reasoning
Let me analyze the problem:
1. We need to implement a 3D convolution + Mish activation + Tanh activation fused into a single CUDA kernel
2. The GPU is an H100 with 3.35 TB/s memory bandwidth and massive compute power
3. The shapes are: batch=16, in_channels=32, out_channels=64, D=32, H=64, W=64, kernel_size=3
4. This is fp32 precision
Compute requirements for conv3d:
- Output size: 16 × 64 × 30 × 62 × 62 ≈ 186M elements
- Per output element: 32 × 3 × 3 × 3 = 864 MACs
- Total MACs: 186M × 864 ≈ 161G FLOPs
- At 67 TFLOPS, this would take ~2.4ms, but this is naive - conv3d in PyTorch uses cudnn which is highly optimized
However, fusing Mish and Tanh can help by:
1. Reducing memory bandwidth (don't need to write conv output then read it back)
2. These activations are memory-bound elementwise ops
For a 3x3x3 conv with stride=1, padding=0, output is:
- D' = 32 - 3 + 1 = 30
- H' = 64 - 3 + 1 = 62
- W' = 64 - 3 + 1 = 62
PyTorch's cudnn conv3d is very fast. The challenge is to fuse activations so we save memory bandwidth.
Actually, for a fused conv+activation kernel, we need to implement conv ourselves or find another approach. Implementing a full 3D conv from scratch that beats cudnn is very hard.
Alternative approach: Use implicit GEMM or a tiled convolution. But given the H100 has tensor cores, we should leverage them if possible. However, `load_inline` makes it tricky to use CUTLASS.
A simpler approach: Can we just fuse the activations into a custom kernel that operates after the conv, saving one memory round-trip? That is:
1. Run cudnn conv3d (writes output)
2. Instead of mish then tanh (which are separate kernels or at least separate reads/writes), fuse them into one kernel
Wait, let me re-read. The question asks us to replace the forward computation with a custom CUDA kernel. We can still call cudnn for the conv part, then do the activations in a fused kernel. Or we can try to do everything in one kernel.
Actually, for maximum performance:
1. Do conv3d (this is compute intensive and cudnn is very good at it)
2. Fuse mish + tanh into a single kernel
But could we also implement conv3d as an implicit GEMM with Tensor Cores? That's very complex.
Let me think about what the PyTorch reference does:
- `self.conv(x)` - cudnn conv3d
- `torch.nn.functional.mish(x)` - mish activation
- `torch.tanh(x)` - tanh activation
PyTorch likely compiles mish as `x * tanh(softplus(x))`. And tanh is another kernel.
- softplus(x) = log(1 + exp(x))
- mish(x) = x * tanh(softplus(x))
- Then tanh(mish(x))
So the reference does: conv → mish → tanh. This is 3-4 kernel launches with multiple memory round trips.
Fusing mish + tanh after conv saves at least one full memory read/write.
If we do conv (cudnn) then fused mish+tanh, that's better.
But can we do better? We could implement a simple explicit 3D conv kernel. Let's check the arithmetic intensity.
Output elements: batch_size × out_channels × D' × H' × W' = 16 × 64 × 30 × 62 × 62 ≈ 118,600,000
Each output needs: in_channels × kD × kH × kW = 32 × 27 = 864 multiply-adds
Total FLOPs ≈ 118.6M × 864 × 2 ≈ 205 GFLOPs (ma = 2 flops)
At 3.35 TB/s bandwidth, if we do a custom kernel that reads input, weights, bias once each:
- Input read: 16 × 32 × 32 × 64 × 64 × 4 bytes ≈ 268MB
- Weight read: 64 × 32 × 27 × 4 bytes ≈ 221KB
- Output write: 16 × 64 × 30 × 62 × 62 × 4 bytes ≈ 475MB
- Bias read: negligible
Total memory traffic ≈ 743MB
Time at bandwidth = 743MB / 3350GB/s = 0.22ms
Theoretical compute time at 67 TFLOPS = 205GF / 67TF = 3.1ms
Wait, the compute is much more than memory bound for this kernel. 205 GFLOPs with 743MB = 276 FLOPs/byte. With 3.35 TB/s, the compute-limited time is 3.1ms. So this is compute-bound.
Cudnn likely runs conv3d in under 1ms (it's very optimized). So implementing conv ourselves would probably be much slower than cudnn.
Let me check: maybe we can use CUTLASS or some other library? But we only have `load_inline`.
Best approach: Call cudnn conv3d inside our kernel wrapper, then fuse the activations.
Wait - can we call cudnn from our custom kernel? No, `load_inline` compiles C++/CUDA, so we could in principle call cudnn, but linking and setup would be complex.
Actually, a simpler approach: In `ModelNew.forward()`:
1. Check if weights are already in some format
2. Use PyTorch's conv3d (which calls cudnn) for the convolution
3. Then apply our custom fused activation kernel
But that's not the spirit of "replace the forward computation with a custom CUDA kernel" - though technically we could have a custom kernel for the activations.
Or, can we use `torch.nn.functional.conv3d` and then our CUDA kernel for the activations?
Actually, the fastest practical approach might be:
1. Keep the conv as a PyTorch operation (cudnn is too good)
2. Write a fused mish+tanh kernel to minimize bandwidth
But for a truly custom solution, we should try implementing the conv ourselves if at all possible. Let me think...
Actually, for 3D conv with the given parameters, explicit GEMM might work if we use tensor cores. We could use the `mma` instructions or even simpler - use CUTLASS-style implicit GEMM.
But that's extremely complex to write inline. Another option is a direct convolution.
For H100, a direct conv with good tiling could potentially match cudnn for small sizes.
Wait, I should check if PyTorch 2.x compile already fuses mish+tanh. Actually, PyTorch's default eager mode does not fuse these. So there's definitely room for improvement.
Let me consider a different approach: can we write a fused kernel using implicit GEMM + activations?
Implicit GEMM for conv3d:
1. Unfold the input into a GEMM matrix (img2col)
2. Multiply with weight matrix
3. Add bias
4. Apply activation
But for conv3d, img2col is expensive. We can do implicit GEMM in shared memory.
Actually, for a 3×3×3 kernel with these dimensions, a naive direct convolution might actually be decent if implemented well with tensor cores.
But let me think about this more carefully. The batch is 16, in_channels is 32, out_channels is 64. We can tile in various ways.
For direct conv, each thread/block could compute a tile of the output. The key is:
- Tile size: maybe 16 or 32 output channels per block, 1×4×4 or 2×4×4 spatial per block
- Load a tile of input into shared memory
- Load weights into shared memory (or registers)
- Compute dot products
With 32 in_channels and 3×3×3 kernel, each output pixel requires 32×27 = 864 input values. That's a lot.
If we use float4 loads and careful tiling, we can get good memory coalescing.
Actually, for this problem, I think the best balance is:
1. Implement a direct 3D convolution in CUDA
2. Fuse mish and tanh into it
3. Use shared memory for input tiling
4. Try to use Tensor Cores if possible, or just do vectorized FMA
Let me first check: could I write an implicit GEMM using `wmma`? With fp32, wmma on Ampere/Hopper supports 16×16×16. For 32 input channels, that's 2 tiles in K dimension. For 64 output channels, that's 4 tiles in M dimension.
Actually, with the constraint of writing a fast kernel inline, let me think about a practical tiling strategy:
Block configuration:
- Consider each block computing a tile of output
- Tile: maybe 8 output channels × 1×4×4 spatial locations = 128 outputs
- Each output needs 864 MACs
- Total work per block: 128 × 864 = 110K FLOPs
With 11+K threads per block (say 256 threads), that's ~430 FLOPs/thread.
For shared memory:
- Input tile for 1×4×4 outputs needs (1×6×6×32) for a 3×3×3 kernel in one dimension? Wait, for 3D conv:
- For 1×4×4 in spatial output, we need 3×6×6 in spatial input
- With 32 channels: 3×6×6×32 = 3456 floats = 13.8KB
- We need weights for 8 output channels: 8×32×3×3×3 = 6912 floats = 27.6KB
- Total shared memory ≈ 42KB. That's reasonable.
Actually, with 256 threads, we can collaboratively load this.
For a full implementation, we need to handle:
- Padding is 0 (no padding needed in the kernel logic, just bounds checking)
- Stride is 1
Let me think about memory access patterns. For direct conv, each thread would compute one or more output elements.
Alternative: Simple tiled convolution where each thread computes one output element.
Actually, maybe I should try a simpler approach first and see the performance. Let me write a basic kernel that:
1. Uses PyTorch conv3d for convolution
2. Fuses the activations
Then check correctness and timing, and iterate.
Actually, when the evaluator runs, it compares against the reference implementation. The reference does `conv → mish → tanh`. If we do `conv → custom_fused_activation`, we should be correct and potentially faster.
But wait - the custom kernel needs to replace *the forward computation*. We can call arbitrary PyTorch ops in forward, then our custom CUDA kernel.
Actually, looking at the instruction more carefully: "Replace the forward computation of the PyTorch model below with a custom CUDA kernel." This suggests the kernel should do the bulk of the work. But we can technically use a combination.
I think the simplest winning strategy is:
1. Host code: compute conv3d using PyTorch's cudnn-backed conv3d
2. Custom CUDA kernel: fused mish + tanh
This saves at least 1 read+write of the full tensor. For 118M elements at 4 bytes, that's ~940MB of memory traffic saved. At 3.35 TB/s, that's 0.28ms saved.
But could we do better by also fusing the conv?
Given the H100's incredible compute power, a hand-written direct conv might not be faster than cudnn. But let me do a quick calculation.
For a well-optimized direct conv:
- On Ampere/H100, with half precision, direct convolutions can be very fast
- But in fp32, tensor core utilization might be harder
Actually, let me try a direct conv approach too. If I can make it competitive with cudnn, the fusion benefit would be larger.
For a direct 3D conv kernel, here's my strategy:
- Each block computes a tile of output
- BlockDim = (32, 8) or similar
- Each thread computes multiple output elements using unrolled loops
Actually, after more thought, I think direct conv with good vectorization and shared memory might work. Let me design:
```cuda
__global__ void conv3d_mish_tanh_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int C, int D, int H, int W,
int K, int KD, int KH, int KW,
int OD, int OH, int OW
) {
// Grid: (OW/TileW, OH/TileH, OD*K*N)
// Or better: (OW, OH, OD*K*N) with block-level tiling
}
```
For 3D conv, the implicit GEMM approach is:
- Input: (N*OD*OH*OW) × (C*KD*KH*KW)
- Weight: (K) × (C*KD*KH*KW)
- Output: (N*OD*OH*OW) × K
If we reshape to 2D, this is a GEMM. But we need img2col for explicit GEMM, which requires extra memory or doing it on-the-fly.
For implicit GEMM without extra memory, we can compute the conv directly using the 6D indexing.
Let me try a simple but well-structured approach:
- Grid decomposition: Each block handles a subset of output channels and spatial locations
- Use shared memory for input and weight tiles
Actually, for this specific problem, let me try to write a cutlass-style kernel or use a simple direct approach with good occupancy.
Wait, I just realized: could we use `cublas` or similar from inside our inline CUDA? If we link against cublas, we could do the conv as a GEMM with on-the-fly im2col in shared memory. But linking and proper setup is tricky with `load_inline`.
Let me go with a practical approach: implement a direct 3D convolution with manual tiling and fusion.
For H100, each SM has a lot of shared memory. Let me maximize shared memory usage.
Tiling strategy:
- Block size: 256 threads
- Each block computes: 64 output channels? No, that's too many.
- Each block computes: 8 out_channels × 2×4×4 spatial = 256 outputs
- Each output needs 864 floats read from input + weight
- But with shared memory reuse, we can share the input tile among output channels
If we compute 8 output channels, the weight needs 8×864 = 6912 floats.
If we compute 4×4×4 spatial output, the input needs (4+2)×(4+2)×(4+2)×32 = 6×6×6×32 = 6912 floats.
Total shared memory ≈ 14KB per tile component. Actually let me recalculate:
- For 4×4×4 output with 3×3×3 kernel: input needs (4+2)×(4+2)×(4+2)×32 = 6^3 × 32 = 6912 floats × 4 = 27.6KB
- For 8 output channels: 8×32×27 = 6912 floats × 4 = 27.6KB
- Total: 55.2KB
This is a bit high but fits in 227KB limit. Actually we can do better.
Alternatively, let me consider a smaller spatial tile. Say 1×4×4 output:
- Input: 3×6×6×32 = 3456 floats = 13.8KB
- Weights for 8 out_ch: 8×32×27 = 6912 floats = 27.6KB
- Total: 41.4KB
With 256 threads, each thread computes 8/256 = 1/32 of an output channel? That's not right.
Better: each thread computes one output element. For 8 out_ch × 1×4×4 spatial = 128 outputs. But we have 256 threads. Each thread computes 1/2 output? No, we can have each thread compute one output for 4 out_ch × 1×4×4 = 64 outputs with 256 threads, meaning 4 threads per output? That doesn't work well.
Let me rethink. Standard CUDA conv approach:
- Each thread computes one output element
- Block size = number of threads = number of outputs computed simultaneously
If we want 256 threads to compute 256 outputs:
- 8 output channels × 8 spatial locations = 64 outputs... that's too few threads.
- Wait, 256 threads can compute 256 different output elements.
For 3D conv, we can decompose:
- BlockDim.x = 32 (W dimension)
- BlockDim.y = 8 (H dimension)
- BlockDim.z = 1 (D dimension)
- But CUDA limits total threads per block to 1024.
Actually, let's use 1D blocks of 256 threads and let each thread compute multiple outputs in some dimensions.
Suppose each thread computes one output channel for one spatial location. Then 256 threads can compute:
- 4 output channels × 64 spatial locations = 256 outputs
Or:
- 8 output channels × 32 spatial locations = 256 outputs
But spatial locality matters for input reuse.
Let me consider a different approach. For 3D conv on large spatial dimensions, we can do:
- Each block handles a 3D spatial tile and a range of output channels
- Within a block, threads cooperate to load input into shared memory
Actually, a very common and effective pattern for direct conv is:
- Each block handles a fixed set of output channels (e.g., 64 or 128) and a spatial tile
- Threads are organized as (spatial_x, spatial_y) with each thread iterating over all output channels in its block's range
But for 64 output channels total, we could have each block handle all 64 output channels for a small spatial region.
Let me try: BlockDim = (32, 8) = 256 threads
- Each block computes 64 output channels × 1 spatial location? No, 256 threads for 64 outputs is too few parallelism.
- Actually, we want high occupancy, so maybe multiple blocks per SM.
What if each block computes 8 output channels × 32 spatial locations = 256 outputs
- 256 threads, each computes 1 output
- 32 spatial locations could be a 2×4×4 or 1×4×8 region
For shared memory, if we compute 8 outputs channels, we load 8×32×27 weights once.
If we compute 32 spatial outputs (say 2×4×4), we need to load a 4×6×6 input window.
But different spatial outputs share some input, so we can load a merged input tile.
For 2×4×4 output with kernel 3:
- D input range: [d, d+4) spanning 4 values
- H input range: [h, h+6) spanning 6 values
- W input range: [w, w+6) spanning 6 values
- Input tile: 4×6×6×32 = 4608 floats = 18.4KB
- Weights for 8 output channels: 8×32×27 = 6912 floats = 27.6KB
- Total: 46KB
This is workable.
Within the block, each thread computes one output. But threads computing different output channels for the same spatial location need the same input but different weights. Threads computing the same output channel for different spatial locations need different input but the same weights.
To maximize reuse, we can:
1. Load the entire input tile into shared memory collaboratively (4608 floats / 256 threads ≈ 18 floats per thread)
2. Load the weight tile into shared memory (6912 floats / 256 threads ≈ 27 floats per thread)
3. Compute: each thread loops over C, KD, KH, KW, reading input and weight from shared memory and accumulating in registers
This approach is straightforward and can be quite effective.
For the spatial dimensions:
- Output D = 30, H = 62, W = 62
- Total output elements = 16 × 64 × 30 × 62 × 62 ≈ 118.6M
- With 256 outputs per block, that's ~463K blocks
- Blocks can be organized as (spatial_blocks, batch×out_ch_blocks)
Actually, let me organize the grid:
- Each block computes: 8 out_ch × 2×4×4 spatial = 256 outputs
- Number of out_ch blocks: 64/8 = 8
- Number of spatial blocks: ceil(30/2) × ceil(62/4) × ceil(62/4) = 15 × 16 × 16 = 3840
- Number of batch blocks: 16
- Total blocks: 16 × 8 × 3840 = 491,520
That seems like a lot of blocks. With 132 SMs on H100 (actually H100 has 132 SMs? Let me check - SXM5 has 132, PCIe has 114). Assuming ~132 SMs, that's ~3700 blocks per SM. At 32 blocks/SM max, that's plenty for occupancy.
Actually, I think H100 SXM has 132 SMS, each with 128 cores. But the exact SM count doesn't matter too much - we have enough blocks.
Let me refine the approach. Actually, I should check if 2×4×4 is optimal. Maybe smaller tiles with more channels or larger tiles.
For H100, with 256 threads per block, we could:
1. Use warp-level primitives for faster accumulation
2. Consider using float4 for loads
Actually, let me simplify first and see what performance we get.
But wait - direct conv in fp32 might be much slower than cudnn if not using tensor cores. Cudnn likely uses tensor cores extensively for this. Without tensor cores, at 67 TFLOPS, 205 GFLOPs would take ~3ms. With cudnn/tensor cores at maybe 300+ TFLOPS effective, it could be ~0.7ms or less.
Hmm, so our custom direct conv might actually be slower.
Alternative idea: Use `wmma` or `mma.sync.aligned.m16n8k8.row.col.f32.tf32.tf32.f32` on H100.
Actually, for fp32 with tensor cores on Hopper, we can use the wmma/mma instructions. But writing a full IMMA/WMMA implicit GEMM is quite complex.
Another thought: On Hopper, `cublas` is available. Can we call cublas from the inline kernel? No, because `load_inline` compiles C++ extensions, not standalone CUDA code for kernels only. If we include `<cublas_v2.h>`, we should be able to link against cublas.
Wait, can we use cublas in a C++ extension compiled via `load_inline`? The extension is compiled as a shared library and loaded into Python. PyTorch already links against cublas. But we can also call PyTorch's own operations.
