reasoning
Good, H100 with lots of memory and high bandwidth. Now let me implement a parallel associative scan kernel.
The algorithm:
1. Up-sweep (reduce) phase: combine pairs in a tree-like fashion
2. Down-sweep phase: propagate partial results
For a sequence of length L, we need O(log L) steps, each step processing all elements.
Key considerations:
- Each batch and channel is independent, so I can process them in parallel
- Use shared memory for intermediate results
- Need to handle arbitrary sequence lengths
Let me write the first version:
```python
import torch
from torch.utils.cpp_extension import load_inline
# CUDA source for parallel associative scan
cuda_source = """
#include <cuda.h>
#include <cuda_runtime.h>
// Parallel associative scan kernel
// Each thread block handles one (batch, channel) pair
// Threads in a block cooperatively perform the scan
__global__ void associative_scan_kernel(
const float* __restrict__ a,
const float* __restrict__ b,
float* __restrict__ output,
int L, int D) {
int batch = blockIdx.x;
int channel = blockIdx.y;
int tid = threadIdx.x;
int block_size = blockDim.x;
extern __shared__ float shared_mem[];
float* a_shared = shared_mem; // size L
float* b_shared = shared_mem + L; // size L
int base_idx = batch * L * D + channel;
// Load data into shared memory (coalesced reads)
for (int t = tid; t < L; t += block_size) {
int idx = base_idx + t * D;
a_shared[t] = a[idx];
b_shared[t] = b[idx];
}
__syncthreads();
// Up-sweep (reduce) phase
for (int stride = 1; stride < L; stride *= 2) {
for (int t = tid; t < L; t += block_size) {
int idx = t;
if (idx >= stride && idx < L) {
int src = idx - stride;
// Combine: (a1, b1) ⊗ (a2, b2) = (a2 * a1, a2 * b1 + b2)
float a_src = a_shared[src];
float b_src = b_shared[src];
float a_dst = a_shared[idx];
float b_dst = b_shared[idx];
// But we need to be careful - we can't modify in place while reading
}
}
__syncthreads();
}
// ...
}
```
Hmm, the issue with the up-sweep/down-sweep approach is that it requires careful synchronization and double buffering. Let me think about a simpler approach.
Actually, for parallel associative scan, a common approach is:
1. Use a work-efficient algorithm that operates in-place
2. Or use Blelloch's algorithm with up-sweep and down-sweep
The issue is that for each step, we need to read and write the same locations, which requires double buffering or careful ordering.
Let me try a different approach - use a more straightforward iterative algorithm where each step doubles the "lookback" distance:
For position t, at step k:
- If t >= 2^k, combine with position t - 2^k
- Result: h_t = a_t * h_{t - 2^k} + b_t (with appropriate accumulations)
Wait, I think I'm overcomplicating this. Let me think about the math more carefully.
For the recurrence h_t = a_t * h_{t-1} + b_t, we can write:
- h_0 = b_0
- h_1 = a_1 * b_0 + b_1
- h_2 = a_2 * (a_1 * b_0 + b_1) + b_2 = a_2 * a_1 * b_0 + a_2 * b_1 + b_2
- h_t = (prod_{i=1}^{t} a_i) * b_0 + sum_{j=1}^{t} (prod_{i=j+1}^{t} a_i) * b_j
This is essentially a prefix sum operation with a special associative operation.
The associative operation is:
- Define element at position t as (a_t, b_t)
- Define combine: (a1, b1) ⊗ (a2, b2) = (a2 * a1, a2 * b1 + b2)
- This represents: if we have prefix up to position t1, and want to add positions t1+1 to t2
- The new a is a_{t2} * a_{t1} (product of all a's)
- The new b is a_{t2} * b_{t1} + b_{t2} (the accumulated b)
Actually wait, let me re-derive:
- At position 0: (a_0, b_0) represents: h_0 = b_0 (a_0 is not used for the first element)
- Hmm, the formulation is a bit tricky for position 0.
