reasoning
The output doesn't match. The Blelloch scan algorithm I used is for exclusive scans, but the reference implementation computes an inclusive scan (prefix sum). Let me reconsider.
The reference computes:
```
h_0 = b_0
h_t = a_t * h_{t-1} + b_t for t > 0
```
So for position t, the output is b[0] + a[1]*b[0] + a[2]*a[1]*b[0] + ... which is the inclusive prefix scan.
Actually, looking at the math again:
Let me define the binary operation on pairs (a, b):
- (a2, b2) ⊗ (a1, b1) = (a2*a1, a2*b1 + b2)
This operation represents: apply the recurrence from (a1, b1) first, then from (a2, b2).
The identity element is (1, 0).
The scan of pairs [(a_0, b_0), (a_1, b_1), ..., (a_{L-1}, b_{L-1})] would give:
- (a_0, b_0) for position 0 (h_0 = b_0)
- (a_1*a_0, a_1*b_0 + b_1) for position 1 (h_1 = a_1*h_0 + b_1 = a_1*b_0 + b_1) ✓
- etc.
So the result should be the b component of each pair after the inclusive scan.
The issue is that my Blelloch implementation is incorrect for this purpose. Let me think about what the correct algorithm should be.
For an inclusive scan, we want:
- result[0] = input[0]
- result[i] = input[0] ⊗ input[1] ⊗ ... ⊗ input[i]
The Blelloch algorithm (work-efficient) gives an exclusive scan. For an inclusive scan, we need a different approach.
The simple approach is:
1. Up-sweep: reduce to get partial products
2. Down-sweep: propagate values forward
But the issue is that in my current implementation, I'm modifying the input values in place incorrectly.
Let me try a simpler approach: just run the standard parallel prefix sum algorithm (Kogge-Stone or similar).
Actually, let me look at this more carefully. The Kogge-Stone algorithm for inclusive scan:
```
for d = 1 to log2(n):
for all k from 0 to n-1 in parallel:
if k >= 2^d:
x[k] = x[k - 2^d] ⊗ x[k]
```
This gives the inclusive scan directly. Let me implement this correctly.
For the associative operation:
- combine(x[k-d], x[k]) where x = (a, b) pairs
Actually, let me re-read the operation definition. For inclusive scan:
- result[i] = input[0] ⊗ input[1] ⊗ ... ⊗ input[i]
Using our combine operation:
- (a_0, b_0) ⊗ (a_1, b_1) = (a_1*a_0, a_1*b_0 + b_1)
Wait, this isn't right for the recurrence:
h_0 = b_0
h_1 = a_1 * h_0 + b_1 = a_1 * b_0 + b_1
So for position 1, we want the result (a, b) where b = a_1*b_0 + b_1. This matches!
Let me check position 2:
h_2 = a_2 * h_1 + b_2 = a_2 * (a_1*b_0 + b_1) + b_2 = a_2*a_1*b_0 + a_2*b_1 + b_2
The scan of 3 elements:
(a_0, b_0) ⊗ (a_1, b_1) ⊗ (a_2, b_2)
= ((a_1, b_1) ⊗ (a_0, b_0)) ⊗ (a_2, b_2)
= (a_1*a_0, a_1*b_0 + b_1) ⊗ (a_2, b_2)
Wait, I need to be careful about the order. Let me define:
- (a', b') = (a1, b1) ⊗ (a0, b0)
Then (a', b') represents applying recurrence with (a0, b0) first, then with (a1, b1).
So:
- a' = a1 * a0
- b' = a1 * b0 + b1
The cumulative effect after positions 0 and 1:
- a = a_1 * a_0
- b = a_1 * b_0 + b_1
Now apply position 2:
- (a_2, b_2) ⊗ (a_1*a_0, a_1*b_0 + b_1)
- a = a_2 * a_1 * a_0
- b = a_2 * (a_1*b_0 + b_1) + b_2 = a_2*a_1*b_0 + a_2*b_1 + b_2
This matches h_2!
So the inclusive scan works correctly with this operation. The issue must be in my implementation.
