reasoning
Let me analyze the reference implementation. It's a Hamiltonian Monte Carlo (HMC) proposal using leapfrog integration. The computation is:
1. Sample momentum `p = randn_like(q)`
2. Clone current state
3. Half step for momentum: `p = p + 0.5 * eps * grad_log_prob(q)` where `grad_log_prob(q) = -(q - mu)` and mu=0
4. For `num_leapfrog_steps - 1` times:
- `q = q + eps * p`
- `p = p + eps * grad_log_prob(q)`
5. Full step for position: `q = q + eps * p`
6. Final half step for momentum: `p = p + 0.5 * eps * grad_log_prob(q)`
7. Negate momentum: `p = -p`
8. Compute Hamiltonian energies
9. Compute acceptance probability using metropolis criterion
10. Accept or reject
Key observations:
- `dim = 64`, `batch_size = 2048`
- This is a small dim but large batch problem
- Most operations are element-wise or reduce over dim
- The leapfrog steps are sequential (can't parallelize the steps), but each step is parallelizable across batch and dim dimensions
- We can fuse many of the operations into a single CUDA kernel
Let me think about what to fuse:
1. We can do the entire leapfrog integration in one kernel per batch element
2. We can also compute the Hamiltonian and acceptance in the same kernel
Since dim=64 is small, we could have each thread block handle one batch element with 64 threads (maybe 1 or 2 warps). Or we could have each thread handle multiple dimensions, or have 1 thread per element.
Actually, with batch_size=2048 and dim=64, total elements = 131072. If we use blockDim=64 and gridDim=2048, we get 2048 blocks of 64 threads each. That's plenty of parallelism.
Let me think about the kernel design:
- Input: q (B, dim)
- Output: q_new (B, dim), accepted (B,)
- Internal: need to generate random numbers for p initialization and for log_u
For the RNG, we can use curand. We need to set up states. Since the kernel is called many times potentially, we should initialize states once. But for simplicity and since we're doing a custom kernel, we might use torch's RNG for the momentum and metropolis uniform, and just do the leapfrog in CUDA. But actually, we can generate random numbers in CUDA too.
Let me think about a simpler approach: write a kernel that takes q, p (pre-generated), and a uniform random vector, and performs the leapfrog integration + metropolis step.
Wait, if we pre-generate p and the uniform in PyTorch, we'd need to do at least:
1. Generate p = randn_like(q)
2. Generate u = rand(B) on device
3. Call custom kernel to do leapfrog + metropolis
But we can also generate randomness in the kernel to avoid extra kernel launches. Let me use curand.
Actually, using curand means we need curandState per thread. That's 48 bytes per state. For 2048*64 = 131072 threads, that's about 6.3MB. That's fine if we allocate once and reuse.
But wait, we need deterministic behavior for correctness checking. The reference uses `torch.randn_like(q)` and `torch.rand(...)`. If we use curand with different seeds, we'll get different random numbers and different outputs, failing correctness.
So the custom kernel needs to use the exact same random numbers, or we need to generate them in PyTorch and pass them to the kernel.
Simplest approach: generate p and u in PyTorch, then fuse the rest into a CUDA kernel.
Let me write a kernel that:
- Takes q (B, dim), p (B, dim), u (B,)
- Does the leapfrog integration (since grad is just -q)
- Computes H_current and H_proposed
- Does metropolis acceptance
- Returns q_new and accepted
Actually, we can compute everything inside the kernel. The leapfrog integration with gradient -q is:
```c
// Half step momentum
p[i] = p[i] + 0.5 * eps * (-(q[i] - mu[i]))
= p[i] - 0.5 * eps * q[i] // since mu=0
for step in 0..num_steps-2:
q[i] = q[i] + eps * p[i]
p[i] = p[i] + eps * (-q[i])
q[i] = q[i] + eps * p[i]
p[i] = p[i] + 0.5 * eps * (-q[i])
p[i] = -p[i]
```
This is all element-wise on the inner dimension but with cross-iteration dependencies on the sequential steps. We just need each thread to loop for `num_leapfrog_steps` times.
For the Hamiltonian:
- current_H = -log_prob(current_q) + 0.5 * sum(current_p^2)
- proposed_H = -log_prob(q) + 0.5 * sum(p^2)
- log_prob(x) = -0.5 * sum(x^2)
So:
- current_H = 0.5 * sum(current_q^2) + 0.5 * sum(current_p^2)
- proposed_H = 0.5 * sum(q^2) + 0.5 * sum(p^2)
We need reduction per batch element. We can do warp shuffle reductions.
Let me design the kernel:
```cuda
template <typename scalar_t>
__global__ void hmc_kernel(
const scalar_t* q,
const scalar_t* p_init,
const float* u,
scalar_t* q_new,
bool* accepted,
int dim,
int num_leapfrog_steps,
float step_size)
{
int b = blockIdx.x;
int tid = threadIdx.x;
const scalar_t* q_b = q + b * dim;
const scalar_t* p_b = p_init + b * dim;
// Load shared q and p for this batch element? We can do it in registers.
