It's not that simple. Finding collisions for md5(f(x)) could be harder than for md5(x) if f is a half-decent hash function. For the former, likely the best you

It's not that simple. Finding collisions for md5(f(x)) could be harder than for md5(x) if f is a half-decent hash function. For the former, likely the best you could do is a (brute-force) brithday attack, with complexity ~64. OTOH, the best known collision attack on SHA-1 has complexity < 63.