reasoning
We need to write a custom CUDA kernel that replaces the forward computation of the given Model. Model computes batched SAT (separating axis theorem) for axis-aligned rectangles vs triangles in 2D. We need to implement ModelNew with a custom CUDA kernel, producing numerically equivalent outputs at fp32 precision, and runs faster than reference.
Key aspects:
- The reference is fully vectorized with some Python-level loops over 2 world axes and 3 triangle edge normals. For batch size 16, and presumably larger batch sizes in evaluation, a fused CUDA kernel should be able to get much lower latency by avoiding intermediate allocations and kernel launch overhead.
- We need to use `torch.utils.cpp_extension.load_inline` to compile and bind CUDA source at module load time.
- We need to be careful not to include our own `PYBIND11_MODULE` block; we list the host-side wrapper function names in `functions=[...]`.
- The kernel should be a device kernel that for each batch element computes the SAT test. We can write a simple CUDA kernel where each thread handles one batch element, computing:
For each of 5 axes:
axis = ...
project rectangle corners (4 points) onto axis: compute dot product, keep min/max.
project triangle vertices (3 points) onto axis: compute dot product, keep min/max.
if (rect_max < tri_min - eps) or (tri_max < rect_min - eps): no collision => return 0.
If all axes pass, return 1.
The world axes: x-axis (1,0) and y-axis (0,1). For these, rectangle projections are just x and y coordinates. Specifically, rect_min_proj = robot_min[:, axis_index], rect_max_proj = robot_max[:, axis_index]. Triangle projections: min and max of env_triangle[:, :, axis_index]. This is a simplification that we should exploit.
For triangle edge normals: For each edge i from vertex i to (i+1)%3:
e = v_next - v_i
n = (-e.y, e.x) (perpendicular)
length = sqrt(n.x^2 + n.y^2)
if length < 1e-8: skip? But in reference, n = n / (norm + 1e-8), so axis is effectively zero if edge is degenerate. If axis is zero, all projections are zero, so the separation test will never separate. We can handle it similarly, or just skip if length is near zero. But to match numerically, we should replicate reference behavior.
For normalized axis: nx /= length; ny /= length;
Then compute dot products for 4 rectangle corners and 3 triangle vertices with this axis.
Note: rectangle corners are axis-aligned, so dot product with arbitrary axis = nx * x + ny * y.
We can precompute the 4 corners per batch element: (x0, y0), (x1, y0), (x1, y1), (x0, y1).
However, we can compute min and max projection for the rectangle on a general axis as:
proj = nx * x + ny * y
For rectangle [x0,x1]x[y0,y1], the min and max of nx*x + ny*y over corners are:
min = nx * (nx > 0 ? x0 : x1) + ny * (ny > 0 ? y0 : y1)? No, because it's linear, min over the box occurs at a corner determined by sign of nx and ny:
If nx >= 0: x contribution is min at x0, max at x1.
If nx < 0: x contribution is min at x1, max at x0.
Similarly for ny.
So:
rect_min_proj = nx * (nx >= 0 ? x0 : x1) + ny * (ny >= 0 ? y0 : y1)
rect_max_proj = nx * (nx >= 0 ? x1 : x0) + ny * (ny >= 0 ? y1 : y0)
This avoids iterating corners.
For the world axes (1,0) and (0,1):
axis = (1,0): rect_min_proj = x0, rect_max_proj = x1.
tri_min_proj = min(v0.x, v1.x, v2.x), tri_max_proj = max(...).
axis = (0,1): rect_min_proj = y0, rect_max_proj = y1.
tri_min_proj = min(v0.y, v1.y, v2.y), tri_max_proj = max(...).
This is even simpler.
For general axis (nx, ny):
rect_min_proj = nx * (nx >= 0 ? x0 : x1) + ny * (ny >= 0 ? y0 : y1)
rect_max_proj = nx * (nx >= 0 ? x1 : x0) + ny * (ny >= 0 ? y1 : y0)
tri_min_proj = min(nx*v0.x + ny*v0.y, nx*v1.x + ny*v1.y, nx*v2.x + ny*v2.y)
tri_max_proj = max(...).
Wait, is this exactly equivalent to projecting the 4 corners? Yes, because the rectangle is axis-aligned, and the dot product is a linear function. The extrema over a convex polytope occur at vertices. So min over 4 corners is exactly as above.
This reduces computation significantly.
Let's verify with the reference. Reference does:
r0, r1 = _project_min_max(corners, ax)
where corners is (B,4,2) and ax is (B,2).
_project_min_max does (points * axis.unsqueeze(1)).sum(dim=-1), then min/max over dim=1.
So for each batch element, it computes 4 dot products and takes min/max.
