then `offset=4` does the first real merge. This is correct but slightly inefficient. We could optimize the loop bounds, but this is fine. Now, there's an issue: we have 1024 rows and block size 256. Grid = (1024, 1). This means 1024 blocks total. On 132 SMs, that's about 7.8 blocks per SM. With 256 threads each, that's about 2000 threads per SM, which is near the max (2048). Good utilization. Actually wait, each block processes ONE row. 1024 blocks for 1024 rows. Each SM can run 8 blocks of 256 threads. So 132 * 8 = 1056 blocks concurrently. That's basically all 1024 blocks running at once with very few waves. This is excellent occupancy! Total concurrent warps = 1024 blocks * 8 warps = 8192 warps. 8192 / 132 SMs = 62 warps per SM. H100 can handle 64 warps per SM. So we're at ~97% occupancy. Great! So the fused kernel should run very efficiently on the memory bandwidth bound portion. Now, the reference uses `self.linear(x)` which calls cuBLAS, then GELU kernel (1024 blocks elementwise), then softmax kernel (~1024 blocks for reduction). The reference likely has 3 kernel launches. We have 2 kernel launches (matmul + fused). We save one kernel launch. Also, for the elementwise portion, PyTorch's GELU and softmax might each have some synchronization or different block configurations. Our fused kernel has high occupancy. Let me also consider whether we should manually handle the linear layer instead of using self.linear. If we use `self.linear(x)`, PyTorch does the matmul and bias addition. But what if TensorFloat32 is enabled? That could cause precision differences. To ensure fp32 precision, we might want to disable TF32 or make sure it's not used. In PyTorch, TF32 is enabled by default on Ampere/Hopper. But the problem asks for fp32 precision outputs. Actually, TF32 is only used inside cuBLAS. The output of cuBLAS with TF32 will have ~3 decimal digits of precision vs fp32's ~7. But 1e-4 tolerance might still pass. Wait, the problem says "Produce numerically equivalent outputs at fp32 precision" with tolerance 1e-4 for fp32. TF32 should be fine for 1e-4 tolerance in many cases. But to be safe, maybe we should use `torch.backends.cuda.matmul.allow_tf32 = False`? But setting global flags might be frowned upon. Actually, the reference implementation also uses `nn.Linear` which calls cuBLAS with TF32 enabled by default (if on H100). So the reference and our code would use the same setting. If we use `self.linear(x)` in our code, it will match the reference. But I'm planning to use `torch.matmul(x, self.weight.t())`. This also calls cuBLAS and will use TF32 in the same way. Wait, but I'm not using `self.linear`. I'm using `torch.matmul`. Both go through cuBLAS, so they should use the same TF32 setting. I could also use `torch.addmm(self.bias, x, self.weight.t())` to fuse bias into the matmul call, which is what `nn.Linear` does internally. This might be slightly more efficient than separate matmul + bias addition. Actually, `F.linear(x, self.weight, self.bias)` does exactly this. Let me code up the full Python module with the hybrid approach and see. But before I do, let me think about whether we should try to also fuse bias into the custom kernel to avoid any potential overhead. Yes, the plan is exactly that. Wait, another optimization: we can use `torch.mm` instead of `torch.matmul` since both inputs are 2D. This avoids some broadcasting overhead. OK let me write the code: ```python import torch import torch.nn as nn from torch.utils.cpp_extension import load_inline cuda_source = r''' #include <torch/extension.h> #include <cuda.h> #include <cuda_runtime.h> #include <math.h> #define WARPS_PER_BLOCK 8 #define THREADS_PER_BLOCK 256 __inline__ __device__ float warp_reduce_max(float val) { for (int offset = 16; offset > 0; offset >>= 1) val = fmaxf(val, __shfl_down_sync(0xFFFFFFFF, val, offset)); return val; } __inline__ __device__ float warp_reduce_sum(float val) { for (int offset = 16; offset > 0; offset >>= 1) val += __shfl_down_sync(0xFFFFFFFF, val, offset); return val; } __global__ void fused_bias_gelu_softmax_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N) { int row = blockIdx.x; int tid = threadIdx.x; int lane = tid & 31; int warp_id = tid >> 5; if (row >= M) return; const float* row_in = input + row * N; float* row_out = output + row * N; const int chunk_size = N / WARPS_PER_BLOCK; const int thread_num_iters = chunk_size >> 5; int warp_start = warp_id * chunk_size; int thread_start = warp_start + lane; float thread_max = -1e30f; float vals[32]; #pragma unroll for (int i = 0; i < thread_num_iters; i++) { int idx = thread_start + (i << 5); float val = row_in[idx] + bias[idx]; // Exact GELU float cdf = 0.5f * (1.0f + erff(val * 0.70710678118654752440f)); val = val * cdf; vals[i] = val; thread_max = fmaxf(thread_max, val); } thread_max = warp_reduce_max(thread_max); __shared__ float shared_max[WARPS_PER_BLOCK]; __shared__ float shared_sum[WARPS_PER_BLOCK]; if (lane == 0) { shared_max[warp_id] = thread_max; } __syncthreads(); float block_max = -1e30f; if (tid < WARPS_PER_BLOCK) { block_max = shared_max[tid]; } block_max = warp_reduce_max(block_max); block_max = __shfl_sync(0xFFFFFFFF, block_max, 0); float thread_sum = 0.0f; #pragma unroll for (int i = 0; i < thread_num_iters; i++) { float exp_val = expf(vals[i] - block_max); vals[i] = exp_val; thread_sum += exp_val; } thread_sum = warp_reduce_sum(thread_sum); if (lane == 0) { shared_sum[warp_id] = thread_sum; } __syncthreads(); float block_sum = 0.0f; if (tid < WARPS_PER_BLOCK) { block_sum = shared_sum[tid]; } block_sum = warp_reduce_sum(block_sum); block_sum = __shfl_sync(0xFFFFFFFF, block_sum, 0); float inv_sum = 1.0f / block_sum; #pragma unroll for (int i = 0; i < thread_num_iters; i++) { int idx = thread_start + (i << 5); row_out[idx] = vals[i] * inv_sum; } } torch::Tensor fused_bias_gelu_softmax(torch::Tensor input, torch::Tensor bias) { int M = input.size(0); int N = input.size(1); auto output = torch::empty_like(input); dim3 grid(M); dim3 block(THREADS_PER_BLOCK); fused_bias_gelu_softmax_kernel<<<grid, block>>>( input.data_ptr<float>(), bias.data_ptr<float>(), output.data_ptr<float>(), M, N ); return output; } ''' cpp_source = r''' torch::Tensor fused_bias_gelu_softmax(torch::Tensor input, torch::Tensor bias); ''' fused_ops = load_inline( name='fused_ops', cpp_sources=cpp_source, cuda_sources=cuda_source, functions=['fused_bias_gelu_softmax'], extra_cuda_cflags=['-O3'], verbose=False, ) class ModelNew(nn.Module): def __init__(self, in_features, out_features): super(ModelNew, self).__init__() self.in_features = in_features self.out_features = out_features self.weight = nn.Parameter(torch.empty(out_features, in_features)) self.bias = nn.Parameter(torch.empty(out_features)) nn.init.kaiming_uniform_(self.weight, a=math.sqrt(5)) fan_in, _ = nn.init._calculate_fan_in_and_fan_out(self.weight) bound = 1 / math.sqrt(fan_in) nn.init.uniform_(self.bias, -bound, bound) def forward(self, x): out = torch.matmul(x, self.weight.t()) return fused_ops.fused_bias_gelu_softmax(out, self.bias) ``` Wait, I need to include `import math`. Also, `load_inline` needs the cpp_sources to only contain declarations (no definitions). And the cuda_source contains the `fused_bias_gelu_softmax` function but the device kernel plus a host wrapper. Actually, in `cpp_sources`, I should only put the declaration: ```cpp torch::Tensor fused_bias_gelu_softmax(torch::Tensor input, torch::Tensor bias); ``` And in `cuda_sources`, I put the full implementation including the host wrapper function. Wait, but `load_inline` auto-generates pybind bindings for the functions listed in `functions=...`. The pybind bindings call the C++ functions. So the C++ functions need to be available. Usually, we put the implementations in `cuda_sources` (which are compiled as `.cu` files and can contain host C++ functions as well as device code). The `cpp_sources` are compiled as `.cpp` files and linked with the `.cu` files. So we can put the host wrapper in `cuda_sources` and declare it in `cpp_sources`. Actually, a common pattern with `load_inline` is to put everything in `cuda_sources` and leave `cpp_sources` empty or just with necessary includes, then declare the function in `cuda_sources`: ```python fused_ops = load_inline( name='fused_ops', cpp_sources='', cuda_sources=cuda_code_with_host_function, functions=['fused_bias_gelu_softmax'], ... ) ``` But the host function needs to be parseable by the cpp compiler for pybind. Since `torch::Tensor` and `ATen` types are used, we need to include `<torch/extension.h>`. Actually, when we don't specify `cpp_sources`, `load_inline` might still include the necessary torch headers. Let me check. In the CUDA source, `#include <torch/extension.h>` is fine because `.cu` files are compiled with nvcc which can also parse C++ headers. The pybind bindings will be generated and linked with the compiled CUDA objects. Actually, I think we can put the entire code including host wrapper in `cuda_sources` and leave `cpp_sources` as minimal includes or empty. The `functions` list tells it which functions to bind. Let me test this approach. I'll write the code and try compiling it. One issue: `erff` in CUDA device code requires `#include <math.h>` or `<cmath>`, and it's available as a device function. But wait, on some CUDA versions, `erff` might not be available in device code. The CUDA math library does include `erff` as an intrinsic since around CUDA 7.0 or earlier, so it should be fine. Wait, I can also use `normcdf` or manually compute. Actually, `erff` is definitely in `<cuda_runtime.h>` and `<math.h>`. Let me make sure. In CUDA device code, we can use `erff`, `erfc`, etc. Another issue: I'm using `load_inline` with `cpp_sources=''` and putting the host function in `cuda_sources`. I think this should work, but let me verify the syntax. Actually, I realize that for `load_inline`, the `cpp_sources` and `cuda_sources` are compiled separately and linked. If we put host C++ code in `cuda_sources`, nvcc will compile it. That's fine. But we don't need any C++ declarations in `cpp_sources` if all our code is in `cuda_sources`. The auto-generated pybind code just needs to know the function signatures, which it gets from the compiled object files. So our setup can be: ```python cuda_source = ''' #include <torch/extension.h> #include <cuda.h> #include <cuda_runtime.h> #include <math.h> ... device kernel ... torch::Tensor fused_bias_gelu_softmax(torch::Tensor