void LUIncPreconditioner::solveMxb(Vector &x, const Vector &b) const
{
#ifdef DEBUG
assert(x.getLength() == b.getLength() );
#endif
Vector y( b.getLength() ); // TEMPORARY VECTOR
forwardSubstitution(y,b);
backwardSubstitution(x,y);
}
static Vec<double> lsSolve(const Mat<double> &L, const Mat<double> &U,
const Vec<double> &b)
{
Vec<double> x(L.ys());
// Solve Ly=b, Here y=x
forwardSubstitution(L, b, x);
// Solve Ux=y, Here x=y
backwardSubstitution(U, x, x);
return x;
}
double * LUSolver(double **A, double *b, int n){
double **L, **U, *c, *x;
L=allocateDoubleMatrix(n, n);
U=allocateDoubleMatrix(n, n);
LUFactotisation(n,A,U,L);
c=forwardSubstitution(n, L, b);
x=backSubstitution(n, U, c);
return x;
}
/*
* Computer problem 4.2 #2
*/
void problem42_2()
{
// variables
struct matrix mat,chol,cholT;
struct matrix4 eq = {{
{0.05,0.07,0.06,0.05},
{0.07,0.10,0.08,0.07},
{0.06,0.08,0.10,0.09},
{0.05,0.07,0.09,0.10}
}};
struct array ans = {{0.23,0.32,0.33,0.31}};
struct array y,x;
// setup
mat.m4 = eq;
mat.size = NUM4;
// print
printf("Problem 4.2 #2\n");
printf("Solve this system by Cholesky method:\n");
printf("0.05x1 + 0.07x2 + 0.06x3 + 0.05x4 = 0.23\n");
printf("0.07x1 + 0.10x2 + 0.08x3 + 0.07x4 = 0.32\n");
printf("0.06x1 + 0.08x2 + 0.10x3 + 0.09x4 = 0.33\n");
printf("0.05x1 + 0.07x2 + 0.09x3 + 0.10x4 = 0.31\n");
// do the work
chol = cholesky(mat);
// transpose
cholT = transposeMatrix(chol);
// solve Ly = b (for y)
y = forwardSubstitution(chol,ans);
// solve L^Tx = y (for x)
x = backSubstitution(cholT,y);
// print out results
printf("\nCholesky\n");
printMatrix(chol);
printf("\nTranspose \n");
printMatrix(cholT);
printf("\nLy = b (for y)\n");
printArray(y,"y");
printf("L^Tx = y (for x)\n");
printArray(x,"x");
printf("\n");
return;
}
void gauss() {
int i, j;
forwardSubstitution();
reverseElimination();
if(isinf(mx[0][n]))
{
printf("hiba");
return;
}
for (i = 0; i < n; ++i) {
// hack! (to avoid -0.00000 and use 0.00000)
if(mx[i][n]<0&&mx[i][n]>-0.00001)
mx[i][n]=0;
if(i+1<n)
printf("%.5f ", mx[i][n]);
else
printf("%.5f", mx[i][n]);
}
}