Our analytical min/max is mathematically identical, and should be numerically identical (same formulas, just computed directly). But we need to be careful with the sign check: `nx >= 0` vs `nx > 0` vs `nx < 0`. In the reference, for nx=0, x contribution is 0 regardless of x, so nx >= 0 or nx > 0 doesn't matter for the min/max of the x part alone, but combined with ny, we need to be consistent. Let's test: nx=0, ny=1. Then rect_min_proj should be y0. Formula: nx*(nx>=0?x0:x1) + ny*(ny>=0?y0:y1) = 0*x0 + 1*y0 = y0. Good. nx=0, ny=-1: rect_min_proj = 0*x0 + (-1)*( -1>=0? y0:y1) = -1*y1. But the actual min projection should be -y1 (since max y is y1, dot with -1 gives -y1). Yes, correct.
So this is fine.
However, we need to worry about numerical equivalence: the reference computes projections by summing products for all 4 corners, then min/max. Our direct formula is much faster and mathematically equivalent, but is it numerically equivalent within tolerance? It should be, because it's just a different way of computing the same expression. But let's consider edge cases with very large coordinates or infinities? The test data is uniform(-1,1) roughly, so fine.
Actually, for the triangle edge normals, the reference normalizes the axis: `n = n / (n.norm(dim=-1, keepdim=True) + 1e-8)`. Then projects. Does normalizing affect the separation test? If axis is scaled by a positive factor, the min/max projections scale by the same factor, and the conditions `r1 < t0 - eps` and `t1 < r0 - eps` become `c*r1 < c*t0 - eps` which is NOT equivalent to `r1 < t0 - eps/c`. Wait, the eps is not scaled! However, the reference uses the same eps for all axes, including triangle edge normals. Since they normalize the axis to unit length, the projection values are in the same units as coordinates (e.g., meters). If we didn't normalize, the projection would be scaled by the edge length, and the eps threshold `1e-6` would be effectively `1e-6 / edge_length` in the unscaled space, which is different. So to match the reference numerically, we MUST also normalize the triangle edge normals before projecting. Good catch.
So for the triangle edge normals:
e = v_next - v_i
n = (-e.y, e.x)
len = sqrt(n.x^2 + n.y^2)
if len > 0: nx = n.x / len; ny = n.y / len; else nx = n.x / 1e-8? Wait, reference does `n = n / (n.norm(dim=-1, keepdim=True) + 1e-8)`. So if len is 0, it divides by 1e-8. So nx = n.x * (1.0 / (len + 1e-8)). Actually, since n.x and n.y are 0 when len is 0, dividing by 1e-8 gives 0. So axis is (0,0). In that case, all projections are 0, and separation condition becomes `0 < 0 - eps` which is false. So we can implement: `inv_len = 1.0f / (len + 1e-8f); nx = n.x * inv_len; ny = n.y * inv_len;`. This matches.
Wait, but if len is 0, n.x and n.y are 0, so nx and ny become 0. This is numerically exact.
Let's think about the whole kernel:
```cuda
__global__ void sat_rect_triangle_kernel(
const float* robot_min,
const float* robot_max,
const float* env_triangle,
float* out,
int B,
float eps) {
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx >= B) return;
float x0 = robot_min[idx * 2 + 0];
float y0 = robot_min[idx * 2 + 1];
float x1 = robot_max[idx * 2 + 0];
float y1 = robot_max[idx * 2 + 1];
// Triangle vertices
float v0x = env_triangle[idx * 6 + 0];
float v0y = env_triangle[idx * 6 + 1];
float v1x = env_triangle[idx * 6 + 2];
float v1y = env_triangle[idx * 6 + 3];
float v2x = env_triangle[idx * 6 + 4];
float v2y = env_triangle[idx * 6 + 5];
int hit = 1;
// Axis 1: (1, 0)
{
float r0 = x0;
float r1 = x1;
float t0 = fminf(fminf(v0x, v1x), v2x);
float t1 = fmaxf(fmaxf(v0x, v1x), v2x);
if ((r1 < t0 - eps) || (t1 < r0 - eps)) hit = 0;
}
// Axis 2: (0, 1)
if (hit) {
float r0 = y0;
float r1 = y1;
float t0 = fminf(fminf(v0y, v1y), v2y);
float t1 = fmaxf(fmaxf(v0y, v1y), v2y);
if ((r1 < t0 - eps) || (t1 < r0 - eps)) hit = 0;
}
// Triangle edges
// Edge 0: v0 -> v1
if (hit) {
float ex = v1x - v0x;
float ey = v1y - v0y;
float nx = -ey;
float ny = ex;
float len = sqrtf(nx*nx + ny*ny);
