pyqubo.Array.dot¶

Array.
dot
(other)[source]¶ Returns a dot product of two arrays.
Parameters: other ( Array
) – Array.Returns: Express
/Array
Example
Dot calculation falls into four patterns.
 If both self and other are 1D arrays, it is inner product of vectors.
>>> from pyqubo import Array, Binary >>> array_a = Array([Binary('a'), Binary('b')]) >>> array_b = Array([Binary('c'), Binary('d')]) >>> array_a.dot(array_b) ((Binary(a)*Binary(c))+(Binary(b)*Binary(d)))
 If self is an ND array and other is a 1D array, it is a sum product over the last axis of self and other.
>>> array_a = Array([[Binary('a'), Binary('b')], [Binary('c'), Binary('d')]]) >>> array_b = Array([Binary('e'), Binary('f')]) >>> array_a.dot(array_b) Array([((Binary(a)*Binary(e))+(Binary(b)*Binary(f))), ((Binary(c)*Binary(e))+(Binary(d)*Binary(f)))])
 If both self and other are 2D arrays, it is matrix multiplication.
>>> array_a = Array([[Binary('a'), Binary('b')], [Binary('c'), Binary('d')]]) >>> array_b = Array([[Binary('e'), Binary('f')], [Binary('g'), Binary('h')]]) >>> array_a.dot(array_b) Array([[((Binary(a)*Binary(e))+(Binary(b)*Binary(g))), ((Binary(a)*Binary(f))+(Binary(b)*Binary(h)))], [((Binary(c)*Binary(e))+(Binary(d)*Binary(g))), ((Binary(c)*Binary(f))+(Binary(d)*Binary(h)))]])
 If self is an ND array and other is an MD array (where N, M>=2), it is a sum product over the last axis of self and the secondtolast axis of other. If N = M = 3, (i, j, k, m) element of a dot product of self and other is:
dot(self, other)[i,j,k,m] = sum(self[i,j,:] * other[k,:,m])
>>> array_a = Array.create('a', shape=(3, 2, 4), vartype='BINARY') >>> array_a.shape (3, 2, 4) >>> array_b = Array.create('b', shape=(5, 4, 3), vartype='BINARY') >>> array_b.shape (5, 4, 3) >>> i, j, k, m = (1, 1, 3, 2) >>> array_a.dot(array_b)[i, j, k, m] == sum(array_a[i, j, :] * array_b[k, :, m]) True
Dot product with list.
>>> array_a = Array([Binary('a'), Binary('b')]) >>> array_b = [3, 4] >>> array_a.dot(array_b) ((Binary(a)*Num(3))+(Binary(b)*Num(4)))