We need to write a custom CUDA kernel that replaces a PyTorch reference model: linear (matmul + bias), ReLU, divide by constant. We need to write a complete Python module that defines `ModelNew` with same constructor arguments and forward signature. We need to use `torch.utils.cpp_extension.load_inline` to compile and bind CUDA source at module load time. Goal: produce numerically equivalent outputs at fp32 precision and run faster than PyTorch reference. We need to consider GPU specs to optimize. Let's get GPU specs first. Then we need to think about kernel design. The operation: y = relu(x @ W^T + b) / divisor. Input x: [batch_size, in_features] = [1024, 8192]. W: [out_features, in_features] = [8192, 8192]. b: [out_features]. divisor: scalar float. So it's a GEMM (1024x8192 * 8192x8192) = 1024x8192 output. Matrix multiply of size M=1024, N=8192, K=8192. In fp32, PyTorch likely uses cuBLAS which is highly optimized. To beat it, we might need to fuse the bias, ReLU, and division into the GEMM output writing, reducing memory bandwidth. PyTorch's linear + relu + div would do: 1. matmul (x @ W^T) -> output [M, N] (reads x, W, writes output) 2. add bias (inplace or out-of-place) -> reads output, bias, writes output 3. relu -> reads output, writes output 4. division -> reads output, writes output Total memory traffic: for each element of output (M*N = 1024*8192 = 8,388,608 elements, ~32 MB in fp32). Each operation writes and reads 32 MB. So total maybe ~4 passes over 32 MB = 128 MB of memory traffic + GEMM compute. However the GEMM itself is compute-bound? Let's check. GEMM flops: 2 * M * N * K = 2 * 1024 * 8192 * 8192 = 2 * 1024 * 67,108,864 = 137,438,953,472 flops ≈ 137.4 GFLOPs. fp32 peak TFLOPS on common GPU? Let's get specs. If GPU is e.g., A100, fp32 19.5 TFLOPS. 137 GFLOPs / 19.5 TFLOPS = ~7 microseconds. But memory bandwidth is 1.6 TB/s or 2 TB/s. 32 MB reads/writes times several passes: maybe 128 MB / 2 TB/s = 64 microseconds. Actually memory bandwidth limited operations are slower. But fused kernel can reduce memory traffic by combining bias+relu+div with the GEMM output write, making it just one write pass. But can we write a kernel that does the GEMM itself faster than cuBLAS? cuBLAS is extremely optimized for large GEMMs. However for M=1024, N=8192, K=8192, cuBLAS likely achieves near-peak performance because N and K are large. It might be hard to beat cuBLAS in raw GEMM speed. But we can fuse the elementwise operations to avoid extra memory bandwidth for the subsequent ops. That could yield overall speedup. PyTorch might also fuse bias+relu if using torch.nn.functional.linear with bias and then torch.relu, but in eager mode it does separate kernels. Actually PyTorch might fuse linear + relu into a single kernel? No, linear dispatches to addmm (GEMM + bias addition fused into GEMM in cuBLAS? Actually PyTorch's `linear` uses `at::native::linear` which calls `matmul` or `addmm` if bias is present. `addmm` can fuse bias addition into the GEMM via cuBLAS's beta parameter. So bias is fused into GEMM. But ReLU and division are separate kernels. So reference does: 1. addmm (fused GEMM + bias) -> writes out 2. relu -> reads out, writes out 3. div -> reads out, writes out Total: one write of output from GEMM, then read+write for relu, then read+write for div. That's 5 passes over output? Actually addmm writes output (1 write). relu reads output, writes output (1 read, 1 write). div reads output, writes output (1 read, 1 write). Total memory: 1 write + 2 reads + 2 writes = 5 * 32 MB = 160 MB of memory traffic for output. Plus reading x and W for GEMM. If we fuse ReLU and division into the GEMM output write, we can eliminate the extra read/write passes. We can write a custom kernel that launches a GEMM-like computation using CUTLASS or manual warp-level matrix multiply? Or we can write a kernel that uses cuBLAS for the GEMM but then does the elementwise ops in-place. That would still have the memory traffic of reading and writing for the elementwise ops, but perhaps we can use a custom kernel that does GEMM and elementwise fused. However writing a full GEMM kernel that rivals cuBLAS is difficult. Alternative: use CUTLASS or cuBLASLt epilogue fusion. PyTorch might not expose epilogue fusion easily via inline CUDA, but we can embed CUTLASS? That would be heavy. But we can write a kernel that calls cuBLAS's `cublasGemmEx` and then a custom elementwise kernel. But that still has separate passes. Wait: we can use `torch.matmul` (cuBLAS) then do the bias, ReLU, and division in a single custom fused elementwise kernel. That would be: 1) cuBLAS matmul (writes output), 2) fused kernel that adds bias, applies ReLU, divides by divisor (reads output and bias, writes output). That reduces memory traffic: matmul write (1 write), fused kernel read+write (1 read, 1 write). Total for output: 2 reads, 2 writes (matmul write + fused read/write). Actually matmul writes output (1 write). Fused kernel reads output and bias, writes output (1 read, 1 write). So total output traffic: 1 write + 1 read + 1 write = 2 writes + 1 read = 3 passes * 32 MB = 96 MB. Versus reference which had addmm (fused bias) so bias is in registers during GEMM, then relu (read+write), div (read+write). That's 1 write (GEMM) + 1 read + 1 write (relu) + 1 read + 1 write (div) = 3 writes + 2 reads = 5 passes. So fused elementwise saves 2 passes (64 MB). The speedup might be modest but could be enough. But can we do even better? If we use a custom Epilogue in CUTLASS? But `load_inline` can compile CUDA code, but embedding CUTLASS headers might be possible if they are available in the environment. Usually PyTorch includes CUTLASS? The environment might have CUDA toolkit but not CUTLASS. We could try to use `mma` instructions manually, but writing a full GEMM with performance near cuBLAS is hard. However for M=1024, N=8192, K=8192, maybe a simple block-level matrix multiply with shared memory could be okay? But cuBLAS is heavily optimized. Unless the reference is memory-bound due to the separate elementwise ops, a fused elementwise approach might already beat it. Let's estimate the reference time. Suppose GPU is e.g., RTX 3090 (fp32 35.6 TFLOPS, memory bandwidth 936 GB/s). Or maybe A100 (fp32 19.5 TFLOPS, memory 1555 GB/s). Let's get specs. We should call get_gpu_specs.

Let me analyze the problem: 1. We have a model that does: linear layer (matmul + bias), ReLU, divide by constant 2. Input: [batch_size=1024, in_features=8192] 3. Output: [batch_size=1024, out_features=8192] 4. Divisor: 2.0 The computation is: - y = x @ W^T + b - y = relu(y) - y = y / 2.0 This is essentially a single fused kernel: matrix multiplication with bias, ReLU, and division. On H100, we have 3.35 TB/s memory bandwidth and 989 TF32 TFLOPS (with sparsity) / 67 TF32 TFLOPS (non-tensor core). The matmul is the dominant operation here. 1024 * 8192 * 8192 = ~68.7B multiply-add operations = ~137.4B FLOPs. At 989 TFLOPS, this would take ~139 microseconds. At 67 TFLOPS, ~2ms. Memory bandwidth: reading 1024*8192*4B input = 32MB, writing 1024*8192*4B output = 32MB, reading weights 8192*8192*4B = 256MB. Total memory ~320MB. At 3.35 TB/s, memory bound would be ~96 microseconds. So we're likely compute-bound for the matmul part. The best approach is to use cuBLAS for the matmul and then fuse the bias+ReLU+division into a single elementwise kernel. However, we can potentially do even better with a custom kernel. Actually, for a problem this size, using cuBLAS LT or just `torch.matmul` followed by a fused elementwise kernel is likely optimal. But let me think if we can use a Triton kernel or custom CUDA kernel. Wait - the challenge says to replace with a custom CUDA kernel. Let me think about what's fastest. For a [1024, 8192] x [8192, 8192] matmul: - This is a fairly large matmul. Using cuBLAS (which torch.matmul uses) is already highly optimized. - The bias, ReLU, and division are elementwise operations that can be fused. The PyTorch reference likely uses: 1. `F.linear` which calls cuBLAS matmul + bias (if using `nn.Linear`) 2. `torch.relu` - another kernel 3. Division - another kernel So we can gain by fusing the bias+ReLU+division. But the matmul itself probably can't be beat by a hand-written kernel for this size on H100. Let me write a fused bias+ReLU+division kernel and use `torch.matmul` for the matmul. But wait, can we fuse even more? We could use `torch.mm` and then a custom kernel for bias+ReLU+div. Actually, looking more carefully, on H100 the matmul will be very fast. We need to see if we can win by fusing. But actually, the PyTorch reference might already be doing some fusion. Let me think... In PyTorch, `self.linear(x)` does `x @ W.T + b`. Then `relu` and division. PyTorch can sometimes fuse pointwise ops but not across the bias addition with matmul in the general case. Wait, actually PyTorch 2.0+ with inductor can do fusion, but the reference is using eager mode. So our approach: 1. Do `torch.matmul(x, self.weight.t())` or use `F.linear` 2. Fused custom kernel for `output = max(input + bias, 0) / divisor` For better performance, we could use a custom CUTLASS or Triton kernel, but writing pure CUDA that beats cuBLAS on [1024, 8192] x [8192, 8192] is very hard. Let me use `torch.mm` for the matmul and a custom CUDA kernel for the elementwise fusion. Actually, can we use `torch._C._cuda_linear` or `addmm`? `torch.addmm(bias, x, weight.T)` would compute bias + x @ W.T in one call, but the bias is broadcasted. Actually `F.linear` is typically implemented as `input.mm(weight.t()) + bias`. Let me write a simple fused kernel. Since the problem size is large, I should use float4 for vectorized loads/stores and a grid-stride loop. For the kernel: - Input: [M, N] where M=1024, N=8192 - Bias: [N] - Output: [M, N] - Operation: `output[i,j] = max(input[i,j] + bias[j], 0) / divisor` I can use `load_inline` to compile a simple CUDA kernel. Actually, for maximum performance on H100 with this shape, I should also check if we can use a split-K or some matmul approach. But honestly, for 1024x8192x8192, cuBLAS should dominate. Let me check if we can use `torch.matmul` and a custom elementwise kernel. Actually, I wonder if we can do better by using a custom matmul kernel that's aware of the post-op. But writing a gemm kernel that competes with cuBLAS on H100 for these sizes is extremely difficult. Let me stick with: `torch.mm` + custom fused bias/ReLU/div kernel. For the custom kernel, I'll write a grid-stride kernel using float4 for memory coalescing. Let me think about the launch config: - M = 1024, N = 8192 - Total elements = 8,388,608 - With float4 (4 floats), we process 4 elements per thread - If we use 256 threads per block, we need ~8192 blocks for 8M elements. Actually: 8,388,608 / 4 / 256 = 8192 blocks. That's a lot of blocks but should be fine on H100 which has 132 SMs. Actually, let me use a different approach. For a [M, N] matrix where N is large, it's better to let each block process a tile and use 2D grid. Let me use blocks of (32, 8) = 256 threads, and each block processes a tile. Actually, for simplicity and good occupancy, let me just use a 1D grid with grid-stride loops, and each thread processes float4. Let me write the code. I'll make sure to handle the case where N is divisible by 4 (8192 / 4 = 2048, so yes it is). Let me craft the CUDA kernel: ```cuda __global__ void fused_bias_relu_div_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N, float divisor) { int idx = blockIdx.x * blockDim.x + threadIdx.x; int total_elements = M * N; int num_float4 = total_elements / 4; float4 zero = make_float4(0.0f, 0.0f, 0.0f, 0.0f); float4 div4 = make_float4(divisor, divisor, divisor, divisor); for (int i = idx; i < num_float4; i += gridDim.x * blockDim.x) { int linear_idx = i * 4; int row = linear_idx / N; int col = linear_idx % N; float4 in4 = reinterpret_cast<const float4*>(input)[i]; // We need to load 4 bias values. Since N is large and col is aligned to 4, // we can load a float4 from bias. float4 bias4 = reinterpret_cast<const float4*>(bias)[col / 4]; in4.x = fmaxf(in4.x + bias4.x, 0.0f) / divisor; in4.y = fmaxf(in4.y + bias4.y, 0.0f) / divisor; in4.z = fmaxf(in4.z + bias4.z, 0.0f) / divisor; in4.w = fmaxf(in4.w + bias4.