Gaussian elimination

From Rosetta Code
Revision as of 17:08, 12 December 2014 by rosettacode>Hajo (→‎{{header|PL/I}}: {{out}})
Task
Gaussian elimination
You are encouraged to solve this task according to the task description, using any language you may know.

Problem: Solve Ax=b using Gaussian elimination then backwards substitution. A being an n by n matrix. Also, x and b are n by 1 vectors. To improve accuracy, please use partial pivoting and scaling.

ALGOL 68

Works with: ALGOL 68 version Revision 1 - extension to language used - "PRAGMA READ" (similar to C's #include directive.)
Works with: ALGOL 68G version Any - tested with release algol68g-2.4.1.

File: prelude_exception.a68<lang algol68># -*- coding: utf-8 -*- # COMMENT PROVIDES

 MODE FIXED; INT fixed exception, unfixed exception;
 PROC (STRING message) FIXED raise, raise value error

END COMMENT

  1. Note: ℵ indicates attribute is "private", and
       should not be used outside of this prelude #

MODE FIXED = BOOL; # if an exception is detected, can it be fixed "on-site"? # FIXED fixed exception = TRUE, unfixed exception = FALSE;

MODE #ℵ#SIMPLEOUTV = [0]UNION(CHAR, STRING, INT, REAL, BOOL, BITS); MODE #ℵ#SIMPLEOUTM = [0]#ℵ#SIMPLEOUTV; MODE #ℵ#SIMPLEOUTT = [0]#ℵ#SIMPLEOUTM; MODE SIMPLEOUT = [0]#ℵ#SIMPLEOUTT;

PROC raise = (#ℵ#SIMPLEOUT message)FIXED: (

 putf(stand error, ($"Exception:"$, $xg$, message, $l$));
 stop

);

PROC raise value error = (#ℵ#SIMPLEOUT message)FIXED:

 IF raise(message) NE fixed exception THEN exception value error; FALSE FI;

SKIP</lang>File: prelude_mat_lib.a68<lang algol68># -*- coding: utf-8 -*- # COMMENT PRELUDE REQUIRES

 MODE SCAL = REAL;
 FORMAT scal repr = real repr
 # and various SCAL OPerators #

END COMMENT

COMMENT PRELUDE PROIVIDES

 MODE VEC, MAT;
 OP :=:, -:=, +:=, *:=, /:=;
 FORMAT sub, sep, bus;
 FORMAT vec repr, mat repr

END COMMENT

  1. Note: ℵ indicates attribute is "private", and
       should not be used outside of this prelude #

INT #ℵ#lwb vec := 1, #ℵ#upb vec := 0; INT #ℵ#lwb mat := 1, #ℵ#upb mat := 0; MODE VEC = [lwb vec:upb vec]SCAL,

    MAT = [lwb mat:upb mat,lwb vec:upb vec]SCAL;

FORMAT sub := $"( "$, sep := $", "$, bus := $")"$, nl:=$lxx$; FORMAT vec repr := $f(sub)n(upb vec - lwb vec)(f(scal repr)f(sep))f(scal repr)f(bus)$; FORMAT mat repr := $f(sub)n(upb mat - lwb mat)(f( vec repr)f(nl))f( vec repr)f(bus)$;

  1. OPerators to swap the contents of two VECtors #

PRIO =:= = 1; OP =:= = (REF VEC u, v)VOID:

 FOR i TO UPB u DO SCAL scal=u[i]; u[i]:=v[i]; v[i]:=scal OD;

OP +:= = (REF VEC lhs, VEC rhs)REF VEC: (

 FOR i TO UPB lhs DO lhs[i] +:= rhs[i] OD;
 lhs

);

OP -:= = (REF VEC lhs, VEC rhs)REF VEC: (

 FOR i TO UPB lhs DO lhs[i] -:= rhs[i] OD;
 lhs

);

OP *:= = (REF VEC lhs, SCAL rhs)REF VEC: (

 FOR i TO UPB lhs DO lhs[i] *:= rhs OD;
 lhs

);

OP /:= = (REF VEC lhs, SCAL rhs)REF VEC: (

 SCAL inv = 1 / rhs; # multiplication is faster #
 FOR i TO UPB lhs DO lhs[i] *:= inv OD;
 lhs

);

SKIP</lang>File: prelude_gaussian_elimination.a68<lang algol68># -*- coding: utf-8 -*- # COMMENT PRELUDE REQUIRES

 MODE SCAL = REAL,
 REAL near min scal = min real ** 0.99,
 MODE VEC = []REAL,
 MODE MAT = [,]REAL,
 FORMAT scal repr = real repr,
 and various OPerators of MAT and VEC

END COMMENT

COMMENT PRELUDE PROVIDES

 PROC(MAT a, b)MAT gaussian elimination;
 PROC(REF MAT a, b)REF MAT in situ gaussian elimination

