reasoning
Good, the kernel is correct. Now I need to optimize it. Let me think about the current kernel:
1. Each thread handles one (b, h, j, i) but i is threadIdx.x and j is blockIdx.z
2. There are B*H*D blocks, each with D threads
3. Each block processes all T timesteps
4. Shared memory is used for reduction across D threads per block
5. The state S is replicated per thread - but actually each thread only computes one (i,j) pair, so we need D*D states per (b,h)
Wait, the state is a D×D matrix. My kernel currently only computes one entry per thread (i,j) but each thread loops over T. For each (b,h), we have D×D threads (organized as D blocks in z-dim × D threads in x-dim), which is actually correct since there are D×D entries.
But the issue is performance:
- Each block processes T timesteps
- Within a block, all threads process the same (b, h, j) but different i
- At each timestep, we need __syncthreads() for the reduction
- The reduction is done by thread 0 in a loop over shared memory
This is very suboptimal because:
1. Only 24 threads per block (head_dim = 24), which is less than a warp
2. Using D shared memory for reduction with only D threads is wasteful
3. The outer product computation is not vectorized
4. Gate and beta are loaded redundantly
5. k and v are loaded from global memory every timestep
Let me think about how to optimize this:
Option 1: Use one thread per (b,h,j), let each thread compute the full inner product over i
- Then each thread needs to maintain D different S values (one for each i)
- At each timestep: S[i] = gate * S[i] + beta * k[t,i] * v[t,j]
- Then output[j] = sum_i q[t,i] * S[i]
- This is nice because each thread computes one output j independently
- No need for __syncthreads() or shared memory
- But each thread has a loop over D for S storage and computation
- We have B*H*D threads, and each thread does O(D) work per timestep
For D=24, this might be fine! Each thread would need an array of 24 floats in registers, and for each timestep it would update all 24 and then sum.
Actually, let me think more carefully. With B=2, H=4, D=24, we have 2*4*24 = 192 threads total. That's very few! We need to increase parallelism.
Wait, each block on Hopper can have up to 1024 threads, and there are many SMs. We should be launching way more threads.
Better approach: Process multiple timesteps in parallel? Or process multiple j's per block?
Actually, let me reconsider. The state S[t] depends on S[t-1], so we can't parallelize over t easily. But we can parallelize over (b, h, j) very well. Currently we have B*H*D = 192 blocks, which is very few for a modern GPU with many SMs.
A better approach might be:
- Assign each block to handle one (b, h)
- Within the block, use all threads to compute different j values
- For D=24, we could use 24 warps, each handling one j
- But that's overkill
Actually, for small D (24), the most efficient approach might be to have each thread handle multiple j values, or use warp-level parallelism.
Let me think about what the reference implementation actually does mathematically:
For each t:
S = gate_t * S + beta_t * outer(k_t, v_t)
y_t = q_t @ S (where @ is matrix multiply)
Wait, `y_t = (q[:, t].unsqueeze(-1) * state).sum(dim=-2)`:
- q[:, t] has shape (B, H, D)
- state has shape (B, H, D, D)
- q.unsqueeze(-1) has shape (B, H, D, 1)
- Multiply: (B, H, D, D) — broadcast over last dim
- sum over dim=-2: sum over D dimension → (B, H, D)
So y_t[b, h, j] = sum_i q[b, h, t, i] * state[b, h, i, j]
And state[b, h, i, j] at time t is:
state_t = gate_t * state_{t-1} + beta_t * k[b,h,t,i] * v[b,h,t,j]
This is exactly what my kernel computes.
Now for optimization:
1. Currently we have too few blocks: B*H*D = 192. For a Hopper H100 with 132 SMs, we want many more.
2. We can process multiple timesteps with fewer blocks, or use more threads per block.
3. Since D is small (24), we can have each block handle multiple (b, h) pairs, or each block handle one (b, h) and all D j's with more threads.
Let me try: each block handles one (b, h). Threads within block handle different j values. If we use 256 threads, each thread handles j = threadIdx.x % D and some other assignment... Or better: we can use 24 warps (768 threads), each warp handling one j. Within each warp, threads could handle nothing (just one thread per j) but that's wasteful.
Actually, for D=24, a cleaner approach:
- Each block: one (b, h)
- Each warp: one j (32 threads per warp, but only 24 are active... or we can have each thread handle one j, and each block has 24 threads)
- But then we're back to small blocks
Better: Have each thread handle one j, but pack many j's per block. With 256 threads and D=24, we can have 256/24 ≈ 10 j's per block with some threads idle, or just use blocks of 32 threads (1 warp) per j, with 24 warps = 768 threads per block.
Actually, the key insight is that for each j, we can compute independently if each thread maintains D states. The update for state[i,j] doesn't depend on other j values. So yes, each j can be handled independently.
Let me try a different approach: use one CUDA block per (b, h, j) as before, but increase block size to use full warps. Actually, for D=24, we need to involve all 32 threads of a warp. We can have thread i compute for... well, there are only 24 i's. So threads 0-23 compute, threads 24-31 are idle. That's not terrible.
Actually, maybe the best approach is to transpose the computation: instead of having threads parallel over i (the sum dimension), have threads parallel over j (the output dimension) and let each thread/vector of threads compute the full inner product.
For D=24, let's have each block = one (b, h), with 24 threads. Each thread handles one j and loops over T. Within each thread, it maintains an array of 24 states (one per i) in registers, and for each t:
1. Load gate, beta, v_j
2. For each i in 0..23: S[i] = gate * S[i] + beta * k_i * v_j
3. Compute y_j = sum_i q_i * S[i]
4. Store y_j
This removes shared memory and syncs entirely! Each thread works independently. But 24 threads per block is still small.
To increase occupancy:
- Each block can handle multiple j values. Actually, a block can have up to 1024 threads. With D=24, we can have 24 threads per j, so 1024/24 = 42 j's... but we only have 24 j's per (b, h).
- So we could handle multiple (b, h) pairs per block! Or assign each block to multiple (b, h, j) combinations.
Actually, let's think about register pressure. If each thread maintains 24 floats = 96 bytes in registers, and we have 1024 threads per block, that's 96KB of registers per block. Each SM has 256KB of registers, so we could fit 2 blocks per SM, which is decent.
But actually, we don't need 24 threads per j. We just need 1 thread per j. If each thread handles 1 j and maintains 24 states, then for one (b, h) we need 24 threads. We can pack 42 (b, h) pairs per block of 1024 threads? No wait, 1024 threads / 24 threads_per_bh = 42.6, so we'd need to handle fractional.
Actually, a simpler approach: one thread handles one (b, h, j), so we have B*H*D threads total. With B=2, H=4, D=24, that's 192 threads. That's very few. To increase parallelism, each thread can handle multiple (b, h, j) combinations? Or we can assign multiple timesteps to different threads using some kind of parallel scan? But the recurrence makes that hard.
Wait, actually for different (b, h) pairs, they are completely independent. So we DO have B*H = 8 independent computations. Each computation involves a D×D state updated over T steps. For each (b, h), there are D output dimensions j that are also somewhat independent (each j only needs the j-th column of the state).
Hmm, with so little parallelism (B*H*D = 192), the GPU is severely underutilized. In such cases, the key is to make each thread do more work efficiently, so that we saturate memory bandwidth.
Actually, maybe I should look at it differently. Each thread can process a "tile" of j values or a tile of timesteps.
For j tiling: a thread handles j_start to j_end, maintaining a (D, tile_j) state. This increases register use but keeps parallelism the same.
Hmm, but what if we vectorize the inner loop? For each j, each thread already does D=24 operations per timestep. If we process 4 j's per thread, that's 96 ops per timestep, which might better utilize the ALUs.
Wait, but even with that, we'd only have 192/4 = 48 threads. Still tiny.
Another idea: process multiple timesteps in a single thread, but "batch" them. For example, a thread could process t=0..T-1 for one j, while another thread processes the same j but... no, the recurrence prevents that.
Let me reconsider: maybe a better strategy is to use one warp (32 threads) per j, and have the 32 threads compute the D=24 values with some threads idle. Or even better, have 24 threads per j, with 24 warps = 576 threads per block, handling all j's for one (b, h). But with B*H = 8 blocks total, that's still only 8 blocks. On Hopper, you want at least 100+ blocks to saturate all SMs.
So the solution is: each block should handle multiple (b,h) pairs, OR we should have many more blocks.
If we have 8 (b,h) pairs and each needs 24 j's, we can launch 8*24 = 192 blocks, each with 24 threads. That's just not enough.
Wait, we can increase blocks by having each block handle only part of the T timesteps! Like, block (b, h, j, chunk) handles timesteps t_start to t_end for (b, h, j). But the state S depends on previous timesteps, so we'd need a scan/reduce to combine chunks.
Actually, a better idea: sequential scan using parallel associative operators? The state update is:
S_t = g_t * S_{t-1} + b_t * k_t ⊗ v_t
This is a linear recurrence. In theory, we can use parallel scan for linear recurrences, but the "state" here is D×D which is 576 floats. Transferring that between threads for a parallel scan would be extremely expensive.
Given the constraints, maybe the best approach is:
1. Have each thread handle one (b, h, j)
2. Have each block handle 32 or 64 such threads (so multiple j's, possibly from different (b,h))
3. Optimize the inner loop with vectorized loads/stores and loop unrolling
For D=24, since 24 is small and doesn't divide 32 evenly, we could:
- Use 32 threads per block, with 8 threads each handling 1 j, giving 8 j's per block. But that wastes 24 threads.
- Or use 24 threads per block, handling 24 j's from different (b,h) pairs.
Actually, the cleanest way is:
- Launch B*H*D blocks (192 blocks)
- Each block has 1 thread? No, that's terrible.
- Each block has D threads (24 threads), processing D different i values for ONE j
- So each block handles one (b, h, j), with threads 0..23 each computing for a different i
- They need __syncthreads() for reduction
This is what I have. The issue is 192 blocks total. But actually, wait! The problem says `seq_len = 128`, `batch_size = 2`, `num_heads = 4`, `head_dim = 24`. But during evaluation, these sizes might vary. Or maybe not - I should check if there are hidden sizes.
Actually, the problem benchmark usually tests with the specified sizes. With B=2, H=4, T=128, D=24, we have only 192 blocks. That's very few, but if each block does a lot of work (T=128 iterations, each doing a few flops), maybe it's OK.
Let me estimate: Each block handles (b, h, j) and has 24 threads. Each thread loops over T=128 and does maybe 10 FLOPs per iteration. Total FLOPs = 192 * 128 * 24 * 10 = ~6M FLOPs. That's tiny. At 100 TFLOPS, this should take 60 nanoseconds. Memory bandwidth is ~3 TB/s, and we're reading q, k, v, gate, beta and writing out. Let's see:
For each timestep, per (b, h):
- We read: q (D), k (D), v (D), gate (1), beta (1) = 3D+2 floats
- And for each j, we need to read the same q, k, gate, beta, but v is the same for all i... wait no.
Actually in the current kernel:
- Each block is (b, h, j) with D threads
- At each t, thread i reads: q_i, k_i, gate, beta (all from global memory, coalesced for k and q since adjacent threads read adjacent addresses)
- Thread i also reads v_j (same for all threads in block, but each block has different j)
- Then computes and writes to shared memory
- Finally thread 0 writes out_j
So per block per timestep:
- Global memory reads: D q's + D k's + 1 gate + 1 beta + 1 v + (D shared mem reads, not global)
- Global memory writes: 1 out
- Total global memory: 2D + 3 reads + 1 write = 51 floats = 204 bytes per block per timestep
- For 192 blocks × 128 timesteps = 24,576 block-timesteps
- Total memory: 24,576 × 204 bytes = ~5 MB
At 3 TB/s, reading 5 MB takes ~1.7 microseconds. Plus compute. The reference is probably much slower because it's a Python loop over T=128.
Actually, my kernel might already be faster than the reference just because the reference is a Python for-loop. Let me check if there are any timing results from the correctness check... no, correctness doesn't show timing.
Let me try to optimize the kernel anyway:
1. **Remove shared memory and __syncthreads()**: Use one thread per (b, h, j), each maintaining D states in registers, and accumulate the output directly. This eliminates sync overhead.
2. **Improve parallelism**: Since we'd only have 192 threads, we should use as many threads per block as possible. If each thread handles one (b, h, j), and a block can have 1024 threads, we can pack multiple (b, h, j) into one block. For example, a block of 192 threads could handle one full (b, h, all j), but 192 doesn't align well with warp size.
3. **Even better**: Since we only have 192 total work units, let's split them into larger blocks. Maybe 64 threads per block handling 64 different (b, h, j) units, giving 3 blocks total. That's even worse.
Actually, to get good occupancy, we want many blocks. 192 blocks is actually not terrible if each block can run fast and we can get multiple blocks per SM. With 132 SMs, that's ~1.45 blocks per SM. But each block only has 24 threads, so we can fit many blocks per SM. On Hopper, max threads per SM is 2048, so we could fit 85 blocks per SM! The issue is each block needs registers and the total work might be too small.
Wait, if we have 192 blocks total, and each SM can run 85 of them concurrently (in terms of thread slots), then we'd need only 3 SMs to hold all blocks. The remaining 129 SMs would be idle. That's terrible utilization.
To fix this, we need to process T in parallel, or use a different parallelization strategy.
Actually, I just realized: for each j, the computation is independent, but for each (b, h, j), the T timesteps are sequential. With only 192 parallel units, we're severely limited.
But wait - the reference is a pure Python loop, which is extremely slow. Even with terrible GPU utilization, the CUDA kernel might be orders of magnitude faster because:
1. Python loop overhead is huge
2. PyTorch creates many intermediate tensors per iteration
3. Each tensor operation has kernel launch overhead
So my kernel with 192 blocks could easily be 100x faster than the reference. Let me verify by running correctness and then submitting. But first, let me try to optimize the kernel to use per-thread accumulation instead of shared memory, which could be faster for this small D case.
Actually, let me try a much cleaner approach:
```cuda
// One thread per (b, h, j)
// Thread maintains state[0..D-1] in registers
// For each t:
// g = gate[b,h,t]
// b = beta[b,h,t]
// vj = v[b,h,t,j]
// for i = 0..D-1:
// state[i] = g * state[i] + b * k[b,h,t,i] * vj
// sum = 0
// for i = 0..D-1:
// sum += q[b,h,t,i] * state[i]
// out[b,h,t,j] = sum
```
For D=24, each thread has 24 registers for state. This is totally fine. And we eliminate all shared memory and syncs.
Now for block/work distribution: we have B*H*D = 192 threads of work. To get good occupancy, we might want to process this as:
- 6 blocks of 32 threads each = 192 threads total
- Or 3 blocks of 64 threads
- Or 1 block of 192 threads
- Or 192 blocks of 1 thread... but that's bad
Actually, 192 blocks of 1 thread is terrible. But 6 blocks of 32 threads = 6 blocks, which is also tiny.
Hmm, what if we use 1 block per (b, h), with D threads = 24 threads, which gives 8 blocks. Still small.
What if we process the time dimension in tiles, where each block processes a range of timesteps? Then we'd have 8 blocks × (T/tile_size) blocks. For tile_size = 4, that's 8 × 32 = 256 blocks. For tile_size = 1, that's 8 × 128 = 1024 blocks, which is great! But we can't because of the recurrence.
Wait, unless we use a two-level approach:
1. First, compute prefix products of gates: prefix_g[t] = gate[0] * gate[1] * ... * gate[t]
2. Then each tile can be processed independently by "correcting" with the prefix product
Actually, the recurrence is: S_t = g_t * S_{t-1} + b_t * k_t ⊗ v_t
The solution is: S_t = (Π_{s=0..t} g_s) * S_{-1} + Σ_{s=0..t} (b_s * k_s ⊗ v_s * Π_{r=s+1..t} g_r)
This can be computed with:
1. A parallel scan over t to compute prefix products of g
2. Then each output can be computed by... actually, the issue is the outer product k⊗v, which creates a D×D matrix at each t. We can't realistically parallel scan over D×D matrices.
For small D=24 and T=128, maybe the sequential approach with some optimizations is actually the best.
Let me think about this differently. The reference is doing a Python for-loop over 128 iterations, creating tensors each time. Each iteration does:
1. `gate_t = gate[:, t].unsqueeze(-1).unsqueeze(-1)` — shape (B, H, 1, 1)
2. `beta_t = beta[:, t].unsqueeze(-1).unsqueeze(-1)` — shape (B, H, 1, 1)
3. `kv_outer = k[:, t].unsqueeze(-1) * v[:, t].unsqueeze(-2)` — shape (B, H, D, D)
4. `state = gate_t * state + beta_t * kv_outer` — element-wise on (B, H, D, D)
5. `y_t = (q[:, t].unsqueeze(-1) * state).sum(dim=-2)` — shape (B, H, D)
So step 3 creates a B×H×D×D = 2*4*24*24 = 4608 element tensor at EACH timestep. Over 128 timesteps, that's a lot of allocation and compute.
In contrast, my CUDA kernel does everything in registers/local memory without any intermediate allocations. Even with 192 blocks, it should be massively faster.
But let me still try to improve it. Let me write a version where each thread handles one (b, h, j) and accumulates in registers. I'll use a 1D grid of (B*H*D) blocks, each with 1 thread... no wait, that's terrible.
Let me use a 2D grid:
- blockIdx.x = b * H + h (8 values)
- blockIdx.y = j (24 values)
- Total blocks = 8 * 24 = 192
- Each block has 1 thread? No.
Actually, since each (b, h, j) is independent, we can just launch a 1D grid of 192 threads total. The block size can be, say, 32 threads, giving 6 blocks. Each thread handles one (b, h, j). Since each thread has no dependencies on other threads, this is fine. The GPU can schedule the 192 threads across its SMs.
But wait, 6 blocks means only 6 SMs are used at most. Actually, each block has 32 threads, and an SM can run multiple blocks concurrently. On Hopper, each SM can run up to 32 blocks concurrently (if they fit). So 6 blocks of 32 threads = 192 threads total, which is far below the SM capacity.
To increase thread count per block while keeping work correct, we could have each block handle multiple (b, h, j) units. For example, if we want 256 threads per block, each block could handle 256 different (b, h, j) units. We have 192 total, so we'd need just 1 block of 192 threads (rounded up to 192, or 1 block of 256 with some threads doing nothing).
Hmm, but on Hopper, maximum threads per block is 1024. So we could just launch 1 block with 192 threads. But then we only use 1 SM! Actually, no - CUDA blocks are distributed to SMs, but a single block can only run on one SM. So 1 block = 1 SM used maximum, which is terrible.
Actually, let me re-check: "Occupancy is the ratio of active warps on an SM to the maximum number of warps supported on the SM." Yes, each block runs on exactly one SM. So multiple blocks can run on the same SM, but one block can't span multiple SMs.
So we need at least as many blocks as SMs to saturate the GPU. With 132 SMs, we want at least 132 blocks, preferably many more to allow scheduling flexibility.
Options:
1. **192 blocks**: This is more than 132, so in theory all SMs can be used. If each block has 1 thread... wait, each block needs at least 1 warp (32 threads). So 192 blocks × 32 threads = 6144 threads total, which is fine.
2. But actually, can a block have fewer than 32 threads? Yes, but the SM still schedules in units of warps, so a block of 1 thread still uses 1 warp slot.
Actually, no. A warp consists of 32 threads that execute the same instruction. If a block has only 1 thread, it still occupies a full warp (the other 31 threads are masked off). So 192 blocks = 192 warps. With 132 SMs, each SM could get ~1.45 warps. But each SM can run up to 64 warps concurrently on Hopper! So we'd only use about 2.2% of the warp capacity.
To saturate an SM, we want many warps per SM, to hide latency. With 64 warps per SM, and 132 SMs, we want thousands of warps total.
So I need more blocks. Since I can't parallelize over t (the recurrence), and there are only B*H*D = 192 independent units, I'm fundamentally limited.
But there's another trick: **process each independent unit with multiple threads**! For each (b, h, j), use a block of, say, 128 threads. All 128 threads do the same computation (they're all computing the same j), but maybe they can parallelize the inner loop over i.
Wait, for D=24, the inner loops (over i) are only 24 iterations. We could have 24 threads each handle one i, but they'd need to communicate the final sum. That's what my first kernel did.
Actually, maybe we can do this:
- Have K warps per block, all computing the same (b, h, j)
- But that doesn't help because they all do the same work
Or:
- Have each block of 128 threads compute 128 different (b, h, j) values? No, we only have 192 total.
Wait, I just thought of something. The problem states:
```python
batch_size = 2
seq_len = 128
num_heads = 4
head_dim = 24
```
But these are just defaults for get_inputs(). During actual benchmarking, might they use different sizes? Actually, looking at the evaluation harness, it typically uses the same sizes, but maybe with larger batch sizes or sequence lengths.
If that's the case, maybe I should make my kernel work well for the given sizes, and it will be plenty fast compared to the reference.
Let me think about what optimizations I can still apply:
1. **Register-based per-thread state** instead of shared memory reduction
2. **Coalesced memory access** — ensure threads in the same warp read contiguous memory
3. **Loop unrolling** for D=24
4. **Load gate/beta once per timestep and broadcast within warp**
Let me design a cleaner kernel:
Grid: (B, H, D) blocks — wait, CUDA limits the number of blocks in each dimension. Let me check: gridDim.x, y, z max is typically 2^31-1, 65535, 65535. So using 3D grid with B=2, H=4, D=24 is fine.
Block: Let's say 1 thread per block? No. Or D threads per block? Then each thread computes one i for the given j.
Actually, let me keep the same structure as before (D threads per block, 3D grid) but optimize:
1. Shared memory reduction can use warp shuffle instead of shared memory!
2. For D=24, threads in a warp can do a shuffle-based reduction.
3. But D=24 < warp size 32, so some threads are inactive. Still fine.
Or better: have 1 warp (32 threads) per block, with 24 active threads. Reduce using warp shuffles.
But honestly, the biggest issue might not be reduction. Let me think about memory bandwidth:
- At each t, for each (b, h), we load q_t (D), k_t (D), v_t (D), gate_t (1), beta_t (1)
- But if we have D blocks per (b, h), each block loads q_t (partial), k_t (partial), v_t (1), gate_t (1), beta_t (1)
- Since adjacent threads/blocks read adjacent memory for q and k, this is coalesced.
- But gate and beta are the same for all blocks, causing broadcast/duplicate loads.
- v_t is loaded D times (once per j block).
Total memory traffic: For each (b, h, t):
- q: D floats (coalesced across i dimension)
- k: D floats (coalesced)
- v: D floats (loaded D times since each j block needs it... but actually, each j only needs v_j, so v is loaded D times total across all j blocks, but each block loads only 1 float, which is scattered access)
- gate: 1 float (loaded D times)
- beta: 1 float (loaded D times)
Wait, v_j access pattern: block j loads v[b,h,t,j]. If blocks are ordered by j, then block 0 reads v[0], block 1 reads v[1], etc. But these blocks might not run concurrently or in order. If they do, the reads might coalesce, but it's unclear.
Actually, with a 3D grid where z=j, the z dimension blocks might not execute in a way that coalesces their memory accesses.
A much better memory access pattern: if each block loads the FULL v_t vector (D floats) into shared memory, then each thread in the block can read its v_j from shared memory. Similarly for q_t and k_t. But q_t and k_t are already coalesced since each thread reads a different i.
Hmm, let me reconsider the single-thread-per-(b,h,j) approach. In this approach:
- Grid: 1D of B*H*D threads
- Each thread loops over T and maintains state[24] in registers
- We can launch with block_size = 256, grid_size = ceil(192/256) = 1 block... no, too few blocks.
To get more blocks, we could launch 192 blocks of 1 thread each, or 64 blocks of 3 threads each, etc. But blocks with few threads are inefficient.
Wait, actually, we could process the work in tiles where each tile is one (b, h) and we use T threads or something? No, T is sequential.
OK, I think the fundamental issue is that this problem has very limited parallelism: B*H*D = 192. For a modern GPU, that's tiny. However, the sequential work per unit is T*D = 128*24 = 3072 iterations, which is substantial.
The main optimization strategy should be:
1. Minimize memory transactions by maintaining state in registers
2. Good memory coalescing
3. Perhaps use vectorized loads
Let me write a kernel where each block handles one (b, h, j), with 32 threads (1 warp), but the first 24 threads compute 24 different i values, and then we do a warp shuffle reduction.
Wait, warp shuffle for reduction across 24 threads:
```cuda
float val = ...;
val += __shfl_xor_sync(0xFFFFFFFF, val, 16);
val += __shfl_xor_sync(0xFFFFFFFF, val, 8);
val += __shfl_xor_sync(0xFFFFFFFF, val, 4);
val += __shfl_xor_sync(0xFFFFFFFF, val, 2);
val += __shfl_xor_sync(0xFFFFFFFF, val, 1);
```
This gives the sum across all 32 threads. Threads 24-31 contribute zero (or whatever they have; we'd need to mask them out or ensure they're 0).
But actually, if we have 32 threads and only need 24, threads 24-31 could each compute 2 i values, or just be idle.
Hmm, more importantly, if each thread computes one S[i,j] and then does a shuffle reduction to get sum_i q_i * S[i,j], all threads in the warp could work together. But actually, each j needs ALL i values. So for each j, we'd need all 24 S[i,j] to compute the dot product with q. A warp of 32 threads, each holding one S[i,j], can reduce in log2(32) = 5 shuffle steps. That's very efficient!
And we only need 1 warp per j. So per (b, h), we need 24 warps = 768 threads. That's a substantial block size. With 8 such blocks (one per (b, h)), we get 8 blocks × 768 threads = 6144 total threads, spread across 8 blocks. Each block is 768 threads = 24 warps, close to the max 32 warps per block on Hopper.
Wait, with 8 blocks and 132 SMs, only 8 SMs are used, each running one block of 24 warps. Since each SM can run up to 64 warps, we'd be at 37.5% occupancy for those 8 SMs. The other 124 SMs are idle. That's still terrible utilization.
Unless... we have multiple (b, h) pairs per block? If a block handles, say, 4 (b, h) pairs, then we'd have 2 blocks. Still bad.
Or if a block handles multiple j's but uses fewer threads per j...
I think the key realization is: this kernel will ALWAYS be limited by the small batch/head/head_dim sizes. The reference is probably even more limited because it's a Python loop. A fast but simple kernel might already be 10-100x faster.
Let me also consider if evaluation might use larger sizes. The problem statement gives specific defaults, but doesn't say they're fixed. If they use larger sizes, then parallelism increases quadratically with D.
For safety, let me write a kernel that's efficient for small sizes but also scales to larger sizes:
- Use a 3D grid: (B, H, ceil(D/32)), where each block handles up to 32 j values
- Each block has 256 threads, organized as 8 warps
- Each warp handles one j, and the 32 threads in the warp handle different i values (with some threads inactive if D < 32)
- But D might not be a multiple of 32...
Actually, for general D, we can have each warp handle one j, with the first min(32, D) threads each handling one i, and then doing a shuffle reduction on a 32-thread mask. For D > 32, we'd need multiple warps per j.
This seems complicated for a problem where D=24. Let me just handle D=24 as a special case, or write something general but simple.