Let me think differently. Define the operation on pairs (coeff, val):
- (c1, v1) represents v1 (c1 is used for future combinations)
- (c2, v2) represents v2
- Combine: (c1, v1) ⊗ (c2, v2) = (c2 * c1, c2 * v1 + v2)
- This represents the accumulated value from combining the two sequences
For the recurrence, the "element" at position t is:
- (a_t, b_t) for t > 0
- (1, b_0) for t = 0 (no previous to multiply)
Wait, looking at the reference code again:
```python
h = torch.zeros(B, D, device=a.device, dtype=a.dtype) # h starts at 0
for t in range(L):
h = a[:, t] * h + b[:, t] # h_t = a_t * h_{t-1} + b_t
outputs.append(h)
```
So h_0 = a[:, 0] * 0 + b[:, 0] = b[:, 0]
h_1 = a[:, 1] * h_0 + b[:, 1] = a[:, 1] * b[:, 0] + b[:, 1]
So the first output h_0 = b_0, and then we apply the recurrence.
For the associative scan, I can define each element as a linear function of the "previous" state:
- Element t transforms previous state h_prev as: h_new = a_t * h_prev + b_t
- This is a linear transformation that can be composed: (a, b) represents the function h -> a * h + b
- Composition: (a2, b2) ∘ (a1, b1) = (a2 * a1, a2 * b1 + b2)
- Apply (a1, b1) then (a2, b2): h -> a1 * h + b1 -> a2 * (a1 * h + b1) + b2 = a2 * a1 * h + a2 * b1 + b2
Now, the prefix scan computes:
- Prefix up to t is the composition of all elements from 0 to t
- Applying this to the identity (h = 0) gives us h_t
So for the scan:
- Each element is (a_t, b_t)
- Identity element is (1, 0) (h -> 1 * h + 0 = h)
- For t = 0: h_0 = a_0 * 0 + b_0 = b_0
- For t > 0: prefix_t = (a_t, b_t) ∘ prefix_{t-1}
Actually, let me just use the standard formulation:
- Identity: (a=1, b=0)
- Elements: e_t = (a_t, b_t)
- Combine: e_1 ⊗ e_2 = (e_2.a * e_1.a, e_2.a * e_1.b + e_2.b)
- Prefix_t = e_0 ⊗ e_1 ⊗ ... ⊗ e_t
- Output h_t = prefix_t.b (when applied to h=0)
Wait, that's not quite right either. Let me be more careful:
- prefix_0 = e_0 = (a_0, b_0)
- h_0 = apply prefix_0 to h=0 = a_0 * 0 + b_0 = b_0 ✓
- prefix_1 = e_1 ⊗ e_0 = (a_1, b_1) ⊗ (a_0, b_0) = (a_1 * a_0, a_1 * b_0 + b_1)
- But wait, we want h_1 = a_1 * h_0 + b_1 = a_1 * b_0 + b_1
- If we apply prefix_1 to h=0: a_1 * a_0 * 0 + a_1 * b_0 + b_1 = a_1 * b_0 + b_1 ✓
So the output h_t = prefix_t.b when the prefix is computed right-to-left (element 0 first, then element 1, etc.)
