We need to write a custom CUDA kernel that performs a parallel associative scan (linear recurrence) over a 1D sequence. The recurrence is: h_0 = b_0 h_t = a_t * h_{t-1} + b_t for t > 0 Given a of shape (B, L, D), b of shape (B, L, D). Output is (B, L, D) where each channel independently has this recurrence. We need to produce numerically equivalent outputs at fp32 precision, and run faster than the reference sequential PyTorch implementation. We need to write a complete Python file defining ModelNew, using torch.utils.cpp_extension.load_inline to compile CUDA source. We need to be careful: the evaluation environment likely runs on a specific GPU. We need to get GPU specs to tailor optimization (e.g., shared memory size, number of SMs, etc.). We should call get_gpu_specs first. The problem: we need to implement a parallel scan. The naive sequential PyTorch loop over t from 0 to L-1, with B=32, D=256, L=4096. That's 32*256=8192 independent scans, each of length 4096. The PyTorch reference does sequential loop in Python, appending to list, which is very slow. Even a simple CUDA kernel that does each scan sequentially within a thread would be much faster because it avoids Python overhead and uses GPU parallelism. But we can do better with a parallel scan within each channel to reduce O(L) to O(log L) steps per channel. However, each scan is only length 4096, and we have 8192 scans. With 4096 threads per block? Let's think about occupancy. A simple approach: each thread handles one (b,d) pair, and loops sequentially over L. Since L=4096, each thread does 4096 multiplications and additions. That's 8192 threads total. GPU can run many threads in parallel. This is very memory bound: reading a and b, writing output. For each thread, memory accesses are coalesced across threads if we store data as (B, L, D) with D as last dimension: a[b, t, d] where d is contiguous. If we launch threads with d as x-dimension and b as y-dimension, then for a fixed t, accesses to a[:, t, :] are contiguous across d. But sequential loop over t means each thread accesses a[:, t, d] for t=0..L-1. The memory layout is (B, L, D). The stride for t is D. So for a fixed b and varying t, the addresses are separated by D * element_size. Within a warp, consecutive d values correspond to consecutive addresses. So for a given t, all threads in a warp read consecutive memory: coalesced. Good. For each thread, we read a and b for each t, compute h, write output. Total memory: B*L*D*3*4 bytes = 32*4096*256*12 bytes = about 402 MB? Let's compute: 32*4096 = 131072; *256 = 33,554,432 elements per tensor. Each tensor is 128 MB (since 33.5M * 4 bytes = 134.2 MB). Three tensors (a, b, out) => 402 MB. At memory bandwidth ~1-2 TB/s, kernel would take ~200-400 microseconds. That's likely fast enough. But perhaps we can do even better using shared memory or warp shuffle to do parallel scan within blocks. However, with B*D = 8192 independent scans, each of length 4096, we can assign one warp (32 threads) per scan, and do a parallel scan using warp shuffle. Then we have 8192 warps = 262144 threads. GPU can handle that. But memory bandwidth is likely the bottleneck anyway. A sequential per-thread scan might already be close to bandwidth limit. However, a warp-scan might be slightly faster because it reduces the number of sequential steps per thread from 4096 to ~128 (if using warp shuffle) or ~log2(4096)=12 steps if full parallel. But each step involves reading memory? Actually, data is already in registers/shared memory after loading. We could load the entire sequence into shared memory per warp/block. 4096 floats = 16 KB per scan. With 8192 scans, that's too much for shared memory. But we can tile: process the sequence in chunks. For example, each block processes a batch of scans, each warp processes one scan in chunks of, say, 256 elements. Use warp shuffle scan within chunk, then apply prefix correction across chunks. That reduces register pressure? Let's think. Alternatively, we could use a block-level scan with 256 threads per scan, but we have 8192 scans; if we use 256 threads per scan, total threads = 2M, which is okay but might have too many blocks? Actually 8192 blocks of 256 threads each is okay. But shared memory per block: storing 4096 elements for 256 threads? If each thread loads 16 elements, total shared memory = 4096*4 = 16KB per block. That's fine. Each block could handle one scan. 8192 blocks, each of 256 threads. The parallel scan within each block would be efficient. However, each block only processes one (b,d) pair. With 8192 blocks, that's fine. But is this more efficient than a sequential thread? Probably yes, because it uses more parallel instructions and shared memory for reduction, but memory bandwidth is same. However, the sequential per-thread scan has simple instruction mix and high memory throughput. The parallel scan might have extra shared memory barriers and bank conflicts. But given the reference is a Python loop, any compiled CUDA kernel will likely beat it by orders of magnitude. We need to be careful about correctness: fp32. The recurrence is h_t = a_t * h_{t-1} + b_t. This is associative? Let's check: the state can be represented as an affine transformation: (h) -> a * h + b. Composition of two steps (a1,b1) and (a2,b2) yields: h' = a2*(a1*h + b1) + b2 = (a2*a1)*h + (a2*b1 + b2). So the associative operator on tuples (a,b) is: (aL, bL) ∘ (aR, bR) = (aL * aR, aL * bR + bL) Wait, careful about order. If we have two segments: first segment maps h -> a1 * h + b1. Second maps h -> a2 * h + b2. Then combined mapping is h -> a2*(a1*h + b1) + b2 = (a2*a1)*h + (a2*b1 + b2). So the tuple for combined is (a2*a1, a2*b1 + b2). So the operator is: pair_left = (a_left, b_left) pair_right = (a_right, b_right) combined = (a_right * a_left, a_right * b_left + b_right) Alternatively, if we think of scan order left to right, we want to combine pair_i (a_i, b_i) such that applying them in order yields the transformation. The scan of tuples (a_t, b_t) with this operator will give at each position the cumulative transformation from start to that position. Then h_t = cum_a_t * h_{-1}? Actually initial h is b_0: h_0 = b_0. For t>0, h_t = a_t * h_{t-1} + b_t. We can define pair_0 = (1, b_0)? Wait, h_0 = b_0 = 1 * 0 + b_0 if we start with h_{-1}=0. But in our recurrence, h_0 is just b_0. Then for t=1, h_1 = a_1 * h_0 + b_1. So if we consider cumulative transformation from h_{-1}=0, we need pair_0 = (0, b_0)? No, h_0 = b_0. That can be seen as applying transformation (a=0? no) to some initial state? Actually we can treat the scan of tuples (a_t, b_t) with identity (1,0). The cumulative tuple after t is (A_t, B_t) such that h_t = A_t * h_{init} + B_t. If h_{init}=0, then h_t = B_t. Let's verify: At t=0: identity combined with (a_0, b_0) = (a_0*1, a_0*0 + b_0) = (a_0, b_0). So (A_0, B_0) = (a_0, b_0). h_0 = A_0 * 0 + B_0 = b_0. Good. At t=1: combine (A_0,B_0) with (a_1,b_1) = (a_1*A_0, a_1*B_0 + b_1) = (a_1*a_0, a_1*b_0 + b_1). h_1 = A_1 * 0 + B_1 = a_1*b_0 + b_1. Correct. At t=2: combine with (a_2,b_2) = (a_2*a_1*a_0, a_2*(a_1*b_0+b_1)+b_2). h_2 = B_2. Correct. So the associative scan on tuples (a_t, b_t) with identity (1,0) and operator combine(left, right) = (right.a * left.a, right.a * left.b + right.b) yields B_t = h_t. So we just need to compute the exclusive or inclusive scan? Actually inclusive scan: the cumulative transformation at position t is the combination of pairs from 0..t. The operator as defined combines left segment then right segment. If we do an inclusive scan over the array of pairs (a_t,b_t) with identity (1,0), the result at position t is the combined transformation for 0..t, and the B component is h_t. So the output is just the B component of the scanned tuples. Alternatively, we can compute h sequentially. For small L and many channels, sequential is fine. Given that we have 8192 independent scans of length 4096, a simple sequential kernel where each thread handles one scan is extremely easy to write and likely near-optimal because it's memory-bound. But we need to ensure we saturate memory bandwidth. With 8192 threads, we only have 8192 * 32 = 262k concurrent operations (the loop is 4096 iterations). But memory latency might be hidden by many warps? Actually each thread loops 4096 times, so each warp issues 4096 loads sequentially. There are 8192/32 = 256 warps. That's not a huge number; on a GPU with 80 SMs, each SM would have ~3 warps. That might not fully hide latency. But each warp has a lot of instruction-level parallelism? Not really, because each load depends on the previous computation (h = a*h + b). So the critical path is long. However, the compiler might unroll and interleave instructions from different iterations? No, because it's a loop with dependency on h. So each thread is effectively sequential. Memory latency per load might not be fully hidden if there aren't enough active warps per SM. If we use a warp-level parallel scan, each warp of 32 threads handles one scan of length 4096. The 4096 elements are processed in chunks. Within a chunk of 32 elements, each thread handles one element, and we use warp shuffle to do a parallel prefix scan. Then across chunks, we maintain a running prefix. This reduces the sequential dependency from 4096 steps per thread to 128 steps (4096/32 = 128 chunks) where each chunk uses warp shuffle (log2(32)=5 steps). Actually within each chunk, each thread loads its element, and then we compute prefix sums within the warp. But the dependency on the previous chunk's final value is still serial across chunks: chunk 0's final value must be known before chunk 1 starts. However, within each chunk, the 32 threads cooperate, so each chunk takes about 5 shuffle steps plus loads/stores. The total number of serial steps is 128 chunks * a few steps. For the sequential per-thread approach, it's 4096 serial steps. So the warp approach reduces serial length by 32x. This is beneficial because it exposes more parallelism and better latency hiding. Alternatively, we can have a block of 256 threads handle one scan. Then each block processes 4096 elements in 16 iterations of 256 threads. Within each iteration, do a block-level scan using shared memory. Serial across 16 iterations. That's even fewer serial steps. But we have 8192 scans, so 8192 blocks. Each block has 256 threads. That is 2M threads, which is fine. Shared memory per block: need to store 256 elements (floats) = 1KB. Also need some reduction values. Very small. So occupancy is high. But launching 8192 blocks might have some overhead, but it's fine. However, the memory bandwidth is still the limit. The sequential per-thread kernel might already achieve near-peak bandwidth if the hardware has enough latency tolerance? Let's think more concretely. Consider a warp handling one scan sequentially. Each thread in the warp loads a different channel (different d). Wait, if one warp handles one scan (one (b,d) pair), then only one thread is active per warp, which is terrible. I meant one warp handles one channel but uses 32 threads to process 32 consecutive time steps in parallel. Let's clarify. Option 1: Each thread handles one (b,d) pair, loops over t sequentially. Launch B*D threads. Each thread processes L elements. Serial length = L. Number of warps = B*D/32 = 8192/32 = 256 warps. On a GPU with 80 SMs, that's 3.2 warps per SM. Memory latency is maybe 400 cycles. To hide latency, we need maybe 10 warps per SM. So 3 warps is insufficient. Therefore memory latency will not be hidden, and the kernel will be slower than peak bandwidth. However, each warp does a lot of instructions (4096 iterations). But each iteration has a dependent load: load a[t], load b[t], fma, store out[t]. The load latency might be ~400 cycles. With only 3 warps per SM, the SM might stall waiting for memory. So throughput will be low. Wait, but there are also many blocks. Each block has 32 threads? If we launch 8192 threads in blocks of 256, that's 32 blocks. Actually B*D=8192. If we use block size 256, then blocks = 8192/256 = 32 blocks. That's even fewer blocks: 32 blocks total. 32 blocks across 80 SMs means some SMs idle. Each block has 256 threads = 8 warps. 8 warps per SM * maybe 1 block per SM if distributed evenly (but only 32 blocks, so maybe 1 block per 2.5 SMs). Actually 32 blocks across 80 SMs: some SMs get 0 blocks. The scheduler can run multiple blocks per SM if resources allow. But with only 32 blocks total, maximum active blocks per SM is limited by the total number. So at most 32 blocks. If each block has 8 warps, total warps = 256. That's exactly the same as before. Wait, B*D=8192 threads total. Whether arranged as 32 blocks of 256 or 256 blocks of 32, total warps = 8192/32 = 256 warps. That's a very small number of warps for a GPU. A100 has 108 SMs. So 256 warps is ~2.4 warps per SM. That's definitely not enough to hide global memory latency. Thus a simple per-thread sequential kernel might be significantly slower than peak bandwidth. We need more warps per scan to increase parallelism. But wait: the problem says B=32, D=256, so B*D=8192. That's fixed by the benchmark. The reference uses a Python loop over L=4096, which is extremely slow. Even a kernel with 256 warps will be much faster than Python. But to maximize speedup, we should try to saturate the GPU. However, maybe the benchmark input sizes are fixed, and we just need to beat the reference. The reference is super slow because it's a Python loop and memory copies? Actually the reference does a Python loop over L, but each iteration does PyTorch ops on (B,D) tensors. Those ops launch kernels each time. So it's launching 4096 kernels, each processing 8192 elements. That's huge overhead. So even a single kernel with 256 warps that processes the whole thing sequentially will be orders of magnitude faster. The speedup might be 100x or more. But maybe the evaluation measures kernel runtime only, not the reference overhead? Actually the reference runtime is the PyTorch implementation. The speedup is reference_time / new_time. The reference time includes all that kernel launch overhead. So we can beat it easily. But we also want to produce a fast kernel for good SOL. The evaluation metrics include speedup and SOL. The problem statement says "Speedup vs the PyTorch reference and SOL (Speed of Light — the higher of DRAM and compute utilization, on a 0–1 scale) are the objectives." So we want high SOL as well. To get high SOL, we need to saturate memory bandwidth or compute. Since the operation is memory-bound (3 floats per element, 2 FLOPs per element), SOL will be determined by memory bandwidth utilization. To maximize memory bandwidth utilization, we need enough concurrent memory requests in flight. With only 8192 threads (256 warps), we might not saturate. But we can process multiple elements per thread. For example, each thread could process a small group of channels, say 4 channels, giving 32768 threads (1024 warps). That's ~10 warps per SM. That might be enough to saturate memory bandwidth? Maybe. Actually typical occupancy needed to saturate memory bandwidth is around 20-40 warps per SM. A100 can have up to 64 warps per SM. So 1024 warps total might still be less than ideal. But we can process, say, 16 channels per thread: 131072 threads = 4096 warps. That's ~38 warps per SM (if 108 SMs). That should be enough. But then each thread loops over L sequentially for its group of channels. The serial length is still L=4096. But memory accesses per iteration are for 16 elements per thread, which are coalesced across threads? Let's think. If each thread handles K channels, then for a given b and t, the thread reads a[b, t, d_start + tid*K + k] for k=0..K-1. If threads in a warp have consecutive d_start values, then the reads for k=0 across the warp are contiguous (since each thread reads a block of K elements, the warp reads K*32 contiguous elements). So coalescing is good as long as K is small enough to fit in one or a few cache lines. Alternatively, we can have each thread handle one (b,d) but process multiple timesteps per iteration with vectorized loads (float4). For example, read 4 timesteps at once: load float4 from a[b, t:t+4, d] and b[b, t:t+4, d]. But the recurrence is sequential across t: h_{t+1} depends on h_t. So we can't process 4 timesteps in parallel within a thread. We still need to process them sequentially. However, we can use vectorized loads and stores to improve memory throughput. For example, load a_t as float, but store out_t. Not much benefit. Another alternative: use a block-level scan where each block processes one (b,d) but has many threads (e.g., 512). The block loads a chunk of a and b into shared memory, computes a parallel prefix scan using shared memory, writes output, then proceeds to the next chunk. The serial steps are L / blockDim.x = 4096 / 512 = 8 steps per block. Each step has a __syncthreads(), but the parallelism within each step is high. The total number of threads is B*D*512 = 8192*512 = 4,194,304 threads. That's 131072 warps. On an A100 with 108 SMs, that's ~1213 warps per SM. That's massive over-subscription, which is great for latency hiding. However, each block only has 512 threads, and there are 8192 blocks. The total number of blocks (8192) is large but manageable. The shared memory per block is small: need to store 512 floats = 2KB. Plus maybe some temporary arrays for scan. That's fine. Register usage might be moderate. Occupancy might be limited by registers per thread, but 512 threads per block is okay. But wait: each block handles one (b,d) pair. So each block processes 4096 floats. The memory bandwidth is the same, but the parallelism is huge. The block-level scan uses many threads to process each scan, but the workload per thread is small. This is good for latency hiding, but there is overhead from __syncthreads() and shared memory operations. However, for memory-bound kernels, the overhead is small if the memory loads are vectorized. But we need to consider that each block processes only 4096 floats, which is very little work. With 8192 blocks, each block is tiny. The overhead of launching 8192 blocks might be non-negligible, but still likely fast. However, if we have 4M threads, the GPU might be heavily loaded. But the kernel will be memory-bound. Actually, maybe the best approach is to use a simple sequential kernel but vectorize and increase occupancy by handling multiple channels per thread. For example, each thread handles 4 or 8 channels. That gives us 8192 * 8 = 65536 threads = 2048 warps. If we use block size 256, that's 256 blocks. Each thread loops over L. That might be enough. But can we do even better with warp-level scan? Let's analyze the block-level scan approach more carefully. Suppose we have 8192 blocks, each of 256 threads. Each block processes one (b,d). We process the sequence in tiles of 256 elements. For each tile: 1. Each thread loads one a and one b: a[global_idx], b[global_idx]. Store in shared memory. 2. Perform a parallel prefix scan on the 256 elements in shared memory. The associative operator is combine(left, right) = (aR*aL, aR*bL + bR). We want inclusive scan. After step 2, each thread has the cumulative pair for its position within the tile. 3. The thread with the last element (or all threads) compute the tile's aggregate pair (the combination of all elements in the tile). Then we need to apply the prefix from previous tiles to this tile. Let's call the running prefix across tiles `running`. Initially `running` = (1, 0). After processing tile i, the aggregate of tile i is `tile_agg`. The cumulative up to tile i is `tile_cum = combine(running, tile_agg)`? Let's check: `running` is the combined transformation of all previous tiles. For each element in tile i, its total transformation is `combine(running, local_cum)`. Because local_cum combines elements within the tile from left to right. The operator combines left segment then right segment. If `running` represents the transformation of everything before the tile, then for an element at position j in the tile, the total transformation is `running` followed by local_cum_j. So total pair = combine(running, local_cum_j) = (local_a * running.a, local_a * running.b + local_b). Then the output h = total.b. After finishing the tile, we update running = combine(running, tile_agg). So algorithm: shared float s_a[256], s_b[256]; running_a = 1.0f; running_b = 0.0f; for tile = 0; tile < num_tiles; ++tile: idx = tile * 256 + threadIdx.x; s_a[threadIdx.x] = a[idx]; s_b[threadIdx.x] = b[idx]; __syncthreads(); // inclusive scan in s_a, s_b float local_a = s_a[threadIdx.x]; float local_b = s_b[threadIdx.x]; // warp-level scan #pragma unroll for offset = 1; offset < 32; offset *= 2: float temp_a = __shfl_up_sync(0xFFFFFFFF, local_a, offset); float temp_b = __shfl_up_sync(0xFFFFFFFF, local_b, offset); if (lane >= offset): // combine temp (left) and local (right) // local = combine(temp, local) = (local.a * temp.a, local.a * temp.b + local.b) local_a = local_a * temp_a; local_b = local_a * temp_b + local_b; // Wait, we need to be careful: the operator is not commutative. // Let's define combine(left, right) = (right_a * left_a, right_a * left_b + right_b). // For prefix scan, we want to accumulate left to right. So at each step, we combine the value from offset positions to the left (temp) with current (local). // So new_local = combine(temp, local) = (local_a * temp_a, local_a * temp_b + local_b). // This is correct. // After warp scan, local is inclusive scan within warp. // Now need to combine across warps. We can have warp leaders write their warp aggregate to shared memory, then do another scan. // Standard block-level scan: // After warp scan, each lane 31 has warp aggregate. // Write warp_agg to shared memory. // __syncthreads(); // Thread 0 does sequential scan of warp aggregates (there are 8 warps for 256 threads). // __syncthreads(); // Each thread reads the prefix from previous warps and combines. // ... // __syncthreads(); // Then apply running prefix. // Write output. // Compute tile_agg from the last thread (or from the scanned aggregates). // Update running = combine(running, tile_agg). // __syncthreads(); But this is quite complex to get right, especially with the non-commutative operator. Alternatively, we can use shared memory with a simple tree-based scan (Blelloch scan) or Hillis-Steele scan. For 256 elements, we can do a Hillis-Steele scan in log2(256)=8 steps. Each step accesses shared memory with stride = 1,2,4,...,128. The operator is combine. Since it's not commutative, we need to ensure the order: for inclusive scan, at step k (stride 2^k), each thread reads element at index i - stride and combines it with current. But combining left to right: the element at i-stride is the cumulative up to i-stride (after previous steps). So we combine that (left) with current (right): new = combine(s[i-stride], s[i])? Let's test. Suppose we want inclusive scan of array p[0..n-1] where p[i] = (a_i, b_i). Inclusive scan s[i] = p[0] ∘ p[1] ∘ ... ∘ p[i]. Hillis-Steele: for d = 1; d < n; d *= 2: if i >= d: temp = s[i - d] s[i] = combine(temp, s[i]) // because temp is left segment, s[i] is right segment. __syncthreads(); Let's verify: after d=1, s[1] = combine(p[0], p[1]) = (a1*a0, a1*b0+b1). s[2] = combine(p[1], p[2])? Wait, i=2 >=1, s[2] = combine(s[1], s[2]) = combine(p[0]∘p[1], p[2]) = correct. But s[1] was updated in this step, and thread 2 reads s[1] which is the updated value. That's fine because we want inclusive scan. For i=3, s[3] = combine(s[2], p[3])? Wait, s[2] before update is p[2]; but thread 3 may read s[2] after thread 2 updated it? In Hillis-Steele inclusive scan, threads read the old values and write new values. This requires double buffering to avoid race conditions. However, if we do it in-place with __syncthreads(), all threads read first, then compute, then write, then sync. So at each step, we read s[i-d] and s[i] from the previous step. That works. But for 256 threads, we need 8 steps, each with a __syncthreads(). That's 8 syncs per tile. With 16 tiles per scan (4096/256=16), that's 128 syncs per block. That's a lot. But maybe still fast? The work per step is small (just a couple of FMAs). The syncs might dominate. However, with 8192 blocks, there are many warps, and the sync cost might be amortized. Alternatively, we can process each scan sequentially but with many threads per block handling many scans. For example, block size 256, each block processes 256 independent channels (or maybe 64 channels if each thread handles 4 channels). That way, we have many blocks (B*D / channels_per_block). Total warps = (B*D)/32 = 256 warps regardless of how we group them! Wait, total warps is fixed by total threads = B*D*K? No, if each thread handles one channel, total threads = B*D = 8192 = 256 warps. The total number of warps cannot exceed total threads / 32. To increase warps, we need to increase total threads, which means each thread handles a fraction of a channel? Not possible. Unless we use multiple threads per channel, which is the block-level parallel scan approach. Ah, right. B*D = 8192 independent recurrences. If each thread handles one recurrence, total threads = 8192. That's 256 warps. That's a hard limit. The only way to get more warps is to use more than one thread per recurrence. So the block-level scan approach is the way to increase total thread count and thus occupancy. Given that the reference is extremely slow, even 256 warps will give huge speedup. But for high SOL, we might want to go further and use block-level scan to have >100k warps. But does block-level scan actually improve bandwidth utilization? The memory accesses are the same total bytes. If the per-thread sequential kernel is limited by latency rather than bandwidth, then increasing warps by using more threads per recurrence could improve throughput. But if the kernel is already bandwidth-limited, adding more threads won't help beyond saturation. However, with only 256 warps, I suspect the sequential kernel is latency-bound, not bandwidth-bound. Because 256 warps is low occupancy. On A100, to saturate memory bandwidth, you typically need at least a few thousand warps active across the chip. For example, each SM can have up to 64 warps. To saturate, you want maybe 20-30 warps per SM. With 108 SMs, that's 2000-3000 warps. We have only 256. So the kernel will be heavily latency-bound. Each warp does 4096 iterations, each with dependent loads. The GPU might still be able to switch to other warps, but with only 2-3 warps per SM, there isn't enough concurrency to hide memory latency. Therefore, to achieve high bandwidth utilization, we need to increase the number of warps by using multiple threads per channel. The block-level scan approach (e.g., 256 threads per channel) yields 8192*8 = 65536 warps if blockDim=256? Wait, each block has 256 threads. There are 8192 blocks (one per channel). Total warps = 8192 * (256/32) = 8192 * 8 = 65536 warps. That's huge occupancy. That should easily saturate memory bandwidth or at least get close. But each block does very little work: 4096 elements processed by 256 threads in 16 tiles. The overhead of 16 __syncthreads() per block might be okay. Also, there are 8192 blocks, which might cause some launch overhead, but load_inline kernels launch quickly. The main concern is that each block is small and the kernel might be limited by block scheduling overhead or the cost of __syncthreads(). Alternatively, we can use a warp-level scan with 32 threads per channel. That gives 8192 blocks of 32 threads. Total warps = 8192 * 1 = 8192 warps. That's also a lot. 8192 warps / 108 SMs = ~76 warps per SM. That's excellent occupancy. And the warp-level scan doesn't need __syncthreads() (except maybe to communicate tile aggregates across warps if we have multiple warps per block). If each block is one warp (32 threads), we don't need inter-warp sync. We process the sequence in chunks of 32 elements per warp. Within each chunk, we do warp shuffle scan. Then we update a running prefix using a single thread and broadcast via shuffle. This is very efficient. Let's design the warp-level scan kernel: Each warp (32 threads) handles one (b,d) pair. So each block can have, say, 8 warps = 256 threads, processing 8 channels. Or each block can have 1 warp = 32 threads, processing 1 channel. Launching 8192 blocks of 32 threads might have higher block overhead, but it's okay. Let's consider block size = 32, grid size = B*D = 8192. For a given channel (b,d), the warp processes L=4096 elements in chunks of 32. Each thread in the warp handles one element per chunk: thread i handles index chunk*32 + i. Algorithm for a chunk: 1. Load a_val = a[b, chunk*32 + lane, d]; b_val = b[b, chunk*32 + lane, d]; 2. Perform inclusive warp scan on the pairs (a_val, b_val) using warp shuffles. The operator: combine(left, right) = (right.a * left.a, right.a * left.b + right.b). Since we are scanning left to right, we can use the standard shuffle-up approach. For offset = 1,2,4,8,16: temp_a = __shfl_up_sync(0xFFFFFFFF, a_val, offset); temp_b = __shfl_up_sync(0xFFFFFFFF, b_val, offset); if (lane >= offset): // a_val is current (right), temp is left. // new = combine(temp, a_val) = (a_val * temp_a, a_val * temp_b + b_val) // Wait, we need to store the result back into a_val, b_val as the inclusive prefix up to this lane. b_val = a_val * temp_b + b_val; a_val = a_val * temp_a; After this, a_val = product of a from start of chunk to lane, b_val = corresponding affine term. But note: this only includes elements within the current chunk. It doesn't include the running prefix from previous chunks. 3. After the warp scan, lane 31 has the aggregate of the chunk: chunk_a = a_val (product of all a in chunk), chunk_b = b_val (the b term after applying all a's). We need to broadcast this to all lanes to apply to the next chunk. But first, we need to combine the running prefix from previous chunks with the current chunk's prefix. Let running_a, running_b be the prefix from all previous chunks. Initially running_a = 1, running_b = 0. For each lane i in the warp, the total prefix up to this element is: total_a = chunk_prefix_a[i] * running_a total_b = chunk_prefix_a[i] * running_b + chunk_prefix_b[i] Because running is left, chunk_prefix is right: combine(running, chunk_prefix) = (chunk_a * running_a, chunk_a * running_b + chunk_b). So we can compute output: out_b = total_b (since h = total_b if initial h was 0). Then we update running for the next chunk: running_a = chunk_a * running_a running_b = chunk_a * running_b + chunk_b But chunk_a is the aggregate a of the chunk (product of all a in chunk), chunk_b is the aggregate b (the b term after full chunk). Actually for the chunk aggregate, we need the result of applying the entire chunk to an input. The chunk's transformation is (chunk_a, chunk_b) where chunk_a = product of a_i, and chunk_b is the result of applying the chunk to initial h=0. That's exactly what the inclusive scan gives at lane 31 if we started the chunk with identity. However, the warp scan as described above computes inclusive scan of the chunk assuming identity before the first element of the chunk. So lane 31's a_val and b_val are the chunk aggregate. Good. But we need to get chunk_prefix_a[i] and chunk_prefix_b[i] for each lane i. After step 2, a_val and b_val already are the chunk prefix for lane i. Then total_b = a_val * running_b + b_val. Then we write output. Then update running: We can get chunk_agg_a from lane 31 via shuffle (broadcast). Similarly chunk_agg_b. chunk_agg_a = __shfl_sync(0xFFFFFFFF, a_val, 31); chunk_agg_b = __shfl_sync(0xFFFFFFFF, b_val, 31); running_b = chunk_agg_a * running_b + chunk_agg_b; running_a = chunk_agg_a * running_a; But wait: running_a is not needed for output (since output is just b). But it's needed for updating running across chunks. So we keep running_a. However, there is a subtlety: the recurrence for h is h_t = a_t * h_{t-1} + b_t. This is an affine map with coefficient a_t and offset b_t. The composition of two affine maps (a1,b1) then (a2,b2) is (a2*a1, a2*b1+b2). So the identity is (1,0). Good. Now, for the chunk scan, each thread loads (a_i, b_i). The inclusive scan within the warp must compute the composition of all (a_j,b_j) for j from 0 to i within the chunk. The warp shuffle code above does exactly that if we define combine(temp, current). Let's verify carefully. Suppose lane 0: a0, b0. After offset=1, lane<1, no change. So lane 0 stays (a0,b0). Good. Lane 1: temp = lane0 (a0,b0). combine(temp, current) = (a1*a0, a1*b0 + b1). Good. Lane 2: offset=1: temp = lane1 (a1,b1). combine(temp, lane2) = (a2*a1, a2*b1+b2). But wait, this is only composition of lane1 and lane2, not lane0. After offset=2: temp = lane0 (a0,b0). combine(temp, current) where current is now (a2*a1, a2*b1+b2). So new = ( (a2*a1)*a0, (a2*a1)*b0 + (a2*b1+b2) ) = (a2*a1*a0, a2*(a1*b0+b1)+b2). Good. So the standard shuffle-up scan works with this non-commutative operator as long as we combine temp (left) with current (right) correctly. Implementation: float a_val = ...; float b_val = ...; #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float temp_a = __shfl_up_sync(0xFFFFFFFF, a_val, offset); float temp_b = __shfl_up_sync(0xFFFFFFFF, b_val, offset); if (lane >= offset) { // combine temp (left) and current (right) b_val = a_val * temp_b + b_val; a_val = a_val * temp_a; } } After this, a_val = product_{j=0}^{i} a_j, b_val = h_i if h_{-1}=0 within the chunk. Then: float total_b = a_val * running_b + b_val; // since running is left, chunk prefix is right out[idx] = total_b; Then update running using chunk aggregate (lane 31): float chunk_agg_a = __shfl_sync(0xFFFFFFFF, a_val, 31); float chunk_agg_b = __shfl_sync(0xFFFFFFFF, b_val, 31); running_b = chunk_agg_a * running_b + chunk_agg_b; running_a = chunk_agg_a * running_a; Wait, running_a is not needed for output but needed for next chunk's total_b? Actually total_b only needs running_b and a_val, b_val. running_a is not used. But to update running_b for the next chunk, we need to know what happens when we apply the chunk aggregate to an arbitrary input. The chunk aggregate transformation is (chunk_agg_a, chunk_agg_b). So if we have a running state h_before (which is the result of applying all previous chunks to 0), then after the chunk, h_after = chunk_agg_a * h_before + chunk_agg_b. So running_b = h_after. Indeed, running_a is not needed because the recurrence always starts from h=0 at the very beginning (since h_0 = b_0). Wait, does the overall scan start with identity (1,0)? Let's see: before processing any elements, the transformation is identity: h = 1 * h_init + 0. But our initial h_init is 0, and h_0 = b_0. So the running state (the result so far) is h = running_b, with running_a implicitly being the product of all a's so far, but we don't need it if we only track the result h. However, for composition of chunks, if we have chunk aggregate (A_chunk, B_chunk) and we want to apply it to the result of previous chunks, we need the previous result h_prev. Then h_new = A_chunk * h_prev + B_chunk. So we only need to keep h_prev (running_b). We don't need running_a. Because running_a is the product of all a's from start to current point, but we never need it except to compute h. Wait, but for the next chunk, when we compute total_b for each element within the next chunk, we need to apply the running prefix to the chunk prefix. For a thread in the next chunk with local prefix (a_local, b_local), total = combine(running, local) = (a_local * running_a, a_local * running_b + b_local). So we do need running_a to compute total_b? Let's check: total_b = a_local * running_b + b_local. Wait, is running_a used? No! Because total_b = a_local * running_b + b_local. The running_a is multiplied with a_local to get total_a, but we don't need total_a except maybe for the chunk aggregate at the end of the next chunk. However, the chunk aggregate's a component is the product of all a_local in the chunk, which is independent of running. So for computing output at each element, we only need running_b. For updating running across chunks, we need to compute the new running_b after the chunk: new_running_b = chunk_agg_a * running_b + chunk_agg_b. So we never need running_a! Great. Wait, let's double-check with algebra. Let running transformation after k chunks be (A_run, B_run). But since initial h_{-1} = 0, the actual h after k chunks is B_run (since h = A_run * 0 + B_run = B_run). So running_h = B_run. Now, local prefix within chunk i for element j is (A_loc_j, B_loc_j). The total transformation up to element j in chunk i is (A_loc_j * A_run, A_loc_j * B_run + B_loc_j). The output h = total_B = A_loc_j * B_run + B_loc_j. So indeed we only need B_run (running_h). For the chunk aggregate (i.e., total transformation of the entire chunk), it's (A_chunk, B_chunk) where A_chunk = product of all a in chunk, B_chunk = result of applying chunk to 0. After the chunk, the new running_h becomes A_chunk * running_h + B_chunk. So we can keep just running_h, and update it with chunk_agg_a and chunk_agg_b. Therefore, we only need to track running_h (float). Initial running_h = 0. For each chunk, after warp scan, each lane computes out = a_val * running_h + b_val. Then lane 31 broadcasts chunk_agg_a and chunk_agg_b. Update running_h = chunk_agg_a * running_h + chunk_agg_b. Proceed to next chunk. This is excellent. It simplifies the kernel. Now, what about the initial condition? The first element of the first chunk (t=0) should output b_0. Let's test: running_h = 0. first chunk, lane 0: a_val = a_0, b_val = b_0. out = a_0 * 0 + b_0 = b_0. Good. lane 1: a_val = a_1*a_0, b_val = a_1*b_0 + b_1. out = a_1*a_0*0 + (a_1*b_0+b_1) = a_1*b_0+b_1. Good. After chunk 0, lane 31's a_val = product of a_0..a_31, b_val = h_31 if start from 0. chunk_agg_a = product a_0..a_31. chunk_agg_b = h_31. running_h updated to h_31. Next chunk, lane 0 (t=32): a_val = a_32, b_val = b_32 (local prefix within chunk). out = a_32 * running_h + b_32 = a_32 * h_31 + b_32 = h_32. Good. So the algorithm is correct. Now, we need to handle 4096 elements. 4096 / 32 = 128 chunks. Each warp processes 128 chunks. Each chunk: - Load a and b. - Warp shuffle scan (5 steps). - Compute out and write. - Broadcast chunk_agg_a and chunk_agg_b (two shuffles). - Update running_h. Total loads: 2 loads per thread per chunk. 128 chunks. So 256 loads per thread. Total memory: B*L*D*3*4 bytes = 402 MB. On A100 (2 TB/s), that's ~200 microseconds. Computation is trivial. Could we vectorize the loads? We can load float2 or float4 to reduce instructions. For example, each thread could process 2 elements per chunk: chunk size = 64 threads? No, warp size is 32. But we can process 2 chunks at once? Or each thread loads a float2 (2 consecutive elements) and then we do a scan across 64 elements using warp shuffles and maybe some extra steps? That might complicate but could reduce load instructions. However, the memory system handles 32-bit loads efficiently. Vectorized loads might improve throughput slightly. Alternatively, we can have each warp process multiple channels. For example, each warp processes 8 channels, with each lane handling one element of each channel? That might increase arithmetic intensity but memory bandwidth is same. Actually, we can have each thread handle one element at a time but the warp processes 32 channels at once. Wait, if we assign one warp per channel, we have 8192 warps. Total threads = 8192*32 = 262144. That's a decent amount. 262144 / 32 = 8192 warps. On 108 SMs, that's ~76 warps per SM. That should be enough to saturate memory bandwidth? Maybe. According to some rules of thumb, 64 warps per SM can saturate bandwidth. With 76 warps per SM, we should be close. So maybe we don't need block-level scan. The warp-level scan gives 8192 warps, which is good. Wait, let me recalculate: B=32, D=256 => 8192 channels. Each warp = 32 threads. If each warp handles one channel, total threads = 8192*32 = 262,144. Warps = total threads / 32 = 8192. Wait, each block can have multiple warps. If block size = 32, then blocks = 8192. Warps per block = 1. Total warps = 8192. Yes. 8192 warps across 108 SMs = 75.9 warps per SM. That's very good. A100 has max 64 warps per SM, but some SMs might have more than 64? Actually A100 supports 2048 threads per SM = 64 warps. So 8192 warps across 108 SMs means if evenly distributed, each SM gets 75.9 warps, which exceeds capacity. So the GPU will be fully occupied. Good. Thus the warp-level scan kernel with block size = 32 (one warp per block) should achieve high occupancy and likely saturate memory bandwidth. But block size 32 might have some overheads. Could we use block size 256 (8 warps) and have each warp in the block handle a different channel? That would still be 8192 warps, but blocks = 8192/8 = 1024. Each block has 8 warps. Total warps = 1024*8 = 8192. The occupancy per SM is the same. But larger blocks might have slightly better scheduling. Also, we can reuse some values? Not really. But we can store running_h in registers per warp. With block size 256, we have 8 warps per block, each processing one channel. We can compute channel index = blockIdx.x * 8 + warpId. This reduces block launch overhead from 8192 blocks to 1024 blocks. That might be slightly better. However, block size 32 is also fine. Let's aim for block size 32 to keep it simple and avoid any shared memory usage. But with block size 32, there are 8192 blocks. Is there any issue with the maximum number of blocks? A100 can handle a lot. The grid dimension can be large. 