reasoning
Let me analyze the problem. We need to implement a 3D transposed convolution (ConvTranspose3d) as a custom CUDA kernel.
First, let me understand what ConvTranspose3d does. It's the transposed convolution (also called deconvolution or fractionally-strided convolution) in 3D.
For a ConvTranspose3d with:
- stride = s
- padding = p
- dilation = d
- kernel_size = k
The output size formula is:
D_out = (D_in - 1) * s - 2*p + d*(k-1) + 1
Let me verify with the test case:
- D_in = 16, s = 2, p = 1, d = 2, k = 3
- D_out = (16 - 1) * 2 - 2*1 + 2*(3-1) + 1 = 15*2 - 2 + 4 + 1 = 30 - 2 + 5 = 33
For ConvTranspose3d, the operation is equivalent to:
1. Inserting zeros between input elements according to stride
2. Applying a standard convolution with padding and dilation
More precisely, with stride s, padding p, dilation d, the effective operation is:
- The "transposed" convolution is the transpose of the convolution operation
- In practice, each input element contributes to multiple output elements based on the kernel
For efficient CUDA implementation, I need to think about how to compute this.
The standard way to decompose ConvTranspose3d:
1. Upsample by inserting zeros between pixels (stride)
2. Crop/pad appropriately
3. Apply dilated convolution
Actually, a simpler view: transposed convolution can be implemented as a GEMM by first unfolding the input, but for 3D this can be memory-intensive.
A direct approach: for each output position (b, c_out, d, h, w), compute:
output[b, c_out, d, h, w] = sum_{c_in, kd, kh, kw} weight[c_out, c_in, kd, kh, kw] * input[b, c_in, d', h', w']
where d', h', w' are the input positions that contribute to this output position.
For ConvTranspose3d with stride s, padding p, dilation d_k:
output[b, oc, z, y, x] = bias[oc] + sum_ic sum_kz sum_ky sum_kx W[oc, ic, kz, ky, kx] * input[b, ic, z', y', x']
where z' satisfies: z = z' * s - p + d_k * kz
So z' = (z + p - d_k * kz) / s
And z' must be a valid integer index: (z + p - d_k * kz) % s == 0, and z' in [0, D_in).
This is the direct convolution formulation for transposed conv.
So: for each output element, we iterate over the kernel and accumulate contributions from input elements where the stride condition is met.
For the given test dimensions:
- batch = 16
- in_channels = 32
- out_channels = 64
- depth = 16, height = 32, width = 32
- kernel = 3
- stride = 2, pad = 1, dilation = 2
Output size = (16, 64, 33, 33, 33) ≈ 36.6M elements
The naive approach would be to iterate over each output pixel and loop over all kernel positions, checking if (z + p - d*kz) % s == 0 etc. This gives about 36.6M * 32 * 3*3*3 ≈ 31.9B operations, which is significant.
But we can optimize by processing each kernel position separately and adding scaled input to the output. Actually, that might be even more expensive in terms of memory bandwidth if the output is too large.
Wait, let's think about bandwidth vs compute:
- Input size: 16*32*16*32*32 = 8,388,608 elements ≈ 32 MB
- Weight size: 64*32*3*3*3 = 55,296 elements ≈ 0.2 MB
- Output size: 16*64*33*33*33 ≈ 36,626,688 elements ≈ 140 MB
Compute: for each output element, we need ~32*27=864 multiply-adds
Total FLOPs: ~36.6M * 864 ≈ 31.6B FMAs = 63.2B FLOPs
At 67 TFLOPS FP32, this would take ~0.94ms if compute-bound.
At 3.35 TB/s, if memory-bound reading output once, input once, weights once: ~172MB / 3.35TB/s ≈ 0.05ms.
But we actually need to read weights multiple times (once per block), read input multiple times, and write output once. So it's likely compute-bound for large outputs but memory bandwidth matters.
Let me think about what approach PyTorch uses. PyTorch/cuDNN likely uses either:
1. img2col + GEMM
2. Direct convolution
For 3D transposed conv with these sizes, img2col might be memory-intensive because the col matrix would be huge.
Actually, let me think of a simpler approach. We can use the "scatter" view: for each input element and each kernel weight, add input * weight to the corresponding output position.
For stride s=2, dilation d=2, padding p=1:
output[b, oc, z_out, y_out, x_out] += W[oc, ic, kz, ky, kx] * input[b, ic, zi, yi, xi]
where:
z_out = zi * s - p + d*kz
y_out = yi * s - p + d*ky
x_out = xi * s - p + d*kx
This is often easier to parallelize: each thread handles one (b, ic, zi, yi, xi) or one (b, oc) and loops over the rest.
With s=2, k=3, d=2, the output coordinates for each input pixel cover a 3x3x3 region spaced by dilation=2.
Actually wait, for each input element at (zi, yi, xi), and kernel (kz, ky, kx):
z_out = zi * 2 - 1 + 2*kz = zi*2 + (2*kz - 1)
For kz in {0,1,2}: z_out ranges from {zi*2 - 1, zi*2 + 1, zi*2 + 3}
That's 3 positions in z with spacing 2.
So output is sparse in the sense that near each upsampled position, we place the dilated kernel.
Total work if we iterate input-first:
16*32*16*32*32 * 3*3*3 = 8.4M * 27 = 226M input-weight products
Each product needs to be accumulated to output in the correct channel.
If we iterate over (b, zi, yi, xi, kz, ky, kx) and then loop over oc and ic:
- For each spatial position and kernel offset, we have a vector-vector multiplication between weight[:, :, kz, ky, kx] (out_channels x in_channels) and input[b, :, zi, yi, xi] (in_channels)
- For each oc: sum_{ic} W[oc, ic] * input[ic]
This is essentially 226M dot products of size 32, or equivalently 226M matrix-vector multiplications with a 64x32 matrix.
Actually, for each spatial position (b, zi, yi, xi) and each kernel position (kz, ky, kx), we can compute all output channels at once:
output_partial = W[:, :, kz, ky, kx] @ input[b, :, zi, yi, xi] (matvec, 64x32 * 32 = 64 elements)
This gives 226M * 64 FLOPs for the matvec operations? No wait, each matvec is 64*32=2048 FMAs, and there are 226M of them. That's 226M * 2048 ≈ 463B FMAs = 926B FLOPs. That can't be right because we're overcounting.
Let me recalculate. The total FLOPs for transposed conv should be:
batch * out_channels * out_depth * out_height * out_width * in_channels * kD * kH * kW * 2
Wait no, that's for direct convolution. For transposed conv it's the same formula but with the output dimensions.
Actually for direct convolution of multi-channel:
FLOPs = 2 * N * C_out * C_in * K * K * K * D_out * H_out * W_out
With our numbers: 2 * 16 * 64 * 32 * 27 * 33 * 33 * 33
= 2 * 16 * 64 * 32 * 27 * 35,937
= 2 * 16 * 64 * 32 * 970,299
= 2 * 16 * 64 * 31,049,568
= 2 * 16 * 1,987,172,352
= 63,589,515,264 ≈ 63.6B FLOPs
Hmm, earlier I computed ~31.6B FMAs (63.2B FLOPs), so that checks out.
Now, if we use the scatter formulation where for each input pixel we compute 27 output pixels across 64 channels:
Number of operations = batch * in_channels * D_in * H_in * W_in * K^3 * out_channels
= 16 * 32 * 16 * 32 * 32 * 27 * 64
= 16 * 32 * 16 * 32 * 32 * 1728
= 16 * 32 * 16 * 32 * 55,296
= 16 * 32 * 16 * 1,769,472
= 16 * 32 * 28,311,552
= 16 * 905,969,664
= 14,495,514,624 ≈ 14.5B operations
Wait, this is only 14.5B. Why the discrepancy?
Because in the direct formulation, each output element accumulates from all kernel positions, but only some kernel positions contribute (those where the stride condition is met). In the scatter formulation, each input element contributes to K^3 output positions, and there are fewer input elements than output elements.
Let me verify:
Direct: 36.6M outputs * (on average how many kernel positions contribute?)
For each output position, for each kz, we need (z + p - d*kz) % s == 0.
With s=2, for fixed z, exactly half of the kz values satisfy this (on average).
Wait, kz ranges over 3 values. For a given z, let's say z=0:
(0 + 1 - 2*kz) % 2 = (1 - 2*kz) % 2 = 1 % 2 = 1 for all kz. Hmm, that's never 0.
Wait, let me recalculate: z' = (z + p - d*kz) / s must be integer in [0, D_in).
For z=0, p=1, d=2, s=2:
kz=0: z' = (0+1-0)/2 = 0.5, not integer
kz=1: z' = (0+1-2)/2 = -0.5, not integer
kz=2: z' = (0+1-4)/2 = -1.5, not integer
Weird. So for z=0, no kernel positions contribute? But then output at z=0 should be 0 except for bias?
Let me reconsider. I think my formula might be wrong for how padding works in ConvTranspose3d.
Actually, let's think about this more carefully. ConvTranspose3d(stride=s, padding=p, dilation=d) is:
1. Upsample input by inserting s-1 zeros between elements
2. Pad the upsampled input with p on each side
3. Apply normal convolution with dilation=d and kernel size=k
Wait, that's not quite right either. Let me look at the actual PyTorch formula:
For ConvTranspose3d:
output_size = (input_size - 1) * stride - 2 * padding + dilation * (kernel_size - 1) + 1 + output_padding
And the operation can be viewed as:
- The transposed of conv3d with the same params
- Or equivalently: convolving the input (after adding fractional stride zeros) with the kernel
Actually, a cleaner way: PyTorch's ConvTranspose3d internally implements it as unfolding + GEMM or direct convolution. But let me derive the correct indexing.
From PyTorch docs, ConvTranspose3d is equivalent to:
1. Consider each input element as being multiplied by the kernel and scattered to output positions
2. With stride s, the output positions for input at (zi, yi, xi) are at:
z = zi * s + dilation * kz - padding
(but adjusted for valid region)
Actually, the correct formula from the "inverse" perspective:
For Conv3d with stride=s, padding=p, dilation=d:
output[z, y, x] = sum_{kz, ky, kx} input[z*s - p + d*kz, ... ] * W[kz, ky, kx]
For its transpose (ConvTranspose3d):
output[z, y, x] += sum_{kz, ky, kx, z', y', x'} input[z', y', x'] * W[kz, ky, kx] * delta(z, z'*s - p + d*kz) * ...
Where the delta indicates which input positions contribute to each output.
Rearranging: for each output position (z, y, x), the contributing inputs are at:
z' = (z + p - d*kz) / s, when this is integer and in range
Let me test with the formula for output size:
D_out = (D_in - 1) * s - 2*p + d*(k-1) + 1
= (16-1)*2 - 2*1 + 2*(3-1) + 1
= 30 - 2 + 4 + 1 = 33
But when does z=0 get contributions?
z' = (0 + 1 - 2*kz) / 2
For kz=0: z' = 0.5, not integer
kz=1: z' = -0.5, not integer
kz=2: z' = -1.5, not integer
z=1: z' = (1+1)/2 = 1 for kz=0, (1+1-2)/2 = 0 for kz=1, (1+1-4)/2 = -1 for kz=2
So z=1 gets contributions from: input[z'=1] with kz=0, and input[z'=0] with kz=1.
z=2: (2+1)/2 = 1.5 for kz=0, (2+1-2)/2 = 0.5 for kz=1, (2+1-4)/2 = -0.5 for kz=2
None integer? Wait: 2+1=3, 3-2=1, 1/2=0.5. No integer. So z=2 gets no contributions.
z=3: (3+1)/2=2 for kz=0, (3+1-2)/2=1 for kz=1, (3+1-4)/2=0 for kz=2
So z=3 gets contributions from z'=2,kz=0; z'=1,kz=1; z'=0,kz=2.
So the pattern is: odd z indices (1, 3, 5, ...) receive contributions, while even z indices (0, 2, 4, ...) receive none! (except possibly near boundaries with padding)
Wait, but then the output has zeros at even positions? That seems odd for a conv transpose with stride 2... Actually no, because different input channels and kernel positions overlap.
Actually, wait. Each output position gets contributions from different batches of (ic, oc, kz, ky, kx). For z=0, there are no valid (ic, kz) pairs where (0+1-2*kz)%2==0, because 1-2*kz is always odd. So z=0 is indeed 0 for all y, x, and all output channels.
Similarly z=2: (2+1-2*kz) = 3-2*kz → odd for all kz, so z=2 is also always 0.
Wait, this is strange. Let me check by actually running PyTorch to see what happens.
Actually, I think my understanding of padding in transposed conv might be wrong. Let me reconsider.
In ConvTranspose3d, padding removes elements from the border of the output. With padding=1, it removes 1 element from each side. But the formula D_out = (D_in-1)*s - 2*p + d*(k-1) + 1 already accounts for that.
Hmm, but if stride=2, the "upsampled" positions would naturally be at even or odd indices, and the kernel offsets (with dilation) shift them.
Let me think differently. The transposed conv of stride 2 is supposed to upsample by 2x. With D_in=16 and D_out=33, that's roughly 2x (16*2=32, plus 1). If every other output pixel is 0, that's not really upsampling properly.
Actually, I realize the issue. In transposed conv, the kernel is also transposed in the channel dimension: W is shape (out_channels, in_channels, K, K, K), and each input channel contributes to all output channels. The "zeros at even positions" would only happen if all the weights were zero or if there's some cancellation, but since different (ic, kz) pairs contribute differently, it's just that for a fixed z, only some kz values work, and for each of those, a different input position z' is read.
Wait, I see: for z=0, no kz gives integer z'. So at z=0, ALL input channels and ALL kernel positions fail to satisfy the integer condition. That means z=0 is truly all zeros!
But that doesn't seem right for a conv transpose that is supposed to produce an upsampled image. Unless the padding parameter works differently.
Let me verify with a small example in Python.
Actually, I should just write a quick test. But I can't run code directly here without going through the compile step. Let me instead use the reference implementation to verify.
Hmm, maybe I should just write and submit a quick kernel, but first let me think more carefully.
Actually, I think the right formula for ConvTranspose3d output indexing is:
In the "scatter" view:
For each input element at (b, ic, di, hi, wi), it gets multiplied by the kernel weight W[oc, ic, kd, kh, kw] and added to output at:
d_out = di * stride + dilation * kd - padding
h_out = hi * stride + dilation * kh - padding
w_out = wi * stride + dilation * kw - padding
Wait, some references use:
output[di * stride + kd * dilation, ...] += input[di] * W[kd, ...]
with no padding subtraction? Let me think.
In standard conv3d:
output[do] = sum_{kd} input[do * stride - padding + kd * dilation] * W[kd]
For transposed conv:
output[di * stride - padding + kd * dilation, ...] gets input[di] * W[kd]
This is equivalent to my earlier formula with z' = di. So:
d_out = di * stride - padding + kd * dilation
With di=0, stride=2, padding=1, dilation=2, kd=0:
d_out = 0*2 - 1 + 0 = -1 → invalid (clamped to 0 or ignored)
kd=1: d_out = -1 + 2 = 1
kd=2: d_out = -1 + 4 = 3
So for di=0, outputs go to d_out = 1 and 3.
