部分选主元的LU分解，处理较为巧妙，性能高效

http://www.johnloomis.org/ece538/notes/Matrix/ludcmp.html

LU Decomposition

Download: matrix.zip.

See: ludcmp.cpp  lusolve.cpp (test program).

LU decomposition is a method of solving linear matrix equations,

A x = b where A is an n by n matrix, b is a n by 1 vector and x is the n by 1 solution.

This program uses the Matrix class.

ludcmp.cpp

---

/*\file ludcmp.cpp
*
*!  Finds solution to set of linear equations A x = b by LU decomposition.
*
*  Chapter 2, Programs 3-5, Fig. 2.8-2.10
*  Gerald/Wheatley, APPLIED NUMERICAL ANALYSIS (fourth edition)
*  Addison-Wesley, 1989
*
*/
#include 
#include 
#include 
#include 
#include "matrix.h"

int ludcmp(Matrix &a, int order[]);
void solvlu(const Matrix &a, const Vec &b, Vec &x, const int order[]);
static int pivot(Matrix &a, int order[], int jcol);

#define TINY 1e-20
//! finds LU decomposition of Matrix
/*!
*   The function ludcmp computes the lower L and upper U triangular
*   matrices equivalent to the A Matrix, such that L U = A.  These
*   matrices are returned in the space of A, in compact form.
*   The U Matrix has ones on its diagonal.  Partial pivoting is used
*   to give maximum valued elements on the diagonal of L.  The order of
*   the rows after pivoting is returned in the integer vector "order".
*   This should be used to reorder the right-hand-side vectors before
*   solving the system A x = b.
*
*
* \param       a     - n by n Matrix of coefficients
* \param      order - integer vector holding row order after pivoting.
*
*/

int ludcmp(Matrix &a, int order[])
{
	int i, j, k, n, nm1;
	int flag = 1;    /* changes sign with each row interchange */
	double sum, diag;

	n = a.rows();
	assert(a.cols()==n);

	/* establish initial ordering in order vector */
	
	for (i=0; i<n; i++) order[i] = i;

	/* do pivoting for first column and check for singularity */

	if (pivot(a,order,0)) flag = -flag;
	diag = 1.0/a[0][0];
	for (i=1; i<n; i++) a[0][i] *= diag;
	
	/* 
	*  Now complete the computing of L and U elements.
	*  The general plan is to compute a column of L's, then
	*  call pivot to interchange rows, and then compute
	*  a row of U's.
	*/
	
	nm1 = n - 1;
	for (j=1; j<nm1; j++) {
		/* column of L's */
		for (i=j; i<n; i++) {
			sum = 0.0;
			for (k=0; k<j; k++) sum += a[i][k]*a[k][j];
			a[i][j] -= sum;
		}
		/* pivot, and check for singularity */
		if (pivot(a,order,j)) flag = -flag;
		/* row of U's */
		diag = 1.0/a[j][j];
		for (k=j+1; k<n; k++) {
			sum = 0.0;
			for (i=0; i<j; i++) sum += a[j][i]*a[i][k];
			a[j][k] = (a[j][k]-sum)*diag;
		}
	}

	/* still need to get last element in L Matrix */

	sum = 0.0;
	for (k=0; k<nm1; k++) sum += a[nm1][k]*a[k][nm1];
	a[nm1][nm1] -= sum;
	return flag;
}

//!  Find pivot element
/*!
*   The function pivot finds the largest element for a pivot in "jcol"
*   of Matrix "a", performs interchanges of the appropriate
*   rows in "a", and also interchanges the corresponding elements in
*   the order vector.
*
*
*  \param     a      -  n by n Matrix of coefficients
*  \param   order  - integer vector to hold row ordering
*  \param    jcol   - column of "a" being searched for pivot element
*
*/
int pivot(Matrix &a, int order[], int jcol)
{
	int i, ipvt,n;
	double big, anext;
	n = a.rows();

	/*
	*  Find biggest element on or below diagonal.
	*  This will be the pivot row.
	*/

	ipvt = jcol;
	big = fabs(a[ipvt][ipvt]);
	for (i = ipvt+1; i<n; i++) {
		anext = fabs(a[i][jcol]);
		if (anext>big) {
			big = anext;
			ipvt = i;
		}
	}
	assert(fabs(big)>TINY); // otherwise Matrix is singular
	
	/*
	*   Interchange pivot row (ipvt) with current row (jcol).
	*/
	
	if (ipvt==jcol) return 0;
	a.swaprows(jcol,ipvt);
	i = order[jcol];
	order[jcol] = order[ipvt];
	order[ipvt] = i;
	return 1;
}