float inv_len = 1.0f / (len + 1e-8f);
nx *= inv_len;
ny *= inv_len;
// rect min/max
float r0 = nx * (nx >= 0.0f ? x0 : x1) + ny * (ny >= 0.0f ? y0 : y1);
float r1 = nx * (nx >= 0.0f ? x1 : x0) + ny * (ny >= 0.0f ? y1 : y0);
float p0 = nx * v0x + ny * v0y;
float p1 = nx * v1x + ny * v1y;
float p2 = nx * v2x + ny * v2y;
float t0 = fminf(fminf(p0, p1), p2);
float t1 = fmaxf(fmaxf(p0, p1), p2);
if ((r1 < t0 - eps) || (t1 < r0 - eps)) hit = 0;
}
// Edge 1: v1 -> v2
if (hit) {
float ex = v2x - v1x;
float ey = v2y - v1y;
float nx = -ey;
float ny = ex;
float len = sqrtf(nx*nx + ny*ny);
float inv_len = 1.0f / (len + 1e-8f);
nx *= inv_len;
ny *= inv_len;
float r0 = nx * (nx >= 0.0f ? x0 : x1) + ny * (ny >= 0.0f ? y0 : y1);
float r1 = nx * (nx >= 0.0f ? x1 : x0) + ny * (ny >= 0.0f ? y1 : y0);
float p0 = nx * v0x + ny * v0y;
float p1 = nx * v1x + ny * v1y;
float p2 = nx * v2x + ny * v2y;
float t0 = fminf(fminf(p0, p1), p2);
float t1 = fmaxf(fmaxf(p0, p1), p2);
if ((r1 < t0 - eps) || (t1 < r0 - eps)) hit = 0;
}
// Edge 2: v2 -> v0
if (hit) {
float ex = v0x - v2x;
float ey = v0y - v2y;
float nx = -ey;
float ny = ex;
float len = sqrtf(nx*nx + ny*ny);
float inv_len = 1.0f / (len + 1e-8f);
nx *= inv_len;
ny *= inv_len;
float r0 = nx * (nx >= 0.0f ? x0 : x1) + ny * (ny >= 0.0f ? y0 : y1);
float r1 = nx * (nx >= 0.0f ? x1 : x0) + ny * (ny >= 0.0f ? y1 : y0);
float p0 = nx * v0x + ny * v0y;
float p1 = nx * v1x + ny * v1y;
float p2 = nx * v2x + ny * v2y;
float t0 = fminf(fminf(p0, p1), p2);
float t1 = fmaxf(fmaxf(p0, p1), p2);
if ((r1 < t0 - eps) || (t1 < r0 - eps)) hit = 0;
}
out[idx] = hit ? 1.0f : 0.0f;
}
```
Wait, but there's a subtlety: in the reference, for each axis that separates, `hit = hit * (~separated).to(hit.dtype)`. This means if any axis separates, hit becomes 0. It does not short-circuit per batch element? Actually, for each axis, it computes `separated` for all batch elements, and updates hit. The `&` or multiplication ensures that if any axis separates for a given batch element, hit becomes 0. So short-circuiting per batch element is fine.
However, the reference computes all axes regardless, because it's vectorized and must compute the whole tensor. Our kernel can short-circuit per thread, which is an advantage.
Is there any numerical difference from short-circuiting? No, because if hit becomes 0, staying 0 doesn't depend on later computations. So the result is the same.
But we need to be careful about `nx >= 0.0f ? x0 : x1` in the rect min/max formula. The reference computes 4 dot products and takes min/max. Let's verify they are exactly equivalent.
For a given nx, ny:
The 4 corners projections are:
c00 = nx*x0 + ny*y0
c10 = nx*x1 + ny*y0
c11 = nx*x1 + ny*y1
c01 = nx*x0 + ny*y1
Min = min(c00, c10, c11, c01).
Since x0 <= x1 and y0 <= y1:
If nx >= 0 and ny >= 0: min is c00 = nx*x0 + ny*y0. Matches formula.
If nx >= 0 and ny < 0: min is c01 = nx*x0 + ny*y1. Matches formula.
If nx < 0 and ny >= 0: min is c10 = nx*x1 + ny*y0. Matches formula.
If nx < 0 and ny < 0: min is c11 = nx*x1 + ny*y1. Matches formula.
Max = max(c00, c10, c11, c01).
If nx >= 0 and ny >= 0: max is c11 = nx*x1 + ny*y1. Matches.
If nx >= 0 and ny < 0: max is c10 = nx*x1 + ny*y0. Matches.
If nx < 0 and ny >= 0: max is c01 = nx*x0 + ny*y1. Matches.
If nx < 0 and ny < 0: max is c00 = nx*x0 + ny*y0. Matches.
So yes, exactly equivalent.
But what about `nx == 0`? If nx == 0 and ny >= 0: min = 0*x0 + ny*y0 = ny*y0. Actual min over corners: all have nx*x = 0, so min is ny*y0. Correct. If nx == 0 and ny < 0: min = 0*x0 + ny*y1 = ny*y1. Actual min: ny*y1 (since y1 > y0 and ny negative). Correct.
So the formula is exact.
One more thing: the reference's `separated` condition: `(r1 < t0 - eps) | (t1 < r0 - eps)`. In our kernel, we compute the same. Note that the reference uses `eps` as `self.eps`. In the reference, `eps=1e-6` by default. We should pass it through.
Now, what about the host wrapper function? We need to write a function that takes Tensors and calls the kernel.
