/// <summary>
/// Solves a linear equation system of the form Ax = b.
/// </summary>
/// <param name="value">The b from the equation Ax = b.</param>
/// <returns>The x from equation Ax = b.</returns>
public Double[] Solve(Double[] value)
{
// Additionally an important property is that if there does not exists a solution
// when the matrix A is singular but replacing 1/Li with 0 will provide a solution
// that minimizes the residue |AX -Y|. SVD finds the least squares best compromise
// solution of the linear equation system. Interestingly SVD can be also used in an
// over-determined system where the number of equations exceeds that of the parameters.
// L is a diagonal matrix with non-negative matrix elements having the same
// dimension as A, Wi ? 0. The diagonal elements of L are the singular values of matrix A.
//singularity threshold
Double e = this.Threshold;
var Y = value;
// Create L*, which is a diagonal matrix with elements
// L*i = 1/Li if Li = e, else 0,
// where e is the so-called singularity threshold.
// In other words, if Li is zero or close to zero (smaller than e),
// one must replace 1/Li with 0. The value of e depends on the precision
// of the hardware. This method can be used to solve linear equations
// systems even if the matrices are singular or close to singular.
int scols = s.Length;
var Ls = new Double[scols, scols];
for (int i = 0; i < s.Length; i++)
{
if (System.Math.Abs(s[i]) <= e)
Ls[i, i] = 0;
else Ls[i, i] = 1 / s[i];
}
//(V x L*) x Ut x Y
var VL = v.Multiply(Ls);
//(V x L* x Ut) x Y
int urows = u.GetLength(0);
int vrows = v.GetLength(0);
var VLU = new Double[vrows, urows];
for (int i = 0; i < vrows; i++)
{
for (int j = 0; j < urows; j++)
{
Double sum = 0;
for (int k = 0; k < scols; k++)
sum += VL[i, k] * u[j, k];
VLU[i, j] = sum;
}
}
//(V x L* x Ut x Y)
return VLU.Multiply(Y);
}