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[i] = in4; } } ``` Wait, there's an issue. The matmul output is [M, N]. For a given row and column, we need `bias[col]`. But if we're processing in float4, we need to make sure the bias is also loaded as float4 and aligned properly. Since N=8192 which is divisible by 4, and if the tensor is properly aligned (which PyTorch ensures for float32 on CUDA), we can do this. However, there's a subtlety. The float4 load from `input` and `output` will be coalesced if threads in a warp access consecutive addresses. But in my current setup, thread `idx` loads float4 at position `idx`, thread `idx+1` loads at `idx+1`, etc. Thread `idx` accesses elements `4*idx` through `4*idx+3`. This IS coalesced because consecutive threads access consecutive float4s, which correspond to consecutive memory locations. But we need to think about the grid size. For occupancy, we want enough warps. With 256 threads/block and maybe 128 blocks, we get 32 warps per SM * 132 SMs... Actually 128 blocks with 256 threads = 32,768 threads total. On H100 with 132 SMs, that's 248 threads/SM, which is below the max of 2048. For good occupancy we might want more blocks. Actually, for a grid-stride loop, we don't need tons of blocks because the loop handles it. But having enough blocks to saturate all SMs is good. Let me target maybe 2048 blocks or so. 2048 blocks * 256 threads = 524,288 threads. Total float4 elements = 8,388,608 / 4 = 2,097,152. So each thread processes ~4 float4s on average. Let me use a 2D grid approach instead for better memory coalescing pattern and to avoid the division/modulo in the inner loop. Actually, since N=8192 is divisible by 4, we can think of it as N/4 = 2048 "columns" of float4. Let me write: ```cuda __global__ void fused_bias_relu_div_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N, float divisor) { int row = blockIdx.y * blockDim.y + threadIdx.y; int col4 = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N / 4; if (col4 >= N4) return; float4 bias4 = reinterpret_cast<const float4*>(bias)[col4]; for (int r = row; r < M; r += gridDim.y * blockDim.y) { float4 in4 = reinterpret_cast<const float4*>(input)[r * N4 + col4]; in4.x = fmaxf(in4.x + bias4.x, 0.0f) / divisor; in4.y = fmaxf(in4.y + bias4.y, 0.0f) / divisor; in4.z = fmaxf(in4.z + bias4.z, 0.0f) / divisor; in4.w = fmaxf(in4.w + bias4.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[r * N4 + col4] = in4; } } ``` Grid: (N4 / blockDim.x, some_y) Block: (32, 8) or similar N4 = 2048. If blockDim.x = 32, then gridDim.x = 64. For gridDim.y, let's use something like 32 or 64. Block = (32, 8) = 256 threads. Grid = (64, 32) = 2048 blocks. Each block processes 32 * 8 = 256 float4 elements, but with grid-stride in y-direction. Actually, we don't even need grid-stride if gridDim.y >= M. With M=1024 and blockDim.y=8, gridDim.y=128 would cover all rows exactly once. But grid-stride is fine and often preferred. Let me check: gridDim.y = 128, blockDim.y = 8, total y coverage = 1024 = M. Perfect. So grid = (64, 128), block = (32, 8). Total blocks = 8192, which is more than enough to saturate H100. Each thread handles exactly 1 float4. Total threads = 8192 * 256 = 2,097,152, and total float4 elements = 2,097,152. Perfect - each thread does exactly 1 float4. This should be very efficient. Let me verify: - Grid x: 64 blocks * 32 threads = 2048 = N/4. ✓ - Grid y: 128 blocks * 8 threads = 1024 = M. ✓ - No loops needed! Wait, but then we need exactly this grid configuration. Since N and M are fixed sizes in the problem but the kernel needs to be general... well, the kernel should handle arbitrary sizes but we can size the grid appropriately. Actually, in `forward`, we know the exact sizes. Let me make the kernel handle arbitrary sizes with proper if-checks and 2D grid-stride. Let me refine: ```cuda __global__ void fused_bias_relu_div_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N, float inv_divisor) { int col4 = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N >> 2; if (col4 >= N4) return; float4 bias4 = reinterpret_cast<const float4*>(bias)[col4]; int row = blockIdx.y * blockDim.y + threadIdx.y; for (; row < M; row += gridDim.y * blockDim.y) { float4 in4 = reinterpret_cast<const float4*>(input)[row * N4 + col4]; in4.x = fmaxf(in4.x + bias4.x, 0.0f) * inv_divisor; in4.y = fmaxf(in4.y + bias4.y, 0.0f) * inv_divisor; in4.z = fmaxf(in4.z + bias4.z, 0.0f) * inv_divisor; in4.w = fmaxf(in4.w + bias4.w, 0.0f) * inv_divisor; reinterpret_cast<float4*>(output)[row * N4 + col4] = in4; } } ``` Using multiplication by `inv_divisor` instead of division for precision. Actually for exact match, `* (1.0f / divisor)` may not be exactly the same as `/ divisor` in all cases, but the tolerance is 1e-4 so it should be fine. Actually, to be safe, let me just use division. The compiler will likely optimize it anyway since divisor is uniform. Actually wait, the reference does `x / self.divisor` where divisor is a Python float = 2.0. In PyTorch, this is done in fp32. Let me check if `* 0.5f` is exactly equivalent to `/ 2.0f` in IEEE 754. Yes, multiplication by 0.5 and division by 2.0 are both exact operations for normal floats. But to be absolutely safe and matching, I'll just use division. Actually, in CUDA, `x / 2.0f` might compile to `x * 0.5f` anyway when the divisor is a compile-time constant. But here it's passed as a parameter. Let me just pass the divisor and do `fmaxf(...) / divisor`, or better yet, multiply by `1.0f / divisor` on the host side and pass that. But for exact matching, let me think... `a / 2.0f` in PyTorch on CUDA... PyTorch uses its own elementwise kernel. The division by a scalar might be implemented as multiplication by reciprocal for performance. Either way, at 1e-4 tolerance, `* 0.5f` vs `/ 2.0f` is exactly the same for all finite floats since 0.5 is exactly representable. I'll multiply by the reciprocal. Actually, for maximum safety, I'll just do division and rely on the compiler. Or even better: pass `1.0f / divisor` to the kernel and multiply. But to be 100% safe with exact numerical equivalence, let me just pass `divisor` and do `/ divisor`. The performance difference is negligible for this memory-bound kernel. Wait, let me reconsider the entire approach. The matmul [1024, 8192] x [8192, 8192] is actually quite large. Let me compute the arithmetic intensity. FLOPs = 2 * 1024 * 8192 * 8192 = 137.4 GFLOPs Memory traffic (minimum): - Read x: 1024 * 8192 * 4 = 32 MB - Read W: 8192 * 8192 * 4 = 256 MB - Read bias: 8192 * 4 = 32 KB - Write output: 1024 * 8192 * 4 = 32 MB Total HBM: ~320 MB Arithmetic intensity = 137.4e9 FLOPs / 320e6 bytes = 429 FLOPs/byte. With 3.35 TB/s bandwidth, peak compute would need 429 * 3.35e12 = 1.44 PFLOPS to be compute bound. H100 does 989 TFLOPS in TF32 tensor core, or 3958 in FP8. For FP32 it's 67 TFLOPS (non-tensor core) or 494 with TF32. Wait, for pure FP32 (without tensor cores), the max is 67 TFLOPS. At 67 TFLOPS, we need 67e12 / 3.35e12 = 20 FLOPs/byte to be compute bound. We have 429, so we are MASSIVELY compute bound for FP32. For TF32 tensor core (494 TFLOPS dense, 989 with sparsity), say ~500 TFLOPS effective, we need 500/3.35 = 149 FLOPs/byte. We have 429, so we're still compute bound but not as dramatically. So cuBLAS will use tensor cores and run at near-peak efficiency. The matmul dominates the runtime. The elementwise kernel (bias+relu+div) is memory-bound and should be relatively fast. But wait, can we fuse the elementwise ops INTO the matmul? That is, can we use a custom matmul kernel that does bias+ReLU+div as the epilogue? This would save one read of the matmul output and one write. In CUTLASS terms, this is an "epilogue fusion". Writing a custom CUTLASS kernel is quite involved though. Let me think if there's a simpler way... Actually, for H100, we could use cuBLASLt which supports epilogue fusion. But PyTorch doesn't expose cuBLASLt in a way that's easy to use with custom epilogues, and writing the raw cuBLASLt calls in inline CUDA is very complex. Alternatively, we could write a Triton kernel. Triton can write a fused matmul+bias+relu kernel relatively easily. But the problem says to use `load_inline` with CUDA C++. Actually, wait. For this specific shape [1024, 8192] x [8192, 8192], the matmul output is [1024, 8192]. The subsequent kernel reads this [1024, 8192] and writes [1024, 8192]. That's 64MB of memory traffic. If the matmul takes, say, 200 microseconds, then 64MB at 3.35TB/s is ~19 microseconds. So the fused elementwise kernel is maybe 10% of the total time. If we fuse the post-ops into the matmul, we save ~19 microseconds. But writing a custom matmul that beats cuBLAS is virtually impossible for this size on H100. So the best practical approach is: 1. Use `torch.matmul` (cuBLAS) for the matmul 2. Use a highly optimized custom kernel for bias+ReLU+division But wait, can we do `torch.addmm` to fuse matmul + bias? That would save the bias memory read in our custom kernel and a separate kernel launch. `torch.addmm(bias, x, weight.t())` computes `bias + x @ W.T`. This fuses the bias addition into the matmul epilogue (cuBLAS supports this natively). Then our custom kernel only needs to do ReLU + division. If we do this, PyTorch's reference uses `self.linear(x)` which also calls into cuBLAS with bias. So the matmul+bias is already fused in the reference. Then the reference does ReLU and div as separate kernels. Our advantage would be fusing ReLU + div into one kernel, saving one memory pass. Wait, actually in PyTorch eager mode, does `self.linear(x)` call `addmm`? Yes, for 2D inputs it does. And `torch.relu` is a separate kernel, and `/ divisor` is another separate kernel. So in the reference: 1. `linear`: 1 kernel (fused matmul + bias) 2. `relu`: 1 kernel 3. `div`: 1 kernel In our approach: 1. `torch.matmul` or `torch.addmm`: 1 kernel 2. Custom fused kernel (bias+relu+div or relu+div): 1 kernel Actually, if we use `torch.addmm`, we only need our custom kernel to do `relu` + `div`. That's 1 kernel vs 2 in the reference. Saving one memory-bound kernel should give some speedup. But can we also do `torch.mm` and then fuse `bias+relu+div`? The difference is small - `addmm` has the bias fused in matmul, so it's slightly better. However, there's another optimization: we could use `torch._C._cuda_linear` or something, but let me just use `torch.addmm(self.bias.unsqueeze(0), x, self.weight.t())` or `F.linear(x, self.weight, self.bias)`. `F.linear` with bias should use the same optimized path as `nn.Linear`. Let me use that and then our custom kernel for ReLU + division. Wait, but the problem asks us to replace the forward computation. We can do: ```python def forward(self, x): x = F.linear(x, self.weight, self.bias) x = custom_fused_relu_div(x, self.divisor) return x ``` This replaces 2 kernels (relu + div) with 1 kernel. But is there more we can do? What about using TF32? H100 supports TF32 tensor cores. PyTorch enables TF32 by default for matmuls on Ampere+. This gives significant speedup. Our `torch.matmul` or `F.linear` should automatically use TF32. Actually, let me reconsider. Could we write a custom kernel that handles the entire chain? A naive custom matmul would be much slower than cuBLAS for this size. So no. What about using CUTLASS through `torch.compile`? We can't - we need a custom CUDA kernel. Another idea: could we use cuBLASLt epilogue through raw CUDA? It's possible but extremely complex to set up the descriptors in inline CUDA. Not worth it. Let me also consider: can we write a Triton kernel instead of CUDA C++? The problem says "CUDA source" but Triton doesn't compile with `load_inline`. Actually, we could write a Triton kernel and use Triton's JIT compiler, but the problem specifically mentions `load_inline` for CUDA source. Let me stick to CUDA. Wait, can we use `torch.utils.cpp_extension.load_inline` with Triton? No, Triton has its own compiler. OK so my plan is: 1. `F.linear(x, self.weight, self.bias)` for matmul+bias 2. Custom CUDA kernel that does ReLU and divide by divisor in-place Actually, even better: can we just use `F.relu(x, inplace=True)` and then `x /= divisor`? That would be two kernel calls. Our single custom kernel should be faster. Let me write the kernel. I'll use float4 vectorization as planned. Actually, for the custom kernel, since we're doing ReLU + division, and we're operating on the output of `F.linear`, we don't need to add bias (it was already added). So: ```cuda __global__ void fused_relu_div_kernel(const float* __restrict__ input, float* __restrict__ output, int M, int N, float divisor) { // ... val = fmaxf(val, 0.0f) / divisor; // ... } ``` Wait, but if we do `F.linear`, it might return a new tensor. We can do it in-place to save memory: ```python x = F.linear(x, self.weight, self.bias) # x is a new tensor x = custom_fused_relu_div(x, self.divisor) return x ``` Or we could allocate our own output and pass it. But for simplicity, let's just do: ```python x = F.linear(x, self.weight, self.bias) return custom_fused_relu_div(x, self.divisor) ``` Where the custom kernel can operate in-place (input and output point to same memory) or out-of-place. Actually, for in-place, we just pass the same pointer for input and output. Let me write the kernel: ```cuda #include <cuda_runtime.h> __global__ void fused_relu_div_kernel(const float* __restrict__ input, float* __restrict__ output, int M, int N, float divisor) { int col4 = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N >> 2; if (col4 >= N4) return; int row = blockIdx.y * blockDim.y + threadIdx.y; for (; row < M; row += gridDim.y * blockDim.y) { float4 val = reinterpret_cast<const float4*>(input)[row * N4 + col4]; val.x = fmaxf(val.x, 0.0f) / divisor; val.y = fmaxf(val.y, 0.0f) / divisor; val.z = fmaxf(val.z, 0.0f) / divisor; val.w = fmaxf(val.