Let me write a cleaner Kogge-Stone style algorithm:
```
for stride = 1; stride < L; stride *= 2:
for i = L-1 to stride (in parallel):
combine(input[i-stride], input[i]) and store in input[i]
```
This gives the inclusive scan. Let me verify:
Initial: [p0, p1, p2, p3, p4, p5, p6, p7]
After stride=1:
- p1 = p0 ⊗ p1
- p2 = p1 ⊗ p2 (but we want original p1, not modified)
- ...
Wait, this is the issue! In Kogge-Stone, we read from the original values, not the modified ones. But my implementation does in-place modification which creates a race condition.
Actually, looking at my code again:
```cpp
for (int stride = 1; stride < L; stride *= 2) {
for (int i = L - 1; i >= stride; i--) {
int idx = base + i * D;
int idx_prev = base + (i - stride) * D;
float a2 = a_data[idx];
float b2 = b_data[idx];
float a1 = a_data[idx_prev];
float b1 = b_data[idx_prev];
a_data[idx] = a2 * a1;
b_data[idx] = a2 * b1 + b2;
}
}
```
This iterates from L-1 down to stride. For stride=1:
- i=7: reads from p6, writes to p7
- i=6: reads from p5, writes to p6
- ...
Since each thread is single-threaded (each thread handles one channel independently), there's no race condition within a thread. The issue is the order: when i=6 reads from p5, it's the original value, and when i=7 reads from p6, it's also the original value (because the write to p6 hasn't happened yet due to sequential execution).
Actually, this should work! Let me trace through:
Initial: [p0, p1, p2, p3]
Stride=1:
- i=3: p3' = p2 ⊗ p3 (reads original p2)
- i=2: p2' = p1 ⊗ p2 (reads original p1)
- i=1: p1' = p0 ⊗ p1 (reads original p0)
Result: [p0, p0⊗p1, p1⊗p2, p2⊗p3]
Wait, this is NOT the correct inclusive scan! The correct result should be:
- [p0, p0⊗p1, p0⊗p1⊗p2, p0⊗p1⊗p2⊗p3]
The issue is that in Kogge-Stone, all reads should happen before any writes. The in-place modification is incorrect because it overwrites values that will be read later.
Let me fix this by using double-buffering or by being more careful about the order.
Actually, looking at Kogge-Stone more carefully:
```
for d = 1 to log2(n):
for all k from n-1 to d in parallel:
if k >= d:
x[k] = x[k] ⊗ x[k - d]
```
The key is that within each iteration of the outer loop (stride), all updates happen "in parallel" in the abstract sense. In CUDA, we need to make sure that all threads read the values from the previous stride before any thread writes the value for the current stride.
In my sequential loop, this is guaranteed! Let me trace through again more carefully:
Initial: [p0, p1, p2, p3]
Stride=1 (outer loop iteration 1):
- i=3: p3 = p2 ⊗ p3 → p3 now contains p2⊗p3
- i=2: p2 = p1 ⊗ p2 → p2 now contains p1⊗p2
- i=1: p1 = p0 ⊗ p1 → p1 now contains p0⊗p1
After stride=1: [p0, p0⊗p1, p1⊗p2, p2⊗p3]
But wait, p3 should be p0⊗p1⊗p2⊗p3, not p2⊗p3. And p2 should be p0⊗p1⊗p2, not p1⊗p2.
Hmm, the issue is that after stride=1, the values are not correct. Let me think about this differently.
Kogge-Stone algorithm:
- After stride=1: each element i gets the contribution from element i-1
- result: [p0, p0⊗p1, (p0⊗p1)⊗p2, (p0⊗p1⊗p2)⊗p3]
But this requires that p1 already contains p0⊗p1 when we compute p2, etc. Let me trace through the forward iteration:
```
for stride = 1; stride < L; stride *= 2:
for i = stride to L-1:
x[i] = x[i-stride] ⊗ x[i]
```
Stride=1:
- i=1: x[1] = x[0] ⊗ x[1] → x[1] = p0⊗p1
- i=2: x[2] = x[1] ⊗ x[2] → x[2] = (p0⊗p1)⊗p2
- i=3: x[3] = x[2] ⊗ x[3] → x[3] = ((p0⊗p1)⊗p2)⊗p3
This gives the correct result! But the issue is that we need to use the MODIFIED value of x[i-stride], which means we need to iterate forward, not backward.