//!  This function is used to find the solution to a system of equations,
/*!   A x = b, after LU decomposition of A has been found.
*    Within this routine, the elements of b are rearranged in the same way
*    that the rows of a were interchanged, using the order vector.
*    The solution is returned in x.
*
*
*  \param  a     - the LU decomposition of the original coefficient Matrix.
*  \param  b     - the vector of right-hand sides
*  \param       x     - the solution vector
*  \param    order - integer array of row order as arranged during pivoting
*
*/
void solvlu(const Matrix &a, const Vec &b, Vec &x, const int order[])
{
	int i,j,n;
	double sum;
	n = a.rows();

	/* rearrange the elements of the b vector. x is used to hold them. */

	for (i=0; i<n; i++) {
		j = order[i];
		x[i] = b[j];
	}

	/* do forward substitution, replacing x vector. */

	x[0] /= a[0][0];
	for (i=1; i<n; i++) {
		sum = 0.0;
		for (j=0; j<i; j++) sum += a[i][j]*x[j];
		x[i] = (x[i]-sum)/a[i][i];
	}

	/* now get the solution vector, x[n-1] is already done */

	for (i=n-2; i>=0; i--) {
		sum = 0.0;
		for (j=i+1; j<n; j++) sum += a[i][j] * x[j];
		x[i] -= sum;
	}
}

---

lusolve.cpp

---

/*\file  lusolve.cpp
*
*   Solve linear equations by LU decomposition
*/
#include 
#include 
#include 
#include "matrix.h"
#include 
using namespace std;

char text[128];
void print_vector(double *v, int n);


void lusolve(Matrix &mx, FILE *in);
int ludcmp(Matrix &a, int order[]);
void solvlu(const Matrix &a, const Vec &b, Vec &x, const int order[]);

int main(int argc, char *argv[])
{
	Matrix mx;
	FILE *in;
	char *filename;

	if (argc>1) {
		filename = argv[1];
		printf("opening: %s\n",filename);
	}
	else {
		printf("This program requires a data file.\n");
		printf("Enter filename: ");
		gets_s(text,72);
		if (*text==0 || *text=='\n') return -1;
		filename = text;
	}
	errno_t nerr = fopen_s(&in, filename,"rt");
	if (nerr) {
		perror(filename);
		return -1;
	}

	string title;
	while (read_matrix(mx,title,in)==0) {
		printf("\n%s\n",title.c_str());
		printf("\nMatrix A\n");
		mx.print();
		lusolve(mx, in);
	}
	return 0;
}

void lusolve(Matrix &mx, FILE *in)
{
	int i, j, k, n, flag, index, nrhs;
	int *ipvt;
	double sum;
	Vec b, x, save;
	double det;
	/*
	*   Create matrix a as a copy of input matrix.
	*/
	assert(mx.rows()==mx.cols()); // input must be square matrix
	n = mx.rows();

	Matrix a = mx;
	ipvt = ivector(n);

	flag = ludcmp(a,ipvt);

	/* Calculate determinant */

	det = flag;
	for (i=0; i<n; i++) det *= a[i][i];

	printf("\ndeterminant: %g\n",det);
	printf("pivot vector: ");
	for (i=0; i<n; i++) printf("%3d",ipvt[i]);
	printf("\n");

	/* print LU Matrix */

	Matrix lower = a;
	Matrix upper = a;
	Matrix product(n, n);

	for (i=0; i<n; i++) {
		for (j=i+1; j<n; j++) lower[i][j] = 0.0;
		for (j=0; j<i; j++) upper[i][j] = 0.0;
		upper[i][i] = 1.0;
	}
	printf("\nLower triangular Matrix\n");
	lower.print();
	printf("\nUpper triangular Matrix\n");
	upper.print();

	/* Multiply the lower and upper matrices */

	for (i=0; i<n; i++) {
		for (j=0; j<n; j++) {
			sum = 0.0;
			for (k=0; k<n; k++) sum += lower[i][k] * upper[k][j];
			product[i][j] = sum;
		}
	}
	printf("\n\nProduct of L U\n\n");
	product.print();