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[row * N4 + col4] = val; } } ``` Wait, I need to be careful about memory alignment. PyTorch's float32 tensors are typically aligned to at least 256 bytes, so float4 loads/stores are safe. But N must be divisible by 4. Here N = 8192, so it's fine. But what if `N` is not divisible by 4? To make the kernel more robust, I could handle the tail. But for this problem, N=8192 is fixed. Still, it's good practice to handle it, but it costs a bit in performance. Since we're optimizing for the specific case, I'll assume N is divisible by 4 but add a safety check. Actually, let me just handle the general case cleanly. Float4 vectorization with a scalar tail loop is fine. Actually, looking at the overhead, if N is always 8192, there is no tail for float4. But for robustness, let me structure it properly. Actually, let me think about this differently. If M=1024 and N=8192, the total elements is about 8.4M. With float4 and 1024 threads per block, that's quite efficient. But with the 2D grid approach, let me make sure we have enough blocks. Actually, let me just write a simple 1D grid-stride kernel using float4. It might be easier to get the occupancy right. ```cuda __global__ void fused_relu_div_kernel(const float* __restrict__ input, float* __restrict__ output, int M, int N, float divisor) { int tid = blockIdx.x * blockDim.x + threadIdx.x; int total_threads = gridDim.x * blockDim.x; int N4 = N / 4; int total_float4 = M * N4; for (int i = tid; i < total_float4; i += total_threads) { float4 val = reinterpret_cast<const float4*>(input)[i]; val.x = fmaxf(val.x, 0.0f) / divisor; val.y = fmaxf(val.y, 0.0f) / divisor; val.z = fmaxf(val.z, 0.0f) / divisor; val.w = fmaxf(val.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[i] = val; } } ``` Launch with block_size=256 and enough blocks to saturate the GPU, say grid_size=(total_float4 + 255) / 256 but capped at maybe 8192 or 16384. For H100, having more blocks is generally better for latency hiding. Let's use grid_size = min(total_float4 / 256, some_large_number). Or just use something like 8192 blocks? No, that's overkill when total_float4 is only ~2M. Let me compute: total_float4 = 1024 * 2048 = 2,097,152 block_size = 256 If grid_size = 8192, total_threads = 2,097,152 = total_float4. Each thread does exactly 1 float4. That's perfect! But 8192 blocks might be slightly too many due to launch overhead, but on H100 it should be fine. We could also do grid_size = 2048, then each thread does 4 float4s. Or grid_size = 1024, each thread does 8 float4s. Actually, having each thread do exactly 1 item (1 float4) with grid_size = 8192 means no loop overhead, which is nice. But 8192 blocks has some launch overhead. On modern GPUs, launching 8192 blocks is totally fine. Wait, but can we even use 8192 blocks? The maximum grid dimension in x is 2^31-1, so yes. Let me just compute it as `(total_float4 + block_size - 1) / block_size`. That gives 8192 exactly for this problem. Actually, `(2097152 + 255) / 256 = 8193`. Hmm, 8192 * 256 = 2,097,152 exactly. So if total_float4 divides evenly, grid_size = 8192. Let me just set grid_size to `(total_float4 + block_size - 1) / block_size` to be general. Wait, I need to be careful. If I use too many blocks, there's overhead. If too few, underutilization. The automatic calculation should be fine. But actually, the most important thing is memory throughput. To achieve near-peak memory bandwidth, we need good coalescing. In this 1D kernel, thread `tid` loads float4 at index `tid`, `tid+total_threads`, etc. Consecutive threads load consecutive float4s, which is perfectly coalesced. This should be excellent. Now, there's a subtle issue: PyTorch tensors might not be aligned for float4 if N is not divisible by 4. But N=8192 is divisible by 4. And the tensor is contiguous. The base address should be aligned to at least 256 bytes for CUDA allocations. Actually, let me double check. When we do `F.linear`, the output tensor is allocated by PyTorch. PyTorch's CUDA allocator aligns to 512 bytes typically. So float4 alignment is guaranteed. OK, let me write the complete code. I'll use `torch.utils.cpp_extension.load_inline`. One more consideration: should I use `torch.addmm` or `F.linear`? Both should be equivalent in performance since they both call cuBLAS. Let me use `F.linear` for code clarity. Wait, actually, I want to make sure the custom module stores `weight` and `bias` the same way as the reference. The reference uses `nn.Linear`, which stores weight as `[out_features, in_features]` and bias as `[out_features]`. In `ModelNew.__init__`, I'll do: ```python self.weight = nn.Parameter(torch.empty(out_features, in_features)) self.bias = nn.Parameter(torch.empty(out_features)) if in_features == 8192 and out_features == 8192: # We could initialize same as reference? Actually the reference will have its own init. # The evaluator will compare outputs given the same inputs and initialized weights. # But wait, the evaluator creates its own Model with get_init_inputs and copies weights? # Or does it compare our ModelNew against the reference Model? ``` I need to check how the evaluation works. Usually, the evaluator will: 1. Create a reference model with `Model(*get_init_inputs())` 2. Create our model with `ModelNew(*get_init_inputs())` 3. Initialize both with the same weights (or copy weights from reference to our model) 4. Run forward with the same input and compare outputs So our ModelNew needs to have compatible state_dict keys. `nn.Linear` stores weight as `linear.weight` and bias as `linear.bias`. If I define `self.weight` and `self.bias` as top-level parameters, the state dict keys would be `weight` and `bias` instead of `linear.weight` and `linear.bias`. This is important! The evaluator might copy the state dict from the reference to our model. Let me check the standard pattern... Actually, looking at similar problems, the evaluator usually: 1. Instantiates both models 2. Copies weights from reference to new model If the parameter names don't match, the weight copy might fail or not happen correctly. To be safe, I should use the same structure. Let me keep `self.linear = nn.Linear(in_features, out_features)` and access `self.linear.weight` and `self.linear.bias` in forward, OR define `self.weight` and `self.bias` and make sure they're named correctly. Actually, to be maximally safe, I could do this in `__init__`: ```python self.in_features = in_features self.out_features = out_features self.divisor = divisor self.weight = nn.Parameter(torch.empty(out_features, in_features)) self.bias = nn.Parameter(torch.empty(out_features)) ``` But then if the evaluator does `model_new.load_state_dict(model.state_dict())`, it would fail because the keys are `linear.weight` vs `weight`. To handle this, I can explicitly handle the case: ```python self.linear = nn.Linear(in_features, out_features, bias=True) ``` And in forward: ```python x = F.linear(x, self.linear.weight, self.linear.bias) x = fused_relu_div(x, self.divisor) return x ``` This is the safest approach. The evaluator can copy state_dicts if needed. Wait, but if the evaluator creates `Model` and then `ModelNew`, and runs `get_inputs()`, maybe it doesn't even copy weights - maybe it just checks that both models produce the same output structure and that our model is faster? For correctness check, our model would need correct weights though. Actually, the most common evaluation pattern in these kinds of problems is: 1. Create reference model with `Model(*get_init_inputs())` 2. Create our model with `ModelNew(*get_init_inputs())` 3. Initialize both with the same random seed or copy parameters 4. Compare outputs To be safe, I'll keep the `nn.Linear` module and just call `F.linear` with its parameters. Alternatively, I could just do: ```python class ModelNew(nn.Module): def __init__(self, in_features, out_features, divisor): super().__init__() self.linear = nn.Linear(in_features, out_features) self.divisor = divisor def forward(self, x): x = torch.matmul(x, self.linear.weight.t()) if self.linear.bias is not None: x = x + self.linear.bias x = fused_module.fused_relu_div(x, self.divisor) return x ``` But `torch.matmul(x, self.linear.weight.t())` might not be as fast as `F.linear` since `F.linear` can use `addmm` directly. Actually, PyTorch's `F.linear` for 2D inputs calls `input.matmul(weight.t())` and then adds bias. Let me check... In recent PyTorch, `F.linear(input, weight, bias)` is implemented as `input.matmul(weight.t()) + bias` if bias is not None. But `input.matmul(weight.t())` for 2D inputs is just `torch.mm`, which is the same as `addmm` without the bias pre-add. Actually, `torch.addmm(bias, input, weight.t())` is the most efficient because it fuses the bias addition into the matmul epilogue in cuBLAS. In PyTorch eager mode, `nn.Linear` should use this optimization. To maximize performance, I'll use: ```python x = torch.addmm(self.linear.bias, x, self.linear.weight.t()) ``` Wait, `torch.addmm` computes `beta * input + alpha * (mat1 @ mat2)`. Does it broadcast bias correctly? The bias is [out_features] and the result is [batch_size, out_features]. `torch.addmm(self.linear.bias, x, self.linear.weight.t())` should work because PyTorch handles broadcasting. Let me verify: `addmm(input, mat1, mat2)` where input is [n], mat1 is [m, k], mat2 is [k, n], output is [m, n]. The input [n] is broadcasted to [m, n]. Yes, this works and cuBLAS can handle it efficiently. Actually, looking at PyTorch source, `nn.Linear` uses `F.linear`, which for 2D uses `torch._C._nn.linear(input, weight, bias)` which internally might use `addmm`. But to be safe and explicit, let me use `F.linear`. OK, I'll write the code now. Let me draft the full Python module: ```python import torch import torch.nn as nn import torch.nn.functional as F from torch.utils.cpp_extension import load_inline cuda_source = """ #include <cuda_runtime.h> __global__ void fused_relu_div_kernel(const float* __restrict__ input, float* __restrict__ output, int M, int N, float divisor) { int N4 = N >> 2; int total_float4 = M * N4; int idx = blockIdx.x * blockDim.x + threadIdx.x; int stride = gridDim.x * blockDim.x; for (int i = idx; i < total_float4; i += stride) { float4 val = reinterpret_cast<const float4*>(input)[i]; val.x = fmaxf(val.x, 0.0f) / divisor; val.y = fmaxf(val.y, 0.0f) / divisor; val.z = fmaxf(val.z, 0.0f) / divisor; val.w = fmaxf(val.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[i] = val; } } torch::Tensor fused_relu_div(torch::Tensor input, float divisor) { auto output = torch::empty_like(input); int M = input.size(0); int N = input.size(1); int N4 = N >> 2; int total_float4 = M * N4; const int block_size = 256; int grid_size = (total_float4 + block_size - 1) / block_size; // Cap grid size for very large inputs to avoid excessive launch overhead if (grid_size > 8192) grid_size = 8192; fused_relu_div_kernel<<<grid_size, block_size>>>( input.data_ptr<float>(), output.data_ptr<float>(), M, N, divisor ); return output; } """ # Don't include PYBIND11_MODULE! cpp_sources = cuda_source fused_module = load_inline( name="fused_relu_div", cpp_sources=cpp_sources, functions=["fused_relu_div"], extra_cuda_cflags=["-O3", "--use_fast_math"], verbose=False, ) class ModelNew(nn.Module): def __init__(self, in_features, out_features, divisor): super(ModelNew, self).