Commit 26114e0d authored by Yuefeng Zhou's avatar Yuefeng Zhou Committed by TensorFlower Gardener
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Use x*x instead of x^2 to calculate square in huber loss implementation. The... 

Use x*x instead of x^2 to calculate square in huber loss implementation. The reason is d(x^y)/dy = x^y * log(x) and when x is zero, it becomes NaN. Even if y is constant, the check op would still report failure.

PiperOrigin-RevId: 185208180
parent 8ced9faa
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