Compatibility and Parsimony

Compatibility, in the binary case at least, is easily shown to be a special case of parsimony. To do that, we will use a slightly different, thresholded version of parsimony, where for each character that changes at least twice we ``charge'' exactly 2 (instead of the real total number of changes). This problem is not more difficult than usual parsimony. To prove this equivalence between thresholded parsimony and compatibility, let us now define n_i to be the number of characters that need i changes. The total score in that case is clearly S(T) = n₁ + 2(n-n₀-n₁) = 2n-2n₀-n₁. The numbers n and n₀ are fixed for the given data, and so minimizing the score S means maximizing n₁, which is exactly the number of compatible characters. Hence we can use the methods described for solving large parsimony, to solve the problem of compatibility. However, this is not of much help, since parsimony is an NP-hard problem.