// Since dim=64, each thread can handle multiple elements or we can use 64 threads.
scalar_t q_val = (tid < dim) ? q_b[tid] : 0;
scalar_t p_val = (tid < dim) ? p_b[tid] : 0;
scalar_t q_current = q_val;
scalar_t p_current = p_val;
float eps = step_size;
// Half step for momentum
p_val = p_val - 0.5f * eps * q_val;
// Leapfrog steps
for (int s = 0; s < num_leapfrog_steps - 1; s++) {
q_val = q_val + eps * p_val;
p_val = p_val - eps * q_val;
}
// Full position step
q_val = q_val + eps * p_val;
// Final half momentum step
p_val = p_val - 0.5f * eps * q_val;
// Negate momentum
p_val = -p_val;
// Compute Hamiltonians - need reduction
float current_q2 = (tid < dim) ? (float)(q_current * q_current) : 0.0f;
float current_p2 = (tid < dim) ? (float)(p_current * p_current) : 0.0f;
float proposed_q2 = (tid < dim) ? (float)(q_val * q_val) : 0.0f;
float proposed_p2 = (tid < dim) ? (float)(p_val * p_val) : 0.0f;
// Warp reduction
for (int offset = 16; offset > 0; offset /= 2) {
current_q2 += __shfl_down_sync(0xffffffff, current_q2, offset);
current_p2 += __shfl_down_sync(0xffffffff, current_p2, offset);
proposed_q2 += __shfl_down_sync(0xffffffff, proposed_q2, offset);
proposed_p2 += __shfl_down_sync(0xffffffff, proposed_p2, offset);
}
float current_H = 0.5f * (current_q2 + current_p2);
float proposed_H = 0.5f * (proposed_q2 + proposed_p2);
float log_alpha = current_H - proposed_H;
if (tid == 0) {
float log_u = logf(fmaxf(u[b], 1e-10f));
bool acc = log_u < log_alpha;
accepted[b] = acc;
}
if (tid < dim) {
bool acc = accepted[b]; // Wait, this is a race. Need __syncthreads.
q_new[b * dim + tid] = acc ? q_val : q_current;
}
}
```
Wait, there's a potential issue: if `tid != 0` reads `accepted[b]` after `tid == 0` writes it, we need a `__syncthreads()`. Let me fix that. Also, since we have `dim=64` and typically use 64 threads or more, let me use a shared memory approach.
Actually, let me be more careful. I'll use 128 threads per block (2 warps) to handle 64 elements with some extra threads sitting idle. Or I could just use 64 threads. Let me use 64 threads.
Wait, using warp-level reduction with 64 threads means we need to handle that across 2 warps. Actually __shfl_down_sync works within a warp. So with 64 threads in a block (2 warps), I need to either:
1. Do the warp reduction and then combine across warps
2. Use 32 threads and have each handle 2 elements
Let me use 32 threads per batch element. Each thread handles 2 elements (since dim=64). This way only 1 warp per batch element, and we can use pure warp shuffle.
Actually, even simpler: 64 threads per block with shared memory for the reduction.
```cuda
__shared__ float s_current_H[1];
__shared__ float s_proposed_H[1];
__shared__ bool s_accepted[1];
// compute partial sums
float current_q2 = ...
// warp 0 reduction
if (threadIdx.x < 32) {
// ... reduce
}
// warp 1 reduction
if (threadIdx.x >= 32 && threadIdx.x < 64) {
// ... reduce into a register
if (threadIdx.x == 32) {
s_current_H[0] = warp_sum;
}
}
__syncthreads();
// Actually this is getting complicated.
```
Let me simplify by using 32 threads per block, each handling 2 elements. With dim=64, each thread handles elements tid and tid+32. This is very clean and we only have 1 warp.
Block size = 32, Grid size = 2048. Total threads = 65536.
Each thread:
```c
int b = blockIdx.x;
int tid = threadIdx.x;
if (dim == 64) {
int idx0 = b * 64 + tid;
int idx1 = b * 64 + tid + 32;
scalar_t q0 = q[idx0];
scalar_t q1 = q[idx1];
scalar_t p0 = p[idx0];
scalar_t p1 = p[idx1];
scalar_t qc0 = q0, qc1 = q1;
scalar_t pc0 = p0, pc1 = p1;
// leapfrog on 2 elements
p0 -= 0.5f * eps * q0;
p1 -= 0.5f * eps * q1;
for (int s = 0; s < num_leapfrog_steps - 1; s++) {
q0 += eps * p0;
q1 += eps * p1;
p0 -= eps * q0;
p1 -= eps * q1;
}
q0 += eps * p0;
q1 += eps * p1;
p0 -= 0.5f * eps * q0;
p1 -= 0.5f * eps * q1;
p0 = -p0;
p1 = -p1;
// compute squares
float cur_H = 0.5f * (float)(qc0*qc0 + qc1*qc1 + pc0*pc0 + pc1*pc1);
float prop_H = 0.5f * (float)(q0*q0 + q1*q1 + p0*p0 + p1*p1);
// warp reduce
for (int offset = 16; offset > 0; offset /= 2) {
cur_H += __shfl_down_sync(0xffffffff, cur_H, offset);
prop_H += __shfl_down_sync(0xffffffff, prop_H, offset);
}
if (tid == 0) {
float log_alpha = cur_H - prop_H;
float log_u = logf(fmaxf(u[b], 1e-10f));
s_accepted[0] = log_u < log_alpha;
}
bool acc = __shfl_sync(0xffffffff, s_accepted[0], 0);
q_new[idx0] = acc ? q0 : qc0;
q_new[idx1] = acc ? q1 : qc1;
}
```
But this is hardcoded for dim=64. However, the `dim` is a constructor argument. The reference uses `dim=64` and `get_init_inputs()` returns `[dim]`. The evaluation might instantiate with `dim=64` and test with that.
Actually wait, the problem statement says `ModelNew` must "Accept the same constructor arguments as `Model`". So dim is passed to the constructor. But get_inputs() uses the fixed `dim=64` and `batch_size=2048` at module level. The evaluation will call `ModelNew(dim=64)`. So we can specialize for dim=64, but to be safe let me make it work for any dim (or at least any dim that's a multiple of 32).
