Gaussian elimination
Solve Ax=b using Gaussian elimination then backwards substitution.
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
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.
- See also
-
- the Wikipedia entry: Gaussian elimination
360 Assembly
<lang 360asm>* Gaussian elimination 09/02/2019 GAUSSEL CSECT
USING GAUSSEL,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
SAVE (14,12) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LA R7,1 j=1
DO WHILE=(C,R7,LE,N) do j=1 to n
LA R9,1(R7) j+1
LR R6,R9 i=j+1
DO WHILE=(C,R6,LE,N) do i=j+1 to n
LR R1,R7 j
MH R1,=AL2(NN) *n
AR R1,R7 +j
BCTR R1,0 j*n+j-1
SLA R1,2 ~
LE F0,A-(NN*4)(R1) a(j,j)
LR R1,R6 i
MH R1,=AL2(NN) *n
AR R1,R7 j
BCTR R1,0 i*n+j-1
SLA R1,2 ~
LE F2,A-(NN*4)(R1) a(i,j)
DER F0,F2 a(j,j)/a(i,j)
STE F0,W w=a(j,j)/a(i,j)
LR R8,R9 k=j+1
DO WHILE=(C,R8,LE,N) do k=j+1 to n
LR R1,R7 j
MH R1,=AL2(NN) *n
AR R1,R8 +k
BCTR R1,0 j*n+k-1
SLA R1,2 ~
LE F0,A-(NN*4)(R1) a(j,k)
LR R1,R6 i
MH R1,=AL2(NN) *n
AR R1,R8 +k
BCTR R1,0 i*n+k-1
SLA R1,2 ~
LE F2,A-(NN*4)(R1) a(i,k)
LE F6,W w
MER F6,F2 *a(i,k)
SER F0,F6 a(j,k)-w*a(i,k)
STE F0,A-(NN*4)(R1) a(i,k)=a(j,k)-w*a(i,k)
LA R8,1(R8) k=k+1
ENDDO , end do k
LR R1,R7 j
SLA R1,2 ~
LE F0,B-4(R1) b(j)
LR R1,R6 i
SLA R1,2 ~
LE F2,B-4(R1) b(i)
LE F6,W w
MER F6,F2 *b(i)
SER F0,F6 b(j)-w*b(i)
STE F0,B-4(R1) b(i)=b(j)-w*b(i)
LA R6,1(R6) i=i+1
ENDDO , end do i
LA R7,1(R7) j=j+1
ENDDO , end do j
L R2,N n
SLA R2,2 ~
LE F0,B-4(R1) b(n)
L R1,N n
MH R1,=AL2(NN) *n
A R1,N n
BCTR R1,0 n*n+n-1
SLA R1,2 ~
LE F2,A-(NN*4)(R1) a(n,n)
DER F0,F2 b(n)/a(n,n)
STE F0,X-4(R2) x(n)=b(n)/a(n,n)
L R7,N n
BCTR R7,0 j=n-1
DO WHILE=(C,R7,GE,=F'1') do j=n-1 to 1 by -1
LE F0,=E'0' 0
STE F0,W w=0
LA R9,1(R7) j+1
LR R6,R9 i=j+1
DO WHILE=(C,R6,LE,N) do i=j+1 to n
LR R1,R7 j
MH R1,=AL2(NN) *n
AR R1,R6 i
BCTR R1,0 j*n+i-1
SLA R1,2 ~
LE F0,A-(NN*4)(R1) a(j,i)
LR R1,R6 i
SLA R1,2 ~
LE F2,X-4(R1) x(i)
MER F0,F2 a(j,i)*x(i)
LE F6,W w
AER F6,F0 +a(j,i)*x(i)
STE F6,W w=w+a(j,i)*x(i)
LA R6,1(R6) i=i+1
ENDDO , end do i
LR R2,R7 j
SLA R2,2 ~
LE F0,B-4(R2) b(j)
SE F0,W -w
LR R1,R7 j
MH R1,=AL2(NN) *n
AR R1,R7 j
BCTR R1,0 j*n+j-1
SLA R1,2 ~
LE F2,A-(NN*4)(R1) a(j,j)
DER F0,F2 (b(j)-w)/a(j,j)
STE F0,X-4(R2) x(j)=(b(j)-w)/a(j,j)
BCTR R7,0 j=j-1
ENDDO , end do j
XPRNT =CL8'SOLUTION',8 print
MVC PG,=CL91' ' clear buffer
LA R6,1 i=1
DO WHILE=(C,R6,LE,N) do i=1 to n
LR R1,R6 i
SLA R1,2 ~
LE F0,X-4(R1) x(i)
LA R0,5 number of decimals
BAL R14,FORMATF edit
MVC PG(13),0(R1) output
XPRNT PG,L'PG print
LA R6,1(R6) i=i+1
ENDDO , end do i
L R13,4(0,R13) restore previous savearea pointer
RETURN (14,12),RC=0 restore registers from calling sav
COPY plig\$_FORMATF.MLC format F13.n
NN EQU (X-B)/4 n N DC A(NN) n A DC E'1',E'0',E'0',E'0',E'0',E'0'
DC E'1',E'0.63',E'0.39',E'0.25',E'0.16',E'0.10'
DC E'1',E'1.26',E'1.58',E'1.98',E'2.49',E'3.13'
DC E'1',E'1.88',E'3.55',E'6.70',E'12.62',E'23.80'
DC E'1',E'2.51',E'6.32',E'15.88',E'39.90',E'100.28'
DC E'1',E'3.14',E'9.87',E'31.01',E'97.41',E'306.02'
B DC E'-0.01',E'0.61',E'0.91',E'0.99',E'0.60',E'0.02' X DS (NN)E x(n) W DS E w PG DC CL91' ' buffer
REGEQU
END GAUSSEL</lang>
- Output:
SOLUTION
-0.00999
1.60279
-1.61322
1.24552
-0.49100
0.06576
Ada
<lang Ada>with Ada.Text_IO; with Ada.Numerics.Generic_Real_Arrays;
procedure Gaussian_Eliminations is
type Real is new Float;
package Real_Arrays is
new Ada.Numerics.Generic_Real_Arrays (Real);
use Real_Arrays;
function Gaussian_Elimination (A : in Real_Matrix;
B : in Real_Vector) return Real_Vector
is
procedure Swap_Row (A : in out Real_Matrix;
B : in out Real_Vector;
R_1, R_2 : in Integer)
is
Temp : Real;
begin
if R_1 = R_2 then return; end if;
-- Swal matrix row
for Col in A'Range (1) loop
Temp := A (R_1, Col);
A (R_1, Col) := A (R_2, Col);
A (R_2, Col) := Temp;
end loop;
-- Swap vector row
Temp := B (R_1);
B (R_1) := B (R_2);
B (R_2) := Temp;
end Swap_Row;
AC : Real_Matrix := A;
BC : Real_Vector := B;
X : Real_Vector (A'Range (1)) := BC;
Max, Tmp : Real;
Max_Row : Integer;
begin
if
A'Length (1) /= A'Length (2) or
A'Length (1) /= B'Length
then
raise Constraint_Error with "Dimensions do not match";
end if;
if
A'First (1) /= A'First (2) or
A'First (1) /= B'First
then
raise Constraint_Error with "First index must be same";
end if;
for Dia in Ac'Range (1) loop
Max_Row := Dia;
Max := Ac (Dia, Dia);
for Row in Dia + 1 .. Ac'Last (1) loop
Tmp := abs (Ac (Row, Dia));
if Tmp > Max then
Max_Row := Row;
Max := Tmp;
end if;
end loop;
Swap_Row (Ac, Bc, Dia, Max_Row);
for Row in Dia + 1 .. Ac'Last (1) loop
Tmp := Ac (Row, Dia) / Ac (Dia, Dia);
for Col in Dia + 1 .. Ac'Last (1) loop
Ac (Row, Col) := Ac (Row, Col) - Tmp * Ac (Dia, Col);
end loop;
Ac (Row, Dia) := 0.0;
Bc (Row) := Bc (Row) - Tmp * Bc (Dia);
end loop;
end loop;
for Row in reverse Ac'Range (1) loop
Tmp := Bc (Row);
for J in reverse Row + 1 .. Ac'Last (1) loop
Tmp := Tmp - X (J) * Ac (Row, J);
end loop;
X (Row) := Tmp / Ac (Row, Row);
end loop;
return X; end Gaussian_Elimination;
procedure Put (V : in Real_Vector) is
use Ada.Text_IO;
package Real_IO is
new Ada.Text_IO.Float_IO (Real);
begin
Put ("[ ");
for E of V loop
Real_IO.Put (E, Exp => 0, Aft => 6);
Put (" ");
end loop;
Put (" ]");
New_Line;
end Put;
A : constant Real_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));
B : constant Real_Vector :=
( -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 );
X : constant Real_Vector := Gaussian_Elimination (A, B);
begin
Put (X);
end Gaussian_Eliminations;</lang>
- Output:
[ -0.010000 1.602774 -1.613148 1.245437 -0.490967 0.065758 ]
ALGOL 68
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
- 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
- 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 := $f(sub)n(upb mat - lwb mat)(f( vec repr)f(nl))f( vec repr)f(bus)$;
- 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
- using Gaussian elimination, find x where A*x = b #
PROC in situ gaussian elimination = (REF MAT a, b)REF MAT: (
- 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;
- 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;
- 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: (
- 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 #
- -*- coding: utf-8 -*- #
PR READ "prelude_exception.a68" PR;
- define the attributes of the scalar field being used #
MODE SCAL = REAL; FORMAT scal repr = $g(-0,real width)$;
- 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>
- include <stdlib.h>
- include <math.h>
- 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) {
- 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); }
- 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
Common Lisp
<lang CommonLisp> (defmacro mapcar-1 (fn n list)
"Maps a function of two parameters where the first one is fixed, over a list" `(mapcar #'(lambda (l) (funcall ,fn ,n l)) ,list) )
(defun gauss (m)
(labels
((redc (m) ; Reduce to triangular form
(if (null (cdr m))
m
(cons (car m) (mapcar-1 #'cons 0 (redc (mapcar #'cdr (mapcar #'(lambda (r) (mapcar #'- (mapcar-1 #'* (caar m) r)
(mapcar-1 #'* (car r) (car m)))) (cdr m)))))) ))
(rev (m) ; Reverse each row except the last element
(reverse (mapcar #'(lambda (r) (append (reverse (butlast r)) (last r))) m)) ))
(catch 'result
(let ((m1 (redc (rev (redc m)))))
(reverse (mapcar #'(lambda (r) (let ((pivot (find-if-not #'zerop r))) (if pivot (/ (car (last r)) pivot) (throw 'result 'singular)))) m1)) ))))
</lang>
- Output:
(setq m1 '((1.00 0.00 0.00 0.00 0.00 0.00 -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 39.90 100.28 0.60)
(1.00 3.14 9.87 31.01 97.41 306.02 0.02) ))
(gauss m1)
=> (-0.009999999 1.6027923 -1.6132091 1.2455008 -0.4909925 0.06576109)
C#
This modifies A and b in place, which might not be quite desirable. <lang csharp> using System;
namespace Rosetta {
internal class Vector
{
private double[] b;
internal readonly int rows;
internal Vector(int rows)
{
this.rows = rows;
b = new double[rows];
}
internal Vector(double[] initArray)
{
b = (double[])initArray.Clone();
rows = b.Length;
}
internal Vector Clone()
{
Vector v = new Vector(b);
return v;
}
internal double this[int row]
{
get { return b[row]; }
set { b[row] = value; }
}
internal void SwapRows(int r1, int r2)
{
if (r1 == r2) return;
double tmp = b[r1];
b[r1] = b[r2];
b[r2] = tmp;
}
internal double norm(double[] weights)
{
double sum = 0;
for (int i = 0; i < rows; i++)
{
double d = b[i] * weights[i];
sum += d*d;
}
return Math.Sqrt(sum);
}
internal void print()
{
for (int i = 0; i < rows; i++)
Console.WriteLine(b[i]);
Console.WriteLine();
}
public static Vector operator-(Vector lhs, Vector rhs)
{
Vector v = new Vector(lhs.rows);
for (int i = 0; i < lhs.rows; i++)
v[i] = lhs[i] - rhs[i];
return v;
}
}
class Matrix
{
private double[] b;
internal readonly int rows, cols;
internal Matrix(int rows, int cols)
{
this.rows = rows;
this.cols = cols;
b = new double[rows * cols];
}
internal Matrix(int size)
{
this.rows = size;
this.cols = size;
b = new double[rows * cols];
for (int i = 0; i < size; i++)
this[i, i] = 1;
}
internal Matrix(int rows, int cols, double[] initArray)
{
this.rows = rows;
this.cols = cols;
b = (double[])initArray.Clone();
if (b.Length != rows * cols) throw new Exception("bad init array");
}
internal double this[int row, int col]
{
get { return b[row * cols + col]; }
set { b[row * cols + col] = value; }
}
public static Vector operator*(Matrix lhs, Vector rhs)
{
if (lhs.cols != rhs.rows) throw new Exception("I can't multiply matrix by vector");
Vector v = new Vector(lhs.rows);
for (int i = 0; i < lhs.rows; i++)
