How to resolve the algorithm Determinant and permanent step by step in the Nim programming language

Published on 12 May 2024 09:40 PM
#Nim

How to resolve the algorithm Determinant and permanent step by step in the Nim programming language

Table of Contents

Problem Statement

For a given matrix, return the determinant and the permanent of the matrix. The determinant is given by while the permanent is given by 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.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Determinant and permanent step by step in the Nim programming language

Source code in the nim programming language

import sequtils, permutationsswap

type Matrix[M,N: static[int]] = array[M, array[N, float]]

proc det[M,N](a: Matrix[M,N]): float =
  let n = toSeq 0..a.high
  for sigma, sign in n.permutations:
    var x = sign.float
    for i in n: x *= a[i][sigma[i]]
    result += x

proc perm[M,N](a: Matrix[M,N]): float =
  let n = toSeq 0..a.high
  for sigma, sign in n.permutations:
    var x = 1.0
    for i in n: x *= a[i][sigma[i]]
    result += x

const
  a = [ [1.0, 2.0]
      , [3.0, 4.0]
      ]
  b = [ [ 1.0,  2,  3,  4]
      , [ 4.0,  5,  6,  7]
      , [ 7.0,  8,  9, 10]
      , [10.0, 11, 12, 13]
      ]
  c = [ [ 0.0,  1,  2,  3,  4]
      , [ 5.0,  6,  7,  8,  9]
      , [10.0, 11, 12, 13, 14]
      , [15.0, 16, 17, 18, 19]
      , [20.0, 21, 22, 23, 24]
      ]

echo "perm: ", a.perm, " det: ", a.det
echo "perm: ", b.perm, " det: ", b.det
echo "perm: ", c.perm, " det: ", c.det

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