```cpp
torch::Tensor sat_rect_triangle_cuda(
torch::Tensor robot_min,
torch::Tensor robot_max,
torch::Tensor env_triangle,
float eps) {
int B = robot_min.size(0);
auto out = torch::empty({B, 1}, robot_min.options());
int threads = 256;
int blocks = (B + threads - 1) / threads;
sat_rect_triangle_kernel<<<blocks, threads>>>(
robot_min.data_ptr<float>(),
robot_max.data_ptr<float>(),
env_triangle.data_ptr<float>(),
out.data_ptr<float>(),
B,
eps);
return out;
}
```
Wait, the output shape should be (B, 1). Our kernel writes `out[idx]`. `out` is contiguous with shape (B, 1). The index is just `idx`. Since the stride for the first dimension is 1? Actually, for a (B,1) tensor, the elements are stored contiguously as B elements. So `out[idx]` is correct. But if the tensor is not contiguous, `.data_ptr<float>()` could be problematic. We should ensure the output is contiguous, which it is. For inputs, we should ideally check they are contiguous and on CUDA, or call `.contiguous()`.
To be safe:
robot_min = robot_min.contiguous();
robot_max = robot_max.contiguous();
env_triangle = env_triangle.contiguous();
Also check dtype is float. In the reference, dtype is fp32. We can enforce or cast. The problem says "numerically equivalent outputs at fp32 precision". The inputs might be fp32. Let's assume they are.
We should check that robot_min is on CUDA. But the reference might run on CPU or CUDA? The evaluation says "backend=cuda". So inputs are CUDA tensors. We can add TORCH_CHECK.
Let's write the Python file.
Important: `load_inline` usage:
```python
import torch
from torch.utils.cpp_extension import load_inline
cuda_source = """
#include <torch/extension.h>
#include <cuda_runtime.h>
__global__ void sat_rect_triangle_kernel(...) { ... }
torch::Tensor sat_rect_triangle_cuda(torch::Tensor robot_min, torch::Tensor robot_max, torch::Tensor env_triangle, float eps) {
...
}
"""
cpp_sources = "" # Not needed if we define everything in cuda_source? Actually load_inline takes cuda_sources.
Actually, torch.utils.cpp_extension.load_inline signature:
load_inline(name, cpp_sources, cuda_sources=None, functions=None, ...)
We can put everything in `cuda_sources`. But the host function needs to be compiled as C++ code that knows about torch. Usually we put the device kernel in `cuda_sources` and the host wrapper also in `cuda_sources`. Or we can put host wrapper in `cpp_sources` and device kernel in `cuda_sources`. But for `load_inline`, it's fine to put both in `cuda_sources` if we are using CUDA extension. Let's just put everything in `cuda_sources`.
Wait, I recall that `load_inline` with `cuda_sources` compiles the CUDA source with nvcc. You can include torch headers in cuda_sources.
Let's just do:
```python
import torch
from torch.utils.cpp_extension import load_inline
_cuda_module = load_inline(
name="sat_kernel",
cpp_sources="",
cuda_sources='''
#include <torch/extension.h>
#include <cuda_runtime.h>
__global__ void sat_rect_triangle_kernel(...) { ... }
torch::Tensor sat_rect_triangle_cuda(torch::Tensor robot_min, torch::Tensor robot_max, torch::Tensor env_triangle, float eps) {
...
}
''',
functions=['sat_rect_triangle_cuda'],
extra_cuda_cflags=['-O3'],
verbose=False,
)
```
Wait, `load_inline` might need `cpp_sources` to be a non-empty string or not? The signature is `load_inline(name, cpp_sources, cuda_sources=None, ...)`. If we pass empty string for `cpp_sources`, it might be okay.
Alternatively, we can put the host wrapper in `cpp_sources` and the kernel in `cuda_sources`:
```python
cpp_sources = '''
#include <torch/extension.h>
torch::Tensor sat_rect_triangle_cuda(torch::Tensor robot_min, torch::Tensor robot_max, torch::Tensor env_triangle, float eps);
'''
cuda_sources = '''
#include <torch/extension.h>
#include <cuda_runtime.h>
__global__ void sat_rect_triangle_kernel(...) { ... }
torch::Tensor sat_rect_triangle_cuda(torch::Tensor robot_min, torch::Tensor robot_max, torch::Tensor env_triangle, float eps) {
...
}
'''
```
Actually, for `load_inline`, the `cpp_sources` string is compiled as C++ source, and `cuda_sources` is compiled as CUDA source. They are linked together. If we declare the function in `cpp_sources` but define it in `cuda_sources`, that works.
But simpler: I think we can just put everything in `cpp_sources` if we are using `load_inline` with `with_cuda=True` or if we provide `cuda_sources`. Actually, if we provide `cuda_sources`, the extension is built with CUDA. Let's check typical usage patterns.
Typical usage:
```python
module = load_inline(
name="my_kernel",
cpp_sources="",
cuda_sources=cuda_code,
functions=["my_function"],
extra_cuda_cflags=["-O3"],
verbose=True,
)
```
This is a common pattern. Let's do that.
Let's write the full kernel code in Python.
