How to implement the Softmax derivative independently from any loss function?

danijar picture danijar · Nov 5, 2015 · Viewed 15.5k times · Source

For a neural networks library I implemented some activation functions and loss functions and their derivatives. They can be combined arbitrarily and the derivative at the output layers just becomes the product of the loss derivative and the activation derivative.

However, I failed to implement the derivative of the Softmax activation function independently from any loss function. Due to the normalization i.e. the denominator in the equation, changing a single input activation changes all output activations and not just one.

Here is my Softmax implementation where the derivative fails the gradient checking by about 1%. How can I implement the Softmax derivative so that it can be combined with any loss function?

import numpy as np


class Softmax:

    def compute(self, incoming):
        exps = np.exp(incoming)
        return exps / exps.sum()

    def delta(self, incoming, outgoing):
        exps = np.exp(incoming)
        others = exps.sum() - exps
        return 1 / (2 + exps / others + others / exps)


activation = Softmax()
cost = SquaredError()

outgoing = activation.compute(incoming)
delta_output_layer = activation.delta(incoming) * cost.delta(outgoing)

Answer

ticcky picture ticcky · Nov 7, 2015

It should be like this: (x is the input to the softmax layer and dy is the delta coming from the loss above it)

    dx = y * dy
    s = dx.sum(axis=dx.ndim - 1, keepdims=True)
    dx -= y * s

    return dx

But the way you compute the error should be:

    yact = activation.compute(x)
    ycost = cost.compute(yact)
    dsoftmax = activation.delta(x, cost.delta(yact, ycost, ytrue)) 

Explanation: Because the delta function is a part of the backpropagation algorithm, its responsibility is to multiply the vector dy (in my code, outgoing in your case) by the Jacobian of the compute(x) function evaluated at x. If you work out what does this Jacobian look like for softmax [1], and then multiply it from the left by a vector dy, after a bit of algebra you'll find out that you get something that corresponds to my Python code.

[1] https://stats.stackexchange.com/questions/79454/softmax-layer-in-a-neural-network