VIEWS: 12 PAGES: 39 POSTED ON: 3/8/2011 Public Domain
CSCI3130 Tutorial 10 Chun-Ho Hung chhung@cse.cuhk.edu.hk Department of Computer Science & Engineering 1 Outline Review P, NP, NPC Polynomial-time Reduction 2 problems Double-SAT Dominating set http://en.wikipedia.org/wiki/Dominating_set_problem 2 P, NP, NPC Polynomial-time Reduction 3 P P is the class of all languages that have poly-time algorithm e.g., Shortest path on a directed graph, Sorting 4 NP NP is the class of all languages that have poly-time verifier A verifier – a Turing Machine, V, s.t. Given a potential x x ∈ L V accepts input <x, s> for some s Solution s V runs in polynomial time 5 NP (con’t) While verifying a solution of a NP problem is easy (in poly-time), finding a solution could be more difficult An 3SAT instance - Find a satisfying assignment for f = (x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1) Verifying Given an assignment, just evaluate the truth value Finding a solution? No efficient algorithm has been discovered yet 6 P versus NP Every language, L in P, L is also in NP Let Verifier = Poly-time TM that solves L Therefore P is contained in NP Note: L in NP does not imply that NP efficient algorithm that decides L does not exist P 7 NPC A language C is NP-complete if: C is in NP Every language L in NP, L poly-time reduces to C What is a reduction…? 8 Reduction The direction of the reduction is very important Saying “A is easier than B” and “B is easier than A” mean different things “A (polynomially) reduces to B” means “B is not easier than A” 9 Reduction (Con’t) Consider 2 problems: 1) BFS on unweighted graph 2) Shortest path on weighted graph Assume we have a TM, V, which solves 2) We can reduce 1) to 2): Given an instance of 1), convert it into an instance of 2): Copy the graph, add weight=1 to every edge in 2) Run this instance on V, output result These two “yes” instances corresponds to each other 10 Poly-Time Reduction How to show that a problem B is not easier than a problem A? Informally, if B can be solved efficiently, we can solve A efficiently Formally, we say A polynomially reduces to B if: 1. Given an instance a of problem, x 2. There is a polynomial time transformation to an instance of B, y = f(x) 3. x is a “yes” instance if and only if y is a “yes” instance 11 Poly-Time Reduction (Con’t) Suppose A poly-time reduces to B Then there exists a poly-time TM, R, s.t., Given an instance of A, x, transforms it to an instance of B, y = f(x), and y is accepted x is accepted 12 Poly-Time Reduction (Implication) Suppose A reduces to B If B is polynomial time solvable, then A is polynomial time solvable If A is not polynomial time solvable, then B is not polynomial time solvable Contrapositive acc x R y TM for L’ rej Poly-time TM 13 Poly-Time Reduction (Implication) Suppose A reduces to B Solving B cannot be easier than solving A Suppose A is “difficult” while B is “easy” However, by this reduction, you find a “easy” way to solve A Consequently, if A is NPC, then B must be NPC acc x R y TM for L’ rej Poly-time TM 14 Poly-Time Reduction - P versus NP To show P = NP, one could try to show that a NPC problem, C, can be solved in polynomial time. Why? Every problem in NP poly-time reduces to C If C can be solved in poly-time, so does each problem in NP Then NP = P!! But this is not that easy and it is counter-intuitive (to most people) too 15 P versus NP (Again) Most believe that P ≠ NP, because intuitively searching for a solution is more difficult than verifying a solution What does P = NP imply? Know how to verify a solution in poly-time Know how to find a solution in poly-time (?!) Indeed we prefer P ≠ NP Encryption algorithms heavily rely on the assumption that P ≠ NP P = NP or P ≠ NP is still an open problem 16 Relations NP-C hard Is there any problem even harder than NP-C? NP Yes! e.g. I-go P easy 17 Methodology To show L is in NP, you can either (i) Show that solutions for L can be verified in polynomial-time, or (ii) Describe a nondeterministic polynomial-time TM for L (Come back to this if we have enough time) To show L is NP-complete Show that L is in NP Poly-time reduce some NPC problem to L i.e., design a polynomial-time reduction from some problem we know to be NP- complete 18 Proving a problem being NPC 19 Double-SAT Problem: Double-SAT = {<φ> | φ is a Boolean formula with at least two satisfying assignments} Goal: Show that Double-SAT is NP-Complete 20 Double-SAT (Proof Sketch) Steps: 1) Show that Double-SAT ∈ NP 2) Show that Double-SAT