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This class

This class. Satisfiability problem NP completeness The complexity class co-NP Some NP-Completeness proofs. CNF-satisfiability  NP. Important problem in logic First problem shown to be NP-Complete by Cook To show CNF-satisfiability  NP :

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This class

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  1. This class • Satisfiability problem • NP completeness • The complexity class co-NP • Some NP-Completeness proofs

  2. CNF-satisfiability  NP • Important problem in logic • First problem shown to be NP-Complete by Cook • To show CNF-satisfiability  NP : • First, Conjunctive Normal Form will be defined • Then, the CNF-satisfiability problem will be defined • Finally, we will show a polynomial boundedverificationalgorithm for the problem

  3. Conjunctive Normal Form (CNF) • A logical (Boolean) variable is a variable that may be assigned the value true or false (for example p, q, r and s) • A literal is a logical variable or the negation of a logical variable (p and Øq are literals) • A clause is a disjunction of literals (for example (pÚqÚs) and (Øq Ú r) are clauses) • A logical expression is in Conjunctive Normal Form if it is a conjunction of clauses. Forexample the expression: (pÚqÚs) Ù(Øq Ú r)Ù(Øp Ú r) Ù(Ør Ú s)Ù(ØpÚØsÚØq)

  4. The CNF-satisfiability problem • Is there a truth assignment for the variables of a CNF expression which causes the CNF expression to be true? • Answer is true, if all clauses evaluate to T (true) • For example, p=T, q=F, r=T and s=T is a truth assignment for (pÚqÚs) Ù(ØqÚ r)Ù(Øp Úr) Ù(Ør Ús)Ù(ØpÚØsÚØq) • Note that if q=F then Øq=T

  5. NP-completeness • A language L (or problem corresponding to the language) is NP-complete if 1. L Î NP and 2. L' µ L for every L'ÎNP. • The class of NP-complete problems (languages) is called NPC.

  6. Theorem 1.If any problem in NPC is polynomial-time solvable, then P = NP. 2.If any problem in NP is not polynomial-time solvable, then all problems in NPC are not polynomial-time solvable.

  7. Proof 1.Given: L ÎP and L ÎNPC For any L'ÎNP L' µ L (definition of NPC). So by last theorem, L' Î P and P = NP. 2.Given: L ÎNP and LÏP Let L' ÎNPC. Assume that L'ÎP. Then L µ L' (definition of NPC) and thus by last theorem, LÎP. Contradiction. So L’ÏP and all NPC problems are not polynomial time solvable

  8. Proving NP completeness • Cook showed CNF-SATISFIABILITY is NP-complete. • To prove any other problem NP-complete, we then just have to reduce CNF- SATISFIABILITY to that problem and show that problem lies in the class NP.

  9. Theorem: polynomial reducibility is transitive If L1  L2 and L2 µ L3 then L1 µ L3 Proof outline: • There is a polynomially computable function • that reduces all inputs x  L1 into an input T1(x)  L2 • that reduces all inputs y  L2 into an input T2(y)  L3 • Applying T1 to x and T2 to T1(x) reduces any instance x  L1 to an instance T2(T1(x))  L3, and this conversion requires a polynomial number of steps

  10. Shortcut for NP-completeness Proofs • To prove a language L is NP-complete: Prove 1. L Î NP. Choose L' Î NPC, and prove 2. L’ L • Why is L is NP-complete? By 1. L Î NP. By definition L Î NPC, if every M Î NP satisfies M  L. We will show that this is correct, and thus L is NP-complete Since L’ Î NPC every M Î NP satisfies M  L’ (definition), and by 2. L’ L. So by transitivity M  L

  11. Complexity Class co-NP • The complexity class co-NP is the class of problems L such that Lbar ÎNP. • An open question: “is NP closed under complement” or is NP = co-NP? • To show that NP=co-NP we would need to show that for every language L Î NP, also LbarÎ NP

  12. TS: Given a set of cities {c1, …, cm} a distance d between every pair, and a bound B. Is it true that there is a TS cycle such that the distance of the cycle is at most B? Certificate is a permutation of the cities Polynomial bound verification algorithm (see slides) TSbar: Given a set of cities {c1, …, cm} a distance d between every pair, and a bound B. Is it true that there is no TS cycle such that the distance of the cycle is at most B? What is the certificate? A polynomial bound verification algorithm? NP=co-NP?