input, torch::Tensor bias) { ... return output; } ''' fused_ops = load_inline( name='fused_ops', cpp_sources='', cuda_sources=cuda_source, functions=['fused_bias_gelu_softmax'], extra_cuda_cflags=['-O3'], verbose=False, ) ``` I think this should work. Wait, I should also check if the input tensors are contiguous and float. We can add checks, but for this problem it's probably safe. Let me also use `torch::Tensor` properly. In the host function: ```cpp torch::Tensor fused_bias_gelu_softmax(torch::Tensor input, torch::Tensor bias) { int M = input.size(0); int N = input.size(1); auto output = torch::empty_like(input); // ... } ``` I need to make sure input is on CUDA and contiguous. The problem says inputs are CUDA tensors. Actually, wait - `torch::empty_like(input)` will create a tensor of the same type. Since the reference uses fp32 and we need fp32 output, this should be float. Let me write the complete code and compile it. One more optimization: in reference, for nn.Linear, the weight is actually transposed for the forward. In nn.Linear, weight is [out_features, in_features]. So `x @ weight.t()` is correct. In ModelNew, if I do `torch.matmul(x, self.weight.t())`, this gives the same as `F.linear(x, self.weight, self.bias)` without the bias. Wait, should I use `torch.addmm(self.bias, x, self.weight.t())` instead? This fuses bias into the cuBLAS call, which is what nn.Linear does internally (well, it often calls addmm). Using addmm might be slightly faster than matmul + custom bias in the kernel because cuBLAS can fuse the bias. But then our custom kernel can't fuse the bias - wait, our custom kernel DOES fuse the bias, so it doesn't matter. Actually, let me think: if I use `F.linear(x, self.weight, self.bias)`, cuBLAS does matmul + bias in one call, writing a temporary. Then our custom fused kernel does... wait, if cuBLAS already added bias, we can't apply GELU and softmax because the output already has bias added. The custom kernel should start from the linear output. If I use `F.linear(x, self.weight, self.bias)`, then the matmul and bias are fused in cuBLAS, and we save the separate bias addition. Then our custom kernel just needs GELU + softmax. If I use `torch.matmul(x, self.weight.t())`, then cuBLAS does only matmul, and our custom kernel needs to add bias, GELU, and softmax. It might not matter much because both approaches have the same total work. But F.linear might be slightly optimized. Actually, keeping bias separate and adding it in our kernel means that the matmul output doesn't go through a separate bias-add kernel. That's already one fewer kernel. And our kernel will fuse bias+GELU+softmax. So using `torch.matmul` and custom bias might be better or equivalent. Wait, but if I use `F.linear(x, self.weight, self.bias)`, PyTorch internally does: ```python output = torch.addmm(self.bias, x, self.weight.t()) ``` This is a single cuBLAS call that does C = alpha * A @ B + beta * C' (where C' is bias broadcasted). cuBLAS is very good at this. The output goes directly to memory. Then our kernel reads it, applies GELU+softmax, writes output. If I use `torch.matmul(x, self.weight.t())`, this does C = A @ B. Then our kernel reads C, adds bias, GELU, softmax, writes output. The difference: in the first case, cuBLAS writes 32MB. Our kernel reads 32MB and writes 32MB. Total extra memory: 64MB. In the second case, cuBLAS writes 32MB. Our kernel reads 32MB and writes 32MB. Total extra memory: 64MB. Same! The bias is either added in cuBLAS (fused) or in our kernel (fused with GELU+softmax). The memory traffic is the same. But adding bias in cuBLAS is very efficient and costs nothing extra in terms of memory. So using `F.linear` is at least as good. Wait, one more thing: if I use `F.linear`, the bias is already in the output. Then my custom kernel would just be GELU + softmax. Let's compare: Approach A: `F.linear` then custom(GELU + softmax) - Memory: read 32MB, write 32MB for custom kernel Approach B: `torch.matmul` then custom(bias + GELU + softmax) - Memory: read 32MB, write 32MB for custom kernel (but reads bias too) The bias vector is 32KB, which is negligible. Both are the same. But if I use F.linear, the custom kernel is simpler: no bias load. Let me stick with just GELU+softmax for the custom kernel, and use F.linear for the matmul+bias. Wait, actually, let me just try both approaches. I'll start with the simpler fused GELU+softmax kernel. Let me write the code: Actually, one more thought. What if I use a custom kernel that does the FULL linear + GELU + softmax, but instead of doing a custom matmul, it uses a different algorithm? For M=1024, K=8192, N=8192, with weight W (256MB), the W is quite large. Maybe we can compute the output in blocks and use some optimized micro-kernel. But honestly, a full custom GEMM is probably too complex and risky. I'll start with the hybrid approach and test. Let me draft the code. Wait, I should also consider whether using `torch.mm` or `F.linear` or `torch.matmul` matters. They should all use cuBLAS/GEMM internally. Let's use `F.linear(x, self.weight, self.bias)` or `torch.addmm(self.bias, x, self.weight.t())`. Actually, `torch.addmm` should be very efficient. And nn.Linear uses it internally. So: ```python out = torch.addmm(self.bias, x, self.weight.t()) ``` Now let me write the code properly. Actually, I realize there might be a subtle issue with how the custom CUDA kernel interacts with PyTorch's autograd. The problem only asks for forward(). So we don't need to worry about backward. Let me write the full file. One more thing: in the CUDA kernel, I'm using `erff`. Let me make sure this is correct. The GELU formula is: `0.5 * x * (1.0 + erf(x / sqrt(2)))` In CUDA device code: ```cpp float cdf = 0.5f * (1.0f + erff(val * 0.70710678118654752440f)); val = val * cdf; ``` This is correct. But I want to double check: PyTorch's `torch.nn.functional.gelu` uses the exact formulation with erf. Let me verify by checking if default approximate is None. In PyTorch documentation: > torch.nn.functional.gelu(input, approximate='none') → Tensor > ... > When the approximate argument is 'tanh', Gelu is estimated with: > $$GELU(x) = 0.5 * x * (1 + tanh(\sqrt(2 / \pi) * (x + 0.044715 * x^3)))$$ The default is 'none', which uses the exact formulation. OK so `erff` is required. Wait, does CUDA have `erff` in device code? Yes, `erff` is a device function in `<math.h>` or `<cuda_runtime.h>`. Actually, in CUDA, `erff` is available in the C runtime math library for both host and device. We can use `erff` directly. One potential issue: `<math.h>` vs `<cmath>`. In CUDA code compiled by nvcc, both work but `<cmath>` might require `std::erff` and is less commonly used. I'll use `<math.h>` and `erff`. Let me write the kernel code. I'll also add `__restrict__` to pointer arguments to help the compiler optimize. Actually, I just realized there's a potential issue with `load_inline`: the generated pybind11 code calls our function. It needs to know about `torch::Tensor`. This is available because `<torch/extension.h>` includes the necessary headers. But if `cpp_sources` is empty, will the compilation still include torch headers? I think when using `load_inline`, if `cpp_sources` is provided, it's compiled as C++. If empty, it might still compile a stub. The `cuda_sources` are also compiled and linked. I think providing `cpp_sources` with just the necessary includes might be safer, or putting everything in `cuda_sources` and keeping `cpp_sources` minimal. Actually, let me check a pattern that I know works: ```python cuda_source = ''' #include <torch/extension.h> #include <cuda.h> #include <cuda_runtime.h> #include <math.h> __global__ void my_kernel(...) { ... } torch::Tensor my_function(torch::Tensor a) { ... return out; } ''' module = load_inline( name='my_module', cpp_sources='', cuda_sources=cuda_source, functions=['my_function'], extra_cuda_cflags=['-O3'], ) ``` This should work because `torch::Tensor` is defined in the headers included via `<torch/extension.h>`, and nvcc can compile this. However, I recall that sometimes it's better to have the function declaration in `cpp_sources` and definition in `cuda_sources`, especially if there are issues with name mangling or linking. But for simplicity, putting everything in `cuda_sources` and nothing in `cpp_sources` should work. Let me also add `-use_fast_math` to the CUDA flags. This might speed up `expf` and `erff`, but `erff` with fast math might be less accurate. For GELU we need precision. Actually, `-use_fast_math` replaces `erff` with a faster but less accurate approximation. Since we need fp32 accuracy to 1e-4, we probably should NOT use fast math for `erff`. Though for expf in softmax, fast math is probably fine. Actually, without `-use_fast_math`, `expf` and `erff` use the accurate implementations from the math library. Let's not use `-use_fast_math` to be safe, or at least test without it first. OK, let me write the code and compile.

The compile failed because I gave an empty string for cpp_sources but needed to pass the CUDA source to cuda_sources and the function was declared in cuda_sources but not accessible to the generated pybind wrapper. Let me check: the issue is that pybind11 needs a declaration of the function signature in cpp_sources or it won't construct the wrapper correctly. Actually, looking at the load_inline docs more carefully: - The `cpp_sources` can contain C++ declarations that are visible to the auto-generated pybind code. - But if I give an empty `cpp_sources`, the generated pybind code can't see the function declaration. Wait, I need to provide a declaration in the `cpp_sources` even though there's no C++ code there. Let me add a proper declaration.

The kernel compiled. Now let me test correctness. But I realize there might be an issue with the kernel - N needs to be exactly divisible by (WARPS_PER_BLOCK * 32) = 256, and 8192 / 256 = 32, which is fine. But let me also think about the GELU implementation. PyTorch uses a specific approximation. Let me check: - PyTorch's GELU: `gelu = input * 0.5 * (1.0 + erf(input / sqrt(2)))` (for `approximate='none'`) - Or with the tanh approximation: `gelu = input * 0.5 * (1 + tanh(sqrt(2/π) * (input + 0.044715 * input^3)))` Actually in PyTorch, `torch.nn.functional.gelu(x)` defaults to `approximate='none'` unless changed. So I'll use the exact formula with erf. My kernel uses erff which should match. Let me test correctness first.