Actually I think the combine order matters. Let me use:
- prefix[t] = e_0 ⊗ e_1 ⊗ ... ⊗ e_t (left-to-right)
- Or equivalently, prefix[t] = e_t ∘ prefix[t-1] for t > 0, and prefix[0] = e_0
With the combine operation:
- (a1, b1) ⊗ (a2, b2) = (a1 * a2, b1 + a1 * b2) -- this is composition: first apply (a2, b2), then (a1, b1)
Hmm, I think I have the order mixed up. Let me just use a straightforward approach:
For the associative scan, I'll use the following:
- Element at position t: (a_t, b_t) representing the transformation h -> a_t * h + b_t
- Combine (a1, b1) after (a2, b2): (a1 * a2, a1 * b2 + b1)
- This means: apply (a2, b2) first, then (a1, b1)
- Result: h -> a2 * h + b2 -> a1 * (a2 * h + b2) + b1 = a1 * a2 * h + a1 * b2 + b1
For prefix scan (inclusive):
- prefix[t] = elements[0] ⊗ elements[1] ⊗ ... ⊗ elements[t]
- This should represent applying elements[0], then elements[1], ..., then elements[t]
With the combine as defined above (left element applied after right):
- prefix[0] = (a_0, b_0)
- prefix[1] = (a_1, b_1) ⊗ (a_0, b_0) = (a_1 * a_0, a_1 * b_0 + b_1)
Hmm, but the order is reversed. Let me define combine differently:
- (a1, b1) ⊗ (a2, b2) where (a1, b1) comes before (a2, b2) in the sequence
- Combined result: (a1 * a2, b1 + a1 * b2)
- This means: first apply (a1, b1), then apply (a2, b2)
- Wait, that's h -> a1 * h + b1 -> a2 * (a1 * h + b1) + b2 = a2 * a1 * h + a2 * b1 + b2
That's not right. Let me try:
- (a1, b1) ⊗ (a2, b2) = (a2 * a1, a2 * b1 + b2)
- This means: first apply (a1, b1), then (a2, b2)
- Check: h -> a1 * h + b1 -> a2 * (a1 * h + b1) + b2 = a2 * a1 * h + a2 * b1 + b2 ✓
So:
- prefix[0] = (a_0, b_0)
- prefix[1] = (a_0, b_0) ⊗ (a_1, b_1) = (a_1 * a_0, a_1 * b_0 + b_1)
- Output h_t = prefix[t].b + prefix[t].a * h_start = prefix[t].b + prefix[t].a * 0 = prefix[t].b
So the output is just the b component of the prefix!
Now let me implement the parallel associative scan. I'll use the standard algorithm:
```
// Up-sweep (reduce)
for d = 0 to log2(n) - 1:
for k = 0 to n-1 by 2^(d+1):
x[k + 2^(d+1) - 1] = x[k + 2^d - 1] ⊗ x[k + 2^(d+1) - 1]
// Down-sweep
x[n-1] = identity
for d = log2(n) - 1 downto 0:
for k = 0 to n-1 by 2^(d+1):
t = x[k + 2^d - 1]
x[k + 2^d - 1] = x[k + 2^(d+1) - 1]
x[k + 2^(d+1) - 1] = t ⊗ x[k + 2^(d+1) - 1]
```
But this is for exclusive scan. For inclusive scan, we need to adjust.
Actually, for inclusive scan, there's a simpler formulation. Let me think...
The inclusive scan can be done as:
```
for d = 0 to log2(n) - 1:
for all k in parallel:
if k >= 2^d:
x[k] = x[k - 2^d] ⊗ x[k]
```
Where ⊗ is the associative operation. At each step, each element k combines with element k - 2^d, effectively "pulling in" contributions from further back.
After log2(n) steps, each element contains the prefix sum up to that position.
This is the algorithm I'll implement!
Let me code it up:
```cpp
__global__ void associative_scan_kernel(
const float* __restrict__ a,
const float* __restrict__ b,
float* __restrict__ output,
int L, int D) {
int batch = blockIdx.x;
int channel = blockIdx.y;
int tid = threadIdx.x;
int block_size = blockDim.x;
// Each block handles one (batch, channel) pair
// Shared memory for double buffering
extern __shared__ float shared_mem[];
float* a_cur = shared_mem; // size L
float* b_cur = shared_mem + L; // size L
float* a_prev = shared_mem + 2 * L; // size L
float* b_prev = shared_mem + 3 * L; // size L
int base_idx = batch * L * D + channel;
// Load data into shared memory (coalesced reads)
for (int t = tid; t < L; t += block_size) {
int idx = base_idx + t * D;
a_cur[t] = a[idx];
b_cur[t] = b[idx];
}
__syncthreads();
// Parallel scan
for (int d = 1; d < L; d *= 2) {
// Swap buffers
float* tmp_a = a_cur; float* tmp_b = b_cur;
a_cur = a_prev; b_cur = b_prev;
a_prev = tmp_a; b_prev = tmp_b;
for (int t = tid; t < L; t += block_size) {
if (t >= d) {
// Combine: (a_prev[t-d], b_prev[t-d]) ⊗ (a_prev[t], b_prev[t])
// = (a_prev[t] * a_prev[t-d], a_prev[t] * b_prev[t-d] + b_prev[t])
a_cur[t] = a_prev[t] * a_prev[t - d];
b_cur[t] = a_prev[t] * b_prev[t - d] + b_prev[t];
} else {
a_cur[t] = a_prev[t];
b_cur[t] = b_prev[t];
}
}
__syncthreads();
}
// Write output (h_t = b_cur[t])
for (int t = tid; t < L; t += block_size) {
int idx = base_idx + t * D;
output[idx] = b_cur[t];
}
}
```
This requires 4 * L * sizeof(float) shared memory per block, which for L=4096 and float is 64KB.