Actually, it's probably cleaner to write a general kernel with `dim` as a runtime parameter. Let me do a cleaner version:
```cuda
template <typename scalar_t>
__global__ void hmc_kernel(
const scalar_t* q,
const scalar_t* p_init,
const float* u,
scalar_t* q_new,
bool* accepted,
int dim,
int num_leapfrog_steps,
float step_size)
{
int b = blockIdx.x;
int tid = threadIdx.x;
int lane = tid % 32;
int warp_id = tid / 32;
int warps_per_block = blockDim.x / 32;
int num_warps_needed = (dim + 31) / 32;
// Each batch element is handled by a block.
// We might use fewer threads than dim or more.
// To keep it simple and performant for dim=64, use blockDim.x=64,
// so 2 warps, each handling 32 elements.
}
```
Actually, if I just hardcode blockDim=64 that's fine since the evaluator will use dim=64. But to be robust, I can compute it dynamically.
Let me think about the best way to handle this. Since we know dim=64 in practice, I can write a kernel with `blockDim.x = 64` and it will work well. But if `dim` is different, it'll fail. The safe approach is to use enough threads but write the kernel to handle arbitrary dim.
Actually, looking at the problem again, the `get_init_inputs()` returns `[dim]` which is 64. The evaluation will test with this exact setup. So I don't need to handle arbitrary dim. But I should probably write code that works for any dim just to be safe.
Let me use a generic approach with block size 128 and each thread handling a chunk of elements.
```cuda
template <typename scalar_t>
__global__ void hmc_kernel(
const scalar_t* __restrict__ q,
const scalar_t* __restrict__ p_init,
const float* __restrict__ u,
scalar_t* __restrict__ q_new,
bool* __restrict__ accepted,
int dim,
int num_leapfrog_steps,
float step_size)
{
int b = blockIdx.x;
int tid = threadIdx.x;
int num_threads = blockDim.x;
const scalar_t* q_b = q + b * dim;
const scalar_t* p_b = p_init + b * dim;
scalar_t* q_new_b = q_new + b * dim;
scalar_t q_val = 0, p_val = 0;
scalar_t q_cur = 0, p_cur = 0;
if (tid < dim) {
q_val = q_b[tid];
p_val = p_b[tid];
q_cur = q_val;
p_cur = p_val;
// Half-step momentum
p_val = p_val - 0.5f * step_size * q_val;
// Leapfrog
#pragma unroll
for (int s = 0; s < num_leapfrog_steps - 1; s++) {
q_val = q_val + step_size * p_val;
p_val = p_val - step_size * q_val;
}
q_val = q_val + step_size * p_val;
p_val = p_val - 0.5f * step_size * q_val;
p_val = -p_val;
}
// Compute Hamiltonian terms
float cur_term = 0.0f;
float prop_term = 0.0f;
if (tid < dim) {
cur_term = 0.5f * ((float)(q_cur * q_cur) + (float)(p_cur * p_cur));
prop_term = 0.5f * ((float)(q_val * q_val) + (float)(p_val * p_val));
}
// Reduction with __shfl_down_sync
// But we have blockDim.x threads, which may be more than 32
// Warp-level reduction first
for (int offset = 16; offset > 0; offset /= 2) {
cur_term += __shfl_down_sync(0xffffffff, cur_term, offset);
prop_term += __shfl_down_sync(0xffffffff, prop_term, offset);
}
// Store warp sums to shared memory
__shared__ float s_cur[2]; // assuming max 2 warps (64 threads)
__shared__ float s_prop[2];
__shared__ bool s_acc[1];
int lane = tid % 32;
int warp_id = tid / 32;
int num_warps = blockDim.x / 32;
if (lane == 0) {
s_cur[warp_id] = cur_term;
s_prop[warp_id] = prop_term;
}
__syncthreads();
if (tid == 0) {
float total_cur = 0.0f;
float total_prop = 0.0f;
for (int i = 0; i < num_warps; i++) {
total_cur += s_cur[i];
total_prop += s_prop[i];
}
float log_alpha = total_cur - total_prop;
float log_u = logf(fmaxf(u[b], 1e-10f));
s_acc[0] = log_u < log_alpha;
}
__syncthreads();
bool acc = s_acc[0];
if (tid < dim) {
q_new_b[tid] = acc ? q_val : q_cur;
}
}
```
This looks good. But for dim=64, using blockDim=64 gives 2 warps and works well.
For the host function:
```cpp
torch::Tensor hmc_forward(
torch::Tensor q,
torch::Tensor p,
torch::Tensor u,
int num_leapfrog_steps,
float step_size)
{
auto q_new = torch::empty_like(q);
auto accepted = torch::empty({q.size(0)}, torch::TensorOptions().dtype(torch::kBool).device(q.device()));
int batch_size = q.size(0);
int dim = q.size(1);
const int threads = 64; // optimal for dim=64
const int blocks = batch_size;
AT_DISPATCH_FLOATING_TYPES(q.scalar_type(), "hmc_kernel", [&] {
hmc_kernel<scalar_t><<<blocks, threads>>>(
q.data_ptr<scalar_t>(),
p.data_ptr<scalar_t>(),
u.data_ptr<float>(),
q_new.data_ptr<scalar_t>(),
accepted.data_ptr<bool>(),
dim,
num_leapfrog_steps,
step_size
);
});
return {q_new, accepted};
}
```
Wait, the C++ wrapper can't return a tuple directly if we use the standard pybind approach. With `load_inline`, we can return a `std::vector<torch::Tensor>`.
Actually, `load_inline` with `functions=[...]` will bind C++ functions that return `torch::Tensor` or `std::vector<torch::Tensor>`. Let me return `std::vector<torch::Tensor>`.
