Does that really follow? The sqrt(N) factor is the overhead of memory access only, so if you're comparing an algorithm where every operation is a memory access

Does that really follow? The sqrt(N) factor is the overhead of memory access only, so if you're comparing an algorithm where every operation is a memory access to another where that is not the case it would seem you could have the inverse relation in Oi and Ot. Edit: A counterexample then: Oi(f(N)) = N^(2/3) (memory accessing operations) Oi(g(N)) = N (non memory accessing operations) Ot(f(N)) = N^(7/6) Ot(g(N)) = N So Oi(f(N)) < Oi(g(N)) but Ot(f(N)) > Ot(g(N))