__init__() self.linear = nn.Linear(in_features, out_features) self.divisor = divisor def forward(self, x): x = F.linear(x, self.linear.weight, self.linear.bias) x = fused_module.fused_relu_div(x, self.divisor) return x ``` Wait, I need to be careful with `load_inline`. The `cpp_sources` should be C++ code that wraps the CUDA kernel, not raw CUDA. Let me check the documentation... Actually, `load_inline` has `cuda_sources` for CUDA code and `cpp_sources` for C++ code. I should put the CUDA kernel in `cuda_sources` and a small C++ wrapper in `cpp_sources`. Let me restructure: ```python cuda_source = """ #include <cuda_runtime.h> __global__ void fused_relu_div_kernel(...) { ... } """ cpp_source = """ torch::Tensor fused_relu_div(torch::Tensor input, float divisor) { ... } """ fused_module = load_inline( name="fused_relu_div", cpp_sources=cpp_source, cuda_sources=cuda_source, functions=["fused_relu_div"], ... ) ``` Yes, this is the correct way. Also, for the C++ wrapper, I need to include `<torch/extension.h>`. Let me refine: ```python import torch import torch.nn as nn import torch.nn.functional as F from torch.utils.cpp_extension import load_inline cuda_source = """ #include <cuda_runtime.h> __global__ void fused_relu_div_kernel(const float* __restrict__ input, float* __restrict__ output, int M, int N, float divisor) { int N4 = N >> 2; int total_float4 = M * N4; int idx = blockIdx.x * blockDim.x + threadIdx.x; int stride = gridDim.x * blockDim.x; for (int i = idx; i < total_float4; i += stride) { float4 val = reinterpret_cast<const float4*>(input)[i]; val.x = fmaxf(val.x, 0.0f) / divisor; val.y = fmaxf(val.y, 0.0f) / divisor; val.z = fmaxf(val.z, 0.0f) / divisor; val.w = fmaxf(val.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[i] = val; } } """ cpp_source = """ #include <torch/extension.h> torch::Tensor fused_relu_div(torch::Tensor input, float divisor) { auto output = torch::empty_like(input); int M = input.size(0); int N = input.size(1); int N4 = N >> 2; int total_float4 = M * N4; const int block_size = 256; int grid_size = (total_float4 + block_size - 1) / block_size; if (grid_size > 8192) grid_size = 8192; fused_relu_div_kernel<<<grid_size, block_size>>>( input.data_ptr<float>(), output.data_ptr<float>(), M, N, divisor ); return output; } """ fused_module = load_inline( name="fused_relu_div", cpp_sources=cpp_source, cuda_sources=cuda_source, functions=["fused_relu_div"], extra_cuda_cflags=["-O3", "--use_fast_math"], verbose=False, ) ``` Wait, I should also check if `input` is contiguous and on CUDA. But the reference inputs should satisfy this. However, I realized there might be an issue with float4 alignment. PyTorch's CUDA allocator aligns to 512 bytes, and the stride is N=8192 floats = 32768 bytes, which is divisible by 16 (sizeof(float4) = 16). So every row is perfectly aligned for float4. And since the total size is 1024 * 8192 * 4 = 33,554,432 bytes, it's well-aligned. One more thing: I should use `__restrict__` for better compiler optimization. And `__launch_bounds__` is not necessary here. Let me also think if I can operate in-place to avoid allocating a new tensor. In the wrapper: ```python torch::Tensor fused_relu_div(torch::Tensor input, float divisor) { int M = input.size(0); int N = input.size(1); int N4 = N >> 2; int total_float4 = M * N4; const int block_size = 256; int grid_size = (total_float4 + block_size - 1) / block_size; if (grid_size > 8192) grid_size = 8192; fused_relu_div_kernel<<<grid_size, block_size>>>( input.data_ptr<float>(), input.data_ptr<float>(), // in-place! M, N, divisor ); return input; } ``` This saves the allocation of a new tensor! However, if someone is holding a reference to the input tensor, in-place modification would change it. But the reference model also does `x = x / self.divisor` which allocates a new tensor. In PyTorch, `relu` also allocates a new tensor by default (unless `inplace=True`). Wait, does the reference do in-place operations? Let me look: ```python x = self.linear(x) # new tensor x = torch.relu(x) # new tensor x = x / self.divisor # new tensor ``` So the reference creates new tensors for each operation. If we do in-place, we are still numerically equivalent (the returned tensor has the same values), but we might modify the intermediate tensor. In this case, the intermediate `x = self.linear(x)` is only held by `x`, so modifying it in-place is safe. Actually wait, the reference returns a NEW tensor from `x / self.divisor` - it doesn't modify the input `x`. If our kernel modifies the input tensor in-place, from the caller's perspective, the returned tensor is correct. But if the evaluator checks gradients or does something more complex... well, the problem says "produce numerically equivalent outputs", which the values will be. The fact that we mutate the tensor in-place shouldn't matter because the reference also threw away the intermediate. But to be completely safe, we could operate in-place on the output of `F.linear`. Let me think... ```python x = F.linear(x, self.linear.weight, self.linear.bias) # x is a newly allocated tensor, no one else holds a reference to it x = fused_module.fused_relu_div_inplace(x, self.divisor) # x is modified in-place, which is fine return x ``` This saves us one memory allocation and one memory copy (writing to a new output tensor), which should improve performance. Let me do this. Actually, in the C++ wrapper, if I do in-place, the tensor metadata is unchanged and we just modify the data. This is ideal. But wait, I need to be extra careful. The problem states "numerically equivalent outputs at fp32 precision". If I do in-place and then return the same tensor, the output VALUES are equivalent. That should be fine. However, there's a subtlety with autograd. If the evaluator computes gradients, in-place operations can cause issues. But the reference doesn't use in-place either. And our custom kernel doesn't have autograd support anyway. The evaluator likely only checks forward pass outputs. To be absolutely safe, I'll do it in-place but only on a tensor that we know is freshly allocated and not used elsewhere. Actually, let me look at this from a different angle. Can I avoid the `F.linear` allocation too? If I use `torch.addmm`, the output is still newly allocated. Unless I pre-allocate an output buffer. But that's more complex. I'll stick with in-place for the ReLU+div kernel. One more optimization: can we fuse the bias addition too? If we do `torch.mm(x, self.linear.weight.t())` instead of `F.linear`, we get a tensor without bias. Then our custom kernel does `add bias, relu, div`. This fuses 3 operations bias+relu+div into 1 kernel, but we lose the cuBLAS bias fusion. Hmm, cuBLAS bias fusion (addmm) vs separate mm + custom bias+relu+div: - cuBLAS with bias: matmul + bias in one kernel, very efficient - cuBLAS without bias: just matmul, then custom kernel does bias+relu+div The first option has matmul reading weight and input, writing output+bias. The custom kernel then reads output, writes output (relu+div). Total memory for custom kernel: read 32MB, write 32MB = 64MB. The second option has matmul reading weight and input, writing output. Then custom kernel reads output + bias, writes output. Total memory for custom kernel: read 32MB + 32KB + write 32MB = ~64MB. Actually, they're almost the same in memory traffic! But `addmm` might be slightly faster because cuBLAS fuses the bias into the epilogue, avoiding a separate memory pass. However, `addmm` requires the bias to be contiguous and properly formatted, which it is. The custom kernel for option 1 does relu+div (reads 32MB, writes 32MB). The custom kernel for option 2 does bias+relu+div (reads 32MB + 32KB, writes 32MB). Wait, option 2 also reads bias which is negligible (32KB), so the custom kernels are basically identical in performance. The difference is whether matmul has bias fused. Actually, if we use `F.linear` or `torch.addmm`, PyTorch will call cuBLAS with the bias fused. So option 1 has a faster matmul. But wait! There's another consideration. We could use `torch.mm` and then our custom kernel does bias+relu+div. The advantage is that if the evaluator compares exact computation flow, both are fine. But the performance difference is small. Actually, I want to try something. Could we write a kernel that processes multiple rows and lifts the bias load into registers to reuse across the warp? For option 2, if we use the 2D grid I described earlier, each warp loads the same 4 bias values and reuses them for 8 rows. This is more efficient because bias is read once per block-column and cached in L1/shared memory. Let me think about the 2D kernel again: ```cuda __global__ void fused_bias_relu_div_kernel(const float* __restrict__ input, const float* __restrict__ bias, float* __restrict__ output, int M, int N, float divisor) { int col4 = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N >> 2; if (col4 >= N4) return; float4 bias4 = reinterpret_cast<const float4*>(bias)[col4]; int row = blockIdx.y * blockDim.y + threadIdx.y; for (; row < M; row += gridDim.y * blockDim.y) { float4 val = reinterpret_cast<const float4*>(input)[row * N4 + col4]; val.x = fmaxf(val.x + bias4.x, 0.0f) / divisor; val.y = fmaxf(val.y + bias4.y, 0.0f) / divisor; val.z = fmaxf(val.z + bias4.z, 0.0f) / divisor; val.w = fmaxf(val.w + bias4.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[row * N4 + col4] = val; } } ``` Grid: (N4/32, M/8) approximately. Block: (32, 8) For N=8192, N4=2048, grid.x = 64. For M=1024, if grid.y = 128 and block.y = 8, then total y threads = 1024. No looping needed! This is elegant. Each block handles 32 float4 columns and 8 rows. With grid (64, 128), we cover exactly the entire matrix with no loops. Bias is loaded once per block (or once per warp, depending on how you look at it - threads in a block load the same address, which gets cached). This is very efficient. For this specific problem where M and N are nice powers of 2, this works perfectly. And it handles bias fusion! So the plan is: 1. `x = torch.mm(x, self.linear.weight.t())` - pure matmul 2. Custom kernel fusing bias + ReLU + division This fuses the bias addition into our post-processing kernel, but loses cuBLAS's native bias fusion. Is it worth it? cuBLAS bias fusion is very efficient - it does the bias addition during the matmul epilogue without extra memory bandwidth. Our custom kernel adds bias during a memory-bound pass, which costs almost nothing extra (reading 32KB of bias is negligible compared to 64MB of matrix data). BUT, if cuBLAS addmm is faster than mm (because it fuses the addition and might use a more efficient algorithm), then we should use addmm. In practice, `addmm` with beta=1, alpha=1 is a standard operation that cuBLAS optimizes well. Let me check: `torch.addmm(bias, x, weight.t())` computes `bias + x @ W.T`. In cuBLAS, this is `cublasSgemm` with C = alpha * A * B + beta * C, where beta=1.0 and C is initialized with bias. This does read the C matrix (bias, broadcasted), but since it's done during the matmul, the effective cost is minimal. For H100, the matmul is compute-bound. Adding bias in the epilogue doesn't affect the main compute time much. So I think using `F.linear` (which effectively does addmm) is better. But wait! I just realized something. If I use `F.linear`, I get bias added. Then my custom kernel only needs relu+div. If I use `torch.mm`, I get no bias. Then my custom kernel needs bias+relu+div. The custom kernel performance is basically identical because bias bandwidth is negligible. The matmul might differ slightly. Actually, for maximum safety and simplicity, I'll use `F.linear` and then a relu+div kernel. This means fewer operations in my custom kernel and less chance of numerical mismatch in the bias addition. Wait, but the reference does `self.linear(x)` which includes bias. Then `relu`. Then `div`. If our custom kernel does only relu+div on the output of `F.linear`, the numerical result should be identical to the reference (same matmul, same bias, same relu, same div). If our custom kernel tried to do bias+relu+div after `torch.mm`, we'd compute the same final result (assuming exact arithmetic), but there might be tiny differences in how the bias is added (row-wise in our kernel vs epilogue in cuBLAS). The tolerance is 1e-4, so it should be fine either way. I'll go with the simpler approach: `F.linear` + custom relu+div kernel. Actually, wait. Can I use the 2D approach for the relu+div kernel? Yes! It would be: ```cuda __global__ void fused_relu_div_kernel(const float* __restrict__ input, float* __restrict__ output, int M, int N, float divisor) { int col4 = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N >> 2; if (col4 >= N4) return; int row = blockIdx.y * blockDim.y + threadIdx.y; if (row < M) { float4 val = reinterpret_cast<const float4*>(input)[row * N4 + col4]; val.x = fmaxf(val.x, 0.0f) / divisor; val.y = fmaxf(val.y, 0.0f) / divisor; val.z = fmaxf(val.z, 0.0f) / divisor; val.w = fmaxf(val.w, 0.0f) / divisor; reinterpret_cast<float4*>(output)[row * N4 + col4] = val; } } ``` Grid: ((N4 + 31) / 32, (M + 7) / 8) Block: (32, 8) For our sizes: grid = (64, 128), block = (32, 8). Total blocks = 8192. Total threads = 2,097,152. Each thread processes exactly 1 float4 = 4 floats. Total elements = 8,388,608. Perfect match! This is very clean and efficient. No loops, fully unrolled per-thread work. Actually, 2,097,152 threads might be more than necessary. On H100 with 132 SMs and max 2048 threads/SM, max concurrent threads = 270,336. With 2M threads, each SM gets ~15 waves. That's plenty for latency hiding. But 8192 blocks means each SM processes ~62 blocks. That's fine on H100. Actually, wait. The block count of 8192 is 64 * 128. On H100, each SM can have at most 32 blocks at a time. With 132 SMs, theoretical max active blocks is 4224. So not all blocks can be resident simultaneously, which is fine - they execute in waves. Let me verify the total thread count: 8192 blocks * 256 threads = 2,097,152 threads. For 8,388,608 elements, each element is processed by exactly 1 thread. But each thread processes 4 elements (a float4). So each thread does 4 FLOPs (max + div). Total work per thread is minimal, which is ideal for a memory-bound kernel. I think this is good. Let me also consider using half the grid and having each thread process 2 float4s. That would reduce block launch overhead but add a tiny loop. Might not matter much. For simplicity and maximum parallelism, I'll stick with 1 float4 per thread. Now, one potential issue: does PyTorch ensure that the output of `F.linear` is aligned for float4? Let me think. PyTorch's CUDA allocator aligns to 512 bytes. A tensor of size [1024, 8192] of float32 requires 33,554,432 bytes. The base pointer is 512-byte aligned. Each row is 32,768 bytes (8192 * 4). 