Merge OpenAI Triton commit c186592
#5480
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…ons in matmul (#8483)
…nerically (#8421) (#8495) This PR relands triton-lang/triton#8386. It depends on triton-lang/triton#8492 to avoid regressing in some workloads.
There's silent data corruption when calling `tl.histogram` with
interpreter.
```python
# test.py
import torch
import ctypes
import triton
import triton.language as tl
@triton.jit
def histogram_kernel(x_ptr, z_ptr):
offset = tl.arange(0, 1)
x = tl.load(x_ptr + offset)
z = tl.histogram(x, 1)
buf = (ctypes.c_int32 * 2).from_address(int(z_ptr))
print(f'before store: {list(buf)}')
tl.store(z_ptr + offset, z) # tl.store treats z values as int64 while they're int32
print(f'after store: {list(buf)}')
device = 'cpu'
torch.manual_seed(17)
x = torch.ones(1, device=device, dtype=torch.int32)
z = torch.ones(2, dtype=torch.int32, device=device)
histogram_kernel[(1, )](x, z)
# Output:
# TRITON_INTERPRET=1 TRITON_TEST_SUITE=interpreter python test.py
# before store: [1, 1]
# after store: [1, 0] <- second element shouldn't be cleared
```
Based on `np.histogram` docs:
https://numpy.org/doc/2.3/reference/generated/numpy.histogram.html
Returned dtype is taken account when optional weights param is passed,
int64 othwerwise.
That leads to `tl.store` thinking it's saving int64 values while there's
int32 in my example tensor passed, so it's writing 8 bytes at once
instead of 4 bytes, leading to writing 4 bytes exceeding it's data range
causing silent data corruption.
```python
import numpy as np
data = np.array([1], dtype=np.int32)
bins = 1
print(f'Data dtype before: {data.dtype}')
histogram = np.histogram(data, bins=bins, range=(0, bins))[0]
print(f'Data dtype after: {histogram.dtype}')
# Data dtype before: int32
# Data dtype after: int64
```
Applying "dummy_weights" fixes returned data type as expected fixing
data corruption.
------------------------------
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- [x] I am not making a trivial change, such as fixing a typo in a
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- [x] I have written a PR description following these
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- [x] This PR does not need a test because np.histogram specific
behavior with interpreter mode.
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… transactions (#8575) `ttg.async_wait` counts the number of outstanding `ttg.commit_groups`. However, when lowering to LLVM on AMD we require the number of outstanding async intrinsics/final assembly instructions. The conversion is already done by `UpdateAsyncWaitCnt` which modifies the `num` of `ttg.async_wait` in place. This PR introduces a new op `amdgpu.async_wait` to make the change in semantics explicit in the IR. `UpdateAsyncWaitCount` is moved to `TTGIR->LLVM` primarily to also include in for `Gluon` kernels and we should always call it since it will only have an effect if there are `ttg.async_wait` ops present in the kernel. To avoid membar changes this also adds a `ttgpu.LocalBarrier` after each `amdgpu.async_wait`. Membar will respect the newly added barrier and behave the same as for `ttg.async_wait`.
Fixes #8578
We're using the wrong output constraint which leads llvm to extend the
fp16 value to 32-bits. Fixing the constraint removes the conversion.
Note that we still end up with a no-op sequence like:
```ptx
mov.b32 {%rs1, %rs2}, %r1
mov.b32 %r2, {%rs1, %rs2}
```
However, `ptxas` is able to optimize these out.
### The Problem with the Original Formula The original formula is: ``` tanh(x) = (e^(2x) - 1) / (e^(2x) + 1) ``` - Issue with large positive x: - When x = 20: e^(40) ≈ 2.4 × 10^17 → manageable - When x = 50: e^(100) ≈ 2.7 × 10^43 → overflow to infinity - Result: (∞ - 1)/(∞ + 1) = NaN x - For negative x: The formula actually works fine because e^(2x) → 0, giving (-1)/(1) = -1 ### The Numerically Stable Solution - For Positive x: Reformulation ``` tanh(x) = (e^(2x) - 1) / (e^(2x) + 1) = (e^(2x) + 1 - 2) / (e^(2x) + 1) = 1 - 2/(e^(2x) + 1) ``` - For Negative x: Using Symmetry ``` tanh(-x) = (e^(-2x) - 1) / (e^(-2x) + 1) = (2/(e^(-2x) + 1) - 1) = -1 × (1 - 2/(e^(2|x|) + 1)) ``` ### Unified formulation: ``` tanh(x) = sign(x) × (1 - 2/(e^(2|x|) + 1)) ```
anmyachev
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Nov 14, 2025
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Signed-off-by: Whitney Tsang <[email protected]>
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This PR changes the Triton base from 3c2e6f8 to c186592 (Oct 29).
Pass rate: 94.91%->94.95%