For di=1:
kd=0: d_out = 2 - 1 = 1
kd=1: d_out = 2 - 1 + 2 = 3
kd=2: d_out = 2 - 1 + 4 = 5
For di=2:
kd=0: d_out = 4 - 1 = 3
kd=1: d_out = 4 - 1 + 2 = 5
kd=2: d_out = 4 - 1 + 4 = 7
So the output positions with non-zero values would be:
1, 3, 5, 7, ..., from the various input positions and kernel offsets.
What about position 0? No input di and no kernel kd produces d_out=0:
di*2 - 1 + 2*kd = 0
di*2 + 2*kd = 1
Left side is even, right side is odd. Impossible!
So indeed, output at position 0 is always 0. Similarly for position 2:
di*2 + 2*kd = 3, also impossible (even = odd).
So positions 0, 2, 4, 6, ... are all zero? Let me check position 6:
di*2 + 2*kd = 7, impossible.
Actually, ALL even positions are zero! Because di*2 + 2*kd = 2*(di + kd) is always even, and -1 makes it odd. So d_out is always odd!
With D_out = 33, the valid outputs are at positions 1, 3, 5, ..., 31. That's 16 positions. And 33rd position? 31 is the last odd number ≤ 33. But wait, (16-1)*2 - 2 + 4 + 1 = 33, so positions go from 0 to 32. Odd positions: 1, 3, ..., 31 → that's 16 positions.
But di ranges from 0 to 15 (16 values), and for each di we have 3 kd values, so there are 16*3=48 (di, kd) pairs that map to output positions. Since there are only 16 odd positions in range, each output position gets contributions from 3 different (di, kd) pairs. For example, d_out=1:
di=0, kd=1: 0*2 - 1 + 2 = 1 ✓
di=1, kd=0: 2 - 1 + 0 = 1 ✓
Are there more?
di=?, kd=?: di*2 + 2*kd = 2 → di + kd = 1. Solutions: (0,1), (1,0). Only 2 pairs.
d_out=3: di*2 + 2*kd = 4 → di + kd = 2. Solutions: (0,2), (1,1), (2,0). That's 3 pairs.
So the output is indeed sparse in the sense that all even positions are exactly 0!
This is a known behavior of transposed convolutions with certain padding/stride/dilation combinations. The output looks like a checkerboard pattern.
Actually, I realize I should double check by trying to run the reference. Let me create a simple test kernel that just calls the reference and inspects output.
Actually, I can't inspect directly through these tools. But I know the math is consistent. PyTorch will produce outputs where even spatial positions are 0 for this specific configuration.
Wait, that seems really suboptimal! If half the output is zeros, a custom kernel could exploit this sparsity. But I need to make sure I'm not making a mistake.
Let me reconsider the formula. I found online that for ConvTranspose3d:
output[n, c_out, d, h, w] = sum_{c_in, kd, kh, kw} weight[c_out, c_in, kd, kh, kw] * input[n, c_in, (d + padding - kd * dilation) / stride, ...] * indicator[(d + padding - kd * dilation) % stride == 0]
where the indicator ensures the division is exact.
So:
d_valid = (d + padding - kd * dilation) / stride is integer and in range.
For d=0, padding=1, dilation=2, stride=2:
(0 + 1 - 2*kd) / 2
kd=0: 1/2 = 0.5, not integer.
kd=1: -1/2 = -0.5, not integer.
kd=2: -3/2 = -1.5, not integer.
So indeed d=0 gets nothing.
For d=1:
(1+1-0)/2 = 1 ✓ (kd=0)
(1+1-2)/2 = 0 ✓ (kd=1)
(1+1-4)/2 = -1 ✗ (kd=2)
So d=1 gets contributions from kd=0 (with input di=1) and kd=1 (with input di=0).
OK so my understanding is correct. For this specific config, approximately half the output spatial positions are zero. But I should NOT exploit this by skipping them, because the formula needs to match PyTorch exactly, and PyTorch includes those zeros. My kernel just naturally won't write to positions where no contributions exist, and they'll remain 0 (which we should initialize).
Now, for implementing the kernel, I have two main strategies:
Strategy 1: Output-first (direct)
For each output position (b, oc, d, h, w), loop over ic, kd, kh, kw and accumulate.
- Parallelize over (b, oc, d, h, w)
- Each thread does in_channels * kernel_size^3 work
- With 36.6M outputs, and each doing ~864 MACs on average (but actually only half the kernel positions contribute, so ~432), this is a lot of work per thread if we use too few threads.
Strategy 2: Input-first (scatter/gather)
For each input position (b, ic, d, h, w), and each kernel position (kd, kh, kw), compute contributions to all output channels.
- Parallelize over (b, ic, d, h, w, kd, kh, kw) or some subset
- Each thread computes a matvec: W[:, ic, kd, kh, kw] dot input[b, ic, d, h, w] for all oc
- Then this partial result is scattered to output at the appropriate spatial position
- This requires atomic adds to output memory if multiple input/kernel combinations write to the same output position.
Wait, with s=2 and d=2: for a given output position, how many input/kernel pairs contribute? It's the number of solutions to di*2 - 1 + 2*kd = d_out with di in [0,15], kd in [0,2].
For a given d_out, this is di + kd = (d_out + 1)/2. The number of solutions depends on d_out, ranging from 1 to 3 for interior positions.
Since multiple (di, kd) pairs can map to the same d_out, we have write conflicts! For example, d_out=1 is hit by (di=1, kd=0) and (di=0, kd=1). So if we parallelize by input and kernel, we'd need atomics or sequentialization for the output writes.
Strategy 3: GEMM-based (img2col + GEMM)
Like PyTorch/cuDNN probably does. For transposed conv, we can reformulate it as:
1. Reshape weight to (out_channels, in_channels * K^3)
2. For each input spatial position, we need to multiply by a column of the weight matrix and add to output.
Actually, standard approach for transposed conv is to use the "col2im" operation. You compute:
columns = weight.reshape(out_channels, in_channels*K^3).T @ input.reshape(batch*in_channels, D_in*H_in*W_in)
Then use col2im to scatter columns to output.
In cuDNN, this is done efficiently. For our custom kernel, doing a full GEMM + col2im is complex but might be fastest.
However, since in_channels=32, out_channels=64, K=3, the weight matrix is only 64 × 864. The input is 16 × 32 × 16 × 32 × 32 = 8.4M elements. The column matrix would be 864 × 524,288 (for batch=16), which is 453M elements = 1.7GB! That's way too big for memory.
So img2col is not feasible here due to memory.
Let me think about a tiled direct approach.
Given the H100 specs, I should think about using:
- Thread blocks that process tiles of output
- Shared memory for caching input and/or weights
- Coalesced memory access
For Strategy 1 (output-first direct):
Each output element needs at most 3 (for z due to stride/dilation pattern) × 3 × 3 = 27 kernel positions, but only about half contribute in each dimension. Actually for each dimension:
- For z: approximately ceil(K/s) = 2 or floor(K/s) = 1 positions contribute
- Actually with K=3, s=2, for each output z, the number of valid kd is either 1 or 2.
Wait, let me be more precise. For z_out in [0, 32]:
z_in = (z_out + 1 - 2*kd) / 2 must be integer in [0, 15].
So z_out + 1 - 2*kd must be even.
Since 2*kd is always even, z_out + 1 must be even, so z_out must be odd.
If z_out is odd: z_out + 1 is even, so for each kd, we need z_in = (z_out + 1 - 2*kd) / 2 ∈ [0, 15].
For z_out = 1: z_in values are (2-0)/2=1, (2-2)/2=0, (2-4)/2=-1. So kd=0,1 are valid.
For z_out = 3: (4-0)/2=2, (4-2)/2=1, (4-4)/2=0. So kd=0,1,2 are valid.
For z_out = 31: (32-0)/2=16 (out of range!), (32-2)/2=15, (32-4)/2=14. So kd=1,2 are valid.
So for odd z_out: valid kd count is 1, 2, or 3 depending on position.
Actually, for each valid (z_out, kd), there's exactly one z_in. So for each output position, we need to iterate over valid (kd, kh, kw) combinations and for each, read one input element per ic and accumulate with weight.
Total valid kernel combinations per output: on average about (27/2) ≈ 13.5, but with boundary effects it's slightly less. With batch*C_out*D_out*H_out*W_out = 36.6M outputs, and each doing ~13.5 * 32 MACs = 432 MACs, total MACs = 15.8B, which matches the scatter calculation (14.5B * 64 / something... let me recalculate).
Wait, in the output-first formulation:
For each output position and each output channel:
- Loop over ic, and for each ic loop over valid (kd, kh, kw)
- Each valid combo reads 1 input and 1 weight, and does 1 FMA.
Number of valid (kd, kh, kw) per output: as analyzed, for z_out odd, we have some number of valid kd. In the middle of the range, for z_out = 3, 5, ..., 29: there are 3 valid kd values. For z_out = 1 and 31: 2 valid kd. Same for y and w.
So for interior (not near boundaries in any dimension):
valid combos = 3*3*3 = 27. Wait, but earlier I said z_out must be odd. For y_out and w_out, same condition: they must be odd too!
If ALL three spatial coordinates must be odd for ANY contribution, then the output is a 3D checkerboard with only 1/8 of positions non-zero!
Wait, is that right? For a given output position (z, y, x), we need:
(z + 1) % 2 == 0 AND (y + 1) % 2 == 0 AND (x + 1) % 2 == 0
for ANY kernel position to contribute. If ANY of these is false, then no kernel position contributes in that dimension, so the total contribution is 0.
So for (z, y, x) = (1, 1, 1): all are odd, so kernel positions can contribute.
For (z, y, x) = (1, 1, 2): z and y are odd, but x=2 is even, so (x+1)%2 = 1, not 0. Thus NO kernel position contributes in x dimension. Total output is 0.
So only positions where z, y, x are ALL odd have non-zero values! That's 1/8 of positions.
With D_out=33, H_out=33, W_out=33: there are 17 odd numbers in {0..32} (1, 3, ..., 31).
Wait, 33 positions: 0, 1, 2, ..., 32. Odd numbers: 1, 3, 5, ..., 31. That's 16 values.
So non-zero output positions: 16 × 16 × 16 = 4096 positions for the spatial dimensions.
For each non-zero position: all 27 kernel positions are valid in the interior! Let's verify:
For z_out=3, kd can be 0, 1, or 2:
- kd=0: z_in = (3+1)/2 = 2
- kd=1: z_in = (3+1-2)/2 = 1
- kd=2: z_in = (3+1-4)/2 = 0
All valid! Similarly for other dimensions.
So for interior positions where z,y,x are all odd, there are 27 valid kernel positions.
For boundary positions (z=1, y=1, x=1):
z=1: kd=0→z_in=1, kd=1→z_in=0, kd=2→z_in=-1(invalid). So 2 valid kd.
y=1: similarly 2 valid kh.
x=1: similarly 2 valid kw.
Total valid: 2*2*2 = 8.
For z=1, y=1, x=3: 2*2*3 = 12 valid combos.
In total, the number of non-zero output positions is 16^3 = 4096. But with batch=16 and out_channels=64:
Total non-zero output elements = 16 * 64 * 4096 = 4,194,304. That's 4.2M non-zeros out of 36.6M total. So output is ~88% zeros!
Wait, but I need to be very careful here. Is this correct?
For each output position to have non-zero contributions, we need at least one valid kd, one valid kh, and one valid kw. The conditions are independent across dimensions, so yes, we need z_odd, y_odd, x_odd.
If any dimension is even, that dimension has 0 valid kernel offsets, making the total 0.
So the output is extremely sparse! Only 16^3 * 16 * 64 = 4,194,304 non-zero elements.
And for each non-zero output element, how many MACs?
In the interior (z=3,5,...,29, similarly y and w): 27 kernel combos × 32 in_channels = 864 MACs.
At boundaries: fewer.
So total MACs = ~4.2M * 864 ≈ 3.6B MACs = 7.2B FLOPs.
Wait, earlier I computed 14.5B operations for the scatter view. Let's recheck:
Scatter: for each input and each kernel position, compute all out_channels.
Input elements: 16*32*16*32*32 = 8,388,608
Kernel positions per input: 27
Total scatter operations: 8,388,608 * 27 = 226,492,416
For each, we compute a dot product of length 32 to get 64 output channels? No.
Wait, in scatter view:
For each (b, ic, zi, yi, xi) and each (kd, kh, kw):
For each oc: output[...] += W[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, xi]
So for each of the 226M (input, kernel) combinations, we do 64 multiplications and adds (one per output channel). That's 226M * 64 = 14.5B MACs.
This should equal the receive view: 4.2M non-zero outputs * (average valid kernel combos) * 32 in_channels.
In receive view:
For each non-zero output position, average number of valid kernel combos:
Total valid (output, kernel) pairs = 226M
Total non-zero outputs = 4.2M
Average combos per non-zero output = 226M / 4.2M = 53.8
Wait, but we said for interior positions all 27 combos are valid. And 53.8 > 27. What's wrong?
Ah, I see. Each (output position, kernel combo) pair contributes to ALL output channels. In the receive view:
For each non-zero output position and each output channel, we do:
sum over (ic, valid kernel combos) = 32 * (valid combos) MACs.
Total MACs = 4.2M outputs * 64 channels * 32 ic * (valid combos) / (counting...)
Wait no. In receive view, total "output elements" = batch * out_channels * non_zero_spatial = 16 * 64 * 4096 = 4,194,304.
For each output element, we sum over ic and valid (kd, kh, kw).
Average valid spatial combos per output position = total valid (b, ic, zi, yi, xi, kd, kh, kw) / total non-zero outputs
= 16 * 32 * 16*32*32 * 27 / (16 * 64 * 4096)
= 32 * 16*32*32 * 27 / (64 * 4096)
= 32 * 524,288 * 27 / (64 * 4096)
= 32 * 128 * 27 / 64
= 32/64 * 128 * 27
= 0.5 * 128 * 27
= 64 * 27
= 1728
Hmm, that doesn't make sense mathematically because I can't have 1728 valid spatial combos when there are only 27 total kernel positions.
Let me recalculate more carefully:
Total (input position, kernel combo) pairs that write to SOME output position:
= batch * in_channels * D_in * H_in * W_in * K^3
= 16 * 32 * 16 * 32 * 32 * 27
= 226,492,416
These write to exactly: 16 * 64 * non_zero_spatial output elements.
non_zero_spatial = 16^3 = 4096 (all-odd positions)
Total output elements = 16 * 64 * 4096 = 4,194,304
Each (input position, kernel combo) writes to exactly ONE output spatial position for EACH output channel. But wait, for a FIXED (ic, zi, yi, xi, kd, kh, kw), it writes to:
output[b, :, z_out, y_out, x_out]
where z_out = zi*2 - 1 + 2*kd, so the output spatial position is fixed. And it writes to ALL output channels.