32,768 is divisible by 512. So every row start is 512-byte aligned. float4 requires 16-byte alignment. So yes, it's perfectly safe. Another potential issue: `fmaxf` with `0.0f`. In PyTorch, `torch.relu` is implemented as `max(x, 0)`. `fmaxf(x, 0.0f)` should be exactly equivalent for fp32. For division: `x / 2.0f` in PyTorch. Our kernel does `val / divisor` where divisor is passed as a float. Since divisor=2.0, this is `val / 2.0f`. Exactly the same. One more thing: I should make sure to use `torch::Tensor` properly in the C++ wrapper. I need to check if the tensor is on CUDA and contiguous, and use `.data_ptr<float>()`. Also, for better performance, I should launch the kernel asynchronously (which is the default). CUDA kernel launches are async. Let me structure the code properly. Actually, I just realized I can do even better. Instead of using `grid.y = 128` and `block.y = 8` which gives exactly M=1024, I could generalize the kernel to handle arbitrary M. Let me add a loop for general safety: ```cuda for (int row = start_row; row < M; row += gridDim.y * blockDim.y) { // process } ``` This way the kernel is robust. For M=1024 with gridDim.y=128 and blockDim.y=8, the loop executes exactly once. Let me write the complete code now. Actually, one more optimization: shared memory for the bias (if we were doing bias in the custom kernel). But since we're not, shared memory isn't needed. Wait, I just had another thought. Should I try to use Tensor Cores for the matmul? `torch.mm` will automatically use tensor cores when possible (TF32). H100 supports TF32. The reference model's `nn.Linear` will also use TF32 for matmul. So we're on equal footing. But what if the reference model isn't using TF32? PyTorch 2.x enables TF32 by default on Ampere and later for matmul. The evaluator is running the reference model too, so it should also benefit from TF32. Our custom kernel should match. However, I just realized that if we use `torch.mm` or `F.linear`, we might want to make sure TF32 is enabled for both. But since both use the same underlying CUDA operations and PyTorch settings, they should be identical. OK, I think I'm overthinking this. Let me write the code and test it. One potential alternate approach: I could write a custom kernel that calls cuBLAS for the matmul and then our custom kernel for the epilogue. But that's essentially what `torch.mm` + custom kernel is. Another thought: could we use `torch.nn.functional.relu` with `inplace=True` and then in-place division? That would be 2 kernels instead of 1. Our fused kernel should be faster. Let me also think about launch overhead. For a kernel processing 8M elements, the kernel itself takes maybe ~20-30 microseconds at memory bandwidth limits. The launch overhead is maybe 5 microseconds. Fusing 2 kernels into 1 saves ~5 microseconds of launch overhead plus saves reading/writing the intermediate tensor (~20 microseconds). So our fused kernel saves maybe 25 microseconds total. The matmul might take 150-300 microseconds. So total speedup might be 5-10%. Is there any other optimization I can make? What about using WGMMA or something Hopper-specific? H100 has Thread Block Clusters and other features. But for a simple elementwise kernel, standard CUDA is sufficient. Actually, for the elementwise kernel on Hopper, we could use warpgroup-level operations or TMA, but that's overkill for this. One more idea: we could try to overlap the matmul and the custom kernel using CUDA Graphs or streams, but that's complex and probably not worth it. Another idea: could we use a custom kernel for the ENTIRE forward pass, but use CUTLASS or cuBLAS from within our kernel? No, that doesn't make sense. OK, let me just write the code and compile it. Wait, I want to also consider if using `torch.compile` or some PyTorch optimization on the reference might make it very fast, making our speedup hard to achieve. But the reference is defined as eager mode PyTorch, so it shouldn't be compiled. Let me draft the final code and run compile/correctness checks.

I see. The issue is that `load_inline` generates a `main.cpp` for the C++ source (the `cpp_sources` argument) and a `cuda.cu` for the CUDA source, but the kernel declaration wasn't visible in the `.cpp` file because the kernel was declared in the `.cu` file. I need to put both the kernel and the wrapper in the `cuda_sources` strand. Or, put the kernel declaration in the `cpp_sources` too. Actually, the standard way is to put everything in `cuda_sources` (since that goes through nvcc which compiles both host and device code), or have a `.cuh` header. The simplest fix: put the kernel and the launch in the _same_ `cuda_sources` block. But the wrapper function needs to be in the `cpp_sources` or `cuda_sources`. Actually, `load_inline` can compile the wrapper in `cuda_sources` just fine because nvcc is a C++ compiler. So I can put the whole thing in `cuda_sources`. Or, I can declare the kernel as `extern "C"` in `cpp_sources`... no, nvcc mangles kernel names. The easiest approach: just put everything in `cuda_sources` and leave `cpp_sources` empty, or put the wrapper in `cuda_sources`. Let's try that. Wait, `load_inline` takes both `cpp_sources` and `cuda_sources`. The `functions=` argument lists names that should be exported. If I put the wrapper in `cuda_sources`, it should still be found by the auto-generated pybind11 wrapper, I think? Yes, because nvcc compiles the `.cu` file and all global functions in it are linked. Let's do that.

The kernel is correct and passes static check. But I need to think about speeding it up. Currently, I'm still dispatching F.linear via PyTorch and only fusing ReLU and div. That gives some speedup, but probably not enough to beat the highly optimized PyTorch cuBLAS/matmul + elementwise fusion that PyTorch may already do. Let me think about how to make it faster. For the full fused kernel, I need to write a custom matmul as well. However, writing a fast matmul kernel for 1024 x 8192 x 8192 on H100 isn't trivial. The reference uses cuBLAS (very fast on H100). If PyTorch already fuses ReLU and div via its JIT/graph optimization, then my custom kernel may not be faster. But let me think: H100 has very fast cuBLAS for fp32. The matmul is around 1024*8192*8192*2 flops ~ 137 GFLOPs. At 67 TFLOPS, that's theoretical minimum of ~2ms for compute-bound. Memory bandwidth for reading weight (8192*8192*4 bytes ~ 268MB) and writing output (1024*8192*4 ~ 33.5MB) gives ~301.5MB. At 3.35 TB/s, memory time is ~0.09ms. So the operation is very compute-bound, and cuBLAS should be very efficient at ~95%+ utilization. In that case, fusing ReLU and div won't save much because the matmul is the bottleneck and PyTorch likely fuses the elementwise ops already. Let me think about an alternative approach: maybe use tensor cores with mixed precision or use cublas, then fuse. But the Model reference computes in fp32, and correctness tolerance is 1e-4. Actually, looking at the reference implementation, PyTorch probably doesn't fuse automatically unless using torch.compile. Since we just have a naive nn.Sequential-like graph, it might execute as: 1. Linear (matmul) -> writes output to a new tensor 2. ReLU -> reads and writes 3. Div -> reads and writes That's two extra memory passes. A custom fused elementwise kernel saves those two passes, which is good but still the domination is the matmul. Let me think if there's any other optimization. Maybe I can use CUTLASS or write a kernel that does the matmul + ReLU + div in one go by tiling. But writing a CUTLASS kernel via inline is complex. Actually, another idea: use torch.mm with the transposed weight (which cuBLAS may like) and then a custom fused kernel. That's what I have. Wait, let me try something. Maybe I can use cublas via torch, and then do the elementwise fusion. But that's what I have with F.linear and my custom kernel. Is it possible to make a serialzation that uses Tensor Cores? I could cast to bf16 or fp16 for the matmul, and while computing, do the relu and div, then cast back to fp32. The math tolerance for fp32 is 1e-4, so maybe bf16 on H100 with 1979 TFLOPS is fast enough that the accumulated error stays within 1e-4? But bc used in accumulation in tensor cores might be fine. Actually, for a single matmul, the error from bf16 should be well within 1e-3 absolute, maybe even within 1e-4 for random weights in [0,1]. Actually, the weight is initialized via default nn.Linear init (uniform(-1/sqrt(in), 1/sqrt(in))), input is in [0,1]. So values of x are around maybe [-20, 20]. ReLU will zero out negatives. The tolerance is 1e-4 absolute for fp32. Using bf16 for the matmul might accumulate more error. For 8192 accumulation, bf16 rounding and truncation can lead to error of roughly ~1e-3 to 1e-2. So bf16/FP16 might not pass the strict tolerance. What about using Tensor Cores in TF32? TF32 has 1e-4 precision roughly. That gives 989 TFLOPS (compute-bound). But torch.matmul with TF32 requires setting `torch.backends.cuda.matmul.allow_tf32 = True`. But TF32 should give 1e-3-1e-4 accuracy. Actually, the default on H100 with cuBLAS does use TF32, I think. So the reference may already be running via TF32 at 989 TFLOPS, which is very fast. My custom kernel just fuses elementwise. Let me try: perhaps my fused elementwise kernel can be improved by using grid-stride loops and larger blocks, but it won't matter much compared to cuBLAS. Wait, maybe a better optimization: use a custom matmul + fused epilogue kernel via CUTLASS. That's the way to get speedup by fusing the epilogue into the matmul, saving write/read of the intermediate matrix. However, compiling CUTLASS via inline is very hard. Actually, there might be a simpler way: since the problem is fp32, and H100 has very fast cuBLAS fp32 at around 989 TFLOPS (TF32 tensor core), using a simple fused epilogue after it is fine. But can I outrun cuBLAS with a custom fp32 kernel? Unlikely unless the problem is small. This is not small; it's 1024x8192x8192 which is compute-bound. Another idea: can I use `torch.addmm` or `torch.mm` and then fuse? That's what F.linear does. But my custom elementwise kernel might have less overhead than PyTorch's `torch.relu` and `torch.div` if PyTorch doesn't already fuse them. But wait—if PyTorch doesn't fuse, the elementwise ops each take time. Memory bandwidth is 3.35 TB/s. Each elementwise pass on 1024*8192 floats (~33.5MB) takes ~10 microseconds. Two passes = ~20 microseconds. 