The task is to replace `torch.cumsum` along dimension 1 with a custom CUDA kernel for a tensor of shape (32768, 32768). Let me analyze this: - Input shape: (32768, 32768) - Operation: cumsum along dim=1 - This means each row of 32768 elements needs a prefix sum For a 2D tensor with cumsum along dim=1, each row is independent. We have 32768 rows, each with 32768 elements. Total elements: 32768 * 32768 = 1,073,741,824 (about 1 billion elements) At fp32, that's about 4 GB of data. A simple approach would be to have each block handle one or more rows, with threads computing the prefix sum within a row. However, for large arrays, a parallel scan algorithm like Hillis-Steele or Brent-Kung is better. But given that each row is 32768 elements and we have 32768 rows, we can have each SM process multiple rows. For the H100: - 132 SMs (H100 has 132 or 144 SMs, let me check - actually H100 SXM has 132 SMs) - We need to process 32768 rows A simple and efficient approach for 1D scan per row: 1. Load a row into shared memory 2. Perform a parallel scan within the block 3. Write back For 32768 elements per row, we can use multiple blocks per row or have each block handle a portion using a multi-level scan. Actually, since we have 32768 rows and 32768 elements per row, a more efficient approach might be to process the transpose or use a block-based scan. Let me think about the memory layout. The tensor is (32768, 32768) with row-major layout. So elements in a row are contiguous in memory. This is good for coalesced access when we operate along dim=1. For a row: - 32768 elements = 131072 bytes - We need to perform prefix sum on each row Strategy 1: One block per row, with threads doing a parallel scan. For 32768 elements, if we use 256 threads per block: - Each thread processes 128 elements (32768/256 = 128) - Load into shared memory - Perform scan But 32768 floats is a lot for shared memory (32K floats = 128KB). The H100 has 228KB shared memory per SM, but max 227KB per block. So we can fit one row in shared memory if we use 128KB per block. Actually wait - 32768 floats * 4 bytes = 131072 bytes = 128KB. This is within limits. But if we want to process multiple rows per SM concurrently, we need less shared memory per block. Alternative: Process chunks of a row. Each block handles a tile of the row, then combine results. But actually, a simple approach might work well enough: - Each block handles one row - 256 threads - Each thread handles 128 elements (coalesced loads) - Tree-based scan in shared memory Wait, but we have 32768 elements and 256 threads, so each thread would need to handle 128 elements sequentially. That might be slow due to the sequential scan part. Better approach for 32768 elements per row: - Use 1024 threads per block (max) - Each thread handles 32 elements - Or use a multi-level approach Actually, for maximum performance on H100, we should consider using warp shuffle operations for intra-warp scan, then combining across warps. Let me think about an efficient 1D scan per row algorithm: For each row (handled by one block): 1. Divide into chunks processed by warps 2. Each warp does a scan of its elements using shuffle 3. The last element of each warp is stored as a warp sum 4. Scan the warp sums 5. Each warp adds its cumulative offset to its elements This is the classic two-level scan. For 32768 elements per row with 1024 threads (32 warps of 32 threads): - Each thread processes 32 elements - Each warp processes 1024 elements - 32 warps process 32768 elements Actually, let's think about this more carefully. If each warp processes 32 elements per thread * 32 threads = 1024 elements per warp. For 32768 elements, we need 32 warps. Wait, 32 warps * 1024 elements = 32768. Perfect! So algorithm per row (one block): 1. Each thread loads 32 elements from global memory (coalesced) 2. Thread does sequential scan on its 32 elements 3. Store last element of each thread in shared memory 4. Warp-level scan using shfl_sync 5. Each thread gets warp prefix, add to elements 6. Write back to global memory But wait, the thread-private scan on 32 elements might limit parallelism. Let me consider if we should reduce elements per thread. Alternative: 512 threads, each handling 64 elements. - 16 warps - Each warp handles 2048 elements - 16 warps * 2048 = 32768 Or: 256 threads, each handling 128 elements. Let me think about what works best. Actually, I think the most efficient on modern GPUs is: - Use as many threads as possible to parallelize the tree scan - For 32768 elements, if we use 1024 threads, each thread processes 32 elements Tree scan approach: 1. Load 32768 elements into shared memory (needs 128KB - too much! Actually it is 128KB which is within 227KB limit, but still high occupancy might be limited) Shared memory per block of 128KB would mean we can only have 1 block per SM at most (228KB/128KB = 1.78, so 1 block). With 132 SMs, only 132 active blocks, but we have 32768 rows. Not great occupancy. Better: Don't load the whole row into shared memory. Process in tiles. Tile-based approach: - Each block processes a tile of, say, 2048 elements - Multiple blocks per row, with block results combined But then we need a second pass to add offsets. This is a multi-pass scan. Actually, let me reconsider. For 32768 rows, if we can process each row with one block, and we have concurrent execution of 32768/132 ~ 248 waves of blocks. That's actually fine for throughput. The key question is: what's the fastest per-row scan? Let me try: 256 threads per block, each thread handles 128 elements. Shared memory: instead of loading the whole row into shared memory, we can use registers for sequential chunks and shared memory only for the warp/thread aggregates. Algorithm (256 threads per block, each thread handles 128 elements): Total elements: 256 * 128 = 32768 ✓ 1. Each thread reads 128 elements sequentially (stride of 256, coalesced across threads in a warp) Wait, for coalesced access: thread i reads elements i, i+256, i+512, ..., i+255*128 Actually, for coalesced: thread 0 reads 0, 256, 512, ... thread 1 reads 1, 257, 513, ... This is NOT coalesced. Better: thread 0 reads 0..127, thread 1 reads 128..255, etc. For coalesced load: thread i reads elements from i*128 to (i+1)*128-1. Then threads in a warp read contiguous memory. Perfect coalescing! 2. Each thread computes the sequential prefix sum of its 128 elements (in registers) Keep track of the total sum of this thread's chunk. 3. Store each thread's chunk sum in shared memory (256 floats = 1KB) 4. Perform a parallel scan on the 256 chunk sums in shared memory - Use Brent-Kung or Kogge-Stone in shared memory - Or use warp shuffles: 8 warps of 32 threads 5. Each thread adds the prefix sum from step 4 to its elements 6. Write back to global memory This uses only 1KB of shared memory! Great occupancy. Let me flesh this out more carefully. In step 4, with 256 threads = 8 warps: - Each warp can perform a scan using shfl - Then scan the 8 warp totals - Then add back For the shared memory scan of 256 thread sums: - 256 elements at 4 bytes = 1024 bytes Actually, we can do this entirely with shuffles since we have 8 warps: Warp-level scan (32 threads per warp): ``` lane = threadIdx.x % 32 warp = threadIdx.x / 32 value = thread_sum for offset in [1, 2, 4, 8, 16]: n = __shfl_up_sync(0xFFFFFFFF, value, offset) if lane >= offset: value += n ``` After this, `value` is the prefix sum within the warp. For cross-warp scan: - The last prefix of each warp needs to be scanned - We can store these 8 values in shared memory - Thread 0 of each warp loads the warp total (before adding to thread_sum?) Actually, in the classic approach: 1. Each warp does scan of its thread sums 2. warp_total[warp_id] = last_element_of_warp_scan 3. Thread 0 scans these 8 warp totals 4. Add the appropriate prefix to each warp's elements But wait, step 2: we have 8 warps, each with thread_sums[0..31]. Each warp does prefix sum. So thread 31 in warp 0 has sum of warp 0's 32 thread sums. We store this. Then threads 0..7 (or just thread 0) scan these 8 warp totals. Then when processing elements, each thread adds: - Its intra-warp prefix (from step 1) - The inter-warp prefix (the sum of all previous warps' totals) This is clean. Let me write this out more carefully: ``` // Each thread handles tid_in_block elements // No, each thread handles a chunk of input // For 32768 elements, 256 threads per block // Each thread handles 128 contiguous elements // Step 1: Read 128 elements and compute prefix sum in registers float local_prefix[128]; float sum = 0.0f; for (int i = 0; i < 128; i++) { float val = input[row * cols + threadIdx.x * 128 + i]; sum += val; local_prefix[i] = sum; } // Now 'sum' is the total sum of this thread's 128 elements // local_prefix[i] is the prefix within this thread's chunk // Step 2: Warp-level scan of thread sums int lane = threadIdx.x & 31; int warp = threadIdx.x >> 5; float warp_prefix = sum; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, warp_prefix, offset); if (lane >= offset) { warp_prefix += other; } } // Step 3: Store warp totals __shared__ float warp_sums[8]; if (lane == 31) { warp_sums[warp] = warp_prefix; } __syncthreads(); // Step 4: Scan of warp totals // Only first warp does this if (warp == 0) { float val = (lane < 8) ? warp_sums[lane] : 0.0f; for (int offset = 1; offset < 8; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) { val += other; } } if (lane < 8) { warp_sums[lane] = val; } } __syncthreads(); // Step 5: Add warp offset to each element float warp_offset = (warp > 0) ? warp_sums[warp - 1] : 0.0f; // Step 6: Write back for (int i = 0; i < 128; i++) { float final_val = local_prefix[i] + warp_offset; output[row * cols + threadIdx.x * 128 + i] = final_val; } ``` Wait, there's an issue. In step 2, `warp_prefix` after the shuffle loop is the prefix including this thread's sum. But for the scan, we want the prefix BEFORE this thread's contribution to add to its elements. Let me reconsider. The classic two-level scan: Level 1 (intra-chunk): Each thread scans its 128 elements. `local_prefix[i]` is the prefix. `sum` is the total of the chunk. Level 2a (intra-warp): Scan the `sum` values within each warp. After the scan, thread with lane `l` has the sum of chunks from lanes 0..l. Let `intra_warp_prefix` = this value. For thread `l`, the offset from previous threads in the SAME WARP is `intra_warp_prefix - sum`. (Because intra_warp_prefix includes this thread's sum, but we want the prefix before this thread's chunk.) Level 2b (inter-warp): Each warp's total. Warp w's total is the last thread's `intra_warp_prefix` in warp w. Let `inter_warp_prefix[w]` be the sum of totals from warps 0..w-1. For a thread in warp w, the offset from previous warps is `inter_warp_prefix[w]`. So total prefix before this thread's chunk: `offset = intra_warp_prefix - sum + inter_warp_prefix[warp]` Then the final prefix for element i in this thread's chunk: `final = local_prefix[i] + offset` Let me verify: - Thread 0, warp 0: intra_warp_prefix = sum0, offset = sum0 - sum0 + 0 = 0. Correct. - Thread 1, warp 0: intra_warp_prefix = sum0+sum1, offset = sum0+sum1 - sum1 + 0 = sum0. Correct. - Thread 0, warp 1: intra_warp_prefix = sum_w1_0, offset = sum_w1_0 - sum_w1_0 + total_warp_0 = total_warp_0. Correct. Great! Let me write corrected code: ``` // After reading and local prefix sum: float thread_sum = sum; // Total of this thread's 128 elements float local_prefix[128]; // local_prefix[i] is prefix within this thread // Intra-warp scan of thread_sum float warp_scan = thread_sum; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, warp_scan, offset); if (lane >= offset) warp_scan += other; } // warp_scan now = sum of thread_sums from lane 0 to this lane // We want prefix before this thread = warp_scan - thread_sum float intra_warp_offset = warp_scan - thread_sum; // Store per-warp totals __shared__ float warp_totals[8]; if (lane == 31) { warp_totals[warp] = warp_scan; } __syncthreads(); // Inter-warp scan if (warp == 0) { float val = (lane < 8) ? warp_totals[lane] : 0.0f; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) val += other; } if (lane < 8) { warp_totals[lane] = val; // Now warp_totals[lane] = sum of warp totals from warp 0 to lane } } __syncthreads(); // For warp w, prefix from previous warps: // warp_totals[w] is the sum of warps 0..w // We want sum of warps 0..w-1 = warp_totals[w] - (last warp's total) // Hmm, that's messy. // Actually, better to store after scan differently. // Alternative: store the scanned prefix at index w // warp_totals[w] = sum of totals from warp 0 to w-1 // Then inter_warp_offset = warp_totals[warp]; ``` Let me redo the inter-warp part more carefully: ``` if (warp == 0) { float val = (lane < 8) ? warp_totals[lane] : 0.0f; // val currently = total of warp lane for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) val += other; } // Now val = sum of warp totals from warp 0 to lane // We want warp_totals[lane] = sum of warp totals from warp 0 to lane-1 if (lane < 8) { warp_totals[lane] = val - ((lane < 8) ? warp_totals[lane] : 0.0f); } } ``` Wait, that's wrong too because warp_totals[lane] was overwritten or not available. Better approach: use a temporary or compute correctly. Actually, better yet: ``` if (warp == 0) { float val = (lane < 8) ? warp_totals[lane] : 0.0f; float original = val; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) val += other; } // val is prefix including this warp // prefix before this warp = val - original if (lane < 8) { warp_totals[lane] = val - original; } } ``` Let's verify: - lane 0: val = total_w0, original = total_w0, result = 0. Correct. - lane 1: val = total_w0 + total_w1, original = total_w1, result = total_w0. Correct. - lane 2: val = total_w0 + total_w1 + total_w2, original = total_w2, result = total_w0 + total_w1. Correct. Great! So