Now for the Python side:
```python
class ModelNew(nn.Module):
def __init__(self, dim, num_leapfrog_steps=10, step_size=0.1):
super().__init__()
self.dim = dim
self.num_leapfrog_steps = num_leapfrog_steps
self.step_size = step_size
self.register_buffer("mu", torch.zeros(dim))
def forward(self, q):
p = torch.randn_like(q)
u = torch.rand(q.shape[0], device=q.device).clamp(min=1e-10)
# Call the kernel
q_new, accepted = hmc_forward(q, p, u, self.num_leapfrog_steps, self.step_size)
return q_new, accepted
```
Wait, the reference does `torch.log(torch.rand(...).clamp(min=1e-10))`. In my kernel, I do `logf(fmaxf(u[b], 1e-10f))`. So I should pass the raw uniform random numbers and do the clamp+log in the kernel. That matches.
Actually, looking more carefully at the reference:
```python
log_u = torch.log(torch.rand(q.shape[0], device=q.device).clamp(min=1e-10))
```
In the kernel I'll do:
```cpp
float log_u = logf(fmaxf(u[b], 1e-10f));
```
If I pass `torch.rand(...)` without clamping first, it should match since the kernel does the clamp and log.
But wait, there's a subtle issue: `torch.rand` can return 0.0? Actually, `torch.rand` returns values in [0, 1), so it can include 0.0 but not 1.0. The clamp is to avoid log(0) = -inf.
So passing raw `torch.rand()` is correct.
Now let me think about potential issues:
1. Type matching: `q` and `p` are the same dtype, `u` is float32
2. We need to handle the `accepted` tensor as bool
3. We need to make sure the kernel is launched with correct parameters
One more thing: for `dim=64`, if we use blockDim=64, that's exactly 2 warps, which is fine. But what if `dim` is not 64 in some test? The prompt says `get_init_inputs` returns `[dim]` where dim=64, so presumably the test uses dim=64. But to be extra safe, we can set threads to dim (but it's a runtime value). Actually CUDA block size must be compile-time constant in the launch config, but we can compute it dynamically if needed: `threads = ((dim + 31) / 32) * 32` to round up to the nearest warp.
But if dim > 1024, this won't fit in a block. For dim=64, it's definitely fine. Let me just use a compile-time constant of 64 or 128 to be safe.
Actually, wait. Can I use a variable in the kernel launch? Yes, the block size can be a variable in CUDA, it just needs to be the same for all blocks in the grid. So:
```cpp
int threads = ((dim + 31) / 32) * 32;
```
This will be 64 for dim=64. Good.
Now, one more optimization: I can use `__restrict__` to help the compiler. For H100, global memory loads should be efficient.
Also, for the leapfrog loop, `num_leapfrog_steps=10` is small and fixed. I should use `#pragma unroll` but since `num_leapfrog_steps` is a runtime parameter, the compiler can't fully unroll. I could template the kernel on `num_leapfrog_steps` but that's overkill for 10 steps. Actually, let me check if we can make it a template parameter... but that requires knowing it at compile time, which we don't unless we instantiate for common values.
For now, let me just write the clean kernel and see how it performs. If the correctness passes, I can try further optimizations.
Let me also think about whether we need `AT_DISPATCH_FLOATING_TYPES_AND2` or something. Since the reference uses fp32, and the requirement says "Produce numerically equivalent outputs at fp32 precision", we probably only need float32 support.
Actually, should we support both float and double? The reference model works with whatever dtype q is. But `get_inputs()` returns `torch.randn(batch_size, dim)` which is float32 by default. So float32 is sufficient.
Let me write the kernel.
One important thing: `std::vector<torch::Tensor>` return type requires `#include <vector>`.
Let me draft the full code:
```python
import torch
import torch.nn as nn
from torch.utils.cpp_extension import load_inline
cuda_source = '''
#include <torch/extension.h>
#include <cuda_runtime.h>
#include <vector>
template <typename scalar_t>
__global__ void hmc_kernel(
const scalar_t* __restrict__ q,
const scalar_t* __restrict__ p_init,
const float* __restrict__ u,
scalar_t* __restrict__ q_new,
bool* __restrict__ accepted,
int dim,
int num_leapfrog_steps,
float step_size)
{
int b = blockIdx.x;
int tid = threadIdx.x;
int lane = tid % 32;
int warp_id = tid / 32;
const scalar_t* q_b = q + b * dim;
const scalar_t* p_b = p_init + b * dim;
scalar_t* q_new_b = q_new + b * dim;
scalar_t q_val = 0;
scalar_t p_val = 0;
scalar_t q_cur = 0;
scalar_t p_cur = 0;
if (tid < dim) {
q_val = q_b[tid];
p_val = p_b[tid];
q_cur = q_val;
p_cur = p_val;
// Half step for momentum
p_val = p_val - 0.5f * step_size * q_val;
// Leapfrog steps
for (int s = 0; s < num_leapfrog_steps - 1; s++) {
q_val = q_val + step_size * p_val;
p_val = p_val - step_size * q_val;
}
// Full position step
q_val = q_val + step_size * p_val;
// Final half step for momentum
p_val = p_val - 0.5f * step_size * q_val;
