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

Published on 12 May 2024 09:40 PM
#Lua

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

Source code in the lua programming language

function ToReducedRowEchelonForm ( M )
    local lead = 1
    local n_rows, n_cols = #M, #M[1]

    for r = 1, n_rows do
        if n_cols <= lead then break end
        
        local i = r
        while M[i][lead] == 0 do
            i = i + 1
            if n_rows == i then
                i = r
                lead = lead + 1
                if n_cols == lead then break end                
            end
        end 
        M[i], M[r] = M[r], M[i]

        local m = M[r][lead]
        for k = 1, n_cols do
            M[r][k] = M[r][k] / m
        end
        for i = 1, n_rows do
            if i ~= r then
                local m = M[i][lead]
                for k = 1, n_cols do
                    M[i][k] = M[i][k] - m * M[r][k]
                end
            end
        end  
        lead = lead + 1     
    end
end

M = { { 1, 2, -1, -4 }, 
      { 2, 3, -1, -11 }, 
      { -2, 0, -3, 22 } }
      
res = ToReducedRowEchelonForm( M )

for i = 1, #M do
    for j = 1, #M[1] do
        io.write( M[i][j], "  " )
    end
    io.write( "\n" )
end

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