the total offset for a thread: ``` float inter_warp_offset = warp_totals[warp]; float offset = intra_warp_offset + inter_warp_offset; ``` And final values: ``` for (int i = 0; i < 128; i++) { output[row * 32768 + threadIdx.x * 128 + i] = local_prefix[i] + offset; } ``` But wait, 128 registers per thread for local_prefix is a lot. At 256 threads, that's 256 * 128 * 4 bytes if stored in registers = 131KB just for the prefix arrays. Plus other state. This might exceed register file limits or cause spilling. Actually, we don't need to store all 128 prefix values in registers simultaneously if we write them out as we compute them. Let me reconsider. Alternative: We can read elements, compute running prefix, add offset, and write back in the same loop. For a sequential chunk of 128 elements: ``` float offset = ... ; // computed as before float prefix = 0.0f; for (int i = 0; i < 128; i++) { int idx = row * 32768 + threadIdx.x * 128 + i; prefix += input[idx]; output[idx] = prefix + offset; } ``` This requires no array in registers! Just a single `prefix` variable. Much better for register pressure. Let's restructure: 1. Read 128 elements, compute sum 2. Perform warp and inter-warp scans to get offset 3. Read 128 elements again, compute running prefix + offset, write back But we're reading the data twice. Is that better? Actually wait, we can do it in one pass if we don't need the data again: ``` float sum = 0.0f; for (int i = 0; i < 128; i++) { sum += input[row * 32768 + threadIdx.x * 128 + i]; } // ... compute offset ... float prefix = offset; for (int i = 0; i < 128; i++) { prefix += input[row * 32768 + threadIdx.x * 128 + i]; output[idx] = prefix; } ``` Two reads but simpler. For compute-bound operations, extra memory reads might hurt. But prefix sum is heavily memory bandwidth bound anyway. Actually, for prefix sum, the arithmetic intensity is very low. Memory bandwidth is the limiting factor. So we should aim to minimize memory traffic. But we need to read each element once and write each element once. Total memory traffic per element is 8 bytes (4 read + 4 write). However, with the two-pass approach above, we read each element twice = 12 bytes per element total. That's 50% more memory traffic. Can we do it in one pass? ``` float prefix = offset; int base = row * 32768 + threadIdx.x * 128; for (int i = 0; i < 128; i++) { prefix += input[base + i]; output[base + i] = prefix; } ``` But then we don't know `offset` until after we've seen all elements. So we need a two-pass approach per row if we want correct prefix sums. Unless we don't need block-level parallelism, which we do. Wait, actually there IS a way. We could process the chunks and write them out, then do a correction pass. But that requires another kernel launch. Or: we store the 128 elements temporarily. We only need 128 floats per thread if we store them in registers. 128 * 4 bytes = 512 bytes. At 256 threads, that's 131072 bytes = 128KB of registers. That might be too much. Actually, in the sequential pass approach: Pass 1: Read 128 elements, compute total sum of chunk Intra/inter warp scan on sums Pass 2: Read 128 elements, compute prefix + offset, write Two passes over the same data, but the data might still be in L2 cache from pass 1, so pass 2 might be fast. For 32768 elements * 4 bytes = 128KB per row. The H100 L2 cache is 50MB. It can hold about 400 rows. With 32768 rows, there's locality benefit but limited. Still, reading from L2 is much faster than HBM. Actually, for 32768 rows and shared memory based approaches, let me think about occupancy more carefully. If each block uses 256 threads and only ~1KB of shared memory, we can have many blocks per SM. With 2048 threads per SM max, that's 8 blocks per SM. With 132 SMs, that's 1056 concurrent blocks. But we have 32768 rows, so that's ~31 waves. That's decent. But wait, 256 threads * 32 registers = 8192 registers. With 64K registers per SM, that's 8 blocks per SM max. So we could have 8 blocks per SM = 1056 concurrent blocks. Each block reads 128KB (one row). At 1056 concurrent blocks = 135MB of reads in flight. That's significant. Actually, let me think about register usage. If each thread uses ~32 registers, we get good occupancy. But if we loop over 128 elements with a couple of variables and index computation, it might be fine. Wait, we might want to consider a different approach entirely. Since the tensor is square (32768, 32768), we could transpose and use a different memory access pattern. But no, the operation is along dim=1 (rows), and row-major means this is contiguous - already optimal. Another consideration: should we use vectorized loads/stores? float4 loads would load 4 floats at once. For 128 elements per thread, that's 32 float4 loads. This could improve memory bandwidth utilization. Let's try to incorporate float4: ``` float4* input_vec = (float4*)(input + row * 32768 + threadIdx.x * 128); float4* output_vec = (float4*)(output + row * 32768 + threadIdx.x * 128); float sum = 0.0f; for (int i = 0; i < 32; i++) { float4 val = input_vec[i]; sum += val.x + val.y + val.z + val.w; } ``` But we can't easily do prefix sums with float4. We'd need to expand them. Still, for the first pass (computing sum), float4 is good. For the second pass: ``` float prefix = offset; for (int i = 0; i < 32; i++) { float4 val = input_vec[i]; prefix += val.x; output_vec[i].x = prefix; prefix += val.y; output_vec[i].y = prefix; prefix += val.z; output_vec[i].z = prefix; prefix += val.w; output_vec[i].w = prefix; } ``` This writes float4 which is coalesced and efficient. For the float4 to work, each thread's chunk must be 128 * 4 = 512 bytes aligned, and thread 0 starts at offset 0, thread 1 at offset 512, etc. Since threadIdx.x * 128 * 4 = threadIdx.x * 512 bytes. For 32 threads in a warp: offsets 0, 512, 1024, ... 15872. Each warp reads 16KB. The load granularity is 32 bytes (one cache line), and consecutive threads in a warp access addresses 512 bytes apart. That's not coalesced into a single transaction - it requires 32 separate 32-byte transactions for 128 floats = 512 bytes per thread = 16 32-byte segments per thread. Wait, that's terrible. For float4 coalescing, threads in a warp should access consecutive float4s. Thread i should access address base + i * sizeof(float4). So if we want float4: - Warp reads 32 * 4 = 128 floats in one instruction (if aligned, all within cache lines) - Total elements per warp: 128 floats - For 32768 elements, with 32 warps: 32 * 128 = 4096 floats per block... wait, that's wrong. Let me recalculate. 32 threads per warp, each reading 1 float4 = 4 floats. Total per warp = 128 floats. For 256 threads = 8 warps. Per block = 1024 floats. But we need 32768 floats per row. So each block only handles 1024 floats = 1/32 of a row. Unless each thread reads many float4s: - Each thread reads 32 float4s = 128 floats - Per warp: 32 * 128 = 4096 floats - Per block (8 warps): 32768 floats ✓ But for coalescing: thread i in a warp reads float4 at indices i*32, i*32+1, ..., i*32+31? No. For coalesced float4: Thread 0 reads float4 at index 0 (floats 0-3) Thread 1 reads float4 at index 1 (floats 4-7) ... Thread 31 reads float4 at index 31 (floats 124-127) Then in the next iteration, thread 0 reads float4 at index 32 (floats 128-131), etc. So each thread's floats are NOT contiguous. They're spread out across the row. For prefix sum, contiguous chunks per thread is much simpler. With float4 and coalesced access, the elements per thread are interleaved with other threads. Can we still compute prefix sum efficiently? For a block of 256 threads handling 32768 elements: - The row is divided into 256 chunks of 128 elements each. - Thread 0 handles elements 0..127 - Thread 1 handles elements 128..255 - etc. With coalesced float4: - Iteration 0: thread i reads elements [i*128 : i*128+3] as a float4 This is NOT coalesced because thread 0 accesses bytes 0..15, thread 1 accesses 512..527, etc. For coalesced access within a warp: - Iteration 0: thread i reads elements [i*4 : i*4+3] Thread 0: 0..3, thread 1: 4..7, ..., thread 31: 124..127 This is coalesced! - Iteration 1: each thread reads the next float4 Thread 0: 128..131, thread 1: 132..135, etc. With this pattern, each thread's elements are at indices: threadIdx.x * 4 + iteration * 128, threadIdx.x * 4 + iteration * 128 + 1, etc. Each thread handles 128 elements, but they're spread across the row in chunks of 4 floats, interleaved by warp size. This is more complex for prefix sum because each thread's elements are not contiguous. We'd need a global reordering or a different algorithm. For simplicity, maybe it's better to stick with contiguous chunks and accept slightly suboptimal coalescing. With 256 threads, consecutive threads access memory 512 bytes apart. For an H100 with 32-bit loads, a warp issues 32 separate 128-byte transactions? Actually, H100 has L2 cache line of 128 bytes. So thread 0 accesses bytes 0..3, thread 1 accesses 512..515, etc. Each thread hits its own cache line (or at least a different sector). So a warp load issues 32 memory requests. That's bad. But wait - consecutive threads in a warp accessing memory 512 bytes apart means each 32-byte sector might be touched by only one thread. Actually, each 128-byte cache line holds 32 floats. Thread 0 touches floats 0..3 (cache line 0), thread 1 touches floats 128..131 (cache line 128), etc. So indeed 32 separate transactions of at least 32 bytes each. Not great for bandwidth. For better coalescing, we should assign elements so consecutive threads access consecutive memory. That means: - Thread 0: elements 0, 256, 512, ... (stride = num_threads) - This is also bad for prefix sum because elements are scattered. Hmm. Actually for prefix sum along a row, the optimal memory pattern is sequential within each thread's chunk, but that hurts coalescing. Wait, let me think again. If each thread processes a contiguous chunk: - Thread 0 reads 0..127 - Thread 1 reads 128..255 - etc. In a warp (threads 0..31), the reads are: - Thread 0: 0..3 (first iteration of 4-float block) - Thread 1: 128..131 - ... - Thread 31: 3968..3971 For a warp, the addresses span 0 to ~4000 bytes in steps of 512 bytes. On H100, memory coalescing groups threads into subgroups. For a warp reading with this pattern, each thread's 4 floats might require separate 32-byte transactions. So the warp issues ~32 transactions for 128 floats = 512 bytes. Efficient utilization: 512 bytes requested, 32*32 = 1024 bytes transferred = 50% efficiency. Not terrible, but not great. Actually, for cache line sizes: if a warp accesses bytes spaced 512 apart, each 128-byte cache line is accessed by at most one thread. The hardware might fetch 32 bytes per thread (minimum granularity). So 32 transactions * 32 bytes = 1024 bytes fetched for 512 bytes requested = 50% efficiency. If we use 32 threads per warp and have each thread stride by 32: - Thread 0: 0, 32, 64, ..., which is 0..3, then 32..35, etc. This is also spread out. For prefix sum, contiguous chunks are algorithmically convenient. Let me see if we can make coalescing work by reversing the assignment. What if we have each warp handle a contiguous region? - Warp 0 (threads 0..31): elements 0..1023 (32 threads * 32 elements each) Thread 0: 0..31 Thread 1: 32..63 ... Thread 31: 992..1023 This is coalesced! Each thread accesses consecutive float4s aligned with warp stride. - Warp 1: elements 1024..2047 - ... - Warp 31: elements 31744..32767 With 1024 threads per block = 32 warps: - Each warp handles 1024 elements - Block handles 32768 elements = 1 row Each thread handles 32 elements (1024 / 32 = 32). This is much better! Let's restructure: - 1024 threads per block (32 warps) - Each warp handles 1024 contiguous elements - Each thread in a warp handles 32 contiguous elements Memory access is coalesced within each warp. Algorithm per block (one row): 1. Each thread reads 32 elements (as float4 * 8 or float * 32) from its position in the warp's region 2. Compute local prefix sum within thread's 32 elements (in registers) 3. Compute thread sum 4. Intra-warp scan of thread sums using shfl 5. Store warp totals in shared memory 6. Inter-warp scan 7. Each thread computes final prefix and writes 32 elements Wait, but if each warp handles 1024 elements, and there are 32 warps in a block, each block handles 32768 elements = 1 row. Each warp handles a contiguous 1024-element chunk. But the chunks are also contiguous across warps. For memory access: - Warp 0 handles elements 0..1023 - Thread 0: 0..31 - Thread 1: 32..63 - ... consecutive and coalesced - Warp 1 handles elements 1024..2047 - Thread 32: 1024..1055 - ... This is coalesced because within a warp, consecutive threads access consecutive memory. But what about shared memory usage? We only need: - warp_totals[32] for inter-warp scan = 128 bytes - Maybe nothing else! That's excellent for occupancy. 1024 threads per block * max 8 blocks per 2048-thread SM = 2 blocks per SM max (due to thread count). 64K registers / 1024 threads = 64 registers per thread. If each thread uses ~32-48 registers, we might have 2 blocks per SM. Actually H100 can have up to 2048 threads per SM and 64K registers. If each thread uses 32 registers, that's 32K registers per block, so 2 blocks per SM. With 132 SMs * 2 blocks = 264 concurrent blocks. 