We need to write a custom CUDA kernel for Softmax along dim=1 on a large matrix: batch_size=4096, dim=393216. The input shape is (4096, 393216). So we need to compute softmax per row. The reference uses torch.softmax(x, dim=1). Since dim is huge (393k), each row is huge. The naive approach: compute max, subtract, compute exp, sum, divide. Since dim is huge and batch_size is moderate (4096), we can process each row with a large number of threads. Due to the large dimension, we need to process elements in chunks. We need to write an efficient CUDA kernel. Let's consider the GPU specs first. Use get_gpu_specs to know what we're dealing with. Then design kernel. Given the large dimension, we can have each block process one row, with blockDim.x maybe 1024 or 512 threads. Each thread loops over the row with stride blockDim.x, computing max in shared memory. Then compute exp and sum in another pass. Then compute output in final pass. But with dim=393216, each block of 1024 threads processes 384 elements per thread (393216/1024 = 384). That's okay. However, 393216 is 3*131072 = 3*2^17. Actually 393216 = 3 * 131072 = 3*2^17. Not a multiple of 1024? 393216 / 1024 = 384 exactly. Yes, 1024*384 = 393216. So it's divisible by 1024. Good. But if we have 4096 rows, and each row gets one block of 1024 threads, total blocks = 4096. With SMs maybe 80-128. That's enough occupancy? Each block has 1024 threads = 32 warps. If register count is low, occupancy might be limited by shared memory or registers. We need to be careful: using shared memory for reduction. For max, we can do a tree reduction. Alternatively, we can use warp-level primitives for reduction. Since 1024 threads = 32 warps, we can have each warp compute its own max and sum, then aggregate across warps using shared memory. But we need to compute max first, then sum of exp(x - max). Two passes over the data (or three passes). But memory bandwidth is likely the bottleneck anyway, so extra passes okay if we keep them fused. Because the dimension is large, the data per row is 393216 * 4 bytes = 1.5 MB. With 4096 rows, total data = 6 GB. That's huge - actually 4096 * 393216 = 1,610,612,736 elements. * 4 bytes = 6.44 GB. That might exceed GPU memory for some GPUs? But the evaluation environment likely has enough memory. The reference will process that too. Memory bandwidth: Let's say the GPU has ~1-2 TB/s. 6.44 GB read + 6.44 GB write = ~12.9 GB. At 1 TB/s, that's ~13 ms. At 2 TB/s, ~6.5 ms. So we need to be close to memory bandwidth. We need to read x once, compute max, then read again? Ideally we want to read once but we need max for exp. However, we can do a two-pass algorithm: first pass compute max (read x). Second pass compute exp and sum (read x again). Third pass write output (or second pass writes output but needs sum, which is not known until after reduction). Actually, after second pass we have sum, then third pass writes output (read x again or reuse data from second pass). That's 3 reads + 1 write = 4 trips over memory. We want to minimize reads. We can do a two-pass algorithm: first pass loads x in tiles, computes per-thread max. Then reduce to block max. Then second pass loads x again, computes exp(x - max), writes to output, and accumulates sum per-thread. Then reduce sum. But after reduction, we need to normalize the output by dividing by sum. That requires a third pass over the output (or we could compute output in shared memory? Not possible due to size). So we need to either: Option A: 3 passes: 1) compute max (read x). 2) compute exp and write unnormalized output (read x, write out). 3) normalize output (read out, write out). That's 2 reads of x + 2 reads/writes of out = 4 memory transfers per element. Option B: Use a two-pass where after computing max, we compute exp and store in output, then after reduction we normalize in place. That's 2 reads of x + 1 read/write of out for normalization? Wait: Pass 1: load x chunks, compute max. (read x) Pass 2: load x chunks, compute exp, write to out. (read x, write out) Pass 3: load out chunks, divide by sum, write back. (read out, write out) Total: 2 reads x + 2 reads out + 2 writes out = 4 memory ops per element. Actually each memory op is 4 bytes. So 16 bytes per element. For 1.6B elements, that's ~25.8 GB. At 1TB/s, 25.8 ms. Hmm. We could possibly combine pass 2 and 3 by doing a prefix? No, sum is global per row. Could compute max, then re-read x, compute exp, store in output, and also store exp in some temporary array? That doesn't help. Alternatively, we can compute max and sum in one pass using online softmax algorithm? Actually, there is an algorithm to compute softmax in a single pass using a running max and sum, but due to numerical stability, you can do it with two passes: first max, then exp and sum. But to write output, you need the final sum. You can write output in a second pass after sum is known, but that still requires reading x again. Wait: Could we compute max and then do a second pass that computes exp, accumulates sum into shared memory, and writes output? But we need sum to normalize. If we do a block-level sum reduction, the sum is known at the end of the block. However, the threads are writing output as they compute exp, but they don't know the final sum yet. Unless we first compute exp and store in shared memory or registers? The dimension is 393216 elements, which is too large for shared memory (max 48KB per block). So we can't store the entire row in shared memory. But we can process the row in tiles: for each tile, we compute max for the tile? No, we need global max for the row to compute exp accurately. Alternatively, we can compute max in pass 1. In pass 2, process tiles, compute exp, but we need the global sum. We can accumulate sum in a register while writing unnormalized output. But to normalize, we need sum after all tiles are processed. That means we need a third pass over the output. Could we use persistent threads or cooperative groups to process the row? Not really. What if we process each row with multiple blocks? For example, each row is split across multiple blocks, each block computes a partial max and partial sum, then we do a cross-block reduction (via global atomic or second kernel), and then a final normalization kernel. This is typical for online softmax in large dimensions. For 4096 rows, we could have e.g., 128 blocks per row? That seems too many blocks (4096*128 = 524k blocks). But with 393216 elements per row, each block would process ~3072 elements. Actually, that's not terrible. But cross-block reduction adds complexity and global atomics. However, given that the dimension is huge, the reference PyTorch kernel probably uses a multi-pass approach as well or perhaps a persistent kernel. PyTorch's native softmax uses optimized kernels from cuDNN or its own. For large inner dimension, it might use the warp or block reduction approach. Actually, PyTorch's CUDA softmax for dim=-1 on a 2D tensor uses a kernel that processes each row with a block of threads. For large dimensions, it uses a loop to compute max and sum. It might also use vectorized loads (float4) to improve bandwidth. Given that we want to beat PyTorch, we need to at least match its bandwidth usage, likely by using vectorized loads/stores (float4) and minimizing overhead. Also note: the input is fp32 and output is fp32. So we can load 4 floats at a time using float4 (128 bits). Since 393216 is divisible by 4 (yes, 393216 / 4 = 98304), we can use float4 loads and stores. We should check GPU specs and perhaps write a kernel using float4 for memory bandwidth. Let's get specs.