Mathematical definition of tensordot operation on TensorFlow tensor

Question

I'm trying to reverse engineer the behavior of tf.tensordot axes parameter, but having a hard time.

Given the following code:

a = tf.constant([[1., 2.], [3., 4.], [4., 5.]])
b = tf.constant([1., 2.])
c = tf.constant([[1., 2.], [2., 3.], [3., 4.]])

print(f'Shape of c: {c.shape}')

ct = tf.transpose(c)

print(f'Shape of ct: {ct.shape}')

print('.................')

d = tf.tensordot(a, ct, axes=1)
print(f'Shape of d: {d.shape}')
print(d)

print('.................')


e = tf.tensordot(a, ct, axes=0)
print(f'Shape of e: {e.shape}')
print(e)


print('.................')


f = tf.tensordot(a, ct, axes=2)
print(f'Shape of f: {f.shape}')
print(e)

I understand how "d" is produced, but I don't understand how "e" and "f" are produced. The TensorFlow Documentation is not sufficient for me to understand.

Answer 1

I post an intermediate answer to a subset of parameters axes can take (when axes is an integer):

axes = 0: No addition is taking place, only multiplication. For example in:

c = tf.tensordot(a, b, axes=0)

This means if a had shape m,n and b had shape o,p, then c would be of shape m,n,o,p because the elements are multiplied, but not summed up

axes = 1:

c = tf.tensordot(a, b, axes=1)

If a had shape m,n and b had shape n,o then c would be of shape m,o because the elements are multiplied, but also summed up

Will update this post once I've learned more on the topic