Matrix arithmetic

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Task
Matrix arithmetic
You are encouraged to solve this task according to the task description, using any language you may know.

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

The determinant is given by

\det(A) = \sum_\sigma\sgn(\sigma)\prod_{i=1}^n M_{i,\sigma_i}

while the permanent is given by

 \operatorname{perm}(A)=\sum_\sigma\prod_{i=1}^n M_{i,\sigma_i}

In both cases the sum is over the permutations σ 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.

Cf.

Contents

[edit] C

C99 code. By no means efficient or reliable. If you need it for serious work, go find a serious library.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
double det_in(double **in, int n, int perm)
{
if (n == 1) return in[0][0];
 
double sum = 0, *m[--n];
for (int i = 0; i < n; i++)
m[i] = in[i + 1] + 1;
 
for (int i = 0, sgn = 1; i <= n; i++) {
sum += sgn * (in[i][0] * det_in(m, n, perm));
if (i == n) break;
 
m[i] = in[i] + 1;
if (!perm) sgn = -sgn;
}
return sum;
}
 
/* wrapper function */
double det(double *in, int n, int perm)
{
double *m[n];
for (int i = 0; i < n; i++)
m[i] = in + (n * i);
 
return det_in(m, n, perm);
}
 
int main(void)
{
double x[] = { 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 };
 
printf("det:  %14.12g\n", det(x, 5, 0));
printf("perm: %14.12g\n", det(x, 5, 1));
 
return 0;
}

A method to calculate determinant that might actually be usable:

#include <stdio.h>
#include <stdlib.h>
#include <tgmath.h>
 
void showmat(const char *s, double **m, int n)
{
printf("%s:\n", s);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
printf("%12.4f", m[i][j]);
putchar('\n');
}
}
 
int trianglize(double **m, int n)
{
int sign = 1;
for (int i = 0; i < n; i++) {
int max = 0;
 
for (int row = i; row < n; row++)
if (fabs(m[row][i]) > fabs(m[max][i]))
max = row;
 
if (max) {
sign = -sign;
double *tmp = m[i];
m[i] = m[max], m[max] = tmp;
}
 
if (!m[i][i]) return 0;
 
for (int row = i + 1; row < n; row++) {
double r = m[row][i] / m[i][i];
if (!r) continue;
 
for (int col = i; col < n; col ++)
m[row][col] -= m[i][col] * r;
}
}
return sign;
}
 
double det(double *in, int n)
{
double *m[n];
m[0] = in;
 
for (int i = 1; i < n; i++)
m[i] = m[i - 1] + n;
 
showmat("Matrix", m, n);
 
int sign = trianglize(m, n);
if (!sign)
return 0;
 
showmat("Upper triangle", m, n);
 
double p = 1;
for (int i = 0; i < n; i++)
p *= m[i][i];
return p * sign;
}
 
#define N 18
int main(void)
{
double x[N * N];
srand(0);
for (int i = 0; i < N * N; i++)
x[i] = rand() % N;
 
printf("det: %19f\n", det(x, N));
return 0;
}

[edit] D

This requires the modules from the Permutations and Permutations by swapping Tasks.

Translation of: Python
import std.algorithm, std.range, std.traits, permutations2,
permutations_by_swapping1;
 
auto prod(Range)(Range r) nothrow @safe @nogc {
return reduce!q{a * b}(ForeachType!Range(1), r);
}
 
T permanent(T)(in T[][] a) nothrow @safe
in {
assert(a.all!(row => row.length == a[0].length));
} body {
auto r = a.length.iota;
T tot = 0;
foreach (const sigma; r.array.permutations)
tot += r.map!(i => a[i][sigma[i]]).prod;
return tot;
}
 