8192 is fine. Alternatively, we could process 2 or 4 elements per thread within the warp using a "virtual" warp size. For example, each warp processes 64 elements (2 per thread) using a sequential scan within thread then warp scan. This would reduce the number of chunks from 128 to 64, saving some shuffle overhead. But it's more complex. Probably unnecessary. Let's think about memory layout and indexing. Input a and b are shape (B, L, D). In PyTorch, the strides are (L*D, D, 1). Actually for a tensor of shape (B, L, D), the default layout is contiguous with stride (L*D, D, 1). So indexing a[b, t, d] = a + b*L*D + t*D + d. Output shape is (B, L, D), same strides. In our kernel, if each warp handles one (b,d), then we can compute: int b = blockIdx.x / D; int d = blockIdx.x % D; Wait, total blocks = B * D. So blockIdx.x corresponds to channel index = b * D + d. Then: int b = blockIdx.x / D; int d = blockIdx.x % D; int lane = threadIdx.x; // 0..31 float running_h = 0.0f; for (int chunk = 0; chunk < L / 32; ++chunk) { int t = chunk * 32 + lane; int idx = b * L * D + t * D + d; float a_val = a[idx]; float b_val = b[idx]; // warp scan #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float temp_a = __shfl_up_sync(0xFFFFFFFF, a_val, offset); float temp_b = __shfl_up_sync(0xFFFFFFFF, b_val, offset); if (lane >= offset) { b_val = a_val * temp_b + b_val; a_val = a_val * temp_a; } } // compute output float out_val = a_val * running_h + b_val; out[idx] = out_val; // broadcast chunk aggregate from lane 31 float chunk_agg_a = __shfl_sync(0xFFFFFFFF, a_val, 31); float chunk_agg_b = __shfl_sync(0xFFFFFFFF, b_val, 31); running_h = chunk_agg_a * running_h + chunk_agg_b; } This is straightforward. But wait: for the chunk scan, the shfl_up_sync will read from lower lanes. For lane < offset, we don't update. That's correct. One concern: performance of __shfl_up_sync inside a loop. With 5 iterations, it's fine. Another concern: the for loop over 128 chunks. It might be unrolled by the compiler partially. We can add #pragma unroll but 128 is large. Might not be fully unrolled, which is fine. Memory coalescing: For a fixed b and d, the warp loads a[b, chunk*32 + lane, d]. The addresses are: base + b*L*D + (chunk*32 + lane)*D + d = (base + b*L*D + chunk*32*D + d) + lane*D. Since D=256, lane*D = lane*256. Consecutive lanes access memory separated by 256 floats = 1024 bytes. That's not coalesced! Because each thread accesses a different t, and the stride between t is D elements. Since D=256, the stride is 1024 bytes. A warp of 32 threads accesses 32 addresses separated by 1024 bytes. That means each thread is in a different cache line (128 bytes). So each load causes 32 cache lines to be fetched. That's terrible for memory bandwidth. We need to fix the memory access pattern. In the reference, a and b are (B, L, D). The natural layout has D as the last dimension. For a fixed b and t, different d are contiguous. So to have coalesced loads, the threads in a warp should iterate over d, not t. That means we should assign each warp to a fixed (b, t_chunk) and have the threads process different d's? But we need a scan over t. If threads in a warp process different d's, they can't do a shuffle scan over t because each thread would be scanning its own d sequentially. Wait, but if we want coalescing, each thread should access contiguous memory. So each thread should handle a different d, and they should load a block of t's? But the scan is along t. How can we use warp shuffles if each thread is scanning a different d? Actually, we can have each thread handle one (b,d) and load multiple t's sequentially. But then the load for a given t across the warp would be for different d, which are contiguous. Yes! If each thread handles one (b,d), then for a fixed t, the 32 threads in a warp correspond to 32 consecutive d's. Then the load a[b, t, d] for all threads in the warp are contiguous! But our warp scan requires the 32 threads to process 32 consecutive t's. So the assignment is: each block/warp processes a fixed b and a range of d's? Wait, if each thread handles one (b,d), then the warp consists of 32 threads each with a different d. To do a scan over t, each thread would have to do its own sequential scan over t. Warp shuffles wouldn't help because each thread's data is for a different d, not different t. However, we could reinterpret the data: each warp processes 32 channels simultaneously, each thread handling one channel's sequence sequentially. In this case, we wouldn't use warp shuffles; each thread would do a sequential loop. But then total threads = B*D = 8192, total warps = 8192/32 = 256. Back to low occupancy. We need to reconcile coalescing with parallelism. If each warp handles one channel, the stride between consecutive t is D. That's a large stride. But we can transpose the data? The input is given as (B, L, D). If we launched threads reading t contiguously, we get coalescing. But the scan is along t. So perhaps we can process the scan with each warp handling a fixed (b, d) but reading t with stride 1? Wait, if each thread in the warp handles a different t, then for a fixed (b,d), the addresses for thread lane are base + b*L*D + (chunk*32+lane)*D + d. Since D is the stride between t, it's D floats. This is a strided access with stride D. If D=256, that's 1024 bytes. This is not coalesced. However, if D is a multiple of warp size, the hardware might still handle it via memory sectors? Actually, for a 32-thread warp, to coalesce, the addresses should fall within a 128-byte segment (or 32 bytes for each bank?). For strided access with 1024 bytes, each thread hits a separate 128-byte cache line, resulting in 32 memory transactions per warp. That's terrible. We need to change the assignment. To get coalesced loads, the threads should read along the D dimension. So we could have each thread handle a different d, and the block/warp processes a single b and a chunk of t. But then each thread does its own scan over t. That's sequential. But we can have each block of, say, 256 threads handle 256 channels for a single b, and each thread loops over t. That gives total blocks = B = 32. Total threads = 32*256 = 8192. Warps = 256. Low occupancy but coalesced memory. However, we can increase occupancy by having each thread process multiple channels or multiple b's. For example, each block could have 256 threads, each thread processes K channels. Then total threads = 8192*K. Warps = 256*K. If K=4, warps=1024. That's better. Coalescing: for a fixed b and t, threads in a warp read d = tid*K + k? If tid is thread index, then for k=0, addresses are contiguous across tid if K=1 and block size = 256. Yes, if each thread handles one d, reading a[b, t, d] across the warp (32 threads) reads 32 contiguous floats. That's coalesced. But then each thread loops over t sequentially. Wait, if each block has 256 threads and handles 256 channels, there are B=32 blocks. Each thread loops over L=4096. So each block has 256 threads = 8 warps. Total warps = 32*8 = 256. That's low. But each thread can handle multiple channels, say 4 channels. Each block still has 256 threads, but now handles 1024 channels. Blocks needed = B * D / 1024 = 8192 / 1024 = 8 blocks. That's even fewer blocks, but same total threads (8*256=2048). Warps = 64. Worse. Actually, total threads is blocks * blockDim.x. If each thread handles M channels, total threads needed = (B*D) / M. To increase total threads, we need to decrease M or increase B, but B and D are fixed. Wait, the total amount of work is B*D*L. If each thread does L steps, total threads = B*D / M. So increasing M decreases total threads. To increase total threads, we need M < 1, i.e., multiple threads per channel. That brings us back to parallel scan per channel. But parallel scan per channel with warp shuffles requires threads to cooperate within a channel, meaning they read different t's for the same channel, causing strided memory access. We can bridge this gap by transposing the data in shared memory: have a warp load a contiguous block of memory (e.g., 32 floats across d) for a fixed t, then use shared memory to exchange data so that each thread gets the t it needs for the scan. But that seems complicated. Alternatively, we can process the data in a transposed manner: treat each channel as independent but have each warp handle 32 channels, each thread handling one channel sequentially. Then to increase warps, we can assign multiple warps per block. But total warps is still B*D/32 = 256. We can't increase warps beyond 256 unless we have multiple threads per channel. So the trade-off is: coalesced sequential kernel (256 warps) vs strided parallel scan kernel (8192 warps). Which is faster? We need to estimate memory bandwidth for the strided kernel. In the strided kernel, each warp reads 32 elements separated by 1024 bytes. That's 32 cache lines per warp load. On A100, L2 cache line size is 128 bytes. So each 4-byte load fetches 128 bytes, but only uses 4 bytes. That's a utilization of 1/32. The effective memory bandwidth is 1/32 of peak. That's terrible. Even with 8192 warps, the memory bandwidth is severely bottlenecked. However, the hardware might have some broadcast/stride caching? For a stride of 1024 bytes, consecutive threads hit different cache lines. The memory controller will coalesce them into multiple transactions. It will likely issue 32 separate 128-byte sectors (or smaller). So the bandwidth is ~1/32. Peak memory bandwidth ~2 TB/s, so effective ~64 GB/s. Total memory moved: 402 MB. So time ~ 6.3 ms. Wait, 402 MB / 64 GB/s = 6.3 ms. Actually, 402MB is total. If each load fetches 128 bytes for a 4 byte value, total bytes moved from DRAM is 402 MB * 32 = ~12.8 GB. At 2 TB/s, that's 6.4 ms. Hmm. 6.4 ms is not super fast, but the reference might be 100 ms or more. So still a big speedup. But we can do much better. Sequential per-thread kernel with coalesced loads: each warp reads 32 contiguous floats (one per d). So loads are fully coalesced into 1 or 2 128-byte transactions. Memory bandwidth near peak. Total memory moved = 402 MB. At 2 TB/s, time ~ 200 microseconds. But with only 256 warps, latency bound. Effective bandwidth might be lower. Let's estimate: 256 warps, 108 SMs, 2.4 warps per SM. Memory latency ~ 400 cycles. To hide latency, need ~20 warps per SM. So effective bandwidth might be ~2.4/20 = 12% of peak = 240 GB/s. Time ~ 1.7 ms. That's still faster than the strided kernel's 6.4 ms. But both are much faster than reference. But can we combine the benefits? We want coalesced loads AND high occupancy. The way to do that is to have multiple threads per channel (for occupancy) but load data in a coalesced way and then transpose in shared memory. Specifically: we can process channels in groups. For a block, suppose we have 32 warps (1024 threads). We can assign each warp to one channel (so block handles 32 channels). But the 32 threads in a warp load 32 consecutive elements along t for the same channel. That's strided. Not good. Alternatively, assign each warp to a channel, but load data transposed: for a chunk of 32 t's, each thread in the warp loads one t. To get coalesced loads, we need the threads to read contiguous memory. That means we need to change the mapping: for a given block, the threads should read a batch of (b, t, d) where d varies across threads. So each thread handles one d. But then how do we do a scan over t for a given d? Each thread could load a block of t's into shared memory, then perform the scan in shared memory across t? But the scan is along t, not d. If each thread handles one d, and we load a tile of t's (e.g., 32 t's) for that d, then we have 32 values per thread (for 32 t's). We could store them in registers and do a sequential scan on them. That doesn't help occupancy. Wait, another idea: process the data in tiles where each tile is a 2D slice (t_chunk, d_chunk). For example, tile size 32x32. Within the tile, we have 32 t's and 32 d's. We can load the entire tile into shared memory in a coalesced manner (threads read along d). Then, within the tile, we need to compute the scan along t for each d. We can have threads in a warp process different d's, but they need to scan along t. Since the data is now in shared memory, and shared memory is banked, we can have each thread sequentially scan its d's t's within the tile. Or we can parallelize across t using multiple warps. But again, total threads per channel would be limited. Let's think about the occupancy vs coalescing trade-off more carefully. Maybe we can use vectorized loads in the strided kernel to reduce the number of memory transactions. For example, each warp could load a float4 (4 consecutive t's) per thread. Wait, if each thread loads 4 consecutive t's for the same d, the addresses are still separated by D floats. So warp loads 128 t's? No, each thread loads 4 t's: addresses for thread lane are base + lane*D*4 + ...? Actually, if we vectorize along t: each thread loads a float4 at a[b, chunk*32*4 + lane*4, d]? Wait, the mapping is still strided by D. If we load float4 per thread, each thread loads 4 contiguous elements along t? No, because the memory layout is (B, L, D). For a fixed b and d, contiguous along t are separated by D. So a[b, t, d] and a[b, t+1, d] are D elements apart. So "vectorizing along t" means loading scattered elements. That's not how vector loads work. Vector loads (float4) require contiguous memory addresses. To load a float4 contiguously, the addresses must be consecutive in memory. That means we need to vary d, not t. So each thread could load a float4 along d: a[b, t, d:d+4] if d is contiguous. That is coalesced. But that loads 4 channels at once. So we could have each thread process 4 channels sequentially. That reduces total threads by 4, but uses vectorized loads. That seems like a good optimization for the sequential kernel. But the fundamental issue remains: to increase occupancy beyond 256 warps, we need multiple threads per channel. Can we have multiple threads per channel and still coalesce loads? Yes, by loading data into shared memory in a transposed manner. Consider a block of size 256. The block processes one channel. We want to load the sequence a[b, :, d] and b[b, :, d] efficiently. This is a 1D strided array (stride D). A warp of 32 threads reading 32 consecutive elements along t would each read with stride D, which is bad. But we can instead have the 256 threads in the block read 256 consecutive elements along t, but each thread reads one element. Still stride D. The memory controller will generate 256 separate transactions. However, we can use the L2 cache: if D is 256, and we read 256 elements with stride 256, the total span is 256*256*4 = 262KB. That's larger than L1 but maybe within L2. But still, 256 separate 128-byte reads = 32KB loaded from memory to read 1KB of data. That's a lot of waste. But there is a trick: reading strided data with many threads is essentially a "gather" pattern. GPUs handle it with many cache lines. However, modern GPUs have L1 cache that can handle some strided accesses if the stride is not too large? Actually, if stride is 1024 bytes, each thread hits a separate cache line. There's no way around it: 32 cache lines per warp. So effective bandwidth is 1/32. Unless... we can use texture memory or constant memory? No. Alternatively, we can transpose the input in memory. But we can't change the input format; it's given as (B, L, D). However, we could launch a kernel that transposes a and b to (B, D, L) first, then do the scan with coalesced loads. But that adds an extra kernel launch and extra memory traffic (2 * B*L*D reads and writes). Total memory traffic becomes larger. But the scan kernel would be much more efficient. Let's estimate: Original sequential kernel (coalesced): reads a and b (2 * 134MB = 268MB), writes out (134MB). Total 402MB. Time maybe 1-2 ms due to low occupancy. Transpose + scan: transpose a: read 134MB, write 134MB. transpose b: same. Then scan on transposed (B, D, L): reads 268MB, writes 134MB. Total memory: 402MB (transposes) + 402MB (scan) = 804MB. If scan kernel has high occupancy and saturates bandwidth, time ~ 400us. Transpose kernels also saturate bandwidth, ~ 400us. Total ~ 800us. That's still faster than the sequential latency-bound kernel. But can we fuse transpose and scan? We could have each block load a tile of a and b from (B, L, D) into shared memory (coalesced across d), then within the block, the threads reorganize and perform scan along t. But scan along t for a given d requires L values. L=4096. If D=256, a tile of (t_chunk=32, d=256) is 32*256 = 8192 floats = 32KB. That's large but fits in shared memory (up to 164KB per SM on A100). We could load a tile of 32 t's and all 256 d's. But then each d's scan is only over 32 t's; we need to combine tiles sequentially. Within a tile, for one d, we have 32 values. We could have one warp per d process those 32 values via shuffle scan. But a block has 256 threads, so it can handle 8 d's at a time (since each warp handles one d). Wait, if we load a tile of 32 t's and 256 d's, that's 32*256 = 8192 values. With 256 threads, each thread can load 32 values? That's too many. Or each thread loads one value: but then we only load 256 values per tile. That would be a tile of 1 t and 256 d's. That's just loading a column. Alternatively, load a tile of 32 t's and 32 d's = 1024 values. With 256 threads, each thread loads 4 values (a float4 along t? No, along d). So 256 threads load 256*4 = 1024 values. The tile is 32 t's x 32 d's. But then each d has 32 t's. We can assign one warp per d within the tile. There are 32 d's in the tile, so 32 warps = 1024 threads. That fits. The block would have 1024 threads. Shared memory for the tile: 1024 floats * 2 tensors = 8192 bytes. Very small. Then for each t-chunk (32 t's), we load the tile into shared memory (coalesced: threads read along d). Then each warp reads the 32 values for its d from shared memory (no bank conflicts if we arrange properly). Then the warp does the shuffle scan on those 32 values. Then writes output back to global memory. But the output is in (B, L, D) layout. Writing 32 values for one d across 32 t's is strided. We can write them from shared memory to global memory in a coalesced way by having all warps write their results for a given t across d's. That is, we could store results in shared memory transposed, then write to global memory coalesced. This is getting into a transpose-scan-transpose approach, but fused. Let's step back. The simplest high-performance approach might be to perform the scan in "registers" using a block-level algorithm on transposed data, but fused with coalesced loads and stores. Given the complexity, maybe a simplest efficient approach is: 1. Transpose a and b from (B, L, D) to (B, D, L) using a fast transpose kernel. 2. Run a simple cooperative scan kernel on (B, D, L) where each warp handles one (b,d) and reads contiguous L values. Since L is now contiguous, memory accesses are fully coalesced. 3. The output is (B, D, L), then we might need to transpose back to (B, L, D). But we can write the output directly in (B, L, D) or do a fused transpose. Actually, the output of scan is (B, L, D) in the original layout. If we transpose a and b to (B, D, L), then scan along the last dimension (L) gives output (B, D, L). Then we need to transpose it back to (B, L, D). That's two extra transpose operations (or one extra if we write the output transposed and then transpose). But the scan kernel on transposed data would be extremely efficient because each warp reads contiguous memory. But perhaps we can avoid the extra transpose by using the fact that the reference returns (B, L, D). The evaluator just checks correctness. It doesn't care about intermediate layout. So we can do: - Transpose a and b to (B, D, L) in-place or to temporary tensors. - Compute scan along L dimension with a simple sequential kernel (each thread handles one (b,d) and loops over L). Now memory is coalesced and we have 8192 threads = 256 warps. Wait, but transposing doesn't increase thread count. We still have only 8192 threads. But memory accesses are now coalesced. So the sequential kernel would be bandwidth-bound? With 256 warps, still low occupancy, but maybe the transpose kernels add more warps and the total time is dominated by the transpose? Actually the sequential scan on transposed data still has 256 warps and might be latency-bound. But if we transpose with a highly parallel kernel (e.g., each block processes a tile), that might saturate the device. However, the scan portion might still be the bottleneck. Alternatively, after transposing, we could run the warp-level parallel scan on (B, D, L). This would have 8192 warps and coalesced loads. The load pattern for the warp-level scan on (B, D, L) is: each warp handles one (b,d), reads a block of 32 contiguous elements along L. Since L is contiguous, it's fully coalesced. Great! So we get both high occupancy (8192 warps) and coalesced loads. The only downside is the need to transpose a and b first, and transpose the output back. But transposes are also bandwidth-saturating and can be fused or done efficiently. Wait, but maybe we can launch a single kernel that does the transposed scan without explicit separate transpose. We can treat the input as (B, D, L) logically by swapping the indexing: idx = b*D*L + d*L + t. But the actual memory layout is (B, L, D): idx = b*L*D + t*D + d. We can't change the layout without copying. So if we want coalesced loads for the scan, we need to copy or transpose. However, note that we can have each warp handle 32 channels (varying d) for a single t-chunk, and use shared memory to reorder data so that each thread gets the values for its channel across t. This is a "swizzle" approach. Specifically, consider a block of 32 warps (1024 threads). The block handles one b and a chunk of 32 t's (t0..t31) and all 256 d's? 1024 threads can handle 1024 values. For 32 t's and 256 d's, that's 8192 values. Too many. But we can process the data in smaller tiles: say block handles one b, 32 t's, and 32 d's. That's 1024 values. 1024 threads can each load one value. So block size = 1024. Each block processes a tile of 32x32. There are B * (L/32) * (D/32) blocks = 32 * 128 * 8 = 32768 blocks. That's a lot. Each block has 1024 threads = 32 warps. Total warps = 32768 * 32 = 1,048,576. Massive occupancy. But each block only processes 1024 values. Is that efficient? The overhead of launching so many blocks might be high, but the memory bandwidth would be saturated. But what do we do inside the block? We load a tile: a[b, t0..t31, d0..d31]. Then we need to compute scan along t for each d. For each d, we have 32 values (one per t). We can assign one warp per d to do a warp shuffle scan on those 32 values. But we have 32 d's per tile, so 32 warps. The block has 32 warps. Perfect! Each warp handles one d within the tile. However, the warp needs the 32 values for its d. After loading the tile into shared memory, the threads of a warp can read the 32 values for its d from shared memory. Since the warp's threads are active, we need to distribute the 32 values to the 32 threads. Thread i reads value at t=i for that d. That's a shared memory read with stride? The shared memory layout could be shmem[t][d] (32x32). Then for a fixed d, the values for different t are in consecutive entries along t dimension. Warp thread i reads shmem[i][d]. That's consecutive addresses in shared memory (if column-major or row-major appropriately). Bank conflicts? If shmem is indexed as [t][d], then each thread i reads shmem[i][d] where d is same across warp. That means all threads in the warp read the same column? Actually if layout is shmem[t*32 + d], then shmem[i*32 + d] for i=0..31. Addresses are separated by 32 floats = 128 bytes. Shared memory has 32 banks, each 4 bytes wide. So shmem[i*32 + d] maps to bank (i*32 + d) % 32 = d % 32. Since d is same for all threads in the warp, all threads access the same bank! That's a 32-way bank conflict. Bad. We can swizzle the layout: shmem[(i%32) * 33 + d] or something. Or we can store as shmem[d][t]. Then thread i reads shmem[d][i] = shmem[d*32 + i]. Addresses are consecutive, bank i. No conflicts! Yes. So layout in shared memory: for each d_offset in 0..31, and t_offset in 0..31, store at index d_offset*32 + t_offset. Then a warp assigned to d_offset reads shmem[d_offset*32 + i] for i=0..31. That's consecutive, no bank conflicts. Good. So the tile approach: - Block size: 1024 threads (32 warps). - Each block processes: one b, one d_tile (32 d's), one t_tile (32 t's). Actually we need to process the entire sequence along t. So we can have the block process one b and one d_tile, and loop over t_tiles sequentially. But each block has 32 warps, each warp handles one d. The 1024 threads load a t_tile of 32 t's and 32 d's. That's 1024 values. Each thread loads one a and one b. Wait, if we loop over t_tiles, we need to maintain running prefix across tiles for each d. That can be stored in registers per warp. Let's design the kernel: Grid: (B, D/32) blocks. Each block handles one b and one group of 32 d's. Block size: 32 warps = 1024 threads. But we only need 32 warps for 32 d's. The 1024 threads are used to load the tile. Inside the kernel: int b = blockIdx.x; int d_tile = blockIdx.y; // d from d_tile*32 to d_tile*32+31 int warp_id = threadIdx.x / 32; int lane = threadIdx.x % 32; // Each warp handles one d: d = d_tile*32 + warp_id. int d = d_tile * 32 + warp_id; float running_h = 0.0f; // Loop over t in chunks of 32 for (int t_tile = 0; t_tile < L/32; ++t_tile) { int t = t_tile * 32 + lane; // Global memory index for a[b, t, d] (original layout) // But loads are coalesced if threads in a warp load for same t and different d? // Actually, if we have 1024 threads, each reading a[b, t, d] where d = d_tile*32 + threadIdx.x % 32? Wait, thread mapping matters. } Wait, we need to load the tile into shared memory. For a tile of 32 t's and 32 d's, we have 1024 elements. If we have 1024 threads, each thread can load one element. To have coalesced loads, the threads should read along d. So we can map threadIdx.x to d_offset and lane? Actually, if we have 1024 threads, we can map: int d_offset = threadIdx.x % 32; // 0..31 within tile int t_offset = threadIdx.x / 32; // 0..31 within tile Then each thread loads: int d = d_tile * 32 + d_offset; int t = t_tile * 32 + t_offset; int idx = b * L * D + t * D + d; Since threads with consecutive threadIdx.x have consecutive d_offset (because threadIdx.x mod 32 increments fastest), for a fixed t_offset, the threads in a warp (32 threads) will have consecutive d_offset and same t_offset. So they load a[b, t, d_offset .. d_offset+31] which are contiguous in memory. Fully coalesced! Great. We load a_val and b_val into shared memory. We need shared memory layout that allows each warp (which handles one d_offset) to read 32 values across t_offset. Shared memory layout: float s_a[32][32]; float s_b[32][32]; But to avoid bank conflicts when reading across t_offset for a fixed d_offset, we should store as s_a[d_offset][t_offset]. Since threads write with t_offset varying and d_offset varying, we need to see bank conflicts on write. Write pattern: thread i writes s_a[t_offset][d_offset]? If we store as s_a[d_offset][t_offset], then thread i writes to s_a[d_offset][t_offset]. d_offset = i%32, t_offset = i/32. So for a fixed t_offset, the 32 threads in a warp have d_offset = 0..31, writing to s_a[d][t] where d varies, t same. That means each thread writes a different d, same t. In shared memory, s_a[d][t] corresponds to address base + d*32 + t. So consecutive d means consecutive addresses, separated by 32 floats = 128 bytes. Bank = (d*32 + t) % 32 = t % 32. Since t is same, all threads write to the same bank! 32-way bank conflict on writes. To avoid bank conflicts on writes, we can pad the inner dimension: e.g., s_a[32][33]. Then address = d*33 + t. Bank = (d*33 + t) % 32. For a fixed t, d varies, bank varies because 33 mod 32 = 1. So banks are different. No conflicts on write. On read, a warp reading s_a[d][lane] for fixed d: address = d*33 + lane. Banks = (d*33 + lane) % 32 = (d + lane) % 32. Lane varies, so banks vary. No conflicts. Great. So shared memory: __shared__ float s_a[32][33]; __shared__ float s_b[32][33]; (or use 1D array with padding). Thus the kernel steps: 1. Determine thread mapping: int tid = threadIdx.x; int d_off = tid % 32; int t_off = tid / 32; 2. Compute global indices: int b = blockIdx.x; int d_tile = blockIdx.y; int d = d_tile * 32 + d_off; 3. Allocate shared memory: __shared__ float s_a[32][33]; __shared__ float s_b[32][33]; 4. For each t_tile from 0 to L/32 - 1: a. Load a and b: int t = t_tile * 32 + t_off; int idx = b * L * D + t * D + d; s_a[t_off][d_off] = a[idx]; // but layout needs to be s_a[d_off][t_off] for later read. Wait, we can store directly into s_a[d_off][t_off] using a transposed store. Let's think about how to store without bank conflicts. If we want final layout in shared memory to be s_a[d][t] (so that warp reading d reads t contiguously), then for the write, threads have t_off varying slowly and d_off varying fast. So each warp has same t_off, varying d_off. As we saw, writing to s_a[d_off][t_off] causes bank conflicts. But with padding s_a[32][33], writing s_a[d_off][t_off] is fine because banks for consecutive d_off differ. Let's verify: address of s_a[d][t] with column size 33: offset = d*33 + t. For a warp with threads i=0..31: d_off = i%32, t_off = i/32. If t_off is constant (e.g., 0 for threads 0..31, 1 for threads 32..63, etc.), then offset = i*33 + t_off. Bank = (i*33 + t_off) % 32 = (i + t_off) % 32. Since i varies, banks vary. So no bank conflicts on writes. Good. For read: a warp assigned to d reads s_a[d][lane]. offset = d*33 + lane. Bank = (d*33 + lane) % 32 = (d + lane) % 32. Lane varies, banks vary. No conflicts. Perfect. So we can use shared memory as float s_a[32][33]; where first index is d (0..31), second is t (0..31). Write: s_a[d_off][t_off] = a[idx]; (d_off varies across warp, t_off constant per warp -> no conflict). Read: each warp (warp_id = threadIdx.x / 32) corresponds to d = d_tile*32 + warp_id. lane = threadIdx.x % 32; float a_val = s_a[warp_id][lane]; // warp_id is block-level, not per warp? Actually warp_id ranges 0..31 (since 1024 threads / 32 = 32 warps). So we can have each warp handle one d. The warp_id corresponds to the d offset within the tile. This is perfect! So block size = 1024 threads (32 warps). Grid = (B, D/32) = (32, 8) = 256 blocks. Total warps = 256 blocks * 32 warps = 8192 warps. That's exactly the same as the one-warp-per-channel kernel, but now memory accesses are coalesced! And we have the same high warp count. This seems ideal. Let's verify: B=32, D=256, D/32=8. So grid size = dim3(B, D/32) = (32,8). Blocks = 256. Each block has 1024 threads. Total threads = 256*1024 = 262,144. Warps = 262,144 / 32 = 8192. Yes. Each block handles one b (blockIdx.x) and one d_tile (blockIdx.y). Within the block, there are 32 warps, each handling one d in the tile. The 1024 threads load a tile of 32 t x 32 d. t varies across threads as t_off = tid / 32 (0..31), d varies as d_off = tid % 32 (0..31). The global load is a[b, t_tile*32 + t_off, d_tile*32 + d_off]. Since threads in a warp have same t_off and consecutive d_off, they read contiguous memory. Coalesced. Then shared memory stores as s_a[d_off][t_off] with padding. Then each warp reads s_a[warp_id][lane] (where warp_id = tid/32). warp_id corresponds to d_off. So warp 0 reads d_off=0 across all t. This is exactly the data for that d's 32 t values. Then the warp does the warp shuffle inclusive scan on those 32 values. Then apply running prefix (register per warp), write output. Then update running prefix with chunk aggregate (from lane 31). Output writing: We need to write out[b, t, d] in original layout. After computing out_val for lane i (t = t_tile*32 + i, d = d_tile*32 + warp_id), we can write directly to global memory: out[idx] = out_val. The write pattern: each warp writes 32 values for the same d and consecutive t. The addresses are out[b, t, d] with t varying, d fixed. Stride = D. So writes are strided! Just like the load in the naive strided kernel. That's bad for output! We will have 32 separate cache lines per warp write. To fix this, we can write results back to shared memory in a transposed layout (or directly write if we accept strided writes). Strided writes are also bad for memory bandwidth. However, we can use shared memory to transpose writes: after computing out_val in registers, store them back to shared memory (s_out[d_off][t_off] maybe) and then have all threads write from shared memory to global memory in a coalesced pattern. That would add some overhead but improve write bandwidth. Alternatively, we can store output in transposed layout if the evaluator expects (B, L, D). It does. So we must write in that layout. But we can write from shared memory to global memory coalesced by having each thread write a contiguous block along d. Let's see: after the warp scan, each lane has out_val for its t and d. We can write to shared memory as s_out[warp_id][lane] = out_val. Then after __syncthreads(), we have the output tile in shared memory. Then we can write to global memory by mapping threads to d_off and t_off again: each thread reads s_out[d_off][t_off] and writes to out[b, t_tile*32 + t_off, d_tile*32 + d_off]. Since t_off and d_off mapping is the same as load, the writes will be coalesced (threads in a warp have same t_off and varying d_off). So the write addresses are contiguous. Yes! So we need an extra shared memory array for output, or we can reuse the shared memory for a and b since they are no longer needed after the scan. We can write out_val into s_b[warp_id][lane] (overwriting b shared mem) or s_a. But we need to be careful: after warp scan, warps read from shared memory. The shared memory arrays are 32x33 floats. We can reuse s_a or s_b for output. Sequence: 1. Load a tile of a and b into smem_a and smem_b. 2. __syncthreads(); 3. Each warp reads its d's values from smem_a and smem_b. 4. Warp shuffle scan, compute out_val. 5. Write out_val to smem_a[warp_id][lane] (overwriting a values). 6. __syncthreads(); 7. Each thread reads out_val = smem_a[d_off][t_off] and writes to global memory out. Wait, note that after step 5, smem_a contains output transposed: smem_a[d_off][t_off] = out for that d and t. But in step 5, warp_id writes to smem_a[warp_id][lane]. warp_id corresponds to d_off. lane corresponds to t_off. So after all warps write, smem_a[d][t] holds the output. Then in step 7, threads read smem_a[d_off][t_off] (where d_off = tid%32, t_off = tid/32). That's the same mapping as the initial load. So threads read the correct output for their global coordinates. Then they write coalesced. This is beautiful. The only extra cost is two __syncthreads() per tile: one after load, one after write to shared mem before global write. And the warp scan. Total tiles per block: L/32 = 128. So 128 iterations of the loop. Each iteration: - Coalesced load 1024 floats from a, 1024 from b. (2*1024*4=8KB) - 2 __syncthreads() - 32 warps each do warp shuffle scan (5 shuffles each) - 32 warps each do 2 shuffles for broadcast aggregate. - Coalesced write 1024 floats to out. (4KB) Total memory traffic: read a: 134MB, read b: 134MB, write out: 134MB. Plus shared memory traffic (negligible). The loads and stores are fully coalesced. The number of warps is 8192, which should saturate the GPU. This kernel should be near-optimal. Let's compute the FLOPs: each element has 2 FLOPs (mul and add). Total FLOPs = B*L*D*2 = 32*4096*256*2 = 67,108,864. ~67 MFLOPs. That's tiny. So the kernel is purely memory-bound. The memory bandwidth utilization (SOL) should be high because we have coalesced accesses and high occupancy. Now, we need to write the CUDA code. Let's flesh out the kernel. Parameters: - B, L, D are known at runtime. They can be passed to the kernel, or we can compute them from tensor sizes. Tensor shapes: a, b are (B, L, D). out is (B, L, D). Kernel signature: __global__ void scan_kernel(const float* a, const float* b, float* out, int B, int L, int D) Block size: 1024. Grid: dim3(B, D/32). Inside kernel: const int tid = threadIdx.x; const int d_off = tid % 32; const int t_off = tid / 32; const int b = blockIdx.x; const int d_tile = blockIdx.y; const int d = d_tile * 32 + d_off; // Shared memory __shared__ float s_a[32][33]; __shared__ float s_b[32][33]; // We only need two arrays. We can reuse s_a for output. // Actually we need to keep a and b for warp scan? No, warps read from s_a and s_b, compute, then write result back to s_a. // So s_a can be overwritten. float running_h = 0.0f; const int num_tiles = L / 32; // L must be multiple of 32. for (int tile = 0; tile < num_tiles; ++tile) { int t = tile * 32 + t_off; int idx = b * L * D + t * D + d; // Load s_a[d_off][t_off] = a[idx]; s_b[d_off][t_off] = b[idx]; __syncthreads(); // Warp scan int warp_id = tid / 32; int lane = tid % 32; // Each warp handles one d: d_idx = warp_id float a_val = s_a[warp_id][lane]; float b_val = s_b[warp_id][lane]; #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float temp_a = __shfl_up_sync(0xFFFFFFFF, a_val, offset); float temp_b = __shfl_up_sync(0xFFFFFFFF, b_val, offset); if (lane >= offset) { b_val = a_val * temp_b + b_val; a_val = a_val * temp_a; } } float out_val = a_val * running_h + b_val; // Write result to shared memory (overwrite s_a) s_a[warp_id][lane] = out_val; // Get chunk aggregate from lane 31 float chunk_agg_a = __shfl_sync(0xFFFFFFFF, a_val, 31); float chunk_agg_b = __shfl_sync(0xFFFFFFFF, b_val, 31); running_h = chunk_agg_a * running_h + chunk_agg_b; __syncthreads(); // Write to global memory coalesced out[idx] = s_a[d_off][t_off]; __syncthreads(); // do we need this? Not before next tile's load? The next iteration loads into s_a[d_off][t_off] and s_b[d_off][t_off]. Since all threads have finished reading s_a for output write, and the next loop overwrites s_a, we need a sync to ensure all writes to global memory are done before next load? Actually writes to global memory don't need sync with shared memory, but we need to ensure that all threads have finished writing out before any thread starts the next iteration's load, because the next iteration's load writes to s_a and s_b, and some threads might still be reading s_a in the previous iteration? No, all threads read s_a in the warp scan step before this write-out. After __syncthreads() after writing s_a, all threads can read s_a[d_off][t_off] for global write. After that, the next iteration will write to s_a again. To avoid race between a slow thread still reading s_a for global write and a fast thread writing s_a in the next iteration, we need a sync at the end of the loop. However, the global write reads s_a[d_off][t_off] using tid mapping. The next iteration's load writes s_a[d_off][t_off] using the same tid mapping. So if a thread is still doing the global write while another thread has moved to the next iteration and writes s_a, it's okay because they are writing to the same location (same d_off, t_off) and same thread writes it. Actually each thread writes s_a[d_off][t_off] and later reads it. Since each thread only accesses its own "slot" (determined by tid), there is no inter-thread conflict after the __syncthreads() following the warp write. Wait, after warp scan, each warp writes to s_a[warp_id][lane]. Then __syncthreads(). Then each thread reads s_a[d_off][t_off] and writes to global memory. Here, thread tid reads s_a[d_off][t_off]. This is a different mapping than write: d_off = tid%32, t_off = tid/32. The write mapping was warp_id = tid/32, lane = tid%32. So they are the same! Because warp_id = tid/32, lane = tid%32. And d_off = tid%32 = lane, t_off = tid/32 = warp_id. So s_a[warp_id][lane] is exactly s_a[d_off][t_off]. So each thread writes to the same location it reads. Therefore, after __syncthreads(), each thread can safely read its own s_a slot without any other thread accessing it. Thus no need for another __syncthreads() before the next iteration's load. However, the next iteration's load writes to s_a[d_off][t_off] again. Since each thread overwrites its own slot, and no other thread reads that slot after the current iteration's global write, it's safe to proceed without an extra sync. But we also have s_b being written in the next iteration. s_b is also per-thread slot. So no race. But what about the warp scan step? Warps read s_a[warp_id][lane]. That is, warp_id = tid/32. Since d_off = tid%32, t_off = tid/32, the warp read uses warp_id = t_off, lane = d_off. So warp 0 reads s_a[0][0..31], which corresponds to t_off=0, all d. These slots were written by threads with t_off=0, d_off=0..31. That's exactly the same warp (warp_id=0). Wait, threads with tid 0..31 have t_off = 0/32? No, tid 0..31: t_off = tid/32 = 0, d_off = tid%32 = 0..31. So warp 0 handles d=0..31? Actually thread 0..31 have same t_off=0 and varying d_off. They belong to the same warp (warp_id=0). When they do the warp scan, warp_id = tid/32 = 0. So they read s_a[0][lane] which corresponds to d_off = lane, t_off = 0. That was written by themselves! But what about warp 1 (tid 32..63)? They have t_off = 1, d_off = 0..31. warp_id = 1. They read s_a[1][lane] which corresponds to d_off=lane, t_off=1. Written by themselves. Good. So within a block, warp_id corresponds to t_off. Wait, that means each warp handles one t_off (one row of t) and scans across d? No! The warp scan is reading s_a[warp_id][lane] where warp_id = t_off, lane = d_off. So the values being scanned are for a fixed t_off and varying d? That would scan across d, not t! That's wrong. We need to be careful: we want each warp to scan along t for a fixed d. In shared memory, the preferred layout for reading a fixed d across t is s_a[d][t]. But our write layout used s_a[d_off][t_off] to avoid bank conflicts. Then d_off is the first index, t_off is the second. So for a fixed d (say d=0), the values across t are s_a[0][0], s_a[0][1], ..., s_a[0][31]. We want a warp to read these 32 values. So the warp should have lane correspond to t_off, and the warp should have a fixed d. That means the warp should consist of threads with the same d_off and varying t_off. But in our block, threads are grouped into warps by tid/32. tid = t_off*32 + d_off. So warp 0 has tid 0..31: t_off=0, d_off=0..31. This warp has varying d_off, fixed t_off. That's the opposite of what we need. We need to restructure the thread mapping. We want each warp to process one d (i.e., one channel) and 32 consecutive t's. So the 32 threads in the warp should have the same d_off and varying t_off. In terms of linear thread index tid (0..1023), we can map: d_off = tid / 32; // 0..31 t_off = tid % 32; // 0..31 Then threads in a warp (tid 0..31) have d_off = 0, t_off = 0..31. Good! Warp 0 handles d_off=0, warp 1 handles d_off=1 (tid 32..63), etc. There are 32 warps, each handling one d. Total threads = 32*32 = 1024. That's perfect. Now, for global memory load: we want coalesced loads. Threads in a warp have same d_off, varying t_off. So they load a[b, t_tile*32 + t_off, d_tile*32 + d_off]. Since d_off is same across warp, but t_off varies, the addresses are separated by D. That's strided! Not coalesced. So this mapping gives good warp scan but bad loads. We need a mapping where threads in a warp have contiguous global memory addresses for load, but after loading into shared memory, we can reorganize so that each warp reads contiguous values for its d. The standard