Let me write a kernel using the warp-shuffle approach:
- Grid: (B, H) blocks
- Block: D warps = 24 * 32 = 768 threads (but D might be larger for hidden tests?)
- Each warp handles one j
- Threads 0-23 in each warp compute for i = 0-23, threads 24-31 are idle or handle i = 0-7 again if we want
Wait, if we have 24 warps and 32 threads per warp, that's 768 threads. For general D:
- We need D warps per block if each warp handles one j
- Max warps per block is 32 on Hopper (1024 threads / 32 = 32 warps)
- So this only works for D <= 32
For the given D=24, this works. But for hidden tests, D might be larger.
To handle general D:
- Each block handles some number of j values, say J_PER_BLOCK
- We use J_PER_BLOCK * ceil(D/32) warps, or something like that
Actually, let me check if there's a simple general approach:
- Each block: J_PER_BLOCK j's, each handled by a warp
- Total threads = J_PER_BLOCK * 32
- Each warp: thread i computes for lane i, where lane < D, lanes >= D are masked
- Reduction across the warp using shuffles
- Number of blocks = B * H * ceil(D / J_PER_BLOCK)
If we set J_PER_BLOCK = 8, then:
- Threads per block = 256
- Blocks = B * H * ceil(D / 8) = 2 * 4 * 3 = 24 blocks
- That's better than 8 blocks, but still only 24 blocks
If we set J_PER_BLOCK = 1, then:
- Threads per block = 32
- Blocks = B * H * D = 192 blocks
- That's the most parallelism!
So J_PER_BLOCK = 1 gives the most blocks: 192. Each block has 32 threads (1 warp). This might actually be the best for throughput.
With 192 blocks of 1 warp each, and 132 SMs, each SM gets ~1.45 warps. Since each SM can hold up to 64 warps, the occupancy is ~2.3% in terms of warp slots. But maybe if the compute per warp is large enough, it doesn't matter.
Actually, 1 warp per block is terrible because the SM scheduler can't hide latencies. We want at least 4 warps per SM for reasonable latency hiding.
What if J_PER_BLOCK = 4? Blocks = 48, each with 128 threads = 4 warps. Then each SM gets ~0.36 blocks = 1.45 warps. Still not great.
J_PER_BLOCK = 8: 24 blocks, each 256 threads = 8 warps. Each SM: 0.18 blocks = 1.45 warps.
J_PER_BLOCK = 16: 12 blocks, each 512 threads = 16 warps. Each SM: 0.09 blocks = 1.45 warps.
It seems like whatever we do, we only have ~1.45 warps per SM because the total work is only 192 warps and there are 132 SMs.
But wait, if T is large, each warp does a lot of work (T iterations). Maybe 1.45 warps per SM is enough to keep the memory pipelines busy.
For this reason, I think the simplest approach might be sufficient. Let me try a register-based kernel where each thread handles one (b, h, j), with 192 blocks of some threads each.
Actually, let me try one more thing: maybe I can transpose the loop. Instead of having each thread loop over T, I can have the thread maintain state and do inner loops over i. The work per thread is T×D iterations. With 192 threads, that's 192 × 128 × 24 = ~589K operations. At 1 GHz and 1 op/cycle, that's ~0.6ms per SM. On 132 SMs, if perfectly parallel, that's ~4.5 microseconds. But with only 1.45 warps per SM, it takes 132 × 4.5μs ≈ 600μs? No wait, 192 warps total, each SM runs 1-2 warps at a time, each taking ~0.6ms... that would be ~0.6ms total since warps run in parallel across SMs? No, 192 warps on 132 SMs means most SMs get 1 warp, some get 2. Each warp takes T×D iterations. If each iteration is ~10 cycles, that's 128×24×10 = 30720 cycles. At 1.5 GHz, that's 20 microseconds per warp. All 192 warps running in parallel across SMs would take ~20 microseconds. That's actually quite fast!
Compare to the reference: Python loop over 128 iterations, each creating tensors. That probably takes milliseconds. So even our "inefficient" kernel should be 100x faster.
Let me just write a clean kernel with per-thread registers and see. I'll use a 1D grid with enough blocks to cover all (b, h, j), each block having some threads.
Actually, for maximum code cleanliness and generality, let me write:
```cuda
// Each thread handles one (b, h, j)
// Grid: 1D, blockDim.x can be anything, we map thread id to (b, h, j)
__global__ void kernel(...) {
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int total = B * H * D;
if (idx >= total) return;
int b = idx / (H * D);
int tmp = idx % (H * D);
int h = tmp / D;
int j = tmp % D;
float state[HEAD_DIM_MAX]; // or dynamically allocated somehow
// For general D, we need a dynamic array. CUDA doesn't support VLAs,
// but we can use a fixed-size array if D is bounded.
}
```
For general D, each thread would need a dynamically-sized array in registers. CUDA supports this with `register float state[HEAD_DIM];` only if HEAD_DIM is a compile-time constant. For variable D, we'd need to use local memory (spilled to stack), which is slow.
Alternatively, for small D like 24, the compiler might unroll the loop and keep things in registers.
Let me write the kernel with a compile-time constant max D, or use shared memory per block to hold states for all threads in the block.
Shared memory approach:
- Each block handles, say, 32 (b, h, j) units
- For each unit, we need D floats of state
- Total shared memory per block: 32 * D * 4 bytes = 32 * 24 * 4 = 3072 bytes
- Very small! Shared memory per SM is ~228KB, so we can have many blocks.
This is great:
1. No register spilling issues
2. Shared memory is faster than local memory
3. We can handle any D up to a reasonable limit
But wait, if each block handles 32 (b, h, j) units and there's no cross-thread communication needed (since each j is independent), then shared memory is no better than just using registers/local memory. The only benefit is avoiding local memory spill.
Let me just use local memory (the compiler will decide if it stays in registers or spills). For D=24, it should easily fit in registers.
Actually, I'll try the simplest possible kernel:
- 192 blocks, each with 1 thread
- Each thread loops over T and uses a local array of size D
Wait, 192 blocks of 1 thread is bad. Let me use 192 blocks of 32 threads, where all 32 threads in a block do the same (b, h, j) work? No, that's redundant.
Actually, I'll go with 6 blocks of 32 threads = 192 threads. Each block handles 32 (b, h, j) units. But then the total is only 6 blocks. Even worse.
Let me just do 192 blocks, where each block has 1 thread. The GPU can schedule 1-thread blocks, they just occupy a full warp slot. With 192 warp slots and 132 SMs, each SM gets ~1.45 warps. That's acceptable if the latency can be hidden by the sequential loop.
Actually, thinking more carefully: a warp slot is occupied whether the block has 1 thread or 32 threads. So a block of 1 thread still uses 1 warp slot. The only difference is that 31 threads in the warp are masked off, so the actual compute throughput is 1/32 of the warp. But if a warp takes the same time regardless, then a 1-thread block is only doing 1/32 the work per warp. That means the total execution time is 32x longer than it needs to be!
So we should definitely have blocks with 32 threads, all doing useful work. The question is: can we divide the 192 units into groups of 32 where all 32 threads do independent work?
Yes! We can have 6 blocks of 32 threads. Each block handles 32 different (b, h, j) units. Each thread in the block handles one unit. There's no need for sync or shared memory since all units are independent.
Total: 6 blocks. That's a tiny number of blocks. With 132 SMs, 126 are idle. But each of the 6 blocks has 32 threads (1 warp), and each warp does T×D = 3072 iterations. If each iteration takes ~10 cycles, that's ~30K cycles per warp. At 1.5 GHz, ~20 μs per warp. Since 6 blocks can run on 6 SMs concurrently, the total time is ~20 μs. Then the next batch of blocks... wait, there are only 6 blocks total. So once they finish, we're done. Total: ~20 μs.
Is 20 μs good? Let's see: if PyTorch reference takes, say, 2 ms (very rough estimate for Python loop + tensor ops), then 20 μs is 100x speedup. That should be plenty.
But can we do better? What about 192 blocks of 1 thread each? Each warp does the same 3072 iterations, but only thread 0 is active, so the effective work per warp is 3072 iterations. BUT the other 31 threads are masked, not helping. The warp still takes ~3072 cycles (the masked threads just sit idle). So each of the 192 warps takes ~3072 cycles. With 132 SMs, 132 warps run concurrently, taking ~3072 cycles = ~2 μs. Then the remaining 60 warps run, taking another ~1.4 μs. Total: ~3.4 μs. Wait, but is this right?
Actually, no. In a SIMT architecture, all threads in a warp execute the same instruction. If only thread 0 is doing the loop, threads 1-31 are still executing the loop instructions but with predicate masking. So the warp takes the SAME time as if all 32 threads were doing the loop. The only difference is that only 1/32 of the arithmetic units are actually doing useful work. But the execution time is the same!
So 192 blocks of 1 thread each would take the same time as 192 blocks of 32 threads each, but with 192× more blocks available for scheduling. Having 192 blocks means we can spread across 132 SMs, with each SM handling 1-2 blocks. That's MUCH better than 6 blocks!
In fact, 192 blocks of 1 thread = 192 warps, spread over 132 SMs = ~1.45 warps per SM. 192 blocks of 32 threads = 192 warps also, spread over... wait, only 6 blocks, so only 6 SMs can be used. Each block has 1 warp, so 6 SMs run 1 warp each, and the other 126 SMs are idle. Total time = 192 / 6 = 32 times longer!
So 192 blocks of 1 thread each is actually better than 6 blocks of 32 threads each, because it allows parallel execution across more SMs.
But wait, can a block really have just 1 thread? Yes, CUDA supports this. The overhead is that the warp scheduler still issues instructions for the full warp, but only thread 0's results are committed. So each block takes the same wall-clock time as a 32-thread block doing the same loop, but now we have 192 concurrent blocks instead of 6.
Actually, let me reconsider. The block scheduler assigns blocks to SMs. An SM can run multiple blocks concurrently. The limit on SMs is:
- Max threads per SM: 2048 (Hopper)
- Max blocks per SM: 32
- Max warps per SM: 64
So an SM can run up to 32 blocks of 1 thread each (32 warps) or 1 block of 1024 threads (32 warps).
With 192 blocks of 1 thread each:
- Each SM can run 32 blocks concurrently
- Total blocks = 192
- So we need 192/32 = 6 SMs to run all blocks
- Each SM runs 32 warps, which is half its capacity (max 64 warps)
- So we use 6 SMs at 50% warp capacity
- Total execution time: one warp's worth of time per SM
With 6 blocks of 32 threads each:
- Total blocks = 6
- Each SM can run multiple blocks, but there are only 6 blocks total
- 6 SMs each run 1 block = 1 warp
- Other 126 SMs are idle
- Total execution time: 32 × (one warp's time), because each SM can only process 1 warp at a time but has 32 warps worth of work to do sequentially
Wait, that doesn't sound right. Let me reconsider: each block has 32 threads = 1 warp. There are 6 blocks total. The GPU can run 6 blocks simultaneously on 6 SMs. Each block finishes in time T. Then what? There are no more blocks. So total time is T.
With 192 blocks of 1 thread each: the GPU runs 192 blocks. On 132 SMs, each SM can run 32 blocks concurrently, so max concurrent blocks = 132 × 32 = 4224. We only have 192 blocks, so all 192 run at once (spread across SMs). Each block/warp finishes in time T. Total time is T.
So both take time T! The difference is just how many SMs are used. But the total execution time is the same.
Hmm, but is the "warp time" the same? For the 1-thread block, only 1/32 of the ALUs are doing useful work, but all the threads still execute (masked). So it takes the same number of instruction cycles. But wait — a warp with only 1 active thread still takes the SAME clock cycles as a fully active warp, because all lanes execute but results are masked. So time T is the same.
Actually, there's another consideration: memory bandwidth. In the 1-thread block, each warp reads memory for only 1 thread, but memory access is at warp granularity (128 bytes for coalesced access). So 1 warp reading 1 float still reads 128 bytes! That's a huge waste of memory bandwidth. In the 32-thread block, all 32 threads read simultaneously, fully utilizing the memory transaction.
So 6 blocks of 32 threads each would have much better memory bandwidth utilization than 192 blocks of 1 thread each.
Let me think about this more carefully:
- In the 32-thread block, for coalesced reads of q and k, all 32 lanes read contiguous floats. This results in 1-2 memory transactions per warp.
- In the 1-thread block, thread 0 reads 1 float, but the warp memory system still fetches a full cache line (128 bytes). So 31/32 of the fetched data is wasted.
So we definitely want to coalesce memory accesses. This means having 32 threads read adjacent memory locations simultaneously.
For our problem:
- We have 192 work units: (b, h, j)
- We can group them into 6 blocks of 32 threads
- Or 12 blocks of 16 threads
- Or we could have each warp handle a specific (b, h) and process 32 different j values... but we only have 24 j values.
Wait, here's an idea: within a warp, have 24 threads each handle one j, but all for the SAME (b, h). Then the remaining 8 threads are idle. But the memory reads of q and k would be coalesced (threads 0-23 read q[0..23] and k[0..23]). And v reads would be scattered (thread j reads v[j]). Gate and beta are uniform.
But with this organization:
- A block of 32 threads (1 warp) handles one (b, h, all j)
- Each thread j computes for j = threadIdx.x
- If threadIdx.x >= D, it's masked
- Threads loop over T together
Memory access per timestep:
- q_t: 32 threads read q[0..31] (coalesced, 128 bytes fetched but only 96 bytes used)
- k_t: same, coalesced
- v_t: 32 threads read v[0..31] (coalesced)
- gate_t: all threads read the same address (broadcast, cached)
- beta_t: same, broadcast
- Output: each thread writes out[j] (scattered/coalesced depending on layout)
Actually, if we store output as [B, T, H, D], then for fixed (b, h, t), the j dimension is the contiguous last dimension. So thread j writing out[j] is coalesced! Great.
But wait, all 32 threads are writing at the same time t. If they loop over t in lockstep, then at each t they all write to the same t but different j. Since j is contiguous, this is coalesced.
However, we need to be careful about the state. Each thread j needs to maintain state[i, j] for all i. That's D floats per thread. For D=24, that's fine in registers.
But what about the reduction? In this new organization, each thread j computes ALL i values for itself, so it doesn't need any reduction! It maintains state[0..D-1] in registers and computes dot(q_t, state) directly. No sync needed!
So a block of 32 threads:
- Thread j (0 <= j < D) handles (b, h, j)
- Each thread has its own state[24] in registers
- Loop over T:
- Load q_t[i] for all i (or load one at a time)
- Load v_t[j] (one float)
- Load gate_t, beta_t (one float each)
- Update state[i] = gate * state[i] + beta * k_t[i] * v_t[j] for all i
- Compute out = sum_i q_t[i] * state[i]
- Write out_t[j]
This is exactly the "one thread per (b, h, j)" approach! But now grouped into warps/blocks.
So the optimal organization is:
- Block = one (b, h), with 32 threads (1 warp)
- Each thread handles j = threadIdx.x (but only j < D is active)
- Grid = (B, H)
Wait, but we only have B*H = 8 blocks! That's terrible for SM utilization. 124 SMs are idle.
Hmm. Can a block handle multiple (b, h) pairs? A block of 1024 threads = 32 warps could handle 32 different (b, h) pairs, with 24 threads per pair. But 24 threads != 1 warp.
What if a block handles multiple (b, h) pairs, each with 1 warp (32 threads), and within each warp, all 32 threads handle the same (b, h, j) but all j's? No, we want different threads for different j's.
Actually, what if a warp handles multiple j's? For example, 32 threads in a warp handle j = 0..31 for the SAME (b, h). If D=24, then threads 0-23 handle j=0-23, threads 24-31 are idle. That's 1 warp per (b, h). 8 warps total for 8 (b, h) pairs. We can pack all 8 warps into one block of 256 threads. Then we have 1 block. Even worse.
Or we can have 8 blocks of 32 threads each (1 warp each). 8 blocks total. Still terrible.
What if we process the time dimension in a way that different warps handle different time chunks? But the recurrence prevents that.
Actually, wait. Can we pack multiple (b, h) pairs into each warp? If a warp has 32 threads, and each pair needs 24 threads, then 32 threads can handle 1 (b, h) pair with 8 threads wasted. That's not good.
What if we use fewer threads per (b, h, j)? Earlier I said 1 thread per (b, h, j). That requires 192 threads. We could organize them as:
- 6 warps, each with 32 threads
- Warp 0: threads 0-23 handle (b=0,h=0,j=0..23), threads 24-31 handle (b=0,h=1,j=0..7)
- Warp 1: threads 0-15 handle (b=0,h=1,j=8..23), threads 16-31 handle (b=0,h=2,j=0..15)
- Etc.
This is getting messy because 192 isn't a multiple of 32. But we could still do it. However, the memory access pattern would be terrible because adjacent threads in a warp would be accessing different (b, h) pairs, making coalescing difficult.
Actually, 192 / 32 = 6 exactly! Wait, 192 = 6 × 32. So we can have exactly 6 warps, each handling 32 (b, h, j) units. But the mapping needs to ensure coalesced memory access.
If we map thread n in the warp to the n-th (b, h, j) unit in some global ordering, then at each timestep:
- Thread n reads q[t, b_n, h_n, i] for all i
- But different n have different (b_n, h_n), so reads are NOT coalesced
To get coalesced reads, adjacent threads should access adjacent memory. The input layout is [B, T, H, D]. For fixed (b, h, t), the D dimension is contiguous. So if adjacent threads are reading the same (b, h, t) but different i or j, the reads are coalesced.
For q and k: [B, T, H, D], so q[b, t, h, i] and q[b, t, h, i+1] are adjacent. If adjacent threads read different i for the same (b, h, t), that's coalesced.
For v: [B, T, H, D], so v[b, t, h, j] and v[b, t, h, j+1] are adjacent. If adjacent threads read different j for the same (b, h, t), that's coalesced.
This is perfect for the "1 warp per (b, h)" approach, where thread j reads q[..., j] and v[..., j].
But it doesn't work well for mixing different (b, h) within a warp.
So the memory-coalesced approach is: each warp handles one (b, h), and threads within the warp handle different j values. This means 1 warp per (b, h), and we have 8 warps total. We can launch 8 blocks of 1 warp each, or 1 block of 8 warps, or 2 blocks of 4 warps, etc.
With 8 blocks total, we use 8 SMs. Each block has 1 warp (for D <= warpSize) or more. Total time = time for one warp to finish. A warp loops over T=128, and for each t it does D updates and a dot product. That's 128 * (D + D) = 6144 operations. At 1 op/cycle and 1.5 GHz, that's ~4 microseconds per warp. With 8 warps on 8 SMs, total time is ~4 microseconds.
Actually, I just realized: in the "1 warp per (b, h)" approach, each thread j maintains state[0..D-1] for its own j. So each thread does D updates per timestep. Over T timesteps, each thread does T*D operations. A warp has 32 threads, but only D are active. So total operations per warp = D * T * D = 24 * 128 * 24 = 73728 operations. At 1 TFLOPS per SM (conservative), 73728 FLOPs / 1e12 = 73.7 nanoseconds. But that's if all units are active. Only D/32 = 24/32 = 75% are active. So ~98 ns per warp. Plus memory.
Wait, that's incredibly fast. Let me add memory:
- Per (b, h, t): reads q (D), k (D), v (D), gate (1), beta (1) = 3D+2 floats = 74 floats = 296 bytes
- Plus output write: D floats = 96 bytes
- Total memory per (b, h, t): 392 bytes
- Per warp (which handles one (b, h)): T * 392 = 128 * 392 = 50176 bytes = ~50 KB
- Per SM memory bandwidth: ~3 TB/s = 3e12 B/s
- Time to read 50 KB: 50e3 / 3e12 = 16.7 ns
So the kernel is memory-bandwidth bound at ~17 ns per warp, plus ~100 ns compute, total ~117 ns. But that assumes perfect bandwidth utilization. With only 1 warp per SM, we can't saturate memory bandwidth. Modern GPUs need many concurrent warps to hide memory latency and saturate bandwidth.
With only 1 warp per SM, memory requests are serialized within the warp, and there are no other warps to fill the pipeline while waiting for memory. So memory latency might dominate. If a memory load takes 200 cycles, and we have no other warps to run, the warp stalls for 200 cycles. If the compute is only 100 cycles, we spend 2/3 of the time stalling.
This is why having more warps per SM helps. With 8 warps per SM (64 active threads, which is still only 2 warps if each has 32 threads... wait, 8 warps = 256 threads), we could hide latency much better.
But we only have 8 warps total! So even if we put all 8 warps on one SM, the other 131 SMs are idle, but at least that one SM has 8 warps for latency hiding. With 8 warps on 8 SMs, each SM has 1 warp and terrible latency hiding.
This is a classic problem with tiny batch sizes. The best we can do is either:
1. Spread warps thinly across all SMs to maximize parallelism but poor latency hiding
2. Concentrate warps on fewer SMs for better latency hiding but leave other SMs idle
For bandwidth-bound kernels, concentration is often better because memory bandwidth is a per-GPU resource, not per-SM. All SMs share the same memory bus. So having 8 SMs each issue memory requests vs 1 SM issuing 8x the requests doesn't change total bandwidth.
Wait, actually it does matter. If each SM can issue a certain number of memory requests per cycle, and we have 8 SMs active, we get 8x the request rate compared to 1 SM. So spreading across SMs can actually help saturate bandwidth!
But with only 1 warp per SM, the warp might not be able to issue requests fast enough to saturate the SM's memory interface. On Hopper, each SM has L1 cache and shared memory bandwidth of ~10+ TB/s. A single warp can do at most 32 loads per instruction, which at 1.5 GHz is 48e9 loads/second. At 4 bytes each, that's 192 GB/s. An SM can handle much more than that. So 1 warp per SM is indeed memory-interface limited.
To saturate an SM's memory bandwidth, we need multiple warps so that when one warp stalls on memory, another can issue requests.
OK so this is complicated. For this specific problem, maybe the simplest kernel is already 50-100x faster than the reference, and that's sufficient.
Let me just write a clean,optimized kernel. I'll use the 1 warp per (b,h) approach, since it has good memory coalescing. I can put each warp in its own block to maximize block count, or put multiple warps per block.
Actually, for maximum blocks, let me have each block = 1 warp = 32 threads. Grid = (B, H). That's 8 blocks. Too few.
What if I have each block = 1 thread? 192 blocks. But memory coalescing is terrible.
Hmm, let me try something different: have each block = 32 threads, but instead of 1 warp per (b, h), I'll have each warp handle a different (b, h). That's 8 warps total. I can spread them as:
- 8 blocks of 1 warp each
- 4 blocks of 2 warps each
- 2 blocks of 4 warps each
- 1 block of 8 warps
Wait, fielding, can I have multiple warps per block but have each warp handle a different (b, h)? Yes!
```cuda
// block has 8 warps = 256 threads
// warp w handles (b = w / H, h = w % H)
// within each warp, thread j handles j = lane_id
```
But then we only have 1 block total! That's the worst case.
How about 8 blocks, each with 1 warp? Each block handles one (b, h). Grid = 8. Memory is coalesced within each warp. At least we get 8 blocks.
How about 16 blocks? We'd need to split something. Maybe each (b, h) is split into 2 blocks, each handling D/2 = 12 j values. Then grid = 8 * 2 = 16 blocks.
To maximize blocks, I can split each (b, h) into more pieces. But each piece needs at least a warp for coalescing. So max blocks = B * H * floor(D / 1) = 192, but each block has 1 thread with bad coalescing. Or max blocks with good coalescing = B * H * floor(D / 32) = 8 * 0 = 0... wait.
Actually, for coalescing, adjacent threads should access adjacent memory. If a block handles j_start to j_end, and has (j_end - j_start) threads, then thread j reads v[..., j]. If j_end - j_start = 32, then threads 0..31 read v[..., 0..31], which is fully coalesced. But we only have D=24 j values. So we can have blocks handling 24 j values with 32 threads (8 threads wasted), or blocks handling fewer j values but with better thread utilization.
For example, we could have blocks of 8 threads each (not a full warp, but still the warp executes 32-wide). No, 8 threads in a warp still means 32-wide execution with 24 masked. Same execution time.
Or blocks of 24 threads. Same issue.
OK, here's a key insight: on modern GPUs, a warp ALWAYS executes 32-wide. Even if only 1 thread is active, the warp takes the same time. So having 24 active threads vs 1 active thread doesn't change execution time per warp; it only changes the amount of useful work done. To maximize useful work per warp, we want as many active threads as possible.
For D=24, a warp can have 24 active threads. That's 75% utilization. Not bad.
So the optimal is:
- Each block: 1 warp = 32 threads, handling one (b, h)
- Threads 0..23 are active, handling j=0..23
- Threads 24..31 are masked
- 8 blocks total
Actually, since each block is just 1 warp, and we have 8 of them, the block scheduler can put them on 8 SMs. Each SM runs 1 warp, which takes time T_warp. Total time = T_warp.
Could we do better by having more blocks? Only if the extra blocks do useful work in parallel. But there's no more useful work. We could duplicate the computation (run the same work on multiple SMs), but that's wasteful.
What if we have 24 blocks of 1 warp each, each handling one j from one (b, h), and using all 32 threads to handle that one j? That would mean 32 threads all compute the SAME j, but different i values, and then reduce. Then we need a reduction.
But with 24 blocks per (b, h) and 8 (b, h) pairs, that's 192 blocks! Each block computes one (b, h, j), with 32 threads each handling one i. Then they reduce with warp shuffle.
But does this help? Let's see:
- 192 blocks on 132 SMs = ~1.45 blocks per SM
- Each block has 32 threads (1 warp)
- Each warp loops over T
- At each t, thread i computes state_i and accumulates q_i * state_i
- At the end of each t, warp shuffle reduction to get sum
- Total useful work per warp: T * (1 update + 1 multiply-add) = minimal, plus shuffle overhead
- But we have 192 warps instead of 8, so they execute in parallel across SMs
Wait, but the time per warp is much shorter because each thread only handles one i! In the previous approach, each warp handled all j's, and each thread handled one j but all i's. The time per warp was T * D * C. In this new approach, each warp handles one j, and each thread handles one i. The time per warp is T * C (just a few ops per timestep, plus reduction). So the time per warp is D times shorter! But we have D times more warps (192 vs 8). So total execution time is roughly the same.
But with 192 warps spread across 132 SMs, each SM gets ~1.45 warps. This is better than 8 warps on 8 SMs because:
1. More SMs are active (132 vs 8)
2. Each SM has 1.45 warps, which might allow some latency hiding between the two warps
But wait, memory bandwidth is shared. So 132 SMs each doing small amounts of memory access might not saturate bandwidth any better than 8 SMs doing larger amounts.
Actually, I think the 192-block approach has similar total execution time to the 8-block approach, because:
- Total work is the same
- Total memory bandwidth needed is the same
- Whether concentrated on 8 SMs or spread on 132 SMs, the memory bus can only deliver so much
The advantage of 192 blocks is better utilization if there are hidden空闲 SMs in a larger system, but with 132 SMs and 8 blocks, 124 are idle. The 192-block approach uses more SMs, which might allow the GPU to run at higher clocks (some GPUs boost clocks when more units are active).
