Routine Name: backSub
Author: Brandon Furman
Language: C++
Description/Purpose: This function computes the solution to the linear system of equations Ax = b where A is a upper triangular coefficient matrix, b is a vector of constant terms, and x is the solution vector.
Input: This function accepts an upper triangular coefficient matrix, A, in the form of a array2D object and a vector of constant terms, b, in the form of a array1D object as inputs. Both inputs are passed by reference.
Output: This function returns a array1D object that is the solution to the linear system of equations.
Usage/Example:
int m, n, l;
m = 4; n = 4; l = 4;
array1D vec, solVec;
array2D mat;
vec = emptyVec(l);
vec(0) = 1; vec(1) = 2; vec(2) = 3; vec(3) = 4;
mat = emptyMat(m,n);
mat(0, 0) = 1; mat(0, 1) = 2; mat(0, 2) = 3; mat(0, 3) = 4;
mat(1, 1) = 5; mat(1, 2) = 6; mat(1, 3) = 7;
mat(2, 2) = 8; mat(2, 3) = 9;
mat(3, 3) = 10;
solVec = backSub(mat, vec);
for (int i = 0; i < l; i++) {
std::cout << solVec(i) << " ";
}
outputs the following to the console
-0.235 -0.07 -0.075 0.4
Implementation/Code:
array1D backSub(array2D& A, array1D& b) {
int m = A.getRows();
int n = A.getCols();
int l = b.getLength();
if (m != n || m != l) {
throw "backSub: Incompatible Sizes";
}
for (int i = 0; i < m; i++) {
if (A(i, i) == 0.0) throw "backSub: Division by zero";
}
array1D x;
x.allocateMem(m);
for (int i = m - 1; i >= 0; i--) {
x(i) = b(i);
for (int j = i + 1; j < m; j++) {
x(i) = x(i) - A(i, j)*x(j);
}
x(i) = x(i) / A(i, i);
}
return x;
}
Last Modified: March/2019