{
double sum = 0;
for (int j = 0; j < rhs.rows; j++)
sum += lhs[i,j]*rhs[j];
v[i] = sum;
}
return v;
}
internal void SwapRows(int r1, int r2)
{
if (r1 == r2) return;
int firstR1 = r1 * cols;
int firstR2 = r2 * cols;
for (int i = 0; i < cols; i++)
{
double tmp = b[firstR1 + i];
b[firstR1 + i] = b[firstR2 + i];
b[firstR2 + i] = tmp;
}
}
//with partial pivot
internal void ElimPartial(Vector B)
{
for (int diag = 0; diag < rows; diag++)
{
int max_row = diag;
double max_val = Math.Abs(this[diag, diag]);
double d;
for (int row = diag + 1; row < rows; row++)
if ((d = Math.Abs(this[row, diag])) > max_val)
{
max_row = row;
max_val = d;
}
SwapRows(diag, max_row);
B.SwapRows(diag, max_row);
double invd = 1 / this[diag, diag];
for (int col = diag; col < cols; col++)
this[diag, col] *= invd;
B[diag] *= invd;
for (int row = 0; row < rows; row++)
{
d = this[row, diag];
if (row != diag)
{
for (int col = diag; col < cols; col++)
this[row, col] -= d * this[diag, col];
B[row] -= d * B[diag];
}
}
}
}
internal void print()
{
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
Console.Write(this[i,j].ToString()+" ");
Console.WriteLine();
}
}
}
} </lang> <lang csharp> using System;
namespace Rosetta {
class Program
{
static void Main(string[] args)
{
Matrix A = new Matrix(6, 6,
new double[] {1.1,0.12,0.13,0.12,0.14,-0.12,
1.21,0.63,0.39,0.25,0.16,0.1,
1.03,1.26,1.58,1.98,2.49,3.13,
1.06,1.88,3.55,6.7,12.62,23.8,
1.12,2.51,6.32,15.88,39.9,100.28,
1.16,3.14,9.87,31.01,97.41,306.02});
Vector B = new Vector(new double[] { -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 });
A.ElimPartial(B);
B.print();
}
}
} </lang>
{{out}}
-0.0597391027501976
1.85018966726278
-1.97278330181163
1.4697587750651
-0.553874184782179
0.0723048745759396
D
<lang d>import std.stdio, std.math, std.algorithm, std.range, std.numeric,
std.typecons;
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]
Delphi
<lang Delphi>program GuassianElimination;
// Modified from: // R. Sureshkumar (10 January 1997) // Gregory J. McRae (22 October 1997) // http://web.mit.edu/10.001/Web/Course_Notes/Gauss_Pivoting.c
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils;
type
TMatrix = class
private
_r, _c : integer;
data : array of TDoubleArray;
function getValue(rIndex, cIndex : integer): double;
procedure setValue(rIndex, cIndex : integer; value: double);
public
constructor Create (r, c : integer);
destructor Destroy; override;
property r : integer read _r;
property c : integer read _c;
property value[rIndex, cIndex: integer]: double read getValue write setValue; default;
end;
constructor TMatrix.Create (r, c : integer);
begin
inherited Create; self.r := r; self.c := c; setLength (data, r, c);
end;
destructor TMatrix.Destroy; begin
data := nil; inherited;
end;
function TMatrix.getValue(rIndex, cIndex: Integer): double; begin
Result := data[rIndex-1, cIndex-1]; // 1-based array
end;
procedure TMatrix.setValue(rIndex, cIndex : integer; value: double); begin
data[rIndex-1, cIndex-1] := value; // 1-based array
end;
// Solve A x = b procedure gauss (A, b, x : TMatrix); var rowx : integer;
i, j, k, n, m : integer; amax, xfac, temp, temp1 : double;
begin
rowx := 0; // Keep count of the row interchanges
n := A.r;
for k := 1 to n - 1 do
begin
amax := abs (A[k,k]);
m := k;
// Find the row with largest pivot
for i := k + 1 to n do
begin
xfac := abs (A[i,k]);
if xfac > amax then
begin
amax := xfac;
m := i;
end;
end;
if m <> k then
begin // Row interchanges
rowx := rowx+1;
temp1 := b[k,1];
b[k,1] := b[m,1];
b[m,1] := temp1;
for j := k to n do
begin
temp := a[k,j];
a[k,j] := a[m,j];
a[m,j] := temp;
end;
end;
for i := k+1 to n do
begin
xfac := a[i, k]/a[k, k];
for j := k+1 to n do
a[i,j] := a[i,j]-xfac*a[k,j];
b[i,1] := b[i,1] - xfac*b[k,1]
end;
end;
// Back substitution
for j := 1 to n do
begin
k := n-j + 1;
x[k,1] := b[k,1];
for i := k+1 to n do
begin
x[k,1] := x[k,1] - a[k,i]*x[i,1];
end;
x[k,1] := x[k,1]/a[k,k];
end;
end;
var A, b, x : TMatrix;
begin
try // Could have been done with simple arrays rather than a specific TMatrix class A := TMatrix.Create (4,4); // Note ideal but use TMatrix to define the vectors as well b := TMatrix.Create (4,1); x := TMatrix.Create (4,1);
A[1,1] := 2; A[1,2] := 1; A[1,3] := 0; A[1,4] := 0; A[2,1] := 1; A[2,2] := 1; A[2,3] := 1; A[2,4] := 0; A[3,1] := 0; A[3,2] := 1; A[3,3] := 2; A[3,4] := 1; A[4,1] := 0; A[3,2] := 0; A[4,3] := 1; A[4,4] := 2;
b[1,1] := 2; b[2,1] := 1; b[3,1] := 4; b[4,1] := 8;
gauss (A, b, x);
writeln (x[1,1]:5:2); writeln (x[2,1]:5:2); writeln (x[3,1]:5:2); writeln (x[4,1]:5:2);
readln;
except
on E: Exception do
Writeln(E.ClassName, ': ', E.Message);
end;
end.
</lang>
- Output:
1.00, 0.00, 0.00, 4.00
F#
The Function
<lang fsharp> // Gaussian Elimination. Nigel Galloway: February 2nd., 2019 let gelim augM=
let f=List.length augM
let fG n (g:bigint list) t=n|>List.map(fun n->List.map2(fun n g->g-n)(List.map(fun n->n*g.[t])n)(List.map(fun g->g*n.[t])g))
let rec fN i (g::e as l)=
match i with i when i=f->l|>List.mapi(fun n (g:bigint list)->(g.[f],g.[n]))
|_->fN (i+1) (fG e g i@[g])
fN 0 augM
</lang>
The Task
This task uses functionality from Continued_fraction/Arithmetic/Construct_from_rational_number#F.23 and Continued_fraction#F.23 <lang fsharp> let test=[[ -6I; -18I; 13I; 6I; -6I; -15I; -2I; -9I; -231I];
[ 2I; 20I; 9I; 2I; 16I; -12I; -18I; -5I; 647I];
[ 23I; 18I; -14I; -14I; -1I; 16I; 25I; -17I; -907I];
[ -8I; -1I; -19I; 4I; 3I; -14I; 23I; 8I; 248I];
[ 25I; 20I; -6I; 15I; 0I; -10I; 9I; 17I; 1316I];
[-13I; -1I; 3I; 5I; -2I; 17I; 14I; -12I; -1080I];
[ 19I; 24I; -21I; -5I; -19I; 0I; -24I; -17I; 1006I];
[ 20I; -3I; -14I; -16I; -23I; -25I; -15I; 20I; 1496I]]
let fN (n,g)=cN2S(π(rI2cf n g)) gelim test |> List.map fN |> List.iteri(fun i n->(printfn "x[%d]=%1.14f " (i+1) (snd (Seq.pairwise n|> Seq.find(fun (n,g)->n-g < 0.0000000000001M))))) </lang>
- Output:
x[1]=12.00000000000000 x[2]=10.00000000000000 x[3]=-20.00000000000000 x[4]=22.00000000000000 x[5]=-1.00000000000000 x[6]=-20.00000000000000 x[7]=-25.00000000000000 x[8]=23.00000000000000
Fortran
Gaussian Elimination with partial pivoting using augmented matrix <lang fortran>
program ge
real, allocatable :: a(:,:),b(:)
a = reshape( &
[1.0, 1.00, 1.00, 1.00, 1.00, 1.00, &
0.0, 0.63, 1.26, 1.88, 2.51, 3.14, &
0.0, 0.39, 1.58, 3.55, 6.32, 9.87, &
0.0, 0.25, 1.98, 6.70, 15.88, 31.01, &
0.0, 0.16, 2.49, 12.62, 39.90, 97.41, &
0.0, 0.10, 3.13, 23.80, 100.28, 306.02], [6,6] )
b = [-0.01, 0.61, 0.91, 0.99, 0.60, 0.02]
print'(f15.7)',solve_wbs(ge_wpp(a,b))
contains
function solve_wbs(u) result(x) ! solve with backward substitution
real :: u(:,:)
integer :: i,n
real , allocatable :: x(:)
n = size(u,1)
allocate(x(n))
forall (i=n:1:-1) x(i) = ( u(i,n+1) - sum(u(i,i+1:n)*x(i+1:n)) ) / u(i,i)
end function
function ge_wpp(a,b) result(u) ! gaussian eliminate with partial pivoting
real :: a(:,:),b(:),upi
integer :: i,j,n,p
real , allocatable :: u(:,:)
n = size(a,1)
u = reshape( [a,b], [n,n+1] )
do j=1,n
p = maxloc(abs(u(j:n,j)),1) + j-1 ! maxloc returns indices between (1,n-j+1)
if (p /= j) u([p,j],j) = u([j,p],j)
u(j+1:,j) = u(j+1:,j)/u(j,j)
do i=j+1,n+1
upi = u(p,i)
if (p /= j) u([p,j],i) = u([j,p],i)
u(j+1:n,i) = u(j+1:n,i) - upi*u(j+1:n,j)
end do
end do
end function
end program
</lang>
FreeBASIC
Gaussian elimination with pivoting. FreeBASIC version 1.05 <lang FreeBASIC>
Sub GaussJordan(matrix() As Double,rhs() As Double,ans() As Double)
Dim As Long n=Ubound(matrix,1)
Redim ans(0):Redim ans(1 To n)
Dim As Double b(1 To n,1 To n),r(1 To n)
For c As Long=1 To n 'take copies
r(c)=rhs(c)
For d As Long=1 To n
b(c,d)=matrix(c,d)
Next d
Next c
#macro pivot(num)
For p1 As Long = num To n - 1
For p2 As Long = p1 + 1 To n
If Abs(b(p1,num))<Abs(b(p2,num)) Then
Swap r(p1),r(p2)
For g As Long=1 To n
Swap b(p1,g),b(p2,g)
Next g
End If
Next p2
Next p1
#endmacro
For k As Long=1 To n-1
pivot(k) 'full pivoting
For row As Long =k To n-1
If b(row+1,k)=0 Then Exit For
Var f=b(k,k)/b(row+1,k)
r(row+1)=r(row+1)*f-r(k)
For g As Long=1 To n
b((row+1),g)=b((row+1),g)*f-b(k,g)
Next g
Next row
Next k
'back substitute
For z As Long=n To 1 Step -1
ans(z)=r(z)/b(z,z)
For j As Long = n To z+1 Step -1
ans(z)=ans(z)-(b(z,j)*ans(j)/b(z,z))
Next j
Next z
End Sub
dim as double a(1 to 6,1 to 6) = { _ {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} _ }
dim as double b(1 to 6) = { -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 }
redim as double result() GaussJordan(a(),b(),result())
for n as long=lbound(result) to ubound(result)
print result(n)
next n sleep </lang>
- Output:
-0.01 1.602790394502115 -1.613203059905572 1.245494121371448 -0.490989719584662 0.06576069617523256
Go
Partial pivoting, no scaling
Gaussian elimination with partial pivoting by pseudocode on WP page Gaussian elimination." <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]
Scaled partial pivoting
Changes from above version noted with comments. For the example data scaling does help a bit. <lang go>package main
import (
"errors" "fmt" "log" "math"
)
type testCase struct {
a [][]float64 b []float64 x []float64
}
var tc = testCase{
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-14
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) {
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
}
for k := range a {
iMax := 0
max := -1.
for i := k; i < m; i++ {
row := a[i]
// compute scale factor s = max abs in row
s := -1.
for j := k; j < m; j++ {
x := math.Abs(row[j])
if x > s {
s = x
}
}
// scale the abs used to pick the pivot.