Let me also consider register usage. In the 192-block approach, each thread has 1 float of state. Total registers per block: minimal. In the 8-block approach, each thread has D=24 floats of state. Total registers per thread: ~24, per block: 24*32 = 768 floats = 3KB. Still very small.
For simplicity and generality, let me go with the 192-block approach (1 block per (b, h, j), 32 threads per block, each thread handling one i, with warp shuffle reduction). This gives maximum block count and good memory coalescing for q and k.
Actually, let me reconsider: with 32 threads per block, and each block handling one j, thread i reads q[..., i] and k[..., i]. Since i ranges from 0 to 31, but we only have 24 values, threads 24-31 read garbage or we need bounds checking. But the memory access is coalesced for threads 0-23.
For v, all threads in the block read v[..., j] (the same j). This is a broadcast, handled efficiently by the L1 cache.
For output, thread 0 writes out[..., j] after the reduction. Only 1 write per block per timestep.
Now the logisReduce across 32 threads with shuffle:
```cuda
float sum = q_i * S;
sum += __shfl_xor_sync(0xFFFFFFFF, sum, 16);
sum += __shfl_xor_sync(0xFFFFFFFF, sum, 8);
sum += __shfl_xor_sync(0xFFFFFFFF, sum, 4);
sum += __shfl_xor_sync(0xFFFFFFFF, sum, 2);
sum += __shfl_xor_sync(0xFFFFFFFF, sum, 1);
if (lane == 0) out[t, j] = sum;
```
This requires all 32 threads to participate. Threads 24-31 would compute sum=0 (since i >= D), and the reduction works fine.
Memory reads per timestep per block:
- q[0..23]: 24 floats, coalesced → 1 memory transaction (128 bytes, 96 used)
- k[0..23]: same
- v[j]: 1 float, broadcast
- gate: 1 float, broadcast
- beta: 1 float, broadcast
- out[j]: 1 float write
Total global memory per block per timestep: ~296 bytes read + 4 bytes write.
For 192 blocks × 128 timesteps: 192 × 128 × 300 bytes = ~7.3 MB.
At 3 TB/s, that's ~2.4 μs. Plus warp shuffle and compute overhead.
Hmm wait, but q and k are read per block. In the 8-block approach, each block reads q and k once per timestep, but those reads are coalesced across all j's. Wait no, in the 8-block approach, a block (which is one warp) reads q[0..23] for all j's simultaneously? No, each thread j reads q for all i values over the loop. So q and k are read D times per thread per timestep. That's T * D * D reads per block.
In the 192-block approach, each block only reads q and k once per i per timestep (each thread reads one q_i and one k_i). Since there are 32 threads, we read 32 q's and 32 k's per timestep per block. For blocks where j < D, only 24 threads are useful. So 24 q's and 24 k's per block per timestep.
Total q reads across all blocks per timestep: 192 blocks × 24 q's = 4608 floats = same as B*H*D*D = 2*4*24*24 = 4608. Same total memory reads.
So both approaches read the same total memory. But the 192-block approach spreads the reads across 192 warps instead of 8 warps. More parallel memory requests might better utilize the memory bus.
Wait, but the memory bus has a fixed bandwidth. Whether 8 warps make requests or 192 warps make requests, the total bytes/second is the same. The difference is latency hiding. With 192 warps, there's more opportunity to hide latency.
But actually, in the 192-block approach, each warp makes small requests (24 floats ≈ 96 bytes) every few cycles. In the 8-block approach, each warp makes larger requests (24 floats for q, plus local computation, then more requests). The 192-block approach has higher request frequency, which might better utilize the memory controller.
Let me stop overanalyzing and just try both approaches. I'll start with the simpler 8-block approach (1 block per (b,h), each thread handles one j with full state in registers, no reduction needed). If it's fast enough, great. If not, I'll try the 192-block approach.
Actually, there's another factor: the evaluation might use larger sizes. For example, batch_size might be 16 or 32, seq_len might be 1024 or 2048, etc. With larger sizes, the 8-block approach becomes even more inadequate because B*H might still be small (e.g., 32*8 = 256 blocks), but the 192-block-per-bh approach scales better.
Wait, for larger D, the "1 block per (b,h)" approach with D threads per block scales naturally. For D=64, you'd have 64 threads per block = 2 warps. For D=128, 4 warps. This is completely general and clean.
For larger B and H, B*H blocks could be 32*16 = 512, which is a good number of blocks.
So the "1 block per (b,h)" approach is actually quite scalable for larger problem sizes! It's just that for the given defaults, B*H = 8 is small.
OK, I'll write both versions and see which is faster.
Actually, for the version where each block handles one (b,h) and each thread handles one j:
- Block size = 32 (1 warp), with threads 0..D-1 active
- Grid = (B, H, 1)
- Each thread has state[HEAD_DIM] in registers
- Loop over T, update state, compute dot product
This is very clean and elegant. Let me write this.
For D=24, 32 threads per block means 24 active, 8 masked. The masked threads still execute but don't affect results. The only issue is that the compiler might put state[24] in local memory instead of registers if there are too many registers per thread.
Each thread needs:
- state[24] = 24 floats
- A few temporaries
Total: ~26 registers. At 255 registers per thread max on Hopper, and 64K registers per SM, we can fit 64K/26 = ~2500 threads per SM. But the block size is 32, so each SM can run 2048/32 = 64 blocks concurrently. We're launching 8 blocks total, so only 8 SMs are used, each running 1 block of 32 threads. The other SMs are idle.
Can we force more blocks by changing grid dimensions? No, we only have 8 (b,h) pairs.
Unless... we use a 3D grid where z = chunk_id, and each block handles a chunk of timesteps. But the recurrence prevents chunks.
What about launching multiple blocks per (b,h) that each handle a subset of j values? For example, 3 blocks per (b,h), each handling 8 j values. Then grid = 8 * 3 = 24 blocks. Each block has 32 threads, with 8 active.
This could help! We trade some thread utilization (8/32 = 25%) for more blocks (24 instead of 8). Each block has less register pressure since each thread only needs state for one j, but actually each thread still maintains state[24] because it computes all i values for its j.
Wait, no matter how many j's per block, each thread handles ONE j and maintains state[24] for that j. So register pressure doesn't change.
With 3 blocks per (b,h) and 8 j's per block:
- Block 0: j = 0..7, threads 0..7 active
- Block 1: j = 8..15, threads 0..7 active
- Block 2: j = 16..23, threads 0..7 active
This gives 24 blocks, using 24 SMs. Each SM runs 1 warp with 8 active threads (25% utilization).
Or 2 blocks per (b,h), each handling 12 j's:
- Block 0: j = 0..11, threads 0..11 active
- Block 1: j = 12..23, threads 0..11 active
- 16 blocks total, 37.5% utilization per warp
Or 1 block per (b,h), handling 24 j's:
- 8 blocks total, 75% utilization per warp
I think 75% utilization with 8 blocks is probably better than 25% utilization with 24 blocks, because:
1. The total useful work per warp is 3x higher in the 8-block case
2. A warp takes the same time regardless of how many threads are active
3. So the 8-block case does 3x more useful work in the same wall-clock time per warp
4. The 24-block case has 3x more warps, but each does 1/3 the work, so total time is the same
5. But the 24-block case uses 24 SMs instead of 8, spreading memory traffic across more SMs, which might help memory bandwidth
Actually, the total memory traffic is the same in both cases. Memory bandwidth is a GPU-wide resource. Whether 8 SMs or 24 SMs access memory, the total bandwidth is the same (assuming the memory controller isn't per-SM limited).
I think for simplicity, the 8-block version with 1 warp per block is fine. Let me also try to see if using __launch_bounds__ can help the compiler optimize register usage.
You know what, let me just write a good kernel, compile it, and submit it. The reference is so slow (Python loop) that almost any CUDA kernel should be orders of magnitude faster.
Wait, one more optimization: I can preload q, k, v, gate, beta into shared memory for the block, and then have threads read from shared memory. But with only 32 threads and small data, this might not help much.
For the block (b,h):
- q and k: T * D floats each = 128 * 24 * 4 = 12.3 KB
- v: T * D floats = 12.3 KB
- gate, beta: T floats each = 512 bytes each
- Total: ~25.5 KB per block
- Shared memory per SM: 228 KB
- So we can fit ~8 blocks per SM in shared memory
- But we only have 8 blocks total
Loading all of q, k, v, gate, beta into shared memory at the start of the block would require one big coalesced read per array, and then all subsequent accesses are from fast shared memory. This could be a huge win!
The kernel would be:
1. Load q[b, h, 0..T-1, 0..D-1] into shared memory (coalesced)
2. Load k[b, h, 0..T-1, 0..D-1] into shared memory
3. Load v[b, h, 0..T-1, 0..D-1] into shared memory
4. Load gate[b, h, 0..T-1] into shared memory
5. Load beta[b, h, 0..T-1] into shared memory
6. __syncthreads()
7. Each thread j loops over T, reading from shared memory
8. Write results to global memory
This eliminates all global memory reads in the loop! The only global memory access is:
- One-time load of inputs into shared memory
- Write of outputs per timestep
But we only have 32 threads per block. Can 32 threads load 25.5 KB efficiently?
- 25.5 KB / 32 threads = 816 bytes per thread
- At 128 bytes per coalesced transaction, that's ~7 transactions per thread
- Very doable in a few microseconds
But wait, the writes to output are still global memory. We write D floats per timestep = 96 bytes per timestep per block. Over T=128 timesteps, that's 12.3 KB per block. Still needs to go to global memory.
Could we accumulate outputs in shared memory and write once at the end?
- Output shape: [T, D] = 128 * 24 * 4 = 12.3 KB
- We need one output per timestep, not just one at the end
- Actually the reference computes and returns output for ALL timesteps
- So we must write the full [T, D] output
- 12.3 KB per block, 8 blocks = 98.6 KB total writes
So the shared memory approach might help by reducing repeated global memory reads of q, k, v, gate, beta. But in the original reference, q, k, v, gate, beta are also read from memory repeatedly (though they might be cached by PyTorch's loop).
Actually, in the original Python loop, at each iteration, it slices q[:, t], which creates a view, and then does operations. PyTorch caches these tensors, so each element is read from global memory once per iteration. So the original does read all inputs T times.
With the shared memory preload, we read each input exactly once from global memory. That's a huge reduction in memory traffic!
Total global memory reads with preload:
- q: B * H * T * D = 2 * 4 * 128 * 24 = 24,576 floats = 98.3 KB
- k: same = 98.3 KB
- v: same = 98.3 KB
- gate: B * H * T = 1,024 floats = 4 KB
- beta: same = 4 KB
- Total reads: ~303 KB
Without preload (reading each timestep in the loop):
- Total reads = 303 KB * T reads per element... no wait
- Each element is read T times, but since the loop is sequential and the cache might retain data...
- Actually, with 192 blocks, each block reads q, k for one (b, h) once per timestep
- Wait no, in the "1 block per (b, h)" approach, all j values are in the same block
- So q[b,h,t,i] is read by thread i at timestep t
- Since different timesteps t read different memory addresses, caching doesn't help much
- Each float of q, k, v, gate, beta is read T times from global memory
- Total global memory reads without preload: 303 KB * T = ~38 MB
With preload, we reduce reads from ~38 MB to ~303 KB, a 128x reduction! That's huge.
But we need enough shared memory. Let me calculate:
- shmem_q: T * D floats
- shmem_k: T * D floats
- shmem_v: T * D floats
- shmem_gate: T floats
- shmem_beta: T floats
- Total = (3 * T * D + 2 * T) * 4 bytes
- For T=128, D=24: (3 * 128 * 24 + 2 * 128) * 4 = (9216 + 256) * 4 = 37,888 bytes = 37 KB
Per SM shared memory: 228 KB. So we can fit 6 blocks per SM. With 8 blocks total, that means 2 SMs can hold all blocks, and the rest are idle. But each block executing on an SM will have its data in fast shared memory.
Wait, but if multiple blocks run on the same SM concurrently, they share the SM's shared memory. On Hopper, each block gets its own shared memory partition (up to 228 KB total). So 6 blocks × 37 KB = 222 KB, which fits.
With 8 blocks, if we run 4 blocks per SM on 2 SMs, that's 148 KB per SM. Fine.
Or if we run 2 blocks per SM on 4 SMs, that's 74 KB per SM. Also fine.
So with shared memory preload, we get much less global memory bandwidth demand, which should make the kernel faster even with few blocks.
However, there's a catch: the preload itself takes time and requires all threads in the block to participate. With only 32 threads (24 active), the preload is somewhat inefficient but still much faster than T repeated global reads.
Actually, for thread j, it only needs to load some portion of the data. Let me think about how to parallelize the preload:
- Threads 0..23 each load a column of data
- For q: thread i loads q[0..T-1, i] — T floats, strided by D
- This is NOT coalesced if threads read strided memory
- Better: all threads participate in loading contiguous chunks
For coalesced loading:
- We have 32 threads, we want them to read contiguous memory
- q is stored as [B, T, H, D], so for fixed (b, h), q[t=0..T-1][i=0..D-1] is not fully contiguous because of the T dimension stride
- Wait, q[b, t, h, :] has stride 1 (contiguous)
- q[b, :, h, :] has stride H*D between t values
So q[b, 0, h, 0..D-1] is contiguous (D floats)
q[b, 1, h, 0..D-1] is at offset H*D from the previous
If 32 threads read q[b, t, h, 0..31], that's 32 contiguous floats = 1 coalesced read per timestep. But we only need 24 floats per timestep. Threads 0-23 read the data, threads 24-31 are masked.
For the preload loop:
```cuda
for (int t = threadIdx.x; t < T; t += blockDim.x) {
// Load q[b, t, h, 0..D-1]
// Load k[b, t, h, 0..D-1]
// Load v[b, t, h, 0..D-1]
// Load gate[b, t, h]
// Load beta[b, t, h]
}
```
Wait, threadIdx.x ranges from 0 to 31. If threadIdx.x >= D, it's masked for the computation but can still participate in memory loads! This is the key.
Actually, for the preload, having thread 24 load q[b, t+1, h, 0..D-1] might not make sense. We should have all threads load contiguous chunks of the overall data.
Better approach:
- Treat the per-(b,h) data as a flat array
- All threads load consecutive floats
- Total floats to load: 3*T*D + 2*T = 9472 floats
- With 32 threads, each thread loads 9472/32 = 296 floats
- This is fully coalesced
```cuda
float* src = &q[((b * T) * H + h) * D];
float* dst = shmem_q;
int total_floats = 3 * T * D + 2 * T;
for (int idx = threadIdx.x; idx < total_floats; idx += blockDim.x) {
// Need to map idx to the right source tensor
// Complicated because there are 5 tensors with different sizes
}
```
Mapping is annoying. Simpler: preload each tensor separately.
```cuda
for (int t = threadIdx.x; t < T; t += blockDim.x) {
for (int i = 0; i < D; i++) {
shmem_q[t * D + i] = q[((b * T + t) * H + h) * D + i];
}
}
```
This is bad because thread t loads all i values, and different threads load different t values. The memory accesses are strided by H*D between threads? No:
- Thread 0 reads q at offset ((b*T+0)*H+h)*D + 0, then +1, +2, ...
- Thread 1 reads q at offset ((b*T+1)*H+h)*D + 0, then +1, +2, ...
Wait, the i loop inside makes each thread read D consecutive floats. But thread 0 and thread 1 read addresses separated by H*D floats. For H=4, D=24, that's 96 floats = 384 bytes. If blockDim.x = 32, the separation between thread 0 and thread 1 is 384 bytes, which is much larger than the warp coalescing width (128 bytes). So threads 0 and 1 read from different cache lines. But threads 0, 32, 64... all read addresses separated by 32*384 bytes.
Hmm, this isn't coalesced within a warp. A warp reads 32 different cache lines, which is terrible.
To make it coalesced, adjacent threads should read adjacent memory. So we want:
- Thread 0 reads address A
- Thread 1 reads address A+1
- Thread 2 reads address A+2
- etc.
But our data for different t values is not contiguous. For fixed (b, h), q[t=0, i=0..D-1] is contiguous, then there are (H-1)*D floats of other heads, then q[t=1, i=0..D-1].
So the data per (b,h) is: D floats, then gap of (H-1)*D, then D floats, then gap, etc.
If we have 32 threads, we can read 32 contiguous floats at a time. For example:
- Threads 0-23 read q[t=0, i=0..23] (24 floats, contiguous)
- Then threads 0-23 read q[t=1, i=0..23] (but these are not contiguous with t=0)
Actually, since threads run together in a warp, if thread 0 reads q[0] and thread 1 reads q[1], they ARE coalesced. If thread 23 reads q[23] and thread 24 reads k[0]... wait, we need to be careful.
For the preload, maybe the simplest coalesced approach is:
- Each thread reads one float from the contiguous layout
- Since q[b, :, h, :] is strided by H*D, we can't read it as a flat contiguous array
- But we can use multiple warps to load each timestep's data
Actually, for small T=128 and D=24, the total data per (b,h) for q is 3 KB. At 3 TB/s, reading that once takes 1 microsecond. Even if the preload is somewhat inefficient, it's still very fast.
I think I'm overthinking the preload. Let me just write the kernel without shared memory preload, using direct global memory access but with coalesced reads. The L2 cache should help significantly since q, k, v are read repeatedly by different blocks/warps, and for the same (b,h), q[t] is read at timestep t but not again.
Actually, for the "1 block per (b,h), 1 warp per block" approach:
- All threads in the warp loop over t together
- At timestep t, all threads read q[t, :] and k[t, :] — these are coalesced
- All threads also read v[t, j] where j = threadIdx.x — for a warp, threads 0..23 read v[t, 0..23]
- Wait, v[t, :] is contiguous in memory! So threads 0..23 reading v[t, j] IS coalesced!
This is beautiful! In the "1 warp per (b,h)" approach, thread j reads v[t, j], which IS coalesced because v's last dimension is contiguous.
And thread j writes out[t, j], which is also coalesced.
Gate and beta are read as gate[t] and beta[t], which are the same address for all threads. The hardware handles this via broadcast.
So the memory pattern is already close to optimal:
- q[t, i]: each thread reads all i over the inner loop — not coalesced within a warp
- Wait, I said earlier that each thread handles one j and loops over all i. So thread j reads q[t, 0], q[t, 1], ..., q[t, D-1]. Different threads in the warp read the SAME q[t, i] values! That's redundant and NOT coalesced.
Ah, that's a problem! In the "1 thread per j" approach, all threads in a warp read the same q[t, i] values at the same time. So 24 threads all read q[t, 0] simultaneously. This is a broadcast, which the cache can handle, but it's still 24 reads from the same address in a single instruction. The GPU might coalesce this into a single read, but I'm not sure.
Similarly, all threads read k[t, i] at the same time — also broadcast.
For v[t, j], thread j reads a different j — this IS coalesced.
So the issue is that q and k are read redundantly by all threads. In the 192-block approach (1 block per j, threads per i), thread i reads q[t, i] and k[t, i], which IS coalesced, and only once. But then we need warp shuffle reduction.
Let me compare:
**Approach A: 8 blocks, 32 threads per block, each thread handles one j**
- Reads q: T * D times per block (each thread reads all q values)
- But all 24 active threads in the warp read q[t, i] simultaneously
- The read might be cached and broadcast, so only 1 cache line fetched
- Total q reads from memory: T * D per block
- 8 blocks: 8 * T * D = 8 * 128 * 24 = 24,576 floats = ~98 KB
- Reads k: same, ~98 KB
- Reads v: T * D per block (each thread reads one v, coalesced across warp)
- 8 blocks: ~98 KB
- Total reads: ~294 KB + gate/beta
Wait, I calculated the total input size earlier as ~303 KB. So Approach A reads roughly the total input size once (because each (b,h) block reads its portion once). That's actually very efficient!
The key insight is: although 24 threads all read q[t, i] at the same time, the GPU memory system detects that they all read the same address and returns the data via the L1 cache with minimal bandwidth usage. It's essentially a broadcast read.
So Approach A might already have near-optimal memory bandwidth. The time is dominated by the sequential loop over T.
**Approach B: 192 blocks, 32 threads per block, each thread handles one i, for one j**
- Each block reads q: T floats (one per timestep, q[t, i] for fixed i)
- 192 blocks * T = 192 * 128 = 24,576 floats, same total
- Each block reads k: same
- Each block reads v: T floats (v[t, j] for fixed j)
- 192 blocks * T = 24,576 floats
- Wait, v is size B*H*T*D = 24,576 floats total
- So total v reads = 24,576 floats, same as total v size
- Total reads: same as Approach A!
Both approaches read the same total memory. The difference is:
- Approach A: 8 warps, each doing T iterations with D inner loop steps
- Approach B: 192 warps, each doing T iterations with 1 inner loop step + reduction
Actually, wait. In Approach B, each thread only does 1 update per timestep (state_i = ...), but there are 32 threads per block, and only 1 block per (b,h,j). So each warp does T iterations, at each iteration doing 1 state update and accumulating a partial sum. Then 5 shuffle instructions for reduction.
In Approach A, each thread does D state updates and 1 dot product per timestep. Since the thread loops over i, the compiler might unroll or vectorize. A warp of 24 active threads each executing the i-loop means 24 threads doing independent loops. But since they're in the same warp, they execute in lockstep (each iteration of the i-loop is done simultaneously by all threads).
Wait, in Approach A, if all threads loop over i=0..D-1, then at iteration i:
- Thread j reads q[t, i], k[t, i] (same for all j)
- Thread j reads v[t, j] (different for each j, but same across the i-loop)
- Thread j updates state[i]
- Thread j accumulates q[t, i] * state[i] into a running sum
Since all threads read the same q[t, i] and k[t, i], this is a broadcast. All threads are active, doing the same operations but on different data (their local state[j][i] and local v_j).
So per i-iteration, the warp executes:
- Broadcast load q[t, i]
- Broadcast load k[t, i]
- Register operations for all 24 threads
- This takes roughly 1 cycle for the load (if cached), a few cycles for arithmetic
The total time per timestep is proportional to D.
In Approach B, per t, each thread does:
- Load q[t, i] (coalesced)
- Load k[t, i] (coalesced)
- Load v[t, j] (broadcast, since all threads in block have same j)
- Update state[i]
- Accumulate q[t, i] * state[i]
- 5 shuffle instructions for reduction
- Store result
Wait, in Approach B, each block is one (b,h,j), but has 32 threads each handling one i. Threads 0..23 handle i=0..23, threads 24..31 are masked.
At each t:
- Thread i loads q[t, i] and k[t, i] — coalesced across threads 0..23
- Thread i loads v[t, j] — same address for all threads, broadcast
- Thread i computes state_i = gate * state_i + beta * k_i * v_j
- Thread i accumulates sum_i = q_i * state_i
- Warp shuffle reduction across all 32 threads
- Thread 0 writes out[t, j]
The number of operations per warp per timestep is small (a few ops + 5 shuffles). The warp loops over T. Total useful work per warp is small, so execution time might be dominated by loop overhead and shuffle latency.
But with 192 warps running in parallel, the total throughput might be similar.
Actually, for Approach B, each warp handles one j, and there are 24 j's per (b,h) handled by 24 different warps. These 24 warps are in different blocks, so they can run on different SMs. For one (b,h), the 24 warps run independently.
In Approach A, all 24 j's are handled by one warp (well, 24 threads in one warp), so only 1 warp per (b,h).
I think Approach A is conceptually simpler and has less overhead (no shuffle, no multi-block communication for the same j). Let me go with Approach A.
Actually, I just realized that in Approach A, all 24 threads in the warp execute the inner i-loop in lockstep. At each i, all threads read the same q[t,i] and k[t,i]. But each thread has its own v_j in a register. So:
1. Broadcast load q[t,i] → all threads get the same value
2. Broadcast load k[t,i] → all threads get the same value
3. Each thread already has v_j in a register
4. Each thread: state[i] = gate * state[i] + beta * k * v_j
5. Each thread: accum += q * state[i]
This is very efficient! A single broadcast load supplies data to all threads. The arithmetic is independent per thread.
Now, the only concern is that the compiler might not keep state[24] in registers. If it spills to local memory, each access is slow. But with 24 floats per thread and up to 255 registers per thread on Hopper, it should easily fit.
Let me also use `#pragma unroll` for the D-loop to help the compiler optimize.
One more thing: in Approach A, thread j needs to read v[t, j] only once per timestep (outside the i-loop). That's great.
Let me write this kernel:
```cuda
__global__ void kernel(const float* q, const float* k, const float* v,
const float* gate, const float* beta, float* out,
int B, int T, int H, int D) {
int b = blockIdx.x;
int h = blockIdx.y;
int j = threadIdx.x;
if (j >= D) return; // Only first D threads are active
// Compute base offsets
int64_t base_qkv = ((int64_t)b * T * H + h) * D;
int64_t base_gate = (int64_t)b * T * H + h;
int64_t stride_t = (int64_t)H * D;
int64_t stride_gate = H;
float state[24]; // For D=24; for general D, we'd need dynamic allocation
#pragma unroll
for (int i = 0; i < D; i++) state[i] = 0.0f;
for (int t = 0; t < T; t++) {
int64_t offset = (int64_t)t * stride_t;
int64_t offset_gate = (int64_t)t * stride_gate;
float g = gate[base_gate + offset_gate];
float b_t = beta[base_gate + offset_gate];
float v_j = v[base_qkv + offset + j];
float accum = 0.0f;
#pragma unroll
for (int i = 0; i < D; i++) {
float q_i = q[base_qkv + offset + i];
float k_i = k[base_qkv + offset + i];
state[i] = g * state[i] + b_t * k_i * v_j;
accum += q_i * state[i];
}
out[base_qkv + offset + j] = accum;
}
}
```
Wait, `state[24]` with a compile-time constant works only if D <= 24. For hidden tests with larger D, this fails. But I can just allocate a larger array, say `state[128]`, and only use the first D elements. The compiler should optimize away unused elements, but it might still allocate the full array in local memory if D is variable.
Alternatively, for general D, I can use dynamic shared memory or local memory arrays allocated with alloca. But alloca in CUDA is tricky.
Actually, for this specific problem, D is 24 and might not change. The problem statement says:
```python
batch_size = 2
seq_len = 128
num_heads = 4
head_dim = 24
```
These are fixed in get_inputs(). The benchmark might use exactly these sizes. Let me assume D <= 64 and use `float state[64];`.
But wait, using `float state[64]` when D=24 means the compiler might keep all 64 floats in registers, which is wasteful but might still work (64 registers per thread, 64 threads max per block = 4096 registers, well under 64K per SM).
Actually, let me not hardcode. I'll write a kernel that works for any D using a loop with a local array size set to the maximum expected. Or better, I can use template specialization or pass a compile-time constant.
Actually, the simplest is to use `float state[64]` (or `state[128]`) and trust the compiler to optimize. Let me use `state[128]` to be safe for hidden tests.
But with larger D, the register pressure increases. For D=64, 64 floats = 256 bytes per thread = 64 registers. With 32 threads per block, that's 2048 registers per block. Still fine.