END COMMENT

  1. using Gaussian elimination, find x where A*x = b #

PROC in situ gaussian elimination = (REF MAT a, b)REF MAT: (

  1. Note: a and b are modified "in situ", and b is returned as x #
 FOR diag TO UPB a-1 DO
   INT pivot row := diag; SCAL pivot factor := ABS a[diag,diag];
   FOR row FROM diag + 1 TO UPB a DO # Full pivoting #
     SCAL abs a diag = ABS a[row,diag];
     IF abs a diag>=pivot factor THEN 
       pivot row := row; pivot factor := abs a diag FI
   OD;
 # now we have the "best" diag to full pivot, do the actual pivot #
   IF diag NE pivot row THEN
   # a[pivot row,] =:= a[diag,]; XXX: unoptimised # #DB#
     a[pivot row,diag:] =:= a[diag,diag:]; # XXX: optimised #
     b[pivot row,] =:= b[diag,] # swap/pivot the diags of a & b #
   FI;
   IF ABS a[diag,diag] <= near min scal THEN 
     raise value error("singular matrix") FI;
   SCAL a diag reciprocal := 1 / a[diag, diag];
   FOR row FROM diag+1 TO UPB a DO
     SCAL factor = a[row,diag] * a diag reciprocal;
   # a[row,] -:= factor * a[diag,] XXX: "unoptimised" # #DB#
     a[row,diag+1:] -:= factor * a[diag,diag+1:];# XXX: "optimised" #
     b[row,] -:= factor * b[diag,]
   OD
 OD;
  1. We have a triangular matrix, at this point we can traverse backwards
 up the diagonal calculating b\A Converting it initial to a diagonal
 matrix, then to the identity.  #
 FOR diag FROM UPB a BY -1 TO 1+LWB a DO
   IF ABS a[diag,diag] <= near min scal THEN 
     raise value error("Zero pivot encountered?") FI;
   SCAL a diag reciprocal = 1 / a[diag,diag];
   FOR row TO diag-1 DO
     SCAL factor = a[row,diag] * a diag reciprocal;
   # a[row,diag] -:= factor * a[diag,diag]; XXX: "unoptimised" so remove # #DB#
     b[row,] -:= factor * b[diag,]
   OD;
  1. Now we have only diagonal elements we can simply divide b
 by the values along the diagonal of A. #
   b[diag,] *:= a diag reciprocal
 OD;
 b # EXIT #

);

PROC gaussian elimination = (MAT in a, in b)MAT: (

  1. Note: a and b are cloned and not modified "in situ" #
 [UPB in a, 2 UPB in a]SCAL a := in a;
 [UPB in b, 2 UPB in b]SCAL b := in b;
 in situ gaussian elimination(a,b)

);

SKIP</lang>File: postlude_exception.a68<lang algol68># -*- coding: utf-8 -*- # COMMENT POSTLUDE PROIVIDES

 PROC VOID exception too many iterations, exception value error;

END COMMENT

SKIP EXIT exception too many iterations: exception value error:

 stop</lang>File: test_Gaussian_elimination.a68<lang algol68>#!/usr/bin/algol68g-full --script #
  1. -*- coding: utf-8 -*- #

PR READ "prelude_exception.a68" PR;

  1. define the attributes of the scalar field being used #

MODE SCAL = REAL; FORMAT scal repr = $g(-0,real width)$;

  1. create "near min scal" as is scales better then small real #

SCAL near min scal = min real ** 0.99;

PR READ "prelude_mat_lib.a68" PR; PR READ "prelude_gaussian_elimination.a68" PR;

MAT a =(( 1.00, 0.00, 0.00, 0.00, 0.00, 0.00),

       ( 1.00, 0.63, 0.39,  0.25,  0.16,   0.10),
       ( 1.00, 1.26, 1.58,  1.98,  2.49,   3.13),
       ( 1.00, 1.88, 3.55,  6.70, 12.62,  23.80),
       ( 1.00, 2.51, 6.32, 15.88, 39.90, 100.28),
       ( 1.00, 3.14, 9.87, 31.01, 97.41, 306.02));

VEC b = (-0.01, 0.61, 0.91, 0.99, 0.60, 0.02);

[UPB b,1]SCAL col b; col b[,1]:= b;

upb vec := 2 UPB a;

printf((vec repr, gaussian elimination(a,col b)));

PR READ "postlude_exception.a68" PR</lang>Output:

( -.010000000000002, 1.602790394502130, -1.613203059905640, 1.245494121371510, -.490989719584686, .065760696175236)

C

This modifies A and b in place, which might not be quite desirable. <lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <math.h>
  1. define mat_elem(a, y, x, n) (a + ((y) * (n) + (x)))

void swap_row(double *a, double *b, int r1, int r2, int n) { double tmp, *p1, *p2; int i;

if (r1 == r2) return; for (i = 0; i < n; i++) { p1 = mat_elem(a, r1, i, n); p2 = mat_elem(a, r2, i, n); tmp = *p1, *p1 = *p2, *p2 = tmp; } tmp = b[r1], b[r1] = b[r2], b[r2] = tmp; }

void gauss_eliminate(double *a, double *b, double *x, int n) {

  1. define A(y, x) (*mat_elem(a, y, x, n))

int i, j, col, row, max_row,dia; double max, tmp;

for (dia = 0; dia < n; dia++) { max_row = dia, max = A(dia, dia);

for (row = dia + 1; row < n; row++) if ((tmp = fabs(A(row, dia))) > max) max_row = row, max = tmp;

swap_row(a, b, dia, max_row, n);

for (row = dia + 1; row < n; row++) { tmp = A(row, dia) / A(dia, dia); for (col = dia+1; col < n; col++) A(row, col) -= tmp * A(dia, col); A(row, dia) = 0; b[row] -= tmp * b[dia]; } } for (row = n - 1; row >= 0; row--) { tmp = b[row]; for (j = n - 1; j > row; j--) tmp -= x[j] * A(row, j); x[row] = tmp / A(row, row); }

  1. undef A

}

int main(void) { double a[] = { 1.00, 0.00, 0.00, 0.00, 0.00, 0.00, 1.00, 0.63, 0.39, 0.25, 0.16, 0.10, 1.00, 1.26, 1.58, 1.98, 2.49, 3.13, 1.00, 1.88, 3.55, 6.70, 12.62, 23.80, 1.00, 2.51, 6.32, 15.88, 39.90, 100.28, 1.00, 3.14, 9.87, 31.01, 97.41, 306.02 }; double b[] = { -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 }; double x[6]; int i;

gauss_eliminate(a, b, x, 6);

for (i = 0; i < 6; i++) printf("%g\n", x[i]);

return 0;

}</lang>
Output:

-0.01 1.60279 -1.6132 1.24549 -0.49099 0.0657607

D

Translation of: Go

<lang d>import std.stdio, std.math, std.algorithm, std.range, std.numeric;

Tuple!(double[],"x", string,"err") gaussPartial(in double[][] a0, in double[] b0) pure /*nothrow*/ in {

   assert(a0.length == a0[0].length);
   assert(a0.length == b0.length);
   assert(a0.all!(row => row.length == a0[0].length));

