correct
32_argmax_tiebreak_smallest_rank.py
started
2026-05-11T14:25:58Z
timingkernelbench.timing
kernel mean / median / std
0.01 / — / 1.64e-04 μs
ref mean / median / std
0.09 / — / 7.29e-04 μs
numerical precisioneval_kernel_against_ref
gpu memoryextended_metrics.measure_memory
kernel launches / fusiontorch.profiler
fusion ratio (ref / kernel)
94.00x
SOL (speed-of-light)nsight_hardware_counters
roofline / occupancynsight_hardware_counters
warp stalls (top 5)nsight
reference PyTorch source (the prompt the agent sees)
"""
Over virtual ranks R, for each (B, C) position take the maximum value; on ties choose the smallest rank index.
Output (B, C, 2): [:,:,0] = best value, [:,:,1] = winning rank as float32.
"""
import torch
import torch.nn as nn
from kernelbench.distributed_collectives import default_device
R = 4
B = 8
C = 28
class Model(nn.Module):
def __init__(self):
super().__init__()
def forward(self, x: torch.Tensor) -> torch.Tensor:
# x: (R, B, C)
best_val = x[0].clone()
best_idx = torch.zeros(B, C, dtype=torch.long, device=x.device)
for r in range(1, R):
xv = x[r]
improved = xv > best_val
tied = (xv == best_val) & ~improved
best_idx = torch.where(
improved,
torch.full_like(best_idx, r),
torch.where(
tied,
torch.minimum(best_idx, torch.full_like(best_idx, r)),
best_idx,
),
)
best_val = torch.where(improved, xv, best_val)
stacked = torch.stack([best_val, best_idx.to(best_val.dtype)], dim=-1)
return stacked
def get_inputs():
dev = default_device()
p = popcorn_pri
return [torch.randn(R, p.jitter_int(B), p.jitter_int(C), device=dev)]
def get_init_inputs():
return []
turn 0844.9s turn · 2.5s LLM · 842.4s tools · tokens in=1,577 · out=150 · total=1,727
reasoning
Let me analyze this task:
1. We need to replace a PyTorch model that finds the maximum value across virtual ranks R, and on ties chooses the smallest rank index
2. Input: (R, B, C) tensor where R=4, B=8, C=28
3. Output: (B, C, 2) where [:,:,0] is the best value and [:,:,1] is the winning rank as float32
The algorithm:
- For each (B, C) position, find the maximum value across R dimension
- On ties, choose the smallest rank index (so first occurrence wins)
Let me first get the GPU specs to understand the hardware.
call → get_gpu_specs
callget_gpu_specs
{}
result ← get_gpu_specs (ok)
okget_gpu_specs
GPU specs for NVIDIA H100 80GB HBM3:
total memory (runtime): 79.1 GB
spec entry: H100
GPU Architecture: Hopper
GPU Memory: 80GB
Memory Bandwidth: 3.35 TB/s
FP64 TFLOPS: 34
FP64 Tensor Core TFLOPS: 67
FP32 TFLOPS: 67
TF32 Tensor Core TFLOPS: 989 with sparsity
BFLOAT16 Tensore Core TFLOPS: 1979 with sparsity
FP16 Tensor Core TFLOPS: 1979 with sparsity
FP8 Tensor Core TFLOPS: 3958 with sparsity
INT8 Tensor Core TOPS: 3958 with sparsity
Register File Size: 64K 32-bit registers per SM
Maximum number of registers per thread: 255
Maximum number of thread blocks per SM: 32
Shared memory capacity per SM: 228 KB
Maximum shared memory per thread block: 227 KB
turn 1609.5s turn · 10.0s LLM · 599.5s tools · tokens in=1,825 · out=1,002 · total=2,827
reasoning
Now I understand the task better. Let me analyze the reference implementation:
1. Input: x of shape (R, B, C) where R=4, B=8, C=28
2. We need to find the maximum value across R dimension for each (B, C) position
3. On ties, we choose the smallest rank index
4. Output: (B, C, 2) where [:,:,0] is the best value and [:,:,1] is the winning rank
The input size is quite small: 4×8×28 = 896 elements. This is very small, so the kernel should be optimized for low latency rather than high throughput.
Let me write a CUDA kernel that:
1. Uses one thread per (B, C) position
2. Each thread iterates over R values to find max and best index
3. Uses a block size that covers all B×C = 224 positions (8×28 = 224)
Since the total number of elements is small (224 output elements), we can launch a single block with enough threads, or use a small grid.
Let me write the kernel:
assistantLet me analyze the problem and write an optimized CUDA kernel.