is not easier than a certain NPC problem For the NPC problem, we choose SAT i.e., we want to poly-time reduce Double-SAT to SAT 3) Show the correspondence of “yes” instance between reduction 21 Double-SAT - (1) NP It is trivial to see that Double-SAT ∈ NP Given 2 assignments for φ, and verify whether both of them satisfy φ We can just evaluate the truth value in poly-time 22 Double-SAT - (2) Reduction Reduction: On input φ(x1, . . . , xn): 1. Introduce a new variable w 2. Output formula φ’(x1, . . . , xn, y) = φ(x1, . . . , xn) ∧ ( w ∨ w ). acc x R y TM for L’ rej x∈L y ∈ L’ TM accepts SAT Double-SAT 23 Double-SAT - (3) Correspondence x ∈ L y ∈ L’ : Suppose there is an satisfying assignment, X, for φ(x1, . . . , xn), we can find two satisfying assignments for φ’(x1, . . . , xn, w): Assignment 1 = {X, w=True} Assignment 2 = {X, w=False} φ’(x1, . . . , xn, w) = φ(x1, . . . , xn) ∧ ( w ∨ w ) For {xi}, assign X, No matter what w is, then this part = True this part = True 24 Double-SAT - (3) Correspondence x ∈ L y ∈ L’ : We use contrapositive i.e., to show x ∉ L ⇒ y ∉ L’ Indeed, if x ∉ L, φ(x1, . . . , xn)=False Then, no matter what the value of y is φ’(x1, . . . , xn, y)=False 25 Dominating Set Problem: Dominating-set = {<G, K> | A dominating set of size K for G exists} Goal: Show that Dominating-set is NP-Complete 26 Dominating Set (Definition) Problem: Dominating-set = {<G, K> | A dominating set of size (at most) K for G exists} Let G=(V,E) be an undirected graph A dominating set D is a set of vertices that covers all vertices i.e., every vertex of G is either in D or is adjacent to at least one vertex from D 27 Dominating Set (Example) Size-2 example : {Yellow vertices} e 28 Dominating Set (Proof Sketch) Steps: 1) Show that Dominating-set ∈ NP. 2) Show that Dominating-set is not easier than a NPC problem We choose this NPC problem to be Vertex cover Reduction from Vertex-cover to Dominating-set 3) Show the correspondence of “yes” instances between the reduction 29 Dominating Set - (1) NP It is trivial to see that Dominating-set ∈ NP Given a vertex set D of size K, we check whether (V-D) are adjacent to D i.e., for each vertex, v, in (V-D), whether v is adjacent to some vertex u in D 30 Dominating Set - (2) Reduction Reduction - Graph transformation Construct a new graph G' by adding new vertices and edges to the graph G as follows: G T G’ <G,k> ∈ L <G’, k> ∈ L’ Vertex-cover Dominating-set 31 Dominating Set - (2) Reduction Reduction - Graph transformation (Con’t) For each edge (v, w) of G, add a vertex vw and the edges (v, vw) and (w, vw) to G' Furthermore, remove all vertices with no incident edges; such vertices would always have to go in a dominating set but are not needed in a vertex cover of G We skip the discussion of this subtle part in the followings G T G’ <G,k> ∈ L <G’, k> ∈ L’ Vertex-cover Dominating-set 32 [Recap] Vertex cover A vertex cover, C, is a set of vertices that covers all edges i.e., each edge is at least adjacent to some node in C 1 2 {2, 4}, {3, 4}, {1, 2, 3} 3 4 are vertex covers 33 Dominating Set – Graph Transformation Example vw v w v w vz wu vu z u z u zu G G' 34 Dominating Set - (3) Correspondence A dominating set of size K in G’ A vertex cover of size K in G Let D be a dominating set of size K in G’ Case 1): D contains only vertices from G Then, all new vertices have an edge to a vertex in D D covers all edges D is a valid vertex cover of G 35 Dominating Set - (3) Correspondence A dominating set of size K in G’ A vertex cover of size K in G Let D be a dominating set of size K in G’ Case 2): D contains some new vertices (vertex in the form of uv) (We show how to construct a vertex cover using only old vertices, otherwise we cannot obtain a vertex cover for G) For each new vertex uv, replace it by u (or v) If u ∈ D, this node is not needed Then the edge u-v in G will be covered After new edges are removed, it is a valid vertex cover of G (of size at most K) 36 Dominating Set - (3) Correspondence A dominating set of size K in G’ A vertex cover of size K in G Let C be a vertex cover of size K in G For an old vertex, v ∈ G’ : By the definition of VC, all edges incident to v are covered v is also covered For a new vertex, uv ∈ G’ : Edge u-v must be covered, either u or v ∈ C This node will cover uv in G’ Thus, C is a valid dominating for G’ (of size at most K) 37 Dominating Set - (3) Correspondence vw v w v w vz wu vu z u z u zu Vertex-cover in G Dominating-set in G' 38 Any questions? (There should be some) 39