} body {

   enum eps = 1e-6;
   immutable m = b0.length;
   // Make augmented matrix.
   //auto a = a0.zip(b0).map!(c => c[0] ~ c[1]).array; // Not mutable.
   auto a = a0.zip(b0).map!(c => [] ~ c[0] ~ c[1]).array;
   // Wikipedia algorithm from Gaussian elimination page,
   // produces row-eschelon form.
   foreach (immutable k; 0 .. a.length) {
       // Find pivot for column k and swap.
       a[k .. m].minPos!((x, y) => x[k] > y[k]).front.swap(a[k]);
       if (a[k][k].abs < eps)
           return typeof(return)(null, "singular");
       // Do for all rows below pivot.
       foreach (immutable i; k + 1 .. m) {
           // Do for all remaining elements in current row.
           a[i][k+1 .. m+1] -= a[k][k+1 .. m+1] * (a[i][k] / a[k][k]);
           a[i][k] = 0; // Fill lower triangular matrix with zeros.
       }
   }
   // End of WP algorithm. Now back substitute to get result.
   auto x = new double[m];
   foreach_reverse (immutable i; 0 .. m)
       x[i] = (a[i][m] - a[i][i+1 .. m].dotProduct(x[i+1 .. m])) / a[i][i];
   return typeof(return)(x, null);

}

void main() {

   // The test case result is correct to this tolerance.
   enum eps = 1e-13;
   // Common RC example. Result computed with rational arithmetic
   // then converted to double, and so should be about as close to
   // correct as double represention allows.
   immutable a = [[1.00, 0.00, 0.00,  0.00,  0.00,   0.00],
                  [1.00, 0.63, 0.39,  0.25,  0.16,   0.10],
                  [1.00, 1.26, 1.58,  1.98,  2.49,   3.13],
                  [1.00, 1.88, 3.55,  6.70, 12.62,  23.80],
                  [1.00, 2.51, 6.32, 15.88, 39.90, 100.28],
                  [1.00, 3.14, 9.87, 31.01, 97.41, 306.02]];
   immutable b = [-0.01, 0.61, 0.91,  0.99,  0.60,   0.02];
   immutable r = gaussPartial(a, b);
   if (!r.err.empty)
       return writeln("Error: ", r.err);
   r.x.writeln;
   immutable result = [-0.01,               1.602790394502114,
                       -1.6132030599055613, 1.2454941213714368,
                       -0.4909897195846576, 0.065760696175232];
   foreach (immutable i, immutable xi; result)
       if (abs(r.x[i] - xi) > eps)
           return writeln("Out of tolerance: ", r.x[i], " ", xi);

}</lang>

Output:
[-0.01, 1.60279, -1.6132, 1.24549, -0.49099, 0.0657607]

Go

Gaussian elimination with partial pivoting by pseudocode on WP page "Gaussian elimination." Scaling not implemented. <lang go>package main

import (

   "errors"
   "fmt"
   "log"
   "math"

)

type testCase struct {

   a [][]float64
   b []float64
   x []float64

}

var tc = testCase{

   // common RC example.  Result x computed with rational arithmetic then
   // converted to float64, and so should be about as close to correct as
   // float64 represention allows.
   a: [][]float64{
       {1.00, 0.00, 0.00, 0.00, 0.00, 0.00},
       {1.00, 0.63, 0.39, 0.25, 0.16, 0.10},
       {1.00, 1.26, 1.58, 1.98, 2.49, 3.13},
       {1.00, 1.88, 3.55, 6.70, 12.62, 23.80},
       {1.00, 2.51, 6.32, 15.88, 39.90, 100.28},
       {1.00, 3.14, 9.87, 31.01, 97.41, 306.02}},
   b: []float64{-0.01, 0.61, 0.91, 0.99, 0.60, 0.02},
   x: []float64{-0.01, 1.602790394502114, -1.6132030599055613,
       1.2454941213714368, -0.4909897195846576, 0.065760696175232},

}

// result from above test case turns out to be correct to this tolerance. const ε = 1e-13

func main() {

   x, err := GaussPartial(tc.a, tc.b)
   if err != nil {
       log.Fatal(err)
   }
   fmt.Println(x)
   for i, xi := range x {
       if math.Abs(tc.x[i]-xi) > ε {
           log.Println("out of tolerance")
           log.Fatal("expected", tc.x)
       }
   }

}

func GaussPartial(a0 [][]float64, b0 []float64) ([]float64, error) {

   // make augmented matrix
   m := len(b0)
   a := make([][]float64, m)
   for i, ai := range a0 {
       row := make([]float64, m+1)
       copy(row, ai)
       row[m] = b0[i]
       a[i] = row
   }
   // WP algorithm from Gaussian elimination page
   // produces row-eschelon form
   for k := range a {
       // Find pivot for column k:
       iMax := k
       max := math.Abs(a[k][k])
       for i := k + 1; i < m; i++ {
           if abs := math.Abs(a[i][k]); abs > max {
               iMax = i
               max = abs
           }
       }
       if a[iMax][k] == 0 {
           return nil, errors.New("singular")
       }
       // swap rows(k, i_max)
       a[k], a[iMax] = a[iMax], a[k]
       // Do for all rows below pivot:
       for i := k + 1; i < m; i++ {
           // Do for all remaining elements in current row:
           for j := k + 1; j <= m; j++ {
               a[i][j] -= a[k][j] * (a[i][k] / a[k][k])
           }
           // Fill lower triangular matrix with zeros:
           a[i][k] = 0
       }
   }
   // end of WP algorithm.
   // now back substitute to get result.
   x := make([]float64, m)
   for i := m - 1; i >= 0; i-- {
       x[i] = a[i][m]
       for j := i + 1; j < m; j++ {
           x[i] -= a[i][j] * x[j]
       }
       x[i] /= a[i][i]
   }
   return x, nil

}</lang>

Output:
[-0.01 1.6027903945020987 -1.613203059905494 1.245494121371364 -0.49098971958462834 0.06576069617522803]

J

%. , J's matrix divide verb, directly solves systems of determined and of over-determined linear equations directly. This example J session builds a noisy sine curve on the half circle, fits quintic and quadratic equations, and displays the results of evaluating these polynomials.

