# jax.numpy.linalg.multi_dot¶

jax.numpy.linalg.multi_dot(arrays, *, precision=None)[source]
Compute the dot product of two or more arrays in a single function call,

while automatically selecting the fastest evaluation order.

LAX-backend implementation of multi_dot(). Original docstring below.

multi_dot chains numpy.dot and uses optimal parenthesization of the matrices 1 2. Depending on the shapes of the matrices, this can speed up the multiplication a lot.

If the first argument is 1-D it is treated as a row vector. If the last argument is 1-D it is treated as a column vector. The other arguments must be 2-D.

Think of multi_dot as:

def multi_dot(arrays): return functools.reduce(np.dot, arrays)

Parameters

arrays (sequence of array_like) – If the first argument is 1-D it is treated as row vector. If the last argument is 1-D it is treated as column vector. The other arguments must be 2-D.

Returns

output – Returns the dot product of the supplied arrays.

Return type

ndarray

See also

dot()

dot multiplication with two arguments.

References

1

Cormen, “Introduction to Algorithms”, Chapter 15.2, p. 370-378

2

https://en.wikipedia.org/wiki/Matrix_chain_multiplication

Examples

multi_dot allows you to write:

>>> from numpy.linalg import multi_dot
>>> # Prepare some data
>>> A = np.random.random((10000, 100))
>>> B = np.random.random((100, 1000))
>>> C = np.random.random((1000, 5))
>>> D = np.random.random((5, 333))
>>> # the actual dot multiplication
>>> _ = multi_dot([A, B, C, D])


instead of:

>>> _ = np.dot(np.dot(np.dot(A, B), C), D)
>>> # or
>>> _ = A.dot(B).dot(C).dot(D)


Notes

The cost for a matrix multiplication can be calculated with the following function:

def cost(A, B):
return A.shape[0] * A.shape[1] * B.shape[1]


Assume we have three matrices $$A_{10x100}, B_{100x5}, C_{5x50}$$.

The costs for the two different parenthesizations are as follows:

cost((AB)C) = 10*100*5 + 10*5*50   = 5000 + 2500   = 7500
cost(A(BC)) = 10*100*50 + 100*5*50 = 50000 + 25000 = 75000