reasoning
This is a straightforward element-wise multiplication problem. The operation is `A.unsqueeze(1) * B`, where A is shape (N,) and B is shape (N, M). This is just broadcasting multiplication.
Since N = M = 4096, the total elements are 16,777,216. Each element is a simple multiplication.
Memory bound: 16M elements * 4 bytes * 3 (2 reads + 1 write) = ~192 MB. At 3.35 TB/s, theoretical minimum is ~57 microseconds. So we need a really efficient kernel.
A simple element-wise kernel using standard grid-stride loop should be close to optimal. But we can try to use vectorized loads/stores for better memory bandwidth.
We can use float4 (or more, like float8) loads to improve memory bandwidth. The H100 has really high bandwidth, so vectorized accesses will help.
Let's write a kernel that processes elements in chunks using float4. N = 4096, M = 4096, so total 16,777,216 elements.
Each thread can load 4 floats at a time, but A needs to be broadcasted across M. Since M is 4096, which is divisible by many powers of 2, we can vectorize.
Layout: B is (N, M) row-major. A is (N,). For row i, we need to multiply every element B[i,j] by A[i].
Since M=4096 is divisible by 4, we can load/store float4. A[i] is a scalar reused for the whole row.
Optimization approach:
- Each thread processes a chunk of 4 elements from B (float4)
- For row i, we read A[i] once
- We read B[i, j:j+4] as float4
- Compute element-wise multiplication
- Write back C[i, j:j+4] as float4
Grid organization:
- Each block handles a tile of rows and columns.
- Let's use 2D blocks. BlockDim (32, 8) = 256 threads.
- Each thread handles a number of float4 elements along M dimension.
- GridDim: along N: ceil(4096 / 8) = 512 blocks. Along M: ceil(4096 / (32 * 4)) = 32 blocks. So 512 * 32 = ~16k blocks. That's fine.
Actually we can also do a 1D grid where each thread handles one element with grid-stride. That might be simpler and still fast due to coalescing. But vectorized is likely better.
Let me think about the thread mapping.
Option 1: 2D kernel
- blockDim.x = 32, blockDim.y = 8
- threadIdx.x handles the x-dimension (M), threadIdx.y handles the y-dimension (N)
- Each thread processes 4 elements in M using float4
- So block covers 32*4 = 128 columns and 8 rows.
- Grid: (32, 512) -> 16,384 blocks. Each block 256 threads. Total threads = 4,194,304. Each thread processes 4 elements = 16,777,216 elements. Good.
But launching many blocks is fine. H100 has many SMs.
Actually a 1D grid-stride loop with float4 might also be good and simpler.
Total float4 elements = 16,777,216 / 4 = 4,194,304.
If we launch 128 blocks * 256 threads = 32,768 threads. Each thread processes 128 float4s. That's a lot of work per thread but fine. Actually we want enough parallelism but not too much overhead.
Wait, for memory bandwidth, the key is just achieving high throughput. A simple 1D kernel:
- Each thread handles a contiguous chunk of float4s.
- Thread i handles indices: i, i+total_threads, i+2*total_threads, ...
- This is fully coalesced if consecutive threads handle consecutive float4s.
Total elements: 16,777,216.
As float4: 4,194,304.
Let's say we launch 128 blocks of 256 threads = 32,768 threads. Each thread processes 128 float4s. That's good.
But wait, how do we handle A? If we use 1D indexing, the row is `idx / M` (integer division). For float4, the base element index is `idx * 4`. The row is `(idx * 4) / M`. Since M = 4096, row = (idx * 4) >> 12.
Or maybe a 2D kernel is more natural for accessing A per row. But 1D with row computation is also fine.
Let me write a 2D kernel. It's cleaner.
```cuda
__global__ void diag_mul_kernel(const float* A, const float* B, float* C, int N, int M) {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col_base = (blockIdx.x * blockDim.x + threadIdx.x) * 4;
if (row < N && col_base < M) {
float a = A[row];
float4 b = *((const float4*)(B + row * M + col_base));
float4 c;
c.x = a * b.x;
c.y = a * b.y;
c.z = a * b.z;
c.w = a * b.w;
*((float4*)(C + row * M + col_base)) = c;
}
}
```
Block size: (32, 8) for 256 threads.
Grid: ((M+127)/128, (N+7)/8) = (32, 512)
128 = 32*4.
M = 4096, so grid.x = 4096 / 128 = 32 exactly.
N = 4096, so grid.y = 4096 / 8 = 512 exactly.
No boundary checks needed for exact divisibility, but we should keep them for safety.
Wait, M = 4096, which is divisible by 4, so float4 alignment is guaranteed. B and C are allocated with M*sizeof(float) which is 16KB per row, so row starts are aligned at least to 16KB? Actually base pointer alignment: PyTorch tensors are typically aligned to at least 512 bytes for CUDA. 4096*4 = 16384 bytes per row, so row offsets are multiples of 16384, which are definitely aligned for float4 (16 bytes). So we can safely use float4.
