For a given matrix, return the determinant and the permanent of the matrix.

The determinant is given by

 



det
(
A
)
=

∑

σ


sgn
⁡
(
σ
)

∏

i
=
1


n



M

i
,

σ

i






{\displaystyle \det(A)=\sum _{\sigma }\operatorname {sgn}(\sigma )\prod _{i=1}^{n}M_{i,\sigma _{i}}}


while the permanent is given by

 



perm
⁡
(
A
)
=

∑

σ



∏

i
=
1


n



M

i
,

σ

i






{\displaystyle \operatorname {perm} (A)=\sum _{\sigma }\prod _{i=1}^{n}M_{i,\sigma _{i}}}


In both cases the sum is over the permutations 



σ


{\displaystyle \sigma }

 of the permutations of 1, 2, ..., n. (A permutation's sign is 1 if there are an even number of inversions and -1 otherwise; see parity of a permutation.)

More efficient algorithms for the determinant are known: LU decomposition, see for example wp:LU decomposition#Computing the determinant. Efficient methods for calculating the permanent are not known.



Related task

 Permutations by swapping


from itertools import permutations
from operator import mul
from math import fsum
from spermutations import spermutations
 
def prod(lst):
    return reduce(mul, lst, 1)
 
def perm(a):
    n = len(a)
    r = range(n)
    s = permutations(r)
    return fsum(prod(a[i][sigma[i]] for i in r) for sigma in s)
 
def det(a):
    n = len(a)
    r = range(n)
    s = spermutations(n)
    return fsum(sign * prod(a[i][sigma[i]] for i in r)
                for sigma, sign in s)
 
if __name__ == '__main__':
    from pprint import pprint as pp
 
    for a in ( 
            [
             [1, 2], 
             [3, 4]], 
 
            [
             [1, 2, 3, 4],
             [4, 5, 6, 7],
             [7, 8, 9, 10],
             [10, 11, 12, 13]],        
 
            [
             [ 0,  1,  2,  3,  4],
             [ 5,  6,  7,  8,  9],
             [10, 11, 12, 13, 14],
             [15, 16, 17, 18, 19],
             [20, 21, 22, 23, 24]],
        ):
        print('')
        pp(a)
        print('Perm: %s Det: %s' % (perm(a), det(a)))