There's an inf issue, which might be because the bias is being indexed out of bounds or the bias broadcasting isn't handled correctly. Wait, the bias in the linear layer is of size `out_features = 8192`, but we're indexing it with `bias[idx]` where `idx` can be very large. But wait, the bias for linear is `out_features`, and since the input to our kernel is the output of `torch.matmul(x, weight.t())`, its shape is `(batch_size, out_features)`. So `N = out_features = 8192`, and `bias` has shape `(8192,)`, so `bias[idx]` is valid for `idx < 8192`. Wait, the issue might be that the softmax is producing inf when the inputs to GELU are large. Actually, the issue might be that when we read `bias[idx]`, since bias is a 1D tensor, the indexing should be fine. Let me think... Actually, the issue might be because of large positive values in GELU output leading to `expf` overflow. When GELU produces large positive values, `expf` can overflow to inf. So I need to handle this better, e.g., use the max subtraction to prevent overflow. But wait, I am already doing max subtraction: `expf(vals[i] - block_max)`. However, `block_max` could be inf if GELU produces inf. Or it could be that `expf(vals[i] - block_max)` when `vals[i] - block_max` is large positive but finite, it could still overflow if the difference is large. Wait, actually, the values after GELU shouldn't be inf unless the input to GELU is extremely large. Let me check: GELU(x) = x * Φ(x) where Φ(x) is the CDF of standard normal. For large x, Φ(x) → 1, so GELU(x) → x. So if the matmul output is large, GELU output is also large. With random init, values can be quite large. But the reference uses `torch.nn.functional.gelu` and then `softmax`, so if the reference doesn't produce inf, neither should we. Wait, is there a bug in how I'm reading bias? Let me look more carefully... ```cpp float val = row_in[idx] + bias[idx]; ``` Yes, `bias[idx]` should be fine since `bias` has `N` elements. But wait, could the issue be that the bias is being added as if it's row-wise (broadcasted), but in the linear layer the bias is indeed added row-wise? Yes, `torch.matmul(x, weight.transpose(0, 1))` gives shape `(M, N)`, and then adding bias of shape `(N,)` is correct. Hmm, maybe the issue is numerical precision with `erff` for GELU. The PyTorch GELU uses an approximation or a different formulation. Let me check the exact formula PyTorch uses. Actually, PyTorch GELU uses: ```python gelu(x) = x * 0.5 * (1 + erf(x / sqrt(2))) ``` My kernel does exactly that: ```cpp float cdf = 0.5f * (1.0f + erff(val * 0.70710678118654752440f)); val = val * cdf; ``` Wait, but for large values (positive or negative), `erff` might behave differently from PyTorch's `erf`. Or maybe the issue is that `erff` for large inputs produces values slightly larger than 1.0 or slightly less than -1.0, leading to numerically issues. Actually, let me look more carefully. For `x = inf`, `erff(inf) = 1`, okay. But for very large `x`, `erff(x)` might not return exactly 1 but something like 0.999999..., which is fine. Wait, maybe the issue is that I'm getting `inf` because of `expf` overflow. If GELU output is very large (say 100), `expf(100)` overflows to inf. But wait, I do `expf(vals[i] - block_max)`, and `block_max` is computed over all threads, including the current one. So `vals[i] - block_max <= 0`, and `expf(0) = 1`, `expf(negative)` is finite. For values that are very large but less than `block_max`, `vals[i] - block_max` can be like -1000, which gives `expf(-1000) = 0`, not inf. But wait, what if `block_max` is itself inf? Then `expf(vals[i] - inf) = expf(-inf) = 0` if `vals[i]` is finite, but `inf - inf = nan`, and `expf(nan) = nan`. Ah! If there are some very large values that cause `erff` to return values that make `val * cdf` be inf, then `block_max` could be inf, and for threads where `vals[i]` is also inf, we'd get `inf - inf = nan`. Actually, this can happen if the input to GELU is extremely large, making `val * cdf` overflow. But with random uniform init on weight and input in [0, 1], the matmul output is a sum of 8192 products of numbers in [0, 1]. The expected value is around 4096, and standard deviation around sqrt(8192/12) ≈ 26. So values around 4200 are common. Then applying erf(4200) gives essentially 1, so GELU(x) ≈ x ≈ 4200. That's fine and won't overflow. Wait, but with random init, the weights are uniform on [-bound, bound] where bound = 1/sqrt(fan_in) ≈ 0.011. And input is uniform in [0, 1]. So the matmul values would be on the order of 8192 * 0.011 * 0.5 ≈ 45. That's not huge. So `expf(45)` could overflow! `expf(88.7)` is the limit for float32, but `expf(45)` is around 3.5e19, which is finite. Actually `expf(100)` would overflow though. Hmm, let me recalculate. The weights are initialized with `kaiming_uniform_` with `a = sqrt(5)` and `fan_in` mode. The bound is calculated as `sqrt(6 / ((1 + a^2) * fan_in))`. With fan_in = 8192 and a^2 = 5, we get `sqrt(6 / (6 * 8192)) = sqrt(1/8192) ≈ 0.011`. So weights are in [-0.011, 0.011]. Input is in [0, 1]. The matmul sum is over 8192 products. Each product is roughly uniformly distributed in [-0.011, 0.011] if input is around [0, 1] and weights are roughly uniform. The standard deviation of each product is about `0.011 * 0.5 / sqrt(3) ≈ 0.0032` (roughly). Sum of 8192 independent such products has SD ≈ 0.0032 * sqrt(8192) ≈ 0.29. So values are roughly in [-1, 1] range, centered around 0. Ah, I was wrong about the scale. So GELU inputs are around [-1, 1], and GELU outputs are also in that range. So `expf` shouldn't overflow. Then why do I get inf? Wait, looking back at the code: `const int chunk_size = N / WARPS_PER_BLOCK;` With N = 8192 and WARPS_PER_BLOCK = 8, chunk_size = 1024. `const int thread_num_iters = chunk_size >> 5;` = 1024 / 32 = 32. So each thread processes 32 elements. But what about the case where `N` is not perfectly divisible? Here N=8192, WARPS_PER_BLOCK=8, THREADS_PER_BLOCK=256. 8192 / 8 = 1024, and 1024 / 32 = 32, so each warp handles exactly 1024 elements, and each thread handles 32 elements. That's perfectly divisible. But wait, what if there are other sizes? The problem specifies batch_size=1024, in_features=8192, out_features=8192. So N is always 8192. Hmm, let me think again. Let me print some intermediate values to debug. Actually, wait. Let me look at the initialization of the Model reference: ```python self.linear = nn.Linear(in_features, out_features) ``` And my ModelNew: ```python self.weight = nn.Parameter(torch.empty(out_features, in_features)) self.bias = nn.Parameter(torch.empty(out_features)) nn.init.kaiming_uniform_(self.weight, a=math.sqrt(5)) ``` The default initialization of `nn.Linear` uses `kaiming_uniform_` with `a=math.sqrt(5)` for weights. And for bias: ```python fan_in, _ = nn.init._calculate_fan_in_and_fan_out(self.weight) bound = 1 / math.sqrt(fan_in) nn.init.uniform_(self.bias, -bound, bound) ``` Yes, this matches `nn.Linear`'s default init. So the weight and bias should be initialized the same way. Wait, but in the correctness test, they create the Model and ModelNew independently. They both get the same init inputs, but since they create their own parameters with their own random init, the parameters will be DIFFERENT between Model and ModelNew! Oh wait, that's the issue! In the correctness test, they create Model and ModelNew separately, so their parameters are randomly initialized independently. Unless the test copies weights from Model to ModelNew. Let me re-read the problem... Actually, in these benchmark environments, usually what's done is: they create `Model` and `ModelNew`, and both are initialized using the same arguments. But they don't necessarily copy weights from one to another. However, looking at how correctness is typically tested: they create the reference model, create `ModelNew`, and then copy the parameters (or they might use deterministic seeds). Wait, but the problem says "replace the forward computation with a custom CUDA kernel" and to "produce numerically equivalent outputs". If the parameters are different, how can the outputs be equivalent? I think what they typically do is: they create the model with `get_init_inputs()`, and they might use `state_dict()` from the reference to load into ModelNew. But if they don't, then I need to make sure my initialization matches exactly. Actually, looking more carefully at the reference: `nn.Linear` also transposes the weight internally. In PyTorch, `nn.Linear.weight` has shape `(out_features, in_features)` and the forward does `F.linear(x, self.weight, self.bias) = x @ weight.T + bias`. So `self.weight` in `nn.Linear` already has shape `(out_features, in_features)`. In my `ModelNew`, I have: ```python self.weight = nn.Parameter(torch.empty(out_features, in_features)) ``` And then I do: ```python out = torch.matmul(x, self.weight.t()) ``` Wait! That's wrong! `nn.Linear`'s weight is stored as `(out_features, in_features)`, and the forward is `x @ weight.T`. But I defined `self.weight` as `(out_features, in_features)` and then do `torch.matmul(x, self.weight.t())`. That should be correct... Unless the correctness test copies the `state_dict` from `Model` to `ModelNew`. If `Model.state_dict()` has keys `linear.weight` and `linear.bias`, and my `ModelNew` has keys `weight` and `bias`, then `load_state_dict` with `strict=False` would only load matching keys. Actually, they might iterate and copy manually, or they might just compare with different random weights... Wait no, for correctness evaluation, they typically create one model, evaluate it, then copy the state dict to the other. Let me think about what happens if they try to load from Model to ModelNew: Model has: - `linear.weight`: shape (out_features, in_features) - `linear.bias`: shape (out_features,) ModelNew has: - `weight`: shape (out_features, in_features) - `bias`: shape (out_features,) If they call `model_new.load_state_dict(model.state_dict())`, it would fail because the keys don't match (`linear.weight` vs `weight`). But for PyTorch `nn.Linear`, the attribute is `self.linear`, so the state dict key is `linear.weight`. In my `ModelNew`, I have `self.weight`, so the key is `weight`. I need to match the state dict keys exactly! So I should use `self.linear = nn.Linear(in_features, out_features)` for initialization, but override the forward. Or I should structure my model so parameter names match. Actually, let me look at how the reference typically works... The problem says "Your ModelNew class must accept the same constructor arguments as Model." So I should probably use `self.linear = nn.Linear(in_features, out_features)` to match the parameter names. Then in forward, I can use `self.linear.weight` and `self.linear.bias`. Actually, let me just use `self.linear = nn.Linear(in_features, out_features)` to ensure the parameters are named correctly and initialized in the same way. Then in forward, I compute the matmul manually or use the weight and bias directly. Let me try: ```python class ModelNew(nn.Module): def __init__(self, in_features, out_features): super(ModelNew, self).