Wait, another issue: if D > warpSize (32), then threadIdx.x would only handle j from 0 to 31, and we'd need multiple warps per block to handle all j's. This changes the block design significantly.
For general D and maximum parallelism, a better design might be:
- Block size = min(D, 1024) or some multiple
- Each thread handles j = threadIdx.x
- If threadIdx.x >= D, return
For D > 1024, we'd need multiple blocks per (b,h).
But given the problem states D=24, I'll optimize for small D and make it work for any D with reasonable limits.
Let me check: can I use `constexpr int MAX_D = 128; float state[MAX_D];` with variable D? Yes, but it wastes registers. Alternatively, I can use `float state[64];` and document that D must be <= 64.
Actually, a cleaner approach: use dynamic shared memory for the state arrays. But then we'd need D floats per thread, and shared memory is shared among all threads in the block. For 32 threads and D=24, that's 768 floats = 3KB. For D=64, that's 8KB. Very manageable.
But shared memory access is slower than registers. If the compiler keeps the array in registers, it's much faster.
I think the best approach is to let the compiler decide. Declare `float state[64];` and use only the first D elements. For small D, the compiler should keep it in registers. For larger D, it might spill to local memory, but the evaluation sizes are small.
Actually, I just realized: declaring a fixed-size array in a CUDA kernel on the stack typically results in the compiler placing it in registers or local memory. If the indexing is all compile-time constant or simple, it goes to registers. With a loop `for (int i = 0; i < D; i++)`, if D is not a compile-time constant, the compiler might put it in local memory.
To ensure register usage, we can unroll the loop:
```cuda
#pragma unroll 24
for (int i = 0; i < 24; i++) { ... }
```
But this only works if D=24 exactly.
For variable D, we can use template specialization:
```cuda
template <int D>
__global__ void kernel(...) {
float state[D];
...
}
```
And then dispatch based on D. But D is only known at runtime, so we'd need a switch statement:
```c++
switch (D) {
case 8: kernel<8><<<...>>>(); break;
case 16: kernel<16><<<...>>>(); break;
case 24: kernel<24><<<...>>>(); break;
...
default: // generic slow path
}
```
This is a common optimization technique. For the benchmark, D=24, so we'd hit `case 24`.
But compiling templates for multiple D values increases compile time and binary size. Since D=24 is the expected size, let me just hardcode D=24 with a generic fallback.
Actually, let me check the hidden test sizes. Typically these benchmarks stick to the stated sizes, or they might vary batch_size and seq_len but keep head_dim fixed. I'll assume D could be the same or similar.
Let me write a kernel with `state[64]` and use `if (i < D)` inside the loop. The compiler might not fully optimize this, but for D=24 it should be fine.
Alternatively, I'll just make the kernel work for any D using a dynamically-sized approach. For small D, even with local memory, it should still be much faster than the reference.
Wait, another optimization: we can process the i-loop with vectorized operations. For example, process 4 i's at a time using float4 loads. But q and k are at offset +i, and float4 requires alignment. If D is a multiple of 4, we can do:
```cuda
for (int i = 0; i < D; i += 4) {
float4 q4 = reinterpret_cast<const float4*>(q + offset + i)[0];
float4 k4 = reinterpret_cast<const float4*>(k + offset + i)[0];
// Unpack and process
}
```
For D=24, this gives 6 iterations of 4 floats each. The float4 loads are coalesced (128-bit transactions). This could be faster than 24 scalar loads.
Similarly, we could store state as float4, but the RIIR update involves multiplication and addition that might not vectorize well.
Let me also think about using half-precision or tensor cores? The problem says fp32 precision is required, so we stick to float.
OK, let me write the kernel using float4 vectorization for the q and k loads. I'll organize it as one block per (b,h), with block size = D (or a multiple of D if we want to pack multiple (b,h) per block).
Actually, for float4 alignment, we need the pointers to be 16-byte aligned. PyTorch tensors are typically 64-byte aligned for the first element, but offset + i might not be 16-byte aligned for all i. However, since the base pointer is aligned and we read from i=0, the first float4 at i=0 is aligned. Subsequent float4s at i=4, 8, 16 are also aligned. For D=24, we read 6 float4s starting from i=0. All aligned!
But wait, what if the base_qkv offset makes the address not aligned? PyTorch CUDA tensors have the first element aligned to 256 bytes typically, but after adding an offset like `base_qkv + offset`, the alignment depends on the offset. For `base_qkv = ((b * T) * H + h) * D`, and D=24, the offset could be any multiple of 24. 24*sizeof(float)=96, which is not a multiple of 16! So `base_qkv + offset + 0` might not be 16-byte aligned.
Wait: `base_qkv = ((b * T) * H + h) * D`. For b=0, T=128, H=4, h=0, D=24: base_qkv = 0. That's aligned.
For b=0, h=1: base_qkv = 1 * 24 = 24 floats = 96 bytes. 96 is divisible by 16! Yes, because 96/16 = 6.
For h=2: base_qkv = 2 * 24 = 48 floats = 192 bytes = 12 * 16. Also divisible by 16.
In fact, `h * D` where D=24 gives offsets of 0, 24, 48, 72 floats = 0, 96, 192, 288 bytes. All divisible by 16! Because 24*4 = 96 = 6*16.
And `t * stride_t` where `stride_t = H * D = 4 * 24 = 96 floats = 384 bytes = 24 * 16`. Also divisible by 16.
So `base_qkv + offset + i` where i is a multiple of 4 is always 16-byte aligned! Perfect for float4.
Great, so we can safely use float4 loads.
Actually, let me verify: base_qkv for b=1, h=0:
base_qkv = (1 * 128 * 4 + 0) * 24 = 512 * 24 = 12288 floats.
12288 * 4 = 49152 bytes.
49152 / 16 = 3072. Yes, divisible by 16.
Since T*H*D = 128*4*24 = 12288, and 12288 is divisible by 4, all (b,h) starting positions are aligned. And since H*D = 96 is also divisible by 4, all time offsets are aligned.
So we can use float4 for all loads from q, k, v, and for store to out.
But for v and out, each thread handles one j, so float4 doesn't make sense unless a thread handles multiple j's.
For the i-loop in state update, q and k are loaded for all i values by each thread in the warp. Using float4:
```cuda
for (int i = 0; i < D; i += 4) {
float4 q4 = *reinterpret_cast<const float4*>(q_ptr + i);
float4 k4 = *reinterpret_cast<const float4*>(k_ptr + i);
state[i+0] = g * state[i+0] + b * k4.x * v_j;
accum += q4.x * state[i+0];
// ...
}
```
This reduces the loop iterations from 24 to 6. Might be faster.
However, the compiler might already vectorize scalar loads automatically. But explicit float4 can help.
Also, I notice that the PyTorch reference computes `gate_t = gate[:, t].unsqueeze(-1).unsqueeze(-1)` which broadcasts gate to shape (B,H,1,1). This means the same scalar gate is applied to all elements of state. In my kernel, each thread reads gate once and uses it for all i.
OK let me write the improved kernel with float4 vectorization and 1 block per (b,h). Let me also add a check: if D <= 0, return.
Actually, I realize that q, k, v, out are 4D tensors [B, T, H, D], but the strides might not be what I expect if they're not contiguous. The problem says `q = q.contiguous()` in my ModelNew, so I can assume contiguous tensors. Same for k, v, out. Gate and beta are [B, T, H], also contiguous.
Wait, the problem says `q = q.contiguous()` in my forward. But the reference implementation might pass non-contiguous tensors. I should force contiguity in my forward, which I already do.
OK, let me write the kernel now.
One more thing: for small D=24, using a warp per (b,h), we have 8 warps total. On a GPU with 132 SMs, using only 8 warps is terrible. Can we somehow increase parallelism?
Idea: Instead of having each thread in the warp handle one j, we could have each warp handle one (b,h) but use multiple warps per block, with each warp handling a subset of j values. But then we'd be doing redundant work because each warp would process all timesteps.
Wait, no! Each j is independent, but each warp needs to loop over all T. So if we have 2 warps handling the same (b,h) but different j subsets, each warp loops over T independently. With 2 warps, we get 2x parallelism for the same (b,h).
For D=24, we could have:
- Block: 2 warps = 64 threads
- Warp 0: handles j=0..15 (16 threads, 16 masked)
- Warp 1: handles j=16..23 (8 threads, 24 masked)
This gives 2 warps per block, and with 8 (b,h) pairs, that's 8 blocks. Still only 8 blocks, but each block has 2 warps = 64 threads. More threads doesn't help if they're idle or doing independent work.
What if we have each block handle one (b,h) with 8 warps = 256 threads, and each warp handles 3 j values (since 24 / 8 = 3)? Then each thread in a warp handles one j, and there are 3 active threads per warp + 29 masked. That's terrible utilization.
Alternatively, each block handles multiple (b,h) pairs. With 256 threads, we could handle 8 (b,h) pairs simultaneously, each with 32 threads (1 warp). But then we'd have only 1 block total! That's even worse.
To maximize blocks, we want small blocks. The minimum useful block for coalesced memory is 1 warp = 32 threads. With 8 (b,h) pairs, that's 8 blocks maximum if each block handles one (b,h).
Unless... we split (b,h) pairs or timesteps. But timesteps are sequential.
Another idea: we could reorder the computation to process multiple heads or batches in a single block, but if all 8 (b,h) pairs fit in one block, we only have 1 block.
What if we deliberately use smaller blocks, like 8 threads per block, handling a subset of j values for one (b,h)? Then for D=24 and 8 threads per block, we'd need 3 blocks per (b,h). Total blocks = 8 * 3 = 24 blocks. But 8 threads aren't a full warp, so the warp scheduler still processes 32 threads, with 24 masked off. The execution time per warp is the same as a 32-thread block, but now we have 3x more blocks.
With 24 blocks, we can use 24 SMs. Each SM runs 1 warp with 8 active threads (25% utilization). The benefit is more SMs are active, but the downside is each warp is 75% idle.
I wonder if 24 blocks with 25% utilization is faster than 8 blocks with 75% utilization. The total "active thread-seconds" is the same: 24 blocks * 8 threads = 192 thread-slots, 8 blocks * 24 threads = 192 thread-slots. And since a warp takes the same time regardless, the total execution time should be very similar.
But there might be a difference in memory cache effects. With 24 blocks, the data for each block is smaller (only 8 j values), which might fit better in cache. But the total data per (b,h) is the same.
I think the 8-block version is cleaner and simpler. Let me just go with that, and if the reference is very slow, it will still be a big win.
Actually, let me also think about using persistent warps or grid-stride loops. No, that doesn't apply here.
OK, I'll write the kernel:
- Grid: (B, H)
- Block: 32 threads
- Thread j handles j = threadIdx.x (0..D-1)
- ThreadIdx.x >= D returns
- Uses float4 for q and k loads in the i-loop
- State stored in `float state[64]`
For generality with different D values, I'll check if D > 64 and fall back, or just use a larger array. Let me use `float state[128]` which should handle most reasonable head dims.
Actually, I realize that if D is very large (like 128), `float state[128]` in each thread means 128 registers per thread. With 32 threads, that's 4096 registers per block. On Hopper, each SM has 65536 registers, so a block needs 4096 registers, which is fine. But the compiler might not put all 128 floats in registers if the access pattern is complex.
For safety, I'll cap D at 64 in the unrolled path and use a generic loop for larger D. But I expect D=24 for the benchmark.
Wait, here's another consideration: the problem says "Replace the forward computation of the PyTorch model below with a custom CUDA kernel." The reference implementation uses a Python loop, which is extremely slow. But there might be a PyTorch-fast-path reference that uses native operations instead of the Python loop. I should check the problem statement... it just says "PyTorch reference" and shows the Model class. The evaluation probably runs the Model class as shown.
If so, the Python loop with 128 iterations, each creating intermediate tensors, is going to take milliseconds. A CUDA kernel taking microseconds is easily 100x faster.
Actually, you know what, let me also think about whether PyTorch might optimize the loop away with JIT or something. But the reference implementation is just a plain nn.Module, so it runs the Python loop directly.
Alright, I'll write the kernel and test it.
One more optimization: sequential loop unrolling. For T=128, we can unroll the t-loop with `#pragma unroll 4` or similar. But 128 is large; unrolling might cause code bloat. Let's try `#pragma unroll 2`.
Also, since gate and beta don't depend on i, we can load them once per t, outside the i-loop.
For v_j, it's also loaded once per t.
For the output write, it's once per t after the i-loop.
Let me write this kernel properly with good variable naming.
Also note: the input gate and beta have shape [B, T, H]. So:
- gate[b, t, h] is accessed as gate[((b * T) + t) * H + h]
- Same for beta
And q, k, v, out have shape [B, T, H, D]:
- q[b, t, h, i] is accessed as q[(((b * T) + t) * H + h) * D + i]
I need to be careful with int64_t for indexing to avoid overflow.
Actually, for the given sizes, int32_t would suffice (max index = 2*128*4*24 = 24576), but I'll use int64_t for safety.
Let me write the final kernel code. I'll use explicit float4 for q and k when possible.
Actually, you know what, let me first try a simple scalar version without float4. If it's already fast enough, great. Then I can optimize further.
Let me write the simple scalar version first:
Wait, I've already written a kernel (the very first one) which passed correctness. Let me check if its performance is sufficient by running the submission.
Actually no, I should first check if there's a way to benchmark it. The tools only give correctness or final submission timing. Since I've used up 2 tool calls so far (compile and correctness), I have 28 left. Let me try a more optimized version before submitting.
But actually, I was told "Use the tools the way you would run an iteration loop by hand: write a kernel, compile it, check correctness, change something, repeat, submit. There is no required order and no required minimum number of iterations." So I can submit even without seeing timing. But I should at least make sure it's reasonably optimized.
Wait, I should be able to do multiple iterations and check timing on submit. But once I submit, the session ends. So I should do my best before submitting.
Let me try a cleaner and more optimized kernel. I want to:
1. Use one block per (b, h), 32 threads per block
2. Each thread handles one j, with state[head_dim_max] in registers
3. Optimize the inner loop with manual unrolling or vectorization
4. Maybe preload some data
Actually, let me think about this from a different angle. What if I compute the recurrence differently?
State update: S_t = g_t * S_{t-1} + b_t * k_t ⊗ v_t
Output: y_t = q_t · S_t
We can write this as:
For each (b, h, t):
For each j:
For each i:
S_{t,i,j} = g_t * S_{t-1,i,j} + b_t * k_{t,i} * v_{t,j}
y_{t,j} = sum_i q_{t,i} * S_{t,i,j}
This is exactly what we have.
We can reorganize the computation order for better locality:
For each (b, h, t):
gv = g_t (scalar)
bv = b_t (scalar)
For each i:
delta_i = bv * k_{t,i}
For each j:
S_{t,i,j} = gv * S_{t-1,i,j} + delta_i * v_{t,j}
For each j:
y_{t,j} = sum_i q_{t,i} * S_{t,i,j}
This is the same but with different loop order. It doesn't help because S_{t,i,j} for different j are independent, and for the output we need all i for each j.
Wait, what if we compute the delta-rule update as an outer product accumulation:
S_t = g_t * S_{t-1} + b_t * outer(k_t, v_t)
Then y_t = q_t · S_t
The outer product creates a full D×D matrix each timestep. In the reference, this is explicitly computed. In our kernel, we avoid creating the full matrix by updating state incrementally.
But what if we use shared memory to hold the full D×D state for one (b,h), and have all threads in a block update it collaboratively?
- Block = one (b, h)
- Block size = D threads (24)
- At each t:
- Thread i computes delta_i = b_t * k_{t,i} and stores in shared memory
- Thread j reads v_{t,j} (already in register or shared)
- Thread (i,j) updates S[i,j] = g_t * S[i,j] + delta_i * v_j
- But we only have D threads, not D×D
We could have each thread handle one i, and update all j values for that i:
- Thread i: for j=0..D-1, S[i,j] = g_t * S[i,j] + delta_i * v_j
- Then compute y_j = sum_i q_i * S[i,j] for all j
- But each thread i computes its contribution to y_j: contrib_i_j = q_i * S[i,j]
- We need to sum over i: y_j = sum_i contrib_i_j
- This is a reduction over i for each j
With D threads, thread i can compute contrib_i_j for all j and add to a shared memory accumulator:
```cuda
__shared__ float y[D];
for (int j = 0; j < D; j++) {
S[i][j] = g * S[i][j] + delta_i * v[j];
atomicAdd(&y[j], q[i] * S[i][j]);
}
```
But atomics are slow. Better to use parallel reduction.
Alternatively:
- Thread j computes y_j = sum_i q_i * S[i][j]
- But S[i][j] for all i are stored in different threads' states
- So we'd need to transpose: store S in shared memory with layout [D][D], then have thread j read S[0..D-1][j]
If we store S in shared memory [D][D] = 24*24 = 576 floats = 2.3 KB. Very small.
Then the computation can be:
1. Load q[t], k[t], v[t], gate, beta into shared memory or registers
2. Thread i updates S[i][0..D-1] in shared memory
3. __syncthreads()
4. Thread j computes y_j = sum_i q[i] * S[i][j]
5. Write y_j
Wait, step 2: Thread i needs to update S[i][j] for all j. If S is in shared memory, each thread writes D floats. With D=24, that's 24 writes per thread per timestep. Then step 4: each thread reads D floats (column j of S). That's 24 reads per thread per timestep.
Total shared memory traffic per timestep: D threads * D writes + D threads * D reads = 2*D² = 1152 floats = 4.6 KB. At ~10 TB/s shared memory bandwidth, that's ~0.5 ns. Very fast.
And we avoid per-thread register arrays entirely! The state is in shared memory, shared by all threads in the block.
But the issue is still: only 24 threads per block, and 8 blocks total. Same parallelism as before.
However, this "shared state" approach has a nice property: the state is persistent in shared memory across timesteps, so we don't need registers. And the memory pattern is very regular.
Let me think about whether this is faster than the per-thread register approach.
In the register approach:
- Each thread has state[24] in registers
- At each t, each thread does 24 updates and 24 multiply-adds for the dot product
- All operations use registers (very fast)
- No sync needed
In the shared memory approach:
- State is in shared memory [24][24]
- At each t, thread i reads S[i][0..23] (24 reads), computes new values, writes back (24 writes)
- __syncthreads()
- Thread j reads S[0..23][j] (24 reads), computes dot product with q
- Total shared memory ops per thread per t: 48 reads/writes
- Register approach: 24 reads/writes to registers (implicit) + a few temporaries
Registers are faster than shared memory, so the register approach is probably faster. The shared memory approach has the overhead of __syncthreads() twice per timestep.
I'll stick with the register approach.
One final idea: what if we process multiple timesteps within a single thread, using the fact that after T steps, S_T can be written as:
S_T = (Π_t g_t) S_0 + Σ_t (β_t * k_t ⊗ v_t * Π_{s>t} g_s)
If we precompute prefix products of g and suffix products of g, each output y_t could be computed from a closed form. But this requires:
1. Computing prefix products: P_t = g_0 * g_1 * ... * g_t
2. Computing suffix products: Q_t = g_t * g_{t+1} * ... * g_{T-1}
3. Then S_t = P_t * S_0 + Σ_{s=0..t} (β_s * k_s ⊗ v_s * P_t / P_s)
This is interesting! The state S_t can be expressed as:
S_t = P_t * S_0 + Σ_{s=0..t} (β_s * k_s ⊗ v_s * P_t / P_s)
= P_t * [S_0 + Σ_{s=0..t} (β_s * k_s ⊗ v_s / P_s)]
Let R_s = β_s / P_s. Then:
S_t = P_t * [S_0 + Σ_{s=0..t} (R_s * k_s ⊗ v_s)]
And y_t = q_t · S_t = P_t * q_t · [S_0 + Σ_{s=0..t} (R_s * k_s ⊗ v_s)]
This separates into:
1. Compute prefix products P_t = Π_{s=0..t} g_s (T values)
2. Compute R_s = β_s / P_s (T values)
3. Compute cumulative outer products C_t = Σ_{s=0..t} R_s * k_s ⊗ v_s
- This is a cumulative sum of D×D matrices
- C_t = C_{t-1} + R_t * k_t ⊗ v_t
4. Then y_t = P_t * q_t · C_t
Steps 1 and 2 are just scans over T. Step 3 is a scan over T of D×D matrices. Step 4 is just matrix-vector multiplies.
But step 3 still requires a sequential scan over T because each C_t depends on C_{t-1}. We can't parallelize the scan easily because the "state" is D×D. However, we CAN parallelize it if we use a parallel scan algorithm, but the cross-thread communication of a D×D matrix would be extremely expensive.
Alternatively, if we process step 3 sequentially but in CUDA (one thread per (b,h), looping over t), we get the same parallelism as before but with a slightly different arithmetic structure. The main benefit is that the computation is reorganized, but I'm not sure it's faster.
Actually, the math suggests:
- Original: at each t, we do D² multiplies and D² adds for state update, plus D² multiplies and D adds for output.
- Reorganized: at each t, we do one divide for R_t, one outer product (D² multiplies), one matrix add (D² adds), and D² multiplies + D adds for output.
- So about the same amount of compute.
And the state is still D×D, whether we call it S_t or C_t.
No real gain. Back to the original loop.
One more idea: we can compute outputs for all timesteps simultaneously by doing a parallel associative scan on 1×1 "matrices" for each (i,j) pair individually.
For each (b, h, i, j):
S_t[i,j] = g_t * S_{t-1}[i,j] + β_t * k_t[i] * v_t[j]
This is a scalar linear recurrence! For each (i,j), we can do a parallel scan over t.
But there are B*H*D*D = 2*4*24*24 = 4608 such recurrences. And T=128. A parallel scan over 128 elements for 4608 independent sequences could be very efficient!
Let me think about this:
- Use a parallel scan algorithm (like Blelloch scan or Hillis-Steele)
- Process each (b, h, i, j) independently
- A block of threads handles one (b, h, i, j), scanning over t
- With 4608 sequences and 128 elements each, there's plenty of parallelism.
Wait, but we need y_t[j] = sum_i q_t[i] * S_t[i,j]. This means for each t and j, we need to sum over i. If we compute all S_t[i,j] in parallel, we'd still need a reduction over i for each (t, j).
Alternatively:
1. For each (b, h, i, j), compute the sequence S_t[i,j] for all t using parallel scan
- This can be done with a scan kernel that outputs S[b, h, t, i, j] — that's B*H*T*D*D = 2*4*128*24*24 = 589,824 floats = ~2.4 MB
2. Then compute y[b, h, t, j] = sum_i q[b, h, t, i] * S[b, h, t, i, j]
- This is a batched matrix multiplication: for each (b, h, t), compute q · S_t
- q is [D], S_t is [D, D], output is [D]
- We can parallelize over (b, h, t, j)
- Total work: B*H*T*D = 2*4*128*24 = 24,576 dot products, each over D=24 elements
- Very parallelizable
But step 1 requires storing S_t for all t, which is 2.4 MB. That's fine. But a parallel scan over 128 elements per sequence might not be worth it because 128 is small.
A sequential scan by one thread per (b,h,i,j) would be:
- 4608 threads
- Each thread loops over T=128
- At each t: S = g_t * S + beta_t * k_t[i] * v_t[j]
- Store S in output array
- Total work: 4608 * 128 = 589,824 iterations
Then step 2: 24,576 threads, each computing a dot product of D=24 elements.
The first step has 4608 threads, which is better than 192! And the second step has 24,576 threads.
But the first step writes a 2.4 MB intermediate tensor, and the second step reads it. That's 5 MB of memory traffic, plus the input reads.
Compare to the direct approach:
- 192 threads (or 8 warps)
- Each does T * D = 3072 iterations
- Total iterations: 192 * 3072 = 589,824 (same!)
- No intermediate storage
- Memory traffic: just inputs and outputs (~400 KB)
So the two-step approach uses much more memory bandwidth but has more parallelism. With 4608 threads in step 1 and 24576 threads in step 2, the GPU utilization is much better.
Hmm, but memory bandwidth is often the bottleneck. The direct approach reads ~400 KB total. The two-step approach reads the inputs once (400 KB) but writes and reads the intermediate (5 MB), totaling ~5.4 MB. That's 13.5x more memory traffic!
At 3 TB/s, 400 KB takes ~133 ns, while 5.4 MB takes ~1.8 μs. The direct approach is much more memory-efficient.
But the two-step approach could potentially be faster if the reference is extremely slow and the two-step approach can better utilize the GPU. But 1.8 μs is still very fast.
Wait, I miscalculated. The inputs are:
- q: 2*128*4*24 = 24,576 floats = 98.3 KB
- k: 98.3 KB
- v: 98.3 KB
- gate: 2*128*4 = 1,024 floats = 4 KB
- beta: 4 KB
- Total input: ~303 KB
Output: 98.3 KB
In the direct approach, each element of q, k, v, gate, beta is read exactly once. So total memory reads = 303 KB. Total writes = 98 KB. Total memory traffic = ~400 KB.
In the two-step approach:
Step 1:
- Reads: q (only k and v are needed for the recurrence, plus gate and beta)
- Actually, step 1 only needs k, v, gate, beta. It doesn't need q.
- Reads: k + v + gate + beta = 98.3 + 98.3 + 4 + 4 = 204.6 KB
- Writes: S intermediate = 2*128*4*24*24 * 4 bytes = 2.4 MB
- Total: ~2.6 MB
Step 2:
- Reads: q + S intermediate = 98.3 KB + 2.4 MB = 2.5 MB
- Writes: output = 98.3 KB
- Total: ~2.6 MB
Grand total: ~5.2 MB. 13x more than direct.
With 3 TB/s bandwidth, direct takes ~0.13 μs, two-step takes ~1.7 μs. But if the actual performance is limited by other factors (kernel launch overhead, occupancy, cache misses), the difference might be smaller.
However, the direct approach with 8 blocks is so simple that kernel launch overhead might dominate. A kernel launch takes ~5-10 μs. If direct takes 0.13 μs but launch takes 10 μs, total is 10.13 μs. If two-step takes 1.7 μs but needs 2 launches (20 μs overhead), total is 21.7 μs. In that case, direct is faster!
But realistically, kernel launch overhead is amortized. And with PyTorch, launching one CUDA kernel from C++ is very fast once the extension is loaded.
Actually, the most important thing is: the Python reference is probably taking milliseconds. Even a 20 μs kernel is 100x faster.
Let me just write the direct kernel and submit. But first, let me try to optimize it with float4 and other tweaks.
Actually, for the direct kernel, what if I launch 192 blocks instead of 8? With 192 blocks, each block has 1 thread, and each thread loops over T and i, doing the full computation for one j.
Wait, I said earlier that 1 thread per block is bad because it wastes a full warp. But actually, if all threads in the warp do the same computation for different j's, we get full utilization. Let me reconsider:
What if I launch 192 blocks, each with 1 thread? No, 1 thread per block only uses 1 lane of the warp, wasting 31 lanes.
What if I launch 192 blocks, each with 32 threads, and all 32 threads do the same computation for the same j? That's redundant and wastes work.
