More NP-Complete Problems Lecture 4 September 5, 2013 Sariel - PowerPoint PPT Presentation

CS 573: Algorithms, Fall 2013 More NP-Complete Problems Lecture 4 September 5, 2013 Sariel (UIUC) CS573 1 Fall 2013 1 / 48

Recap NP : languages that have polynomial time certifiers/verifiers. A language L is NP-Complete ⇐ ⇒ L is in NP for every L ′ in NP , L ′ ≤ P L L is NP-Hard if for every L ′ in NP , L ′ ≤ P L . . Theorem (Cook-Levin) . Circuit-SAT and SAT are NP-Complete . . Sariel (UIUC) CS573 2 Fall 2013 2 / 48

Recap NP : languages that have polynomial time certifiers/verifiers. A language L is NP-Complete ⇐ ⇒ L is in NP for every L ′ in NP , L ′ ≤ P L L is NP-Hard if for every L ′ in NP , L ′ ≤ P L . . Theorem (Cook-Levin) . Circuit-SAT and SAT are NP-Complete . . Sariel (UIUC) CS573 2 Fall 2013 2 / 48

Recap NP : languages that have polynomial time certifiers/verifiers. A language L is NP-Complete ⇐ ⇒ L is in NP for every L ′ in NP , L ′ ≤ P L L is NP-Hard if for every L ′ in NP , L ′ ≤ P L . . Theorem (Cook-Levin) . Circuit-SAT and SAT are NP-Complete . . Sariel (UIUC) CS573 2 Fall 2013 2 / 48

Recap NP : languages that have polynomial time certifiers/verifiers. A language L is NP-Complete ⇐ ⇒ L is in NP for every L ′ in NP , L ′ ≤ P L L is NP-Hard if for every L ′ in NP , L ′ ≤ P L . . Theorem (Cook-Levin) . Circuit-SAT and SAT are NP-Complete . . Sariel (UIUC) CS573 2 Fall 2013 2 / 48

Recap contd . Theorem (Cook-Levin) . Circuit-SAT and SAT are NP-Complete . . Establish NP-Complete ness via reductions: SAT ≤ P 3-SAT and hence 3-SAT is NP -complete 3-SAT ≤ P Independent Set (which is in NP ) and hence Independent Set is NP-Complete Vertex Cover is NP-Complete Clique is NP-Complete Set Cover is NP-Complete Sariel (UIUC) CS573 3 Fall 2013 3 / 48

Today Prove Hamiltonian Cycle Problem is NP-Complete . 3-Coloring is NP-Complete . Subset Sum . Sariel (UIUC) CS573 4 Fall 2013 4 / 48

Part I . NP-Completeness of Hamiltonian Cycle . Sariel (UIUC) CS573 5 Fall 2013 5 / 48

Directed Hamiltonian Cycle Input Given a directed graph G = ( V , E ) with n vertices Goal Does G have a Hamiltonian cycle? A Hamiltonian cycle is a cycle in the graph that visits every vertex in G exactly once Sariel (UIUC) CS573 6 Fall 2013 6 / 48

Directed Hamiltonian Cycle Input Given a directed graph G = ( V , E ) with n vertices Goal Does G have a Hamiltonian cycle? A Hamiltonian cycle is a cycle in the graph that visits every vertex in G exactly once Sariel (UIUC) CS573 6 Fall 2013 6 / 48

Directed Hamiltonian Cycle is NP-Complete Directed Hamiltonian Cycle is in NP Certificate: Sequence of vertices Certifier: Check if every vertex (except the first) appears exactly once, and that consecutive vertices are connected by a directed edge Hardness: We will show 3-SAT ≤ P Directed Hamiltonian Cycle Sariel (UIUC) CS573 7 Fall 2013 7 / 48

Reduction Given 3-SAT formula φ create a graph G φ such that G φ has a Hamiltonian cycle if and only if φ is satisfiable G φ should be constructible from φ by a polynomial time algorithm A Notation: φ has n variables x 1 , x 2 , . . . , x n and m clauses C 1 , C 2 , . . . , C m . Sariel (UIUC) CS573 8 Fall 2013 8 / 48