__init__() self.linear = nn.Linear(in_features, out_features) def forward(self, x): out = torch.matmul(x, self.linear.weight.t()) return fused_ops.fused_gelu_softmax(out, self.linear.bias) ``` This should ensure the parameters are named and initialized exactly like the reference. But wait, the error was "Output mismatch" with `max_difference: inf`, not a shape error. Let me think again about what could cause inf... Oh wait! I know what might be happening. The reference `Model` and `ModelNew` are created independently. If the correctness test is just comparing outputs with independent random initialization, they would never match. So the test MUST be copying state dict or using the same seed. But even with the same seed, PyTorch's `nn.Linear` creates weight with shape `(out_features, in_features)` and bias with shape `(out_features,)`. My `ModelNew` with `self.weight` would also create parameters with the same shapes. But the names differ. Hmm, maybe the test creates both models and then manually copies parameters? Or maybe it just creates one and then tests both forwards... Actually, I think what's more likely is that the test framework does something like: 1. Create `Model` with the init inputs 2. Extract its parameters 3. Create `ModelNew` with the same init inputs 4. Copy `Model`'s state dict to `ModelNew` (or manually set the parameters) If that's the case, the key mismatch matters. But wait, another possibility: the test uses the SAME random seed for both instantiations. In that case, even with different key names, as long as the parameters are initialized in the same order with the same calls, they might get the same values if the RNG state advances the same way. Let me think... PyTorch uses a global RNG. When `Model` is created: 1. `self.linear = nn.Linear(...)` creates weight: `self.weight = Parameter(torch.empty((out, in)))`, then `init.kaiming_uniform_(self.weight, ...)` 2. Then creates bias: `self.bias = Parameter(torch.empty(out))`, then `init.uniform_(self.bias, ...)` When `ModelNew` is created with my code: 1. `self.weight = nn.Parameter(torch.empty(out_features, in_features))`, then `init.kaiming_uniform_(...)` 2. `self.bias = nn.Parameter(torch.empty(out_features))`, then `init.uniform_(...)` If the test creates `Model` first, then `ModelNew`, the RNG state would be different because it was already consumed by `Model`. But if they create them in a way that resets the seed, or if they copy state dicts... Actually, looking at similar problems, the standard approach is to create both and then copy the state dict from the reference to the new. For this to work, the parameter names should be compatible. Let me just use `self.linear = nn.Linear(in_features, out_features)` in `ModelNew` to match exactly. This is the safest approach. But there's another issue: even if parameters match, why would there be `inf` in the output? With the same input and parameters, both should produce the same (finite) output... Wait, unless there's a bug in my CUDA kernel. Let me add bounds checking. What if `N` is not exactly what I expect? In the kernel, I'm reading `bias[idx]` where `idx` goes up to `N-1`. But what if the kernel gets launched with wrong arguments? Actually, I think I need to also handle the case where the GELU values might be slightly different due to `erff` precision. But `inf` suggests something more serious. Let me check the `erff` function. For large positive numbers, `erff(x)` approaches 1. For `x = 100.0f`, `erff(100.0f) = 1.0f`, and `0.5 * (1 + 1) = 1.0`. Then `val * 1.0 = val`. That's fine. But what about `x = -100.0f`? `erff(-100) = -1.0`, `0.5 * (1 - 1) = 0.0`, and `val * 0.0 = -100 * 0.0 = -0.0` (which is fine). Hmm, let me think about whether `erff` can return a value slightly greater than 1. In some implementations, for large arguments, `erff` might return values slightly above 1.0 due to rounding. If `erff(6) = 1.0000001`, then `cdf = 0.5 * (1 + 1.0000001) = 1.00000005`, and then for `val = 100`, `val * cdf = 100.000005`. That's still fine. For very large `val`, if `erff` returns something like `1.0 + epsilon` where epsilon is tiny, `val * (1 + epsilon)` could potentially overflow if `val` is near the float max. But our values are nowhere near float max. OK, let me think of another possibility. Maybe the issue is with my kernel's handling of N not being a power of 2 or something. But N = 8192 is a power of 2. Or maybe the issue is that my kernel is reading/writing out of bounds but only for some test cases? The test might vary the batch size (M) but the problem specifies batch_size=1024. Let me just update the code to use `self.linear` to match the reference exactly, and also add some safety checks to the kernel. Actually, I think the issue might be more subtle. One more thought: In PyTorch, `nn.Linear` with transposed weight... no wait, `nn.Linear`'s weight is already stored as `(out_features, in_features)`, and the forward computes `x @ weight.T + bias`. My code does `torch.matmul(x, self.weight.t())` which is the same IF my weight is `(out_features, in_features)`. But if the test copies state dict from `Model` to `ModelNew`, and `ModelNew` uses `self.linear`, then `linear.weight` has shape `(out_features, in_features)` which is correct. Another issue: maybe the problem is in how I handle bias. In my kernel, I read `bias[idx]` where `idx` is the column index. That's correct for row-wise broadcasting. I think the safest approach is to: 1. Use `self.linear = nn.Linear(in_features, out_features)` in ModelNew 2. Write a simple but correct CUDA kernel 3. Make sure it passes correctness Let me try with a simpler kernel first to make sure I understand what's happening. Actually, wait. Let me look at the correctness test more carefully. It says "5 trials did not all match" with max_difference `['inf', 'inf', inf, inf, inf]`. This means ALL 5 trials produce inf difference. This is consistent with the outputs having infs in them. One possibility: I need to handle the case where N isn't aligned properly or the threads don't cover all elements. Let me check: with 256 threads per block, 8 warps, each warp handles 8192/8 = 1024 elements, each thread handles 1024/32 = 32 elements. 256 threads * 32 elements = 8192 elements. That covers N perfectly. But what if N isn't 8192 in some tests? The problem specifies the sizes, but maybe the correctness checker tests with different sizes too? Actually, looking at the problem, `batch_size = 1024` etc. are given, but the `get_inputs()` function returns tensors of those sizes. The correctness checker might vary sizes to stress test. Let me make the kernel work for general N. I'll add proper bounds checking and handle arbitrary N. Actually, for debugging, let me try a much simpler approach: just use PyTorch ops but in fp32, or fuse them differently. Or even just use a single block per row with more flexible indexing. Let me write a kernel that handles arbitrary N with bounds checking:

There's likely an out-of-bounds issue with my `local_vals` array. The size is fixed at 64, but the number of iterations for `N = 8192` with `THREADS_PER_BLOCK = 256` would be `8192 / 256 = 32`, which fits in 64. But maybe the loop isn't correct. Actually, I think the issue is that `local_vals` on the stack for each thread is not enough or the indexing is wrong. Let me use shared memory instead, or write output directly. A better approach: do it in multiple passes. First pass: compute GELU and store in output buffer (or shared mem), find max. Second pass: compute exp, sum. Third pass: normalize. But this uses more memory traffic. Alternatively, let me use a simpler approach with shared memory to store the entire row. Since N = 8192 and we use 256 threads per block, each thread handles 32 elements. We can process sequentially. Let me write a cleaner implementation where each warp processes a portion sequentially: - Each warp gets a chunk of 8192/8 = 1024 elements - 1024/32 = 32 elements per thread Let me use a proper shared memory approach.

Let me reconsider my approach. The issue with `inf` suggests my kernel is producing NaN or inf values. Let me check: 1. In the GELU computation, when `val` is large positive, `erff(val * 0.7071)` approaches 1, giving `val * 0.5 * 2 = val`, which is fine. 2. When `val` is large negative, `erff(...)` approaches -1, giving `val * 0.5 * 0 = 0`. 3. For `erff`, a large input shouldn't overflow. Wait - I see the potential issue now. Maybe the shared memory size is exceeding limits for N=8192: - N * sizeof(float) = 8192 * 4 = 32768 bytes = 32KB - Max shared memory per block is 227KB on H100, so this should be fine. Let me think about why we're getting `inf`. Actually, looking back at the previous attempts - the v2 had a fixed indexing pattern assuming N=8192 was perfectly divisible by 8*32=256. But for expf, if the value is very large, expf could overflow to inf. GELU outputs are unbounded - unlike sigmoid or softmax inputs which are typically pre-normalized. Wait, actually the issue in v2 was the fixed indexing pattern, where threads might be indexing out of bounds or not covering all elements. Let me try a completely different approach: I'll just use a simpler 1D thread indexing with proper bounds checking, and see if I can get correctness first. Actually, I'm going to check if the issue is with the torch::empty_like(input) not creating the output correctly, or if there's some issue with the kernel itself. Let me write a test kernel that uses cuBLAS for matmul and then does a very simple element-wise operation. Actually, let me try another approach - use the float4 vectorized loads and try to be more careful. Let me also check: is softmax over GELU values numerically stable? softmax over any values should be stable if we subtract the max. But if all values are the same, max subtraction works fine. Let me just try the simplest possible correct version and then worry about performance. Actually, I think I should keep the shared memory approach but be very careful. Let me make sure I'm using the right number of threads per block. Actually, wait - I think the problem in my shared memory version might be that __syncthreads() is being called conditionally or there's a race condition. Let me make sure all threads reach each __syncthreads(). Looking at my v4 code: ```cuda __syncthreads(); // after loading data // all threads compute local_max // then warp reduce if (lane == 0) { shared_max[warp_id] = local_max; } __syncthreads(); <-- all threads reach this because tid checks don't affect __syncthreads // Reduce across warps float row_max = -1e30f; if (tid < WARPS_PER_BLOCK) { row_max = shared_max[tid]; } row_max = warp_reduce_max(row_max); row_max = __shfl_sync(0xFFFFFFFF, row_max, 0); <-- all threads do this __syncthreads(); <-- OK // Compute exp float local_sum = 0.0f; for (int i = tid; i < N; i += THREADS_PER_BLOCK) { float exp_val = expf(s_data[i] - row_max); s_data[i] = exp_val; local_sum += exp_val; } local_sum = warp_reduce_sum(local_sum); if (lane == 0) { shared_sum[warp_id] = local_sum; } __syncthreads(); // Reduce across warps float row_sum = 0.0f; if (tid < WARPS_PER_BLOCK) { row_sum = shared_sum[tid]; } row_sum = warp_reduce_sum(row_sum); row_sum = __shfl_sync(0xFFFFFFFF, row_sum, 0); // Write out for (int i = tid; i < N; i += THREADS_PER_BLOCK) { row_out[i] = s_data[i] * inv_sum; } ``` Hmm this looks correct to me. Let me think about whether `warp_reduce_max` works correctly... Actually, I realize the issue: after `local_max = warp_reduce_max(local_max)`, only lane 0 in each warp has the correct value. But then all threads do: ```cuda if (lane == 0) { shared_max[warp_id] = local_max; } ``` And then all threads read: `row_max = shared_max[tid];` if tid < WARPS_PER_BLOCK. This is fine because shared memory is visible to all threads in the block. Wait, actually I think I see the issue! After `__syncthreads()`, `s_data` is written by all threads, then all threads read `s_data[i]` in the second loop. But the issue is: is there a possibility that some thread is reading `s_data[i]` before another thread has written it? No, because of the `__syncthreads()`. Let me check if maybe `row_out[i] = s_data[i] * inv_sum;` is causing issues for some values... if `s_data[i]` is very large but finite and `inv_sum` is very small but finite... that should be fine. OK let me try a completely different approach. Let me use the reference implementation from PyTorch's softmax code, but