if abs := math.Abs(row[k]) / s; abs > max {
iMax = i
max = abs
}
}
if a[iMax][k] == 0 {
return nil, errors.New("singular")
}
a[k], a[iMax] = a[iMax], a[k]
for i := k + 1; i < m; i++ {
for j := k + 1; j <= m; j++ {
a[i][j] -= a[k][j] * (a[i][k] / a[k][k])
}
a[i][k] = 0
}
}
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 1.6027903945021131 -1.6132030599055596 1.245494121371436 -0.49098971958465754 0.065760696175232]
Haskell
From scratch
<lang Haskell> isMatrix xs = null xs || all ((== (length.head $ xs)).length) xs
isSquareMatrix xs = null xs || all ((== (length xs)).length) xs
multiply:: Num a => a -> a -> a multiply us vs = map (mult [] vs) us
where mult xs [] _ = xs mult xs _ [] = xs mult [] (zs:zss) (y:ys) = mult (map (\v -> v*y) zs) zss ys mult xs (zs:zss) (y:ys) = mult (zipWith (\u v -> u+v*y) xs zs) zss ys
gauss::Double -> Double -> Double gauss xs bs = map (map fromRational) $ solveGauss (toR xs) (toR bs)
where toR = map $ map toRational
solveGauss:: (Fractional a, Ord a) => a -> a -> a solveGauss xs bs | null xs || null bs || length xs /= length bs || (not $ isSquareMatrix xs) || (not $ isMatrix bs) = []
| otherwise = uncurry solveTriangle $ triangle xs bs
solveTriangle::(Fractional a,Eq a) => a -> a -> a solveTriangle us _ | not.null.dropWhile ((/= 0).head) $ us = [] solveTriangle ([c]:as) (b:bs) = go as bs [map (/c) b]
where val us vs ws = let u = head us in map (/u) $ zipWith (-) vs (head $ multiply [tail us] ws) go [] _ zs = zs go _ [] zs = zs go (x:xs) (y:ys) zs = go xs ys $ (val x y zs):zs
triangle::(Num a, Ord a) => a -> a -> (a,a) triangle xs bs = triang ([],[]) (xs,bs)
where
triang ts (_,[]) = ts
triang ts ([],_) = ts
triang (os,ps) zs = triang (us:os,cs:ps).unzip $ [(fun tus vs, fun cs es) | (v:vs,es) <- zip uss css,let fun = zipWith (\x y -> v*x - u*y)]
where ((us@(u:tus)):uss,cs:css) = bubble zs
bubble::(Num a, Ord a) => (a,a) -> (a,a) bubble (xs,bs) = (go xs, go bs)
where idmax = snd.minimum.flip zip [0..].map (negate.abs.head) $ xs go ys = let (us,vs) = splitAt idmax ys in vs ++ us
main = do
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]]
mapM_ print $ gauss a b
</lang>
- Output:
[-1.0e-2] [1.6027903945021098] [-1.6132030599055482] [1.245494121371424] [-0.49098971958465265] [6.576069617523134e-2]
Another example
We use Rational numbers for having more precision. a % b is the rational a / b. <lang Haskell> foldlZipWith::(a -> b -> c) -> (d -> c -> d) -> d -> [a] -> [b] -> d foldlZipWith _ _ u [] _ = u foldlZipWith _ _ u _ [] = u foldlZipWith f g u (x:xs) (y:ys) = foldlZipWith f g (g u (f x y)) xs ys
foldl1ZipWith::(a -> b -> c) -> (c -> c -> c) -> [a] -> [b] -> c foldl1ZipWith _ _ [] _ = error "First list is empty" foldl1ZipWith _ _ _ [] = error "Second list is empty" foldl1ZipWith f g (x:xs) (y:ys) = foldlZipWith f g (f x y) xs ys
multAdd::(a -> b -> c) -> (c -> c -> c) -> a -> b -> c multAdd f g xs ys = map (\us -> foldl1ZipWith (\u vs -> map (f u) vs) (zipWith g) us ys) xs
mult:: Num a => a -> a -> a mult xs ys = multAdd (*) (+) xs ys
bubble::([a] -> c) -> (c -> c -> Bool) -> a -> b -> (a,b) bubble _ _ [] ts = ([],ts) bubble _ _ rs [] = (rs,[]) bubble f g (r:rs) (t:ts) = bub r t (f r) rs ts [] []
where
bub l k _ [] _ xs ys = (l:xs,k:ys)
bub l k _ _ [] xs ys = (l:xs,k:ys)
bub l k m (u:us) (v:vs) xs ys = ans
where
mu = f u
ans | g m mu = bub l k m us vs (u:xs) (v:ys)
| otherwise = bub u v mu us vs (l:xs) (k:ys)
pivot::Num a => [a] -> [a] -> a -> a -> (a,a) pivot xs ks ys ls = go ys ls [] []
where
x = head xs
fun r = zipWith (\u v -> u*r - v*x)
val rs ts = let f = fun (head rs) in (tail $ f xs rs,f ks ts)
go [] _ us vs = (us,vs)
go _ [] us vs = (us,vs)
go rs ts us vs = go (tail rs) (tail ts) (es:us) (fs:vs)
where (es,fs) = val (head rs) (head ts)
triangle::(Num a,Ord a) => a -> a -> (a,a) triangle as bs = go (as,bs) [] []
where go ([],_) us vs = (us,vs) go (_,[]) us vs = (us,vs) go (rs,ts) us vs = ans where (xs:ys,ks:ls) = bubble (abs.head) (>=) rs ts ans = go (pivot xs ks ys ls) (xs:us) (ks:vs)
solveTriangle::(Fractional a,Eq a) => a -> a -> a solveTriangle [] _ = [] solveTriangle _ [] = [] solveTriangle as _ | not.null.dropWhile ((/= 0).head) $ as = [] solveTriangle ([c]:as) (b:bs) = go as bs [map (/c) b]
where val us vs ws = let u = head us in map (/u) $ zipWith (-) vs (head $ mult [tail us] ws) go [] _ zs = zs go _ [] zs = zs go (x:xs) (y:ys) zs = go xs ys $ (val x y zs):zs
solveGauss:: (Fractional a, Ord a) => a -> a -> a solveGauss as bs = uncurry solveTriangle $ triangle as bs
matI::(Num a) => Int -> a matI n = [ [fromIntegral.fromEnum $ i == j | j <- [1..n]] | i <- [1..n]]
task::Rational -> Rational -> IO() task a b = do
let x = solveGauss a b let u = map (map fromRational) x let y = mult a x let identity = matI (length x) let a1 = solveGauss a identity let h = mult a a1 let z = mult a1 b putStrLn "a =" mapM_ print a putStrLn "b =" mapM_ print b putStrLn "solve: a * x = b => x = solveGauss a b =" mapM_ print x putStrLn "u = fromRationaltoDouble x =" mapM_ print u putStrLn "verification: y = a * x = mult a x =" mapM_ print y putStrLn $ "test: y == b = " print $ y == b putStrLn "identity matrix: identity =" mapM_ print identity putStrLn "find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveGauss a identity =" mapM_ print a1 putStrLn "verification: h = a * a1 = mult a a1 =" mapM_ print h putStrLn $ "test: h == identity = " print $ h == identity putStrLn "z = a1 * b = mult a1 b =" mapM_ print z putStrLn "test: z == x =" print $ z == x
main = do
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]]
task a b
</lang>
- Output:
a = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [1 % 1,63 % 100,39 % 100,1 % 4,4 % 25,1 % 10] [1 % 1,63 % 50,79 % 50,99 % 50,249 % 100,313 % 100] [1 % 1,47 % 25,71 % 20,67 % 10,631 % 50,119 % 5] [1 % 1,251 % 100,158 % 25,397 % 25,399 % 10,2507 % 25] [1 % 1,157 % 50,987 % 100,3101 % 100,9741 % 100,15301 % 50] b = [(-1) % 100] [61 % 100] [91 % 100] [99 % 100] [3 % 5] [1 % 50] solve: a * x = b => x = solveGauss a b = [(-1) % 100] [655870882787 % 409205648497] [(-660131804286) % 409205648497] [509663229635 % 409205648497] [(-200915766608) % 409205648497] [26909648324 % 409205648497] u = fromRationaltoDouble x = [-1.0e-2] [1.602790394502114] [-1.6132030599055613] [1.2454941213714368] [-0.4909897195846576] [6.5760696175232e-2] verification: y = a * x = mult a x = [(-1) % 100] [61 % 100] [91 % 100] [99 % 100] [3 % 5] [1 % 50] test: y == b = True identity matrix: identity = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,1 % 1] find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveGauss a identity = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [(-1373267314900) % 409205648497,2792895413400 % 409205648497,(-2539722499600) % 409205648497,1620086418000 % 409205648497,(-593562467900) % 409205648497,93570451000 % 409205648497] [1683936576500 % 409205648497,(-5515373801600) % 409205648497,7425272193600 % 409205648497,(-5318952383900) % 409205648497,2060945510400 % 409205648497,(-335828095000) % 409205648497] [(-955389934100) % 409205648497,3910562856500 % 409205648497,(-6532196158200) % 409205648497,5493636552500 % 409205648497,(-2312764532500) % 409205648497,396151215800 % 409205648497] [253880215500 % 409205648497,(-1187959549100) % 409205648497,2281116328400 % 409205648497,(-2180688584400) % 409205648497,1021846842100 % 409205648497,(-188195252500) % 409205648497] [(-25558559000) % 409205648497,131101344100 % 409205648497,(-277605537500) % 409205648497,292380217600 % 409205648497,(-151287558900) % 409205648497,30970093700 % 409205648497] verification: h = a * a1 = mult a a1 = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,1 % 1] test: h == identity = True z = a1 * b = mult a1 b = [(-1) % 100] [655870882787 % 409205648497] [(-660131804286) % 409205648497] [509663229635 % 409205648497] [(-200915766608) % 409205648497] [26909648324 % 409205648497] test: z == x = True
Determinant and permutation matrix are given
<lang Haskell> foldlZipWith::(a -> b -> c) -> (d -> c -> d) -> d -> [a] -> [b] -> d foldlZipWith _ _ u [] _ = u foldlZipWith _ _ u _ [] = u foldlZipWith f g u (x:xs) (y:ys) = foldlZipWith f g (g u (f x y)) xs ys
foldl1ZipWith::(a -> b -> c) -> (c -> c -> c) -> [a] -> [b] -> c foldl1ZipWith _ _ [] _ = error "First list is empty" foldl1ZipWith _ _ _ [] = error "Second list is empty" foldl1ZipWith f g (x:xs) (y:ys) = foldlZipWith f g (f x y) xs ys
multAdd::(a -> b -> c) -> (c -> c -> c) -> a -> b -> c multAdd f g xs ys = map (\us -> foldl1ZipWith (\u vs -> map (f u) vs) (zipWith g) us ys) xs
mult:: Num a => a -> a -> a mult xs ys = multAdd (*) (+) xs ys
triangle::(Fractional a, Ord a) => a -> a -> (a,[(([a],[a]),Int)]) triangle as bs = pivot 1 [] $ zipWith3 (\x y i -> ((x,y),i)) as bs [(0::Int)..]