T determinant(T)(in T[][] a) nothrow
in {
assert(a.all!(row => row.length == a[0].length));
} body {
immutable n = a.length;
auto r = n.iota;
T tot = 0;
//foreach (sigma, sign; n.spermutations) {
foreach (const sigma_sign; n.spermutations) {
const sigma = sigma_sign[0];
immutable sign = sigma_sign[1];
tot += sign * r.map!(i => a[i][sigma[i]]).prod;
}
return tot;
}
 
void main() {
import std.stdio;
 
foreach (/*immutable*/ const a; [[[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]]]) {
writefln("[%([%(%2s, %)],\n %)]]", a);
writefln("Permanent: %s, determinant: %s\n",
a.permanent, a.determinant);
}
}
Output:
[[ 1,  2],
 [ 3,  4]]
Permanent: 10, determinant: -2

[[ 1,  2,  3,  4],
 [ 4,  5,  6,  7],
 [ 7,  8,  9, 10],
 [10, 11, 12, 13]]
Permanent: 29556, determinant: 0

[[ 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]]
Permanent: 6778800, determinant: 0

[edit] Fortran

Please find the compilation and example run at the start of the comments in the f90 source. Thank you.

 
!-*- mode: compilation; default-directory: "/tmp/" -*-
!Compilation started at Sat May 18 23:25:42
!
!a=./F && make $a && $a < unixdict.txt
!f95 -Wall -ffree-form F.F -o F
! j example, determinant: 7.00000000
! j example, permanent: 5.00000000
! maxima, determinant: -360.000000
! maxima, permanent: 900.000000
!
!Compilation finished at Sat May 18 23:25:43
 
 
 
! NB. example computed by J
! NB. fixed seed random matrix
! _2+3 3?.@$5
! 2 _1 1
!_1 _2 1
!_1 _1 _1
!
! (-/ .*)_2+3 3?.@$5 NB. determinant
!7
! (+/ .*)_2+3 3?.@$5 NB. permanent
!5
 
!maxima example
!a: matrix([2, 9, 4], [7, 5, 3], [6, 1, 8])$
!determinant(a);
!-360
!
!permanent(a);
!900
 
 
! compute permanent or determinant
program f
implicit none
real, dimension(3,3) :: j, m
data j/ 2,-1, 1,-1,-2, 1,-1,-1,-1/
data m/2, 9, 4, 7, 5, 3, 6, 1, 8/
write(6,*) 'j example, determinant: ',det(j,3,-1)
write(6,*) 'j example, permanent: ',det(j,3,1)
write(6,*) 'maxima, determinant: ',det(m,3,-1)
write(6,*) 'maxima, permanent: ',det(m,3,1)
 
contains
 
recursive function det(a,n,permanent) result(accumulation)
! setting permanent to 1 computes the permanent.
! setting permanent to -1 computes the determinant.
real, dimension(n,n), intent(in) :: a
integer, intent(in) :: n, permanent
real, dimension(n-1, n-1) :: b
real :: accumulation
integer :: i, sgn
if (n .eq. 1) then
accumulation = a(1,1)
else
accumulation = 0
sgn = 1
do i=1, n
b(:, :(i-1)) = a(2:, :i-1)
b(:, i:) = a(2:, i+1:)
accumulation = accumulation + sgn * a(1, i) * det(b, n-1, permanent)
sgn = sgn * permanent
enddo
endif
end function det
 
end program f
 

[edit] FunL

From the task description:

def sgn( p ) = product( (if s(0) < s(1) xor i(0) < i(1) then -1 else 1) | (s, i) <- p.combinations(2).zip( (0:p.length()).combinations(2) ) )
 
def perm( m ) = sum( product(m(i, sigma(i)) | i <- 0:m.length()) | sigma <- (0:m.length()).permutations() )
 
def det( m ) = sum( sgn(sigma)*product(m(i, sigma(i)) | i <- 0:m.length()) | sigma <- (0:m.length()).permutations() )

Laplace expansion (recursive):

def perm( m )
| m.length() == 1 and m(0).length() == 1 = m(0, 0)
| otherwise = sum( m(i, 0)*perm(m(0:i, 1:m.length()) + m(i+1:m.length(), 1:m.length())) | i <- 0:m.length() )
 
def det( m )
| m.length() == 1 and m(0).length() == 1 = m(0, 0)
| otherwise = sum( (-1)^i*m(i, 0)*det(m(0:i, 1:m.length()) + m(i+1:m.length(), 1:m.length())) | i <- 0:m.length() )