But wait, I'm handling each (batch, channel) as a separate block. With B=32 and D=256, that's 32 * 256 = 8192 blocks. That's a lot of blocks, but they should queue up on the GPU.
Actually, there's an issue: I'm using the wrong indexing for shared memory and the output. Let me fix that.
Actually wait, for L=4096, each block needs 64KB of shared memory, and the H100 has 228KB per SM. So we can fit at most 3 blocks per SM (228/64 = 3.56). With 132 SMs on H100, that's 396 blocks running concurrently. With 8192 total blocks, this should still be efficient.
But the bigger issue is that the block indexing is wrong. Let me reconsider the launch configuration:
- Grid: (B, D) -> blockIdx.x = batch, blockIdx.y = channel
- Block: (blockDim.x,) -> threadIdx.x handles elements along L
Hmm, but if L=4096, we might have issues with max threads per block (1024). Let me use a loop within the thread.
Actually, let me reconsider the whole approach. Maybe it's better to have each thread handle multiple channels or something different.
Let me try a different approach: each thread handles one output element, and we use a different scan strategy.
Actually, let me just try the straightforward approach first and see how it performs:
```cpp
__global__ void associative_scan_kernel(
const float* __restrict__ a,
const float* __restrict__ b,
float* __restrict__ output,
int B, int L, int D) {
// Each block handles one (batch, channel) pair
int batch = blockIdx.x;
int channel = blockIdx.y;
int tid = threadIdx.x;
int n_threads = blockDim.x;
// Shared memory: need to store a and b for all L elements, double buffered
extern __shared__ float smem[];
float* a_buf[2] = {smem, smem + L};
float* b_buf[2] = {smem + 2*L, smem + 3*L};
int cur = 0;
// Load data
for (int t = tid; t < L; t += n_threads) {
int idx = batch * L * D + t * D + channel;
a_buf[cur][t] = a[idx];
b_buf[cur][t] = b[idx];
}
__syncthreads();
// Parallel inclusive scan
for (int stride = 1; stride < L; stride *= 2) {
int prev = cur;
cur = 1 - cur;
for (int t = tid; t < L; t += n_threads) {
if (t >= stride) {
float a_prev = a_buf[prev][t - stride];
float b_prev = b_buf[prev][t - stride];
float a_cur_val = a_buf[prev][t];
float b_cur_val = b_buf[prev][t];
a_buf[cur][t] = a_cur_val * a_prev;
b_buf[cur][t] = a_cur_val * b_prev + b_cur_val;
} else {
a_buf[cur][t] = a_buf[prev][t];
b_buf[cur][t] = b_buf[prev][t];
}
}
__syncthreads();
}
// Write output
for (int t = tid; t < L; t += n_threads) {
int idx = batch * L * D + t * D + channel;
output[idx] = b_buf[cur][t];
}
}
```
Actually, I realize there might be an issue with shared memory size. For L=4096, we need 4 * 4096 * 4 = 64KB. The H100 supports up to 227KB per block, so this should be fine.
Let me also think about the number of threads per block. With L=4096, we want enough threads to efficiently process the elements. Let's use 256 or 512 threads per block.