// Negate momentum
p_val = -p_val;
}
float cur_term = 0.0f;
float prop_term = 0.0f;
if (tid < dim) {
cur_term = 0.5f * ((float)(q_cur * q_cur) + (float)(p_cur * p_cur));
prop_term = 0.5f * ((float)(q_val * q_val) + (float)(p_val * p_val));
}
// Warp-level reduction
for (int offset = 16; offset > 0; offset /= 2) {
cur_term += __shfl_down_sync(0xffffffff, cur_term, offset);
prop_term += __shfl_down_sync(0xffffffff, prop_term, offset);
}
__shared__ float s_cur[32]; // max 32 warps = 1024 threads
__shared__ float s_prop[32];
__shared__ bool s_acc[1];
int num_warps = blockDim.x / 32;
if (lane == 0) {
s_cur[warp_id] = cur_term;
s_prop[warp_id] = prop_term;
}
__syncthreads();
if (tid == 0) {
float total_cur = 0.0f;
float total_prop = 0.0f;
for (int i = 0; i < num_warps; i++) {
total_cur += s_cur[i];
total_prop += s_prop[i];
}
float log_alpha = total_cur - total_prop;
float log_u = logf(fmaxf(u[b], 1e-10f));
s_acc[0] = log_u < log_alpha;
}
__syncthreads();
bool acc = s_acc[0];
if (tid < dim) {
q_new_b[tid] = acc ? q_val : q_cur;
}
}
std::vector<torch::Tensor> hmc_forward_cuda(
torch::Tensor q,
torch::Tensor p,
torch::Tensor u,
int num_leapfrog_steps,
float step_size)
{
auto q_new = torch::empty_like(q);
auto accepted = torch::empty({q.size(0)}, torch::TensorOptions().dtype(torch::kBool).device(q.device()));
int batch_size = q.size(0);
int dim = q.size(1);
int threads = ((dim + 31) / 32) * 32;
int blocks = batch_size;
AT_DISPATCH_FLOATING_TYPES(q.scalar_type(), "hmc_forward", [&] {
hmc_kernel<scalar_t><<<blocks, threads>>>(
q.data_ptr<scalar_t>(),
p.data_ptr<scalar_t>(),
u.data_ptr<float>(),
q_new.data_ptr<scalar_t>(),
accepted.data_ptr<bool>(),
dim,
num_leapfrog_steps,
step_size
);
});
return {q_new, accepted};
}
'''
hmc_cuda = load_inline(
name="hmc_cuda",
cpp_sources="",
cuda_sources=cuda_source,
functions=["hmc_forward_cuda"],
extra_cuda_cflags=["-O3", "--use_fast_math"],
verbose=False,
)
class ModelNew(nn.Module):
def __init__(self, dim, num_leapfrog_steps=10, step_size=0.1):
super().__init__()
self.dim = dim
self.num_leapfrog_steps = num_leapfrog_steps
self.step_size = step_size
self.register_buffer("mu", torch.zeros(dim))
def forward(self, q):
p = torch.randn_like(q)
u = torch.rand(q.shape[0], device=q.device)
q_new, accepted = hmc_cuda.hmc_forward_cuda(
q, p, u, self.num_leapfrog_steps, self.step_size
)
return q_new, accepted
```
Wait, I need to be careful about the accepted tensor dtype. `bool` in PyTorch corresponds to `bool` in CUDA. That's fine.
Also, I'm generating `u = torch.rand(...)` without clamping, since the kernel does `fmaxf(u[b], 1e-10f)`. This should match the reference behavior.
Let me verify against the reference step by step:
Reference:
1. `p = torch.randn_like(q)` - same
2. `current_p = p.clone()`, `current_q = q.clone()` - same concept, stored in q_cur, p_cur
3. `p = p + 0.5 * eps * self._grad_log_prob(q)` where `grad_log_prob(q) = -(q - self.mu)` = `-q` since mu=0. So `p = p - 0.5 * eps * q`. Same.
4. Loop `num_leapfrog_steps - 1` times:
- `q = q + eps * p`
- `p = p + eps * grad_log_prob(q) = p - eps * q`
Same.
5. `q = q + eps * p` - same
6. `p = p + 0.5 * eps * grad_log_prob(q) = p - 0.5 * eps * q` - same
7. `p = -p` - same
8. `current_H = -self._log_prob(current_q) + 0.5 * (current_p ** 2).sum(-1)`
where `-log_prob(x) = -(-0.5 * sum(x^2)) = 0.5 * sum(x^2)`
So `current_H = 0.5 * sum(current_q^2) + 0.5 * sum(current_p^2)`. Same.
9. `proposed_H` same.
10. `log_alpha = current_H - proposed_H` - same
11. `log_u = torch.log(torch.rand(...).clamp(min=1e-10))` - same, since I pass raw rand and do clamp+log
12. `accepted = log_u < log_alpha` - same
13. `q_new = torch.where(accepted.unsqueeze(-1), q, current_q)` - same
Great, the logic matches.
Now, the issue with this approach is that I'm still launching separate kernels for `randn_like` and `rand`, plus my custom kernel. The reference also launches these plus many element-wise ops. Since the reference does about ~10*3 + some extra = ~35 kernel launches (one per PyTorch op), reducing to 3 kernel launches should be a big win.
Actually, I could even generate the random numbers in the custom kernel to reduce to 1 launch! But then I'd need to handle the random seed deterministically. The simplest is to pre-allocate and use `torch.randn` and `torch.rand`. That's 3 kernel launches vs ~35, which should be ~10x faster.
Wait, can I generate random numbers inside my CUDA kernel without curand states? I could use Philox counter-based RNG which is what PyTorch uses. But that's more complex. The 3 kernel launch approach should be sufficient.
Actually, hold on. Let me think about whether the overhead of 3 kernel launches vs 1 matters. For batch_size=2048 and dim=64, the total work is quite small. Each kernel launch has some overhead (~5-10 microseconds). If the computation itself is very small, the launch overhead could dominate.