Test using the first set of definitions (from task description):

matrices = [
( (1, 2),
(3, 4)),
( (-2, 2, -3),
(-1, 1, 3),
( 2, 0, -1)),
( ( 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)) ]
 
for m <- matrices
println( m, 'perm: ' + perm(m), 'det: ' + det(m) )
Output:
((1, 2), (3, 4)), perm: 10, det: -2
((-2, 2, -3), (-1, 1, 3), (2, 0, -1)), perm: 10, det: 18
((1, 2, 3, 4), (4, 5, 6, 7), (7, 8, 9, 10), (10, 11, 12, 13)), perm: 29556, det: 0
((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)), perm: 6778800, det: 0

[edit] Go

Importing the permute packge from the Permutations by swapping task.

package main
 
import (
"fmt"
"permute"
)
 
func determinant(m [][]float64) (d float64) {
p := make([]int, len(m))
for i := range p {
p[i] = i
}
it := permute.Iter(p)
for s := it(); s != 0; s = it() {
pr := 1.
for i, σ := range p {
pr *= m[i][σ]
}
d += float64(s) * pr
}
return
}
 
func permanent(m [][]float64) (d float64) {
p := make([]int, len(m))
for i := range p {
p[i] = i
}
it := permute.Iter(p)
for s := it(); s != 0; s = it() {
pr := 1.
for i, σ := range p {
pr *= m[i][σ]
}
d += pr
}
return
}
 
var m2 = [][]float64{
{1, 2},
{3, 4}}
 
var m3 = [][]float64{
{2, 9, 4},
{7, 5, 3},
{6, 1, 8}}
 
func main() {
fmt.Println(determinant(m2), permanent(m2))
fmt.Println(determinant(m3), permanent(m3))
}
Output:
-2 10
-360 900

[edit] J

J has a conjunction for defining verbs which can act as determinant. This conjunction is symbolized as a space followed by a dot. (See also: "definitions" for some details on how to read that definition.)

For people who do not care to sort out what "recursive expansion by minors" means: -/ .* y gives the determinant of y and +/ .* y gives the permanent of y.

For example, given the matrix:

   i. 5 5
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

Its determinant is 0. When we use IEEE floating point, we only get an approximation of this result:

   -/ .* i. 5 5
_1.30277e_44

If we use exact (rational) arithmetic, we get a precise result:

   -/ .* i. 5 5x
0

The permanent does not have this problem in this example (the matrix contains no negative values and permanent does not use subtraction):

   +/ .* i. 5 5
6778800

As an aside, note that for specific verbs (like -/ .*) J uses an algorithm which is more efficient than the brute force approach.

[edit] Julia

The determinant of a matrix A can be computed by the built-in function

det(A)
Translation of: Python

The following function computes the permanent of a matrix A from the definition:

function perm(A)
m, n = size(A)
if m != n; throw(ArgumentError("permanent is for square matrices only")); end
sum(σ -> prod(i -> A[i,σ[i]], 1:n), permutations(1:n))
end

Example output:

julia> A = [2 9 4; 7 5 3; 6 1 8]
julia> det(A), perm(A)
(-360.0,900)

[edit] МК-61/52

П4	ИПE	П2	КИП0	ИП0	П1	С/П	ИП4	/	КП2
L1	06	ИПE	П3	ИП0	П1	Сx	КП2	L1	17
ИП0	ИП2	+	П1	П2	ИП3	-	x#0	34	С/П
ПП	80	БП	21	КИП0	ИП4	С/П	КИП2	-	*
П4	ИП0	П3	x#0	35	Вx	С/П	КИП2	-	<->
/	КП1	L3	45	ИП1	ИП0	+	П3	ИПE	П1
П2	КИП1	/-/	ПП	80	ИП3	+	П3	ИП1	-
x=0	61	ИП0	П1	КИП3	КП2	L1	74	БП	12
ИП0	<->	^	КИП3	*	КИП1	+	КП2	->	L0
82	->	П0	В/О

This program calculates the determinant of the matrix of order <= 5. Prior to startup, РE entered 13, entered the order of the matrix Р0, and the elements are introduced with the launch of the program after one of them, the last on the screen will be determinant. Permanent is calculated in this way.