where good rs ts = (abs.head.fst.fst $ ts) <= (abs.head.fst.fst $ rs) go (us,vs) ((os,ps),i) = if o == 0 then ((rs,f vs ps),i) else ((f us rs,f vs ps),i) where (o,rs) = (head os,tail os) f = zipWith (\x y -> y - x*o) change i (ys:zs) = map (\xs -> if (==i).snd $ xs then ys else xs) zs pivot d ls [] = (d,ls) pivot d ls zs@((_,j):ys) = if u == 0 then (0,ls) else pivot e (ps:ls) ws where e = if i == j then u*d else -u*d ws = map (go (map (/u) us,map (/u) vs)) $ if i == j then ys else change i zs ps@((u:us,vs),i) = foldl1 (\rs ts -> if good rs ts then rs else ts) zs
-- ((det,sol),permutation) = gauss as bs -- det = determinant as -- sol is solution of: as * sol = bs -- perm is a permutation with: (matPerm perm) * as * sol = (matPerm perm) * bs gauss::(Fractional a,Ord a) => a -> a -> ((a,a),[Int]) gauss as bs = if 0 == det then ((0,[]),[]) else solveTriangle ms
where (det,ms) = triangle as bs solveTriangle ((([c],b),i):sys) = go sys [map (/c) b] [i] where val us vs ws = let u = head us in map (/u) $ zipWith (-) vs (head $ mult [tail us] ws) go [] zs is = ((det,zs),is) go (((x,y),i):sys) zs is = go sys ((val x y zs):zs) (i:is)
solveGauss::(Fractional a,Ord a) => a -> a -> a solveGauss as = snd.fst.gauss as
matI::Num a => Int -> a matI n = [ [fromIntegral.fromEnum $ i == j | i <- [1..n]] | j <- [1..n]]
matPerm::Num a => [Int] -> a matPerm ns = [ [fromIntegral.fromEnum $ i == j | (j,_) <- zip [0..] ns] | i <- ns]
task::Rational -> Rational -> IO() task a b = do
let ((d,x),perm) = gauss a b let ps = matPerm perm let u = map (map fromRational) x let y = mult a x let identity = matI (length x) let a1 = solveGauss a identity let h = mult a a1 let z = mult a1 b putStrLn "d = determinant a =" print d putStrLn "a =" mapM_ print a putStrLn "b =" mapM_ print b putStrLn "solve: a * x = b => x = solveGauss a b =" mapM_ print x putStrLn "u = fromRationaltoDouble x =" mapM_ print u putStrLn "verification: y = a * x = mult a x =" mapM_ print y putStrLn $ "test: y == b = " print $ y == b putStrLn "ps is the permutation associated to matrix a and ps =" mapM_ print ps putStrLn "identity matrix: identity =" mapM_ print identity putStrLn "find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveGauss a identity =" mapM_ print a1 putStrLn "verification: h = a * a1 = mult a a1 =" mapM_ print h putStrLn $ "test: h == identity = " print $ h == identity putStrLn "z = a1 * b = mult a1 b =" mapM_ print z putStrLn "test: z == x =" print $ z == x
main = do
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]]
task a b
</lang>
- Output:
d = determinant a = 409205648497 % 10000000000 a = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [1 % 1,63 % 100,39 % 100,1 % 4,4 % 25,1 % 10] [1 % 1,63 % 50,79 % 50,99 % 50,249 % 100,313 % 100] [1 % 1,47 % 25,71 % 20,67 % 10,631 % 50,119 % 5] [1 % 1,251 % 100,158 % 25,397 % 25,399 % 10,2507 % 25] [1 % 1,157 % 50,987 % 100,3101 % 100,9741 % 100,15301 % 50] b = [(-1) % 100] [61 % 100] [91 % 100] [99 % 100] [3 % 5] [1 % 50] solve: a * x = b => x = solveGauss a b = [(-1) % 100] [655870882787 % 409205648497] [(-660131804286) % 409205648497] [509663229635 % 409205648497] [(-200915766608) % 409205648497] [26909648324 % 409205648497] u = fromRationaltoDouble x = [-1.0e-2] [1.602790394502114] [-1.6132030599055613] [1.2454941213714368] [-0.4909897195846576] [6.5760696175232e-2] verification: y = a * x = mult a x = [(-1) % 100] [61 % 100] [91 % 100] [99 % 100] [3 % 5] [1 % 50] test: y == b = True ps is the permutation associated to matrix a and ps = [1,0,0,0,0,0] [0,0,0,0,0,1] [0,0,1,0,0,0] [0,0,0,0,1,0] [0,1,0,0,0,0] [0,0,0,1,0,0] identity matrix: identity = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,1 % 1] find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveGauss a identity = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [(-1373267314900) % 409205648497,2792895413400 % 409205648497,(-2539722499600) % 409205648497,1620086418000 % 409205648497,(-593562467900) % 409205648497,93570451000 % 409205648497] [1683936576500 % 409205648497,(-5515373801600) % 409205648497,7425272193600 % 409205648497,(-5318952383900) % 409205648497,2060945510400 % 409205648497,(-335828095000) % 409205648497] [(-955389934100) % 409205648497,3910562856500 % 409205648497,(-6532196158200) % 409205648497,5493636552500 % 409205648497,(-2312764532500) % 409205648497,396151215800 % 409205648497] [253880215500 % 409205648497,(-1187959549100) % 409205648497,2281116328400 % 409205648497,(-2180688584400) % 409205648497,1021846842100 % 409205648497,(-188195252500) % 409205648497] [(-25558559000) % 409205648497,131101344100 % 409205648497,(-277605537500) % 409205648497,292380217600 % 409205648497,(-151287558900) % 409205648497,30970093700 % 409205648497] verification: h = a * a1 = mult a a1 = [1 % 1,0 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,0 % 1,0 % 1,1 % 1] test: h == identity = True z = a1 * b = mult a1 b = [(-1) % 100] [655870882787 % 409205648497] [(-660131804286) % 409205648497] [509663229635 % 409205648497] [(-200915766608) % 409205648497] [26909648324 % 409205648497] test: z == x = True
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>
JavaScript
From Numerical Recipes in C: <lang javascript>// Lower Upper Solver function lusolve(A, b, update) { var lu = ludcmp(A, update) if (lu === undefined) return // Singular Matrix! return lubksb(lu, b, update) }
// Lower Upper Decomposition function ludcmp(A, update) { // A is a matrix that we want to decompose into Lower and Upper matrices. var d = true var n = A.length var idx = new Array(n) // Output vector with row permutations from partial pivoting var vv = new Array(n) // Scaling information
for (var i=0; i<n; i++) { var max = 0 for (var j=0; j<n; j++) { var temp = Math.abs(A[i][j]) if (temp > max) max = temp } if (max == 0) return // Singular Matrix! vv[i] = 1 / max // Scaling }
if (!update) { // make a copy of A var Acpy = new Array(n) for (var i=0; i<n; i++) { var Ai = A[i] Acpyi = new Array(Ai.length) for (j=0; j<Ai.length; j+=1) Acpyi[j] = Ai[j] Acpy[i] = Acpyi } A = Acpy }
var tiny = 1e-20 // in case pivot element is zero for (var i=0; ; i++) { for (var j=0; j<i; j++) { var sum = A[j][i] for (var k=0; k<j; k++) sum -= A[j][k] * A[k][i]; A[j][i] = sum } var jmax = 0 var max = 0; for (var j=i; j<n; j++) { var sum = A[j][i] for (var k=0; k<i; k++) sum -= A[j][k] * A[k][i]; A[j][i] = sum var temp = vv[j] * Math.abs(sum) if (temp >= max) { max = temp jmax = j } } if (i <= jmax) { for (var j=0; j<n; j++) { var temp = A[jmax][j] A[jmax][j] = A[i][j] A[i][j] = temp } d = !d; vv[jmax] = vv[i] } idx[i] = jmax; if (i == n-1) break; var temp = A[i][i] if (temp == 0) A[i][i] = temp = tiny temp = 1 / temp for (var j=i+1; j<n; j++) A[j][i] *= temp } return {A:A, idx:idx, d:d} }
// Lower Upper Back Substitution function lubksb(lu, b, update) { // solves the set of n linear equations A*x = b. // lu is the object containing A, idx and d as determined by the routine ludcmp. var A = lu.A var idx = lu.idx var n = idx.length
if (!update) { // make a copy of b var bcpy = new Array(n) for (var i=0; i<b.length; i+=1) bcpy[i] = b[i] b = bcpy }
for (var ii=-1, i=0; i<n; i++) { var ix = idx[i] var sum = b[ix] b[ix] = b[i] if (ii > -1) for (var j=ii; j<i; j++) sum -= A[i][j] * b[j] else if (sum) ii = i b[i] = sum } for (var i=n-1; i>=0; i--) { var sum = b[i] for (var j=i+1; j<n; j++) sum -= A[i][j] * b[j] b[i] = sum / A[i][i] } return b // solution vector x }
document.write( lusolve( [ [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]
],
[-0.01, 0.61, 0.91, 0.99, 0.60, 0.02]
) )</lang>
- Output:
-0.01000000000000004, 1.6027903945021095, -1.6132030599055475, 1.2454941213714232, -0.4909897195846526, 0.06576069617523138
Julia
Using built-in LAPACK-based linear solver (which employs partial-pivoted Gaussian elimination): <lang julia>x = A \ b</lang>
Klong
<lang K> elim::{[h m];h::*m::x@>*'x;
:[2>#x;x;(,h),0,:\.f({1_x}'{x-h**x%*h}'1_m)]}
subst::{[v];v::[];
{v::v,((*x)-/:[[]~v;[];v*x@1+!#v])%x@1+#v}'||'x;|v}
gauss::{subst(elim(x))} </lang>
Example, matrix taken from C version:
<lang K>
gauss([[1.00 0.00 0.00 0.00 0.00 0.00 -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 39.90 100.28 0.60]
[1.00 3.14 9.87 31.01 97.41 306.02 0.02]]
[-0.00999999999999981
1.60279039450211414 -1.6132030599055625 1.24549412137143782 -0.490989719584658025 0.0657606961752320591]
</lang>
Kotlin
<lang scala>// version 1.1.51
val ta = arrayOf(
doubleArrayOf(1.00, 0.00, 0.00, 0.00, 0.00, 0.00), doubleArrayOf(1.00, 0.63, 0.39, 0.25, 0.16, 0.10), doubleArrayOf(1.00, 1.26, 1.58, 1.98, 2.49, 3.13), doubleArrayOf(1.00, 1.88, 3.55, 6.70, 12.62, 23.80), doubleArrayOf(1.00, 2.51, 6.32, 15.88, 39.90, 100.28), doubleArrayOf(1.00, 3.14, 9.87, 31.01, 97.41, 306.02)
)
val tb = doubleArrayOf(-0.01, 0.61, 0.91, 0.99, 0.60, 0.02)
val tx = doubleArrayOf(
-0.01, 1.602790394502114, -1.6132030599055613, 1.2454941213714368, -0.4909897195846576, 0.065760696175232
)
const val EPSILON = 1e-14 // tolerance required
fun gaussPartial(a0: Array<DoubleArray>, b0: DoubleArray): DoubleArray {
val m = b0.size
val a = Array(m) { DoubleArray(m) }
for ((i, ai) in a0.withIndex()) {
val row = ai.copyOf(m + 1)
row[m] = b0[i]
a[i] = row
}
for (k in 0 until a.size) {
var iMax = 0
var max = -1.0
for (i in k until m) {
val row = a[i]
// compute scale factor s = max abs in row
var s = -1.0
for (j in k until m) {
val e = Math.abs(row[j])
if (e > s) s = e
}
// scale the abs used to pick the pivot
val abs = Math.abs(row[k]) / s
if (abs > max) {
iMax = i
max = abs
}
}
if (a[iMax][k] == 0.0) {
throw RuntimeException("Matrix is singular.")
}
val tmp = a[k]
a[k] = a[iMax]
a[iMax] = tmp
for (i in k + 1 until m) {
for (j in k + 1..m) {
a[i][j] -= a[k][j] * a[i][k] / a[k][k]
}
a[i][k] = 0.0
}
}
val x = DoubleArray(m)
for (i in m - 1 downTo 0) {
x[i] = a[i][m]
for (j in i + 1 until m) {
x[i] -= a[i][j] * x[j]
}
x[i] /= a[i][i]
}
return x
}
fun main(args: Array<String>) {
val x = gaussPartial(ta, tb)
println(x.asList())
for ((i, xi) in x.withIndex()) {
if (Math.abs(tx[i] - xi) > EPSILON) {
println("Out of tolerance.")