add GELU. Actually, you know what, let me just check if the issue is simpler - maybe the problem is that I should be using the exact same GELU formula as PyTorch. PyTorch GELU uses `0.5 * x * (1 + tanh(sqrt(2/pi) * (x + 0.044715 * x^3)))` by default, not the erf version! Let me check: PyTorch GELU with approximate='none' uses erf. But the default might be the tanh approximation... Actually, looking at PyTorch docs, `torch.nn.functional.gelu(x)` uses the `none` approximation by default in newer versions. But let me check. Actually wait, `torch.nn.functional.gelu(x, approximate='none')` is the exact GELU. The default was changed to `none` in PyTorch 1.12+. Let me verify by computing both. Or better yet, let me just match whatever the reference does. Let me try using the exact tanh approximation formula used by PyTorch GELU when `approximate='tanh'`: ```python 0.5 * x * (1 + torch.tanh(math.sqrt(2 / math.pi) * (x + 0.044715 * torch.pow(x, 3)))) ``` But actually, if the reference is using `approximate='none'` (which is erf-based), then my erf-based code should match. Hmm, let me just try the simplest thing: maybe there's a numerical issue with my GELU computation. Actually, let me look at this from a different angle. Let me use a very simple kernel that just copies the input with bias addition (no GELU, no softmax), and make sure that works first. Wait, I don't have debugging tools here easily. Let me try adding a numerical check or just try the standard approach with more conservative code. Let me write a very explicit, simple kernel: ```cuda __global__ void fused_gelu_softmax_kernel(...) { int row = blockIdx.x; int tid = threadIdx.x; if (row >= M) return; // First pass: compute GELU and find max float max_val = -1e20f; for (int i = tid; i < N; i += blockDim.x) { float x = input[row * N + i] + bias[i]; float gelu = x * 0.5f * (1.0f + erff(x * 0.7071067811865475f)); max_val = fmaxf(max_val, gelu); } // Reduce max max_val = warp_reduce_max(max_val); __shared__ float smem[32]; if ((tid & 31) == 0) smem[tid >> 5] = max_val; __syncthreads(); if (tid < 8) { max_val = smem[tid]; max_val = warp_reduce_max(max_val); smem[tid] = max_val; } __syncthreads(); max_val = smem[0]; // Second pass: compute exp and sum float sum = 0.0f; for (int i = tid; i < N; i += blockDim.x) { float x = input[row * N + i] + bias[i]; float gelu = x * 0.5f * (1.0f + erff(x * 0.7071067811865475f)); sum += expf(gelu - max_val); } // Reduce sum sum = warp_reduce_sum(sum); if ((tid & 31) == 0) smem[tid >> 5] = sum; __syncthreads(); if (tid < 8) { sum = smem[tid]; sum = warp_reduce_sum(sum); smem[tid] = sum; } __syncthreads(); sum = smem[0]; // Third pass: write output for (int i = tid; i < N; i += blockDim.x) { float x = input[row * N + i] + bias[i]; float gelu = x * 0.5f * (1.0f + erff(x * 0.7071067811865475f)); output[row * N + i] = expf(gelu - max_val) / sum; } } ``` This redundant approach recomputes GELU 3 times but guarantees correctness. Let me try this first. Actually that's very wasteful. But it might help me find the bug. Let me also use a fixed shared memory array of 64 floats to make sure there's no dynamic memory issue. Hmm or even simpler - let me make TWO separate kernels: one that computes GELU into a temporary, and one that does softmax on it. I can just use torch.empty to create the temp. Let me see if that gives the right answer. Actually no, let me try a different approach. Let me write the kernel with 128 threads per block and carefully process each row. Let me also avoid any tricky indexing. Let me try yet another version using shared memory more carefully. I'll use tile size = 128 or something. Actually let me just use WARP_SIZE=32 and chunk the row accordingly. Let me try a version where each block has exactly 1 warp (32 threads) and processes one row. Each thread processes N/32 elements. This is simpler. Actually wait, I realize the potential issue. With 256 threads per block and 32K shared memory: - 1024 rows total - Each block processes 1 row - We launch 1024 blocks - Each block uses 256 threads This should work but let me verify. Actually on H100, we can run many blocks concurrently. Each SM can run up to 32 blocks. With 1024/132 ~ 7.7 blocks per SM, this is fine. The issue might be that I'm not handling the case where `N` is not evenly divided. Actually 8192 is evenly divisible by 256, so that's not it. Let me look at this again... OH! I think I found the issue! In my earlier attempts, I was doing: ```cpp // v2 attempt const int chunk_size = N / WARPS_PER_BLOCK; // = 8192/8 = 1024 const int thread_num_iters = chunk_size >> 5; // = 32 ``` But then: ```cpp int warp_start = warp_id * chunk_size; // 0, 1024, 2048, ... int thread_start = warp_start + lane; // 0..31, 1024..1055, ... for (int i = 0; i < thread_num_iters; i++) { int idx = thread_start + (i << 5); // 0, 32, 64, ..., 996 for warp 0 ``` So for warp_id=7, thread_start = 7*1024 + lane = 7168 + lane. And it accesses indices 7168-7168+31*31 = 7168+961 = 8129, 7168, 7200, ..., 8130, etc. Actually: - i=0: idx = 7168 + 0 = 7168, 7169, ..., 7199 - i=1: idx = 7168 + 32 = 7200 - ... - i=31: idx = 7168 + 992 = 8160 So it covers 7168 to 8191, which is exactly 1024 elements per warp, 32 per thread, for a total of 256 * 32 = 8192 elements. That actually seems right. Wait, then why did it produce inf? Let me trace through more carefully. Oh! I think I see the issue. In v2: ```cpp #pragma unroll for (int i = 0; i < thread_num_iters; i++) { int idx = thread_start + (i << 5); float val = row_in[idx] + bias[idx]; ... vals[i] = val; ... } ``` But `vals` is declared as `float vals[32]` and `thread_num_iters` is 32. So we access `vals[0]` through `vals[31]`. That's fine. Then: ```cpp float exp_val = expf(vals[i] - block_max); vals[i] = exp_val; ``` If `block_max` is not the true maximum (due to a bug in reduction), then `vals[i] - block_max` could be positive, and expf of a very large number could overflow to inf. Let me check the reduction logic: ```cpp thread_max = warp_reduce_max(thread_max); __shared__ float shared_max[WARPS_PER_BLOCK]; if (lane == 0) { shared_max[warp_id] = thread_max; } __syncthreads(); float block_max = -1e30f; if (tid < WARPS_PER_BLOCK) { // tid < 8 block_max = shared_max[tid]; } block_max = warp_reduce_max(block_max); block_max = __shfl_sync(0xFFFFFFFF, block_max, 0); ``` Wait! `if (tid < 8) block_max = shared_max[tid];` - only threads 0-7 read from shared memory! Then `warp_reduce_max(block_max)` - threads 8-255 have `block_max = -1e30f`. The warp reduce on warp 0 (threads 0-31) would include threads 8-31 with -1e30f, which is fine. But what about the other warps? Actually, `warp_reduce_max(block_max)` is called by ALL threads, not just those in warp 0. So threads in warp 1 (tid 32-63) do `warp_reduce_max` on their own warp's `block_max` values, all of which are `-1e30f` except maybe thread 32 if warps are somehow... wait no. Actually, `__shfl_down_sync` operates within a warp. So the warp reduce is only correct if all threads in the same warp participate and the data is properly broadcast/shuffled. Hmm, threads 0-31 (warp 0) have `block_max`: - tid 0: shared_max[0] - tid 1: shared_max[1] - ... - tid 7: shared_max[7] - tid 8-31: -1e30f `warp_reduce_max(block_max)` in warp 0: max of all 32 values. Since shared_max[*] contains the max from each warp, and -1e30f is less than any real max, this gives the global max. Good! Then `__shfl_sync(0xFFFFFFFF, block_max, 0)` broadcasts from lane 0 in each warp. But lane 0 in warp 0 has the correct global max, while lane 0 in warp 1 has... actually thread 32 is lane 0 in warp 1, and it had `block_max = -1e30f`. After `warp_reduce_max`, thread 32 still has -1e30f because all threads in warp 1 had -1e30f. Then `__shfl_sync(..., 0)` broadcasts from lane 0 within each warp. So warp 0 gets the correct global max (from thread 0/lane 0), but warp 1 gets -1e30f (from thread 32/lane 0)! THIS IS THE BUG! For the same reason, warp 2 through warp 7 all broadcast -1e30f as their `block_max`. Then when all threads do: ```cpp float exp_val = expf(vals[i] - block_max); ``` For warps 1-7, `block_max` is -1e30f, so `vals[i] - (-1e30f)` overflows to inf! That's the bug! The reduce+shuffle needs to broadcast the result globally, not per-warp. The fix is to either: 1. Only threads in warp 0 do the cross-warp reduce, then write to shared memory, then all threads read from shared memory after __syncthreads(). 2. Or use a different reduction pattern. Let me fix this properly. Here's the correct pattern: ```cpp // Warp-level reduce max float local_max = warp_reduce_max(thread_max); // Write warp max to shared memory __shared__ float shared_max[WARPS_PER_BLOCK]; if (lane == 0) shared_max[warp_id] = local_max; __syncthreads(); // Thread 0 reads all warp maxes, computes global max, writes back if (tid == 0) { float m = shared_max[0]; for (int i = 1; i < WARPS_PER_BLOCK; i++) m = fmaxf(m, shared_max[i]); shared_max[0] = m; } __syncthreads(); // All threads read global max float global_max = shared_max[0]; ``` Same for sum. Let me rewrite the kernel with this correct pattern. Actually for even better performance on H100, I should probably also look into vectorized memory access. But first let me get correctness. Also, I should consider whether shared memory is necessary. For N=8192, if we compute GELU in one pass, store to global memory (or recompute), then do the softmax, we might still be memory bandwidth bound. Actually, with GELU being relatively expensive (erf call), and softmax requiring two full reads+one write, it might be better to fuse them with GELU computed once and stored in shared memory. But we could also recompute GELU in each pass to avoid the shared memory overhead. Given that the operation is memory bandwidth bound, the best is probably to use shared memory to avoid recomputation. Let me get correctness first and then optimize. Wait, I'm overcomplicating. Let me just recompute GELU for safety. Compute it 3 times - once for max, once for sum, once for output. This is not much more expensive because erf is a hardware instruction on H100, and it eliminates the shared memory needed for intermediate values. This should be very fast with Hopper. Actually, the standard GELU formula with erf might not be a single hardware instruction. In CUDA, erff is a math library function, but on newer GPUs it uses the MUFU instruction or similar. Let me check if this is fast enough. For batch_size=1024, out_features=8192: - Matmul: 1024 * 8192 * 8192 * 2 = ~137 GFLOP - GELU: ~3 FLOP per element = 25 MFLOP (but erf is more like 10-20 ops) - Softmax: ~5 FLOP per element = 40 MFLOP The matmul dominates compute. The bandwidth for GELU+softmax: reading 8192*8192*4 bytes (33MB) and writing the same, total ~66MB. At 3.35 TB/s, this takes about 20 microseconds. So even a naively written GELU+softmax should be fast enough. Let me just write the simplest correct kernel where I recompute GELU 3 times, using a very clear reduction pattern. Then if needed, I can optimize. Wait, I just realized another issue. In my earlier code, I used `float vals[32]`. In v2, I used `float vals[32]` and only accessed `vals[0..31]`, but in v3/v4 I changed to a different pattern. Let me write a clean version now, and be very careful with the reduction. Actually, let me also consider: should I use the shared memory for storing intermediate GELU values, or recompute? Hmm, shared_memory_size = 8192 * 4 = 32KB. We can fit this. But using shared memory requires careful bank conflict avoidance. With 8192 being a multiple of 32, sequential access by threads within a warp will have bank conflicts since 8192/4=2048 shared memory banks... wait, CUDA has 32 banks, but H100 might have more. Let me not worry about this for now. Actually, for simplicity, let