approach is to use the "swizzle" mapping: - For loading: map threads so that consecutive threads have consecutive d_off. So d_off = tid % 32, t_off = tid / 32. This gives coalesced loads. - Store into shared memory in a transposed layout (or padded layout) such that the data is arranged by d then t. - For reading: we need warps to read data for one d across t. So warp_id should correspond to d, and lane to t. Since warps are tid/32, we could have a second phase where threads are remapped. But we can't change warp composition. However, we can read shared memory with a different mapping: each thread can read any element. If we want warp 0 (tid 0..31) to read d=0 across t=0..31, then thread i (lane i) reads s_a[0][i]. That means thread i reads the element for d=0, t=i. That works! The warp doesn't need to consist of threads with same d; it just needs to read the appropriate values. We can have each warp handle one d, and each thread in the warp read the element for that d and its lane as t. Since threads in a warp are tid=0..31, we can set: int lane = tid % 32; int d_for_warp = tid / 32; // but tid/32 is warp_id, so each warp has a unique d_for_warp. Wait, if tid 0..31: d_for_warp = 0. lane = 0..31. So warp 0 reads s_a[0][lane]. That's correct. If tid 32..63: d_for_warp = 1. lane = 0..31. Warp 1 reads s_a[1][lane]. Correct. So we can keep the block mapping as: For load: int d_off = tid % 32; int t_off = tid / 32; For warp scan: int warp_id = tid / 32; // 0..31 int lane = tid % 32; // 0..31 float a_val = s_a[warp_id][lane]; // read the d corresponding to warp_id, t corresponding to lane. This works! Because s_a[d][t] stores values for d across t. The load stored s_a[d_off][t_off] = a[idx] (with d_off = tid%32, t_off=tid/32). Then warp_id = tid/32 reads s_a[warp_id][lane] = s_a[tid/32][tid%32]? Wait, warp_id = tid/32, lane = tid%32. For tid 0..31, warp_id=0, lane=0..31. They read s_a[0][0], s_a[0][1], ..., s_a[0][31]. These were written by threads with d_off=0? Let's see: s_a[0][0] was written by thread with d_off=0, t_off=0. That's tid = t_off*32 + d_off = 0*32 + 0 = 0. s_a[0][1] was written by thread d_off=0? Wait, the write is s_a[d_off][t_off] = a[idx]. So s_a[0][1] is written by thread with d_off=0, t_off=1. That's tid = 1*32 + 0 = 32. So s_a[0][1] is written by thread 32, which is in warp 1. That's fine. s_a[0][31] is written by thread with d_off=0, t_off=31: tid = 31*32 + 0 = 992. That is warp 31. So all threads across the block contributed to writing d=0 column. Good. Then warp 0 (tid 0..31) reads s_a[0][0..31]. These values are all for d=0, t=0..31. Exactly what warp 0 needs. Similarly, warp 1 (tid 32..63) reads s_a[1][0..31]. These are for d=1, t=0..31. Good. So the mapping works. The load is coalesced because threads in a warp have same t_off and varying d_off (consecutive). The read for warp scan accesses different banks because s_a[warp_id][lane] is stored with padding [32][33], and warp_id is constant within a warp, lane varies. Banks = (warp_id*33 + lane) % 32 = (warp_id + lane) % 32. Lane varies, so no bank conflicts. After warp scan, each warp writes result back to s_a[warp_id][lane] (overwrite). Then __syncthreads(). Then for global write, we want coalesced writes. We need threads in a warp to write consecutive addresses. Consecutive addresses in output correspond to varying d and fixed t. So we need threads in a warp to have fixed t_off and varying d_off. That's the same as the load mapping. Since after __syncthreads(), s_a[d_off][t_off] holds the output for d=d_tile*32+d_off, t=tile*32+t_off. Wait, does it? After warp scan and write, s_a[warp_id][lane] = out_val for warp_id's d and lane's t. So s_a[d][t] = out for that d and t. Then if we have each thread read s_a[d_off][t_off] with d_off = tid%32, t_off = tid/32, it will read the output for its original d and t. Then write to global memory: out[b, t, d]. Since threads in a warp have same t_off and varying d_off, writes are consecutive in d. Coalesced! Perfect. So the kernel structure is: - tid = threadIdx.x - d_off = tid % 32 - t_off = tid / 32 - b = blockIdx.x - d_tile = blockIdx.y - d_global = d_tile * 32 + d_off - shared float s_a[32][33]; - shared float s_b[32][33]; - (maybe reuse s_a for output) Loop over tiles: t_global = tile*32 + t_off idx = b*L*D + t_global*D + d_global s_a[d_off][t_off] = a[idx] s_b[d_off][t_off] = b[idx] __syncthreads() // warp scan int warp_id = tid / 32; // 0..31 int lane = tid % 32; // 0..31 float a_val = s_a[warp_id][lane]; float b_val = s_b[warp_id][lane]; float ... warp shuffle ... float out_val = a_val * running_h + b_val; // write back to smem s_a[warp_id][lane] = out_val; // update running_h using lane 31 aggregate float chunk_a = __shfl_sync(..., a_val, 31); float chunk_b = __shfl_sync(..., b_val, 31); running_h = chunk_a * running_h + chunk_b; __syncthreads() // write to global memory out[idx] = s_a[d_off][t_off]; // optional __syncthreads() at end of loop to avoid next iteration load overwriting before all reads done? But as argued, after writing s_a[warp_id][lane], __syncthreads(), then reading s_a[d_off][t_off] (which is same location as written by this thread), and then next iteration writes s_a[d_off][t_off] again. Since no other thread reads this slot after __syncthreads(), it's safe. // However, the next iteration's load also writes to s_b[d_off][t_off], which is independent. // So we don't need an extra __syncthreads() at the very end. // But to be safe, we can add one. It costs a little. Wait, is there any cross-thread read after the second __syncthreads()? The threads read s_a[d_off][t_off] and write to global. No other thread reads that s_a slot after that. The next iteration's load writes to s_a[d_off][t_off] and s_b[d_off][t_off]. Since it's the same thread writing to the same slot, it's safe even without sync. But the previous iteration's warp scan reads s_a[warp_id][lane] which might be written by other threads (since warp_id corresponds to t_off in write mapping). For tile N, the load writes s_a[d_off][t_off]. Then __syncthreads. Then warp scan reads s_a[warp_id][lane]. warp_id = tid/32, lane = tid%32. So s_a[tid/32][tid%32] is read. This location was written by a thread with d_off = tid%32, t_off = tid/32. That is exactly the same thread (tid). Wait, no: for warp scan, warp 0 reads s_a[0][0] (tid 0), s_a[0][1] (tid 1), ..., s_a[0][31] (tid 31). s_a[0][1] was written by thread with d_off=0, t_off=1. Which thread is that? tid = t_off*32 + d_off = 1*32 + 0 = 32. That is a different thread! So warp 0 reading s_a[0][1] reads data written by thread 32. That's fine, because we have __syncthreads() after the load. Good. For the output write, each thread reads s_a[d_off][t_off] = s_a[tid%32][tid/32]. This location was written by warp scan write: s_a[warp_id][lane] = out_val. warp_id = tid/32? No, warp_id here is the warp id of the thread that wrote it. Which thread wrote s_a[d_off][t_off]? Let's find: we need a thread with warp_id_w = ? and lane_w = ? such that s_a[warp_id_w][lane_w] corresponds to s_a[d_off][t_off]. The write mapping is warp_id_w = some_t_off, lane_w = some_d_off. Wait, warp_id in the write step is tid/32. For a given output at (d_off, t_off), which thread wrote it? The thread that processed d = d_off in the warp scan. That thread belongs to warp with warp_id = d_off? Actually, in the warp scan step, each warp handles one d. Which warp handles d_off? warp_id = d_off? Let's check: warp 0 handles d=0 (global d = d_tile*32 + 0). warp 1 handles d=1, etc. So warp_id corresponds to d_off. The lane corresponds to t_off. So the write is s_a[d_off][t_off] = out_val, performed by thread in warp d_off, lane t_off. What is the tid of that thread? tid = warp_id*32 + lane = d_off*32 + t_off. That's not the same as our load mapping tid = t_off*32 + d_off (unless d_off = t_off). So the output for (d_off, t_off) is written by thread with tid = d_off*32 + t_off. Then in the global write step, the thread with tid = t_off*32 + d_off reads s_a[d_off][t_off]. Since these are different threads, we need a __syncthreads() between the write to s_a and the read from s_a for global output. We have __syncthreads() before the warp scan, and after the warp scan write, we also have __syncthreads(). Let's check: Sequence: 1. Load: s_a[d_off][t_off] = a[idx]. (mapping M1: tid = t_off*32 + d_off) 2. __syncthreads(); 3. Warp scan: read s_a[warp_id][lane] where warp_id = tid/32, lane = tid%32 (mapping M2: tid = warp_id*32 + lane). Wait, here tid is the same linear thread index. For tid = t_off*32 + d_off (M1), warp_id = tid/32 = t_off, lane = tid%32 = d_off. So warp scan reads s_a[t_off][d_off]. But the desired data for scan is for a fixed d across t. This reads s_a[t_off][d_off] which is for fixed t_off across d? That's scanning across d, not t! Oh no! I made a mistake. Let's re-evaluate. We said earlier: warp_id = tid/32, lane = tid%32. For load mapping: tid = t_off*32 + d_off. So for tid 0..31: t_off = 0, d_off = 0..31. warp_id = 0, lane = 0..31. They read s_a[warp_id][lane] = s_a[0][0..31]. In shared memory, s_a[d][t] holds data for d and t. So s_a[0][0..31] holds data for d=0, t=0..31. But these tid 0..31 have d_off = 0..31, t_off = 0. They are the load threads for different d's, same t=0. Now they read s_a[0][lane] = data for d=0, t=lane. So they all read data for d=0? Yes! All 32 threads in warp 0 read data for d=0 but different t. That's exactly what we want for a warp scan along t for d=0. Even though the threads originally loaded data for different d's, they now collectively read data for d=0 across t. That's fine. Similarly, warp 1 (tid 32..63) has t_off = 1, d_off = 0..31. They read s_a[1][lane] = data for d=1, t=lane. So warp 1 scans along t for d=1. Good. Now after warp scan, each warp writes result back to s_a[warp_id][lane]. Warp 0 writes to s_a[0][lane] (d=0, t=lane). Warp 1 writes to s_a[1][lane] (d=1, t=lane). ... Warp 31 writes to s_a[31][lane] (d=31, t=lane). Now, for global write, we need coalesced writes. We want threads in a warp to write consecutive d's for a fixed t. Which threads should do that? Threads with same t_off and varying d_off. That is tid = t_off*32 + d_off. For a given t_off = k, the 32 threads with tid = k*32 + d_off (d_off=0..31) form a warp! Wait, tid = k*32 + d_off. For fixed k, these are consecutive tids: k*32 to k*32+31. That's exactly a warp! Because warps are groups of 32 consecutive threads. So the same group of threads that loaded coalescedly (warp with fixed t_off) can also write coalescedly. They just need to read the correct output from shared memory. Specifically, thread with tid = t_off*32 + d_off needs to read the output for its own d_off and t_off. Where is that stored in shared memory? It's stored at s_a[d_off][t_off] (since warp d_off wrote to s_a[d_off][t_off] because warp_id = d_off, lane = t_off). Wait, warp_id = d_off? Let's verify: the warp that wrote s_a[d_off][t_off] is the warp with warp_id = d_off. In that warp, the thread with lane = t_off wrote s_a[warp_id][lane] = s_a[d_off][t_off]. The tid of that thread is warp_id*32 + lane = d_off*32 + t_off. Now, the thread with tid = t_off*32 + d_off wants to read s_a[d_off][t_off]. Since these are different threads (unless d_off = t_off? No, in general tid differs), we need a __syncthreads() after the warp scan write, and before the global write read. Let's trace for a specific element: d=0, t=1. - It was loaded by thread with t_off=1, d_off=0: tid = 1*32 + 0 = 32. - After __syncthreads(), warp 0 (tid 0..31) reads s_a[0][lane] for d=0. The thread in warp 0 with lane=1 (tid=1) reads this value, does scan, and writes back to s_a[0][1]. So the output is at s_a[0][1], written by tid=1. - Now for global write, we need thread with t_off=1, d_off=0 (tid=32) to read s_a[0][1] and write it out. Since tid=32 is in a different warp (warp 1) than tid=1 (warp 0), we need __syncthreads() to ensure tid=1's write is visible to tid=32. - Then tid=32 reads s_a[0][1]. But tid=32's warp_id = 1, lane = 0. We can just read s_a[d_off][t_off] = s_a[0][1]. So each thread computes d_off = tid % 32, t_off = tid / 32, reads s_a[d_off][t_off]. This works! Let's verify for all tids: - tid = t_off*32 + d_off. - It reads s_a[d_off][t_off]. - This slot was written by the warp scan of warp d_off, lane t_off. That writer had tid_w = d_off*32 + t_off. - Since tid and tid_w are different threads in general, __syncthreads() is required between warp write and global read. So the kernel flow is: for tile: // Load phase: threads mapped by (t_off, d_off) = (tid/32, tid%32) int d_off = tid % 32; int t_off = tid / 32; int idx = b*L*D + (tile*32 + t_off)*D + d_tile*32 + d_off; s_a[d_off][t_off] = a[idx]; s_b[d_off][t_off] = b[idx]; __syncthreads(); // Scan phase: each warp scans along t for one d int warp_id = tid / 32; // this is t_off from load mapping! Wait, for tid = t_off*32 + d_off, tid/32 = t_off. So warp_id = t_off. int lane = tid % 32; // = d_off. // But we want each warp to handle one d. Currently warp_id = t_off. // So with this mapping, each warp handles one t and scans across d. That's not what we want. Hold on. The load mapping groups threads into warps by t_off. Warp 0: t_off=0, d_off=0..31. Warp 1: t_off=1, d_off=0..31, etc. So warp_id = t_off. If we then do warp scan with warp_id = tid/32, we are scanning across d for each t. That's wrong. We need to remap threads for the scan phase. We can't change warp composition, but we can have each warp read data that corresponds to one d across t, even though the warp's threads have varying d_off and fixed t_off. Is that possible? Let's see: warp 0 has threads with d_off = 0..31, t_off = 0. If we want warp 0 to scan along t for d=0, we need thread 0 to read t=0, thread 1 to read t=1, ..., thread 31 to read t=31 for d=0. But in shared memory, s_a[d=0][t=0..31] is the data for d=0. The threads in warp 0 have lane = d_off. So thread i (lane=i) can read s_a[0][i]. That gives each thread in warp 0 a value for d=0, t=i. Then warp 0 can do shuffle scan across lanes (which correspond to t). Yes! This is exactly what I described earlier but got confused. Let's formalize: - Load phase: thread tid computes t_off = tid / 32, d_off = tid % 32. Loads a[b, tile*32+t_off, d_tile*32+d_off]. Writes to s_a[d_off][t_off]. - __syncthreads(). - Scan phase: warp_id = tid / 32 (which is t_off). lane = tid % 32 (which is d_off). But we don't use warp_id for indexing d. Instead, we can have each warp handle a specific d. Which warp handles which d? If we have 32 warps and 32 d's, we can assign warp i to handle d=i. But warp i consists of threads with t_off = i, d_off = 0..31. So warp i's threads have fixed t_off=i, varying d_off. If they read s_a[d_fixed][lane], that means each thread reads a different t (lane) for the same d. That works! We just need to pick d_fixed for each warp. Since there are 32 warps and 32 d's, we can assign warp i to read d=i. That is: int d_for_warp = warp_id; // warp_id = tid / 32 = 0..31. Since there are 32 warps. float a_val = s_a[d_for_warp][lane]; // lane = tid%32 = 0..31. Then the warp does shuffle scan on a_val. After scan, each thread has the prefix for d=d_for_warp, t=lane within tile. Then warp i writes back result to s_a[d_for_warp][lane]. Then update running_h for warp i. This works! Because warp_id = tid/32 uniquely identifies the warp within the block. With 1024 threads, there are 32 warps, numbered 0..31. Each warp handles one d (d_for_warp = warp_id). The threads in the warp read s_a[warp_id][lane]. Since lane = tid%32 varies from 0..31, they read t=0..31 for that d. Now, does this cause shared memory bank conflicts on read? - s_a is declared as float s_a[32][33]. - Read: s_a[d_for_warp][lane], where d_for_warp is constant across the warp, lane varies 0..31. - Address = d_for_warp * 33 + lane. Banks = (d_for_warp*33 + lane) % 32 = (d_for_warp + lane) % 32. Since lane varies, banks vary across threads. No bank conflicts. Great. On write: - s_a[d_for_warp][lane] = out_val. Same pattern. No bank conflicts. Now, after all warps write results back, __syncthreads(). Then for global write, we need coalesced writes: threads in a warp should write consecutive d for same t. So we use the original load mapping: - d_off = tid % 32 - t_off = tid / 32 - Read from shared memory: out_val = s_a[d_off][t_off]. - Write to out[idx] with idx = b*L*D + (tile*32 + t_off)*D + d_tile*32 + d_off. Does this read cause bank conflicts? - s_a[d_off][t_off] with d_off = tid%32, t_off = tid/32. - Address = d_off*33 + t_off. For a warp (tid 0..31): d_off = 0..31, t_off = 0. Addresses = 0*33+0, 1*33+0, ..., 31*33+0. Banks = (0,1,2,...,31). No conflicts. For warp 1 (tid 32..63): d_off = 0..31, t_off = 1. Banks = (0+1, 1+1, ..., 31+1) % 32. No conflicts. So reads are bank-conflict-free. Great! Thus the full algorithm is correct and efficient. Let's write the kernel carefully. One subtlety: s_a and s_b. We need s_a for loading a and later for output. We can reuse s_a for output. s_b is only needed for the warp scan. So we can declare: __shared__ float smem_a[32][33]; __shared__ float smem_b[32][33]; We can also pack them in a single array to reduce shared memory use, but 2*32*33*4 = ~8.4KB. That's fine. Implementation details: - L is assumed to be a multiple of 32. In the benchmark, L=4096, which is a multiple of 32. We can assert or just assume. - D is 256, a multiple of 32. B=32. - We need to handle arbitrary D? The constructor d_model is passed, but inputs might have that dimension. We should make the kernel generic, handling D not necessarily a multiple of 32. But since the benchmark likely uses d_model=256, we can optimize for multiples of 32 but also handle remainders. Simpler: we can write a generic kernel that handles arbitrary D by processing full tiles of 32 and a remainder. For D not multiple of 32, the last tile has fewer than 32 d's. We can still launch blocks for D/32, and for the last block, only process the remaining d's. However, the block always has 32 warps and loads 32x32 tiles. But if the remaining d's < 32, some warps would be out of bounds. That's a bit more code. Since the problem states d_model=256, we can assume D is a multiple of 32. But to be safe, we can add a generic fallback or handle it. Actually, the model's forward gets a and b of shape (B, L, D). D can be any value. The reference code works generically. We should handle generically, but optimized for D multiple of 32 and L multiple of 32. However, the benchmark only tests with D=256, L=4096. But it's safer to write a generic kernel. Generic approach: process D in tiles of 32. For each block (b, d_tile), d_base = d_tile * 32. The block handles up to 32 d's, but we must check bounds: if d_base + d_off >= D, the thread should not load/store. In the warp scan, only warps with d_base + warp_id < D should participate; others can be inactive. But all threads in the block load/store anyway, with boundary checks. Let's design the generic kernel: __global__ void scan_kernel(const float* a, const float* b, float* out, int B, int L, int D) { const int tid = threadIdx.x; const int d_off = tid % 32; const int t_off = tid / 32; const int b_idx = blockIdx.x; const int d_tile = blockIdx.y; const int d = d_tile * 32 + d_off; __shared__ float smem_a[32][33]; __shared__ float smem_b[32][33]; float running_h = 0.0f; const int num_tiles = (L + 31) / 32; for (int tile = 0; tile < num_tiles; ++tile) { int t = tile * 32 + t_off; bool valid = (b_idx < B) && (t < L) && (d < D); int idx = b_idx * L * D + t * D + d; // Load if (valid) { smem_a[d_off][t_off] = a[idx]; smem_b[d_off][t_off] = b[idx]; } else { smem_a[d_off][t_off] = 1.0f; // identity for a smem_b[d_off][t_off] = 0.0f; // identity for b } __syncthreads(); // Warp scan int warp_id = tid / 32; // 0..31 int lane = tid % 32; int d_scan = d_tile * 32 + warp_id; float a_val = 1.0f; float b_val = 0.0f; if (d_scan < D) { a_val = smem_a[warp_id][lane]; b_val = smem_b[warp_id][lane]; } // Only perform scan if this warp corresponds to a valid d if (d_scan < D) { #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float temp_a = __shfl_up_sync(0xFFFFFFFF, a_val, offset); float temp_b = __shfl_up_sync(0xFFFFFFFF, b_val, offset); if (lane >= offset) { b_val = a_val * temp_b + b_val; a_val = a_val * temp_a; } } float out_val = a_val * running_h + b_val; // Write output to smem_a smem_a[warp_id][lane] = out_val; float chunk_agg_a = __shfl_sync(0xFFFFFFFF, a_val, 31); float chunk_agg_b = __shfl_sync(0xFFFFFFFF, b_val, 31); running_h = chunk_agg_a * running_h + chunk_agg_b; } __syncthreads(); // Write to global memory if (valid) { out[idx] = smem_a[d_off][t_off]; } __syncthreads(); // needed for next iteration to avoid overwriting smem before all threads read } } Wait, the if (d_scan < D) inside the loop means some warps are idle. That's fine. The __syncthreads() must be called by all threads, which it is. The running_h is per warp (register). For invalid warps, running_h is unused. But there is a problem: the output write reads smem_a[d_off][t_off]. This will read the output written by the warp scan. However, if d_scan = d_tile*32 + warp_id >= D, then that warp didn't write to smem_a[warp_id][lane]. So for invalid d, smem_a contains the identity values (1,0) from load. But if the output for invalid d is not written to global memory, it's fine. However, valid threads might read smem_a[d_off][t_off] where d_off corresponds to an invalid d if d_tile*32 + d_off >= D. But we guard the global write with valid. So it's okay. But there is a subtle issue with the __syncthreads() at the end of the loop. In the next iteration, we overwrite smem_a[d_off][t_off] with new a values. Since all threads in the block participate in the load, and the previous iteration's global write is guarded by valid, some invalid threads might proceed to the next load immediately while valid threads are still writing global memory? No, all threads reach the __syncthreads() at the end of the loop before proceeding. So after that sync, it's safe to overwrite smem. Wait, the last __syncthreads() in the loop ensures all threads have finished writing global memory before any thread starts the next iteration. Good. However, there is another subtlety: in the warp scan, we have: if (d_scan < D) { ... smem_a[warp_id][lane] = out_val; ... } __syncthreads(); If d_scan >= D, the warp does not write to smem_a. But other warps might write, and then all threads read smem_a for global write. For valid d, the write happened. For invalid d, the smem still holds identity from load. Since we don't write to global for invalid d, it's fine. One more issue: in the generic kernel, t can go up to L. For the last tile, t may exceed L for some t_off. We set smem to identity for invalid t. In the warp scan, we read smem_a[warp_id][lane] for all lane 0..31. If t >= L for some lane, we loaded identity (1,0). The warp scan will include these identity elements at the end. Does that affect the scan? Yes, because inclusive scan over the tile includes elements beyond L, which would corrupt the prefix for the valid elements. We need to handle the last tile correctly. For the last tile, only the first (L % 32) elements are valid. The rest should be treated as identity so that they don't affect the prefix of valid elements. However, in an inclusive scan, if the invalid elements are at the end of the tile and have identity (1,0), then including them after the valid elements will not change the prefix for the valid elements because identity is (1,0): combine(valid_prefix, identity) = (1*valid_a, 1*valid_b + 0) = (valid_a, valid_b). So the prefix for the last valid element is unchanged. For elements after the valid range (invalid), we don't care because they won't be written to output (valid flag false). So loading identity for invalid elements is correct. Wait, need to check: the warp scan is inclusive. The elements are ordered by lane 0..31. If t goes from tile*32 to tile*32+31, and only the first M are valid, then lanes 0..M-1 have real data, lanes M..31 have identity. The inclusive scan will compute prefix for lane M-1 as combination of 0..M-1. For lane M, it combines prefix up to M-1 with identity, resulting in same prefix. For lane M+1, same, etc. So the prefix for valid lanes is correct. The invalid lanes get the same prefix as the last valid lane, but since we don't write them, it's fine. However, the chunk aggregate (lane 31) for the last tile will be the prefix up to lane 31, which is same as prefix up to M-1 (since rest are identity). So chunk_agg is correct. running_h update is correct. Great. Similarly for D: if a warp corresponds to an invalid d (d_scan >= D), we load identity and skip scan, leaving smem_a with identity. But we don't write output. And other warps are unaffected. Good. Now, what about the first tile? running_h = 0. The first element (t=0) should output b_0. Let's verify with identity: smem for d=0, t=0..31: lane 0: a_0, b_0. lanes 1..31: a_1..a_31, b_1..b_31. Warp scan inclusive: lane 0: a_0, b_0. out = a_0*0 + b_0 = b_0. lane 1: a_1*a_0, a_1*b_0+b_1. out = (a_1*a_0)*0 + ... = a_1*b_0+b_1. Correct. running_h after tile = chunk_agg_b (lane 31) = h_31. Correct. Now, about block/grid dimensions. Grid: dim3(B, (D+31)/32). For D=256, that's (32,8). Blocks = 256. Each block 1024 threads. Total threads = 262144. Warps = 8192. Shared memory per block: 2 * 32*33 * 4 = ~8448 bytes. That's well within limits. Register usage: each thread holds a few floats: a_val, b_val, running_h, temp vars. Probably < 32 registers. Occupancy will be high. Compute: O(B*L*D) work. Memory fully coalesced. Should be very fast. One concern: using two __syncthreads() per tile (128 tiles) = 256 syncs per block. Is that a lot? On A100, __syncthreads() cost is small (~10-20 cycles) if all warps are resident. With 32 warps per block, it's okay. But 256 syncs * 256 blocks * maybe some overhead. Still, with high occupancy, memory bandwidth dominates. Could we do the warp scan without storing back to shared memory and then reading again for output? If we write output directly to global memory from each warp, the writes would be strided (since warp writes out for fixed d and varying t). That is bad. The extra shared memory step ensures coalesced writes. We could also have each warp write its output to a temporary buffer in global memory with stride D, and then have a separate transpose kernel. But that adds another kernel and extra memory traffic. The fused approach is better. Alternatively, we could store output in registers, then use warp shuffle to exchange data among warps? No, warp shuffle only within a warp. We need to transpose data across warps to write coalescedly. Shared memory is the right tool. Another optimization: we can avoid the second __syncthreads() inside the loop by using double buffering in shared memory. But 128 iterations is fine. Wait, we can perhaps reduce the second __syncthreads() by noting that after the first __syncthreads() within the loop, we load from smem, compute, write back, then __syncthreads(). Then we read from smem for output and write to global. The next iteration overwrites smem_a and smem_b. To ensure that all threads have finished reading smem for output before the next iteration's load overwrites, we need a sync at the end of the loop. However, we could avoid the extra sync if we use two buffers and swap them each iteration. But shared memory is small, 2 buffers = 16KB. Still fine. But the __syncthreads() cost is likely small compared to memory loads. Let's estimate: each tile reads 8KB, writes 4KB. At 2TB/s, that's 6 microseconds per tile. __syncthreads() is ~0.05 microseconds. So negligible. Thus the kernel should be very efficient. Now, let's write the host-side wrapper in C++. We need to define a function that takes at::Tensor a, at::Tensor b, and returns at::Tensor out. The tensors are float (fp32). We can use a.options() to create output. CUDA code: ```cpp torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b) { auto B = a.size(0); auto L = a.size(1); auto D = a.size(2); auto out = torch::empty_like(a); const int tile_d = 32; const int tile_t = 32; dim3 block(1024); dim3 grid(B, (D + tile_d - 1) / tile_d); scan_kernel<<<grid, block>>>( a.data_ptr<float>(), b.data_ptr<float>(), out.data_ptr<float>(), B, L, D ); return out; } ``` Wait, block size is 1024. But each block will have 1024 threads. However, if D < 32, we still launch 1024 threads but some warps are idle. That's okay. But if D is small, many warps are idle. For D=1, only warp 0 is active, 31 warps idle. 1024 threads is overkill. But again, the benchmark uses D=256. We can also launch block size = 32 * min(32, D). For generic, we could compute blockDim.x = 32 * ((D+31)/32) * 32? Let's see: we need one warp per d in the tile. The tile has up to 32 d's. So we need up to 32 warps = 1024 threads. But if the tile has fewer d's (e.g., last tile), we still launch 1024 threads, which is fine. The extra warps just do nothing. Actually they still load data (identity) and skip scan. That's fine. But if D is less than 32, say D=16, then we still launch 1024 threads, but only 16 warps are useful. The other 16 warps load identity and skip. That's fine. One subtlety: if D is not a multiple of 32, then for the last tile, some warps are invalid. Our kernel handles that. Now, about the input shapes: a and b are (B, L, D). We assume they are contiguous? The reference code uses standard shape. In PyTorch, tensors might not be contiguous, but given the get_inputs function creates contiguous tensors (randn), they are contiguous. However, to be safe, we could check or call .contiguous(). But it's overhead. Usually it's fine. We should also handle non-contiguous by using strides. But for simplicity and speed, we can assume contiguous. Since the benchmark likely uses contiguous tensors generated by get_inputs, it's okay. But to be robust, we might want to check and make contiguous. However, copying would add overhead. Better: in the wrapper, we can check if a is contiguous, if not call contiguous(). Same for b. But making contiguous might allocate and copy, which is expensive. Since the benchmark passes them directly, and they are contiguous, we can skip. Alternatively, we can pass strides to the kernel and do strided indexing. But that adds a bit of arithmetic. Since tensors are contiguous, it's fine. Let's write the full inline code. One more thought: we need to ensure that the CUDA code compiles with the right architecture. load_inline will compile with the default arch. We can add extra_cuda_cflags if needed, but not necessary. Also, we need to include <cuda_runtime.h> and <cuda_fp16.h>? Not needed since we use float. We need to include <torch/extension.h>. Let's draft the code. ```python import torch from torch.utils.cpp_extension import load_inline cpp_source = """ #include <torch/extension.h> #include <cuda_runtime.h> __global__ void scan_kernel(const float* __restrict__ a, const float* __restrict__ b, float* __restrict__ out, int B, int L, int D) { const int tid = threadIdx.x; const int d_off = tid % 32; const int t_off = tid / 32; const int b_idx = blockIdx.x; const int d_tile = blockIdx.y; __shared__ float smem_a[32][33]; __shared__ float smem_b[32][33]; float running_h = 0.0f; const int num_tiles = (L + 31) / 32; for (int tile = 0; tile < num_tiles; ++tile) { int t = tile * 32 + t_off; int d = d_tile * 32 + d_off; bool valid = (b_idx < B) && (t < L) && (d < D); int idx = b_idx * L * D + t * D + d; // Coalesced load into shared memory (transposed layout) float av = valid ? a[idx] : 1.0f; float bv = valid ? b[idx] : 0.0f; smem_a[d_off][t_off] = av; smem_b[d_off][t_off] = bv; __syncthreads(); // Warp scan: each warp handles one d within the tile int warp_id = tid / 32; // 0..31 corresponds to one d offset int lane = tid % 32; int d_scan = d_tile * 32 + warp_id; float a_val = 1.0f; float b_val = 0.0f; if (d_scan < D) { a_val = smem_a[warp_id][lane]; b_val = smem_b[warp_id][lane]; } if (d_scan < D) { #pragma unroll for (int offset = 1; offset < 32; offset <<= 1) { float temp_a = __shfl_up_sync(0xFFFFFFFF, a_val, offset); float temp_b = __shfl_up_sync(0xFFFFFFFF, b_val, offset); if (lane >= offset) { b_val = a_val * temp_b + b_val; a_val = a_val * temp_a; } } float out_val = a_val * running_h + b_val; smem_a[warp_id][lane] = out_val; float chunk_a = __shfl_sync(0xFFFFFFFF, a_val, 31); float chunk_b = __shfl_sync(0xFFFFFFFF, b_val, 31); running_h = chunk_a * running_h + chunk_b; } __syncthreads(); // Coalesced write to global memory if (valid) { out[idx] = smem_a[d_off][t_off]; } __syncthreads(); } } torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b) { int B = a.size(0); int L = a.size(1); int D = a.size(2); auto out = torch::empty_like(a); dim3 block(1024); dim3 grid(B, (D + 31) / 32); scan_kernel<<<grid, block>>>( a.data_ptr<float>(), b.data_ptr<float>(), out.data_ptr<float>(), B, L, D ); return out; } """ scan = load_inline( name="scan_cuda", cpp_sources=cpp_source, functions=["scan_forward"], extra_cuda_cflags=["-O3", "--use_fast_math"], verbose=False, ) ``` Wait, the `cpp_sources` should include only the C++ code. But we need to ensure that `load_inline` uses `cpp_sources` and `cuda_sources`? Actually `load_inline` takes `cpp_sources` and `cuda_sources`. The CUDA kernel code should be in `cuda_sources` if it's `.cu` code. But we can put it all in `cpp_sources`? Actually `load_inline` compiles `cpp_sources` as C++ and `cuda_sources` as CUDA. If we include CUDA-specific syntax like `__global__` in `cpp_sources`, it might fail because the C++ compiler doesn't understand it. We need to put the CUDA kernel in `cuda_sources`. The wrapper function `scan_forward` can be in `cpp_sources` or `cuda_sources`. Typically, we put everything in `cuda_sources` if it contains CUDA code. Let's check `torch.utils.cpp_extension.load_inline` signature: `torch.utils.cpp_extension.load_inline(name, cpp_sources, cuda_sources=None, functions=None, extra_cuda_cflags=None, extra_cflags=None, verbose=False, ...)` So we should put the wrapper in `cpp_sources` and the kernel in `cuda_sources`. However, the wrapper calls the kernel, so they need to be linked together. If we declare the kernel in `cuda_sources` and use it in `cpp_sources`, we might need a forward declaration. Actually, we can put everything in `cuda_sources`, because `scan_forward` can also be compiled as CUDA (it's just C++ with some CUDA calls). There's no issue putting host code in `cuda_sources`. Many examples put all code in `cuda_sources`. Let's check typical usage: ```python scan = load_inline( name="scan", cpp_sources="torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b);", cuda_sources=cuda_source, functions=["scan_forward"], extra_cuda_cflags=["-O3"], ) ``` But if we put everything in `cuda_sources`, does it need a C++ wrapper? Actually, `load_inline` generates a pybind11 wrapper for the C++ functions declared in `cpp_sources` or `cuda_sources`. The `functions` list should contain the names of functions to expose. If we put `scan_forward` in `cuda_sources`, that's fine. We can set `cpp_sources=""` and `cuda_sources=cuda_source`. Or we can put the wrapper in `cpp_sources` and the kernel declaration in `cuda_sources`. To simplify, I'll put everything in `cuda_sources`. Actually, the wrapper needs `torch::empty_like` etc. That requires `<torch/extension.h>`. That can be included in `cuda_sources`. Let's do: ```python cuda_source = r''' #include <torch/extension.h> #include <cuda_runtime.h> __global__ void scan_kernel(...) { ... } torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b) { ... } ''' ``` And then `load_inline(name="scan", cpp_sources="", cuda_sources=cuda_source, functions=["scan_forward"], ...)`. But I recall that `cpp_sources` cannot be empty? It might be okay. Let's check typical patterns. Usually both are provided. If `cpp_sources` is empty string, it should be fine. Let's just put everything in `cuda_sources`. Or put the wrapper in `cpp_sources` and kernel in `cuda_sources`. The latter is more standard. Let's do: ```python cuda_source = r''' #include <cuda_runtime.h> __global__ void scan_kernel(const float* __restrict__ a, const float* __restrict__ b, float* __restrict__ out, int B, int L, int D) { ... } ''' cpp_source = r''' #include <torch/extension.h> torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b) { int B = a.size(0); int L = a.size(1); int D = a.size(2); auto out = torch::empty_like(a); dim3 block(1024); dim3 grid(B, (D + 31) / 32); scan_kernel<<<grid, block>>>( a.data_ptr<float>(), b.data_ptr<float>(), out.data_ptr<float>(), B, L, D ); return out; } ''' ``` This way, `cpp_sources` contains the host code, `cuda_sources` contains the device code. The linker will combine them. This is safe. Now, we need to ensure `extra_cuda_cflags` are set. Use `["-O3", "--use_fast_math"]`. `-O3` is standard. `--use_fast_math` might slightly affect precision, but the operator is just multiply-add. Fast math shouldn't matter much. But to be safe for correctness tolerance 1e-4, we might avoid `--use_fast_math` because it can affect precision of certain ops (like division, sqrt). For multiply and add, fast math is fine. But to be safe, we can omit `--use_fast_math`. Also, we should consider using `__launch_bounds__(1024)` on the kernel to help the compiler. Now, `ModelNew` class: ```python class ModelNew(nn.Module): def __init__(self, d_model: int): super().__init__() self.d_model = d_model def forward(self, a: torch.Tensor, b: torch.Tensor) -> torch.Tensor: return scan.scan_forward(a, b) ``` But we need to make sure `scan` module is available. The `load_inline` call returns a module with the exposed functions. We can assign it to a global variable in the Python file. Potential issue: `load_inline` caches the compiled extension based on source code. If we change the source, it recompiles. The first compilation might take time, but that's okay. Now, we need to test correctness. Let's call compile_kernel first. Wait, the compile_kernel tool expects the full Python source. So we need to write the complete file as a string. Let's draft the full Python source. One concern: the `scan_kernel` uses 1024 threads per block. The grid dimension y is (D+31)/32. For D=256, that's 8. So grid is (32,8). Blocks = 256. That is fine. But what if D is not a multiple of 32 and the last tile has fewer d's? For example, D=1. Then grid.y = 1. Block has 1024 threads. warp_id = tid/32 ranges 0..31. d_scan = 0*32 + warp_id = warp_id. So only warp_id=0 has d_scan < D (d_scan=0). Warps 1..31 are idle. But all threads still load data (for d_off = tid%32). For D=1, d = 0*32 + d_off. Only d_off=0 is valid. So valid threads are those with d_off=0. They load into smem_a[0][t_off] and smem_b[0][t_off]. The other threads load identity into smem_a[d_off][t_off] for d_off>0. Then warp 0 reads smem_a[0][lane] and does scan. Others skip. Then warp 0 writes result to smem_a[0][lane]. Then threads with d_off=0 read smem_a[0][t_off] and write output. All other threads skip write. This works! Good. Similarly for L not multiple of 32: last tile, some t_off are invalid. They load identity. Warp scan includes them, but they don't affect prefix for valid lanes. But wait, if L=33, last tile (tile=1) has t_off=0 valid, t_off=1..31 invalid. For warp 0 (d_scan=0), it reads smem_a[0][lane]. lane=0 gets a_32, b_32. lanes 1..31 get identity (1,0). Inclusive scan: lane 0: (a_32, b_32). lane 1: combine((a_32,b_32), identity) = (a_32, b_32). So lane 0 output is a_32*running + b_32. That's correct (h_32). lane 1 output is same as h_32, but we don't write because t=33 is invalid. Valid threads only write lane 0. So output is correct. Wait, but what about lanes >0 in the warp that are invalid? They get h_32 as out_val. Since they don't write, fine. But the chunk aggregate from lane 31 will be h_32 (since identity doesn't change it). So running_h update is correct. Good. Thus the kernel is generic. One minor issue: `__syncthreads()` inside a loop with conditional execution can cause deadlock if some threads in the block take a different branch that skips the __syncthreads(). In our kernel, the __syncthreads() calls are outside the `if (d_scan < D)` condition. All threads in the block reach them. The only conditional is around the warp scan body, but __syncthreads() is outside. So no deadlock. However, there is a conditional around the global write `if (valid)`. The __syncthreads() after it is reached by all threads because it's outside the if. Actually the code places __syncthreads() after the if block? In my draft: ```cpp if (valid) { out[idx] = smem_a[d_off][t_off]; } __syncthreads(); ``` All threads reach the __syncthreads() after the if. Since the if doesn't contain a return or break, all threads reach it. Good. But wait: after the warp scan, we have: ```cpp if (d_scan < D) { // ... smem_a[warp_id][lane] = out_val; // ... } __syncthreads(); ``` All threads reach the __syncthreads(). Good. Now, let's think about performance. The kernel has 256 blocks, each with 32 warps. 8192 warps total. Each warp processes L/32 = 128 tiles. So each warp executes 128 iterations. Each iteration does: - 2 loads (a,b) per thread from global memory (coalesced). - 2 stores to shared memory. - __syncthreads() - 1 load from shared memory per thread. - 5 shfl_up + 5 shfl (broadcast) + arithmetic. - 1 store to shared memory. - __syncthreads() - 1 load from shared memory per thread. - 1 store to global memory (coalesced). - __syncthreads() Actually, the first load from global and store to shared is done by all threads. Then __syncthreads(). Then warp scan loads from shared. Then writes to shared. Then __syncthreads(). Then loads from shared and writes to global. Then __syncthreads(). Total global memory transactions: 3 per thread per tile (read a, read b, write out). With 1024 threads per block and 256 blocks, that's 262144 threads. Each tile touches 32*32=1024 elements. Total elements = B*L*D = 33,554,432. Each element touched once per tile. Total global memory bytes = 33.5M * 4 * 3 = 402MB. All coalesced. With 8192 warps, memory bandwidth should be near peak. The __syncthreads() and shuffle overhead is small. Potential improvement: we can remove the final __syncthreads() by unrolling or restructuring. But it's fine. One more thing: `__shfl_up_sync` and `__shfl_sync` require compute capability >= 3.0. That's fine. Now, we should also consider half-precision? The problem says "produce numerically equivalent outputs at fp32 precision." The inputs in get_inputs are default float32 (randn returns float32). So our kernel uses float. Good. We need to compile and test. Let's write the Python source and call compile_kernel.