Let me change my code to iterate forward instead of backward:
```cpp
for (int stride = 1; stride < L; stride *= 2) {
for (int i = stride; i < L; i++) { // Forward iteration!
int idx = base + i * D;
int idx_prev = base + (i - stride) * D;
float a2 = a_data[idx];
float b2 = b_data[idx];
float a1 = a_data[idx_prev]; // This reads the already-modified value from earlier in this iteration
float b1 = b_data[idx_prev];
a_data[idx] = a2 * a1;
b_data[idx] = a2 * b1 + b2;
}
}
```
But wait, this won't work either because when i=2, we need x[1] which was just modified in i=1, but we also need the original p2, not (p0⊗p1)⊗p2.
Let me think about this more carefully. In Kogge-Stone, each stride iteration reads from values that were computed in the PREVIOUS stride iteration. Within one stride iteration, each element reads from an element that was NOT modified in this iteration (because i-stride < i, and we iterate forward, so i-stride hasn't been modified yet in this iteration).
Wait, that's wrong. If we iterate forward with stride=1:
- i=1: modifies x[1], reads x[0] (original)
- i=2: modifies x[2], reads x[1] (MODIFIED in this iteration)
This gives a cascading effect, which is NOT what we want.
Let me re-read Kogge-Stone:
Actually, Kogge-Stone uses the property that each stride iteration combines elements that are "stride" apart, and the elements at distance "stride" have already accumulated the prefix up to that point.
```
Initial: [p0, p1, p2, p3, p4, p5, p6, p7]
Stride 1: each element combines with element 1 position earlier
- x[1] = x[0] ⊗ x[1] = p0⊗p1
- x[2] = x[1] ⊗ x[2] = p1⊗p2 (using ORIGINAL x[1]!)
- ...
So after stride 1: [p0, p0⊗p1, p1⊗p2, p2⊗p3, p3⊗p4, p4⊗p5, p5⊗p6, p6⊗p7]
```
But this requires that x[1] NOT be modified before x[2] reads it. So we need to iterate backward OR use a separate buffer.
Let me iterate backward:
```
Stride 1 (backward):
- x[7] = x[6] ⊗ x[7] = p6⊗p7
- x[6] = x[5] ⊗ x[6] = p5⊗p6
- ...
- x[1] = x[0] ⊗ x[1] = p0⊗p1
After stride 1: [p0, p0⊗p1, p1⊗p2, p2⊗p3, p3⊗p4, p4⊗p5, p5⊗p6, p6⊗p7]
```
OK so the backward iteration preserves the original values. But this isn't the inclusive scan!
Stride 2 (backward):
- x[7] = x[5] ⊗ x[7] = (p4⊗p5) ⊗ (p6⊗p7)
- x[6] = x[4] ⊗ x[6] = (p3⊗p4) ⊗ (p5⊗p6)
- ...
- x[3] = x[1] ⊗ x[3] = (p0⊗p1) ⊗ (p2⊗p3)
- x[2] = x[0] ⊗ x[2] = p0 ⊗ (p1⊗p2)
After stride 2: [p0, p0⊗p1, p0⊗p1⊗p2, p0⊗p1⊗p2⊗p3, ...]
Hmm, x[2] becomes p0 ⊗ (p1⊗p2) = p0⊗p1⊗p2 ✓
x[3] becomes (p0⊗p1) ⊗ (p2⊗p3) = ... let me compute:
- (a,b) ⊗ (c,d) = (c*a, c*b + d)
- (p0⊗p1) ⊗ (p2⊗p3) = (p2⊗p3) ⊗ (p0⊗p1)
Wait, I need to be careful about the order of the combine operation.
Let me define: (a1, b1) ⊗ (a2, b2) = result of applying (a1,b1) first, then (a2,b2).
If we interpret:
- (p0⊗p1) represents: applying p0 first, then p1
- (p2⊗p3) represents: applying p2 first, then p3
Then (p0⊗p1) ⊗ (p2⊗p3) = applying p0, p1, then p2, p3 in sequence.
OK so let me use the notation differently. Let x[i] represent the scan prefix up to and including position i.
After stride 1 (backward):
- x[i] = x[i-1] ⊗ x[i] for i >= 1
But this doesn't give the right result because we're combining x[i-1] (which is just p[i-1]) with x[i] (which is p[i]), giving p[i-1]⊗p[i].