Wait, I think the simplest and most practical approach that has a good chance of winning is:
1. In `forward()`, call `F.conv3d` for the actual convolution (uses cudnn)
2. Then call our custom fused kernel for mish + tanh
But this only optimizes the activation part. For large convolutions, the conv dominates.
Actually, let me reconsider. Is there a way to call cudnn from C++ and get the conv result, then apply our activation in the same kernel?
In PyTorch C++ extension, we can use `at::conv3d` which calls cudnn internally, and then write our fused activation kernel. But then it's just two operations, not a single kernel.
For competitions like this, a true single fused kernel is often better, but if cudnn conv3d is 10× faster than any direct conv we can write, then using cudnn + fused activation is still the winning strategy.
Actually, I wonder if we can use Flash Attention-style approach for 3D conv? No, that doesn't make sense.
Let me try a different approach. Let me see if there's a CUTLASS path. We can't easily integrate CUTLASS via load_inline. But we might be able to write a reasonably optimized kernel.
Actually, for this specific shape, let me check the arithmetic density again more carefully.
Input size: 16 × 32 × 32 × 64 × 64 = 67,108,864 elements ≈ 256 MB
Output size: 16 × 64 × 30 × 62 × 62 ≈ 118,601,000 elements ≈ 452 MB
Wait, my output calculation: 16 × 64 = 1024, 1024 × 30 = 30720, 30720 × 62 = 1,904,640, 1,904,640 × 62 = 118,087,680
So output is ~118M elements ≈ 473 MB in fp32.
For the conv itself:
- Memory reads: input (256MB) + weights (~0.2MB) + bias (~0.00025MB)
- Memory writes: output (473MB)
- Total memory: ~729MB
- Compute: 118M outputs × 32 × 27 MACs = 102 GFMACs = 204 GFLOPs
- Arithmetic intensity: 204G / 729M = 280 FLOPs/byte
At 3.35 TB/s, the memory bandwidth bound would be 204G / 3350G = 0.061 seconds? No wait:
- Time = max(compute_time, memory_time)
- Compute_time = 204G / 67T = 3.0ms
- Memory_time = 729MB / 3350GB/s = 0.22ms
- So compute bound.
But with tensor cores at effective 500+ TFLOPS for fp32? Wait, H100 fp32 is 67 TFLOPS without tensor cores, and with TF32 tensor cores it's 494 TFLOPS (no sparsity) according to some specs? Actually the spec says TF32 Tensor Core is 494 TFLOPS (without sparsity), but TF32 is not true fp32. True FP32 Tensor Core is 67 TFLOPS.
Actually, the spec says:
- FP32 TFLOPS: 67
- TF32 Tensor Core TFLOPS: 494 without sparsity
TF32 is sufficient for many ML tasks, and cudnn uses it. So cudnn could be running at ~400-500 TFLOPS effective.
At 494 TFLOPS, compute time = 204G / 494T = 0.41ms. That's very fast.
A hand-written direct conv without tensor cores would take 3ms, which is 7× slower.
So using cudnn for conv + custom kernel for activations is the way to go.
Wait, but PyTorch probably already fuses some operations in its ATen backend. Does it fuse conv + mish + tanh? Probably not in eager mode. With torch.compile, it might. But the reference implementation is eager mode.
So our approach:
1. `ModelNew` uses `torch.nn.Conv3d` or `F.conv3d` for the conv
2. A custom fused kernel for `mish + tanh`
But wait, can we do better by fusing `mish` into a single kernel rather than two? Yes. Mish(x) = x * tanh(softplus(x)). Then tanh(mish(x)).
Actually, the reference does:
```
x = conv(x)
x = F.mish(x)
x = torch.tanh(x)
```
This is at least 3 kernel launches. Mish itself contains tanh and exp, so it's already non-trivial.
Our fused kernel would be one launch:
```cpp
y = tanh(x * tanh(log(1 + exp(x))))
```
For each element, that's:
1. exp(x)
2. 1 + exp(x)
3. log(...)
4. tanh(...)
5. x * tanh(...)
6. tanh(...)
So per element: 1 exp, 1 log, 2 tanh, some arithmetic.
For 118M elements at 4 bytes each, reading input and writing output:
- Memory: read 473MB + write 473MB = 946MB
- Time: 946MB / 3350GB/s = 0.28ms
With compute overhead, maybe 0.4-0.5ms.
Reference would be:
- Conv: ~0.5ms (if cudnn TF32)
- Mish: ~0.3ms (read+compute+write)
- Tanh: ~0.25ms (read+compute+write)
- Total: ~1.05ms
Our fused: ~0.5ms + 0.3ms = 0.8ms → ~1.3× speedup. Not great.
But wait, the activations might be fused by PyTorch into a single kernel already? Let me check. Actually, in newer PyTorch versions, `F.mish` and `torch.tanh` are separate ATen operations.
If we just fuse them, the speedup is moderate. What if we write a much better activation kernel than PyTorch's? PyTorch's tanh and mish are well-optimized already, but a custom kernel might have slightly less overhead or better instruction scheduling.
Actually, to get a really good speedup, we probably need to fuse everything. But can we beat cudnn in conv3d?
Wait, I just thought of something. Could we use the fact that the activations are applied elementwise after conv to do a partial fusion? Or use CUTLASS through some clever means?
Actually, let me think about whether we can beat cudnn with a custom kernel that uses shared memory optimally. For 3D conv, cudnn uses various algorithms including Winograd and FFT for certain sizes. For 3×3×3 conv, Winograd might not be as common. But implicit GEMM is very effective.
Can we write an implicit GEMM kernel using shared memory and load the img2col transform on-the-fly? Yes!
For implicit GEMM without extra memory:
- We reshape the conv into a GEMM: M = N × OD × OH × OW, K = C × KD × KH × KW, N = K_out (64)
- Each block computes a tile of the output matrix
- For each tile in the K dimension, we:
1. Load the weight tile into shared memory (easy, just linear access)
2. Load the corresponding "img2col" input into shared memory by reading from the 3D input tensor
3. Multiply the tiles using FMAs
This is essentially what optimized conv kernels do. If we implement this well, we might approach cudnn performance, especially with the fusion benefit.
For the input tile loading: given a position in the M dimension (which corresponds to a batch and spatial output location), and a range in the K dimension (which corresponds to input channels and kernel spatial offsets), we need to load input elements. We can collaborate among threads to load them.
Let me design this more carefully.
GEMM dimensions:
- M = N × OD × OH × OW = 16 × 30 × 62 × 62
- K = C × KD × KH × KW = 32 × 3 × 3 × 3 = 864
- N = K_out = 64 (output channels)
Wait, in GEMM notation, output matrix is M × N where:
- M = number of output positions = N × OD × OH × OW
- N = number of output channels = 64
- K = number of input features per position = C × KD × KH × KW = 864
Each block computes a tile of size BM × BN.
- Let's say BM = 128, BN = 64
- Then each block computes 128 × 64 outputs
- For K = 864, if we tile BK = 32, we need 864/32 = 27 iterations
Per iteration, we load:
- Input tile: BM × BK = 128 × 32 floats = 16KB
- Weight tile: BK × BN = 32 × 64 floats = 8KB
- Total SMEM: 24KB (plus double buffering = 48KB)
Wait, but the "input tile" is not contiguous. For implicit GEMM, the input matrix has rows indexed by (n, od, oh, ow) and columns by (c, kd, kh, kw). When we load a tile in the K dimension, the M×K input matrix elements for a given M row and K column correspond to a specific input position.
For loading the input tile efficiently:
- We need to load elements where row ∈ [m_start, m_start + BM) and col ∈ [k_start, k_start + BK)
- Each column in K dimension maps to (c, kd, kh, kw)
- Each row in M dimension maps to (n, od, oh, ow)
- The value at (m, k) is input[n, c, d, h, w] where d = od + kd, etc.
So for a block of M rows and K columns, the input access pattern is fairly scattered across the input tensor. This makes implicit GEMM with on-the-fly loading quite challenging for good memory coalescing.
For explicit GEMM (with actual img2col), we'd first materialize the M×K matrix, then call cublas. But that requires M×K elements = 118M × 864 ≈ 102B elements = 400GB. That's way too much memory.
Wait, img2col doesn't materialize the full matrix for the entire tensor at once. In practice, batched GEMM is used or the matrix is materialized in chunks.
For cudnn's implicit GEMM, the kernel effectively does the img2col transform in registers/shared memory without global memory traffic.
For our custom kernel, one option is to do the img2col in shared memory. Each thread computes one output and loops over K, reading weights and doing the img2col indexing to read input.
Let's reconsider: maybe a direct convolution is easier and potentially competitive for fp32 if we have good register usage.
For direct conv, each thread computes one output element. The thread loops over all in_channels and kernel spatial dimensions. With 864 iterations and good instruction-level parallelism, we can hide memory latency.
If each thread accesses input in a coalesced way (e.g., threads in a warp accessing adjacent input channels), memory bandwidth should be well utilized.
For 256 threads in a block computing 256 outputs:
- Let's organize as 4 warps × 64 threads
- Each warp could compute a 2×4×8 spatial region for 1 output channel
Wait, each output has 64 channels. If each block computes all 64 channels for a small spatial region:
- 64 channels × 1×4×4 spatial = 256 outputs
- 256 threads, each computes 1 output
- Thread i computes output channel (i % 64), spatial location (i / 64)
For thread i where i < 64, they're computing the same spatial location but different output channels. For spatial location (od, oh, ow), all 64 threads need the same input window (the 3×3×3 region around it).
- Input window: 3×3×3×32 = 864 floats
- But each thread needs different weights: weight[out_ch, :, :, :, :] for 64 output channels = 64 × 32 × 27 = 55,296 floats
The weights are too large for shared memory (55K floats = 221KB > 48KB L1 cache). But if threads in the same block read the same weights, maybe we can cache them in L2 or use a clever approach.
Actually, for threads computing different output channels, we can load weights in a cooperative way - each thread loads some weights and shares them. But with 64 output channels, if each warp handles 32 channels and we use register broadcasting, it's complex.
A simpler approach: each warp computes one output channel for 32 spatial locations.
- Warp size: 32 threads
- Each thread computes one spatial location for one output channel
- 32 threads = 32 spatial locations (e.g., 1×4×8 or 1×8×4 or 2×4×4)
For 64 output channels, we need 64 warps = 2048 threads per block. That's too many.
For 2 warps (64 threads) per block:
- Each warp computes 32 outputs = 1 output channel × 32 spatial locations
- 2 warps = 2 output channels × 32 spatial locations = 64 outputs
- To cover 64 output channels, we'd need 32 such blocks per spatial tile.
This is getting complicated. Let me think about what NVIDIA's recommended approach is.
Actually, for direct conv on Ampere/Hopper, a common pattern is:
- Each block handles a set of output channels (e.g., 64 or 128) and a 2D spatial tile
- Each warp handles a subset of output channels and a 1D strip of the spatial tile
- Threads use vectorized loads (float4) for input and weights
- Unroll the kernel loops
For 3D conv, we can extend this to three spatial dimensions.
One efficient pattern for our sizes:
- BlockDim = (16, 4, 4) = 256 threads? (16*4*4 = 256, but CUDA supports up to 3D blocks with max 1024 total)
- Actually, 16×16 = 256 threads in 2D
- Grid: (OW/16, OH/16, OD×N×(K/BLK_K))
This is getting too abstract. Let me try something concrete and optimize from there.
Actually, I just realized: can we use `wmma` on Ampere/Hopper for this?
For fp32, Hopper has wmma.mma.sync.aligned.m16n8k8.row.col.tf32.tf32.f32 which uses tf32 (10-bit mantissa). If fp32 tolerance is 1e-4, tf32 might be borderline. But we can use `mma.sync.aligned.m16n8k8.row.col.f32.f32.f32` for true fp32.
True fp32 mma on Hopper: I believe there are instructions like `mma.sync.aligned.m16n8k8.row.col.f32.f32.f32.f32` or similar.
Actually, let me try a different strategy that doesn't require writing a GEMM from scratch but still gets significant speedup.
Idea: Use PyTorch's conv3d, then implement the activations using a custom kernel with aggressive vectorization and warp-level optimizations.
But for real speedup, we should try to write a direct conv kernel. Let me start simple.
Simple direct conv kernel design:
```cuda
__global__ void conv3d_mish_tanh_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int C, int D, int H, int W,
int K, int OD, int OH, int OW
) {
int tid = blockIdx.x * blockDim.x + threadIdx.x;
int total_outputs = N * K * OD * OH * OW;
if (tid >= total_outputs) return;
// Decode tid into n, k, od, oh, ow
int ow = tid % OW;
int tmp = tid / OW;
int oh = tmp % OH;
tmp /= OH;
int od = tmp % OD;
tmp /= OD;
int k = tmp % K;
int n = tmp / K;
float sum = bias[k];
for (int c = 0; c < C; c++) {
for (int kd = 0; kd < 3; kd++) {
for (int kh = 0; kh < 3; kh++) {
for (int kw = 0; kw < 3; kw++) {
int d = od + kd;
int h = oh + kh;
int w = ow + kw;
float in_val = input[((n * C + c) * D + d) * H * W + h * W + w];
float w_val = weight[((k * C + c) * 3 + kd) * 3 * 3 + kh * 3 + kw];
sum += in_val * w_val;
}
}
}
}
// Apply mish + tanh
float x = sum;
float sp = logf(1.0f + expf(x));
float mish = x * tanhf(sp);
float out = tanhf(mish);
output[((n * K + k) * OD + od) * OH * OW + oh * OW + ow] = out;
}
```
This naive kernel would be terrible because:
1. No coalescing (adjacent threads access different output channels, same spatial location, but the memory layout is NCHW, so adjacent output channels are far apart in memory)
2. Massive register pressure from the 4 nested loops
3. No shared memory reuse
Wait, in PyTorch, the default format is NCDHW for 3D conv? Let me check. PyTorch conv3d uses NCDHW (batch, channel, depth, height, width). Yes.
So output[((n * K + k) * OD + od) * OH * OW + oh * OW + ow].
Adjacent threads with adjacent `k` would stride by OD*OH*OW, which is 30*62*62 = 115,320 elements. That's not coalesced at all!
To fix coalescing, adjacent threads should compute adjacent spatial locations for the same output channel, OR we should change how threads are mapped.
Better mapping:
- Group threads by output channel first, then spatial
- Within a warp, threads should compute adjacent spatial locations (varying `ow` primarily)
Let's make each warp compute one output channel, with 32 threads computing 32 adjacent `ow` values for fixed `n, k, od, oh`.
Block configuration:
- BlockDim.x = 32 (W dimension)
- BlockDim.y = 8 (H dimension)
- BlockDim.z = 1 (or 2 for D dimension, but 32*8*2 = 512 > 256 if we want specific block size)
Let's do BlockDim = (32, 8) = 256 threads.
Each block computes:
- 1 output channel (k)
- 1 batch (n)
- 1 D position (od)
- 8 H positions (oh)
- 32 W positions (ow)
So each block computes 1 × 1 × 1 × 8 × 32 = 256 outputs.
Grid configuration:
- x = ceil(OW / 32) = ceil(62/32) = 2
- y = ceil(OH / 8) = ceil(62/8) = 8
- z = OD × K × N = 30 × 64 × 16 = 30720
Total blocks = 2 × 8 × 30720 = 491,520. That's plenty.
For coalescing: within a warp (32 threads in x-dimension), threads compute adjacent `ow` values. Since `ow` is the fastest varying dimension in memory, adjacent threads access adjacent memory locations. Perfect coalescing!
For each output channel k:
- Threads load weights for channel k. Since all threads in a block use the same output channel, the same weights are loaded repeatedly. We can use shared memory or L1 cache for weights.
- For input: each thread needs a 3×3×3 window around its spatial position. Threads computing adjacent `ow` values need overlapping input windows. Good spatial locality.
For the weights: with 32 input channels and 3×3×3 kernel, weights per output channel = 32 × 27 = 864 floats = 3.4KB. For 64 output channels total, that's 221KB. For one block (one output channel), it's 3.4KB. We can load weights into shared memory!
For the input: each block computes 8 oh × 32 ow for one od. The input needed is:
- D: od to od+2 (3 values)
- H: oh_start to oh_start+8+2 (10 values, but block computes 8 consecutive oh values starting at oh_start*8)
- W: ow_start to ow_start+32+2 (34 values)
Wait, each thread in the block computes one (oh, ow) pair. For BlockDim.y = 8, block computes oh in [oh_block*8, oh_block*8+7]. For BlockDim.x = 32, block computes ow in [ow_block*32, ow_block*32+31].
The input window needed:
- D: [od, od+3) = 3 values
- H: [oh_block*8, oh_block*8+7+2) = [oh_block*8, oh_block*8+10) = 10 values
- W: [ow_block*32, ow_block*32+31+2) = [ow_block*32, ow_block*32+34) = 34 values
For 32 input channels: input window = 3 × 10 × 34 × 32 floats ≈ 32,640 floats = 130KB.
130KB > 48KB shared memory limit per block (on Ampere). On Hopper, shared memory limit is up to 227KB per block, but that's only if we configure it, and it reduces the number of blocks per SM.
Actually, Hopper has 228KB shared memory per SM. If one block uses 130KB + 3.4KB = 133KB, we can only have 1 block per SM. With 132 SMs, that's 132 blocks in parallel. Since each block has 256 threads, that's 33,792 threads total. Total blocks are 491k, so it would take ~15 waves. That might still be okay, but occupancy would be low (1 block per SM, or potentially 2 if we reduce shared memory). 256 threads per block means 256/1536 = 1/6th occupancy per block. With 1 block per SM, it's 1/6th occupancy. On H100, 1536 threads per SM is the max? Actually H100 has 1536 CUDA cores per SM but max threads per SM is 2048. So 256 threads per block with 1 block per SM = 256/2048 = 12.5% occupancy. That might be low.
We can reduce shared memory usage:
Option 1: Use registers instead of shared memory for input. Each thread loads its own 3×3×3 input window. That's 27 floats per thread = 108 bytes. For 256 threads, the block loads 256 × 27 × 32 channels values over time. But we load them sequentially, so no shared memory benefit. L1 cache might help.
Option 2: Load input on-the-fly from global memory for each output. Each thread needs to load 864 input values (32 channels × 27 spatial). For 256 threads, each loading 864 values, that's a lot of global memory traffic. But if well-coalesced, it might be okay.
Wait, actually, if each thread loads its input window on-the-fly, the total input reads per block:
- 256 outputs × 864 input values = 221,184 values
- But outputs share input. How much sharing?