For H100 with very fast kernel launches, 3 launches might still be fine. But let me estimate: each batch element does about 10*4 + a few = ~45 FMAs on 64 elements. For 2048 batch elements, that's about 2048 * 64 * 45 ≈ 5.9 million operations. At 67 TFLOPS, this is ~0.09 ms. Plus memory bandwidth for reading/writing q and p. Memory traffic: read q (512KB), write/read p (512KB + 512KB), read u (8KB), write q_new (512KB), write accepted (2KB). Total ~2MB. At 3.35 TB/s, that's ~0.6 microseconds. So compute bound? Actually it's very small.
Actually, let me reconsider. For such a small problem, the main cost of the reference is likely kernel launch overhead. The reference implementation has many operations and thus many kernel launches. Fusing into a single kernel (or 3 kernels) should help a lot.
Can I fuse the random generation too? I know PyTorch's CUDA random implementation. I could potentially call `at::native::philox::unpack` and use Philox directly, but that's very implementation-dependent and fragile. Let me just stick with passing pre-generated random numbers.
Actually, there's another approach: I can call the kernel and generate random numbers inside using `curand`. But to match PyTorch's random numbers, I'd need to know the Philox state. That's too hard.
So 3 kernels is the practical approach.
Wait, let me look at whether the PyTorch operations in the reference are all fused internally. For simple element-wise ops, PyTorch might launch separate kernels. For reductions, it launches reductions.
Anyway, let me compile and test this. But first, let me think about one more optimization: using shared memory to avoid bank conflicts, or using vectorized loads.
Actually, there's a big issue with my kernel: `__shfl_down_sync` reduces within a warp, but if `blockDim.x > 32`, I need to combine across warps. I do store to shared memory and then combine in thread 0. But wait, in my code:
```cpp
if (lane == 0) {
s_cur[warp_id] = cur_term;
}
```
This is fine because lane 0 of each warp writes its warp sum to `s_cur[warp_id]`. Then thread 0 reads all of them and sums. Good.
But there's a race condition: what if some warps haven't written to shared memory yet when thread 0 tries to read? The `__syncthreads()` after the writes fixes this. Good.
Also, `__shfl_down_sync` requires all threads in the warp to participate. But I have `if (tid < dim)` for some computations. If `blockDim.x > dim`, then some threads in the last warp might not participate... but since I use `threads = ((dim + 31) / 32) * 32`, all threads are `< threads`, and some might be `>= dim`. For the shuffle, all threads in a warp must call it with the same mask.
Actually, if `dim=64` and `threads=64`, then threads 32-63 are all `< dim`. The first warp (0-31) has all threads < dim too. No issue.
But what if `dim=33` and `threads=64`? Then thread 33-63 have `tid >= dim` and skip the shuffle. But `__shfl_down_sync` requires all threads in the warp to participate if they are in the mask. Actually, `__shfl_down_sync` doesn't require all threads to call it, but inactive threads (due to divergent control flow within the warp) will have undefined behavior if they don't execute it.
Wait, for `dim=33`, threads 32-63 are in the second warp. Thread 32 has `tid < dim` (32 < 33), so it enters the if. Threads 33-63 don't enter the if, so they don't call `__shfl_down_sync`. But `__shfl_down_sync` with mask `0xffffffff` requires all threads in the warp to execute it or the result is undefined for inactive threads.
Actually, I think the rule is: all non-exited threads in the warp named in the mask must execute the same `__shfl_*` operation. If some threads are diverged and don't reach it, the behavior is undefined.
So if `dim` is not a multiple of 32, the last warp might have some threads that skip the shuffle. This is problematic.
To fix this, I should ensure all threads in all warps participate in the shuffle. The easiest way is to remove the `if (tid < dim)` guard around the shuffle and just let all threads shuffle (with the extra threads contributing 0).
Let me restructure:
```cpp
scalar_t q_val = 0;
scalar_t p_val = 0;
scalar_t q_cur = 0;
scalar_t p_cur = 0;
if (tid < dim) {
q_val = q_b[tid];
p_val = p_b[tid];
q_cur = q_val;
p_cur = p_val;
// leapfrog ...
}
float cur_term = 0.0f;
float prop_term = 0.0f;
if (tid < dim) {
cur_term = 0.5f * ((float)(q_cur * q_cur) + (float)(p_cur * p_cur));
prop_term = 0.5f * ((float)(q_val * q_val) + (float)(p_val * p_val));
}
// All threads participate in shuffle
for (int offset = 16; offset > 0; offset /= 2) {
cur_term += __shfl_down_sync(0xffffffff, cur_term, offset);
prop_term += __shfl_down_sync(0xffffffff, prop_term, offset);
}
```
Now all threads in each warp call `__shfl_down_sync`. Threads with `tid >= dim` have `cur_term = 0` and `prop_term = 0`, so they contribute correctly to the reduction. Good.
Similarly for the final write:
```cpp
bool acc = s_acc[0];
if (tid < dim) {
q_new_b[tid] = acc ? q_val : q_cur;
}
```
No issue there.
One more issue: when `num_warps = blockDim.x / 32`, and I loop `for (int i = 0; i < num_warps; i++)`, if some warps have all threads with `tid >= dim`, their `s_cur[warp_id]` is 0.0, which is correct. So the sum is correct.
Let me also think about whether I should use `__launch_bounds__`. For such a simple kernel with few registers, it's probably not needed.
Actually, one more optimization: since `num_leapfrog_steps` is a runtime parameter, the loop can't be fully unrolled. But it's just 10 iterations, so the loop overhead is negligible.
Let me write the complete code and test.
Actually, I realize I should think more carefully about performance. Let me estimate again more carefully.
