Numpy matrix power/exponent with modulo?

Question

Is it possible to use numpy\'s linalg.matrix_power with a modulo so the elements don\'t grow larger than a certain value?

Answer 1

Using the implementation from Numpy:

https://github.com/numpy/numpy/blob/master/numpy/matrixlib/defmatrix.py#L98

I adapted it by adding a modulo term. HOWEVER, there is a bug, in that if an overflow occurs, no OverflowError or any other sort of exception is raised. From that point on, the solution will be wrong. There is a bug report here.

Here is the code. Use with care:

from numpy.core.numeric import concatenate, isscalar, binary_repr, identity, asanyarray, dot
from numpy.core.numerictypes import issubdtype    
def matrix_power(M, n, mod_val):
    # Implementation shadows numpy's matrix_power, but with modulo included
    M = asanyarray(M)
    if len(M.shape) != 2 or M.shape[0] != M.shape[1]:
        raise ValueError("input  must be a square array")
    if not issubdtype(type(n), int):
        raise TypeError("exponent must be an integer")

    from numpy.linalg import inv

    if n==0:
        M = M.copy()
        M[:] = identity(M.shape[0])
        return M
    elif n<0:
        M = inv(M)
        n *= -1

    result = M % mod_val
    if n <= 3:
        for _ in range(n-1):
            result = dot(result, M) % mod_val
        return result

    # binary decompositon to reduce the number of matrix
    # multiplications for n > 3
    beta = binary_repr(n)
    Z, q, t = M, 0, len(beta)
    while beta[t-q-1] == '0':
        Z = dot(Z, Z) % mod_val
        q += 1
    result = Z
    for k in range(q+1, t):
        Z = dot(Z, Z) % mod_val
        if beta[t-k-1] == '1':
            result = dot(result, Z) % mod_val
    return result % mod_val

Answer 2

I had overflow issues with all the previous solutions, so I had to write an algorithm that accounts for overflows after every single integer multiplication. This is how I did it:

def matrix_power_mod(x, n, modulus):
    x = np.asanyarray(x)
    if len(x.shape) != 2:
        raise ValueError("input must be a matrix")
    if x.shape[0] != x.shape[1]:
        raise ValueError("input must be a square matrix")
    if not isinstance(n, int):
        raise ValueError("power must be an integer")

    if n < 0:
        x = np.linalg.inv(x)
        n = -n
    if n == 0:
        return np.identity(x.shape[0], dtype=x.dtype)
    y = None
    while n > 1:
        if n % 2 == 1:
            y = _matrix_mul_mod_opt(x, y, modulus=modulus)
        x = _matrix_mul_mod(x, x, modulus=modulus)
        n = n // 2
    return _matrix_mul_mod_opt(x, y, modulus=modulus)


def matrix_mul_mod(a, b, modulus):
    if len(a.shape) != 2:
        raise ValueError("input a must be a matrix")
    if len(b.shape) != 2:
        raise ValueError("input b must be a matrix")
    if a.shape[1] != a.shape[0]:
        raise ValueError("input a and b must have compatible shape for multiplication")
    return _matrix_mul_mod(a, b, modulus=modulus)


def _matrix_mul_mod_opt(a, b, modulus):
    if b is None:
        return a
    return _matrix_mul_mod(a, b, modulus=modulus)


def _matrix_mul_mod(a, b, modulus):
    r = np.zeros((a.shape[0], b.shape[1]), dtype=a.dtype)
    bT = b.T
    for rowindex in range(r.shape[0]):
        x = (a[rowindex, :] * bT) % modulus
        x = np.sum(x, 1) % modulus
        r[rowindex, :] = x
    return r

Answer 3

In order to prevent overflow, you can use the fact that you get the same result if you first take the modulo of each of your input numbers; in fact:

(M**k) mod p = ([M mod p]**k) mod p,

for a matrix M. This comes from the following two fundamental identities, which are valid for integers x and y:

(x+y) mod p = ([x mod p]+[y mod p]) mod p  # All additions can be done on numbers *modulo p*
(x*y) mod p = ([x mod p]*[y mod p]) mod p  # All multiplications can be done on numbers *modulo p*

The same identities hold for matrices as well, since matrix addition and multiplication can be expressed through scalar addition and multiplication. With this, you only exponentiate small numbers (n mod p is generally much smaller than n) and are much less likely to get overflows. In NumPy, you would therefore simply do

((arr % p)**k) % p

in order to get (arr**k) mod p.

If this is still not enough (i.e., if there is a risk that [n mod p]**k causes overflow despite n mod p being small), you can break up the exponentiation into multiple exponentiations. The fundamental identities above yield

(n**[a+b]) mod p = ([{n mod p}**a mod p] * [{n mod p}**b mod p]) mod p

and

(n**[a*b]) mod p = ([n mod p]**a mod p)**b mod p.

Thus, you can break up power k as a+b+… or a*b*… or any combination thereof. The identities above allow you to perform only exponentiations of small numbers by small numbers, which greatly lowers the risk of integer overflows.

Answer 4

What's wrong with the obvious approach?

E.g.

import numpy as np

x = np.arange(100).reshape(10,10)
y = np.linalg.matrix_power(x, 2) % 50