public static void SubstituteLU(SFix[,] LU, SFix[] b, SFix[] y, SFix[] x)
{
int n = LU.GetLength(0);
for (int i = 0; i < n; i++)
{
x[i] = b[i];
}
/* do forward substitution, replacing x vector. */
x[0] /= LU[0, 0];
for (int i = 1; i < n; i++)
{
SFix sum = SFix.FromLong(0, 1);
for (int j = 0; j < i; j++)
{
sum += LU[i, j] * x[j];
}
x[i] = (x[i] - sum) / LU[i, i];
}
/* now get the solution vector, x[n-1] is already done */
for (int i = n - 2; i >= 0; i--)
{
SFix sum = SFix.FromLong(0, 1);
for (int j = i + 1; j < n; j++)
{
sum += LU[i, j] * x[j];
}
x[i] -= sum;
}
}
/*
* Code partially ported from:
*
* Chapter 2, Programs 3-5, Fig. 2.8-2.10
* Gerald/Wheatley, APPLIED NUMERICAL ANALYSIS (fourth edition)
* Addison-Wesley, 1989
*/
public static void ComputeLU(SFix[,] A, int iw, int fw)
{
int n = A.GetLength(0);
SFix sum;
SFix diag = (SFix.FromLong(1, iw, fw) / A[0, 0]).Resize(iw, fw);
for (int i = 1; i < n; i++)
{
A[0, i] = (A[0, i] * diag).Resize(iw, fw);
}
/*
* 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.
*/
int nm1 = n - 1;
for (int j = 1; j < nm1; j++)
{
/* column of L's */
for (int i = j; i < n; i++)
{
sum = SFix.FromLong(0, iw, fw);
for (int k = 0; k < j; k++)
{
sum = (sum + A[i, k] * A[k, j]).Resize(iw, fw);
}
A[i, j] = (A[i, j] - sum).Resize(iw, fw);
}
/* row of U's */
diag = (SFix.FromLong(1, iw, fw) / A[j, j]).Resize(iw, fw);
for (int k = j + 1; k < n; k++)
{
sum = SFix.FromLong(0, iw, fw);
for (int i = 0; i < j; i++)
{
sum = (sum + A[j, i] * A[i, k]).Resize(iw, fw);
}
A[j, k] = ((A[j, k] - sum) * diag).Resize(iw, fw);
}
}
/* still need to get last element in L Matrix */
sum = SFix.FromLong(0, iw, fw);
for (int k = 0; k < nm1; k++)
{
sum = (sum + A[nm1, k] * A[k, nm1]).Resize(iw, fw);
}
A[nm1, nm1] = (A[nm1, nm1] - sum).Resize(iw, fw);
}