- For the same output channel, all threads need the same weights
- For adjacent spatial positions, threads share input spatial locations
- Maximum reuse: the input tile has 3×10×34×32 = 32,640 values
- Each is used by multiple outputs: an input value at (d,h,w) is used by up to 27 outputs (all kernel offsets that include it)
- So total "useful" memory bandwidth without reuse would still be the input tile size, which is 32,640 values
- But each thread loading independently would read the input tile multiple times, wasting bandwidth.
If we don't use shared memory, each thread reads its own window. The total reads = 221,184 values (per channel? No, this includes all channels). Wait:
- 256 threads × 32 channels × 27 kernel positions = 221,184 values
- Input tile size = 3 × 10 × 34 × 32 = 32,640 values
- Overread factor = 221k / 32k = 6.8×
- That means 6.8× more memory traffic than necessary.
With shared memory, we load 32,640 values once, and all threads read from shared memory.
Without shared memory, we load 221,184 values from global memory.
221,184 × 4 bytes = 885KB per block. With 491k blocks, total input traffic = 434GB. That's okay at 3.35TB/s (129ms total for all blocks? No, blocks run in parallel). Wait, total memory traffic for the whole kernel = all input reads + all weight reads + all output writes.
If we calculate total memory traffic without shared memory:
- Input: each output position requires 864 input reads. Total reads = 118M × 864 = 102B values = 408GB. This is the theoretical minimum without reuse.
- But in practice, L1/L2 cache provides some reuse. Adjacent blocks computing adjacent spatial tiles might reuse some input through L2 cache.
- With good L2 cache reuse (e.g., blocks processing adjacent tiles run on the same SM consecutively), overread might be 2-3×.
Actually, the theoretical minimum input traffic (accounting for all reuse) is just reading the input tensor once plus overlaps at tile boundaries. Input size is 16 × 32 × 32 × 64 × 64 = 67M floats = 256MB. With overlaps due to kernel size, each spatial position is read multiple times, but total unique input reads is still bounded by a small factor.
The question is whether L1 cache is sufficient. For a block of 256 threads computing 1×8×32 outputs, the input footprint per channel is 3×10×34 = 1020 floats ≈ 4KB. For 32 channels, it's 130KB. This fits in L1 cache (128-256KB on H100 per SM, I think). Actually, H100 L1 cache is tied to shared memory, up to 228KB. The L1 cache itself might be 16-32KB.
Let me check H100 architecture: Hopper has 128KB L1 cache per SM that can be partitioned with shared memory. If we use 0 bytes shared memory, we get 128KB L1. Our input footprint is 130KB, which barely doesn't fit!
We can reduce the spatial tile. If we compute 1×4×16 outputs per block:
- Needed input: 3 × 6 × 18 × 32 = 10,368 floats = 41KB
- Plus weights: 3.4KB
- Total: 44KB - this fits in L1 nicely!
Block size: 64 threads (1×4×16 = 64). Too small? We want at least 128 or 256 threads per block for good parallelism.
If we compute 1×8×16 = 128 outputs per block, with 128 threads:
- Input: 3 × 10 × 18 × 32 = 17,280 floats = 69KB
- Still somewhat large for L1.
Alternatively, compute 2×4×16 = 128 outputs:
- Input: 4 × 6 × 18 × 32 = 13,824 floats = 55KB
Or 2×4×8 = 64 outputs with 64 threads. Let's try to scale up.
Actually, we can have each thread compute multiple outputs! This is a common optimization.
BlockDim = 256 threads, each thread computes 1 output. Block computes 256 outputs.
But we can also have each thread compute 2 outputs, block computes 512 outputs with 256 threads.
If each thread computes outputs at different spatial locations, it needs different input windows. Unless the outputs are far apart, they might not share input. But if each thread computes 2 adjacent outputs in W dimension, the input windows overlap significantly.
Let's say each thread computes 2 adjacent ow values for the same (n, k, od, oh):
- Block computes: 1 k × 1 n × 1 od × 8 oh × 64 ow = 512 outputs? Wait, 256 threads × 2 = 512 outputs.
- Spatial: 8 × 64 = 512. Input window: 3 × 10 × 66 = 1980 per channel × 32 = 63,360 floats ≈ 253KB. Too large.
Let's try smaller: each thread computes 2 outputs, block computes 256 outputs total.
- Spatial tile: 4 × 32 = 128 outputs (or 2×4×16 = 128, times 2 = 256 with 128 threads, no let's think differently).
If we use 256 threads and each computes 1 output, but adjust the tile dimensions:
- Tile: 1×4×64 outputs = 256 outputs
- Input: 3 × 6 × 66 × 32 = 38,016 floats = 152KB. Too large.
What about 4×4×16 = 256 outputs?
- Input: 6 × 6 × 18 × 32 = 20,736 floats = 83KB. This might barely fit in 128KB L1 cache.
Hmm. What if we split by output channel too?
Tile: 4 output channels × 4×4×4 spatial = 256 outputs
- 256 threads, each computes one output
- Input window: 6 × 6 × 6 × 32 = 6,912 floats = 27.6KB (shared by all threads in block!)
- Weight window: 4 × 32 × 27 = 3,456 floats = 13.8KB (for 4 output channels)
- Total shared memory: 41.4KB
- This is excellent and fits perfectly!
Now, with 4 output channels, the weights are small enough for shared memory, and the input is also reasonable.
Grid configuration:
- Spatial blocks: ceil(30/4) × ceil(62/4) × ceil(62/4) = 8 × 16 × 16 = 2048
- Channel blocks: ceil(64/4) = 16
- Batch blocks: 16
- Total blocks: 2048 × 16 × 16 = 524,288
- Very parallelizable!
Block mapping:
- blockIdx.z could cover batch (16) and channel blocks (16) = 256
- blockIdx.y covers D blocks (8) and H blocks (16) = 128
- blockIdx.x covers W blocks (16)
Wait, CUDA grids have limited dimensions (max 2^31-1 for x, 65535 for y and z on older GPUs, but modern is 2^31-1 for x, and quite large for y/z). Let's keep it simple:
- GridDim.x = W_blocks = 16
- GridDim.y = H_blocks × D_blocks = 16 × 8 = 128
- GridDim.z = K_blocks × N = 16 × 16 = 256
Total blocks = 16 × 128 × 256 = 524,288
BlockDim = 256 threads. We need to map threadIdx to (k_offset, d_offset, h_offset, w_offset) within the tile.
Inside the block, we can map:
- threadIdx.x = 0..255
- Decode to: w = tid % 4, h = (tid/4) % 4, d = (tid/16) % 4, k = tid / 64
So thread 0 computes k=0, d=0, h=0, w=0
Thread 1 computes k=0, d=0, h=0, w=1
...
Thread 63 computes k=0, d=3, h=3, w=3
Thread 64 computes k=1, d=0, h=0, w=0
etc.
This means:
- Threads with the same k but different spatial compute adjacent w first, then h, then d.
- For the same k, adjacent threads compute adjacent w values. Good for output coalescing!
- For different k, threads compute different output channels. Output memory is laid out as [N][K][D][H][W], so adjacent k values are separated by D×H×W elements. Not coalesced if adjacent threads have different k!
Wait, we have a problem. If thread 0 computes k=0 and thread 64 computes k=1, their outputs are 30×62×62 = 115,320 floats apart. That's terrible for coalescing.
To fix this, we need to make sure threads writing to adjacent memory locations run simultaneously in a warp. We want threads in the same warp to have the same k but different (d, h, w), or at least adjacent memory positions.
Since output is NCHW (NCDHW), adjacent w values for the same k are adjacent in memory. So we want a warp to process the same k, varying w primarily.
With warp size 32, we can have:
- 32 threads all computing k=0, varying d/h/w
- Next 32 threads computing k=1, etc.
For 4 output channels, that's 4 warps.
Within a warp of 32 threads computing the same k:
- We can assign w = tid % 4, h = (tid / 4) % 4, d = (tid / 16) % 2?
Wait, 4×4×2 = 32. Yes!
- w in [0,3], h in [0,3], d in [0,1] for a total of 32 positions.
But our spatial tile is 4×4×4 = 64 positions. So one warp can compute half the spatial tile (32 positions).
For 256 threads:
- 8 warps, each of 32 threads
- Warp 0-3: k = warp_id, each computes 32 spatial positions
- Wait, 4 k values × 8 warps = 32 warps, but we only have 8 warps.
- 8 warps / 4 k values = 2 warps per k value.
So: 2 warps per k, each warp computes 32 of the 64 spatial positions.
Total: 4 k × 2 warps × 32 positions = 256 outputs. Perfect!
Memory coalescing:
- Within a warp, all threads have the same k but different (w, h, d).
- For the same k, output memory is [n][k][d][h][w]. Adjacent w is adjacent in memory.
- If we assign w = lane_id % 4, then threads 0,1,2,3 compute w=0,1,2,3 for the same (d,h). These are adjacent memory locations! Coalescing is excellent.
- Threads 4,5,6,7 compute w=0,1,2,3 for the next h value. Memory is still adjacent within the warp... wait.
Actually, for fixed n, k, d, h, the memory layout is contiguous in w. So for 4 consecutive w values, they occupy 4 floats = 16 bytes.
If lane 0 computes (d=0, h=0, w=0) and lane 1 computes (d=0, h=0, w=1), they access adjacent floats. Coalesced.
But lane 4 computes (d=0, h=1, w=0). The offset from lane 0's address is (h_diff × W + w_diff) = 1 × 62 + 0 = 62 floats = 248 bytes. This is still within a 256-byte cache line, so it's still efficiently coalesced! The warp will load a few cache lines that cover all the needed addresses.
Actually, for a warp computing 32 positions in a 4×4×4 tile, the memory locations span from (0,0,0) to (0,3,3) to (1,0,0) to (1,3,3) to (2,0,0) etc. The maximum distance in the tile is from (0,0,0) to (3,3,3) which is (3×62 + 3)×62 + 3 = 192 × 62 + 3 = 11,907 floats. That's about 47KB apart. A warp accessing 47KB span cannot be fully coalesced into a single cache line.
But this is just true for a naive mapping. We can improve it.
For 32 threads in a warp computing the same k:
- Divide into 8 groups of 4 threads
- Each group computes one (d, h) combination, with w=0,1,2,3
- e.g., group 0: (d=0,h=0,w=0..3), group 1: (d=0,h=1,w=0..3), ..., group 3: (d=0,h=3,w=0..3), group 4: (d=1,h=0,w=0..3), etc.
Within the warp:
- Lanes 0-3: (d=0,h=0,w=0..3) - addresses 0, 1, 2, 3
- Lanes 4-7: (d=0,h=1,w=0..3) - addresses 62, 63, 64, 65
- Lanes 8-11: (d=0,h=2,w=0..3) - addresses 124, 125, 126, 127
- ...
- Lanes 28-31: (d=1,h=3,w=0..3) - addresses 115320+186, etc.
Wait, addresses for group (d, h, w=0..3) = base_k + (d × OH × OW + h × OW + w) = base_k + d × 3844 + h × 62 + w.
For group 0: base_k + 0 to 3
For group 1: base_k + 62 to 65
For group 2: base_k + 124 to 127
For group 3: base_k + 186 to 189
For group 4: base_k + 3844 to 3847
The differences between addresses:
- Within a group: 1 float apart (perfect coalescing for 4 threads)
- Between groups: 58, 59, or 3779 floats apart
A warp load instruction will try to coalesce. For 32 addresses with this pattern, the memory controller will load multiple cache lines. But since the stride is 62 floats = 248 bytes, which is larger than the 128-byte cache line, each group of 4 threads might need a separate cache line for w=0..3 (except for w=0,1 which might be on the same line if aligned). Actually, 4 floats = 16 bytes, so w=0..3 fits in one cache line.
For 8 groups, we need 8 cache lines (if perfectly aligned). But the addresses are at offsets 0, 62, 124, 186, 248, 310, 372, 434 from base_k (for d=0). Then for d=1, offsets 3844, 3906, etc.
Wait, I miscalculated. 62 floats × 4 bytes = 248 bytes. So groups are 248 bytes apart. A cache line is 128 bytes. So groups 0 and 1 overlap in cache lines slightly (if a group spans 16 bytes starting at offset 0, and next is at offset 248, they're in different lines). Actually, 248 / 128 = 1.94, so they're almost 2 cache lines apart.
For groups in the same d but different h, we're loading 8 cache lines for the 8 groups. That's up to 8 × 128 = 1024 bytes per warp. But a warp executes a single instruction and can load up to 128 bytes per instruction per thread? Actually, modern GPUs can load multiple cache lines per warp instruction. But it's not as efficient as a perfectly contiguous access.
Still, 8 cache lines for 32 threads (averaging 32 bytes/thread effective bandwidth) is reasonable.
For the second warp computing the remaining 32 positions of the same k:
- It might compute (d=2,h=0..3,w=0..3) and (d=3,h=0..3,w=0..3)
- These are even farther apart.
So for 2 warps computing the same k, they access very different memory regions.
Actually, the real issue is: can the memory system handle this efficiently? With 32 threads requesting 32 addresses spanning ~16KB, the L1 cache might struggle. But if we use vectorized stores (e.g., each thread writes 1 float), the memory subsystem should still be fairly efficient.
Alternatively, we could restructure so each warp writes contiguous 128 bytes. For 32 threads, if each writes 1 float4 (4 floats), that's 128 bytes. Perfect!
But our output tile is 4×4×4 = 64 floats per k. For 4 k values, that's 256 floats total. If each thread writes 1 float, that's 256 floats = 1KB.
For vectorized writes: if we make tile dimensions such that each warp processes a contiguous chunk...
- e.g., warp processes 32 floats in W dimension (for fixed n,k,d,h). But OW=62, and our tile is only 4×4×4.
- For one k and one d,h, there are only 4 w values. That's 16 bytes. Too small for a warp.
I think the basic tiled approach with shared memory will work reasonably well. Let me not overthink the coalescing too much. Modern GPUs handle scattered accesses much better than older ones, especially within L1 cache.
Let me design the kernel:
```cuda
template <int BLOCK_SIZE, int TILE_K, int TILE_D, int TILE_H, int TILE_W>
__global__ void conv3d_fused_kernel(...) {
// Block computes TILE_K out_ch × TILE_D×TILE_H×TILE_W spatial for one batch
int n = blockIdx.z / K_BLOCKS;
int k_block = blockIdx.z % K_BLOCKS;
int d_block = blockIdx.y / H_BLOCKS;
int h_block = blockIdx.y % H_BLOCKS;
int w_block = blockIdx.x;
int k_start = k_block * TILE_K;
int d_start = d_block * TILE_D;
int h_start = h_block * TILE_H;
int w_start = w_block * TILE_W;
// Thread mapping
int tid = threadIdx.x;
int w = tid % TILE_W;
int h = (tid / TILE_W) % TILE_H;
int d = (tid / (TILE_W * TILE_H)) % TILE_D;
int k = tid / (TILE_W * TILE_H * TILE_D);
if (k >= TILE_K || d >= TILE_D || h >= TILE_H || w >= TILE_W) return;
int k_out = k_start + k;
int d_out = d_start + d;
int h_out = h_start + h;
int w_out = w_start + w;
if (k_out >= K || d_out >= OD || h_out >= OH || w_out >= OW) return;
// Compute one output element
float sum = bias[k_out];
// To use shared memory: load input tile and weight tile
// We'll need to handle this carefully...
}
```
Actually, let me simplify. Instead of trying to perfectly map threads to outputs, I'll use a simpler approach where each thread computes exactly one output, and I'll use shared memory for both input and weights.
But loading shared memory cooperatively with the 4D mapping is tricky. Let's use a 1D thread indexing for the block:
```cuda
#define TILE_K 4
#define TILE_D 4
#define TILE_H 4
#define TILE_W 4
#define TILE_SPATIAL (TILE_D * TILE_H * TILE_W) // 64
#define TILE_OUTPUTS (TILE_K * TILE_SPATIAL) // 256
__shared__ float smem_input[TILE_D + KD - 1][TILE_H + KH - 1][TILE_W + KW - 1][C]; // 6*6*6*32 = 6912 floats
__shared__ float smem_weight[TILE_K][C][KD][KH][KW]; // 4*32*27 = 3456 floats
```
Wait, static shared memory with variable C is not allowed in standard CUDA unless we use template parameters. We can pass C as a template parameter or use dynamic shared memory.
For our specific problem: C=32, K=64, KD=KH=KW=3.
Input shared: 6×6×6×32 = 6912 floats = 27648 bytes
Weight shared: 4×32×3×3×3 = 3456 floats = 13824 bytes
Total: 41472 bytes = 40.5KB
Actually, the shared memory layout should be [channel][d][h][w] for better access when threads read the same channel but different spatial locations. Or perhaps [d][h][w][channel] for coalesced channel reads.
When threads read input for computation, all threads in a warp might process different spatial locations, so accessing different (d,h,w) but same channel. If layout is [channel][d][h][w], then same channel means same outer index, and different spatial means adjacent or nearby memory. This might be okay.
Actually, for threads computing different spatial for the same k_out, they all iterate over c, kd, kh, kw in the same order. At a given (c,kd,kh,kw), each thread reads input[c][d+kd][h+kh][w+kw]. If c is the fastest varying index in memory, then threads reading different c would be coalesced, but they all read the same c at a given loop iteration. So they read different spatial locations for the same c.
If memory is [c][d][h][w], then for fixed c, spatial locations vary. Threads reading different spatial are NOT coalesced (they access strided locations). But if memory is [d][h][w][c], then for fixed spatial, different c are coalesced.
Wait, in our loop, at each step, all threads read the SAME (c,kd,kh,kw)? No! In direct conv, we loop over c, kd, kh, kw. At each iteration, all threads read one specific input value for their output.
For iteration (c=0, kd=0, kh=0, kw=0):
- Thread 0 (computing k=0, d=0,h=0,w=0) reads input[n][0][0][0][0]
- Thread 1 (computing k=0, d=0,h=0,w=1) reads input[n][0][0][0][1]
- Thread 2 reads input[n][0][0][0][2]
- Thread 3 reads input[n][0][0][0][3]
These are adjacent in the W dimension! If layout is NCDHW (N, C, D, H, W), then for fixed N,C,D,H, W is the fastest dimension. So threads 0,1,2,3 access adjacent memory! Excellent coalescing!
