NP Completeness. CS 374: Algorithms & Models of Computation, Spring Lecture 23. November 19, 2015
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1 CS 374: Algorithms & Models of Computation, Spring 2015 NP Completeness Lecture 23 November 19, 2015 Chandra & Lenny (UIUC) CS374 1 Spring / 37
2 Part I NP-Completeness Chandra & Lenny (UIUC) CS374 2 Spring / 37
3 P and NP and Turing Machines 1 P: set of decision problems that have polynomial time algorithms. 2 NP: set of decision problems that have polynomial time verification algorithms. Many natural problems we would like to solve are in NP. Every problem in NP has an exponential time algorithm (try verifying each possible certificate). P NP So some problems in NP are in P (example, shortest path problem) Big Question: Does every problem in NP have an efficient algorithm? Same as asking whether P = NP. Chandra & Lenny (UIUC) CS374 3 Spring / 37
4 Hardest Problems Question What is the hardest problem in NP? How do we define it? Towards a definition 1 Hardest problem must be in NP. 2 Hardest problem must be at least as difficult as every other problem in NP. Chandra & Lenny (UIUC) CS374 4 Spring / 37
5 NP-Complete Problems Definition A problem X is said to be NP-Complete if 1 X NP, and 2 (Hardness) For any Y NP, Y P X. Chandra & Lenny (UIUC) CS374 5 Spring / 37
6 NP-Complete Problems Definition A problem X is said to be NP-Complete if 1 X NP, and 2 (Hardness) For any Y NP, Y P X. Recall reduction: Y P X means that an instance of Y can be efficiently modeled as an instance of X. Chandra & Lenny (UIUC) CS374 5 Spring / 37
7 Solving NP-Complete Problems Proposition Suppose X is NP-Complete. Then X can be solved in polynomial time if and only if P = NP. Proof. Suppose X can be solved in polynomial time 1 Let Y NP. We know Y P X. 2 We showed that if Y P X and X can be solved in polynomial time, then Y can be solved in polynomial time. 3 Thus, every problem Y NP is such that Y P; NP P. 4 Since P NP, we have P = NP. Since P = NP, and X NP, we have a polynomial time algorithm for X. Chandra & Lenny (UIUC) CS374 6 Spring / 37
8 NP-Hard Problems Definition A problem X is said to be NP-Hard if 1 (Hardness) For any Y NP, we have that Y P X. An NP-Hard problem need not be in NP! Example: Halting problem is NP-Hard (why?) but not NP-Complete. Chandra & Lenny (UIUC) CS374 7 Spring / 37
9 Consequences of proving NP-Completeness If X is NP-Complete 1 Since we believe P NP, 2 and solving X implies P = NP. Chandra & Lenny (UIUC) CS374 8 Spring / 37
10 Consequences of proving NP-Completeness If X is NP-Complete 1 Since we believe P NP, 2 and solving X implies P = NP. X is unlikely to be efficiently solvable. Chandra & Lenny (UIUC) CS374 8 Spring / 37
11 Consequences of proving NP-Completeness If X is NP-Complete 1 Since we believe P NP, 2 and solving X implies P = NP. X is unlikely to be efficiently solvable. At the very least, many smart people before you have failed to find an efficient algorithm for X. Chandra & Lenny (UIUC) CS374 8 Spring / 37
12 Consequences of proving NP-Completeness If X is NP-Complete 1 Since we believe P NP, 2 and solving X implies P = NP. X is unlikely to be efficiently solvable. At the very least, many smart people before you have failed to find an efficient algorithm for X. (This is proof by mob opinion take with a grain of salt.) Chandra & Lenny (UIUC) CS374 8 Spring / 37
13 NP-Complete Problems Question Are there any natural problems that are NP-Complete? Answer Yes! Many, many important problems are NP-Complete. Chandra & Lenny (UIUC) CS374 9 Spring / 37
14 Cook-Levin Theorem Theorem (Cook-Levin) SAT is NP-Complete. Chandra & Lenny (UIUC) CS Spring / 37
15 Cook-Levin Theorem Theorem (Cook-Levin) SAT is NP-Complete. Need to show 1 SAT is in NP. 2 every NP problem X reduces to SAT. Will see proof in next lecture. Steve Cook won the Turing award for his theorem. Chandra & Lenny (UIUC) CS Spring / 37