This kernel should be very fast. Very low arithmetic intensity, purely memory bandwidth bound. Vectorized loads/stores should get close to peak bandwidth.
Let me write the full code.
Actually, we might want to consider using `__restrict__` and `__ldg` for A and B. But `__ldg` is mostly useful for read-only data that is broadcast or has locality. Here A has temporal locality within a warp (all threads in a warp read the same A[row]). But A is read once per row and reused 4 times. Actually each thread reads one A value and uses it for 4 multiplications. So we don't necessarily need `__ldg`, but it can help.
For B and C, it's just streaming, so normal loads/stores are fine.
Actually, since M=4096 and N=4096, there might be an even simpler approach: Triton or CUTLASS... but we need raw CUDA.
Another optimization: we can use half4 or something? No, the task says fp32 precision. So float4 is the way.
Actually, I should think about using `stream` or something. But PyTorch handles streams.
What if we try to use more vectorization? Like float8? CUDA doesn't have built-in float8, but we can do two float4 per thread.
Or we can transpose and use vector loads differently? No, row-major with contiguous M dimension is already optimal.
Let's try the kernel. I'll also add a fallback for when M is not divisible by 4, though for this specific problem it is. But to be safe, we can write a general kernel, or just write the vectorized one and a scalar tail. But for the benchmark, M=N=4096 so it will always hit the vectorized path.
Actually, the current kernel assumes M%4==0 and alignment, which is true for the benchmark. But we should make it general.
General approach: handle float4 for main body, scalar for tail. But for simplicity and since it will only be run with M=4096, I'll just write the vectorized kernel. Actually to be safe, let's do the tail too.
Wait, we need to be careful about `blockDim.x`. If we set blockDim.x = 32 and each thread handles 4 elements, then each block handles 128 elements. With grid.x = 32 and grid.y = 512, total float4s covered = 32*128*512 = 2,097,152 float4s. Wait, no:
Total work per block in x: blockDim.x * 4 = 128 columns.
Grid.x = ceil(M / 128) = 32.
Total columns covered = 32 * 128 = 4096. Correct.
Total rows per block: blockDim.y = 8.
Grid.y = ceil(N / 8) = 512.
Total rows covered = 512 * 8 = 4096. Correct.
Total float4s processed per block = 8 rows * 128 cols = 1024 float4s.
Total blocks = 32 * 512 = 16384.
Total float4s = 16384 * 1024 = 16,777,216 floats / 4 = 4,194,304 float4s. Correct.
Each block has 32*8 = 256 threads. Each thread handles exactly 1 float4.
So each thread does: 1 load from A (but actually from cache since same warp threads share it), 1 float4 load from B, 4 FMAs/muls, 1 float4 store to C.
Wait, A is loaded by every thread in the block. But threads in the same warp that have the same row will all load the same A value. Actually, threads with the same threadIdx.y (within a warp) have the same row. Warp has 32 threads in x. So all 32 threads in a warp have the same row! That means all 32 threads in a warp will try to load the same A[row]. This will be cached and broadcast within the warp, so it's fine (only 1 cache line fetched, or cache hit after first thread). But N=4096, each row is 4096 floats = 16KB. A whole cache line is 128 bytes = 32 floats. So A is very compact.
Actually, each warp loads A[row] redundantly, but it hits L1 cache and is broadcast. No issue.
What if we load A into shared memory? Each block processes 8 rows. We could load A[rows] into SMEM (8 floats = 32 bytes). This would guarantee the loads are coalesced and reused efficiently, but it's probably overkill. However, it might reduce the "global loads" counter and help occupancy slightly. Wait, for H100, L1 cache can easily handle this. But let's think: each block does 8 rows. It needs 8 floats from A. If we put them in shared memory:
- Thread 0 loads 8 contiguous floats? No, block handles rows `blockIdx.y*8` to `blockIdx.y*8+7`. These are contiguous in A. A block of 256 threads can load these 8 floats in one coalesced transaction easily.
- But SMEM is 32 bytes. That saves 255 redundant global loads per block. Since there are 16,384 blocks, that's about 4M redundant loads eliminated. Each redundant load goes through cache anyway, so it might be fast. But it does consume load/store bandwidth.
However, the main bandwidth is B and C. B is 64MB, C is 64MB, A is 16KB. So A is negligible. We don't need SMEM for A.
What if we use more vectorization per thread to reduce instruction overhead? Like 2 float4s per thread?
- Block size (32, 8): 256 threads.
- Each thread handles 2 float4s (8 floats) along M.
- Block covers 32*8 = 256 columns, 8 rows.
- Grid: (16, 512).
- This halves the number of blocks to 8192. Still plenty.
- Each thread does more work: 2 A loads (redundant), 2 float4 B loads, 8 muls, 2 float4 stores.
Actually, the A load is the same value, compiler should keep it in a register.
Maybe we should use a 1D kernel. It's often better because it avoids 2D grid overhead and integer division.