But wait, our block computes 4 output channels (k=0..3) and 4×4×4 spatial. Thread 0 computes (k=0,d=0,h=0,w=0), thread 1 computes (k=0,d=0,h=0,w=1), ..., thread 63 computes (k=0,d=3,h=3,w=3), thread 64 computes (k=1,d=0,h=0,w=0).
At iteration (c=0, kd=0, kh=0, kw=0):
- Thread 0: input[n][0][0][0][0] (offset = 0)
- Thread 1: input[n][0][0][0][1] (offset = 1)
- ...
- Thread 63: input[n][0][3][3][3] (offset = 3*62*62 + 3*62 + 3)
- Thread 64: input[n][0][0][0][0] (same as thread 0!)
Threads 0-63 access a wide range of spatial locations. The range is from offset 0 to offset ~11,000. These are NOT all on the same cache line. But thread 0 and thread 64 access the same address.
For the weight read at this iteration (c=0, kd=0, kh=0, kw=0):
- Thread 0 reads weight[k=0][0][0][0][0]
- Thread 1 reads weight[k=0][0][0][0][0] (same!)
- Thread 64 reads weight[k=1][0][0][0][0]
So all threads with the same k read the same weight. This is good for shared memory broadcast.
For the input read: threads 0-63 read different spatial locations for the same channel. With NCDHW layout, these are strided.
Actually, this is where a shared memory transpose or clever layout helps. If we load the input tile into shared memory in NCDHW order, then threads reading from shared memory don't need to be coalesced (shared memory is much faster).
But loading the input tile INTO shared memory requires global memory reads. We can do this collaboratively:
- Each thread loads some input elements from global memory to shared memory.
- We ensure the global memory reads are coalesced.
For loading the input tile:
The input tile covers [d_start, d_start+6), [h_start, h_start+6), [w_start, w_start+6), and all 32 channels.
We have 256 threads. Total elements to load: 6×6×6×32 = 6912.
Each thread loads 6912 / 256 = 27 elements.
To coalesce: thread i loads element i, i+256, i+512, etc. We need to map linear index to (c, d, h, w).
If we store in smem as [c][d][h][w], then linear index = ((c * 6 + d) * 6 + h) * 6 + w.
For thread i = 0: c=0,d=0,h=0,w=0
Thread i = 1: c=0,d=0,h=0,w=1
...
Thread i = 5: c=0,d=0,h=0,w=5
Thread i = 6: c=0,d=0,h=1,w=0
Adjacent threads read adjacent w values for fixed c,d,h. This is perfectly coalesced in global memory (since W is the fastest dimension)!
Similarly, for weights, linear index = ((k * 32 + c) * 3 + kd) * 9 + kh * 3 + kw.
Thread 0: k=0,c=0,kd=0,kh=0,kw=0
Thread 1: k=0,c=0,kd=0,kh=0,kw=1
Adjacent threads read adjacent kw values, which is perfectly coalesced in weight memory!
Great! So the collaborative load is naturally coalesced.
Now for the compute phase: each thread loops over c, kd, kh, kw and reads from shared memory.
Shared memory layout: smem_input[c][d_local][h_local][w_local] and smem_weight[k_local][c][kd][kh][kw].
For a given (c, kd, kh, kw), a warp of threads reads:
- smem_input[c][d_thread][h_thread][w_thread] - each thread reads a different spatial for the same c.
- If smem layout is [c][d][h][w], then for fixed c, different (d,h,w) map to different banks.
Shared memory has 32 banks (on modern GPUs). Each bank is 4 bytes.
Address = ((c * 6 + d) * 6 + h) * 6 + w
For fixed c, the stride between adjacent w is 1 float = 4 bytes = 1 bank.
For adjacent h: stride is 6 floats = 24 bytes = 6 banks.
For adjacent d: stride is 36 floats = 144 bytes = 36 mod 32 = 4 banks.
For threads in a warp computing the same k but different spatial:
- Thread 0: (d=0,h=0,w=0) → bank 0
- Thread 1: (d=0,h=0,w=1) → bank 1
- ...
- Thread 31: (d=0,h=0,w=31) wait, w only goes up to 5 in the tile!
Our tile W is only 4. With w wrapping to next h:
- Thread 0: (0,0,0) bank 0
- Thread 1: (0,0,1) bank 1
- Thread 2: (0,0,2) bank 2
- Thread 3: (0,0,3) bank 3
- Thread 4: (0,1,0) bank 6
- Thread 5: (0,1,1) bank 7
- ...
- Thread 7: (0,1,3) bank 9
- Thread 8: (0,2,0) bank 12
- Thread 9: (0,2,1) bank 13
- Thread 11: (0,2,3) bank 15
- Thread 12: (0,3,0) bank 18
- Thread 15: (0,3,3) bank 21
- Thread 16: (1,0,0) bank 36 mod 32 = 4
- Thread 17: (1,0,1) bank 5
- ...
- Thread 31: (1,3,3) bank 36+21 = 57 mod 32 = 25
All 32 threads access different banks! No bank conflicts! This is great.
For the weight shared memory access:
- All threads in a warp read smem_weight[k_local][c][kd][kh][kw] for their respective k_local.
- But in our block, k_local ranges from 0 to 3, and threads are grouped by k.
- Warp 0: threads 0-31 all have k_local=0,1, or maybe mixed.
- Actually, with 256 threads and k_local = tid / 64, threads 0-63 have k_local=0, threads 64-127 have k_local=1, etc.
- So warp 0 (threads 0-31) all have k_local=0. They all read smem_weight[0][c][kd][kh][kw]. This is a broadcast - all threads read the SAME address. On modern GPUs, this is handled with no bank conflicts (broadcast capability).
- Warp 1 also has k_local=0. Same broadcast.
- Warp 2 has k_local=1. Broadcast for k=1.
- etc.
So weight access has no bank conflicts either!
This is excellent. The shared memory approach looks very promising.
Now let me design the full kernel:
```cuda
#include <cuda_runtime.h>
#include <cuda_fp16.h>
#define KD 3
#define KH 3
#define KW 3
template <int TILE_K, int TILE_D, int TILE_H, int TILE_W>
__global__ void conv3d_fused_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int C, int D, int H, int W,
int K, int OD, int OH, int OW
) {
const int INPUT_TILE_D = TILE_D + KD - 1;
const int INPUT_TILE_H = TILE_H + KH - 1;
const int INPUT_TILE_W = TILE_W + KW - 1;
const int INPUT_TILE_SIZE = INPUT_TILE_D * INPUT_TILE_H * INPUT_TILE_W * C;
const int WEIGHT_TILE_SIZE = TILE_K * C * KD * KH * KW;
__shared__ float smem_input[INPUT_TILE_SIZE];
__shared__ float smem_weight[WEIGHT_TILE_SIZE];
// Block indices
int w_block = blockIdx.x;
int h_block = blockIdx.y % ((OH + TILE_H - 1) / TILE_H);
int d_block = blockIdx.y / ((OH + TILE_H - 1) / TILE_H);
int k_block = blockIdx.z % ((K + TILE_K - 1) / TILE_K);
int n = blockIdx.z / ((K + TILE_K - 1) / TILE_K);
int w_start = w_block * TILE_W;
int h_start = h_block * TILE_H;
int d_start = d_block * TILE_D;
int k_start = k_block * TILE_K;
// Thread index within block
int tid = threadIdx.x;
int nthreads = blockDim.x;
// Load input tile into shared memory
for (int idx = tid; idx < INPUT_TILE_SIZE; idx += nthreads) {
int w = idx % INPUT_TILE_W;
int tmp = idx / INPUT_TILE_W;
int h = tmp % INPUT_TILE_H;
tmp /= INPUT_TILE_H;
int d = tmp % INPUT_TILE_D;
int c = tmp / INPUT_TILE_D;
int d_in = d_start + d;
int h_in = h_start + h;
int w_in = w_start + w;
float val = 0.0f;
if (d_in < D && h_in < H && w_in < W && c < C) {
int in_idx = ((n * C + c) * D + d_in) * H * W + h_in * W + w_in;
val = input[in_idx];
}
smem_input[idx] = val;
}
// Load weight tile into shared memory
for (int idx = tid; idx < WEIGHT_TILE_SIZE; idx += nthreads) {
int kw = idx % KW;
int tmp = idx / KW;
int kh = tmp % KH;
tmp /= KH;
int kd = tmp % KD;
tmp /= KD;
int c = tmp % C;
int k = tmp / C;
int k_out = k_start + k;
float val = 0.0f;
if (k_out < K && c < C) {
int w_idx = ((((k_out * C + c) * KD + kd) * KH + kh) * KW + kw);
val = weight[w_idx];
}
smem_weight[idx] = val;
}
__syncthreads();
// Compute output
int outers = TILE_K * TILE_D * TILE_H * TILE_W;
int out_tid = tid;
if (out_tid < outers) {
int w = out_tid % TILE_W;
int tmp = out_tid / TILE_W;
int h = tmp % TILE_H;
tmp /= TILE_H;
int d = tmp % TILE_D;
int k = tmp / TILE_D;
int k_out = k_start + k;
int d_out = d_start + d;
int h_out = h_start + h;
int w_out = w_start + w;
if (k_out < K && d_out < OD && h_out < OH && w_out < OW) {
float sum = (k_out < K) ? bias[k_out] : 0.0f;
for (int c = 0; c < C; c++) {
for (int kd = 0; kd < KD; kd++) {
for (int kh = 0; kh < KH; kh++) {
for (int kw = 0; kw < KW; kw++) {
int in_idx = ((c * INPUT_TILE_D + (d + kd)) * INPUT_TILE_H + (h + kh)) * INPUT_TILE_W + (w + kw);
int w_idx = ((k * C + c) * KD + kd) * KH * KW + kh * KW + kw;
sum += smem_input[in_idx] * smem_weight[w_idx];
}
}
}
}
// Mish + Tanh
float sp = logf(1.0f + expf(sum));
float mish = sum * tanhf(sp);
float out_val = tanhf(mish);
int out_idx = (((n * K + k_out) * OD + d_out) * OH + h_out) * OW + w_out;
output[out_idx] = out_val;
}
}
}
```
Wait, there's a problem. I mapped 256 threads to 256 outputs (4×4×4×4). But what if blockDim.x is 256 and there are only 128 valid outputs due to boundary conditions? The boundary threads do nothing, but we still load the full input/weight tiles, which is fine.
But `outers = 256`, so `out_tid = tid` for tid=0..255, and `out_tid < outers` is always true. The boundary check `if (k_out < K && ...)` handles the actual output bounds.
Actually, I realize that with this design, we can increase the tile sizes to use all 256 threads more efficiently. Since 256 threads = 256 outputs, tile dimensions must multiply to 256. Let me check valid combinations where K<=64, D<=30, H<=62, W<=62.
4×4×4×4 = 256 ✓ (what we have)
8×4×4×2 = 256 ✓
16×4×4×1 = 256 ✓
For spatial dimensions, we want to maximize W tile to get output coalescing. So maybe:
- TILE_K = 4
- TILE_D = 2
- TILE_H = 4
- TILE_W = 8
Product: 4×2×4×8 = 256 ✓
With W=8, output memory for a warp (assuming same k, different w) would be more contiguous per warp.
Or:
- TILE_K = 2
- TILE_D = 2
- TILE_H = 8
- TILE_W = 8
Product: 2×2×8×8 = 256
Or even better for coalescing:
- TILE_K = 1
- TILE_D = 2
- TILE_H = 8
- TILE_W = 16
Product: 1×2×8×16 = 256
But if TILE_K = 1, then each block computes only 1 output channel. The weight shared memory is tiny (32×27 = 864), but we need more blocks overall (64× more in the K dimension). That could be okay.
Actually, with TILE_K = 1 and tile spatial = 2×8×16 = 256, the grid would be:
- N = 16
- K = 64 (each block computes 1)
- Spatial: 15 × 8 × 4 = 480
- Total blocks: 16 × 64 × 480 = 491,520. Still fine.
With TILE_K = 1, all threads in a block compute the same output channel, which means they all read the same weight. For shared memory, this is a tiny weight tile. For compute, all threads have perfect weight reuse via shared memory broadcast.
And output coalescing: all 256 threads compute different spatial for the same k. If we store in NCDHW order, they write to scattered locations. But with TILE_W=16, threads 0..15 write adjacent w for same d,h. Threads 16..31 write adjacent w for next h, etc. Within a warp of 32 threads, they'd write to 2 different h values × 16 w values, or some pattern.
Actually, let me think about access locality. With TILE_K=1, output for a block is [n][k][d..d+2][h..h+8][w..w+16]. This is a 2×8×16 subtensor of the output. Threads within the block vary in d,h,w. Adjacent threads in a warp: let's say warp computes tid 0..31. If mapping is w=tid%16, h=(tid/16)%8, d=(tid/128)%2:
Lane 0: w=0, h=0, d=0
Lane 1: w=1, h=0, d=0
...
Lane 15: w=15, h=0, d=0
Lane 16: w=0, h=1, d=0
...
Lane 31: w=15, h=1, d=0
For fixed n,k,d=0, the output addresses are:
- lane i (i<16): offset = h×OW + w = 0 + i = i
- lane 16+i: offset = 1×OW + i = 62 + i
So within the warp, we have two 16-float contiguous segments separated by 62-16 = 46 floats. These segments are 186 bytes apart. A single warp store might touch 2-3 cache lines. Acceptable.
For the next warp (lanes 32-63):
- d=0, h=2..3, w=0..15
Similar pattern, next 2 cache lines per 16-float segment.
If we vectorize writes using float4, we could write 4 floats per thread. But our W tile is only 16, so each h row has 16 floats = 4 float4s. If we use 4 threads to write a float4 each, they cover one h row.
But implementing float4 with boundary checks in our general tiling is complex. Let's stick to float stores for now.
Wait, actually we can optimize the inner loops. The 4 nested loops (c, kd, kh, kw) with C=32 and K-spatial=27 can be unrolled.
CUDA compiler might unroll small loops automatically, but we can help by using `#pragma unroll` or template-based unrolling.
For C=32, the compiler can definitely unroll if we use a constant. But C is passed as a runtime parameter. If we assume C is always 32 and K-spatial is always 3, we can use template specialization or just hardcode 32 and 3.
But the kernel should be general enough. We can pass C as a template parameter, or just assume the specific values and hardcode them.
For the purpose of this competition, the shapes are fixed. So we can hardcode C=32, K_out=64, D=32, H=64, W=64, KD=KH=KW=3, etc.
Hardcoding allows the compiler to fully unroll loops and optimize register usage.
Let me think about whether we can use constant memory for weights. PyTorch weight tensors are in global memory. We could copy them to constant memory, but that's only 64KB total. Our weights are 64×32×27×4 = 221KB. Too large for constant memory.
But we can use shared memory as already planned.
Let me reconsider whether we should try to use Tensor Cores. For HMMA with fp32 on Hopper... Actually, fp32 mma on Hopper:
- `mma.sync.aligned.m16n8k8.row.col.f32.f32.f32.f32` works on Ampere/Hopper
- But we'd need to organize our data into fragments. This is quite complex for a direct conv.
Alternatively, we can try to achieve good performance without tensor cores. At 67 TFLOPS and compute-bound conv, if our direct conv achieves say 50% of peak, that's 33 TFLOPS. Compute time = 204G / 33T = 6.1ms. Wait, that's much slower than cudnn at 494 TFLOPS (0.41ms).
Actually, I realize that without tensor cores, our custom conv kernel will likely be much slower than cudnn. Even if we get 100% utilization of non-tensor FP32 units (67 TFLOPS), it would take 3ms. With cudnn using tensor cores at ~300-500 TFLOPS effective, it takes <1ms.
So a pure custom conv kernel without tensor cores is probably not going to win.
Unless... can we use tensor cores from a custom inline kernel? Yes, if we write the mma instructions directly. But it's very complex.
Another idea: use CUTLASS. Can we include CUTLASS headers in `load_inline`? Potentially yes, if we set the include paths. But CUTLASS is a massive codebase and `load_inline` has limited include paths. PyTorch itself includes some CUTLASS for Flash Attention, but getting it working via `load_inline` is non-trivial.
Actually, I think the most reliable approach for this competition is:
1. Keep PyTorch's conv3d
2. Write a highly optimized fused activation kernel
But will this be enough to beat the reference?
Wait, let me re-read the reference. The reference model is:
```python
class Model(nn.Module):
def __init__(self, in_channels, out_channels, kernel_size, stride=1, padding=0):
super(Model, self).__init__()
self.conv = nn.Conv3d(in_channels, out_channels, kernel_size, stride=stride, padding=padding)
def forward(self, x):
x = self.conv(x)
x = torch.nn.functional.mish(x)
x = torch.tanh(x)
return x
```
In PyTorch eager mode, `self.conv(x)` goes through aten::conv3d which calls cudnn.
Then `F.mish(x)` is a separate kernel.
Then `torch.tanh(x)` is another kernel.
But wait! PyTorch recently added support for fusing conv + activation through the cudnn frontend API. Does `nn.Conv3d` automatically fuse with subsequent activations? In eager mode, no. In `torch.compile`, yes. But since the reference is eager mode, it's three separate operations.
Moreover, does PyTorch have a fused mish + tanh? No.
So our fused mish+tanh kernel saves:
1. One read of conv output (473MB)
2. One write of mish output (473MB)
3. One read of mish output for tanh (473MB)
Total saved: ~1.4GB of memory traffic.
At 3.35TB/s, 1.4GB = 0.42ms.
If conv takes 0.5ms, mish 0.3ms, tanh 0.3ms → total 1.1ms.
Our version: conv 0.5ms + fused 0.3ms = 0.8ms.
Speedup: 1.375x. Not huge, but might be enough.
But wait, can we also save the write-to-memory after conv by doing conv + fused activation in a single kernel? Only if we implement conv ourselves or use some fused API.
Actually, cudnn 8+ supports fused conv + activation. But PyTorch's eager mode doesn't automatically fuse conv + mish + tanh.
What if our approach is:
1. Check if PyTorch's conv3d supports some fused activation?
2. No, mish and tanh are not standard conv fusion targets.
Another idea: Can we replace the conv3d with a grouped/depthwise equivalent? No, it's a regular conv.
What if we write a custom kernel that does just the activation part, but we implement it in a way that matches or exceeds PyTorch's elementwise kernel performance?
PyTorch's elementwise kernels use the TensorIterator and are already quite optimized. But our fused kernel might have lower launch overhead and better cache behavior.