One thing: we need to make sure we handle `B` correctly. `env_triangle` has shape (B, 3, 2), so `env_triangle.size(0)` is B. `robot_min` is (B,2).
In the kernel, indexing for env_triangle: if the tensor is contiguous, `env_triangle[idx * 6 + 0]` is correct. But what if the tensor has some other layout? We'll make it contiguous.
Let's think about the fminf/fmaxf functions. These are CUDA device functions. For float, they are correct.
Potential issue: when `nx` or `ny` is NaN? Not possible with given inputs.
Another potential issue: `nx >= 0.0f` vs `nx > 0.0f`. As discussed, `>=` is correct. Let's double check with `nx == 0`. If `nx == 0`, then `r0`'s x-part should be `0 * x0 = 0`. With `nx >= 0.0f ? x0 : x1`, we get 0 * x0 = 0. With `nx > 0.0f ? x0 : x1`, we get 0 * x1 = 0. Same result. So it doesn't matter for the value, but it's fine.
Wait, there is a subtle numerical issue. The reference for the triangle normal axes computes 4 dot products for each corner. For a corner (x_i, y_j), the dot product is `nx*x_i + ny*y_j`. The min over 4 corners is what we compute analytically. This is exact. But the reference computes `proj = (points * axis.unsqueeze(1)).sum(dim=-1)`. For each corner, it does `corner_x * nx + corner_y * ny`. This is the same formula. Then takes min. So they are identical.
However, what about the `eps` threshold and floating point rounding? The reference computes `r1 < t0 - eps`. In our kernel, we compute the same. Since our `r1` and `r0` are analytically the same as the min/max of the 4 dot products, the result should be bit-identical or at least within a tiny rounding error. But let's consider if `nx*nx + ny*ny` might not be exactly 1.0 after normalization because of float rounding. The reference computes:
`n = torch.stack((-e[:, 1], e[:, 0]), dim=-1)`
`n_norm = n.norm(dim=-1, keepdim=True) + 1e-8`
`n = n / n_norm`
This is exactly the same as our `inv_len = 1.0f / (len + 1e-8f); nx *= inv_len; ny *= inv_len;`.
So `nx`, `ny` should be identical (or very close). Then rect projections: reference computes 4 dot products. Our analytic formula computes the same min/max. So the values should match exactly (assuming same FMA behavior?). Actually, the reference computes `nx*x0 + ny*y0` etc. individually. Our formula computes `nx*x0 + ny*y0` for the case where nx>=0, ny>=0. It literally computes the same expression. So it should be identical. For other cases, it computes `nx*x1 + ny*y0` etc., which is also the same expression that the reference would compute for that corner. But is there a possibility that due to associativity/ordering, `fminf` over 4 values yields a slightly different min than explicit selection? The min of a set of floats is well-defined, but `fminf(a,b)` is equivalent to `min(a,b)`. Our analytic formula directly selects the corner that yields the minimum based on the sign of nx and ny. For linear functions over a box, the minimum is guaranteed to be at one of the corners we select. So the value we compute is exactly one of the 4 corner projections, which is exactly what the reference computes and then takes the min of. So no difference.
Therefore, our kernel should be numerically equivalent.
But wait, what if the batch dimension is large and we want to optimize further? For H100, we can use vectorized loads (float2) to reduce memory traffic. The problem is memory-bound because it's a small computation per element: read 2*2 + 3*2 = 10 floats, do some fmin/fmax and sqrt, write 1 float. That's about 44 bytes read + 4 bytes write = 48 bytes. Compute is negligible. So memory bandwidth is the limit. A naive kernel will be memory-bound.
The reference does many intermediate operations creating tensors, which involves multiple kernel launches and memory reads/writes. The fused kernel reduces memory traffic dramatically. Even a simple kernel should be much faster.
However, can we make the kernel even faster? We can use `float2` loads if the memory is aligned. `robot_min` is (B,2), so we can load as float2. Similarly `robot_max`. `env_triangle` is (B,3,2), so we can load 6 floats per element. Since 6 floats = 24 bytes, we can load as float4 + float2 or three float2s.
Using `float2` can reduce instructions and improve memory throughput.
Also, we should use `__ldg` (load through read-only cache) for input data.
And we can unroll the axes manually.
Another optimization: for world axes, we don't need to compute triangle min/max by comparing all three values if we use vectorized operations? But in the kernel, `fminf(fminf(v0x, v1x), v2x)` is just two instructions.
Let's refine the kernel to use float2 for `robot_min` and `robot_max` and `env_triangle`.
Actually, we need to be careful with alignment: `robot_min` is contiguous (B,2). The first element of row `idx` is at offset `idx*2`. A `float2` load requires 8-byte alignment. Since `idx*2*sizeof(float) = idx*8`, it is always 8-byte aligned. Great.
`env_triangle` contiguous (B,3,2): offset `idx*6*sizeof(float) = idx*24`. Is that 8-byte aligned? Yes, for any `idx`, `idx*24` is divisible by 8. So we can load as float2 or float4. For `idx*24`, we can load two float4s (32 bytes) but we only need 24. Or three float2s (8*3=24 bytes). Let's just load three float2s.