println("Expected values are ${tx.asList()}")
return
}
}
}</lang>
- Output:
[-0.01, 1.6027903945021138, -1.6132030599055616, 1.2454941213714392, -0.49098971958465953, 0.06576069617523238]
M2000 Interpreter
Faster, with accuracy of 25 decimals <lang M2000 Interpreter> module checkit {
Dim Base 1, a(6, 6), b(6)
a(1,1)= 1.00, 0.00, 0.00, 0.00, 0.00, 0.00, 1.00, 0.63, 0.39, 0.25, 0.16, 0.10, 1.00, 1.26, 1.58, 1.98, 2.49, 3.13, 1.00, 1.88, 3.55, 6.70, 12.62, 23.80, 1.00, 2.51, 6.32, 15.88, 39.90, 100.28, 1.00, 3.14, 9.87, 31.01, 97.41, 306.02
\\ remove \\ to feed next array
\\ a(1,1)=1.1,0.12,0.13,0.12,0.14,-0.12,1.21,0.63,0.39,0.25,0.16,0.1,1.03,1.26,1.58,1.98,2.49,3.13, 1.06,1.88,3.55,6.7,12.62,23.8, 1.12,2.51,6.32,15.88,39.9,100.28,1.16,3.14,9.87,31.01,97.41,306.02
for i=1 to 6 : for j=1 to 6 : a(i,j)=val(a(i,j)->Decimal) :Next j:Next i
b(1)=-0.01, 0.61, 0.91, 0.99, 0.60, 0.02
for i=1 to 6 : b(i)=val(b(i)->Decimal) :Next i
function GaussJordan(a(), b()) {
cols=dimension(a(),1)
rows=dimension(a(),2)
\\ make augmented matrix
Dim Base 1, a(cols, rows)
\\ feed array with rationals
Dim Base 1, b(Len(b()))
for diag=1 to rows {
max_row=diag
max_val=abs(a(diag, diag))
if diag<rows Then {
for ro=diag+1 to rows {
d=abs(a(ro, diag))
if d>max_val then max_row=ro : max_val=d
}
}
\\ SwapRows diag, max_row
if diag<>max_row then {
for i=1 to cols {
swap a(diag, i), a(max_row, i)
}
swap b(diag), b(max_row)
}
invd= a(diag, diag)
if diag<=cols then {
for col=diag to cols {
a(diag, col)/=invd
}
}
b(diag)/=invd
for ro=1 to rows {
d1=a(ro,diag)
d2=d1*b(diag)
if ro<>diag Then {
for col=diag to cols {a(ro, col)-=d1*a(diag, col)}
b(ro)-=d2
}
}
}
=b()
}
Function ArrayLines$(a(), leftmargin=6, maxwidth=8,decimals$="") {
\\ defualt no set decimals, can show any number
ex$={
}
const way$=", {0:"+decimals$+":-"+str$(maxwidth,"")+"}"
if dimension(a())=1 then {
m=each(a())
while m {ex$+=format$(way$,array(m))}
Insert 3, 2 ex$=string$(" ", leftmargin)
=ex$ : Break
}
for i=1 to dimension(a(),1) {
ex1$=""
for j=1 to dimension(a(),2 ) {
ex1$+=format$(way$,a(i,j))
}
Insert 1,2 ex1$=string$(" ", leftmargin)
ex$+=ex1$+{
}
}
=ex$
}
mm=GaussJordan(a(), b())
c=each(mm)
while c {
print array(c)
}
\\ check accuracy
link mm to r()
\\ prepare output document
Document out$={Algorithm using decimals
}+"Matrix A:"+ArrayLines$(a(),,,"2")+{
}+"Vector B:"+ArrayLines$(b(),,,"2")+{
}+"Solution: "+{
}
acc=25
for i=1 to dimension(a(),1)
sum=a(1,1)-a(1,1)
For j=1 to dimension(a(),2)
sum+=r(j)*a(i,j)
next j
p$=format$("Coef. {0::-2}, rounding to {1} decimal, compare {2:-5}, solution: {3}", i, acc, round(sum-b(i),acc)=0@, r(i) )
Print p$
Out$=p$+{
}
next i
Report out$
clipboard out$
} checkit </lang>
slower with accuracy of 26 decimals <lang M2000 Interpreter> Module Checkit2 {
Dim Base 1, a(6, 6), b(6)
\\ a(1,1)= 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
a(1,1)=1.1,0.12,0.13,0.12,0.14,-0.12,1.21,0.63,0.39,0.25,0.16,0.1,1.03,1.26,1.58,1.98,2.49,3.13, 1.06,1.88,3.55,6.7,12.62,23.8, 1.12,2.51,6.32,15.88,39.9,100.28,1.16,3.14,9.87,31.01,97.41,306.02
for i=1 to 6 : for j=1 to 6 : a(i,j)=val(a(i,j)->Decimal) :Next j:Next i
b(1)=-0.01, 0.61, 0.91, 0.99, 0.60, 0.02
for i=1 to 6 : b(i)=val(b(i)->Decimal) :Next i
\\ modules/function to use rational nymbers
Module Global subd(m as array, n as array) { ' change m
link m to m()
link n to n()
if m(0)=0 then return m, 0:=-n(0), 1:=n(1) : exit
if n(0)=0 then exit
return m, 0:=m(0)*(n(1)/m(1))-n(0), 1:=n(1)
}
Function Global Inv(m as array){
link m to m()
if m(0)=0@ then =m : exit
=(m(1), m(0))
}
Function Global mul(m as array, n as array){' nothing change
link m to m()
link n to n()
if n(0)=0 or n(1)=0 then =(0@,0@) : exit
=((m(0)/n(1))*n(0),m(1))
}
Module Global mul(m as array, n as array) { ' change m
link m to m()
link n to n()
if n(0)=0 or n(1)=0 then m=(0@,0@) : exit
return m, 0:=(m(0)/n(1))*n(0)
}
Function Global Res(m as array) {
link m to m()
if m(0)=0@ then =0@: exit
=m(0)/m(1)
}
\\ GaussJordan get arrays byvalue
function GaussJordan(a(), b()) {
Function copypointer(m) { Dim a() : a()=m:=a()}
\\ we can use : def copypointer(a())=a(0),a(1)
cols=dimension(a(),1)
rows=dimension(a(),2)
Dim Base 1, a(cols, rows)
for i=1 to cols : for j=1 to rows : a(i, j)=(a(i, j), 1@) : next j : next i
def d as decimal
for j=1 to rows : b(j)=(b(j), 1@) : next j
for diag=1 to rows {
max_row=diag
max_val=abs(Res(a(diag, diag)))
if diag<rows Then {
for ro=diag+1 to rows {
d=abs(Res(a(ro, diag)))
if d>max_val then max_row=ro : max_val=d
}
}
\\ SwapRows diag, max_row
if diag<>max_row then {
for i=1 to cols {
swap a(diag, i), a(max_row, i)
}
swap b(diag), b(max_row)
}
invd= Inv(a(diag, diag))
if diag<=cols then {
for col=diag to cols {
mul a(diag, col), invd
}
}
mul b(diag), invd
for ro=1 to rows {
\\ work also d1=(a(ro,diag)(0), a(ro,diag)(1))
d1=copypointer(a(ro, diag))
if ro<>diag Then {
for col=diag to cols {subd a(ro, col), mul(d1, a(diag, col))}
subd b(ro), mul(d1, b(diag))
}
}
}
dim base 1, ans(len(b()))
for i=1 to cols {
ans(i)=res(b(i)) \\ : Print b(i) ' print pairs
}
=ans()
}
Function ArrayLines$(a(), leftmargin=6, maxwidth=8,decimals$="") {
\\ defualt no set decimals, can show any number
ex$={
}
const way$=", {0:"+decimals$+":-"+str$(maxwidth,"")+"}"
if dimension(a())=1 then {
m=each(a())
while m {ex$+=format$(way$,array(m))}
Insert 3, 2 ex$=string$(" ", leftmargin)
=ex$ : Break
}
for i=1 to dimension(a(),1) {
ex1$=""
for j=1 to dimension(a(),2 ) {
ex1$+=format$(way$,a(i,j))
}
Insert 1,2 ex1$=string$(" ", leftmargin)
ex$+=ex1$+{
}
}
=ex$
}
mm=GaussJordan(a(), b())
c=each(mm)
while c {
print array(c)
}
\\ check accuracy
link mm to r()
for i=1 to dimension(a(),1)
sum=a(1,1)-a(1,1)
For j=1 to dimension(a(),2)
sum+=r(j)*a(i,j)
next j
Print round(sum-b(i),26), b(i)
next i
\\ check accuracy
Document out$={Algorithm using pair of decimals as rational numbers
}+"Matrix A:"+ArrayLines$(a(),,,"2")+{
}+"Vector B:"+ArrayLines$(b(),,,"2")+{
}+"Solution: "+{
}
acc=26
for i=1 to dimension(a(),1)
sum=a(1,1)-a(1,1)
For j=1 to dimension(a(),2)
sum+=r(j)*a(i,j)
next j
p$=format$("Coef. {0::-2}, rounding to {1} decimal, compare {2:-5}, solution: {3}", i, acc, round(sum-b(i),acc)=0@, r(i) )
Print p$
Out$=p$+{
}
next i
Report out$
clipboard out$
} Checkit2 </lang>
- Output:
Algorithm using decimals
Matrix A:
1,10, 0,12, 0,13, 0,12, 0,14, -0,12
1,21, 0,63, 0,39, 0,25, 0,16, 0,10
1,03, 1,26, 1,58, 1,98, 2,49, 3,13
1,06, 1,88, 3,55, 6,70, 12,62, 23,80
1,12, 2,51, 6,32, 15,88, 39,90, 100,28
1,16, 3,14, 9,87, 31,01, 97,41, 306,02
Vector B:
-0,01, 0,61, 0,91, 0,99, 0,60, 0,02
Solution:
Coef. 1, rounding to 26 decimal, compare True, solution: -0,0597391027501962649904316335
Coef. 2, rounding to 26 decimal, compare True, solution: 1,8501896672627829700670299288
Coef. 3, rounding to 26 decimal, compare True, solution: -1,9727833018116428175300387318
Coef. 4, rounding to 26 decimal, compare True, solution: 1,4697587750651240151384675034
Coef. 5, rounding to 26 decimal, compare True, solution: -0,5538741847821888403564152897
Coef. 6, rounding to 26 decimal, compare True, solution: 0,0723048745759411900531809852
Algorithm using pair of decimals as rational numbers
Matrix A:
1,10, 0,12, 0,13, 0,12, 0,14, -0,12
1,21, 0,63, 0,39, 0,25, 0,16, 0,10
1,03, 1,26, 1,58, 1,98, 2,49, 3,13
1,06, 1,88, 3,55, 6,70, 12,62, 23,80
1,12, 2,51, 6,32, 15,88, 39,90, 100,28
1,16, 3,14, 9,87, 31,01, 97,41, 306,02
Vector B:
-0,01, 0,61, 0,91, 0,99, 0,60, 0,02
Solution:
Coef. 1, rounding to 26 decimal, compare True, solution: -0,0597391027501962649904316335
Coef. 2, rounding to 26 decimal, compare True, solution: 1,8501896672627829700670299288
Coef. 3, rounding to 26 decimal, compare True, solution: -1,9727833018116428175300387317
Coef. 4, rounding to 26 decimal, compare True, solution: 1,4697587750651240151384675034
Coef. 5, rounding to 26 decimal, compare True, solution: -0,5538741847821888403564152897
Coef. 6, rounding to 26 decimal, compare True, solution: 0,0723048745759411900531809852
Algorithm using decimals
Matrix 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
Vector B:
-0,01, 0,61, 0,91, 0,99, 0,60, 0,02
Solution:
Coef. 1, rounding to 25 decimal, compare True, solution: -0,01
Coef. 2, rounding to 25 decimal, compare True, solution: 1,6027903945021139442641548525
Coef. 3, rounding to 25 decimal, compare True, solution: -1,6132030599055614189052834829
Coef. 4, rounding to 25 decimal, compare True, solution: 1,2454941213714367443882298102
Coef. 5, rounding to 25 decimal, compare True, solution: -0,4909897195846576129526569211
Coef. 6, rounding to 25 decimal, compare True, solution: 0,0657606961752320046201065486
Algorithm using pair of decimals as rational numbers
Matrix 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
Vector B:
-0,01, 0,61, 0,91, 0,99, 0,60, 0,02
Solution:
Coef. 1, rounding to 26 decimal, compare True, solution: -0,01
Coef. 2, rounding to 26 decimal, compare True, solution: 1,6027903945021139442641548522
Coef. 3, rounding to 26 decimal, compare True, solution: -1,6132030599055614189052834817
Coef. 4, rounding to 26 decimal, compare True, solution: 1,2454941213714367443882298085
Coef. 5, rounding to 26 decimal, compare True, solution: -0,4909897195846576129526569203
Coef. 6, rounding to 26 decimal, compare True, solution: 0,0657606961752320046201065485
Mathematica / Wolfram Language
<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>
Modula-3
This implementation defines a generic Matrix type so that the code can be used with different types. As a bonus, we implemented it to work with rings rather than fields, and tested it on two rings: the ring of integers and the ring of integers modulo 46. We include the interface of a ring modulo 46 below; the project's m3makefile (not included) is set up to automatically generates an interface and module for a matrix over each ring.
- requirements of the generic type
The Matrix needs its generic type to implement the following:
- It must have a type
T, as per Modula-3 convention. - It must have procedures
Nonzero(a: T): BOOLEAN, which indicates whetherais nonzero;Minus(a, b: T): TandTimes(a, b: T): T, which return the results of the procedures' names; andPrint(a: T)which does what the name implies.
- Matrix interface
<lang modula3>GENERIC INTERFACE Matrix(RingElem);
(* "RingElem" must export the following: - a type T; - procedures
+ "Nonzero(a: T): BOOLEAN", which indicates whether "a" is nonzero; + "Minus(a, b: T): T" and "Times(a, b: T): T", which return the results you'd guess from the procedures' names; and + "Print(a: T)", which does what the name implies.
- )
TYPE
T <: Public;
Public = OBJECT
METHODS
init(READONLY data: ARRAY OF ARRAY OF RingElem.T): T;
(* use this to copy the entries in "data"; returns "self" *)
initDimensions(m, n: CARDINAL): T;
(* use this for an mxn matrix of random entries *)
num_rows(): CARDINAL;
(* returns the number of rows in "self" *)
num_cols(): CARDINAL;
(* returns the number of columns in "self" *)
entries(): REF ARRAY OF ARRAY OF RingElem.T;
(* returns the entries in "self" *)
triangularize();
(*
Performs Gaussian elimination in the context of a ring.
We can add scalar multiples of rows,
and we can swap rows, but we may lack multiplicative inverses,
so we cannot necessarily obtain 1 as a row's first entry.
