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Getting around tf.argmax which is not differentiable

I've written a custom loss function for my neural network but it can't compute any gradients. I thinks it is because I need the index of the highest value and are therefore using argmax to get this index.

As argmax is not differentiable I to get around this but I don't know how it is possible.

Can anyone help?

like image 823
Martin Bagge Avatar asked Oct 25 '17 07:10

Martin Bagge


2 Answers

As aidan suggested, it's just a softargmax stretched to the limits by beta. We can use tf.nn.softmax to get around the numerical issues:

def softargmax(x, beta=1e10):
  x = tf.convert_to_tensor(x)
  x_range = tf.range(x.shape.as_list()[-1], dtype=x.dtype)
  return tf.reduce_sum(tf.nn.softmax(x*beta) * x_range, axis=-1)
like image 147
Nova Avatar answered Sep 17 '22 18:09

Nova


If you are cool with approximates,

import tensorflow as tf
import numpy as np

sess = tf.Session()
x = tf.placeholder(dtype=tf.float32, shape=(None,))
beta = tf.placeholder(dtype=tf.float32)

# Pseudo-math for the below
# y = sum( i * exp(beta * x[i]) ) / sum( exp(beta * x[i]) )
y = tf.reduce_sum(tf.cumsum(tf.ones_like(x)) * tf.exp(beta * x) / tf.reduce_sum(tf.exp(beta * x))) - 1

print("I can compute the gradient", tf.gradients(y, x))

for run in range(10):
    data = np.random.randn(10)
    print(data.argmax(), sess.run(y, feed_dict={x:data/np.linalg.norm(data), beta:1e2}))

This is using a trick that computing the mean in low temperature environments gives to the approximate maximum of the probability space. Low temperature in this case correlates with beta being very large.

In fact, as beta approaches infinity, my algorithm will converge to the maximum (assuming the maximum is unique). Unfortunately, beta can't get too large before you have numerical errors and get NaN, but there are tricks to solve that I can go into if you care.

The output looks something like,

0 2.24459
9 9.0
8 8.0
4 4.0
4 4.0
8 8.0
9 9.0
6 6.0
9 8.99995
1 1.0

So you can see that it messes up in some spots, but often gets the right answer. Depending on your algorithm, this might be fine.

like image 45
ignorance Avatar answered Sep 21 '22 18:09

ignorance