Let me write the complete kernel code:
call → compile_kernel
callcompile_kernel
{
"kernel_code": "import torch\nfrom torch.utils.cpp_extension import load_inline\n\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n\n// Parallel associative scan kernel\n// Each block handles one (batch, channel) pair\n// Uses O(log L) iterations with the associative combine operation\n\n__global__ void associative_scan_kernel(\n const float* __restrict__ a,\n const float* __restrict__ b,\n float* __restrict__ output,\n int B, int L, int D) {\n \n int batch = blockIdx.x;\n int channel = blockIdx.y;\n int tid = threadIdx.x;\n int n_threads = blockDim.x;\n \n // Shared memory: double buffered a and b arrays\n extern __shared__ float smem[];\n float* a0 = smem;\n float* b0 = smem + L;\n float* a1 = smem + 2 * L;\n float* b1 = smem + 3 * L;\n \n // Load data into shared memory\n for (int t = tid; t < L; t += n_threads) {\n int idx = batch * L * D + t * D + channel;\n a0[t] = a[idx];\n b0[t] = b[idx];\n }\n __syncthreads();\n \n // Pointers for double buffering\n float* a_cur = a0;\n float* b_cur = b0;\n float* a_prev = a1;\n float* b_prev = b1;\n \n // Parallel inclusive scan: each step doubles the lookback distance\n // Combine: (a_src, b_src) \u2297 (a_dst, b_dst) = (a_dst * a_src, a_dst * b_src + b_dst)\n for (int stride = 1; stride < L; stride *= 2) {\n // Swap buffers\n float* tmp_a = a_cur; float* tmp_b = b_cur;\n a_cur = a_prev; b_cur = b_prev;\n a_prev = tmp_a; b_prev = tmp_b;\n \n for (int t = tid; t < L; t += n_threads) {\n if (t >= stride) {\n float a_src = a_prev[t - stride];\n float b_src = b_prev[t - stride];\n float a_dst = a_prev[t];\n float b_dst = b_prev[t];\n a_cur[t] = a_dst * a_src;\n b_cur[t] = a_dst * b_src + b_dst;\n } else {\n a_cur[t] = a_prev[t];\n b_cur[t] = b_prev[t];\n }\n }\n __syncthreads();\n }\n \n // Write output: h_t = b component of prefix\n for (int t = tid; t < L; t += n_threads) {\n int idx = batch * L * D + t * D + channel;\n output[idx] = b_cur[t];\n }\n}\n\ntorch::Tensor associative_scan_cuda(torch::Tensor a, torch::Tensor b) {\n TORCH_CHECK(a.dim() == 3, \"a must be 3D\");\n TORCH_CHECK(b.dim() == 3, \"b must be 3D\");\n TORCH_CHECK(a.sizes() == b.sizes(), \"a and b must have same shape\");\n \n int B = a.size(0);\n int L = a.size(1);\n int D = a.size(2);\n \n auto output = torch::empty_like(a);\n \n // Shared memory: 4 * L floats (double buffered a and b)\n size_t shared_mem_size = 4 * L * sizeof(float);\n \n // Launch configuration\n int threads_per_block = 256;\n dim3 grid(B, D);\n dim3 block(threads_per_block);\n \n associative_scan_kernel<<<grid, block, shared_mem_size>>>(\n a.data_ptr<float>(),\n b.data_ptr<float>(),\n output.data_ptr<float>(),\n B, L, D\n );\n \n return output;\n}\n\"\"\"\n\n# Load the CUDA extension\nscan_extension = load_inline(\n name='associative_scan',\n cpp_sources='',\n cuda_sources=cuda_source,\n functions=['associative_scan_cuda'],\n extra_cuda_cflags=['-O3'],\n verbose=False\n)\n\nclass ModelNew:\n \"\"\"\n Performs a parallel associative scan (linear recurrence) over a 1D sequence.\n Uses a custom CUDA kernel with O(log L) parallel algorithm.\n \"\"\"\n def __init__(self, d_model: int):\n self.d_model = d_model\n \n def forward(self, a: torch.Tensor, b: torch.Tensor) -> torch.Tensor:\n return scan_extension.associative_scan_cuda(a, b)"
}