<lang J>

  f=: 6j2&":   NB. formatting verb
  sin=: 1&o.   NB. verb to evaluate circle function 1, the sine
  add_noise=: ] + (* (_0.5 + 0 ?@:#~ #))   NB. AMPLITUDE add_noise SIGNAL
  f RADIANS=: o.@:(%~ i.@:>:)5  NB. monadic circle function is  pi times
 0.00  0.63  1.26  1.88  2.51  3.14
  f SINES=: sin RADIANS
 0.00  0.59  0.95  0.95  0.59  0.00
  f NOISY_SINES=: 0.1 add_noise SINES
_0.01  0.61  0.91  0.99  0.60  0.02
  A=: (^/ i.@:#) RADIANS  NB. A is the quintic coefficient matrix
  NB. display the equation to solve
  (f A) ; 'x' ; '=' ; f@:,. NOISY_SINES

┌────────────────────────────────────┬─┬─┬──────┐ │ 1.00 0.00 0.00 0.00 0.00 0.00│x│=│ _0.01│ │ 1.00 0.63 0.39 0.25 0.16 0.10│ │ │ 0.61│ │ 1.00 1.26 1.58 1.98 2.49 3.13│ │ │ 0.91│ │ 1.00 1.88 3.55 6.70 12.62 23.80│ │ │ 0.99│ │ 1.00 2.51 6.32 15.88 39.90100.28│ │ │ 0.60│ │ 1.00 3.14 9.87 31.01 97.41306.02│ │ │ 0.02│ └────────────────────────────────────┴─┴─┴──────┘

  f QUINTIC_COEFFICIENTS=: NOISY_SINES %. A   NB. %. solves the linear system
_0.01  1.71 _1.88  1.48 _0.58  0.08
  quintic=: QUINTIC_COEFFICIENTS&p.  NB. verb to evaluate the polynomial
  NB. %. also solves the least squares fit for overdetermined system
  quadratic=: (NOISY_SINES %. (^/ i.@:3:) RADIANS)&p.  NB. verb to evaluate quadratic.
  quadratic

_0.0200630695393961729 1.26066877804926536 _0.398275112136019516&p.

  NB. The quintic is agrees with the noisy data, as it should
  f@:(NOISY_SINES ,. sin ,. quadratic ,. quintic) RADIANS
_0.01  0.00 _0.02 _0.01
 0.61  0.59  0.61  0.61
 0.91  0.95  0.94  0.91
 0.99  0.95  0.94  0.99
 0.60  0.59  0.63  0.60
 0.02  0.00  0.01  0.02
  f MID_POINTS=: (+ -:@:(-/@:(2&{.)))RADIANS
_0.31  0.31  0.94  1.57  2.20  2.83
  f@:(sin ,. quadratic ,. quintic) MID_POINTS
_0.31 _0.46 _0.79
 0.31  0.34  0.38
 0.81  0.81  0.77
 1.00  0.98  1.00
 0.81  0.83  0.86
 0.31  0.36  0.27

</lang>

Julia

Using built-in LAPACK-based linear solver (which employs partial-pivoted Gaussian elimination): <lang julia>x = A \ b</lang>

Mathematica

<lang Mathematica>GaussianElimination[A_?MatrixQ, b_?VectorQ] := Last /@ RowReduce[Flatten /@ Transpose[{A, b}]]</lang>

MATLAB

<lang MATLAB> function [ x ] = GaussElim( A, b)

% Ensures A is n by n sz = size(A); if sz(1)~=sz(2)

   fprintf('A is not n by n\n');
   clear x;
   return;

end

n = sz(1);

% Ensures b is n x 1. if n~=sz(1)

   fprintf('b is not 1 by n.\n');
   return

end

x = zeros(n,1); aug = [A b]; tempmatrix = aug;

for i=2:sz(1)


   % Find maximum of row and divide by the maximum
   tempmatrix(1,:) = tempmatrix(1,:)/max(tempmatrix(1,:));
   
   % Finds the maximum in column
   temp = find(abs(tempmatrix) - max(abs(tempmatrix(:,1))));
   if length(temp)>2
       for j=1:length(temp)-1
           if j~=temp(j)
               maxi = j; %maxi = column number of maximum
               break;
           end
       end
   else % length(temp)==2
       maxi=1;
   end
   
   % Row swap if maxi is not 1
   if maxi~=1
       temp = tempmatrix(maxi,:);
       tempmatrix(maxi,:) = tempmatrix(1,:);
       tempmatrix(1,:) = temp;
   end    
   
   % Row reducing
   for j=2:length(tempmatrix)-1
       tempmatrix(j,:) = tempmatrix(j,:)-tempmatrix(j,1)/tempmatrix(1,1)*tempmatrix(1,:);
       if tempmatrix(j,j)==0 || isnan(tempmatrix(j,j)) || abs(tempmatrix(j,j))==Inf
           fprintf('Error: Matrix is singular.\n');
           clear x;
           return
       end
   end
   aug(i-1:end,i-1:end) = tempmatrix;
   
   % Decrease matrix size
   tempmatrix = tempmatrix(2:end,2:end);

end

% Backwards Substitution x(end) = aug(end,end)/aug(end,end-1); for i=n-1:-1:1

   x(i) = (aug(i,end)-dot(aug(i,1:end-1),x))/aug(i,i);

end

end </lang>

OCaml

The OCaml stdlib is fairly lean, so these stand-alone solutions often need to include support functions which would be part of a codebase, like these... <lang OCaml> module Array = struct

 include Array
 (* Computes: f a.(0) + f a.(1) + ... where + is 'g'. *)
 let foldmap g f a =
   let n = Array.length a in
   let rec aux acc i =
     if i >= n then acc else aux (g acc (f a.(i))) (succ i)
   in aux (f a.(0)) 1
 (* like the stdlib fold_left, but also provides index to f *)
 let foldi_left f x a =
   let r = ref x in
   for i = 0 to length a - 1 do
     r := f i !r (unsafe_get a i)
   done;
   !r

end

let foldmap_range g f (a,b) =

 let rec aux acc n =
   let n = succ n in
   if n > b then acc else aux (g acc (f n)) n
 in aux (f a) a

let fold_range f init (a,b) =

 let rec aux acc n =
   if n > b then acc else aux (f acc n) (succ n)
 in aux init a