  13. P=co-P • The class of languages in P are closed under union, intersection, concatenation, Kleene star and complement • Proof for P=co-P • Let L  P. There exists a polynomial bound algorithm A that decides L (i.e, answers yes or no). • By changing A to answer no instead of yes and to yes instead of no we have a polynomial time algorithm B for Lbar. So Lbar  P and P=co-P

  14. SpanTree= {G|G is an undirected connected graph that contains a spanning tree of length at most B} Algorithm for SpanTree: L=Kruskal(G) //find length of MST if (L<=B) return true else return false SpanTreebar= {G|G is an undirected connected graph that does not contain a spanning tree of length at most B} Algorithm for SpanTreebar: L=Kruskal(G) //find length of MST if (L<=B) return false else return true Example

  15. P  (NP  co-NP) • Show P  co-NP. We need to show that for every language L, LP  Lco-NP Proof: Since LP and P = co-P  LbarP Since LbarP and P  NP  LbarNP So (Lbar)bar=L co-NP • Since P  NP and P  co-NP P  (NP co-NP)

  16. Theorem If NP co-NP then P  NP ProofMain idea: Use contradiction. Assume P=NP We know P=co-P. We will show that P=NP  NP=co-NP Contradicting the assumption NP co-NP

  17. Detailed Proof (cont.) • Given : P = NP, and P = co-P • We will show NP= co-NP 1. X  NP  X  P Xbar  P  Xbar  NP X  co-NP NP co-NP 2. X  co-NP Xbar  NP Xbar  P  X  P X  NP co-NP NP • From 1. And 2. NP= co-NP

  18. The four possibilities • P= NP = co-NP • P  NP and NP = co-NP • P = (NP co-NP) NP co-NP • P  (NP co-NP), NP co-NP P= NP=co-NP co-NP NPco-NP P= NPco-NP NP NP P co-NP NP=co-NP P

  19. NP-completeness Proofs in detail • To prove a language L is NP-complete: Step 1. Prove LÎ NP. Step 2. Choose a known language L' Î NPC. Step 3. Describe T mapping every instance of L' to an instance of L.

  20. NP-completeness Proofs in detail Step 4. Show that x Î L' iff T(x) Î L for all x. Step 5. Show that T , the algorithm that computes T(x) , runs in polynomial time.

  21. 3-CNF-SAT • Every clause has exactly 3 literals. • 3-CNF-SAT is NP-Complete • It can be shown that 3-CNF-SAT Î NP and CNF-SAT µ 3-CNF-SAT

  22. The Clique Problem • A clique is a complete undirected graph --- every vertex is connected to every other vertex. CLIQUE Input: An undirected graph G and a positive integer k. Output: YES iff a clique of size k exists in G.

  23. Clique example 1 2 8 3 7 4 • G contains a clique of 4 (with vertices 3, 5, 6, 7) • The 4 people 3, 5, 6, 7 “know” (can work with each other) each other 6 5

  24. The Clique Problem • Theorem: CLIQUE is NP-complete. • Proof: • Step 1. CLIQUE Î NP Given a set of k vertices V' ÍV, we can check if V' forms a clique by checking for every pair of nodes u, v ÎV’ that (u,v) Î E • Clearly, this can be done in polynomial time.

  25. The Reduction Step 2. Selection 3-CNF-SATwhich is NP-Complete. Step 3. Mapping For a formula C1Ç... Ç Ck such that Cr = l1,r+ l2,r+l3,r we construct a graph G with vertices v1,r v2,r v3,r for r = 1,…, k, where vi,r represents the literal li,r

  26. The Reduction We put an edge between vi,r and vj,s if both of the following hold: 1. r ¹ s and 2. li,r is not the negation of lj,s.

  27. v1,1 v1,2 k=2 The graph has 6 cliques of size 2 v2,1 v2,2 v3,2 v3,1

  28. Step 4a. Yes for 3-Sat implies yes for clique • Assume formula satisfiable. • With the satisfying assignment each clause contains at least 1 literal that is assigned 1. • Since each literal from each clause is a vertex in the graph, if we pick out a literal that is assigned 1 from each of the k clauses, we get k vertices in the graph.

  29. Step 4a. Yes for 3-Sat implies yes for clique • This set of k vertices is a clique. • For any two vertices, the corresponding literals are from different clauses, and are both assigned 1, so they cannot be complements of a single variable • Thus there is an edge between any two such vertices.