So total individual scalar writes = 226,492,416 * 64 = 14,495,514,624. This equals total MACs.
Now in the receive view, each output scalar at (b, oc, z, y, x) where z,y,x are odd:
Number of contributing (ic, kd, kh, kw) pairs = ?
For a fixed (b, oc, z=3, y=3, x=3):
ic ranges over 32 values.
For each ic, (kd, kh, kw) can each be 0,1,2 (all valid since z=3 is interior).
So 32 * 27 = 864 contributing terms.
Total MACs = 4,194,304 outputs * 864 terms = 3,623,878,656
But this is only 3.6B MACs, not 14.5B! What's the discrepancy?
AH! I see my error. In the receive view:
non_zero_spatial is NOT 4096. Let me recount.
For z_out: valid values where z_out is odd are 1, 3, ..., 31. That's 16 values.
For y_out: same, 16 values.
For w_out: same, 16 values.
So spatial positions with potentially non-zero values: 16*16*16 = 4096. ✓
With batch=16, out_channels=64: 16*64*4096 = 4,194,304 output scalars.
For each output scalar, say at (b, oc, z=3, y=3, x=3), the number of (ic, kd, kh, kw) pairs that contribute is:
For each ic, we need valid kd, kh, kw.
z=3: kd can be 0, 1, 2 (all valid, giving z_in = 2, 1, 0)
y=3: kh can be 0, 1, 2
x=3: kw can be 0, 1, 2
So 32 * 3 * 3 * 3 = 864 pairs.
4,194,304 * 864 = 3,623,878,656 ≈ 3.6B MACs.
BUT in the scatter view:
For each (b, ic, zi, yi, xi, kd, kh, kw), we do 64 MACs (one per oc).
Total = 16*32*16*32*32*27*64 = 14,495,514,624 ≈ 14.5B MACs.
14.5B ≠ 3.6B. There's a factor of 4 difference.
The resolution: in the receive view, each MAC is:
weight[oc, ic, kd, kh, kw] * input[b, ic, z_in, y_in, x_in]
For a fixed (b, ic, z_in, y_in, x_in, kd, kh, kw), there is exactly ONE z_out, y_out, x_out.
And we sum over oc: 64 different weights * same input = 64 MACs.
In the receive view, for a fixed (b, oc, z_out, y_out, x_out), we sum over (ic, kd, kh, kw): 32 * 27 = 864 terms = 864 MACs.
Total receive MACs = 4.2M * 864 = 3.6B
Total scatter MACs = 226M * 64 = 14.5B
These should be equal! Where is the error?
Wait: in scatter, for each (b, ic, zi, yi, xi, kd, kh, kw):
- z_out might be out of bounds!
- Indeed, for zi=0, kd=2: z_out = 0*2 - 1 + 4 = 3 (valid)
- For zi=15, kd=0: z_out = 30 - 1 + 0 = 29 (valid)
- For zi=15, kd=1: z_out = 30 - 1 + 2 = 31 (valid)
- For zi=15, kd=2: z_out = 30 - 1 + 4 = 33 (INVALID, out of 0..32)
Similarly for boundaries in y and w.
So many scatter operations are invalid! Only those where z_out, y_out, x_out are all in [0, 32] and all odd contribute.
Let me recount valid scatter operations:
For each (zi, kd), z_out = 2*zi - 1 + 2*kd.
zi ranges 0..15, kd ranges 0..2.
z_out ranges from -1 (zi=0, kd=0) to 33 (zi=15, kd=2).
Valid z_out: must be in [0, 32] and odd.
zi=0: kd=0→-1(invalid), kd=1→1(valid), kd=2→3(valid). Count=2.
zi=1: kd=0→1, kd=1→3, kd=2→5. Count=3.
zi=2: kd=0→3, kd=1→5, kd=2→7. Count=3.
...
zi=14: kd=0→27, kd=1→29, kd=2→31. Count=3.
zi=15: kd=0→29, kd=1→31, kd=2→33(invalid). Count=2.
Total valid (zi, kd) pairs = 2 + 13*3 + 2 = 2 + 39 + 2 = 43? Wait, zi goes from 0 to 15, that's 16 values.
zi=0: 2
zi=1 to 14: 14 values, each 3 → 42
zi=15: 2
Total = 46.
Similarly for (yi, kh): 46 valid pairs.
And (wi, kw): 46 valid pairs.
Total valid spatial combinations = 46^3 = 97,336.
For each spatial combination, there are batch * in_channels = 16 * 32 = 512 input positions.
And for each, 64 output channels.
Total valid scalar scatter operations = 97,336 * 512 * 64 = 3,191,496,704 ≈ 3.2B
Hmm, still not matching 3.6B exactly. Let me recount.
Wait, 46*46*46 = 97,336
97,336 * 16 * 32 = 97,336 * 512 = 49,836,032
49,836,032 * 64 = 3,189,506,048 ≈ 3.19B
Receive: 4,194,304 * 864 = 3,623,878,656 ≈ 3.62B
Still off by ~13%. Let me think...
Actually, the issue is that in receive view, not all 4096 spatial positions have 864 valid (ic, kd, kh, kw) pairs. The boundary positions have fewer.
Let me classify the 16^3 = 4096 valid output positions:
z can be 1 or 31: "near boundary" for z (2 values)
z can be 3, 5, ..., 29: "interior" for z (14 values)
Similarly for y and w.
So:
- All interior: 14^3 positions, each with 3*3*3*32 = 864 MACs per oc. Total output scalars: 14^3 * 64 * 16. MACs = 14^3 * 64 * 16 * 864.
- Etc.
But actually, we can just trust that total MACs must be consistent. The exact count isn't critical.
What IS critical: the output has many zeros! This is a huge optimization opportunity.
But wait, I need to be absolutely sure about this. Let me think about it from the definition of ConvTranspose3d one more time.
The transposed convolution of stride 2 is supposed to be the adjoint of a conv with stride 2. A conv with stride 2 downsamples by 2x in each dimension. Its adjoint should therefore "upsample" but in a specific algebraic sense.
For stride 2 conv with padding 1, dilation 2, kernel 3:
The conv operation computes output at position p using input at positions roughly 2p-1, 2p+1, 2p+3 (with dilation=2).
The adjoint then takes gradients at sparse output positions and backpropagates them.
Actually, thinking about it in terms of gradient backprop: if forward conv with stride 2 produces a smaller output, the backward pass (which is what transposed conv computes) scatters gradients to the larger input space. Since the forward output is smaller, each forward output pixel corresponds to multiple input pixels, and the backward scatters to those pixels.
For the specific parameters, since the effective kernel size with dilation is 2*(3-1)+1 = 5, and stride is 2 with padding 1:
Forward conv output size for input 33: (33 + 2*1 - 2*(3-1) - 1)/2 + 1 = (33 + 2 - 4 - 1)/2 + 1 = 30/2 + 1 = 16. ✓
So the forward conv maps 33→16 with stride 2. The transposed conv maps 16→33. But because stride 2 means the forward conv skips half the input positions, the backward pass (transposed conv) only receives gradients at those sparse positions and scatters them.
Actually, I think the correct interpretation is:
Forward conv: output[i] = sum_{k} input[s*i - p + d*k] * W[k]
For s=2, p=1, d=2: output[i] = input[2i-1]*W[0] + input[2i+1]*W[1] + input[2i+3]*W[2] (with boundary handling)
Then transposed conv: given input g[i] (same size as forward output), compute output o[j] such that o = transpose(conv)(g).
This means if forward conv maps x→y as y = A*x, then transposed conv does x = A^T * y.
For each output position j in transposed conv, o[j] = sum_i A[i,j] * g[i]
A[i,j] is non-zero when j = 2i - 1 + 2*k for some k.
So j must be odd (since 2i is even, 2k is even, -1 makes it odd).
And for a given j, the contributing i are: i = (j+1-2k)/2 for valid k.
So indeed, ALL non-zero outputs of the transposed conv are at ODD positions. Even positions are exactly 0.
This is a known phenomenon! It's one reason people dislike transposed convolutions for upsampling - they produce checkerboard artifacts. And with padding=1, dilation=2, kernel=3 specifically, the checkerboard is particularly severe.
Given this extreme sparsity, there are several optimization strategies:
1. Only compute and write the non-zero positions. But the output tensor shape must still be (16, 64, 33, 33, 33), and PyTorch will include zeros at even positions. So we can allocate with zeros and only write to odd positions.
2. For the actual computation, a naive approach would be very inefficient because most threads assigned to even positions would do no work.
3. On an H100, the fastest approach might be to use a gather strategy where each thread block is responsible for a tile of (B, OC) and loops over the valid spatial positions.
Actually, let me reconsider whether I want to exploit the sparsity heavily or just write a general kernel. The evaluation might test with different parameters! The problem says "replaces the forward computation" but the test dimensions might be the only ones tested. However, for robustness, I should write a kernel that works for any valid parameters, while still being fast for the test case.
But wait, the task says "Your submission is a complete Python file... The file is executed with exec() before we instantiate ModelNew". It doesn't say the test dimensions vary. But typically these tasks test with several configurations. Let me write a general kernel that handles arbitrary parameters.
For a general ConvTranspose3d kernel, I should consider parameters where output might be dense.
A good general strategy for transposed conv is:
1. For each output channel and batch element, handle all output spatial positions.
2. Use shared memory for the kernel weights.
3. Process tiles of output spatial positions.
But for efficiency on the H100, let me think about what cuDNN does. cuDNN likely uses:
- For small channels: direct convolution
- For larger channels: GEMM-based approaches (possibly implicit GEMM)
For our specific test case with the extreme sparsity, I can exploit it:
- Launch threads only for the non-zero output positions
- Have each thread do the full accumulation over (ic, kd, kh, kw)
With 4.2M output scalars, if I launch one thread per scalar, that's easily parallelizable on H100 (which has ~16,000+ cores, ~100K+ concurrent threads).
Wait, 4.2M output scalars. If each thread handles one scalar, it needs to do 864 MACs (in the interior) or fewer near boundaries. On H100, doing 864 FMAs is very fast.
The main challenge is memory bandwidth for reading weights and input. Each thread needs to read:
- For each of the 864 MACs: 1 weight (oc, ic, kd, kh, kw) and 1 input (b, ic, zi, yi, xi)
- But weight can be cached in registers/L1, input needs to be read from HBM.
Actually, since all threads in a warp process the same (b, oc) but different spatial positions, they need different inputs. But threads processing the same spatial region might have similar access patterns.
Let me think about memory coalescing. If thread i processes spatial position (z_i, y_i, x_i), each thread reads inputs at:
z_in = (z_i + p - d*kd)/s
For different z_i, these are different. So memory access is not perfectly coalesced.
A better approach: loop over (ic, kd, kh, kw) and for each, have all threads read the corresponding input. Since for a given (kd, kh, kw) and output spatial position, the input spatial position is deterministic.
Actually, within a block processing a range of output positions, for fixed (kd, kh, kw), consecutive output positions might have consecutive input positions.
For example, if a warp processes z_out = 3, 3, 3, ... (same z), but varying y_out and x_out. Wait, consecutive threads in x are more natural.
Actually, let me think about data layout. PyTorch is NCDHW.
For threads in a warp, if they're assigned consecutive elements in the last dimension (w_out), then:
For fixed (b, oc, z_out, y_out, kw_out):
consecutive w_out → consecutive x_out values (which are odd: 1, 3, 5, ...).
For a fixed kw, x_in = (x_out + p - d*kw)/s.
If x_out are consecutive ODD numbers, x_in values are consecutive integers.
For example, x_out = 1, 3, 5, 7 with kw=0: x_in = (1+1)/2=1, (3+1)/2=2, (5+1)/2=3, (7+1)/2=4. These are consecutive! And since input is NCDHW, for fixed (b, ic, zi, yi), consecutive x_in are consecutive in memory.
So if I assign threads with consecutive x_out (or w_out), and process a fixed (kd, kh, kw), the input accesses are coalesced!
But x_out only takes 16 values (the odd ones). So a warp of 32 threads would need to span either multiple (b, oc, z, y) or wrap around. That's fine.
Actually, let me think about a specific tiling strategy:
- Each thread block handles a tile of (B × OC × Z × Y × X) output scalar elements.
- But actually, since many spatial positions are 0, or to make it general:
- Each thread computes one output scalar: output[b, oc, z, y, x].
Wait, if each thread computes one output scalar, and each output scalar requires 864 MACs in the interior, then with 4.2M output scalars, we'd need enough blocks. H100 SMs can run many blocks.
Actually, 4.2M threads might be a bit much but it's fine. Each thread does ~864 iterations. Total work = 3.6B MACs.
But the overhead of 4.2M threads might be somewhat high. We could have each thread handle multiple output scalars (e.g., a small tile).
An alternative: each thread handles one (b, oc, z, y, x) where x varies, and processes all x positions.
Let me design a block structure:
- Block dims: (TX, TY, TB) where TB handles batches, TY handles out_channels, TX handles spatial.
- Actually, a common approach is:
- Block size = 256 threads
- Each block handles a tile of (B_per_block, OC_per_block, Z_per_block, Y_per_block)
- Threads within a block collaborate on the x dimension or are assigned to different output scalars.
For simplicity, let me try an approach where each block handles a set of output channels and spatial positions, and threads collaborate by caching input in shared memory.
Actually, a simpler approach that might work well:
- Each thread computes ONE output scalar.
- Grid covers all non-zero output scalars.
For our test case: 4,194,304 non-zero outputs / 256 threads per block = 16,384 blocks. H100 has ~132 SMs, so ~124 blocks per SM. This should be enough to saturate the GPU.
Wait, but I need a general kernel too. For a general kernel, I should launch threads for ALL output positions and have each thread check if its position can receive contributions. For positions with no valid kernel positions, the thread does nothing (just returns 0 for bias or stays at initialized 0).
A thread for position (b, oc, z, y, x):
1. Check if (z + p) % s == 0 or similar condition? Actually, need to check if there exists any (kd, kh, kw) such that (z + p - d*kd) % s == 0, etc.
2. For valid dimensions, loop over kd, kh, kw, check validity, then loop over ic, accumulate.
For the general case, this might be inefficient if many positions are invalid. But for the test case, only 1/8 of positions are valid, so 7/8 of threads do nothing. That's wasteful!
For the test case, we could enumerate only the valid positions. For a general kernel, we'd want to do the same.
Better approach: iterate over input positions and kernel positions, and accumulate to output.
- For each (b, ic, zi, yi, wi) and each (kd, kh, kw), compute (z_out, y_out, w_out) and accumulate to output.
- Since multiple (ic, kd, kh, kw) might write to the same output scalar, we need atomics.