*)
END;
PROCEDURE PrintMatrix(m: T); (* prints the matrix row-by-row; sorry, no special padding to line up columns *)
END Matrix.</lang>
- Matrix implementation
<lang modula3>GENERIC MODULE Matrix(RingElem);
IMPORT IO;
TYPE
REVEAL T = Public BRANDED OBJECT rows, cols: CARDINAL; data: REF ARRAY OF ARRAY OF RingElem.T; OVERRIDES init := Init; initDimensions := InitDimensions; num_rows := Rows; num_cols := Columns; entries := Entries; triangularize := Triangularize; END;
PROCEDURE Init(self: T; READONLY d: ARRAY OF ARRAY OF RingElem.T): T = BEGIN
self.rows := NUMBER(d);
self.cols := NUMBER(d[0]);
self.data := NEW(REF ARRAY OF ARRAY OF RingElem.T, self.rows, self.cols);
FOR i := FIRST(d) TO LAST(d) DO
FOR j := FIRST(d[0]) TO LAST(d[0]) DO
self.data[i-FIRST(d)][j-FIRST(d[0])] := d[i][j];
END;
END;
RETURN self;
END Init;
PROCEDURE InitDimensions(self: T; r, c: CARDINAL): T = BEGIN
self.rows := r; self.cols := c; self.data := NEW(REF ARRAY OF ARRAY OF RingElem.T, r, c); RETURN self;
END InitDimensions;
PROCEDURE Rows(self: T): CARDINAL = BEGIN
RETURN self.rows;
END Rows;
PROCEDURE Columns(self: T): CARDINAL = BEGIN
RETURN self.cols;
END Columns;
PROCEDURE Entries(self: T): REF ARRAY OF ARRAY OF RingElem.T = BEGIN
RETURN self.data;
END Entries;
PROCEDURE SwapRows(VAR data: ARRAY OF ARRAY OF RingElem.T; i, j: CARDINAL) = (* swaps rows i and j of data *) VAR
a: RingElem.T;
BEGIN
WITH Ai = data[i], Aj = data[j], m = FIRST(data[0]), n = LAST(data[0]) DO
FOR k := m TO n DO
a := Ai[k];
Ai[k] := Aj[k];
Aj[k] := a;
END;
END;
END SwapRows;
PROCEDURE PivotExists(
VAR data: ARRAY OF ARRAY OF RingElem.T; r: CARDINAL; VAR i: CARDINAL; j: CARDINAL
): BOOLEAN = (*
Returns true iff column j of data has a pivot in some row at or after r. The row with a pivot is stored in i.
- )
VAR
searching := TRUE; result := LAST(data) + 1;
BEGIN
i := r;
WHILE searching AND i <= LAST(data) DO
IF RingElem.Nonzero(data[i,j]) THEN
searching := FALSE;
result := i;
ELSE
INC(i);
END;
END;
RETURN NOT searching;
END PivotExists;
PROCEDURE Pivot(VAR data: ARRAY OF ARRAY OF RingElem.T; i, j, k: CARDINAL) = (*
Pivots on row i, column j to eliminate row k, column j.
- )
BEGIN
WITH n = LAST(data[0]), Ai = data[i], Ak = data[k] DO
VAR a := Ai[j]; b := Ak[j];
BEGIN
FOR l := j TO n DO
IF RingElem.Nonzero(Ai[l]) THEN
Ak[l] := RingElem.Minus(
RingElem.Times(Ak[l], a),
RingElem.Times(Ai[l], b)
);
ELSE
Ak[l] := RingElem.Times(Ak[l], a);
END;
END;
END;
END;
END Pivot;
PROCEDURE Triangularize(self: T) = VAR
i: CARDINAL; r := FIRST(self.data[0]);
BEGIN
WITH data = self.data, m = FIRST(data[0]), n = LAST(data[0]) DO
FOR j := m TO n DO
IF PivotExists(data^, r, i, j) THEN
IF i # j THEN
SwapRows(data^, i, r);
END;
FOR k := r + 1 TO LAST(data^) DO
IF RingElem.Nonzero(data[k][j]) THEN
Pivot(data^, r, j, k);
END;
END;
INC(r);
END;
END;
END;
END Triangularize;
PROCEDURE PrintMatrix(self: T) = BEGIN
WITH data = self.data DO
FOR i := FIRST(data^) TO LAST(data^) DO
IO.Put("[ ");
WITH Ai = data[i] DO
FOR j := FIRST(Ai) TO LAST(Ai) DO
RingElem.Print(Ai[j]);
IF j # LAST(Ai) THEN
IO.PutChar(' ');
END;
END;
END;
IO.Put(" ]\n");
END;
END;
END PrintMatrix;
BEGIN END Matrix.</lang>
- interface for the ring of integers modulo an integer
<lang modula3>INTERFACE ModularRing;
(* Implements arithmetic modulo a nonzero integer. Assertions check that the modulus is nonzero.
- )
TYPE
T = RECORD value, modulus: CARDINAL; END;
PROCEDURE Init(VAR a: T; value: INTEGER; modulus: CARDINAL); (* initializes a to the given value and modulus *)
PROCEDURE Nonzero(n: T): BOOLEAN;
PROCEDURE Plus(a, b: T): T;
PROCEDURE Minus(a, b: T): T;
PROCEDURE Times(a, b: T): T;
PROCEDURE Print(a: T; withModulus := FALSE); (*
when "withModulus" is "TRUE", this adds after "a" the letter "m", followed by the modulus
- )
END ModularRing.</lang>
- test implementation
It's fairly easy to initialize an array of types in Modula-3, but it can get cumbersome with structured types, so we wrote a procedure to convert an integer matrix to a matrix of integers modulo a number.
<lang modula3>MODULE GaussianElimination EXPORTS Main;
IMPORT IO, ModularRing AS MR, IntMatrix AS IM, ModMatrix AS MM;
CONST
(* data to set up the matrices *)
A1 = ARRAY OF INTEGER { 2, 1, 0 };
A2 = ARRAY OF INTEGER { 1, 2, 0 };
A3 = ARRAY OF INTEGER { 0, 3, 0 };
A = ARRAY OF ARRAY OF INTEGER { A1, A2, A3 };
B1 = ARRAY OF INTEGER { 4, 8, 0, -4, 0 };
B2 = ARRAY OF INTEGER { -3, -6, 0, 9, 0 };
B3 = ARRAY OF INTEGER { 1, 3, 5, 7, 2 };
B4 = ARRAY OF INTEGER { 7, 5, 3, 1, 2 };
B = ARRAY OF ARRAY OF INTEGER { B1, B2, B3, B4 };
PROCEDURE IntToModArray(READONLY A: IM.T; VAR B: MM.T; mod: CARDINAL) = (*
copies a two-dimensional array of integers to a two-dimension array of integers modulo "mod"
- )
BEGIN
B := NEW(MM.T).initDimensions(A.num_rows(), A.num_cols());
WITH Adata = A.entries(), Bdata = B.entries() DO
FOR i := FIRST(Adata^) TO LAST(Adata^) DO
WITH Ai = Adata[i], Bi = Bdata[i] DO
FOR j := FIRST(Ai) TO LAST(Ai) DO
MR.Init(Bi[j], Ai[j], mod);
END;
END;
END;
END;
END IntToModArray;
VAR
M: IM.T; N: MM.T;
BEGIN
(* triangularize the data in A *)
M := NEW(IM.T).init(A);
IO.Put("Initial A:\n");
IM.PrintMatrix(M);
IO.PutChar('\n');
M.triangularize();
IO.Put("Final A:\n");
IM.PrintMatrix(M);
IO.PutChar('\n');
IO.PutChar('\n');
(* triangularize the data in B, all computations modulo 46 *)
M := NEW(IM.T).init(B);
IntToModArray(M, N, 46);
IO.Put("Initial B:\n");
MM.PrintMatrix(N);
IO.PutChar('\n');
N.triangularize();
IO.Put("Final B:\n");
MM.PrintMatrix(N);
IO.PutChar('\n');
END GaussianElimination.</lang>
- Output:
Initial A: [ 2 1 0 ] [ 1 2 0 ] [ 0 3 0 ] Final A: [ 2 1 0 ] [ 0 3 0 ] [ 0 0 0 ] Initial B: [ 4 8 0 42 0 ] [ 43 40 0 9 0 ] [ 1 3 5 7 2 ] [ 7 5 3 1 2 ] Final B: [ 4 8 0 42 0 ] [ 0 4 20 32 8 ] [ 0 0 32 38 44 ] [ 0 0 0 24 0 ]
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>
- 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>
- 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
<lang Perl>use Math::Matrix; my $a = Math::Matrix->new([0,1,0],
[0,0,1],
[2,0,1]);
my $b = Math::Matrix->new([1],
[2],
[4]);
my $x = $a->concat($b)->solve; print $x;</lang>
Math::Matrix solve() expects the column vector to be an extra column in the matrix, hence concat(). Putting not just a column there but a whole identity matrix (making Nx2N) is how its invert() is implemented. Note that solve() doesn't notice singular matrices and still gives a return when there is in fact no solution to Ax=B.
Perl 6
Gaussian elimination results in a matrix in row echelon form. Gaussian elimination with back-substitution (also known as Gauss-Jordan elimination) results in a matrix in reduced row echelon form. That being the case, we can reuse much of the code from the Reduced row echelon form task. Perl 6 stores and does calculations on decimal numbers within its limit of precision using Rational numbers by default, meaning the calculations are exact.
<lang perl6>sub gauss-jordan-solve (@a, @b) {
@b.kv.map: { @a[$^k].append: $^v };
@a.&rref[*]»[*-1];
}
- reduced row echelon form (Gauss-Jordan elimination)
sub rref (@m) {
return unless @m; my ($lead, $rows, $cols) = 0, +@m, +@m[0];
for ^$rows -> $r {
$lead < $cols or return @m;
my $i = $r;
until @m[$i;$lead] {
++$i == $rows or next;
$i = $r;
++$lead == $cols and return @m;
}
@m[$i, $r] = @m[$r, $i] if $r != $i;
my $lv = @m[$r;$lead];
@m[$r] »/=» $lv;
for ^$rows -> $n {
next if $n == $r;
@m[$n] »-=» @m[$r] »*» (@m[$n;$lead] // 0);
}
++$lead;
}
@m
}
sub rat-or-int ($num) {
return $num unless $num ~~ Rat; return $num.narrow if $num.narrow.WHAT ~~ Int; $num.nude.join: '/';
}
sub say-it ($message, @array, $fmt = " %8s") {
say "\n$message"; $_».&rat-or-int.fmt($fmt).put for @array;
}
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-it 'A matrix:', @a, "%6.2f"; say-it 'or, A in exact rationals:', @a; say-it 'B matrix:', @b, "%6.2f"; say-it 'or, B in exact rationals:', @b; say-it 'x matrix:', (my @gj = gauss-jordan-solve @a, @b), "%16.12f"; say-it 'or, x in exact rationals:', @gj, "%28s"; </lang>
- Output:
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
or, A in exact rationals:
1 0 0 0 0 0
1 63/100 39/100 1/4 4/25 1/10
1 63/50 79/50 99/50 249/100 313/100
1 47/25 71/20 67/10 631/50 119/5
1 251/100 158/25 397/25 399/10 2507/25
1 157/50 987/100 3101/100 9741/100 15301/50
B matrix:
-0.01
0.61
0.91
0.99
0.60
0.02
or, B in exact rationals:
-1/100
61/100
91/100
99/100
3/5
1/50
x matrix:
-0.010000000000
1.602790394502
-1.613203059906
1.245494121371
-0.490989719585
0.065760696175
or, x in exact rationals:
-1/100
655870882787/409205648497
-660131804286/409205648497
509663229635/409205648497
-200915766608/409205648497
26909648324/409205648497
Phix
<lang Phix>function gauss_eliminate(sequence a, b)
integer n = length(b)
atom tmp
for col=1 to n do
integer m = col
atom mx = a[m][m]
for i=col+1 to n do
tmp = abs(a[i][col])
if tmp>mx then
{m,mx} = {i,tmp}
end if
end for
if col!=m then
{a[col],a[m]} = {a[m],a[col]}
{b[col],b[m]} = {b[m],b[col]}
end if
for i=col+1 to n do
tmp = a[i][col]/a[col][col]
for j=col+1 to n do
a[i][j] -= tmp*a[col][j]
end for
a[i][col] = 0
b[i] -= tmp*b[col]
end for
end for
sequence x = repeat(0,n)
for col=n to 1 by -1 do
tmp = b[col]
for j=n to col+1 by -1 do
tmp -= x[j]*a[col][j]
end for
x[col] = tmp/a[col][col]
end for
return x
end function
constant 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}},
b = {-0.01, 0.61, 0.91, 0.99, 0.60, 0.02}
pp(gauss_eliminate(a, b))</lang>
- Output:
{-0.01,1.602790395,-1.61320306,1.245494121,-0.4909897196,0.06576069618}
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
PowerShell
Gauss
<lang PowerShell> function gauss($a,$b) {
$n = $a.count
for ($k = 0; $k -lt $n; $k++) {
$lmax, $max = $k, [Math]::Abs($a[$k][$k])
for ($l = $k+1; $l -lt $n; $l++) {
$tmp = [Math]::Abs($a[$l][$k])
if($max -lt $tmp) {
$max, $lmax = $tmp, $l
}
}
if ($k -ne $lmax) {
$a[$k], $a[$lmax] = $a[$lmax], $a[$k]
$b[$k], $b[$lmax] = $b[$lmax], $b[$k]
}
$akk = $a[$k][$k]
for ($i = $k+1; $i -lt $n; $i++){
$aik = $a[$i][$k]
for ($j = $k; $j -lt $n; $j++) {
$a[$i][$j] = $a[$i][$j]*$akk - $a[$k][$j]*$aik
}
$b[$i] = $b[$i]*$akk - $b[$k]*$aik
}
}
for ($i = $n-1; $i -ge 0; $i--) {
for ($j = $i+1; $j -lt $n; $j++) {
$b[$i] -= $b[$j]*$a[$i][$j]
}
$b[$i] = $b[$i]/$a[$i][$i]
}
$b
} function show($a) {
if($a) {
0..($a.Count - 1) | foreach{ if($a[$_]){"$($a[$_][0..($a[$_].count -1)])"}else{""} }
}
} $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) ) "a =" show $a "" $b = @(-0.01, 0.61, 0.91, 0.99, 0.60, 0.02) "b =" $b "" "x =" gauss $a $b
</lang> Output:
a = 1 0 0 0 0 0 1 0.63 0.39 0.25 0.16 0.1 1 1.26 1.58 1.98 2.49 3.13 1 1.88 3.55 6.7 12.62 23.8 1 2.51 6.32 15.88 39.9 100.28 1 3.14 9.87 31.01 97.41 306.02 b = -0.01 0.61 0.91 0.99 0.6 0.02 x = -0.01 1.60279039450213 -1.6132030599056 1.24549412137148 -0.490989719584674 0.0657606961752342
Gauss-Jordan
<lang PowerShell> function gauss-jordan($a,$b) {
$n = $a.count
for ($k = 0; $k -lt $n; $k++) {
$lmax, $max = $k, [Math]::Abs($a[$k][$k])
for ($l = $k+1; $l -lt $n; $l++) {
$tmp = [Math]::Abs($a[$l][$k])
if($max -lt $tmp) {
$max, $lmax = $tmp, $l
}
}
if ($k -ne $lmax) {
$a[$k], $a[$lmax] = $a[$lmax], $a[$k]
$b[$k], $b[$lmax] = $b[$lmax], $b[$k]
}
$akk = $a[$k][$k]
for ($j = $k; $j -lt $n; $j++) {$a[$k][$j] /= $akk}
$b[$k] /= $akk
for ($i = 0; $i -lt $n; $i++){
if ($i -ne $k) {
$aik = $a[$i][$k]
for ($j = $k; $j -lt $n; $j++) {
$a[$i][$j] = $a[$i][$j] - $a[$k][$j]*$aik
}
$b[$i] = $b[$i] - $b[$k]*$aik
}
}
}
$b
} function show($a) {
if($a) {
0..($a.Count - 1) | foreach{ if($a[$_]){"$($a[$_][0..($a[$_].count -1)])"}else{""} }
}
} $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) ) "a =" show $a "" $b = @(-0.01, 0.61, 0.91, 0.99, 0.60, 0.02) "b =" $b "" "x =" gauss-jordan $a $b </lang> Output:
a = 1 0 0 0 0 0 1 0.63 0.39 0.25 0.16 0.1 1 1.26 1.58 1.98 2.49 3.13 1 1.88 3.55 6.7 12.62 23.8 1 2.51 6.32 15.88 39.9 100.28 1 3.14 9.87 31.01 97.41 306.02 b = -0.01 0.61 0.91 0.99 0.6 0.02 x = -0.01 1.60279039450211 -1.61320305990556 1.24549412137144 -0.490989719584659 0.0657606961752323
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.
