INFR 11102 : Computational Complexity 01 / 02 / 2018 Lecture 6 : Reductions , NP - completeness

We will not prove the theorem here, but we should step back and take a minute to think about the meaning of this theorem. A naive NL algorithm for Conn is to take the same NTM for Conn but flip its answers. Thus, we will accept, if there exists a wrong guess connecting s to t. However, for any (s, t), there exists a wrong guess, and we will end up accepting everything! Essentially, if (G, s, t) is in Conn, then for any sequence of vertices of G, it cannot connect s to t. There is a universal quantifier “for any” in the statement. However, recall the verification definition of NP, what an NTM is capable of doing is to state an existential quantifier “there exists ...”. Thus, Theorem 1 is non-trivial in that it implicitly transforms the universal quantifier into an existential one. Without the space requirement, this can actually be done easily. For example, if there exists a cut1 (X,Y ) of G, such that s ∈ X and t ∈ Y , then we know that s and t are disconnected. However, a cut cannot be stored in logarithmic space. The proof of Theorem 1 uses a method called inductive counting. The idea is that, if we know how many vertices are reachable from s, then we can simply guess connectivity for every vertex, verify the guess, and then verify the total number is correct. The ability to verify the guess makes sure that we never overestimate the number of reachable vertices. Hence, if the total number in the end is correct, then we know all our guesses are correct, and thus know whether t is among reachable vertices. To find out this number, let Ck be the number of vertices that can be reached within k steps. To compute Cn, we do it inductively from C1 to Cn. If we know Ck, then we can reconstruct all vertices