P = NP; P NP. Intuition of the reduction idea:

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1 1 Polynomial Time Reducibility The question of whether P = NP is one of the greatest unsolved problems in the theoretical computer science. Two possibilities of relationship between P and N P P = NP; P NP. Intuition of the reduction idea: If a problem A is reducible to a problem B, an algorithm forsolvingb canbeusedtoconstructanalgorithmfora. If A is polynomially reducible to B, then a polynomial algorithm for B can be used to construct a polynomial algorithm for A. 1

2 Definition 1 Language A is polynomial time (mapping) reducible to language B, written A P B, if a polynomial time computable function exists, where for every w, f : Σ Σ w A iff f(w) B. A f B A f B 2

3 Theorem 1 If A P B and B P C, then A P C. Definition 2 A language L is called NP-complete iff 1. L NP; 2. for every language L NP, there is a polynomialtime mapping from L to L. Proposition 1 Let L be an NP-complete language. Then P = NP iff L P. Theorem 2 SAT P 3SAT. Theorem 3 3SAT p CLIQUE. Corollary 1 SAT P CLIQUE. 3

4 2 More Computational Classes Class P of problems that can be decided in polynomial time on a deterministic Turing machine. Class NP of problems that can be decided in polynomial time on a non-deterministic Turing machine. Class CNP of NP-complete problems; a language A NP is called NP-complete if B NP B P A. Class co-np of problems {L} such that L NP. 4

5 Definition 3 Two graphs G 1 (V 1,E 1 ) and G 2 (V 2,E 2 ) are called isomorphic if there is a one-to-one mapping f from V 1 onto V 2 such that (x,y) E 1 iff (f(x),f(y)) E 2. Graph Isomorphism Problem: Given two graphs G and H, is it true that they are isomorphic? This is a decision problem with a polynomial verifier. A polynomial verifier is known to check if graphs are isomorphic. 5

6 Graph Non-isomorphism Problem: Given two graphs G and H, is it true that they are not isomorphic? This is a decision problem for which no polynomial verifier was found. The complement to this problem belongs to NP; thus, the problem is in co-np. No polynomial verifier is known to check if two graphs are non-isomorphic. 6

7 3 Decision problems vs Optimization Problems Example: Clique (decision version): Given a graph G(V, E) and an integer k > 0, is there a clique of size k in G? L = { G(V,E),k : G contains a clique of size k}. Clique (optimization version): Given a graph G(V, E), find a clique in G of the maximal size. Proposition 2 If A is an algorithm for the decision version of the Clique-problem, then O(log n) applications of A solves the optimization version of the Clique-problem. 7

8 Problem 1 If A is an algorithm for the decision version of the Clique-problem, and A accepted G(V, E), k, how to find a clique of size k in G? Solution idea: Repeat 1. for every vertex v of G, apply A to H,k where H is the subgraph induced on the set containing v and its neighbors; 2. include into the clique the first v for which the application of A above yielded acceptance; Reset G to be the subgraph induced on the neighbors of v and k = k 1. 8

9 Problem 2 If A is an algorithm for the decision version of the Partitioning-problem, and A accepted S, C, how to find a subset T S for which a i = C? a i T Solution idea: Let S = {a 1,a 2,...,a n }. Repeat 1. find a i S for which A accepts S {a i },C a i ; include a i into T. 2. resets andc bys = S {a i }andc = C a i. 9

10 4 Vertex Coloring A k-vertex-coloring of a graph G(V,E) is a functioon f : V [1,k], such that for any edge xy E, f(x) f(y). Vertex Color (decision version): Given a graph G(V, E) and an integer k > 0, is there a vertex coloring of G which uses k colors? L = { G(V,E),k : G : there exists k vertex coloring.} Vertex Color (optimization version): Given a graph G(V,E), find a vertex-coloring of G which uses the smallest number of colors. Proposition 3 If A is an algorithm for the decision version of Vertex Color, then O(log n) applications of A are sufficient to determine the smallest number k for which there exists a k-vertex- coloring of G. Question: How to discover an optimal vertex-coloring? 10

11 Given an algorithm A for the decision version of Vertex_Color, we design a coloring algorithm Construct (G) which for every graph G outputs a k coloring, where k is such that A accepted <G,k> <G(V,E); k> is G complete? yes color G no select non adjacent pair xy. Form graph G[x,y] by fusing x and y apply A is G k colorable? yes Construct (G[x,y]) no Form graph H by adding xy to G Use k coloring of G[x,y] to create k coloring of G Construct (H) Use k coloring of H to create k coloring of G 11

12 Proposition 4 Let G be k-colorable and complete(any two vertiices are adjacent). Then, a k-coloring is obtained by the following rule: every vertex gets its own color. Proposition 5 Let G be k-colorable and not complete, and let x and y be two non-adjacent vertices in G. Form two graphs G[x,y] and H as follows: G[x,y]: remove x and y and add a new vertex [x,y] making it adjacent to all vertices in G that were adjacent to x or to y; H: add edge xy to G. Then there exists a k-coloring f of G iff at least one of the two options is correct: f(x) = f(y), in which case there is a k-coloring g of G[x,y] which preserves all colors and uses color f(x) for vertex [x,y]; f(x) f(y), in which case there is a k-coloring h on H which preserves all colors of f on G. Theorem 4 Assume that every application of procedure A is executed in one time unit. Prove that under this assumption, there is a polynomial algorithm for k-vertex-coloring. 12