Thus we can rewrite to Here • (AB) is independent of φ since ∀f ∈Ã, f ∈ {n + 1, n + 3} : ∀f ∈Ã : • (BA) is independent of φ since ∀f ∈Ã, f ∈B : • (BB) is independent of φ.
with some constant c, we can finally rewrite Equation 28 as Inserting the definition of N f f and S f f (simplified using constants c , and c ) we obtain which is equivalent to the original quadratic assignment problem we have shown that a solution to the quadratic assignment problem in Equation 34 is a solution to the decipherment problem in Equation 20 and vice versa.