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As noted above, for the factorization we only need to store the pivots, so it suffices to allocate a pivot array of length n ( pivots(1:n)). In fact, we will store the inverses of the pivots rather than the pivots themselves. This implies that during the system solution no divisions have to be performed.
Additionally, we assume that an extra integer array diag_ptr(1:n) has been allocated that contains the column (or row) indices of the diagonal elements in each row, that is, .
The factorization begins by copying the matrix diagonal
for i = 1, n pivots(i) = val(diag_ptr(i)) end;Each elimination step starts by inverting the pivot
for i = 1, n pivots(i) = 1 / pivots(i)For all nonzero elements with , we next check whether is a nonzero matrix element, since this is the only element that can cause fill with .
for j = diag_ptr(i)+1, row_ptr(i+1)-1 found = FALSE for k = row_ptr(col_ind(j)), diag_ptr(col_ind(j))-1 if(col_ind(k) = i) then found = TRUE element = val(k) endif end;If so, we update .
if (found = TRUE) val(diag_ptr(col_ind(j))) = val(diag_ptr(col_ind(j))) - element * pivots(i) * val(j) end; end;