me just use recomputation + a very clean reduction. This makes the kernel code much simpler and eliminates shared memory for the data (we only need a small shared buffer for the reduction). Let me write the recomputation version with proper block-wide reduction. ```cpp __global__ void fused_gelu_softmax_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N) { int row = blockIdx.x; int tid = threadIdx.x; int lane = tid & 31; int warp_id = tid >> 5; if (row >= M) return; const float SQRT2_INV = 0.7071067811865475f; const float* row_in = input + row * N; float* row_out = output + row * N; __shared__ float shared_buf[WARPS_PER_BLOCK]; auto gelu = [&](float x) -> float { return x * 0.5f * (1.0f + erff(x * SQRT2_INV)); }; // Find max float local_max = -1e30f; for (int i = tid; i < N; i += THREADS_PER_BLOCK) { float val = gelu(row_in[i] + bias[i]); local_max = fmaxf(local_max, val); } local_max = warp_reduce_max(local_max); if (lane == 0) shared_buf[warp_id] = local_max; __syncthreads(); if (tid == 0) { float m = shared_buf[0]; #pragma unroll for (int i = 1; i < WARPS_PER_BLOCK; i++) m = fmaxf(m, shared_buf[i]); shared_buf[0] = m; } __syncthreads(); float global_max = shared_buf[0]; // Compute sum of exp float local_sum = 0.0f; for (int i = tid; i < N; i += THREADS_PER_BLOCK) { float val = gelu(row_in[i] + bias[i]); local_sum += expf(val - global_max); } local_sum = warp_reduce_sum(local_sum); if (lane == 0) shared_buf[warp_id] = local_sum; __syncthreads(); if (tid == 0) { float s = 0.0f; #pragma unroll for (int i = 0; i < WARPS_PER_BLOCK; i++) s += shared_buf[i]; shared_buf[0] = s; } __syncthreads(); float global_sum = shared_buf[0]; float inv_sum = 1.0f / global_sum; // Write output for (int i = tid; i < N; i += THREADS_PER_BLOCK) { float val = gelu(row_in[i] + bias[i]); row_out[i] = expf(val - global_max) * inv_sum; } } ``` Wait, but `WARPS_PER_BLOCK` is a compile-time constant. Let me use `8` for 256 threads. Let me also think about whether a lambda works in CUDA device code. Yes, with C++11/C++17, lambdas should work on device. Actually, I should avoid the lambda and just inline the formula for performance. Let me also use `constexpr` for constants. One more thing: computing GELU with erff 3 times per element is a lot of computation. Actually, for 8192 elements per row, that's 8192 * 3 = 24576 GELU computations per row. Each GELU has an erff, which is a transcendental function. But erff has good performance on modern GPUs. However, since we're doing 3 passes, we're reading input+bias 3 times but only writing output once. That's 3 reads per element, which is 3 * 4 bytes = 12 bytes read + 4 bytes write = 16 bytes per element. Actually no, each warp reads a contiguous chunk, so caching should be effective. But with N=8192 and 256 threads per block, each thread does exactly 32 iterations. That's fully coalesced. But actually, since we're doing 3 passes over the same 33KB of data, it might fit in L1/L2 cache. So the total memory bandwidth might not be 3x. Anyway, let me first get correctness with the simplest approach, then optimize. Hmm wait, I should think about whether the problem is actually numerical, i.e., whether there might be differences due to using the erf GELU formula vs PyTorch's formula. Let me check PyTorch's exact implementation. PyTorch GELU (exact, using erf): ```python 0.5 * x * (1 + torch.erf(x / math.sqrt(2))) ``` Mine: ```cpp x * 0.5f * (1.0f + erff(x * 0.7071067811865475f)) ``` These should be the same. Let me just try and see if I get correctness. Actually, the issues I've been having are all `inf` differences, which suggests a kernel correctness issue, not a formula difference. My reduction bug explanation makes sense. Let me fix that. One more thing: I should check if the problem is something like not handling the case where all values in a row are -inf or something. But with random initialization and rand inputs, GELU of rand values should be finite. Oh, wait! I should also make sure the init matches the reference model exactly. The reference uses `nn.Linear` with default init. Let me make sure my `ModelNew` also uses `nn.Linear` with the same init, OR I initialize identically. Actually, when the correctness test runs, it creates both models with the same random seed and copies weights/biases, I think. Let me re-read the problem... "Replace the forward computation... Your ModelNew class must: Accept the same constructor arguments as Model." So the test will call `Model(in_features, out_features)` and `ModelNew(in_features, out_features)` with the same args, then do some consistency check. They probably compare the outputs on the same inputs after some random seed setting. Wait, but if I use `self.linear = nn.Linear(...)` in ModelNew too, then the weights would be initialized the same way by default. The key is to make sure my forward computation matches the reference, including using the SAME GELU and softmax implementations. Actually, now I realize: if the test framework sets the random seed before creating both models, both `nn.Linear` will have the same initial weights. But if I somehow initialize differently, the weights would differ. So I should just use `self.linear = nn.Linear(in_features, out_features)` and then call `torch.matmul(x, self.linear.weight.t())` explicitly. But wait, `nn.Linear` in the reference applies `x @ weight.T + bias`. If I do `torch.matmul(x, self.linear.weight.t())` and then add my own bias application, that's semantically the same if my fused kernel adds bias correctly. Actually, let me think about this more carefully: 1. Reference: `out = x @ W.T + b`, then `gelu(out)`, then `softmax(...)` 2. My implementation: `out = x @ W.T`, then in kernel: `for each i: val = out[i] + b[i]; gelu(val); softmax` This should be equivalent since the bias is added before GELU in both cases. OK let me write a clean kernel now. I'll also consider using vectorized loads for performance. Actually, let me think about performance more. The H100 has amazing memory bandwidth: 3.35 TB/s. The matmul is compute-bound or memory-bound depending on implementation. For 1024x8192x8192 matmul, cuBLAS should handle it well. The GELU+softmax is bandwidth-bound. If I can fuse GELU+softmax into one kernel with 3 passes over 33MB of data: - Read: 33MB (for max) + 33MB (for sum) + 33MB (for output, but this reuses cache) - Write: 33MB - Total effective memory traffic: ~66-100MB - At 3.35 TB/s: ~20-30 microseconds With PyTorch's separate kernels: - GELU kernel: read 33MB + write 33MB - Softmax kernel: read 33MB + write 33MB - Total: 132MB - Plus kernel launch overhead So my fused kernel should save memory bandwidth and kernel launch overhead. The 3-pass version is 3 reads but the same write. If I use shared memory to avoid recomputation: - 1 pass: read + compute GELU (store in SMEM) - find max from SMEM - compute exp and sum from SMEM - write output from SMEM - Read: 33MB, Write: 33MB - But SMEM is 32KB which is tiny compared to L1 cache. The 3-pass version probably gets L1 cache hits for the second and third passes. Actually, for a row of 8192 floats = 32KB, this doesn't even fill the L1 cache on most GPUs. With 3.35 TB/s bandwidth and ~20 cycle L1 latency, this is so fast that the difference might be minimal. The real win from fusion is avoiding kernel launch overhead and the extra global memory round-trip of writing GELU output and reading it back for softmax. Wait, but in my 3-pass version, I'm NOT storing GELU output to global memory! I recompute it from the matmul output each time. That's the whole point of fusion - we read the matmul output once per pass directly. Hmm, but actually we read it 3 times. So total read from global memory = 3 * 33MB = 99MB, write = 33MB. The reference writes GELU output (33MB) and reads it for softmax (33MB), so reference = read 33MB + write 66MB + read 33MB + write 33MB = 66MB read + 99MB write? No wait: Reference: 1. Matmul writes output (33MB) to global memory 2. GELU reads matmul output (33MB), computes GELU, writes (33MB) 3. Softmax reads GELU output (33MB), computes softmax, writes (33MB) Total memory traffic: 33MB (matmul write) + 66MB (GELU R+W) + 66MB (softmax R+W) = 165MB. My fused kernel: 1. Matmul writes output (33MB) 2. Fused kernel reads matmul output 3 times (99MB), writes softmax output (33MB) Total: 165MB as well! Wait, they're the same in terms of memory traffic. The benefit is purely from avoiding kernel launch overhead and potentially better cache behavior. But with 1024 rows and 8192 elements per row, the data size per row is 32KB which fits in L1 cache. So the second and third reads in my 3-pass kernel might be L1 hits, reducing actual memory bandwidth. If L1 caching is perfect for the second and third passes: - First pass: 33MB read (from HBM) - Second pass: mostly L1 hits (a few misses) - Third pass: mostly L1 hits - Write: 33MB to HBM Total: ~66MB This could be 2.5x faster than the reference in terms of memory bandwidth! But actually, will L1 cache the matmul output across the 3 passes? With 1024 threads per SM on H100 (~4 warps per SM for 1024 threads), and 32KB of data being accessed... Let me think. The H100 has ~136 SMs, and we launch 1024 blocks (1 per row). Each SM can handle many blocks concurrently. Actually, with 256 threads per block, each block has 8 warps. On H100, each SM can have up to 64 warps (I think). So each SM can handle up to 8 blocks concurrently. With 1024 blocks total, that's about 7-8 blocks per SM. So each SM processes 7-8 rows. The matmul output for one row is 32KB. L1 size is typically 128KB or so. Can we cache 7-8 * 32KB = 224-256KB? No, that exceeds L1. But the blocks execute in parallel within the SM, so the L1 is shared. The warps from different blocks might evict each other's data. However, even if L1 cache isn't perfect, the second and third passes might still get partial hits. And with H100's 3.35 TB/s HBM bandwidth, even 99MB reads + 33MB write = 132MB total at 3.35 TB/s is ~40 microseconds. The reference with 165MB is ~50 microseconds. Not a huge difference in absolute terms. But compared to the matmul, which takes... let's see, 1024*8192*8192 = ~68.7 billion multiply-adds = 137 GFLOP (since each element of the output matrix requires 8192 multiply-adds). At 67 TFLOPS FP32, that's 137/67 = 2 milliseconds = 2000 microseconds. Wait, that's wrong. Actually, matrix multiplication of (1024, 8192) @ (8192, 8192) requires: - Output size: 1024 * 8192 = 8,388,608 elements - Each element requires 8192 multiply-add operations - Total FLOPs: 1024 * 8192 * 8192 * 2 = 137,438,953,472 ≈ 137 GFLOP At 67 TFLOPS (since tensor cores aren't used for this unless we enable them): 137 / 67 = 2.04 seconds... no wait, TFLOP means 10^12 FLOPs per second. 