16 Proving that a problem X is NP-Complete To prove X is NP-Complete, show 1 Show that X is in NP. 2 Give a polynomial-time reduction from a known NP-Complete problem such as SAT to X Chandra & Lenny (UIUC) CS Spring / 37
17 Proving that a problem X is NP-Complete To prove X is NP-Complete, show 1 Show that X is in NP. 2 Give a polynomial-time reduction from a known NP-Complete problem such as SAT to X SAT P X implies that every NP problem Y P X. Why? Chandra & Lenny (UIUC) CS Spring / 37
18 Proving that a problem X is NP-Complete To prove X is NP-Complete, show 1 Show that X is in NP. 2 Give a polynomial-time reduction from a known NP-Complete problem such as SAT to X SAT P X implies that every NP problem Y P X. Why? Transitivity of reductions: Y P SAT and SAT P X and hence Y P X. Chandra & Lenny (UIUC) CS Spring / 37
19 3-SAT is NP-Complete 3-SAT is in NP SAT P 3-SAT as we saw Chandra & Lenny (UIUC) CS Spring / 37
20 NP-Completeness via Reductions 1 SAT is NP-Complete due to Cook-Levin theorem 2 SAT P 3-SAT 3 3-SAT P Independent Set 4 Independent Set P Vertex Cover 5 Independent Set P Clique 6 3-SAT P 3-Color 7 3-SAT P Hamiltonian Cycle Chandra & Lenny (UIUC) CS Spring / 37
21 NP-Completeness via Reductions 1 SAT is NP-Complete due to Cook-Levin theorem 2 SAT P 3-SAT 3 3-SAT P Independent Set 4 Independent Set P Vertex Cover 5 Independent Set P Clique 6 3-SAT P 3-Color 7 3-SAT P Hamiltonian Cycle Hundreds and thousands of different problems from many areas of science and engineering have been shown to be NP-Complete. A surprisingly frequent phenomenon! Chandra & Lenny (UIUC) CS Spring / 37
22 NP-Completeness via Reductions Chandra & Lenny (UIUC) CS Spring / 37
23 Part II Reducing 3-SAT to Independent Set Chandra & Lenny (UIUC) CS Spring / 37
24 Independent Set Problem: Independent Set Instance: A graph G, integer k. Question: Is there an independent set in G of size k? Chandra & Lenny (UIUC) CS Spring / 37
25 3SAT P Independent Set The reduction 3SAT P Independent Set Input: Given a 3CNF formula ϕ Goal: Construct a graph G ϕ and number k such that G ϕ has an independent set of size k if and only if ϕ is satisfiable. Chandra & Lenny (UIUC) CS Spring / 37
26 3SAT P Independent Set The reduction 3SAT P Independent Set Input: Given a 3CNF formula ϕ Goal: Construct a graph G ϕ and number k such that G ϕ has an independent set of size k if and only if ϕ is satisfiable. G ϕ should be constructable in time polynomial in size of ϕ Chandra & Lenny (UIUC) CS Spring / 37
27 3SAT P Independent Set The reduction 3SAT P Independent Set Input: Given a 3CNF formula ϕ Goal: Construct a graph G ϕ and number k such that G ϕ has an independent set of size k if and only if ϕ is satisfiable. G ϕ should be constructable in time polynomial in size of ϕ Importance of reduction: Although 3SAT is much more expressive, it can be reduced to a seemingly specialized Independent Set problem. Notice: We handle only 3CNF formulas reduction would not work for other kinds of boolean formulas. Chandra & Lenny (UIUC) CS Spring / 37
28 Interpreting 3SAT There are two ways to think about 3SAT Chandra & Lenny (UIUC) CS Spring / 37
29 Interpreting 3SAT There are two ways to think about 3SAT 1 Find a way to assign 0/1 (false/true) to the variables such that the formula evaluates to true, that is each clause evaluates to true. Chandra & Lenny (UIUC) CS Spring / 37
30 Interpreting 3SAT There are two ways to think about 3SAT 1 Find a way to assign 0/1 (false/true) to the variables such that the formula evaluates to true, that is each clause evaluates to true. 2 Pick a literal from each clause and find a truth assignment to make all of them true Chandra & Lenny (UIUC) CS Spring / 37