Let me analyze the problem. I need to implement a parallel associative scan (linear recurrence) on GPU. The recurrence is: ``` h_0 = b_0 h_t = a_t * h_{t-1} + b_t for t > 0 ``` This is a classic prefix scan that can be parallelized using Blelloch scan or similar approaches. For the associative scan, we need to define an associative binary operator. For two operations at positions i and j (where j comes after i): - If we have `(a_i, b_i)` and `(a_j, b_j)`, the combined operation is: - `a_combined = a_j * a_i` - `b_combined = a_j * b_i + b_j` This is because: `h_j = a_j * h_i + b_j = a_j * (a_i * h_{i-1} + b_i) + b_j = (a_j * a_i) * h_{i-1} + (a_j * b_i + b_j)` So the binary operator is: `(a2, b2) ∘ (a1, b1) = (a2 * a1, a2 * b1 + b2)` Wait, let me be careful about ordering. If we process elements in order, and we want to combine segment [0:i] with [i:j]: - Segment [0:i] is represented by `(a_left, b_left)` where applying to initial state gives `h_i = a_left * h_{-1} + b_left` (with `h_{-1} = 0`, but `h_0 = b_0`). - Actually, let's think of each element as an affine transformation: `f_t(x) = a_t * x + b_t`. - The composition is: `f_j(f_i(x)) = a_j * (a_i * x + b_i) + b_j = (a_j * a_i) * x + (a_j * b_i + b_j)`. - So the composed transformation is `(a_j * a_i, a_j * b_i + b_j)`. For a scan, we want to apply this composition left-to-right. A common efficient parallel scan on GPU uses a two-pass approach (Blelloch or Hillis-Steele), or a block-based scan with shared memory. For good performance on H100, I should use a work-efficient parallel scan implemented in CUDA. Key considerations: 1. The sequence length is 4096 2. Batch size is 32, d_model is 256 3. Total elements: 32 * 4096 * 256 = 33,554,432 This is a moderate problem size. The main challenge is that we have B*D independent scans (32*256 = 8192 scans), each of length 4096. Since each scan is relatively short (4096), but there are many of them (8192), we have good parallelism. Strategy: - Launch one block per scan (one per channel per batch), so 8192 blocks. - Each block handles a sequence of length 4096. - Use shared memory for the scan within a block. - Since 4096 is relatively small, we can do a synchronous scan within a block. For a block scan with 4096 elements: - Load data into shared memory - Perform parallel reduction scan - Write back Actually, 4096 elements per block might be too many threads. Let's think about thread count. - With 256 threads per block, each thread handles 16 elements (4096/256 = 16). - We can use a sequential scan per thread on 16 elements, then a parallel scan on the 256 partial results, then add back. This is the classic "double-buffered" or "chunked" scan: 1. Each thread sequentially scans its 16 elements, producing a local result and a local aggregate `(a_local, b_local)`. 2. Do a parallel block-level scan on the 256 local aggregates. 3. Each thread combines its local scan with the prefix aggregate from previous threads. For the block-level scan on 256 items with 256 threads: - Use the Kogge-Stone (Hillis-Steele) scan or Brent-Kung ( Blelloch) scan. - With 256 threads and 256 aggregates, Kogge-Stone is simple: log2(256) = 8 steps. - Brent-Kung is more efficient (2*log2(n) - 1 steps, but with better lane utilization). Actually, for our binary operator, the scan is: ```cuda struct Pair { float a; float b; }; __device__ Pair op(Pair x, Pair y) { // Combine: first apply x, then y // y ∘ x return {y.a * x.a, y.a * x.b + y.b}; } ``` Wait, careful. In a scan, we usually want `scan[i] = op(elements[0], elements[1], ..., elements[i])`. If each element is an affine transform `(a_i, b_i)`, and composing `f_j ∘ f_i` gives `(a_j * a_i, a_j * b_i + b_j)`, then: - `scan[0] = (a_0, b_0)` - `scan[1] = (a_1 * a_0, a_1 * b_0 + b_1)` - `scan[2] = (a_2 * (a_1 * a_0), a_2 * (a_1 * b_0 + b_1) + b_2)` Let's verify: if initial state is `h_{-1} = 0`, then: - `h_0 = a_0 * 0 + b_0 = b_0` ✓ - `h_1 = a_1 * h_0 + b_1 = a_1 * b_0 + b_1` ✓ So the scan gives us `(A_i, B_i)` where `h_i = A_i * 0 + B_i = B_i`. Wait, that's interesting. Let me check: - For element 0: `(a_0, b_0)`, and `h_0 = b_0`. So the scan result's B component is `h_0`. - For element 1: `(a_1 * a_0, a_1 * b_0 + b_1)`, and `h_1 = a_1 * b_0 + b_1`. ✓ - For element 2: `(a_2 * a_1 * a_0, a_2 * (a_1 * b_0 + b_1) + b_2)`, and `h_2 = ...` ✓ So we don't even need the A component for the output! But we need it for the scan computation. Actually, we can compute the scan of just the B values by maintaining both. But we can optimize: once we have the scan result, we only need the `b` component. Implementation plan: 1. Flatten batch and channel dimensions into one dimension: `N = B * D = 8192` 2. Launch `N` blocks, each with `THREADS` threads (e.g., 256) 3. Each block reads `L = 4096` elements 4. Each thread handles `ITEMS_PER_THREAD = L / THREADS = 16` elements 5. Sequential scan within each thread, storing results in shared memory and computing aggregate `(a_prod, b_acc)` 6. Parallel scan on the `THREADS` aggregates using shared memory 7. Each thread combines its local scan with the scanned aggregate from previous threads 8. Write back only the `b` values (which are the `h` values) Let me think about the memory access pattern: - Input `a` is shape `[B, L, D]`. We need to access it as `a[b, t, d]`. - If we group by `(b, d)` with each block handling one `(b, d)` pair, then within a block, threads access `a[b, t, d]` for consecutive `t`. - For coalesced access, consecutive threads should access consecutive `t` values. - So thread `tid` accesses `t = tid * ITEMS_PER_THREAD + i`. - But this means thread 0 accesses t=0,1,2,...,15 and thread 1 accesses t=16,17,...,31. - In memory, these are consecutive. So threads in a warp access consecutive memory - this is coalesced. Wait, actually `a[b, t, d] = a[(b * L + t) * D + d]`. For fixed `b` and `d`, consecutive `t` values are separated by `D` in memory. So if thread 0 reads `t=0` and thread 1 reads `t=1`, they access addresses separated by `D*sizeof(float) = 1024 bytes`. This might not be perfectly coalesced, but it's strided. Actually, if we instead have each block handle a subsequence and threads handle interleaved elements... For strided access by `D`, we need to think about memory coalescing. In CUDA, a warp of 32 threads accesses memory. If thread `i` accesses address `base + i * stride`, and `stride` is large, this results in 32 separate cache lines if `stride * sizeof(float)` exceeds cache line size (128 bytes). Here `stride = D * sizeof(float) = 1024 bytes`. So each thread in a warp accesses a different cache line. This is not ideal. To improve coalescing, we could: 1. Have each warp work on adjacent channels rather than adjacent timesteps 2. Or transpose the data first 3. Or use vectorized loads Actually, wait. Let me reconsider the memory layout. PyTorch default is row-major: `a[b, t, d] = a.base + (b * L * D + t * D + d) * sizeof(float)` Within a block handling `(b, d)`, thread `tid` accesses `a[b, tid, d]` for various `t`. If threads in a warp have consecutive `tid` and access the same `t` but different `d`... no, in our case each block handles fixed `(b, d)` and different `t`. So in a warp of 32 threads, each thread accesses: - Thread 0: `a[b, t0, d]` at offset `(b * L * D + t0 * D + d) * 4` - Thread 1: `a[b, t1, d]` at offset `(b * L * D + t1 * D + d) * 4` If `t0, t1, ..., t31` are consecutive, then the addresses differ by `D * 4 = 1024` bytes. Each warp accesses 32 cache lines. This is bad. If `t0 = 0, t1 = 16, t2 = 32, ...` (for ITEMS_PER_THREAD=16, THREADS=256), then thread 0 accesses t=0, thread 1 accesses t=16, etc. The difference is `16 * D * 4 = 16384` bytes, which is even worse. To get coalesced access, we need consecutive threads in a warp to access consecutive memory addresses. That means we need consecutive threads to vary `d`, not `t`. So perhaps a better approach is: - Each block handles a range of timesteps for a batch element - Or each block handles multiple `(b, d)` pairs and threads in a warp vary `d` Actually, an even better approach for this problem: given that we have 8192 scans, and each scan has 4096 elements, and the memory layout groups `d` together (since `a[b, t, d]`), the most natural way to get coalescing is to have threads in a warp work on different `d` but the same `t`. Let me think about an alternative layout: - Each block handles one timestep `t` for all `d` in a batch, or something like that. - But then scan within a block doesn't make sense because scans are along `t`. Actually, the best approach might be to transpose the data conceptually, or to change our thread layout. Consider: process the scans in a different way. Since PyTorch stores `a` as `[B, L, D]`, consecutive `d` values are adjacent in memory. So for a fixed `b` and `t`, `a[b, t, 0], a[b, t, 1], ..., a[b, t, D-1]` are contiguous. An approach that achieves coalescing: - Each block processes a batch element `b` and a range of `d` values. - For example, a block has `THREADS` threads, and each thread handles one `d` value. - Then when loading `a[b, t, d]` for all threads in the block, consecutive threads access consecutive `d` values → coalesced! - Each thread then scans along `t` for its `d` value. But if each thread scans sequentially along `t`, we get no parallelism within a scan. However, we have 256 parallel scans per block (if block size is 256), and with 8192 total scans, we'd have 32 blocks (8192/256 = 32). Each block does a sequential scan of 4096 elements per thread. That's 4096 * 32 blocks = not much parallelism, but on H100 with many SMs, 32 blocks is too few. Actually, with B=32 and D=256, if each block handles one batch and 8 channels (256 channels / 8 = 32 threads? No...) Let me reconsider. We want: 1. Good coalescing 2. Enough parallelism 3. Efficient scan Option A: Warp-level scan, each warp handles multiple channels - A warp has 32 threads. If each thread handles one `d`, we get coalesced loads per warp. - Each warp could handle 32 channels for one batch element. - With D=256, we need 8 warps per batch element. - With B=32, we have 32 batch elements * 8 warps = 256 warps total. - Each warp does a sequential scan of 4096 elements: 4096 iterations per warp. This might be slow. Option B: A block handles multiple batch elements, with threads varying over both - This gets complicated. Option C: Load transposed into shared memory, then do scan - First, load data in a coalesced way, then reorganize in shared memory, then do the scan. - For example, have a block of 256 threads load 256 consecutive `d` values for multiple `t` values. - Actually, we could have a block handle a sub-sequence of `t` values for all `d` in one batch element, but with a thread layout where thread `i` handles `d = i`. Wait, I think the cleanest solution is: - Use 256 threads per block - Each block handles one batch element `b` and 8 channels... no, that's not enough. Let me recalculate: - B = 32, L = 4096, D = 256 - Total work: 32 * 4096 * 256 = 33.5M multiply-add operations - The naive PyTorch implementation is sequential per scan but there's a Python loop overhead. - On H100, we can easily get good speedup. If we process by having each block handle one `(b, d)` pair, that's 8192 blocks. With 132 SMs on H100, that's 62 blocks per SM. With 256 threads per block, that's 16K threads per SM (up to 2048 max per SM), so we can have up to 8 blocks per SM concurrently, meaning about 1056 blocks active across all SMs. Wait, actually 8192 blocks is plenty - it's way more than the hardware can run concurrently, which is good for occupancy. But the memory access is strided by D=256. Each thread in a warp accesses memory stride 256*4=1024 bytes apart. This is not ideal but maybe the L2 cache can help. Actually, H100 has very good L2 cache bandwidth (3.35 TB/s effective). With 8192 blocks, each block accesses 4096*4 bytes = 16KB per input. But these 16KB chunks are strided throughout memory. For `a[b, :, d]`, the memory is at `base + (b * L * D + t * D + d) * 4`. For consecutive `t`, the addresses differ by `D * 4 = 1024`. So each access is to a different cache line. But the accesses within a block (4096 elements) cover a range of `4096 * 1024 = 4MB` in memory for each input. With two inputs (a and b), that's 8MB. L2 cache on H100 is 50MB, so all batch elements' data might fit in L2 or at least be cache-friendly. For `a[0, :, 0]`, addresses are `base + t * 1024`. A warp of 32 threads accesses 32 different cache lines. This results in 32 cache line loads. If each cache line is 128 bytes, that's 4096 bytes loaded for 128 bytes used (32 floats * 4 bytes). Only 3.125% bandwidth utilization! This is terrible. But wait - we can use vectorized loads! Instead of each thread loading one float, each thread loads a float4. Then for a warp loading 32*4=128 floats, we still access 32 cache lines, but each cache line provides 4 floats. So 512 bytes loaded for 512 bytes used... no, still 32 cache lines * 128 bytes = 4096 bytes for 512 bytes of data. Still only 12.5% utilization. The problem is the stride. Each float we want is separated by 1024 bytes. To fix this, we need consecutive threads to access consecutive memory. That means consecutive threads must access consecutive `d` values. Approach: Change our parallelism dimension. Instead of having one block per `(b, d)`, have one block per `b`, and within a block, threads handle different `d` values. All threads in a block load the same `t` but different `d` values → coalesced. So: - Block dimensions: 256 threads (or maybe 128 for register pressure) - Each block handles one batch element `b` - Thread `tid` handles channel `d = tid` (if blockDim.x = D = 256, which it is!) - There are 32 blocks (one per batch element) Now each thread does a sequential scan over L=4096 timesteps: ```cuda h = 0; for (t = 0; t < L; t++) { float a_val = a[b * L * D + t * D + tid]; float b_val = b[b * L * D + t * D + tid]; h = a_val * h + b_val; out[b * L * D + t * D + tid] = h; } ``` Wait! This is exactly equivalent to the Python loop but parallelized over channels. Since we have 256 threads per block and 32 blocks, total threads = 8192. Each thread does 4096 iterations. The memory access for loads and stores is coalesced because at each timestep `t`, threads with consecutive `tid` access consecutive memory addresses. But is this faster than PyTorch? The reference is: ```python for t in range(L): h = a[:, t] * h + b[:, t] outputs.append(h) return torch.stack(outputs, dim=1) ``` In PyTorch, the operations `a[:, t] * h + b[:, t]` operate on all B*D elements at once and can use vectorized kernels. The Python loop overhead is small for 4096 iterations, but there might be kernel launch overhead for each step (4096 kernel launches for elementwise ops), plus intermediate tensor allocations. Our single kernel approach eliminates all the Python/kernel launch overhead and does everything in one kernel with coalesced memory access. But wait - 32 blocks * 256 threads = only 8192 threads total. On H100 with 132 SMs, we only use 8192 / 2048 = 4 SMs at most (assuming 2048 threads per SM). That's very low occupancy! H100 can run up to 2048 threads per SM, so 32*256 = 8192 threads means we could theoretically use 4 SMs fully, leaving 128 SMs idle. This is a problem. We need more blocks to utilize all SMs. To increase block count while maintaining coalescing, we could: 1. Have each block handle a subset of timesteps for a batch element, with threads varying over `d`. - Block handles `b` and `t_range`. Threads handle different `d`. - But then we need to communicate the initial `h` value across blocks, which is complicated. 2. Have each block handle multiple batch elements, with threads varying over `d`. - Block 0 handles b=0 and b=1, with first warp handling b=0, second warp handling b=1. - But with 256 threads and D=256, we can't easily fit multiple batches in one block unless we reduce threads per block. 3. Reduce threads per block to 128 or 64, increasing block count. - If each block has 128 threads handling d=0..127 for one batch, we need 2 blocks per batch to cover all D=256. - Total blocks: 32 * 2 = 64. Still not enough. - If 64 threads per block, we need 4 blocks per batch, total 128 blocks. Still only about 1 block per SM. 4. Interleave batch elements across blocks differently. - Actually, we can just make the grid larger. What if we have multiple blocks per `(b, d)` pair, each handling a chunk of timesteps? But we need the scan state to propagate between chunks. Actually, the scan propagation issue is solvable with a two-level scan: - Level 1: Each block scans its chunk, outputs local results and block aggregate. - Level 2: Scan the block aggregates. - Level 3: Each block adds the prefix aggregate to its local results. This is the standard work-efficient scan. But for a recurrence where the operator depends on all previous values, we still need the two-level approach. However, let's think about whether simpler is better. With 8192 threads doing 4096 iterations each, each thread does 4096 loads, 4096 mul-adds, and 4096 stores. The main bottleneck will likely be memory bandwidth if not compute. Wait, but H100 can handle much more parallelism. If we only use 4 SMs, we're leaving a lot of performance on the table. We need more blocks. Alternative: What if we transpose the data in a preprocessing kernel? Then we can have many blocks per scan. Or, what if we use a different kernel where: - We have 8192 blocks (one per scan, like originally planned) - But within each block, we load data into shared memory in a coalesced way and then do a parallel scan. Actually, for the one-block-per-scan approach, we can still have coalesced loads if we structure it right. But with 8192 blocks and only 132 SMs, we get good occupancy. For coalesced loads in the one-block-per-scan approach... we already determined that consecutive threads accessing consecutive `t` gives strided memory. But what if we use fewer threads and have them load multiple elements? Actually, another idea: Compute the transpose on-the-fly in shared memory. - For a block handling scan `(b, d)`, we have 256 threads. - First, let's load a chunk of `a` and `b` into shared memory. - But we can't load coalesced for the same scan because the memory is strided. Unless... we load in a different pattern. What if we have threads in a block work on adjacent timesteps for adjacent scans? Like: - Block works on scans `(b, d=0..255)` for a range of `t`. - Then we can load `a[b, t, d=0..255]` coalesced! - But the issue is that each scan only needs one `d`, so we'd be loading 256 times more data than we use. That doesn't help. Let me reconsider the simple sequential approach with 32 blocks. Even though it only uses 4 SMs, it might still be faster than PyTorch because: 1. PyTorch reference does 4096 kernel calls (each doing elementwise ops on 32*256 elements) 2. Our kernel does everything in one call 3. Memory is coalesced 4. No intermediate allocations Actually, looking more carefully at the PyTorch reference: ```python for t in range(L): h = a[:, t] * h + b[:, t] outputs.append(h) ``` `a[:, t]` returns a view, so `a[:, t] * h + b[:, t]` is a single elementwise kernel call. It should be fused into one kernel by PyTorch's JIT or eager mode might launch 2-3 kernels (mul, add). But for 4096 iterations, that's still 4096 * 2 = ~8192 kernel launches. Kernel launch latency is ~5-10 microseconds, so 8192 * 5us = 40ms just in launch overhead. The actual computation is negligible. Our single kernel with 8192 threads and 4096 sequential steps per thread will: - Read 2 * 32*4096*256*4 = 256MB of input - Write 32*4096*256*4 = 128MB of output - Total memory traffic: ~384MB - At 3.35 TB/s, that's ~115 microseconds just for memory bandwidth. - With 4096 * 1 FMA per step, and FP32 throughput of 67 TFLOPS, compute is ~2.2 TFLOPS, which is ~3% of peak, so memory bound. If we could use all SMs, we'd still be memory bandwidth bound at ~115us. But with only 4 SMs active, we'd get at most 4/132 of peak bandwidth, which is ~101 GB/s, so ~3.8ms. That's much slower! Wait, that's a good point. If we're memory bandwidth bound and only use 4 SMs, we'll be limited by those 4 SMs' memory bandwidth. Each SM on H100 has some fraction of the total bandwidth. We need more parallelism to use all SMs. So I definitely need more blocks. How about: use 1024 threads per block, 32 blocks per batch element (1024 threads / 32 warps, but D=256). Hmm. Actually, wait. D=256. If we use 64 threads per block (two warps), each block handles 64 channels. For one batch element, we need 4 blocks. Total blocks: 32 * 4 = 128. Still not enough. If we use 32 threads per block, each block handles 32 channels. For one batch element, we need 8 blocks. Total blocks: 32 * 8 = 256. Still only 2 per SM. If we use 16 threads per block: 32 * 16 = 512 blocks. 4 per SM. Getting better but still low. If we use 8 threads per block: 32 * 32 = 1024 blocks. ~8 per SM. That's better, but 8 threads leaves a lot of warp capacity unused. And we also need to handle the case where each thread does sequential work. With fewer threads per block, each thread does more work. Actually, the occupancy isn't just about block count. It's about threads per SM. If we have 512 blocks of 32 threads, that's 16,384 threads total. At 2048 threads/SM, that's 8 SMs, still leaving 124 SMs idle. To use all 132 SMs with 2048 threads each, we need ~270,000 threads total. With 8192 scans, that means each scan would need ~33 threads. But each scan is sequential. Wait, we don't need all SMs to be fully occupied. But we do want enough warps to hide memory latency. With sequential execution per thread, memory latency is the main concern. If a thread is doing 4096 loads sequentially, and each load has ~400 cycle latency, the thread would stall a lot unless the warp has other threads to execute. But in a warp, all 32 threads execute the same instruction, so if all 32 threads are doing the same sequential loop, they all stall together. So we really want memory-level parallelism within a warp. That means either: 1. Issuing loads for multiple elements before using them (software pipelining / loop unrolling), or 2. Having enough warps per SM so that when one warp stalls on memory, another warp can execute. For option 2, with 32 threads per block, we'd need many blocks per SM. For example, 32 blocks of 32 threads = 1024 threads per SM. If each SM can handle 2048 threads, we could run 64 blocks per SM. But we don't have enough blocks unless we process multiple batch elements per block. So let's think about having each block process multiple batch elements. Block structure: 256 threads. Each warp (32 threads) processes one batch element, with each thread handling 8 channels. So: - Warp 0: batch b0, channels d=0..31 (each thread handles 1 channel... actually 32 threads * 1 channel = 32 channels) - Warp 1: batch b0, channels d=32..63 - ... - Warp 7: batch b0, channels d=224..255 - Warp 8: batch b1, channels d=0..31 - ... With 256 threads (8 warps), we can handle 1 batch element fully (8 warps * 32 channels = 256 channels). To handle more batch elements, we'd need more warps. We have exactly 8 warps. So one block handles 1 batch element with 256 threads (1 thread per channel). What if each block handles 2 batch elements? We'd need 512 threads, which is possible but less efficient. Or: each block handles 1 batch element, but we process multiple timesteps per thread to increase the block count? No, that changes the grid dimensions. Actually, what if we process the sequence in chunks and use many blocks? For example: - Each block handles one `(b, d)` pair and a chunk of 128 timesteps. - For a scan of length 4096, we'd have 32 chunks. - Total blocks: 8192 scans * 32 chunks = 262,144 blocks! That's enough to saturate all SMs. But the problem is: chunk k needs the final state from chunk k-1. So we'd need a multi-pass approach: - Pass 1: Each block scans its chunk and outputs the chunk's aggregate `(a_agg, b_agg)`. - Pass 2: Scan the aggregates to get prefix states for each chunk. - Pass 3: Each block uses the prefix state to finalize its local scan. This is the classic three-pass parallel scan, also known as "reduce-then-scan" or "Blelloch scan variant." For our affine recurrence: - Pass 1: For each chunk of length C, local scan and output aggregate `(A_chunk, B_chunk)` where `A_chunk = prod(a_i)` and `B_chunk = scan result`. - Pass 2: For each position, scan the aggregates. Actually, each chunk's aggregate is a single pair. We need to scan these aggregates across chunks. - Pass 3: Apply the scanned prefix to each local scan. This gives us lots of blocks! And we can make each chunk small enough that it fits in registers or shared memory. Let me design this: **Pass 1: Local scan per chunk** - Grid: `(num_chunks, num_scans)` = `(32, 8192)` = 262,144 blocks - Each block has 32 threads (or 128, etc.), processing `C` elements per scan - Each thread sequentially processes `C / threads_per_block` elements - Store local scan results in output array - Compute aggregate and store in a separate `aggregates` array of shape `[num_scans, num_chunks, 2]` Wait, output from pass 1 is incomplete. We could write local results to output and then update them in pass 3. Actually, let me think about whether we actually need pass 3 to write to output, or if we can keep local results in shared memory. Given the problem size (4096 timesteps, 8192 scans), if we use chunks of 128: - num_chunks = 4096 / 128 = 32 - Blocks in pass 1: 8192 * 32 = 262,144 - Each block scans 128 elements. With 32 threads, each thread scans 4 elements sequentially. - Pass 2: We need to scan 32 chunks for each of 8192 scans. So we have 8192 scans, each with 32 aggregates. We can launch 8192 blocks, each scanning 32 aggregates using 32 threads (1 thread per chunk, or parallel scan with 32 threads). - Pass 3: 262,144 blocks again, each applying one prefix to its local scan. This is complex but very parallel. However, the extra passes add overhead. But wait - for a scan of only 32 elements (the 32 chunks), pass 2 could be incorporated differently. We can do a block-level scan with 32 threads in Kogge-Stone style over 32 elements. Very fast. Is this worth it? Let's compare: - Sequential approach (per thread): 1 pass, 8192 threads, but only 32 blocks. Bad occupancy. - Multi-pass approach: 3 passes, 262k blocks, great occupancy. Actually, there's an intermediate approach. We can use the sequential approach but launch many more blocks by having each block handle one `(b, d)` pair but process the data in a way that increases block count... no, that's not how it works. Wait! What if we use a hybrid approach: - Have 8192 blocks (one per scan) - Each block has 256 threads, but the threads don't scan sequentially. Instead, they do a parallel scan on 4096 elements within the block. - 4096 elements / 256 threads = 16 elements per thread. - We use the Brent-Kung or Kogge-Stone scan in shared memory. This gives us 8192 blocks. Each block uses 256 threads. Total threads = 2,097,152. That's about 15,900 threads per SM, or about 8 blocks per SM (at 2048 threads/SM limit, actually 2048/256 = 8 blocks per SM). Total active blocks = 132 * 8 = 1056 blocks concurrently. But we have 8192 blocks total, so they queue up. This is actually pretty good occupancy! And each block handles 4096 elements with 256 threads. The shared memory approach will be efficient. The main issue is memory access pattern: non-coalesced loads because threads in a warp access strided memory. But maybe we can fix the memory access pattern with shared memory! Here's how: - Instead of having consecutive threads access consecutive `t` (which is strided in memory), we can have the block load a tile of data in a coalesced way. - For example, if each block handles scan `(b, d)`, and we have 256 threads, we could first transpose the data so that we can read it more efficiently. Actually no - the fundamental problem is that for a single scan `(b, d)`, the elements are stored with stride `D=256`. We can't change that without transposing first. But we can issue 4096 loads per block, and with good caching... let's think. Each block loads 4096 elements at stride 1024 bytes. On H100, the L2 cache line is 128 bytes. So each load pulls in 128 bytes, but we only use 4 bytes. The cache bandwidth would need to be 32x higher than useful bandwidth. However, H100 has 50MB L2 cache and 3.35 TB/s memory bandwidth. With 8192 blocks, each accessing 4096 * 4 = 16KB of strided data (but actually the total span is 4096 * 1024 = 4MB per scan), the cache might not help much. Wait, no: for `a[0, :, 0]`, we access `base + 0*D*4, base + 1*D*4, ..., base + 4095*D*4`. That's scattered across a 4MB range with 1024-byte stride. Not cache friendly at all. So the strided access is indeed a problem. Let me reconsider. What if we transpose `a` and `b` first? - `a_t = a.permute(0, 2, 1)` gives shape `[B, D, L]`. Now `a_t[b, d, t]` has consecutive `t` values adjacent in memory. - This is a memory copy/transpose operation that PyTorch can do efficiently. - After transpose, each scan `(b, d)` is contiguous. - Then our kernel can easily access it with fully coalesced loads if consecutive threads access consecutive `t`. Wait, but permute in PyTorch is just a view operation! It doesn't copy data. So `a.permute(0, 2, 1)` gives a tensor with new strides but same underlying storage. The actual data is still stored in `[B, L, D]` order. If we want contiguous memory per scan, we need to actually transpose/copy the data. That's an extra kernel call. But... what if our custom CUDA kernel first does an efficient transpose into shared memory, then does the scan? Or, even simpler: transpose on the fly using shared memory. For the sequential per-thread approach where each block handles one batch element and threads handle channels: - At each timestep `t`, all threads in a block (256 threads) load `a[b, t, 0..255]` and `b[b, t, 0..255]`. - These loads are coalesced! Consecutive threads access consecutive memory. - Each thread does `h = a_val * h + b_val`. - Then threads move to `t+1` and repeat. - For each timestep, the loads are coalesced. This is 4096 iterations per block, with each iteration doing one coalesced load of 256 floats (1024 bytes) for `a` and one for `b`. With 32 blocks total (one per batch), this is 32 blocks. At 256 threads per block, we have 8192 threads. But 32 blocks is only 32/132 = 24% of SMs even at 1 block per SM. Actually, can we have multiple blocks per batch element? If we split channels across blocks: - Block 0 handles batch b, channels 0..127 (128 threads) - Block 1 handles batch b, channels 128..255 (128 threads) - Total blocks: 32 * 2 = 64 Or with 64 threads per block: - 4 blocks per batch, 128 blocks total With 32 threads per block: - 8 blocks per batch, 256 blocks total With 16 threads per block: - 16 blocks per batch, 512 blocks total But smaller block sizes mean less ability to hide latency. With 32 threads (1 warp) per block, if the warp stalls, the SM has no other warps in that block to switch to. But other blocks on the same SM can run. For 256 blocks of 32 threads = 8192 threads total. About 62 threads per SM. With 2048 threads capacity per SM, that's 3% occupancy. Not great. What about 128 blocks of 64 threads? Still 8192 threads. 15 blocks per SM... wait, 128 blocks / 132 SMs ≈ 1 block per SM for the first wave, then a few more. Still low thread count per SM. Hmm. 8192 threads is just not enough to saturate H100. Unless... we process one batch element per warp, and have many warps per block. No, a warp is 32 threads, and D=256 channels means 8 warps per batch element. A block can have 8 warps (256 threads) and handle 1 batch element. To handle more, we'd need larger blocks. What about a block handling 4 batch elements? We'd need 1024 threads. Possible but might have issues. Actually, let me check: can a thread block on H100 have 1024 threads? Yes, max is 1024. Can it have 2048? No, 1024 is the max. So a block of 1024 threads can handle 4 batch elements (4 * 256 = 1024 threads). Total blocks: 32 / 4 = 8 blocks. That's even worse! What about having each thread handle multiple channels? A thread could process 2 channels, so a block of 256 threads handles 2 batch elements. Total blocks: 16. Worse. This seems to contradict my earlier calculation. Let me recalculate. We have 8192 independent scans (B*D = 32*256 = 8192). We need to assign these to threads. Total threads needed if each thread handles 1 scan: 8192 threads. H100 has 132 SMs * 2048 threads/SM = 270,336 thread capacity. We need 8192 threads. That's only 3% of the capacity. If we can't create more threads, then we can't increase occupancy by just having more threads. Instead, we need each thread to do more work and somehow hide latency. But wait... we CAN create more blocks/threads if we parallelize within a scan! Parallel scan on 4096 elements with 256 threads per block gives 16x more threads per scan. So for the parallel scan approach: - 8192 blocks - 256 threads per block - Total threads: 2,097,152 - Thread capacity utilization: 2M / 270k ≈ 7.75x over-subscription. Good! But the memory access was strided. Can we do something about that? Idea: Use vectorized loads from strided memory! Each thread loads 16 elements with stride 1024 bytes between consecutive warps. This will use cache efficiently if we process in tiles. Actually, H100 has L1 cache per SM (128KB or 256KB?). With 256 threads each loading 16 floats = 4096 floats = 16KB per block. Wait, we need to load 4096 floats for `a` and 4096 for `b` = 32KB per block. This fits in L1 cache! And with 8 blocks per SM (2048 threads / 256 threads = 8 blocks), that's 8 * 32KB = 256KB of L1 cache used. H100 L1 is 256KB per SM (I think; actually it's 128KB shared+L1 configurable, but maybe 256KB on some configs). This might be tight but workable. Wait, actually each block doesn't need to keep all data in L1 simultaneously. The scan can stream through it. If we load 4096 elements of a single scan, and they're spaced 1024 bytes apart, they'll map to different cache lines. In L1 cache (128KB = 1024 cache lines of 128 bytes each), 4096 elements would need 4096 cache lines if they don't alias. But there are only 1024 cache lines! So there will be heavy conflict misses. 4096 elements, each in its own cache line (since stride 1024 > cache line size 128). In a 128KB cache with 1024 cache lines, and 4096 accesses, we're way over. Conflict misses will be terrible. BUT - the data for `a` and `b` also need to be loaded, so 8192 cache lines accessed. With 1024 cache lines, each cache line can hold 32 floats (128 bytes / 4 bytes = 32 floats), but our accesses are 1024 bytes apart, so each float gets its own cache line. This is the worst possible access pattern. So the one-block-per-scan with strided access is fundamentally bad for cache. The only way to get good memory performance is: 1. Transpose the data first (expensive extra pass), or 2. Use the sequential per-channel approach where consecutive threads access consecutive memory (coalesced), and try to hide latency through instruction-level parallelism. For option 2, even with low occupancy, let's think about whether it's fast enough. - We have 32 blocks, each with 256 threads (or more blocks with fewer threads). - At each timestep, a block loads 256 floats for `a` and 256 for `b`. That's 2048 bytes. This fits in one or two cache lines if coalesced. - Wait, coalesced means consecutive threads access consecutive memory. So threads 0..255 access `a[b, t, 0..255]`, which is 256*4 = 1024 contiguous bytes. This is exactly 8 cache lines (128 bytes each), or one 1024-byte sector. Actually on H100, memory is accessed in 128-byte cache lines, so 1024 bytes is 8 consecutive cache lines. A warp of 32 threads accesses 128 bytes (1 cache line) since 32*4 = 128 bytes. So warp 0 accesses cache line 0, warp 1 accesses cache line 1, etc. This is perfectly coalesced! A block of 8 warps accesses 8 consecutive cache lines. Beautiful. So the memory bandwidth efficiency is near 100% for loads and stores. But we only have 32 blocks. With 132 SMs, most SMs are idle. However, each block does 4096 iterations. The memory bandwidth per SM is total_bw / num_SMs = 3.35 TB/s / 132 ≈ 25.4 GB/s per SM. At 100% efficiency, each block consumes: - Per iteration: 2 loads + 1 store = 3 * 256 * 4 = 3072 bytes - 4096 iterations: 12,582,912 bytes ≈ 12 MB per block - Time at 25.4 GB/s per SM: 12MB / 25.4GB/s = 0.47ms per block on one SM. - With 32 blocks running on 32 different SMs: 0.47ms total. - But with SMs running multiple blocks sequentially: 32 blocks / 132 SMs ≈ 0.24 blocks per SM. Actually, SMs can run multiple blocks concurrently. With 256 threads per block, each SM can run 8 blocks (2048/256 = 8). So 32 blocks occupy 4 SMs fully, leaving 128 SMs idle. - The 4 occupied SMs process 8 blocks each. Total time: 8 * 0.47ms = 3.76ms on those 4 SMs. Hmm, wait. Memory bandwidth is shared across the whole GPU, not per-SM in a hard-partitioned way. The memory controller can serve requests from all SMs. With only 4 SMs active, can they saturate the full memory bandwidth? Actually no - there's a limit to how many memory requests each SM can issue. With 8 warps per SM (256 threads), each warp can have outstanding memory requests. The memory subsystem can handle requests from multiple SMs concurrently. But with only 4 SMs active vs 132 total, we're using only about 3% of the memory request generation capacity. However, each SM on H100 is very powerful. Can 4 SMs saturate the memory bus? Let's think. A100 has 108 SMs and memory bandwidth of 2 TB/s. One SM can do ~20-30 GB/s? Actually, at peak occupancy, all SMs together saturate the bus. With H100 having 132 SMs at 3.35 TB/s, each SM needs to generate about 25 GB/s of traffic to saturate its share. Does 1 SM with 8 warps generate enough memory traffic to saturate 25 GB/s? Probably yes if the warps are memory-bound. But with only 4 SMs, that's potentially 100 GB/s total, which is only 3% of peak. So we'd be severely bandwidth-starved at the SM level. But I might be wrong about this. Let me think about actual kernel execution. Each warp does: ```cuda for (int t = 0; t < 4096; t++) { float av = a[b, t, tid]; float bv = b[b, t, tid]; h = av * h + bv; out[b, t, tid] = h; } ``` With 8 warps per SM, and each warp doing this sequentially, each warp issues one load every ~4 cycles (L2 hit latency). With warp switching, SM can handle multiple warps. 8 warps might generate enough concurrent memory requests. But even if 1 SM can generate 100 GB/s with good occupancy, 4 SMs generate 400 GB/s, and 32 SMs generate 3.2 TB/s. We need at least ~32 SMs to approach memory bandwidth saturation. Since we only use 4 SMs, we're likely limited to ~400-800 GB/s. This means our kernel would take 384MB / 600GB/s ≈ 640 microseconds. Is that faster than PyTorch? The PyTorch reference with 4096 iterations of Python overhead and kernel launches might take 20-40ms. So 640us would be a 60x speedup. Let me verify by implementing the simple approach first and testing it! Actually, let me also consider a hybrid: process multiple batch elements per block with persistent threads or something. But I think the simple approach is the best starting point. Wait, let me re-examine the PyTorch reference. It does: ```python for t in range(L): h = a[:, t] * h + b[:, t] outputs.append(h) ``` Each iteration: 1. `a[:, t]` extracts a [B, D] slice. This creates a view. 2. `a[:, t] * h` does elementwise multiplication on [B, D] tensors. This is a single kernel call that processes 8192 elements. 3. `+ b[:, t]` is another elementwise kernel call. So per iteration: 2 kernel calls * 4096 = 8192 kernel calls. Plus the final `torch.stack`. Kernel launch overhead on H100 is about 3-5 microseconds in ideal cases. So 8192 * 4us = 32.7ms just in launch overhead. The actual compute is tiny (8K FMAs per kernel, so ~16K FLOPs per iteration, negligible). Even with our simple 32-block kernel that takes 640us, we'd be 50x faster. But can we do even better? If we could use more SMs, we might approach the ~110us memory bandwidth limit, another 6x improvement. Actually, I realize I miscounted. The grid can have more than 32 blocks. What if I make each warp handle one channel, and I have multiple warps in a block, but also launch many blocks? For example, what if I use 1D grid where: - Block has 32 threads (1 warp) - Grid has 8192 blocks (one per scan) Then each block/warp handles one scan sequentially. - Memory access is strided (bad) - But we have 8192 blocks! Good occupancy. What if I use a clever memory access pattern? Like, each warp loads a float4 or something? No, the stride is still bad. What if I first pool memory from multiple scans so that threads in a warp load coalesced data, then redistribute? That sounds like a transpose. Actually, there's a well-known trick for this kind of problem. We can have a block read a tile of the input into shared memory in a coalesced way, then process it. For example, consider reading a tile of shape `[B_tile, T_tile, D_tile]` where threads load along D. But our scan is along T for fixed (B, D). Another idea: Load a chunk of 32 timesteps for 32 channels into shared memory in a coalesced way, but this doesn't help the scan because the scan is along T for each channel separately. We'd need to synchronize across blocks or something. Let me think about this differently. What if we use a register-based warp shuffle scan? For each channel `d`, we could assign it to a warp and do the scan using warp shuffles. But we need 256 channels, so 8 warps per batch element. What if we use a 2D grid? - Blocks of 256 threads (8 warps) - Grid: [B, D/256] = [32, 1] = 32 blocks... no. What about [B, D/32] = [32, 8] = 256 blocks. Each block of 32 threads (1 warp) handles one channel? No, each block of 32 threads handles 32 channels for one batch element. Then threads in a warp access different channels at the same timestep - coalesced! Wait! Let me think about this carefully: - Grid: [32, 8] = 256 blocks - Block: 32 threads (1 warp) - Block (b, c) handles batch element b, channels d = c*32 + tid (where tid is 0..31) - Each thread scans 4096 timesteps sequentially. - At timestep t, thread tid loads `a[b, t, c*32 + tid]`. Threads in the warp access consecutive channels → coalesced! Blocks: 256 blocks. 256/132 ≈ 2 blocks per SM. At 32 threads per block, that's 64 threads per SM. Very low occupancy. But better than 32 blocks. Can we make block dimensions [32, 16] = 512 blocks? Each block would need 16 warps (512 threads), and handle 32 channels. Total threads per block = 512. Each SM can handle 4 blocks of 512 threads. 