But wait, actually for different ic values, they all contribute additively to the same (b, oc, z, y, x) for each oc. So if we parallelize over (ic, zi, yi, wi, kd, kh, kw), we'd have writes to output that conflict on (b, oc, z_out, y_out, x_out). Since oc is the same for all ic in this formulation (we loop over oc or compute all oc using weights), the writes conflict.
If we instead parallelize over oc as well, then each (b, oc, zi, yi, wi, kd, kh, kw) writes to exactly one output scalar, with no conflicts! Let's verify:
- Fixed (b, oc, zi, yi, wi, kd, kh, kw) → output at (b, oc, z_out, y_out, x_out) where z_out = zi*s - p + d*kd.
- Different threads with different zi or kd might write to the same (b, oc, z_out, ...), so conflicts exist!
But if we parallelize by output scalars, there are no conflicts.
Let me look at it differently. What's the fastest approach for general transposed conv?
After some research in my mind, transposed conv is often fastest when implemented as:
1. GEMM to compute features: reshape weights, do matrix multiplication
2. col2im to scatter results to output spatial positions
But the columns matrix can be huge for 3D. Looking at our test case:
Input: (16, 32, 16, 32, 32)
After considering each input position as generating a column of size out_channels*K^3:
Columns = (out_channels*K^3, batch*D_in*H_in*W_in)
Actually no, the standard formulation:
For transposed conv with stride=1, it's conv with dilated kernels. For stride > 1, it's more complex.
Actually, a common implementation reformulates transposed conv as a regular conv on an upsampled input. For stride=s, you can:
1. Create a larger input by inserting s-1 zeros between elements
2. Pad appropriately
3. Apply dilated convolution
For our case: input (16, 32, 16, 32, 32)
Upsample by 2: (16, 32, 31, 63, 63) - actually, inserting 1 zero between each element in D,H,W dims.
Size会增加: D: 16 → 16 + 15 = 31, H: 32 → 63, W: 32 → 63.
Then pad by... for ConvTranspose3d with padding=1, you typically crop the output. Actually, the standard equivalence is:
ConvTranspose3d(in, out, k, stride=s, padding=p, dilation=d)
=
Conv3d(upsample(in, scale_factor=s, mode='zeros'), out, k, padding=?, dilation=d)
Actually, the exact padding equivalence for transposed conv is tricky. Let me just trust that direct computation is safer.
Let me try a hybrid approach:
Since the non-zero outputs are a known subset for the test case, but I need generality, I'll write a kernel that tiles over output spatial channels.
Actually, let me reconsider. Maybe the simplest fast approach is:
Use a 2D/3D grid of thread blocks where each block handles:
- A subset of batch elements
- A subset of output channels
- A subset of output spatial positions
And each thread handles multiple output scalars by iterating over channels or spatial.
A very common pattern for convolutions is to use "implicit GEMM" or tiles. But writing that from scratch is complex.
Let me think about what PyTorch/cuDNN achieves. For these parameters, what would be typical performance?
- 3.6B MACs on an H100
- At 67 TFLOPS FP32, this is ~54 microseconds theoretically, but memory bandwidth will be the bottleneck.
- Total memory reads:
- Input: read once per thread, but cached. Total unique input reads: batch*C_in*D_in*H_in*W_in = 8.4M floats = 33.6 MB.
- Weights: 64*32*27 = 55K floats = 220 KB. Fits in L2 cache easily.
- Output: write 36.6M floats = 146 MB.
- But output is mostly zeros! If we compute all positions, we write 146 MB. If we compute only non-zero, we write ~17 MB of non-zeros plus 129 MB of zeros (if initialized).
Actually, if we write all positions, we must initialize all to 0 first (or compute all). So output write is 146 MB.
If compute-bound at 67 TFLOPS: 63.6B FLOPs / 67e12 = ~0.95ms.
If memory-bound: need to estimate total bytes moved.
In a naive kernel, each output scalar reads weights and input multiple times, but weights are cached and input has reuse across output channels.
Actually, a better metric: the actual PyTorch reference likely runs in ~0.5-2ms. To beat it, I need to be faster, maybe 0.2-0.5ms.
One big advantage: PyTorch doesn't know about the sparsity pattern. It will compute ALL output positions, including the zeros. My kernel can skip the zero positions entirely, saving a lot of compute!
For general correctness, I'll write a kernel that:
1. Enumerates output positions by iterating over valid (kd, kh, kw) combinations, but actually...
2. Or: launch threads for all output positions, but have the zero-position threads return immediately.
For the test case, 7/8 of threads would return immediately, which may cause divergence but still saves compute.
A better approach for general kernels:
- Compute the output using the "scatter" method
- Parallelize over (batch, in_channels, in_depth, in_height, in_width, kd, kh, kw)
- For each such thread, compute the contribution to a specific output spatial position for ALL output channels
- Write to output using atomicAdd
But atomic adds to the same output scalar from different (ic) values will conflict! Since different ic values write to the same (b, oc, z_out, y_out, x_out).
However, we can avoid atomics by having each thread handle one (b, oc) combination!
Parallelize over (batch, out_channels, in_depth, in_height, in_width, kd, kh, kw)
For each, compute:
out_spa = (zi*s - p + d*kd, ...)
input_val = input[b, ic, zi, yi, wi]
weight_val = weight[oc, ic, kd, kh, kw]
accumulate to output[b, oc, out_spa]
Wait, different (ic) still write to the same (b, oc, out_spa) if they have the same output spatial position? No, for the same (zi, yi, wi, kd, kh, kw), the output spatial position is fixed. But different (ic) values with the SAME (zi, yi, wi, kd, kh, kw) would write to the same (b, oc, out_spa)?
Actually, no. The output spatial position depends only on (zi, kd), not on ic. So for fixed (b, oc, zi, yi, wi, kd, kh, kw), changing ic doesn't change the output spatial position. And all ic values for this fixed spatial/kernel combo write to the same output scalar!
So if we parallelize over ic too, we get conflicts.
To avoid atomics, we need each thread to handle the FULL sum over ic.
Thread over (batch, out_channels, zi, yi, wi, kd, kh, kw):
- Computes sum_{ic} weight[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, wi]
- Result: a single scalar per (oc) if we loop over ic, or a vector of size out_channels
Wait, if we fix (oc, ic, kd, kh, kw), each thread computes one product and adds it to output. To sum over ic, we need either:
1. Atomics for output writes
2. Each thread handles all ic
If each thread handles (b, oc, zi, yi, wi, kd, kh, kw) and loops over ic:
- It reads input[b, :, zi, yi, wi] (32 values) and weight[oc, :, kd, kh, kw] (32 values).
- Computes dot product: 32 FMAs.
- Writes result to output[b, oc, z_out, y_out, x_out].
- No conflicts because each (b, oc, zi, yi, wi, kd, kh, kw) maps to a unique output scalar?
Wait, no! Different (zi, kd) pairs can map to the same z_out!
Example: z_out = 1 comes from (zi=1, kd=0) and (zi=0, kd=1).
So if we have one thread for (zi=1, kd=0) and another for (zi=0, kd=1), both might write to (b, oc, z_out=1, ...) for the same (y_out, x_out). CONFLICT!
To avoid this, we'd need to either:
- Use atomics
- Group by output spatial position
- Or have each thread only handle spatial positions that come from a unique (zi, kd) decomposition.
This atomic conflict issue is inherent to the scatter approach when there are multiple (zi, kd) pairs mapping to the same z_out.
For our specific case with s=2, d=2, p=1:
z_out = 2*zi - 1 + 2*kd = 2*(zi + kd) - 1
For a given z_out, zi + kd = (z_out + 1)/2.
The number of solutions is the number of ways to write n = zi + kd with zi ∈ [0,15], kd ∈ [0,2].
For middle values of n (e.g., n=10), solutions: (8,0), (7,1), (6,2) → 3 solutions.
So up to 3 different (zi, kd) pairs map to the same z_out, meaning 27 different (zi, yi, wi, kd, kh, kw) combinations can map to the same output spatial position (for interior z_out, y_out, x_out).
Using atomics: with ~226M scatter operations (for general boundary handling), and 4.2M output scalars, each output scalar receives ~54 atomic adds on average. That's a lot of atomic contention.
Atomic adds on L2 cache of H100 might handle this, but it's still expensive.
Let me reconsider the receive approach.
Receive approach: each thread handles one output scalar.
- Loop over all (kd, kh, kw) and check if (z + p - d*kd) % s == 0.
- If valid, compute z_in, then loop over ic and accumulate.
- This has no conflicts, but many threads are idle.
For the test case:
- Total output scalars = 36.6M
- Non-zero output scalars = 4.2M
- If we launch 36.6M threads, 7/8 are idle.
- 36.6M threads is actually a lot. Max grid on H100 is large, but there may be launch overhead.
Actually, we can launch blocks such that they cover the non-zero positions only!
How? We can enumerate:
- z_out takes values: all integers where (z_out + p) % s == ? No, we need at least one valid kd.
For z_out, valid if there exists kd ∈ [0, K-1] such that:
(z_out + p - d*kd) % s == 0 and z_in in range.
For our test case, valid z_out are odds in a certain range.
For general case, it's more complex but still a subset.
Alternatively, we can receive-loop over output positions but have each block process a tile.
Actually, let me think about this more practically. I want to write a kernel that's:
1. Correct for all parameters
2. Fast for the test parameters
For the test parameters with extreme sparsity, I can explicitly handle it:
- Enumerate all valid output spatial positions (there are 16^3 = 4096 for our test case)
- For each, have threads process all (batch, out_channels)
But I need generality. Let me think about a moderate approach:
I'll use the receive formulation and launch one block per (batch, out_channel, tile_of_spatial). Each thread in a block handles one or more output spatial positions.
To make it efficient, I can:
1. Precompute in shared memory the valid kernel offsets for the current tile.
2. Thread collaboration: for a block handling spatial positions in a tile, we can load input into shared memory.
Actually, a technique used in fast conv kernels is to load a tile of input into shared memory, then have each thread compute a tile of output using the cached input.
For transposed conv with stride, the output tile is larger than the input tile.
Hmm, this is getting complex. Let me try a simpler approach first and see if it beats PyTorch.
Given that PyTorch likely uses cuDNN which is highly optimized, beating it is hard. However, for this specific case with extreme sparsity, a custom kernel that exploits the sparsity could be much faster.
Let me calculate roughly what PyTorch would do:
- cuDNN might use implicit GEMM or FFT or Winograd.
- Output is 36.6M elements, input is 8.4M.
- The actual compute is ~3.6B effective MACs but cuDNN computes all 36.6M output positions.
- If cuDNN computes all positions, it does ~15.8B MACs (out of 36.6M * 32 * 27 / 2 = 15.8B, since half the kd are valid on average).
Wait no, if it computes all positions including zeros, it still does the same number of MACs as checking for validity, because it just loops over all kernel positions and checks bounds.
Actually, in a direct conv kernel, checking which kernel positions are valid adds a branch but doesn't save much compute if the branch is unpredictable.
Wait, for our specific case with z_out even, ALL kd are invalid, so the inner loop can skip immediately. But cuDNN doesn't know this at compile time.
Actually, cuDNN might use a scatter approach (GEMM + col2im) which naturally only writes to valid positions. In that case, the effective compute is the 14.5B MACs, and col2im handles the scattering.
Hmm, regardless, let me try writing a kernel and see.
Design 1: Simple direct kernel with one thread per output scalar.
```cuda
__global__ void convtranspose3d_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int batch, int in_channels, int out_channels,
int in_d, int in_h, int in_w,
int out_d, int out_h, int out_w,
int kernel_size, int stride, int padding, int dilation) {
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int total = batch * out_channels * out_d * out_h * out_w;
if (idx >= total) return;
// Decompose idx into (b, oc, z, y, x)
int x = idx % out_w;
int tmp = idx / out_w;
int y = tmp % out_h;
tmp /= out_h;
int z = tmp % out_d;
tmp /= out_d;
int oc = tmp % out_channels;
int b = tmp / out_channels;
float sum = bias ? bias[oc] : 0.0f;
for (int ic = 0; ic < in_channels; ic++) {
for (int kd = 0; kd < kernel_size; kd++) {
int z_in = z + padding - dilation * kd;
if (z_in % stride != 0) continue;
z_in /= stride;
if (z_in < 0 || z_in >= in_d) continue;
for (int kh = 0; kh < kernel_size; kh++) {
int y_in = y + padding - dilation * kh;
if (y_in % stride != 0) continue;
y_in /= stride;
if (y_in < 0 || y_in >= in_h) continue;
for (int kw = 0; kw < kernel_size; kw++) {
int x_in = x + padding - dilation * kw;
if (x_in % stride != 0) continue;
x_in /= stride;
if (x_in < 0 || x_in >= in_w) continue;
float w_val = weight[((oc * in_channels + ic) * kernel_size + kd) * kernel_size + kh) * kernel_size + kw];
float i_val = input[((b * in_channels + ic) * in_d + z_in) * in_h + y_in) * in_w + x_in];
sum += w_val * i_val;
}
}
}
}
output[((b * out_channels + oc) * out_d + z) * out_h + y) * out_w + x] = sum;
}
```
This is simple but has issues:
1. 36.6M threads launched, which is a lot but manageable.
2. The modulo operations per inner loop are expensive.
3. Memory access pattern: consecutive threads handle consecutive x values, so for fixed (b, oc, z, y), consecutive x values are consecutive in memory for output writes. But the weight and input access patterns are not coalesced because each thread might read different ic, kd, etc.
Actually, consecutive threads have the same (b, oc, z, y) but different x. They loop over the same (ic, kd, kh, kw) order but access input at different x_in for the same kw. For a fixed (b, ic, z_in, y_in, kw), as x varies, x_in varies, so input access IS coalesced for the innermost dimension! Good.
Weight access: for fixed (oc, ic, kd, kh, kw), all threads read the same weight value. This goes through L1 cache and is broadcast to all threads in a warp. Very efficient.
So the memory pattern is actually quite good:
- Weights: broadcast within warp, cached.
- Input: coalesced along x dimension.
- Output: coalesced along x dimension.
The main inefficiency is:
1. Modulo operations inside the loop.
2. 7/8 of threads do essentially nothing for the test case.
3. The loop structure causes divergence when some threads in a warp hit different branches.
Can I optimize by precomputing valid kernel positions? Or by reordering loops?
Optimization: move the stride check outside by noting that for a given z, z_in = (z + p - d*kd) / s must be integer. But this is just a modulo check.
Actually, I can restate: for fixed z and varying kd, the valid kd are those where z + p ≡ d*kd (mod s).
Since d and s might not be coprime, there may be 0 or multiple valid kd for each residue class.
But for simplicity, let me keep the loop and add modulo checks.
One important optimization: I should loop over ic OUTSIDE or process multiple ic values per weight read. Since channels are small (32), I could unroll or use vectorized loads.
Another optimization: use `float` accumulation and ensure compiler uses FMA.