32768 rows / 264 ≈ 124 waves. Still decent. Actually, I think 256 or 512 threads per block might be better for occupancy. Let me reconsider. With 512 threads (16 warps): - Each warp handles 2048 elements? No, 32768 / 512 = 64 elements per thread. - Wait, let's recalculate. If block handles 1 row = 32768 elements: - 512 threads * 64 elements/thread = 32768. ✓ With 64 elements per thread, stored as 16 floats. Or as float4 * 16. Within a warp of 32 threads: - Each thread handles 64 elements - Warp handles 2048 elements - Thread 0: elements 0..63 - Thread 1: elements 64..127 - ... Each thread's memory access is 64 contiguous floats = 256 bytes. Consecutive threads access consecutive 256-byte regions. This is still very coalesced. For warp-level scan: - 32 threads in warp, each with a sum of 64 elements - Intra-warp scan of 32 sums - Warp totals: 16 warps = 16 values - Inter-warp scan of 16 values This is manageable. Actually, the warp assignment is: warp 0 handles elements 0..2047, warp 1 handles 2048..4095, etc. Wait, no. With 512 threads = 16 warps. Each warp has 32 threads. Each thread handles 64 elements. So each warp handles 32*64 = 2048 elements. 16 warps * 2048 = 32768. Warp 0: elements 0..2047 Warp 1: elements 2048..4095 ... Warp 15: elements 30720..32767 Within warp 0: Thread 0: 0..63 Thread 1: 64..127 ... Thread 31: 1984..2047 This is coalesced. For the inter-warp scan, we have 16 warp totals. We need to scan them. Let me think about whether 512 threads or 1024 threads is better. With 1024 threads: - 32 warps - Each thread handles 32 elements - 2 blocks per SM max (thread limited) - But register pressure: each thread needs fewer registers With 512 threads: - 16 warps - Each thread handles 64 elements - 4 blocks per SM max (thread limited) - But each thread needs more registers for its loop I think 512 threads with 64 elements each might be a sweet spot. But actually, since the operation is memory bandwidth bound, having more blocks (higher occupancy) helps hide latency. Let me try 512 threads first. Actually, I want to also consider: what if we process the transpose? No, dim=1 is contiguous, so that's already optimal. Another thought: H100 has 3.35 TB/s bandwidth. Our data is 4GB input + 4GB output = 8GB of memory traffic for the theoretical minimum (1 read + 1 write per element). At 3.35 TB/s, 8GB takes 2.38 ms. Reference cumsum might be doing something less optimal, maybe multiple passes or atomic operations. A clean custom kernel should be competitive. Actually, `torch.cumsum` on CUDA might be implemented with a parallel scan algorithm, perhaps using CUB or similar. For large 2D arrays, it could be efficient. To beat it, we need to be clever. One thing: maybe `torch.cumsum` along dim=1 for large arrays does a transpose, cumsum on dim=0, then transpose back? Or it might use a library that's well-optimized. Let me think about register pressure more carefully. If we process 64 elements per thread with float4: - 16 float4 reads per thread - Sum them up in a loop - Second loop: read 16 float4, compute prefix + offset, write 16 float4 We need variables: - `thread_sum` (1 float) - `base_idx` (1 int) - Loop counter (1 int) - `prefix` / running sum (1 float) - For float4: `val` (4 floats, but typically in 4 registers or a vector register) So maybe ~10-15 registers per thread. Plus some for the warp scan. That's very light. Let me write out the kernel for 512 threads per block, 64 elements per thread, using float4. Wait, 64 elements = 16 float4s. For coalescing, within a warp: Thread i in warp w accesses: Base for warp w: w * 2048 Offset for thread i: i * 64 = address of (w * 2048 + i * 64) In float4 indices: (w * 2048 + i * 64) / 4 = w * 512 + i * 16 So thread i loads float4 at index `w * 512 + i * 16 + j` for j = 0..15. Wait, consecutive threads i and i+1: - Thread i: base + i*64 to base + i*64 + 3 (first float4) - Thread i+1: base + (i+1)*64 to base + (i+1)*64 + 3 The difference is 64 floats = 256 bytes. So consecutive threads access memory 256 bytes apart. In a warp, threads 0..31 access bytes 0..15, 256..271, 512..527, ..., 7936..7951. This spans from byte 0 to byte 7951 + 16 = 7968 in a block. The H100 coalescing unit groups threads by addresses. For addresses spaced by 256 bytes, a warp might issue multiple transactions. Actually, coalescing depends on whether the addresses fall within the same cache line or can be grouped. With 128-byte cache lines: - Thread 0: line 0 (bytes 0-127) - Thread 1: line 2 (bytes 256-383) - Thread 2: line 4 (bytes 512-639) - ... So each thread hits its own cache line, but the load granularity is 32 bytes. So 32 transactions of 32 bytes each = 1024 bytes fetched for 512 bytes requested (32 threads * 16 bytes per float4). That's 50% efficiency. If we use 1024 threads per block: - Each warp handles 1024 elements = 4096 bytes - Thread i handles 32 elements = 128 bytes - Thread 0: 0..3, thread 1: 32..35, etc. - Difference = 32 floats = 128 bytes - Thread 0: line 0, thread 1: line 1, thread 2: line 2, etc. - Consecutive threads access consecutive cache lines! - A warp accesses 32 cache lines = 4096 bytes, exactly matching 32 threads * 128 bytes. How does the memory unit handle this? For a warp reading 128 bytes per thread = 4096 bytes total, with addresses perfectly strided by 128 bytes... Actually, in modern GPUs, consecutive threads accessing consecutive cache lines can be coalesced into fewer transactions if they fall within the same segment. But 128-byte stride means each thread touches a different cache line. The memory subsystem might still issue ~32 transactions. But these can be executed in parallel if the memory channels allow. Hmm, 50% efficiency isn't amazing, but it might still be OK. Let me see if there's a better way. What if threads in a warp are assigned elements with stride 1? - Thread 0: 0, 32, 64, ... (no, that's not coalesced either) Wait. An important insight: for prefix sum, the data intra-dependency within a thread's chunk matters. If we assign elements so that consecutive threads access consecutive memory, then each thread's elements are spread throughout the warp's region. For warp 0 handling 1024 elements: - Thread 0: 0, 32, 64, ..., 992 (32 elements spaced by 32) - Thread 1: 1, 33, 65, ..., 993 - ... This is coalesced! Each warp load fetches 32 consecutive floats = 128 bytes with each thread getting one float. For vectorized loads, we can load 128-byte chunks per instruction. Then each thread's elements are at indices: `warp_offset + threadIdx.x + i * 32` for i = 0..31. To compute the prefix sum of a thread's elements, we'd process them in order of the thread index, not memory order. But for prefix sum of the whole row, we need the elements in index order. Actually, we could process them in memory order if we restructure. Each thread handles every 32nd element starting from its thread ID. So thread 0 handles elements 0, 32, 64, ... These are not contiguous in index, so computing prefix sum across them doesn't make sense for the row-wise scan. The issue is that prefix sum requires elements in index order, but coalescing requires memory access by thread order. These are in tension. For a row-major scan, there's a fundamental tradeoff: either you assign contiguous chunks to threads (good for the algorithm, bad for coalescing within a warp's first loads) or you interleave elements (good for coalescing, bad for the algorithm). Actually, wait. I think I was wrong about coalescing with contiguous chunks. Let me think about it from a different angle. If we have threads in a warp load contiguous elements: - Warp loads elements [0..31] - Then [32..63] (if each thread did 1 element) - For our case with 32 elements per thread, if we organize it as: Approach A (interleaved within warp): - On iteration i, thread j loads element `i * 32 + j` - This is perfectly coalesced - After all iterations, thread j has elements j, j+32, j+64, ..., j+31*32 - These elements are in the row but not contiguous Can we compute prefix sum with this layout? The row is divided into 32-element groups. On iteration i, all threads load group i. Thread j loads element i*32 + j, which is the j-th element of group i. To compute the prefix sum: - First, each thread computes the prefix of its own elements (which are scattered). But the prefix of scattered elements isn't useful for the overall prefix sum. Unless we transpose the thinking: process the array in groups of 32, where each group is processed by all threads in a warp. Actually, this is a well-known approach for scan algorithms. Let me think about warp-level scan with this layout. For a block of 1024 threads handling 32768 elements: - Process elements in tiles - Each warp handles 1024 elements in 32 iterations - On iteration k, all 32 threads in warp load elements warp_start + k*32 + lane - The warp has loaded a contiguous 32-element chunk What can the warp do with this? The warp can compute the sum of these 32 elements using warp reduction. But for prefix sum, we'd need a different approach. There's a known parallel scan algorithm that operates on this layout: 1. Load chunk (e.g., 32 contiguous elements, one per thread) 2. Warp-level scan of the loaded values (using shfl) 3. Now each thread has the prefix within this 32-element chunk 4. The last value of the chunk is the chunk sum 5. After processing all chunks, scan the chunk sums 6. Add chunk prefix offsets Wait, this is interesting! On each iteration, the warp loads a contiguous chunk: - Iteration 0: thread j loads element j. Now warp has elements 0..31. - Do a warp-level scan using shfl. Thread j now has prefix[0..j]. - Store prefix[0..j] back? But each thread only needs to store its own prefix. - The chunk sum is element 0..31 = prefix[31] at thread 31. - Save chunk sum. Then do this for all 32 chunks in the warp's region: - Iteration 0: elements 0..31, chunk sum = s0 - Iteration 1: elements 32..63, chunk sum = s1 - ... - Iteration 31: elements 992..1023, chunk sum = s31 After the loop, do a scan of chunk sums s0..s31. This can be done in shared memory. Then on a second pass (or in the same pass if we store values), add the chunk prefix to each element. This approach is: - Coalesced on both reads and writes - Uses warp shuffles efficiently - Requires storing intermediate values For 32 chunks per warp and 32 warps per block: - 1024 chunk sums total... wait. Let me think again. With 1024 threads = 32 warps: - Each warp handles 1024 elements = 32 chunks of 32 elements - For each warp, we have 32 chunk sums - 32 warps * 32 chunk sums = 1024 chunk sums per block - Each chunk sum is one float - We need to scan these 1024 chunk sums - Then propagate offsets to all chunks This requires more shared memory but is still manageable (1024 * 4 = 4096 bytes). Actually, let me reconsider the scale. If each block handles one row of 32768 elements: - 1024 threads = 32 warps - Each warp handles 1024 elements - Each warp processes 32 chunks of 32 elements - Total chunks = 32 * 32 = 1024 chunks of 32 elements each = 32768 elements ✓ Algorithm: 1. For each warp, for each chunk: - Load 32 elements (coalesced, one per thread) - Warp-level scan using shfl - Write back the prefix for this chunk (coalesced) - Save chunk sum 2. After all chunks are processed, scan the 1024 chunk sums - We have 1024 floats in shared memory - 1024 threads can do a parallel scan - Use the classic block scan algorithm on chunk sums 3. For each chunk, add its block-offset to all elements - On second pass, or if we stored elements... The problem is we write back in step 1 without the chunk offsets. Then we need a second pass to add offsets. Or we could buffer the elements. If we buffer elements, each thread needs 32 floats for one chunk, but we process chunks sequentially. So we don't need to buffer all elements at once. Alternative: two-pass approach: Pass 1: For all chunks, load, compute prefix within chunk, write prefix back. Also compute and store chunk sums in shared memory. After pass 1: scan chunk sums in shared memory. Pass 2: For all chunks, load the written values, add chunk offset, write back. This requires reading and writing the data twice. Total memory traffic: 16 bytes per element (2 reads + 2 writes). That's double the minimum! Not great. Can we do single pass? If we process chunks and the chunk offsets are known before we write... they aren't, unless we pre-scan all chunk sums. But chunk sums are computed as we go. What if we use a two-level approach where: - Level 1: process all chunks, store local prefix and chunk sum - Level 2: scan chunk sums - Level 3: write final prefix using chunk offsets For Level 1, we need to store the local prefixes. 32768 floats per row = 128KB. That's too much for shared memory, but we can write to global memory. Then in Level 3, we read back from global memory, add offset, and write back. That still requires two passes over global memory (3 reads + 2 writes = 20 bytes per element). Actually, for bandwidth-bound algorithms, multiple passes can sometimes still be fast if the working set fits in cache. For 128KB per row, with L2 cache of 50MB, we can hold ~400 rows. But with 32768 rows, that doesn't help much. Hmm. Let me think differently. What if we don't process one row per block? For very large rows (32768 elements), a single block per row with two passes might be fine if the passes are efficient. The reference implementation might already do something similar. Actually, wait. The cuDUB/cuFFT libraries' scan implementation for large arrays might use a recursive approach or multi-block scan. For PyTorch, `torch.cumsum` likely delegates to a Thrust or Cub implementation which is quite optimized. A simple per-row block scan with sequential per-thread chunks might already be competitive or even slower than the reference. We need to think about why our custom kernel could be faster. One advantage of a custom kernel: we can fuse operations, avoid temporary allocations, and tailor the algorithm to this specific shape. Another idea: process multiple rows per block, or use a different tiling strategy. What about using shared memory more aggressively with tile decomposition? Tile size = 1024 elements (one warp's worth) 32 tiles per row 32 warps per block Each warp handles one tile. Warps can work in parallel. Actually, I just realized that for single-pass prefix sum within a block, if we store thread-level aggregates and do a hierarchical scan, we can avoid multiple passes IF we can compute the final offset before writing. But we can't compute the final offset for later elements without knowing the sums of earlier elements. Unless... What if we process the row from left to right within the block? We could have 32 warps each handle a tile, but we need to propagate the prefix from tile 0 to tile 1, etc. This is sequential across tiles, which defeats parallelism. OK so we need multiple passes or storage. Let me consider the two-pass approach with good caching. Actually, there's a known technique: store the row in shared memory and perform a Brent-Kung scan in shared memory. But 32768 elements * 4 bytes = 128KB, which exceeds 48KB / 1024 shared memory in older GPUs, but H100 has 228KB per SM and allows up to 227KB per block! So we CAN fit 128KB in shared memory! 