1D float4 kernel:
```cuda
__global__ void diag_mul_kernel(const float* A, const float* B, float* C, int N, int M) {
int total_float4 = N * M / 4;
int idx = blockIdx.x * blockDim.x + threadIdx.x;
for (int i = idx; i < total_float4; i += gridDim.x * blockDim.x) {
int elem_idx = i * 4;
int row = elem_idx / M;
int col = elem_idx % M;
float a = A[row];
float4 b = *((const float4*)(B + row * M + col));
float4 c = make_float4(a * b.x, a * b.y, a * b.z, a * b.w);
*((float4*)(C + row * M + col)) = c;
}
}
```
But `elem_idx / M` and `elem_idx % M` are expensive integer divisions. Since M=4096, we can use shifts: row = elem_idx >> 12. But if we want generality, division is needed.
However, we can compute row more efficiently:
Since i is the float4 index, and each row has M/4 float4s = 1024 float4s per row.
So `row = i / (M/4)`, `col_offset = (i % (M/4)) * 4`.
Since M/4 = 1024, division and modulo by 1024 are just shifts/ands! That's perfect.
So for this specific case where M is a power of 2 and we divide by M/4 (also power of 2), it's very fast.
General code:
`int float4_per_row = M / 4;`
`int row = i / float4_per_row;` // expensive if not power of 2
`int col4 = i % float4_per_row;`
But we can write it as:
`int row = i / (M / 4);`
`int col_base = (i % (M / 4)) * 4;`
If M is known to be divisible by 4. The compiler might not optimize `/` and `%` unless M is a compile-time constant or known to be power of 2. We can pass M_div_4 as an argument, or just trust that with M=4096 the division is fast... but actually, if M is a runtime variable, thecompiler doesn't know it's a power of 2.
However, we can compute it using a parameter `int M4 = M / 4;` passed to the kernel? Or we can just precompute on the host and pass `M_div_4`.
Actually, for the 2D kernel, we avoid all division. The 2D kernel has no division inside. `blockIdx.y * blockDim.y + threadIdx.y` is simple arithmetic. So the 2D kernel is likely faster because it has fewer integer instructions.
2D kernel each thread does:
`int row = blockIdx.y * blockDim.y + threadIdx.y;` -> mad
`int col_base = (blockIdx.x * blockDim.x + threadIdx.x) * 4;` -> mad, shift
1D kernel each thread does (inside loop):
`int i = ...;`
`int row = i / M4;` -> div (even if shift, extra op)
`int col_base = (i % M4) * 4;` -> mod, mul
And the 1D kernel has a loop. But grid-stride loops sometimes help occupancy.
For a bandwidth-bound kernel, both should be fine. 2D is simpler and no loops. Let's go with 2D, but we can also try both.
Let me think about whether we should unroll or do multiple float4s per thread.
If each thread does 1 float4, with 16384 blocks and 256 threads each = 4.2M threads. H100 has 132 SMs. So ~31 blocks per SM. Each block has very little work. Is that okay? Yes, but the block launch overhead might be slightly higher than necessary.
If we do 2 float4s per thread:
- Block size: (16, 8) = 128 threads. Each thread does 2 float4s contiguous?
- Then block covers 16*2*4 = 128 columns, 8 rows.
- Grid: (32, 512). Blocks = 16384. Threads per block = 128. Total threads = 2.1M.
- Wait, still a lot of blocks.
Actually, we can make each thread handle a whole "tile" of data. Let's say each block is (32, 8) and each thread handles 2 float4s along the M dimension.
Grid.x = M / (32 * 2 * 4) = 4096 / 256 = 16.
Grid.y = N / 8 = 512.
Total blocks = 16 * 512 = 8192.
Total threads = 8192 * 256 = 2,097,152.
Each thread does 2 float4 operations (2 loads, 2 stores, 8 muls).
This reduces block count by half but keeps the same total thread count. It adds a tiny loop (iterations=2) but reduces grid launch overhead. Probably not a big difference either way.
Actually, for an H100, launching 16k blocks is instant. The overhead is negligible. What's more important is memory coalescing and occupancy.
Occupancy: 2D kernel with (32, 8) -> 256 threads per block. Registers: each thread needs ~10-20 regs. So 256*20 = 5120 regs per block. Limit is 64K per SM. So ~12 blocks per SM. Limit is 32 blocks per SM. So occupancy is high.
Wait, we need enough warps to hide latency. 256 threads = 8 warps per block. 12 blocks = 96 warps per SM. That's plenty.
Let's just go with the simplest 2D vectorized kernel. It should be very close to peak bandwidth.
Actually, is there any reason to use `__launch_bounds__`? Not really needed here.
Also, should we check alignment before using float4? We can assert M % 4 == 0 in the host wrapper.
What about using `reinterpret_cast` vs explicit pointer cast? Let's use C-style cast or `reinterpret_cast<const float4*>`. In CUDA C, both work.
Actually, I want to make sure the compiler generates good code. Maybe mark pointers as `__restrict__` and `const` where applicable.