- 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)
- 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]]
- 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]]
- 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>
Using numpy
<lang python3> $ python3 Python 3.6.0 |Anaconda custom (64-bit)| (default, Dec 23 2016, 12:22:00) [GCC 4.4.7 20120313 (Red Hat 4.4.7-1)] on linux Type "help", "copyright", "credits" or "license" for more information. >>> # https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.solve.html >>> import numpy.linalg >>> a = [[2, 9, 4], [7, 5, 3], [6, 1, 8]] >>> b = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] >>> numpy.linalg.solve(a,b) array([[-0.10277778, 0.18888889, -0.01944444],
[ 0.10555556, 0.02222222, -0.06111111],
[ 0.06388889, -0.14444444, 0.14722222]])
>>> </lang>
R
Here 'b' is a matrix. Partial pivoting is used, and the determinant of 'a' is returned as well.
<lang R>gauss <- function(a, b) {
n <- nrow(a) det <- 1
for (i in seq_len(n - 1)) {
j <- which.max(a[i:n, i]) + i - 1
if (j != i) {
a[c(i, j), i:n] <- a[c(j, i), i:n]
b[c(i, j), ] <- b[c(j, i), ]
det <- -det
}
k <- seq(i + 1, n)
for (j in k) {
s <- aj, i / ai, i
a[j, k] <- a[j, k] - s * a[i, k]
b[j, ] <- b[j, ] - s * b[i, ]
}
}
for (i in seq(n, 1)) {
if (i < n) {
for (j in seq(i + 1, n)) {
b[i, ] <- b[i, ] - ai, j * b[j, ]
}
}
b[i, ] <- b[i, ] / ai, i
det <- det * ai, i
}
list(x=b, det=det)
}
a <- matrix(c(2, 9, 4, 7, 5, 3, 6, 1, 8), 3, 3, byrow=T) gauss(a, diag(3))</lang>
- Output:
$x
[,1] [,2] [,3]
[1,] -0.10277778 0.18888889 -0.01944444
[2,] 0.10555556 0.02222222 -0.06111111
[3,] 0.06388889 -0.14444444 0.14722222
$det
[1] -360
Racket
<lang racket>
- 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>
- <array
'#(6 1) #[-0.01 1.602790394502109 -1.613203059905556 1.2454941213714346 -0.4909897195846582 0.06576069617523222]>
</lang>
REXX
version 1
<lang rexx>/* REXX ---------------------------------------------------------------
- 07.08.2014 Walter Pachl translated from PL/I)
- improved to get integer results for, e.g. this input:
-6 -18 13 6 -6 -15 -2 -9 -231 2 20 9 2 16 -12 -18 -5 647 23 18 -14 -14 -1 16 25 -17 -907 -8 -1 -19 4 3 -14 23 8 248 25 20 -6 15 0 -10 9 17 1316 -13 -1 3 5 -2 17 14 -12 -1080 19 24 -21 -5 -19 0 -24 -17 1006 20 -3 -14 -16 -23 -25 -15 20 1496
- --------------------------------------------------------------------*/
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-1
ma=a.j.j
Do ja=j+1 To n
mb=a.ja.j
Do i=1 To n
new=a.j.i*mb-a.ja.i*ma
a.ja.i=new
End
b.ja=b.j*mb-b.ja*ma
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
version 2
(Data was placed into a file instead of placing the data into the REXX program.)
Programming note: with the large precision (numeric digits 1000), the residuals were insignificant.
Only 8 (fractional) decimal digits were used for the output display. <lang rexx>/*REXX program solves Ax=b with Gaussian elimination and backwards substitution. */ parse arg iFID . /*obtain optional argument from the CL.*/ numeric digits 1000 /*heavy─duty decimal digits precision. */ if iFID== | iFID=="," then iFID= 'GAUSS_E.DAT' /*Not specified? Then use the default.*/
do rec=1 while lines(iFID) \== 0 /*read the equation sets. */
#=0 /*the number of equations (so far). */
do $=1 while lines(iFID) \== 0 /*process the equation. */
z=linein(iFID); if z= then leave /*Is this a blank line? end─of─data.*/
if $==1 then do; say; say center(' equations ', 75, "▓"); say
end /* [↑] if 1st equation, then show hdr.*/
say z /*display an equation to the terminal. */
if left(space(z), 1)=='*' then iterate /*Is this a comment? Then ignore it.*/
#=# + 1; n=words(z) - 1 /*assign equation #; calculate # items.*/
do e=1 for n; a.#.e= word(z, e)
end /*e*/ /* [↑] process A numbers. */
b.#=word(z, n + 1) /* ◄─── " B " */
end /*$*/
if #\==0 then call Gauss_elim /*Not zero? Then display the results. */
end /*rec*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Gauss_elim: do j=1 for n; jp=j + 1
do i=jp to n; _=a.j.j / a.i.j
do k=jp to n; a.i.k=a.j.k - _ * a.i.k
end /*k*/
b.i=b.j - _ * b.i
end /*i*/
end /*j*/
x.n=b.n / a.n.n
do j=n-1 to 1 by -1; _=0
do i=j+1 to n; _=_ + a.j.i * x.i
end /*i*/
x.j=(b.j - _) / a.j.j
end /*j*/ /* [↑] uses backwards substitution. */
say
numeric digits 8 /*for the display, only use 8 digits. */
say center('solution', 75, "═"); say /*a title line for articulated output. */
do o=1 for n; say right('x['o"] = ", 38) left(, x.o>=0) x.o/1
end /*o*/
return</lang>
- input file : GAUSS_E.DAT
* a1 a2 a3 b
* ─── ─── ─── ───
1 2 3 14
2 1 3 13
3 -2 -1 -4
* a1 a2 a3 a4 a5 a6 b
* ─────── ─────── ─────── ─────── ─────── ─────── ───────
1 0 0 0 0 0 -0.01
1 0.63 0.39 0.25 0.16 0.10 0.61
1 1.26 1.58 1.98 2.49 3.13 0.91
1 1.88 3.55 6.70 12.62 23.80 0.99
1 2.51 6.32 15.88 39.90 100.28 0.60
1 3.14 9.87 31.01 97.41 306.02 0.02
- output when using the default input file:
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓ equations ▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓
* a1 a2 a3 b
* ─── ─── ─── ───
1 2 3 14
2 1 3 13
3 -2 -1 -4
═════════════════════════════════solution══════════════════════════════════
x[1] = 1
x[2] = 2
x[3] = 3
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓ equations ▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓
* a1 a2 a3 a4 a5 a6 b
* ─────── ─────── ─────── ─────── ─────── ─────── ───────
1 0 0 0 0 0 -0.01
1 0.63 0.39 0.25 0.16 0.10 0.61
1 1.26 1.58 1.98 2.49 3.13 0.91
1 1.88 3.55 6.70 12.62 23.80 0.99
1 2.51 6.32 15.88 39.90 100.28 0.60
1 3.14 9.87 31.01 97.41 306.02 0.02
═════════════════════════════════solution══════════════════════════════════
x[1] = -0.01
x[2] = 1.6027904
x[3] = -1.6132031
x[4] = 1.2454941
x[5] = -0.49098972
x[6] = 0.065760696
version 3
This is the same as version 2, but in addition, it also shows the residuals.
Code was added to this program version to keep a copy of the original A.i.k and B.# arrays (for calculating the residuals).
Also added was rounding the residual numbers to zero if the number of significant decimal digits was
less or equal to 5% of the number of significant fractional decimal digits (in this case, 5% of 1,000 digits for the decimal fraction).