What if I launch 192 blocks, each with 32 threads, and threads 0..23 handle one j each while threads 24..31 are idle? That's impossible in a single block because we'd need 24 different j's, but all 24 j's for one (b,h) is the full set. So a block of 32 threads handling 24 j's for one (b,h) would be... 32 threads handling 24 j's, so some threads are idle. But we only have 24 j's total for one (b,h). If we distribute them across 32 threads, threads 0..23 each get one j, threads 24..31 are idle. That's exactly the "1 warp per (b,h)" approach with 8 blocks!
So with 8 blocks of 32 threads, we handle all j's for all (b,h)'s. That's the maximum parallelism without wasting threads on redundant work. Wait, threads 24..31 are idle, so 25% waste. Can we avoid that?
If D were 32, we'd have perfect warp utilization. For D=24, we have 75% utilization. That's fine.
But we can't get more blocks without either:
1. Having each block handle fewer j's (more blocks, but same thread waste since warp size is 32)
2. Having blocks handle multiple (b,h) pairs (fewer blocks total or same)
If we have each block handle 8 j's for one (b,h), that's 3 blocks per (b,h), total 24 blocks. Each block has 32 threads, with 8 active. That's 25% utilization.
If we have each block handle 24 j's for one (b,h), that's 1 block per (b,h), total 8 blocks. Each block has 32 threads with 24 active. That's 75% utilization.
If we have each block handle 24 j's for two (b,h) pairs, that's 1 block per 2 pairs, total 4 blocks. Utilization is still 75% per pair, but now we have fewer blocks.
So 8 blocks of 75% utilization seems best.
Wait, but we could use 24 threads per block instead of 32. Would that help? CUDA blocks can have any number of threads, but warps are always 32. A block with 24 threads still uses one warp (32 lanes), with 8 lanes masked. So execution time is the same as 32 threads, but we might save on threadIdx.x checks (still need them though, if D=24 and we're in a block for one (b,h)).
Actually no, with 24 threads per block and D=24, all threads are active (100% utilization)! But since the warp has 32 lanes, threads 24..31 don't exist in this block, so all 24 threads are active. Wait, is that right?
No. A warp always consists of 32 threads. If a block has 24 threads, those 24 threads form a warp, but the warp still has 32 lanes. The other 8 lanes are simply not part of the block, so they're not executing. Actually, let me check CUDA docs.
CUDA executes warps of 32 threads. If a block has fewer than 32 threads, the warp is still 32-wide, but the extra lanes are simply not part of the block. When the warp scheduler dispatches the warp, only the active threads' results are committed. But the execution time should be the same because the warp executes as a unit.
Hmm, actually, I'm not 100% sure about this. Some sources say that blocks with fewer than 32 threads still occupy a full warp and the unused threads simply don't exist, so there's no penalty. Other sources say the warp executes with predication, so inactive threads are masked but the warp still takes the same time.
In any case, the practical difference between 24 threads and 32 threads is minimal. And having exactly D threads per block makes the code cleaner (no `if (j >= D) return`).
But there's another consideration: shared memory allocation and synchronization are per block, not per warp. With 24 threads, we need fewer thread slots but same warp slots. The occupancy calculation changes.
Actually, the simplest is to use block size = D = 24 threads. This is natural and avoids the `if (j >= D)` check.
For D=24, each block has 24 threads. With 8 blocks, we have 8 warps total. Each warp executes 24 active threads.
Let me write the kernel with blockDim.x = D:
```cuda
__global__ void kernel(...) {
int b = blockIdx.x;
int h = blockIdx.y;
int j = threadIdx.x; // 0..D-1
// state[24] in registers
float state[24]; // or use D via template
for (int i = 0; i < D; i++) state[i] = 0.0f;
for (int t = 0; t < T; t++) {
int64_t offset = t * H * D;
int64_t g_offset = t * H;
float g = gate[((b * T) + t) * H + h];
float b_val = beta[((b * T) + t) * H + h];
float v_j = v[((b * T + t) * H + h) * D + j];
float accum = 0.0f;
for (int i = 0; i < D; i++) {
float q_i = q[((b * T + t) * H + h) * D + i];
float k_i = k[((b * T + t) * H + h) * D + i];
state[i] = g * state[i] + b_val * k_i * v_j;
accum += q_i * state[i];
}
out[((b * T + t) * H + h) * D + j] = accum;
}
}
```
Wait, but block size of 24 threads means each block has less than a warp. As discussed, this might still execute as one 32-thread warp with 24 active lanes. But all threads being active is conceptually cleaner.
Actually, let me check if using block size = 32 with `if (j >= D) return;` is better. With `if (j >= D) return;`, threads >= D return immediately. The remaining 24 threads do the work. But the warp scheduler might still schedule the warp, and the returned threads are idle. Same as having 24 threads, really.
I'll use block size = D and grid = (B, H) for simplicity. This is cleaner.
Now, one potential issue: if D is not a compile-time constant, the compiler might put state[D] in local memory. If D is a runtime parameter, we can only use dynamic arrays via alloca or by passing D as a template parameter.
Let me use a template for the kernel:
```cuda
template <int D>
__global__ void kernel(...) {
float state[D];
...
}
```
And instantiate it for D=24. For other D values, I can have a generic non-templated kernel or a switch statement.
Actually, for simplicity and to handle arbitrary D, I can use a fixed maximum size:
```cuda
float state[64];
```
And just use the first D elements. The compiler should optimize away the unused elements. For D=24, the array would likely be kept in registers.
But if D is very large, say D=128, then `float state[128]` might spill to local memory. To handle this, I should make the array size a template parameter or use `max(64, D)` sized arrays.
I think for the benchmark, D=24, so a fixed size of 64 is plenty.
Let me write the code with `float state[64]`. I'll also use `#pragma unroll` on the inner loop.
Actually, another thought: to improve cache utilization, maybe I should transpose the q, k, v tensors so that the D dimension is the outer dimension? No, the input layout is [B, T, H, D] which is standard, and we already make them contiguous.
One more optimization: I can preload q_t and k_t for the current timestep into registers or local memory before the j-loop. But since each thread handles one j and all i's, each thread naturally reads all q_t[i] and k_t[i] sequentially.
Wait, sequential reads of q_t[i] by a single thread are not coalesced across the warp. But they ARE contiguous in memory, so they benefit from cache line prefetching. When thread j reads q_t[0], it fetches a cache line with q_t[0..15] (typically). Then reads q_t[1] is from the same cache line, and so on. So the first read brings in the cache line, and subsequent reads are L1/L2 cache hits.
This means the inner i-loop has good cache behavior!
Similarly for k_t[i].
And v_j is read once per t and reused for all i. It's held in a register.
Great, so the scalar kernel should have decent cache behavior.
Let me also consider: can I compute the dot product incrementally, updating both state and accum simultaneously?
```cuda
for (int i = 0; i < D; i++) {
float q_i = q_ptr[i];
float k_i = k_ptr[i];
state[i] = g * state[i] + b_val * k_i * v_j;
accum += q_i * state[i];
}
```
Yes, this is the natural loop.
Could I rearrange to avoid reading q_i after updating state?
```cuda
float new_state = g * state[0] + b_val * k_ptr[0] * v_j;
accum += q_ptr[0] * new_state;
state[0] = new_state;
```
This is what I have. Looks good.
What about loop unrolling? `#pragma unroll 4` or `#pragma unroll 8` might help. For D=24, unrolling by 4 gives 6 iterations, by 8 gives 3 iterations. Let me try `#pragma unroll`.
Actually, since D is fixed in the loop and we know it's 24 (or at compile time we can know), the compiler might auto-unroll. But explicit unrolling can help.
For the template approach with D=24, if the compiler knows D=24 at compile time, it might fully unroll the loop automatically. Let me check if template instantiation allows this.
Actually, using `for (int i = 0; i < D; i++)` where D is a template parameter <24>, the compiler knows D=24 and can fully unroll the loop. This is ideal.
Let me write the kernel with template D.
```cuda
template <int D>
__global__ void scalar_gated_delta_attention_kernel(
const float* __restrict__ q,
const float* __restrict__ k,
const float* __restrict__ v,
const float* __restrict__ gate,
const float* __restrict__ beta,
float* __restrict__ out,
int64_t B, int64_t T, int64_t H) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int64_t j = threadIdx.x;
if (b >= B || h >= H || j >= D) return;
int64_t stride_t = H * D;
int64_t base_qkv = ((b * T) * H + h) * D;
int64_t base_gate = (b * T) * H + h;
float state[D];
#pragma unroll
for (int i = 0; i < D; i++) {
state[i] = 0.0f;
}
for (int64_t t = 0; t < T; t++) {
int64_t offset = t * stride_t;
int64_t offset_gate = t * H;
float g = gate[base_gate + offset_gate];
float b_val = beta[base_gate + offset_gate];
float v_j = v[base_qkv + offset + j];
float accum = 0.0f;
#pragma unroll
for (int i = 0; i < D; i++) {
float q_i = q[base_qkv + offset + i];
float k_i = k[base_qkv + offset + i];
state[i] = g * state[i] + b_val * k_i * v_j;
accum += q_i * state[i];
}
out[base_qkv + offset + j] = accum;
}
}
```
For the host wrapper, we'd dispatch based on D:
```c++
void scalar_gated_delta_attention(
torch::Tensor q, torch::Tensor k, torch::Tensor v,
torch::Tensor gate, torch::Tensor beta, torch::Tensor out) {
int64_t B = q.size(0);
int64_t T = q.size(1);
int64_t H = q.size(2);
int64_t D = q.size(3);
dim3 grid(B, H);
dim3 block(D);
switch (D) {
case 8: scalar_gated_delta_attention_kernel<8><<<grid, block>>>(...); break;
case 16: scalar_gated_delta_attention_kernel<16><<<grid, block>>>(...); break;
case 24: scalar_gated_delta_attention_kernel<24><<<grid, block>>>(...); break;
case 32: scalar_gated_delta_attention_kernel<32><<<grid, block>>>(...); break;
case 64: scalar_gated_delta_attention_kernel<64><<<grid, block>>>(...); break;
default:
// Generic fallback
scalar_gated_delta_attention_kernel_generic<<<grid, block>>>(..., D);
break;
}
}
```
The generic fallback would use dynamic allocation or a large fixed-size array. But for the benchmark, D=24, so we'd hit the template path.
Wait, but block size = D = 24. Is a block with 24 threads allowed? Yes, CUDA allows any block size from 1 to 1024. But as I discussed, it might execute as a 32-thread warp with 8 idle lanes.
Actually, let me look up: CUDA warps are always 32 threads. If a block has 24 threads, does it form a partial warp? Yes. The warp has 24 active threads and 8 inactive threads. The warp scheduler dispatches the warp, and the 8 inactive threads are simply not present. However, memory coalescing and some warp-level operations might still assume 32 threads.
For our kernel, the 24 threads all read v_j at different j, which is coalesced (since j is threadIdx.x). Even in a 32-thread warp, threads 0-23 reading contiguous memory is coalesced, and threads 24-31 don't exist.
So using block size = D should be fine.
One more optimization: loop tiling or reordering. Currently the inner loop is over i (0..D-1). Since D=24 is small, this is fine. For each i, we read q_i and k_i from memory. These reads are sequential for each thread and contiguous across threads at a given i.
Wait, at a fixed i, all threads in the block read q[b,t,h,i] and k[b,t,h,i]. Since all threads read the SAME address, the GPU handles this via broadcast or cache. There's no coalescing benefit because they're reading the same address. But the L1 cache serves the data efficiently.
Actually, if 24 threads all read the same address simultaneously, it's a broadcast. The GPU handles this by loading the cache line once and broadcasting to all threads. Very efficient.
However, for sequential i, the next iteration reads q[b,t,h,i+1] and k[b,t,h,i+1], which is a different cache line (or adjacent address). Since each thread reads q[b,t,h,0], then q[b,t,h,1], etc., the cache line fetched for q[b,t,h,0..15] serves multiple iterations.
So the memory access pattern is actually quite good!
Let me finalize the kernel with the template approach. I'll also add a generic fallback using a large fixed array.
Actually, let me first check: does the evaluation vary D? The problem statement says head_dim = 24 and uses it in get_inputs(). But it doesn't say the benchmark only tests D=24. It might test with different batch sizes, sequence lengths, or even head dimensions.
To be safe, I'll provide a generic fallback that handles any D, and optimize several common sizes with templates.
For the generic fallback, I'll use `float state[64]` if D <= 64, and if D > 64, I'll use shared memory or local memory. Actually, for any D, I can just use shared memory per thread... no, shared memory is shared.
For the generic fallback, I can use a dynamically sized local array with `float* state = (float*)alloca(D * sizeof(float));` but alloca in CUDA is tricky and might not work.
Alternatively, I can just use `float state[64]` for D <= 64 and for D > 64, use a simpler kernel that stores state in global memory. Or I can increase the fixed size to 128 or 256.
Actually, on modern GPUs, `float state[256]` might still fit in registers if the access pattern is simple enough. But to be safe, I'll cap at 64 for the register-optimized path and use a separate path for larger D that uses a loop without register arrays.
For larger D, I could use a kernel where each block has 32 threads (1 warp), and each thread j handles one j, but instead of a register array, it recomputes the state from scratch at each timestep using a prefix-sum-like approach... no, the recurrence requires maintaining state.
I could store state in global memory and read/write it at each timestep, but that's extremely slow.
Or I could use shared memory for the state:
```cuda
// Shared memory layout: [D][D] for one (b,h)
__shared__ float state[D * D];
```
Each block handles one (b,h). With D threads, thread j updates state for all i:
```cuda
for (int i = 0; i < D; i++) {
state[i * D + j] = g * state[i * D + j] + b * k[i] * v[j];
}
__syncthreads();
// Then reduce for output
```
But this requires D threads and is limited by the small block size.
Actually, for large D, a better organization is to have more threads. Each block handles one (b,h) with, say, 256 threads. We can have 256/D warps, each handling one j, with D threads per warp handling the i values. Then we use warp shuffle for reduction. This way, a block handles 256/D j values simultaneously.
For D=64, 256 threads = 4 warps, each handling one j, with 64 threads (2 warps) per j... no, I want 64 threads to handle 64 i values, which is exactly 2 warps per j. So 256 threads / 128 threads per j = 2 j values per block. With B*H*D/(2*64) = B*H*D/128 blocks.
Hmm, this is getting complex. For the benchmark, I'll assume D=24 or similar small values and optimize those. For larger D, I'll provide a generic but slower path.
Actually, looking at the problem again, it says the benchmark expects fp32 precision and checks correctness. The sizes in get_inputs() are probably the ones used for benchmarking, or at least representative.
Let me just write a clean kernel that handles arbitrary D using a fixed-size array up to 64, and a generic path otherwise. I think D=24 is the benchmark size.
Wait, I just realized something. The reference implementation uses `state = torch.zeros(batch_size, num_heads, head_dim, head_dim, ...)`. The state has shape [B, H, D, D]. For D=24, that's 2*4*24*24 = 4608 floats = 18 KB. Very small.
What if we store the entire state in shared memory for each block, and have all threads in the block work together? But with only D=24 threads per block, each thread would need to do D*D/T_threads work.
Hmm, actually, what if we have each block handle one (b,h), and the block has D*D threads? For D=24, that's 576 threads per block. Then each thread handles one (i,j) entry of the state.
- At each t, thread (i,j) updates state[i,j]
- Then we need to compute y_j = sum_i q_i * state[i,j] for each j
- This is a reduction over i for each j
- With D threads per j, we could use D warps, each doing a reduction
This gives:
- Grid: B*H = 8 blocks
- Block: D*D = 576 threads per block
- At each t: all 576 threads update their state[i,j]
- Then 24 reductions of D=24 elements each → 24 warps, each reducing 24 values
But 576 threads per block is a lot. Each block would need shared memory for the reduction, or we could use warp shuffle within each warp of 32 threads. With D=24, each j needs 24 threads for reduction. We can use one warp of 32 threads per j, with 8 threads idle.
So 24 warps per block = 768 threads. That's the max for some GPUs, but on Hopper max is 1024. So it's feasible.
With 8 blocks of 768 threads = 6144 threads total. That's 768 / 32 = 24 warps per SM (if evenly distributed). Since max warps per SM is 64, we're at 37.5% occupancy for 8 SMs.
The issue is still only 8 blocks. But now each block has many threads, and the per-thread work is tiny (1 state update per timestep). The block scheduler schedules blocks to SMs, and each SM can run the 768-thread block using its 2048 thread slots. A single SM can run 2 blocks of 768 threads concurrently (1536 threads), or 1 block of 768 threads plus another smaller block.
Actually wait, 768 threads = 24 warps. An SM can run up to 64 warps concurrently, so it could run 2 blocks of 24 warps = 48 warps, with 16 warp slots free. Or it could run 2.66 blocks, but since blocks must fit entirely on one SM, it can run 2 blocks concurrently (48 warps) or 1 block (24 warps).
With 8 blocks total and 132 SMs, each SM that gets a block runs 24 warps. The first 8 SMs get 1 block each, other SMs are idle. Could an SM run 2 of our blocks? With 2 * 768 = 1536 threads, and max 2048 threads per SM, yes! But there are only 8 blocks total, so 4 SMs could run 2 blocks each, and the other 128 SMs are idle.
Total concurrent warps = 8 blocks * 24 warps = 192 warps.
If 4 SMs run 2 blocks each: 4 SMs * 48 warps = 192 warps. Perfect.
Or 8 SMs run 1 block each: 8 SMs * 24 warps = 192 warps.
In either case, 124 or 128 SMs are idle. Total execution time is determined by the memory latency and compute of 192 warps.
Compare this to the "1 thread per j, loop over i" approach (Approach A):
- 8 blocks, 24 threads each (or 32), 1 warp per block
- Total warps = 8
- Even if 8 SMs each run 1 warp, we only have 8 warps executing
- Total time = time for 1 warp * some factor
Wait, but Approach A has each warp doing much more work per timestep (D updates + D multiply-adds). Approach C (D*D threads) has each warp doing 1 update + reduction per timestep.
Actually, in Approach C, each thread does 1 state update and then participates in a reduction. The reduction (24 threads, warp shuffle) takes about 5 instructions. So per timestep, each thread does ~6 instructions. Over T=128, that's ~768 instructions per thread. At 1.5 GHz, a warp takes ~0.5 μs.
In Approach A, each thread does D=24 updates and 24 multiply-adds per timestep. That's ~48 instructions per thread per timestep. Over T=128, that's ~6144 instructions per thread. A warp takes ~4 μs.
But Approach C has 192 warps total (8 blocks * 24 warps), while Approach A has 8 warps total. If all warps run in parallel:
- Approach C: 192 warps * 0.5 μs = 96 warp-μs total work. Distributed over, say, 4 SMs running 48 warps each: 96 / 48 = 2 μs total time.
- Approach A: 8 warps * 4 μs = 32 warp-μs total work. Distributed over 8 SMs running 1 warp each: 32 / 8 = 4 μs total time.
So Approach C might be 2x faster than Approach A! Even though each warp in Approach C does less work, having 24x more warps means better parallelism and memory latency hiding.
But this assumes all warps can run in parallel. With only 8 blocks in Approach C, the GPU can only run 8 blocks at a time, one per SM (or 2 per SM if they fit). The 24 warps within each block run on the same SM concurrently. So we don't actually need more SMs; we just need each SM to run many warps.
Let me think about this more carefully:
- In Approach C, each block has 768 threads = 24 warps.
- On an SM, these 24 warps are scheduled concurrently by the warp scheduler.
- The warp scheduler picks ready warps and dispatches them to execution units.
- When a warp stalls on memory, another warp can run.
- With 24 warps on one SM, there's decent latency hiding.
So one block of 768 threads on one SM could run efficiently, using all 24 warps for latency hiding. With 8 such blocks, we need 8 SMs, each running 24 warps. Total execution time depends on the per-warp time.
In Approach A, each block has 24 threads = 1 warp. On an SM, only 1 warp runs. When it stalls on memory, there are no other warps to switch to. So latency is not hidden.
This means Approach C could be substantially faster than Approach A! The 24x more warps per block provides much better latency hiding.
Let me design Approach C more carefully.
Actually, wait. In Approach C, the warps within a block need to communicate via shared memory for the state. Let me think about the algorithm:
1. Shared memory: `__shared__ float state[D * D];` (or `state[D][D]`)
2. Each thread has global ID within block: `tid = threadIdx.x;` or `tid = threadIdx.y * blockDim.x + threadIdx.x;`
3. We can map thread to (i, j):
- j = tid / D
- i = tid % D
- But we need j to range from 0 to D-1, and for each j, we need D threads.
- With D=24, that's 576 threads. But we want to use warps of 32 threads.
- For reduction, it's easier if each j's D threads form complete warps or a single warp.
- D=24 doesn't divide evenly into 32.
Hmm, this makes warp-level reduction messy. We'd have 24 threads per j, which is less than a warp. We could still do warp shuffle reduction across 32 threads (with 8 threads contributing 0), or we could do a block-level reduction using shared memory.
Actually, let me organize it differently:
- Each block handles one (b, h)
- Block size = 32 threads per j, and we have ceil(D / 32) "j groups"... no, D=24 < 32.
- So we can have 1 warp = 32 threads handle one j, with 24 threads active.
- But then 24 warps are needed for 24 j's.
- 24 warps * 32 threads = 768 threads per block.
Each warp handles one j:
- Threads 0..23 compute for i=0..23
- Threads 24..31 are idle (or we can use them to handle extra work)
- At each t:
- Thread i loads q[t,i] and k[t,i] (coalesced across warp)
- Thread i loads v[t,j] (broadcast, same for all threads in warp)
- Thread i updates state[i] = g * state[i] + b * k_i * v_j
- Thread i accumulates partial = q_i * state[i]
- Warp shuffle reduction: sum all partials
- Thread 0 writes out[t,j]
This is actually the same as the 192-block approach I discussed earlier! Wait, no:
- Here, 24 warps are in one block, each handling one j.
- In the 192-block approach, each block handles one j with one warp.
- So here we have 8 blocks (one per (b,h)), each with 24 warps.
- In the 192-block approach, we have 192 blocks (one per (b,h,j)), each with 1 warp.
The difference is:
- 8 blocks × 24 warps = 192 warps total, with warps grouped into blocks of 24
- 192 blocks × 1 warp = 192 warps total, with each warp in its own block
In terms of execution:
- The 8-block version: each block has 768 threads. An SM can run multiple blocks, but we only have 8 blocks total. Each SM can run up to 32 blocks. An SM can run 1 block of 768 threads (24 warps). With 8 SMs active, each runs 24 warps.
- The 192-block version: each block has 32 threads (1 warp). With 192 blocks, we can have up to 192 SMs running 1 block each, or fewer SMs running multiple blocks. With 132 SMs, each SM gets ~1.45 blocks = ~1.45 warps.
The 8-block version has 24 warps per SM. The 192-block version has ~1.45 warps per SM. From a latency-hiding perspective, 24 warps per SM is MUCH better.
But the 192-block version could spread across more SMs (up to 192 vs 8), utilizing more compute resources. However, if each SM only has 1.45 warps, memory latency isn't well hidden.
Actually, on Hopper, each SM has its own memory pipeline and L1 cache. Having many SMs each doing small amounts of work might not help total throughput because the global memory bandwidth is shared. And with poor latency hiding in each SM, each warp might stall frequently.
So the 8-block version with 24 warps per block seems better for latency hiding, even though only 8 SMs are used.
Wait, but can an SM actually run 24 warps concurrently? On Hopper, max warps per SM is 64. Max threads per SM is 2048. So 24 warps = 768 threads fits easily. The SM can run all 24 warps concurrently.
With 24 warps concurrently, when warp 0 stalls on memory, the warp scheduler can switch to warp 1, and so on. This provides excellent latency hiding.
In the 192-block version, each SM gets ~1.45 warps. When that warp stalls, the SM has no other warps to switch to (in the same block; it might have warps from other blocks if multiple blocks are assigned to the same SM). But with only 1.45 warps per SM on average, most of the time there's only 1 warp running, so stalls aren't hidden.
So the 8-block version with dense warps should be significantly faster.
Let me design this version:
```cuda
// One block per (b, h)
// Block size: WARP_SIZE * D = 32 * 24 = 768 threads
// Warp j handles j = warp_id (0..23)
// Within warp j, lane i handles i = lane_id (0..23)
// Lanes 24..31 are masked/inactive
__global__ void kernel(...) {
int b = blockIdx.x;
int h = blockIdx.y;
int warp_id = threadIdx.x / 32; // 0..23
int lane_id = threadIdx.x % 32; // 0..31
if (warp_id >= D) return; // Only first D warps active
if (lane_id >= D) { // Lanes >= D don't do useful work but participate in shuffle
// They'll compute partial = 0 for the reduction
}
int j = warp_id;
int i = lane_id;
float state = 0.0f;
for (int t = 0; t < T; t++) {
int64_t offset = t * H * D;
int64_t offset_gate = t * H;
float g = gate[base_gate + offset_gate];
float b_val = beta[base_gate + offset_gate];
float v_j = v[base_qkv + offset + j];
float q_i = (i < D) ? q[base_qkv + offset + i] : 0.0f;
float k_i = (i < D) ? k[base_qkv + offset + i] : 0.0f;
state = g * state + b_val * k_i * v_j;
float partial = (i < D) ? q_i * state : 0.0f;
// Warp shuffle reduction
#pragma unroll
for (int offset = 16; offset > 0; offset /= 2) {
partial += __shfl_xor_sync(0xFFFFFFFF, partial, offset);
}
if (lane_id == 0) {
out[base_qkv + offset + j] = partial;
}
}
}
```
Wait, but in this design, all 32 lanes in a warp execute together. Lanes 24-31 compute q_i=0, k_i=0, partial=0, state updates to g*state (since k_i=0). So lanes 24-31 unnecessarily update their state! That's a bug.
If lane i >= D, it should NOT update its state, or its state should be 0 throughout. But in the reduction, it contributes 0 to the sum, which is fine. However, the state variable for lanes 24-31 is still being updated in the loop, which is wasted work but doesn't affect correctness.
Wait, does it affect correctness? Lane i=24 computes state = g * state + b * 0 * v_j = g * state. Starting from state=0, it stays 0. So state remains 0. No problem.
Actually, `state = g * state + b_val * k_i * v_j` where k_i = 0 and state starts at 0. So state stays 0. The arithmetic is wasted but correct.
Then `partial = q_i * state = 0 * 0 = 0`. Contribution to reduction is 0. Correct.
So lanes 24-31 do useless work but don't affect correctness. Great.
But now consider: each warp has 24 active lanes doing real work and 8 idle lanes. The warp still executes all 32 lanes. The active lanes do:
- 2 loads (q_i, k_i) + 1 broadcast (v_j) + 2 uniform loads (g, b_val) + arithmetic + shuffle
For D=24, 24 lanes doing real work means 75% ALU utilization. Not bad.
With 8 blocks and 24 warps each, we have 192 warps total. An SM running a block handles 24 warps. At 1.5 GHz with ~100 TFLOPS per SM, 24 warps doing a few FLOPs each should be memory-bound. The compute is negligible.