```cuda
__global__ void diag_mul_kernel(const float* __restrict__ A, const float* __restrict__ B, float* __restrict__ C, int N, int M) {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col_base = (blockIdx.x * blockDim.x + threadIdx.x) * 4;
if (row < N && col_base < M) {
float a = A[row];
const float4* b_ptr = reinterpret_cast<const float4*>(B + row * M + col_base);
float4 b = *b_ptr;
float4 c;
c.x = a * b.x;
c.y = a * b.y;
c.z = a * b.z;
c.w = a * b.w;
*reinterpret_cast<float4*>(C + row * M + col_base) = c;
}
}
```
Should we use `make_float4`?
`float4 c = make_float4(a * b.x, a * b.y, a * b.z, a * b.w);`
This might be cleaner, but direct assignment is fine.
One more optimization: use `__ldg` for B since it's read-only and linearly accessed. On H100, this doesn't hurt.
`float4 b = __ldg(reinterpret_cast<const float4*>(B + row * M + col_base));`
Actually, `__ldg` can help with caching behavior for read-only data. For streaming data, it's generally recommended to use regular loads to allow the L2 cache to optimize (e.g., allocate via evict_first). But `__ldg` goes through read-only cache/texture cache. On modern GPUs, regular loads are usually fine. I'll skip `__ldg`.
What about TMA? Overkill.
Another thought: can we get PyTorch to optimize this automatically? The operation `A.unsqueeze(1) * B` is a simple element-wise multiply with broadcasting. PyTorch likely dispatches to a standard element-wise kernel. A custom CUDA kernel might not be much faster if PyTorch is already using vectorized loads. But PyTorch's general-purpose element-wise kernel incurs some overhead for indexing, broadcasting, and dispatch. A custom kernel that knows the exact shapes and memory layout might be slightly faster.
Let me check actual PyTorch performance. For a simple `A[:, None] * B` with A (4096,) and B (4096, 4096), the memory traffic is:
- Read A: 4096 * 4 = 16KB
- Read B: 4096*4096*4 = 64MB
- Write C: 64MB
- Total ~128MB + 16KB.
At 3.35 TB/s, theoretical time is ~38 microseconds for 128MB. Wait, memory bandwidth is usually quoted for the highest achievable. Effective bandwidth might be lower.
Wait, actually: reads + writes = 1 read A + 1 read B + 1 write C = 128MB + 16KB. But in GPU, writes often go through write-combine or are counted differently. The theoretical minimum based on bytes transferred is 128MB / 3.35TB/s ≈ 38.2 microseconds.
But PyTorch's kernel probably has some overhead. A custom kernel should achieve similar memory bandwidth, so the speedup might be small (maybe 1.1x - 1.2x). But any speedup counts.
Wait, could we do better? Not really, it's memory bound. But we can try to fuse some operation. There are no other ops to fuse.
One trick: if we use vectorized float4, we effectively transfer the same amount of data but with fewer instructions, which can help instruction throughput and reduce overhead.
Let's code it up and test.
I need to write the full Python file with `ModelNew` class and `load_inline`.
```python
import torch
from torch.utils.cpp_extension import load_inline
cuda_source = '''
#include <torch/extension.h>
__global__ void diag_mul_kernel(const float* __restrict__ A, const float* __restrict__ B, float* __restrict__ C, int N, int M) {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col_base = (blockIdx.x * blockDim.x + threadIdx.x) * 4;
if (row < N && col_base < M) {
float a = A[row];
const float4 b = *reinterpret_cast<const float4*>(B + row * M + col_base);
float4 c;
c.x = a * b.x;
c.y = a * b.y;
c.z = a * b.z;
c.w = a * b.w;
*reinterpret_cast<float4*>(C + row * M + col_base) = c;
}
}
torch::Tensor diag_mul_forward(torch::Tensor A, torch::Tensor B) {
int N = A.size(0);
int M = B.size(1);
auto C = torch::empty_like(B);
dim3 block(32, 8);
dim3 grid((M + block.x * 4 - 1) / (block.x * 4), (N + block.y - 1) / block.y);
diag_mul_kernel<<<grid, block>>>(
A.data_ptr<float>(),
B.data_ptr<float>(),
C.data_ptr<float>(),
N,
M
);
return C;
}
'''
cpp_sources = '''
torch::Tensor diag_mul_forward(torch::Tensor A, torch::Tensor B);
'''
module = load_inline(
name='diag_mul',
cpp_sources=cpp_sources,
cuda_sources=cuda_source,
functions=['diag_mul_forward'],
extra_cuda_cflags=['-O3'],
extra_include_paths=[],
)
```
Wait, `load_inline` doesn't use `cpp_sources` argument like that. Let me check the signature.
`load_inline(name, cpp_sources, cuda_sources, functions, extra_cuda_cflags=None, extra_cflags=None, extra_include_paths=None, ...)`
Actually `cpp_sources` is the string for C++ code? Or is it `cpp_sources` and `cuda_sources`?