512 blocks / 132 SMs ≈ 4 blocks per SM. Up to 2048 threads per SM. Good! Wait, blockDim.x would be 32 (for channels), and we'd have 16 "sub-blocks" per batch? No, in a 2D grid, the total block count is gridDim.x * gridDim.y. If gridDim.x = 32 (batch) and gridDim.y = 16 (channel groups), and blockDim.x = 32 (threads per channel group), then: - Each block handles batch element b and channel group g. Thread tid handles channel d = g*32 + tid. - There are 32 * 16 = 512 blocks. - Each block has 32 threads. - Total threads = 512 * 32 = 16,384. - Threads per SM = 16,384 / 132 ≈ 124 threads. - Blocks per SM = 124 / 32 ≈ 4 blocks. - This is still very low occupancy. What if blockDim.x = 32 and we have more channel groups? D=256, so with groups of 32 we have 8 groups. Can't have more. What if each block handles MULTIPLE batch elements? Block of 256 threads. - Threads 0-31 handle b=0, channels 0-31 - Threads 32-63 handle b=0, channels 32-63 - ... - Threads 224-255 handle b=0, channels 224-255 - That's one batch element. We can't fit two batch elements in 256 threads if each batch needs 256 threads. Unless... each block has 512 threads! - First 256 threads: batch 0, all channels - Next 256 threads: batch 1, all channels - Grid: 16 blocks (batches 0-1, 2-3, ..., 30-31) - Total blocks: 16. Worse! Or each block handles multiple batches with fewer threads per batch: - Block of 256 threads, each handling 2 channels for 4 batches? No, 256 threads / (4 batches * 2 channels) = 32 threads... getting messy. Actually, what about persistent kernels? Launch a 1D grid of many blocks, and each block loops over multiple scans. - Launch 4096 blocks with 256 threads each. - Each block handles `ceil(8192 / 4096) = 2` scans. - Threads within a block: for scan 0, thread tid handles channel tid of batch b0. For scan 1, thread tid handles channel tid of batch b1. - But then within a warp, threads are accessing different batches at the same time. Let me check memory layout. If block 0 handles (b=0, all d) and (b=1, all d), then: - Thread 0 handles d=0 for b=0 - Thread 1 handles d=1 for b=0 - ... - Thread 255 handles d=255 for b=0? No, we said each block handles 2 scans. If block 0 handles scan (b=0, d=0..127) and (b=1, d=0..127): - Thread 0: (b=0, d=0) - Thread 1: (b=0, d=1) - ... - Thread 127: (b=0, d=127) - Thread 128: (b=1, d=0) - Thread 129: (b=1, d=1) - ... At timestep t: - Threads 0-127 load `a[0, t, 0..127]` — coalesced! - Threads 128-255 load `a[1, t, 0..127]` — also coalesced! This works! With 4096 blocks and each handling 2 scans of 128 channels each, total scans = 8192. But each block has 256 threads. Total blocks = 4096. This uses 4096 / 132 ≈ 31 blocks per SM. At 256 threads each, that's 7936 threads per SM, but max is 2048, so only 8 blocks can run per SM concurrently. Total active blocks = 1056, with 3038 blocks waiting. Good! But we need to figure out the indexing. Each block handles 2 batches × 128 channels = 256 "channel-instances". That's exactly 256 threads. So: - Block `bid` handles: batch = bid * 2 / (D/blockChannels), channel_start = (bid * 2) % (D/blockChannels) * blockChannels... wait. Actually, let's simplify: - Each block handles `B_PER_BLOCK` batch elements and `D_PER_BLOCK` channels per batch. - Total "work items" per block = B_PER_BLOCK * D_PER_BLOCK. - This must equal blockDim.x (e.g., 256). So B_per_block * D_per_block = 256. Options: - B_pb=1, D_pb=256 (one batch, all channels) — 32 blocks - B_pb=2, D_pb=128 — 16 blocks! Worse. - B_pb=4, D_pb=64 — 8 blocks - B_pb=8, D_pb=32 — 4 blocks This goes in the wrong direction! More batch elements per block means fewer blocks. Wait, I think I confused myself. Total scans = B * D = 32 * 256 = 8192. If each block handles `B_pb * D_pb` scans, then number of blocks = 8192 / (B_pb * D_pb). If B_pb=1, D_pb=256, blocks = 32. If B_pb=2, D_pb=128, blocks = 8192 / 256 = 32. If B_pb=4, D_pb=64, blocks = 8192 / 256 = 32. Oh! The number of blocks is always the same if B_pb * D_pb = 256! Because total scans / scans_per_block = 8192 / 256 = 32. So with fixed block size of 256 threads and each thread handling exactly 1 scan, we always get 32 blocks. This is because 8192 / 256 = 32. Therefore, to get more blocks, we MUST have more than 8192 total threads, which means either: 1. Multiple threads per scan (parallel scan), or 2. More threads than total scans, meaning some threads are idle or help with other scans. Actually, we could have each thread handle a FRACTION of a scan. Like 2 threads per scan, each doing half the work. But then we need parallel scan communication between the threads. Or, each block has 128 threads and handles 128 scans. Total blocks = 8192 / 128 = 64. With 64 threads per block: 8192 / 64 = 128 blocks. With 32 threads per block: 8192 / 32 = 256 blocks. With 16 threads per block: 8192 / 16 = 512 blocks. With 8 threads per block: 8192 / 8 = 1024 blocks. With 4 threads per block: 8192 / 4 = 2048 blocks. With 2 threads per block: 8192 / 2 = 4096 blocks. With 1 thread per block: 8192 blocks. With 1 thread per block and 8192 blocks, we have great block count but terrible thread count per block. A warp has 32 threads, and a block of 1 thread is terrible for the warp scheduler. With 32 threads per block (1 warp), we get 256 blocks. That's 256/132 ≈ 2 blocks per SM. But each SM can run up to 32 blocks and up to 64 warps. With 32 threads = 1 warp per block, 256 blocks = 256 warps. 256/132 ≈ 2 warps per SM. Very low occupancy. So the trade-off is: - Smaller block size → more blocks, but fewer warps per SM - Larger block size → fewer blocks, but more warps per block The product is constant: 8192 total warps if each thread handles 1 scan. Wait, 8192 scans * 1 thread/scan = 8192 threads = 256 warps. That's not enough to saturate the GPU no matter how we divide them into blocks! This is the fundamental insight. With 1 thread per scan, we can only create 256 warps total. H100 has 132 SMs * 64 warps/SM = 8448 warp capacity. We can only fill 3% of the warp capacity. So we MUST use more than 1 thread per scan. We need parallel scan algorithms within each scan. For a scan of length 4096 with 256 threads, we get 16x more warps. Total warps = 8192 scans * 16 warps/scan = 131,072 warps. That's 15x over-subscription! This is great for occupancy. But we had the strided memory access problem. How to fix strided memory access in the parallel scan approach? Idea: Use a preprocessing step to transpose the data. Not a full transpose, but just copy to a contiguous buffer scan-by-scan. But that's an extra kernel + memory pass. What if the parallel scan kernel does the transpose on-the-fly into shared memory first? Consider one block per scan, 256 threads. Load 4096 elements. For scan `(b, d)`, element at `t` is at `addr = base + ((b * L + t) * D + d) * 4`. Stride between consecutive `t` is `D * 4 = 1024`. To load this into shared memory with coalesced access... we can't, because only one block is loading this scan's data. Unless we use vectorized memory operations that load multiple strided elements at once? Like a texture fetch or L2 cache bypass? No, L2 cache is what we use. Actually, NVPTX has `.global` loads with any stride, and the hardware will cache lines. But with 1024-byte stride, each float gets its own cache line. The L2 cache bandwidth is 3.35 TB/s, but only a fraction of that is useful. Wait, I just realized: if we're launching 8192 blocks, each loading 4096 elements with 256 threads, the GPU scheduler distributes these blocks across SMs. Even with terrible cache behavior per block, with 8 blocks per SM active concurrently, the memory system is servicing many requests. But each float loaded wastes 31/32 of the cache line bandwidth (since each cache line is 128 bytes but we only use 4 bytes). So effective bandwidth is 3.35 TB/s * (4/128) = 104 GB/s. That's terrible. Total data to move: 2 inputs + 1 output = 384MB. At 104 GB/s, that's 3.7ms. Still faster than PyTorch's 32ms, but far from optimal. However, we can actually improve this! If we use `float4` loads or gather instructions... no, the data layout defeats us. Let me think about whether we can process the data in a different order. What if we transpose the problem: instead of having each block process one scan, what if we have blocks that process data in tiles and multiple scans share the loaded data? For example, tile the (B, D) space. A block handles 32 batch elements × 8 channels = 256 threads. Each thread handles one (b, d). But instead of one scan each, 256 threads cooperate... no, each scan is independent. Wait, ALL scans are independent! They don't communicate. So we can't parallelize within a scan and load data efficiently unless we load extra data and share it. What if a block handles a tile of 8 timesteps × 32 channels × 1 batch element? The block has 256 threads. We can assign: - Threads vary over channels (32 values) and timesteps (8 values) = 256 combinations. - So each block processes 8 timesteps for 32 channels of 1 batch. - Total blocks: (4096/8) * (256/32) * 32 = 512 * 8 * 32 = 131,072 blocks. Now within a block, threads access `a[b, t0+dt, d0+dd]` where `dt` is 0..7 and `dd` is 0..31. - Thread with `(dt, dd)` accesses address `base + ((b * L + t0 + dt) * D + d0 + dd) * 4`. - Consecutive `dd` gives consecutive memory (coalesced!). - But consecutive `dt` gives strided memory. If we organize threads so that consecutive thread IDs vary `dd` first, then `dt`: - Thread 0: dt=0, dd=0 - Thread 1: dt=0, dd=1 - ... - Thread 31: dt=0, dd=31 - Thread 32: dt=1, dd=0 - etc. Then within a warp (threads 0-31), all threads have dt=0 and dd=0..31. They access `a[b, t0, d0..d0+31]`. This is 128 contiguous bytes = 1 cache line. Coalesced! Now, the question is: can we do the scan recurrence with this tile structure? - Each tile handles 8 consecutive timesteps for 32 channels. - The scan for each channel requires the result from the previous tile. - So we need a multi-pass approach: - Pass 1: For each tile, compute local scan and aggregate. - Pass 2: For each channel, scan the tile aggregates. - Pass 3: For each tile, apply the prefix and finalize. This is exactly the work-efficient scan but now tiles are 2D (time × channels) and scans are along time. Let me work this out: - Tile size: T_tile = 8 timesteps, D_tile = 32 channels. - Grid: [B=32, num_tiles_t=512, num_tiles_d=8] = 32 * 512 * 8 = 131,072 blocks. - Each block has 256 threads = T_tile * D_tile = 8 * 32. Pass 1 (local scan): - Block (b, tt, dt) loads `a[b, tt*8:(tt+1)*8, dt*32:(dt+1)*32]` into shared memory. - Wait, shared memory layout: we can store as [8][32] or [32][8]. - For coalesced loads, we load into shared memory directly. - Then for each of the 32 channels in the tile, the 8 threads with that channel do a sequential scan over the 8 timesteps in the tile. - We also compute the aggregate for each channel over the tile: `a_prod = prod(a_t)` and `b_effect = local_scan_result_for_last_timestep` (which is the result of scanning the 8 timesteps starting from h=0). - Actually, the aggregate that propagates to the next tile is just the final `h` value if the initial state was 0, along with the product of `a` values. Hmm, we need both because the next tile's prefix depends on both. Wait, for a tile from t_start to t_end, the relation between input `h_in` (state before t_start) and output `h_out` (state after t_end) is: `h_out = A_tile * h_in + B_tile` where `A_tile = prod(a_t for t in tile)` and `B_tile = scan_of_tile_with_h_in=0`. Actually, `B_tile` is exactly what the local scan computes as the final value when initialized with 0. So the aggregate for a tile is `(A_tile, B_tile)`. Pass 2 (scan aggregates): - For each channel, we have 512 tiles along time. - We need to scan 512 aggregates per channel. - There are 32 * 256 = 8192 channels total. - We can launch 8192 blocks, each handling one channel and scanning 512 aggregates. - Or we can launch blocks that scan multiple channels in parallel. With 8192 blocks scanning 512 items each, and block size 512 threads... or maybe 256 threads. With 256 threads doing parallel scan on 512 elements, each thread handles 2 elements. This gives 8192 blocks. Actually, 8192 blocks for pass 2. Each block scans 512 items. Total work: 8192 * 512 = 4M items to scan, but each scan is independent. That's similar to our original total element count. Hmm, the multi-pass with tiles of [8, 32] is getting complex. Let me think if there's a simpler way. Actually, what if we use T_tile = 4096/32 = 128 and D_tile = 256/... wait, a block of 256 threads can handle 128 timesteps × 2 channels = 256 work items. But then loads would be: - Thread (dt, dd): access `a[b, t0+dt, d0+dd]` - With dt=0..127, dd=0..1. - Thread 0: t0, d0 - Thread 1: t0, d0+1 - Thread 2: t0+1, d0 - etc. For a warp of 32 threads, if we arrange as dd first: - Threads 0-1: dt=0, dd=0,1 - Threads 2-3: dt=1, dd=0,1 - etc. Warp accesses: for each thread pair, two consecutive dd values. Not ideal coalescing because adjacent threads access different timesteps. Let me think more carefully. If we want a warp to access consecutive memory, we need all threads in the warp to access the same timestep but consecutive channels. So in the tile, the fastest-varying dimension should be channels (d), not time (t). So a tile should be T_tile timesteps × D_tile channels, where D_tile ≥ 32 (warp size). Then: - Thread (dt, dd) where dd varies fastest. - Threads in a warp have different dd but same dt → same timestep, consecutive channels → coalesced! So we need D_tile ≥ 32. Since D = 256, we can choose D_tile = 32, 64, 128, or 256. If D_tile = 256 (all channels), then each block handles T_tile timesteps × 256 channels. With block size 256, T_tile = 1. Not useful. If D_tile = 128, T_tile = 2. Warp accesses 128 consecutive floats = 512 bytes = 4 cache lines. Coalesced. Each block handles 2 timesteps for 128 channels. Not enough work per block. If D_tile = 64, T_tile = 4. Block handles 4 timesteps for 64 channels. If D_tile = 32, T_tile = 8. Block handles 8 timesteps for 32 channels. Let me try D_tile = 32, T_tile = 8: - Each block processes 8 timesteps × 32 channels for one batch. - Number of blocks: B * (L/8) * (D/32) = 32 * 512 * 8 = 131,072 blocks. - Each block has 256 threads. - Total warps = 131,072 * 8 = 1,048,576 warps. This is excellent occupancy! - Blocks per SM = 131,072 / 132 ≈ 992. But each SM can only hold 32 blocks concurrently. So lots of waves but full utilization. Now the question is: can we perform the scan efficiently with this tiling? Pass 1: Local scan within each tile - Load 8 timesteps × 32 channels into shared memory. Coalesced loads! - For each channel (32 channels), 8 threads cooperate to scan the 8 timesteps. - Wait, but threads in a warp are spread across channels, not across timesteps within a channel. - For a single channel `d`, which threads handle it? Threads with `threadIdx.x % 32 == d`?? No, if we arrange as `(dt, dd)` with `dd` fastest, then: - Thread index = dt * 32 + dd - For a fixed channel dd, threads are: dd, dd+32, dd+64, ..., dd+224. That's 8 threads with consecutive dt values. - These 8 threads are NOT in the same warp! Warp 0 has threads 0-31, which all have dt=0. So warp 0 handles all 32 channels at dt=0. Warp 1 handles all 32 channels at dt=1. etc. This is a problem for doing a scan along dt, because the threads handling a single channel's 8 timesteps are in different warps (they're at positions dd, dd+32, dd+64, ..., dd+224 in the thread block). Can we do a 32-channel × 8-timestep scan where warps are organized by dt? Actually, each warp handles all channels for a single dt. We have 8 warps handling dt=0,1,2,...,7. For a scan along dt for a fixed channel dd, we need communication between warps. This is easy with shared memory. Algorithm for Pass 1: 1. Load `a[b, t0+dt, d0+dd]` and `b[b, t0+dt, d0+dd]` for all dt in 0..7, dd in 0..31. Use coalesced loads. Store in shared memory as `smem_a[8][32]` and `smem_b[8][32]`. 2. For each channel dd (0..31), scan along dt (0..7) using shared memory: - We can have warp 0 (threads 0..31) process all channels simultaneously. - Each thread in warp 0 handles one channel dd, and sequentially processes dt=0..7. - For channel dd, thread dd reads `smem_a[0][dd]`, `smem_b[0][dd]`, computes `h = a*h + b`, then `smem_h[0][dd] = h`. - Continues for dt=1..7. - After processing all 8 timesteps, the aggregate for the tile is stored: - `agg_a[dd] = prod(smem_a[dt][dd])` for dt=0..7 (can be accumulated during the scan) - `agg_b[dd] = h_final` (with h_initial=0) Wait, thread dd in warp 0 can do this sequentially for dt=0..7. Other warps are idle during this step. But that's fine because we only need 1 warp * 32 threads = 32 threads, and our block has 8 warps. We could use multiple warps to process multiple channels, but warp 0 handles ALL 32 channels sequentially. Actually, 32 channels × 8 timesteps = 256 operations. One warp can do this sequentially in about 256 * (2 loads + 1 mul + 1 add + 1 store) cycles ≈ 1280 cycles. With 8 warps in the block, if they're all doing this, we finish 8x faster. But as I said, each warp handles a different dt. Hmm. Alternative: After loading data to shared memory, have warp `w` handle channels `w*4` to `w*4+3` for all 8 timesteps sequentially. So warp 0 handles channels 0-3, warp 1 handles channels 4-7, etc. Wait, this doesn't help because a warp's threads need to access shared memory without bank conflicts. Shared memory layout `smem_a[8][32]`: - Thread in warp w accesses `smem_a[dt][dd]`. - If warp w loads all 32 channels for dt=w (as in the original layout), then threads 0..31 access `smem_a[w][0..31]`. These are 32 consecutive words in the same row → bank conflicts! Because shared memory is banked by 32-bit words, and consecutive words are in different banks. So no bank conflict for a row access! Actually, accessing 32 consecutive 32-bit words maps to 32 different banks. No conflicts. But for the scan, thread dd needs to access `smem_a[0][dd]`, `smem_a[1][dd]`, etc. These are in the same column, separated by 32 words (the row pitch). `smem_a[dt][dd]` and `smem_a[dt+1][dd]` are 32 words apart. Since there are 32 banks, these map to the SAME bank! So all 32 threads in a warp accessing column dd for different dt values would cause 32-way bank conflicts. So `smem_a[8][32]` has column-major access pattern causing bank conflicts. Solution: Pad the shared memory. `smem_a[8][33]` or `smem_a[8][36]`. Then `smem_a[dt][dd]` and `smem_a[dt+1][dd]` are 33 or 36 words apart, avoiding bank conflicts. With padded shared memory and `smem_a[8][33]`: - Thread dd accesses `smem_a[0][dd]`, `smem_a[1][dd]`, etc. - These are at offsets `0*33 + dd = dd`, `1*33 + dd`, `2*33 + dd`, etc. - Offset difference = 33, which is not a multiple of 32. So bank(dd) != bank(dd + 33). No conflicts. Now, for the scan along dt in shared memory: - We can have all 32 threads in the block work on different channels. - But each thread needs to sequentially scan 8 timesteps. - Threads 0..31 handle channels 0..31. Each thread does: ```cuda float h = 0; for (int dt = 0; dt < 8; dt++) { float av = smem_a[dt][threadIdx.x]; float bv = smem_b[dt][threadIdx.x]; h = av * h + bv; smem_h[dt][threadIdx.x] = h; } ``` - After this, `smem_h` contains the local scan (assuming h_in=0). - And `h` contains the tile aggregate B_tile. - We also need A_tile = product of `a` values, which we can compute during the loop. But wait, all 256 threads try to run this simultaneously. Thread 0 handles channel d=0, thread 1 handles d=1, etc. But the shared memory can handle 32 concurrent accesses to different columns without conflict (with padding). So 32 threads can do this in parallel. But a warp has 32 threads, and all 32 threads access different columns but they're spread across... actually, threads 0..31 are one warp. Warp 0 handles channels 0..31. The other warps (1..7) also try to handle channels... but in our setup, we said threads are `(dt, dd)` with `dt` varying by warp. That's not the right assignment for this phase. Let me redefine. After loading shared memory, we can repurpose the threads: - Thread index `tid` (0..255) handles channel `tid % 32`... no, we have 32 channels in the tile, so we can have 8 threads per channel. - Each group of 8 threads handles one channel. Thread 0,1,2,...,7 handle channel 0 for dt=0,1,2,...,7. But they don't need to cooperate for a sequential scan! - For a sequential scan, only 1 thread per channel is needed. The other 7 threads per channel are wasted. So with D_tile = 32 and 256 threads, we have 8 threads per channel, but only 1 is needed for sequential scan. 87.5% of threads are wasted in the scan phase. This wastes the GPU. Is there a way to use all threads? Option: Each thread could handle multiple channels. Thread `w` in the 8-warps-per-channel setup could handle 4 channels, one per warp. Actually, each of the 8 warps could process 4 channels (32 channels / 8 warps = 4 channels per warp). Each warp has 32 threads, but only 1 is needed per channel for sequential scan. So within a warp, 4 threads are active and 28 are idle. This is still very wasteful. The fundamental issue is: a sequential scan along 8 elements only needs 1 thread. A parallel scan on 8 elements might use 8 threads, but with such small size, the overhead of parallel coordination (barriers, reductions) is comparable to just doing it sequentially. What if we use a larger tile? Like D_tile = 32, T_tile = 128. Then each block handles 128 timesteps × 32 channels. Block has 256 threads. So 2 threads per (dt, dd) pair, or threads handle a subset of 64 timesteps each. Wait, with 128 timesteps and 32 channels = 4096 work items. We have 256 threads. Each thread handles 4096/256 = 16 work items. If we arrange by channel, each thread handles 16 timesteps for one channel. But we have 32 channels and 256 threads, so 8 threads per channel, each handling 16 timesteps. That means a channel's 128 timesteps are split among 8 threads. This brings us back to the multi-threaded parallel scan, but at the tile level. OK, I think the tiled approach with small tiles leads to thread underutilization in the sequential phase. Let me reconsider. Maybe I should just implement the simple 32-block kernel and see how it performs. If PyTorch takes ~30ms and our kernel takes ~600us, that's already a 50x win. The challenge says "run faster than the PyTorch reference," not "achieve peak bandwidth." But I suspect the reference is actually much faster than I calculated because PyTorch might JIT compile the loop or fuse operations. Let me look at the reference more carefully. Actually, in PyTorch eager mode: ```python for t in range(L): h = a[:, t] * h + b[:, t] outputs.append(h) ``` This involves: 1. `a[:, t]` indexing: cheap, returns a view 2. `a[:, t] * h`: element-wise mul, launches a kernel 3. `+ b[:, t]`: element-wise add, might be fused with mul if using NNC or if PyTorch uses a fused mul-add kernel. In eager mode without JIT, it's typically separate kernels or one fused kernel. 4. `outputs.append(h)`: stores a reference, no copy If PyTorch uses ATen for broadcasting element-wise ops, `a[:, t] * h + b[:, t]` probably compiles to a single element-wise kernel that reads `a[:, t]`, `h`, `b[:, t]` and writes to a new `h`. Or if it's in-place, it reads and writes to `h`. In any case, each iteration involves at least 1 kernel call (if fused) or 2-3 kernel calls. Let's assume 1 fused kernel per iteration. That's 4096 kernel launches. Each launch: CPU overhead ~3-5us, GPU kernel execution for 8192 elements ~1-2us. Total per iteration: ~5us. 4096 * 5us = 20.5ms. So the reference might take ~20ms. Our simple kernel at ~600us would be 33x faster. That's a huge win. But what if PyTorch has better compilation/optimization? The evaluation framework runs the model in some mode. Let me not worry too much. I want to try a hybrid approach that gets good coalescing and decent occupancy: - Use 256 threads per block - Grid: 128 blocks (4 batch elements per block? No...) Actually, I just realized: we can have each block handle a subset of sequences from DIFFERENT batch elements. For example: - Block has 256 threads. - Each thread handles 1 scan from any batch/channel. - The 256 scans in a block are chosen so that at each timestep, the threads access memory close together. If we put scans with the same batch element but consecutive channels in the same block, consecutive threads access consecutive memory at each timestep. Coalesced! And we can launch more than 32 blocks by splitting each batch element's channels across multiple blocks. With D=256 and block size 256, one block handles all 256 channels of one batch element. That's 32 blocks. If we use block size 128, each block handles 128 channels of one batch element. That's 64 blocks. If we use block size 64, that's 128 blocks. But block size also affects occupancy. With block size 64 and 128 blocks: - Total threads = 8192 - Threads per SM = 8192 / 132 ≈ 62 - Blocks per SM = 62 / 64 < 1... wait, each SM can run multiple blocks. - 128 blocks / 132 SMs ≈ 1 block per SM. Each SM has 2048 thread slots, so 64 threads is only 3% occupancy. Other warps on the SM can help hide latency, but we only have 2 warps per block (64 threads = 2 warps). Hmm. With 1 block per SM and only 2 warps, hiding latency is hard. But consecutive warps from other blocks on the same SM can help. I think a block size of 128 might be a sweet spot: - 128 blocks - 128 threads per block - 16,384 total threads - 128 blocks / 132 SMs ≈ 1 block per SM initially - But each SM can hold 2048/128 = 16 blocks concurrently! - So initially 128 blocks fill 128 SMs with 1 block each. But as blocks finish, more blocks from the queue are assigned. Total waves: ceil(128 / 132) = 1 wave, then maybe some stragglers. - Actually, 128 blocks for 132 SMs means almost all SMs get 1 block. With 16 blocks per SM capacity, after the first 128 blocks, there are no more blocks to fill the remaining capacity. So occupancy is still limited by total thread count, not block size. Total threads = 8192. Divide by 132 SMs = 62 threads per SM. H100 needs maybe 512-1024 threads per SM to saturate memory bandwidth. So we're still under-utilized. Is there any way to increase the total thread count without using multi-threaded scan? Wait, I just realized something. Do we really have only 8192 independent work items (one per scan)? For the sequential approach, yes. But what if we process the sequence in chunks and assign chunks to different threads? Example: Sequence length 4096. We split into 16 chunks of 256 timesteps. - Each scan of length 4096 becomes 16 sub-scans. - But we need to chain them together. - Pass 1: Do 8192 * 16 = 131,072 independent scans of length 256. - Each scan of length 256 can be done by one thread sequentially. - Total threads: 131,072. Threads per SM: 131,072 / 132 ≈ 992. That's 50% occupancy. Good! - Memory access: if thread handles one sub-scan of length 256, and the sub-scan is for a fixed (b, d) and chunk c, then memory is still strided by D within a chunk. But consecutive threads in a warp accessing DIFFERENT sub-scans might not be coalesced. Can we assign threads in a warp to handle sub-scans from the same batch but consecutive channels? Yes! - Warp 0: batch 0, channels 0..31, chunk 0 (timesteps 0..255) - Warp 1: batch 0, channels 0..31, chunk 1 (timesteps 256..511) - ... At timestep t in chunk c, thread tid accesses `a[b, c*256 + t, d]`. Consecutive tid (channels 0..31) access consecutive memory. Coalesced! And with chunks, each thread does a sequential scan of 256 timesteps. Number of blocks: if each block has 256 threads, and each thread handles one sub-scan, then we get 131,072 blocks. Great! But each sub-scan of 256 timesteps produces an output of 256 values and an aggregate `(A_chunk, B_chunk)` for the next chunk. We need to chain the chunks. This is the three-pass scan again: 1. Local scan on each chunk, output local results and aggregates 2. Scan aggregates across chunks (for each channel) 3. Apply prefix aggregates to local results Pass 1: 131,072 blocks of 256 threads - Each block handles 256 timesteps for 1 scan (or multiple scans) - Let's say each block handles 1 scan with 256 threads. Wait, 256 threads for 256 timesteps means each thread does 1 timestep? That's parallel, not sequential. - Or each block handles 256 scans, each thread handling 1 scan of 256 timesteps sequentially. Total blocks = 131,072 / 256 = 512 blocks. That's 512 blocks of 256 threads = 131,072 threads. Memory access for pass 1: - Block 0: batch 0, channels 0..255, chunk 0. Thread tid handles channel tid, timesteps 0..255. - At timestep t, thread tid accesses `a[0, t, tid]`. Coalesced! - Number of blocks: B * (D/256) * num_chunks = 32 * 1 * 16 = 512 blocks. Each block has 256 threads. - Total threads = 131,072. - Blocks per SM = 512 / 132 ≈ 4 blocks per SM. - Threads per SM = 131,072 / 132 ≈ 992 threads/SM ≈ 50% occupancy. Pass 1 output: - local_results: [B, L, D] same layout as input - aggregates: [B, num_chunks, D, 2] where each entry is (A_chunk, B_chunk) Pass 2: Scan aggregates across chunks - For each scan (b, d), we have 16 aggregates. - We need to scan these 16 values to get prefix aggregates. - Launch 8192 blocks, each with 16 threads (scanning 16 items). Or 8192 blocks with 32 threads. - Actually, 16 is very small. We could do a sequential scan with 1 thread per scan. - 8192 threads for pass 2. That's too few. Or we can combine this differently. Instead of 16 chunks, do 32 chunks of 128. Then pass 2 needs 8192 threads scanning 32 items. Or, we can do a block-level scan where one block handles 32 channels and all 16 chunks. Block size 256 = 32 channels * 16 chunk positions (but chunk positions don't work that way). Actually, for pass 2, the data is much smaller. 8192 scans × 16 chunks × 2 values = 262,144 values. At 4 bytes each, ~1MB. We can process this efficiently with many threads. For example, each block handles 32 scans (batch 0, channels 0..31, all 16 chunks). 256 threads per block. Each thread handles one scan's 16 aggregates sequentially. Number of blocks: 8192 / 32 = 256 blocks. Same as before. Hmm, it seems like we always end up with limited blocks when we process channels sequentially within a block. Wait! For pass 2, we have 8192 scans, each needing a scan of 16 values. We can launch 8192 blocks of 1 thread each. Each thread sequentially scans 16 values. Number of blocks: 8192. Good! Pass 3: Apply prefix to local results - 512 blocks of 256 threads, same as pass 1. - Each thread handles one scan over 256 timesteps. - For each timestep, apply the prefix aggregate. Total kernel count: 3. Total memory traffic: - Pass 1: read 2*128MB, write 128MB + small aggregates - Pass 2: read 1MB, write 1MB - Pass 3: read 128MB + 1MB prefixes, write 128MB - Total: ~640MB of memory traffic. - At 3.35 TB/s = 0.19ms. Realistically with coalescing and overhead, maybe 0.5ms. But 3 kernel launches add a bit of overhead. Still very fast. However, is this possible to implement correctly? Yes, but it's more complex. What if we simplify even further: instead of chunks, use ONE block per scan but process 4096 elements sequentially with 1 thread, and bank on the fact that 8192 blocks with 1 thread each is enough to get some parallelism? Wait, a CUDA block cannot have fewer than 32 threads and run efficiently. Actually, it can have 1 thread, but it's terrible. And with 8192 scans, if we use 1 thread per scan in a block of 32 threads (where only 1 thread does work and 31 are idle), we'd launch 8192/32 = 256 blocks. Each block has 1 active thread and 31 idle threads. This is terrible. What if each block handles 32 scans with 32 threads, and each thread sequentially processes 4096 timesteps? - Blocks: 8192 / 32 = 256 blocks - Each block: 32 threads - Total threads: 8192 - At each timestep, 32 threads access 32 consecutive channels of the same batch and timestep. Coalesced! - Memory access is perfect. - But only 256 blocks = ~2 blocks per SM. Wait, I keep coming back to this. The total number of warps is 8192/32 = 256 warps. No matter how we arrange them, we have 256 warps total if each thread handles exactly one scan. This is because there are 8192 scans and each scan needs exactly 1 thread for sequential processing. To get more warps, we MUST use multiple threads per scan. So the real question becomes: what's the most efficient multiple-threads-per-scan approach that doesn't have terrible memory access? Let me reconsider the parallel scan per block with the tile-based transpose idea. The key insight from cutlass and modern scan kernels is to do a tiling where you load data in a coalesced way, then transpose in shared memory. For our problem: - We have [B, L, D] data. - We want to scan along L for each (B, D) independently. - The memory layout makes scanning along L strided by D. If we create a kernel where a block of 256 threads works on a tile of [B_tile, L_tile, D_tile], and loads data in a coalesced way... Actually, the most common GPU scan implementation (like the one in CUB or thrust) for segmented scan or array-of-scans typically assumes the scan dimension is the innermost (contiguous) dimension. If it's not, they might transpose first. For our case, maybe the simplest high-performance solution is: 1. Write a kernel that transposes `a` from [B, L, D] to [B, D, L] (contiguous along L) 2. Write a kernel that does parallel scan on [B, D, L] with contiguous L 3. Transpose output back from [B, D, L] to [B, L, D] Wait, but PyTorch's permute/reshape doesn't copy. We need an actual transpose copy kernel. Transpose [B, L, D] → [B, D, L]: - This is a standard matrix transpose for each batch element. - Each batch element is L×D = 4096×256 = 1M elements = 4MB. - 32 batch elements = 128MB per tensor. - Two tensors = 256MB. - Transpose bandwidth is typically 70-90% of peak. - Time: ~100-150us. Then scan on contiguous data [B, D, L]: - For each (b, d), scan over L=4096 contiguous elements. - We can use a very efficient parallel scan kernel. - With B*D = 8192 scans, each of 4096 elements. - CUB-style block scan with 256 threads per block can scan 4096 elements per block (16 elements per thread). - Blocks: 8192. Very high occupancy. - Time: For well-optimized scan on contiguous data, maybe 50-100us. Then transpose output back: - Another 100-150us. Total: ~300-400us. Very fast! But is this faster than the simple sequential approach? Let me estimate the simple sequential approach more carefully. Simple sequential approach (32 blocks of 256 threads, coalesced memory): - 32 blocks, each does 4096 iterations. - At each iteration, each block reads 2048 bytes and writes 1024 bytes = 3072 bytes. - Total memory per block: 3072 * 4096 = 12,582,912 bytes ≈ 12 MB. - Total for all blocks: 32 * 12 MB = 384 MB. - With coalesced access, we can approach memory bandwidth. - BUT: with only 32 blocks, we only use ~4 SMs fully. How much bandwidth can 4 SMs generate? With 8 warps each (256 threads = 8 warps), that's 32 warps total. Each warp can have outstanding memory requests. But at 100% utilization of 4 SMs, maybe we get 4/132 of 3.35 TB/s = 100 GB/s. Actually, I don't think it's linearly partitioned like that. Memory bandwidth is more about how many requests reach the memory controllers. Actually, on modern GPUs, even a single SM can generate a lot of memory traffic. With many concurrent warps, memory latency is hidden. With 8 warps per SM (256 threads), we might hide some latency but not saturate the bus. Let me think about it differently. At each timestep, a block issues: - Load from `a`: 256 threads × 4 bytes = 1024 bytes, coalesced into 8 cache lines (128 bytes each). Or maybe 1 sector of 1024 bytes on H100. - Load from `b`: same. - Store to `out`: 1024 bytes. - FMA: 256 FMAs. The memory bandwidth for 1024 bytes at 3.35 TB/s is ~300ns. The FMAs take negligible time. So each timestep takes ~300ns if memory-bound. 4096 timesteps * 300ns = 1.23ms per block. 32 blocks running on SMs. If 4 SMs run 8 blocks each sequentially: 8 * 1.23ms = 9.84ms per SM, but they run in parallel. So total ~9.84ms. Wait, that can't be right. At 3.35 TB/s total, 384MB should take ~115us. But with only 4 SMs active, we can't use full bandwidth. If 1 SM can use, say, 200 GB/s (which is likely achievable with sufficient warps), 4 SMs use 800 GB/s. 384MB / 800GB/s = 480us. That's reasonable. But with 32 warps total (8 per SM x 4 SMs), can we generate 200 GB/s per SM? 200 GB/s / 8 warps = 25 GB/s per warp. Each warp moves 128 bytes per load (1 cache line). To get 25 GB/s, a warp needs to issue a load every 128 / 25GB = 5.1ns. At 2GHz clock, that's every 10 cycles. Is that achievable? With 32 threads doing back-to-back loads, maybe. But there are only 4096 iterations, and each iteration does 3 memory ops. With loop overhead, maybe 20-30 cycles per iteration. That gives 128 bytes / 25 cycles = 5 bytes/cycle. At 2GHz, that's 10 GB/s per warp. 8 warps = 80 GB/s per SM. 4 SMs = 320 GB/s. 384MB / 320GB/s = 1.2ms. Hmm, so maybe the simple approach takes ~1ms. Still faster than 20ms PyTorch. But the transpose + scan approach might do 384MB in ~200us total. That's 5x faster. However, the transpose + scan approach requires 3 kernel calls and careful implementation. Let me think if there's a simpler approach that achieves similar performance. Actually, I just thought of another approach. We can use the tile-sequential approach but with more tiles by having each block handle one batch element and a small number of channels, but process the channels in parallel using warp shuffles. Wait, no. We need more blocks. What if we make the sequence length smaller by having each thread process a chunk? Actually, the chunked scan with 3 passes can be implemented with: - Chunk size = 256 timesteps - Number of chunks = 16 - Pass 1: 512 blocks of 256 threads (coalesced, sequential within chunk) - Pass 2: 8192 blocks of 1 thread (sequential scan of 16 aggregates) — but 1 thread per block is terrible Alternative for Pass 2: Use 256 blocks of 32 threads. Each block handles 32 channels from one batch, scanning 16 chunk aggregates. 32 threads can do a parallel scan on 16 items with Kogge-Stone algorithm in 4 steps. This is efficient! Or even simpler, each thread does sequential scan of 16 items, but that's 32 sequential scans in parallel. Wait, for each channel, scanning 16 aggregates sequentially is trivial. A block of 32 threads can process 32 channels simultaneously, each thread scanning 16 items. At 256 batch elements (32 batches × 8 channel groups of 32), we need 32 * 8 = 256 blocks. Each block has 32 threads. Total threads for Pass 2 = 8192. But we can do better. Each block can handle 256 channels from one batch (256 threads), scanning 16 aggregates per channel. That's 32 blocks total. Same issue as before. For Pass 2, if each block handles 32 channels and we need 32 batches * 8 groups = 256 blocks, the total is 256 blocks of 32 threads = 8192 threads. Same low thread count. The problem is that once we chunk into 16 pieces, Pass 2 still only has 8192 independent scans (of length 16), which again limits us to 8192 threads if each scan is done sequentially. For a scan of 16 elements, using 1 thread is actually fine and we can't do much better. But 8192 threads total is still low. What if we use a parallel algorithm for the 16-element scan? 16 threads can scan 16 elements in log(16) = 4 steps. That's 16x more parallelism. But the problem size is too small to justify the overhead. Wait, I think I see the way. In Pass 2, one block can handle MULTIPLE channels' aggregate scans simultaneously using a single warp per channel group, or by having all threads in a block work on one channel scan at a time and loop over channels. For example, a block of 256 threads could scan the aggregates for one batch element's 256 channels: - Use shared memory to hold 256 channels × 16 aggregates = 4096 pairs. - Perform a parallel scan over 16 values for each channel. But 256 threads and 16 values per channel... we could process 16 channels at a time with 16 threads each. - 256 threads / 16 threads per channel = 16 channels processed in parallel. We have 256 channels, so 16 iterations. - Total blocks for Pass 2: 32 (one per batch). Again, only 32 blocks. What if a block processes 4 batch elements? 4 * 256 = 1024 channels. If the block has 256 threads, each thread processes 4 channels. Still total threads = 8192. It seems impossible to use more than 8192 threads for the sequential scan of aggregates unless we parallelize within each aggregate scan. For a 16-element scan, using 16 threads per scan and processing many simultaneously: - 256 threads can process 16 scans simultaneously (16 threads per scan). - 8192 scans / 16 = 512 blocks. - Each block has 256 threads. - Total threads = 131,072. Great occupancy! - Each block loads 16 aggregates for 16 channels into shared memory, does parallel scan, writes back. But wait, for a scan of 16 items, if we use 16 threads per item... Kogge-Stone scan on 16 items with 16 threads: - Step 1: thread i reads item i - Step 2: thread i adds item i-1 (if i >= 1) - Step 3: thread i adds item i-2 (if i >= 2) - Step 4: thread i adds item i-4 (if i >= 4) - Step 5: thread i adds item i-8 (if i >= 8) For our operator, each step involves a binary op. The scan completes in 5 steps. Very fast! But we need 16 threads per channel scan. With 16 channels being processed simultaneously in a block, we need 16 * 16 = 256 threads. Perfect! So for Pass 2: - Each block handles 16 channels from one batch. - Block has 256 threads arranged as [16 channels, 16 threads_per_scan]. - Thread (c, t) handles aggregate position t of channel c. - Use warp shuffles or shared memory to do Kogge-Stone scan. - Wait, Kogge-Stone with 16 threads means we need communication between threads. If we use 16 threads per channel, and we have 16 channels, that's 256 threads. A block has 8 warps. We can organize so that each channel's 16 threads are in one warp (half a warp... no, 16 threads is half a warp). Actually, a warp is 32 threads, so we can fit 2 channels per warp (16 threads each). But warp shuffle operates on 32 threads. We could use a different arrangement. Actually, for 16 threads, we could either: 1. Use one warp for 2 channels (16 threads each), and use shared memory for scan because warp shuffle across lanes 0-15 and 16-31 separately is tricky (possible though with masks). 2. Use one warp for 1 channel but only use 16 threads, leaving 16 idle. Wasteful but simple. Or use 32 threads per channel scan. Then 256 threads / 32 = 8 channels per block. For 8192 channel scans, we need 8192 / 8 = 1024 blocks. Each block has 256 threads. Total threads = 262,144. Excellent occupancy! And scanning 16 items with 32 threads is even easier: just have threads 0..15 active, or use all 32 threads with some doing nothing. Kogge-Stone on 16 items with 32 threads still takes 5 steps (log2(32)=5). Actually, with 32 threads, we can have each thread handle one item (items 0..15), and threads 16..31 are idle or handle duplicate items. Or we can be smarter and process 8 channels per block with 32 threads per channel. Wait, but Kogge-Stone scan of 16 items doesn't need 32 threads. 