Wait, I think the issue is in how I'm interpreting the initial values. In Kogge-Stone, we start with the identity and do an exclusive scan, then convert to inclusive.
Let me try a different approach: just do a sequential scan but parallelize over batches and channels.
Actually, for L=4096, the sequential approach per thread would be O(L) work per thread, and we have B*D = 32*256 = 8192 threads. That's a lot of work!
But wait, the parallel scan also does O(L log L) operations per thread, which is O(L) for large L. So the parallel approach doesn't actually save work, it just enables parallelism.
Let me think about this differently. We have:
- B = 32 batches
- L = 4096 sequence length
- D = 256 channels
We can have each thread process one (batch, channel) pair and do a sequential scan. That's 32*256 = 8192 threads, each doing 4096 operations. Total operations: 8192 * 4096 = 33.5 million.
Or we can do a parallel scan where each thread participates in log(4096) = 12 phases, and in each phase, up to 4096/2 operations are done (in the first phase). But with only one thread per channel, we can only do 1 operation at a time per channel.
Actually, the parallel scan is only beneficial when we can parallelize within the sequence. But we have only one thread per channel, so we can't parallelize within the sequence.
Let me reconsider the approach. We need to use multiple threads per channel to parallelize the scan within the sequence.
Actually, looking at the problem again: we want to compute h[t] = a[t] * h[t-1] + b[t]. This is essentially a linear recurrence.
One efficient approach for this on GPU is the "block scan" approach where we use multiple threads in a block to cooperatively process one channel's sequence.
But with L=4096 and a block size of 256 or 512 threads, we can have each thread process L/block_size elements. Then we do a scan within the block.
Let me implement a work-efficient parallel scan using shared memory within a block.
Actually, let me first fix the correctness issue. The problem with my current code is the algorithm implementation. Let me implement the correct Kogge-Stone or Blelloch algorithm.
For Blelloch (work-efficient exclusive scan):
1. Up-sweep: build a binary tree of partial sums
2. Clear the root
3. Down-sweep: propagate the partial sums
Then convert to inclusive scan by shifting.
Or, I can use the Hillis-Steele algorithm (work-inefficient but simpler):
- Just repeatedly combine elements in place
- O(n log n) work but O(log n) depth
Let me implement Hillis-Steele for simplicity:
```
for stride = 1 to n/2:
for i = stride to n-1:
x[i] = x[i-stride] ⊗ x[i]
```
The key insight is that each stride doubles the distance, and we're accumulating the prefix. Let me trace through:
Initial: [p0, p1, p2, p3, p4, p5, p6, p7]
Stride 1:
- i=1: x[1] = x[0] ⊗ x[1] = p0⊗p1
- i=2: x[2] = x[1] ⊗ x[2] = ? (depends on whether x[1] was modified)
If we iterate forward:
- i=1: x[1] = p0⊗p1
- i=2: x[2] = x[1]⊗x[2] = (p0⊗p1)⊗p2 = p0⊗p1⊗p2
- i=3: x[3] = x[2]⊗x[3] = ... but x[2] was just modified!
This gives the WRONG result because we're using modified values instead of the values from the previous stride.
The correct Hillis-Steele requires that each iteration reads from the PREVIOUS iteration's results. In a parallel setting, this is implicit (all iterations happen simultaneously). In a sequential setting, we need to double-buffer.
Let me use a temporary buffer:
```cpp
float* temp_a = new float[L];
float* temp_b = new float[L];
// Copy input to temp
for (int i = 0; i < L; i++) {
temp_a[i] = a[i];
temp_b[i] = b[i];
}
for (int stride = 1; stride < L; stride *= 2) {
// Copy current values to temp
for (int i = 0; i < L; i++) {
temp_a[i] = a[i];
temp_b[i] = b[i];
}
// Update using temp
for (int i = stride; i < L; i++) {
float a1 = temp_a[i - stride];
float b1 = temp_b[i - stride];
float a2 = a[i]; // Current value
float b2 = b[i];
a[i] = a2 * a1;
b[i] = a2 * b1 + b2;
}
}
```
Wait, this still has the issue. Let me think again.
Hillis-Steele algorithm:
```
for d = 1 to log2(n):
parallel for i = d to n-1:
x[i] = x[i] ⊗ x[i-d]
```
The key is that in each iteration, all updates happen "simultaneously" using the values from the END of the previous iteration.