Actually, let me think about this differently. The input to activations is 118M elements. PyTorch's elementwise kernels typically achieve close to bandwidth. So:
- Mish: reads 473MB, computes, writes 473MB. Time ≈ 946MB / 3.35TB/s = 0.28ms + compute
- Tanh: same, 0.28ms + compute
- Fused: read 473MB, compute both, write 473MB. Time ≈ 946MB / 3.35TB/s = 0.28ms + compute
Wait, for fused, it's one read + one write = 946MB total traffic.
For separate: conv write (473MB) + mish read+write (946MB) + tanh read+write (946MB) = 2365MB.
But conv also reads input and weights during its computation.
For the activation part specifically:
- Reference: mish reads output(473) + writes temp(473) + tanh reads temp(473) + writes final(473) = 1892MB activation traffic
- Fused: reads conv output(473) + writes final(473) = 946MB activation traffic
- Saved: 946MB = 0.28ms at 3.35TB/s
This is decent but not amazing.
Actually, can we also eliminate the conv output write by doing conv in-place or using a temporary buffer? No, we need the conv output as input to our kernel.
But wait, what if we allocate the output tensor first, then have the conv write directly into it? That's what PyTorch already does.
Hmm, I think for this to be a really compelling win, we need to implement the conv ourselves using an approach that can match or exceed cudnn for this specific shape, AND fuse the activations.
For smallish conv3d like this (3×3×3, 32→64 channels), maybe a direct conv with shared memory can actually be quite competitive with cudnn in fp32. Cudnn has overhead and might not be perfectly optimized for every shape. For 3D convs, cudnn support might not be as mature as 2D conv.
Also, the batch size is small (16). 3D convs are less commonly benchmarked and optimized in cudnn.
Let me actually try to write the direct conv and see how it performs. If it's terrible, I can fall back to conv + fused activation.
Let me write the kernel with the shared memory tiling approach I designed.
Actually, there's one more thing. I can use `cp.async` (Hopper/Ampere) for asynchronous loads to shared memory. This can help overlap memory and compute.
But let's start simple. I'll write a clean direct conv kernel with shared memory tiles, then compile and benchmark.
Wait, let's also check if there's any possibility to use the CUTLASS library. PyTorch has `torch.backends.cuda.cutlass` in newer versions, but not directly accessible for custom kernels.
One more idea: can we use Triton? But `load_inline` expects C++/CUDA code, not Triton.
Alright, let me start writing the kernel. I'll use the tile approach with TILE_K=4, TILE_D=4, TILE_H=4, TILE_W=4, block=256 threads.
For performance, I'll unroll the inner spatial loops since KD=KH=KW=3 are constants.
Let me write the code:
```python
kernel_code = """
import torch
import torch.nn as nn
from torch.utils.cpp_extension import load_inline
cpp_source = '''
#include <cuda_runtime.h>
#include <cuda_fp16.h>
#include <math.h>
#define KD 3
#define KH 3
#define KW 3
#define TILE_K 4
#define TILE_D 4
#define TILE_H 4
#define TILE_W 4
#define BLOCK_SIZE 256
__global__ void conv3d_mish_tanh_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int C, int D, int H, int W,
int K, int OD, int OH, int OW
) {
const int IN_TILE_D = TILE_D + KD - 1;
const int IN_TILE_H = TILE_H + KH - 1;
const int IN_TILE_W = TILE_W + KW - 1;
const int IN_TILE_SIZE = IN_TILE_D * IN_TILE_H * IN_TILE_W * C;
const int W_TILE_SIZE = TILE_K * C * KD * KH * KW;
extern __shared__ float smem[];
float* smem_input = smem;
float* smem_weight = smem + IN_TILE_SIZE;
int n = blockIdx.z;
int k_block = blockIdx.y;
int d_block = blockIdx.x / ((OH + TILE_H - 1) / TILE_H * (OW + TILE_W - 1) / TILE_W);
int h_block = (blockIdx.x / ((OW + TILE_W - 1) / TILE_W)) % ((OH + TILE_H - 1) / TILE_H);
int w_block = blockIdx.x % ((OW + TILE_W - 1) / TILE_W);
int k_start = k_block * TILE_K;
int d_start = d_block * TILE_D;
int h_start = h_block * TILE_H;
int w_start = w_block * TILE_W;
// Collaborative load of input tile
for (int idx = threadIdx.x; idx < IN_TILE_SIZE; idx += BLOCK_SIZE) {
int w = idx % IN_TILE_W;
int tmp = idx / IN_TILE_W;
int h = tmp % IN_TILE_H;
tmp /= IN_TILE_H;
int d = tmp % IN_TILE_D;
int c = tmp / IN_TILE_D;
int d_in = d_start + d;
int h_in = h_start + h;
int w_in = w_start + w;
float val = 0.0f;
if (d_in < D && h_in < H && w_in < W && c < C) {
val = input[((n * C + c) * D + d_in) * H * W + h_in * W + w_in];
}
smem_input[idx] = val;
}
// Collaborative load of weight tile
for (int idx = threadIdx.x; idx < W_TILE_SIZE; idx += BLOCK_SIZE) {
int kw = idx % KW;
int tmp = idx / KW;
int kh = tmp % KH;
tmp /= KH;
int kd = tmp % KD;
tmp /= KD;
int c = tmp % C;
int k = tmp / C;
int k_out = k_start + k;
float val = 0.0f;
if (k_out < K && c < C) {
val = weight[((((k_out * C + c) * KD + kd) * KH + kh) * KW + kw)];
}
smem_weight[idx] = val;
}
__syncthreads();
// Compute outputs
int outers = TILE_K * TILE_D * TILE_H * TILE_W;
int out_tid = threadIdx.x;
if (out_tid < outers) {
int w = out_tid % TILE_W;
int tmp = out_tid / TILE_W;
int h = tmp % TILE_H;
tmp /= TILE_H;
int d = tmp % TILE_D;
int k = tmp / TILE_D;
int k_out = k_start + k;
int d_out = d_start + d;
int h_out = h_start + h;
int w_out = w_start + w;
if (k_out < K && d_out < OD && h_out < OH && w_out < OW) {
float sum = bias[k_out];
for (int c = 0; c < C; c++) {
#pragma unroll
for (int kd = 0; kd < KD; kd++) {
#pragma unroll
for (int kh = 0; kh < KH; kh++) {
#pragma unroll
for (int kw = 0; kw < KW; kw++) {
int in_idx = ((c * IN_TILE_D + (d + kd)) * IN_TILE_H + (h + kh)) * IN_TILE_W + (w + kw);
int w_idx = ((k * C + c) * KD + kd) * KH * KW + kh * KW + kw;
sum += smem_input[in_idx] * smem_weight[w_idx];
}
}
}
}
float sp = logf(1.0f + expf(sum));
float mish = sum * tanhf(sp);
float out_val = tanhf(mish);
output[(((n * K + k_out) * OD + d_out) * OH + h_out) * OW + w_out] = out_val;
}
}
}
torch::Tensor conv3d_mish_tanh_cuda(
torch::Tensor input,
torch::Tensor weight,
torch::Tensor bias,
int stride,
int padding
) {
int N = input.size(0);
int C = input.size(1);
int D = input.size(2);
int H = input.size(3);
int W = input.size(4);
int K = weight.size(0);
int OD = (D + 2 * padding - KD) / stride + 1;
int OH = (H + 2 * padding - KH) / stride + 1;
int OW = (W + 2 * padding - KW) / stride + 1;
auto output = torch::empty({N, K, OD, OH, OW}, input.options());
int w_blocks = (OW + TILE_W - 1) / TILE_W;
int h_blocks = (OH + TILE_H - 1) / TILE_H;
int d_blocks = (OD + TILE_D - 1) / TILE_D;
int k_blocks = (K + TILE_K - 1) / TILE_K;
dim3 grid(w_blocks * h_blocks * d_blocks, k_blocks, N);
dim3 block(BLOCK_SIZE);
int shared_mem_size = (IN_TILE_D * IN_TILE_H * IN_TILE_W * C +
TILE_K * C * KD * KH * KW) * sizeof(float);
conv3d_mish_tanh_kernel<<<grid, block, shared_mem_size>>>(
input.data_ptr<float>(),
weight.data_ptr<float>(),
bias.data_ptr<float>(),
output.data_ptr<float>(),
N, C, D, H, W, K, OD, OH, OW
);
return output;
}
'''
conv3d_mish_tanh = load_inline(
name='conv3d_mish_tanh',
cpp_sources=cpp_source,
functions=['conv3d_mish_tanh_cuda'],
extra_cuda_cflags=['-O3', '--use_fast_math'],
verbose=False
)
class ModelNew(nn.Module):
def __init__(self, in_channels, out_channels, kernel_size, stride=1, padding=0):
super(ModelNew, self).__init__()
self.conv = nn.Conv3d(in_channels, out_channels, kernel_size, stride=stride, padding=padding)
self.stride = stride
self.padding = padding
def forward(self, x):
w = self.conv.weight
b = self.conv.bias
return conv3d_mish_tanh.conv3d_mish_tanh_cuda(x, w, b, self.stride, self.padding)
"""
```
Wait, there are several issues:
1. `blockIdx.x` mapping - I used a complex 1D grid x mapping. That might be okay, but let me make sure it's correct.
2. Grid dimensions: `dim3 grid(w_blocks * h_blocks * d_blocks, k_blocks, N)`. CUDA limits: x can be very large, y and z up to 65535 on some devices, but on modern GPUs they're much larger. For our case: w*h*d = 16*16*8 = 2048, k_blocks = 16, N = 16. So grid = (2048, 16, 16). This is fine.
3. The grid decomposition might not be optimal. Let me reconsider.
Actually, using `(blockIdx.x, blockIdx.y, blockIdx.z)` for `(spatial, k, n)` is okay, but it might be better to put the largest dimension in x for better hardware scheduling. `w_blocks * h_blocks * d_blocks = 2048` is the largest.
But wait, I need to be careful with `extern __shared__ float smem[];`. The size is passed at kernel launch. But inside the kernel, I declare `float* smem_input = smem; float* smem_weight = smem + IN_TILE_SIZE;`. This is fine.
Wait, `IN_TILE_SIZE` is not a compile-time constant in a way that works with array declarations? Actually, `IN_TILE_D`, etc. are `#define` macros, so `IN_TILE_SIZE` is a compile-time constant. Good.
But there might be a problem: `extern __shared__ float smem[];` followed by `float* smem_input = smem;` is the standard pattern for dynamic shared memory. But wait, I could also use static shared memory since the sizes are known at compile time for our specific values. But `C` is a runtime parameter. So `smem_input[IN_TILE_SIZE]` cannot be static because `IN_TILE_SIZE = IN_TILE_D * IN_TILE_H * IN_TILE_W * C` and C is a runtime variable.
Wait, but IN_TILE_SIZE uses `C` which is a function parameter! That's illegal for static shared memory. So `extern __shared__` with dynamic allocation is the right approach.
But there's an issue: the shared memory size I pass to the kernel launch should account for the actual sizes. However, if C is a runtime variable, then IN_TILE_SIZE is computed at runtime. But I pass `shared_mem_size` which is computed in host code using `C`. The kernel also computes `IN_TILE_SIZE` using the parameter `C`. These should match.
Actually, I just realized: in our specific problem, C=32, which is fixed. So I could make it a template parameter or just hardcode it. Hardcoding C=32 would allow the compiler to unroll loops better.
Let me hardcode the specific dimensions to maximize compiler optimization. But the kernel should still accept the parameters for generality.
Wait, if I hardcode C=32, and the compiler sees `for (int c = 0; c < 32; c++)`, it can fully unroll. This is a huge win.
Similarly, KD=KH=KW=3 can be hardcoded.
Let me revise the kernel to use template parameters for C, KD, KH, KW, or just hardcode the constants.
Actually, since the shapes are fixed, I can write:
```cpp
__global__ void conv3d_mish_tanh_kernel(...) {
const int C_FIXED = 32;
const int KD_FIXED = 3;
...
}
```
But to get the compiler to unroll, the loop bound needs to be a compile-time constant. If it's `const int`, the compiler might or might not unroll depending on optimization level. `#pragma unroll` helps, but the compiler still needs to know the bound. Actually, with `-O3`, `const int` loop bounds inside a kernel are typically treated as compile-time constants if they're initialized with literals.
Wait, but `C` is a function parameter. The compiler won't know C=32 at compile time. It will generate a generic loop.
For maximum performance, I should use template parameters:
```cpp
template <int C, int KD, int KH, int KW>
__global__ void conv3d_mish_tanh_kernel(...)
```
Then in the host wrapper:
```cpp
conv3d_mish_tanh_kernel<32, 3, 3, 3><<<grid, block, shared_mem_size>>>(...)
```
Yes! This ensures the compiler knows all these dimensions at compile time.
Similarly, we could template TILE sizes if we want, but `#define` macros also work as compile-time constants.
Let me also template the tile sizes for experimentation, but for now, I'll use macros.
Actually, one concern: the grid mapping. With `dim3 grid(w_blocks * h_blocks * d_blocks, k_blocks, N)`, blockIdx.x is very large. In the kernel, I compute:
```cpp
int d_block = blockIdx.x / (h_blocks * w_blocks);
int h_block = (blockIdx.x / w_blocks) % h_blocks;
int w_block = blockIdx.x % w_blocks;
```
I need `h_blocks` and `w_blocks` inside the kernel. But these are not passed as parameters! I only passed N, C, D, H, W, K, OD, OH, OW. I'd need to compute h_blocks and w_blocks from OH, OW, etc.
```cpp
int w_blocks_grid = (OW + TILE_W - 1) / TILE_W;
int h_blocks_grid = (OH + TILE_H - 1) / TILE_H;
```
Yes, I can compute these at runtime.
Let me refine the code structure. Actually, I realize that since the block decomposition uses runtime values (OH, OW), but the tile sizes are compile-time constants, this is fine.
Here's my refined plan for the kernel:
```cpp
template <int C, int KD, int KH, int KW, int TILE_K, int TILE_D, int TILE_H, int TILE_W, int BLOCK_SIZE>
__global__ void conv3d_mish_tanh_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int D, int H, int W,
int K, int OD, int OH, int OW
) {
// In our case: C=32, KD=3, KH=3, KW=3, TILE_K=4, etc.
...
}
```
But passing many template parameters is verbose. Since the shapes are fixed, I can just `#define` them all.
Actually, let me think about whether we need dynamic shared memory at all if we hardcode everything. If C=32, then IN_TILE_SIZE = 6*6*6*32 = 6912. This is a compile-time constant! So we can use static shared memory:
```cpp
__shared__ float smem_input[6912];
__shared__ float smem_weight[3456];
```
This is simpler and might be slightly faster.
Let me count: TILE_K=4, C=32, KD=3, KH=3, KW=3. Weight tile = 4*32*27 = 3456 floats.
Static shared memory total = 6912 + 3456 = 10368 floats = 41,472 bytes = 40.5KB.
This is well under the 48KB default limit, so we don't even need to call `cudaFuncSetAttribute` to increase shared memory.
On Hopper, we could use up to 227KB, but 40.5KB is plenty and allows multiple blocks per SM.
With 40.5KB per block and 228KB per SM, we can have up to 5 blocks per SM. With 256 threads per block, that's 1280 threads per SM, yielding ~62.5% occupancy. Pretty good.
Alright, let me write the code using static shared memory with hardcoded constants.
Actually, wait. The weights for 4 output channels are 3456 floats. In shared memory, we store them as `smem_weight[k][c][kd][kh][kw]`. When threads read from this, for a fixed (c,kd,kh,kw) and varying k, they read:
- k=0: offset 0
- k=1: offset 32*27 = 864
- k=2: offset 1728
- k=3: offset 2592
These are nicely spread across banks. Each k's weights for a given c are 27 floats = 108 bytes = 27 banks, so no bank conflicts.
For input shared memory: `smem_input[c][d][h][w]`. As I analyzed before, for threads reading the same c but different (d,h,w), there's good bank distribution.
One more optimization: instead of 4 nested loops with `#pragma unroll`, I can manually unroll or use a single linear loop over the 864 combinations. But the compiler should handle unrolling well for loops with compile-time bounds.
Let me write the full kernel now. I'll carefully compute indices to avoid integer division inside the inner loop.
Actually, I realize that with C=32 and kernel 3×3×3, we can restructure the computation to be more efficient:
- For each c (32 iterations), we need 27 input values and 27 weight values (per output channel).
- We can compute partial sums for all output channels simultaneously.
- For smem_input, we read input_val = smem_input[c][d+kd][h+kh][w+kw]
- For smem_weight, we read weight_val[k] = smem_weight[k][c][kd][kh][kw]
- Sum[k] += input_val * weight_val[k]
This is what the nested loops already do.
Let me also think about register usage. We have TILE_K=4 output channels. Each thread computes 1 output (1 of the 4 channels, or 1 spatial). So each thread only needs 1 accumulator register. The k-loop in the 4D decomposition handles which channel the thread computes.
Wait, with 256 threads and 256 outputs (4 channels × 4×4×4 spatial), each thread computes exactly 1 output. So each thread has 1 accumulator. The thread reads the appropriate weight for its channel. No vector accumulation across channels.
This is fine. With 4 nested loops of 32×3×3×3 = 864 iterations, and each iteration doing 1 FMA, that's 864 FMAs per thread. At 2 FLOPs per FMA, 1728 FLOPs per thread. For 256 threads per block, that's 442K FLOPs per block.
With 491k blocks, total FLOPs = 217 GFLOPs for the MACs + overhead.
Wait, I calculated 205 GFLOPs earlier. The difference might be due to tiling overhead or my arithmetic.
Actually, outputs = 16*64*30*62*62 = 118,087,680
MACs per output = 32*27 = 864
Total MACs = 118,087,680 * 864 = 102,027,755,520
In FLOPs (multiply-add = 2): 204 GFLOPs. OK.
Now, per block, valid outputs might be fewer due to boundary effects. Since 62/4 = 15.5, the last block in H and W dimension has some invalid threads.
Actually, in our mapping, we have `if (k_out < K && d_out < OD && h_out < OH && w_out < OW)`. This means boundary threads just return early, but they still participated in loading shared memory.
Total blocks if no boundaries: 16*64*30*62*62 / 256 = exactly 118,087,680 / 256 = 461,280 blocks. But with boundary rounding:
- W: ceil(62/4) = 16, actual W positions covered = 64, but we only need 62
- H: ceil(62/4) = 16, actual = 64, need 62
- D: ceil(30/4) = 8, actual = 32, need 30
- K: ceil(64/4) = 16, actual = 64
Total blocks launched = 16*16*8*16*16 = 524,288.
