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authorsrcarroll <50210727+srcarroll@users.noreply.github.com>2024-03-17 11:23:30 -0500
committerGitHub <noreply@github.com>2024-03-17 11:23:30 -0500
commitd39ac3a8e010582c25e5d7e193ad3153402b1c4f (patch)
treee2c2f5beeda656c7bb1feccdf18a64e1c5d4cf6d /llvm/lib/CodeGen/ModuloSchedule.cpp
parent2c6fb7ce99dfad615ccf64464c79e65b26c1bc93 (diff)
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[mlir][math] Reland 58ef9bec071383744fb703ff08df9806f25e4095 (#85436)
The previous implementation decomposes tanh(x) into `(exp(2x) - 1)/(exp(2x)+1), x < 0` `(1 - exp(-2x))/(1 + exp(-2x)), x >= 0` This is fine as it avoids overflow with the exponential, but the whole decomposition is computed for both cases unconditionally, then the result is chosen based off the sign of the input. This results in doing two expensive exp computations. The proposed change avoids doing the whole computation twice by exploiting the reflection symmetry `tanh(-x) = -tanh(x)`. We can "normalize" the input to be positive by setting `y = sign(x) * x`, where the sign of `x` is computed as `sign(x) = (float)(x > 0) * (-2) + 1`. Then compute `z = tanh(y) `with the decomposition above for `x >=0` and "denormalize" the result `z * sign(x)` to retain the sign. The reason it is done this way is that it is very amenable to vectorization. This method trades the duplicate decomposition computations (which takes 5 instructions including an extra expensive exp and div) for 4 cheap instructions to compute the signs value `arith.cmpf `(which is a pre-existing instruction in the previous impl) `arith.sitofp` `arith.mulf` `arith.addf` and 1 more instruction to get the right sign in the result 5. `arith.mulf`. Moreover, numerically, this implementation will yield the exact same results as the previous implementation. As part of the relanding, a casting issue from the original commit has been fixed, i.e. casting bool to float with `uitofp`. Additionally a correctness test with `mlir-cpu-runner` has been added.
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