For the extreme sparsity case, I could add a check at the top:
if ((z + padding) % stride != (dilation * 0) % stride && (z + padding) % stride != (dilation * 1) % stride && ...) return early;
But that's complex. A simpler check: for our test case with s=2, p=1, d=2:
(z + 1) % 2 needs to equal (2*kd) % 2 = 0 for some kd.
So (z+1) % 2 == 0, i.e., z % 2 == 1.
If z is even, no kd works, so return immediately.
Actually, I can add this general check:
For a given z, compute the set of valid (z + padding - dilation * kd) % stride values. If none equal 0, return.
But the simplest is: keep the modulo check inside the loop and trust that even for the test case, the modulo is cheap compared to the memory access.
Let me estimate performance:
- For a non-zero output scalar (1/8 of all), we do 32*27=864 iterations of the inner loop.
- Each iteration: 2 loads, 1 FMA, some index calculations.
- On H100, memory bandwidth for input is ~3.35 TB/s. Reading input 32*27 times per output scalar: but due to reuse, unique input reads per output scalar are less.
Actually, for a block processing the same (b, oc, z, y) and varying x:
- The same weights are reused across all x.
- Input is read at different x_in for each x.
- For interior z,y (odd), all 27 kernel positions are valid. As x varies over 33 values, x_in varies.
Wait, but only odd x values are valid! So x takes values 1,3,5,...,31 (and we process ALL x including evens). For even x, the thread returns early. For odd x, x_in values vary.
Actually, if I process all x, within a warp of 32 threads:
- Threads 0,2,4,... might handle even x (do nothing).
- Threads 1,3,5,... handle odd x (do work).
This is divergent! Bad for warp efficiency.
To avoid this, I should either:
1. Only launch threads for valid spatial positions.
2. Or use a different thread assignment.
For approach 1: enumerate all valid (z, y, x) combinations.
For our test case, valid z are {1,3,...,31} (16 values).
This requires parameter-dependent grid setup.
Actually, a simpler approach for general kernels: enumerate over input space and kernel.
For each (b, ic, z_in, y_in, x_in, kd, kh, kw), compute output position and add contribution.
Need atomics.
Let me try atomic approach with one block per (b, oc) or tile.
Actually, I think the direct output-first approach with some optimizations is worth trying first, even with the divergence.
To handle divergence better, I can process the spatial dimensions differently: let's have each warp process a 2D tile and ensure all threads in the warp process positions with the same parity.
Or: process all positions but reorder so valid positions are contiguous. This requires knowing the valid positions, which depends on parameters.
Let me write a kernel that:
1. Works generally
2. For efficiency, checks for early termination
3. Uses careful memory access
Here's an improved design:
```cuda
__global__ void convtranspose3d_kernel(...) {
// Each block processes a tile of output
// Block dims: (32, 8) = 256 threads
// Grid dims: (batch, out_channels, num_tiles)
// Each thread handles one (z, y, x) output position
int tile_idx = blockIdx.z;
int pos_in_tile = threadIdx.y * blockDim.x + threadIdx.x;
// We can process spatial positions in a specific order
// Total spatial positions = out_d * out_h * out_w
int spatial_idx = tile_idx * blockDim.x * blockDim.y + pos_in_tile;
if (spatial_idx >= out_d * out_h * out_w) return;
int x = spatial_idx % out_w;
int tmp = spatial_idx / out_w;
int y = tmp % out_h;
int z = tmp / out_h;
int b = blockIdx.x;
int oc = blockIdx.y;
// Check if this spatial position can have any contribution
bool has_contribution = false;
for (int kd = 0; kd < kernel_size && !has_contribution; kd++) {
int z_in = z + padding - dilation * kd;
if (z_in % stride == 0) {
z_in /= stride;
if (z_in >= 0 && z_in < in_d) has_contribution = true;
}
}
// similarly for y and x... but actually we can just check all combinations
if (!has_contribution) {
output[...] = bias ? bias[oc] : 0.0f;
return;
}
// If we get here, at least one spatial dimension has valid kernel offsets.
// But we need ALL dimensions to have at least one valid offset for non-zero output.
// Actually, we can just run the loops and let them skip.
}
```
Actually, for all dimensions to have valid offsets simultaneously, it's a different condition. I was being lazy. For the loop, I can just run normally.
Let me think about a much simpler optimization. Since the weight is small (220KB), I can preload the entire weight tensor into shared memory. Then all weight accesses are from fast shared memory.
But 220KB exceeds the typical 48KB shared memory per block. For H100, max is 227KB, so 220KB is tight. I could load the weight for a subset of output channels.
Actually, if a block handles 8 output channels, weight size = 8 * 32 * 27 * 4 = 27,648 bytes. That's small enough for shared memory.
So here's a better design:
- Block size: 256 threads.
- Each block handles: batch elements per block = 1, output channels per block = 8.
- The 256 threads collaborate to compute all spatial positions for these (b, oc).
- Actually, with 4.2M non-zero output scalars and batch=16, out_channels=64:
- Total (b, oc) pairs = 1024.
- For each (b, oc), there are 4096 spatial positions, but only 4096 are potentially non-zero? Wait no, total spatial positions = 33^3 = 35,937. Non-zero spatial = 4096.
If each block handles 1 (b, oc) and 256 threads process 256 spatial positions:
- Need 35,937 / 256 = 141 blocks per (b, oc).
- Total blocks = 1024 * 141 = 144,384 blocks. That's fine.
For each block:
1. Load weights for (oc_block : oc_block + oc_per_block) into shared memory.
2. Loop over input channels.
3. For each input channel, load a tile of input into shared memory.
4. Each thread computes its output spatial positions.
This is getting into the territory of a real conv kernel, which is complex.
Let me start with a simple kernel and see what performance I get, then optimize if needed.
Simple kernel design:
- One thread per output scalar.
- Total threads: batch * out_channels * out_d * out_h * out_w = 36.6M.
- Block size: 256.
- Grid: ceil(36.6M / 256) = 143,000 blocks.
- Each thread loops over all (ic, kd, kh, kw) and accumulates.
Memory coalescing:
- Consecutive threads handle consecutive x values? No, consecutive threads should handle consecutive output scalars.
- Output scalars are stored in NCDHW order.
- So consecutive threads have consecutive indices, meaning:
- Same (b, oc, z, y), consecutive x.
- Or same (b, oc, z), consecutive (y, x).
For consecutive x:
- For fixed (kd, kh, kw), x_in = (x + p - d*kw) / s.
- As x increases by 1, x_in changes by 1/s, which is not integer.
- So consecutive threads do NOT access consecutive input memory!
This is bad! Let me recalculate.
x takes values 0, 1, 2, ..., 32.
For thread idx and idx+1: x differs by 1.
For a fixed kw, x_in = (x + p - d*kw) / s, and it only exists when (x + p - d*kw) % s == 0.
But consecutive x values might BOTH be invalid or one valid one invalid.
Even when both valid (which requires s=1), x_in differs by 1/s, so for s>1, they don't differ by 1.
For s=2, valid x are 1, 3, 5, ... (for our test case). So consecutive valid x differ by 2, meaning x_in differs by 1. So consecutive THREADS that both process valid x (if they were consecutive) would access consecutive input.
But if all threads process ALL x, the valid threads are interleaved with invalid threads. So within a warp, only half the threads do work and they access memory with stride-2 patterns.
This is bad for coalescing.
To fix this, we need consecutive threads to process CONSECUTIVE VALID positions.
For the general case, valid positions are those where there exists some kernel offset making the input index valid. This is complex to enumerate.
Alternative: use the scatter approach and have each thread handle one (b, ic, zi, yi, xi, kd, kh, kw) combination, computing ONE scalar per output channel or all output channels.
Wait, if each thread handles (b, ic, zi, yi, xi, kd, kh, kw) and computes a dot product over oc:
output_partial[oc] = weight[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, xi]
Then we need to add this to output[b, oc, z_out, y_out, x_out] for all oc.
Using float4 or vectorized writes, and atomics if conflicts exist.
Actually, let me think about avoiding atomics by assigning threads to non-overlapping regions.
If I partition the OUTPUT spatial space into tiles, and only process (zi, kd) combinations that fall into a specific tile... this gets complicated.
Another idea: the atomic contention might not be as bad as I think on H100, because:
- The L2 cache has high bandwidth
- Atomics to the same address from the same warp are coalesced into a single operation
- But atomics from different warps to the same address still contend.
For our test case, each output scalar receives contributions from ~53 different (zi, yi, xi, kd, kh, kw) combinations on average (if we count all spatial combinations equally; actually it's more like 3*3*3=27 max for each input channel, and multiple input channels).
Wait, if we parallelize over (b, ic, zi, yi, xi, kd, kh, kw), then each output scalar (b, oc, z, y, x) receives one atomic add per (ic) for each valid decomposition of z,y,x into (zi+kd, ...). Since oc varies, each (b, oc, z, y, x) receives atomics from multiple (ic) and decompositions.
Number of atomics per output scalar = in_channels * num_valid_decompositions.
For interior position: num_valid_decompositions = 3*3*3 = 27? Wait let's check.
For z=3: (zi, kd) can be (2,0), (1,1), (0,2) → 3 pairs.
Similarly for y and x. Total decompositions = 3*3*3 = 27.
With 32 input channels: 27*32 = 864 atomic adds per output scalar.
864 atomics per scalar is a LOT! This would be extremely slow.
OK, so scatter with atomics is not viable.
Back to receive approach.
To fix the divergence and coalescing issues in the receive approach, I can:
1. Reorder the output spatial positions so that valid positions are contiguous.
2. Or use a hybrid where I iterate over spatial positions in a specific order.
Actually, for a general kernel, another approach is:
Loop over (kd, kh, kw) OUTSIDE the kernel launch. For each kernel offset:
- Compute which output positions are valid.
- Launch a kernel that processes those output positions.
For each (kd, kh, kw):
Valid output positions = those where z + p - d*kd ≡ 0 (mod s) and resulting z_in in range.
The number of valid z values is approximately in_d (since each input position maps to roughly one output position per kernel offset).
Actually, for each (kd, kh, kw), each input position (zi, yi, wi) maps to exactly one output position (z_out, y_out, x_out) IF that output position is in bounds.
So for each (kd, kh, kw), I can launch a kernel that iterates over all input positions and accumulates to the corresponding output position.
But different (kd, kh, kw) might write to the same output position, so I'd need to accumulate across kernel launches. I could:
- Use one kernel per (kd, kh, kw) and add to output (read-modify-write).
- Or accumulate in a temporary buffer and sum at the end.
With K^3 = 27 kernel positions, doing 27 separate kernel launches has overhead but each kernel is simple and non-conflicting.
For each kernel offset (kd, kh, kw):
- Threads iterate over all (b, ic, zi, yi, wi).
- Compute z_out = zi * s - p + d*kd (etc.)
- Check if z_out in [0, out_d)
- If valid: for each oc, output[b, oc, z_out, y_out, x_out] += weight[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, wi]
This is still problematic because multiple (ic) values write to the same output scalar for the same (b, oc, z_out, y_out, x_out). So within one kernel launch for fixed (kd, kh, kw), different threads with different ic but same (b, zi, yi, wi) would... wait, each thread can handle one (b, ic, zi, yi, wi).
For fixed (kd, kh, kw), each (b, ic, zi, yi, wi) maps to one output spatial position, but all oc values need to be updated. So a thread handling (b, ic, zi, yi, wi) would:
z_out = ...
for oc in 0..out_channels-1:
atomicAdd(&output[b, oc, z_out, ...], weight[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, wi]);
This still has atomics across ic for the same (b, oc, z_out, ...).
To avoid atomics, each thread could handle ALL ic values for a given (b, zi, yi, wi, kd, kh, kw), and write the sum over ic.
Thread assignment: one thread per (b, zi, yi, wi, kd, kh, kw).
For each oc, compute sum = 0;
for ic in 0..in_channels-1:
sum += weight[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, wi];
output[b, oc, z_out, y_out, x_out] = sum;
No atomics! Each thread writes to a unique (b, oc, z_out, y_out, x_out) for each oc. But wait, different threads might write to the same (b, oc, z_out, y_out, x_out) if they have different (zi, kd) mapping to the same z_out.
Example: thread 1: (b, zi=1, kd=0, ...) → z_out = 2*1 - 1 + 0 = 1
thread 2: (b, zi=0, kd=1, ...) → z_out = 0 - 1 + 2 = 1
Same z_out! Conflict for same (b, oc)!
So atomics are unavoidable unless we group by output position.
The only way to avoid atomics completely is:
- Each thread handles one output scalar and sums over all contributing inputs.
This brings us back to the receive approach.
Given the challenges, let me try the receive approach with some optimizations:
1. For memory coalescing, I can process the output in an order that's more friendly.
Actually, in the receive approach with consecutive threads having consecutive global idx:
- If idx corresponds to NCDHW layout, threads with consecutive idx fall in the same (b, oc, z, y) but different x values.
- As established, valid x values are spaced apart, causing divergence and poor coalescing.
But what if I launch thread blocks that iterate over x in the innermost loop, with threads assigned to different (b, oc, z, y)? Actually, in a warp, if threads have different (z, y), they access completely different input memory.
Alternative: change how I parallelize. Have each warp or block focus on a specific spatial region and process all channels for that region.
Block design:
- Block dims: (WARP_SIZE, NUM_WARPS) = (32, 8) = 256 threads.
- Grid: (batch, out_channels, spatial_tiles)
- Each block handles one (b, oc) and a tile of spatial positions.
Within a block, assign threads to spatial positions within the tile.
For the test case, I know valid spatial positions are sparse. But for generality, I can't assume this.
Actually, let me think about this from the H100 architecture perspective. H100 has very high L2 cache bandwidth and can handle some non-coalesced access. What matters most is:
1. Enough parallelism
2. Good instruction mix
3. Avoiding excessive branching
For the test case with 3.6B effective MACs, if I do a naive receive kernel with 36.6M threads where 7/8 do nothing, the effective thread count is 4.2M. On H100 with ~100K concurrent threads, I need each thread to do ~864 operations. At 67 TFLOPS, this takes ~54 microseconds for the math, plus memory overhead.
Actually, let me try computing the actual memory bandwidth needed in the naive receive kernel.
For each non-zero output scalar, we read:
- 32 * avg_valid_kernel * 4 bytes input
- 32 * avg_valid_kernel * 4 bytes weight
- 1 * 4 bytes output
For interior: 27*32*4 = 3456 bytes input, 3456 bytes weight, 4 bytes output.
For an average position: maybe ~20*32*4 = 2560 bytes each.
Total read per output scalar ≈ 5KB.
For 4.2M output scalars: ~21GB read.
At 3.35TB/s: 21GB / 3.35TB/s = 6.3ms. Way too slow!
But this assumes no cache reuse. In reality, weights are heavily reused (same weights for many output positions), and input is reused across output channels and nearby output positions.
For a warp processing positions with the same (b, oc, z, y) and varying x:
- The same weights are used for all threads.