16 threads is enough (log2(16)=4 steps). Using 32 threads means half are idle but we process fewer channels per block. Channels per block with 32 threads per channel: 256/32 = 8 channels. Blocks needed: 8192 / 8 = 1024 blocks. With 16 threads per channel and 2 channels per warp: 256/16 = 16 channels per block. Blocks needed: 8192 / 16 = 512 blocks. With 8 threads per channel and using Brent-Kung scan: 256/8 = 32 channels per block. Blocks needed: 8192 / 32 = 256 blocks. I think 16 threads per channel with 16 channels per block (256 blocks total) is a good trade-off. But we can still do better by having each channel use fewer threads. Actually, for 16 items, just using 1 thread and doing it sequentially might be faster than using 16 threads with 4 barriers and warp shuffles! The overhead of parallel scan can dominate for small arrays. With 1 thread per channel and 256 threads per block, we process 256 channels per block. - Blocks: 32 (one per batch). - Threads: 8192. - Each thread sequentially scans 16 aggregates. - Time: very fast, probably dominated by memory bandwidth. - Memory: 8192 scans * 16 items * 2 values * 4 bytes ≈ 1MB of reads, 1MB of writes. - At full bandwidth, ~0.6us. Realistically, maybe 10-20us. So for Pass 2 with sequential scan on 1 thread per channel, the time is negligible (~20us). Pass 1: 512 blocks of 256 threads, each thread scans 256 timesteps sequentially. - Coalesced memory access. - Total memory: read a (~128MB), read b (~128MB), write local (~128MB), write aggregates (~1MB). - Time: With good memory utilization, maybe 400-500us. Pass 3: 512 blocks of 256 threads, applying prefix aggregates. - Read local results (~128MB), read prefixes (~1MB), write final (~128MB). - Time: ~300-400us. Total: ~800us + overhead. That's competitive with the transpose approach and might be simpler to implement. But wait, can we do Pass 1 and Pass 3 in fewer passes or combine them? Actually, note that in Pass 1, each block writes the local scan results AND the aggregate. In Pass 3, we read the local scan results and the prefix aggregate, then write the final result. Can we avoid writing local results to global memory and keep them in shared memory while also doing the prefix application? Not easily, because the prefix from Pass 2 isn't available until all blocks finish Pass 1. However, if we use a persistent kernel or cooperative groups, we might be able to do more. But that's overkill. Let me reconsider whether a single-pass parallel scan might be better after all. For a parallel scan of 4096 elements per scan with 8192 scans: - We can use 8192 blocks, 256 threads each. - Each block does a parallel scan of 4096 elements. - Total threads: 2,097,152 (good occupancy). - Memory access is strided (bad). But what if we use a very clever memory access pattern? What if we use `float4` and the compiler does some magic? No. What about texture memory or __ldg (load global through read-only cache)? The read-only data cache (L2) still loads cache lines. The bandwidth waste (31/32) is still there. But L2 bandwidth is high. Actually, let me compute: with L2 bandwidth of 3.35 TB/s, and loading 128MB of useful data but wasting 31/32 of each cache line, we need 128MB * 32 = 4GB of L2 bandwidth. At 3.35 TB/s, that's 1.2ms. Plus stores of 128MB * 32 = 4GB of write bandwidth, total 8GB at 3.35 TB/s = 2.4ms. Plus the fact that we only use 4GB out of the 8GB moved (the cache line waste). Still much faster than PyTorch. But with 2M threads active, can we sustain L2 bandwidth? The L2 is shared by all SMs. With all 132 SMs active, L2 bandwidth is 3.35 TB/s. With 2M threads, there are plenty of requests. Wait, for the parallel scan approach, the memory footprint per block is: - Load 4096 floats for a: 16KB - Load 4096 floats for b: 16KB - Store 4096 floats for output: 16KB - But due to strided access, each float load fetches a 128-byte cache line. So actual L2 traffic per block: 4096 * 128 * 3 = 1.5MB. - For 8192 blocks: 12.3GB of L2 traffic. - At 3.35 TB/s: 3.7ms. Hmm, that's not great. The sequential coalesced approach only moves 384MB total, which is much less. So despite lower occupancy, the sequential coalesced approach is more bandwidth-efficient and might actually be faster! For sequential coalesced: - Total data moved: 384MB (all useful, no waste) - Time at 500 GB/s (conservative with 4-8 SMs active): 768us. - Time at 1 TB/s (optimistic): 384us. For parallel scan with strided access: - Total data moved: ~12GB (384MB useful, 11.6GB wasted) - Time at 3.35 TB/s: 3.6ms. So sequential coalesced is likely 4-10x faster than parallel strided, despite using far fewer SMs. This is because coalesced access is SO MUCH more efficient. Now, can we improve the sequential coalesced approach to use more SMs and get closer to memory bandwidth saturation? We need more warps while keeping coalesced access. This means we need more blocks where each block's threads access consecutive memory. Since we have 32 batch elements and 256 channels, and coalescing works when threads in a warp access consecutive channels at the same timestep, we need warps that work on SAME batch, SAME timestep, DIFFERENT consecutive channels. In the sequential approach, each warp iterates over timesteps. At timestep t, warp w accesses channels (w*32) to (w*32+31) of batch b. These are consecutive. To get more warps active, we can have each block process MULTIPLE batch elements, and within each batch, have multiple warps processing different channel ranges. But with 256 threads per block: - Warp 0: batch 0, channels 0-31 - Warp 1: batch 0, channels 32-63 - ... - Warp 7: batch 0, channels 224-255 - Warp 8: batch 1, channels 0-31 - ... - Warp 15: batch 1, channels 224-255 This block has 512 threads (16 warps)! But CUDA max threads per block is 1024, and H100 can handle 1024. So a block of 512 threads is fine. With 512 threads, the block handles 2 batch elements. - Total blocks: 32/2 = 16 blocks. - Warps: 16 * 16 = 256 warps. - Still only 256 warps total. To get more warps, we need to process the SAME scan (batch, channel) with multiple warps, i.e., parallelize the scan of 4096 timesteps. But then we're back to needing parallel scan and its strided access issues. OR, we can process the sequences in smaller chunks where each chunk is assigned to a different warp, like the 3-pass chunked scan. In that approach: - Pass 1: 512 blocks of 256 threads, each thread handles 256 timesteps of one scan. Coalesced! Total warps: 512 * 8 = 4096 warps. - Pass 2: 32 blocks of 256 threads, each thread handles 16 aggregate values of one scan. Sequential but tiny data. Could use more blocks for this. - Pass 3: 512 blocks of 256 threads. Total warps: 4096. Total warps for the main passes: 4096 + 4096 = 8192 warps. That's 10x more than the simple 32-block approach! And the memory access in Pass 1 and Pass 3 is coalesced. The overhead of the extra passes is small because Pass 2 is tiny. Let me calculate the time for this chunked approach more carefully. Assume chunk size = 256 (16 chunks of 4096). Pass 1: - 512 blocks, 256 threads each. - Each thread processes 256 timesteps sequentially. - Per thread: 256 loads from a, 256 loads from b, 256 FMAs, 256 stores to local output, compute aggregate. - Per block: 256 threads * 256 timesteps = 65,536 timesteps = 65K elements. - Total blocks: B * (D/256) * num_chunks = 32 * 1 * 16 = 512 blocks. - Memory traffic: read a and b for all elements = 256MB, write local output = 128MB, write aggregates = ~1MB. - With coalesced access, effective bandwidth maybe 1-2 TB/s (using many SMs better now). - Time: ~200-400us. Pass 2: - 32 blocks of 256 threads. Each thread scans 16 aggregates. - Or to get more parallelism, use 256 blocks of 32 threads. - Actually, let's use 256 blocks of 32 threads. Each block handles 32 channels, one batch. - Memory: read 1MB, write 1MB. - Time: ~10us. Pass 3: - 512 blocks, 256 threads each. - Each thread reads 256 local outputs + 1 aggregate prefix, writes 256 final outputs. - Memory: read 128MB + 1MB, write 128MB. - Time: ~150-300us. Total: ~400-800us. This is better than the simple approach's ~600-1000us estimate. And it's relatively simple to implement. Can we do even better? What if chunk size = 128 (32 chunks)? - Pass 1: 32 * 1 * 32 = 1024 blocks. Better occupancy. - Each thread processes 128 timesteps. - Pass 2: 256 blocks of 32 threads, scanning 32 aggregates. More work in Pass 2 but still small. - Pass 3: 1024 blocks. Even more parallelism. What if chunk size = 64 (64 chunks)? - Pass 1: 2048 blocks. - Pass 2: 256 blocks, scanning 64 aggregates. - Each thread in Pass 2 scans 64 values. That's more sequential work but manageable. - Pass 3: 2048 blocks. What if chunk size = 32 (128 chunks)? - Pass 1: 4096 blocks. - Pass 2: 256 blocks, scanning 128 aggregates per thread. That's a lot of sequential work per thread for Pass 2. A chunk size of 64 or 128 seems like a sweet spot. Actually, let me think about whether I can merge Pass 1 and Pass 2 more efficiently. For chunk size = 64: - Pass 1: 2048 blocks. - Pass 2: For each scan, scan 64 aggregates. - Using 1 thread per scan: 8192 threads total. - We can launch 256 blocks of 32 threads, where each thread handles 1 scan's 64 aggregates sequentially. - Each thread does 64 loads and 64 FMAs. Very fast. - Total time for Pass 2: negligible (~10-20us). Actually, since Pass 2 is so fast, the overhead of launching the extra kernel is comparable to the kernel execution time. We might want to combine kernels or increase chunk size to make Pass 2 more parallel. Chunk size = 256: - Pass 2 is 16 values per scan, still fast. - But 512 blocks in Pass 1 and 3 might already give decent occupancy. Let me calculate: with block size 256, 512 blocks in Pass 1 means 131,072 threads. 131K / 132 SMs = 992 threads per SM. That's 50% occupancy. Not bad. With 1024 blocks, 1984 threads per SM = 97% occupancy! So let's use chunk size = 128 (32 chunks): - Pass 1: 1024 blocks of 256 threads = 262,144 threads. - 262K / 132 = 1986 threads per SM. Near 100% occupancy! - Each thread handles 128 timesteps sequentially. - Pass 2: 256 blocks of 32 threads = 8192 threads, scanning 32 aggregates each. Or 128 blocks of 64 threads. Let's use 256 blocks of 32 threads. - Pass 3: 1024 blocks of 256 threads = 262,144 threads. This looks very promising. But I want to see if we can do even better with warp-level reductions or shared memory within each chunk. For Pass 1, each thread handles 128 sequential timesteps. It computes: ```cuda float h = 0; float a_prod = 1; for (int i = 0; i < 128; i++) { int t = chunk_start + i; float av = a[b, t, d]; float bv = b[b, t, d]; h = av * h + bv; out_local[b, t, d] = h; // write local result a_prod *= av; } agg_a[b, chunk, d] = a_prod; agg_b[b, chunk, d] = h; ``` Wait, `a_prod` should be the product of all `a` values in the chunk. For the forward recurrence, this is correct. For Pass 2, for each (b, d): ```cuda float prefix_a = 1; float prefix_b = 0; for (int c = 0; c < 32; c++) { float av = agg_a[b, c, d]; float bv = agg_b[b, c, d]; // The aggregate of chunk c transforms h_in to h_out = av * h_in + bv // When chaining: prefix_{c} applied to h_{-1} gives prefix up to chunk c // new_prefix_a = av * prefix_a // new_prefix_b = av * prefix_b + bv prefix_a = av * prefix_a; prefix_b = av * prefix_b + bv; prefix_a_arr[b, c, d] = prefix_a; prefix_b_arr[b, c, d] = prefix_b; } ``` Wait, what exactly is the prefix we need for each chunk? If chunk c covers timesteps [c*128, (c+1)*128), and we want to apply the effect of all previous chunks to the local scan within chunk c. Let's define: - `local_h[c][i]` = result of scanning chunk c starting from h=0, at position i within chunk. - `agg_a[c]` = product of a over chunk c. - `agg_b[c]` = final result of scanning chunk c starting from h=0. If the input state to chunk c is `h_in`, then the output state is: - `h_out = agg_a[c] * h_in + agg_b[c]` For the scan, we need the input state to each chunk. Define: - `chunk_prefix[c]` = state before chunk c (i.e., after all previous chunks). - `chunk_prefix[0] = 0` (since h_0 = b_0, and the first element is in chunk 0). Actually, wait. The local scan of chunk 0 starting from 0 gives `h_0, h_1, ..., h_127` where these are the actual values if h_before_chunk0 = 0. But in the actual recurrence, h_before_chunk0 is well-defined based on prior elements. For chunk 0, there are no prior elements, so `h_before_chunk0` should be... hmm. Actually, the recurrence is: - `h_0 = b_0` - `h_t = a_t * h_{t-1} + b_t` For chunk 0 covering t=0..127, `h_before_chunk0` should be 0, because `h_0 = a_0 * 0 + b_0 = b_0`. So using h_in=0 for chunk 0 is correct! For chunk 1 covering t=128..255, `h_before_chunk1 = h_127` (the actual value). - `h_127_actual = agg_a[0] * 0 + agg_b[0] = agg_b[0]` - So `h_before_chunk1 = agg_b[0]` For chunk 2, `h_before_chunk2 = agg_a[1] * h_before_chunk1 + agg_b[1]`. In general, the prefix for chunk c is: - `prefix[0] = 0` - `prefix[c+1] = agg_a[c] * prefix[c] + agg_b[c]` And the actual value at position i in chunk c is: - `actual_h[c][i] = local_a[c][i] * prefix[c] + local_b[c][i]` where `local_a[c][i]` is the product of a from chunk start to position i, and `local_b[c][i]` is the local scan result at i starting from 0. Wait, I previously thought we only needed to store `local_h` (the local scan result). But to apply the prefix, we actually need to know the product of a up to each position. Hmm, if we have `local_h[c][i]` = scan result starting from 0, and `prefix[c]` = correct input state for chunk c, then: - `actual_h[c][i] = local_a[c][i] * prefix[c] + local_h[c][i]` But `local_a[c][i]` is the product of a from chunk start to position i. Do we have this? Actually, can we derive `actual_h[c][i]` from `local_h[c][i]` and `prefix[c]`? For a chunk starting at t_start: - `h[t] = a[t] * h[t-1] + b[t]` - Let `h[t] = local_h[t]` when `h[t_start-1] = 0`. - Let `h_actual[t] = a[t] * h_actual[t-1] + b[t]` with the true initial state. The affine transform from `h[t_start-1]` to `h[t]` is: - `h[t] = A[t] * h[t_start-1] + B[t]` where `A[t] = prod(a[t_start..t])` and `B[t] = local_h[t]`. So `h_actual[t] = A[t] * prefix + local_h[t]`. We need both `A[t]` and `local_h[t]`. If we only stored `local_h[t]`, we'd need to recompute `A[t]`. But we can store both, or we can store `local_h` and then for Pass 3, recompute or load `a` again. Options for Pass 3: 1. Recompute `A[t]` on-the-fly from `a`. This means loading `a` again and doing a sequential prefix product within each thread's 128 timesteps. That's additional memory traffic. 2. Store `A[t]` in Pass 1. This doubles the intermediate storage from 128MB to 256MB. 3. Note that `A[t] = a_prod_up_to_t`. We can compute this during Pass 1 and store it. Or we can compute it during Pass 3. Actually, wait. Let me think more carefully. During Pass 1, each thread scans 128 timesteps. It can compute and store the local scan result for each timestep. For Pass 3, to apply the prefix, it needs `A[t]` for each timestep. Option: In Pass 1, store only the final `local_h[127]` as `agg_b`. Don't store per-timestep `local_h` at all. Instead, in Pass 3, re-scan the chunk from scratch, loading `a` and `b` again, but this time apply the prefix. This means: - Pass 1: Scan chunk, compute aggregate, write aggregate to global memory. Do NOT write per-timestep results. - Pass 2: Scan aggregates to get prefixes. - Pass 3: Scan chunk again from scratch, applying prefix, write final output. This avoids storing intermediate per-timestep results! It does re-read `a` and `b` in Pass 3, but that's fine because memory bandwidth is the main resource. Wait, but we need the local scan without prefix to... no, we don't! We only need the final output with the correct prefix. In Pass 3: ```cuda float h = prefix_b; // prefix for this chunk // prefix_b is the correct state BEFORE the chunk starts // Actually, prefix_b should be applied as the initial state // So we scan the chunk starting from h = prefix_b instead of h = 0 for (int i = 0; i < 128; i++) { int t = chunk_start + i; float av = a[b, t, d]; float bv = b[b, t, d]; h = av * h + bv; out[b, t, d] = h; } ``` Wait, is this correct? - If prefix_b is the state before the chunk starts (i.e., after all previous chunks), then scanning this chunk starting with h = prefix_b gives the correct results. Yes! Because the recurrence `h = a * h + b` is the same regardless of initial state. So we don't need to store per-timestep intermediate results at all! We only need the chunk aggregates. This simplifies things: - Pass 1: For each chunk, scan sequentially, compute agg_a and agg_b. Write agg_a and agg_b to global memory. - Pass 2: For each scan, scan the chunk aggregates to compute prefix states for each chunk. - Pass 3: For each chunk, scan sequentially starting from the prefix state, write final output. Memory traffic: - Pass 1: read a (128MB), read b (128MB), write aggregates (1MB) - Pass 2: read aggregates (1MB), write prefixes (1MB) - Pass 3: read a (128MB), read b (128MB), read prefixes (1MB), write output (128MB) - Total: read a+b twice = 512MB, write output + intermediates = 258MB. Total: ~770MB. At 3.35 TB/s and good coalescing: ~230us. But realistically with overhead and not achieving peak, maybe 400-600us. And we get great parallelism with 1024 blocks in Pass 1 and 3, each with 256 threads. That's 524,288 total threads in the main passes. Excellent occupancy. Actually, we can further optimize by fusing the aggregate computation and having each block process multiple chunks or multiple scans in a tiled way. But let's also consider: can we avoid the second read of a and b by having a different structure? What if in Pass 1, we write the per-timestep results to a workspace, and in Pass 3, we read them and apply the prefix? Workspace size: 128MB for local results. Pass 3 would read 128MB workspace + 1MB prefix, write 128MB output. Total memory: Pass 1: 256MB read + 128MB write + 1MB write = 385MB. Pass 2: 2MB. Pass 3: 129MB read + 128MB write = 257MB. Total: 644MB. Compared to the recompute approach's 770MB, this saves about 126MB of reads but costs 128MB of workspace writes and reads. So it's roughly a wash. The recompute approach is simpler! No workspace allocation. However, there is another consideration: we could potentially overlap Pass 1 for some chunks with Pass 2 for others, or use a single kernel launch with persistent blocks. But for simplicity, let's stick with 3 kernel calls. Wait, can we at least combine Pass 2 into a smaller, more efficient kernel? For chunk size 128, num_chunks = 32. Pass 2 scans 32 aggregates per scan. With 8192 scans and each scan taking 32 steps, we can process this very efficiently. One option: launch 8192 blocks of 1 thread each. That's a lot of blocks, but each block has only 1 thread which is terrible. Better: launch 256 blocks of 32 threads. Each block handles 32 scans (one batch element's worth, but we only have 32 batches... wait, 32 batches * 256 channels / 32 scans per block = 256 blocks). Each thread in a block handles one scan (one channel), scanning 32 aggregates sequentially. That gives 256 blocks of 32 threads = 8192 threads. Blocks per SM: 256/132 ≈ 2. Threads per SM: 62. Occupancy: low, but the work is tiny. Another option: launch 1024 blocks of 8 threads. 1024 blocks is better distributed. Each block handles 32 channels of 1 batch? No, 8 threads can't handle 32 channels well. Actually, for 32 items per scan, we might be able to use parallel scan within a warp! - A warp of 32 threads can scan 32 items in parallel. - We have 8192 scans. Each warp scans 32 items for 1 scan? No, each warp has 32 threads. - If each thread handles 1 item, a warp can scan the 32 aggregates of ONE channel using Kogge-Stone in 5 steps. - 32 warps can scan 32 channels simultaneously. - 8192 channels / 32 = 256 blocks of 32 warps each = 1024 threads? No, 32 warps = 1024 threads. That's 256 blocks of 1024 threads. - Total threads: 262,144. Great! - But each block only processes 32 channels... For 8192 channels, 256 blocks. - Blocks per SM: 2. Not great. Wait, we can pack more into each block. A block of 1024 threads = 32 warps. - Warp 0: channel 0, aggregates 0..31 (using 32 threads) - Warp 1: channel 1, aggregates 0..31 - ... - Warp 31: channel 31, aggregates 0..31 - So one block handles 32 channels. - But we need inter-warp communication to write results, and each warp operates independently. - Total blocks: 8192 / 32 = 256 blocks. Same issue: 256 blocks of 1024 threads = 262,144 threads total, but 256 blocks spread across 132 SMs gives only ~2 blocks per SM initially. Actually, the issue is that we only have 256 independent work units (32 batches × 8 groups of 32 channels). Unless we split batches further. What if a block handles 8 channels from 4 different batches? - 32 warps in a block. - Warp 0: batch 0, channel 0 - Warp 1: batch 0, channel 1 - ... - Warp 7: batch 0, channel 7 - Warp 8: batch 1, channel 0 - ... - Warp 31: batch 3, channel 7 - Block handles 4 batches × 8 channels = 32 channel-scans. - Still 256 blocks total. I think 256 blocks is the minimum for any workload that processes batch elements and channels together with 32 as a group size. Because 32 batches × 256 channels / 32 channels per block = 256 blocks. If we process 16 channels per block (half a warp per channel scan), we get 512 blocks. If we process 8 channels per block (quarter warp), we get 1024 blocks. If we process 4 channels per block, we get 2048 blocks. With 2048 blocks, we get 15-16 blocks per SM. That's much better! For Pass 2 with 32 aggregates per scan, and 4 channels per block: - Block handles 4 channels. - For each channel, 8 threads scan 32 aggregates... that's weird. - Or, we can use 1 thread per channel and have 4 threads per block. Block size 4 is terrible. Alternatively, process 32 channels per block but have the block only use 1 warp (32 threads), and launch 8x more blocks by having blocks from different "waves" execute the same workload... no, CUDA doesn't work that way. Actually, we CAN launch more blocks than the grid size I computed. I was dividing work as if each block had to handle a contiguous subset. But we can have each block handle a smaller subset, or even better, use a 3D grid. The total number of independent units to process in Pass 2 is 8192 scans × 32 chunk boundaries = 262,144 items. But these are not independent; they need to be scanned per channel. Since the scan per channel has 32 items and is independent, we have 8192 independent scans of length 32. We need to assign these to blocks. Standard CUB/thrust approach for batch scan: launch many blocks, each handling a portion of the batch. For 8192 scans of length 32, we can launch, say, 4096 blocks where each block handles 2 scans (2 channels) and does sequential scan on 32 items per scan. With block size 2 (1 thread per scan): terrible block size. With block size 64 (32 threads per scan, handling 2 scans): 64 threads. Blocks: 8192 / 2 = 4096 blocks. Each block uses 64 threads = 2 warps. Total warps: 8192. Blocks per SM: 4096 / 132 = 31. Threads per SM: 4096 * 64 / 132 = 1986. Near 100% occupancy! And memory access: each block loads 32 aggregates for 2 channels = 64 pairs = 512 bytes. With 2 warps, loads might be coalesced if arranged properly. Wait, if each block handles 2 channels, thread 0 handles channel 0 aggregate 0, thread 1 handles channel 0 aggregate 1, etc. That's a scan within a channel using multiple threads. Or thread 0 handles all 32 aggregates for channel 0 sequentially, and thread 1 handles all 32 for channel 1. Then at each aggregate position, thread 0 and thread 1 access different channels. If arranged properly, consecutive threads access consecutive channels, which is contiguous in the aggregate tensor. If `agg_a` is stored as `[B, num_chunks, D]`, then `agg_a[b, c, d]` and `agg_a[b, c, d+1]` are adjacent. So if thread 0 accesses `d=0` and thread 1 accesses `d=1` at the same chunk `c`, the accesses are adjacent and coalesced! Wait, but in a sequential scan, thread 0 iterates over `c=0..31` for `d=0`, then thread 1 does the same for `d=1`. They don't access the same `c` at the same time. So memory access within a warp is strided by `D * num_chunks * sizeof(float)`. If D=256 and num_chunks=32, the stride is 32KB! This is terrible. To fix this, we need the memory layout of aggregates to be something like `[B, D, num_chunks]` so that consecutive chunks of the same channel are adjacent. Then thread 0 accessing `c=0,1,2,...,31` for `d=0` accesses contiguous memory. If we store aggregates as `[B, D, num_chunks, 2]` (two values a and b), then thread 0 accesses contiguous memory for its scan. Thread 1 accesses a different contiguous block for `d=1`. Within a warp, threads 0 and 1 access different 128-byte ranges. This is fine - each warp accesses 32 cache lines (one per thread), each in a different address. The total range is 32 * 128 bytes = 4KB. But the lines are not contiguous (they're for different channels). The memory controller can handle this, but it's not perfectly coalesced. Actually, for the aggregate tensor, if we arrange as `[B, D, num_chunks, 2]` with `B` varying slowest and `num_chunks` varying fastest within a channel, then thread 0 accesses `base + 0*32*2 + chunk*2 + ab`, thread 1 accesses `base + 1*32*2 + chunk*2 + ab`. The addresses are separated by 64*4 = 256 bytes. Not contiguous, but within the same general area. A warp loads 32 cache lines, each separated by 256 bytes. This is somewhat scattered but better than 32KB apart. To make it coalesced, we'd need threads in a warp to access the same channel, different chunks. But then 32 threads would all be working on the same scan, doing a parallel scan. That brings us back to parallel scan. For simplicity, let's just store aggregates as `[B, D, num_chunks, 2]` and do sequential scan per thread. The aggregate data is tiny (1MB), so even with bad coalescing, it won't be a bottleneck. OK, let me now outline the kernel design I'll implement: **Aggregate Tensor Layout**: `[B, D, num_chunks, 2]` where the last dimension stores (agg_a, agg_b). **Pass 1: Compute chunk aggregates** - Chunk size: 256 (so 16 chunks for L=4096) - Grid: `dim3(B, D / 256, num_chunks)`? No, CUDA only supports up to 3D grid. - Better: `dim3(B * (D / CHUNK_THREADS), num_chunks)` where CHUNK_THREADS = 256. - Actually, if each block has 256 threads and handles 256 channels from one batch, then: - Grid.x = B = 32 - Grid.y = num_chunks = 16 - Each block: 256 threads handling channels 0..255 - Thread tid handles channel tid. - Total blocks: 32 * 16 = 512 blocks. - In the kernel: ```cuda int b = blockIdx.x; int chunk = blockIdx.y; int d = threadIdx.x; int t_start = chunk * 256; float h = 0; float a_prod = 1; for (int i = 0; i < 256; i++) { int t = t_start + i; float av = a[b * L * D + t * D + d]; float bv = b[b * L * D + t * D + d]; h = av * h + bv; a_prod *= av; } agg[b * D * num_chunks + d * num_chunks + chunk] = a_prod; agg[b * D * num_chunks + d * num_chunks + chunk + offset] = h; ``` Wait, the memory access pattern for `a` and `b` in this kernel: - Each block handles batch `b` and chunk `c`. - Threads 0..255 access channels 0..255 at timestep `t_start + i`. - These are contiguous! Consecutive threads (varying `d`) access consecutive memory addresses. - Per load instruction for a warp: 32 consecutive floats = 128 bytes = 1 cache line. - Coalesced! And we have 512 blocks. Each block does 256 iterations. Total threads: 512 * 256 = 131,072. Blocks per SM: 512 / 132 ≈ 4. Threads per SM: ~992. 50% occupancy. Decent. Actually, wait. 512 blocks × 256 threads = 131,072 threads. If each SM handles 2048 threads, that's 64 SMs worth of threads. With 132 SMs, we have plenty of room. 512 blocks / 132 SMs = ~4 blocks per SM. But each SM can hold up to 32 blocks. So we process 4 blocks per SM, then move on. All blocks start fairly quickly. Total active threads at once: 132 SMs × 4 blocks × 256 threads = 135,168 threads. That's basically all of them at once! Because 512 blocks / 32 blocks_per_SM = 16 SMs to hold all blocks? No wait, 512 blocks spread across 132 SMs means 3.88 blocks per SM concurrently. With 256 threads each, that's 992 threads per SM. So about half capacity. The other half of thread slots are empty. Is 50% occupancy enough? Probably yes. Each thread does 256 iterations with memory-bound behavior. 992 threads per SM generating memory requests should be enough to saturate a good fraction of memory bandwidth. **Pass 2: Scan chunk aggregates** - Input: `agg` of shape `[B, D, num_chunks, 2]` - Output: `prefix` of same shape - For each scan (b, d), we scan `num_chunks` aggregates. - Launch 8192 blocks of 1 thread each? No. - Launch 256 blocks of 32 threads. Each block handles 32 channels of one batch. - Grid: `dim3(B, D/32)` = `dim3(32, 8)` = 256 blocks - Block: 32 threads - Thread tid handles channel blockIdx.y * 32 + tid - In the kernel: ```cuda int b = blockIdx.x; int d = blockIdx.y * 32 + threadIdx.x; float prefix_a = 1; // Actually, the prefix starts as applying no chunks float prefix_b = 0; // h_before_chunk0 = 0 for (int c = 0; c < num_chunks; c++) { float av = agg_a[b, d, c]; float bv = agg_b[b, d, c]; // Compose: new_prefix = (av, bv) ∘ (prefix_a, prefix_b) // new_prefix_a = av * prefix_a // new_prefix_b = av * prefix_b + bv float new_a = av * prefix_a; float new_b = av * prefix_b + bv; prefix_a = new_a; prefix_b = new_b; prefix_a_out[b, d, c] = prefix_a; prefix_b_out[b, d, c] = prefix_b; } ``` Wait, I need to be careful about what `prefix` means. For chunk c, we want the state BEFORE applying chunk c. That is, the result of scanning chunks 0..c-1. - prefix for chunk 0: state before chunk 0 = 0 (no prior elements) - After processing chunk 0: state = agg_a[0] * 0 + agg_b[0] = agg_b[0] - prefix for chunk 1: state before chunk 1 = agg_b[0] - After processing chunk 1: state = agg_a[1] * agg_b[0] + agg_b[1] - prefix for chunk 2: state before chunk 2 = agg_a[1] * agg_b[0] + agg_b[1] In the loop: ```cuda float state_a = 1; // Applying state_a to h gives state_a * h + state_b float state_b = 0; // Initial state is 0 for (int c = 0; c < num_chunks; c++) { // state is the prefix before chunk c prefix_a[b, d, c] = state_a; prefix_b[b, d, c] = state_b; // Update state with chunk c float av = agg_a[b, d, c]; float bv = agg_b[b, d, c]; state_b = av * state_b + bv; state_a = av * state_a; } ``` Yes! This stores the prefix before each chunk, which is what Pass 3 needs. But wait, do we really need both `state_a` and `state_b`? Let's check. For Pass 3, we want to start scanning chunk c with the correct initial state. That initial state is the result of applying all previous chunks to h=0. Let's call it `prefix_h`. Actually, if we only store `prefix_h` for each chunk, is that enough for Pass 3? In Pass 3, we scan chunk c starting from `h = prefix_h`. We don't need `prefix_a` because `prefix_h` already encodes the combined effect. We just need the scalar value `prefix_h`. Wait, that's right! The prefix state is just a scalar float, not a pair. Because when we apply the previous chunks to h=0, we get a scalar value. So we only need to store `prefix_h` (state_b) for each chunk! Not state_a. Let me verify: - For chunk 0: prefix_h[0] = 0. We scan starting from h=0. Correct. - For chunk 1: prefix_h[1] = agg_b[0]. We scan starting from h=agg_b[0]. - First element of chunk 1: actual h = a[t1] * agg_b[0] + b[t1]. Correct. - For chunk 2: prefix_h[2] = agg_a[1] * agg_b[0] + agg_b[1]. - First element of chunk 2: actual h = a[t2] * prefix_h[2] + b[t2]. Correct. So `prefix_h[c] = result of applying chunks 0..c-1 to initial h=0`. The update rule is: ```cuda float prefix_h = 0; for (int c = 0; c < num_chunks; c++) { prefix_out[b, d, c] = prefix_h; prefix_h = agg_a[b, d, c] * prefix_h + agg_b[b, d, c]; } ``` We store `prefix_h` for each chunk. That's one float per chunk per scan. Memory for prefixes: 32 * 256 * 16 * 4 = 512KB. Tiny! This simplifies Pass 2 significantly. **Pass 3: Final scan with prefix** - Grid: same as Pass 1. `dim3(B, num_chunks)` with 256 threads per block. - Thread tid handles channel tid. - Load prefix for this chunk. - Scan chunk sequentially, starting from h = prefix. - Write output. ```cuda int b = blockIdx.x; int chunk = blockIdx.y; int d = threadIdx.x; int t_start = chunk * 256; float prefix_h = prefix[b * D * num_chunks + d * num_chunks + chunk]; float h = prefix_h; for (int i = 0; i < 256; i++) { int t = t_start + i; float av = a[b * L * D + t * D + d]; float bv = b[b * L * D + t * D + d]; h = av * h + bv; out[b * L * D + t * D + d] = h; } ``` Memory access: coalesced! Good. Total kernel launches: 3. Now let's think about whether we can do better. One optimization: instead of 3 separate kernel launches, can we combine Pass 1 for all chunks and then do the prefix propagation within the same kernel using shared memory or inter-block communication? Not really for general case. Another optimization: can we avoid writing the aggregates to global memory and instead use shared memory within a block? Not without processing all chunks of a scan in the same block, which would mean 16 chunks × 256 timesteps = 4096 timesteps per block. That might work! Wait! What if a block handles ONE scan (one b, one d) for ALL 4096 timesteps? But with 256 threads per block, we could have 256 blocks per batch element (one per channel). Total 8192 blocks. Each block handles 4096 timesteps using 256 threads in parallel! Or better yet: a block handles 256 channels from one batch for ALL 4096 timesteps. But with block size 256, and 4096 timesteps, that's 4096 timesteps / 256 threads = 16 timesteps per thread. This is the parallel scan approach, and memory is strided. What if the block first loads a tile of 256 timesteps for 256 channels into shared memory in a coalesced way, then processes it, then loads the next tile? Let's think about a streaming approach: - Block: 256 threads. - Handles: batch b, channels 0..255. - We process the sequence in tiles of, say, 32 timesteps. - For each tile: 1. Load 32 timesteps × 256 channels from `a` and `b` into shared memory. Coalesced! 2. Wait, we need to store 32 × 256 = 8192 floats per input. That's 32KB for `a` and 32KB for `b` = 64KB shared memory. Plus output. H100 has up to 228KB shared memory per block, so this fits. 3. Process the 32 timesteps: perform the scan for each channel using the state carried over from the previous tile. Processing the tile: - After loading, for each channel d (0..255), we need to scan 32 timesteps starting from the carried state. - We have 256 threads and 256 channels. We could assign each thread to one channel and have it sequentially scan the 32 timesteps. - But the data is in shared memory. Thread d accesses `smem_a[tile_t][d]` for t=0..31. - With `smem_a[32][256+padding]`, thread d accesses column d. Consecutive threads access different columns → no bank conflicts (each column is a different bank, and with padding if needed, horizontal accesses also have no conflicts). - Wait, `smem_a[t][d]` and `smem_a[t+1][d]` are separated by 256+padding. If padding=1, offset difference = 257, not multiple of 32. So no bank conflict. - Thread d scans 32 timesteps sequentially: ```cuda for (int i = 0; i < 32; i++) { float av = smem_a[i][d]; float bv = smem_b[i][d]; h = av * h + bv; smem_out[i][d] = h; } ``` - Then write `smem_out` back to global memory. Coalesced! This is brilliant! It combines: - Coalesced global memory loads/stores (because we load/store tiles with consecutive channels accessed by consecutive threads) - Sequential scan within each thread (efficient, no synchronization needed) - Streaming through the sequence in tiles - Only ONE kernel launch! Memory traffic: - Read `a` once: 128MB - Read `b` once: 128MB - Write `out` once: 128MB - Total: 384MB. Perfect! No extra reads or writes. And with 32 blocks (one per batch element) and 256 threads each, total threads = 8192. But each block now processes 4096 timesteps in 128 tiles of 32 timesteps. Or maybe 64 tiles of 64 timesteps. Wait, 32 blocks is still only 32 blocks. Can we increase the number of blocks by having each block handle fewer channels? If each block handles 64 channels, block size = 64 threads. - Total blocks: 32 batches * (256/64) = 32 * 4 = 128 blocks. - Each block streams through 4096 timesteps. - Total threads: 128 * 64 = 8192. Same. If each block handles 32 channels, block size = 32 threads. - Total blocks: 32 * 8 = 256 blocks. - Total threads: 256 * 32 = 8192. Same. The total thread count is always 8192 because that's the number of scans. And each scan needs only 1 thread for sequential processing. BUT: we're now streaming tiles through shared memory. The tile approach makes the memory access coalesced. And since shared memory is used as a buffer, the performance might be much less sensitive to block count because each block does a lot of work (4096 iterations). Actually, with the tile approach, each block does: - num_tiles = 4096 / TILE_T = say 128 tiles (for TILE_T=32) - For each tile: - Load 32*channels floats from a and b - Sequential scan of 32 steps - Store 32*channels floats to output - Total memory: 384MB, all coalesced - With 32 blocks, each accesses 12MB of data. But, the key insight: we can use ANY block size and ANY number of blocks as long as each scan is assigned to exactly one thread. The shared memory tiling works regardless of how many channels per block, as long as all threads in a block work on channels for the same batch and tile. Wait, actually the shared memory size scales with channels per block. If block handles 256 channels, shared memory needs 32 * 256 floats = 32KB for a and 32KB for b = 64KB for TILE_T=32. With TILE_T=64, that's 128KB. Still under 228KB. With TILE_T=128, that's 256KB. Too much for 256 channels. So for 256 channels per block, max TILE_T = 64 (maybe 128 if we only keep a and b in shared memory and compute output directly). Actually, we don't need to keep output in shared memory. We can compute and write directly to global memory. But we DO need both `a` and `b` in shared memory for the sequential scan. Or do we? We could load one tile of `a` into shared memory and `b` into shared memory, then each thread reads from shared memory sequentially. For 256 channels and TILE_T=64: - `smem_a`: 64 * 256 * 4 = 64KB - `smem_b`: 64KB - Total: 128KB. Fits! For TILE_T=128: - Total: 256KB. Might be tight but possible on H100 (max 227KB). Actually 128 * 256 * 4 * 2 = 256KB. Too much. So TILE_T=64 with 256 channels per block gives 32 blocks and 128KB shared memory. TILE_T=32 with 128 channels per block gives 64 blocks and 32KB shared memory. What about TILE_T=16 with 64 channels per block? - 128 blocks, 8KB shared memory. I think 32 blocks with 256 channels and TILE_T=64 is a good starting point. But we can also try 64 blocks with 128 channels and TILE_T=32. Actually, the memory access pattern for loading tiles: - A warp loads `a[b, t_start + tile_t, d_start + 0..31]` for tile_t = 0..TILE_T-1. - We can load in various patterns. Let's think about the optimal tile load. We want consecutive threads to access consecutive memory. - Thread `tid` in the block handles channel `d_start + tid`. - To load the tile, we iterate over `tile_t` from 0 to TILE_T-1. - At each `tile_t`, all threads load `a[b, t_start + tile_t, d_start + tid]`. - These loads are coalesced and can be issued simultaneously for all threads in the block. - So for TILE_T=64, we issue 64 coalesced loads per block, each loading 256 floats. With 32 blocks, we're issuing 32 * 64 = 2048 coalesced loads total, plus the stores. This is very efficient! Now the question is: can 32 blocks saturate H100 memory bandwidth if the loads are coalesced? With 32 blocks and 256 threads each, we have 8192 threads. At each load, each block issues a request for 256 floats = 1KB. With 32 blocks, that's 32KB per load wave. At 2GHz and memory latency of ~400 cycles, to hide latency we need many waves in flight. With 64 loads per block, and assuming the hardware overlaps them, we might only have 1-2 waves per SM. But with 256 threads = 8 warps per block, and 32 blocks occupying up to 32 SMs (1 block per SM), each SM has 8 warps. The SM can switch between warps to hide latency. With 8 warps, can it hide 400-cycle latency? Barely, if each warp has independent instructions. But all warps in the block are doing the same loads and will likely stall together. Actually, because the loads are coalesced, the memory controller handles them efficiently. And modern GPUs (Hopper) have very good memory subsystems. I suspect even 8 warps per SM can generate enough memory traffic. But to be safe, let's increase the number of blocks by having each block handle fewer channels. Option A: 128 channels per block, 64 blocks total. - Block size: 128 threads (4 warps) - Shared memory: TILE_T * 128 * 4 * 2. For TILE_T=32: 32KB. - 64 blocks distributed across 132 SMs. Some SMs get 1 block, some get 0 initially. - Threads per SM: 64 * 128 / 132 ≈ 62. Same thread count. Actually, the total thread count is always 8192. The thread count per SM is 62 whether we have 32 blocks of 256 or 64 blocks of 128. So occupancy is the same. What matters is whether the SM has enough active warps to hide memory latency. With 8 warps per block (256 threads), each SM running 1 block has 8 warps. With 4 warps per block (128 threads), each SM running 1 block has 4 warps. More blocks means more blocks per SM, but each block has fewer warps. The key question: does 8 warps in one block generate more concurrent memory requests than 4 warps in one block + 4 warps in another block? On modern GPUs, memory requests come from warps across the whole SM, and the L1 and L2 cache are shared. So 8 warps on an SM should generate twice as many concurrent requests as 4 warps, assuming the instruction mix allows it. But the issue with 32 blocks is that only ~32 SMs have blocks initially. The other 100 SMs are idle! With 64 blocks, ~64 SMs have blocks. With 256 blocks, all 132 SMs have blocks and there are 124 blocks in the queue. So more blocks is better for SM utilization, even if total warps is the same. The blocks are distributed across more SMs, and each SM processes blocks one after another. With 32 blocks on 132 SMs: 32 SMs process 1 block each. The remaining 100 SMs are idle. The 32 SMs process their block, then they finish. Total time = time for 1 block. If a block takes 500us, total time is 500us regardless of SM count. Wait, that's the key! With 32 blocks, each block runs on its own SM (assuming enough SMs). They ALL run concurrently! So total time is the time to process one block, not 32 sequential blocks. We have 132 SMs, so 32 blocks run fully in parallel on 32 SMs, with 100 SMs idle. So total time = (work per block) / (memory bandwidth of 32 SMs). If 32 SMs provide 32/132 of total memory bandwidth = ~800 GB/s. And each block moves 12MB. Time = 12MB / 800GB/s = 15us. That seems too fast. Let me recalculate. Actually, memory bandwidth isn't partitioned by SM. All SMs share the memory bus. The question is: can 32 blocks generate enough requests to saturate the bus? I think 32 blocks with 256 threads each (8192 threads total) might not saturate the memory bus because only 32 * 8 = 256 warps are generating requests. For comparison, to saturate an H100's memory, we might need thousands of warps issuing requests concurrently. If a warp can have, say, 100 outstanding memory requests, and there are 256 warps, that's 25,600 outstanding requests. That might be enough to saturate the bus. Actually, H100 has a very sophisticated memory system. With 8192 threads doing coalesced loads, we might achieve 60-80% of peak bandwidth. Let me estimate: Total data: 384MB. At 80% of 3.35 TB/s = 2.68 TB/s. Time = 384MB / 2.68TB/s = 143us. At 50% bandwidth: 230us. At 30% bandwidth: 384us. Even at 10% bandwidth (with only 32 blocks), we'd get 1.15ms. Still faster than PyTorch. So I think the streaming tiled sequential approach with 32 blocks is the simplest and likely very effective. But... can we get closer to bandwidth saturation with more blocks? Yes, if we somehow have more blocks. But we don't want to use multiple threads per scan because that complicates things. Wait, actually we CAN have more blocks! The constraint was that each scan needs 1 thread. But we can split each scan into sub-scans and have multiple blocks process them, like the 3-pass chunk approach. OR, we can have blocks work on different batches and channels, and also split the sequence so that one block processes half the sequence and another block processes the other half. But then we need a prefix pass. Let me compare: - Simple tiled: 1 kernel, 32 blocks, coalesced, ~200-400us. - 3-pass chunked: 3 kernels, 512+256+512 blocks, coalesced, ~400us. The simple tiled has fewer kernel launches and less synchronization overhead. The 3-pass chunked has more blocks but more overhead. I think I'll start with the simple tiled approach (1 kernel, 32 blocks, streaming through shared memory). Let me implement it and test. Wait, actually there's a concern. With 32 blocks and 8192 threads, will the GPU actually run all blocks concurrently? With 132 SMs, 32 blocks is only 24% of SMs. But each block uses 256 threads, which might be small enough that the warp scheduler can handle it well. The issue is not the number of SMs but whether the memory system is saturated. Actually, I want to try something: what if each thread handles MULTIPLE channels? Then we can have more blocks. For example: - Each block has 128 threads. - Each thread handles 2 channels (e.g., d = d_start + threadIdx.x and d = d_start + threadIdx.x + 128). - Each block handles 256 channels. - Total blocks: 32. - Total threads: 32 * 128 = 4096. This halves the thread count. Worse. What if each warp (32 threads) handles 32 channels, and we have one block per 32 channels per batch? - Total blocks: 32 batches * 8 channel groups = 256 blocks. - Each block has 32 threads (1 warp). - Total threads: 256 * 32 = 8192. Same. - But now we have 256 blocks! Much better distribution across SMs! - Each block processes all 4096 timesteps with 1 warp. - Memory access is still coalesced! - But with only 1 warp per block, warp switching (to hide latency) is limited. There are no other warps in the same block. However, with 256 blocks on 132 SMs, most SMs get 2 blocks. The SM can switch between the 2 warps from the 2 blocks. But that's only 2 warps per SM. Is that enough to hide memory latency? Probably not well. But the GPU might still run at decent speed. Actually, we can have even more blocks by processing fewer channels per block. With 16 channels per block: - Blocks: 32 * 16 = 512 blocks. - Threads per block: 16... but CUDA warps are 32 threads. A block of 16 threads still occupies a full warp, so effectively 32 threads with 16 idle. Not good. With 32 channels per block (1 warp): - Blocks: 32 * 8 = 256 blocks. - Threads: 32. - This is natural warp size. Can we use thread blocks of 64 threads (2 warps) handling 64 channels? - Blocks: 32 * 4 = 128 blocks. - Threads: 64. - This might be a sweet spot. Or 128 threads handling 128 channels: - Blocks: 32 * 2 = 64 blocks. I think I'll go with 64 blocks of 128 threads each. Each block handles 128 channels of one batch. Total 64 blocks × 128 threads = 8192 threads. And I'll stream tiles of 64 timesteps through shared memory. - For TILE_T=64, shared memory needs 64 * 128 * 4 * 2 = 64KB. Very comfortable. - Number of tiles: 4096 / 64 = 64 tiles per block. - Each block does 64 tiles. Wait, but each block now handles only 128 channels, so half the channels of a batch. 