128KB < 227KB limit. This is huge! With 128KB of shared memory per block: - We can load the entire row into shared memory - Do a parallel scan in shared memory - Write back 128KB per block means we can have at most floor(228KB/128KB) = 1 block per SM. With 132 SMs, only 132 blocks in flight. But we have 32768 rows, so each block handles one row, and we have 32768 / 132 = 248 waves. With 1024 threads per block and 1 block per SM: 1024 threads per SM. That's only 50% thread occupancy. But shared memory is the limiter. For Brent-Kung scan with 32768 elements: - Up-sweep phase: log2(32768) = 15 steps - Down-sweep phase: 15 steps - Total: 30 steps of __syncthreads() Each step does adds and syncs. 30 syncs per row. Is this fast? The memory accesses from shared memory are very fast (~20 TB/s). The compute is minimal. The syncs might be expensive. But with only 1024 threads per SM and 30 syncs, the latency hiding might be poor. What about a hybrid approach: use shared memory for tiles and a multi-level scan. Actually, let me look at this from the perspective of what PyTorch does. For `torch.cumsum` with dim=1 on a large 2D tensor, PyTorch might: 1. Call into ATen, which uses TensorIterator 2. TensorIterator might parallelize over the outer dimension (dim=0) 3. Each thread/row does a sequential scan If each row is handled by a separate thread or warp with sequential scan, then: - 32768 rows * 32768 elements = 1B elements - Each row takes 32768 serial adds - With 132 SMs * 2048 threads = 270K threads, we have 32768 rows / 270K threads ≈ 0.12 rows per thread... wait, that's not how it works. If each thread handles one row, we have 32768 threads' worth of work. With 270K threads, we'd have 270K concurrent threads (if we launched that many). But 32768 < 270K, so all rows could be processed concurrently! Each thread does a sequential scan of 32768 elements. A serial loop of 32768 float adds. At ~1 cycle per fma maybe a few cycles per add. 32768 adds ≈ 32768 * 4 cycles? Actually memory bound. If each thread reads and writes sequentially, memory access is coalesced across threads? No, if thread 0 accesses row 0 elements 0..32767, thread 1 accesses row 1 elements 0..32767, etc. Across a warp, at any point in time, threads are reading different rows but the same column. If rows are 32768 * 4 = 131072 bytes apart, then threads in a warp are accessing memory 131KB apart. That's terrible coalescing! A warp of 32 threads would issue 32 transactions to 32 different memory locations, possibly all in different cache lines. Very inefficient. So if PyTorch uses TensorIterator with thread-per-row, its memory bandwidth utilization might be terrible. Our custom kernel can do much better by processing a row with multiple threads and maintaining coalescing. Let me reconsider the coalescing for multi-thread-per-row. If we have one block per row, with threads processing the row cooperatively: - Thread j in block processes elements j * chunk_size to (j+1) * chunk_size - 1 - For a warp of 32 threads, consecutive threads access memory chunk_size * 4 bytes apart - If chunk_size = 1, threads access consecutive elements (coalesced!) - If chunk_size = 128, threads access memory 512 bytes apart So we should use chunk_size = 1! Each thread processes every N-th element, where N is block size. But then each thread's elements are strided by N. For prefix sum, a thread needs to process elements in order, so it would process indices: threadIdx.x, threadIdx.x + blockDim.x, threadIdx.x + 2*blockDim.x, etc. This strided access is perfect for coalescing! On each iteration, all threads in a warp load consecutive elements from memory. BUT: the prefix sum of strided elements within a thread is NOT the global prefix sum. The global prefix spans across all threads. For example, with 1024 threads: - Thread 0: 0, 1024, 2048, ... - Thread 1: 1, 1025, 2049, ... ... - Thread 1023: 1023, 2047, 3071, ... If thread 0 processes its elements: prefix[0] = x[0], prefix[1024] = x[0] + x[1024], etc. But the global prefix at element 1024 should be sum of x[0..1024], not just x[0] + x[1024]. So this strided assignment doesn't directly work for prefix sum. However, we CAN use it if we reorder: 1. Load elements in strided order (coalesced) 2. Compute local prefix within each thread's strided set 3. Transpose/reorder so that elements are in index order 4. Do cross-thread scan 5. Write back Steps 3-4 are tricky and require shared memory communication. Actually, there's a known parallel scan algorithm called the "decoupled look-back" scan that might help here, but for a block-level scan... Another well-known approach is to use shared memory as an intermediary: 1. Load strided elements into shared memory in a blocked layout 2. Transpose in shared memory so elements are in index order 3. Perform scan in shared memory (Brent-Kung or similar) 4. Write back coalesced For 1024 elements: - 1024 threads, each loads 1 element in strided fashion (perfect coalescing) - Store in shared memory at index threadIdx.x - Then thread i accesses shared memory element i to do scan - Scan in shared memory - Write back element i in strided fashion (perfect coalescing) This is the classic "shfl or shared memory scan" and works perfectly! But we have 32768 elements per row, not 1024. And a block has at most 1024 threads. If we use 1024 threads per block, we could process the row in tiles of 1024 elements: - Tile 0: elements 0..1023 - Tile 1: elements 1024..2047 - ... - Tile 31: elements 31744..32767 For each tile: 1. Load 1024 elements into shared memory (coalesced, strided) 2. Brent-Kung scan in shared memory 3. Read scan result, write back (coalesced) After processing all tiles, we have computed the prefix within each tile, but tiles are independent. We need to add the total of previous tiles to each tile. So we need a second pass or a precomputation of tile sums. If we process tiles left to right within a block, we CAN carry over the prefix: - Process tile 0: scan, write back. Total of tile 0 = last element. - Process tile 1: load, add tile_0_total to all elements, scan, write back. - Process tile 2: add tile_0_total + tile_1_total, etc. This is sequential across tiles! But within each tile, 1024 threads work in parallel. For 32768 elements / 1024 per tile = 32 tiles. With 32 sequential tiles, each with ~log2(1024) ≈ 10 steps of shared memory scan plus sync. About 320 steps total. Each step is a shared memory access (fast). At 20 TB/s shared memory bandwidth and ~2 accesses per element per step... Hmm, this might be slow because of the sequentialism. What if we launch with multiple blocks per row, and use an inter-block prefix sum algorithm? But then we need inter-block communication, which is expensive. Actually, for 32768 rows and our GPU, I wonder if a per-row sequential approach in global memory might be fine if we ensure good memory access. What about: have each thread process one row sequentially, but organize blocks so that threads in the same warp process the same row? No, that doesn't help. Wait, what if each block processes multiple rows? Say 4 rows per block with 256 threads: - 256 threads / 4 = 64 threads per row - Each row is 32768 elements - 64 threads per row, each thread handles 512 elements sequentially - Consecutive threads in a warp access memory 512*4 = 2048 bytes apart Then we need cross-thread scan per row. With 64 threads, we can use shuffle. For coalescing: 64 threads per row in a warp of 32... a warp might have threads from different rows if we're not careful. Actually, if a block processes 4 rows, and we assign threads to rows by contiguous groups: - Threads 0..63: row 0 - Threads 64..127: row 1 - etc. A warp has 32 threads. Warp 0 has threads 0..31 = all from row 0. Warp 1 has threads 32..63 = all from row 0. Warp 2 has threads 64..95 = all from row 1. This means the first 2 warps handle row 0. Within warp 0, each thread handles 512 elements. Consecutive threads access memory 2048 bytes apart. Coalescing is poor. What if we interleave threads across rows? - Thread 0: row 0 - Thread 1: row 1 - Thread 2: row 2 - Thread 3: row 3 - Thread 4: row 0 - ... Then in a warp, 8 threads handle row 0, 8 handle row 1, etc. When reading element 0 of their respective rows, threads in a warp access 4 different memory addresses (one per row). Still poor coalescing. Hmm. The fundamental issue is that for good coalescing, threads in a warp need to access consecutive memory in the same row. But for a row scan, we want threads in a warp to cooperate on the same row. The solution: have multiple warps all work on the same row, with each warp accessing a different part of the row. But ensure that within each warp, threads access consecutive memory. Let me go back to the earlier approach but optimize coalescing by making chunk_size = 1. For a block of 1024 threads processing 1 row: - The row has 32768 elements - We can think of the row as 32 groups of 1024 elements - On iteration i, all 1024 threads load element i * 1024 + threadIdx.x - This is perfectly coalesced - We store these 1024 elements in shared memory at index threadIdx.x - Now perform a block scan on the 1024 elements in shared memory - Then write back: output[i * 1024 + threadIdx.x] = shared_mem[threadIdx.x] + global_offset For the block scan on 1024 elements in shared memory: - We can use Brent-Kung: O(2 * log N) = 20 steps - Or Kogge-Stone: log N = 10 steps, but more writes Brent-Kung up-sweep: For d = 512, 256, ..., 1: if threadIdx.x < d: shared[2*(threadIdx.x+1)*stride - 1] += shared[2*threadIdx.x*stride + stride - 1] Then down-sweep: Set root to 0 For d = 1, 2, ..., 512: if threadIdx.x < d: swap and add For 1024 threads, the up-sweep and down-sweep each take 10 steps. Total 20 __syncthreads() calls per tile. For 32 tiles: 640 syncs! That's way too many. What about using warp shuffles for the 1024-element scan? With 32 warps of 32 threads: 1. Load into shared memory at threadIdx.x (coalesced) 2. Each warp does a scan of 32 elements using shfl - After this, warp has prefix sums for elements warp*32 to warp*32+31 3. Last element of each warp is the warp total 4. Store warp totals in shared memory (32 values) 5. Thread 0 scans these 32 warp totals 6. Add warp offset back This takes minimal syncs: - shfl sync (implicit in warp) - One __syncthreads after step 3 - Step 5 by thread 0 - One __syncthreads after step 5 For each tile of 1024 elements, only 2 __syncthreads! And we need to process 32 tiles per row. If we do them sequentially with prefix carry: ``` float carry = 0.0f; for (int tile = 0; tile < 32; tile++) { // Load tile into shared mem (coalesced) shared[threadIdx.x] = input[row * 32768 + tile * 1024 + threadIdx.x]; __syncthreads(); // Warp-level scan float val = shared[threadIdx.x]; int lane = threadIdx.x & 31; int warp = threadIdx.x >> 5; // Intra-warp scan for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) val += other; } // Store warp totals __shared__ float warp_totals[32]; if (lane == 31) warp_totals[warp] = val; __syncthreads(); // Scan warp totals if (warp == 0) { float wval = (lane < 32) ? warp_totals[lane] : 0.0f; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, wval, offset); if (lane >= offset) wval += other; } if (lane < 32) warp_totals[lane] = wval; } __syncthreads(); // Add offset float warp_offset = (warp > 0) ? warp_totals[warp - 1] : 0.0f; float final_val = val - shared[threadIdx.x] + warp_offset + carry; // Wait, val includes this thread's element. We want: // val = prefix including this thread's element // intra_prefix = val - shared[threadIdx.x] = prefix before this thread // final = intra_prefix + warp_offset + carry + shared[threadIdx.x] // = val - shared[threadIdx.x] + warp_offset + carry + shared[threadIdx.x] // = val + warp_offset + carry - wait, no. // Let me recompute. // shared[tidx] = original value at this position. // After intra-warp scan, val = sum of shared[warp*32 .. warp*32+lane] // For the element at position tidx, the prefix should be sum of all elements before tidx plus this element. // Before this thread in the warp: val - shared[tidx] // Before this warp: warp_offset // Before this tile: carry // So total prefix = (val - shared[tidx]) + warp_offset + carry + shared[tidx] = val + warp_offset + carry // Wait, is val the sum including this element? After shfl scan, yes. // val = sum_{j=0}^{lane} shared[warp*32+j] // We want prefix for element at warp*32+lane: // prefix = sum_{k=0}^{warp*32+lane} shared[k] + carry // = sum_{k=0}^{warp*32-1} shared[k] + sum_{j=0}^{lane} shared[warp*32+j] + carry // = warp_offset + val + carry // YES! Because warp_offset is defined as sum of all previous warps. // But warp_totals[warp] after scan = sum of warps 0..warp. // So warp_offset should be sum of warps 0..warp-1. if (warp == 0 && lane == 0) { warp_totals[0] = 0; // No, wait. } } ``` Let me redo the inter-warp scan more carefully. After storing warp totals where each warp's total is the last element of its scan (which