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FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

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NP = NTIME(n k ) k N

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Theorem: L NP if there exists a poly-time Turing machine V(erifier) with L = { x | y (witness ) |y| = poly(|x|) and V(x,y) accepts } Proof: (1) If L = { x | y |y| = poly(|x|) and V(x,y) accepts } then L NP Because we can guess y and then run V (2) If L NP then L = { x | y |y| = poly(|x|) and V(x,y) accepts } Let N be a non-deterministic poly-time TM that decides L and define V(x,y) to accept if y is an accepting computation history of N on x

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A language is in NP if and only if there exist polynomial-length certificates for membership to the language SAT is in NP because a satisfying assignment is a polynomial-length certificate that a formula is satisfiable

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NP = all the problems for which once you have the answer it is easy (i.e. efficient) to verify

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P = NP?

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If P = NP… Cryptography as we know it would not be possible (e.g. RSA) Mathematicians would be out of a job AI program become perfect as exhaustive search is efficient

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If P = NP… Writing symphonies is as easy as listening to them. Being a chef is as easy as eating. Writing Shakespeare is as easy as recognizing Shakespeare. Generation is as easy as recognition:

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POLY-TIME REDUCIBILITY A language A is polynomial time reducible to language B, written A P B, if there is a polynomial time computable function f : Σ* Σ*, where for every w, w A f(w) B f is called a polynomial time reduction of A to B

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NP Complete Problem: hardest problem in NP Intuitively, L is harder than L’ if L’ is polynomial reducible to L. NP Complete Problem: If such a problem has an efficient algorithm (in P), then every other problem also has efficient algorithm.

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Theorem (Cook-Levin): SAT is NP-complete Corollary: SAT P if and only if P = NP

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P NP SAT Any thing in NP P 3SAT You can think of -> as “easier than”. SAT is the hardest problem in NP.

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NP-Complete Problems 3SAT k-Clique Vertex Cover Independent Set

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3SAT Problem: Given a CNF where each clause has 3 variables, decide whether it is satisfiable or not. (x 1 x 1 x 2 ) ( x 1 x 2 x 2 ) ( x 1 x 2 x 2 )

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3SAT = { | y such that y is a satisfying assigment to and is in 3cnf } 3SAT is NP Complete How about 2SAT?

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P NP 3SAT SAT What we want to prove?

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How to prove? 1.We can convert (in polynomial time) a given SAT instance S into a 3SAT instance S’ such that If S is satisfiable, then S’ is satisfiable. If S’ is satisfiable, then S is satisfiable.

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Key Observation x 1 x 2 x 3 x 4 x 5 is satisfiable if and only if (x 1 x 2 z 1 ) ( z 1 x 3 z 2 ) ( z 2 x 4 z 3 ) ( z 3 x 4 x 5 ) is satisfiable.

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Polynomial Time Reduction Clause in SAT x 1 x 1 x 2 x 1 x 2 x 3 x 1 x 2 x 3 x 4 x 1 x 2 x 3 x 4 x 5 Clauses in 3SAT x 1 x 1 x 1 x 1 x 1 x 2 x 1 x 2 x 3 (x 1 x 2 z 1 ) ( z 1 x 3 x 4 ) (x 1 x 2 z 1 ) ( z 1 x 3 z 2 ) ( z 2 x 4 z 3 ) ( z 3 x 4 x 5 )

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CLIQUE b a e c d f g k-clique = complete subgraph of k nodes

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K-Cliques A K-clique is a set of K nodes with all K(K- 1)/2 possible edges between them This graph contains a 4-clique

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CLIQUE = { (G,k) | G is an undirected graph with a k-clique } Theorem: CLIQUE is NP-Complete (1) CLIQUE NP (2) 3SAT P CLIQUE Assume a reasonable encoding of graphs (example: the adjacency matrix is reasonable) Brute Force Algorithm: Try out all {n choose k} possible locations for the k clique

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P NP CLIQUE 3SAT CLIQUE is NP-Complete

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3SAT P CLIQUE We transform a 3-cnf formula into (G,k) such that 3SAT (G,k) CLIQUE The transformation can be done in time that is polynomial in the length of

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3SAT P CLIQUE We transform a 3-cnf formula into (G,k) such that 3SAT (G,k) CLIQUE If has m clauses, we create a graph with m clusters of 3 nodes each, and set k=m Each cluster corresponds to a clause. Each node in a cluster is labeled with a literal from the clause. We do not connect any nodes in the same cluster We connect nodes in different clusters whenever they are not contradictory

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(x 1 x 1 x 2 ) ( x 1 x 2 x 2 ) ( x 1 x 2 x 2 ) x1x1 x1x1 x1x1 x2x2 x2x2 x2x2 x2x2 x2x2 x1x1 k = #clauses clauseclause #nodes = 3(# clauses)

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(x 1 x 1 x 1 ) ( x 1 x 1 x 2 ) (x 2 x 2 x 2 ) ( x 2 x 2 x 1 ) x1x1 x1x1 x2x2 x2x2 x1x1 x1x1 x2x2 x2x2 x1x1 x2x2 x2x2 x1x1

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This graph contains an independent set of size 3 Independent Set An independent set is a set of nodes with no edges between them

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Independent Set G(V,E) Problem: Given a graph G and k, is there a size k independent set?

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Complement of G Given a graph G, let G*, the complement of G, be the graph such that two nodes are connected in G* if and only if the corresponding nodes are not connected in G G G*

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Key Observation For a graph G, vertex set S is an independent set if and only if S is clique in G*.

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Let G be an n-node graph Is there size clique in G? Is there size k independent set in G*?

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VERTEX COVER b a e c d b a e c d vertex cover = set of nodes that cover all edges

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Vertex Cover is NP Complete Given a Graph G(V,E), decide if there is k vertex such that every edge is covered by one of them?

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K-Indep Set P Vertex Cover

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Key Observation For a graph G(V,E), S is a independent set if and only if V-S is a vertex cover.

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Other NP-Complete Problems Travelling Salesman Problem Hamiltonian Path Max Cut Subset Sum Integer Programming ….

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Other Problems in NP Graph Isomorphism Factoring Number We don’t know if they are NP Complete or not.

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