	/*
	*   Read in the right-hand side vectors.
	*/
	fscanf_s(in," %d",&nrhs);
	fgets(text,72,in);
	x.create(n);
	b.create(n);
	index = 0;
	while (nrhs--) {
		for (i=0; i<n; i++) fscanf_s(in," %lf", &b[i]);
		if (feof(in)) break;
		printf("\nRight-hand-side number %d\n\n",++index);
		b.print();
		save = b;

		if (fabs(det)<1e-10) {
			printf("\nCoefficient Matrix is singular.\n");
			continue;
		}
		solvlu(a,b,x,ipvt);
		printf("\nSolution vector\n\n");
		x.print();
				
		k = 0;
		for (i=0; i<n; i++) {
			sum = 0.0;
			for (j=0; j<n; j++) sum += mx[i][j]*x[j];
			if (fabs(sum-save[i])>1e-8) k++;
		}
		printf("\nsolution %s.\n",(k? "does not check": "checks"));
	}
	free(ipvt);
}

void print_vector(double *v, int n)
{
	while (n--) printf("%10.4f", *v++);
	printf("\n");
}

---

Results

The following output was produced from the command:

    lusolve eqn5.dat > output.txt

Here are the results:

opening: eqn5.dat

  Example set of equations used in text, beginning on p. 113

Matrix A
         4        -2         1
        -3        -1         4
         1        -1         3

determinant: -18
pivot vector:   0  1  2

Lower triangular matrix
         4         0         0
        -3      -2.5         0
         1      -0.5       1.8

Upper triangular matrix
         1      -0.5      0.25
         0         1      -1.9
         0         0         1


Product of L U

         4        -2         1
        -3        -1         4
         1        -1         3

Right-hand-side number 1

   15.0000    8.0000   13.0000

Solution vector

    2.0000   -2.0000    3.0000

solution checks.

  Example set of equations used in text, beginning on p. 120

Matrix A
         0         2         0         1
         2         2         3         2
         4        -3         0         1
         6         1        -6        -5

determinant: -234
pivot vector:   3  2  1  0

Lower triangular matrix
         6         0         0         0
         4    -3.667         0         0
         2     1.667     6.818         0
         0         2     2.182      1.56

Upper triangular matrix
         1    0.1667        -1   -0.8333
         0         1    -1.091    -1.182
         0         0         1    0.8267
         0         0         0         1


Product of L U

         6         1        -6        -5
         4        -3         0         1
         2         2         3         2
         0         2         0         1

Right-hand-side number 1

    0.0000   -2.0000   -7.0000    6.0000

Solution vector

   -0.5000    1.0000    0.3333   -2.0000

solution checks.

 Test data used in Chapter 2, Program 2.1  See page 184 (Fig 2.4).

Matrix A
         6         1        -6        -5
         2         2         3         2
         4        -3         0         1
         0         2         0         1

determinant: 234
pivot vector:   0  2  1  3

Lower triangular matrix
         6         0         0         0
         4    -3.667         0         0
         2     1.667     6.818         0
         0         2     2.182      1.56

Upper triangular matrix
         1    0.1667        -1   -0.8333
         0         1    -1.091    -1.182
         0         0         1    0.8267
         0         0         0         1


Product of L U

         6         1        -6        -5
         4        -3         0         1
         2         2         3         2
         0         2         0         1

Right-hand-side number 1

    6.0000   -2.0000   -7.0000    0.0000

Solution vector

   -0.5000    1.0000    0.3333   -2.0000

solution checks.

   Set of equations for problem 2-5 (page 201)

Matrix A
         2         1         1        -2
         4         0         2         1
         3         2         2         0
         1         3         2         0

determinant: 11
pivot vector:   1  3  2  0

Lower triangular matrix
         4         0         0         0
         1         3         0         0
         3         2      -0.5         0
         2         1      -0.5    -1.833

Upper triangular matrix
         1         0       0.5      0.25
         0         1       0.5  -0.08333
         0         0         1     1.167
         0         0         0         1


Product of L U

         4         0         2         1
         1         3         2         0
         3         2         2         0
         2         1         1        -2

Right-hand-side number 1

    0.0000    8.0000    7.0000    3.0000

Solution vector

    3.1818    2.3636   -3.6364    2.5455

solution checks.

   Set of equations given for problem 2-11 (page 201)

Matrix A
         3         2        -1        -4
         1        -1         3        -1
         2         1        -3         0
         0        -1         8        -5

determinant: -1.33227e-014
pivot vector:   0  1  3  2

Lower triangular matrix
         3         0         0         0
         1    -1.667         0         0
         0        -1         6         0
         2   -0.3333        -3-4.441e-016

Upper triangular matrix
         1    0.6667   -0.3333    -1.333
         0         1        -2      -0.2
         0         0         1   -0.8667
         0         0         0         1


Product of L U

         3         2        -1        -4
         1        -1         3        -1
         0        -1         8        -5
         2         1        -3         0

Right-hand-side number 1

   10.0000   -4.0000   16.0000    3.0000

Coefficient matrix is singular.

---

Maintained by John Loomis,updated Wed Feb 07 23:01:57 2007