is sum of all elements in that warp): ``` if (lane == 31) warp_totals[warp] = val; ``` So warp_totals[warp] = sum of elements in warp `warp`. Then scanning warp_totals: ``` if (warp == 0) { float wval = (lane < 32) ? warp_totals[lane] : 0.0f; float woriginal = wval; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, wval, offset); if (lane >= offset) wval += other; } // wval = sum of warp_totals[0..lane] // We want prefix before warp lane = sum of warp_totals[0..lane-1] if (lane < 32) { warp_totals[lane] = wval - woriginal; } } ``` So warp_totals[warp] now = sum of all elements in warps 0..warp-1. Then: ``` float warp_offset = warp_totals[warp]; float final_val = val + warp_offset + carry; ``` And for the carry to the next tile: ``` if (threadIdx.x == 1023) { // val is sum of this warp (warp 31), warp_offset is sum of warps 0..30 // So val + warp_offset = total of this tile carry = val + warp_offset; } __syncthreads(); ``` Wait, we need another __syncthreads() after carry update? Actually, all threads need to see the updated carry for the next iteration. So we should do the carry computation and then __syncthreads() before the next iteration begins. Actually, since `carry` is in a register and only thread 1023 computes it, we need to communicate it to all threads. Options: 1. Store in shared memory: __shared__ float tile_carry; if (tidx == 1023) tile_carry = val + warp_offset; __syncthreads(); then each thread adds tile_carry to its next iteration... wait, no. We want `carry` to accumulate across tiles. Actually, the carry should be accumulated. Let me restructure: ``` __shared__ float sdata[1024]; __shared__ float warp_totals[32]; __shared__ float tile_sum; float carry = 0.0f; for (int tile = 0; tile < 32; tile++) { int idx = row * 32768 + tile * 1024 + threadIdx.x; sdata[threadIdx.x] = input[idx]; __syncthreads(); float val = sdata[threadIdx.x]; // ... intra-warp scan ... // ... warp totals and inter-warp scan ... float result = val + warp_totals[warp] + carry; output[idx] = result; if (threadIdx.x == 1023) { tile_sum = val + warp_totals[warp]; // Total of this tile = last thread's prefix } __syncthreads(); carry += tile_sum; __syncthreads(); } ``` Wait, this has 4 __syncthreads per tile: after load, after warp scan, after tile_sum store, and after carry update. That's 128 syncs for 32 tiles. Plus the warp scan is within warp so doesn't need sync. Actually, the intra-warp and inter-warp scans using shfl don't need __syncthreads. Only stores/loads to shared memory need sync for visibility. Let me restructure to minimize syncs: 1. Load into sdata, __syncthreads() 2. Read val from sdata, do shfl scan (no sync needed within warp) 3. Store warp total (one thread per warp), __syncthreads() 4. Thread 0 does inter-warp scan using shfl (data already in shared mem, but threads in warp 0 read from shared mem - no additional sync needed since step 3 had a sync) 5. __syncthreads() so all threads see updated warp_totals 6. Compute result = val + warp_totals[warp] + carry 7. Write output 8. Thread 1023 writes tile_sum, __syncthreads() 9. carry += tile_sum, __syncthreads() Actually, steps 8 and 9 can be combined. Since `carry` is the same for all threads, we can compute it once and broadcast. Optimization: since warp_totals is only 32 floats, and after step 5 all threads read warp_totals[warp], we might be able to avoid some syncs. Let me think: in step 4, warp 0 reads warp_totals and writes back. Do we need a sync before step 5? Well, step 3 already had a sync after writing warp_totals. Then warp 0 reads and writes warp_totals. Other warps also read warp_totals in step 5. So we need a sync between step 4 and 5. Actually, we could have warp 0 read the 32 values into registers, scan them, and write back. Then __syncthreads(). Then all threads read their warp prefix. Total syncs per tile: 4. For 32 tiles: 128 syncs per row. With 32768 rows and limited occupancy due to shared memory (128KB per block, so 1 block per SM = 132 concurrent blocks), this might take a while. But shared memory scan is very fast. Each sync is maybe ~20-30 cycles. 128 * 30 = 3840 cycles per row. At 1.5 GHz, that's 2.5 microseconds per row. For 32768 rows with 132 concurrent: 32768/132 * 2.5us = 620us. The memory bandwidth bound is 2.38ms for 8GB. So this could be within memory bound. But the limited occupancy (1 block per SM) means we're using only 1024 out of 2048 threads per SM. That's 50% thread occupancy, which might not hide all the latency. But the heavy use of shared memory within the block provides some latency hiding. Actually, I want to try a different approach that's simpler and might have better occupancy. What about 256 threads per block, each handling 128 elements sequentially? - Block handles 1 row = 32768 elements - 256 threads * 128 = 32768 ✓ - Shared memory: only for warp totals (8 floats = 32 bytes) and maybe tile sum - Register usage: low For memory access: Thread j accesses elements [j * 128, j * 128 + 127] In a warp (threads 0..31): offsets 0, 512, 1024, ..., 15872 bytes. Each thread loads 128 floats = 512 bytes. This is NOT coalesced. But maybe it doesn't matter if we use vector loads. Actually, with float4 loads (16 bytes each), each thread does 32 float4 loads. Thread 0: bytes 0..15, 16..31, ..., 496..511 (all in the first 512 bytes) Thread 1: bytes 512..527, 528..543, ..., 1008..1023 For a warp, threads access memory in chunks of 16 bytes at offsets 0, 16, 32, ..., 496 (thread 0), then 512, 528, ..., 1008 (thread 1), etc. The memory coalescing unit sees a warp trying to load bytes 0..15, 512..527, 1024..1039, ..., 15872..15887 all at once. That's 32 transactions of 16 bytes each. The memory subsystem might not coalesce these well. But H100 has very high L2 bandwidth and multiple memory channels. It might still be OK. Actually, the H100 memory controller can handle many concurrent transactions. With 32 transactions per warp and each warp processing 128*32=4096 floats=16KB, maybe it's fine. But let me think about another approach: 256 threads, but each thread doesn't handle contiguous elements. Instead, each warp handles a contiguous region. 256 threads = 8 warps. Each warp handles 32768 / 8 = 4096 elements. Within warp 0: threads 0..31 handle 4096 elements total = 128 elements per thread. Thread 0: 0..127 Thread 1: 128..255 ... Thread 31: 3968..4095 For thread 0, accesses are 0..3, 4..7, ..., 124..127 (as float4). Thread 1: 128..131, 132..135, ..., 252..255. This is still 128 bytes apart between threads. Hmm, it seems difficult to get perfect coalescing for multi-thread row scan without strided assignment, which complicates the algorithm. Let me look for another angle. What if we transpose the problem? If we transpose the 32768x32768 matrix, cumsum along dim=1 becomes cumsum along dim=0 of the transpose. But PyTorch already handles this. What does our custom kernel do differently? Actually, I just checked: for a 32768x32768 float tensor, transposing is expensive (needs to read and write the whole tensor). PyTorch's cumsum might or might not transpose internally. For large 2D cumsum, the fastest approach on GPU is often the "Blelloch scan" or "decoupled lookback" scan. NVIDIA CUB has highly optimized implementations. Since we have to beat PyTorch, and PyTorch likely uses CUB (which is quite optimized), we need to be very clever. Wait, but PyTorch uses tensoriterator which might not always dispatch to CUB for all cases. Let me check if torch.cumsum uses CUB... For CUDA, `at::cuda::cub::cumulative_sum` probably exists. Or it might be a custom implementation. But I recall that PyTorch's cumsum internally does use efficient parallel algorithms. Let me consider whether our kernel can be competitive. The key advantage of a custom kernel is we can avoid some overheads. For this specific shape (32768, 32768) and dimension 1, a very simple approach might be: - Each block handles one row - Each thread in the block does a sequential scan of its chunk - Use warp shuffles for cross-thread scan - Use shared memory only for minimal communication This has very low overhead and fits well in registers. I think the 256-thread block with 128 elements per thread is worth trying. The memory access pattern isn't ideal, but it might still be fast. And the register pressure is very low, allowing high occupancy (multiple blocks per SM). Actually, let me think about this more carefully. With 256 threads per block handling one row, each thread handles 128 elements. In the first pass, each thread reads 128 floats to compute a sum. In the second pass, each thread reads 128 floats, adds the offset, and writes 128 floats. Memory traffic: 32768 rows * 32768 floats/row * (2 reads + 1 write) * 4 bytes = 32768 * 32768 * 12 = ~13 GB. Wait, 2 reads + 1 write = 12 bytes per element. Total: 1B elements * 12 bytes = 12 GB. For bandwidth analysis: if we achieve 50% of peak bandwidth (due to poor coalescing), that's ~1.67 TB/s. 12GB / 1.67 TB/s = 7.2 ms. Reference time... I don't know exactly, but if the reference achieves better coalescing, it might be faster. Let me think about improving coalescing for the 256-thread approach. What if we have each WARP handle a contiguous region, but instead of 256 threads = 8 warps with 4096 elements each, we have the threads within a warp access the warp's region with stride 1? In warp 0 (128 elements per thread): Thread 0: 0, 1, 2, ..., 127? No, that doesn't work with 32 threads and 4096 elements. Wait, 4096 elements / 32 threads = 128 elements per thread. If thread j accesses elements j, j+32, j+64, ..., j+127*32. That's 128 elements per thread, spread across the 4096-element region with stride 32. On iteration k, all threads in warp load element k*32 + lane. This is coalesced! And consecutive iterations load consecutive 32-element chunks. After loading a 32-element chunk (one per thread), the warp can do a scan within that chunk using shfl. Then the chunk prefix can be accumulated. This is promising! Let me work it out. For a row of 32768 elements, with 256 threads = 8 warps: - Warp w handles elements [w * 4096, (w+1) * 4096 - 1] = 4096 elements - Within warp, on iteration k (0..127), thread j loads element w*4096 + k*32 + j - This is coalesced (consecutive threads access consecutive elements) - 32 elements per iteration, 128 iterations, 4096 elements per warp For the scan within warp: At iteration k, the warp loads elements [k*32, k*32+31] of its region. Thread j has value input[w*4096 + k*32 + j]. We want the prefix sums for these elements. After all iterations, thread j needs the prefix at global index w*4096 + k*32 + j for all k. This is the entire region. Algorithm: For each warp handling 4096 elements: 1. Have an array of 128 floats per thread (register). But that's 512 bytes per thread, too much. Alternative: process in chunks and accumulate a running prefix, writing results as we go. But we need the warp-level and inter-warp offsets before we can write the final prefix. Two-pass approach per warp's region: Pass 1: Compute the sum of each 32-element chunk. For k = 0..127: Load chunk k (coalesced) Compute chunk_sum[k] using warp reduction Now warp has 128 chunk sums (one per iteration) Pass 2: Scan the 128 chunk sums. Since all threads in a warp have the same 128 chunk sums if we stored them... but storing 128 floats per warp requires shared memory or warp shuffles over 128 items. Actually, after pass 1, each thread can have a local copy of chunk sums. Then they all do the same scan computation. Wait, on iteration k, all threads in the warp load the 32-element chunk. A warp reduction gives the chunk sum, which all threads can get. ``` float chunk_sums[128]; for (int k = 0; k < 128; k++) { float val = input[w*4096 + k*32 + lane]; // Warp reduction for sum float chunk_sum = val; for (int offset = 16; offset > 0; offset >>= 1) { chunk_sum += __shfl_down_sync(0xFFFFFFFF, chunk_sum, offset); } chunk_sum = __shfl_sync(0xFFFFFFFF, chunk_sum, 0); chunk_sums[k] = chunk_sum; } ``` This uses 128 registers per thread just for chunk_sums. Too much. Alternative: compute running totals and store prefixes in shared memory temporarily. Actually, let me think about this more simply. What if we do a two-level scan: - Level 1: each warp processes 4096 elements in 128 chunks of 32. For each chunk, load, do warp scan, write back the local prefix. Also store the chunk sum in shared memory. - Level 2: scan the 128 chunk sums for this warp (using shared memory or shfl). - Level 3: read back the local prefixes and add chunk offsets. But this requires shared memory for 128 chunk sums per warp, and 8 warps = 1024 floats = 4KB. Plus we need to write and read back the 4096 local prefixes = another pass. Hmm, this is getting complex. And we already saw that two-pass approaches double memory traffic. Let me reconsider: maybe we can get decent performance with the "poorly coalesced" contiguous chunk assignment. For the H100 with its massive memory bandwidth and multi-channel memory, strided accesses might not be as bad as on older GPUs. In fact, for 256 threads, each thread's chunk is 128 contiguous elements. A warp of 32 threads accesses 32 chunks = 4096 elements = 16KB. The memory controller serves these from different channels. At 128-byte cache line granularity, each thread's 128 floats span 4 cache lines. But consecutive threads read non-consecutive cache lines. Still, the H100 has many memory channels and can handle this. Let me just try a simple approach first and see how it performs. Simplest approach: - 256 threads per block - Each block handles 1 row - Each thread handles 128 contiguous elements - Two passes: first computes chunk sum, second applies offsets and writes Kernel structure: ``` __global__ void cumsum_kernel(const float* input, float* output, int rows, int cols) { int row = blockIdx.x; if (row >= rows) return; int tid = threadIdx.x; int chunk_size = 128; int base = row * cols + tid * chunk_size; // Pass 1: compute chunk sum float sum = 0.0f; for (int i = 0; i < chunk_size; i++) { sum += input[base + i]; } // Intra-warp scan of sums int lane = tid & 31; int warp = tid >> 5; float warp_scan = sum; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, warp_scan, offset); if (lane >= offset) warp_scan += other; } __shared__ float warp_totals[8]; if (lane == 31) { warp_totals[warp] = warp_scan; } __syncthreads(); // Inter-warp scan if (warp == 0) { float val = (lane < 8) ? warp_totals[lane] : 0.0f; float original = val; for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) val += other; } if (lane < 8) { warp_totals[lane] = val - original; } } __syncthreads(); float offset = warp_totals[warp] + warp_scan - sum; // Pass 2: write prefix with offset float prefix = offset; for (int i = 0; i < chunk_size; i++) { prefix += input[base + i]; output[base + i] = prefix; } } ``` Wait, `offset = warp_totals[warp] + warp_scan - sum;` - warp_totals[warp] = sum of all previous warps' totals - warp_scan - sum = sum of all previous threads' sums in this warp - So offset = sum of all previous threads' sums in all previous warps and this warp - This is the prefix before this thread's chunk. Then in pass 2, `prefix = offset`, and we add each input element to prefix. That gives the global prefix. Correct! Memory traffic: each element is read twice, written once. 12 bytes/element. Register usage: sum (1), offset (1), prefix (1), loop counter (1), plus some temporaries = maybe 16-20 registers. Very low. With 256 threads and ~20 registers: 5120 registers per block. At 64K registers per SM, we can have 12 blocks per SM (register-limited). But thread-limited: 256 threads * 32 blocks = 8192 threads > 2048 max. So thread-limited to 8 blocks per SM (256*8=2040 threads). Shared memory: 8 floats = 32 bytes. Negligible. So we can have 8 blocks per SM. With 132 SMs: 1056 concurrent blocks. 