  30. Step 4b. Yes for Clique implies yes for 3-Sat • Assume G has a clique V' of size k. • No edge connects vertices in the same triple, so each of k triples has exactly one vertex in V'. • Assign 1 to each literal in V' without getting an inconsistent assignment (why?) and arbitrary values to the rest of the variables. • For this assignment each clause is satisfied and thus the answer for 3-SAT is yes.

  31. Step 5. Reduction is polynomial • Step 5. The reduction is polynomial. • The formula is read and 3k vertices are generated in O(k) steps. Then, each pair of literals ( ) from two different clauses is checked and an edge is added if the literals are not complimentary. • The reduction is O(k2)

  32. The vertex-cover problem • A vertex cover of an undirected graph is a set of vertices V' such that for every edge (u,v), either u or v or both are in V'. The problem is to find a cover of minimum size. • VERTEX-COVER • Input: A graph G and a number k. • Output: YES iff G has a vertex cover of size k.

  33. 1 2 3 5 4 Example of a vertex cover problem G k=2

  34. Application of vertex cover • What is the fewest # of guards we need to place in a museum to cover all the corridors? An airport to cover all the main walkways Hall 1 Hall 5 Hall 2 Hall 3 Hall 6 Hall 4

  35. The vertex-cover problem • Theorem: VERTEX-COVER is NP-complete. • Proof: Step 1. VERTEX-COVER ÎNP (obvious algorithm, given a subset of vertices). • Step 2. We select CLIQUE (will show that CLIQUE µ VERTEX-COVER)

  36. The reduction • Step 3. The mapping. • Given an instance of the CLIQUE problem <G, k> we output an instance <G’, |V|-k> of the VERTEX-COVER problem. • G’ has the same vertices as G and exactly those edges that are not in G. • It is easy to show the reduction is polynomial (step 5)

  37. Reduction Example G G’ 1 1 2 2 3 3 5 5 4 4 Clique {1,2} of size 2 Cover {3,4,5} of size 3

  38. Step 4. Correctness of the reduction • Assume G has a clique C of size k. • In G’ there are no edges between any pair of vertices in C G G’ 1 1 2 2 3 3 5 5 4 4

  39. Step 4 cont • So all edges in G’ are between a node in C and a node in V-C, or two nodes in V-C. • So V-C is a vertex cover for G’. G G’ 1 1 2 2 3 3 5 5 4 4

  40. Step 4. Correctness of the reduction • Assume G’=(V, E’) has a vertex cover V'Í V, where |V'| = |V|-k. • Thus for all u, v ÎV-V‘ (not in the cover), (u,v) E’ and thus (u,v) ÎE • V-V' is thus a clique. G G’ 1 1 2 2 3 3 5 5 4 4

  41. Hamiltonian Cycle • A Hamiltonian cycle of a graph G is a cycle that contains each vertex in V exactly once. A graph is Hamiltonian if it has a Hamiltonian cycle. • HAM-CYCLE • Input: A graph G. • Output: YES iff G is Hamiltonian.

  42. Hamiltonian Cycle • Theorem: HAM-CYCLE is NP-complete. • 3-CNF-SATµ HAM-CYCLE (proof omitted).

  43. Traveling Salesman • A tour is a Hamiltonian cycle in a graph. We want the minimum cost tour in a weighted graph. • TSP: • Input: A graph G, weights c for edges and a positive integer k. • Output: YES iff G with weights c has a TS tour of cost at most k.

  44. Traveling Salesman • Theorem: TSP is NP-complete. • Proof: Step 1: TSP is in NP • A certificate is a permutation of the cities. • This certificate can easily be verified by checking that all cities are included exactly once and that the sum of the distances is k or less. • This can be done in polynomial time, so TSP ÎNP.

  45. The reduction • Step 2: Select HAM-CYCLE (We will show that HAM-CYCLE µ TSP). • Step 3: The reduction • Given an instance G of HAM-CYCLE, we construct a graph G' = (V,E'). G' is a complete graph and c(i,j) = 0 if (i,j) is an edge and 1 otherwise. • The instance of TSP is then (G',c,0) where 0 is the bound on the cost of the tour. This conversion can be done in polynomial time (step 5).

  46. The reduction (example) G 0, G’ 1 2 1 2 0 0 1 1 0 0 3 3 1 5 0 5 0 4 4

  47. The reduction (step 4) • If G has a Hamiltonian cycle h, each edge in h belongs to E and thus has no cost in G'. Thus h is a tour with cost 0. • If G' has a tour of cost 0, the tour must have edges from E (since any edge not in E adds 1 to the cost). Thus, the tour must be a Hamiltonian cycle in G.

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