Valid outputs per block: some blocks on the boundary have fewer than 256 valid outputs.
The extra work is small. But it's fine.
I should also consider whether to use `float4` for loads. For the input tile loading with 6912 elements and 256 threads, we can use `float4` to load 4 floats per thread in a single instruction. This reduces instruction count and might help.
For loading input tile: each thread loads 6912/256 = 27 values. Not divisible by 4. We could have some threads load float4 and some load singles.
Actually, let me just use float loads for simplicity. The compiler might vectorize automatically.
Wait, for the output write, can we use a single store instruction per thread? Yes, `output[idx] = out_val;` is a single float store.
Actually, before I finalize the kernel, let me think about whether I should use a different tile shape.
The current tile: 4 channels × 4×4×4 spatial = 256 outputs.
Alternative: 8 channels × 2×4×4 spatial = 256 outputs.
Input for 2×4×4 spatial: 4×6×6 = 144 positions × 32 channels = 4608 floats = 18KB.
Weight for 8 channels: 8×32×27 = 6912 floats = 27KB.
Total SMEM: 45KB. Still under 48KB.
With 8 channels per block, the weight tile is larger (27KB), but the input tile is smaller (18KB). Total is actually slightly larger (45KB vs 40.5KB), but still under 48KB.
But with 8 channels, there are fewer channel blocks (8 instead of 16), so total blocks = 16 * 8 * (ceil(30/2)*ceil(62/4)*ceil(62/4)) = 16 * 8 * 15 * 16 * 16 = 491,520. Wait, that's 16×8×15×16×16. The spatial blocks changed because D tile is 2 now.
Actually, with TILE_D=2, we have d_blocks = ceil(30/2) = 15.
With TILE_H=4, h_blocks = ceil(62/4) = 16.
With TILE_W=4, w_blocks = ceil(62/4) = 16.
Spatial blocks = 15*16*16 = 3840.
Total blocks = 3840 * 8 * 16 = 491,520.
With TILE_D=4, TILE_H=4, TILE_W=4: spatial blocks = 8*16*16 = 2048, k_blocks = 16, N = 16.
Total = 2048 * 16 * 16 = 524,288.
More blocks is generally better for occupancy and scheduling, but both are plenty.
I think both configurations would work similarly. Let me stick with 4×4×4×4 for balanced memory usage.
One concern: with 4 output channels per block and 256 threads, threads are grouped as 64 threads per channel. But a warp has 32 threads. So warp 0-1 compute channel 0, warp 2-3 compute channel 1, etc.
For weight reads: warp 0 reads weight for k=0. Broadcast. Warp 1 also reads weight for k=0. Broadcast. Good.
For input reads: warp 0 threads compute 32 different spatial positions for channel 0. They read input[c][d_thread][h_thread][w_thread] for varying spatial. Bank distribution is fine as analyzed.
For output writes: warp 0 writes 32 outputs for channel 0 at different spatial positions. As discussed, these are somewhat scattered but within L1 cache range.
I think this is solid. Let me write the complete module code.
Actually, let me also think about activation precision. Mish involves `exp` and `log` and `tanh`. Using `expf`, `logf`, `tanhf` should be precise enough. But we need to match PyTorch's Mish and Tanh within 1e-4. PyTorch uses `std::tanh` and its own softplus. On GPU, PyTorch likely uses `tanh` and `log1p(exp(x))` from libdevice.
For numerical equivalence, I should be careful with large values. The `expf` can overflow for large positive values. In PyTorch, mish uses `softplus(x) = log(1 + exp(x))`. For large x, `log1p(exp(x))` might overflow if exp(x) > max_float. PyTorch probably handles this with a stable implementation.
Let's check PyTorch's mish implementation. In ATen, Mish is implemented as:
```cpp
auto softplus = at::softplus(self, 1.0, 20.0);
auto tanh_softplus = at::tanh(softplus);
auto mish = self * tanh_softplus;
```
And softplus uses a threshold:
```cpp
template <typename T>
C10_HOST_DEVICE T softplus(T x, T beta, T threshold) {
if (x * beta > threshold) {
return x;
} else {
return (std::log1p(std::exp(x * beta))) / beta;
}
}
```
Wait, I need to verify. For default parameters, beta=1.0, threshold=20.0. So if x > 20, softplus(x) = x.
So mush(x) = x * tanh(softplus(x)).
For x > 20: softplus(x) = x, mish(x) = x * tanh(x). As x → ∞, tanh(x) → 1, so mish(x) → x.
For x very negative: exp(x) → 0, softplus(x) → 0, tanh(0) = 0, mish(x) → x * 0 = 0.
My kernel computes `logf(1.0f + expf(x))` directly. For x > 20, `expf(x)` might overflow (exp(88) ≈ 1.7e38 is near float max). For x around 89, expf overflows to inf, then log(1+inf) = inf, then tanh(inf) = 1, and x * 1 = x. But the intermediate inf might cause issues with numerical correctness if PyTorch handles it differently.
Wait, if x=90 and we compute `logf(1.0f + expf(90.0f))`:
- expf(90) = inf (since max float32 exp is ~88.7)
- log1p(inf) = inf
- tanh(inf) = 1.0
- mish = 90 * 1 = 90
But PyTorch's implementation: since x > 20, softplus(x) = x = 90, tanh(90) = 1, mish = 90 * 1 = 90.
Same result.
For x=89:
- Our code: expf(89) = inf, mish = 89 * 1 = 89
- PyTorch: softplus(89) = 89, mish = 89 * tanh(89) ≈ 89 * 1 = 89
Same.
For x = 100:
- Our code: expf(100) = inf, tanh(inf) = 1, mish = 100
- PyTorch: same
So overflow in expf doesn't actually cause incorrect results for the final mish computation, because tanh(inf) = 1 anyway.
But what about intermediate values where x is large negative? expf(x) underflows to 0, log1p(0) = 0, tanh(0) = 0, mish = x * 0 = 0. For very negative x, PyTorch gives approximately the same.
However, PyTorch uses `log1p(exp(x))` which for very negative x computes `log1p(underflow) = log1p(0) = 0`. If we use `logf(1.0f + expf(x))`, for x < -103, expf(x) = 0, log1p(0) = 0. Same result.
What about x near -1? Our `logf(1.0f + expf(x))` vs PyTorch's `log1p(exp(x))`. These should be very close.
For the `tanh(mish(x))` part, we're applying `tanh` to the result of `mish`. This is straightforward.
I think using `--use_fast_math` might affect precision. The fast math flag makes `expf`, `logf`, and `tanhf` use faster but slightly less accurate intrinsic functions. For `atan2` and division, precision might drop. But for exp/log/tanh on H100, the difference should be within 2-3 ULPs.
Given tolerance is 1e-4 for fp32, and our input values are from `torch.rand` (0 to 1), the conv output will be moderate. Actually, conv output could be larger depending on weights. But weights are initialized by PyTorch (default is Kaiming uniform). Output magnitude is usually O(1) to O(sqrt(C*K^3)).
Wait, with He initialization, weight variance is proportional to 2/(C*K^3) = 2/864 ≈ 0.0023. Standard deviation ≈ 0.048. With 864 inputs, conv output std ≈ sqrt(864) * 0.048 ≈ 1.4. And with 32 input channels averaging ~0.5, the mean might be around 0. The output of conv3d is typically in the range [-5, 5] or so.
At this range, exp, tanh, log are well-behaved and fast-math should be accurate to within 1e-5 or better. So `atol=rtol=1e-4` should be fine.
Actually, one thing to check: PyTorch computes mish first, then tanh:
`mish = x * tanh(softplus(x))`
`out = tanh(mish)`
So `out = tanh(x * tanh(softplus(x)))`.
Our kernel computes exactly this. Good.
Let me now finalize the code. I want to make sure the PyTorch module works correctly.
One important detail: `nn.Conv3d` stores weights in the order [out_channels, in_channels, kD, kH, kW]. Our kernel reads weight[k_out][c][kd][kh][kw]. Let me verify PyTorch's weight format.
In PyTorch, `conv3d` weight shape is `(out_channels, in_channels/groups, kD, kH, kW)`. With groups=1, it's `(K, C, KD, KH, KW)`. And the layout is dense in that order.
So our indexing:
```cpp
weight[((((k_out * C + c) * KD + kd) * KH + kh) * KW + kw)]
```
This matches the `(K, C, KD, KH, KW)` layout. Good.
Input layout is `(N, C, D, H, W)`.
Our input indexing:
```cpp
input[((n * C + c) * D + d_in) * H * W + h_in * W + w_in]
```
This matches NCDHW. Good.
Output layout:
```cpp
output[(((n * K + k_out) * OD + d_out) * OH + h_out) * OW + w_out]
```
This is also NCDHW. Good.
For the `ModelNew` class, I need to properly initialize the conv weights and make sure they're accessible.
```python
class ModelNew(nn.Module):
def __init__(self, in_channels, out_channels, kernel_size, stride=1, padding=0):
super(ModelNew, self).__init__()
self.in_channels = in_channels
self.out_channels = out_channels
self.kernel_size = kernel_size
self.stride = stride
self.padding = padding
# Initialize weights and bias the same way as nn.Conv3d
self.weight = nn.Parameter(torch.empty(out_channels, in_channels, kernel_size, kernel_size, kernel_size))
self.bias = nn.Parameter(torch.empty(out_channels))
# Reset parameters using same init as Conv3d
nn.init.kaiming_uniform_(self.weight, a=math.sqrt(5))
if self.bias is not None:
fan_in, _ = nn.init._calculate_fan_in_and_fan_out(self.weight)
bound = 1 / math.sqrt(fan_in)
nn.init.uniform_(self.bias, -bound, bound)
def forward(self, x):
return conv3d_mish_tanh_cuda(x, self.weight, self.bias, self.stride, self.padding)
```
Wait, that's not quite right. The evaluator will compare `ModelNew` against `Model` on randomly initialized weights. We need to ensure our module is compatible with how `Model` is initialized.
Actually, looking at the evaluator code pattern (not shown, but standard), it typically:
1. Creates `model = Model(*get_init_inputs())`
2. Creates `model_new = ModelNew(*get_init_inputs())`
3. Copies weights from `model` to `model_new` or initializes them the same way
But the problem statement says: "Your `ModelNew` class must: 1. Accept the same constructor arguments as `Model`."
We need to match PyTorch's Conv3d behavior. But we can't easily extract the weights from a separate Conv3d layer if we're copying parameters.
Actually, the safest approach is to have `ModelNew` contain an `nn.Conv3d` layer and use its weights, OR implement our own weight initialization that matches `nn.Conv3d`.
If the evaluator does `model_new.load_state_dict(model.state_dict())`, then we need to have matching parameter names.
Let me use the approach where `ModelNew` has the same structure as `Model` but calls our custom kernel:
```python
class ModelNew(nn.Module):
def __init__(self, in_channels, out_channels, kernel_size, stride=1, padding=0):
super(ModelNew, self).__init__()
self.conv = nn.Conv3d(in_channels, out_channels, kernel_size, stride=stride, padding=padding)
def forward(self, x):
return conv3d_mish_tanh_cuda(x, self.conv.weight, self.conv.bias,
self.conv.stride[0], self.conv.padding[0])
```
This way, `state_dict` is compatible, and initialization is automatic.
Wait, but `self.conv.stride` is a tuple `(stride, stride, stride)`. And `self.conv.padding` is also a tuple.
For our kernel, we need `stride=1` and `padding=0`. If the constructor can receive different values, we should handle them. But the reference uses defaults.
In the problem, `get_init_inputs()` returns `[in_channels, out_channels, kernel_size]`. So stride defaults to 1 and padding to 0. Our kernel should handle these via the `stride` and `padding` parameters.
But wait, for a general stride and padding, our kernel's output size formula might work, but our direct conv implementation ONLY works for stride=1 (since we directly add spatial indices). For stride > 1, the kernel would need to multiply spatial indices by stride, which is easy to add.
For padding > 0, our current kernel adds 0-padding by checking bounds and loading 0 when out of bounds. This already handles padding implicitly IF the input dimensions include the padding. But since the input tensor doesn't include padding, we'd need to offset the input reads.
Actually, in our input loading:
```cpp
if (d_in < D && h_in < H && w_in < W && c < C) {
val = input[...];
} else {
val = 0.0f;
}
```
This works for padding=0. For padding>0, the input coordinates would need to be shifted by -padding:
`d_in = d_start + d - padding`
And the bounds check should be:
`if (d_in >= 0 && d_in < D && h_in >= 0 && h_in < H && w_in >= 0 && w_in < W && c < C)`
Also, the output formula is:
`OD = (D + 2*padding - KD) / stride + 1`
And `d_start = d_block * TILE_D` gives the output d index. For stride > 1, the input starting position should be `d_start * stride`.
Let me update the kernel to properly support stride and padding.
For output position `d_out`, the corresponding input start is `d_out * stride - padding`.
The input window covers `[d_out * stride - padding, d_out * stride - padding + KD)`.
For loading input into shared memory:
```cpp
int d_in = d_start * stride - padding + d;
int h_in = h_start * stride - padding + h;
int w_in = w_start * stride - padding + w;
```
And the output bounds:
```cpp
int d_out = d_start + d;
int h_out = h_start + h;
int w_out = w_start + w;
```
Wait, actually, for stride, the loop should be:
`d_out` ranges from `d_start` to `d_start + TILE_D - 1`.
The input corresponding to `d_out` is at `d_out * stride`.
So for the shared memory tile starting at `d_start * stride - padding`, the element at offset `d` in shared memory corresponds to input position `d_start * stride - padding + d`.
For the convolution, output `d_out` uses input starting at `d_out * stride - padding`.
Within the shared memory tile, output `d_out = d_start + d_offset` needs input at shared memory position `d_offset * stride` relative to the start... wait, this simple indexing only works for stride=1.
For stride > 1, the relationship between output spatial and input spatial is nonlinear. For example, with stride=2:
- output d=0 uses input d=0,1,2
- output d=1 uses input d=2,3,4
This means adjacent outputs share some inputs, but the pattern is strided. For direct conv with shared memory, this complicates the tile loading because different outputs in the same tile need DIFFERENT input windows if stride > 1.
For a tile of outputs with stride=2:
- output 0 needs input [0, 2) (with padding=0, kernel=3)
- output 1 needs input [2, 4)
- output 2 needs input [4, 6)
- etc.
These windows don't overlap in a simple way like they do with stride=1. The total input span for TILE_D outputs is `(TILE_D - 1) * stride + KD`. For TILE_D=4, stride=2, KD=3: `(3)*2 + 3 = 9`. But the overlap is minimal.
Loading all needed input into shared memory is still possible, but the indexing from output to input becomes more complex and might lose some coalescing.
However, the task says the reference uses `stride=1, padding=0` by default. And `get_init_inputs()` only passes `in_channels, out_channels, kernel_size`. So for the actual evaluation, stride=1 and padding=0.
To be safe, I can add stride support. But to keep the kernel simple and performant, I'll focus on stride=1 and handle other values by falling back to PyTorch. Wait, no fallback. The task says not to use try/except fallbacks.
I can either:
1. Support general stride and padding in the CUDA kernel
2. Only support stride=1, padding=0 and rely on the fact that tests use these defaults
Since the evaluation likely only tests with the default values (as shown in `get_init_inputs`), option 2 is safer for performance. But to be robust, let me add support for arbitrary stride and padding.
For stride, I'll adjust the input loading:
```cpp
int d_in = d_start * stride - padding + d;
// etc.
```
And for the convolution indexing within shared memory:
If stride=1: output at local offset `d` reads input at local offset `d + kd` from shared memory.
If stride=2: output at local offset `d` reads input at local offset `d * stride + kd` from shared memory.
Wait, actually the shared memory contains all input values from `d_start * stride - padding` to some end. For output `d_out = d_start + d`, the corresponding input window in global terms starts at `(d_start + d) * stride - padding`. In shared memory local terms (where shared memory starts at `d_start * stride - padding`), the offset is `((d_start + d) * stride - padding) - (d_start * stride - padding) = d * stride`.
So within shared memory, output at local offset `d` needs input at local offset `d * stride + kd`.
For stride=1: `d + kd`. Correct.
For stride=2: `2*d + kd`.
Similarly for h and w.
This changes the indexing in the inner loop:
```cpp
int in_idx = ((c * IN_TILE_D + (d * stride_d + kd)) * IN_TILE_H + (h * stride_h + kh)) * IN_TILE_W + (w * stride_w + kw);
```
Wait, but `d * stride_d` might exceed the shared memory bounds for the tile! We need `IN_TILE_D = (TILE_D - 1) * stride_d + KD`. And similarly for H and W.
For stride=1: IN_TILE_D = TILE_D + KD - 1. Correct.
For stride=2, TILE_D=4, KD=3: IN_TILE_D = (4-1)*2 + 3 = 9. We need 9 values in D dimension.
This is getting more complex. Since our test only uses stride=1, padding=0, I'll just hardcode for that case but keep the parameters in the interface.
Actually, let me just support it properly. It's not that complicated.
Let me redesign:
```cpp
const int IN_TILE_D = (TILE_D - 1) * stride_d + KD;
const int IN_TILE_H = (TILE_H - 1) * stride_h + KH;
const int IN_TILE_W = (TILE_W - 1) * stride_w + KW;
```
But wait, `stride_d`, `stride_h`, `stride_w` need to be passed to the kernel or constant. We can pass them as parameters.
For simplicity, since we're using templates for compile-time constants, we could template `stride_d` etc. But they might be runtime values.
Since the problem uses stride=1 and padding=0, I'll just assume that and document it. If the evaluator tries other strides, our kernel would break. But based on the problem statement, it seems like only the defaults are used.
Actually, looking at the constructor signature, stride and padding are configurable. We should handle arbitrary values correctly. Let me do it properly.
For the input loading with arbitrary stride and padding:
```cpp
int d_in = d_start * stride_d - padding_d + d;
int h_in = h_start * stride_h - padding_h + h;
int w_in = w_start * stride_w - padding_w + w;
```
And output computation:
```cpp
int d_out = d_start + d;
int h_out = h_start + h;
int w_out = w_start + w;
```
And the convolution index:
```cpp
int in_smem_d = d * stride_d + kd;
int in_smem_h = h * stride_h + kh;
int in_smem_w = w * stride_w + kw;
```
This is correct. For stride=1, it reduces to `d + kd`, which is our original formula.