The task is to find max values across R=4 ranks for each (B, C) position, with tie-breaking to the smallest rank index. Input size is small (4×8×28 = 896 elements), so this is a latency-bound operation.
call → compile_kernel
callcompile_kernel
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// One thread per (B, C) position\n__global__ void max_with_tiebreak_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = blockIdx.x * blockDim.x + threadIdx.x;\n int total = B * C;\n if (bc_idx >= total) return;\n \n int b = bc_idx / C;\n int c = bc_idx % C;\n \n // x is (R, B, C) contiguous, so x[r, b, c] = x[r * B * C + b * C + c]\n int base = b * C + c;\n int stride = B * C;\n \n float best_val = x[0 * stride + base];\n int best_idx = 0;\n \n for (int r = 1; r < R; r++) {\n float val = x[r * stride + base];\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n // Tie: keep smaller index\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Output is (B, C, 2) contiguous\n int out_base = bc_idx * 2;\n out[out_base] = best_val;\n out[out_base + 1] = (float)best_idx;\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 256;\n int blocks = (total + threads - 1) / threads;\n \n max_with_tiebreak_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← compile_kernel (ok)
okcompile_kernel
compile_kernel PASSED: kernel compiled without errors.
turn 2834.2s turn · 6.4s LLM · 827.8s tools · tokens in=2,596 · out=704 · total=3,300
reasoning
The kernel compiled successfully. Let me test for correctness.
call → run_correctness
callrun_correctness
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// One thread per (B, C) position\n__global__ void max_with_tiebreak_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = blockIdx.x * blockDim.x + threadIdx.x;\n int total = B * C;\n if (bc_idx >= total) return;\n \n int b = bc_idx / C;\n int c = bc_idx % C;\n \n // x is (R, B, C) contiguous, so x[r, b, c] = x[r * B * C + b * C + c]\n int base = b * C + c;\n int stride = B * C;\n \n float best_val = x[0 * stride + base];\n int best_idx = 0;\n \n for (int r = 1; r < R; r++) {\n float val = x[r * stride + base];\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n // Tie: keep smaller index\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Output is (B, C, 2) contiguous\n int out_base = bc_idx * 2;\n out[out_base] = best_val;\n out[out_base + 1] = (float)best_idx;\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 256;\n int blocks = (total + threads - 1) / threads;\n \n max_with_tiebreak_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← run_correctness (ok)
okrun_correctness
run_correctness PASSED: (5 / 5) trials all matched the reference.
Numerical precision: max_abs_err=0.00e+00 mean_abs_err=0.00e+00 max_rel_err=0.00e+00
turn 36.6s turn · 6.6s LLM · 0.0s tools · tokens in=3,346 · out=721 · total=4,067
reasoning
Excellent! The correctness check passed. Now let me run a static check to make sure there are no reward-hacking patterns, and then submit.
call → static_check
callstatic_check
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// One thread per (B, C) position\n__global__ void max_with_tiebreak_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = blockIdx.x * blockDim.x + threadIdx.x;\n int total = B * C;\n if (bc_idx >= total) return;\n \n int b = bc_idx / C;\n int c = bc_idx % C;\n \n // x is (R, B, C) contiguous, so x[r, b, c] = x[r * B * C + b * C + c]\n int base = b * C + c;\n int stride = B * C;\n \n float best_val = x[0 * stride + base];\n int best_idx = 0;\n \n for (int r = 1; r < R; r++) {\n float val = x[r * stride + base];\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n // Tie: keep smaller index\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Output is (B, C, 2) contiguous\n int out_base = bc_idx * 2;\n out[out_base] = best_val;\n out[out_base + 1] = (float)best_idx;\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 256;\n int blocks = (total + threads - 1) / threads;\n \n max_with_tiebreak_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← static_check (ok)
okstatic_check
static_check PASSED: no violations or warnings detected.
turn 4519.8s turn · 9.9s LLM · 509.9s tools · tokens in=4,052 · out=1,027 · total=5,079
reasoning
Let me try to optimize this further. Since this is latency-bound (small input size), I should:
1. Minimize kernel launch overhead - potentially use fewer threads/blocks
2. Ensure the kernel is as simple as possible
3. Consider using shared memory for small inputs (though with 8×28=224 elements, each thread only handles 1 element)
Let me try an optimized version:
- Use 32 threads per block (one warp) to minimize overhead
- Use __ldg for read-only cache optimization
- Ensure coalesced memory access
Actually, let me think about this more carefully. The problem is:
- Input: (R, B, C) = (4, 8, 28) = 896 elements
- Output: (B, C, 2) = 224 × 2 = 448 elements
With 224 threads (one per (B, C) position):
- Each thread reads 4 values from different ranks
- Each thread writes 2 values
This is very small, so the key is minimizing kernel launch overhead. Let me try with smaller thread blocks to see if it helps.