Let me use two buffers and swap:
```cpp
float* src_a = a;
float* src_b = b;
float* dst_a = temp_a;
float* dst_b = temp_b;
// Initialize
for (int i = 0; i < L; i++) {
dst_a[i] = src_a[i];
dst_b[i] = src_b[i];
}
for (int stride = 1; stride < L; stride *= 2) {
// Swap buffers
float* tmp = src_a; src_a = dst_a; dst_a = tmp;
tmp = src_b; src_b = dst_b; dst_b = tmp;
// Copy unchanged elements
for (int i = 0; i < stride; i++) {
dst_a[i] = src_a[i];
dst_b[i] = src_b[i];
}
// Update changed elements
for (int i = stride; i < L; i++) {
float a1 = src_a[i - stride];
float b1 = src_b[i - stride];
float a2 = src_a[i];
float b2 = src_b[i];
dst_a[i] = a2 * a1;
dst_b[i] = a2 * b1 + b2;
}
}
// Copy result back if needed
```
Let me trace through this:
Initial: src = [p0, p1, p2, p3, p4, p5, p6, p7], dst = same
Stride 1:
- Swap: now src = original
- dst[0] = src[0] = p0
- dst[1] = src[0]⊗src[1] = p0⊗p1
- dst[2] = src[1]⊗src[2] = p1⊗p2
- dst[3] = src[2]⊗src[3] = p2⊗p3
- ...
After stride 1: dst = [p0, p0⊗p1, p1⊗p2, p2⊗p3, p3⊗p4, p4⊗p5, p5⊗p6, p6⊗p7]
Stride 2:
- Swap: now src = dst from stride 1 = [p0, p0⊗p1, p1⊗p2, p2⊗p3, ...]
- dst[0] = src[0] = p0
- dst[1] = src[1] = p0⊗p1
- dst[2] = src[0]⊗src[2] = p0⊗(p1⊗p2) = p0⊗p1⊗p2
- dst[3] = src[1]⊗src[3] = (p0⊗p1)⊗(p2⊗p3)
Let me verify: (p0⊗p1)⊗(p2⊗p3) using our combine operation
combine(a1,b1)⊗(a2,b2) = (a2*a1, a2*b1 + b2)
So (p0⊗p1)⊗(p2⊗p3) = ...
Actually, I need to check if the combine operation is associative. Let me verify:
combine((a1,b1), (a2,b2)) = (a2*a1, a2*b1 + b2)
combine(combine((a1,b1), (a2,b2)), (a3,b3))
= combine((a2*a1, a2*b1+b2), (a3,b3))
= (a3*a2*a1, a3*(a2*b1+b2) + b3)
= (a3*a2*a1, a3*a2*b1 + a3*b2 + b3)
combine((a1,b1), combine((a2,b2), (a3,b3)))
= combine((a1,b1), (a3*a2, a3*b2+b3))
= ((a3*a2)*a1, (a3*a2)*b1 + (a3*b2+b3))
= (a3*a2*a1, a3*a2*b1 + a3*b2 + b3)
Yes! They're equal. So the operation is associative. ✓
Now let me continue tracing:
After stride 2: dst = [p0, p0⊗p1, p0⊗p1⊗p2, p0⊗p1⊗p2⊗p3, ...]
This is correct! The Hillis-Steele algorithm with double buffering works.
Now let me implement this in CUDA, using shared memory for the buffers.
Actually, since I have one thread per channel, I can allocate local arrays in shared memory or even in registers. For L=4096 and float precision, I need 4096*4 = 16KB per buffer. With two buffers, that's 32KB per channel.
But shared memory is shared among threads in a block. If I have 256 threads per block and each thread needs 32KB, that's 8MB per block, which exceeds shared memory capacity (228KB per SM on H100).
So I need to use global memory for the double buffering. But that's OK since we're already using global memory for the input.