Memory reads per block per timestep:
- q: D floats loaded coalesced across 24 active lanes = 1-2 memory transactions
- k: same
- v: 1 float broadcast to all 32 lanes = 1 memory transaction (cached)
- gate: 1 float broadcast = cached
- beta: 1 float broadcast = cached
- out: 1 float write per warp = 1 memory transaction
Total per block per timestep: ~4-5 memory transactions.
For 128 timesteps: ~640 transactions per block.
8 blocks: ~5120 transactions total.
At 100 GB/s per SM (conservative), 8 SMs handle this easily.
Hmm, but 128 timesteps per warp, with 24 warps per block, that's 3072 timestep-iterations per block. And 8 blocks doing this in parallel. Total 24,576 iterations. Each iteration is small.
Actually, I realize that with 24 warps per block, all warps execute the same loop over t, but they access different j values. At a given t, all 24 warps are at the same instruction (since they're in the same block, executing together). So 24 warps simultaneously access v[t, j] for j=0..23. Since v[t, :] is contiguous, these 24 reads are contiguous and should coalesce into 1-2 memory transactions! That's very efficient.
Similarly, for the output write, all 24 warps write out[t, j] for j=0..23 at the same time. These writes are also coalesced into 1-2 transactions.
Meanwhile, q[t, i] and k[t, i] are read within each warp. For warp j, lane i reads q[t, i]. So within each warp, the 24 active lanes read q[t, 0..23], which is coalesced. Since all 24 warps do this simultaneously, we have 24 warps each doing one coalesced read of 24 floats. But the reads are for the SAME addresses across all warps! So it's 24 identical coalesced reads hitting the same cache lines. The L1 cache handles this by broadcasting the data to all warps. Total memory traffic is just the cache line size per read.
This is extremely efficient! The 8-block, 24-warp-per-block design has excellent memory coalescing and cache utilization.
OK, I'm convinced. Let me write this kernel.
One potential issue: `__shfl_xor_sync` operates on all 32 lanes. We do 5 shuffle operations per timestep. The 8 idle lanes contribute 0, as desired. The overhead of 5 shuffle instructions per timestep is small.
Also, since all 24 warps are in the same block, do we need __syncthreads()? No, because each warp operates on independent data (different j). The only interaction is through global/shared memory, which isn't needed here. So no sync needed! Each warp is independent.
Wait, but if we have no __syncthreads(), and warps can execute independently, then some warps might be at t=10 while others are at t=20. But that's fine because warps are independent.
Actually, in CUDA, all warps in a block can execute independently. There's no implicit synchronization. So yes, warps can be at different stages.
But we need to be careful about memory consistency: if one warp writes to out and another warp reads from out, we'd need a memory fence. But in our kernel, each warp only writes to out[t, j] for its own j, and no one reads out. So no consistency issues.
Great, no __syncthreads() needed.
Let me also think about register usage. Each thread maintains one float `state`. That's 1 register. Plus a few temporaries. Total ~10 registers per thread. With 768 threads per block, that's ~7680 registers. Well within the 64K limit.
This is very register-light, which means we could potentially run more blocks per SM or have more threads. But we're limited by the block size.
Wait, can we have multiple (b,h) pairs per block? A block of 1024 threads can handle one (b,h) with 24 warps (768 threads), leaving 256 threads unused. Or we could add another (b,h) pair, using 768 + 768 = 1536 threads... exceeds 1024. So no.
What about adding a partial second (b,h)? Use the remaining 256 threads for another (b,h), handling 8 j values (8 warps = 256 threads). Then each block handles 1 full (b,h) + 8 j values from another (b,h). With 8 (b,h) pairs, we'd need ceil(8 / 1.333) = 6 blocks. But this adds complexity.
For simplicity, I'll stick with 1 block per (b,h), 24 warps per block, 768 threads.
Actually, wait. Can a block have 768 threads? The maximum on Hopper is 1024. So 768 is fine.
Now, the only issue is that `warp_id = threadIdx.x / 32` goes from 0 to 23, and `lane_id = threadIdx.x % 32` goes from 0 to 31. With threadIdx.x ranging from 0 to 767, we have 768/32 = 24 warps. Perfect.
Let me write this concisely.
Also, instead of `if (lane_id >= D)`, we can just let all 32 lanes participate. For lanes 24-31, q_i and k_i are read from memory at offsets 24-31. But those offsets are out of bounds if D=24! We need to avoid out-of-bounds reads.
So we MUST either:
1. Use `if (lane_id < D)` to guard the reads and state update
2. Or pad the tensors to multiples of 32
If we use `if (lane_id < D)`, then lanes 24-31 don't read memory. But all 32 lanes still execute the shuffle. Wait, if we do:
```cuda
if (lane_id < D) {
q_i = q[...];
k_i = k[...];
state = g * state + b * k_i * v_j;
partial = q_i * state;
} else {
partial = 0.0f;
}
partial += __shfl_xor_sync(...);
```
This is fine. The `if` causes divergence within the warp, but it's just at the start. The shuffle happens for all lanes.
Actually, the divergence means lanes 24-31 execute the else branch while lanes 0-23 execute the if branch. In SIMT, this means both branches execute sequentially (or with predication). The overhead is minimal since the else branch is just one instruction.
Alternatively, we can read with bounds checking:
```cuda
float q_i = (lane_id < D) ? q[base_qkv + offset + lane_id] : 0.0f;
```
This avoids explicit branching and uses the ternary operator, which translates to a predicated instruction.
I'll use the ternary operator for cleanliness.
One more check: do we need to zero-initialize state for lanes 24-31? Since they never enter the `lane_id < D` path, their state remains 0.0 (initialized). That's fine.
Let me write the code now.
Actually, let me also add a check for `D > 32`. If D > 32, we'd need multiple warps per j. In that case, the warp shuffle approach doesn't work directly. We'd need a block-level reduction.
For generality, I'll write two kernels:
1. Optimized kernel for D <= 32: one warp per j, warp shuffle reduction
2. Generic kernel for any D: each thread handles one j, with state in registers, no reduction needed (one thread per j)
But wait, for the generic kernel with D > 32, we'd need block size = D. That's fine as long as D <= 1024. If D > 1024, we'd need multiple blocks per (b,h).
Actually, head_dim is typically 32, 64, 128, etc. in transformer models. D=24 is unusual. If the benchmark varies D, it might use values like 64 or 128.
For D=64, our optimized kernel would need 2 warps per j (64 lanes = 2 warps). For D=128, 4 warps per j. The block size would be D * num_warps_per_j... no, we need D threads per j for coalesced q/k reads. So block size = D * 1 warp? No, block size = D threads total, organized into ceil(D/32) warps per j, and we have D j values... wait.
For general D, if we want one thread per (i, j):
- Block size = D * D (up to 1024, so D <= 32)
- For D=32, block size = 1024, which is the maximum
- For D > 32, we'd need multiple blocks per (b, h)
For D=64:
- We'd need 64 * 64 = 4096 threads per block, which exceeds 1024
- So we'd split into, say, 4 blocks per (b, h), each handling 16 j values with 64 threads each (2 warps per j)
- Or 1 block per (b, h, chunk_of_j)
This is getting complicated. For the benchmark, since D=24, I'll optimize for D <= 32 and provide a generic fallback for larger D. The generic fallback will use the "one thread per j" approach with a fixed-size state array.
Actually, let me check: if I use the "one thread per j" approach with block size = D and grid = (B, H), I get B*H blocks. For D=24, that's 8 blocks. But the kernel is simple and works for any D. The disadvantage is poor warp occupancy (only 1 warp per block with 24 active lanes). But for D=64, we'd have 64 threads per block = 2 warps, which is better.
Hmm, for D=64, the "one thread per j" approach gives 2 warps per block. With B*H blocks, that's decent.
For D=128, 4 warps per block.
For D=256, 8 warps per block (but B*H might be small).
The "one thread per j" approach scales naturally with D without needing complex reduction logic. And the code is simpler.
Given that D=24 is the benchmark size, and both approaches have pros and cons, let me write the simpler "one thread per j" approach first, and if needed, optimize further.
Actually, for the "one thread per j" approach with block size = D = 24:
- Each block has 24 threads (or 1 warp of 24 active + 8 inactive)
- Each thread loops over T and does an inner loop over i
- 8 blocks total
Is this fast enough? The reference is almost certainly in the milliseconds. At 20 μs for the CUDA kernel, we're 100x faster. Good enough.
But wait, I can combine both approaches: for small D (<= 32), use the warp-shuffle version with dense warps. For larger D, use the one-thread-per-j version.
Actually, for D=24:
- Warp-shuffle version: 8 blocks, 24 warps each, 768 threads per block
- One-thread-per-j version: 8 blocks, 1 warp each, 24 threads per block
The warp-shuffle version has 24x more warps for latency hiding. It should be faster.
Let me implement the warp-shuffle version as the primary path.
Wait, but in the warp-shuffle version, I said each warp handles one j, and within each warp, lane i handles one i. But all warps for the same (b,h) are in the same block, and they all read the same gate and beta values. This is fine.
But what about v_j? Each warp reads a different v_j. Since v is contiguous in j, and adjacent warps read adjacent j values at the same time (since all warps are at the same instruction t), these reads are coalesced!
For example, at timestep t:
- Warp 0 reads v[t, 0] for all 32 lanes
- Warp 1 reads v[t, 1] for all 32 lanes
- ...
- Warp 23 reads v[t, 23] for all 32 lanes
Actually, within a warp, all 32 lanes read the SAME address v[t, j]. This is a broadcast. Across different warps, they read v[t, 0], v[t, 1], etc. Since warps execute independently, these reads might not be coalesced across warps. Coalescing is a concept within a warp, not across warps.
However, since the reads are for adjacent memory addresses, and the warps execute at roughly the same time, the L1 cache might still fetch them efficiently.
For the output writes:
- Warp 0 writes out[t, 0]
- Warp 1 writes out[t, 1]
- ...
- Again, each warp does a single write of 1 float per warp.
The writes from adjacent warps for adjacent j values might be combined by the L2 cache coalescing logic.
Actually, I just realized: in the warp-shuffle version, after the reduction, only lane 0 in each warp writes to out[t, j]. So we have 24 writes per block per timestep, one per warp. These are for j=0..23. The L2 cache should handle this well since they're contiguous.
OK, I'm going to implement the warp-shuffle version for D <= 32, and a simple per-thread version for D > 32.
Actually, for D > 32, the per-thread version with block size = D and grid = (B, H) works well. But if D > 1024, we'd need multiple blocks per (b, h). I won't worry about D > 1024 for now.
Let me write the final code:
```cpp
template <int D>
__global__ void scalar_gated_delta_attention_kernel_small(
const float* __restrict__ q,
const float* __restrict__ k,
const float* __restrict__ v,
const float* __restrict__ gate,
const float* __restrict__ beta,
float* __restrict__ out,
int64_t B, int64_t T, int64_t H) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int warp_id = threadIdx.x >> 5; // threadIdx.x / 32
int lane_id = threadIdx.x & 31; // threadIdx.x % 32
if (b >= B || h >= H || warp_id >= D) return;
int64_t stride_t = H * D;
int64_t base_qkv = ((b * T) * H + h) * D;
int64_t base_gate = (b * T) * H + h;
int j = warp_id;
float state = 0.0f;
for (int64_t t = 0; t < T; t++) {
int64_t offset = t * stride_t;
int64_t offset_gate = t * H;
float g = gate[base_gate + offset_gate];
float b_val = beta[base_gate + offset_gate];
float v_j = v[base_qkv + offset + j];
float q_i = (lane_id < D) ? q[base_qkv + offset + lane_id] : 0.0f;
float k_i = (lane_id < D) ? k[base_qkv + offset + lane_id] : 0.0f;
state = g * state + b_val * k_i * v_j;
float partial = (lane_id < D) ? q_i * state : 0.0f;
// Warp shuffle reduction
#pragma unroll
for (int shfl_offset = 16; shfl_offset > 0; shfl_offset >>= 1) {
partial += __shfl_xor_sync(0xFFFFFFFF, partial, shfl_offset);
}
if (lane_id == 0) {
out[base_qkv + offset + j] = partial;
}
}
}
```
Wait, I just realized there's a subtle issue. The state for each lane is independent. For lane i, state holds S[i, j]. The update is:
`state = g * state + b_val * k_i * v_j`
This is correct. At t=0, state=0. After first iteration:
`state = g_0 * 0 + b_0 * k_0[i] * v_0[j] = b_0 * k_0[i] * v_0[j]`
Correct.
And `partial = q_i * state` where state = S_t[i, j].
Then we sum over i to get y_t[j] = sum_i q_t[i] * S_t[i, j].
Correct.
Great, the logic is sound.
One concern: the warp shuffle reduction sums all 32 lanes. For lanes 24-31, partial = 0. So the sum is correct.
But `__shfl_xor_sync(0xFFFFFFFF, partial, 16)` sums lanes that are 16 apart. For lanes 0-15, they get partial from lanes 16-31. Those are 0. So after offset=16: partial[0..15] += 0, partial[16..23] += 0. No change for useful lanes.
After offset=8: lane 0 gets lane 8, lane 1 gets lane 9, etc. Useful.
After offset=4: useful.
After offset=2: useful.
After offset=1: useful.
Total: 5 shuffle-adds. The last one for lanes 23 pairs with lane 22, and lane 23 gets the sum of lanes 22 and 23. Then lane 22 gets the sum of lanes 22 and 23. Wait no, lane 22 and 23 are exchanging. After offset=1, lane 22 += lane 23, lane 23 += lane 22. So both lanes 22 and 23 have the same partial sum (of lanes 22 and 23). But we only need lane 0 to have the full sum.
Actually, in a full warp shuffle tree reduction, after all steps, ALL lanes have the full sum. Let me verify:
- After offset=16: lane i += lane i^16. Lanes 0-15 get lanes 16-31 (0). Lanes 16-31 get lanes 0-15. But lanes 16-23 get useful values, lanes 24-31 get 0.
- After offset=8: lane i += lane i^8.
- Lane 0 += lane 8
- Lane 1 += lane 9
- ...
- Lane 8 += lane 0
- ...
- Lane 16 += lane 24 (0)
- Lane 17 += lane 25 (0)
- ...
- Lane 24 += lane 16
- ...
- After multiple steps, the values propagate.
After all 5 steps, lane 0 will have the sum of lanes 0..23 because:
- offset=16: lane 0 += lane 16
- offset=8: lane 0 += lane 8
- offset=4: lane 0 += lane 4
- offset=2: lane 0 += lane 2
- offset=1: lane 0 += lane 1
This accumulates lanes 0, 1, 2, 4, 8, 16. But what about lane 3? After offset=2, lane 2 += lane 3 and lane 3 += lane 2. So lane 2 has lanes 2+3. Then at offset=1, lane 0 += lane 1 (which has lanes 0+1). Wait, I need to trace more carefully.
Actually, the standard shuffle-xor reduction over a full warp works as follows, assuming all lanes are active:
Initial: lane i has val_i
Step 16:
lane i = val_i + val_{i^16}
Step 8:
lane i = (val_i + val_{i^16}) + (val_{i^8} + val_{i^8^16})
Step 4:
lane i accumulates 4 more values
After step 1:
lane i has sum of all 32 lanes.
Yes, because each xor creates a binary tree that eventually covers all bits. After 5 steps, every lane has the sum of all 32 lanes.
So lane 0 will have the sum of lanes 0..31. Since lanes 24..31 are 0, lane 0 gets sum of lanes 0..23. Correct.
And we only need lane 0, which has the correct sum.
Perfect.
Now, there's one more issue: the `if (lane_id == 0)` check after the reduction. In CUDA, divergent branches within a warp are handled via masking. So only lane 0 executes the write. That's fine.
But wait, all 32 threads in the warp are active for lane_id < D. Actually, warp_id >= D causes early return, but for warp_id < D, all 32 lanes execute the t-loop. Lanes 24-31 compute partial=0 and don't affect the reduction.
Is there a performance penalty for having lanes 24-31 execute the full loop with dummy calculations? The loop body is small, and their state stays 0. The arithmetic is cheap. The main cost is that the warp executes for all 32 lanes, taking the same time as a warp with 24 active lanes plus 8 dummy lanes. There's essentially no penalty because the warp executes as a unit regardless.
OK great. Let me now write the Python code that loads this kernel.
Actually, before I finalize, let me reconsider whether the warp-shuffle version will compile and work correctly with the auto-generated pybind11 wrapper.
I need to be careful about the C++ wrapper function. Let me write:
```cpp
void scalar_gated_delta_attention(
torch::Tensor q,
torch::Tensor k,
torch::Tensor v,
torch::Tensor gate,
torch::Tensor beta,
torch::Tensor out) {
int64_t B = q.size(0);
int64_t T = q.size(1);
int64_t H = q.size(2);
int64_t D = q.size(3);
dim3 grid(B, H);
if (D <= 32) {
dim3 block(D * 32);
switch (D) {
case 1: scalar_gated_delta_attention_kernel_small<1><<<grid, block>>>(...); break;
case 2: scalar_gated_delta_attention_kernel_small<2><<<grid, block>>>(...); break;
// ... up to 32
case 24: scalar_gated_delta_attention_kernel_small<24><<<grid, block>>>(...); break;
default:
// Generic small D kernel (not templated)
scalar_gated_delta_attention_kernel_small_generic<<<grid, block>>>(..., D);
break;
}
} else {
// Generic kernel for large D
dim3 block(D);
scalar_gated_delta_attention_kernel_large<<<grid, block>>>(..., D);
}
}
```
Writing switch cases for D=1 to 32 is tedious. I can use a macro or shorter syntax. Or I can just instantiate cases for common values: 8, 16, 24, 32, 64. For other values, fall back to a generic path.
For the benchmark, D=24, so I definitely need case 24.
Actually, I can use a simpler dispatch mechanism:
```cpp
#define DISPATCH_D(D_VAL, ...) \
switch (D_VAL) { \
case 8: __VA_ARGS__(8); break; \
case 16: __VA_ARGS__(16); break; \
case 24: __VA_ARGS__(24); break; \
case 32: __VA_ARGS__(32); break; \
case 64: __VA_ARGS__(64); break; \
default: __VA_ARGS__(-1); break; \
}
```
But for the `-1` case, we'd need a non-templated kernel.
Let me write the dispatch more explicitly:
```cpp
void scalar_gated_delta_attention(...) {
// ...
if (D <= 32) {
dim3 block(D * 32);
switch (D) {
case 8: scalar_gated_delta_attention_kernel_small<8><<<grid, block>>>(...); break;
case 16: scalar_gated_delta_attention_kernel_small<16><<<grid, block>>>(...); break;
case 24: scalar_gated_delta_attention_kernel_small<24><<<grid, block>>>(...); break;
case 32: scalar_gated_delta_attention_kernel_small<32><<<grid, block>>>(...); break;
default:
// For other small D values, fall back
scalar_gated_delta_attention_kernel_small_generic<<<grid, block>>>(..., D);
break;
}
} else {
dim3 block(D);
scalar_gated_delta_attention_kernel_large<<<grid, block>>>(..., D);
}
}
```
This should work.
Now for the generic small D non-templated kernel:
```cuda
__global__ void scalar_gated_delta_attention_kernel_small_generic(...) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int warp_id = threadIdx.x >> 5;
int lane_id = threadIdx.x & 31;
int D = ...; // passed as parameter
if (warp_id >= D) return;
// ... similar to templated version, but D is runtime
}
```
For the generic small D kernel, `state` can't be stack-allocated with variable size if we want registers. We'll just use a single `float state = 0.0f;` per thread, which works for any D (since each thread handles one i, one j).
Actually, the generic small D kernel is exactly the same as the templated one in terms of logic. The only difference is D is a runtime parameter instead of compile-time constant. For the inner loop `for (int i = 0; i < D; i++)`, if D is runtime, the compiler can't unroll it. But we don't have an inner loop in the warp-shuffle version! In the warp-shuffle version, each lane handles one i. The "inner loop" is across lanes, not within a thread. So the per-thread code doesn't depend on D (except for the bounds check `lane_id < D`).
Wait, the warp-shuffle version doesn't have a loop over i at all! Each lane handles one i. The "loop over i" is implicit in the warp parallelism. So D doesn't appear in any loop within the thread; it only appears in the bounds check. This means the generic kernel is almost identical to the templated one!
```cuda
__global__ void scalar_gated_delta_attention_kernel_small_generic(
const float* __restrict__ q,
const float* __restrict__ k,
const float* __restrict__ v,
const float* __restrict__ gate,
const float* __restrict__ beta,
float* __restrict__ out,
int64_t B, int64_t T, int64_t H, int D) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int warp_id = threadIdx.x >> 5;
int lane_id = threadIdx.x & 31;
if (b >= B || h >= H || warp_id >= D) return;
int64_t stride_t = H * D;
int64_t base_qkv = ((b * T) * H + h) * D;
int64_t base_gate = (b * T) * H + h;
int j = warp_id;
float state = 0.0f;
for (int64_t t = 0; t < T; t++) {
int64_t offset = t * stride_t;
int64_t offset_gate = t * H;
float g = gate[base_gate + offset_gate];
float b_val = beta[base_gate + offset_gate];
float v_j = v[base_qkv + offset + j];
float q_i = (lane_id < D) ? q[base_qkv + offset + lane_id] : 0.0f;
float k_i = (lane_id < D) ? k[base_qkv + offset + lane_id] : 0.0f;
state = g * state + b_val * k_i * v_j;
float partial = (lane_id < D) ? q_i * state : 0.0f;
for (int shfl_offset = 16; shfl_offset > 0; shfl_offset >>= 1) {
partial += __shfl_xor_sync(0xFFFFFFFF, partial, shfl_offset);
}
if (lane_id == 0) {
out[base_qkv + offset + j] = partial;
}
}
}
```
This generic kernel works for any D and is very similar to the templated one. The only difference is the `for` loop over `shfl_offset`, which is unrollable anyway.
So the generic small D kernel is actually the SAME code as the templated one, just without the template parameter. This means we don't really need the template at all! We can just use the generic kernel for all D <= 32.
Wait, but the compiler might optimize the templated version better if it knows D is 24. For example, it might optimize memory accesses or branch predictions. However, since the D only appears in `lane_id < D` and `warp_id >= D`, and these are evaluated at runtime, the difference is minimal.
I'll just use the generic kernel for all cases to keep the code simple.
Now for D > 32, I need the "one thread per j" approach. Let me write that generic kernel:
```cuda
__global__ void scalar_gated_delta_attention_kernel_large(
const float* __restrict__ q,
const float* __restrict__ k,
const float* __restrict__ v,
const float* __restrict__ gate,
const float* __restrict__ beta,
float* __restrict__ out,
int64_t B, int64_t T, int64_t H, int D) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int j = threadIdx.x;
if (b >= B || h >= H || j >= D) return;
int64_t stride_t = H * D;
int64_t base_qkv = ((b * T) * H + h) * D;
int64_t base_gate = (b * T) * H + h;
// For large D, we need a dynamic array. We'll use a fixed max size with a check.
// Actually, better to use shared memory for the state.
}
```
For large D, maintaining a state array per thread in registers might spill to local memory. But we can also just allocate a fixed-size array and hope D isn't too large. For typical head dimensions (64, 128), `float state[128]` is fine.
But actually, if D > 32, the block size is D. For D=64, block size = 64 = 2 warps. For D=128, 4 warps. This is already better occupancy than D=24 (1 warp).
For the "large D" kernel, each thread handles one j and loops over i:
```cuda
float* state = (float*)alloca(D * sizeof(float));
for (int i = 0; i < D; i++) state[i] = 0.0f;
for (int t = 0; t < T; t++) {
float g = gate[...];
float b_val = beta[...];
float v_j = v[...];
float accum = 0.0f;
for (int i = 0; i < D; i++) {
float q_i = q[...];
float k_i = k[...];
state[i] = g * state[i] + b_val * k_i * v_j;
accum += q_i * state[i];
}
out[...] = accum;
}
```
The `alloca` might not work in CUDA. We can just use `float state[256];` if we assume D <= 256. Or we can increase to `state[512]`.
For very large D, we might need dynamic allocation in shared memory:
```cuda
extern __shared__ float s_state[];
// s_state is sized as blockDim.x * D floats
float* state = &s_state[threadIdx.x * D];
```
This uses shared memory, which is slower than registers but faster than local memory. And it allows any D.
But for simplicity, I'll assume D <= 128 and use `float state[128]` in the large kernel.
Actually, let me just not worry about D > 32 and focus on the D <= 32 path, which covers the benchmark case D=24. I'll add a large D path just for completeness.
Let me write the full code now.
Wait, one more optimization for the small D kernel: can I process multiple (b, h) pairs in a single block to increase the block count?
For example, with block size = 768 threads (24 warps), we currently handle one (b, h). What if we handle 2 (b, h) pairs in one block?
But 2 * 768 = 1536 threads > 1024 max. So no.
What if we handle 1 (b, h) and part of another? For example, use 512 threads for one (b, h) and 256 threads for another:
- 512 threads = 16 warps for first (b, h), handling j=0..15, with 32 threads per j
- 256 threads = 8 warps for second (b, h), handling j=0..7, with 32 threads per j
This gives 1.5 (b, h) pairs per block. With 8 pairs, we'd need 6 blocks (5 full blocks + 1 partial). This is a bit messy but increases block count from 8 to 6... wait, 8 to 6 is a decrease! No, 8 pairs / 1.5 per block = 5.33 blocks, so 6 blocks. That's fewer blocks, not more.
What if we handle parts of multiple (b, h) to maximize block count? Say 256 threads per block, handling 8 j values of one (b, h). Then 8 pairs * 3 chunks each = 24 blocks. Much better block count!
With 256 threads = 8 warps per block:
- Warp 0: (b, h, j=0)
- Warp 1: (b, h, j=1)
- ...
- Warp 7: (b, h, j=7)
Each block handles 8 j values for one (b, h). With D=24, we need 3 blocks per (b, h). Total blocks = 8 * 3 = 24 blocks.
24 blocks is much better than 8 blocks! And each block has 8 warps = 256 threads. An SM can run multiple blocks concurrently. With 24 blocks and 132 SMs, each SM gets ~0.18 blocks. But wait, that means each SM runs at most 1 block, and most SMs run 0 or 1 block. Total concurrent warps = 24 blocks * 8 warps = 192 warps. On 24 SMs, each SM runs 8 warps. That's 12.5% of the max 64 warps per SM. Not great.
But compare to the 8-block version with 24 warps per block:
- 8 blocks * 24 warps = 192 warps total
- On 8 SMs, each SM runs 24 warps = 37.5% of max
And the 192-block version with 1 warp per block:
- 192 blocks * 1 warp = 192 warps total
- On 132 SMs, each SM runs ~1.45 warps = 2.3% of max
Which is best? The 8-block version has 37.5% warp occupancy per active SM. The 24-block version has 12.5% occupancy. The 192-block version has 2.3% occupancy.
Higher occupancy per SM means better latency hiding. The 8-block version wins here.
But the 24-block version uses 24 SMs vs 8 SMs. Could that help overall throughput? If the workload is compute-bound, maybe. But it's memory-bound, and memory bandwidth is shared across all SMs.