</lang> The solver: <lang OCaml> (* Some less-general support functions for 'solve'. *) let swap_elem m i j = let x = m.(i) in m.(i) <- m.(j); m.(j) <- x let maxtup a b = if (snd a) > (snd b) then a else b let augmented_matrix m b =

 Array.(init (length m) ( fun i -> append m.(i) [|b.(i)|] ))

(* Solve Ax=b for x, using gaussian elimination with scaled partial pivot,

* and then back-substitution of the resulting row-echelon matrix. *)

let solve m b =

 let n = Array.length m in
 let n' = pred n in (* last index = n-1 *)
 let s = Array.(map (foldmap max abs_float) m) in  (* scaling vector *)
 let a = augmented_matrix m b in
 for k = 0 to pred n' do
   (* Scaled partial pivot, to preserve precision *)
   let pair i = (i, abs_float a.(i).(k) /. s.(i)) in
   let i_max,v = foldmap_range maxtup pair (k,n') in
   if v < epsilon_float then failwith "Matrix is singular.";
   swap_elem a k i_max;
   swap_elem s k i_max;
   (* Eliminate one column *)
   for i = succ k to n' do
     let tmp = a.(i).(k) /. a.(k).(k) in
     for j = succ k to n do
       a.(i).(j) <- a.(i).(j) -. tmp *. a.(k).(j);
     done
   done
 done;
 (* Backward substitution; 'b' is in the 'nth' column of 'a' *)
 let x = Array.copy b in (* just a fresh array of the right size and type *)
 for i = n' downto 0 do
   let minus_dprod t j = t -. x.(j) *. a.(i).(j) in
   x.(i) <- fold_range minus_dprod a.(i).(n) (i+1,n') /. a.(i).(i);
 done;
 x

</lang> Example data... <lang OCaml> let a =

 [| [| 1.00; 0.00; 0.00;  0.00;  0.00; 0.00 |];
    [| 1.00; 0.63; 0.39;  0.25;  0.16; 0.10 |];
    [| 1.00; 1.26; 1.58;  1.98;  2.49; 3.13 |];
    [| 1.00; 1.88; 3.55;  6.70; 12.62; 23.80 |];
    [| 1.00; 2.51; 6.32; 15.88; 39.90; 100.28 |];
    [| 1.00; 3.14; 9.87; 31.01; 97.41; 306.02 |] |]

let b = [| -0.01; 0.61; 0.91; 0.99; 0.60; 0.02 |] </lang> In the REPL, the solution is: <lang OCaml>

  1. let x = solve a b;;

val x : float array = [|-0.0100000000000000991; 1.60279039450210536; -1.61320305990553226;

 1.24549412137140547; -0.490989719584644546; 0.0657606961752301433|]

</lang> Further, let's define multiplication and subtraction to check our results... <lang OCaml> let mul m v =

 Array.mapi (fun i u ->
   Array.foldi_left (fun j sum uj ->
     sum +. uj *. v.(j)
   ) 0. u
 ) m

let sub u v = Array.mapi (fun i e -> e -. v.(i)) u </lang> Now 'x' can be plugged into the equation to calculate the residual: <lang OCaml>

  1. let residual = sub b (mul a x);;

val residual : float array =

 [|9.8879238130678e-17; 1.11022302462515654e-16; 2.22044604925031308e-16;
   8.88178419700125232e-16; -5.5511151231257827e-16; 4.26741975090294545e-16|]

</lang>

PARI/GP

If A and B have floating-point numbers (t_REALs) then the following uses Gaussian elimination: <lang parigp>matsolve(A,B)</lang>

If the entries are integers, then p-adic lifting (Dixon 1982) is used instead.

Perl 6

Multi-dimensional arrays are not fully implemented in rakudo yet, and that makes Linear Algebra and matrix manipulation quite painful currently. However, it's always possible to proceed as in C with one-dimensional arrays and some index reordering tricks.

Translation of: C

<lang perl6>sub mat_elem(@a, $y, $x, $n) is rw { @a[ $y * $n + $x ] } sub swap_row(@a, @b, $r1, $r2, $n) {

   return if $r1 == $r2;
   for ^$n -> $i {

( mat_elem(@a, $r1, $i, $n), mat_elem(@a, $r2, $i, $n) ).=reverse;

   }
   @b[$r1, $r2].=reverse;

}

sub gauss_eliminate(@a, @b, $n) {

   sub A($y, $x) is rw { mat_elem(@a, $y, $x, $n) }
   my ($i, $j, $col, $row, $max_row, $dia);
   my ($max, $tmp);
   for ^$n -> $dia {

for $dia ^..^ $n -> $row { swap_row @a, @b, $dia, max(:by({ abs(A($_, $dia)) }), $dia ^..^ $n), $n; $tmp = A($row, $dia) / A($dia, $dia); for $dia ^..^ $n -> $col { A($row, $col) -= $tmp * A($dia, $col); } A($row, $dia) = 0; @b[$row] -= $tmp * @b[$dia]; }

   }
   my @x;
   for $n - 1, $n - 2 ... 0 -> $row {

$tmp = @b[$row]; for $n - 1, $n - 2 ...^ $row -> $j { $tmp -= @x[$j] * A($row, $j); } @x[$row] = $tmp / A($row, $row);

   }
   return @x;