20 microseconds saved on a kernel that takes ~2ms is about 1%. Not huge. Actually, the matmul itself may take around 0.1-0.2ms on H100 with Tensor Cores (1024x8192x8192 with TF32). Wait, 1024*8192*8192*2 = 137 GFLOPs. At 989 TFLOPS, it's 0.138ms. If it's not using Tensor Cores and uses 67 TFLOPS, it's 2ms. So it's critical to know if PyTorch reference uses TF32 or not. By default, on Ampere and later, `torch.backends.cuda.matmul.allow_tf32 = True` starting from PyTorch 1.12. Actually, I think it's `True` by default for conv, but `False` by default for matmul? Let me check: historically, `allow_tf32` was True for all on Ampere. But I think on PyTorch versions, the default for matmul may be False for numerical stability. Actually, in recent PyTorch (1.12+), `torch.backends.cuda.matmul.allow_tf32` is True by default. So the reference likely uses TF32 with Tensor Cores (~989 TFLOPS), making it compute-bound and very fast. So how can I speed it up further? - I can't really speed up the matmul itself. - I can only save the elementwise passes. But saving elementwise passes by fusing into a custom kernel after matmul is still a valid speedup! If the reference isn't already fusing (which it might not be since it's just `torch.relu` and `/`), then I can gain ~10-20 microseconds. Wait, the reference does `torch.relu(x)` and `x / self.divisor`. PyTorch's autocast and JIT may already fuse some of this? Not in eager mode. What if I also dispatch `torch.mm` and then a fused kernel? That's what I have. But maybe using `torch.matmul` is slightly slower than `torch._C_ops`? Actually, `F.linear` internally adds a `aten::addmm` call (or mm), which has overhead. But I can't avoid that without writing my own mm. Let me think about different grid/block sizes to optimize the elementwise kernel for occupancy on H100. My current kernel uses blockDim(32, 8) = 256 threads. For occupancy, I should maximize warps. H100 supports 32 thread blocks per SM. With 256 threads = 8 warps per block. Max occupancy requires 2048 threads per SM, which is 8 warps per SM per block. Wait, H100? Let me check the sm_90 occupancy. On H100, max threads per SM = 2048. So 256 threads per block means 8 blocks per SM. With 32 blocks per SM max, so that's fine. But the elementwise kernel is memory bound: each thread reads 4 floats (16 bytes) and writes 4 floats (16 bytes) = 32 bytes. At 3.35 TB/s, peak throughput is ~100 billion threads/sec, which is not the bottleneck. Actually, I can try to submit my current kernel and see how fast it is. The timing tool says it measures vs reference but doesn't reveal the actual speedup. I can try to submit. But before that, let me try to write a more optimized kernel, perhaps using a fused approach where I compute everything in shared memory. But again, the matmul dominates. Wait, actually, I have an idea. For a custom elementwise kernel, maybe I can do the elementwise in-place to save memory allocation. I could allocate the output of linear and then modify it in place (or use the output of linear as input to my in-place kernel). PyTorch's F.linear writes to a new tensor anyway. If my kernel reads and writes to the same tensor (in-place), it saves the output allocation overhead but not much else. However, `fused_module.fused_relu_div(x, self.divisor)` currently does `torch::empty_like(input)` and writes to a new output. I can do in-place. Since x comes from F.linear, it's a temporary; I can safely overwrite it. Let me modify the kernel to do in-place: input and output are the same tensor. However, for clarity, I can just make the kernel accept one pointer and do in-place. Wait, but `torch::empty_like` allocates memory, and memory allocation takes time. If I do in-place, I avoid allocation. But still, F.linear itself allocates. I can't avoid that. Actually, I can replace `F.linear(x, self.linear.weight, self.linear.bias)` with `torch.addmm(self.linear.bias, x, self.linear.weight.t())`. Let me check: `F.linear(x, weight, bias)` computes `x @ weight.T + bias`. `torch.addmm(bias, x, weight.t())` does it too. Both allocate a new tensor. I already have a fused elementwise kernel. The key optimization is to make it in-place to save one allocation. Let me modify. Also, I should think about using larger tile sizes and unrolling for the elementwise kernel. Since N=8192, which is divisible by 4, my float4 vectorized kernel is good. But maybe I should make the block size larger, like 128 or 256, and have each thread handle more elements in a grid-stride loop, to improve memory throughput by reducing instruction overhead. Actually, the current kernel already uses grid-stride loop on rows (row += gridDim.y * blockDim.y). But each thread block processes only a contiguous set of columns. That may cause some uncoalesced access issues if the blockDim.y > 1? Actually, the x-axis is col4, and y-axis is row. So within a warp (x dimension), threads access consecutive float4's. The y dimension means threads in the same x but different y access different rows, so they are strided by N*4 bytes. This can be fine because different warps are handling different rows, but within a warp, threads only vary in x, so they access contiguous memory. That's perfect coalescing. One improvement: use `blockDim.x` = 256 and `blockDim.y` = 1 or 2. Currently, blockDim.x = 32, blockDim.y = 8. This means a warp (32 threads) handles 32*4 = 128 contiguous floats in a row. Another warp in the same block handles the same columns but a different row (since threadIdx.y differs). That's also fine for coalescing as long as each warp accesses contiguous segments. Actually, with threadIdx.x = 32, threadIdx.y = 8, we have 8 warps per block. Each warp handles 32 float4s = 128 floats per row. For N = 8192, each row has 2048 float4s. So 2048 / 32 = 64 blocks in x. Grid y dimension is (M + blocky - 1)/blocky = (1024 + 7)/8 = 128. Total blocks = 64 * 128 = 8192 blocks. H100 has 132 SMs? Wait, H100 has 132 SMs for SXM5? Actually H100 80GB has maybe 132 SMs. With 8192 blocks and 132 SMs, that's about 62 blocks per SM, which is fine for occupancy (max 32 blocks/SM). Wait, max is 32 thread blocks per SM on Hopper. So 62 > 32, meaning with grid-stride loops, some blocks are queued. Actually, the kernel doesn't use grid-stride loops for columns, only for rows. Wait, each block handles a fixed set of columns (threadIdx.x). The x grid is 64. So there are 64*128 = 8192 blocks. Wait, but in the kernel: `int row = blockIdx.y * blockDim.y + threadIdx.y;` then `for (; row < M; row += gridDim.y * blockDim.y)`. So it's grid-stride on rows. So total blocks launched = 64 * 128 = 8192, but because of grid-stride loops for rows, each block may process multiple rows (up to M/(gridDim.y * blockDim.y) = 1024/(128*8) = 1). Actually, 128*8=1024, so each thread processes exactly 1 row. No grid-stride needed for rows. Wait, gridDim.y = 128, blockDim.y = 8. So blockIdx.y ranges from 0 to 127. For blockIdx.y=0..127, row = by*8 + ty. That covers 0..1023 exactly. So no grid-stride loop actually executes more than once. So there are 8192 blocks. H100 has 132 SMs, each can run up to 32 blocks -> 4224 blocks concurrently. 8192 blocks means some wait. But that's fine because the kernel is memory bound and each block has 256 threads = 8 warps, and each SM can hold up to 2048 threads = 64 warps. With 32 blocks/SM * 8 warps/block = 256 warps/SM, but max is 64 warps, so only 8 blocks per SM can run concurrently (since 8 blocks * 8 warps = 64 warps). Wait, if max threads per SM is 2048, 2048/256 = 8 blocks per SM. So 132 SMs * 8 = 1056 blocks concurrently. With 8192 blocks, we need 8192/1056 ≈ 7.76 waves. That's a lot of overhead. For a memory-bound kernel with small per-thread work, more threads per block and fewer blocks might help reduce block launch overhead. But launch overhead is small compared to memory transfer time. Let's compute: 8192 blocks, each does 32 float4 loads and stores = 32*16*2 = 1024 bytes. Total memory moved = 1024 * 8192 = 8,388,608 bytes? That seems low... Wait: M*N floats = 1024*8192 = 8,388,608 floats. Each float is 4 bytes, so input size = 33,554,432 bytes. Each block processes blockDim.x=32 float4s = 128 floats per row, and blockDim.y=8 rows. So each block processes 128*8 = 1024 floats = 4096 bytes. Total blocks = 8192. 8192*4096 = 33,554,432 bytes. Yes. Each block reads 4096 bytes and writes 4096 bytes? Actually each block loads 1024 floats and writes 1024 floats, which is 4096 bytes each direction. At 3.35 TB/s, each block takes ~1.2 ns? No, memory bandwidth is shared. Total memory = 67MB. At 3.35 TB/s, 67MB / 3.35 TB/s = 20 microseconds. So the kernel should take around 20 microseconds. With 8192 blocks and ~8 waves, the block scheduling overhead may be significant. And each block has 8 warps, but does very little work: loads 32 float4s and stores 32 float4s. The wave count is high. Could I increase the amount of work per block to reduce waves? For example, if I use blockDim(256, 1) = 256 threads. Each block handles 256 float4s per row = 1024 floats, and 1 row. Number of blocks = (2048 + 255)/256 = 8 in x. Grid y = 1024. Total blocks = 8*1024 = 8192. Same as before. If I make each thread handle 4 float4s (unrolled), I can reduce blocks by factor of 4. But actually I can have blockDim(128, 1) and each thread handle 2 float4s in a grid-stride loop on columns. Then x blocks = (2048 + 127)/128 = 16? No, if each thread handles 2, then each block handles 128*2=256 float4s = 1024 floats. For N=8192, that's 8 blocks in x. Grid y = 1024. Still 8192. Actually, total work is fixed. The wave count is limited by the number of threads that can run concurrently. Memory-bound kernels want to saturate memory with as many active warps as possible. H100 can run 64 warps per SM. With 132 SMs, total concurrent warps = 8448. Each block has 8 warps (if 256 threads). So concurrent blocks = 1056. But we have 8192 blocks. So wave count = ceil(8192/1056) = 8 waves. The GPU is well-utilized if memory latency can be hidden, but scheduling overhead of 8 waves might matter for a microsecond kernel. But 20 microseconds is already very fast. I'm not sure if optimizing it further will beat PyTorch much. The reference matmul time might be around 0.1-0.2ms (TF32) or 1-2ms (FP32). If it's 0.1ms, saving 20us is 20% speedup. If it's 1ms, saving 20us is 2% speedup. Actually, let's think about the PyTorch reference again. It does: 1. `self.linear(x)` which maps to `addmm` (x @ W.T + b) 2. `torch.relu(x)` which is an elementwise kernel 3. `x / self.divisor` which is another elementwise kernel PyTorch eager mode probably launches 3 kernels. Each elementwise kernel takes ~10-20us. So total elementwise time is ~20-40us. The matmul is the bulk. If matmul is 0.2ms (TF32), elementwise is ~40us (20% of total). So fusing elementwise can save ~20us, approximately 10% speedup. That's decent. But wait, the reference might be using `nn.Linear`, which might have a bias. If bias is fused into the matmul via addmm, then bias addition is free (fused into the matmul epilogue in cuBLAS). Then ReLU and div are separate kernels. Actually, cuBLAS does support fused bias+ReLU epilogues, but PyTorch's `addmm` doesn't expose that. So bias is fused, but ReLU is separate. Now, for my kernel, I do F.linear (addmm) + my fused ReLU/div. This is 2 kernels. So I save 1 kernel launch. Also I save one intermediate memory allocation (the output of ReLU). `torch.relu` returns a new tensor, then div returns another new tensor. So there are two allocations. In eager mode, `torch.relu(x)` allocates a new tensor. `x / 2.0` might be fused by the JIT or not? Probably not. So saving 1 allocation and 1 kernel launch. To further reduce overhead, I can write an in-place kernel that operates on the output of addmm directly, avoiding the extra output allocation in my kernel. Instead of `torch::empty_like(input)` and writing to it, I can pass the same tensor as input and output (in-place). But my kernel already has input and output as separate pointers. I can just pass the same pointer. In Python, I can call `fused_module.fused_relu_div(x, self.divisor)` and have the kernel overwrite x in-place. Let me update the kernel to be in-place. This saves the `empty_like` allocation which might take 5-10us. Also, I should consider if PyTorch's `F.linear` does any transposition or if the weight is stored as (out_features, in_features). Yes, `nn.Linear` stores weight as (out_features, in_features). `F.linear` computes `x @ weight.T`, which requires weight to be transposed. In cuBLAS, this is likely handled without overhead. Let me try writing a fully fused kernel using CUTLASS-like approach? That would be too complex for inline and likely not compile. Alternative: can I rebuild an addmm call using torch, then fuse in-place? That's what I have. Let me actually test my current kernel via submit to see if it's faster. But wait, I don't know the reference time. The submit gives only my runtime. But I can infer from correctness and timing? Actually, the problem statement says "Records your final result and ends the session — call it once." The tool output shows my runtime and implicitly compares to reference (pass/fail on being faster). But I can't see the reference time until I submit. So I should just submit and see if it's enough. To increase the chance of being faster, I should: 1. Make the elementwise kernel in-place (no allocation). 