This is done via linear algebra, yes? As in, you have a 27xk (for some k) matrix with the one's of the l'th row corresponding to letters of l and zeros for the

This is done via linear algebra, yes? As in, you have a 27xk (for some k) matrix with the one's of the l'th row corresponding to letters of l and zeros for the rest of indices. Reduce the matrix to reduced row echelon form. Increase k until you have an over-constrained equation. Repeat with multiple languages to have high success values.