32768 rows / 1056 = 31 waves. Very nice parallelism! And each block does 2 passes over its 128 floats. The memory accesses from different blocks are interleaved and can hide latency. I think this approach could work well. The main concern is memory coalescing. Let me think about whether 128 elements per thread is the right number. If we use more elements per thread: - Fewer threads per block - More blocks per SM (thread-limited) - But each thread does more sequential work If we use 64 elements per thread: - 512 threads per block - 4 blocks per SM (thread limited) - Still good occupancy If we use 32 elements per thread: - 1024 threads per block - 2 blocks per SM - Less occupancy I think 128 or 256 elements per thread gives the best balance. Let me check 256 elements per thread: - 128 threads per block - 16 blocks per SM (thread limited: 128*16=2048) - 16 concurrent blocks per SM * 132 SMs = 2112 concurrent blocks - 32768 rows / 2112 = 15.5 waves - Even fewer waves but more parallelism For coalescing, with 128 threads per block: - Thread 0: 0..255 - Thread 1: 256..511 - ... - In a warp of 32 threads, offsets = 0, 1024, 2048, ..., 31744. Even worse striding. So more elements per thread means worse coalescing within a warp. With 512 threads per block and 64 elements per thread: - Warp 0: threads 0..31, offsets 0, 256, 512, ..., 7936. Better than 128. Actually, 64 elements = 256 bytes. Thread i accesses bytes i*256 to i*256+255. In a warp, addresses are 256 bytes apart. H100 memory transactions are at least 32 bytes. So each thread in the warp needs 8 transactions of 32 bytes. A warp needs 256 transactions for 32 threads * 64 floats = 2048 floats = 8KB of data. 256 transactions * 32 bytes = 8KB fetched. Wait, that's 100% efficiency! Each thread accesses 64 floats = 256 bytes. 256 / 32 = 8 transactions of 32 bytes per thread. 32 threads * 8 = 256 transactions. Total fetched: 256 * 32 = 8192 bytes = 8KB. Total requested: 32 threads * 256 bytes = 8192 bytes. 100%! Wow. The transaction size depends on the request pattern. For a warp requesting contiguous addresses within a cache line, the hardware fetches in 32-byte sectors. For strided accesses, each thread's requests might be independent. Actually, I think my calculation is wrong. Let me reconsider. For coalesced access (consecutive threads access consecutive memory), a warp of 32 threads requesting 32-bit values issues 4 transactions of 32 bytes each to cover 128 bytes (assuming perfect coalescing). For our strided case: thread i accesses bytes [i*256, i*256+255] = [i*256, i*256+255]. Each thread accesses 64 floats = 256 bytes. This spans 2 cache lines of 128 bytes each. For a single float load (not float4), thread i loads float at i*256 + j for j=0..63. - When j=0: thread i accesses byte i*256. Threads in warp access bytes 0, 256, 512, ..., 7936. Each in a different 32-byte sector. So 32 transactions for 32 floats. - When j=1: same pattern, 32 transactions. - Total: 32*64 = 2048 transactions for 2048 floats. Very inefficient. If we use float4 loads: thread i loads float4 at address i*256 + j*16 for j=0..15. - j=0: addresses 0, 256, 512, ..., 7936. 32 transactions of 16 bytes each? Actually minimum transaction might be 32 bytes. So some efficiency loss. - Each float4 is 16 bytes. H100 might fetch 32 bytes and throw away half. Let me check if we can use 32-byte loads. float8 doesn't exist, but we could do float4x2 or use 64-bit loads of double2. Actually, for H100, the minimum sector size is 32 bytes. A float4 is 16 bytes, so loading float4 wastes 50% if the memory isn't cached for nearby threads. What if each thread uses two float4 loads adjacent to each other? That makes float8 = 32 bytes. Thread i loads 32 bytes at i*256 + j*32. Or what if we use __ldg for cached loads? With L1 cache, repeated accesses to the same cache line by different threads might be coalesced? No, __ldg is for read-only data and might help with cache, but doesn't change coalescing. OK, let me think about this differently. Maybe the memory access pattern with 256 threads and 128 elements each is actually not as bad as I think for the following reason: The L2 cache on H100 is 50MB. A row is 128KB. There are 32768 rows = 4GB total. The L2 can hold about 400 rows. But with 1056 concurrent blocks (8 per SM * 132 SMs), each block reads 128KB. So the working set in flight is about 132MB of row data. That's well within L2 cache! Wait, but blocks are processing DIFFERENT rows. Each block reads from a different 128KB chunk. With 1056 concurrent blocks, that's 132MB of unique data. The L2 is 50MB. So there WILL be cache misses and HBM traffic. But actually, for 8 blocks per SM and 132 SMs, not all blocks run simultaneously within the same wave due to memory latency. The memory system pipelines requests from many warps. Given the complexity, let me just try something and see. I think the 256-thread approach with sequential per-thread chunks and warp shuffles is worth testing. If it's not fast enough, I'll iterate. But before writing code, I had another idea: what if we process transpose-style? Instead of having blocks process rows, what if we have blocks process COLUMNS and accumulate? No, that doesn't make sense for row-wise scan. Or: transpose the input in shared memory, scan in shared memory, transpose back. But with 128KB per row, transposing in shared memory per block is impossible (only 128KB per block total). Actually, what if we process tiles that are square-ish? For example, a tile of 128x128 elements. But the scan needs to be along rows... Let me reconsider the problem size. (32768, 32768) is very large. Each row is 32768 floats. With 132 SMs and 64K registers each, we have about 8.4M registers total. At 32 registers per thread, that's 262K threads. With 32768 rows, that's 8 threads per row on average. But since we're assigning one block per row, we have 32768 blocks. What if we assign 8 threads per row? 32768 * 8 = 262K threads total. At 2048 threads per SM, that's 128 blocks of some size, or with 8 threads per block, that's 32768 blocks. With max 32 blocks per SM, we can have at most 32*8 = 256 threads per SM. Not enough to hide latency. What if we assign more threads per row to improve per-row parallelism but process multiple rows per block? For example, a block of 1024 threads processes 4 rows: - 256 threads per row - Each thread handles 128 elements - Warps are split across rows (4 consecutive warps per row) Within a warp of 32 threads, all 32 threads work on the same row. Consecutive threads handle chunks 128 elements apart. This is the same memory pattern but with the benefit that a block only needs 4 rows' worth of memory in flight. Actually, I just realized: with 1024 threads = 32 warps, and 4 rows per block, each row gets 8 warps = 256 threads. Same as one row per block with 256 threads. The memory pattern and coalescing characteristics are identical. So there's no benefit from processing multiple rows per block in terms of coalescing. Unless... we interleave the rows. What if threads 0..7 work on row 0, threads 8..15 on row 1, etc.? No, warps would then have threads from multiple rows, making coalescing even worse. I think the cleanest approach is one block per row. Let me just write and test it. Before writing code, I want to consider one more optimization: vectorized memory access. Even though the addresses are strided, if we use float4 loads, consecutive float4 loads by the same thread hit consecutive cache lines. And different threads hit different cache lines, but in parallel. Actually, for the float2 or float4 loads, the hardware might still need to issue multiple transactions. But the key is whether the memory channels can handle many concurrent requests. H100 has 12 memory channels (or similar), and each can handle many requests. For 256 threads per block = 8 warps, and 8 blocks per SM = 64 warps per SM. At 132 SMs: 8448 warps. Each warp might issue ~8-32 memory transactions per load. The memory system can pipeline these. Actually, let me try to compute more carefully. For 128 floats per thread with float4 loads: - 32 float4 loads per thread - 256 threads per block = 8192 float4 loads - Per warp (32 threads): 1024 float4 loads = 4096 floats = 16KB memory traffic per pass - 2 passes = 32KB + 16KB writes = 48KB per block - At 1056 concurrent blocks: 48MB of memory traffic in flight - At 3.35 TB/s, 48MB takes 14 microseconds Hmm, 14us for memory, plus compute. For the whole problem (32768 blocks in 31 waves): 31 * 14us = 434us. Very fast! But this assumes perfect bandwidth utilization. With poor coalescing, maybe we get 1 TB/s effective. Then 48MB / 1TB/s = 48us per wave, total ~1.5ms. Still, that's probably competitive with the reference. Wait, I just thought of something. What if we launch 32 blocks per row (one per chunk of 1024 elements), and use an algorithm to combine the chunks? Ala CUB's multi-block scan: - Each block scans its 1024-element tile - Store tile sum in a global array - Scan the global array of tile sums (can be done with one block or recursively) - Each block reads its tile offset and adds to all elements For 32768 rows and 32 tiles per row: - 32768 * 32 = 1,048,576 blocks in pass 1 - That's a LOT of blocks. But H100 handles millions of blocks fine. - Each block: 256 threads, 1024 elements = 4 elements per thread Actually with 256 threads and 1024 elements, each thread handles 4 elements. Wait, with 1024 elements and 256 threads, each thread handles only 4 elements. That's not enough to amortize the scan overhead. Better to use fewer threads. With 128 threads and 1024 elements: 8 elements per thread. With 64 threads: 16 elements per thread. For CUB-style multi-block scan: Pass 1: Each tile block loads its 1024 elements, computes local prefix and tile sum. - 1024 threads is natural for this (1 element per thread, shared memory scan) - Or 256 threads with 4 elements per thread (sequential within thread, then warp scan) Pass 2: Scan the tile sums. For 32768 rows * 32 tiles = 1,048,576 tile sums. We can scan each row's 32 tile sums independently. - Launch 32768 blocks, each scanning 32 floats. Pass 3: Each tile block loads its local prefix, adds tile offset, writes back. This is 3 passes over the data. But CUB has optimized this to minimize passes and use clever synchronization. For a custom kernel, 3 kernel launches might have overhead. And the extra passes might hurt. I think the single-block-per-row approach with 2 internal passes is simpler and might be more efficient due to avoiding kernel launch overhead and global communication. Let me try to write a kernel with 256 threads and 128 elements per thread, using float4 for memory access. Actually, wait. Let me verify float4 alignment. With 256 threads per block and each thread handling 128 contiguous elements: - Thread 0: elements 0..127 - Thread 1: elements 128..255 - ... Thread 0 begins at element 0, which is aligned to float4 (element 0 is address 0). Thread 1 begins at element 128, which is 512 bytes, also aligned to 16 bytes. Great! All threads start at float4-aligned addresses. For loading 128 elements as float4: each thread does 32 float4 loads. Addresses: base, base+4, base+8, ..., base+124 (in terms of floats). In terms of bytes: base*4, base*4+16, base*4+32, ... For thread 0: bytes 0, 16, 32, ..., 496, 512, ..., 1008... wait, 128 floats = 512 bytes. base = 0. float4 indices: 0, 1, 2, ..., 31. Addresses: 0, 16, 32, ..., 496. Actually, 32 float4 loads covers 32 * 4 = 128 floats. Perfect. For thread 1: base = 128. float4 indices: 32, 33, ..., 63. Addresses: 512, 528, ..., 1008. For a warp of 32 threads, on the first float4 load: - Thread 0: address 0..15 - Thread 1: address 512..527 - Thread 2: address 1024..1039 - ... - Thread 31: address 15872..15887 These are spaced by 512 bytes (or 128 floats). The L2 cache line is 128 bytes. So each thread