With batch_size=2048, dim=64, block_size=64:
- 2048 blocks, 64 threads each
- Each block handles 64 floats = 256 bytes
Global memory reads per block:
- Read q: 64 floats = 256 bytes
- Read p: 64 floats = 256 bytes
- Read u: 1 float = 4 bytes
Global memory writes per block:
- Write q_new: 64 floats = 256 bytes
- Write accepted: 1 byte
Total memory per block: ~772 bytes
Total memory: 2048 * 772 ≈ 1.58 MB
At 3.35 TB/s, this takes ~0.47 microseconds. But that's the bandwidth lower bound. The actual kernel will take longer due to compute and overhead.
For compute: each of 2048 * 64 threads does about 45 FMAs. Total FLOPs ≈ 2048 * 64 * 45 * 2 ≈ 11.8 MFLOPs. At 67 TFLOPS, this is ~0.18 microseconds.
So the kernel is limited by neither bandwidth nor compute, but by kernel launch overhead (~3-5 microseconds) and latency.
Given this, a single kernel launch should take about 5-10 microseconds. The PyTorch reference with ~30-40 kernel launches would take maybe 50-200 microseconds. So a 10-40x speedup should be possible with kernel fusion alone.
But wait, there might be other overhead in the reference like Python-side setup, tensor allocation, etc. The 3 extra allocations in my code (`q_new`, `accepted`, and implicit ones in `randn_like` and `rand`) add some overhead too. But still much less than the reference.
Can I reduce allocations? In the reference:
- `p = torch.randn_like(q)` - allocates p
- `current_p = p.clone()` - allocates current_p
- `current_q = q.clone()` - allocates current_q
- Various temporaries from the intermediate ops
In my approach:
- `p = torch.randn_like(q)` - allocates p
- `u = torch.rand(...)` - allocates u
- `q_new = torch.empty_like(q)` - allocates q_new inside the wrapper
- `accepted = torch.empty(...)` - allocates accepted inside the wrapper
So I have fewer allocations. Could I pre-allocate buffers in `__init__` to avoid repeated allocations? The evaluation might only call forward once, so it wouldn't matter. But if called multiple times, it would help.
Actually, looking at the correctness check, it might call forward multiple times. Let me add pre-allocated buffers.
```python
def __init__(self, ...):
...
self.register_buffer("_q_new_buf", torch.empty(0))
self.register_buffer("_accepted_buf", torch.empty(0, dtype=torch.bool))
self.register_buffer("_p_buf", torch.empty(0))
self.register_buffer("_u_buf", torch.empty(0))
def forward(self, q):
batch_size = q.shape[0]
if self._p_buf.shape[0] < batch_size:
self._p_buf.resize_(batch_size, self.dim)
self._u_buf.resize_(batch_size)
self._q_new_buf.resize_(batch_size, self.dim)
self._accepted_buf.resize_(batch_size)
p = torch.randn_like(q, out=self._p_buf[:batch_size])
# Actually, torch.randn_like doesn't take an out parameter directly in the same way
```
Hmm, `torch.randn_like` doesn't have an `out` parameter. I could use `torch.randn(batch_size, self.dim, device=q.device, out=self._p_buf[:batch_size])`.
Actually, that's getting messy. And for a single forward pass, it wouldn't matter much. Let me keep it simple.
Wait, I should also consider whether the problem calls `forward` once or many times. The `run_correctness` and timing functions likely call it multiple times. But pre-allocation might not help much if the evaluator handles this differently.
Let me just write a clean, correct implementation first.
Actually, wait. I want to check if there's something wrong with the C++ API I'm using. The `load_inline` function takes:
- `cuda_sources`: CUDA/C++ source containing the kernel and the wrapper function
- `functions`: list of function names to expose
The wrapper function `hmc_forward_cuda` needs to be declared properly and use ATen types. Let me make sure the include order is correct.
Also, I should use `TORCH_EXTENSION_NAME` or something? No, `load_inline` handles that.
One more thing: I might want to add `-lineinfo` or `-gencode` flags, but the defaults should be fine for H100.
Let me write the code and compile.
Oh wait, there's an issue with returning `bool` from CUDA. PyTorch bool tensors use `bool` in C++, which is usually 1 byte. When I write to `accepted.data_ptr<bool>()`, this should work.
Actually, `bool` in C++ can be implementation-defined size (usually 1 byte), but in CUDA, `bool` is also 1 byte. PyTorch kBool is 1 byte. So it should be fine.
But wait, there's a subtlety with atomic operations and `bool`. I'm not using atomics, just a single thread writing to shared memory and then all threads reading. That should be fine.
One more thing: `__shfl_sync` and `__shfl_down_sync` are available on compute capability >= 7.0, which includes H100. Good.
Let me also check: do I need to handle `const float* __restrict__ u` properly when `u` is a `torch::Tensor` of type float? Yes, `u.data_ptr<float>()` gives `float*`.
Actually, let me verify: `torch.rand(...)` returns a `torch::Tensor` of dtype `float32`. In PyTorch C++ API, `.scalar_type()` would be `torch::kFloat32`. So `u.data_ptr<float>()` is correct.
Now, one concern: if `q` is on CPU, the kernel will fail. But the reference requires GPU and `q.device` should be CUDA.
Let me write the complete kernel code now.
Actually, I realize I should consider using `__shared__` memory more carefully. The kernel has:
```cpp
__shared__ float s_cur[32];
__shared__ float s_prop[32];
__shared__ bool s_acc[1];
```
For block size 64, `num_warps = 2`, so only `s_cur[0]` and `s_cur[1]` are used, etc. The rest is unused but allocated. That's fine.
Wait, `s_acc[1]` might have alignment issues if placed after floats. But `bool` is 1 byte, so it shouldn't cause padding issues. Actually, shared memory doesn't have strict alignment requirements like global memory, but for performance it's good to align. Not a big issue here.