64 blocks across 132 SMs, each SM gets ~0.5 blocks initially. But as SMs finish, they pick up more blocks. Total waves: ceil(64 / 132) = 1 wave. So all 64 blocks run concurrently or nearly so. With 64 blocks, we're using about half the SMs. Each SM runs at most 1 block (with 128 threads = 4 warps). 4 warps per SM might generate decent memory traffic. Alternatively, use 32 channels per block (1 warp = 32 threads). 256 blocks. Each SM gets ~2 blocks. With 32 threads per block = 1 warp, each SM gets ~2 warps. Total warps: 256. Same as before, but spread across more SMs. I think the sweet spot might actually be 128 threads per block (32 batches × 2 channel groups = 64 blocks), or 256 threads per block (32 batches × 1 channel group = 32 blocks). Let me try both approaches: 32 blocks and 64 blocks. Actually, I'll implement a flexible kernel that can use any block size, and tune the TILE_T and channels_per_block. For simplicity, I'll use one block per batch element (32 blocks), 256 threads per block, and TILE_T=32 or 64. This is the simplest and most natural given D=256. Let me draft the kernel: ```cuda __global__ void scan_kernel(const float* a, const float* b, float* out, int B, int L, int D) { // blockIdx.x = batch index (0..B-1) // threadIdx.x = channel index (0..D-1) int b = blockIdx.x; int d = threadIdx.x; // Shared memory for a tile of TILE_T x D (but D is blockDim.x = 256) extern __shared__ float smem[]; // size = 2 * TILE_T * D float* smem_a = smem; float* smem_b = smem + TILE_T * D; float h = 0; for (int tile = 0; tile < L / TILE_T; tile++) { int t_base = tile * TILE_T; // Load tile into shared memory for (int i = 0; i < TILE_T; i++) { int t = t_base + i; smem_a[i * D + d] = a[b * L * D + t * D + d]; smem_b[i * D + d] = b[b * L * D + t * D + d]; } __syncthreads(); // Sequential scan within the tile for (int i = 0; i < TILE_T; i++) { float av = smem_a[i * D + d]; float bv = smem_b[i * D + d]; h = av * h + bv; out[b * L * D + (t_base + i) * D + d] = h; } __syncthreads(); } } ``` Wait, the load loop: ```cuda for (int i = 0; i < TILE_T; i++) { int t = t_base + i; smem_a[i * D + d] = a[...]; } ``` All 256 threads run this loop simultaneously. At each `i`, thread `d` loads `a[b, t_base+i, d]`. This is coalesced! Good. But we have `__syncthreads()` after the load and after the store. With 32 blocks, this should be fine. However, there's a problem: the `__syncthreads()` after the sequential scan is not needed for correctness within the block, because each thread only accesses its own `d` column in shared memory, and global memory writes are independent. BUT, we need to make sure all threads finish the store before we overwrite shared memory in the next tile. Actually, since each thread writes to a different global memory location, we don't need a block-wide barrier between tiles. We only need to make sure the shared memory loads for the next tile happen after all threads finish reading shared memory from the current tile. Since each thread finishes reading shared memory before writing to global memory, and the loads for the next tile don't depend on the previous tile's shared memory contents, do we even need `__syncthreads()` between tiles? Actually, yes we do. Without `__syncthreads()`, some fast threads might start loading the next tile into shared memory while slower threads are still reading the current tile. This causes data corruption. So we DO need `__syncthreads()` between tiles. Wait, actually, if each thread only reads column `d` and no thread ever reads another thread's column, then the shared memory slots for column `d` in the next tile are `smem_a[i * D + d]` which are the same addresses as the current tile (since `i` ranges 0..TILE_T-1 in both). If a fast thread overwrites `smem_a[0][d]` with the next tile's data while a slow thread is still reading `smem_a[0][d]`, there's a race. So yes, we need `__syncthreads()` between the sequential scan and the next tile's loads. Can we eliminate shared memory and just load directly? We COULD have each thread load directly from global memory: ```cuda float h = 0; for (int t = 0; t < L; t++) { float av = a[b * L * D + t * D + d]; float bv = b[b * L * D + t * D + d]; h = av * h + bv; out[b * L * D + t * D + d] = h; } ``` This is the simple approach without shared memory. It does coalesced loads and stores directly. No need for shared memory or barriers. Wait! Why do we need shared memory at all? The loads and stores are already coalesced. The sequential scan doesn't need shared memory because each thread operates independently. Let me verify memory access for the direct global memory approach: - Thread `d` accesses `a[b * L * D + t * D + d]` for varying `t`. - At a given `t`, threads 0..31 in a warp access `a[b, t, 0..31]`. These are contiguous. - The warp issues a single coalesced load for 128 bytes. - Perfect! And without shared memory, there's no need for `__syncthreads()`. The kernel is dead simple. So why did I think about shared memory? Because I was worried about cache performance, but for this access pattern, the L1/L2 cache should handle it well because consecutive warps in the block access the same cache lines (channels 0..31, 32..63, etc., at the same timestep). Actually, at timestep `t`: - Warp 0 accesses `a[b, t, 0..31]` → addresses `base + t*D + 0..31`. This is one 128-byte cache line. - Warp 1 accesses `a[b, t, 32..63]` → addresses `base + t*D + 32..63`. This is another 128-byte cache line. - ... - Warp 7 accesses `a[b, t, 224..255]` → 8th cache line. - Total for the block: 8 cache lines. And this happens for `a` and `b` (16 cache lines total). Then stores for `out` (another 8 cache lines). Total 24 cache lines per timestep per block. With 32 blocks: 768 cache lines per timestep. Each cache line is 128 bytes, so about 96KB loaded per timestep across all blocks. Over 4096 timesteps: 384MB. No waste! This is exactly the theoretical minimum memory traffic. And the loads are perfectly coalesced. So the simplest kernel is actually the best one! No shared memory needed. ```cuda __global__ void scan_kernel(const float* a, const float* b, float* out, int B, int L, int D) { int b = blockIdx.x; int d = threadIdx.x; float h = 0; for (int t = 0; t < L; t++) { float av = a[b * L * D + t * D + d]; float bv = b[b * L * D + t * D + d]; h = av * h + bv; out[b * L * D + t * D + d] = h; } } ``` With `gridDim.x = B = 32` and `blockDim.x = D = 256`. Now, the only concern is occupancy. With 32 blocks × 256 threads = 8192 threads, which is only 8192/2048/132 ≈ 3% of H100's thread capacity. BUT each block does a LOT of work (4096 iterations). And memory access is perfect. I think this might still be very fast for this problem size. Let me quickly think about if there are any compiler optimizations we can add: 1. `#pragma unroll` on the loop? 4096 is too large for full unrolling, but we can unroll by a factor. 2. Use `__restrict__` for pointers. 3. Use `float` consistently (inputs are fp32). Actually, let's estimate performance: - Per iteration per thread: 2 loads, 1 FMA, 1 store. - The loop has a data dependency: `h` depends on the previous iteration. - The compiler can pipeline the loads and FMAs using loop unrolling. - With `pragma unroll 4`, the compiler can interleave loads from iterations i, i+1, i+2, i+3. Let me add `#pragma unroll 8` to help with instruction-level parallelism. But there's a dependency chain: ```cuda h_0 = a_0 * h_{-1} + b_0 (4-6 cycle latency for FMA) h_1 = a_1 * h_0 + b_1 (waits for h_0) ``` Each FMA has ~4 cycle latency on Hopper. So iterations can't overlap much. The loop is latency-bound to the FMA chain. Wait, if the loop is FMA-latency bound, memory bandwidth doesn't matter! The FMAs can't be overlapped because of the dependency. FP32 FMA latency on Hopper is about 4 cycles. Throughput per SM is 128 FMAs per clock (with Tensor Cores, but for regular ALU... actually Hopper has 128 FP32 FMAs per SM per clock? No, that's too high. Let me think.) For Ampere, each SM has 4 warp schedulers, each can issue 1 FP32 FMA per clock, so 4 FMAs per clock per SM per warp, but overall the SM can issue up to 128 FP32 ops per clock? Actually, A100 SM can do 64 FP32 FMAs per clock (128 FLOPS). H100 is similar or higher. With a 4-cycle latency and 64 FMAs/clock throughput, the SM needs 4 * 64 = 256 independent FMAs in flight to saturate throughput. In our kernel, each thread does one FMA per iteration, and iterations are dependent. So each thread can issue one FMA every 4 cycles. To saturate the SM, we need 64 threads issuing FMAs every 4 cycles. With 256 threads per block, that's 4x more than needed. But the FMAs from different threads are independent! Thread 0's FMA at iteration i is independent of thread 1's FMA at iteration i. So the SM can interleave FMAs from all 256 threads. With 256 threads doing back-to-back dependent FMAs, and each FMA taking 4 cycles, the throughput across 256 threads is 256 FMAs / 4 cycles = 64 FMAs per cycle. This is exactly the SM's FP32 throughput! Wait, is that right? Each thread issues 1 FMA every 4 cycles (due to latency). With 256 threads, that's 64 FMAs per cycle. If the SM can do 64 FMAs per cycle, then we're compute-bound, not memory-bound! Actually, modern GPUs can issue multiple independent instructions per cycle per thread. If a thread has other work (like the next load) while waiting for the FMA, the scheduler might be able to overlap. But in our case, the loop is: ```cuda load a load b fma (depends on a, b, and previous h) store h ``` The loads can be pipelined. The FMA depends on the loads and previous FMA. The store depends on the FMA. With unrolling, we could have: ```cuda // iteration i load a_i load b_i // iteration i+1 (independent loads) load a_{i+1} load b_{i+1} // FMA for iteration i fma_i // store h_i store_i fma_{i+1} store_{i+1} ``` If the compiler can schedule this, loads from future iterations can overlap with FMAs from current iterations. But the FMA chain is still the critical path. Each FMA takes ~4 cycles, and we do one per iteration. Can we overlap FMAs from different iterations of the SAME thread? No, because of data dependency. Can we overlap FMAs from different threads? Yes, but they already run in parallel on different warp schedulers. Hmm, actually, with 256 threads = 8 warps, and 4 warp schedulers per SM Hopper, each scheduler handles 2 warps. At each cycle, each scheduler can issue an instruction from one warp. If we have enough warps to cover the 4-cycle FMA latency, the SM can issue 1 FMA per cycle per scheduler × 4 schedulers = 4 FMAs per cycle. But the throughput is 64 FMA/cycle, which suggests 16x more capacity. Wait, I need to get the actual SM throughput. H100 SM has 128 FP32 CUDA cores per SM (same as A100). Actually, I think A100 has 64 FP32 cores per SM that can do FMAs (2 FLOPS each), so 128 FLOPS per clock. H100 has 128 FP32 cores per SM, giving 256 FLOPS per clock? Or maybe 64 cores but higher clock. Actually, looking it up: A100 has 64 FP32 cores per SM (Ampere architecture). H100 has 128 FP32 cores per SM (Hopper architecture). At 2GHz, that's 256 FLOPS per clock per SM. For 132 SMs, that's ~33.8 TFLOPS at 2GHz, but spec says 67 TFLOPS. So either it has 128 cores and higher clock, or the FMA counts as 2 FLOPS. Anyway, if each SM can issue 64 or 128 FMAs per clock, and our 8 warps (256 threads) can only provide 64 FMAs per 4 cycles = 16 FMAs per cycle, then the SM is underutilized! We'd need many more warps to saturate the compute. Wait, 256 threads in one block, but if we only have 1 block per SM, that's 256 threads. To saturate, we need maybe 512 or 1024 threads per SM. But even without saturation, the FMA throughput might be sufficient that memory bandwidth becomes the bottleneck. Or vice versa. Let's do a roofline analysis for this kernel on one SM: - Arithmetic intensity: 1 FMA per 12 bytes (2 loads + 1 store of 4 bytes each = 12 bytes, but coalesced with perfect reuse). Actually, each element is loaded once, so 2 loads + 1 store = 12 bytes per FMA. - FMA throughput: let's say 64 FMA/clock. - Memory bandwidth: let's say per-SM effective bandwidth is 25 GB/s at 2GHz = 12.5 bytes/clock. - At memory bandwidth bound: 12.5 bytes/clock / 12 bytes/FMA = 1.04 FMA/clock. - At compute bound: 64 FMA/clock. The kernel needs 64x more compute than bandwidth, so it is unambiguously MEMORY BOUND. Wait, but earlier I calculated that with 256 threads (=8 warps) on one SM, FMA throughput is only 16 FMA/clock (256 threads * 1 FMA / 4 cycles = 64 FMA / 4 cycles = 16 FMA/clock). At 16 FMA/clock with memory delivering 12.5 bytes/clock, the kernel would need 16 * 12 = 192 bytes/clock. But memory only provides 12.5. So we're even more memory bound: 192 / 12.5 = 15.4x over-subscribed on memory. Actually wait, if the FMA throughput from 256 threads is 16/cycle, and memory bandwidth is 12.5 bytes/cycle, the bottleneck is memory at 12.5 bytes/cycle. The FMAs are not the limiting factor because we could theoretically do more (up to 64/cycle) but memory can't keep up. So the kernel is memory bandwidth bound. And since we only have 8 warps per SM (256 threads), can those 8 warps generate enough memory requests to saturate the per-SM bandwidth? At 12.5 bytes/cycle per SM, and 8 warps, each warp needs to generate 12.5/8 = 1.56 bytes/cycle. Since each warp moves 128 bytes per coalesced load, a warp needs to issue a load every 128/1.56 = 82 cycles. In our loop, a warp does 1 load, 1 load, 1 FMA, 1 store per iteration. That's maybe 4-8 instructions. At 2GHz, an instruction takes ~0.5ns. So 8 instructions = 4ns. 82 cycles at 2GHz is 41ns. So we need to sustain one iteration per 41ns, or ~10 iterations per warp per 400ns memory latency. Hmm, this seems achievable. But what's the actual time? If memory is saturated, and we need to move 384MB at 3.35 TB/s = 115us. But with only 32 blocks running on ~32 SMs, can we use 3.35 TB/s? If 32 SMs out of 132 are active, and each SM generates 12.5 bytes/cycle, total is 400 bytes/cycle = 800 GB/s at 2GHz. This is only 24% of peak bandwidth. Is it because 32 SMs only have 24% of the L1-to-L2 bandwidth? Or is L2 bandwidth shared among all SMs? Actually, the memory bandwidth is a global resource. Even if only 32 SMs are active, they can potentially saturate the memory bus if they generate enough requests. Each SM on H100 can have many outstanding memory requests (maybe 1000+). So 32 SMs * 1000 requests = 32,000 outstanding requests. The memory controllers can handle this. I think 32 SMs CAN saturate or nearly saturate the memory bus if each SM is fully loaded with memory requests. But with only 8 warps per SM, and each warp hitting memory every few cycles, we might not have enough in-flight requests per SM to saturate its request pipeline. Each warp can have ~64 outstanding loads. With 8 warps, that's 512 outstanding requests per SM. At 32 SMs, that's 16,384 outstanding requests. The global memory system can probably handle tens of thousands. So we might actually get close to saturation! Wait, I think this is promising. With 32 blocks, 256 threads each, perfectly coalesced memory access, and streaming through 4096 timesteps, the kernel might achieve a large fraction of peak bandwidth. But there is one more thing: the FMA latency bottleneck. Even though memory is the theoretical bottleneck, with only 256 threads per SM, and each thread having a dependency chain of FMAs, the SM might be stalling waiting for FMA results, even if memory is ready. Actually, the warp scheduler can execute instructions from different warps while one warp waits for FMA results. With 8 warps per SM and 4 warp schedulers, at any given cycle, 4 warps can be issuing instructions. If a warp is waiting for FMA results (4 cycles), the scheduler issues instructions from other warps. With 8 warps and 4 schedulers, there are always warps ready to issue (since each warp only stalls for 4 cycles after an FMA, but memory loads take much longer). Actually, in this kernel, the critical path per thread is: 1. Load a (stalls ~400 cycles for L2 miss, but might hit L1) 2. Load b 3. FMA (stalls 4 cycles) 4. Store The loads at iteration i can be issued while the FMA from iteration i-1 is still in flight. With good compiler scheduling, the loads for iteration i might start right after issuing the FMA for iteration i-1, giving ~4 cycles for the FMA to complete before the next FMA is ready. Wait, but the FMA at iteration i needs the result of the FMA from iteration i-1, AND the loads from iteration i. So as long as loads finish before the FMA needs them, there's no stall. Loads take 400 cycles (L2 miss), but they can be pipelined. If we issue loads for iterations i, i+1, i+2, i+3 in advance, by the time we need them, they might be ready. This is where `#pragma unroll` really helps. If we unroll by 8: ```cuda #pragma unroll 8 for (int t = 0; t < L; t += 8) { float a0 = a[...t+0]; float b0 = b[...t+0]; float a1 = a[...t+1]; float b1 = b[...t+1]; ... float a7 = a[...t+7]; float b7 = b[...t+7]; h0 = a0 * h_prev + b0; h1 = a1 * h0 + b1; ... h7 = a7 * h6 + b7; out[...t+0] = h0; ... out[...t+7] = h7; h_prev = h7; } ``` With 8 loads from `a` and 8 from `b` happening first, the compiler can pipeline them. Then 8 dependent FMAs. But the dependency chain is still 8 FMAs long. This doesn't help latency hiding. Actually, unrolling doesn't help with the FMA dependency chain within a thread. But it does help with memory latency hiding by batching loads. For better performance, we should process the loop as simply as possible and let the compiler handle it. OK, I'm going to implement this kernel exactly as described: 32 blocks, 256 threads per block, direct global memory access with coalesced loads. I'll add `#pragma unroll 4` and see how it performs. Wait, but I want to test both: the simple kernel and a version with more blocks (e.g., 256 blocks of 32 threads). The version with more blocks should have better SM distribution. Actually, let me think about 256 blocks of 32 threads: - Grid: `dim3(B, D / 32)` = `dim3(32, 8)` = 256 blocks. - Block: 32 threads (1 warp). - Thread `tid` handles channel `blockIdx.y * 32 + threadIdx.x`. - Each warp processes 4096 timesteps sequentially. - At each timestep, the warp accesses 32 consecutive channels. Coalesced! - 256 blocks on 132 SMs. ~2 blocks per SM. 1 warp per block, so 2 warps per SM. Total warps = 256. - With only 2 warps per SM, latency hiding is poor. 256 warps vs 256 warps in the 32-block case (8 warps per block * 32 blocks = 256 warps). The total number of warps is the same! We just distributed them differently. In the 32-block case: 32 blocks * 8 warps = 256 warps. Some SMs have 1 block (8 warps), some have 0. In the 256-block case: 256 blocks * 1 warp = 256 warps. Most SMs have 2 blocks (2 warps). With 8 warps on an SM, the scheduler can switch between them. With 2 warps, switching is limited. I think the 32-block case with 8 warps per SM is BETTER than 256 blocks with 2 warps per SM, even though fewer SMs are used. What about 64 blocks of 128 threads = 4 warps per block? - 64 blocks * 4 warps = 256 warps total. - 64 blocks / 132 SMs = ~0.5 blocks per SM. - Some SMs have 1 block (4 warps), most have 0. This is worse than 32 blocks with 8 warps. What about 128 blocks of 64 threads = 2 warps per block? - 128 blocks * 2 warps = 256 warps. - Worse. So among the options where 1 thread = 1 scan, the 32-block with 256 threads per block maximizes warps per active SM (8 warps), which should help with latency hiding. But 256 warps total is still only 3% of H100's warp capacity. If each warp is memory-bound and stalls often, 8 warps per SM might not be enough to keep the execution units busy. Actually, on H100, each SM can execute up to 64 warps concurrently. With 256 threads = 8 warps per SM, the SM has plenty of capacity for more warps. But we don't have more warps. Hmm. I think for this problem, the memory throughput might be bounded by the per-SM memory request generation rate. If only 8 warps per SM are issuing loads, the SM might not generate enough requests to fully utilize its L2 bandwidth. One way to artificially increase warp count: use thread blocks that process FAKE scans or have idle threads. But they would just add overhead, not throughput. Another way: split the sequence into chunks and have multiple blocks process each scan. This gives more total warps without requiring multiple threads per scan in a parallel algorithm. Specifically, the 3-pass chunked approach with chunk size 256 gives: - Pass 1: 512 blocks of 256 threads = 4096 warps. - That's 16x more warps! And memory is still coalesced. - With 4096 warps on 132 SMs, that's 31 warps per SM. Now we can really saturate the memory system! And the overhead of Pass 2 and 3 is small. This is starting to look like the winner. Let me calculate the total memory traffic: - Pass 1: reads a and b (256MB), writes aggregates (1MB). Total: 257MB. - Pass 2: reads and writes aggregates (2MB). Total: 2MB. - Pass 3: reads a and b (256MB), reads prefixes (0.5MB), writes output (128MB). Total: 384.5MB. - Overall: ~644MB. At 3.35 TB/s: 192us. Realistically with overhead, maybe 300-400us. Compare to the single-kernel approach: - 384MB traffic. - At say 1 TB/s (due to low occupancy): 384us. - At 2 TB/s (if we're lucky): 192us. They're competitive. The single-kernel is simpler. The 3-pass has more parallelism. I'll implement both and test! Actually, let me do a quick calculation for the 3-pass approach with optimal bandwidth. 644MB at 2 TB/s = 322us. Plus 3 kernel launch overheads (~10us each). Total ~350us. For the single-kernel approach, if we get 1.5 TB/s (good but not peak due to occupancy): 384MB / 1.5TB = 256us. If the single-kernel can achieve >1 TB/s, it might actually be faster than the 3-pass approach due to less total memory traffic. I'll start with the single-kernel approach because it's much simpler to implement correctly. Let me also consider: what if I launch the single kernel with 32 blocks but each block handles multiple batch elements via loop? For example, block 0 handles batch 0 and batch 16, block 1 handles batch 1 and batch 17, etc. This doubles the work per block but keeps the same number of blocks. Not helpful. What if I use a 1D grid of many blocks where each block loops over multiple batch elements? Like 256 blocks, each handling 1/8th of the channels for 1 batch element, and there are enough batch elements to go around... no, we have exactly 32 batch elements. I think the single-kernel approach is simple and worth trying first. If it's not fast enough, I'll switch to the 3-pass approach. One more idea: what if we use a "grid-stride loop" pattern where each thread handles multiple channels? No, that just reduces parallelism. Wait, what about using warp-level primitives to scan 32 timesteps in parallel, and then having the warp iterate over 128 groups of 32? This way, each thread is active more of the time. For a warp of 32 threads, we could scan 32 timesteps using warp shuffles in log(32) = 5 steps. Then iterate 128 times for the full 4096 timesteps. But each warp handles only 1 channel, and 32 threads in the warp work on that one channel. For 256 channels, we need 256 warps. With 32 warps per block (1024 threads), we need 8 blocks (one per batch). Only 8 blocks. Terrible. If we have 32 threads per block (1 warp), and each block handles 1 channel, then 8192 blocks. Great block count! But each block has only 1 warp. And memory access: the 32 threads all load data for the SAME channel at DIFFERENT timesteps. That's strided memory access! Non-coalesced! To fix memory access, we could reorganize: have a block of 32 threads handle 32 channels at the same timestep, and use warp shuffles to do... no, warp shuffles work on one channel's scan, not across channels. This parallel scan within a warp for one channel requires strided access. Back to the same problem. So any approach that parallelizes within a scan has strided memory access. Approaches that process each scan sequentially have good memory access but limited parallelism. The 3-pass chunk approach is the only one that gives both good memory access (because chunks are processed sequentially with coalesced access) AND good parallelism (because each chunk is assigned to a different block). Actually wait! In the 3-pass approach, Pass 1 assigns each chunk of each scan to a thread. If chunk size is 256 and each block handles 256 channels, then a block handles 1 chunk of 1 batch. Threads in the block access different channels at the same timesteps. Coalesced! And we get 512 blocks. Much better. OK, I'm now quite confident the 3-pass approach is superior. Let me design it carefully. **Design for 3-pass approach:** **Pass 1: Compute chunk aggregates** - Chunk size: 256 - Grid: `dim3(B, num_chunks)` = `dim3(32, 16)` = 512 blocks - Block: 256 threads - Each block handles 1 batch, 1 chunk, 256 channels. - Thread `tid` handles channel `tid`. - Kernel: ```cuda __global__ void aggregate_kernel(const float* a, const float* b, float* agg_a, float* agg_b, int B, int L, int D, int chunk_size, int num_chunks) { int b = blockIdx.x; int chunk = blockIdx.y; int d = threadIdx.x; int t_start = chunk * chunk_size; float h = 0; float a_prod = 1; for (int i = 0; i < chunk_size; i++) { int t = t_start + i; int idx = b * L * D + t * D + d; float av = a[idx]; float bv = b[idx]; h = av * h + bv; a_prod *= av; } int agg_idx = b * D * num_chunks + d * num_chunks + chunk; agg_a[agg_idx] = a_prod; agg_b[agg_idx] = h; } ``` Wait, here the memory access for `agg_a` and `agg_b`: - Thread `d` writes to offset `d * num_chunks + chunk`. - For a warp of 32 threads with consecutive `d`, they write to addresses separated by `num_chunks` (16). So offsets are `chunk, chunk+16, chunk+32, ...`. Each address is 64 bytes apart (16 * 4). Not perfectly coalesced, but we can transpose the aggregate output. To make aggregate writes coalesced, we should store as `[B, num_chunks, D]` instead of `[B, D, num_chunks]`: - `agg_idx = b * num_chunks * D + chunk * D + d;` - Then threads 0..31 write to `chunk*D + 0..31`, which is contiguous. Coalesced! Let's use this layout: - `agg_a` shape: `[B, num_chunks, D]` - `agg_b` shape: `[B, num_chunks, D]` Actually, we can combine them: `agg` shape `[B, num_chunks, D, 2]` or two separate arrays. Two separate arrays are simpler. So Pass 1 writes: - `agg_a[b * num_chunks * D + chunk * D + d] = a_prod;` - `agg_b[b * num_chunks * D + chunk * D + d] = h;` And Pass 2 reads from this layout. **Pass 2: Scan aggregates** - Input: `[B, num_chunks, D]` for agg_a and agg_b - Output: `[B, num_chunks, D]` for prefix - For each (b, d), scan num_chunks aggregates. - We want to launch many blocks. Option A: one block per batch element, 256 threads. - 32 blocks. Each thread handles 1 channel, scanning 16 chunks sequentially. - Memory access: threads read `agg_a[b * num_chunks * D + c * D + d]`. - At a fixed `c`, threads 0..31 read contiguous memory. Coalesced! - Number of blocks: 32. Low. Option B: one block per channel group of 32, for each batch. - 32 * 8 = 256 blocks, each with 32 threads. - Thread `tid` handles channel `blockIdx.y * 32 + tid`. - Memory access: coalesced at each chunk iteration. - Better block count! ```cuda __global__ void prefix_kernel(const float* agg_a, const float* agg_b, float* prefix, int B, int num_chunks, int D) { int b = blockIdx.x; int d_group = blockIdx.y; // 0..7 int d = d_group * 32 + threadIdx.x; // 0..255 int base = b * num_chunks * D; float h = 0; for (int c = 0; c < num_chunks; c++) { int idx = base + c * D + d; float av = agg_a[idx]; float bv = agg_b[idx]; prefix[idx] = h; // prefix before applying chunk c h = av * h + bv; } } ``` Wait, the loop has `prefix[idx] = h;` followed by `h = av * h + bv;`. At each iteration `c`, the prefix stored is the state before chunk `c`. For chunk 0, prefix = 0. Correct. For chunk 1, prefix = agg_b[0]. Correct. This stores a scalar prefix per chunk per channel. Can we make this more parallel? We could process the sequential scan on 16 items with a warp, but 16 items is too small. Or we could have each block handle 32 channels and all 32 threads process them in a way that... no, sequential scan is fine for 16 items. Total work for Pass 2: 8192 scans × 16 items = 131,072 items. At 4 bytes each for read and write, about 1MB of traffic. Even with low occupancy, this will be very fast. With 256 blocks of 32 threads, total threads = 8192. Same as before, but much less work per thread. I think this is fine. **Pass 3: Final scan with prefix** - Grid: same as Pass 1. `dim3(B, num_chunks)` = 512 blocks, 256 threads. - Thread `tid` handles channel `tid`. - Load prefix for this chunk. - Scan chunk sequentially, write output. ```cuda __global__ void final_kernel(const float* a, const float* b, const float* prefix, float* out, int B, int L, int D, int chunk_size, int num_chunks) { int b = blockIdx.x; int chunk = blockIdx.y; int d = threadIdx.x; int t_start = chunk * chunk_size; int prefix_idx = b * num_chunks * D + chunk * D + d; float h = prefix[prefix_idx]; for (int i = 0; i < chunk_size; i++) { int t = t_start + i; int idx = b * L * D + t * D + d; float av = a[idx]; float bv = b[idx]; h = av * h + bv; out[idx] = h; } } ``` Memory access: coalesced! Now let me think about whether we can optimize further. For Pass 3, we read `a`, `b`, and `prefix`, then write `out`. `prefix` is only 1 float per thread. We can load it into a register. The main traffic is `a`, `b`, and `out`. For Pass 1, we read `a` and `b`, write `agg_a` and `agg_b`. One optimization: instead of writing both `agg_a` and `agg_b`, can we compute the prefix on-the-fly and avoid writing `agg_a`? No, because we need all aggregates to compute the scan in Pass 2. Unless... we do a hierarchical scan within the same kernel invocation using cooperative groups. But that's more complex. Actually, there's another clever approach. We can process the chunks in a tree-like fashion within a block: 1. Divide the sequence into chunks. 2. Have all threads in a block (256 threads) compute local aggregates for their chunks. 3. Do a parallel scan on the aggregates within the block using warp shuffles or shared memory. 4. Apply the prefixes and do the final scan within each chunk. But each block would need to handle all chunks of one scan, which means 256 threads for 16 chunks. Each chunk could be handled by 16 threads... no, a scan is within a channel, not across channels. Wait, what if a block handles one channel scan, and 256 threads process the 4096 timesteps in parallel? That's the parallel scan with strided memory access. Not good. What if a block handles multiple channels (say 16 channels) and all their chunks (16 chunks per channel, 16 channels = 256 chunks). Each thread handles one chunk. After computing local aggregates, we need to scan 16 aggregates for each of 16 channels. We could use shared memory for this. Let's design a "mega-block" approach: - Block size: 256 threads. - Handles: 16 channels of 1 batch, all 4096 timesteps divided into 16 chunks of 256. - Total blocks: 32 * (256/16) = 32 * 16 = 512 blocks. Same as before! But instead of 3 kernel calls, we do it all in one kernel with shared memory communication: Phase 1: Each thread handles 1 chunk (256 timesteps) for 1 channel. Compute local aggregate. Write to shared memory. - 256 threads = 16 channels × 16 chunks. - After Phase 1, shared memory has 16 channels × 16 aggregates. Phase 2: For each channel, scan the 16 aggregates. We can have 16 threads per channel do this sequentially or in parallel. - If sequential: 16 threads per channel, 16 channels = 256 threads. Each thread scans 16 items sequentially. Same work as Pass 2. - Store prefixes in shared memory. Phase 3: Each thread re-scans its chunk with the prefix and writes output. Can we do all 3 phases in one kernel? Yes! Let's define the thread mapping: - Channels per block: 16 - Threads per channel: 16 - Total threads: 16 * 16 = 256 - Thread mapping: `threadIdx.x = channel_id * 16 + chunk_id` where `channel_id = 0..15` and `chunk_id = 0..15`. Phase 1: Each thread scans chunk `chunk_id` of channel `channel_id`. - Sequential scan of 256 timesteps. - Memory access: at each timestep `t`, thread `(channel, chunk)` accesses `a[b, t, channel]`. - Threads in the same warp: warp has 32 threads. With our mapping, threads 0..15 are channel 0, chunks 0..15. Threads 16..31 are channel 1, chunks 0..15. - At a given timestep `t` (within a chunk), thread `(0, 0)` accesses `a[b, t, 0]`, thread `(0, 1)` accesses `a[b, t+256, 0]`, etc. - Wait, different threads in the same warp are in the same channel but different chunks. They access different timesteps for the SAME channel. These are NOT coalesced! Memory addresses are separated by `256 * D * 4 = 256KB`! This is terrible memory access. We'd need to change the thread mapping. Alternative thread mapping for Phase 1: - Consecutive threads in a warp should access consecutive channels at the same timestep. - So for Phase 1, we want thread `(chunk, channel)` with `channel` varying fastest. - Thread `tid = chunk * 16 + channel` where chunk = 0..15, channel = 0..15. - Warp 0 has threads with chunk=0, channels 0..15 and chunk=1, channels 0..15? No, with 16 channels, threads 0..15 are chunk 0, channels 0..15. Threads 16..31 are chunk 1, channels 0..15. - At timestep `t` within chunk 0, only threads 0..15 are active (processing chunk 0). Threads 16..31 are processing chunk 1 at a different `t`. They don't access memory at the same time because they're in the same warp executing sequentially? No, all threads in a warp execute the same instruction. If threads 0..15 load at `t=0` (chunk 0), threads 16..31 load at `t=0` (chunk 1, which is `t=256`). The warp issues a single load instruction, and threads 0..15 and 16..31 all load simultaneously. Memory access pattern for the warp: - Thread (0, 0): `a[b, 0, 0]` - Thread (0, 1): `a[b, 0, 1]` - ... - Thread (0, 15): `a[b, 0, 15]` - Thread (1, 0): `a[b, 256, 0]` - Thread (1, 1): `a[b, 256, 1]` - ... - Thread (1, 15): `a[b, 256, 15]` The first 16 threads access channels 0..15 at t=0. The next 16 threads access channels 0..15 at t=256. These are NOT contiguous! The first 16 addresses span 64 bytes. The next 16 addresses start at `t=256`, which is `256 * D * 4 = 256 * 256 * 4 = 262,144` bytes away. The warp loads 32 cache lines, some near t=0 and some near t=256. This is bad. To fix this, we need to process one chunk at a time within a warp. But threads in a warp are supposed to execute together. Actually, if we divide into warps where each warp handles one chunk: - Warp 0 handles chunk 0, channels 0..31. It accesses `a[b, t, 0..31]` for t=0..255. At each t, this is coalesced! - Warp 1 handles chunk 1, channels 0..31. - etc. With 16 chunks and 16 channels per block, and each warp handling 1 chunk for all 16 channels... wait, a warp has 32 threads. If it handles 16 channels, that's 2 threads per channel, or we could handle 32 channels per warp but we only have 16 channels per block. If each block handles 32 channels (1 warp per chunk, 16 chunks = 16 warps = 512 threads). That's a large block. Or each block handles 16 channels, 16 chunks = 256 threads = 8 warps. - Warp 0: chunk 0, channels 0..31? No, we only have 16 channels. - Warp 0: chunk 0, channels 0..15 + chunk 1, channels 0..15? That gives 32 threads. But then memory access is bad. What if each block handles 32 channels, 16 chunks, and has 512 threads (16 warps)? - Warp 0: chunk 0, channels 0..31. Coalesced! - Warp 1: chunk 1, channels 0..31. Coalesced! - ... - Warp 15: chunk 15, channels 0..31. Coalesced! - Total blocks: 32 batches * (256/32) = 32 * 8 = 256 blocks. - Each block: 512 threads = 16 warps. - Total warps: 256 * 16 = 4096 warps! - Blocks per SM: 256/132 ≈ 2. Threads per SM: 256 * 512 / 132 ≈ 992 threads. ~50% occupancy. Good! - Shared memory for Phase 2: 32 channels × 16 aggregates × 2 values × 4 bytes = 4KB. Very small! - Actually, each warp computes one chunk's aggregate for 32 channels. The writes to shared memory can be coalesced within the warp. Phase 1 with 256 blocks of 512 threads: - 16 warps per block, each handling one chunk for 32 channels. - Each warp loads coalesced data and sequentially scans 256 timesteps for 32 channels. - Actually, wait. A warp has 32 threads. If it handles 32 channels for one chunk, then thread `tid` handles channel `tid`. That makes 32 threads scanning 256 timesteps each. Coalesced loads! Phase 2: Scan the 16 chunk aggregates for each of the 32 channels. - We have 32 channels and 16 aggregates per channel. - 512 threads in the block. How do we organize? - We could have 16 threads per channel scan the 16 aggregates sequentially or use parallel scan. - 32 channels * 16 threads = 512 threads. - Thread mapping: `tid = channel * 16 + aggregate_idx`. But then threads in a warp are mixed across channels. - Alternatively, we could use a different approach: have 512 threads all participate in a parallel scan across all channels simultaneously. Actually, for Phase 2, since we only have 16 aggregates per channel, and 32 channels, we can have each channel's scan done by 1 thread using a sequential loop. That's 32 threads active and 480 threads idle. Not good. But Phase 2 is tiny (16 steps * 32 channels = 512 operations), so even with 1 thread per channel, it takes negligible time. Or we can use warp shuffles! Each warp of 32 threads handles one channel's 16 aggregates? No, warp has 32 threads and we only have 16 aggregates. With 16 aggregates, we can have 2 channels per warp (16 threads per channel). Use warp shuffles for parallel scan on 16 items. This is efficient! Let's map: - Warp 0: channel 0, aggregates 0..15 (using threads 0..15, threads 16..31 idle) - Warp 1: channel 1, aggregates 0..15 - ... - Warp 31: channel 31, aggregates 0..15 - Total 32 warps. Perfect! Each warp does a parallel scan of 16 aggregates using warp shuffles. The scan of 16 items with 16 threads takes 4 steps. Very fast! Phase 3: Apply prefixes and write final output. - Same structure as Phase 1. 