}

sub MAIN {

   my @a = <
       1.00 0.00 0.00  0.00  0.00   0.00
       1.00 0.63 0.39  0.25  0.16   0.10
       1.00 1.26 1.58  1.98  2.49   3.13
       1.00 1.88 3.55  6.70 12.62  23.80
       1.00 2.51 6.32 15.88 39.90 100.28
       1.00 3.14 9.87 31.01 97.41 306.02
   >;
   my @b = <
       -0.01 0.61 0.91 0.99 0.60 0.02
   >;
   .say for gauss_eliminate(@a, @b, 6);

}</lang>

Output:
-0.01
1.60279039450211
-1.61320305990556
1.24549412137144
-0.490989719584658
0.0657606961752

PHP

<lang php>function swap_rows(&$a, &$b, $r1, $r2) {

   if ($r1 == $r2) return;
   $tmp = $a[$r1];
   $a[$r1] = $a[$r2];
   $a[$r2] = $tmp;
   $tmp = $b[$r1];
   $b[$r1] = $b[$r2];
   $b[$r2] = $tmp;

}

function gauss_eliminate($A, $b, $N) {

   for ($col = 0; $col < $N; $col++)
   {
       $j = $col;
       $max = $A[$j][$j];
       for ($i = $col + 1; $i < $N; $i++)
       {
           $tmp = abs($A[$i][$col]);
           if ($tmp > $max)
           {
               $j = $i;
               $max = $tmp;
           }
       }
       swap_rows($A, $b, $col, $j);
       for ($i = $col + 1; $i < $N; $i++)
       {
           $tmp = $A[$i][$col] / $A[$col][$col];
           for ($j = $col + 1; $j < $N; $j++)
           {
               $A[$i][$j] -= $tmp * $A[$col][$j];
           }
           $A[$i][$col] = 0;
           $b[$i] -= $tmp * $b[$col];
       }
   }
   $x = array();
   for ($col = $N - 1; $col >= 0; $col--)
   {
       $tmp = $b[$col];
       for ($j = $N - 1; $j > $col; $j--)
       {
           $tmp -= $x[$j] * $A[$col][$j];
       }
       $x[$col] = $tmp / $A[$col][$col];
   }
   return $x;

}

function test_gauss() {

   $a = array(
       array(1.00, 0.00, 0.00,  0.00,  0.00, 0.00),
       array(1.00, 0.63, 0.39,  0.25,  0.16, 0.10),
       array(1.00, 1.26, 1.58,  1.98,  2.49, 3.13),
       array(1.00, 1.88, 3.55,  6.70, 12.62, 23.80),
       array(1.00, 2.51, 6.32, 15.88, 39.90, 100.28),
       array(1.00, 3.14, 9.87, 31.01, 97.41, 306.02)
   );
   $b = array( -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 );
   $x = gauss_eliminate($a, $b, 6);
   ksort($x);
   print_r($x);

}

test_gauss();</lang>

Output:
Array
(
    [0] => -0.01
    [1] => 1.6027903945021
    [2] => -1.6132030599055
    [3] => 1.2454941213714
    [4] => -0.49098971958463
    [5] => 0.065760696175228
)

PL/I

<lang pli>Solve: procedure options (main); /* 11 January 2014 */

  declare n fixed binary;
  put ('Program to solve n simultaneous equations of the form Ax = b. Please type n:' );
  get (n);

begin;

  declare (A(n, n), b(n), x(n)) float(18);
  declare (SA(n,n), Sb(n)) float (18);
  declare i fixed binary;
  put skip list ('Please type A:');
  get (a);
  put skip list ('Please type the right-hand sides, b:');
  get (b);
  SA = A; Sb = b;
  put skip list ('The equations are:');
  do i = 1 to n;
     put skip edit (A(i,*), b(i)) (f(5), x(1));
  end;
  call Gauss_elimination (A, b);
  call Backward_substitution (A, b, x);
  put skip list ('Solutions:'); put skip data (x);
  /* Check solutions: */
  put skip list ('Residuals:');
  do i = 1 to n;
     put skip list (sum(SA(i,*) * x(*)) - Sb(i));
  end;

end;

Gauss_elimination: procedure (A, b) options (reorder); /* Triangularise */

  declare (A(*,*), b(*)) float(18);
  declare n fixed binary initial (hbound(A, 1));
  declare (i, j, k) fixed binary;
  declare t float(18);
  do j = 1 to n;
     do i = j+1 to n; /* For each of the rows beneath the current (pivot) row. */
        t = A(j,j) / A(i,j);
        do k = j+1 to n; /* Subtract a multiple of row i from row j. */
           A(i,k) = A(j,k) - t*A(i,k);
        end;
        b(i) = b(j) - t*b(i); /* ... and the right-hand side. */
     end;
  end;

end Gauss_elimination;

Backward_substitution: procedure (A, b, x) options (reorder);

  declare (A(*,*), b(*), x(*)) float(18);
  declare t float(18);
  declare n fixed binary initial (hbound(A, 1));
  declare (i, j) fixed binary;
  x(n) = b(n) / a(n,n);
  do j = n-1 to 1 by -1;
     t = 0;
     do i = j+1 to n;
        t = t + a(j,i)*x(i);
     end;
     x(j) = (b(j) - t) / a(j,j);
  end;

end Backward_substitution;

end Solve;</lang>

Output:
Program to solve n simultaneous equations of the form Ax = b. Please type n: 

Please type A: 

Please type the right-hand sides, b: 

The equations are: 
    1     2     3    14
    2     1     3    13
    3    -2    -1    -4
Solutions: 
X(1)= 1.00000000000000000E+0000                 X(2)= 2.00000000000000000E+0000
X(3)= 3.00000000000000000E+0000;
Residuals: 
 0.00000000000000000E+0000 
 0.00000000000000000E+0000 
 0.00000000000000000E+0000 

Python

<lang python># The 'gauss' function takes two matrices, 'a' and 'b', with 'a' square, and it return the determinant of 'a' and a matrix 'x' such that a*x = b.