2. Make sure F.linear uses the fastest path. 3. Consider using `torch.matmul(x, self.linear.weight.t()) + self.linear.bias` instead of `F.linear`. They do the same. But maybe `addmm` is faster because it fuses bias. In PyTorch, `F.linear` calls `torch._C._nn.linear`, which maps to `aten::linear`, which under the hood uses `addmm` if bias is present. So it's the same. 4. Consider using `torch.compile`? But we're replacing the forward with a custom CUDA kernel. `torch.compile` would likely re-optimize the graph, maybe fusing ReLU and div. But the reference isn't compiled. So we just need to be faster than the eager reference. Let me optimize the kernel further. How about I use CUB or something? No. Actually, another idea: since `divisor = 2.0`, dividing by 2.0 is equivalent to multiplying by 0.5. I can use multiplication instead of division. The compiler might already optimize this if divisor is a compile-time constant, but here it's passed as a runtime float. I can pass it as 1.0f/divisor from the CPU side and do multiplication inside the kernel. Or I can precompute the reciprocal in Python and multiplicatively apply. But wait, since divisor comes from the model init, it's stored in the model. I can compute `self.inv_divisor = 1.0 / divisor` in `__init__` and multiply. That might be slightly more efficient, though fmaxf and division with a constant may already be optimized. But changing to multiplication can help because fdiv is more expensive than fmul, even with fast_math. I'll do `factor = 1.0 / self.divisor` in `__init__` and multiply. And I'll make the kernel in-place. Let me rewrite. Also, let me consider shared memory for elementwise? For a simple in-place elementwise op, shared memory is unnecessary and would just add latency. What about using vectorized loads of float4 and ensuring we compile for sm_90? The compilation already has `-gencode=arch=compute_90,code=sm_90`. One more optimization: use `__ldg` for load-through-L1? On older GPUs, `__ldg` uses read-only cache. On H100, L1/texture are unified, but `__ldg` can still help. I'll try `const float4 val = __ldg(reinterpret_cast<const float4*>(input) + row * N4 + col4);`. But since the pointer is marked `__restrict__`, the compiler can already optimize. Adding `__ldg` explicitly might not help much, but I can try. Another idea: `launch bounds` isn't needed because occupancy is not limiting. What about using 1D blocks and 1D grids and letting each thread handle multiple columns? Actually, the current 2D grid is fine. Wait, I realize: for the CUDA kernel, if I make it in-place, I don't need two pointers. I can just do: ```cpp __global__ void fused_relu_mul_kernel(float* __restrict__ data, int M, int N, float factor) { ... val = __ldg(data_ptr); // no, load must be from same pointer } ``` Better to just load and store to the same pointer. Let me rewrite the full kernel code: cpp_source = r""" #include <torch/extension.h> void fused_relu_mul_inplace(torch::Tensor tensor, float factor); """ cuda_source = r""" #include <cuda_runtime.h> #include <torch/extension.h> __global__ void fused_relu_mul_kernel(float* __restrict__ data, int M, int N, float factor) { int col4 = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N >> 2; if (col4 >= N4) return; int row = blockIdx.y * blockDim.y + threadIdx.y; int row_stride = gridDim.y * blockDim.y; for (; row < M; row += row_stride) { float4 val = reinterpret_cast<float4*>(data)[row * N4 + col4]; val.x = fmaxf(val.x, 0.0f) * factor; val.y = fmaxf(val.y, 0.0f) * factor; val.z = fmaxf(val.z, 0.0f) * factor; val.w = fmaxf(val.w, 0.0f) * factor; reinterpret_cast<float4*>(data)[row * N4 + col4] = val; } } void fused_relu_mul_inplace(torch::Tensor tensor, float factor) { int M = tensor.size(0); int N = tensor.size(1); TORCH_CHECK(tensor.is_cuda(), "tensor must be CUDA"); TORCH_CHECK(tensor.is_contiguous(), "tensor must be contiguous"); TORCH_CHECK(tensor.scalar_type() == at::kFloat, "tensor must be float32"); TORCH_CHECK((N % 4) == 0, "N must be divisible by 4"); int N4 = N >> 2; const int blockx = 128; const int blocky = 2; dim3 block(blockx, blocky); dim3 grid((N4 + blockx - 1) / blockx, (M + blocky - 1) / blocky); fused_relu_mul_kernel<<<grid, block>>>( tensor.data_ptr<float>(), M, N, factor); } """ Python: class ModelNew(nn.Module): def __init__(...): ... self.factor = 1.0 / divisor def forward(self, x): x = F.linear(x, self.linear.weight, self.linear.bias) fused_module.fused_relu_mul_inplace(x, self.factor) return x Here I changed blockx to 128 and blocky to 2. With blockDim(128,2)=256 threads. Grid x = (2048+127)/128 = 16. Grid y = (1024+1)/2 = 512. Total blocks = 16*512 = 8192. Same total blocks as before. But now each block handles 128*2=256 rows? No, each block handles 128 float4s per row (512 floats) and 2 rows. So each block handles 1024 floats. Total blocks = (8192 floats per row / 512 floats per block per row) * 1024 rows / 2 rows per block = 16 * 512 = 8192. Same. The wave count is the same. But using blockx=128 means each warp accesses 32 float4s in a row, which is much more contiguous per warp. This might be fine. Actually, I want to INCREASE the amount of work per block to reduce block count and wave count. How about each thread handles 8 float4s (32 floats) in a column-stride loop? Or I can make block larger. For example, blockDim(256, 1) = 256 threads. Each thread handles 1 float4. Block handles 256 float4s = 1024 floats, 1 row. Grid x = (2048+255)/256 = 8. Grid y = 1024. Total blocks = 8192. Still same. To reduce blocks, each thread should process multiple float4s. Let's have each thread process 4 float4s. BlockDim(128, 1). Each block handles 128*4 = 512 float4s = 2048 floats per row. For N=8192, that's 4 blocks in x. Grid y = 1024. Total blocks = 4096. Wave count roughly halved. Let's do that. ```cpp __global__ void fused_relu_mul_kernel(float* __restrict__ data, int M, int N, float factor) { int tid = blockIdx.x * blockDim.x + threadIdx.x; int N4 = N >> 2; int total_elems4 = M * N4; int num_threads = gridDim.x * blockDim.x; for (int idx = tid; idx < total_elems4; idx += num_threads) { float4 val = reinterpret_cast<float4*>(data)[idx]; val.x = fmaxf(val.x, 0.0f) * factor; val.y = fmaxf(val.y, 0.0f) * factor; val.z = fmaxf(val.z, 0.0f) * factor; val.w = fmaxf(val.w, 0.0f) * factor; reinterpret_cast<float4*>(data)[idx] = val; } } ``` Wait, this is a 1D grid. Each thread processes elements in a strided fashion across all rows. The memory access pattern: threads in a warp access consecutive float4s (idx, idx+1, idx+2...). That is perfect coalescing. For idx = tid; idx < total_elems4; idx += num_threads. Total_elems4 = 1024 * 2048 = 2,097,152. If blockDim.x = 256, and gridDim.x = 132 * 8 = 1056 (to fill all SMs), total threads = 270,336. Each thread processes 2097152 / 270336 ≈ 7.75 elements. So 8 elements per thread. That's a good amount. We can set gridDim.x = num_SMs * occupancy. H100 has 132 SMs. Max threads per SM = 2048. With 256 threads per block, 8 blocks per SM. 132*8 = 1056 blocks concurrently. To saturate memory, we need many warps. We can launch e.g., gridDim.x = 132 * 16 = 2112 blocks. Or just use a heuristic. For a simple elementwise kernel, the number of blocks isn't critical as long as enough to fill SMs. Let's use a large gridDim.x to ensure the GPU is fully occupied. For example: ```cpp const int block_size = 256; int num_blocks = (total_elems4 + block_size - 1) / block_size; ``` num_blocks = (2097152 + 255) / 256 = 8192. That's the same as before. But since it's 1D, each thread in the block processes elements in a grid-stride loop. Actually, if we use the standard 1D grid-stride loop, each block handles a contiguous chunk? No, in the code above, each thread does `idx += num_threads`. This means block 0 thread 0 processes 0, 8192, 16384... This scatters the blocks across the array, which may hurt L2 locality if too many blocks. But actually, since memory is streaming, L2 locality might not matter much. However, it could be better to have each warp process contiguous elements to maximize cache line utilization. Let's use a different approach: block-strided loop where each thread processes contiguous elements with stride `blockDim.x * gridDim.x`. ```cpp int idx = blockIdx.x * blockDim.x + threadIdx.x; int stride = blockDim.x * gridDim.x; for (int i = idx; i < total_elems4; i += stride) { ... } ``` This is what I wrote above. Within a warp, thread t accesses element `idx + t`, which are contiguous. So it's perfectly coalesced. Total threads = 8192 * 256 = 2,097,152 threads. Wait, if num_blocks = 8192 and blockDim=256, total threads = 2,097,152, which equals total_elems4. So each thread does exactly 1 element! No grid-stride loop needed. But then block count is 8192. Same as before. If I reduce num_blocks to e.g., 4096, blockDim=256, total threads = 1,048,576. Each thread processes 2 elements. That halves block count. Or use blockDim=512, num_blocks = 4096 -> total threads = 2,097,152, again 1 element each. The key to reducing block count is to have each thread process multiple elements, but the launch configuration `num_blocks = (total_elems4 + block_size - 1) / block_size` inherently creates at least `total_elems4 / block_size` blocks. To have fewer blocks, I need to not launch enough blocks to cover the whole array. Then the grid-stride loop kicks in. For example, I can launch `num_blocks = min(8192, some_sm_based_number)`. A common heuristic for memory-bound kernels is `num_blocks = number_of_SMs * 8` or something. For H100, 132 * 8 = 1056 blocks (since max 8 blocks per SM with 256 threads). Total threads = 1056 * 256 = 270,336. Each thread processes 2097152 / 270336 ≈ 7.75 float4s. So I could launch 1056 blocks and the grid-stride loop handles the rest. This would reduce block count from 8192 to 1056, which is much better for launch overhead. But the total kernel time is only ~20 microseconds. Block launch overhead per block on H100 is maybe 1-2 us? No, block launch overhead is amortized; the GPU can launch blocks very fast in hardware. But there is still some host-side overhead for each block. Actually, on modern GPUs, the block scheduler can launch blocks very quickly. 8192 blocks for a tiny kernel might be okay. But if each block does tiny work, the per-block fixed overhead (synchronizing warps, etc.) can dominate. With only 1 float4 per thread and 256 threads per block, each block does 256 float4 loads and stores = 4096 bytes. There are 8192 blocks. If each block does very small work, the time to issue all blocks and the tail latency of the last waves becomes significant. So it's better to have larger granularities. I think launching ~1056 blocks with ~8 float4s per thread is better. Let me use a heuristic: number of SMs * 8. But I don't know the SM count at runtime in pure CUDA easily. But I can compute: `int num_blocks = 132 * 8;` since I know it's H100. But for portability, I'll compute based on device or use `cudaOccupancyMaxActiveBlocksPerMultiprocessor`. That's too much for inline. A simpler heuristic: (total_elems4 + block_size - 1) / block_size is standard. 8192 blocks might be fine. PyTorch launches similar numbers of blocks for elementwise kernels. So maybe it's okay. Michelangelo's razor: since the elementwise part is a tiny fraction of total time, and launching more blocks is fine, I'll just use the standard `num_blocks` and keep it simple. The main speedup comes from fusing the two elementwise ops into one kernel and doing it in-place. Let me use the simpler 2D grid kernel but make it in-place. Actually, I'll go back to the 2D grid but with blockDim(128, 2) = 256 threads. Grid x = (2048+127)/128 = 16. Grid y = (1024+1)/2 = 512. Total blocks = 8192. Each block handles 2 rows * 128*4 floats = 1024 floats. Total work per block is same. But the access pattern within a warp: warp 0 has threads (0..31) in x, same y. They access contiguous float4s. Good. Another thing: I should use `fmaxf` and multiplication by factor. Also, I should mark the kernel with `__forceinline__` but it's already a `__global__` and the compiler will inline anyway. Let me also consider using `torch.relu_` and `x.mul_(1/divisor)` in PyTorch to do in-place operations. That would fuse? No, `torch.relu_` is in-place ReLU, then `x.div_(self.divisor)` is in-place division. That still launches two kernels but avoids allocations. But my custom fused kernel is still more efficient because it only launches one kernel. Actually, PyTorch might fuse `relu_` and `div_` via the NNC or NVFuser if it's in a single graph? In eager mode, probably not. What if I just do `x = F.linear(...); x = torch.relu_(x); x.div_(self.divisor); return x`? That might already be close to my custom kernel, since it avoids allocations and uses PyTorch's optimized elementwise kernels. PyTorch's elementwise kernels are very well optimized (vectorized, etc.). Fusing into one kernel only saves one kernel launch (maybe 5us). So it's not a huge difference. Hmm, then maybe my custom kernel isn't enough to beat the reference significantly if I only fuse two elementwise ops. The reference has 3 kernels (addmm, relu, div). Mine has 2 kernels (addmm, fused_relu_div). The difference is one kernel launch and one memory pass. In many benchmark settings, that's enough if the reference isn't already doing it, and the tolerance is tight. But on H100 with very fast memory, saving 20us on a ~200us total might give 10% speedup. The problem says "faster than the reference." 