How to resolve the algorithm Reduced row echelon form step by step in the Ring programming language

Published on 12 May 2024 09:40 PM
#Ring

How to resolve the algorithm Reduced row echelon form step by step in the Ring programming language

Table of Contents

Problem Statement

Show how to compute the reduced row echelon form (a.k.a. row canonical form) of a matrix. The matrix can be stored in any datatype that is convenient (for most languages, this will probably be a two-dimensional array). Built-in functions or this pseudocode (from Wikipedia) may be used: For testing purposes, the RREF of this matrix: is:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Reduced row echelon form step by step in the Ring programming language

Source code in the ring programming language

# Project : Reduced row echelon form

matrix = [[1, 2, -1, -4],
              [2, 3, -1, -11],
              [ -2, 0, -3, 22]]
ref(matrix)
for row = 1 to 3
     for col = 1 to 4
           if matrix[row][col] = -0
              see "0 " 
           else
              see "" + matrix[row][col] + " "
           ok
     next
     see nl
next
 
func ref(m)
nrows = 3
ncols = 4
lead = 1
for r = 1 to nrows
      if lead >= ncols
         exit
      ok
      i = r
      while m[i][lead] = 0
                i = i + 1
                if i = nrows
                   i = r
                   lead = lead + 1
                   if lead = ncols
                      exit 2
                   ok
                ok
      end
      for j = 1 to ncols
           temp = m[i][j]
           m[i][j] = m[r][j]
           m[r][j] = temp 
      next
      n = m[r][lead]
      if n != 0
         for j = 1 to ncols
              m[r][j] = m[r][j] / n 
         next
      ok
      for i = 1 to nrows
           if i != r 
              n = m[i][lead]
              for j = 1 to ncols
                   m[i][j] = m[i][j] - m[r][j] * n
              next
           ok
      next
     lead = lead + 1
next

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