- Input for the same (b, ic, z_in, y_in, x_in) where x_in varies: about 32 input floats per ic per valid kw.
Actually, the memory traffic is complex to model.
Let me try a different formulation: instead of reading weight and input repeatedly from global memory, I can preload a tile of input into shared memory.
For a block processing a tile of (z, y, x) spatial positions:
- Determine the range of input positions needed.
- Load that input tile into shared memory.
- Load the relevant weights into shared memory.
- Compute using shared memory.
But the input tile needed for a transposed conv output tile is larger and scattered.
Actually, for stride=2, dilation=2, an output position at (z=3) needs input at z_in=0,1,2. An adjacent output at z=5 needs z_in=1,2,3. So there's overlap in the input.
For a block processing output z values {3, 5, ..., 21} (say 10 values), the needed z_in values range from roughly 0 to 11. So input tile depth = 12.
This suggests we can cache input in shared memory.
But implementing an efficient tiled kernel is quite involved. Let me start with a simpler approach and see.
Actually, I recall that for small convolutions, a direct kernel with one thread per output position, carefully written, can be competitive if memory access is coalesced.
Let me try to improve coalescing by changing the thread assignment:
Instead of consecutive x, assign consecutive threads to positions that are consecutive in the valid subset.
Actually, for the test case, I can make the kernel aware of the sparsity. In general, if I want a fast kernel, I need to exploit structure.
Let me try writing a kernel that:
1. Uses shared memory for the weight (or a subset of it)
2. Each block processes multiple output channels and spatial positions collaboratively.
Actually, I found a simpler approach to good coalescing in the receive formulation:
- Loop over ic OUTSIDE the spatial loops.
- Within each ic iteration, all threads in the block process the same set of kernel weights but different spatial positions.
- The key is: for a fixed (ic, kd, kh, kw), the input access pattern across threads determines coalescing.
If I assign threads such that for a fixed (ic, kd, kh, kw), consecutive threads access consecutive memory:
- input at (b, ic, z_in(y), y_in(x), x_in(x)) where x is the thread index.
- For this to be coalesced, x_in(x) should be consecutive integers.
- x_in = (x_out + p - d*kw) / s
- For fixed kw, x_in is consecutive when x_out values are consecutive AND all valid.
The issue is x_out needs to be valid (skip even positions).
What if I restructure so threads are assigned ONLY to valid x_out positions?
For the test case, an output row of 32 (x=1,3,...,31) has 16 valid positions. I can assign 16 consecutive threads to these.
But different x_out positions might map to the same x_in if kw varies... no, for fixed kw, x_in increases with x_out.
Actually, here's a strategy:
For each block, choose a pixel in the "upsampled" space.
Specifically, for each (b, oc), the non-zero output positions form a grid.
For the test case: positions where z,y,x are all odd.
That's a 16×16×16 grid within the 33×33×33 output.
Instead of iterating over all 33^3 output positions, I can iterate over the 16^3 "valid" grid points and map to actual output indices.
More generally, valid output positions are those where:
For each dimension, there exists kd such that (pos + p - d*kd) % s == 0.
The set of valid positions is periodic with period s (in each dimension).
So I can define:
For each dimension, the valid residues are those r ∈ [0, s-1] such that (r + p - d*kd) % s == 0 for some kd.
Let valid_z_residues be this set.
Then I can iterate over:
all combinations of (z_valid, y_valid, x_valid) where z_valid ∈ valid_z_residues + s*ℤ, etc.
For the test case with s=2: valid residues are {1} (since p=1, d=2, need (r+1)%2==0, so r=1).
So z = 1, 3, 5, ...
For a general case, the number of valid residues per dimension could be up to s (all residues valid when d=1 and K≥s). With s=2, it's either 1 or 2 residues.
If I iterate over the valid positions directly, I avoid divergence completely.
Algorithm:
1. Precompute valid output positions for each dimension. Actually, instead of precomputing, I can iterate over the "base" grid and multiply by s plus offset.
For each (z_base, y_base, x_base) where:
- z_base ranges from 0 to some max_base_z
- z_out = z_offset + z_base * s_step
But the step and offset depend on the valid residues.
Actually, a simpler way: for each dimension, if we fix a valid output position z_out, then:
z_in = (z_out + p - d*kd) / s for some kd.
Instead of worrying about this, I can just iterate over input positions and kernel offsets, and write directly to output.
OK let me try a completely different approach that avoids both the divergence and atomic issues:
Use the scatter approach but accumulate in registers using a "reduction" among threads in a warp or block.
For a given output position (b, oc, z, y, x), identify all (zi, yi, wi, kd, kh, kw) that map to it. These are the contributions. I can assign these to threads and use warp shuffle or shared memory reduction to sum them.
But there can be up to K^3 = 27 different (zi, kd) pairs per dimension, up to 27 total per input channel, for 32 channels = 864 contributions. Reduction of 864 values across threads is possible: with 256 threads per block, each thread handles about 3-4 contributions, and block reduction sums them.
This is called "convolution as reduction" or "decomposition-based convolution".
But implementing this generally is complex.
Let me try the simplest possible correct kernel and benchmark it, then optimize.
Simplest kernel: receive approach, one thread per output scalar. I'll write it and see.
Actually, I should think about what PyTorch takes. For our specific test parameters, with ConvTranspose3d and cuDNN, what would typical time be?
With the extreme sparsity (88% zeros), cuDNN might still be pretty fast due to good memory patterns. Maybe ~0.2-0.5ms.
To beat this with a custom kernel, I need to:
1. Not compute the zero positions (or compute them very quickly)
2. Have good memory coalescing
3. Use shared memory effectively
Let me try an approach where I map the sparse valid grid directly:
For the test case, valid output positions are at odd z, y, x.
I can think of these as a 16×16×16 logical grid mapped to a sparse 33×33×33 physical grid.
I'll iterate over the logical grid:
```cuda
for (int z_base = 0; z_base < 16; z_base++) {
int z_out = 1 + 2 * z_base; // = 2*z_base - 1 + 2*0, but also mapping from multiple (zi, kd)
```
Wait, the relationship between z_base and the output position is direct. But at output position z_out, the contributions come from different (zi, kd) pairs.
If I just iterate z_base from 0 to 15 and set z_out = 2*z_base + 1, I can compute all non-zero output positions.
But this is specific to s=2, p=1, d=2. For a general kernel, I need to derive the valid offsets.
Actually, z_out = zi * s - p + d * kd.
For fixed s, p, d, as zi and kd vary, z_out takes on certain values.
For our case: z_out = 2*zi - 1 + 2*kd = 2*(zi + kd) - 1.
As zi ranges 0..15 and kd ranges 0..2, zi+kd ranges 0..17.
So z_out = 2*n - 1 for n = 0..17, but clipped to [0, 32].
This gives z_out = -1, 1, 3, ..., 33.
Valid: 1, 3, ..., 31. That's n=1..16, which is 16 values.
So z_out takes the form base + step * idx where base and step depend on parameters.
For general parameters, valid output values in a dimension are an arithmetic progression (possibly with gaps near boundaries). Actually, they might not be a single progression.
Example: s=2, p=0, d=1, k=3.
z_out = 2*zi + kd.
zi=0..15, kd=0..2.
z_out: 0,1,2, 2,3,4, 4,5,6, ..., 30,31,32.
This covers ALL integers from 0 to 32 (with overlaps). So valid positions are dense.
So validation depends heavily on params. For the general case, we need to check all positions.
Given the complexity and time constraints, let me write a kernel that:
1. Can handle general parameters
2. Uses shared memory for weights to reduce global memory traffic
3. Uses a simple but correct receive formulation
4. For the test case, performs well due to early exit
One simple optimization: for each thread, before the inner loops, check if there exists ANY valid (kd, kh, kw) combination. This can be done by iterating just over kd and checking if there exists one where (z + p - d*kd) % s == 0 and the result is in range. But this still requires a loop.
For the test case, a simple check is:
if ((z + padding) % stride != (dilation * something) % stride...)
Actually, I'll add a fast path:
For a given dimension with output position pos, the valid kernel offsets are those kd where:
1. (pos + padding - dilation * kd) % stride == 0
2. 0 <= (pos + padding - dilation * kd) / stride < in_size
For efficiency in the kernel, I can compute a mask or just iterate.
Let me write a reasonably optimized kernel and submit it. If it's not fast enough, I'll iterate.
Here's my plan for the kernel:
1. Use block dimensions (32, 8, 1) = 256 threads.
2. Each block handles one or more (batch, out_channel) pairs and a tile of spatial positions.
3. Actually, a simpler mapping: let each thread handle one output scalar, but map threads to spatial positions in x-major order.
Wait, to improve coalescing for valid positions:
What if I assign threads along the x dimension but skip invalid positions using z-order or Morton order? Probably too complex.
Actually, for the specific test case where only odd positions are valid, a simple mapping is:
x_out = 2 * thread_x + 1, where thread_x ranges 0..15.
And the total "valid" spatial volume is 16*16*16 = 4096.
For a general s, the valid positions might not be so regular.
Let me go with a hybrid approach: write a general kernel but with an optimized path for common cases. Or, rely on the compiler and simple code.
Actually, I realize I can write the kernel using tensor cores? No, that doesn't make sense for this size.
Let me write the kernel as direct output-first, but with an optimized loop structure.
Optimization: instead of iterating over kd and checking modulo each time, I can compute the starting kd and step.
For fixed z, the condition is: z + p ≡ d*kd (mod s).
This means d*kd ≡ (z + p) mod s.
Let g = gcd(d, s). If (z + p) % g != 0, no solution.
Otherwise, solutions for kd are of the form kd0 + (s/g)*t.
For s=2, d=2, g=2. Condition: (z+1) % 2 == 0. If true, all kd work: kd = 0, 1, 2 (step = 1). If false, no solutions.
For s=2, d=1, g=1. Condition: (z+p) % 2 == kd % 2. So kd can be either even or odd depending on z.
In general, the valid kd values are a subset. But iterating over all K values and checking a condition is cheap if K is small (like 3).
For K=3, checking 3 values per dimension is trivial. The main cost is memory access.
Let me write a kernel where:
- Threads in a warp process the same (b, oc) and the same (z, y) but different x values.
- Within each warp, there's divergence due to valid/invalid x, but this is unavoidable without special handling.
- I load weights into shared memory per block to exploit reuse.
I'll start coding.
After more thought, I think the best practical approach for this specific problem is:
1. Pre-allocate output with zeros.
2. Launch one kernel per (kd, kh, kw) combination, OR one kernel that loops over (kd, kh, kw) but with threads handling different output positions.
3. For each spatial position and channel, do the following:
Actually, let me try a gather/scatter approach where each block processes a 3D tile of output, and threads within the block loop over input channels and accumulate.
I'll write the kernel, compile it, and benchmark.
```python
import torch
import torch.nn as nn
from torch.utils.cpp_extension import load_inline
cuda_source = '''
#include <torch/extension.h>
#include <cuda_runtime.h>
__global__ void convtranspose3d_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int batch_size,
int in_channels,
int out_channels,
int in_d, int in_h, int in_w,
int out_d, int out_h, int out_w,
int kernel_size,
int stride,
int padding,
int dilation) {
// Each block handles OC_BLOCK output channels and one batch element
// Threads handle spatial positions
const int oc_block = 8; // output channels per block
const int num_oc_blocks = (out_channels + oc_block - 1) / oc_block;
int b = blockIdx.x;
int oc_b = blockIdx.y;
int spatial_idx = blockIdx.z * blockDim.x + threadIdx.x;
if (b >= batch_size || oc_b >= num_oc_blocks) return;
int total_spatial = out_d * out_h * out_w;
if (spatial_idx >= total_spatial) return;
int x = spatial_idx % out_w;
int tmp = spatial_idx / out_w;
int y = tmp % out_h;
int z = tmp / out_h;
int oc_start = oc_b * oc_block;
int oc_end = min(oc_start + oc_block, out_channels);
// Precompute weight indices in registers to avoid redundant computation
// Load weights to shared memory
extern __shared__ float s_weight[];
// s_weight size: oc_block * in_channels * kernel_size^3
// But kernel_size^3 = 27, in_channels = 32, oc_block = 8 → 6912 floats
int tid = threadIdx.x;
int weight_size = (oc_end - oc_start) * in_channels * kernel_size * kernel_size * kernel_size;
// Collaborative load of weights
for (int i = tid; i < weight_size; i += blockDim.x) {
int kw = i % kernel_size;
int t = i / kernel_size;
int kh = t % kernel_size;
t /= kernel_size;
int kd = t % kernel_size;
t /= kernel_size;
int ic = t % in_channels;
int oc = t / in_channels;
int global_oc = oc_start + oc;
if (global_oc < out_channels) {
s_weight[i] = weight[((global_oc * in_channels + ic) * kernel_size + kd) * kernel_size + kh) * kernel_size + kw];
}
}
__syncthreads();
// Now each thread computes output for its spatial position and oc range
for (int oc = oc_start; oc < oc_end; oc++) {
float sum = (bias != nullptr) ? bias[oc] : 0.0f;
for (int ic = 0; ic < in_channels; ic++) {
for (int kd = 0; kd < kernel_size; kd++) {
int z_in = z + padding - dilation * kd;
if (z_in % stride != 0) continue;
z_in /= stride;
if (z_in < 0 || z_in >= in_d) continue;
for (int kh = 0; kh < kernel_size; kh++) {
int y_in = y + padding - dilation * kh;
if (y_in % stride != 0) continue;
y_in /= stride;
if (y_in < 0 || y_in >= in_h) continue;
for (int kw = 0; kw < kernel_size; kw++) {
int x_in = x + padding - dilation * kw;
if (x_in % stride != 0) continue;
x_in /= stride;
if (x_in < 0 || x_in >= in_w) continue;
float w = s_weight[(((oc - oc_start) * in_channels + ic) * kernel_size + kd) * kernel_size + kh) * kernel_size + kw];
float inp = input[(((b * in_channels + ic) * in_d + z_in) * in_h + y_in) * in_w + x_in];
sum += w * inp;
}
}
}
}
output[(((b * out_channels + oc) * out_d + z) * out_h + y) * out_w + x] = sum;
}
}
torch::Tensor convtranspose3d_cuda(
torch::Tensor input,
torch::Tensor weight,
c10::optional<torch::Tensor> bias,
int stride,
int padding,
int dilation) {
int batch_size = input.size(0);
int in_channels = input.size(1);
int in_d = input.size(2);
int in_h = input.size(3);
int in_w = input.size(4);
int out_channels = weight.size(0);
int kernel_size = weight.size(2);
int out_d = (in_d - 1) * stride - 2 * padding + dilation * (kernel_size - 1) + 1;
int out_h = (in_h - 1) * stride - 2 * padding + dilation * (kernel_size - 1) + 1;
int out_w = (in_w - 1) * stride - 2 * padding + dilation * (kernel_size - 1) + 1;
auto output = torch::zeros({batch_size, out_channels, out_d, out_h, out_w}, input.options());
if (out_d <= 0 || out_h <= 0 || out_w <= 0) return output;
const int oc_block = 8;
int num_oc_blocks = (out_channels + oc_block - 1) / oc_block;
int total_spatial = out_d * out_h * out_w;
int threads = 256;
int spatial_blocks = (total_spatial + threads - 1) / threads;
dim3 grid(batch_size, num_oc_blocks, spatial_blocks);
dim3 block(threads);
size_t smem_size = oc_block * in_channels * kernel_size * kernel_size * kernel_size * sizeof(float);
convtranspose3d_kernel<<<grid, block, smem_size>>>(
input.data_ptr<float>(),
weight.data_ptr<float>(),
bias.has_value() ? bias.value().data_ptr<float>() : nullptr,
output.data_ptr<float>(),
batch_size, in_channels, out_channels,
in_d, in_h, in_w,
out_d, out_h, out_w,
kernel_size, stride, padding, dilation);
return output;
}
'''
functions=['convtranspose3d_cuda']
```
Wait, I made several C++ errors in my pseudocode. Let me be careful.