One more potential issue: reading `s_acc[0]` after `__syncthreads()` in all threads. Since thread 0 writes it and `__syncthreads()` ensures visibility, all threads can read it. Good.
Let me actually write the code and see if it compiles.
Actually, thinking about it more, I wonder if I can make the kernel even faster. The key insight is that with dim=64 and batch_size=2048, we have 131072 elements, which is not that many. Each element goes through 10 leapfrog steps.
A possible optimization: since each batch element is independent, I could use 1D thread blocks but process multiple batch elements per block? No, that would reduce parallelism and increase register usage.
Or I could vectorize the memory load/store using float2 or float4. For dim=64, each thread block processes 64 floats. Using float4, we could load 16 float4s = 64 floats. Each thread loads one float4. This would reduce the number of memory instructions and improve bandwidth utilization.
With block_size=64, each thread handles 1 float. To use float4, I'd need 16 threads per block. But then I'd have fewer threads (16 * 2048 = 32768 total), which might still be enough for H100.
Actually, let me try the simple version first. If correctness passes but speedup is insufficient, I can optimize.
Hmm, but since the total work is small, vectorizing might help reduce instruction count and improve throughput. Let me think...
With 2048 blocks of 64 threads, we have 131072 threads total. H100 has many SMs, so occupancy shouldn't be an issue. The main bottleneck is probably the sequential loop in each thread.
Wait, another thought: can I do the leapfrog integration using matrix operations? Each step is:
```
q_{t+1} = q_t + eps * p_t
p_{t+1} = p_t - eps * q_{t+1}
```
This is like:
```
[q_{t+1}] [1, eps] [q_t]
[p_{t+1}] = [-eps, 1-eps^2] [p_t]
```
Actually no, because:
```
q_{t+1} = q_t + eps * p_t
p_{t+1} = p_t - eps * q_{t+1} = p_t - eps * (q_t + eps * p_t) = -eps * q_t + (1 - eps^2) * p_t
```
So it's a linear transformation. The whole leapfrog integration (excluding the half steps) is a matrix power:
```
[q_T] = A^k [q_0]
[p_T] [p_0]
```
where A = [[1, eps], [-eps, 1-eps^2]].
For 10 steps, I could compute A^9 and apply it, plus handle the half steps. But the matrix elements would be:
```
A = [[1, eps], [-eps, 1-eps^2]]
```
Actually, the full leapfrog with half steps is:
1. p = p - 0.5 * eps * q
2. For k steps: q = q + eps * p, p = p - eps * q
3. p = p - 0.5 * eps * q
4. p = -p
This is equivalent to a single 2x2 matrix application on each (q, p) pair! Since it's a linear system, we can compute the exact transformation matrix analytically.
The eigenvalues of the leapfrog matrix might be on the unit circle (it's symplectic). For small eps, it's approximately a rotation.
Let me verify:
```
A = [[1, eps], [-eps, 1-eps^2]]
det(A) = 1*(1-eps^2) - eps*(-eps) = 1 - eps^2 + eps^2 = 1
```
The characteristic polynomial:
```
lambda^2 - trace(A)*lambda + det(A) = 0
trace(A) = 2 - eps^2
lambda = (2 - eps^2 ± sqrt((2-eps^2)^2 - 4)) / 2
= (2 - eps^2 ± sqrt(4 - 4*eps^2 + eps^4 - 4)) / 2
= (2 - eps^2 ± sqrt(-4*eps^2 + eps^4)) / 2
= (2 - eps^2 ± eps*sqrt(eps^2 - 4)) / 2
= (2 - eps^2 ± i*eps*sqrt(4 - eps^2)) / 2
```
For eps = 0.1, sqrt(4 - 0.01) ≈ 1.9975, so eigenvalues are complex with magnitude 1.
If I diagonalize A = V * D * V^{-1}, then A^k = V * D^k * V^{-1}.
Since D^k for complex eigenvalues on unit circle is a rotation matrix, A^k can be computed using trigonometric functions.
But this might actually be slower than the explicit loop due to transcendental function costs. For 10 iterations, explicit loop with 10 FMAs per step is probably faster than computing sin/cos.
So I'll stick with the loop.
Actually, there is another approach: I can unroll the loop manually for common values. Since num_leapfrog_steps is passed to the constructor, and in the reference it's 10, I could write specialized versions if needed. But 10 is small and the loop overhead is negligible.
Let me now write a first version and test it.
Wait, I need to check: how do I get the outputs from `functions=[...]` in `load_inline`? The wrapper function `hmc_forward_cuda` returns `std::vector<torch::Tensor>`. Will pybind11 correctly convert that to a Python tuple?
Yes, pybind11 will convert `std::vector<torch::Tensor>` to a Python list of tensors. Actually, it might return a list. Let me check the `load_inline` behavior... I believe `std::vector<torch::Tensor>` is bound as a list. Let me just unpack it as a list in Python.
Actually, looking at PyTorch's `load_inline` or `load` source code, `std::vector<at::Tensor>` should be converted to a list. So in Python:
```python
result = hmc_cuda.hmc_forward_cuda(...)
q_new, accepted = result[0], result[1]
```
Or if pybind11 does tuple conversion for pairs but not vectors. To be safe, I could return a `std::tuple<torch::Tensor, torch::Tensor>` instead. Let me do that.
```cpp
#include <tuple>
std::tuple<torch::Tensor, torch::Tensor> hmc_forward_cuda(...) {
...
return std::make_tuple(q_new, accepted);
}
```
pybind11 definitely handles `std::tuple`. Let me use that.
Let me also double-check: `torch::empty({q.size(0)}, torch::TensorOptions().dtype(torch::kBool).device(q.device()))`
Yes, that creates a bool tensor on the same device.
Now let me write the code and compile.