<lang rexx>/*REXX program solves Ax=b with Gaussian elimination and backwards substitution. */
numeric digits 1000 /*heavy─duty decimal digits precision. */
parse arg iFID . /*obtain optional argument from the CL.*/
if iFID== | iFID=="," then iFID= 'GAUSS_E.DAT' /*Not specified? Then use the default.*/
pad=left(, 23) /*used for indenting residual numbers. */
do rec=1 while lines(iFID) \== 0 /*read the equation sets. */
#=0 /*the number of equations (so far). */
do $=1 while lines(iFID) \== 0 /*process the equation. */
z=linein(iFID); if z= then leave /*Is this a blank line? end─of─data.*/
if $==1 then do; say; say center(' equations ', 75, "▓"); say
end /* [↑] if 1st equation, then show hdr.*/
say z /*display an equation to the terminal. */
if left(space(z), 1)=='*' then iterate /*Is this a comment? Then ignore it.*/
#=# + 1; n=words(z) - 1 /*assign equation #; calculate # items.*/
do e=1 for n; a.#.e= word(z, e); oa.#.e= a.#.e
end /*e*/ /* [↑] process A numbers; save orig.*/
b.#=word(z, n + 1); ob.#=b.# /* ◄─── " B " " " */
end /*$*/
if #\==0 then call Gauss_elim /*Not zero? Then display the results. */
say
do i=1 for n; r=0 /*display the residuals to the terminal*/
do j=1 for n; r=r + oa.i.j * x.j /* ┌───◄ don't display a fraction if */
end /*j*/ /* ↓ res ≤ 5% of significant digs.*/
r=format(r - ob.i, , digits() - digits() * 0.05 % 1 , 0) / 1 /*should be tiny*/
say pad 'residual['right(i, length(n) )"] = " left(, r>=0) r /*right justify.*/
end /*i*/
end /*rec*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Gauss_elim: do j=1 for n; jp=j + 1
do i=jp to n; _=a.j.j / a.i.j
do k=jp to n; a.i.k=a.j.k - _ * a.i.k
end /*k*/
b.i=b.j - _ * b.i
end /*i*/
end /*j*/
x.n=b.n / a.n.n
do j=n-1 to 1 by -1; _=0
do i=j+1 to n; _=_ + a.j.i * x.i
end /*i*/
x.j=(b.j - _) / a.j.j
end /*j*/ /* [↑] uses backwards substitution. */
say
numeric digits 8 /*for the display, only use 8 digits. */
say center('solution', 75, "═"); say /*a title line for articulated output. */
do o=1 for n; say right('x['o"] = ", 38) left(, x.o>=0) x.o/1
end /*o*/
return</lang>
- output when using the same default input file as for version 2:
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓ equations ▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓
* a1 a2 a3 b
* ─── ─── ─── ───
1 2 3 14
2 1 3 13
3 -2 -1 -4
═════════════════════════════════solution══════════════════════════════════
x[1] = 1
x[2] = 2
x[3] = 3
residual[1] = 0
residual[2] = 0
residual[3] = 0
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓ equations ▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓
* a1 a2 a3 a4 a5 a6 b
* ─────── ─────── ─────── ─────── ─────── ─────── ───────
1 0 0 0 0 0 -0.01
1 0.63 0.39 0.25 0.16 0.10 0.61
1 1.26 1.58 1.98 2.49 3.13 0.91
1 1.88 3.55 6.70 12.62 23.80 0.99
1 2.51 6.32 15.88 39.90 100.28 0.60
1 3.14 9.87 31.01 97.41 306.02 0.02
═════════════════════════════════solution══════════════════════════════════
x[1] = -0.01
x[2] = 1.6027904
x[3] = -1.6132031
x[4] = 1.2454941
x[5] = -0.49098972
x[6] = 0.065760696
residual[1] = 0
residual[2] = 0
residual[3] = 0
residual[4] = 0
residual[5] = 0
residual[6] = 0
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
Rust
<lang rust> // using a Vec<f32> might be a better idea // for now, let us create a fixed size array // of size: const SIZE: usize = 6;
pub fn eliminate(mut system: [[f32; SIZE+1]; SIZE]) -> Option<Vec<f32>> {
// produce the row reduced echelon form
//
// for every row...
for i in 0..SIZE-1 {
// for every column in that row...
for j in i..SIZE-1 {
if system[i][i] == 0f32 {
continue;
} else {
// reduce every element under that element to 0
let factor = system[j + 1][i] as f32 / system[i][i] as f32;
for k in i..SIZE+1 {
// potential optimization: set every element to zero, instead of subtracting
// i think subtraction helps showcase the process better
system[j + 1][k] -= factor * system[i][k] as f32;
}
}
}
}
// produce gaussian eliminated array
//
// the process follows a similar pattern
// but this one reduces the upper triangular
// elements
for i in (1..SIZE).rev() {
if system[i][i] == 0f32 {
continue;
} else {
for j in (1..i+1).rev() {
let factor = system[j - 1][i] as f32 / system[i][i] as f32;
for k in (0..SIZE+1).rev() {
system[j - 1][k] -= factor * system[i][k] as f32;
}
}
}
}
// produce solutions through back substitution
let mut solutions: Vec<f32> = vec![];
for i in 0..SIZE {
if system[i][i] == 0f32 {
return None;
}
else {
system[i][SIZE] /= system[i][i] as f32;
system[i][i] = 1f32;
println!("X{} = {}", i + 1, system[i][SIZE]);
solutions.push(system[i][SIZE])
}
}
return Some(solutions);
}
- [cfg(test)]
mod tests {
use super::*;
// sample run of the program
#[test]
fn eliminate_seven_by_six() {
let system: [[f32; SIZE +1]; SIZE] = [
[1.00 , 0.00 , 0.00 , 0.00 , 0.00 , 0.00 , -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 , 39.90 , 100.28 , 0.60 ] ,
[1.00 , 3.14 , 9.87 , 31.01 , 97.41 , 306.02 , 0.02 ]
] ;
let solutions = eliminate(system).unwrap();
assert_eq!(6, solutions.len());
let assert_solns = vec![-0.01, 1.60278, -1.61320, 1.24549, -0.49098, 0.06576];
for (ans, key) in solutions.iter().zip(assert_solns.iter()) {
if (ans - key).abs() > 1E-4 { panic!("Test Failed!") }
}
}
} </lang>
Sidef
Uses the rref(A) function from Reduced row echelon form.
<lang ruby>func gauss_jordan_solve (a, b) {
var A = gather {
^b -> each {|i| take(a[i] + b[i]) }
}
rref(A).map{ .last }
}
var 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 ],
]
var b = [ -0.01, 0.61, 0.91, 0.99, 0.60, 0.02 ]
var G = gauss_jordan_solve(a, b) say G.map { "%27s" % .as_rat }.join("\n")</lang>
- Output:
-1/100 655870882787/409205648497 -660131804286/409205648497 509663229635/409205648497 -200915766608/409205648497 26909648324/409205648497
Stata
Gaussian elimination
This implementation computes also the determinant of the matrix A, as it requires only a few operations. The matrix B is overwritten with the solution of the system, and A is overwritten with garbage.
<lang stata>void gauss(real matrix a, real matrix b, real scalar det) { real scalar i,j,n,s real vector js
det = 1 n = rows(a) for (i=1; i<n; i++) { maxindex(abs(a[i::n,i]), 1, js=., .) j = js[1]+i-1 if (j!=i) { a[(i\j),i..n] = a[(j\i),i..n] b[(i\j),.] = b[(j\i),.] det = -det } for (j=i+1; j<=n; j++) { s = a[j,i]/a[i,i] a[j,i+1..n] = a[j,i+1..n]-s*a[i,i+1..n] b[j,.] = b[j,.]-s*b[i,.] } }
for (i=n; i>=1; i--) { for (j=i+1; j<=n; j++) { b[i,.] = b[i,.]-a[i,j]*b[j,.] } b[i,.] = b[i,.]/a[i,i] det = det*a[i,i] } }</lang>
LU decomposition and backsubstitution
<lang stata>void ludec(real matrix a, real matrix l, real matrix u, real vector p) { real scalar i,j,n,s real vector js
l = a n = rows(a) p = 1::n for (i=1; i<n; i++) { maxindex(abs(l[i::n,i]), 1, js=., .) j = js[1]+i-1 if (j!=i) { l[(i\j),.] = l[(j\i),.] p[(i\j)] = p[(j\i)] } for (j=i+1; j<=n; j++) { l[j,i] = s = l[j,i]/l[i,i] l[j,i+1..n] = l[j,i+1..n]-s*l[i,i+1..n] } }
u = uppertriangle(l) l = lowertriangle(l, 1) }
void luback(real matrix l, real matrix u, real vector p, real matrix y) { real scalar i,j,n
n = rows(y) y = y[p,.] for (i=1; i<=n; i++) { for (j=1; j<i; j++) { y[i,.] = y[i,.]-l[i,j]*y[j,.] } /*y[i,.] = y[i,.]/l[i,i]*/ }
for (i=n; i>=1; i--) { for (j=i+1; j<=n; j++) { y[i,.] = y[i,.]-u[i,j]*y[j,.] } y[i,.] = y[i,.]/u[i,i] } }</lang>
Example
Here we are computing the inverse of a 3x3 matrix (which happens to be a magic square), using both methods.
<lang stata>: gauss(a=(2,9,4\7,5,3\6,1,8),b=I(3),det=.)
- b
1 2 3 +----------------------------------------------+ 1 | -.1027777778 .1888888889 -.0194444444 | 2 | .1055555556 .0222222222 -.0611111111 | 3 | .0638888889 -.1444444444 .1472222222 | +----------------------------------------------+
- ludec(a=(2,9,4\7,5,3\6,1,8),l=.,u=.,p=.)
- luback(l,u,p,y=I(3))
- y
1 2 3 +----------------------------------------------+ 1 | -.1027777778 .1888888889 -.0194444444 | 2 | .1055555556 .0222222222 -.0611111111 | 3 | .0638888889 -.1444444444 .1472222222 | +----------------------------------------------+</lang>
Tcl
<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
TI-83 BASIC
The rref() function performs reduced row-echelon form using Gaussian elimination
on a n*(n+1) matrix. The (n+1)th column receives the resulting vector.
The n*n maxtrix is set to 0 and the pivots are set to 1.
The Matr>List() subroutine extracts the (n+1)th column to a list.
The matrix can be more easily entered by the matrix editor.
On TI-83 or TI-84, another way to solve this task is to use the PlySmlt2 internal apps and choose
"simult equ solver" with 6 equations and 6 unknowns.
<lang ti83b>[[ 1.00 0.00 0.00 0.00 0.00 0.00 -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 39.90 100.28 0.60] [ 1.00 3.14 9.87 31.01 97.41 306.02 0.02]]→[A]
Matr>List(rref([A]),7,L1) L1</lang>
- Output:
{-.01 1.602790395 -1.61320306 1.245494121 -.4909897196 .0657606962}
VBA
<lang vb>'Option Base 1
Private Function gauss_eliminate(a As Variant, b As Variant) As Variant
Dim n As Integer: n = UBound(b)
Dim tmp As Variant, m As Integer, mx As Variant
For col = 1 To n
m = col
mx = a(m, m)
For i = col + 1 To n
tmp = Abs(a(i, col))
If tmp > mx Then
m = i
mx = tmp
End If
Next i
If col <> m Then
For j = 1 To UBound(a, 2)
tmp = a(col, j)
a(col, j) = a(m, j)
a(m, j) = tmp
Next j
tmp = b(col)
b(col) = b(m)
b(m) = tmp
End If
For i = col + 1 To n
tmp = a(i, col) / a(col, col)
For j = col + 1 To n
a(i, j) = a(i, j) - tmp * a(col, j)
Next j
a(i, col) = 0
b(i) = b(i) - tmp * b(col)
Next i
Next col
Dim x() As Variant
ReDim x(n)
For col = n To 1 Step -1
tmp = b(col)
For j = n To col + 1 Step -1
tmp = tmp - x(j) * a(col, j)
Next j
x(col) = tmp / a(col, col)
Next col
gauss_eliminate = x
End Function Public Sub main()
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}]
b = [{-0.01, 0.61, 0.91, 0.99, 0.60, 0.02}]
Dim s() As String, x() As Variant
ReDim s(UBound(b)), x(UBound(b))
Debug.Print "(";
x = gauss_eliminate(a, b)
For i = 1 To UBound(x)
s(i) = CStr(x(i))
Next i
t = Join(s, ", ")
Debug.Print t; ")"
End Sub</lang>
- Output:
(-0.01, 1.60279039450209, -1.61320305990548, 1.24549412137136, -0.490989719584628, 0.065760696175228)
VBScript
<lang vb>' Gaussian elimination - VBScript
const n=6
dim a(6,6),b(6),x(6),ab
ab=array( 1 , 0 , 0 , 0 , 0 , 0 , -0.01, _
1 , 0.63, 0.39, 0.25, 0.16, 0.10, 0.61, _
1 , 1.26, 1.58, 1.98, 2.49, 3.13, 0.91, _
1 , 1.88, 3.55, 6.70, 12.62, 23.80, 0.99, _
1 , 2.51, 6.32, 15.88, 39.90, 100.28, 0.60, _
1 , 3.14, 9.87, 31.01, 97.41, 306.02, 0.02)
k=-1
for i=1 to n
buf=""
for j=1 to n+1
k=k+1
if j<=n then
a(i,j)=ab(k)
else
b(i)=ab(k)
end if
buf=buf&right(space(8)&formatnumber(ab(k),2),8)&" "
next
wscript.echo buf
next
for j=1 to n
for i=j+1 to n
w=a(j,j)/a(i,j)
for k=j+1 to n
a(i,k)=a(j,k)-w*a(i,k)
next
b(i)=b(j)-w*b(i)
next
next
x(n)=b(n)/a(n,n)
for j=n-1 to 1 step -1
w=0
for i=j+1 to n
w=w+a(j,i)*x(i)
next
x(j)=(b(j)-w)/a(j,j)
next
wscript.echo "solution"
buf=""
for i=1 to n
buf=buf&right(space(8)&formatnumber(x(i),2),8)&vbcrlf
next
wscript.echo buf</lang>
- Output:
-0,01
1,60
-1,61
1,25
-0,49
0,07
zkl
Using the GNU Scientific Library: <lang zkl>var [const] GSL=Import("zklGSL"); // libGSL (GNU Scientific Library) a:=GSL.Matrix(6,6).set(
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:=GSL.VectorFromData(-0.01, 0.61, 0.91, 0.99, 0.60, 0.02); x:=a.AxEQb(b); x.format(8,5).println();</lang>
- Output:
-0.01000, 1.60279,-1.61320, 1.24549,-0.49099, 0.06576
Or, using lists:
<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)