Reduction: First Ideas Viewing SAT: Assign values to n variables, and each clauses has 3 ways in which it can be satisfied. Construct graph with 2 n Hamiltonian cycles, where each cycle corresponds to some boolean assignment. Then add more graph structure to encode constraints on assignments imposed by the clauses. Sariel (UIUC) CS573 9 Fall 2013 9 / 48

The Reduction: Phase I Traverse path i from left to right if and only if x i is set to true. Each path has 3( m + 1) nodes where m is number of clauses in φ ; nodes numbered from left to right ( 1 to 3 m + 3 ) x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 10 Fall 2013 10 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ”Buffer” vertices ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

The Reduction: Phase II Add vertex c j for clause C j . c j has edge from vertex 3 j and to vertex 3 j + 1 on path i if x i appears in clause C j , and has edge from vertex 3 j + 1 and to vertex 3 j if ¬ x i appears in C j . x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 48

Correctness Proof . Proposition . φ has a satisfying assignment ⇐ ⇒ G φ has a Hamiltonian cycle. . . Proof. . ⇒ Let α be the satisfying assignment for φ . Define Hamiltonian cycle as follows If α ( x i ) = 1 then traverse path i from left to right If α ( x i ) = 0 then traverse path i from right to left. For each clause, path of at least one variable is in the “right” direction to splice in the node corresponding to clause. . Sariel (UIUC) CS573 12 Fall 2013 12 / 48

Hamiltonian Cycle ⇒ Satisfying assignment . Proof continued . Suppose Π is a Hamiltonian cycle in G φ If Π enters c j (vertex for clause C j ) from vertex 3 j on path i then it must leave the clause vertex on edge to 3 j + 1 on the same path i If not, then only unvisited neighbor of 3 j + 1 on path i is 3 j + 2 Thus, we don’t have two unvisited neighbors (one to enter from, and the other to leave) to have a Hamiltonian Cycle Similarly, if Π enters c j from vertex 3 j + 1 on path i then it must leave the clause vertex c j on edge to 3 j on path i . Sariel (UIUC) CS573 13 Fall 2013 13 / 48

Example x 1 ∨ ¬ x 2 ∨ x 4 ¬ x 1 ∨ ¬ x 2 ∨ ¬ x 3 x 1 x 2 x 3 x 4 Sariel (UIUC) CS573 14 Fall 2013 14 / 48

Hamiltonian Cycle = ⇒ Satisfying assignment (contd) Thus, vertices visited immediately before and after C i are connected by an edge We can remove c j from cycle, and get Hamiltonian cycle in G − c j Consider Hamiltonian cycle in G − { c 1 , . . . c m } ; it traverses each path in only one direction, which determines the truth assignment Sariel (UIUC) CS573 15 Fall 2013 15 / 48

(Undirected) Hamiltonian Cycle . Problem ( Undirected Hamiltonian Cycle ) . Input: Given undirected graph G = ( V , E ) . Goal: Does G have a Hamiltonian cycle? That is, is there a cycle that visits every vertex exactly one (except start and end vertex)? . Sariel (UIUC) CS573 16 Fall 2013 16 / 48

NP -Completeness . Theorem . Hamiltonian cycle problem for undirected graphs is NP-Complete . . . Proof. . The problem is in NP ; proof left as exercise. Hardness proved by reducing Directed Hamiltonian Cycle to this problem. . Sariel (UIUC) CS573 17 Fall 2013 17 / 48

Reduction Sketch Goal: Given directed graph G , need to construct undirected graph G ′ such that G has Hamiltonian Path if and only if G ′ has Hamiltonian path . Reduction . Replace each vertex v by 3 vertices: v in , v , and v out A directed edge ( a , b ) is replaced by edge ( a out , b in ) a c v b d . Sariel (UIUC) CS573 18 Fall 2013 18 / 48

Reduction Sketch Goal: Given directed graph G , need to construct undirected graph G ′ such that G has Hamiltonian Path if and only if G ′ has Hamiltonian path . Reduction . Replace each vertex v by 3 vertices: v in , v , and v out A directed edge ( a , b ) is replaced by edge ( a out , b in ) a c v b d . Sariel (UIUC) CS573 18 Fall 2013 18 / 48