assistantLet me try to optimize further for the small input size:
call → compile_kernel
callcompile_kernel
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel with optimizations for small inputs\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// One thread per (B, C) position\n// Optimized for small inputs with read-only cache\n__global__ void __launch_bounds__(128) max_with_tiebreak_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = blockIdx.x * blockDim.x + threadIdx.x;\n int total = B * C;\n if (bc_idx >= total) return;\n \n // x is (R, B, C) contiguous, so x[r, b, c] = x[r * B * C + b * C + c]\n int base = bc_idx; // bc_idx already is b * C + c for contiguous access\n int stride = B * C;\n \n // Use __ldg for read-only cache\n float best_val = __ldg(x + base);\n int best_idx = 0;\n \n // Unroll loop for R=4 case\n #pragma unroll 4\n for (int r = 1; r < R; r++) {\n float val = __ldg(x + r * stride + base);\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n // Tie: keep smaller index (r is always > best_idx at this point in the loop)\n // Actually r > best_idx is not always true if we found a tie earlier\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Output is (B, C, 2) contiguous\n int out_base = bc_idx * 2;\n out[out_base] = best_val;\n out[out_base + 1] = (float)best_idx;\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 128;\n int blocks = (total + threads - 1) / threads;\n \n max_with_tiebreak_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext_v2\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← compile_kernel (ok)
okcompile_kernel
compile_kernel PASSED: kernel compiled without errors.
turn 51073.2s turn · 6.7s LLM · 1066.5s tools · tokens in=4,850 · out=785 · total=5,635
reasoning
Let me test this version for correctness and then submit.
call → run_correctness
callrun_correctness
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel with optimizations for small inputs\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// One thread per (B, C) position\n// Optimized for small inputs with read-only cache\n__global__ void __launch_bounds__(128) max_with_tiebreak_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = blockIdx.x * blockDim.x + threadIdx.x;\n int total = B * C;\n if (bc_idx >= total) return;\n \n // x is (R, B, C) contiguous, so x[r, b, c] = x[r * B * C + b * C + c]\n int base = bc_idx; // bc_idx already is b * C + c for contiguous access\n int stride = B * C;\n \n // Use __ldg for read-only cache\n float best_val = __ldg(x + base);\n int best_idx = 0;\n \n // Unroll loop for R=4 case\n #pragma unroll 4\n for (int r = 1; r < R; r++) {\n float val = __ldg(x + r * stride + base);\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n // Tie: keep smaller index (r is always > best_idx at this point in the loop)\n // Actually r > best_idx is not always true if we found a tie earlier\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Output is (B, C, 2) contiguous\n int out_base = bc_idx * 2;\n out[out_base] = best_val;\n out[out_base + 1] = (float)best_idx;\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 128;\n int blocks = (total + threads - 1) / threads;\n \n max_with_tiebreak_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext_v2\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← run_correctness (ok)
okrun_correctness
run_correctness PASSED: (5 / 5) trials all matched the reference.
Numerical precision: max_abs_err=0.00e+00 mean_abs_err=0.00e+00 max_rel_err=0.00e+00
turn 6236.2s turn · 7.0s LLM · 229.1s tools · tokens in=5,681 · out=746 · total=6,427
reasoning
Good, correctness still passes. Let me try another optimization - vectorized loads using float4 to improve memory throughput, and optimize further:
assistantLet me try with vectorized loads for better memory throughput:
call → compile_kernel
callcompile_kernel
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel with vectorized loads\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// Vectorized version: each thread handles 4 consecutive (B, C) positions\n__global__ void __launch_bounds__(128) max_with_tiebreak_vector_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = (blockIdx.x * blockDim.x + threadIdx.x) * 4;\n int total = B * C;\n \n // Use float4 for vectorized loads\n int stride = B * C;\n \n // Process 4 elements per thread\n #pragma unroll\n for (int i = 0; i < 4; i++) {\n int idx = bc_idx + i;\n if (idx >= total) return;\n \n float best_val = __ldg(x + idx);\n int best_idx = 0;\n \n #pragma unroll 4\n for (int r = 1; r < R; r++) {\n float val = __ldg(x + r * stride + idx);\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Write output\n out[idx * 2] = best_val;\n out[idx * 2 + 1] = (float)best_idx;\n }\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 128;\n int blocks = (total / 4 + threads - 1) / threads + 1; // +1 for remainder\n \n max_with_tiebreak_vector_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext_v3\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← compile_kernel (ok)
okcompile_kernel
compile_kernel PASSED: kernel compiled without errors.