Looking at PyTorch docs:
```python
torch.utils.cpp_extension.load_inline(name, cpp_sources, cuda_sources, functions, extra_cflags=None, extra_cuda_cflags=None, extra_ldflags=None, extra_include_paths=None, build_directory=None, verbose=False)
```
So `cpp_sources` is the C++ host wrapper code, and `cuda_sources` is the CUDA device code. But they are concatenated together. Actually, in `load_inline`, both strings are written to temporary files and compiled. We can just put everything in `cuda_sources` if it contains CUDA code? No, typically the function definition goes in cpp_sources (or one of them), and the declaration is used for the pybind module.
Wait, the docs say: "cpp_sources – C++ sources containing the functions to bind." and "cuda_sources – CUDA sources to compile." But since CUDA files can contain host code too, we can just put everything in `cpp_sources`? Actually no, `load_inline` separates them: `.cpp` for cpp_sources and `.cu` for cuda_sources. The functions in `cpp_sources` are compiled with the C++ compiler and linked. The `cuda_sources` with nvcc.
But we want the kernel and the host wrapper to be compiled together by nvcc. Typically, for `load_inline`, we put the host wrapper in `cpp_sources` and the kernel in `cuda_sources`, but we need a way for the host wrapper to call the kernel. We can declare the kernel as `extern "C"` or just forward-declare it. Wait, actually `cpp_sources` is compiled with g++ and `cuda_sources` with nvcc, then linked. So the kernel needs to be declared as `extern` or forward-declared in the cpp file, but since g++ doesn't understand CUDA syntax like `<<<>>>`, the host wrapper that launches the kernel MUST be in the `.cu` file (i.e., `cuda_sources`).
Wait, that means we should put the host wrapper `diag_mul_forward` in `cuda_sources`, and put only a C++-compatible declaration in `cpp_sources`? Or actually, `cpp_sources` is where the pybind module looks for functions. The auto-generated pybind code just calls the functions. If `cpp_sources` contains only declarations and `cuda_sources` contains the definitions, the linker should resolve them... but the `.cpp` file won't compile if it sees declarations of functions returning `torch::Tensor`? Actually declarations are fine in C++.
Let me think about how `load_inline` works. It auto-generates `torch/extension.h` includes and a pybind module. The pybind module calls the functions. The functions should be declared in `cpp_sources` or `cuda_sources`? Actually, the pybind module is generated based on `functions` and is compiled with the `cpp_sources` and `cuda_sources` linked together. If the function definition is in `cuda_sources`, and there's no declaration in `cpp_sources`, will the pybind-generated code find it? The pybind generated code is usually compiled as part of the cpp extension, so as long as the function is declared or defined in a header included by the pybind code... Actually, `load_inline` might just generate:
```cpp
#include <pybind11/pybind11.h>
#include <torch/extension.h>
// forward declarations from cpp_sources/cuda_sources?
pybind11::def("diag_mul_forward", ...);
```
Wait, no. `load_inline` generates the entire module code? Let me recall.
I think `load_inline` works by:
1. Writing `cpp_sources` to a `.cpp` file and `cuda_sources` to a `.cu` file.
2. Adding `#include <torch/extension.h>` and other stuff.
3. Auto-generating a pybind module that imports the functions listed in `functions`.
The auto-generated pybind code probably looks like:
```cpp
#include <pybind11/pybind11.h>
// forward declare or include declarations
PYBIND11_MODULE(...) {
m.def("diag_mul_forward", &diag_mul_forward, ...);
}
```
For this to compile, `diag_mul_forward` must be declared before the `PYBIND11_MODULE`. This declaration can come from `cpp_sources`. But if we define it in `cuda_sources` and only declare it in `cpp_sources`, that should work.
However, a simpler approach is to put the declaration AND definition all in `cuda_sources`, and just have `cpp_sources` empty or minimal? Wait, the docs say `cpp_sources` is a string. Can we pass an empty string or just the declaration?
Actually, looking at common usage:
```python
module = load_inline(
name='my_module',
cpp_sources='',
cuda_sources=cuda_code,
functions=['my_function'],
extra_cuda_cflags=['-O3'],
)
```
Does this work? I think yes, because `load_inline` might extract the necessary signatures or just generate the pybind code and rely on the fact that the `.cu` file is compiled with nvcc (which can compile host code too) and linked. The pybind generated code might be in its own `.cpp` file and it forward-declares the functions? Or maybe it includes the `cpp_sources` string, so if it's empty, the declarations must be in `cuda_sources`, but the pybind generated code doesn't see them because it's in a separate compilation unit...
Actually, in `load_inline`, the pybind11 module code is usually appended to `cpp_sources`. So if the function is defined in `cuda_sources`, there's a link-time dependency. The `cpp_sources` (including the generated pybind code) just need a declaration of the function. But `load_inline` doesn't automatically add declarations. However, since nvcc compiles `.cu` files and gcc compiles `.cpp` files, and they are linked together, the function must be declared in the `.cpp` file.