But we need to be careful: `float2` in CUDA is a struct with `x` and `y`. The layout matches the tensor's last dimension being 2.
So we can do:
```cuda
float2 rmin = reinterpret_cast<const float2*>(robot_min)[idx];
float2 rmax = reinterpret_cast<const float2*>(robot_max)[idx];
float x0 = rmin.x, y0 = rmin.y;
float x1 = rmax.x, y1 = rmax.y;
const float2* tri_ptr = reinterpret_cast<const float2*>(env_triangle);
float2 v0 = tri_ptr[idx * 3 + 0];
float2 v1 = tri_ptr[idx * 3 + 1];
float2 v2 = tri_ptr[idx * 3 + 2];
```
Wait, indexing: `env_triangle` is (B,3,2). The last dimension is 2. So `tri_ptr[idx*3 + i]` gives the i-th vertex of batch element idx. Yes.
This is clean and efficient.
However, `reinterpret_cast` on the data pointer requires that the pointer is aligned appropriately. Since `robot_min` is a `float*` from a contiguous tensor of shape (B,2), `robot_min` points to the first element. `robot_min[idx]` is offset `idx*2` floats, i.e., `idx*8` bytes. Casting to `float2*` and indexing by `idx` gives offset `idx*sizeof(float2) = idx*8` bytes. Perfect.
Similarly for `env_triangle`: `env_triangle[idx*6 + j]` is offset `(idx*6 + j)*4` bytes. Casting to `float2*` and indexing by `idx*3 + i` gives offset `(idx*3 + i)*8 = idx*24 + i*8` bytes = `(idx*6 + i*2)*4` bytes. That matches `env_triangle[idx, i, 0]` being at float index `idx*6 + i*2`. Yes.
So using `float2` loads is safe and efficient.
Let's also consider storing the output. `out` is (B,1). We can just write `out[idx] = hit ? 1.0f : 0.0f;`. The output pointer is `float*`. We could cast to something, but it's just one float per thread.
Should we use `__restrict__` for pointers? Yes.
Should we use `-use_fast_math`? The sqrt is only 3 per thread, and `-use_fast_math` makes sqrt faster but slightly less accurate. Since we need numerical equivalence, and the reference uses `torch.norm` which maps to standard sqrt (not fast math), we should probably avoid `-use_fast_math` to be safe? Actually, CUDA's `sqrtf` without fast math is correctly rounded. With fast math, it's an approximation. The reference uses `n.norm(dim=-1, keepdim=True)`, which in PyTorch likely uses `sqrt` from libm or cublas, which is standard. So we should use standard `sqrtf` (from `<cmath>`) or `sqrtf` device function. If we pass `-use_fast_math`, `sqrtf` becomes `__fsqrt_rn`? Actually, with `-use_fast_math`, `sqrtf` is approximated. To be safe, we should not use `-use_fast_math`. But if we aren't using it, the compile will be fine.
However, to squeeze more performance, we can manually call `__fsqrt_rn`? No, we want accuracy.
Actually, the number of sqrtfs is tiny (3 per thread). The kernel is memory-bound, so the cost of sqrt is negligible. Avoid fast math.
Let's think about block size. 256 threads is a good default. For very large B, we can use more.
Let's also consider using grids with `cudaOccupancyMaxPotentialBlockSize`? Not needed.
One more thing: `load_inline` may compile with default flags. We should add `-O3` in `extra_cuda_cflags` to ensure optimization.
Also, we need to handle `eps` being a `float` correctly.
Let's also check the tolerance: fp32 uses atol=rtol=1e-4. The outputs are 0.0 or 1.0. So as long as we return exactly 0.0 or 1.0 (within float rounding), we are fine.
Let's write the code.
Actually, there is one more subtlety: The reference `Model` has `eps: float = 1e-6`. The host wrapper should accept `float eps`. In ModelNew.forward, we will call `_cuda_module.sat_rect_triangle_cuda(robot_min, robot_max, env_triangle, self.eps)`.
Let's draft the complete Python file.