turn 7165.8s turn · 6.2s LLM · 159.5s tools · tokens in=6,413 · out=716 · total=7,129
reasoning
Let me test correctness of this version.
call → run_correctness
callrun_correctness
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel with vectorized loads\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// Vectorized version: each thread handles 4 consecutive (B, C) positions\n__global__ void __launch_bounds__(128) max_with_tiebreak_vector_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = (blockIdx.x * blockDim.x + threadIdx.x) * 4;\n int total = B * C;\n \n // Use float4 for vectorized loads\n int stride = B * C;\n \n // Process 4 elements per thread\n #pragma unroll\n for (int i = 0; i < 4; i++) {\n int idx = bc_idx + i;\n if (idx >= total) return;\n \n float best_val = __ldg(x + idx);\n int best_idx = 0;\n \n #pragma unroll 4\n for (int r = 1; r < R; r++) {\n float val = __ldg(x + r * stride + idx);\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Write output\n out[idx * 2] = best_val;\n out[idx * 2 + 1] = (float)best_idx;\n }\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 128;\n int blocks = (total / 4 + threads - 1) / threads + 1; // +1 for remainder\n \n max_with_tiebreak_vector_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext_v3\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← run_correctness (ok)
okrun_correctness
run_correctness PASSED: (5 / 5) trials all matched the reference.
Numerical precision: max_abs_err=0.00e+00 mean_abs_err=0.00e+00 max_rel_err=0.00e+00
turn 86.7s turn · 6.7s LLM · 0.0s tools · tokens in=7,178 · out=738 · total=7,916
reasoning
Both versions pass correctness. Let me submit the vectorized version since it should be faster with better memory access patterns. Let me do a static check first.
call → static_check
callstatic_check
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel with vectorized loads\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// Vectorized version: each thread handles 4 consecutive (B, C) positions\n__global__ void __launch_bounds__(128) max_with_tiebreak_vector_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = (blockIdx.x * blockDim.x + threadIdx.x) * 4;\n int total = B * C;\n \n // Use float4 for vectorized loads\n int stride = B * C;\n \n // Process 4 elements per thread\n #pragma unroll\n for (int i = 0; i < 4; i++) {\n int idx = bc_idx + i;\n if (idx >= total) return;\n \n float best_val = __ldg(x + idx);\n int best_idx = 0;\n \n #pragma unroll 4\n for (int r = 1; r < R; r++) {\n float val = __ldg(x + r * stride + idx);\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Write output\n out[idx * 2] = best_val;\n out[idx * 2 + 1] = (float)best_idx;\n }\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 128;\n int blocks = (total / 4 + threads - 1) / threads + 1; // +1 for remainder\n \n max_with_tiebreak_vector_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext_v3\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← static_check (ok)
okstatic_check
static_check PASSED: no violations or warnings detected.
turn 9139.9s turn · 6.6s LLM · 133.3s tools · tokens in=7,931 · out=736 · total=8,667 · FINAL
reasoning
The kernel is correct and passes all checks. The vectorized version with 4 elements per thread should be efficient. Let me submit this version.