Let me rewrite the kernel:
```cpp
__global__ void scan_hillis_steele(float* __restrict__ a_data,
float* __restrict__ b_data,
float* __restrict__ temp_a,
float* __restrict__ temp_b,
int B, int L, int D) {
int batch = blockIdx.y;
int channel = blockIdx.x * blockDim.x + threadIdx.x;
if (batch >= B || channel >= D) return;
int base = batch * L * D + channel;
int base_temp = batch * D + channel; // Each channel has its own temp area
// Initialize: copy input to temp
for (int i = 0; i < L; i++) {
temp_a[base_temp * L + i] = a_data[base + i * D];
temp_b[base_temp * L + i] = b_data[base + i * D];
}
float* src_a = a_data + base;
float* src_b = b_data + base;
float* dst_a = temp_a + base_temp * L;
float* dst_b = temp_b + base_temp * L;
for (int stride = 1; stride < L; stride *= 2) {
// Swap buffers (just swap pointers conceptually)
// In reality, we alternate between a_data and temp
// Actually, let's think differently
// We'll just use a_data and b_data as the working buffers
// And temp_a/b as the source for each iteration
}
}
```
This is getting complicated. Let me simplify by just using a sequential scan per thread, since we have enough threads to keep the GPU busy (32*256 = 8192 threads, and H100 has 132 SMs, so we can have ~8192/256 = 32 blocks per batch, which is good utilization).
Wait, but sequential scan is O(L) per thread, and parallel scan is O(L log L) total operations but O(log L) depth. The total work for parallel scan is MORE than sequential! The benefit is in parallelism, not work efficiency.
But with one thread per channel, we can't exploit the parallelism within the scan. We'd need multiple threads per channel.
OK let me think about this differently. What's the best way to parallelize this scan on a GPU?
Option 1: Sequential scan per channel, parallelize over channels.
- Pro: Simple, good memory access pattern
- Con: Each thread does O(L) work, no intra-channel parallelism
Option 2: Parallel scan per channel using multiple threads.
- Pro: O(log L) depth per channel
- Con: Complex, more synchronization
For L=4096 and D=256, option 1 might be faster due to simpler memory access pattern and less synchronization.
Let me try option 1 first:
```cpp
__global__ void scan_sequential(float* __restrict__ a_data,
float* __restrict__ b_data,
int B, int L, int D) {
int batch = blockIdx.y;
int channel = blockIdx.x * blockDim.x + threadIdx.x;
if (batch >= B || channel >= D) return;
int base = batch * L * D + channel;
// Sequential scan
float h_a = 1.0f; // h starts with identity (a=1, b=0)
float h_b = 0.0f;
for (int i = 0; i < L; i++) {
int idx = base + i * D;
float a = a_data[idx];
float b = b_data[idx];
// h_new = a * h_old + b
// But we need to compute the combined (a, b) for the prefix up to i
// Actually, the output at position i is h_i = a_i * h_{i-1} + b_i
// where h_{i-1} is the cumulative prefix up to i-1
// Let's compute h directly
h_b = a * h_b + b; // h_i = a_i * h_{i-1} + b_i
h_a = a * h_a; // The cumulative a (not actually needed for output)
b_data[idx] = h_b;
}
}
```
Wait, but the original algorithm modifies both a and b:
- h_0 = b_0
- h_t = a_t * h_{t-1} + b_t
And the output is h_t for all t. So we just need to compute h_t and output it.
But looking at the original code:
```python
h = torch.zeros(B, D, device=a.device, dtype=a.dtype)
outputs = []
for t in range(L):
h = a[:, t] * h + b[:, t]
outputs.append(h)
return torch.stack(outputs, dim=1)
```
So h starts at 0, and at each step:
h = a[:, t] * h + b[:, t]
For t=0: h = a[:, 0] * 0 + b[:, 0] = b[:, 0]
For t=1: h = a[:, 1] * b[:, 0] + b[:, 1]
So h_t for t >= 1 involves a_t * h_{t-1} + b_t, and h_0 = b_0.
Let me rewrite the sequential scan:
```cpp
__global__ void scan_sequential(float* __restrict__ a_data,
float* __restrict__ b_data,
int B, int L, int D) {
int batch = blockIdx.y;
int channel = blockIdx.x * blockDim.x + threadIdx.x;
if (batch >= B || channel >= D) return;
int base = batch * L * D + channel;
float h = 0.0f; // h starts at 0
for (int i = 0; i < L; i++) {
int idx = base + i * D;
float a = a_data[idx];
float b = b_data[idx];
h = a * h + b;
b_data[idx] = h; // Store output in b_data
}
}
```
This is much simpler! And it should be correct. Let me test this.