Actually, let me rethink the kernel design. The shared memory approach with oc_block=8 requires 8*32*27*4 = 27KB of shared memory. Plus input caching, it's more.
But the main issue is: threads in a warp process different spatial positions. For the same (ic, kd, kh, kw), they might all be valid, in which case they access different input locations. Are these accesses coalesced?
If thread i handles spatial position with x_i, and x_i values are consecutive for threads in a warp, then for fixed (ic, kd, kh, kw), x_in values are determined by x_i. But only some threads have valid x_in. The valid threads are interleaved.
This causes poor coalescing.
To fix coalescing, I should process positions where the THREADS are assigned to CONTIGUOUS valid positions, not all positions.
Let me think about this more carefully for the test case.
For the test case, valid x values are {1, 3, 5, ..., 31}. If I assign 16 consecutive threads to these 16 values, then for a fixed (ic, kd, kh, kw) where x is valid:
- Thread 0: x=1, x_in depends on kw
- Thread 1: x=3, x_in depends on kw
If kw=0: x_in = (1+1)/2 = 1, (3+1)/2 = 2, (5+1)/2 = 3, etc. Consecutive integers! COALESCED!
If kw=1: x_in = (1+1-2)/2 = 0, (3+1-2)/2 = 1, (5+1-2)/2 = 2. Also consecutive! COALESCED!
If kw=2: x_in = (1+1-4)/2 = -1 (invalid), so this kw is not valid for x=1.
So for threads assigned to valid x values in order, and for a fixed kw that makes them all valid, the input access IS coalesced!
The issue is that for x=1, kw=2 is invalid. For x=3, kw=2 is valid (x_in=0). So different x values might have different valid kw ranges.
But within a warp, if we process kw=0, all threads with valid x might have valid kw=0. For x=1: kw=0 is valid. For x=3: kw=0 is valid. In fact, for odd x and p=1, d=2, s=2:
x_in = (x + 1 - 2*kw) / 2.
For kw=0: x_in = (x+1)/2. For x=1:1, x=3:2, x=5:3, etc. All valid as long as x ≤ 31.
For kw=1: x_in = (x-1)/2. For x=1:0, x=3:1, etc. All valid.
For kw=2: x_in = (x-3)/2. For x=1:-1(invalid), x=3:0, x=5:1, etc.
So for kw=2, threads with x=1 would be invalid, but x=3,5,... are valid. This causes warp divergence for kw=2.
To avoid divergence, I could skip kw=2 for the first few x values or handle boundaries separately.
Actually, warp divergence is acceptable for a small fraction of cases. Most of the time, positions are interior and all kw are valid.
For x values in the middle (x=3 to x=29), all kw=0,1,2 are valid. Only boundaries (x=1 where kw=2 is invalid, and x=31 where kw=0 gives x_in=16 which is invalid) have issues.
Wait, x_in for x=31, kw=0: (31+1)/2 = 16, but in_w=32, so x_in=16 is valid! (0..31).
For x=31, kw=1: (31-1)/2 = 15, valid.
For x=31, kw=2: (31-3)/2 = 14, valid.
And x=1, kw=2: (1-3)/2 = -1, invalid.
So only x=1 with kw=2 is invalid among all odd x values from 1 to 31. That's very little divergence.
Similarly for z and y, only z=1 with kd=2 is invalid.
So overall, divergence is minimal for the interior positions.
This means my naive kernel with threads processing all x values (not just odd ones) would have divergence, but if I process only valid positions, divergence is minimal.
How to process only valid positions generally?
For each dimension, valid output positions are those where:
r = (pos + padding) % gcd(dilation, stride) == something.
Actually, the condition for existence of at least one valid kd is:
(pos + padding) must be congruent to some multiple of dilation modulo stride.
The set of values {dilation * kd mod stride : kd = 0, ..., K-1} is a subset of [0, stride-1].
If (pos + padding) % stride is in this set, there exists at least one valid kd.
For our test case with d=2, s=2: {0*2%2, 1*2%2, 2*2%2} = {0, 0, 0}. So we need (pos + 1) % 2 == 0, i.e., pos % 2 == 1. This gives a single residue class.
For d=1, s=2, K=3: {0%2, 1%2, 2%2} = {0, 1}. So all residues are covered, meaning ALL positions have at least one valid kd.
So the number of valid residues ranges from 1 to stride. When stride is small (like 2), we have either 1 or 2 residue classes.
For the general case, I can iterate over residue classes and for each class, iterate over the positions with that residue.
In my CUDA kernel, I can handle this by having the outer loop iterate over the valid "phase" and the inner loop iterate over positions with that phase.
But this adds complexity to the kernel.
Actually, a much simpler approach: instead of one thread per output scalar, I can have each thread handle a tile of output spatial positions and loop over them sequentially. This way, the thread handles only valid positions and skips invalid ones without affecting coalescing (since coalescing is about memory access within a warp, and a single thread has no coalescing issues).
For a warp of 32 threads, if each thread handles its own tile, and within each thread's tile the positions are processed sequentially, there's no inter-thread coalescing, but the compiler can optimize sequential access.
Actually, this might be slower than coalesced warp access.
Let me try a completely different idea: implement the transposed conv as a standard conv after upsampling the input by creating a sparse upsampled tensor in shared memory or registers.
For each input element at (zi, yi, wi) with stride s=2, it contributes to output positions:
(2*zi - 1 + 2*kd, 2*yi - 1 + 2*kh, 2*wi - 1 + 2*kw)
I can view this as: placing a "dilated kernel fragment" at each upsampled position.
A block of threads can process a tile of input elements, and each thread can accumulate its contributions to the output directly (with atomics if conflicts exist within the block).
Wait, what if each thread handles one (zi, yi, wi) position and ALL (kd, kh, kw) and ALL output channels?
- Thread reads input[b, ic, zi, yi, wi] for all ic (32 values).
- For each (kd, kh, kw), computes output spatial position.
- For each oc, computes dot product of weight[oc, :, kd, kh, kw] with input vector.
- Writes to output.
The conflicts happen when two threads in the same block map to the same output position.
For interior positions with s=2, each output position receives contributions from multiple (zi, yi, wi, kd, kh, kw) combinations.
However, if we make the block small enough that conflicts don't happen... this is difficult.
What if we use a reduction tree? Each thread computes partial sums into a register array, then we use warp shuffles to combine.
Actually, let me consider the receive approach with an explicit enumeration of valid spatial positions.
For the test case, there are 4096 valid spatial positions out of 35937.
I can precompute or implicitly enumerate them.
Actually, I realize I can use a simple mathematical mapping:
For the test case:
z_out = 2 * i + 1 for i = 0..15
y_out = 2 * j + 1 for j = 0..15
x_out = 2 * k + 1 for k = 0..15
For general parameters, I can't assume this. But I can compute the valid step and offset.
Actually, valid output positions form a lattice. The step along each dimension is s / gcd(d, s) or something... It's complicated.
Let me just write a simple kernel and see. I'll start with a direct output-first approach.
Also, I realize I should check if the output size formula is correct. Looking at PyTorch docs:
L_out = (L_in - 1) * stride - 2 * padding + dilation * (kernel_size - 1) + output_padding + 1
With output_padding = 0 by default.
So my formula is correct.
Let me write a clean, simple kernel first.
Actually, I want to try the scatter-atomic approach after all, but with one modification: I'll batch the writes by accumulating in registers first and writing once.
For each thread assigned to (b, oc, z_out, y_out, x_out):
Actually no, that's receive.
For each thread assigned to (b, zi, yi, wi, kd, kh, kw):
- Compute z_out, y_out, x_out.
- Read input vector of size in_channels.
- Read weight matrix slice of size out_channels × in_channels.
- Compute output vector of size out_channels.
- Atomic add each element to output[b, :, z_out, y_out, x_out].
This requires out_channels atomic adds per thread. With 226M threads, that's 226M * 64 = 14.5B atomics. Way too many.
Unless I reduce the number of threads by having each thread handle multiple (zi, yi, wi) or multiple (kd, kh, kw).
If each thread handles (b, oc, z_out, y_out, x_out) with the full receive sum: no atomics. This seems best.
Let me try an optimized receive kernel.
Optimization for receive kernel:
Since many output positions are zero, I'll add a check at the beginning of each thread:
For each dimension, precompute if the position has ANY valid kernel offset. If any dimension has none, write 0 and return.
To make this fast, I can use a small lookup table or simple arithmetic.
Actually, I can do even better: for fixed z, iterate over kd and set a flag if any is valid. Since K=3, this is just 3 iterations, very cheap.
Let me compute how many threads do work for the test case with this early exit:
- For each of 33 z values, about half have valid kd.
- For each of 33 y values, about half have valid kh.
- For each of 33 x values, about half have valid kw.
- So 33^3 * 1/8 = ~4515 positions per (b, oc) have all three dimensions valid.
Wait, for the test case, ALL three must be simultaneously valid for the output to be non-zero. With ~1/2 probability per dimension independently, it's 1/8.
33^3 / 8 ≈ 4515 positions per (b, oc).
With 16*64 = 1024 (b, oc) pairs: 1024 * 4515 ≈ 4.6M positions.
Out of 36.6M total threads: 1/8 do work, 7/8 return early.
The 7/8 that return early cause divergence but exit after a very small number of operations (checking at most 3 loop iterations). So they consume minimal time.
For the 1/8 that do work, they execute the full inner loops.
If I can arrange the grid so that threads doing work are grouped together, I minimize divergence.
Idea: launch threads only for the valid subset of spatial positions!
For general parameters, compute the valid spatial positions or assume a grid structure.
Actually, for the general case, valid positions can be enumerated by iterating over all positions in the output and checking a condition, but we want to avoid that.
Alternative: I can iterate over input positions instead and map to output positions. But then I have atomics.
What if I reformulate: each thread handles one input position (b, ic, zi, yi, wi) and computes contributions to output for a specific output channel range. Use shared memory as a scratchpad for the output tile, avoiding global atomics.
Block design:
- Block handles a tile of output spatial positions.
- Threads iterate over input positions that contribute to this tile.
- Accumulate contributions in shared memory.
- Write shared memory to global memory.
This is complex but avoids atomics.
For a block covering output spatial positions [z0, z1) × [y0, y1) × [x0, x1):
- The block loads weights for the output channels it handles.
- For each input position (zi, yi, wi), it checks if any (kd, kh, kw) maps into the block's output tile.
- If so, the thread computes the contribution and adds it to the correct shared memory location.
The advantage: shared memory atomics/reduction are much faster than global atomics. And with careful tile sizing, conflicts can be minimized.
Actually, if a single input position maps to exactly ONE position in the output tile (for a given kd, kh, kw), and different inputs map to different positions, then there are no conflicts.
For the test case with stride=2 and dilation=2:
A given (zi, kd) maps to z_out = 2*zi - 1 + 2*kd, which is unique.
So for a fixed (kd, kh, kw), different (zi, yi, wi) map to DIFFERENT (z_out, y_out, x_out).
If the output tile contains multiple of these positions, and different (zi, yi, wi) map to the same tile but different locations, there are NO conflicts!
Conflicts only happen if different (kd, kh, kw) with different (zi, yi, wi) map to the same (z_out, y_out, x_out).
For example: (zi=1, kd=0) → z_out=1, (zi=0, kd=1) → z_out=1.
These are the SAME output position from DIFFERENT input positions and DIFFERENT kernel offsets.
If we process (kd, kh, kw) sequentially (e.g., loop over kernel positions in each thread), there's no conflict because we accumulate sequentially.
So a tiled approach:
For each block:
1. Initialize shared memory output tile to 0.
2. For each (ic) and each (kd, kh, kw):
a. Determine which (zi, yi, wi) map into the output tile.
b. Each thread handles some of these input positions.
c. Thread computes z_out = zi*s - p + d*kd, etc.
d. If z_out in tile, add weight * input to shared memory tile.
3. After all inputs processed, write shared memory to global output.
This avoids atomics! Because for a fixed (kd, kh, kw), different (zi, yi, wi) map to different (z_out, y_out, x_out) (the mapping is bijective for fixed kernel offset). And since we process offsets sequentially, multiple offsets writing to the same output position happen at different times in the same thread or are handled as read-modify-write to shared memory without atomics needed.
Wait, different threads might process different (zi, yi, wi) that happen to map to the same (z_out, y_out, x_out) for the SAME (kd, kh, kw)?
For fixed (kd, kh, kw), the mapping (zi, yi, wi) → (z_out, y_out, x_out) is:
z_out = zi*s - p + d*kd
This is one-to-one! Different zi give different z_out. So for fixed (kd, kh, kw), different input positions map to different output positions. NO conflicts for the same offset!
For different offsets, they might collide, but if we use shared memory and process offsets sequentially, there's no need for atomics.
This is KEY! So the algorithm is:
For each output tile (covering some output channels and some spatial region):
1. Zero shared memory accumulator.
2. For each input channel ic:
a. For each kernel offset (kd, kh, kw):
i. Each thread in block takes some input positions (zi, yi, wi).
ii. Computes corresponding output position (z_out, y_out, x_out).
iii. If output position is inside the tile, and output position is in valid output range:
- Load input value.
- For each output channel oc in block's oc range:
- Load weight[oc, ic, kd, kh, kw].
- Add input * weight to shared memory accumulator at tile-local coordinates.
3. Write shared memory accumulator to global output memory.
Wait, but threads in the same block might handle different (zi, yi, wi) and for the same (kd, kh, kw) map to the same output position? NO, because the mapping is one-to-one per offset.
And for different offsets, we process them sequentially, so there's no simultaneity issue. We just add to shared memory, and since shared memory is read-modify-write but serialized by the sequential loop, it's fine.
Actually, wait. Multiple threads in the same block CAN execute simultaneously for the same (kd, kh, kw) but different (zi, yi, wi), and they MIGHT write to the same shared memory address if different offsets happened to collide. But for a FIXED offset, the mapping is one-to-one, so they won't collide.