31 Interpreting 3SAT There are two ways to think about 3SAT 1 Find a way to assign 0/1 (false/true) to the variables such that the formula evaluates to true, that is each clause evaluates to true. 2 Pick a literal from each clause and find a truth assignment to make all of them true. You will fail if two of the literals you pick are in conflict, i.e., you pick x i and x i We will take the second view of 3SAT to construct the reduction. Chandra & Lenny (UIUC) CS Spring / 37
32 The Reduction 1 G ϕ will have one vertex for each literal in a clause x 1 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 4 Figure: Graph for ϕ = ( x 1 x 2 x 3 ) (x 1 x 2 x 3 ) ( x 1 x 2 x 4 ) Chandra & Lenny (UIUC) CS Spring / 37
33 The Reduction 1 G ϕ will have one vertex for each literal in a clause 2 Connect the 3 literals in a clause to form a triangle; the independent set will pick at most one vertex from each clause, which will correspond to the literal to be set to true x 1 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 4 Figure: Graph for ϕ = ( x 1 x 2 x 3 ) (x 1 x 2 x 3 ) ( x 1 x 2 x 4 ) Chandra & Lenny (UIUC) CS Spring / 37
34 The Reduction 1 G ϕ will have one vertex for each literal in a clause 2 Connect the 3 literals in a clause to form a triangle; the independent set will pick at most one vertex from each clause, which will correspond to the literal to be set to true x 1 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 4 Figure: Graph for ϕ = ( x 1 x 2 x 3 ) (x 1 x 2 x 3 ) ( x 1 x 2 x 4 ) Chandra & Lenny (UIUC) CS Spring / 37
35 The Reduction 1 G ϕ will have one vertex for each literal in a clause 2 Connect the 3 literals in a clause to form a triangle; the independent set will pick at most one vertex from each clause, which will correspond to the literal to be set to true 3 Connect 2 vertices if they label complementary literals; this ensures that the literals corresponding to the independent set do not have a conflict x 1 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 4 Figure: Graph for ϕ = ( x 1 x 2 x 3 ) (x 1 x 2 x 3 ) ( x 1 x 2 x 4 ) Chandra & Lenny (UIUC) CS Spring / 37
36 The Reduction 1 G ϕ will have one vertex for each literal in a clause 2 Connect the 3 literals in a clause to form a triangle; the independent set will pick at most one vertex from each clause, which will correspond to the literal to be set to true 3 Connect 2 vertices if they label complementary literals; this ensures that the literals corresponding to the independent set do not have a conflict 4 Take k to be the number of clauses x 1 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 4 Figure: Graph for ϕ = ( x 1 x 2 x 3 ) (x 1 x 2 x 3 ) ( x 1 x 2 x 4 ) Chandra & Lenny (UIUC) CS Spring / 37
37 Correctness Proposition ϕ is satisfiable iff G ϕ has an independent set of size k (= number of clauses in ϕ). Proof. Let a be the truth assignment satisfying ϕ Chandra & Lenny (UIUC) CS Spring / 37
38 Correctness Proposition ϕ is satisfiable iff G ϕ has an independent set of size k (= number of clauses in ϕ). Proof. Let a be the truth assignment satisfying ϕ 1 Pick one of the vertices, corresponding to true literals under a, from each triangle. This is an independent set of the appropriate size. Why? Chandra & Lenny (UIUC) CS Spring / 37
39 Correctness (contd) Proposition ϕ is satisfiable iff G ϕ has an independent set of size k (= number of clauses in ϕ). Proof. Let S be an independent set of size k 1 S must contain exactly one vertex from each clause 2 S cannot contain vertices labeled by conflicting literals 3 Thus, it is possible to obtain a truth assignment that makes in the literals in S true; such an assignment satisfies one literal in every clause Chandra & Lenny (UIUC) CS Spring / 37
40 Part III NP-Completeness of Hamiltonian Cycle Chandra & Lenny (UIUC) CS Spring / 37
41 Directed Hamiltonian Cycle Input Given a directed graph G = (V, E) with n vertices Goal Does G have a Hamiltonian cycle? Chandra & Lenny (UIUC) CS Spring / 37