16 warps, each handling one chunk for 32 channels. - Each warp reads its prefix from shared memory and scans 256 timesteps, writing output. This single-kernel approach with 256 blocks of 512 threads gives: - 4096 warps total - ~50% occupancy - Coalesced memory access - No extra global memory traffic - One kernel launch! Wait, but there's a problem. In Phase 1, each warp handles one chunk. We have 16 warps. Warp 0 handles chunk 0, warp 1 handles chunk 1, etc. They all execute concurrently. At the end, each warp writes its aggregate to shared memory. Then Phase 2: warp `w` scans the aggregates for channel `w` (if we map 1 warp per channel for the prefix scan). But we need to scan aggregates for ALL 32 channels. And we have 32 warps (with 16 channels... no, we have 16 warps in the block and 32 channels). This doesn't match. Let me restructure. Block: 512 threads = 16 warps. Channels per block: 32. Chunks per block: 16. Phase 1: - We need 32 channels × 16 chunks = 512 aggregates. - Each thread handles 1 channel and 1 chunk. Thread scans 256 timesteps sequentially. - But how to organize threads for coalesced memory access? - We want consecutive threads in a warp to access consecutive channels at the same timestep. - But different threads handle different chunks, so they access different timesteps. - This is the fundamental conflict: coalescing requires same timestep, different channels. But we want to process different chunks in parallel. SOLUTION: Process chunks sequentially within each warp, not in parallel. In Phase 1: - Warp 0 handles channels 0..31 for chunk 0. - Warp 1 handles channels 0..31 for chunk 1. - ... - Warp 15 handles channels 0..31 for chunk 15. - Each warp has 32 threads, each handling one channel. - A warp processes its chunk by iterating over 256 timesteps. - At each timestep, all 32 threads in the warp access consecutive channels. Coalesced! - After finishing chunk 0, warp 0 computes its aggregate and writes to shared memory. - Then it proceeds to chunk 1? No, we already assigned chunk 1 to warp 1. Wait, if warp 0 only handles chunk 0, what does it do while other warps handle chunks 1-15? It finishes chunk 0 and then sits idle until Phase 2. That's fine! The block has 16 warps. Each warp handles one chunk for all 32 channels. They all start together. They finish at roughly the same time (since each does the same amount of work). Then all warps participate in Phase 2. Phase 2: - We have 32 channels, each with 16 aggregates. - We need to scan the 16 aggregates for each channel. - We have 16 warps, each with 32 threads. - One approach: warp 0 handles channel 0, warp 1 handles channel 1, ..., warp 15 handles channel 15. But we have 32 channels, not 16. - So this doesn't work directly. Wait, we can have 2 blocks per batch element. Each block handles 16 channels. - Block 0: channels 0..15, all 16 chunks. - Block 1: channels 16..31, all 16 chunks. - Each block has 256 threads = 8 warps. - Total blocks: 32 batches * 2 = 64 blocks. Phase 1: - Block handles 16 channels, 16 chunks. - 8 warps. Each warp handles 2 chunks? Or 1 chunk but then we don't have enough warps for all channels. If 8 warps and 16 chunks, each warp handles 2 chunks. Warp 0: chunks 0 and 1, channels 0..15. Wait, a warp has 32 threads. If it handles 16 channels, that's 2 threads per channel. Not ideal. Alternatively, each block handles 8 channels, 16 chunks. Total blocks: 32 * 32 = 1024 blocks. Each block: 128 threads = 4 warps. - Warp 0: chunk 0, channels 0..7 (but warp has 32 threads, so 4 threads per channel? Or warp handles 32 channels?) This gets messy. Let me step back. The simplest correct and fast implementation is the 3-pass approach with: - Pass 1: 512 blocks of 256 threads - Pass 2: 32 blocks of 256 threads (or 256 blocks of 32 threads) - Pass 3: 512 blocks of 256 threads This avoids the complexity of the single-kernel multi-phase block. And 3 kernel launches are not that expensive. Let me verify Pass 2 more carefully. With 32 blocks of 256 threads: - Each block handles 1 batch, 256 channels. - Each thread handles 1 channel, scans 16 chunks sequentially. - Total blocks: 32. - Memory access: threads read `agg_a[b, c, d]` where `c` varies in the loop, `d` is fixed per thread. - For `c=0`, threads 0..31 read `agg_a[b, 0, 0..31]`. Coalesced! - For `c=1`, threads 0..31 read `agg_a[b, 1, 0..31]`. Coalesced! - Perfect. The only issue is 32 blocks. But Pass 2 is tiny work, so 32 blocks is fine. Actually, wait. With 32 blocks, 256 threads each, total threads = 8192. Each thread does 16 iterations. Total iterations = 131,072. Each iteration reads 2 floats and writes 1 float = 12 bytes. Total memory = 1.5MB. At ANY bandwidth, this takes ~1us-10us. The kernel launch overhead dominates. So Pass 2 being only 32 blocks is completely fine. Now, the main work is Pass 1 and Pass 3: - Pass 1: 512 blocks, 256 threads each. Total threads = 131,072. - Pass 3: same. - Each thread does 256 iterations. - Coalesced memory access. - Total memory traffic: Pass 1 reads 256MB, writes 1MB. Pass 3 reads 257MB, writes 128MB. Total 642MB. At 2 TB/s (achievable with good occupancy): 321us. At 3 TB/s: 214us. Plus kernel launch overhead: ~20us. Total time: 240-340us. For comparison, the simple 1-kernel approach: - 32 blocks, 256 threads. - Total memory: 384MB. - If bandwidth is limited to, say, 1 TB/s (due to occupancy): 384us. - If 1.5 TB/s: 256us. The 3-pass approach has more total memory traffic (642MB vs 384MB) but potentially higher bandwidth utilization. If it achieves 2.5 TB/s while the 1-kernel only achieves 1 TB/s, the 3-pass wins: 642/2.5 = 257us vs 384/1 = 384us. But there's also the overhead of 3 kernel launches. On H100, launch overhead might be ~5-10us per kernel, so 15-30us total. Negligible. I'll implement the 3-pass approach because: 1. It's still relatively simple. 2. It has much better parallelism (512 blocks vs 32 blocks). 3. It should achieve much higher memory bandwidth fraction. But actually, I want to test the 1-kernel simple approach first, because it's genuinely the simplest. If it already gives a huge speedup over PyTorch, maybe it's enough. Then I can optimize further if needed. Actually, looking at the GPU specifications again, H100 has 3.35 TB/s memory bandwidth. For 384MB of data, the absolute minimum time is 114us. If the single-kernel approach achieves, say, 50% of that (due to occupancy or other bottlenecks), it takes 228us. If the 3-pass achieves 80% of peak (because it has 512 blocks), it moves 642MB in 239us. They're very close! And the 3-pass has more overhead (multiple kernels, extra memory traffic). The single-kernel might actually be competitive or even better because it has less total traffic. I'll test the single-kernel first. Wait, but I keep going back and forth. Let me look at this from a different angle. What do state-of-the-art scan libraries do? CUB's DeviceScan handles this by: 1. If the array is large, it uses a two-level or three-level approach. 2. It assigns multiple blocks to the array. For our case, with 4096 elements per scan and 8192 scans, CUB would likely use: - A block per segment with multiple items per thread (exactly our simple approach!) - Or it might tile the segments. Actually, I think Nvidia's CUB uses a single kernel for segmented scan when the segment sizes are uniform and not too large? Or it might use a multi-pass approach. In any case, for our problem, the PyTorch reference is so slow that almost any CUDA kernel will beat it by a huge margin. The competition is about absolute speedup. Let me look at the actual kernel I'll write. I'll implement BOTH the simple 1-kernel version and the 3-pass version, and see which is faster. Wait, I need to write one Python file. Let me write the 3-pass version because I'm more confident it will perform well. Actually, no. Let me write the 1-kernel version first because it's simpler and test it with the correctness tool. Then test performance (via submit, or just check if it passes and hope for the best). I can iterate. Actually, the `submit_kernel` tool does timing measurement but doesn't tell me the reference time. So I'll know my kernel's absolute time but not the speedup ratio. I need to make it as fast as possible. Let me think about what might limit the single kernel: - 32 blocks. That's very few. - But each block does 4096 * 256 = 1,048,576 iterations. - With 256 threads, each thread does 4096 iterations. - The loop has a recurrence dependency on `h`. - The FMA latency is about 4 cycles. - The memory loads have ~400 cycle latency for L2 miss. Because of the `h` dependency, the FMA from iteration `i` must complete before iteration `i+1` can start. This means iterations execute sequentially. Within one thread, the throughput is 1 iteration per max(FMA latency, memory latency). If memory latency is 400 cycles, and we can hide it by overlapping loads from future iterations, then the throughput per thread might be closer to 1 iteration per FMA latency (4 cycles). But can we overlap loads from future iterations? The compiler can software pipeline the loop. With sufficient unrolling, loads for iterations i+1, i+2, etc., can be issued while waiting for the FMA from iteration i. However, the address calculation for each load depends on `t`, which is a loop counter. This is predictable, so the load unit can prefetch. But the actual value loaded has no dependency on previous loads (except that we need it for the FMA). So loads from iterations i and i+1 are independent and can be pipelined. If we issue loads for iterations i, i+1, i+2, i+3 before doing any FMAs, by the time we need them, they might be in registers or L1 cache. But the FMA chain is still the bottleneck if memory is resolved quickly. Actually, I realize that with coalesced access and the L2 cache, the loads might be quite fast. At t=0, we load a cache line containing 32 floats. These 32 floats are used by 32 threads in the warp. For t=1, we load the next cache line for the same 32 channels. Wait, a cache line is 128 bytes = 32 floats. For channels 0..31 at t=0, they are in one cache line. For channels 0..31 at t=1, they are `D * 4 = 1024` bytes away. So a different cache line. But since we're accessing them sequentially, and the cache line from t=0 was already loaded, the cache might prefetch t=1, t=2, etc. Actually, with a stride of 1024 bytes, adjacent timesteps for the same channel are 1024 bytes apart. The memory controller might detect this stride and prefetch. On H100, the L2 cache has stride-aware prefetching. A sequence of loads with constant stride might be prefetched efficiently! This could dramatically reduce effective memory latency. If the prefetcher works well, the loads appear as L2 hits with ~20-40 cycle latency. Then the FMA latency of 4 cycles becomes the bottleneck: 1 iteration every 4 cycles. With 256 threads per block, that's 256 iterations every 4 cycles = 64 iterations per cycle. Per SM: if running 1 block (256 threads), 64 iters/cycle. Total for 32 SMs: 2048 iters/cycle. Total iterations: 32 * 256 * 4096 = 33,554,432. Time: 33,554,432 / 2048 = 16,384 cycles. At 2GHz: 8.2 microseconds. Wait, that's impossibly fast. Let me recalculate. 33,554,432 iterations / 2048 iterations per cycle = 16,384 cycles. Yes. 16,384 cycles at 2GHz = 8.2us. Is that achievable? Only if the FMA is the bottleneck AND we can sustain 64 FMAs per cycle per SM AND all SMs are active. But wait, we only have 32 blocks. They run on 32 SMs. 32 SMs * 64 FMAs/cycle = 2048 FMAs/cycle. Total FMAs = 33,554,432. Time = 16,384 cycles = 8.2us. This assumes perfect FMA throughput and no memory bottleneck. If memory is actually the bottleneck, time = 384MB / bandwidth. At 3.35 TB/s = 114us. So the theoretical minimum is 8.2us (compute bound) or 114us (memory bound). The actual time will be whichever is larger, plus overheads. Can we actually achieve compute-bound behavior? For that, memory needs to be "free" (prefetched into cache so load latency < FMA latency). If L2 prefetching works perfectly, each load is an L2 hit with ~20 cycle latency. Since FMAs are only 4 cycles apart, and we can overlap loads from multiple iterations, it's possible that memory doesn't bottleneck the inner loop. But to hide 20-cycle load latency, we need ~5 independent FMA chains. We only have 1 per thread. But across 256 threads, we have 256 independent chains. The warp scheduler can execute FMAs from different threads to hide latency. Actually, with 256 threads = 8 warps, and 4 warp schedulers, each scheduler handles 2 warps. While one warp waits for FMA results, the scheduler executes another warp. But all FMAs take 4 cycles, and after issuing an FMA, a warp can't issue another for 4 cycles. Since each warp has 32 threads, and each thread issues 1 FMA per iteration, a warp takes 32 cycles to issue all FMAs? No, modern GPUs issue instructions per warp, with 32 threads executing in SIMD. A warp executes one instruction for all 32 threads simultaneously. So a warp issues one FMA instruction per cycle (covering all 32 threads). Wait, actually, each SM has multiple execution units. For FP32, there are enough units to execute 32 FMAs per cycle across a warp. So one warp can issue one FMA instruction per cycle, and it executes for all 32 threads. Then with 8 warps and 4 schedulers, each scheduler can issue one warp instruction per cycle. So we can issue 4 warp FMAs per cycle. But the SM throughput might be, say, 4 warp FMAs per cycle (matching the scheduler count). So 4 warps can execute FMAs every cycle, and the other 4 warps are waiting. After 4 cycles, the first warp's FMA is done, and it can issue the next one. So throughput: 4 warps * 32 FMAs = 128 FMAs per cycle. Wait, but the SM might only have enough FP32 units for 1 warp FMA per cycle or 2. This is getting complicated. I think a more practical approach is to just look at how fast this kernel could run based on similar problems. From experience, a kernel that loops over 4096 elements per thread with FMAs and coalesced loads, with 8192 threads total, typically runs in 0.2-0.5ms on an A100/H100 for ~30M elements. For example, a simple elemenwise kernel over 33M elements takes ~30us on H100. Our kernel has a loop with a dependency, so it's about 5-10x slower, maybe 150-300us. I think both the 1-kernel and 3-pass approaches will be in the 100-500us range. Given the PyTorch reference might take 20-100ms, either will give a 100x+ speedup. Let me just implement the 1-kernel approach first because of its simplicity. Then if there's time, I'll implement the 3-pass approach. Actually, I realize I should look at whether there is a way to use `torch.utils.cpp_extension.load_inline` effectively and whether there are any PyTorch-specific optimizations. Let me draft the complete Python file: ```python import torch from torch.utils.cpp_extension import load_inline cuda_source = ''' __global__ void scan_kernel(const float* __restrict__ a, const float* __restrict__ b, float* __restrict__ out, int B, int L, int D) { int b = blockIdx.x; int d = threadIdx.x; if (b >= B || d >= D) return; float h = 0.0f; for (int t = 0; t < L; t++) { int idx = b * L * D + t * D + d; float av = a[idx]; float bv = b[idx]; h = av * h + bv; out[idx] = h; } } ''' cpp_source = ''' torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b) { int B = a.size(0); int L = a.size(1); int D = a.size(2); auto out = torch::empty_like(a); scan_kernel<<<B, D>>>(a.data_ptr<float>(), b.data_ptr<float>(), out.data_ptr<float>(), B, L, D); return out; } ''' # Compile scan_cuda = load_inline( name="scan_cuda", cpp_sources=cpp_source, cuda_sources=cuda_source, functions=["scan_forward"], extra_cuda_cflags=["-O3"], verbose=False, ) class ModelNew(nn.Module): def __init__(self, d_model: int): super().__init__() self.d_model = d_model def forward(self, a: torch.Tensor, b: torch.Tensor) -> torch.Tensor: return scan_cuda.scan_forward(a, b) ``` Wait, does `a.data_ptr<float>()` work? Yes, if `a` is contiguous float tensor. The memory layout of PyTorch tensor `[B, L, D]` is row-major: the rightmost index `D` is contiguous. So `a[b, t, d]` maps to `base + (b * L * D + t * D + d)`. This is what I used. Correct. And `blockIdx.x = b`, `threadIdx.x = d`. At each timestep `t`, threads 0..255 access `t*D + 0..255`, which is contiguous. Coalesced! Actually, `blockDim.x = D = 256`. With `threadIdx.x` from 0 to 255, all threads in a block are in the same warp? No, 256 threads = 8 warps. Warp 0 has threads 0-31, warp 1 has 32-63, etc. Each warp accesses 32 consecutive channels. So 8 warps access 8 consecutive segments of 32 channels each. This is perfectly coalesced. One concern: `if (b >= B || d >= D) return;` — with `gridDim.x = B = 32` and `blockDim.x = D = 256`, neither condition can be true. So I'll remove it for slight overhead reduction. Also, I should use `const float* __restrict__` to help the compiler optimize. And `#pragma unroll 4` in the loop might help. Let me add it. ```cuda #pragma unroll 4 for (int t = 0; t < L; t++) { ... } ``` Actually, unrolling by 4 with 4096 iterations is fine. The compiler might do it automatically with `-O3`. Let me also consider using `float4` loads. Since consecutive threads access consecutive floats, we can have each thread load a `float4` and process 4 timesteps. But the recurrence `h = a * h + b` has a dependency across timesteps, so we can't process 4 timesteps in parallel within a thread. We'd still need a sequential loop, but each iteration would process 4 timesteps. This reduces loop overhead but the memory access pattern becomes `float4` loads. Since the data is contiguous and aligned (assuming D is a multiple of 4, and `t*D + d` might not be aligned for `float4` unless `d` is a multiple of 4). Actually, for thread `d`, it accesses `t*D + d`, and `d` varies from 0 to 255. So `float4` loads would require each thread to load 4 consecutive floats starting from its `d`. But `d` is different for each thread. Thread 0 loads `t*D + 0,1,2,3`. Thread 1 loads `t*D + 1,2,3,4`. These overlap! Not good for coalescing. If we have threads access in groups of 4: thread 0 handles d=0,1,2,3; thread 1 handles d=4,5,6,7; etc. Then `float4` loads are coalesced. But we'd need 64 threads per block instead of 256. Each thread handles 4 channels sequentially. Then we have 32 blocks * 64 threads = 2048 threads. Worse parallelism. Or thread 0 handles d=0..3 at t=0, then d=0..3 at t=1, etc. Same thing. Not better. So `float4` doesn't help much here because of the recurrence granularity. Let me just write the simple kernel and test. Actually, wait. I want to consider one more optimization: loop unrolling with manual software pipelining. ```cuda float h = 0; float av0 = a[base + 0 * D + d]; float bv0 = b[base + 0 * D + d]; for (int t = 0; t < L - 1; t++) { float av1 = a[base + (t + 1) * D + d]; float bv1 = b[base + (t + 1) * D + d]; h = av0 * h + bv0; out[base + t * D + d] = h; av0 = av1; bv0 = bv1; } h = av0 * h + bv0; out[base + (L - 1) * D + d] = h; ``` This might overlap the loads of iteration t+1 with the FMA/store of iteration t. The compiler might do this already, but explicit might help. Hmm, actually the compiler with `-O3` is very smart. Let me just write the simple version and let the compiler optimize. Another thought: can we use `bfloat16` or `float16`? No, the reference is fp32, and we need fp32 precision. OK, I'm going to write the kernel. Let me make sure the Python syntax for `load_inline` is correct. The key is: 1. `cuda_sources` contains the CUDA code with `__global__` function. 2. `cpp_sources` contains the C++ wrapper that calls the CUDA kernel. 3. `functions` lists the C++ wrapper function names. Also, I need to handle CUDA streams properly? `load_inline` binds the function to accept Tensors. The C++ code should probably use `a.data_ptr<float>()` and check that the tensor is contiguous. And for CUDA kernel launch, there's no stream argument, so it uses the default stream. That's fine. Wait, will `load_inline` work with a function that takes `torch::Tensor` and returns `torch::Tensor`? Yes. Let me check the exact signature: the C++ function should be callable from Python. Since we use `functions=["scan_forward"]`, `load_inline` generates a pybind11 binding for `scan_forward`. In Python, we call `scan_cuda.scan_forward(a, b)`. One issue: `scan_cuda` is a module. We need to make sure it's accessible from `ModelNew.forward`. We'll store it as a module-level variable. Let me write the complete code. Actually, I realize there might be a subtle issue. The reference implementation initializes `h = torch.zeros(B, D)` and then iterates. Our kernel initializes `h = 0.0f` per thread. That's correct because at t=0, `h = a[:,0] * 0 + b[:,0] = b[:,0]`, which matches the reference. For t > 0, `h = a[:,t] * h + b[:,t]`. Our kernel does the same. So correctness should hold. One more thing: floating point accumulation. The reference does the operations in a specific order. Our kernel does the same order (t=0,1,2,...,L-1). And within each timestep, all channels are independent. So there should be no numerical difference. However, the parallel scan algorithms might have slightly different accumulation order due to associativity. Since our 1-kernel sequential approach preserves the exact same order, it will produce identical results (bit-exact). The 3-pass approach might have slightly different rounding because the prefix is applied differently: `actual_h = local_a * prefix + local_h` vs `h = a * h + b` sequentially. With floating point, these are NOT exactly the same because of rounding. For example, if prefix = 0.1, and local_a = 0.2, local_h = 0.3: - Actual sequential: h = 0.2 * 0.1 + 0.3 = 0.32 (with rounding) - Prefix formula: 0.2 * 0.1 = 0.020000001, + 0.3 = 0.32000002 Wait, floating point multiplication and addition are not associative. The 3-pass approach might give slightly different results than the exact sequential scan. The tolerance is 1e-4 for fp32, so it should be fine. But the 1-kernel sequential approach is guaranteed to be exact! This is a strong reason to prefer the 1-kernel sequential approach: exact numerical match. For the 3-pass approach, we'd need to check if the error is within 1e-4. Since we're doing ~4096 operations, error accumulation might be small but let's be safe. The exact match approach is safer. So I'll implement the 1-kernel sequential approach. Even though it only uses 32 blocks, I'll see how it performs. If it's not fast enough, I can optimize. Wait, actually, can I increase the number of blocks without changing the algorithm? What if I set `gridDim.x = B * 2` and each block handles D/2 channels? Then: - 64 blocks, 128 threads each. - Threads 0..127 handle channels 0..127 if block is even, 128..255 if block is odd. - Each thread's `d` is `blockIdx.y * 128 + threadIdx.x` or something. Let me define: ```cuda int b = blockIdx.x / 2; int d_group = blockIdx.x % 2; int d = d_group * 128 + threadIdx.x; ``` With `gridDim.x = B * 2 = 64`, `blockDim.x = 128`. - Block 0: batch 0, channels 0..127 - Block 1: batch 0, channels 128..255 - Block 2: batch 1, channels 0..127 - ... - Total: 64 blocks. Threads per SM: 64 * 128 / 132 ≈ 62. Same total thread count, more blocks, fewer threads per block. Not necessarily better for throughput, but better block distribution. What about `gridDim.x = B * 4 = 128`, `blockDim.x = 64`? - 128 blocks, 64 threads each. - Total threads: 8192. - At each load, 64 threads access 64 consecutive floats = 256 bytes = 2 cache lines. - Still coalesced. - 128 blocks on 132 SMs. Almost 1 block per SM. Better distribution! What about `gridDim.x = B * 8 = 256`, `blockDim.x = 32`? - 256 blocks, 32 threads each (1 warp). - Total threads: 8192. - At each load, 32 threads access 32 floats = 128 bytes = 1 cache line. - Perfect coalescing. - 256 blocks on 132 SMs. ~2 blocks per SM. - Each SM runs 2 warps concurrently. - Can 2 warps per SM hide latency? Not as well as 8 warps, but the memory system is shared. I think 64 blocks of 128 threads or 128 blocks of 64 threads might be a good middle ground. What about 256 blocks of 32 threads? With only 1 warp per block, there are no opportunities for intra-block warp scheduling. But the GPU can schedule other blocks on the same SM. With 2 blocks per SM, there are 2 warps per SM. The key question is whether 2 warps per SM can generate enough concurrent memory requests. Each warp can have ~64 outstanding loads. With 2 warps, that's 128 outstanding requests per SM. At 32 SMs (64 warps active), total outstanding = 4096. The memory system can handle much more. Actually, H100 has 132 SMs. With 256 blocks, most SMs get 2 blocks (2 warps). Some SMs get 1 block. Total warps active at once = ~250 warps (since 256 total, minus a few waiting). Each SM has ~2 warps. Total outstanding memory requests ≈ 2 * 64 * 132 = 16,896. This might be enough to saturate L2. But the issue is not just outstanding requests; it's also the rate at which new requests are issued. If a warp only issues a load every ~10 cycles (due to loop body), 2 warps issue loads every 5 cycles. The memory subsystem wants more frequent requests. With 8 warps per SM (from larger blocks), loads are issued every ~1.25 cycles. Better. So I think larger blocks are better, even if there are fewer of them. 32 blocks of 256 threads gives 8 warps per active SM. 64 blocks of 128 threads gives 4 warps per active SM (assuming each SM runs 1 block). 128 blocks of 64 threads gives 2 warps. 256 blocks of 32 threads gives 1 warp. I think 32 blocks of 256 threads or 64 blocks of 128 threads are the best options. I'll try 32 blocks first, then 64 if needed. Actually, there's one more idea: can I use block size `D=256` but launch more blocks by having each block handle a RANGE of timesteps and loop over batches? No, that changes the structure. Or launch `B * num_tiles` blocks where each block handles 256 channels and a subset of timesteps, but then I'd need a prefix pass. That's the 3-pass approach. Let me implement the 1-kernel approach with `blockDim.x = 256` and `gridDim.x = B = 32`. I'll add `#pragma unroll 4` and use `__restrict__`. Actually, I want to make it more flexible so I can easily tune block size. I can pass block dimensions as template parameters or just check at runtime. But for simplicity, I'll use `blockDim.x` dynamically. Wait, `D=256` is fixed by the problem, but the kernel should work for any `d_model`. So using `threadIdx.x` with a check `if (d < D)` is good practice, but since `D=256` and we use `blockDim.x <= 256`, the check might not be needed if we handle padding. For simplicity, I'll use: ```cuda int b = blockIdx.x; int d = threadIdx.x; if (d >= D) return; ``` And launch with `blockDim.x = ((D + 31) / 32) * 32` to round up to warp size. For D=256, that's 256. OK, let me write this and test. I'll do compile, then correctness. ```python import torch import torch.nn as nn from torch.utils.cpp_extension import load_inline cuda_source = """ __global__ void scan_kernel(const float* __restrict__ a, const float* __restrict__ b, float* __restrict__ out, int B, int L, int D) { int b = blockIdx.x; int d = threadIdx.x; if (b >= B || d >= D) return; float h = 0.0f; int base = b * L * D + d; #pragma unroll 4 for (int t = 0; t < L; t++) { int idx = base + t * D; float av = a[idx]; float bv = b[idx]; h = av * h + bv; out[idx] = h; } } """ cpp_source = """ torch::Tensor scan_forward(torch::Tensor a, torch::Tensor b) { int B = a.size(0); int L = a.size(1); int D = a.size(2); auto out = torch::empty_like(a); const int threads = D; // 256 scan_kernel<<<B, threads>>>( a.data_ptr<float>(), b.data_ptr<float>(), out.data_ptr<float>(), B, L, D ); return out; } """ scan_cuda = load_inline( name="scan_cuda", cpp_sources=cpp_source, cuda_sources=cuda_source, functions=["scan_forward"], extra_cuda_cflags=["-O3", "--use_fast_math"], verbose=False, ) class ModelNew(nn.Module): def __init__(self, d_model: int): super().__init__() self.d_model = d_model def forward(self, a: torch.Tensor, b: torch.Tensor) -> torch.Tensor: return scan_cuda.scan_forward(a, b) ``` Wait, I need to make sure the tensors are contiguous. I'll add a check or make them contiguous. Also, I should check that `a` and `b` are CUDA float tensors. Let me refine: ```python def forward(self, a: torch.Tensor, b: torch.Tensor) -> torch.Tensor: a = a.contiguous() b = b.contiguous() return scan_cuda.scan_forward(a, b) ``` Let me submit this for compilation and correctness testing.

It seems the cpp_source and cuda_source are being compiled separately and the CUDA kernel isn't visible from the C++ code. I need to put both in the same file, or use a single source. The `load_inline` function typically combines them. Let me check the proper way to do this. For `load_inline`, the `cpp_sources` should contain the host code, and `cuda_sources` contains the device code. But in order for the C++ compiler to see the kernel declaration, we need to provide it with the `extern "C"` declaration or a forward declaration. Actually, looking at the PyTorch docs, typically you put everything in a single string via `cpp_sources + cuda_sources`. The issue here is that the C++ compiler sees the .cpp file and doesn't know about `scan_kernel`. I should add a forward declaration for it in the cpp source, or include the kernel in a header that's included in the cpp source. Let me add an `extern "C"` forward declaration. Actually, for CUDA kernels, they have C++ linkage. I can declare them in the C++ source.

The basic sequential kernel passed correctness. Now let me optimize it. The current kernel launches B blocks with D threads each, and each thread loops over L sequentially. This is memory-bound since each thread reads memory coalesced for D strides but not optimal for L. Let me think about better approaches: 1. **Memory Coalescing**: Currently, each thread accesses memory with stride D between timesteps. This is bad because neighboring threads access neighboring elements at the same time, but within a thread the accesses are strided by D*L or D. Actually, let me recalculate: - Thread d accesses idx = batch*L*D + t*D + d - At time t, threads d and d+1 access adjacent memory → coalesced! - But for a single thread, consecutive accesses are D elements apart → not cache friendly 2. **Better approach - Work per thread**: Each thread can process multiple D elements. Currently D=256 threads per block, which is fine. But we could process with vectorized loads (float4) to improve memory bandwidth. 3. **Parallel scan using shared memory**: We could use a parallel associative scan algorithm within a block. But for L=4096 and D=256, a basic sequential scan per thread might be close to memory bound anyway. Actually, let me think more carefully. The kernel reads 2 inputs and writes 1 output per element. Total memory traffic = 3 * B * L * D * 4 bytes = 3 * 32 * 4096 * 256 * 4 = ~402 MB per forward pass. At 3.35 TB/s, this would take ~120 microseconds just for memory. The reference PyTorch implementation does Python-level loops, so it's much slower. But can we do better? Let's think about a more parallel approach using the associative scan to exploit more parallelism. Since B=32 and D=256, we have 8192 independent scans of length 4096. With 8192 threads and we want to parallelize each scan too. For a parallel scan of length 4096, we could use a block-level scan with ~512 threads, each processing 8 elements. This would require shared memory for the associative operation. The associative operation is: given (a1, b1) and (a2, b2), the combined state is (a2*a1, a2*b1 + b2). Let me write a more optimized kernel using warp-level or block-level parallel scan. But actually, for L=4096, a sequential loop per thread is only 4096 iterations. With 8192 threads, that's 33M operations total. At 67 TFLOPS, that's 0.5 microseconds compute bound. So this is definitely memory bound. The main optimization would be: 1. Use float4 vectorized loads to maximize memory bandwidth 2. Better use of shared memory / L2 cache 3. Larger tiles/ blocks Let me try a float4 vectorized version first. We can process 4 elements per thread. Actually, let me also consider that we can transpose the memory layout: if we store data as (B, D, L) instead of (B, L, D), then each thread accesses consecutive memory. But the input comes as (B, L, D), so we'd need to transpose first. Alternatively, we can have each thread process a contiguous chunk of L, but loads from (B, L, D) with stride D would be strided. Not good. Wait - if we process with each warp handling a block of D and a chunk of L, we could use shared memory to transpose reads. Actually, a simpler approach: since D=256, we have exactly 256 threads = 8 warps. If we tile the L dimension and have each warp process a tile: For a given batch and d, the memory accesses are at: batch*L*D + t*D + d for t = 0, 1, 2, ... If D=256, then consecutive t accesses are 256 floats = 1024 bytes apart. That's exactly the size of a cache line. So a single thread's accesses skip cache lines, but they're perfectly coalesced within a warp at each timestep. Actually let me reconsider: at timestep t, warp accesses addresses: batch*L*D + t*D + [0..31] (for warp 0) These are consecutive floats = 128 bytes = 2 cache lines per warp. Then at timestep t+1, warp accesses: batch*L*D + (t+1)*D + [0..31] These are 256 floats = 1024 bytes after the previous accesses. So each warp reads 2 cache lines per timestep, then jumps 1024 bytes. This means we only utilize 2/64 of each cache line for scalar data in L1, but L2 should be fine since it's sequential at the warp level across time. A better approach would be to have each thread process multiple D elements, or process L in chunks where threads collaborate to load a tile into shared memory. Let me try a tiled approach: - Each block processes one batch and a tile of D (e.g., D=256 all of them) - For L, we process in chunks of say 128 or 256 - Within each chunk, we can use shared memory for the scan - But the scan is inherently sequential in the dependency chain, so parallel scan doesn't help the critical path. Actually for a linear recurrence h_t = a_t * h_{t-1} + b_t, the operation is sequential. Within a chunk, we can't parallelize the scan unless we use the associative scan algorithm, but that's only worthwhile if individual elements are expensive to compute. Let me try vectorized memory access with float4. Each thread can process 4 consecutive D elements... wait, actually the data is in (B, L, D) layout. If thread d processes element d, then to process 4 elements, thread d would process d, d+1, d+2, d+3. But these are adjacent in memory. Wait, if a thread processes elements d, d+1, d+2, d+3 at the same t, then for each t it reads a float4. Consecutive t for the same (d,d+1,d+2,d+3) are 1024 bytes apart. Alternatively, if we process 4 timesteps per thread, we'd read at d, d+D, d+2D, d+3D. These are 256 floats apart and NOT contiguous. So float4 wouldn't help there. Actually let me think about transposition. If we transpose the input from (B, L, D) to (B, D, L), then each thread processing a single d would access consecutive memory. Let me try this: 1. Transpose a and b from (B, L, D) to (B, D, L) 2. Each thread scans its L elements sequentially with perfect memory coalescing 3. Optionally transpose back, or keep in (B, D, L) and transpose at the end The transpose step can be done efficiently with shared memory. Let's try this approach. Actually, wait. Let me reconsider the original kernel. The memory layout is (B, L, D). Thread d accesses: base + 0*D + d, base + 1*D + d, base + 2*D + d, ... This is strided by D=256 floats = 1024 bytes. On H100, L2 cache lines are likely 128 bytes. So each access brings in 128 bytes but we only use 4 bytes. That's poor L2 utilization. But L2 still has good spatial locality because the warp accesses 128 consecutive bytes. So L2 prefetches might help across warps and timesteps. Actually, on Hopper/H100, the L2 cache is 50MB. Our working set is B*L*D*4*2 = 256MB (for both a and b). This is larger than L2, so we'll be memory bound to HBM. The raw memory coalescing for the warp is good (consecutive addresses), so we should get good HBM bandwidth utilization. The poor spatial locality within a thread doesn't matter that much for HBM since warp-level coalescing determines throughput. Let me check if float4 can help by having each thread process 4 adjacent D elements. Thread d would process elements d, d+1, d+2, d+3. For each t, it reads a float4 starting at base + t*D + d. The address is aligned if d is a multiple of 4. With D=256, we have 64 float4s. We launch 64 threads per block, each processing 4 D channels. Wait, but then we'd need 64 blocks per batch instead of 1. Let me think about occupancy. 64 threads per block might be too few. Actually, let me try a different approach: process 4 timesteps per float4 read. Since the data for consecutive t is D floats apart, we can't read it as float4 directly. But we can use shared memory to transpose tiles. Let me design a tiled kernel: - Tile size: D=256 by T=... say T=32 or T=64 - Each block processes one batch - Load a T×D tile from global memory into shared memory in a transposed manner - Then each thread can scan its D dimension locally across multiple tiles - Wait, but the recurrence spans across tiles. Hmm, the cross-tile dependency means thread d needs the final h value from tile [0:T-1] to start tile [T:2T-1]. So within a block, threads could do a parallel scan across the tile boundaries, but that requires communication. Let me try a simpler optimization: just use the vectorized float4 approach for the original layout. Actually, let me benchmark the current kernel more carefully. The reference PyTorch implementation uses a Python loop, which is extremely slow. Our kernel might already be 100x+ faster. But to maximize speed, we need to think about memory bandwidth. A key optimization: since we're memory-bound, the important thing is to saturate HBM bandwidth. Let me try different thread configurations and see. Actually, let me try processing multiple D elements per thread with float4 but keeping 256 threads. If thread d processes elements d, d+64, d+128, d+192 (splitting D=256 into 64 groups of 4), that doesn't help with memory. Let me try: each block handles one batch and all D=256. We have 256 threads. If we unroll the loop more aggressively and use LDG (load global) with good pipelining, we might get better performance. Given that D=256, and we want to use as many SMs as possible: B=32 batches → 32 blocks. H100 has 132 SMs (I think H100 has 132 or 144 SMs for SXM5). So 32 blocks means we only use ~32 SMs. We'd want more parallelism. We could split D into multiple blocks per batch. For example, 4 blocks per batch, each handling D/4 = 64 channels. Then we'd have 32*4 = 128 blocks, which is much better SM utilization. Let me try this approach: each block has 64 threads (one warp), processes all L for 64 D channels. This gives us 128 blocks total, much better parallelism. Actually wait, 32 batches × 4 blocks/batch = 128 blocks. And with 64 threads per block, each warp processes 64 D channels across L timesteps. The memory accesses within a warp are still consecutive (channels d to d+63 at each t), so coalescing is good. But we'd need many more blocks to saturate H100. Let me try 8 or 16 blocks per batch: - 16 blocks/batch × 32 batches = 512 blocks - Each block: 16 threads (half warp), processes 16 D channels × L timesteps With 512 blocks, we get much better SM utilization. However, 16 threads per block is below the minimum recommended. Let me try a configuration. Actually, the simplest good configuration: - Blocks: B × (D/32) = 32 × 8 = 256 blocks - Threads per block: 32 (one warp) - Each warp processes 32 D channels × L timesteps 256 blocks × 32 threads = 8192 threads total, which is exactly the number of independent recurrences. This maximizes parallelism. Let me try this. Memory access pattern per warp: - Thread i in warp processes channel d+i for all timesteps - At each t, warp reads 32 consecutive floats = coalesced 128 bytes This is pretty good. Let me implement this. Actually, I realize I should also consider using larger blocks with fewer blocks total. The issue is occupancy. On H100, each SM can have up to 32 blocks. With 132 SMs, we can have up to 4224 blocks resident. So 256 blocks is fine, but each block only has 1 warp. The occupancy might be limited by registers or other resources. Actually let me try another approach. Let me launch more fine-grained parallelism by also parallelizing over L. We can split the scan into chunks, compute prefix sums within chunks in parallel, then resolve cross-chunk dependencies. For a scan with associative operation, we can do: 1. Split each (batch, d) scan of length L into chunks of size C 2. Each thread block processes one chunk, computing local prefix sums 3. Then compute the "carry" across chunks using another kernel or within the same kernel For the operation (a, b) ∘ (a', b') = (a'*a, a'*b + b'): - The local scan in each chunk gives (a_local, b_local) at each position - The cumulative effect of a chunk is: a_chunk = product of all a in the chunk, b_chunk = the final h value if starting from h=0 - Then we propagate the chunk carries across chunks This is the classic two-level scan. It allows parallel processing of chunks. For L=4096 and C=256, we'd have 16 chunks per (batch, d). With B=32, D=256, that's 8192 scans × 16 chunks = 131072 chunks. We can process all chunks in parallel! Let me implement this. This will be much faster because we parallelize over both the scan dimension and the time chunks. Actually, there's an even simpler approach: since B=32, D=256 gives 8192 independent scans, and L=4096, we can just have one thread per (batch, d) and do a sequential scan. With 8192 threads and 132 SMs, each SM gets ~62 threads. If each thread does 4096 iterations, the total work per SM is reasonable. But the difference is: with chunk-based parallel scan, we can launch 8192 × 16 = 131072 threads, which fully saturates the GPU. However, we then need a second step to combine the chunk results, and a third step to apply the carries. Let me think about whether the overhead is worth it. For L=4096, each thread does 4096 iterations in the sequential approach. At memory-bound speeds, this might be ~200-400 microseconds. The sequential kernel passed correctness, so it should already be much faster than PyTorch. Let me try to optimize the sequential kernel first with better parallelism configuration. Actually, you know what, let me try a vectorized load approach first, then if needed move to parallel scan.

The compilation passed. Let's run correctness check.