  1. If 'b' is the identity, then 'x' is the inverse of 'a'.

import copy from fractions import Fraction

def gauss(a, b):

   a = copy.deepcopy(a)
   b = copy.deepcopy(b)
   n = len(a)
   p = len(b[0])
   det = 1
   for i in range(n - 1):
       k = i
       for j in range(i + 1, n):
           if abs(a[j][i]) > abs(a[k][i]):
               k = j
       if k != i:
           a[i], a[k] = a[k], a[i]
           b[i], b[k] = b[k], b[i]
           det = -det
           
       for j in range(i + 1, n):
           t = a[j][i]/a[i][i]
           for k in range(i + 1, n):
               a[j][k] -= t*a[i][k]
           for k in range(p):
               b[j][k] -= t*b[i][k]
               
   for i in range(n - 1, -1, -1):
       for j in range(i + 1, n):
           t = a[i][j]
           for k in range(p):
               b[i][k] -= t*b[j][k]
       t = 1/a[i][i]
       det *= a[i][i]
       for j in range(p):
           b[i][j] *= t
   return det, b

def zeromat(p, q):

   return [[0]*q for i in range(p)]

def matmul(a, b):

   n, p = len(a), len(a[0])
   p1, q = len(b), len(b[0])
   if p != p1:
       raise ValueError("Incompatible dimensions")
   c = zeromat(n, q)
   for i in range(n):
       for j in range(q):
               c[i][j] = sum(a[i][k]*b[k][j] for k in range(p))
   return c


def mapmat(f, a):

   return [list(map(f, v)) for v in a]

def ratmat(a):

   return mapmat(Fraction, a)
  1. As an example, compute the determinant and inverse of 3x3 magic square

a = [[2, 9, 4], [7, 5, 3], [6, 1, 8]] b = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] det, c = gauss(a, b)

det -360.0

c [[-0.10277777777777776, 0.18888888888888888, -0.019444444444444438], [0.10555555555555554, 0.02222222222222223, -0.061111111111111116], [0.0638888888888889, -0.14444444444444446, 0.14722222222222223]]

  1. Check product

matmul(a, c) [[1.0, 0.0, 0.0], [5.551115123125783e-17, 1.0, 0.0], [1.1102230246251565e-16, -2.220446049250313e-16, 1.0]]

  1. Same with fractions, so the result is exact

det, c = gauss(ratmat(a), ratmat(b))

det Fraction(-360, 1)

c [[Fraction(-37, 360), Fraction(17, 90), Fraction(-7, 360)], [Fraction(19, 180), Fraction(1, 45), Fraction(-11, 180)], [Fraction(23, 360), Fraction(-13, 90), Fraction(53, 360)]]

matmul(a, c) [[Fraction(1, 1), Fraction(0, 1), Fraction(0, 1)], [Fraction(0, 1), Fraction(1, 1), Fraction(0, 1)], [Fraction(0, 1), Fraction(0, 1), Fraction(1, 1)]]</lang>

Racket

<lang racket>

  1. lang racket

(require math/matrix) (define A

 (matrix [[1.00  0.00  0.00  0.00  0.00   0.00]
          [1.00  0.63  0.39  0.25  0.16   0.10]
          [1.00  1.26  1.58  1.98  2.49   3.13]
          [1.00  1.88  3.55  6.70 12.62  23.80]
          [1.00  2.51  6.32 15.88 39.90 100.28]
          [1.00  3.14  9.87 31.01 97.41 306.02]]))

(define b (col-matrix [-0.01 0.61 0.91 0.99 0.60 0.02]))

(matrix-solve A b) </lang> Output: <lang racket>

  1. <array
 '#(6 1)
 #[-0.01
  1.602790394502109
  -1.613203059905556
  1.2454941213714346
  -0.4909897195846582
  0.06576069617523222]>

</lang>

REXX

<lang rexx>/* REXX ---------------------------------------------------------------

  • 07.08.2014 Walter Pachl translated from PL/I)
  • --------------------------------------------------------------------*/
 Numeric Digits 20
 Parse Arg t
 n=3
 Parse Value '1  2  3 14' With a.1.1 a.1.2 a.1.3 b.1
 Parse Value '2  1  3 13' With a.2.1 a.2.2 a.2.3 b.2
 Parse Value '3 -2 -1 -4' With a.3.1 a.3.2 a.3.3 b.3
 If t=6 Then Do
   n=6
   Parse Value '1.00 0.00 0.00  0.00  0.00 0.00  ' With a.1.1 a.1.2 a.1.3 a.1.4 a.1.5 a.1.6 .
   Parse Value '1.00 0.63 0.39  0.25  0.16 0.10  ' With a.2.1 a.2.2 a.2.3 a.2.4 a.2.5 a.2.6 .
   Parse Value '1.00 1.26 1.58  1.98  2.49 3.13  ' With a.3.1 a.3.2 a.3.3 a.3.4 a.3.5 a.3.6 .
   Parse Value '1.00 1.88 3.55  6.70 12.62 23.80 ' With a.4.1 a.4.2 a.4.3 a.4.4 a.4.5 a.4.6 .
   Parse Value '1.00 2.51 6.32 15.88 39.90 100.28' With a.5.1 a.5.2 a.5.3 a.5.4 a.5.5 a.5.6 .
   Parse Value '1.00 3.14 9.87 31.01 97.41 306.02' With a.6.1 a.6.2 a.6.3 a.6.4 a.6.5 a.6.6 .
   Parse Value '-0.01 0.61 0.91 0.99 0.60 0.02'    With b.1 b.2 b.3 b.4 b.5 b.6 .
   End
 Do i=1 To n
   Do j=1 To n
     sa.i.j=a.i.j
     End
   sb.i=b.i
   End
 Say 'The equations are:'
 do i = 1 to n;
   ol=
   Do j=1 To n
     ol=ol format(a.i.j,4,4)
     End
   ol=ol'  'format(b.i,4,4)
   Say ol
   end
 call Gauss_elimination
 call Backward_substitution
 Say 'Solutions:'
 Do i=1 To n
   Say 'x('i')='||x.i
   End
 /* Check solutions: */
 Say 'Residuals:'
 do i = 1 to n
   res=0
   Do j=1 To n
     res=res+(sa.i.j*x.j)
     End
   res=res-sb.i
   Say 'res('i')='res
   End