accesses its own cache line. And the minimum transaction might indeed be 32 bytes. So each float4 load by a thread requires one or two 32-byte transactions. But here we have 32 threads in a warp, each trying to load from a different cache line. The warp issues 32 separate memory requests. The memory subsystem processes them in parallel if possible. For the next float4 load (j=1): - Thread 0: address 16..31 - Thread 1: address 528..543 - ... Now thread 0 accesses bytes 16..31, which might be in the same cache line as its previous load (bytes 0..15). Same for other threads. So for j=0: threads load cache lines 0, 4, 8, ..., 124 (in 128-byte cache line units). Actually, 512 bytes = 4 cache lines. Thread i accesses cache line i*4. For j=1: thread i accesses cache line i*4, bytes 16..31. Same cache line as j=0! Oh! This means for j=0 and j=1, the threads access the SAME cache lines, just different offsets within those lines. The cache lines were fetched on j=0 and are still in L1/L2 for j=1! If we do sequential float4 loads within each thread's chunk, the first load (j=0) fetches the cache line. Loads j=0 to j=7 (covering bytes 0..127 of each thread's region) all hit the same cache line! So we're getting good cache reuse within each thread. Wait, but different threads in a warp access DIFFERENT cache lines. Thread 0 touches cache line 0, thread 1 touches cache line 4, etc. So the warp touches 32 different cache lines. The first warp load fetches 32 cache lines. The next 3 warp loads (j=1,2,3) hit the same 32 cache lines. Then j=4 touches cache line 1 (for thread 0), which is a new cache line. And so on. So within a warp, threads don't share cache lines. But each thread's sequential loads DO reuse its own cache lines. For thread 0: cache lines 0, 1, 2, 3 (for j=0..31) For thread 1: cache lines 4, 5, 6, 7 ... For thread 31: cache lines 124, 125, 126, 127 A warp touches cache lines 0..127. That's 128 cache lines * 128 bytes = 16KB. L1 cache per SM is 256KB (I think for H100). L1 can easily hold 16KB. So if the warp is resident on the same SM, the L1 cache retains these 128 cache lines across the 32 load instructions. But wait, with 8 concurrent warps per SM (if 8 blocks per SM and 1 warp per block... no, 256 threads = 8 warps per block). 8 blocks per SM * 8 warps = 64 warps. They all compete for L1. Actually, each block handles a different row. Different blocks on the same SM might read different rows, causing L1 cache thrashing. But L1 can hold 256KB, and a block reads 128KB. So 2 blocks' worth fits in L1. With 8 blocks competing, there's thrashing. This is getting complex. Let me just write the kernel and test it. Actually, I want to try another approach. What about using shared memory to improve coalescing? We can load data into shared memory in a coalesced pattern, then process it from shared memory. For 256 threads and 128 elements per thread: we want each thread to read 128 elements. Total elements per block = 32768 = 128KB of shared memory. We have 227KB max per block. So we COULD allocate 128KB of shared memory per block. Algorithm: 1. Load row into shared memory in a coalesced pattern - On iteration j, all 256 threads load element threadIdx.x + j*256 - Store at shared_mem[threadIdx.x + j*256] - This is perfectly coalesced! 2. After loading the entire row into shared memory, do the scan in shared memory 3. Write back in a coalesced pattern But step 2 requires scanning 32768 elements in shared memory. With 256 threads doing a Brent-Kung scan: log2(32768) = 15 steps up, 15 steps down. Each step is a few shared memory accesses and a sync. With 30 syncs and shared memory operations... on H100 shared memory is very fast. But 30 syncs for 32768 elements = only ~1000 elements processed per sync, which seems like overhead. Wait, but since we can do warp-level scans without sync, we can reduce the number of syncs. Hybrid approach: 1. Load row into shared memory (coalesced, 128 iterations) 2. Each thread reads its 128 elements from shared memory (contiguous) and computes local prefix 3. Store thread sum in shared memory 4. Warp scan of thread sums using shfl 5. Inter-warp scan using shared memory 6. Write back from shared memory (coalesced) Wait, the shfl scan uses registers, not shared memory. But we already have the entire row in shared memory! We could just do a scan directly in shared memory. Actually, let me think about the simplest correct approach: 1. Load entire row into shared memory (coalesced) 2. Do a scan in shared memory using the classic algorithm 3. Write back (coalesced) For step 2, with 32768 elements and 256 threads: Brent-Kung up-sweep: ``` for (int stride = 32768/2; stride > 0; stride /= 2) { __syncthreads(); if (threadIdx.x < stride) { int idx = 2 * (threadIdx.x + 1) * some_factor - 1; // Actually with 256 threads and 32768 elements, each thread needs to work on 128 elements } } ``` This is complicated because 256 threads can't directly operate on 32768 elements in a standard Brent-Kung where each thread operates on one element. We'd need a loop over elements. Alternatively, use Kogge-Stone style: ``` float val = shared[tid]; for (int offset = 1; offset < 32768; offset *= 2) { __syncthreads(); shared[tid] = val; __syncthreads(); if (tid >= offset) val += shared[tid - offset]; } ``` But with 32768 elements, that's 15 iterations, each with 2 syncs. 30 syncs total. And `tid` is only 0..255. We'd need to loop: ``` for (int base = 0; base < 32768; base += 256) { int idx = base + tid; float val = shared[idx]; for (int offset = 1; offset < 32768; offset *= 2) { // This doesn't work because offset spans the whole array } } ``` No, a Kogge-Stone scan on 32768 elements would need all 32768 threads active simultaneously, then reduce to 256. That doesn't work. What about a hierarchical scan using the fact that 32768 = 128 * 256? Level 1: 256 threads, each scans 128 contiguous elements (sequential) Level 2: Scan the 256 thread sums Level 3: Add offsets to the 128 elements per thread This is exactly what we described earlier, but we can do it with shared memory instead of global memory for the intermediate data! Actually, we don't even need shared memory for the elements if they fit. But they do fit in shared memory. Wait, but if the elements are already in shared memory, the sequential scan within each thread's 128 elements is fast because shared memory is fast. And we don't need to reload from global memory. Algorithm using shared memory: 1. Coalesced load of 32768 floats into shared memory - 128 iterations, each thread loads one float per iteration - All loads are perfectly coalesced - 128 syncs? No, we can do: ``` for (int i = 0; i < 128; i++) { int global_idx = row * 32768 + i * 256 + threadIdx.x; int shared_idx = i * 256 + threadIdx.x; shared[shared_idx] = input[global_idx]; } __syncthreads(); ``` This loop has no sync inside it! Each thread does 128 loads. The loads are coalesced across the warp because on each iteration, thread j accesses global address `base + j`, which is consecutive. 2. Scan in shared memory. Now we have 32768 elements in shared memory. Each element is at index `i * 256 + threadIdx.x` where i=0..127. We need to compute prefix sums along the row. The row is laid out consecutively in shared memory: elements 0..32767 are at shared[0..32767]. Each thread can scan its 128 elements sequentially: ``` float prefix = 0.0f; for (int i = 0; i < 128; i++) { int idx = i * 256 + threadIdx.x; prefix += shared[idx]; shared[idx] = prefix; } ``` But this is NOT a parallel scan. Each thread only scans elements at indices `threadIdx.x, threadIdx.x + 256, threadIdx.x + 512, ...`. These elements are scattered throughout the row, not contiguous. The result would be wrong. To do a correct scan, we need elements in order. Thread 0 should process elements 0..127, thread 1 should process 128..255, etc. But in shared memory, elements are stored at `i * 256 + threadIdx.x`. Thread 0's elements are at indices `0, 256, 512, ...`. NOT contiguous. So if we store in this strided layout, a sequential scan by each thread is scanning strided elements, which is wrong for row order. We need the shared memory layout to match the row order: elements 0..32767 linearly. On load: ``` for (int i = 0; i < 128; i++) { int global_idx = row * 32768 + i * 256 + threadIdx.x; int shared_idx = i * 256 + threadIdx.x; shared[shared_idx] = input[global_idx]; } ``` This stores elements as: - shared[0] = element 0 (thread 0) - shared[1] = element 1 (thread 1) - ... - shared[255] = element 255 (thread 255) - shared[256] = element 256 (thread 0) - shared[257] = element 257 (thread 1) - ... So the shared memory IS in row order! Elements 0..32767 are stored at shared[0..32767] in order. Wait, `shared_idx = i * 256 + threadIdx.x`. - i=0: indices 0..255 store elements 0..255 - i=1: indices 256..511 store elements 256..511 - ... Yes! The elements are stored consecutively in shared memory in row order. Now for scan, thread j normally handles elements j*128 to j*128+127. But in shared memory, these are contiguous. So thread j CAN scan them sequentially in shared memory. ``` int start = threadIdx.x * 128; int end = start + 128; float prefix = 0.0f; for (int i = start; i < end; i++) { prefix += shared[i]; // But we can't write back yet because other threads need original values } ``` Actually, we can use a two-pass approach: Pass 1: Each thread reads 128 elements and computes sum. ``` float sum = 0.0f; for (int i = start; i < end; i++) { sum += shared[i]; } ``` Pass 2: Cross-thread scan of sums using warp shuffles. Pass 3: Each thread scans its 128 elements again and writes prefix + offset. ``` float prefix = offset; for (int i = start; i < end; i++) { prefix += shared[i]; shared[i] = prefix; } ``` After this, shared memory contains the prefix sums. Then write back: ``` for (int i = 0; i < 128; i++) { int shared_idx = i * 256 + threadIdx.x; int global_idx = row * 32768 + i * 256 + threadIdx.x; output[global_idx] = shared[shared_idx]; } ``` Coalesced write! Total shared memory used: 32768 floats = 128KB. With H100 allowing 227KB per block, we can have 1 block per SM (shared-memory limited). Threads: 256 threads per block. Thread-limited occupancy: 2048 / 256 = 8 blocks per SM. But shared memory limits to 1 block per SM. So actual occupancy: 1 block per SM = 256 threads per SM. With 132 SMs: 132 concurrent blocks. 32768 rows / 132 = 248 waves. Per block work: - Load: 32768 coalesced loads (128 iterations * 256 threads, perfect coalescing) - Pass 1: 128 reads from shared memory per thread = 32768 reads (fast, no bank conflicts if we handle offsets carefully... wait, threads read contiguous memory. Thread 0 reads shared[0..127], thread 1 reads shared[128..255], etc. These are in different shared memory banks. On H100, shared memory has 32 banks. Address i is in bank i % 32. Thread 0 accesses banks 0..127 % 32, which is no issue. But multiple threads reading from shared memory: thread 0 reads shared[0], thread 1 reads shared[128] = bank 0. So threads 0 and 1 both access bank 0! Bank conflict! Actually, let me check. For Pass 1, thread j reads shared[j*128 + i] for i=0..127. - Thread 0: shared[0], shared[1], ..., shared[127] (banks 0..127 % 32) - Thread 1: shared[128], shared[129], ..., shared[255] (banks 0..127 % 32, since 128 % 32 = 0) So on the first access (i=0): - Thread 0 reads bank 0 - Thread 1 reads bank 0 - Thread 2 reads bank 0 (shared[256] = 256 % 32 = 0) - ... ALL threads access bank 0! 32-way bank conflict! This is TERRIBLE for shared memory performance. So accessing contiguous 128-element chunks from shared memory causes bank conflicts on H100 (32 banks). To avoid bank conflicts, we need threads in a warp to access different banks. With 32 banks and 4 bytes per float, consecutive floats in shared memory are in consecutive banks. If thread j accesses shared[j], then threads 0..31 access banks 0..31: no conflict. If thread j accesses shared[j * 128], then thread j accesses bank (j * 128) % 32 = 0 (since 128 = 4 * 32). All threads access bank 0: 32-way conflict. Solution: transpose the data or use a different assignment. If thread j accesses shared[j * 4], banks are (j*4)%32 = 0, 4, 8, ..., 28 for j=0..7. Then j=8: bank 0 again. So 4-way conflict. Actually, the access pattern `shared[j * stride]` has bank conflicts when `stride % 32 == 0` for 32-way, `stride % 8 == 0` for 4-way, etc. For our case, stride = 128, and 128 % 32 = 0, 128 % 8 = 0. 128 / 32 = 4. So each bank gets 4 accesses per cycle? Actually, for 32 threads and stride 128: thread j accesses bank (j*128)%32 = 0. All 32 threads access bank 0. On H100, each bank can serve 1 request. So 32 requests to 1 bank = 32 cycles to serve. This is terrible. To fix this, we could use shared memory with padding. If the row is 32768 elements, we could pad to 32768 + some amount. But that increases shared memory usage. Or we can change the thread assignment so threads in a warp access different banks. Instead of thread j accessing elements j*128 to j*128+127, what if thread j accesses elements starting at j*4 + (j/8)*128? That's getting complicated. Actually, if we want thread j to read a contiguous chunk, and avoid bank conflicts when multiple threads read simultaneously, we need the starting addresses of different threads' chunks to be in different banks. Bank of shared[j*chunk_size] = (j*chunk_size*4) % 128 / 4 = (j*chunk_size) % 32. For no conflict: j*chunk_size should be different mod 32 for j=0..31. This requires chunk_size to be coprime with 32. If chunk_size = 128: 128 % 32 = 0. Not coprime. Terrible. If chunk_size = 127: 127 % 32 = 31. gcd(127,32)=1. Good! Thread j accesses bank (j*127)%32, which cycles through all banks. If chunk_size = 129: 129 % 32 = 1. Also good! So using chunk sizes like 129, 127, etc. avoids bank conflicts. But 32768 / 127 = 257.6, not integer. 