[edit] Maple

M:=<<2|9|4>,<7|5|3>,<6|1|8>>:
 
with(LinearAlgebra):
 
Determinant(M);
Permanent(M);

Output:

                                    -360
                                     900

[edit] Mathematica

Determinant is a built in function Det

Permanent[m_List] :=
With[{v = Array[x, Length[m]]},
Coefficient[Times @@ (m.v), Times @@ v]
]

[edit] Maxima

a: matrix([2, 9, 4], [7, 5, 3], [6, 1, 8])$
 
determinant(a);
-360
 
permanent(a);
900

[edit] PARI/GP

The determinant is built in:

matdet(M)

and the permanent can be defined as

matperm(M)=my(n=#M,t);sum(i=1,n!,t=numtoperm(n,i);prod(j=1,n,M[j,t[j]]))

[edit] Perl 6

Works with: Rakudo version 2013.05

Uses the permutations generator from the Permutations by swapping task. This implementation is naive and brute-force (slow) but exact.

sub insert( $x, @xs) { [@xs[0..$_-1], $x, @xs[$_..*]] for 0..@xs }
sub order ($sg, @xs) { $sg > 0 ?? @xs !! @xs.reverse }
 
multi σ_permutations ([]) { [] => 1 }
 
multi σ_permutations ([$x, *@xs]) {
σ_permutations(@xs).map({ order($_.value, insert($x, $_.key)) }) Z=> (1,-1) xx *
}
 
sub m_arith ( @a, $op ) {
note "Not a square matrix" and return
if [||] map { @a.elems cmp @a[$_].elems }, ^@a;
[+] map {
my $permutation = .key;
my $term = $op eq 'perm' ?? 1 !! .value;
for $permutation.kv -> $i, $j { $term *= @a[$i][$j] };
$term
}, σ_permutations [^@a];
}
 
########### Testing ###########
 
my @tests = (
[
[ 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 ]
]
);
 
sub dump (@matrix) {
say $_».fmt: "%3s" for @matrix, '';
}
 
for @tests -> @matrix {
say 'Matrix:';
@matrix.&dump;
say "Determinant:\t", @matrix.&m_arith: <det>;
say "Permanent: \t", @matrix.&m_arith: <perm>;
say '-' x 25;
}
 

Output

Matrix:
  1   2
  3   4
   
Determinant:	-2
Permanent:  	10
-------------------------
Matrix:
  1   2   3   4
  4   5   6   7
  7   8   9  10
 10  11  12  13
   
Determinant:	0
Permanent:  	29556
-------------------------
Matrix:
  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
   
Determinant:	0
Permanent:  	6778800
-------------------------

[edit] Python

Using the module file spermutations.py from Permutations by swapping. The algorithm for the determinant is a more literal translation of the expression in the task description and the Wikipedia reference.

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)))
Sample output
[[1, 2], [3, 4]]
Perm: 10 Det: -2

[[1, 2, 3, 4], [4, 5, 6, 7], [7, 8, 9, 10], [10, 11, 12, 13]]
Perm: 29556 Det: 0

[[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]]
Perm: 6778800 Det: 0

The second matrix above is that used in the Tcl example. The third matrix is from the J language example. Note that the determinant seems to be 'exact' using this method of calculation without needing to resort to other than Pythons default numbers.

