1. 3SAT Independent Set Hamiltonian Cycle Vertex Cover CLIQUE Traveling-Salesman Problem(TSP) Subset-Sum NP-complete Problems SAT Stephen Cook 1971 Richard Karp 1972 …… …… About 1000 NP-complete problems have been discovered since.

2. e b d a c f g Independent Set Let G be an undirected graph. A V(G) is an independent set if no two vertices in A share an edge.  u, v  A, (u, v)  E(G) e Ex. d a g A = { a, d, e, g } is an independent set. B = { a, c, f } is not

3. 2 deterministicO(K )time The Independent-Set Problem IND-SET Input:Graph G, integer K  1. Q:Does G have an independent set of size  K ? Theorem IND-SET is NP-complete. Proof  IND-SET  NP. Below is a non-deterministic algorithm: a) Pick K vertices from V(G) to form a subset A V. non-deterministicO(K)time certificate b) Check if  u, v  A, (u, v)  E(G). If so answer Yes; otherwise, answer No.

4. Construct (G , K) such that   3SAT  (G , K)  IND-SET.   P 3SAT ≤ IND SET Proof (cont’d)  We show that 3SAT is polynomial-time reducible to IND-SET. This will follow from Lemma 1. Therefore IND-SET is NP-complete.

5. Ex.  = ( x  x  x )  ( x  x  x )  ( x  x   x ) x x 1 2 3 1 3 4 2 3 4 x 1 2 1 x x x x x x 2 3 3 4 4 3  ConstructingG One vertex for each literal.  Literals in the same clause form a triangle.  Opposite literals share edges.

6. Satisfiability vs. Independent Set Lemma 1 satisfiable  G has an independent set of size K (= #clauses in ). Proof (  ) Suppose  is satisfiable. Then at least one literal from every clause is true. Pick exactly one such literal from each clause and pick its corresponding vertex.  K vertices are picked  For two of these vertices to share an edge, the corresponding literals would be  either in the same clause Impossible given the way these vertices are picked  or opposite literals. Impossible given that  is satisfiable. So none of the K vertices are adjacent. Thus the K vertices form an independent set.

7. (  ) Suppose G has an independent set of size K.  Then the vertices in the set  must be in different triangles. Cont’d  do not include a pair of opposite literals. Now we assign T to the corresponding literals in , which will be true under the induced truth value assignment to the variables.

8. C V(G) is a vertex cover if for every edge (u, v)  E(G) either u C or v  C b a G = K n e c d Vertex Cover Let G be an undirected graph. a |C| |V| – 1 complete graph with n vertices d { a, d } is a vertex cover.

9. b b a a e e c c d d The Vertex Cover Problem Vertex Cover (VC) Input:Graph G = (V, E), positiveinteger K  | V |. Q:Does G have a vertex cover of size  K ? LemmaC is a vertex cover  V – C is an independent set. Proof () Let C be a vertex cover. Suppose V – C is not an independent set. Then there exists two vertices u, v V – C such that (u, v)  E. {a, d}: vertex cover So edge (u, v) has both vertices not in C and C is not a vertex cover. ( ) Similarly. {b, c, e}: independent set

10. VC is NP-complete Corollary IND-SET  VC P It is easy to show that VC  NP. TheoremVC  NPC.

11. c b e a d f g CLIQUE A cliqueQ is a subset of vertices such that (u, v) is an edge for every u, v Q. cliques: { a, b, c, d } { e, f, g } { d, e } { g }, … The subgraph induced by Q is a complete graph.

12. c b e c a b d f g e a d f g Complement of a Graph The complementG of a graph G = (V, E) has  the same vertex set V edge set E such that (u, v)  E  (u, v)  E Complement: Original graph:

13. IND-SET  CLIQUE P The CLIQUE Problem Input:Graph G = (V, E), positiveinteger K  | V |. Q:Does G have a clique of size  K ? LemmaQ is a clique in G Q is an independent set in the complement G . Theorem CLIQUE  NPC.