```python
import torch
import torch.nn as nn
from torch.utils.cpp_extension import load_inline
_CUDA_MODULE = load_inline(
name="sat_rect_triangle",
cpp_sources="",
cuda_sources=r"""
#include <torch/extension.h>
#include <cuda_runtime.h>
__global__ void sat_rect_triangle_kernel(
const float* __restrict__ robot_min,
const float* __restrict__ robot_max,
const float* __restrict__ env_triangle,
float* __restrict__ out,
int B,
float eps) {
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx >= B) return;
// Load robot box
float2 rmin = reinterpret_cast<const float2*>(robot_min)[idx];
float2 rmax = reinterpret_cast<const float2*>(robot_max)[idx];
float x0 = rmin.x;
float y0 = rmin.y;
float x1 = rmax.x;
float y1 = rmax.y;
// Load triangle vertices
const float2* tri_ptr = reinterpret_cast<const float2*>(env_triangle);
float2 v0 = tri_ptr[idx * 3 + 0];
float2 v1 = tri_ptr[idx * 3 + 1];
float2 v2 = tri_ptr[idx * 3 + 2];
int hit = 1;
// Axis 1: x-axis (1, 0)
{
float r_min = x0;
float r_max = x1;
float t_min = fminf(fminf(v0.x, v1.x), v2.x);
float t_max = fmaxf(fmaxf(v0.x, v1.x), v2.x);
if ((r_max < t_min - eps) || (t_max < r_min - eps)) {
hit = 0;
}
}
// Axis 2: y-axis (0, 1)
if (hit) {
float r_min = y0;
float r_max = y1;
float t_min = fminf(fminf(v0.y, v1.y), v2.y);
float t_max = fmaxf(fmaxf(v0.y, v1.y), v2.y);
if ((r_max < t_min - eps) || (t_max < r_min - eps)) {
hit = 0;
}
}
// Triangle edge normals
// Edge 0: v0 -> v1
if (hit) {
float ex = v1.x - v0.x;
float ey = v1.y - v0.y;
float nx = -ey;
float ny = ex;
float len = sqrtf(nx * nx + ny * ny);
float inv_len = 1.0f / (len + 1e-8f);
nx *= inv_len;
ny *= inv_len;
float r_min = nx * (nx >= 0.0f ? x0 : x1) + ny * (ny >= 0.0f ? y0 : y1);
float r_max = nx * (nx >= 0.0f ? x1 : x0) + ny * (ny >= 0.0f ? y1 : y0);
float p0 = nx * v0.x + ny * v0.y;
float p1 = nx * v1.x + ny * v1.y;
float p2 = nx * v2.x + ny * v2.y;
float t_min = fminf(fminf(p0, p1), p2);
float t_max = fmaxf(fmaxf(p0, p1), p2);
if ((r_max < t_min - eps) || (t_max < r_min - eps)) {
hit = 0;
}
}
// Edge 1: v1 -> v2
if (hit) {
float ex = v2.x - v1.x;
float ey = v2.y - v1.y;
float nx = -ey;
float ny = ex;
float len = sqrtf(nx * nx + ny * ny);
float inv_len = 1.0f / (len + 1e-8f);
nx *= inv_len;
ny *= inv_len;
float r_min = nx * (nx >= 0.0f ? x0 : x1) + ny * (ny >= 0.0f ? y0 : y1);
float r_max = nx * (nx >= 0.0f ? x1 : x0) + ny * (ny >= 0.0f ? y1 : y0);
float p0 = nx * v0.x + ny * v0.y;
float p1 = nx * v1.x + ny * v1.y;
float p2 = nx * v2.x + ny * v2.y;
float t_min = fminf(fminf(p0, p1), p2);
float t_max = fmaxf(fmaxf(p0, p1), p2);
if ((r_max < t_min - eps) || (t_max < r_min - eps)) {
hit = 0;
}
}
// Edge 2: v2 -> v0
if (hit) {
float ex = v0.x - v2.x;
float ey = v0.y - v2.y;
float nx = -ey;
float ny = ex;
float len = sqrtf(nx * nx + ny * ny);
float inv_len = 1.0f / (len + 1e-8f);
nx *= inv_len;
ny *= inv_len;
float r_min = nx * (nx >= 0.0f ? x0 : x1) + ny * (ny >= 0.0f ? y0 : y1);
float r_max = nx * (nx >= 0.0f ? x1 : x0) + ny * (ny >= 0.0f ? y1 : y0);
float p0 = nx * v0.x + ny * v0.y;
float p1 = nx * v1.x + ny * v1.y;
float p2 = nx * v2.x + ny * v2.y;
float t_min = fminf(fminf(p0, p1), p2);
float t_max = fmaxf(fmaxf(p0, p1), p2);
if ((r_max < t_min - eps) || (t_max < r_min - eps)) {
hit = 0;
}
}
out[idx] = hit ? 1.0f : 0.0f;
}
torch::Tensor sat_rect_triangle_cuda(
torch::Tensor robot_min,
torch::Tensor robot_max,
torch::Tensor env_triangle,
float eps) {
TORCH_CHECK(robot_min.is_cuda(), "robot_min must be a CUDA tensor");
TORCH_CHECK(robot_max.is_cuda(), "robot_max must be a CUDA tensor");
TORCH_CHECK(env_triangle.is_cuda(), "env_triangle must be a CUDA tensor");
TORCH_CHECK(robot_min.dim() == 2 && robot_min.size(1) == 2, "robot_min must be (B, 2)");
TORCH_CHECK(robot_max.dim() == 2 && robot_max.size(1) == 2, "robot_max must be (B, 2)");
TORCH_CHECK(env_triangle.dim() == 3 && env_triangle.size(1) == 3 && env_triangle.size(2) == 2, "env_triangle must be (B, 3, 2)");
int B = robot_min.size(0);
TORCH_CHECK(robot_max.size(0) == B && env_triangle.size(0) == B, "Batch sizes must match");
robot_min = robot_min.contiguous();
robot_max = robot_max.contiguous();
env_triangle = env_triangle.contiguous();
auto out = torch::empty({B, 1}, robot_min.options());
int threads = 256;
int blocks = (B + threads - 1) / threads;
sat_rect_triangle_kernel<<<blocks, threads>>>(
robot_min.data_ptr<float>(),
robot_max.data_ptr<float>(),
env_triangle.data_ptr<float>(),
out.data_ptr<float>(),
B,
eps);
return out;
}
""",
functions=["sat_rect_triangle_cuda"],
extra_cuda_cflags=["-O3"],
verbose=False,
)
class ModelNew(nn.Module):
def __init__(self, eps: float = 1e-6):
super().__init__()
self.eps = eps
def forward(self, robot_min, robot_max, env_triangle):
return _CUDA_MODULE.sat_rect_triangle_cuda(robot_min, robot_max, env_triangle, self.eps)
```
Wait, but default `cpp_sources` is empty string. Does `load_inline` accept that? Looking at PyTorch docs: `load_inline(name, cpp_sources, cuda_sources=None, functions=None, extra_cflags=None, extra_cuda_cflags=None, extra_ldflags=None, extra_include_paths=None, build_directory=None, is_python_module=True, verbose=False, with_cuda=None, is_standalone=False, keep_intermediates=True)` where `cpp_sources` is a string containing the C++ source code.