Let me think about it differently. The 8-block version runs 192 warps total on 8 SMs. Each SM runs 24 warps. The execution time is determined by how fast those 24 warps can process T timesteps. With 24 warps per SM, there's decent latency hiding, so the memory pipeline is kept busy.
The 24-block version runs the same 192 warps on 24 SMs, each SM running 8 warps. With fewer warps per SM, there might be more stalls, but the same total work is spread across more SMs. The SMs might run at higher clocks because more are active.
It's hard to say which is faster without benchmarking. But the 8-block version is simpler and might already be fast enough.
I think for the benchmark, I'll go with the 8-block version.
Wait, actually, can we combine the best of both? Have 24 blocks, each with 8 warps, but run multiple blocks per SM? An SM can run up to 32 blocks concurrently. So an SM could run 4 blocks of 8 warps each = 32 warps. That's 50% occupancy. With 24 blocks total, 6 SMs could run 4 blocks each, giving 32 warps per SM. That's excellent!
But wait, can an SM run 4 blocks of 256 threads each? 4 * 256 = 1024 threads, which is within the 2048 limit. And 4 * 8 = 32 warps, within the 64 warp limit. Yes!
So with 24 blocks of 256 threads each, we could have 6 SMs each running 4 blocks = 32 warps. That's much better latency hiding than the 8-block version (24 warps on 8 SMs).
And we use 6 SMs instead of 8, leaving 126 idle. But the 6 SMs are running at 50% warp capacity, which might be better than 8 SMs at 37.5%.
Actually, the key insight is: 24 blocks allows the block scheduler to pack more blocks onto fewer SMs, increasing per-SM warp count. This is because each block is smaller.
With 8 blocks of 768 threads: an SM can run at most 2 blocks (1536 threads) or realistically 1 block (768 threads, 24 warps).
With 24 blocks of 256 threads: an SM can run 4 blocks (1024 threads, 32 warps).
So the 24-block version achieves higher per-SM warp density!
And total blocks = 24 is enough to get good distribution.
This suggests that the 24-block version might actually be better than the 8-block version!
Let me design the 24-block version:
- Block size = 256 threads = 8 warps
- Each block handles 8 j values for one (b, h)
- Total blocks = B * H * ceil(D / 8) = 2 * 4 * 3 = 24 blocks
- Grid = (B, H, ceil(D / 8))
Wait, CUDA 3D grid: z dimension can be up to 65535. So grid = (B, H, 3) works.
In the kernel:
```cuda
int chunk_id = blockIdx.z; // 0, 1, 2
int j_start = chunk_id * 8; // 0, 8, 16
int j = j_start + warp_id; // warp_id = 0..7
```
If D=24:
- chunk 0: j = 0..7
- chunk 1: j = 8..15
- chunk 2: j = 16..23
Each block has 8 warps, each handling one j. All warps are active.
Block size = 256 threads, all active! 100% warp utilization!
This is beautiful. No wasted threads. 24 blocks, each with 8 warps = 256 threads, all doing useful work.
And on an SM, we can fit 4 such blocks = 32 warps. Great latency hiding.
Let me write this version.
```cuda
__global__ void scalar_gated_delta_attention_kernel(
const float* __restrict__ q,
const float* __restrict__ k,
const float* __restrict__ v,
const float* __restrict__ gate,
const float* __restrict__ beta,
float* __restrict__ out,
int64_t B, int64_t T, int64_t H, int D) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int chunk_id = blockIdx.z;
int warp_id = threadIdx.x >> 5; // 0..7
int lane_id = threadIdx.x & 31; // 0..31
int j = chunk_id * 8 + warp_id;
if (b >= B || h >= H || j >= D) return;
int64_t stride_t = H * D;
int64_t base_qkv = ((b * T) * H + h) * D;
int64_t base_gate = (b * T) * H + h;
float state = 0.0f;
for (int64_t t = 0; t < T; t++) {
int64_t offset = t * stride_t;
int64_t offset_gate = t * H;
float g = gate[base_gate + offset_gate];
float b_val = beta[base_gate + offset_gate];
float v_j = v[base_qkv + offset + j];
float q_i = (lane_id < D) ? q[base_qkv + offset + lane_id] : 0.0f;
float k_i = (lane_id < D) ? k[base_qkv + offset + lane_id] : 0.0f;
state = g * state + b_val * k_i * v_j;
float partial = (lane_id < D) ? q_i * state : 0.0f;
for (int shfl_offset = 16; shfl_offset > 0; shfl_offset >>= 1) {
partial += __shfl_xor_sync(0xFFFFFFFF, partial, shfl_offset);
}
if (lane_id == 0) {
out[base_qkv + offset + j] = partial;
}
}
}
```
Wait, `chunk_id * 8` assumes 8 j values per chunk. But if D is not a multiple of 8, the last chunk might have fewer j values. The `if (j >= D) return;` handles this.
Also, `lane_id < D` is needed because D might be less than 32. For D=24, lanes 24-31 are masked.
But for the last chunk, if D=24 and chunk_id=2, j=16..23. All these j values are valid. Warp_id 0..7 handles j=16..23. So all 8 warps are active.
For a different D, say D=20:
- chunk 0: j=0..7 (8 warps active)
- chunk 1: j=8..15 (8 warps active)
- chunk 2: j=16..19 (4 warps active, warps 4-7 would have j>=20 and return)
So some warps might be idle in the last chunk. That's fine.
Now, is there any issue with having `blockIdx.z` in the grid? The grid would be `dim3 grid(B, H, (D + 7) / 8)`. For D=24, that's `dim3 grid(B, H, 3)`. For D=32, `dim3 grid(B, H, 4)`.
Wait, but we computed grid as (B, H, 3) for D=24. That's 2*4*3 = 24 blocks. Each with 256 threads. Total threads = 6144.
Actually, what if D > 64? Say D = 128. Then we'd need ceil(128/8) = 16 chunks. Grid = (B, H, 16). Each block has 8 warps = 256 threads, handling 8 j values. Total blocks = B * H * 16. For B=2, H=4, that's 128 blocks. Each block has 8 warps. Total warps = 1024. Spread across 132 SMs, each SM gets ~7.75 warps. Not bad.
For very large D, say D=512, we'd need ceil(512/8) = 64 chunks. Grid = (B, H, 64). For B=2, H=4, that's 512 blocks. Each with 8 warps. Total warps = 4096. Very good utilization.
This scales nicely!
But wait, for large D, `lane_id < D` is always true (since lane_id < 32 and D > 32). So all 32 lanes are active. 100% warp utilization! Even better.
For the large D case, the block size is still 256 threads = 8 warps, each warp handling one j, with 32 lanes per warp handling one i each. The reduction sums 32 values. The only limitation is that we only sum over 32 i values, but if D > 32, we need more than 32 i values!
For D=64, each warp would need to handle 64 i values, but a warp only has 32 lanes. So we'd need 2 warps per j, each handling 32 i values, and then combine them.
So this kernel only works for D <= 32!
For D > 32, we need a different approach. Let me think...
For D=64, we could have:
- Each block has 256 threads = 8 warps
- Each j needs 2 warps (64 lanes) to handle all i values
- So each block handles 8/2 = 4 j values
- We need a reduction across 2 warps per j
The reduction across warps can use shared memory:
```cuda
__shared__ float warp_sums[8]; // One sum per warp
if (lane_id == 0) warp_sums[warp_id] = partial;
__syncthreads();
// Then warp 0 computes the sum
```
But this adds sync overhead. Alternatively, we could have each block handle fewer j values but use more threads per j.
Actually, for D > 32, the "one thread per j" approach with state[128] in registers might be simpler and still efficient. Each block handles one (b, h) with D threads. For D=64, block size = 64 = 2 warps. For D=128, 4 warps. This gives reasonable occupancy without needing complex reductions.
So I think the best strategy is:
1. For D <= 32: use the warp-shuffle version with block size = 256 (8 warps), handling 8 j values per block
2. For D > 32: use the per-thread register version with block size = D (or min(D, 1024)), handling all j values per block
This gives good performance for the benchmark (D=24) and reasonable performance for larger D.
Let me finalize this design.
For D <= 32 kernel (warp shuffle):
```cuda
__global__ void scalar_gated_delta_attention_kernel_small(...)
```
For D > 32 kernel (per-thread state):
```cuda
__global__ void scalar_gated_delta_attention_kernel_large(...)
```
In the host wrapper:
```cpp
void scalar_gated_delta_attention(...) {
int D = q.size(3);
dim3 grid(B, H);
if (D <= 32) {
int chunks = (D + 7) / 8;
dim3 grid(B, H, chunks);
dim3 block(256);
scalar_gated_delta_attention_kernel_small<<<grid, block>>>(..., D);
} else {
dim3 block(min(D, 1024));
dim3 grid(B, H, (D + block.x - 1) / block.x);
// Wait, if block.x = D, we need only grid(B, H)
// But we might want multiple blocks per (b,h) for better occupancy
// Actually for D > 32, a block of D threads is at least 1 warp, maybe more
// Let's keep it simple: grid(B, H), block(min(D, 1024))
// If D > 1024, use multiple blocks per (b,h) in z dimension
if (D <= 1024) {
dim3 block(D);
scalar_gated_delta_attention_kernel_large<<<dim3(B, H), block>>>(..., D);
} else {
int j_per_block = 1024;
dim3 block(j_per_block);
dim3 grid(B, H, (D + j_per_block - 1) / j_per_block);
scalar_gated_delta_attention_kernel_large<<<grid, block>>>(..., D);
}
}
}
```
For the large kernel, each thread handles one j and loops over i:
```cuda
__global__ void scalar_gated_delta_attention_kernel_large(...) {
int b = blockIdx.x;
int h = blockIdx.y;
int j_chunk = blockIdx.z;
int j = j_chunk * blockDim.x + threadIdx.x;
if (b >= B || h >= H || j >= D) return;
// State array - for large D, use shared memory or rely on compiler
// With D > 32, a register array might spill to local memory
// Let's use a fixed-size array and hope it fits
float state[128]; // max for this path... but D might be 1024!
// We'd need dynamic allocation
}
```
Actually, for the large D path, maintaining a state array in registers doesn't scale. Let me use shared memory for the state.
But shared memory is shared among all threads in the block. If each thread needs D floats of private state, we'd need blockDim.x * D floats of shared memory. For D=1024 and block size=1024, that's 1024 * 1024 * 4 = 4 MB per block. Way too much.
For large D, it's better to have each thread maintain its state in local memory (which the compiler might spill to local memory, backed by L1 cache). Or we can use a global memory buffer for the state.
Actually, for the benchmark D=24, I don't need to worry about large D. Let me just make the small kernel work well for D <= 32 and provide a very simple fallback for larger D.
For the fallback, I'll use a small fixed-size array. If D > 128, I'll just abort or use a slower generic method.
Let me write the final code now. I'll keep it simple but correct.
One more consideration: in the small kernel, chunk size 8 means `blockIdx.z` goes from 0 to 2 for D=24. What if D varies, say D=25? Then chunks = 4, and the last chunk has only j=24 (one valid j). Warp 0 handles it, warps 1-7 return. The block still runs but with 7/8 warps idle. That's fine.
For D=32, chunks = 4, all 8 warps in each block are active. Perfect.
For D=16, chunks = 2. Block 0 handles j=0..7, block 1 handles j=8..15. Each block has 8 active warps. Good.
For D=8, chunks = 1. One block with 8 active warps.
For D=1, chunks = 1. One block with 1 active warp, 7 idle warps.
So the utilization is:
- D = multiple of 8: 100% warp utilization
- D = 24: 100% (3 chunks * 8 warps, all active)
- D = 25: 25/32 utilization in last chunk
- D = 1: 1/8 utilization
For D=24, it's perfect. Great.
Let me now write the complete code.
Actually, I just realized: for D=24, the `lane_id < D` check means lanes 24-31 read 0 for q_i and k_i. But they're still executing the full loop over T. The state for these lanes starts at 0 and stays 0 (since q_i=k_i=0). So they don't affect correctness.
But wait, do they affect the warp shuffle reduction? As I showed earlier, they contribute 0 to the sum, so the reduction is correct.
However, for D=24, `lane_id < D` means threads 24-31 are effectively idle but still executing instructions. This is fine.
Could I eliminate this waste by using `blockDim.x = D * 1` instead of `blockDim.x = 32`? No, because a warp is 32 threads. Even if I only launch D threads per j, the hardware still schedules a 32-thread warp. The extra threads would just be "not present" in the block, but from the warp scheduler's perspective, the warp still has 32 lanes... I think?
Actually, in CUDA, if a block has fewer than 32 threads, those threads still form a partial warp, and the warp scheduler dispatches the full warp with some lanes inactive. The difference is that the inactive lanes are not part of the block, so they don't use register file space or thread slots. But the warp still executes as a 32-wide unit, with the extra lanes simply masked off.
So from a performance perspective, launching a block with 24 threads vs 32 threads is essentially the same: the warp still takes the same time to execute. The only difference might be register allocation: with 24 threads, the block allocates registers for 24 threads, not 32.
But in our design, we're launching blocks with 256 threads (8 warps of 32 threads each). For D=24, all 8 warps are active, but within each warp, 8 lanes are masked. The total block size is still 256 threads. If we could somehow use only 24 lanes per warp, we'd still have 256 threads because warps are 32-wide. Unless we launch blocks with 24 * 8 = 192 threads. But then we'd have 6 warps of 32 threads... wait no, 192 / 32 = 6 warps.
If we launch blocks with 192 threads = 6 warps:
- Warp 0: j=0, lanes 0-23 active
- Warp 1: j=1, lanes 0-23 active
- ...
- Warp 5: j=5, lanes 0-23 active
Then we'd need 4 blocks per (b,h) to handle all 24 j values. Total blocks = 8 * 4 = 32 blocks. Each block has 6 warps.
With 32 blocks, each SM could run 1 block (6 warps) or multiple. Better block count than 24, but fewer warps per block.
Hmm, 32 blocks of 6 warps each = 192 warps total.
24 blocks of 8 warps each = 192 warps total.
Same total warps!
So it doesn't matter whether we have 32 blocks of 6 warps or 24 blocks of 8 warps. The total work is the same.
What matters is the per-SM warp density. With 32 blocks of 6 warps:
- An SM can run floor(64/6) = 10 blocks? No, max blocks per SM is 32, but 10*6=60 warps, close to max.
- Actually, can an SM run 10 blocks? Only if they fit in registers and shared memory.
- More realistically, an SM might run 4 blocks of 6 warps = 24 warps, or 5 blocks = 30 warps.
- With 32 blocks total, if each SM runs 4 blocks, we need 8 SMs.
- If each SM runs 5 blocks, we need 6.4 SMs, say 7.
- Warps per SM: 24-30. Similar to before.
With 24 blocks of 8 warps:
- An SM can run 4 blocks = 32 warps, or 3 blocks = 24 warps, or 2 blocks = 16 warps.
- With 24 blocks, if each SM runs 4 blocks, we need 6 SMs.
- Warps per SM: 32. Slightly better than 24-30.
So the 24-block version with 8 warps each might achieve slightly better per-SM warp density (32 warps vs 24-30). But the difference is small.
For simplicity, I'll stick with 24 blocks of 8 warps each. The chunk size of 8 gives a nice round number and ensures that for D=24, all warps are active.
Actually, one more thought: for the 24-block version, the grid is (B, H, 3). With B=2, H=4, that's 24 blocks total. But in what order are the blocks scheduled? The CUDA runtime typically schedules blocks in a linear order, but the exact mapping to SMs is not specified.
In general, having more blocks gives the scheduler more flexibility to balance load across SMs.
OK, I'll finalize with the 24-block version.
Wait, I just realized that if D is much smaller than 8, like D=1, we'd have chunks=1, and only warp 0 would be active, with warps 1-7 returning immediately. The block still launches with 256 threads. 7/8 of the warps do nothing. That's wasteful.
For very small D, we could handle multiple (b,h) pairs in one block to fill the warps. But D=24 is our target, so it's fine.
One more optimization: should I store the state in a local array or a single float?
In the warp-shuffle version, each lane maintains one float `state` for its i value. Over T timesteps, this float is updated and reused. That's optimal.
For the output write, only lane 0 writes. But could I have all lanes write to different outputs? No, because each j only gets one output value per timestep.
What about using `__shfl_sync` instead of `__shfl_xor_sync`? `__shfl_sync` (or `__shfl_down_sync`) can do a tree reduction that only accumulates to lane 0:
```cuda
for (int offset = warpSize/2; offset > 0; offset /= 2) {
partial += __shfl_down_sync(0xFFFFFFFF, partial, offset);
}
```
Wait, `__shfl_down_sync` works like: lane i gets value from lane i+offset.
- After offset=16: lane 0 += lane 16, lane 1 += lane 17, etc.
- After offset=8: lane 0 += lane 8, etc.
- After offset=1: lane 0 += lane 1.
- At the end, lane 0 has the sum of lanes 0..31. Other lanes have partial sums.
This is actually more efficient than `__shfl_xor_sync` because:
1. Fewer total shuffle operations needed? Actually, still 5 operations.
2. But the final result is only in lane 0, which is exactly what we need.
3. With `__shfl_xor_sync`, all lanes have the full sum at the end, which we don't need.
Wait, both take 5 operations and have the same latency. The difference is that `__shfl_down_sync` puts the result in lane 0 (and lane i has sum of lanes i..i+15, etc.), but `__shfl_xor_sync` puts the full sum in all lanes.
Since we only need lane 0, `__shfl_down_sync` is conceptually better. But performance-wise, they're identical.
Actually, let me verify:
`__shfl_down_sync(mask, var, delta)` returns the value of `var` held by the thread whose lane ID is `lane_id + delta`. If lane_id + delta >= warpSize, it returns the original `var`.
So after the loop, lane 0 has the sum of all lanes. Perfect.
I'll use `__shfl_down_sync`. But I need to be careful about the boundary case: for D=24, lanes 24-31 have partial=0. `__shfl_down_sync` works fine with this.
```cuda
#pragma unroll
for (int offset = 16; offset > 0; offset >>= 1) {
partial += __shfl_down_sync(0xFFFFFFFF, partial, offset);
}
if (lane_id == 0) {
out[...] = partial;
}
```
Wait, `__shfl_down_sync` for lane 0 at offset=16 gets lane 16. Then at offset=8, gets lane 8. etc. This accumulates lanes 0, 1, 2, 4, 8, 16. But does it accumulate ALL lanes? Let me trace:
Start: p_i = partial from lane i
offset=16:
p_0 += p_16
p_1 += p_17
...
p_15 += p_31
p_16 += p_16 (since 16+16=32 >= 32, returns p_16)
offset=8:
p_0 += p_8
p_1 += p_9
...
p_7 += p_15
p_8 += p_16 (wait, p_16 was updated at offset=16 to p_16+p_32_invalid=p_16)
Hmm, actually, after offset=16:
- p_0 now has p_0 + p_16
- p_8 now has p_8 + p_24
At offset=8:
- p_0 += p_8, so p_0 gets (p_0 + p_16) + (p_8 + p_24) = p_0+p_8+p_16+p_24
- p_1 += p_9, so p_1 gets p_1+p_9+p_17+p_25
- etc.
At offset=4:
- p_0 += p_4, so p_0 gets p_0+p_4+p_8+p_12+p_16+p_20+p_24+p_28
At offset=2:
- p_0 += p_2, so p_0 gets more terms
At offset=1:
- p_0 += p_1, so p_0 gets ALL terms p_0..p_31
Yes! `__shfl_down_sync` with a tree reduction correctly accumulates all lanes into lane 0.
And since lanes 24-31 have p=0, lane 0 gets sum of lanes 0..23. Correct.
Great, I'll use `__shfl_down_sync`.
One more thing: I should use `__launch_bounds__` to tell the compiler the maximum number of threads per block, which can help it optimize register usage.
```cuda
__launch_bounds__(256)
__global__ void kernel(...) { ... }
```
This tells the compiler the block size is at most 256 threads, which can help with register allocation.
Let's write the final code. I'll handle D up to 32 with the warp-shuffle version, and D > 32 with a simple per-thread version. For the benchmark D=24, the warp-shuffle version will be used.
Actually, I just want to make sure about one thing: the `__shfl_down_sync` operates on the warp level. Within a block, each warp does its own reduction independently. Since each warp handles a different j, there's no cross-warp communication needed. Perfect.
Let me also think about whether I should use `__restrict__` for the pointers. Yes, to tell the compiler there are no aliasing issues.
And should I mark the pointers as `__align__(16)`? PyTorch tensors are aligned, but the base offset might change alignment for sub-tensors. However, for our indexing calculations, the addresses we access might not be 16-byte aligned if the offset in elements is not a multiple of 4. But I showed earlier that the offsets should be 16-byte aligned for all (b,h) and all t because T*H*D and H*D are multiples of 4 (in fact, multiples of 16/4 = 4, but we showed they're multiples of 6, which is more than enough for 16-byte alignment since 96 bytes = 6 * 16 bytes).
Actually, let me verify again:
- D = 24, sizeof(float) = 4, so D * sizeof(float) = 96 bytes
- 96 / 16 = 6, so it's divisible by 16.
- base_qkv = ((b * T) * H + h) * D
- b * T * H * D = b * (128 * 4 * 24) = b * 12288 floats = b * 49152 bytes
- 49152 / 16 = 3072, divisible by 16.
- h * D = h * 24 floats = h * 96 bytes
- 96 / 16 = 6, divisible by 16.
- t * H * D = t * 96 floats = t * 384 bytes
- 384 / 16 = 24, divisible by 16.
- So any address q[base_qkv + offset + i] where i is a multiple of 4 is 16-byte aligned.
Great, so float4 loads are possible and safe in this specific case. But since our warp-shuffle kernel doesn't use float4 (each lane reads one float), the alignment doesn't matter as much.
However, for the warp-shuffle kernel, at each t, all 32 lanes in a warp attempt to read q[base + lane_id] for lane_id = 0..31. For D=24, lanes 24-31 read out of bounds. We prevent this with the conditional.
But for lanes 0-23, the reads are at base + 0, base + 1, ..., base + 23. These are consecutive floats, forming a coalesced read of 24 floats. The memory system will fetch this as 2 128-byte cache lines (or 1.5 cache lines, rounded up to 2). So it's efficient.
If D=24 and lane_id ranges from 0 to 23, we do 24 coalesced reads. But we're reading via 32 threads where 24 are active. The 24 reads by the active threads are coalesced. The 8 inactive threads don't issue reads.
Wait, but in my kernel, I have:
```cuda
float q_i = (lane_id < D) ? q[base_qkv + offset + lane_id] : 0.0f;
```
This means lanes 24-31 evaluate the else branch and don't read memory. Lanes 0-23 read memory. The reads from lanes 0-23 are coalesced. Great.
But does the `?:` operator actually prevent the memory read for lanes 24-31? In CUDA, the compiler typically converts this to predicated loads, where inactive lanes don't issue the load. So yes, memory access is only from active lanes.
Actually, I should double-check this. If the compiler doesn't predicate the load, lanes 24-31 might read q[base + 24..31], which is out of bounds! That would be a memory error.
To be safe, I should either:
1. Ensure that reading q[base + 24..31] is in bounds (i.e., the tensor has padding)
2. Or explicitly guard with `if (lane_id < D)` to prevent memory access
PyTorch tensors might have some padding, but relying on that is unsafe. The safe approach is:
```cuda
float q_i = 0.0f;
if (lane_id < D) {
q_i = q[base_qkv + offset + lane_id];
}
```
With an explicit `if`, the compiler generates predicated memory accesses for each branch. This is safe.
But `__shfl_down_sync` requires all lanes in the warp to be convergent, right? Actually, `__shfl_down_sync` with mask 0xFFFFFFFF requires all lanes in the warp to participate. If some lanes are in a divergent branch, the behavior is undefined unless the mask correctly identifies the active lanes.
From CUDA docs: "The new *_sync shfl intrinsics take in a mask indicating which threads in the warp should participate in the shuffle."
If we use `__shfl_down_sync(0xFFFFFFFF, partial, offset)`, it requires all lanes in the warp to be active. But if some lanes took the `else` branch of the `if`, they might not be converged.
Wait, after the `if (lane_id < D) { ... }`, the threads reconverge. The `__shfl_down_sync` is executed after the `if` block, when all threads are converged again. So it should work.
Actually, I should use `__shfl_down_sync` with a mask that includes only the active lanes. For D=24, the active lanes are 0..23. But the CUDA shuffle functions don't allow non-contiguous masks easily.
Looking at the CUDA docs: `__shfl_down_sync(unsigned mask, T var, unsigned int delta, int width=warpSize)`. The `mask` specifies which threads are participating. But `__shfl_down_sync` is a warp-collective operation, and the mask should include all threads that are supposed to participate.
For a full warp reduction, we want all 32 threads to participate, even if some have partial=0. The mask 0xFFFFFFFF works as long as all threads are converged when `__shfl_down_sync` is called.
After the `if` block, all threads reconverge. So `__shfl_down_sync(0xFFFFFFFF, ...)` should work.
But wait, I read somewhere that `__shfl_down_sync` with inactive threads can result in incorrect values if the inactive threads are not at the same program counter. However, after branch reconvergence, all threads are at the same PC.
To be extra safe, I can avoid the `if` by using a bounded read. But for D=24 and the tensor being contiguous, `base + 24..31` might be in bounds if there's enough padding. I shouldn't rely on that.
Alternatively, I can process exactly D warps with exactly D threads each, with no inactive lanes. But that requires a block with D threads, which as discussed is essentially the same as 32 threads with D active lanes.
For safety, let me use the explicit `if` and then assume reconvergence works. If there are issues, I can try a different approach.
Actually, there's another approach: launch exactly D warps per block, each warp having exactly D threads (well, a warp always has 32 threads, but we can have each warp start with D threads... no).
Or, we can use the "one thread per j" kernel initially, which passed correctness and is simpler. Let me compare the performance risk.
Actually, I'd like to try both approaches and see. But since I can't easily benchmark without submitting, I'll make an educated guess.
The simpler "one thread per j" approach already passed correctness. It uses 192 blocks of 32 threads if I structure it that way, or 8 blocks of 24 threads. Let me use 8 blocks of 32 threads (matching the warp size) and see how it performs.
Wait, the "one thread per j" approach in my very first submission used:
- Grid: (B, H, D)
- Block: (D)
- Total blocks = 192
And each block had D threads = 24 threads. That's less than a warp.
What if I change the first kernel to grid=(B*H*D), block=(1)? No, terrible.
What if I change it to grid=(B*H*D), block=(32), where all 32 threads in a block do the same work for one j? That's redundant.
What if I change it to grid=(B*H), block=(D), and use the simple per-thread approach? That's 8 blocks of 24 threads. This was my original plan.
With 8 blocks of 24 threads, each block has less than a warp. The execution time is 1 warp per block. So 8 warps total, on 8 SMs (1 warp each). Not great for latency hiding.
The warp-shuffle version gives 24 blocks of 8 warps = 192 warps total. This is 24x more warps, spread across more SMs with better per-SM density. Much better potential.
I think the warp-shuffle version has a much better chance of being fast. Let me write it carefully.
One concern: will the `if (lane_id < D)` before the memory read cause issues with `__shfl_down_sync`?
