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NP-complete and NP-hard problems Transitivity of polynomial-time many-one reductions Definition of complexity class NP –Nondeterministic computation –Problems that can be verified The P = NP Question –Concept of NP-hard and NP-complete problems

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Π 1 ≤ p Π 2 & Π 2 ≤ p Π 3 Π 1 ≤ p Π 3 Let R 1 be the reduction used to prove Π 1 ≤ p Π 2 Let R 2 be the reduction used to prove Π 2 ≤ p Π 3 Let x be an input to Π 1 Define R 3 (x) to be R 2 (R 1 (x))

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Answer-preserving argument Because R 1 is a reduction between Π 1 and Π 2, we know that R 1 (x) is a yes input instance of Π 2 iff x is a yes input instance of Π 1 Because R 2 is a reduction between Π 2 and Π 3, we know that R 2 (R 1 (x)) is a yes input instance of Π 3 iff R 1 (x) is a yes input instance of Π 2 Applying transitivity of iff, we get that R 3 (x) is a yes input of Π 3 iff x is a yes input instance of Π 1

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Polynomial-time Argument Let R 1 take time n c1 Let R 2 take time n c2 Let n be the size of x Then the R 1 call of R 3 takes time at most n c1 Furthermore, R 1 (x) has size at most max(n,n c1 ) Therefore, the R 2 call of R 3 takes time at most max(n c2, (n c1 ) c2 ) = max (n c2, n c1 c2 ) In either case, the total time taken by R 3 is polynomial in n

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NP-complete and NP-hard problems Transitivity of polynomial-time many-one reductions Definition of complexity class NP –Nondeterministic computation –Problems that can be verified The P = NP Question –Concept of NP-hard and NP-complete problems

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Traditional definition of NP Turing machine model of computation –Simple model where data is on an infinite capacity tape –Only operations are reading char stored in current tape cell, writing a char to current tape cell, moving tape head left or right one square Deterministic versus nondeterministic computation –Deterministic: At any point in time, next move is determined –Nondeterministic: At any point in time, several next moves are possible NP: Class of problems that can be solved by a nondeterminstic turing machine in polynomial time

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Turing Machines A Turing machine has a finite-state-control (its program), a two way infinite tape (its memory) and a read-write head (its program counter) 01111001001 Head Tape …. Finite State Control

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Nondeterministic Running Time We measure running time by looking at height of computation tree, NOT number of nodes explored Both computation have same height 4 and thus same running time Deterministic Computation Nondeterministic Computation

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ND computation returning yes If any leaf node returns yes, we consider the input to be a yes input. If all leaf nodes return no, then we consider the input to be a no input. Yes ResultNo Result

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Showing a problem is in NP Hamiltonian Path –Input: Undirected graph G = (V,E) –Y/N Question: Does G contain a HP? Nondeterministic polynomial-time solution –Guess a hamiltonian path P (ordering of vertices) V! possible orderings For binary tree, V log V height to generate all guesses –Verify guessed ordering is correct –Return yes/no if ordering is actually a HP

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Illustration 1 2 3 Yes input graph 123132 213 231 312 321 Guess Phase Nondeterministic --------------- Verify Phase Deterministic 1 2 3 No input graph 123132 213 231 312 321 Guess Phase Nondeterministic --------------- Verify Phase Deterministic

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Alternate definition of NP Preliminary Definitions –Let Π be a decision problem –Let I be an input instance of Π –Let Y(Π) be the set of yes input instances of Π –Let N(Π) be the set of no input instances of Π Π belongs to NP iff For any I ε Y(Π), there exists a “certificate” [solution] C(I) such that a deterministic algorithm can verify I ε Y(Π) in polynomial time with the help of C(I) For any I ε N(Π), no “certificate” [solution] C(I) will convince the algorithm that I ε Y(Π).

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Connection 1 2 3 Yes input graph 123132 213 231 312 321 Guess Phase Nondeterministic --------------- Verify Phase Deterministic 1 2 3 No input graph 123132 213 231 312 321 Guess Phase Nondeterministic --------------- Verify Phase Deterministic Certificate [Solution] A Hamiltonian Path C(I 1 ): 123 or 321 C(I 2 ): none Verification Alg: Verify certificate is a possible HP Check for edge between all adjacent nodes in path

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Example: Clique Problem Clique Problem –Input: Undirected graph G = (V,E), integer k –Y/N Question: Does G contain a clique of size ≥ k? Certificate –A clique C of size at least k Verification algorithm –Verify this is a potential clique of size k –Verify that all nodes in C are connected in E