Common practice is to put the host wrapper in `cpp_sources` and the kernel declaration in `cuda_sources`? No, you can't put the host wrapper in `cpp_sources` because it contains `<<< >>>`. So the host wrapper must be in `cuda_sources`.
Ah, but you CAN have a declaration of the CUDA kernel in `cpp_sources`? No, g++ can't parse `__global__`.
Solution: Put the host wrapper function in `cuda_sources`, and also put a C++ declaration of it (just the signature) in `cpp_sources`.
Wait, but we can also just put everything in `cuda_sources` and make `cpp_sources` contain:
```cpp
torch::Tensor diag_mul_forward(torch::Tensor A, torch::Tensor B);
```
I think that works. Let me check if `load_inline` adds the generated pybind code to `cpp_sources` or to a separate file. It probably writes:
```cpp
// contents of cpp_sources string
#include <pybind11/pybind11.h>
...
PYBIND11_MODULE(name, m) {
m.def("diag_mul_forward", &diag_mul_forward, ...);
}
```
Wait, if `cpp_sources` contains the declaration, then the generated pybind code can find `diag_mul_forward` because it links with the `.cu` object which defines it. The declaration in `cpp_sources` is enough. That's the standard pattern.
So:
```python
cpp_sources = 'torch::Tensor diag_mul_forward(torch::Tensor A, torch::Tensor B);'
```
Wait, but `cpp_sources` might need `#include <torch/extension.h>`? `load_inline` might add it automatically. Actually, it usually does add necessary includes. But to be safe, we can include it.
Actually, the docs say `load_inline` automatically adds `#include <torch/extension.h>`. So we probably don't need includes in `cpp_sources`. But adding them doesn't hurt.
Let me just write it and test compile.
Another approach: some people put the entire code in `cuda_sources` and leave `cpp_sources` empty. Does that work? Only if `load_inline` also parses `cuda_sources` for the function signatures to generate pybind code, or if it treats `cuda_sources` similarly. I'm not sure.
Actually, I just remembered: `load_inline` generates the pybind module and appends it to the `cpp_sources`. The `cuda_sources` is kept separate. The compiler for `cpp_sources` is the C++ compiler. So if we define `diag_mul_forward` only in `cuda_sources`, the `cpp_sources` must have a declaration.
OK let's write the full script and compile it.
Wait, one potential issue: we need to make sure the input tensors are contiguous and on CUDA. The reference implementation `A.unsqueeze(1) * B` handles broadcasting and non-contiguous if needed. But since the benchmark provides standard inputs, we can assume they are contiguous and on the same device.
Actually, to be safe, we should check device and make sure inputs are contiguous. But the benchmark probably provides contiguous CPU or CUDA inputs. `get_inputs` in the problem doesn't specify `.cuda()`, but likely they are moved to CUDA before calling. The reference uses PyTorch ops which handle device automatically. Our custom kernel should probably ensure tensors are contiguous and on CUDA.
Let's add `.contiguous()` and check device. Or simpler, just assume. For safety:
```python
A = A.contiguous().cuda()
B = B.contiguous().cuda()
```
But if A and B are already on CPU, `.cuda()` will move them. The benchmark might time the `.cuda()` calls too. Hmm. Better not to move unless necessary. Actually, the `forward` signature just receives tensors. The benchmark probably provides CUDA tensors or moves them before timing. Let's just call `.contiguous()`.
Wait, if we use `A.data_ptr<float>()`, we need `A` to be contiguous? Not necessarily if we access by row, but since it's 1D, any layout is contiguous effectively. For B, we assume it's contiguous in memory with shape (N, M). If it's not contiguous, our linear indexing `row * M + col_base` is wrong.
So to be safe:
```python
B = B.contiguous()
```
And we should probably create the output on the same device as B.
```python
C = torch.empty_like(B)
```
`empty_like` preserves device.
That's fine.
Another concern: should we use `at::cuda::CUDAStreamGuard` or something? PyTorch sets the current stream for the kernel. When we launch `<<<>>>`, it goes to the default stream or current stream. Since we're in PyTorch, we should use the current CUDA stream. We can get it via `at::cuda::getCurrentCUDAStream()`.
So the host wrapper should be:
```cpp
torch::Tensor diag_mul_forward(torch::Tensor A, torch::Tensor B) {
TORCH_CHECK(A.is_cuda(), "A must be a CUDA tensor");
TORCH_CHECK(B.is_cuda(), "B must be a CUDA tensor");
TORCH_CHECK(A.dim() == 1, "A must be 1D");
TORCH_CHECK(B.dim() == 2, "B must be 2D");
int N = A.size(0);
int M = B.size(1);
TORCH_CHECK(B.size(0) == N, "B must have size N in first dimension");
TORCH_CHECK(M % 4 == 0, "M must be divisible by 4");
auto C = torch::empty_like(B);
dim3 block(32, 8);
dim3 grid((M + block.x * 4 - 1) / (block.x * 4), (N + block.y - 1) / block.y);
auto stream = at::cuda::getCurrentCUDAStream();
diag_mul_kernel<<<grid, block, 0, stream.stream()>>>(
A.data_ptr<float>(),
B.data_ptr<float>(),
C.data_ptr<float>(),
N,
M
);
return C;
}
```
Wait, `stream.stream()` returns the raw `cudaStream_t`. Or we can use `c10::cuda::CUDAStream stream = c10::cuda::getCurrentCUDAStream();` and pass `stream.stream()`. Actually `at::cuda::getCurrentCUDAStream()` returns a `CUDAStream` which has `.stream()`.