Looking at the PTX ISA, `shfl.sync` requires all non-exited threads in the warp to participate. If a thread is in a divergent branch, it still participates as long as it's not exited. The value it provides for the shuffle is the value of the register at the point of the shuffle instruction.
With the `if` statement, after the `if` block, threads reconverge. At the point of `__shfl_down_sync`, all threads are at the same instruction. The threads that took the else branch have `partial = 0.0f`. So they contribute 0 to the shuffle. This is correct.
So I think the warp-shuffle version is safe and correct.
Let me write it.
Actually, one more thing: in my kernel I use `float state = 0.0f;` and update it in the loop. For lanes >= D, state is never updated (since they're in the else branch). After the if block, they have state = 0.0f. Then `partial = (lane_id < D) ? q_i * state : 0.0f`. For lanes >= D, partial = 0.0f. Then shuffle. Correct.
Wait, what if I write it as:
```cuda
float state = 0.0f;
float partial = 0.0f;
if (lane_id < D) {
for (int64_t t = 0; t < T; t++) {
// ... read g, b, v_j, q_i, k_i ...
state = g * state + b_val * k_i * v_j;
partial = q_i * state;
// shuffle
for (int offset = 16; offset > 0; offset >>= 1) {
partial += __shfl_down_sync(0xFFFFFFFF, partial, offset);
}
if (lane_id == 0) {
out[...] = partial;
}
}
}
```
This places the ENTIRE t-loop inside the `if (lane_id < D)`. For lanes >= D, they skip the entire loop. After the if block, they just return.
Wait, but `__shfl_down_sync` inside the `if` block might have issues! The CUDA docs say that `__shfl_down_sync` requires a convergent set of threads. If it's inside an `if` block, only some threads execute it, which violates the convergence requirement.
The `__shfl_*` intrinsics can be called from within conditional code as long as all threads in the warp participate in the shuffle or have exited. But if only a subset of threads execute the shuffle, it's undefined behavior.
From the CUDA C Programming Guide:
> "The __shfl_*sync() intrinsics are available when the compute capability is 6.1 or higher. The mask argument indicates which threads in the warp should participate in the shuffle. Each calling thread must have its own bit set in the mask and all non-exited threads named in mask must execute the same intrinsic with the same mask, or the result is undefined."
So ALL threads in the warp must execute `__shfl_down_sync`. If I put it inside `if (lane_id < D)`, then threads with lane_id >= D do NOT execute it, which is undefined behavior!
To fix this, I must ensure all threads in the warp execute the shuffle. So the shuffle must be outside the `if` block.
But I can safely put the memory reads inside the `if` block, and then reconverge before the shuffle. The partial for threads outside the `if` should be 0.
Let me restructure:
```cuda
float state = 0.0f;
for (int64_t t = 0; t < T; t++) {
float q_i = 0.0f;
float k_i = 0.0f;
float v_j = 0.0f;
float g = 0.0f;
float b_val = 0.0f;
if (lane_id < D) {
int64_t offset = t * stride_t;
int64_t offset_gate = t * H;
g = gate[base_gate + offset_gate];
b_val = beta[base_gate + offset_gate];
v_j = v[base_qkv + offset + j];
q_i = q[base_qkv + offset + lane_id];
k_i = k[base_qkv + offset + lane_id];
state = g * state + b_val * k_i * v_j;
}
float partial = (lane_id < D) ? q_i * state : 0.0f;
for (int offset = 16; offset > 0; offset >>= 1) {
partial += __shfl_down_sync(0xFFFFFFFF, partial, offset);
}
if (lane_id == 0) {
out[base_qkv + t * stride_t + j] = partial;
}
}
```
Wait, but `g` and `b_val` are only loaded by lanes < D, and then used by lanes >= D in the `if` condition `(lane_id < D) ? q_i * state : 0.0f`. Actually, `(lane_id < D) ? q_i * state : 0.0f` uses q_i and state, which are 0 for lanes >= D. So partial = 0. That's correct.
But `state` for lanes >= D: it's initialized to 0.0f. In the loop, for lanes >= D, the `if (lane_id < D)` block is skipped, so state remains 0.0f. Correct.
For lanes < D, the state is updated. But `g` and `b_val` are only loaded once per thread per timestep. Since all threads in a warp read the same gate and beta values, but we only load them in the `if` block for lanes < D, only some threads load them. Wait, for a warp handling one j, all active lanes (0..23) load the same g and b_val. That's redundant but safe.
Actually, could I load g and b_val outside the if block to avoid redundant loads? Since all active threads load the same value, and the inactive threads don't need it:
```cuda
for (int64_t t = 0; t < T; t++) {
int64_t offset = t * stride_t;
int64_t offset_gate = t * H;
float g = gate[base_gate + offset_gate];
float b_val = beta[base_gate + offset_gate];
float v_j = (lane_id < D) ? v[base_qkv + offset + j] : 0.0f;
float q_i = 0.0f;
float k_i = 0.0f;
if (lane_id < D) {
q_i = q[base_qkv + offset + lane_id];
k_i = k[base_qkv + offset + lane_id];
state = g * state + b_val * k_i * v_j;
}
float partial = (lane_id < D) ? q_i * state : 0.0f;
for (int shfl_offset = 16; shfl_offset > 0; shfl_offset >>= 1) {
partial += __shfl_down_sync(0xFFFFFFFF, partial, shfl_offset);
}
if (lane_id == 0) {
out[base_qkv + offset + j] = partial;
}
}
```
This is cleaner. `g` and `b_val` are loaded by all threads (redundant but coalesced/broadcast). `v_j` is loaded only by active threads. Because all threads in a warp have the same j, but only lanes < D are "active" for this computation. Wait, for a given warp, j is fixed. All lanes in the warp have the same j. But all lanes in the warp will be active if j < D.
Actually, in our block structure, warp_id determines j. Warp 0 has j=chunk_id*8+0. Warp 1 has j=chunk_id*8+1, etc. For D=24, all j values are < 24, so all warps are active. But within each warp, lanes >= D (lanes 24-31) are inactive for the i-dimension but the warp itself is active because j < D.
So in the loop:
- `g` and `b_val` are loaded by all lanes (broadcast, cached)
- `v_j` is loaded by all lanes (same address, broadcast)
- `q_i` and `k_i` are loaded only by lanes < D
For the conditional `if (lane_id < D)`, lanes 24-31 don't execute the body. But they need to wait for lanes 0-23 to finish before reconverging.
The `__shfl_down_sync` is executed by all lanes after reconvergence. This is safe.
Actually, is the `if` even necessary for q_i and k_i? What if D were exactly 32? Then lane_id < D is always true, and all 32 lanes are active.
For D < 32, lanes >= D would read out of bounds. So we need the guard.
Could I pad the q, k, v tensors to multiples of 32? No, I can't modify the input tensors.
Could I change the layout so that the "extra" lanes read valid memory? If the tensor has extra elements beyond the D dimension, those reads might be valid but we shouldn't use them. For contiguous tensors, q[base + offset + 24] is q[b, t, h, 24], which is out of bounds for D=24.
So we DO need the guard.
One optimization: since g and b_val are the same for all lanes, and v_j is the same for all lanes in a warp, I could load them once per warp to reduce register pressure. But with auto-broadcasting in the memory system, all lanes reading the same address doesn't waste bandwidth. And the compiler likely optimizes this well.
OK, I'll finalize the code now.
Actually, one last optimization: loop unrolling for the t-loop. Since T is a runtime parameter, we can't fully unroll it. But we can use `#pragma unroll 4` to unroll by a factor of 4.
Wait, but T could be any value. Partial unrolling by 4 should still help.
For D <= 32, the inner loop is gone (it's parallelized across the warp). The only loop is over t. Unrolling by 2 or 4 might reduce loop overhead.
Let me add `#pragma unroll 4` to the t-loop.
But 128 is divisible by 4, so unrolling by 4 is fine.
Actually, for the memory-bound kernel, unrolling the t-loop might not help much because each iteration depends on the previous state. There's no independent work to overlap.
I'll still add `#pragma unroll 2` to be safe.
Now, one more thing: I should use `const` wherever possible to help the compiler optimize.
And I should use 64-bit math for indexing to avoid overflow, even though the sizes are small.
Let me put it all together.
Wait, actually, I realize I should also handle the "large D" path just in case. For D > 32, I'll write a kernel where each block handles one (b, h), with block size = min(D, 1024). Each thread handles one j and loops over i.
For the large D path, I'll allocate `float state[256]` on the stack. If D > 256, I'll abort or use a slower path. But since the benchmark D=24, this won't matter.
Let me write the code. I'll be careful with syntax.
Actually, I want to use `__align__(16)` for the kernel parameters? No, that's not needed.
Let me also check: the CUDA grid can be up to 2^31-1 in x, 65535 in y, 65535 in z. With B=2, H=4, chunks=3, the grid is (2, 4, 3). That's well within limits.
OK, here's my plan for the code:
```python
import torch
import torch.nn as nn
import torch.utils.cpp_extension as ext
cpp_source = "..."
cuda_source = "..."
_cpp_ext = ext.load_inline(...)
class ModelNew(nn.Module):
def __init__(self):
super().__init__()
def forward(self, q, k, v, gate, beta):
q = q.contiguous()
k = k.contiguous()
v = v.contiguous()
gate = gate.contiguous()
beta = beta.contiguous()
out = torch.empty_like(q)
_cpp_ext.scalar_gated_delta_attention(q, k, v, gate, beta, out)
return out
```
Now let me write the exact CUDA source.
In the C++ wrapper function, I'll get the tensor sizes and launch the appropriate kernel.
One important thing: in `load_inline`, the `cpp_sources` should declare the function, and `cuda_sources` should define it. The `functions` parameter lists the host-side wrapper function names.
Also, I should NOT write a `PYBIND11_MODULE` block. `load_inline` generates it automatically.
The C++ wrapper function signature should match what pybind11 can infer from the torch::Tensor types. A function taking torch::Tensor arguments should work.
Let me write the code carefully.
For the switch dispatch, I could write a helper template to avoid repetition:
```cpp
template <int D>
__global__ void kernel(...) { ... }
template <int D>
void launch_kernel(dim3 grid, dim3 block, ...) {
kernel<D><<<grid, block>>>(...);
}
void scalar_gated_delta_attention(...) {
// ...
switch (D) {
case 8: launch_kernel<8>(grid, block, ...); break;
case 16: launch_kernel<16>(grid, block, ...); break;
case 24: launch_kernel<24>(grid, block, ...); break;
case 32: launch_kernel<32>(grid, block, ...); break;
default:
// generic path
break;
}
}
```
But to keep it simple, I'll just inline the dispatch without a helper.
Actually, for the small D path with generic fallback, I can avoid templates entirely and just pass D as a runtime parameter. Since the only place D matters is in the `lane_id < D` check and `warp_id >= D` check, a runtime D is fine.
But I want the compiler to optimize the loop unrolling and bounds checks. With a template, the compiler knows D at compile time. With a runtime parameter, it doesn't. For the `lane_id < D` check, the compiler can still optimize if D is passed as a constant via template.
However, the performance impact might be minimal because `lane_id < D` is evaluated once per timestep? No, it's evaluated for memory access at each timestep, but the warp-level instruction masking handles it efficiently.
Actually, looking at my kernel again, there are no loops over D in the per-thread code. The only D-dependent checks are `warp_id >= D` (once) and `lane_id < D` (per timestep, but it's a simple comparison). So the template isn't critical.
I'll use a non-templated kernel for simplicity. This way I don't need the switch statement, and the code works for any D.
Wait, but for the small kernel, the block size is 256 (8 warps), and chunks = (D + 7) / 8. For D > 256*8 = 2048, we'd have a huge grid. But that's fine.
Actually, if D is very large, say D=1024, chunks = 128. Grid = (B, H, 128). With B=2, H=4, that's 1024 blocks. Each with 256 threads. Total threads = 262,144. That's great utilization!
But wait, the small kernel assumes D <= 32 for the `lane_id < D` check. If D=1024, then lane_id < 1024 is always true. Good.
And the check `warp_id >= D` is `warp_id >= 1024`. But warp_id ranges from 0 to 7 (8 warps per block). So `warp_id >= D` is never true for D > 7. Wait, D=1024, warp_id is 0..7, so warp_id < 1024 always. Good.
So the small kernel actually works for ANY D! Because `lane_id < D` is always true for D >= 32 (all 32 lanes active), and `warp_id >= D` is never true for any reasonable D when we have only 8 warps per block.
Wait, what if D=8? Then `warp_id >= 8` would cause warps 8..31 to return... but we only have 8 warps! So all are active. But we only want j=0..7. For chunk=0, j=chunk*8+warp_id=0..7. For D=8, j=0..7 is valid. For chunk=1, j=8..15, but D=8 so j >= 8 is invalid. But chunks = (8+7)/8 = 1, so chunk only goes from 0 to 0. So only j=0..7 are created. Good.
What if D=16? chunks=2. chunk=0: j=0..7. chunk=1: j=8..15. All valid.
What if D=20? chunks=3. chunk=0: j=0..7. chunk=1: j=8..15. chunk=2: j=16..19. Warps 0..3 active, warps 4..7 return due to `j >= D`. Good.
So the small kernel with chunk_size=8 and block_size=256 works for ALL D >= 1!
And for D=24, chunk=2 gives j=16..23, all valid, all 8 warps active.
For D=64, chunk=0: j=0..7, chunk=1: j=8..15, ..., chunk=7: j=56..63. All valid, all 8 warps active per block. Grid = (B, H, 8).
For D=128, grid = (B, H, 16). 16 * 8 = 128 j values. All good.
Wow, this kernel is actually very general! It scales to any D by adjusting the grid z-dimension.
And for D < 32, some lanes are masked in each warp, but for D >= 32, all lanes are active.
This is a very clean and scalable design.
One issue: for very large D, say D=10,000, grid z-dimension = 1250. That's within the 65535 limit.
For D=100,000, grid z = 12,500. Still within limit.
For D=1,000,000, grid z = 125,000. Exceeds 65535 limit.
But D won't be that large in practice. Head dimensions are typically <= 512.
So the small kernel actually works for all practical D values. And for D < 32, it has some wasted lanes, but that's fine.
Now, the only thing missing is: for the `__shfl_down_sync` reduction, we sum 32 values. But the mathematical sum is over i=0..D-1. For D < 32, we correctly sum lanes 0..D-1 plus zeros for lanes D..31. For D > 32, we only sum 32 values, but D might be 64, 128, etc.! For D > 32, the reduction is incomplete!
Oh! This is critical! For D=64, each warp has 32 lanes, but there are 64 i values. We can't sum 64 values with a 32-lane warp shuffle. We'd need multiple warps per j, each handling 32 i values, and then combine them.
So the kernel as designed only works correctly for D <= 32.
I was tricked into thinking it scales to any D because the block count increases, but the reduction within each warp is capped at 32 elements.
So for D > 32, we need a different approach.
Options for D > 32:
1. Use multiple warps per j, handling chunks of 32 i values each. Then combine warp sums using shared memory.
2. Use the "one thread per j" approach, where each thread loops over all i values, no reduction needed.
For D > 32, the "one thread per j" approach is simpler. Each block handles one (b, h), with block size = min(D, 1024). Each thread loops over all i values.
Let me add this as the fallback.
Actually, for the benchmark D=24 <= 32, the small kernel is sufficient. I'll just implement the small kernel and provide a generic large kernel for completeness.
For the large kernel, I'll use `float state[256]` (or a dynamically sized approach).
OK, let me write the final code. I want to keep it simple and correct.
For the host wrapper:
```cpp
void scalar_gated_delta_attention(
torch::Tensor q, torch::Tensor k, torch::Tensor v,
torch::Tensor gate, torch::Tensor beta, torch::Tensor out) {
int64_t B = q.size(0);
int64_t T = q.size(1);
int64_t H = q.size(2);
int64_t D = q.size(3);
dim3 grid(B, H);
const int CHUNK_SIZE = 8;
int chunks = (int)((D + CHUNK_SIZE - 1) / CHUNK_SIZE);
if (D <= 32) {
dim3 grid(B, H, chunks);
dim3 block(CHUNK_SIZE * 32);
scalar_gated_delta_attention_kernel_small<<<grid, block>>>(
q.data_ptr<float>(), k.data_ptr<float>(), v.data_ptr<float>(),
gate.data_ptr<float>(), beta.data_ptr<float>(), out.data_ptr<float>(),
B, T, H, (int)D);
} else {
// Generic path for D > 32
// ...
}
}
```
For the generic path, I'll write a simple kernel:
```cuda
__global__ void scalar_gated_delta_attention_kernel_large(...) {
int64_t b = blockIdx.x;
int64_t h = blockIdx.y;
int j = blockIdx.z * blockDim.x + threadIdx.x;
if (b >= B || h >= H || j >= D) return;
// State in local memory (may spill to registers or local mem)
float state[128]; // Assuming D <= 128 for this path
// If D > 128, we'd need dynamic allocation
for (int i = 0; i < D; i++) state[i] = 0.0f;
for (int64_t t = 0; t < T; t++) {
int64_t offset = t * H * D;
int64_t g_offset = t * H;
float g = gate[(b * T * H + h) + g_offset];
float b_val = beta[(b * T * H + h) + g_offset];
float v_j = v[((b * T + t) * H + h) * D + j];
float accum = 0.0f;
for (int i = 0; i < D; i++) {
float q_i = q[((b * T + t) * H + h) * D + i];
float k_i = k[((b * T + t) * H + h) * D + i];
state[i] = g * state[i] + b_val * k_i * v_j;
accum += q_i * state[i];
}
out[((b * T + t) * H + h) * D + j] = accum;
}
}
```
Wait, the indexing in the large kernel is slightly wrong. Let me be consistent:
```cuda
int64_t base_qkv = ((b * T) * H + h) * D;
int64_t base_gate = (b * T) * H + h;
int64_t stride_t = H * D;
int64_t stride_gate = H;
```
Then:
- `gate[base_gate + t * stride_gate]`
- `beta[base_gate + t * stride_gate]`
- `v[base_qkv + t * stride_t + j]`
- `q[base_qkv + t * stride_t + i]`
- `k[base_qkv + t * stride_t + i]`
- `out[base_qkv + t * stride_t + j]`
This is consistent.
For the large kernel, I'll assume D <= 128 (which should cover most cases). If D > 128, I'll just clamp or fall back to a global memory state. But since the benchmark D=24, it doesn't matter.
Actually, I can make the state allocation dynamic by passing the maximum size as a compile-time constant, or by checking D at runtime. But `float state[128]` should be fine.
For D > 128, the compiler would definitely spill to local memory. But local memory is cached in L1, which is still much faster than global memory. For the very rare case where D > 128, performance might degrade, but correctness is maintained.
Let me finalize the code now. I'll make sure the large kernel uses the correct indexing.
Also, for the large kernel, I'll use a 3D grid: (B, H, chunks) where chunks = ceil(D / block_size). Block size = min(D, 1024). But actually, for simplicity, I'll just set block size = 256 and chunks = ceil(D / 256). Or block size = 1024.
For the large kernel, the block size choice affects performance. With D=64, block size=64 is natural. But 64 threads per block might not saturate the SM. So block size = 256 or 512 might be better, allowing multiple j values to be processed concurrently within a block.
Actually, in the large kernel, all threads in a block are independent. They just happen to share the same (b, h). So there are no cross-thread dependencies. A larger block size means more threads per block, which can increase per-SM warp density.
But the total number of blocks is B*H*ceil(D/block_size). To maximize block count, we want smaller block sizes. To maximize per-block parallelism, we want larger block sizes.
For the generic large kernel, I'll just use block_size = 256 for simplicity. This gives decent occupancy and is easy to implement.
Wait, for D=64 and block_size=256, we'd have 1 block per (b,h) because 64 < 256. The block would have 64 threads, not 256. That's fine.
Actually, `dim3 block(min((int)D, 1024))` is better.
Let me just write:
```cpp
dim3 block(min((int)D, 256));
int chunks = (int)((D + block.x - 1) / block.x);
dim3 grid_large(B, H, chunks);
scalar_gated_delta_attention_kernel_large<<<grid_large, block>>>(...);
```
For the large kernel, `block.x` threads handle `block.x` different j values for one (b,h). Since all threads are independent, this works.
But wait, for the large kernel, I need each thread to loop over all i from 0 to D-1. If D is large (like 256), each thread does a lot of work. And the memory reads of q and k are sequential for each thread, with good cache behavior.
But all threads in a block read the same q[t, i] values at the same i iteration. This is a broadcast access pattern. The L1 cache handles it well.
For v and out, each thread accesses a different j value, which is coalesced.
So the large kernel has decent memory behavior too.
One issue: state[128] per thread. With 256 threads, that's 256 * 128 * 4 bytes = 131 KB per block. If this spills to local memory, it could be significant. But for D <= 128, the compiler might keep some or all in registers.
Actually, for D=64, 64 floats = 256 bytes per thread. With 64 threads, that's 16 KB. The compiler should be able to keep this in local memory (which is backed by L1 cache) or spill minimally.
Let me just write the code.
Actually, one final optimization for the large kernel: loop tiling or strip-mining. We can process the i-loop in chunks that fit in registers.
But I'll keep it simple.
Let me write the complete kernel code now.
I need to be very careful with the Python string for cpp_sources and cuda_sources. It's a raw string, so I need to escape backslashes properly, or use triple quotes.
Actually, looking at my first working attempt, the CUDA source was included as a raw string in Python. That worked. I'll do the same.
One important note: in `ext.load_inline`, the `cpp_sources` parameter should contain the C++ wrapper function declaration. Since our wrapper is defined in `cuda_sources`, I need to make sure it's declared in `cpp_sources` or defined there.
Actually, `load_inline` puts the cpp_sources in a .cpp file and cuda_sources in a .cu file. The C++ file can call functions defined in the CUDA file if they're declared in the C++ file.
So I need to provide a declaration in `cpp_sources`:
```cpp
void scalar_gated_delta_attention(
torch::Tensor q, torch::Tensor k, torch::Tensor v,
torch::Tensor gate, torch::Tensor beta, torch::Tensor out);
```
Or I can put everything in `cuda_sources` and leave `cpp_sources` empty. In my first working attempt, I had:
```python
cpp_sources="",
cuda_sources=CUDA_SOURCE,
```
And it worked because all functions were in the CUDA source. But the pybind11 auto-generated code calls `scalar_gated_delta_attention`, which is defined in the CUDA source. Since the pybind11 code is generated in a .cpp file, it needs the declaration.
Hmm, but in my first attempt, I provided `cpp_sources=""` and it compiled successfully (after I fixed the missing declaration issue). Wait, my first attempt actually had the declaration in cpp_sources! Looking back:
```python
CPP_SOURCE = r"""
#include <torch/extension.h>
void scalar_gated_delta_attention(
torch::Tensor q,
torch::Tensor k,
torch::Tensor v,
torch::Tensor gate,
torch::Tensor beta,
torch::Tensor out);
"""
CUDA_SOURCE = r"""
#include <torch/extension.h>
#include <cuda_runtime.h>
... definition of scalar_gated_delta_attention ...
"""
```
So the declaration is in cpp_sources, and the definition is in cuda_sources. This works because the CUDA compiler processes the .cu file, generating a .o file with the symbol, and the C++ linker links it all together.
I'll follow the same pattern.
Let me write the final code carefully.
One more thing: I should add `TORCH_CHECK` in the host wrapper to ensure tensors are contiguous and have the right dtype/device. But the problem statement says my forward already makes them contiguous. And the evaluation might check correctness directly. So I'll keep it simple.
Also, I should check that q, k, v are on CUDA. Since the reference implementation uses CUDA tensors (from `get_inputs`), they should be. But it's good practice to add a check.
Actually, let me look at the reference again:
```python
q = torch.randn(batch_size, seq_len, num_heads, head_dim, dtype=torch.float32)
```
This creates a CPU tensor by default! But in `get_inputs()`, they're probably moved to CUDA in the evaluation harness. The evaluation typically calls `.cuda()` on inputs before passing them to the model. So my kernel should handle CUDA tensors.
I should check `q.is_cuda()` in the wrapper.
Let me add:
```cpp
TORCH_CHECK(q.is_cuda(), "q must be a CUDA tensor");
TORCH_CHECK(k.is_cuda(), "k must be a CUDA tensor");
// ...
TORCH_CHECK(q.dtype() == torch::kFloat32, "q must be float32");
```
But `load_inline` might compile without issues. I'll add these checks.
One issue: `TORCH_CHECK` is available from `<ATen/core/TensorBody.h>` or `<torch/extension.h>`. I'll include `<torch/extension.h>` which should bring it in.
Alright, I'll write the complete code now. Let me assemble it carefully.
For the CUDA kernel, I'll write both the small D path and the large D path.
Let me also think about whether I need `cudaDeviceSynchronize()` after the kernel launch. Typically not, since PyTorch's CUDA stream synchronization handles it implicitly when accessing the output tensor.
Actually, when we call `torch.empty_like(q)` and then pass it to the CUDA kernel, the kernel executes asynchronously on the CUDA stream. PyTorch operations will synchronize appropriately. So no explicit sync is needed.
But in the C++ wrapper, if we're going to return immediately after the kernel launch, we don't need a sync either. The tensor will be returned to Python, and PyTorch will handle stream dependencies.
OK, let me write the code.
Actually, before I write the large kernel, let me think about whether I can just use the small kernel for ALL D by modifying the reduction.
For D > 32, each j needs more than 32 i values. I can still use one warp per j, but each warp handles only 32 i values, and we'd need ceil(D/32) warps per j. The warps for the same j would need to combine their partial sums.
For example, D=64:
- For each j, we need 2 warps (32 threads each) to handle i=0..31 and i=32..63
- After each warp computes its partial sum, warp 0 and warp 1 need to combine
- We can store partial sums in shared memory and have warp 0 read and combine
This is doable but complex. For the large kernel, the simple per-thread loop is cleaner.
OK, I'll write both kernels.
Wait, for D > 32, the per-thread loop essentially does:
```cuda
for (int i = 0; i < D; i++) {
state[i] = g * state[i] + b_val * k[t,i] * v_j;
accum += q[t,i] * state[i];
}
```
I can optimize this with `#pragma unroll` if D is known at compile time. But for the generic path, D is runtime.
One optimization: precompute `b_val * v_j` outside the i-loop:
```cuda
float bvj = b_val * v_j;
for (int i = 0; i < D; i++) {
state[i] = g * state[i] + bvj * k_i;
accum += q_i * state[i];
}
```
This saves one multiply per iteration. Small but easy.
Similarly, I could precompute `g * state[i]` and reuse... but state[i] changes each iteration.
Another optimization: process 4 i values at a time to enable instruction-level parallelism.
But again, for D=24, we don't need the large kernel.
Let me write the code now. I'll keep the large kernel simple.
Actually, one more thought: for the large kernel, instead of `float state[128]`, I could use `float state[64]` for D <= 64. And for larger D, I can use the same trick as the small kernel but with multiple warps per j.
Since the benchmark D=24, I'll just focus on making the small kernel correct and fast. The large kernel is a backup.
Let me assemble the complete code string.