Reputation: 87
I want to create a symmetric matrix of n*n
and train this matrix in TensorFlow. Effectively I should only train (n+1)*n/2
parameters. How should I do this?
I saw some previous threads which suggest do the following:
X = tf.Variable(tf.random_uniform([d,d], minval=-.1, maxval=.1, dtype=tf.float64))
X_symm = 0.5 * (X + tf.transpose(X))
However, this means I have to train n*n
variables, not n*(n+1)/2
variables.
Even there is no function to achieve this, a patch of self-written code would help!
Thanks!
Upvotes: 7
Views: 2734
Reputation: 41
gdelab's answer is correct and will work, since a neural network can adjust the 0.5 factor by itself. I aimed for a solution, where the neural network actually only has (n+1)*n/2
output neurons. The following function transforms these into a symmetric matrix:
def create_symmetric_matrix(x,n):
x_rev = tf.reverse(x[:, n:], [1])
xc = tf.concat([x, x_rev], axis=1)
x_res = tf.reshape(xc, [-1, n, n])
x_upper_triangular = tf.linalg.band_part(x_res, 0, -1)
x_lower_triangular = tf.linalg.set_diag( tf.transpose(x_upper_triangular, perm=[0, 2, 1]), tf.zeros([tf.shape(x)[0], n], dtype=tf.float32))
return x_upper_triangular + x_lower_triangular
with x
as a vector of rank [batch,n*(n+1)/2]
and n
as the rank of the output matrix.
The code is inspired by tfp.math.fill_triangular.
Upvotes: 0
Reputation: 55
Referring to answer of gdelab: in Tensorflow 2.x, you have to use following code.
X_upper = tf.linalg.band_part(X, 0, -1)
Upvotes: 1
Reputation: 6220
You can use tf.matrix_band_part(input, 0, -1)
to create an upper triangular matrix from a square one, so this code would allow you to train on n(n+1)/2
variables although it has you create n*n
:
X = tf.Variable(tf.random_uniform([d,d], minval=-.1, maxval=.1, dtype=tf.float64))
X_upper = tf.matrix_band_part(X, 0, -1)
X_symm = 0.5 * (X_upper + tf.transpose(X_upper))
Upvotes: 8