Gradient for tf.multinomial

[Sicheng Wang]

I am a user of Keras with TF backend. When implementing a custom layer in keras, I did something like this:

def build(self, input_shape):

    self.weight = self.add_weight(shape=(1,self.num_classes),trainable=True,name='prob')

def call(self,x):

   x_shape = K.shape(x)[0] 

   weight = K.softmax(self.weight)

   samples = tf.multinomial(tf.log(weight), x_shape,output_dtype='int32')

   return K.cast(K.transpose(SB),dtype='float32')

def compute_output_shape(self, input_shape):

   return (input_shape[0],1)

The gist is to get x_shape number of samples. However, after compiling and trying to train the layer, It raised the following error:

    raise ValueError('An operation has `None` for gradient. '

ValueError: An operation has `None` for gradient. Please make sure that all of your ops have a gradient defined (i.e. are differentiable). Common ops without gradient: K.argmax, K.round, K.eval.

Does that mean gradient is not implemented for  tf.multinomial?

Thank you.

[Martin Wicke]

Can you remove cast? Cast may not have a gradient.

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[Sicheng Wang]

Actually if I do

x = tf.Variable([[5,3,2]])

y = tf.cast(x,dtype=tf.float32)

z = tf.multinomial(x,10,output_dtype=tf.int32)

Then both gradients yield None:

tf.gradients(y,x)

[None]

tf.gradients(z,x)

[None]

So it seems there is no way out?

Sicheng

[Martin Wicke]

If the output type is int, the gradient will be none. Ints are not typically differentiable. 

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