Now, for the tile sizes IN_TILE_D etc.:
```cpp
const int IN_TILE_D = (TILE_D - 1) * stride + KD;
const int IN_TILE_H = (TILE_H - 1) * stride + KH;
const int IN_TILE_W = (TILE_W - 1) * stride + KW;
```
Since stride is a runtime variable, these are runtime constants inside the kernel, but the compiler won't know them at compile time for array sizing with `__shared__`.
For dynamic shared memory, we calculate `IN_TILE_SIZE = ((TILE_D - 1) * stride + KD) * ... * C` in host code. Inside the kernel, we also compute it. But we only need the sizes for indexing, not for declaring arrays if we use dynamic shared memory.
With `extern __shared__ float smem[];`, we don't need compile-time sizes for the array itself. We just need them for indexing.
But wait, we still need `smem_input` and `smem_weight` to have compile-time bounds? No, with dynamic shared memory, `smem` is just a pointer. We partition it manually.
So dynamic shared memory works perfectly for runtime stride.
But for performance, the compiler needs to unroll the loops. With runtime stride, the inner loop bounds are still constants (KD=3, KH=3, KW=3), but the shared memory indexing involves multiplication by stride. The compiler can still unroll KD=3 etc., but the array index will have runtime stride in it.
For stride=1, `d * stride + kd` compiles to `d + kd` if the compiler inlines and constant-propagates. But with runtime stride, it would be `d * stride + kd`.
Since we expect stride=1, the performance difference might be negligible. But to be sure, let me hardcode that we only optimize for stride=1 and handle general stride by... hmm, no fallback.
Actually, I'll pass stride as a parameter but structure the code so that if stride=1 (as expected), the compiler can still optimize well. Or I can use `if (stride == 1)` inside the kernel to have two paths, but that creates divergence.
Since the task strongly implies stride=1, padding=0, let me just hardcode stride=1 and padding=0 in the optimized path. But I'll still pass the parameters and use them in the formulas.
Actually, no. Let me just support it generically. The overhead of a few multiplications by `stride` is tiny compared to the 864 FMAs per thread.
Alright, let me write the complete code now.
One more thing: for grids with many blocks, we want to make sure our occupancy is good. With 40.5KB shared memory and 256 threads, let's see how many blocks can run concurrently on an SM.
H100 has 228KB shared memory per SM. 228KB / 40.5KB = 5.6, so maximum 5 blocks per SM.
Each block has 256 threads. 5 blocks = 1280 threads.
Max threads per SM on Hopper = 2048.
So occupancy = 1280 / 2048 = 62.5%. Good.
Can we reduce shared memory to increase occupancy? If we use a smaller tile...
Say TILE_K=2, TILE_D=4, TILE_H=4, TILE_W=8:
- Outputs = 2*4*4*8 = 256. Same thread count.
- Input: (4-1)*1+3=6 D, (4-1)*1+3=6 H, (8-1)*1+3=11 W, plus channels = 6*6*11*32 = 12,672 floats = 50.8KB.
- Weights: 2*32*27 = 1,728 floats = 6.9KB.
- Total: 57.7KB > 48KB. Not good for default config, but we can use dynamic shared memory with larger limit.
On Hopper, we can configure up to 227KB. So 57.7KB is fine, but it reduces blocks per SM to 228/57.7 = 3.9 → 3 blocks = 768 threads = 37.5% occupancy.
So our original 40.5KB with 62.5% occupancy is better.
What about TILE_K=4, TILE_D=2, TILE_H=4, TILE_W=8:
- Outputs = 4*2*4*8 = 256
- Input: 4*6*11*32 = 8,448 = 33.8KB
- Weights: 4*32*27 = 13.8KB
- Total: 47.6KB → still under 48KB!
- Blocks per SM = 4 (if using default shared memory), threads = 1024, occupancy = 50%.
- Or on Hopper with dynamic, 228/47.6 = 4.7 → 4 blocks = 1024 threads.
Occupancy is slightly worse than our 40.5KB config. But output coalescing for W=8 might be better than W=4.
Let me think about memory access patterns. With TILE_W=8:
- Warp of 32 threads computing same k: need to map 32 threads to the 2*4*8=64 spatial positions.
- Half warp (16 threads) could compute 2×4×2 = 16 positions?
- If we map w = tid % 8, h = (tid / 8) % 4, d = (tid / 32) % 2:
- Threads 0-7: d=0, h=0, w=0..7
- Threads 8-15: d=0, h=1, w=0..7
- Threads 16-23: d=0, h=2 or 3? tid=16: d=0, h=2, w=0
- Threads 24-31: d=0, h=3, w=0..7
Wait, 32 threads for same k: if we have 4 k values, total positions = 64 per k.
One warp = 32 threads. If we assign warp 0 to positions 0..31 and warp 1 to positions 32..63 for k=0.
For positions 0..31: could be d=0, h=0..3, w=0..7? That's 32 positions exactly!
So warp 0 for k=0, d=0 (fixed), h=0..3, w=0..7.
Output addresses for tid:
tid=0: (d=0,h=0,w=0) offset=0
tid=1: (d=0,h=0,w=1) offset=1
...
tid=7: (d=0,h=0,w=7) offset=7
tid=8: (d=0,h=1,w=0) offset=62
tid=9: offset=63
...
tid=15: offset=69
tid=16: offset=124
...
tid=23: offset=131
tid=24: offset=186
...
tid=31: offset=193
Within a warp, we're accessing offsets 0..7, 62..69, 124..131, 186..193. These are interleaved across cache lines. A single warp instruction would need to access 8 cache lines for w=0..7 at h=0, another 8 for h=1, etc. Actually, each group of 8 threads accesses 32 bytes (8 floats). With 128-byte cache lines, each group touches 1 cache line. And groups are 62 floats apart. So 4 groups × 1 cache line each = 4 cache lines. But they are spread out in memory.
For L1 cache: if the output subtensor from (d=0, h=0..3, w=0..7) is cached, the L1 cache might hold these cache lines.
With TILE_W=8, each "row" (fixed d,h) has 8 floats = 32 bytes. With 4 rows (h=0..3), total 128 bytes per warp store. These 128 bytes are spread across 4 cache lines (each 128 bytes but starting at different offsets). So it's not perfectly contiguous, but still reasonable.
I think W=4 or W=8 will both work fine. Since the L1 cache on H100 is large, the scattered accesses within our tile will hit cache.
Let me stick with my original TILE_K=4, TILE_D=4, TILE_H=4, TILE_W=4 for balanced memory usage.
Actually, wait. I want to double check one thing. For the warp-level coalescing of output writes, would it be better to have threads in a warp write to consecutive memory locations?
With TILE_K=4 and 256 threads per block, if we assign 64 threads per K (as I did), then warp 0 (threads 0-31) writes K=0 at various spatial positions. Warp 1 (threads 32-63) also writes K=0. Warps 2-3 write K=1, etc.
For K=0, the 64 threads write to a 4×4×4 spatial cube. In memory, these are at offsets:
`base + d*OH*OW + h*OW + w`
If we want a warp to write contiguous memory, we'd need 32 adjacent positions. But a 4×4×4 cube only has 4 adjacent W values. So a warp writes 32 scattered positions.
With TILE_K=1 and 256 threads for spatial, we'd have 256 threads all writing to the same K. We could map them to a 2×8×16 spatial tile. Then warp 0 could write 32 positions in a 1×4×8 or similar contiguous(ish) pattern.
Let me try TILE_K=1, TILE_D=2, TILE_H=8, TILE_W=16:
- 256 threads, all for k=0
- Warp 0: 32 positions
- Mapping: w=tid%16, h=(tid/16)%8, d=(tid/128)%2
- Warp 0 (tid 0-31): d=0, h=0..1, w=0..15
- Lane 0-15: h=0, w=0..15 → offsets 0..15. Contiguous 16 floats = 64 bytes.
- Lane 16-31: h=1, w=0..15 → offsets 62..77. Another contiguous 16 floats.
- These two segments are 62 floats apart, so 2 cache lines.
- Warp 1 (tid 32-63): d=0, h=2..3, w=0..15
- Offsets 124..139 and 186..201.
- Etc.
With 8 warps per block (256 threads), writing 256 positions for k=0. The span is from d=0,h=0,w=0 to d=1,h=7,w=15 = offsets 0 to 3844+7*62+15 = 3844+449 = 4293. That's about 17KB.
A warp touching 2 cache lines is great. But with 8 warps, we'd touch 16 cache lines, spread across 17KB. The L1 cache can hold this.
The shared memory for TILE_K=1:
- Input: 4*10*18*32 = 23,040 floats = 92KB. Wait, for 2×8×16 spatial with kernel 3:
- IN_D = (2-1)*1 + 3 = 4
- IN_H = (8-1)*1 + 3 = 11
- IN_W = (16-1)*1 + 3 = 18
- Total: 4*11*18*32 = 25,344 floats = 101KB
- Weights: 1*32*27 = 864 floats = 3.5KB
- Total: 104.5KB
This exceeds the 48KB default shared memory limit but fits in Hopper's 227KB. We'd need to configure the kernel to use 104.5KB shared memory. With `cudaFuncSetAttribute` for `cudaFuncAttributeMaxDynamicSharedMemorySize`, we can request up to 227KB.
But with 104.5KB per block, blocks per SM = 228/104.5 = 2.18 → 2 blocks = 512 threads = 25% occupancy. That's low.
On the other hand, the smaller shared memory config (40.5KB) gives 62.5% occupancy. Higher occupancy might be better for latency hiding and throughput, even if each warp does slightly more scattered stores.
I think 40.5KB shared memory with TILE_K=4 is a good balanced choice.
Let me think about one more optimization: using `__ldg` for read-only data. The input, weight, and bias are read-only from the kernel's perspective. We can mark them with `__restrict__` and even use `__ldg` for cache loading. Actually, `const float* __restrict__` already tells the compiler to use LDG if available.
For the compute loop, we're reading shared memory, not global memory, so LDG doesn't apply there.
One more thing: I can try to use `mma` instructions for the K=4 channels by having each thread compute 4 output channels for 1 spatial position. Then a warp of 32 threads computes 32 spatial positions × 4 channels = 128 outputs. With 2 warps = 256 outputs.
If thread computes 4 channels for 1 spatial, it needs 4 accumulators. The thread loops over c,kd,kh,kw. For each iteration, it reads 1 input value and 4 weight values. The 4 weight values can be loaded and multiplied sequentially, or if we use float4, we can load 4 weights at once.
But loading weights from shared memory: if weights are stored as `smem_weight[k][c][kd][kh][kw]`, then for fixed c,kd,kh,kw and k=0..3, the 4 weight values are at offsets 0, 32*27, 64*27, 96*27 in a flat array. That's not contiguous in memory (they are TILE_K apart). Wait, with our mapping:
```cpp
int w_idx = ((k * C + c) * KD + kd) * KH * KW + kh * KW + kw;
```
For k=0: offset = c*27 + kd*9 + kh*3 + kw
For k=1: offset = (1*32 + c)*27 + kd*9 + kh*3 + kw = 864 + c*27 + ...
For k=2: offset = 1728 + ...
For k=3: offset = 2592 + ...
So the 4 weight values for the same (c,kd,kh,kw) but different k are 864 floats apart! That's 3456 bytes. They do NOT share cache lines. Loading them together is not cache-friendly.
If we reorder weight storage to `smem_weight[c][kd][kh][kw][k]`, then for fixed c,kd,kh,kw, the 4 k values are contiguous. This would let us load them with float4 or at least within 1-2 cache lines.
But then the thread mapping to weight load needs to change. Currently, each thread loads weights independently during the collaborative load phase. If we change storage order, the collaborative load also changes.
Let me reconsider the weight storage order. Currently:
Collaborative load linear index maps to `(k, c, kd, kh, kw)` in row-major order.
This means in memory: all c,kd,kh,kw for k=0, then all for k=1, etc.
If we instead map linear index to `(c, kd, kh, kw, k)` in row-major order:
- Index = ((((c * KD + kd) * KH + kh) * KW + kw) * TILE_K + k)
- Thread 0: c=0,kd=0,kh=0,kw=0,k=0
- Thread 1: c=0,kd=0,kh=0,kw=0,k=1
- Thread 2: k=2
- Thread 3: k=3
- Thread 4: c=0,kd=0,kh=0,kw=1,k=0
Adjacent threads read adjacent k for the same c,kd,kh,kw. These are contiguous in GLOBAL memory if the global weight tensor has format [K, C, KD, KH, KW].
In global memory, weight[k][c][kd][kh][kw]:
- For k=0, offset = 0
- For k=1, offset = C*KD*KH*KW = 32*27 = 864
So k=0 and k=1 are 864 floats apart! They are NOT contiguous in global memory.
Aha! That's why our original load order `(k, c, kd, kh, kw)` with k varying fastest would have terrible coalescing. Because adjacent threads reading different k would access memory 864 floats apart.
In my original design, I used:
```cpp
int kw = idx % KW;
tmp = idx / KW;
int kh = tmp % KH;
tmp /= KH;
int kd = tmp % KD;
tmp /= KD;
int c = tmp % C;
int k = tmp / C;
```
So k is the slowest varying dimension, and kw is fastest. Adjacent threads (idx and idx+1) have the same k,c,kd,kh but different kw. These ARE contiguous in global memory! Because layout is [k][c][kd][kh][kw], and kw is the innermost dimension.
Excellent! So the collaborative load for weights IS coalesced.
But if we want to read 4 weights for different k from shared memory efficiently, we'd want them contiguous in shared memory. Since the global memory has k as the outermost dimension, there's a trade-off.
If we load weights in the same order as global memory `(k, c, kd, kh, kw)`, then adjacent threads in the load are coalesced. But in shared memory, different k are far apart.
If we transpose in shared memory to `(c, kd, kh, kw, k)`, we'd lose coalescing in the load unless we do a transpose during the load.
One approach: load with coalesced pattern into a temporary, then transpose. But that requires extra shared memory or complex indexing.
Alternatively, accept non-contiguous weight reads from shared memory. Since shared memory is much faster than global memory, and we're doing 4 reads per thread per iteration, bank conflicts might be an issue.
With `smem_weight[k][c][kd][kh][kw]`:
Thread computing output for k=0 reads from bank based on `k=0`.
Thread computing output for k=1 reads from bank based on `k=1`.
Wait, the shared memory address is `((k * C + c) * 27 + kd*9 + kh*3 + kw)`. For fixed c,kd,kh,kw and varying k:
- k=0: offset = c*27 + rest
- k=1: offset = 864 + c*27 + rest
- k=2: offset = 1728 + c*27 + rest
- k=3: offset = 2592 + c*27 + rest
Bank = (offset * 4) % 32 = (offset % 8) since 32 banks / 4 bytes = 8.
For k=0: bank = (c*27 + rest) % 8
For k=1: bank = (864 + c*27 + rest) % 8 = (c*27 + rest) % 8 (since 864 % 8 = 0)
For k=2: bank = (1728 % 8 = 0) → same
For k=3: same
All 4 k values map to the SAME bank! This means if a warp contains threads computing different k values, they experience bank conflicts when reading weights.
In our configuration, threads 0-63 compute k=0, threads 64-127 compute k=1, etc. So warp 0 has all threads with k=0. No bank conflicts for weight reads (broadcast). Warp 2 has all threads with k=1. Also no conflicts.
The bank conflict issue arises only if a single warp reads multiple k values simultaneously. With 64 threads per k and warp size 32, each warp has uniform k. So NO bank conflicts. This is great!
If we wanted each thread to compute multiple k values (e.g., for better register usage), then a warp would have mixed k and we'd get conflicts. But our current design avoids this.
So our shared memory layout with k as the outermost dimension is actually optimal for our thread mapping.
This confirms that TILE_K=4 with 64 threads per K is a good design.
One more check: we have 256 threads. 64 per K. That's 2 warps per K. With 4 K values, that's 8 warps total.
For output writes: warp 0 and warp 1 both write k=0 at different spatial. The L2 cache will handle coherence between warps writing to nearby addresses.
Alright, I'm confident in this design. Let me write the full Python module.
Actually, I realize I should test whether my `blockIdx` mapping is correct. With `dim3 grid(w_blocks * h_blocks * d_blocks, k_blocks, N)`, I need:
In kernel:
```cpp
int spatial_id = blockIdx.x;
int k_block = blockIdx.y;
int n = blockIdx.z;
int w_block = spatial_id % w_blocks;
int tmp = spatial_id / w_blocks;
int h_block = tmp % h_blocks;
int d_block = tmp / h_blocks;
```
For spatial_id from 0 to w_blocks*h_blocks*d_blocks - 1, this correctly maps to (w_block, h_block, d_block).
Wait, the divisor in the second step: `int tmp = spatial_id / w_blocks;` gives a value from 0 to h_blocks*d_blocks - 1.
Then `h_block = tmp % h_blocks;` gives 0 to h_blocks - 1.
And `d_block = tmp / h_blocks;` gives 0 to d_blocks - 1.
Yes, this is correct.
Let me also make sure `k_block` ranges from 0 to `k_blocks - 1`, and `k_start = k_block * TILE_K`.
Alright here's my draft code. I'll compile and test it.
Actually, let me first check if there's an issue with using `nn.Conv3d` inside `ModelNew` but not calling it in forward. When the evaluator creates `Model` and `ModelNew`, it might want them to have compatible state dicts. If `ModelNew` has `self.conv` which is an `nn.Conv3d`, its state dict will have keys like `conv.weight` and `conv.bias`. `Model` also has these keys. So state dict loading will work.
But wait, we also need `ModelNew` to work with `grad` during backward? The evaluator only tests forward, based on the problem description. So we don't need backward.
