Slide 1

CSCI3130 Tutorial 10 Chun-Ho Hung chhung@cse.cuhk.edu.hk Department of Computer Science & Engineering 1 Slide 2 Outline Review P, NP, NPC Polynomial-time Reduction 2 problems Double-SAT Dominating set http://en.wikipedia.org/wiki/Dominating_set_problem 2 Slide 3 P, NP, NPC Polynomial-time Reduction 3 Slide 4 P P is the class of all languages that have poly-time algorithm e.g., Shortest path on a directed graph, Sorting 4 Slide 5 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 for some s Solution s V runs in polynomial time 5 Slide 6 NP (cont) 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 Verifying Given an assignment, just evaluate the truth value Finding a solution? No efficient algorithm has been discovered yet 6 (x 1 x 2 ) (x 2 x 3 x 4 ) (x 1 ) = Slide 7 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 efficient algorithm that decides L does not exist 7 NP P Slide 8 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 Slide 9 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 Slide 10 Reduction (Cont) 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 Slide 11 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 Slide 12 Poly-Time Reduction (Cont) 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 Slide 13 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 13 x y R TM for L acc rej Poly-time TM Slide 14 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 14 x y R TM for L acc rej Poly-time TM Slide 15 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 Slide 16 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 Slide 17 Relations NP P NP-C easy hard Is there any problem even harder than NP-C? Yes! e.g. I-go 17 Slide 18 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 Slide 19 Proving a problem being NPC 19 Slide 20 Double-SAT Problem: Double-SAT = { | is a Boolean formula with at least two satisfying assignments} Goal: Show that Double-SAT is NP-Complete 20 Slide 21 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 Slide 22 Double-SAT - (1) NP It is trivial to see that Double-SAT NP Given 2 assignments for , and verify whether both of them satisfy We can just evaluate the truth value in poly-time 22 Slide 23 Double-SAT - (2) Reduction Reduction: On input (x 1,..., x n ): 1. Introduce a new variable w 2. Output formula (x 1,..., x n, y) = (x 1,..., x n ) ( w w ). 23 x y x Lx L y Ly L R TM for L acc rej TM accepts SATDouble-SAT Slide 24 Double-SAT - (3) Correspondence x L y L : Suppose there is an satisfying assignment, X, for (x 1,..., x n ), we can find two satisfying assignments for (x 1,..., x n, w): Assignment 1 = {X, w=True} Assignment 2 = {X, w=False} (x 1,..., x n, w) = (x 1,..., x n ) ( w w ) 24 For {x i }, assign X, then this part = True No matter what w is, this part = True Slide 25 Double-SAT - (3) Correspondence x L y L : We use contrapositive i.e., to show x L y L Indeed, if x L, (x 1,..., x n )=False Then, no matter what the value of y is (x 1,..., x n, y)=False 25 Slide 26 Dominating Set Problem: Dominating-set = { | A dominating set of size K for G exists} Goal: Show that Dominating-set is NP-Complete 26 Slide 27 Dominating Set (Definition) Problem: Dominating-set = { | 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 Slide 28 Dominating Set (Example) Size-2 example : {Yellow vertices} e 28 Slide 29 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 Slide 30 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 Slide 31 Dominating Set - (2) Reduction Reduction - Graph transformation Construct a new graph G' by adding new vertices and edges to the graph G as follows: 31 G G L L T Vertex-cover Dominating-set Slide 32 Dominating Set - (2) Reduction Reduction - Graph transformation (Cont) 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 32 G G L L T Vertex-cover Dominating-set Slide 33 [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 33 {2, 4}, {3, 4}, {1, 2, 3} are vertex covers 12 3 4 Slide 34 Dominating Set Graph Transformation Example vw zu G wv zu vzwu vw zu vu G' 34 Slide 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 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 Slide 36 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 Slide 37 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 Slide 38 Dominating Set - (3) Correspondence vw zu wv zu vzwu vw zu vu Vertex-cover in G Dominating-set in G' 38 Slide 39 Any questions? (There should be some) 39