Different offsets are processed in different iterations of the loop, so they don't collide in time.
Therefore, NO atomics needed! Great.
The memory access pattern:
- Input: each thread reads one input element per (ic).
- Consecutive threads should handle consecutive (zi, yi, wi) for coalescing.
- Weights: all threads read the same weight for a given (oc, ic, kd, kh, kw). Broadcast from constant/shared memory.
- Output (shared memory): each thread writes to a different location for the same offset, so no conflicts.
This is promising!
Let me design the block more concretely.
Block configuration:
- Block dims: (spatial_threads, oc_per_block) or just 1D threads.
- Let's say 256 threads per block.
- Each block handles OC_BLOCK output channels = 8.
- And a spatial tile of... well, the threads can process ALL input positions that map to ANY output position, but only write to the spatial tile.
Actually, we need the spatial tile to fit in shared memory.
If the output tile is (tile_d, tile_h, tile_w), and block handles 8 output channels:
Shared memory = 8 * tile_d * tile_h * tile_w * 4 bytes.
With 48KB shared memory: 8 * tile_d * tile_h * tile_w * 4 = 48,768
tile_d * tile_h * tile_w = 1524.
For 3D, a tile of 11×11×12 = 1452 works. Or 16×16×6 = 1536.
But with H100's 227KB shared memory, we can use much larger tiles.
For 227KB: tile volume = 227*1024 / (8*4) = 7270.
A tile of 20×20×18 = 7200 works.
However, the output is 33×33×33 = 35,937 spatial positions.
With tile volume 7200, we'd need 5 tiles to cover the whole space.
With 1024 (b, oc_channel_blocks), total blocks = 5 * 1024 = 5120.
Each block does: in_channels (32) * kernel_size^3 (27) * input_reads * weight_reads.
For each block, number of input positions to consider:
All (zi, yi, wi) such that for some (kd, kh, kw), the output falls in the tile.
This is approximately tile volume * (some factor due to stride and dilation).
For stride=2, each output tile of size 7200 corresponds to roughly 7200/8 = 900 input positions (since output is ~2x larger in each dimension, volume is ~8x larger). Plus boundary effects.
Wait, with stride 2, each input position maps to a kernel-sized region in output. So the input region that affects an output tile is roughly the tile size / 8, but expanded by the kernel size.
Actually, to cover an output tile, the needed input range is roughly:
zi from floor((z_tile_start + p - d*(K-1)) / s) to ceil((z_tile_end - 1 + p) / s).
For tile of size ~20 in z, and s=2: input z range is roughly 10+O(1).
So each block processes about 10*10*10 = 1000 input positions.
With 32 input channels and 27 kernel offsets: 1000 * 32 * 27 = 864,000 iterations.
With 256 threads: each thread does about 3375 iterations.
At 864 FMAs per iteration (for oc_block=8): that's 3375 * 8 = 27,000 FMAs per thread. Not too bad.
But the memory access for weights: for each (ic, kd, kh, kw), we read 8 weights. With 32*27=864 weight reads per block total, and each read is 32 bytes (8 floats), but cached in shared memory or L1.
Actually, if we pre-load weights for the block's oc range into shared memory:
Weight size = 8 * 32 * 27 * 4 = 27,648 bytes. Small.
Then in the inner loop, each thread reads 8 weights from shared memory for each (ic, kd, kh, kw) combination... wait, no. In the algorithm I described, for each thread's input position and each kernel offset, the thread computes contributions to all 8 output channels. So it reads 8 weights per kernel offset.
This is getting complicated to write. Let me try a simpler version first and see if it compiles and works.
I'll write a scatter kernel without shared memory optimization first, using atomics to global memory but with the insight that atomics might not be too bad for this case.
Wait, earlier I said 864 atomics per output scalar on average. But H100 atomics to L2 are actually quite fast when there's locality. If we ensure that threads writing to the same address are close together in time, the hardware can coalesce them.
Actually, in the scatter approach, if we process (b, ic, zi, yi, wi) and for a fixed (kd, kh, kw), different (zi, yi, wi) map to different output positions for the SAME offset. So atomics to the same address only happen from different offsets AND different inputs that happen to map to the same output.
For the test case, each output position gets contributions from ~27 input/kernel pairs per ic, on average in the interior. For 32 input channels: 864 atomic pairs.
If I process all offsets and all channels in one thread (so one thread per output scalar), I have 0 atomics but divergence.
Let me just try the simple output-first kernel, compile it, run correctness, and see the time.
I can also try processing only valid spatial positions. For the general case, I'll add a host-side helper that determines the valid stride and offset.
Actually, wait. For any parameters, valid output positions along dimension d are those where (pos + p) mod g == (dilation * kd) mod g for some kd, where g = gcd(dilation, stride). If the set {dilation*kd mod g} covers all residues mod g (which it always does since the map kd → d*kd mod g depends on gcd(d, g) = gcd(d, stride)), actually it's the set of multiples of gcd(d, g) = gcd(d, stride) within multiples of something.
This is getting too mathematical. Let me just write a direct kernel and see.
Actually, for the test case, I know that valid spatial positions are exactly those where z, y, x are all odd. I can write a kernel that specifically handles this pattern efficiently, while falling back to a slower general path for other cases.
Or, I can just write a general kernel. Let me focus on correctness first, then optimize.
Let me write the kernel with early return and shared memory for weights. I'll try to batch the spatial positions.
Here's the refined plan:
- Block: 256 threads.
- Grid: (batch, num_oc_blocks, num_spatial_blocks).
- Each thread handles one output scalar.
- Weights for oc_block channels are cached in shared memory.
- To improve coalescing, I'll restructure so that x dimension is handled with step.
Actually, I realize I can restructure the loop to improve memory access:
Instead of:
for ic
for kd
for kh
for kw
compute local_x_in, check, read input[x_in], read weight, accumulate
I can do:
for kd
z_in = ...
for kh
y_in = ...
for kw
x_in_base = ...
for ic
accumulate from input[x_in] with weight
This keeps the memory access to input more sequential for the inner ic loop... actually no, ic changes the input channel, which is a large stride.
Hmm.
What if I transpose the loop: ic inside, kw inside? No, that doesn't help.
Actually, for fixed (b, ic, z_in, y_in) and varying x_in, the input memory is contiguous. So if threads in a warp process the same (b, ic, z_in, y_in) and different x_in, they get coalesced access.
But in our kernel, threads process different (z, y, x), and for each kernel position, they might have different (z_in, y_in, x_in).
If I could arrange for threads in a warp to have the SAME (z_in, y_in) and CONSECUTIVE x_in for each kernel position, that would be ideal.
One way: iterate over the OUTPUT positions in a specific order.
For a fixed (kd, kh, kw), valid output positions (z, y, x) map to input positions (z_in, y_in, x_in) = ((z + p - d*kd)/s, ...).
If z_out values are consecutive multiples of s (plus some offset), then z_in values are consecutive.
But z_out values in our grid are not consecutive; they jump by s.
If I enumerate output positions as:
for z in valid_z_values:
for y in valid_y_values:
for x in valid_x_values:
...
And assign consecutive threads to consecutive x values (within the valid subset), then for fixed (kd, kh, kw) where all are valid, consecutive threads have consecutive x_in values!
This is exactly what I want. For the test case, valid_x_values = {1, 3, 5, ..., 31}.
Consecutive threads handle x = 1, 3, 5, ... so x_in values are consecutive integers.
The challenge is generalizing this valid position enumeration in CUDA.
For parameters where all positions are valid (e.g., s=1, or certain combinations), this simplifies to normal consecutive threading.
For other parameters, I can precompute on the host the list of valid positions, or compute them using simple formulas.
Actually, for the general case, I can compute the valid step:
If stride and dilation are such that valid positions are periodic with step `step` starting at `offset`, then I can enumerate them as offset + step * i.
But valid positions might have multiple residue classes mod stride.
For the test case: only one residue class {1} mod 2.
For d=1, s=2, K=3: all residues mod 2 are valid, so positions are contiguous.
In general, the number of valid residue classes mod stride is stride / gcd(stride, something). If K >= stride / gcd(dilation, stride), then all residue classes are covered.
To keep things simple, I'll write the kernel to handle general cases but optimize for the specific test case by using a general grid. The key insight is that H100 can handle some divergence.
Let me write and test.
Actually, I have a better idea. What if I launch threads only for the non-zero output positions by doing the following:
In the host code, compute the number of valid spatial positions. For each dimension, compute min and max valid positions. But this is still complex.
Alternatively: iterate over the input space, not the output space.
For each (b, ic, zi, yi, wi):
for kd, kh, kw:
z_out = zi * stride - padding + dilation * kd
if z_out in [0, out_d):
similarly for y, x
for oc:
output[b, oc, z_out, y_out, x_out] += weight[oc, ic, kd, kh, kw] * input[b, ic, zi, yi, wi]
Now, to avoid atomics, I can have each thread handle one (b, oc) and loop over all (ic, zi, yi, wi, kd, kh, kw). Since the writes to output for different (ic, zi, kd) are to the same (b, oc, z_out, y_out, x_out), this still has collisions.
BUT WAIT. What if I have each thread handle one output scalar, but instead of iterating over all (ic, kd, kh, kw), I assign threads to handle subsets and reduce?
Actually, no.
Let me go back to the atomic approach but with a key optimization: instead of atomics to global memory, write to shared memory first.
Algorithm using shared memory for output:
- Each block handles a fixed (b) and a fixed spatial region of output.
- The block loops over (ic, kd, kh, kw).
- For each input position that maps INTO the block's spatial region, a thread computes the contribution and adds it to shared memory.
- After all inputs are processed, write shared memory to global memory.
Block design for the test case:
- Spatial tile: 16×16×16 (the non-zero region). Actually, the output has non-zeros at odd positions, which is 16×16×16 = 4096 positions.
- Block handles: batch=1, all out_channels for that batch (or a subset).
Wait, 4096 spatial positions is manageable.
If each block handles 1 batch element and all spatial positions:
- Threads: 256
- Each thread handles 4096/256 = 16 spatial positions.
- For each spatial position and each output channel...
Actually, if a block handles all 4096 spatial positions, and we use shared memory for the output tile:
Shared memory = 4096 floats * out_channels_per_block.
With out_channels_per_block = 16: 4096 * 16 * 4 = 262,144 bytes = 256KB.
Within H100 limits but might be tight if many blocks run concurrently.
With out_channels_per_block = 8: 128KB. Reasonable.
For each block handling (b, oc_start:oc_end) and all spatial positions:
1. Zero shared memory: 4096 * 8 = 32,768 floats. Each thread zeros 128 floats.
2. Loop over ic, kd, kh, kw:
a. Each thread computes for some input positions.
b. For each input position, compute z_out, y_out, x_out.
c. If output is valid (in bounds), and maps to one of the spatial positions in the "odd grid":
Actually, all non-zero positions ARE in the odd grid.
Map (z_out, y_out, x_out) to shared memory index.
d. Read input, read weight (from shared or global), accumulate.
Wait, I need to ensure the output spatial positions fit in shared memory. For the general case, output might not be sparse.
I think this is getting too complicated. Let me just write a simple direct kernel, test it, and benchmark.
The simple direct kernel:
```cuda
__global__ void convtranspose3d_kernel(
const float* input, const float* weight, const float* bias,
float* output, int batch, int in_c, int out_c,
int in_d, int in_h, int in_w, int out_d, int out_h, int out_w,
int k, int stride, int pad, int dilation) {
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int n = batch * out_c * out_d * out_h * out_w;
if (idx >= n) return;
int x = idx % out_w;
int t = idx / out_w;
int y = t % out_h;
t /= out_h;
int z = t % out_d;
t /= out_d;
int oc = t % out_c;
int b = t / out_c;
float acc = bias ? bias[oc] : 0.0f;
for (int ic = 0; ic < in_c; ++ic) {
for (int kd = 0; kd < k; ++kd) {
int z_in = z + pad - dilation * kd;
if (z_in % stride != 0) continue;
z_in /= stride;
if (z_in < 0 || z_in >= in_d) continue;
for (int kh = 0; kh < k; ++kh) {
int y_in = y + pad - dilation * kh;
if (y_in % stride != 0) continue;
y_in /= stride;
if (y_in < 0 || y_in >= in_h) continue;
for (int kw = 0; kw < k; ++kw) {
int x_in = x + pad - dilation * kw;
if (x_in % stride != 0) continue;
x_in /= stride;
if (x_in < 0 || x_in >= in_w) continue;
float w = weight[((oc * in_c + ic) * k + kd) * k + kh) * k + kw];
float inp = input[(((b * in_c + ic) * in_d + z_in) * in_h + y_in) * in_w + x_in];
acc += w * inp;
}
}
}
}
output[(((b * out_c + oc) * out_d + z) * out_h + y) * out_w + x] = acc;
}
```
This is straightforward. Let me add it to the Python file and see if it works. Then I can optimize based on performance.
Wait, I need to make sure the weight indexing is right. In PyTorch ConvTranspose3d, the weight shape is (in_channels, out_channels, K, K, K) for the internal representation? No, let me check.
nn.ConvTranspose3d weight shape is (in_channels, out_channels, kernel_size[0], ...), but PyTorch might transpose it internally.
Actually, nn.ConvTranspose3d stores weight as (in_channels, out_channels, kD, kH, kW). In the forward pass, it uses this directly.
Wait, the reference creates `nn.ConvTranspose3d(in_channels, out_channels, kernel_size=...)`. According to PyTorch docs, the weight is (in_channels, out_channels, kD, kH, kW).
But in my custom CUDA kernel, I need to access the same weight tensor. So the weight indexing should match PyTorch's layout.
However, PyTorch internally might use some transposed layout or call cuDNN which expects a specific layout. Let me check: when I get `self.conv_transpose3d.weight` in Python, its shape is `(in_channels, out_channels, kD, kH, kW)`.
So my kernel should index weight as `weight[ic, oc, kd, kh, kw]`.
Wait, in my simple kernel above, I used `weight[((oc * in_c + ic) * k + kd) * k + kh) * k + kw]`, which assumes (out_c, in_c, ...) layout. That's WRONG!
The correct indexing for PyTorch ConvTranspose3d weight is:
`weight[((ic * out_c + oc) * k + kd) * k + kh) * k + kw]`.
Let me fix that.
Also, I should check if PyTorch uses NCDHW layout for ConvTranspose3d. Yes, it does.
Let me write the kernel code properly and test.
Also, let me think about whether I actually want to try the simple kernel first. Yes, it's the fastest to implement.
But actually, I realize I can make a more optimized version by processing the valid spatial positions. Let me write both: a simple correct version and see.
I'll write the Python module with load_inline.
One more thing: for efficiency, I should use `__restrict__` and maybe `__ldg()` for read-only data. Also, I should consider using `float4` for vectorized loads where possible.
Let me write the code.