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Proving a problem is in NP You need to describe what the certificate C(I) will be for any input instance I You need to describe the verification algorithm –usually trivial You need to argue that all yes input instances and only yes input instances have an appropriate certificate C(I) –also usually trivial (typically do not require)

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Example: Vertex Cover Vertex Cover –Input: Undirected graph G = (V,E), integer k –Y/N Question: Does G contain a vertex cover of size ≤ k? Vertex cover: A set of vertices C such that for every edge (u,v) in E, either u is in C or v is in C (or both are in C) Certificate –A vertex cover C of size at most k Verification algorithm –Verify C is a potential vertex cover of size at most k –Verify that all edges in E contain a node in C

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Example: Satisfiability Satisfiability –Input: Set of variables X and set of clauses C over X –Y/N Question: Is there a satisfying truth assignment T for the variables in X such that all clauses in C are true? Certificate? Verification algorithm?

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Example: Unsatisfiability Unsatisfiability –Input: Set of variables X and set of clauses C over X –Y/N Question: Is there no satisfying truth assignment T for the variables in X such that all clauses in C are true? Certificate and Verification algorithm? Negative certificate and Negative verification algorithm?

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Example: Exact Vertex Cover Exact Vertex Cover –Input: Undirected graph G = (V,E), integer k –Y/N Question: Does the smallest vertex cover in G have size exactly k? Vertex cover: A set of vertices C such that for every edge (u,v) in E, either u is in C or v is in C (or both are in C) Certificate and Verification algorithm? Negative certificate and Negative verification algorithm?

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NP-complete and NP-hard problems Transitivity of polynomial-time many-one reductions Definition of complexity class NP –Nondeterministic computation –Problems that can be verified The P = NP Question –Concept of NP-hard and NP-complete problems

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Definition of NP-hard A problem П is NP-hard if –for all П’ ε NP П’ ≤ p П holds. Intuitively, an NP-hard problem П is at least as hard (defined by membership in P) as any problem in NP

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Definition of NP-complete A problem П is NP-complete if –П is NP-hard and –П is in NP Intuitively, an NP-complete problem П is the hardest problem in NP –That is, if П is in P, then P=NP –If P ≠ NP, then П is not in P P=NP=NP-complete OR NP PNP-complete

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Importance of NP-completeness Importance of “Is P=NP” Question Practitioners view –There exist a large number of interesting and seemingly different problems which have been proven to be NP-complete –The P=NP question represents the question of whether or not all of these interesting and different problems belong to P –As the set of NP-complete problems grows, the question becomes more and more interesting

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List of Problem Types from Garey & Johnson, 1979 Graph Theory Network Design Sets and Partitions Storage and Retrieval Sequencing and Scheduling Mathematical Programming Algebra and Number Theory Games and Puzzles Logic Automata and Languages Program Optimization Miscellaneous

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Importance of NP-completeness Importance of “Is P=NP” Question Theoretician’s view –NP is exactly the set of problems that can be “verified” in polynomial time –Thus “Is P=NP?” can be rephrased as follows: Is it true that any problem that can be “verified” in polynomial time can also be “solved” in polynomial time? Hardness Implications –It seems unlikely that all problems that can be verified in polynomial time also can be solved in polynomial time –If so, then P≠NP –Thus, proving a problem to be NP-complete is a hardness result as such a problem will not be in P if P≠NP.

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Proving a problem П is NP-complete –Show П is in NP (usually easy step) –Prove for all П’ ε NP П’ ≤ p П holds. Show that П’ ≤ p П for some NP-hard problemП’ –This only works if we have a known NP-hard problem П’ to reduce from –Also depends on transitivity property proven earlier We need to prove the existence of a first NP-hard problem –Cook-Levin Thm –Developing new reductions is a skill or art form Over time, it gets easier

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Select the right source problem 3-SAT: The old reliable. When none of the other problems seem to work, this is the one to come back to. Integer Partition: A good choice for number problems. 3-Partition: A good choice for proving “strong” NP- completeness for number problems. Vertex Cover: A good choice for selection problems. Hamiltonian Path: A good choice for ordering problems.

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Some history Cook: “The complexity of theorem-proving procedures” STOC 1971, pp. 151-158 –Polynomial-time reductions –NP complexity class –SAT is NP-complete Levin: “Universal sorting problems”, Problemi Peredachi Informatsii 9:3 (1973), pp. 265-266 –Independent discovery of many of the same ideas Karp: “Reducibility among combinatorial problems”, in Complexity of Computer Computations, 1972, pp. 85- 103 –Showed 21 problems from a wide variety of areas are NP- complete

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