42 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 Chandra & Lenny (UIUC) CS Spring / 37
43 Is the following graph Hamiltonian? (A) Yes. (B) No. Chandra & Lenny (UIUC) CS Spring / 37
44 Directed Hamiltonian Cycle is NP-Complete Directed Hamiltonian Cycle is in NP: Why? Hardness: We will show 3-SAT P Directed Hamiltonian Cycle Chandra & Lenny (UIUC) CS Spring / 37
45 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. Chandra & Lenny (UIUC) CS Spring / 37
46 Reduction: First Ideas Viewing SAT: Assign values to n variables, and each clause has multiple 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. Chandra & Lenny (UIUC) CS Spring / 37
47 The Reduction: Phase I Traverse path i from left to right iff 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 3m + 3) x 1 x 2 x 3 x 4 Chandra & Lenny (UIUC) CS Spring / 37
48 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
49 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
50 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
51 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
52 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
53 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
54 The Reduction: Phase II Add vertex c j for clause C j. c j has edge from vertex 3j and to vertex 3j + 1 on path i if x i appears in clause C j, and has edge from vertex 3j + 1 and to vertex 3j 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 Chandra & Lenny (UIUC) CS Spring / 37
55 Correctness Proof Proposition ϕ has a satisfying assignment iff G ϕ has a Hamiltonian cycle. Proof. Let a be the satisfying assignment for ϕ. Define Hamiltonian cycle as follows If a(x i ) = 1 then traverse path i from left to right If a(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 Chandra & Lenny (UIUC) CS Spring / 37
56 Hamiltonian Cycle Satisfying assignment Suppose Π is a Hamiltonian cycle in G ϕ If Π enters c j (vertex for clause C j ) from vertex 3j on path i then it must leave the clause vertex on edge to 3j + 1 on the same path i If not, then only unvisited neighbor of 3j + 1 on path i is 3j + 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 3j + 1 on path i then it must leave the clause vertex c j on edge to 3j on path i Chandra & Lenny (UIUC) CS Spring / 37
57 Example x 1 x 2 x 3 x 4 Chandra & Lenny (UIUC) CS Spring / 37
58 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 Chandra & Lenny (UIUC) CS Spring / 37
59 Hamiltonian Cycle Problem 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)? Chandra & Lenny (UIUC) CS Spring / 37
60 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 Chandra & Lenny (UIUC) CS Spring / 37
61 Reduction Sketch Goal: Given directed graph G, need to construct undirected graph G such that G has Hamiltonian cycle iff G has Hamiltonian cycle Chandra & Lenny (UIUC) CS Spring / 37
62 Reduction Sketch Goal: Given directed graph G, need to construct undirected graph G such that G has Hamiltonian cycle iff G has Hamiltonian cycle Reduction Replace each vertex v by 3 vertices: v in, v, and v out a b v c d Chandra & Lenny (UIUC) CS Spring / 37
63 Reduction Sketch Goal: Given directed graph G, need to construct undirected graph G such that G has Hamiltonian cycle iff G has Hamiltonian cycle Reduction Replace each vertex v by 3 vertices: v in, v, and v out a b v c d a o v i v v o b o c i d i Chandra & Lenny (UIUC) CS Spring / 37
64 Reduction Sketch Goal: Given directed graph G, need to construct undirected graph G such that G has Hamiltonian cycle iff G has Hamiltonian cycle Reduction Replace each vertex v by 3 vertices: v in, v, and v out A directed edge (x, y) is replaced by edge (x out, y in ) a b v c d a o v i v v o b o c i d i Chandra & Lenny (UIUC) CS Spring / 37
65 Reduction: Wrapup The reduction is polynomial time (exercise) The reduction is correct (exercise) Chandra & Lenny (UIUC) CS Spring / 37
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