Let me check: in `torch/extension.h`, `at::cuda::getCurrentCUDAStream()` returns `at::cuda::CUDAStream`? Or `c10::cuda::CUDAStream`? Actually `at::cuda` is usually an alias. The method is `stream()`. But we can also just use:
```cpp
diag_mul_kernel<<<grid, block, 0, at::cuda::getCurrentCUDAStream().stream()>>>(...);
```
Or we can just omit the stream argument and it uses the default stream... but that might cause synchronization issues if PyTorch uses non-default streams. Better to use the current stream.
Actually, the simplest and most reliable is:
```cpp
at::cuda::CUDAStream stream = at::cuda::getCurrentCUDAStream();
diag_mul_kernel<<<grid, block, 0, stream>>>(...);
```
Wait, `stream` is an object. Can we pass it directly? I think `CUDAStream` has an implicit conversion or `stream()` method. Let me just use `.stream()`.
`at::cuda::getCurrentCUDAStream().stream()` should work.
Actually, I'm not 100% sure about the API. Let me think of a safer alternative. In many PyTorch inline CUDA extensions, people just use:
```cpp
cudaStream_t stream = at::cuda::getCurrentCUDAStream().stream();
```
Or they just launch without a stream, which uses the legacy default stream. But since PyTorch may set the current stream, and the kernel might need to be in that stream...
If we look at `aten/src/ATen/cuda/CUDAContext.h`, `at::cuda::getCurrentCUDAStream()` returns `CUDAStream` which is implicitly convertible to `cudaStream_t`? Actually, in old versions it was convertible. In newer versions, there's `.stream()`. I think `.stream()` is safe.
Alternatively, we can use `c10::cuda::getCurrentCUDAStream()` which returns a `c10::cuda::CUDAStream`, and it has a `stream()` method.
To be safe across versions, maybe just skip specifying the stream. The `<<< >>>` syntax without stream uses the per-thread default stream in modern CUDA if compiled with `--default-stream per-thread`, which is what PyTorch does? Actually, PyTorch's extension build might use the default stream settings. Hmm.
Actually, looking at PyTorch's own code, they almost always pass the stream explicitly. Let me check a common pattern in tutorials:
```cpp
void my_kernel_launch(...) {
my_kernel<<<blocks, threads>>>(...);
}
```
Many tutorials do NOT pass a stream, and it works fine because the kernel goes to the current stream automatically? No, `<<< >>>` without stream argument goes to stream 0 (legacy default stream), which synchronizes with all other streams. This can hurt performance if PyTorch uses stream parallelism, but for a single op it usually doesn't matter. However, it's not "correct" in the sense that it might introduce unwanted synchronization.
But since this is a simple op, let's pass the stream. I'll use `c10::cuda::getCurrentCUDAStream().stream()`.
```cpp
#include <c10/cuda/CUDAStream.h>
```
`torch/extension.h` already includes everything.
Actually, `at::cuda::getCurrentCUDAStream()` returns an `at::cuda::CUDAStream` (or `CUDAStreamGuard`). And `.stream()` gives `cudaStream_t`.
Let me just write: `at::cuda::getCurrentCUDAStream().stream()`.
One more thing: do we need to handle non-contiguous B? For simplicity, I can call `B = B.contiguous();` in the host wrapper. That would allocate a new contiguous tensor if needed. But `B.contiguous()` returns a new tensor. I should do:
```cpp
auto B_contig = B.contiguous();
auto C = torch::empty_like(B_contig);
```
But wait, `empty_like(B)` where B is non-contiguous creates a non-contiguous output? No, `empty_like` creates a tensor with the same size and strides? Actually `torch.empty_like(tensor)` creates a tensor with the same size and device, but strided like the input? Wait, no: `empty_like(x)` returns a new tensor with the same size, dtype, device, and memory layout (strides) as `x`. But we want C to be contiguous so we can write with float4.
If B is non-contiguous, we cannot easily use our simple indexing. We should handle it or make it contiguous.
Actually, the reference op `A.unsqueeze(1) * B` handles broadcasting and arbitrary strides. But for the benchmark inputs, B is standard random tensor, so it's contiguous. We can just add a check/convert.
In the host wrapper:
```cpp
auto B_contig = B.contiguous();
auto A_contig = A.contiguous();
auto C = torch::empty_like(B_contig);
```
This is safe.
