As per following the inital thread make efficient the copy of symmetric matrix in c-sharp from cMinor.
I would be quite interesting with some inputs in how to build a symmetric square matrix multiplication with one line vector and one column vector by using an array implementation of the matrix, instead of the classical
long s = 0;
List<double> columnVector = new List<double>(N);
List<double> lineVector = new List<double>(N);
//- init. vectors and symmetric square matrix m
for (int i=0; i < N; i++)
{
for(int j=0; j < N; j++){
s += lineVector[i] * columnVector[j] * m[i,j];
}
}
Thanks for your input !
The line vector times symmetric matrix equals to the transpose of the matrix times the column vector. So only the column vector case needs to be considered.
Originally the i-th element of y=A*x is defined as
y[i] = SUM( A[i,j]*x[j], j=0..N-1 )
but since A is symmetric, the sum be split into sums, one below the diagonal and the other above
y[i] = SUM( A[i,j]*x[j], j=0..i-1) + SUM( A[i,j]*x[j], j=i..N-1 )
From the other posting the matrix index is
A[i,j] = A[i*N-i*(i+1)/2+j] // j>=i
A[i,j] = A[j*N-j*(j+1)/2+i] // j< i
For a N×N symmetric matrix A = new double[N*(N+1)/2];
In C# code the above is:
int k;
for(int i=0; i<N; i++)
{
// start sum with zero
y[i]=0;
// below diagonal
k=i;
for(int j=0; j<=i-1; j++)
{
y[i]+=A[k]*x[j];
k+=N-j-1;
}
// above diagonal
k=i*N-i*(i+1)/2+i;
for(int j=i; j<=N-1; j++)
{
y[i]+=A[k]*x[j];
k++;
}
}
Example for you to try:
| -7 -6 -5 -4 -3 | | -2 | | -5 |
| -6 -2 -1 0 1 | | -1 | | 21 |
| -5 -1 2 3 4 | | 0 | = | 42 |
| -4 0 3 5 6 | | 1 | | 55 |
| -3 1 4 6 7 | | 7 | | 60 |
To get the quadratic form do a dot product with the multiplication result vector x·A·y = Dot(x,A*y)
You could make matrix multiplication pretty fast with unsafe code. I have blogged about it.
Making matrix multiplication as fast as possible is easy: Use a well-known library. Insane amounts of performance work has gone into such libraries. You cannot compete with that.
