CSCI 3130: Formal languages and automata theory

Fall 2011 The Chinese University of Hong Kong CSCI 3130: Formal languages and automata theory More on NP-completeness Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

NP problems 1 1 1 2 2 2 3 3 3 4 4 4 clique independent set vertex cover 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 and 3SAT SAT = {〈f〉: fis a satisfiable Boolean formula} ((x1∨x2 ) ∧ (x1∨x2)) ∨ ((x1∨(x2 ∧ x3)) ∧x3) 3SAT = {〈f〉: fis a satisfiable 3CNF formula} (x1∨x2∨x2 ) ∧ (x2∨x3∨x4) ∧ (x2∨x3∨x5) ∧ (x1∨x5∨x5)

P, NP, and reductions reduction from CLIQUE to IS IS reduces to VC CLIQUE 3SAT SAT (G, k) (G’, k’) G has a k-clique G’ has a k’-ind set P NP (easy to solve) (easy to verify solution)

The plan for today Why? IS reduces to VC CLIQUE Because everything in NP reduces to SAT 3SAT SAT NP-complete (the Cook-Levin theorem) … so all these are NP-complete P NP (easy to solve) (easy to verify solution)

Reducing IS to VC • Proof: We describe a reduction from IS to VC • Example 2 1 R (G, k) (G’, k’) 4 3 G has an IS of size k G’ has a VC of size k’ vertex covers independent sets {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4} ∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}

Reducing IS to VC • Claim • Proof 2 1 S is an independent set of G if and only if S is a vertex cover of G 4 3 VC IS ∅ {1} {2} {3} {4} {1, 2} {1, 3} {2, 4} {3, 4} {1, 2, 3} {1, 2, 4} {1, 3, 4} {2, 3, 4} {1, 2, 3, 4} S is an independent set of G no edge has both endpoints in S every edge has an endpoint in S S is a vertex cover of G

Reducing IS to VC VC R (G, k) (G’, k’) ✓ IS ✓ CLIQUE R: On input (G, k), Output (G, n – k). 3SAT SAT G has an IS of size k G has a VC of size n – k

Reducing 3SAT to CLIQUE • Proof: We give a reduction from 3SAT to CLIQUE 3SAT = {f: f is a satisfiable Boolean formula in 3CNF} CLIQUE = {(G, k): G is a graph with a clique of k vertices} R 3CNF formula f (G, k) G has a cliqueof size k f is satisfiable

Reducing 3SAT to CLIQUE • Example: f = (x1∨x1∨x2 ) ∧ (x1∨x2∨x2) ∧ (x1∨x2∨x3) x1 x1 x1 x1 x2 x2 x2 x2 x3 Put a vertex for every literal Put an edge for every consistent pair

Reducing 3SAT to CLIQUE R 3CNF formula f (G, k) R: On input f, where f is a 3CNF formula with m clauses Construct the following graph G: G has 3m vertices, divided into m groups, one for each literal in f If a and b are in different groups and a ≠ b, put an edge (a, b) Output(G, m)

Reducing 3SAT to CLIQUE R 3CNF formula f (G, m) G has a cliqueof size m f is satisfiable x1 x1 x1 x1 x2 x2 x2 x2 x3 f =(x1∨x1∨x2 ) ∧ (x1∨x2∨x2) ∧ (x1∨x2∨x3) T T F F F T F F T

Reducing 3SAT to CLIQUE R 3CNF formula f (G, m) G has a cliqueof size m f is satisfiable x1 x1 x1 x1 x2 x2 x2 x2 x3 f =(x1∨x1∨x2 ) ∧ (x1∨x2∨x2) ∧ (x1∨x2∨x3) F F T T F F T T T

Reducing 3SAT to CLIQUE • Every satisfying assignment of f gives a clique of size m in G • Conversely, every clique of size m in Ggives a consistent satisfying assignment of f. R 3CNF formula f (G, m) f is satisfiable G has a clique of size m VC ✓ IS ✓ CLIQUE ✓ 3SAT SAT

SAT and 3SAT SAT = {〈f〉: fis a satisfiable Boolean formula} ((x1∨x2 ) ∧ (x1∨x2)) ∨ ((x1∨(x2 ∧ x3)) ∧x3) 3SAT = {〈f〉: fis a satisfiable 3CNF formula} (x1∨x2∨x2 ) ∧ (x2∨x3∨x4) ∧ (x2∨x3∨x5) ∧ (x1∨x5∨x5)

Reducing SAT to 3SAT • Example: f = (x2∨(x1∧x2 ))∧(x1∧(x1∨x2 )) x4x5 x7 x7 = x4 ∧ x5 x10 T T T TT T F FT F T FT F F TF T T FF T F TF F T FF F F T AND (x4∨x5∨x7) x8 x9 (x4∨x5∨x7) OR NOT x6 x7 (x4∨x5∨x7) AND AND (x4∨x5∨x7) x4 x5 x3 NOT NOT OR (x4∨x5∨x7) ∧(x4∨x5∨x7) x2 x1 x2 x1 x1 x2 ∧(x4∨x5∨x7) ∧(x4∨x5∨x7) We add extra variables for every “wire”

Turning gates into 3CNFs z z z Gj: AND OR NOT x y x y x x y z = y ∧ x x y z = y ∨ x x z = x z z z T T T TT T F FT F T FT F F TF T T FF T F TF F T FF F F T T T T TT T F FT F T TT F F FF T T TF T F FF F T FF F F T T T FT F T F T TF F F (x∨z)∧(x∨z) (x∨y∨z)∧(x∨y∨z) ∧(x∨y∨z)∧(x∨y∨z) (x∨y∨z)∧(x∨y∨z) ∧(x∨y∨z)∧(x∨y∨z) fj: (x∨x∨z)∧(x∨x∨z)

Reducing SAT to 3SAT R Boolean formula f 3CNF formula f’ R:= On input f, where f is a boolean formula Construct and output the following 3CNF formula f’: f’ has extra variables xn+1, ..., xn+t one for each gate Gj in f For each gate Gj, construct the formulafj Output f’ = fn+1 ∧ fn+2 ∧ ... ∧ ft ∧ (xn+t ∨xn+t ∨xn+t ) requires thatoutput of f is true

Reducing SAT to 3SAT • Every satisfying assignment of fextends uniquely to a satisfying assignment of f’ • In the other direction, in every satisfying assignment of f’ the part x1, ..., xnis satisfies f R Boolean formula f 3CNF formula f’ ✔ f’ is satisfiable f is satisfiable

CSCI 3130: Formal languages and automata theory