32768 / 129 = 253.6, not integer. What about chunk_size = 64? 64 % 32 = 0. Still bad. chunk_size = 65: gcd(65,32) = 1. Good! 32768 / 65 = 504.1, not integer. chunk_size = 33: gcd(33,32) = 1. Good! 32768 / 33 = 992.96, not integer. chunk_size = 31: gcd(31,32) = 1. 32768 / 31 = 1057.03. chunk_size = 1: No bank conflict. But then we need 32768 threads, too many. What chunk size divides 32768 and is coprime to 32? 32768 = 2^15. Any divisor of 32768 is a power of 2. Any power of 2 > 1 shares a factor with 32. So NO power-of-2 chunk size is coprime to 32. This means if we assign contiguous chunks of size power-of-2 to threads, the starting addresses will have bank conflicts. But wait, in practice, threads don't all start reading simultaneously if they're in a sequential loop. The bank conflict only matters when threads in a warp access shared memory in the same instruction. In our Pass 1 and Pass 3, each thread is in a for loop reading shared memory: ``` for (int i = 0; i < chunk_size; i++) { float val = shared[start + i]; } ``` Threads in a warp are at possibly different `i` values (if they diverge) or the same `i` value if they're in lockstep. Since all threads execute the same loop body and there's no divergence, at instruction `i`, all threads read `shared[start + i]`. For `i=0`: threads read `shared[start]`, `shared[start+128]`, `shared[start+256]`, ... As we established, these all map to bank 0. So 32-way bank conflict. To avoid this, we'd need threads in a warp to access elements that are in different banks. This means either: 1. The chunk size is not a multiple of 32 2. Or we use a different access pattern What if we pad shared memory? Allocate 32768 + 32 floats, and shift each thread's chunk by 1 float: Thread j accesses shared[j*129 + i] for i=0..127. Total shared memory: 256 * 129 = 33024 floats = 132KB. Still within limits. Bank of shared[j*129] = (j*129) % 32 = j*1 % 32 = j. So thread j starts in bank j, no bank conflict! On load, we write strided but with padding: ``` for (int i = 0; i < 128; i++) { int global_idx = row * 32768 + i * 256 + threadIdx.x; int shared_idx = i * (256 + 1) + threadIdx.x; // Padded layout shared[shared_idx] = input[global_idx]; } ``` Wait, this stores elements in a padded strided layout, not row order. And then for scan, we'd need to read in order... it gets complicated. Actually, maybe bank conflicts in shared memory aren't a huge deal for this stage because shared memory is so fast. A 32-way bank conflict might take ~32 cycles, but that's still only about 30ns. For 128 iterations, that's 3.8 microseconds per pass. For 3 passes: ~11us. Still very fast compared to global memory. And actually, for the load phase, we're reading from global memory which is the bottleneck. The compute and shared memory phases are much faster. Let me go back to the simpler approach without loading everything into shared memory. ### Simplest viable approach 256 threads per block. Each thread handles 128 contiguous elements. Compute chunk sum, do warp/inter-warp scan, write prefix. This uses minimal shared memory, good register efficiency, and processes 1 row per block. Memory is read twice per element. Global memory bandwidth is the bottleneck. Let's write this kernel and see what performance we get. ```cuda __global__ void cumsum_kernel(const float* input, float* output, int rows, int cols) { int row = blockIdx.x; if (row >= rows) return; int tid = threadIdx.x; const int chunk_size = 128; int base = row * cols + tid * chunk_size; // Pass 1: compute chunk sum float sum = 0.0f; #pragma unroll for (int i = 0; i < chunk_size; i++) { sum += input[base + i]; } // Intra-warp scan of chunk sums int lane = tid & 31; int warp = tid >> 5; float warp_scan = sum; #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, warp_scan, offset); if (lane >= offset) warp_scan += other; } __shared__ float warp_totals[8]; if (lane == 31) { warp_totals[warp] = warp_scan; } __syncthreads(); // Inter-warp scan if (warp == 0) { float val = (lane < 8) ? warp_totals[lane] : 0.0f; float original = val; #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float other = __shfl_up_sync(0xFFFFFFFF, val, offset); if (lane >= offset) val += other; } if (lane < 8) { warp_totals[lane] = val - original; } } __syncthreads(); float offset = (warp > 0 ? warp_totals[warp] : 0.0f) + warp_scan - sum; // Pass 2: write prefix with offset float prefix = offset; #pragma unroll for (int i = 0; i < chunk_size; i++) { prefix += input[base + i]; output[base + i] = prefix; } } ``` Wait, there's an issue. `warp_scan - sum` is the intra-warp prefix before this thread's chunk. `warp_totals[warp]` is the inter-warp prefix (sum of all previous warps). For warp 0, thread 0: warp_scan = sum0, warp_totals[0] = 0. offset = 0 + sum0 - sum0 = 0. Correct. For warp 0, thread 1: warp_scan = sum0 + sum1. offset = 0 + sum0 + sum1 - sum1 = sum0. Correct. For warp 1, thread 0: warp_scan = sum_w1_0. warp_totals[1] = sum of warps before warp 1 = total of warp 0. offset = total_warp_0 + sum_w1_0 - sum_w1_0 = total_warp_0. Correct. Great! This looks correct. One issue: `input[base + i]` is loaded twice. For cache efficiency, maybe we should use `__ldg()` for the read-only loads to use the read-only cache. But `__ldg` requires const __restrict__ pointers and specific compilation flags. Actually, in CUDA, `const float* __restrict__ input` allows `__ldg`. But `load_inline` might not support `__restrict__` easily. Let me just use regular loads. The compiler might optimize them. For reducing register pressure, we could also try to coalesce loads manually with float4. Let me add some optimizations: 1. Use float4 for loading in pass 1 and pass 2 to reduce instruction count 2. Unroll the inner loops With float4, each thread handles 128 elements = 32 float4s. Thread j accesses float4 at base/4 + i for i=0..31. For coalescing of float4: thread 0 accesses address 0, thread 1 accesses 512, etc. Like we discussed, this is not coalesced across the warp. But it might still be OK. Actually, let me not overcomplicate with float4 for now. Just write the simple version and test. Let me write the complete Python module with ModelNew class and the kernel.

The basic correctness check passes. Now let me make sure the static check passes and then submit the kernel. But wait, the numerical tolerance for fp32 is 1e-4. The max_abs_err=8.20e-02 seems high. Let me think about whether this is a real issue. Actually, looking more carefully, cumsum in fp32 can accumulate error. PyTorch's cumsum might use different algorithms (thread-level scan can have different rounding). With 32768 elements being summed, the accumulated rounding error can be quite high. Let me check if the error is within tolerance still. The max_abs_err is 8.20e-02 which is greater than 1e-4. Wait, but the test said it passed. Let me re-read the output: "run_correctness PASSED: (5 / 5) trials all matched the reference." It says it passed! But the max_abs_err is 8.20e-02. Maybe the tolerance for this specific check is different, or perhaps I should still be concerned. Actually, let me think about this more carefully. The tolerance for fp32 is atol=rtol=1e-4. If the max absolute error is 0.082, that's 820x the tolerance. But the test still says it passed. Hmm, actually the tool might use a different tolerance in its internal check, or perhaps I should just proceed. Let me re-examine my kernel to make sure it's correct. Actually, wait - the cuda correctness tests do match numerically, the report shows the error. Let me proceed with the static check and submit. Maybe the test framework has a looser tolerance or uses relative error. Actually, I realize the tolerance check is: if torch.allclose(a, b, atol=1e-4, rtol=1e-4). For large tensors with cumulative sums around 15000, the relative error times magnitude could be: 1e-4 * 15000 = 1.5. And max_abs_err = 0.082 < 1.5, so it should pass using atol directly as well. Actually, let me also think about potential issues: our kernel should handle cases where dim is not 1. But based on the problem, dim = 1 and the input is 2D. Still, let me make the kernel handle general cases or at least work for dim=1. Since the input is (32768, 32768) and dim=1, my kernel should handle this. Let me also think about optimization: 1. Each row of 32768 elements is processed by 256 threads, each handling 128 elements. 2. This makes 32768/256 = 128 chunks. Each chunk is 128 elements. 3. Wait, 256 * 128 = 32768. That's correct! Now let me also consider if we should handle some edge cases or make the kernel a bit more robust. Let me also think about whether there are performance issues. The kernel launches 32768 blocks, one per row. Each block has 256 threads. On H100, each SM can theoretically handle many blocks. With 32768 blocks, we have enough parallelism. However, one potential optimization: each thread accesses memory in a strided pattern (tid * chunk_size to (tid+1) * chunk_size - 1). This pattern might hurt memory coalescing. Let me think about whether this matters. Actually for the sequential scan within a chunk, strided access by thread index is bad for coalescing. But for prefix sum, this might be the best approach anyway since each thread processes a contiguous chunk. Let me think about memory access pattern: - Thread 0 accesses elements 0..127 - Thread 1 accesses elements 128..255 - etc. When warp 0 executes, threads 0..31 are accessing elements 0, 128, 256, ..., 3968. This is stride-128 access which is terrible for coalescing! Actually, I should redesign for better coalescing. A better approach: - Have threads load data in a coalesced manner first - Each thread loads consecutive elements (thread i accesses elements i, i+32, i+64, ... in coalesced groups within the warp) - But for prefix sum, we want each thread to work on a contiguous range Actually, let me think about this more carefully. With global memory on H100: - Each warp executes a 128B load transaction when threads access consecutive floats - For 32 threads, if each thread loads a float4, that's 128*4 = 512 bytes, requiring 4 transactions If threads access with stride 128: - Thread 0 accesses 0, thread 1 accesses 128, etc. - This would be much worse for cache lines But let me think about whether the memory bandwidth is actually a bottleneck or not. For the 2-pass approach: - Pass 1: Read input, compute chunk sums (read 32768*32768*4 bytes ~ 4GB, write nothing) - Pass 2: Read input again, write output (read 4GB, write 4GB) - Total memory: ~12GB At 3.35TB/s, this would take ~3.6ms. If my kernel takes much longer, it's computation bound. If it's close to this, it's bandwidth bound. Actually, let me just submit and see. But first let me run the static check. Actually, I want to think about one more optimization: warp-level optimization. With 256 threads and 32768 elements per row, we can also increase block size or change chunk size. But the fastest configuration might just vary. Let me also think about using shared memory more efficiently. Currently, inter-warp communication uses shared memory for only 8 floats (warp_totals). This is very efficient. Now, one concern: the problem asks for numerical equivalence at fp32 precision. My algorithm accumulates the prefix sum in the order: 1. Sum each chunk (128 elements) 2. Compute prefix of chunk sums 3. Use those prefixes to compute element-level prefix This is different from PyTorch's cumsum which computes a single sequential scan in some order. The numerical results can differ due to floating point non-associativity. But the test passed, so it should be fine. Let me just submit. Actually, let me first do the static check to make sure there are no issues.

The static check passed. Now let me submit the kernel. The current implementation uses a two-pass algorithm with 256 threads per block, where each thread handles a chunk of 128 elements. The first pass computes chunk sums, does a hierarchical scan (intra-warp and inter-warp), then the second pass writes prefix sums with the correct offset. The correctness check passed with max absolute error 8.20e-02 which is within the 1e-4 tolerance for fp32? Wait, that seems high. But actually, cumsum accumulates floating point errors naturally. The reference torch.cumsum also accumulates errors. Let me re-check what the tolerance is... The task says fp32 uses atol=rtol=1e-4. But the error reported is 8.20e-02 which is much larger than 1e-4. Yet the run_correctness says "PASSED". So the evaluation might be using a different tolerance, or the comparison is against torch.cumsum with same accumulation order but my implementation matches it closely. Wait, actually the error is 8.20e-02 for absolute error on a tensor of size 32K x 32K with random values between 0 and 1. The expected cumsum values can be up to ~32768. On such large values, 0.08 absolute error is about 2.4e-6 relative, which is well within fp32 precision. So it's fine. Let me submit this kernel. But wait - I should consider if there's a performance issue. The kernel launches 32768 blocks with 256 threads each. That's 8,388,608 threads total. On H100 with 132 SMs, that's about 63,552 threads per SM, which seems reasonable (2048 threads per SM max on Hopper? Actually H100 has 2048 threads per SM). So we're using about 31 blocks per SM, which is a decent amount of parallelism but might not fully saturate the GPU. Actually, H100 SMs can handle many concurrent thread blocks. But the overall memory bandwidth is what matters. Since each element is read twice and written once (2 reads + 1 write = 12 bytes per element), at 3.35 TB/s that's a lower bound of 3.75ms. My kernel might be around that. One concern: I use contiguous #pragma unroll with 128 iterations, which might increase register pressure and reduce occupancy. But since we only have 256 threads, it should be fine. Let me submit and see the timing.