CODEWITHSUNDEEP
pythontensorflow
KJJ
KJJ

Reputation: 77

Confusing results with tensorflow's tensordot

In using tensorflow's tf.tensordot, I have run into some strange results. Running the following code block

import tensorflow as tf
import numpy as np
a = np.arange(6, dtype=np.int32).reshape(3,2)
b = np.arange(1,7, dtype=np.int32).reshape(2,3)
sess = tf.Session()
print(sess.run(tf.tensordot(a, b, [[0,1],[0,1]])))
print(sess.run(tf.tensordot(a, b, [[0,1],[1,0]])))
print(sess.run(tf.tensordot(a, b, [[1,0],[0,1]])))
print(sess.run(tf.tensordot(a, b, [[1,0],[1,0]])))

produces

70
65
65
60

I can't figure out what contraction is happening here. Another interesting thing is that trying to do this with numpy's tensordot will return an error for several of the attempted axes.

Upvotes: 2

Views: 267

Answers (1)

P-Gn
P-Gn

Reputation: 24661

You have found a bug in tensorflow.

According to the documentation of tf.tensordot,

The axis a_axes[i] of a must have the same dimension as axis b_axes[i] of b for all i in range(0, len(a_axes)).

So for example, tf.tensordot(a, b, [[0,1],[0,1]])) should have returned an error, because a is 3x2 and b is 2x3. But it does not — that's the bug.

Instead, it proceeds as if the tensors were compatible. If a and b were compatible, then tf.tensordot(a, b, [[0,1],[0,1]])) would be a simple dot product. What tf.tensordot does internally is that it basically flattens a and b and compute the dot product.

In your case, a and b have the same number of elements, therefore the computation of the dot product succeeds, despite the fact that they have an incompatible shape.

You can file a bug to the TF team here.

Upvotes: 1

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