10% faster should be enough. But I can try to squeeze more: use `torch.matmul` instead of `F.linear`? `F.linear` may have some overhead (e.g., dtype checking). But it's minimal. Another idea: use `torch.addmm(self.linear.bias, x, self.linear.weight.t())` but with `out=` kwarg to a pre-allocated buffer? I can't easily pre-allocate because the input shape may vary per forward call. Wait, in the problem, batch_size is fixed at 1024. So I could allocate a persistent buffer in `__init__`! That's a huge win because `addmm` won't need to allocate its output tensor in `forward`, which removes allocation overhead and the associated CUDA synchronization. Let me do that. In `__init__`: ```python self.register_buffer('output_buf', torch.empty(batch_size, out_features, dtype=torch.float32)) ``` But wait, `batch_size` is not passed to `__init__` in `get_init_inputs`. Only `in_features`, `out_features`, `divisor`. So I can allocate a buffer of the right shape. However, `forward` receives `x` which is torch.rand(batch_size, in_features). The batch size may be fixed in this benchmark but could vary in other calls? Let's check get_inputs: it returns `[torch.rand(batch_size, in_features)]`. So batch_size is fixed. And `forward` gets `x` as that tensor. Actually, the reference implementation creates a new tensor on every forward pass. If I pre-allocate a persistent buffer and use `out=` parameter for `torch.addmm`, I save the allocation overhead. But `torch.addmm(bias, x, weight.t(), out=output_buf)` writes to output_buf. Then I apply my in-place fused kernel on output_buf and return it. Wait, `torch.addmm` with `out=` requires the output tensor to be the correct shape. If I pre-allocate `self.output_buf` in `__init__`, I need to know the batch size. The problem gives `batch_size = 1024` globally. But is `batch_size` accessible in the module? I can hardcode it if it's defined in the global scope, but better to be flexible. Actually, when the file is executed, `batch_size` is in the global namespace, but when `get_init_inputs` is called and `ModelNew` is instantiated, does the file have `batch_size`? The reference code is: ```python batch_size = 1024 in_features = 8192 out_features = 8192 divisor = 2.0 def get_inputs(): return [torch.rand(batch_size, in_features)] def get_init_inputs(): return [in_features, out_features, divisor] ``` So `batch_size` is a global in the module. My `ModelNew` is defined in the same module (since it's exec'd), so I can reference `batch_size` in `__init__`. However, to be safe, I can allocate the buffer lazily on first forward call based on the input shape, or just allocate it based on the known batch_size. I'll allocate based on input shape to be robust. Alternatively, I can allocate in forward if the buffer doesn't exist or is the wrong shape. But actually, the biggest overhead might not be allocation. PyTorch uses a caching allocator, so allocation is often just looking up a cached block. But for a new tensor of size 1024*8192*4 = 33MB, it might take some time to find/allocate, but likely not much. Still, pre-allocation is a good optimization. Wait, but I need to return the buffer from forward. However, if I use an in-place operation on the buffer, and then later forward calls also use the same buffer, the autograd graph might get screwed up if gradients are tracked. But this is inference (no gradients?), and the model is a forward-only bench. But even for autograd, modifying an in-place buffer can be problematic if gradients are needed. However, in this benchmark, we only care about forward speed and numerical correctness of output. Since the output is consumed immediately, using a persistent buffer is okay for inference but might break autograd. Since the problem says "Replace the forward computation", and doesn't mention training, forward-only is fine. But to be safe, I should not break autograd if possible. Actually, if I do `out = self.output_buf.clone()` that defeats the purpose. But if I just return `self.output_buf`, and the next forward call overwrites it, the previous output tensor gets mutated. This is okay if the caller doesn't hold onto the tensor. In a benchmark, they probably do hold it for correctness check. For correctness check, they call forward once per trial and compare. The reference output is held, my output is held. So returning a persistent buffer is fine as long as each forward call writes fresh values. But if they run multiple in parallel (unlikely), there would be a race. I think it's safer to allocate on-the-fly. PyTorch's caching allocator is fast. Let me go with the fused in-place kernel and see if it's enough. Actually, I want to see if I can speed up the matmul too. Could I use CUTLASS or custom MMA? Let's think if there's any other way. Since the matrix multiplication is M=1024, K=8192, N=8192, it's a classic GEMM. cuBLAS on H100 is highly optimized for this size. It's unlikely I can beat it with a custom kernel unless I fuse the epilogue, which saves memory traffic. Fusing epilogue into matmul is the only meaningful optimization beyond what I have. But custom GEMM is very complex and risky for correctness. What about using `torch.matmul` with a custom epilogue via torchinductor/triton? I can't use Triton because the instruction says to use `load_inline` with CUDA C++. Wait, actually, I could write a much simpler tiled matmul kernel specifically for this problem size that uses shared memory, but it would still be far slower than cuBLAS for GEMM of this size. For H100, cuBLAS uses specialized kernels with warp-specialization and async copy, achieving near-peak performance. I cannot compete. So my strategy is: 1. Keep `F.linear` (cuBLAS addmm). 2. Fuse ReLU and div into a single in-place CUDA kernel. 3. Use multiplication by reciprocal instead of division. 4. Optimize the elementwise kernel launch config and vectorization. I think that's the best I can do. Let me also try to use `torch.nn.functional.relu` with inplace=True? No, that still launches a separate kernel. What if I use `x = F.linear(x, self.linear.weight, self.linear.bias)` then `x = fused_module.fused_relu_mul_inplace(x, self.factor)`? That's my plan. Let me write the final code. I'll also add ` torch.backends.cuda.matmul.allow_tf32 = True` in the module init to ensure matmul uses TF32? Wait, doing that might make my result numerically different from the reference if the reference also uses TF32. Actually, the reference uses PyTorch default. If the default is True, then both will use TF32. If False, then my setting it to True will make my results slightly different (more TF32 error) but potentially faster. However, I need numerical equivalence within 1e-4. TF32 vs FP32 difference is typically within 1e-3 for large GEMMs. So enabling TF32 if reference doesn't use it would likely cause the correctness test to fail. So I should NOT change it. I should use whatever PyTorch default is. Actually, the reference implementation just uses `self.linear(x)`. It doesn't explicitly set any backend flag. So I should just do the same. One more thought: maybe the reference implementation is slow because it uses `nn.Linear` which is not as optimized as directly using cuBLAS with transposed weight, and PyTorch's elementwise ops might not be vectorized properly for this size? No, PyTorch elementwise ops are very well vectorized. But let's trust that fusing ReLU + div and doing it in-place with vectorized float4 gives enough speedup. I will also try to tune the block size. With a 1D block of 256 threads and using a grid-stride loop over the entire flat array, and launching a smaller number of blocks (like 132*16=2112), each thread processes ~4 float4s. Let's code that. ```cpp __global__ void fused_relu_mul_kernel(float* __restrict__ data, int count4, float factor) { int idx = blockIdx.x * blockDim.x + threadIdx.x; int stride = gridDim.x * blockDim.x; for (int i = idx; i < count4; i += stride) { float4 val = reinterpret_cast<float4*>(data)[i]; val.x = fmaxf(val.x, 0.0f) * factor; val.y = fmaxf(val.y, 0.0f) * factor; val.z = fmaxf(val.z, 0.0f) * factor; val.w = fmaxf(val.w, 0.0f) * factor; reinterpret_cast<float4*>(data)[i] = val; } } ``` Launch: choose num_blocks to saturate SMs. A good heuristic for elementwise kernels: `num_blocks = min(65535, (count4 + block_size - 1) / block_size)`? Actually on modern GPUs we can launch up to 2^31-1 blocks in x. So we can just launch `(count4 + block_size - 1) / block_size` which is 8192. But we said reducing blocks might help. For a very fast kernel, 8192 blocks might be okay. Actually, maybe I should launch exactly enough blocks to keep all SMs busy but not too many. Since max active blocks per SM is 32, but limited by threads to 8 for 256-thread blocks. So 132 * 8 = 1056 blocks concurrently. If I launch 8192 blocks, there are ~8 waves. Wave scheduling overhead is tiny on H100. But each wave has to finish before the next? No, blocks are scheduled as SMs become free. For a memory-bound kernel, SMs become free very quickly (since each block is short). So waves are pipelined. The overhead of many waves is the tail: the last wave has fewer blocks, so some SMs are idle. With 8192 blocks and 1056 concurrent, tail has 8192 - 7*1056 = 8192 - 7392 = 800 blocks. 800 blocks spread over 132 SMs ≈ 6 blocks per SM, which is fine. But each block only has 256 threads doing 1 float4 each. That's maybe 10-20 instructions. The block launch overhead might be comparable to execution time. So fewer blocks with more work per thread is better. I'll launch a heuristic number of blocks. Since I don't know the SM count at runtime without a cudaGetDeviceProperties call, I can hardcode it for H100? That's brittle. Or I can just use `min(8192, 2048)` or something. Actually, many optimized elementwise kernels in PyTorch use `num_blocks = (N + block_size - 1) / block_size`. So 8192 blocks is standard and shouldn't be a problem for PyTorch's performance. Let me just use the simple flat 1D approach with 8192 blocks or 2D approach. I'll go with the 2D approach because it was already correct and compiled. But wait, I can improve the 2D kernel by having each thread process multiple float4s along the x-axis. This reduces total blocks. For example, each thread processes 4 float4s (stride by gridDim.x * blockDim.x in the x-direction). Let's do that in a 2D grid. ```cpp __global__ void fused_relu_mul_kernel(float* __restrict__ data, int M, int N, float factor) { int col4_start = blockIdx.x * blockDim.x * 4 + threadIdx.x * 4; // no, messy } ``` Maybe flat 1D is simpler and more efficient. Let's use flat 1D with `block_size = 256` and `num_blocks = (count4 + block_size - 1) / block_size`. For count4 = 2,097,152, num_blocks = 8192. Actually, I can also vectorize by float4 and further unroll by 2 or 4 inside the loop: ```cpp for (int i = idx; i < count4; i += stride) { float4 val = __ldg(reinterpret_cast<const float4*>(data) + i); ... reinterpret_cast<float4*>(data)[i] = val; } ``` If I use `__ldg`, I need a `const float*` for loading. But since it's in-place, `data` is non-const. I can use `__ldg((const float4*)(data + i*4))`. But I can just load normally; `__restrict__` will help the compiler optimize. One more micro-optimization: compute `one_over_factor` in C++? No, multiplication by `factor` where factor = `1.0/divisor`. Wait, if `divisor = 2.0`, factor = 0.5. The compiler may already optimize `val.x / 2.0f` to `val.x * 0.5f`, but only if divisor is a compile-time constant. Since it's passed as an argument, I'll compute `inv_divisor = 1.0 / divisor` in Python and pass that to the kernel. I already planned that. Alright, let me write the final improved kernel with in-place update and try to compile + correctness check.