Let me write the code:
```python
import torch
import torch.nn as nn
import math
from torch.utils.cpp_extension import load_inline
cpp_source = """
#include <torch/extension.h>
#include <cuda_runtime.h>
#include <math.h>
#define KD 3
#define KH 3
#define KW 3
#define TILE_K 4
#define TILE_D 4
#define TILE_H 4
#define TILE_W 4
#define BLOCK_SIZE 256
__global__ void conv3d_mish_tanh_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int C, int D, int H, int W,
int K, int OD, int OH, int OW
) {
const int IN_TILE_D = (TILE_D - 1) + KD;
const int IN_TILE_H = (TILE_H - 1) + KH;
const int IN_TILE_W = (TILE_W - 1) + KW;
const int IN_TILE_SIZE = IN_TILE_D * IN_TILE_H * IN_TILE_W * C;
const int W_TILE_SIZE = TILE_K * C * KD * KH * KW;
extern __shared__ float smem[];
float* smem_input = smem;
float* smem_weight = smem + IN_TILE_SIZE;
int w_blocks = (OW + TILE_W - 1) / TILE_W;
int h_blocks = (OH + TILE_H - 1) / TILE_H;
int d_blocks = (OD + TILE_D - 1) / TILE_D;
int spatial_id = blockIdx.x;
int k_block = blockIdx.y;
int n = blockIdx.z;
int w_block = spatial_id % w_blocks;
int tmp = spatial_id / w_blocks;
int h_block = tmp % h_blocks;
int d_block = tmp / h_blocks;
int w_start = w_block * TILE_W;
int h_start = h_block * TILE_H;
int d_start = d_block * TILE_D;
int k_start = k_block * TILE_K;
// Load input tile
for (int idx = threadIdx.x; idx < IN_TILE_SIZE; idx += BLOCK_SIZE) {
int w = idx % IN_TILE_W;
int tmp2 = idx / IN_TILE_W;
int h = tmp2 % IN_TILE_H;
tmp2 /= IN_TILE_H;
int d = tmp2 % IN_TILE_D;
int c = tmp2 / IN_TILE_D;
int d_in = d_start + d;
int h_in = h_start + h;
int w_in = w_start + w;
float val = 0.0f;
if (d_in < D && h_in < H && w_in < W && c < C) {
val = input[(((n * C + c) * D + d_in) * H + h_in) * W + w_in];
}
smem_input[idx] = val;
}
// Load weight tile
for (int idx = threadIdx.x; idx < W_TILE_SIZE; idx += BLOCK_SIZE) {
int kw = idx % KW;
int tmp2 = idx / KW;
int kh = tmp2 % KH;
tmp2 /= KH;
int kd = tmp2 % KD;
tmp2 /= KD;
int c = tmp2 % C;
int k = tmp2 / C;
int k_out = k_start + k;
float val = 0.0f;
if (k_out < K && c < C) {
val = weight[((((k_out * C + c) * KD + kd) * KH + kh) * KW + kw)];
}
smem_weight[idx] = val;
}
__syncthreads();
int outers = TILE_K * TILE_D * TILE_H * TILE_W;
int out_tid = threadIdx.x;
if (out_tid < outers) {
int w = out_tid % TILE_W;
int tmp2 = out_tid / TILE_W;
int h = tmp2 % TILE_H;
tmp2 /= TILE_H;
int d = tmp2 % TILE_D;
int k = tmp2 / TILE_D;
int k_out = k_start + k;
int d_out = d_start + d;
int h_out = h_start + h;
int w_out = w_start + w;
if (k_out < K && d_out < OD && h_out < OH && w_out < OW) {
float sum = bias[k_out];
#pragma unroll
for (int c = 0; c < C; c++) {
#pragma unroll
for (int kd = 0; kd < KD; kd++) {
#pragma unroll
for (int kh = 0; kh < KH; kh++) {
#pragma unroll
for (int kw = 0; kw < KW; kw++) {
int in_idx = (((c * IN_TILE_D + (d + kd)) * IN_TILE_H) + (h + kh)) * IN_TILE_W + (w + kw);
int w_idx = (((k * C + c) * KD + kd) * KH + kh) * KW + kw;
sum += smem_input[in_idx] * smem_weight[w_idx];
}
}
}
}
float sp = logf(1.0f + expf(sum));
float mish = sum * tanhf(sp);
float out_val = tanhf(mish);
output[((((n * K + k_out) * OD + d_out) * OH + h_out) * OW + w_out)] = out_val;
}
}
}
torch::Tensor conv3d_mish_tanh_cuda(
torch::Tensor input,
torch::Tensor weight,
torch::Tensor bias,
int stride,
int padding
) {
int N = input.size(0);
int C = input.size(1);
int D = input.size(2);
int H = input.size(3);
int W = input.size(4);
int K = weight.size(0);
int OD = (D + 2 * padding - KD) / stride + 1;
int OH = (H + 2 * padding - KH) / stride + 1;
int OW = (W + 2 * padding - KW) / stride + 1;
auto output = torch::empty({N, K, OD, OH, OW}, input.options());
int w_blocks = (OW + TILE_W - 1) / TILE_W;
int h_blocks = (OH + TILE_H - 1) / TILE_H;
int d_blocks = (OD + TILE_D - 1) / TILE_D;
int k_blocks = (K + TILE_K - 1) / TILE_K;
dim3 grid(w_blocks * h_blocks * d_blocks, k_blocks, N);
dim3 block(BLOCK_SIZE);
int IN_TILE_SIZE = ((TILE_D - 1) * stride + KD) * ((TILE_H - 1) * stride + KH) * ((TILE_W - 1) * stride + KW) * C;
int W_TILE_SIZE = TILE_K * C * KD * KH * KW;
int shared_mem_size = (IN_TILE_SIZE + W_TILE_SIZE) * sizeof(float);
conv3d_mish_tanh_kernel<<<grid, block, shared_mem_size>>>(
input.data_ptr<float>(),
weight.data_ptr<float>(),
bias.data_ptr<float>(),
output.data_ptr<float>(),
N, C, D, H, W, K, OD, OH, OW
);
return output;
}
"""
conv3d_mish_tanh = load_inline(
name='conv3d_mish_tanh',
cpp_sources=cpp_source,
functions=['conv3d_mish_tanh_cuda'],
extra_cuda_cflags=['-O3', '--use_fast_math'],
verbose=False
)
class ModelNew(nn.Module):
def __init__(self, in_channels, out_channels, kernel_size, stride=1, padding=0):
super(ModelNew, self).__init__()
self.conv = nn.Conv3d(in_channels, out_channels, kernel_size, stride=stride, padding=padding)
def forward(self, x):
return conv3d_mish_tanh.conv3d_mish_tanh_cuda(
x, self.conv.weight, self.conv.bias,
self.conv.stride[0] if isinstance(self.conv.stride, tuple) else self.conv.stride,
self.conv.padding[0] if isinstance(self.conv.padding, tuple) else self.conv.padding
)
"""
```
Wait, there's an issue with `stride` in the kernel. I'm computing `IN_TILE_SIZE` in host code as:
```cpp
int IN_TILE_SIZE = ((TILE_D - 1) * stride + KD) * ((TILE_H - 1) * stride + KH) * ((TILE_W - 1) * stride + KW) * C;
```
But inside the kernel, I'm using:
```cpp
const int IN_TILE_D = (TILE_D - 1) + KD;
```
This assumes stride=1. If stride>1, the shared memory size allocated would be larger, but the kernel would still only use `(TILE_D - 1) + KD` elements. This would cause some threads to write out of bounds in shared memory if we use the smaller indexing, or we'd need to fix the kernel indexing.
Actually, looking at my kernel code more carefully:
```cpp
const int IN_TILE_D = (TILE_D - 1) + KD;
```
But I should compute it based on stride:
```cpp
const int IN_TILE_D = (TILE_D - 1) * stride + KD;
```
And similarly for IN_TILE_H and IN_TILE_W.
But wait, I don't pass `stride` to the kernel! The kernel signature doesn't include `stride` or `padding`. I need to add them.
Hmm, but for stride > 1, the inner loop indexing changes too:
```cpp
int in_idx = (((c * IN_TILE_D + (d * stride + kd)) * IN_TILE_H) + (h * stride + kh)) * IN_TILE_W + (w * stride + kw);
```
Wait, but inside the kernel, d, h, w are the output offsets within the tile (0 to TILE_D-1, etc.). For output at global position `d_start + d`, the corresponding input window starts at `(d_start + d) * stride - padding`. In shared memory, which starts at `d_start * stride - padding`, the offset for this output is `d * stride - padding - (d_start * stride - padding) = d * stride`... wait, no.
Actually, shared memory element at index `d` corresponds to global input at `d_start * stride - padding + d`.
For output element `d_out = d_start + d`, the `kd`-th kernel offset needs global input at `(d_start + d) * stride - padding + kd`.
In shared memory local coordinates, this is:
`((d_start + d) * stride - padding + kd) - (d_start * stride - padding) = d * stride + kd`
So the shared memory index is indeed `d * stride + kd`.
For stride=1, this is `d + kd`. For stride=2, it's `2*d + kd`.
I need to pass `stride` and `padding` to the kernel to compute this correctly. Actually, wait - the input loading already accounts for stride and padding:
```cpp
int d_in = d_start * stride - padding + d;
```
But in the compute phase, I need to know `stride` to index into shared memory. And `padding` doesn't matter for shared memory indexing because the shared memory is already aligned to the padded input.
Let me add `stride` to the kernel parameters. Padding is not needed in the compute phase if we loaded the padded values.
Also, I need to handle `padding` in the output size calculation and in the input loading.
Wait, there's an issue. PyTorch's Conv3d with padding `p` pads all sides equally. The output formula is:
`OD = floor((D + 2*p - KD) / stride) + 1`
And in the reference, `padding=0`, `stride=1`. So `OD = D - KD + 1 = 30`.
My code computes this correctly:
```cpp
int OD = (D + 2 * padding - KD) / stride + 1;
```
For loading input with padding, I do:
```cpp
int d_in = d_start + d - padding;
```
Wait, no. I wrote:
```cpp
int d_in = d_start + d;
```
This is wrong! For padding=0, `d_start` is the output start. The input needed is `d_start + d` for the top-left corner. But with padding, the actual input coordinate should be shifted.
Actually, in my current code, I didn't account for padding at all in the input loading. Let me fix it.
For the shared memory load, the first element (d=0) should correspond to the first input position needed by the block. For output `d_start`, the first input position is `d_start * stride - padding`. Then subsequent elements are `d_start * stride - padding + 1`, `d_start * stride - padding + 2`, etc.
So:
```cpp
int d_in = d_start * stride - padding + d;
```
For the compute phase:
```cpp
int in_smem_d = d * stride + kd;
```
This doesn't depend on padding because shared memory is pre-aligned.
For the `if` check in input loading:
```cpp
if (d_in >= 0 && d_in < D && h_in >= 0 && h_in < H && w_in >= 0 && w_in < W && c < C) {
val = input[...];
} else {
val = 0.0f;
}
```
This handles padding correctly.
And for the output bounds:
```cpp
if (k_out < K && d_out < OD && h_out < OH && w_out < OW) {
// compute
}
```
This is already correct.
Let me update the kernel to pass `stride` and handle padding properly.
But wait, I also need to verify that `bias` might be None if `bias=False` is set in Conv3d. In the reference, `nn.Conv3d` defaults to `bias=True`. But `ModelNew` should match. Since we're using `self.conv.bias`, it will be a tensor if bias=True, or None if bias=False.
My kernel requires bias. Let me check if `nn.Conv3d` defaults to `bias=True`. Yes, it does. And the reference doesn't pass `bias=False`. So we're safe.
But to be robust, I should handle the case where bias is None by passing a zero tensor or adding a code path. For simplicity, I'll assume bias exists.
Let me now update the kernel. I'll add `stride` parameter.
Also, I notice a potential issue with `IN_TILE_D` in the compute loop. Currently I have:
```cpp
const int IN_TILE_D = (TILE_D - 1) + KD;
```
If I want to support general stride, this should be:
```cpp
const int IN_TILE_D = (TILE_D - 1) * stride + KD;
```
But `stride` is a runtime parameter now. That's fine since `IN_TILE_D` is just a local variable, not used for array sizing (arrays are dynamically sized).
Wait, but I'm using `IN_TILE_D` in `smem_input` indexing inside the loop:
```cpp
int in_idx = (((c * IN_TILE_D + (d + kd)) * IN_TILE_H) + (h + kh)) * IN_TILE_W + (w + kw);
```
For stride > 1, this should be:
```cpp
int in_idx = (((c * IN_TILE_D + (d * stride + kd)) * IN_TILE_H) + (h * stride + kh)) * IN_TILE_W + (w * stride + kw);
```
Let me fix this.
Also, I realize that `IN_TILE_D` in the host code and kernel should match. Let me make them consistent.
Actually, instead of computing `IN_TILE_SIZE` on host and inside kernel, I'll just pass the needed sizes. But for simplicity, I can compute consistently.
Let me write the kernel one more time with all these considerations. But I want to keep it simple since only stride=1, padding=0 is used.
Actually, here's what I'll do: I'll write the kernel to be correct for general stride and padding, but I'll add a branch in the host wrapper that calls a specialized version when stride=1 and padding=0. But since we can't easily have two kernels without extra complexity, I'll just write one correct general kernel.
For the specific test case, the compiler might not be able to optimize `stride=1` as well since it's a runtime parameter. Should I template on `stride_d`, `stride_h`, `stride_w`?
For the evaluation, stride is always 1. But to keep it simple and avoid template explosion, I'll just use runtime stride. The extra multiplication `d * stride` is trivial overhead compared to the rest of the computation.
Wait, but I also need `padding` for the input bounds check. I need to make sure `d_in = d_start * stride - padding + d >= 0`.
Let me update the kernel:
```cpp
__global__ void conv3d_mish_tanh_kernel(
const float* __restrict__ input,
const float* __restrict__ weight,
const float* __restrict__ bias,
float* __restrict__ output,
int N, int C, int D, int H, int W,
int K, int OD, int OH, int OW,
int stride, int padding
) {
const int IN_TILE_D = (TILE_D - 1) * stride + KD;
const int IN_TILE_H = (TILE_H - 1) * stride + KH;
const int IN_TILE_W = (TILE_W - 1) * stride + KW;
const int IN_TILE_SIZE = IN_TILE_D * IN_TILE_H * IN_TILE_W * C;
const int W_TILE_SIZE = TILE_K * C * KD * KH * KW;
extern __shared__ float smem[];
float* smem_input = smem;
float* smem_weight = smem + IN_TILE_SIZE;
// ... block decomposition ...
// Load input
for (int idx = threadIdx.x; idx < IN_TILE_SIZE; idx += BLOCK_SIZE) {
int w = idx % IN_TILE_W;
int tmp2 = idx / IN_TILE_W;
int h = tmp2 % IN_TILE_H;
tmp2 /= IN_TILE_H;
int d = tmp2 % IN_TILE_D;
int c = tmp2 / IN_TILE_D;
int d_in = d_start * stride - padding + d;
int h_in = h_start * stride - padding + h;
int w_in = w_start * stride - padding + w;
float val = 0.0f;
if (d_in >= 0 && d_in < D && h_in >= 0 && h_in < H && w_in >= 0 && w_in < W && c < C) {
val = input[(((n * C + c) * D + d_in) * H + h_in) * W + w_in];
}
smem_input[idx] = val;
}
// ... load weights ...
__syncthreads();
// Compute
if (out_tid < outers) {
// ... decode out_tid ...
if (k_out < K && d_out < OD && h_out < OH && w_out < OW) {
float sum = bias[k_out];
#pragma unroll
for (int c = 0; c < C; c++) {
#pragma unroll
for (int kd = 0; kd < KD; kd++) {
#pragma unroll
for (int kh = 0; kh < KH; kh++) {
#pragma unroll
for (int kw = 0; kw < KW; kw++) {
int in_idx = (((c * IN_TILE_D + (d * stride + kd)) * IN_TILE_H) + (h * stride + kh)) * IN_TILE_W + (w * stride + kw);
int w_idx = (((k * C + c) * KD + kd) * KH + kh) * KW + kw;
sum += smem_input[in_idx] * smem_weight[w_idx];
}
}
}
}
// ... activation and store ...
}
}
}
```
Wait, the `d_out` should be computed as `d_start + d`, but `d_start` is the output start. Is `d_start` computed correctly?
`d_start = d_block * TILE_D;`
`d_out = d_start + d;`
Yes, and the corresponding input in global memory starts at `d_out * stride - padding`.
And the shared memory covers from `d_start * stride - padding` to `d_start * stride - padding + IN_TILE_D - 1`.
For output `d_out`, the needed input range is `[d_out * stride - padding, d_out * stride - padding + KD - 1]`.
In shared memory local index: `[d * stride, d * stride + KD - 1]`.
The max local index used is `(TILE_D - 1) * stride + KD - 1 = IN_TILE_D - 1`. This matches the allocated size. Good.
I think this is correct.
One more thing: I should make sure the grid size computation uses the updated formulas:
```cpp
int OD = (D + 2 * padding - KD) / stride + 1;
```
For negative values of `(D + 2 * padding - KD)`, integer division in C++ truncates towards zero, but PyTorch would raise an error or produce 0/negative sizes. For valid conv parameters, this should be non-negative.
Alright, now let me also consider whether I should use `math.h` or `<cmath>`. In CUDA, `tanhf`, `expf`, `logf` are from CUDA math library. Using `<cmath>` in C++ with `std::tanh` might use double precision on host. In device code, we should use `tanhf`, `expf`, `logf` directly, which are float versions.
Also, I should check if I need `ATen` or `torch/extension.h` headers. `torch::Tensor`, `torch::empty`, `torch::TensorOptions` are in `torch/extension.h`. Yes.
Let me now compile and test. I'll create the module code and use `compile_kernel`.
Actually, wait. When I call `load_inline` inside the Python module, it compiles at module load time. But since our submission is a Python file that gets `exec()`'d, the compilation will happen during `exec()`. This is fine.
But there might be compilation issues. Let me first use `compile_kernel` to check for syntax errors. However, `compile_kernel` requires a Python file that defines `ModelNew`. Let me just write the full code and try to compile it.
Wait, I should use `compile_kernel` first with a simple version to check syntax, then iterate.
Actually, looking at the tools, `compile_kernel` compiles the kernel but doesn't run correctness. Let me write the code and try compile_kernel.
One potential issue: if bias is None, `bias.data_ptr<float>()` will crash. Let me check: `nn.Conv3d` defaults to `bias=True`, so bias is always a tensor. Good.
Another potential issue: `self.conv.stride` and `self.conv.padding` might be tuples like `(1, 1, 1)`. I should extract the first element or all elements. For uniform stride/padding, `self.conv.stride[0]` works. But what if stride is an int? PyTorch internally converts it to tuple. So `self.conv.stride` should be a tuple. Same for padding.
Actually, `nn.Conv3d` accepts int or tuple for stride and padding. If int, it's internally converted to tuple of that int. So `self.conv.stride[0]` is safe.
Let me write the code and try compiling.