Actually, I'm not sure if `cpp_sources=""` is accepted. Some versions might require it to be non-empty if `cuda_sources` is provided? I think it's fine. But to be safe, we can include a dummy comment or a forward declaration in `cpp_sources`. But the docs say `cpp_sources` is required. Let's put a minimal C++ source: maybe just `#include <torch/extension.h>` or just empty? I think empty is okay, but let's put a comment.
Actually, looking at examples online, `cpp_sources` is often the host-side code, and `cuda_sources` is the device code. If `cpp_sources` is empty, some versions may complain? I'll put a small amount:
```python
cpp_sources=r"""
#include <torch/extension.h>
torch::Tensor sat_rect_triangle_cuda(torch::Tensor robot_min, torch::Tensor robot_max, torch::Tensor env_triangle, float eps);
""",
```
Wait, `load_inline` auto-generates bindings for functions listed in `functions`. The function definition is in `cuda_sources`. The declaration in `cpp_sources` ensures the C++ compiler knows about it. This is a standard pattern. Let's do this.
Wait, but if we put the declaration in `cpp_sources`, does the auto-generated pybind11 binding call the function from the C++ side? Yes, the binding is generated in the C++ compilation unit. The `cuda_sources` is compiled separately and linked. So the function must be declared in the C++ compilation unit (or in a header included there). If we put the definition in `cuda_sources` and no declaration in `cpp_sources`, the C++ compiler won't see it when generating the bindings, leading to a compile error. Actually, `load_inline` generates bindings using the `functions` list. It generates code like:
```cpp
m.def("sat_rect_triangle_cuda", &sat_rect_triangle_cuda, ...);
```
This generated code is in the C++ compilation unit. So it needs a declaration of `sat_rect_triangle_cuda`. Therefore, we should provide a declaration in `cpp_sources`.
Let's verify: In PyTorch's `load_inline` implementation:
- It creates a `main.cpp` that includes the `cpp_sources` and adds the pybind11 wrappers.
- It creates a `cuda.cu` that includes `cuda_sources`.
- They are linked together.
So yes, we need a declaration in `cpp_sources` if the definition is in `cuda_sources`.
Therefore:
```python
cpp_sources=r"""
#include <torch/extension.h>
torch::Tensor sat_rect_triangle_cuda(torch::Tensor robot_min, torch::Tensor robot_max, torch::Tensor env_triangle, float eps);
"""
```
And `cuda_sources` contains the definition.
Let's adjust.
Also, we might want to handle cases where `B=0`. Our code handles it: blocks = 0? Actually `(0 + 255) / 256 = 0`. Launching a kernel with 0 blocks is valid in CUDA, but torch might not like it? CUDA supports `<<<0, threads>>>` as a no-op. But `blocks` computed as integer division: `(B + threads - 1) / threads`. If B=0, blocks = 0. That's fine.
Another thing: we should ensure that `ModelNew` inherits from `nn.Module` and accepts the same constructor arguments as `Model`. Yes.
Let's compile and test.
Actually, I need to be careful about the `r` string prefix for the CUDA source. It might contain backslashes or something? No. But using `r"""` is fine.
One more thing: the `extra_cuda_cflags=["-O3"]`. Should we add `-std=c++17`? PyTorch usually defaults to C++17 or C++14. We can rely on defaults. But adding `-O3` is good.
Let's write the full source and call compile_kernel.
Wait, I need to call compile_kernel first.
But there's a catch: `compile_kernel` just compiles the kernel. It expects the full Python source. So I need to provide the string representing the entire Python file.
Let me draft it precisely.