NP-completeness The Chinese University of Hong Kong Fall 2008 CSC 3130: Automata theory and formal languages NP-completeness Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

P and NP P = languages that can be decided on a TM with polynomial running time (problems that admit efficient algorithms) NP = languages decidable on a nondeterministic TM with polynomial running time (solutions can be checked efficiently) decidable We believe that NP is bigger than P, but we are not 100% sure NP P

Evidence that NP is bigger than P CLIQUE = {(G, k): G is a graph with a clique of k vertices} IS = {(G, k): G is a graph with an independent set of k vertices} VC = {(G, k): G is a graph with a vertex cover of k vertices} SAT = {f: f is a satisfiable Boolean formula} These (and many others) are in NP Their solutions, once found, are easy to verify But no efficient algorithms are known for any of them The fastest known algorithms take time ≈2n

Equivalence of certain NP languages We strongly suspect that problems like CLIQUE, SAT, etc. require time ≈2n to solve We do not know how to prove this, but what we can prove is that If any one of them is tractable (by a polynomial- time algorithm), then all of them are tractable

Equivalence of certain NP languages All these problems are as hard as one another Moreover, they are at the “frontier” of NP They are at least as hard as any problem in NP NP clique independent set vertex-cover P satisfiability

Polynomial-time reductions What do we mean when we say, for example, We mean that Or, we can convert an imaginary poly-time algorithm for IS into one for CLIQUE “CLIQUE is at least as hard as IS” If CLIQUE has a polynomial-time algorithm, so does IS

Polynomial-time reductions CLIQUE = {(G, k): G is a graph with a clique of k vertices} IS = {(G, k): G is a graph with an independent set of k vertices} Theorem 1 2 If CLIQUE has a polynomial-time algorithm, so does IS 3 4 {1, 4}, {2, 3, 4}, {1} are cliques {1, 2}, {1, 3}, {4} are independent sets

Polynomial-time reductions If CLIQUE has a polynomial-time algorithm, so does IS Proof: Suppose CLIQUE has an algorithm A We want to use it to solve IS G, k reject if not accept if G has IS of size k reject if not accept if G’ has clique of size k A G’, k’

Reducing IS to CLIQUE We look for a polynomial-time transformation s.t.: If G’ has a clique of size k’, then G has an IS of size k R G, k G’, k’ If G’ does not have a clique of size k’, then G has no IS of size k 1 2 3 4 1 2 3 4 flip all edges G G’ set k’ = k ISs of size k cliques of size k’

✓ ✓ Reducing IS to CLIQUE R On input (G, k) Construct G’ by flipping all edges of G Set k’ = k Output (G’, k’) R G, k G’, k’ independent sets in G cliques in G’ If G’ has a clique of size k’, then G has an IS of size k ✓ If G’ does not have a clique of size k’, then G has no IS of size k ✓

Reduction recap We showed that by converting an imaginary algorithm for CLIQUE into one for IS To do this, we came up with a reduction that transforms instances of IS into ones for CLIQUE If CLIQUE has a polynomial-time algorithm, so does IS

Polynomial-time reductions Language L polynomial-time reduces to L’ if there exists a polynomial-time computable map R that takes an instance x of L into instance y of L’ s.t. x ∈ L if and only if y ∈ L’ L (IS) L’ (CLIQUE) x (G, k) R y (G’, k’) x ∈ L y ∈ L’ (G has IS of size k) (G’ has clique of size k’)

Meaning of poly-time reductions Saying L reduces to L’ means L is easier than L’ In other words, if we can solve L’, then we can also solve L Theorem Proof If L reduces to L’ and L’ ∈ P, then L ∈ P acc R algorithm for L’ x y rej x ∈ L y ∈ L’ algorithm accepts

Notes about reductions The direction of the reduction is very important This makes sense, because “A is easier than B” and “B is easier than A” mean different things However, it is possible that L reduces to L’ and L’ reduces to L This means that L and L’ are as hard as one another For example, IS and CLIQUE reduce to one another

The Cook-Levin Theorem Every L ∈ NP reduces to SAT SAT = {f: f is a satisfiable Boolean formula} So every problem in NP is easier than SAT But SAT itself is in NP, so SAT must be the “hardest problem” in NP: If SAT ∈ P, then P = NP

Interpretation of Cook-Levin Theorem Optimistic view: Pessimistic view: If we manage to solve SAT, then we can also solve CLIQUE, scheduling, and almost anything Since we do not believe P = NP, it is unlikely that we will ever have a fast algorithm for SAT