Exit

Gauss_elimination:

 do j = 1 to n
    do i = j+1 to n /* For each of the rows beneath the current (pivot) row. */
       t = a.j.j / a.i.j
       do k = j+1 to n /* Subtract a multiple of row i from row j. */
          a.i.k = a.j.k - t*a.i.k
       end
       b.i = b.j - t*b.i /* ... and the right-hand side. */
    end
 end
 Return

Backward_substitution:

 x.n = b.n / a.n.n
 do j = n-1 to 1 by -1
    t = 0
    do i = j+1 to n
       t = t + a.j.i*x.i
    end
    x.j = (b.j - t) / a.j.j
 end
 Return</lang>
Output:
The equations are:
    1.0000    2.0000    3.0000    14.0000
    2.0000    1.0000    3.0000    13.0000
    3.0000   -2.0000   -1.0000    -4.0000
Solutions:
x(1)=1
x(2)=2
x(3)=3
Residuals:
res(1)=0
res(2)=0
res(3)=0

and with test data from PHP

The equations are:
    1.0000    0.0000    0.0000    0.0000    0.0000    0.0000    -0.0100
    1.0000    0.6300    0.3900    0.2500    0.1600    0.1000     0.6100
    1.0000    1.2600    1.5800    1.9800    2.4900    3.1300     0.9100
    1.0000    1.8800    3.5500    6.7000   12.6200   23.8000     0.9900
    1.0000    2.5100    6.3200   15.8800   39.9000  100.2800     0.6000
    1.0000    3.1400    9.8700   31.0100   97.4100  306.0200     0.0200
Solutions:
x(1)=-0.01
x(2)=1.6027903945021139463
x(3)=-1.6132030599055614262
x(4)=1.2454941213714367527
x(5)=-0.49098971958465761669
x(6)=0.065760696175232005188
Residuals:
res(1)=0
res(2)=0.00000000000000000001
res(3)=-0.00000000000000000016
res(4)=0
res(5)=-0.0000000000000000017
res(6)=0.000000000000000001

Ruby

<lang ruby> require 'bigdecimal/ludcmp' include LUSolve

BigDecimal::limit(30)

a = [1.00, 0.00, 0.00, 0.00, 0.00, 0.00,

    1.00, 0.63, 0.39, 0.25, 0.16, 0.10,
    1.00, 1.26, 1.58, 1.98, 2.49, 3.13,
    1.00, 1.88, 3.55, 6.70, 12.62, 23.80,
    1.00, 2.51, 6.32, 15.88, 39.90, 100.28,
    1.00, 3.14, 9.87, 31.01, 97.41, 306.02].map{|i|BigDecimal(i,16)}

b = [-0.01, 0.61, 0.91, 0.99, 0.60, 0.02].map{|i|BigDecimal(i,16)}

n = 6 zero = BigDecimal("0.0") one = BigDecimal("1.0")

lusolve(a, b, ludecomp(a, n, zero,one), zero).each{|v| puts v.to_s('F')[0..20]}</lang>

Output:
-0.01
1.6027903945021135753
-1.613203059905560094
1.2454941213714351826
-0.490989719584656871
0.0657606961752318825

Tcl

Library: Tcllib (Package: math::linearalgebra)

<lang tcl>package require math::linearalgebra

set A {

   {1.00  0.00  0.00  0.00  0.00   0.00}
   {1.00  0.63  0.39  0.25  0.16   0.10}
   {1.00  1.26  1.58  1.98  2.49   3.13}
   {1.00  1.88  3.55  6.70 12.62  23.80}
   {1.00  2.51  6.32 15.88 39.90 100.28}
   {1.00  3.14  9.87 31.01 97.41 306.02}

} set b {-0.01 0.61 0.91 0.99 0.60 0.02} puts -nonewline [math::linearalgebra::show [math::linearalgebra::solveGauss $A $b] "%.2f"]</lang>

Output:
-0.01
1.60
-1.61
1.25
-0.49
0.07

zkl

Translation of: C

<lang zkl>fcn gaussEliminate(a,b){ // modifies a&b --> vector

  n:=b.len();
  foreach dia in ([0..n-1]){
     maxRow:=dia; max:=a[dia][dia];
     foreach row in ([dia+1 .. n-1]){
        if((tmp:=a[row][dia].abs()) > max){ maxRow=row; max=tmp; }
     }
     a.swap(dia,maxRow); b.swap(dia,maxRow);  // swap rows
     foreach row in ([dia+1 .. n-1]){
        ar:=a[row]; ad:=a[dia]; tmp:=ar[dia] / ad[dia];

foreach col in ([dia+1 .. n-1]){ ar[col]-=tmp*ad[col]; } ar[dia]=0.0; b[row]-=tmp*b[dia];

     }
  }
  x:=(0).pump(n,List().write);  // -->list filled with garbage
  foreach row in ([n-1 .. 0,-1]){
     tmp:=b[row]; ar:=a[row];
     foreach j in ([n-1 .. row+1,-1]){ tmp-=x[j]*ar[j]; }
     x[row]=tmp/a[row][row];
  }
  x

}</lang> <lang zkl>a:=List( List(1.00, 0.00, 0.00, 0.00, 0.00, 0.00,),

        List(1.00, 0.63, 0.39,  0.25,  0.16, 0.10,),
        List(1.00, 1.26, 1.58,  1.98,  2.49, 3.13,),
        List(1.00, 1.88, 3.55,  6.70, 12.62, 23.80,),
        List(1.00, 2.51, 6.32, 15.88, 39.90, 100.28,),
        List(1.00, 3.14, 9.87, 31.01, 97.41, 306.02) );

b:=List( -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 ); gaussEliminate(a,b).println();</lang>

Output:
L(-0.01,1.60279,-1.6132,1.24549,-0.49099,0.0657607)