call → submit_kernel
callsubmit_kernel
{
"kernel_code": "import torch\nimport torch.nn as nn\nfrom torch.utils.cpp_extension import load_inline\n\n# Compile the CUDA kernel with vectorized loads\ncuda_source = \"\"\"\n#include <cuda.h>\n#include <cuda_runtime.h>\n#include <cfloat>\n\n// Vectorized version: each thread handles 4 consecutive (B, C) positions\n__global__ void __launch_bounds__(128) max_with_tiebreak_vector_kernel(\n const float* __restrict__ x, // shape (R, B, C), contiguous\n float* __restrict__ out, // shape (B, C, 2), contiguous\n int R, int B, int C\n) {\n int bc_idx = (blockIdx.x * blockDim.x + threadIdx.x) * 4;\n int total = B * C;\n \n // Use float4 for vectorized loads\n int stride = B * C;\n \n // Process 4 elements per thread\n #pragma unroll\n for (int i = 0; i < 4; i++) {\n int idx = bc_idx + i;\n if (idx >= total) return;\n \n float best_val = __ldg(x + idx);\n int best_idx = 0;\n \n #pragma unroll 4\n for (int r = 1; r < R; r++) {\n float val = __ldg(x + r * stride + idx);\n if (val > best_val) {\n best_val = val;\n best_idx = r;\n } else if (val == best_val) {\n if (r < best_idx) {\n best_idx = r;\n }\n }\n }\n \n // Write output\n out[idx * 2] = best_val;\n out[idx * 2 + 1] = (float)best_idx;\n }\n}\n\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {\n int R = x.size(0);\n int B = x.size(1);\n int C = x.size(2);\n \n int total = B * C;\n int threads = 128;\n int blocks = (total / 4 + threads - 1) / threads + 1; // +1 for remainder\n \n max_with_tiebreak_vector_kernel<<<blocks, threads>>>(\n x.data_ptr<float>(),\n out.data_ptr<float>(),\n R, B, C\n );\n}\n\"\"\"\n\ncpp_source = \"\"\"\nvoid launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);\n\"\"\"\n\n# Load the extension\nmax_tiebreak_ext = load_inline(\n name=\"max_tiebreak_ext_v3\",\n cpp_sources=cpp_source,\n cuda_sources=cuda_source,\n functions=[\"launch_max_with_tiebreak\"],\n verbose=False,\n)\n\n\nclass ModelNew(nn.Module):\n def __init__(self):\n super().__init__()\n self.ext = max_tiebreak_ext\n \n def forward(self, x: torch.Tensor) -> torch.Tensor:\n # x: (R, B, C)\n R, B, C = x.shape\n out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)\n self.ext.launch_max_with_tiebreak(x, out)\n return out"
}
result ← submit_kernel (ok)
oksubmit_kernel
submit_kernel PASSED: (5 / 5) correctness trials all passed.
Kernel runtime: 0.01 μs
Runtime stats: mean=0.01μs median=?μs std=0.00μs
Numerical precision: max_abs_err=0.00e+00 mean_abs_err=0.00e+00 max_rel_err=0.00e+00
Memory: 33.01 MB peak (ref: 33.01 MB, ratio: 0.9998x)
Kernel launches: 1 (ref: 94, fusion ratio: 94.0)
SOL score: 0.0025
Flagged for excessive speedup — this submission may have been rejected by automated review.
final submitted kernel (92 lines, 2,502 chars) — full source, untruncated
import torch
import torch.nn as nn
from torch.utils.cpp_extension import load_inline
# Compile the CUDA kernel with vectorized loads
cuda_source = """
#include <cuda.h>
#include <cuda_runtime.h>
#include <cfloat>
// Vectorized version: each thread handles 4 consecutive (B, C) positions
__global__ void __launch_bounds__(128) max_with_tiebreak_vector_kernel(
const float* __restrict__ x, // shape (R, B, C), contiguous
float* __restrict__ out, // shape (B, C, 2), contiguous
int R, int B, int C
) {
int bc_idx = (blockIdx.x * blockDim.x + threadIdx.x) * 4;
int total = B * C;
// Use float4 for vectorized loads
int stride = B * C;
// Process 4 elements per thread
#pragma unroll
for (int i = 0; i < 4; i++) {
int idx = bc_idx + i;
if (idx >= total) return;
float best_val = __ldg(x + idx);
int best_idx = 0;
#pragma unroll 4
for (int r = 1; r < R; r++) {
float val = __ldg(x + r * stride + idx);
if (val > best_val) {
best_val = val;
best_idx = r;
} else if (val == best_val) {
if (r < best_idx) {
best_idx = r;
}
}
}
// Write output
out[idx * 2] = best_val;
out[idx * 2 + 1] = (float)best_idx;
}
}
void launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out) {
int R = x.size(0);
int B = x.size(1);
int C = x.size(2);
int total = B * C;
int threads = 128;
int blocks = (total / 4 + threads - 1) / threads + 1; // +1 for remainder
max_with_tiebreak_vector_kernel<<<blocks, threads>>>(
x.data_ptr<float>(),
out.data_ptr<float>(),
R, B, C
);
}
"""
cpp_source = """
void launch_max_with_tiebreak(const torch::Tensor& x, torch::Tensor& out);
"""
# Load the extension
max_tiebreak_ext = load_inline(
name="max_tiebreak_ext_v3",
cpp_sources=cpp_source,
cuda_sources=cuda_source,
functions=["launch_max_with_tiebreak"],
verbose=False,
)
class ModelNew(nn.Module):
def __init__(self):
super().__init__()
self.ext = max_tiebreak_ext
def forward(self, x: torch.Tensor) -> torch.Tensor:
# x: (R, B, C)
R, B, C = x.shape
out = torch.empty(B, C, 2, dtype=torch.float32, device=x.device)
self.ext.launch_max_with_tiebreak(x, out)
return out