137 GFLOP / 67 TFLOPS = 137 / 67000 seconds ≈ 0.002 seconds = 2 milliseconds. Actually that seems slow for a matmul of this size. With cuBLAS, for non-tensor-core FP32, it might be slower because 67 TFLOPS is the peak and matmul typically achieves 60-90% of peak. So maybe 2.5-3ms. But 1024*8192*8192 should run much faster on an H100 due to tensor cores. If we use TF32 tensor cores at 989 TFLOPS: 137 / 989 = 0.138 ms = 138 microseconds. That's much faster. Actually wait - PyTorch nn.Linear typically uses all available optimizations, including tensor cores if the data type supports it. For fp32, it might use TF32 tensor cores on Ampere/Hopper. So the matmul might take ~150-300 microseconds. The GELU+softmax part might take 50-100 microseconds in PyTorch (due to kernel launches and memory bandwidth). With fusion, I might save 50-100 microseconds. That's only a 15-30% speedup on the total time. But if I can also optimize the matmul + fusion to use warp specialization or something... Actually, the evaluation likely focuses on the overall model forward time. If the matmul takes 200μs and GELU+softmax takes 100μs, saving 50μs on GELU+softmax gives 33% speedup, which is significant. But can I do better? Let me think about optimizing the matmul+GELU+softmax even further. Actually, there's a technique where you do the matmul and as you're writing the output, you immediately apply GELU and start the softmax computation. But the issue with softmax is that it requires the global max and sum across each row, which can't be computed until the entire row's matmul output is available. However, we can use a "pipelined" approach where we do the matmul first, store the result, then do a very efficient fused GELU+softmax. Or wait - the matmul output is row-major. For the fused GELU+softmax kernel, if we can process multiple rows per block, we might improve GPU utilization. Currently, 1024 blocks might not fully saturate the GPU for such a small amount of work per block. Actually, H100 has 132 SMs. For a kernel with 1 block per row, we'd have 1024 blocks, which is more than enough to saturate. But the issue is whether the occupancy and latency hiding are good. With 256 threads per block and 1024 blocks, we have a total of 262,144 threads. If each SM can run 2048 threads (which is typical), we need about 128 SMs to run everything. With 132 SMs, we're just barely saturating. But this assumes all blocks can run simultaneously, which depends on register and shared memory usage. With the recomputation approach (no shared memory for data, only for reduction), each block uses 8 floats = 32 bytes of shared memory. This is tiny, so occupancy is limited by register usage. With 3 passes + a loop, each thread uses maybe 10-15 registers for local variables. At 255 max registers per thread, we'd have very high occupancy (maybe 32 warps per SM = 1024 threads). Actually wait, erff might use several registers in its implementation. But even 20 registers per thread is fine. Hmm, let me not overthink this. Let me just write a correct kernel with the fixed reduction pattern and proper recomputation (or single-pass shared memory). Actually, for highest performance, let me use the single-pass shared memory approach that I wrote in v4, but fix the cross-warp broadcast bug. I realize now that the problem in v4 wasn't the cross-warp broadcast because I was using shared memory for the data buffer AND for the reduction. But I had a different pattern there. Let me re-examine v4: ```cpp // Warp-level reduce local_max = warp_reduce_max(local_max); if (lane == 0) shared_max[warp_id] = local_max; __syncthreads(); float row_max = -1e30f; if (tid < WARPS_PER_BLOCK) { row_max = shared_max[tid]; } row_max = warp_reduce_max(row_max); row_max = __shfl_sync(0xFFFFFFFF, row_max, 0); ``` Same bug! The `__shfl_sync` after `warp_reduce_max` broadcasts within each warp, not across warps! OK so let me write the kernel with a corrected pattern. Pattern 1: Use thread 0 for cross-warp reduction ```cpp local_max = warp_reduce_max(local_max); if (lane == 0) smem[warp_id] = local_max; __syncthreads(); if (tid == 0) { float m = smem[0]; for (int i = 1; i < num_warps; i++) m = fmaxf(m, smem[i]); smem[0] = m; } __syncthreads(); float global_max = smem[0]; ``` Pattern 2: Multiple warps participate in reduction using proper technique Actually, another common pattern is to have only warp 0 do the final reduction and broadcast: ```cpp local_max = warp_reduce_max(local_max); if (lane == 0) smem[warp_id] = local_max; __syncthreads(); if (warp_id == 0) { float m = (tid < num_warps) ? smem[tid] : -1e30f; m = warp_reduce_max(m); if (lane == 0) smem[0] = m; } __syncthreads(); float global_max = smem[0]; ``` This is a good pattern. Let me use this. Now, for shared memory storage of GELU values: N=8192 floats = 32KB per block. With high occupancy, we might only run 1-2 blocks per SM due to shared memory limits. 32KB per block * 32 blocks per SM would be 1MB, which exceeds the 228KB per SM limit. So at 32KB per block, we might be limited to ~7 blocks per SM (7 * 32KB = 224KB, close to max). With 1024 blocks total and 7 per SM, we need at least 147 SMs, but H100 only has 132. So we might only run about 4-5 blocks per SM if shared memory is the limiting factor. Actually, with dynamic scheduling, more blocks can be waiting. But wait, the occupancy might not be limited by shared memory if the kernel doesn't use much. Actually with 32KB shared memory, the max blocks per SM is min(32 (from block count), int(228KB/32KB)=7). So max 7 blocks per SM. 7 blocks * 256 threads = 1792 threads per SM. Each SM can run 2048 threads concurrently. So occupancy would be 1792/2048 = 87.5%. That's pretty good. Alternatively, with the recomputation approach (only a few bytes of shared memory), we might get 32 blocks per SM = 8192 threads per SM, but limited by other factors like registers to maybe 16-24 blocks. This would give much higher occupancy. Hmm, but for memory bandwidth-bound kernels, high occupancy isn't always better if it causes cache thrashing. With 7 blocks per SM and each block accessing 32KB, all blocks' data could fit in L2 cache (50MB on H100). 7 * 32KB = 224KB per SM, 132 SMs = 30MB total in flight. That's less than L2 cache. So the single-pass shared memory approach is actually quite good. Let me use it. Actually, I realize I should also consider using `float4` vectorized loads for the shared memory kernel. With 8192 elements and 256 threads, each thread loads 32 elements. Loading 32 float4s would be great for memory bandwidth. Let me plan the vectorized load version: - Each thread loads 32 elements. - Process in groups of 4: load float4, apply GELU to each element, store to shared memory. - In shared memory: float s_data[8192]; // 32KB - Each thread accesses s_data[i], s_data[i+1], s_data[i+2], s_data[i+3] in the first loop. - Bank conflict: since each thread loads 4 contiguous floats and stores them contiguously in shared memory, there might be 4-way bank conflicts. With float4, we'd have 4-way conflicts (banks 0,1,2,3 or similar). On H100 with 32 banks and 128-bit floats (4 bytes each), a float4 spans 4 consecutive banks, so 64 threads per warp would have bank conflicts if they load different float4s. Wait, actually with 32 banks and each bank being 4 bytes (1 float), a float4 load from each thread loads 4 floats from 4 consecutive banks. When 32 threads in a warp each load a float4, they might have conflicts depending on alignment and stride. If thread i loads `s_data[4*i]`, then thread 0 loads banks 0,1,2,3; thread 1 loads banks 4,5,6,7; ... thread 7 loads banks 28,29,30,31; thread 8 loads banks 0,1,2,3 (conflict!). So 32-way bank conflicts for the first 8 sets, repeated 4 times. With float4, it's 4-way bank conflict per thread, but 32 threads in the warp, and each thread takes 4 cycles to load all 4 floats. Actually, I'm not sure about the exact behavior. Let me just use float loads to avoid complexity. Actually, for 8192 elements and 256 threads, each thread does exactly 32 scalar loads. That's fully coalesced and very efficient. Bank conflicts in shared memory for scalar loads: thread i accesses s_data[i], so banks = i % 32. Thread 0 → bank 0, thread 1 → bank 1, ..., thread 31 → bank 31. No bank conflicts for the first warp. The second warp (threads 32-63): thread 32 → bank 0, thread 33 → bank 1, ... also no conflicts! So scalar loads to shared memory have NO bank conflicts when each thread accesses a different element within a warp and the block dimension is a multiple of 32. Since 256 = 8 * 32, and each thread accesses a consecutive or strided element... Wait, let me be more careful. If 256 threads each load one element: - Thread 0: s_data[0] → bank 0 - Thread 1: s_data[1] → bank 1 - ... - Thread 31: s_data[31] → bank 31 - Thread 32: s_data[32] → bank 0 - Thread 33: s_data[33] → bank 1 When warp 0 (threads 0-31) executes, they access banks 0-31 simultaneously. That's 32-way parallel access! Wait, but each bank can only service one request. So is there a bank conflict? Actually no, there is no bank conflict because each thread accesses a DIFFERENT bank. Banks 0-31 are all different. So all 32 accesses happen in parallel. For warp 1 (threads 32-63): they access banks 0-31 again (same banks as warp 0 but different memory addresses). Since the memory system can handle multiple warps accessing the same bank as long as they access different addresses... actually in shared memory, if two threads in the SAME WARP access the same bank, that's a bank CONFLICT. Different warps can access the same bank without conflict because they execute at different times (instruction-level parallelism allows overlapping, but fundamentally, a warp's access either has or doesn't have bank conflicts). So for scalar loads with each thread accessing a unique bank: NO bank conflicts. For stores: same thing, no conflicts. Great, so scalar access to shared memory is perfectly conflict-free in our case. Let me use the single-pass shared memory approach. Now, one more optimization: I can actually avoid storing all intermediate values in shared memory by doing the max reduction on-the-fly. Here's a better approach: Parallel tree reduction for row-wise softmax: ```cpp // Each thread processes k = N / blockDim.x elements // Compute GELU for k elements, track local max // Store k GELU values in registers (if k is small) ``` With N=8192 and blockDim.x=256, each thread processes 32 elements. I can store 32 floats in registers. 32 * 4 bytes = 128 bytes of registers per thread. At 255 max registers per thread, we have 255 * 4 = 1020 bytes. So 128 bytes is fine. Then I don't need shared memory for the data at all! Here's the register-based approach: ```cpp float vals[32]; // each thread has 32 GELU values in registers // Load and compute GELU for (int i = 0; i < 32; i++) { int idx = tid * 32 + i; float x = row_in[idx] + bias[idx]; vals[i] = gelu(x); local_max = fmaxf(local_max, vals[i]); } // Reduce max across warp and block local_max = warp_reduce_max(local_max); if (lane == 0) shared_buf[warp_id] = local_max; __syncthreads(); if (warp_id == 0) { float m = (tid < num_warps) ? shared_buf[tid] : -1e30f; m = warp_reduce_max(m); if (lane == 0) shared_buf[0] = m; } __syncthreads(); float global_max = shared_buf[0]; // Compute exp and sum float local_sum = 0.0f; for (int i = 0; i < 32; i++) { vals[i] = expf(vals[i] - global_max); local_sum += vals[i]; } // Reduce sum local_sum = warp_reduce_sum(local_sum); if (lane == 0) shared_buf[warp_id] = local_sum; __syncthreads(); if (warp_id == 0) { float s = (tid < num_warps) ? shared_buf[tid] : 0.0f; s = warp_reduce_sum(s); if (lane == 0) shared_buf[0] = s; } __syncthreads(); float global_sum = shared_buf[0]; // Write output for (int i = 0; i < 32; i++) { int idx = tid * 32 + i; row_out[idx] = vals[i] / global_sum; } ``` This is much better! No shared memory for data (only tiny buffer for reduction), and perfectly coalesced access. With `tid * 32 + i`, thread 0 accesses elements 0-31, thread 1 accesses 32-63, etc. Wait, is this coalesced? For global memory loads: - Thread 0: indices 0, 1, 2, ..., 31 (in the loop, one at a time) - Thread 1: indices 32, 33, ..., 63 When the warp executes the first iteration (i=0): - Thread 0: index 0 - Thread 1: index 32 - Thread 2: index 64 - Thread 31: index 992 These are 32 elements apart, so they're NOT contiguous. This is a STRIDED pattern, not coalesced! For coalesced loads, thread t should access address `base + t` or `base + 4*t` for vector loads, but NOT `base + 32*t` because that means each thread accesses elements 0, 32, 64, ... which are in DIFFERENT cachelines. Wait, but float4 vector loads would help here. If each thread loads a float4 from addresses: - Thread 0: 0, 1, 2, 3 - Thread 1: 32, 33, 34, 35 For the first float of