[edit] Racket

 
#lang racket
(require math)
(define determinant matrix-determinant)
 
(define (permanent M)
(define n (matrix-num-rows M))
(for/sum ([σ (in-permutations (range n))])
(for/product ([i n] [σi σ])
(matrix-ref M i σi))))
 

[edit] REXX

/* REXX ***************************************************************
* Test the two functions determinant and permanent
* using the matrix specifications shown for other languages
* 21.05.2013 Walter Pachl
**********************************************************************/

Call test ' 1 2',
' 3 4',2
 
Call test ' 1 2 3 4',
' 4 5 6 7',
' 7 8 9 10',
'10 11 12 13',4
 
Call test ' 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',5
 
Exit
 
test:
/**********************************************************************
* Show the given matrix and compute and show determinant and permanent
**********************************************************************/

Parse Arg as,n
asc=as
Do i=1 To n
ol=''
Do j=1 To n
Parse Var asc a.i.j asc
ol=ol right(a.i.j,3)
End
Say ol
End
Say 'determinant='right(determinant(as),7)
Say ' permanent='right(permanent(as),7)
Say copies('-',50)
Return
/* REXX ***************************************************************
* determinant.rex
* compute the determinant of the given square matrix
* Input: as: the representation of the matrix as vector (n**2 elements)
* 21.05.2013 Walter Pachl
**********************************************************************/

Parse Arg as
n=sqrt(words(as))
Do i=1 To n
Do j=1 To n
Parse Var as a.i.j as
End
End
Select
When n=2 Then det=a.1.1*a.2.2-a.1.2*a.2.1
When n=3 Then det= a.1.1*a.2.2*a.3.3,
+a.1.2*a.2.3*a.3.1,
+a.1.3*a.2.1*a.3.2,
-a.1.3*a.2.2*a.3.1,
-a.1.2*a.2.1*a.3.3,
-a.1.1*a.2.3*a.3.2
Otherwise Do
det=0
Do k=1 To n
det=det+((-1)**(k+1))*a.1.k*determinant(subm(k))
End
End
End
Return det
 
subm: Procedure Expose a. n
/**********************************************************************
* compute the submatrix resulting when row 1 and column k are removed
* Input: a.*.*, k
* Output: bs the representation of the submatrix as vector
**********************************************************************/

Parse Arg k
bs=''
do i=2 To n
Do j=1 To n
If j=k Then Iterate
bs=bs a.i.j
End
End
Return bs
 
sqrt: Procedure
/**********************************************************************
* compute and return the (integer) square root of the given argument
* terminate the program if the argument is not a square
**********************************************************************/

Parse Arg nn
Do n=1 By 1 while n*n<nn
End
If n*n=nn Then
Return n
Else Do
Say 'invalid number of elements:' nn 'is not a square.'
Exit
End
/* REXX ***************************************************************
* permanent.rex
* compute the permanent of a matrix
* I found an algorithm here:
* http://www.codeproject.com/Articles/21282/Compute-Permanent-of-a-Matrix-with-Ryser-s-Algorit
* see there for the original author.
* translated it to REXX (hopefully correctly) to REXX
* and believe that I can "publish" it here, on rosettacode
* when I look at the copyright rules shown there:
* http://www.codeproject.com/info/cpol10.aspx
* 20.05.2013 Walter Pachl
**********************************************************************/

Call init arg(1) /* initialize the matrix (n and a.* */
sum=0
rowsumprod=0
rowsum=0
chi.=0
c=2**n
Do k=1 To c-1 /* loop all 2^n submatrices of A */
rowsumprod = 1
chis=dec2binarr(k,n) /* characteristic vector */
Do ci=0 By 1 While chis<>''
Parse Var chis chi.ci chis
End
Do m=0 To n-1 /* loop columns of submatrix #k */
rowsum = 0
Do p=0 To n-1 /* loop rows and compute rowsum */
mnp=m*n+p
rowsum=rowsum+chi.p*A.mnp
End
rowsumprod=rowsumprod*rowsum /* update product of rowsums */
/* (optional -- use for sparse matrices) */
/* if (rowsumprod == 0) break; */
End
sum=sum+((-1)**(n-chi.n))*rowsumprod
End
Return sum
/**********************************************************************
* Notes
* 1.The submatrices are chosen by use of a characteristic vector chi
* (only the columns are considered, where chi[p] == 1).
* To retrieve the t from Ryser's formula, we need to save the number
* n-t, as is done in chi[n]. Then we get t = n - chi[n].
* 2.The matrix parameter A is expected to be a one-dimensional integer
* array -- should the matrix be encoded row-wise or column-wise?
* -- It doesn't matter. The permanent is invariant under
* row-switching and column-switching, and it is Screenshot
* - per_inv.gif .
* 3.Further enhancements: If any rowsum equals zero,
* the entire rowsumprod becomes zero, and thus the m-loop can be broken.
* Since if-statements are relatively expensive compared to integer
* operations, this might save time only for sparse matrices
* (where most entries are zeros).
* 4.If anyone finds a polynomial algorithm for permanents,
* he will get rich and famous (at least in the computer science world).
**********************************************************************/