What if A is shape (N, 1) but we expect (N,)? The reference uses unsqueeze. Our kernel expects 1D A. The benchmark gives 1D A, so fine.
Now, about block size. (16, 16) gives 256 threads. (32, 8) gives 256. (64, 4) gives 256. Which is best?
For memory coalescing, we want consecutive thread IDs to access consecutive memory. In a 2D block, threads are ordered with x varying fastest. Thread (x, y) has linear ID = y * blockDim.x + x.
Our memory access is along M (columns). Thread `threadIdx.x` handles different columns. Since columns are contiguous, as `threadIdx.x` increases, the column increases. So `threadIdx.x` varying fastest means consecutive threads access consecutive float4s. That's perfect coalescing for any blockDim.x.
So `blockDim.x` should be at least 32 (warp size). Let's use 32.
`blockDim.y` can be 8 for 256 threads total.
Could we use 16 x 16? Thread order: (0,0), (1,0), ..., (15,0), (0,1), ..., (15,15).
So threads 0-15 access columns 0-3, 16-31 access columns 4-7, etc.
Wait, with blockDim.x=16, each thread handles 4 columns. So thread (0,0) handles cols 0-3, thread (1,0) handles cols 4-7, ..., thread (15,0) handles cols 60-63. Thread (0,1) handles cols 0-3 in the next row.
Consecutive threads (in linear ID) handle consecutive float4s. So coalescing is still perfect.
But using 32 in x means 32*4 = 128 columns per row per block. With 8 in y, 8 rows per block.
Using 16 in x means 16*4 = 64 columns per row per block. With 16 in y, 16 rows per block.
Total elements per block: same (128*8 = 64*16 = 1024 floats? No, wait: 128 cols * 8 rows * 4 floats/col = 4096 floats. 64 cols * 16 rows * 4 = 4096 floats). Same amount of data.
The memory access pattern for A is one value per row. With blockDim.y=16, a warp contains 16 threads from different rows? Actually, warp size is 32. With block (16, 16):
Linear IDs 0-15 are row 0, columns 0-15*4.
Linear IDs 16-31 are row 1, columns 0-15*4.
So a warp covers 2 rows (or 1 row if blockDim.x >= 32, but here it's 16, so 2 rows).
With block (32, 8):
Linear IDs 0-31 are row 0, columns 0-31*4.
Linear IDs 32-63 are row 1, columns 0-31*4.
A warp covers 1 row.
Loading A: with block (32,8), each warp accesses 1 A element (all 32 threads need the same A[row]). This is broadcast within the warp, ideal.
With block (16,16), each warp accesses 2 A elements (for 2 rows). Also fine.
I think (32, 8) is a very standard choice. Let's go with that.
Now, let's also consider using `__launch_bounds__(256)` for the kernel. It tells the compiler the max threads per block, allowing better register allocation. Not strictly necessary but might help.
```cuda
__global__ void __launch_bounds__(256) diag_mul_kernel(...)
```
One more optimization: use half2 or something? No, fp32 required.
Actually, what if we use loads of 8 or 16 bytes? float4 is 16 bytes. H100 supports 128-bit loads. That's good.
Could we do 2 float4s per thread to increase ILP?
```cuda
__global__ void diag_mul_kernel(...) {
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col_base0 = (blockIdx.x * blockDim.x + threadIdx.x) * 8;
if (row < N && col_base0 < M) {
float a = A[row];
const float4 b0 = *reinterpret_cast<const float4*>(B + row*M + col_base0);
const float4 b1 = *reinterpret_cast<const float4*>(B + row*M + col_base0 + 4);
float4 c0 = make_float4(a*b0.x, a*b0.y, a*b0.z, a*b0.w);
float4 c1 = make_float4(a*b1.x, a*b1.y, a*b1.z, a*b1.w);
*reinterpret_cast<float4*>(C + row*M + col_base0) = c0;
*reinterpret_cast<float4*>(C + row*M + col_base0 + 4) = c1;
}
}
```
Grid: (16, 512) blocks.
This halves the number of blocks but doubles work per thread. Since memory is streaming, ILP might help if the GPU can issue multiple loads in flight. However, warps already have lots of threads to hide latency. For pure bandwidth, it probably doesn't matter much.
I think 1 float4 per thread is already good. Let's start with that.
Let's write the full Python module and compile.
Wait, the problem says "Do not write your own `PYBIND11_MODULE(...)` block". `load_inline` handles it. We just need to specify `functions=['diag_mul_forward']`.
Actually, I want to double check: `load_inline`'s `functions` argument is a list of function names to expose. The signatures are automatically inferred? No, `load_inline` auto-generates the bindings using some heuristics or template magic. Wait, actually in PyTorch, `load_inline` might not auto-infer signatures in the latest versions. Let me check.
In many examples, `functions` is indeed a list of strings, and `load_inline` generates a pybind module with those functions. This has worked in PyTorch for many versions.
OK, let me write the code.