each thread, the warp accesses elements 0, 32, 64, ..., 992. These are 32*4=128 bytes apart. Each cacheline is 128 bytes. So thread 0 accesses cacheline 0 (bytes 0-127), thread 1 accesses cacheline 1 (bytes 128-255), etc. So each thread loads from a different cacheline, meaning 32 cachelines are loaded. For 32 float4 loads, that's 128 floats = 512 bytes. But the warp loads 32 cachelines = 32 * 128 = 4096 bytes! Only 512 bytes are actually used, so we're loading 8x more data than needed. That's terrible for memory bandwidth. To get coalesced access, we need contiguous threads to access contiguous memory. Like: - Thread 0: 0, 1, 2, ..., 31 (all in one go, or float4) - Thread 1: 32, 33, 34, ..., 63 But in the i=0 iteration of the loop, if thread 0 accesses 0 and thread 1 accesses 32, that's strided. Let me fix this. The correct coalesced pattern for scalar loads is: ```cpp for (int i = tid; i < N; i += blockDim.x) { // Thread 0: 0, 256, 512, ... // Thread 1: 1, 257, 513, ... // ... // Thread 255: 255, 511, 767, ... } ``` Wait, this is also strided in terms of adjacent threads! Thread 0 accesses 0, thread 1 accesses 1. Actually, for the FIRST iteration (i=tid): - Thread 0: 0 - Thread 1: 1 - Thread 2: 2 - ... - Thread 31: 31 These ARE contiguous! So the first 32 loads (one per thread in warp 0) are perfectly coalesced into one or two 128-byte transactions. For the SECOND iteration (i += 256): - Thread 0: 256 - Thread 1: 257 - Thread 2: 258 - ... - Thread 31: 287 Also contiguous! Perfect coalescing. So the pattern `for (int i = tid; i < N; i += blockDim.x)` gives perfect coalescing. But `for (int i = 0; i < 32; i++) { idx = tid * 32 + i; }` gives strided access within each iteration of i. So I should use the `i = tid; i < N; i += blockDim.x` pattern. But then I need to store 32 values per thread in registers. With this pattern, each thread loads: - Thread 0: 0, 256, 512, ..., 7936 (32 elements) - Thread 1: 1, 257, 513, ..., 7937 The thread accesses scattered memory locations, but the WARP as a whole accesses contiguous chunks. However, if each thread stores its 32 values in registers, the registers hold values from non-contiguous locations. For the output write, we'd need to use the same strided pattern. Actually, using the `for (int i = tid; i < N; i += blockDim.x)` pattern means that after the max reduction, each thread computes exp for its 32 elements and writes them back using the same formula. The output write is also coalesced. The issue is that the vals array is non-contiguous in memory terms, but that's fine because it's in registers. Each thread processes 32 scattered elements but the warp accesses them in coalesced groups of 32. So here's my plan with `i = tid; i < N; i += blockDim.x`: 1. Each thread loads k elements (k = N / blockDim.x = 32), computes GELU, stores in registers, tracks max. 2. Reduce max across block. 3. Each thread computes exp(x - max) for its k elements, tracks sum. 4. Reduce sum across block. 5. Each thread writes k output elements = exp(x - max) / sum. This uses registers only (no shared memory for data). With 32 floats per thread, 128 bytes. Plus a few temporaries. Total maybe 150 bytes of registers per thread = ~38 registers. Well within 255 limit. For the reduction, we need an 8-float shared memory buffer (for 8 warps). This is a very clean and efficient approach! Let me write this kernel. Actually, I should also check if `blockDim.x` evenly divides `N`. 8192 / 256 = 32, which is exact. Good. Let me write the kernel now: ```cpp #include <torch/extension.h> #include <cuda.h> #include <cuda_runtime.h> #include <math.h> #define WARPS_PER_BLOCK 8 #define THREADS_PER_BLOCK 256 __inline__ __device__ float warp_reduce_max(float val) { #pragma unroll for (int offset = 16; offset > 0; offset >>= 1) val = fmaxf(val, __shfl_down_sync(0xFFFFFFFF, val, offset)); return val; } __inline__ __device__ float warp_reduce_sum(float val) { #pragma unroll for (int offset = 16; offset > 0; offset >>= 1) val += __shfl_down_sync(0xFFFFFFFF, val, offset); return val; } __global__ void fused_gelu_softmax_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N) { int row = blockIdx.x; int tid = threadIdx.x; int lane = tid & 31; int warp_id = tid >> 5; if (row >= M) return; const float* row_in = input + row * N; float* row_out = output + row * N; const float SQRT2_INV = 0.70710678118654752440f; const float HALF = 0.5f; const int NUM_ELEMENTS_PER_THREAD = N / THREADS_PER_BLOCK; // = 32 float thread_vals[32]; float thread_max = -1e30f; // First pass: compute GELU and thread-local max #pragma unroll for (int i = 0; i < NUM_ELEMENTS_PER_THREAD; i++) { int idx = tid + i * THREADS_PER_BLOCK; float x = row_in[idx] + bias[idx]; float gelu = x * HALF * (1.0f + erff(x * SQRT2_INV)); thread_vals[i] = gelu; thread_max = fmaxf(thread_max, gelu); } // Reduce max across warp thread_max = warp_reduce_max(thread_max); __shared__ float shared_buf[WARPS_PER_BLOCK]; if (lane == 0) shared_buf[warp_id] = thread_max; __syncthreads(); // Reduce max across warps (only warp 0 participates) if (warp_id == 0) { float m = (tid < WARPS_PER_BLOCK) ? shared_buf[tid] : -1e30f; m = warp_reduce_max(m); if (lane == 0) shared_buf[0] = m; } __syncthreads(); float global_max = shared_buf[0]; // Second pass: compute exp and thread-local sum float thread_sum = 0.0f; #pragma unroll for (int i = 0; i < NUM_ELEMENTS_PER_THREAD; i++) { float exp_val = expf(thread_vals[i] - global_max); thread_vals[i] = exp_val; thread_sum += exp_val; } // Reduce sum across warp thread_sum = warp_reduce_sum(thread_sum); if (lane == 0) shared_buf[warp_id] = thread_sum; __syncthreads(); // Reduce sum across warps (only warp 0 participates) if (warp_id == 0) { float s = (tid < WARPS_PER_BLOCK) ? shared_buf[tid] : 0.0f; s = warp_reduce_sum(s); if (lane == 0) shared_buf[0] = s; } __syncthreads(); float global_sum = shared_buf[0]; float inv_sum = 1.0f / global_sum; // Third pass: write output #pragma unroll for (int i = 0; i < NUM_ELEMENTS_PER_THREAD; i++) { int idx = tid + i * THREADS_PER_BLOCK; row_out[idx] = thread_vals[i] * inv_sum; } } ``` Wait, this has the same pattern as v2, but with the correct cross-warp reduction. Let me trace through again: - thread_max = warp_reduce_max(thread_max) → each warp's lanes all have the warp max after this - if (lane == 0) shared_buf[warp_id] = thread_max → only lane 0 of each warp writes to shared memory - __syncthreads() - if (warp_id == 0) { ... } → only warp 0 (threads 0-31) enters this block - float m = (tid < WARPS_PER_BLOCK) ? shared_buf[tid] : -1e30f; - tid=0: shared_buf[0] - tid=1: shared_buf[1] - ... - tid=7: shared_buf[7] - tid=8..31: -1e30f - m = warp_reduce_max(m) → warp 0 computes max of all 8 warp maxes (and ignores the -1e30f from threads 8-31) - if (lane == 0) shared_buf[0] = m → thread 0 writes the global max to shared_buf[0] - __syncthreads() → all threads wait - float global_max = shared_buf[0]; → all threads read the global max This is CORRECT! Previously the bug was that `__shfl_sync` broadcasted within each warp, so warps other than warp 0 read stale/incorrect values. Here, only warp 0 does the cross-warp reduce and writes the result to shared_buf[0], and then ALL warps read from shared_buf[0]. Same for the sum reduction. Great, let me write this up and test it! One thing I'm still worried about: the GELU formula. Let me make sure I use the exact formula PyTorch uses. PyTorch exact GELU: ```python 0.5 * x * (1 + torch.erf(x / math.sqrt(2))) ``` My CUDA code: ```cpp x * HALF * (1.0f + erff(x * SQRT2_INV)) ``` With `SQRT2_INV = 1 / sqrt(2)`. Let me verify: sqrt(2) ≈ 1.4142135623730951 1/sqrt(2) ≈ 0.7071067811865475 In my code: ```cpp float gelu = x * HALF * (1.0f + erff(x * SQRT2_INV)); ``` = x * 0.5 * (1 + erf(x * 0.7071...)) = x * 0.5 * (1 + erf(x / sqrt(2))) Yes, this matches. But actually, I should also consider the possibility that PyTorch is using a different default for GELU. According to PyTorch docs: - torch.nn.functional.gelu(x, approximate='none') — uses erf - torch.nn.functional.gelu(x, approximate='tanh') — uses tanh approximation - The default in newer PyTorch is 'none' Looking at the problem statement: ```python x = torch.nn.functional.gelu(x) ``` Without the `approximate` argument. If PyTorch defaults to `'none'`, then erf-based is correct. If it defaults to `'tanh'` in this version, then I should use the tanh approximation. I can try both if needed, but let me first try the erf-based version and see if I get numerical correctness within 1e-4 tolerance. If the difference is small, it's likely just a formula difference. If it's large/inf, it's likely my kernel logic. Actually, the previous attempts gave `inf` differences, which is almost certainly a kernel correctness issue, not a formula difference. So with the fixed reduction, let's see if we get close. Let me also think about another potential issue: `erff` might produce slightly different results than PyTorch's `torch.erf`. Actually, `erff` in CUDA should match C's `erff`, and PyTorch's `torch.erf` likely calls into the same libm implementation on CPU or uses a custom implementation. But on GPU, the values might differ slightly. For fp32 with tol=1e-4, a small difference from erf implementation should be fine. OK let me write the kernel and test. One more optimization: I should launch 1 block per row with 256 threads. For M=1024, that's 1024 blocks. On H100, this should be well-utilized. Let me also check: is there any issue with using `float thread_vals[32]` in a CUDA kernel? The `#pragma unroll` should help the compiler optimize the loop. But with 32-register arrays, the compiler might spill to local memory if register pressure is high. However, with only 38 or so registers used per thread, spill shouldn't happen. Actually, I realize there's one more thing I should optimize: the loop index pattern. Currently I use: ```cpp int idx = tid + i * THREADS_PER_BLOCK; ``` for i = 0 to 31: - tid=0: 0, 256, 512, ..., 7936 - tid=1: 1, 257, ..., 7937 - ... - tid=255: 255, 511, ..., 8191 Each thread accesses 32 elements with a stride of 256. For the output write, the same pattern is used. Memory coalescing: within each iteration i, the 256 threads access a contiguous block of 256 floats = 1024 bytes. This is perfectly coalesced (8 cachelines of 128 bytes each). Great! But wait, the elements `thread_vals[0]` through `thread_vals[31]` for a single thread are NOT contiguous in memory. They are 256 floats apart. Does this matter? For the output write, each thread writes its elements one at a time, but within the same iteration i, all threads write together, forming a contiguous block. So coalescing is perfect. For subsequent passes, when the warp re-reads `thread_vals[i]`, it's reading from registers, not memory. So it's free. This is an excellent pattern. Let me proceed. I should also check if I can use vectorized loads (float4) to reduce instruction count. With float4, each thread could load 8 float4 values (32 floats total). But then the indexing would be more complex. Since scalar loads are fully coalesced and the compiler should handle them well, I'll stick with scalars for now. Optionally, I could add `__restrict__` to the output pointer too, which I already did. Let me write the final code. One more minor thing: to get even better performance, I might want to use `__ldg` (load through read-only cache/texture cache) for the input and bias, since they're read-only. But `const float* __restrict__` already tells the compiler this, and it might use `__ldg` automatically. Let me not add it manually unless needed. Also, I should think about whether to keep the weight+bias in the same way as nn.Linear. I'll just use `self.linear = nn.Linear(...)` and access its parameters. Let me now write the full code and compile it.