/**********************************************************************
* At first, we need to transform a decimal to a binary array
* with an additional element
* (the last one) saving the number of ones in the array:
**********************************************************************/

dec2binarr: Procedure
Parse Arg n,dim
ol='n='n 'dim='dim
res.=0
pos=dim-1
Do While n>0
res.pos=n//2
res.dim=res.dim+res.pos
n=n%2
pos=pos-1
End
res_s=''
Do i=0 To dim
res_s=res_s res.i
End
Return res_s
 
init: Procedure Expose a. n
/**********************************************************************
* a.* (starting with index 0) contains all array elements
* n is the dimension of the square matrix
**********************************************************************/

Parse Arg as
n=sqrt(words(as))
a.=0
Do ai=0 By 1 While as<>''
Parse Var as a.ai as
End
Return
 
sqrt: Procedure
/**********************************************************************
* compute and return the (integer) square root of the given argument
* terminate the program if the argument is not a square
**********************************************************************/

Parse Arg nn
Do n=1 By 1 while n*n<nn
End
If n*n=nn Then
Return n
Else Do
Say 'invalid number of elements:' nn 'is not a square.'
Exit
End

Output:

   1   2
   3   4
determinant=     -2
  permanent=     10
--------------------------------------------------
   1   2   3   4
   4   5   6   7
   7   8   9  10
  10  11  12  13
determinant=      0
  permanent=  29556
--------------------------------------------------
   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
determinant=      0
  permanent=6778800
--------------------------------------------------

[edit] Ruby

Matrix in the standard library provides a method for the determinant, but not for the permanent.

require 'matrix'
 
class Matrix
# Add "permanent" method to Matrix class
def permanent
r = (0...row_count).to_a # [0,1] (first example), [0,1,2,3] (second example)
r.permutation.inject(0) do |sum, sigma|
sum += sigma.zip(r).inject(1){|prod, (row, col)| prod *= self[row, col] }
end
end
end
 
m1 = Matrix[[1,2],[3,4]] # testcases from Python version
 
m2 = Matrix[[1, 2, 3, 4], [4, 5, 6, 7], [7, 8, 9, 10], [10, 11, 12, 13]]
 
m3 = Matrix[[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]]
 
[m1, m2, m3].each do |m|
puts "determinant:\t #{m.determinant}", "permanent:\t #{m.permanent}"
puts
end
Output:
determinant:	 -2
permanent:	 10

determinant:	 0
permanent:	 29556

determinant:	 0
permanent:	 6778800

[edit] Tcl

The determinant is provided by the linear algebra package in Tcllib. The permanent (being somewhat less common) requires definition, but is easily described:

Library: Tcllib (Package: math::linearalgebra)
Library: Tcllib (Package: struct::list)
package require math::linearalgebra
package require struct::list
 
proc permanent {matrix} {
for {set plist {};set i 0} {$i<[llength $matrix]} {incr i} {
lappend plist $i
}
foreach p [::struct::list permutations $plist] {
foreach i $plist j $p {
lappend prod [lindex $matrix $i $j]
}
lappend sum [::tcl::mathop::* {*}$prod[set prod {}]]
}
return [::tcl::mathop::+ {*}$sum]
}

Demonstrating with a sample matrix:

set mat {
{1 2 3 4}
{4 5 6 7}
{7 8 9 10}
{10 11 12 13}
}
puts [::math::linearalgebra::det $mat]
puts [permanent $mat]
Output:
1.1315223609263888e-29
29556
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