Design and Analysis of Algorithms

Transcription

1 CSE 0, Winter 208 Design and Analsis of Algorithms Lecture 6: NP-Completeness, Part 2 Class URL:

2 The Classes P and NP Classes of Decision Problems P: Problems for which there eists a deterministic polnomial-time algorithm NP: Problems for which there eists a nondeterministic polnomial-time algorithm What is non-deterministic polnomial-time? Solutions are small (== polnomial-size) Guessing is free, but solutions must be checkable in polnomial time Succinct Certificate P NP, but whether P = NP is not known Most people believe P!= NP, which would impl P NP

3 NP-Hard, NP-Complete NP-Hard: Problem X is NP-hard if ever problem in NP is polnomiall reducible to X NP-Complete: Problem X is NP-complete if: X belongs to NP, and X is NP-hard I.e., () Ever problem in NP is pol-time reducible to X ; (2) X is as hard as an problem in NP ( as hard as it gets ) Also, Y is NP-complete if Y NP and some NPcomplete problem X is polnomiall reducible to Y NP-Complete problems are the hardest problems in NP

4 Picture of P, NP, NP-Hard, NP-Complete Graph Non- Isomorphism Shortest Path Graph Isomorphism Halting Problem SAT, 3SAT, Verte Cover, Independent Set, Dominating Set, Hamilton Ccle, CLIQUE, NumPart,.

5 SAT is NP-Complete (Cook, 97) Cook s theorem (97): SAT is NP-complete Levin (972)) Once we know one problem to be NP-complete, proving other problems NP-complete becomes easier A new problem Y is NP-complete if: () Y is in NP (2) SAT or an other NP-complete problem is polnomiall reducible to Y I.e., reduce from SAT to Y

6 Two Parts To An NP-Completeness Proof Problem is in NP A known NP-complete problem reduces to it

7 NP-Complete Problems 0, 0 Number Partition (NumPart): Given a set of numbers S = {, 2,, n }, is there a disjoint partition of S into S, S 2 such that sum of numbers in S = sum of numbers of S 2? disjoint partition: S S 2 = S, and S S 2 = k-tolerant Number Partition (ktnp): Given a set of numbers S = {, 2,, n } and a number k, is there a disjoint partition of S into S, S 2 such that (sum of numbers in S ) (sum of numbers of S 2 ) k?

8 NP-Complete Problems 0, 0 Number Partition (NumPart): Given a set of numbers S = {, 2,, n }, is there a disjoint partition of S into S, S 2 such that sum of numbers in S = sum of numbers of S 2? disjoint partition: S S 2 = S, and S S 2 = k-tolerant Number Partition (ktnp): Given a set of numbers S = {, 2,, n } and a number k, is there a disjoint partition of S into S, S 2 such that (sum of numbers in S ) (sum of numbers of S 2 ) k? NumPart p ktnp

9 NP-Complete Problems 0, 0 Number Partition (NumPart): Given a set of numbers S = {, 2,, n }, is there a disjoint partition of S into S, S 2 such that sum of numbers in S = sum of numbers of S 2? disjoint partition: S S 2 = S, and S S 2 = k-tolerant Number Partition (ktnp): Given a set of numbers S = {, 2,, n } and a number k, is there a disjoint partition of S into S, S 2 such that (sum of numbers in S ) (sum of numbers of S 2 ) k? ktnp p NumPart

10 NP-Complete Problems 0, 0 Reduction of k-tolerant Number Partition (ktnp) to Number Partition (NumPart)

11 NumPart, Bin-Packing NumPart(S): Given a set of numbers S = {, 2,, n } with 0 < i, there a disjoint partition of S into S, S 2 (i.e., where S S 2 = S, and S S 2 = ) such that i S = j S 2? Bin-Packing(L,B): Given a list of items L = {, 2,, n } with 0 < i, and a number B, can L be packed into B unit-capacit bins? Given that NumPart is NP-complete, prove that Bin-Packing is NP-complete

12 NP-Complete Problems #, #2, #3 The following five problems are NP-complete (Manber, Introduction to Algorithms: A Creative Approach) Definition: A verte cover of G =( V, E) is V V such that ever edge in E is incident to some v V. VC(G,k): Given undirected G = (V, E) and integer k, does G have a verte cover with k vertices? CLIQUE(G,k): Does G contain a clique of size k? 3COLOR(G): Is G properl colorable with 3 colors?

13 NP-Complete Problems #4, #5 Definition: A dominating set D of G = (V, E) is D V such that ever v V is either in D or adjacent to at least one verte of D DS(G,k): given G, k, does G have a dominating set of size k? 3SAT(F): Given Boolean epression F in CNF such that each clause has eactl 3 variables, is F satisfiable?

14 NP-Completeness Proofs: Reductions All NP problems Clique Verte Cover SAT 4 3 3SAT 5 3-Colorabilit 2 Dominating Set To prove NP-completeness of a new problem: Prove that the problem belongs to NP Show polnomial reducibilit from a known NPcomplete problem This slide shows the order in which classic NP-completeness reductions are often presented

15 NP-Completeness Proof: Verte Cover (VC) Problem: Given undirected G = (V, E) and integer k, does G have a verte cover with k vertices? Theorem: The VC problem is NP-complete. Proof: Approach: Reduction from CLIQUE I.e., given that CLIQUE is NP-complete, we will prove that VC is NP-complete VC is in NP. This is true since we can guess a cover of size k and check it in polnomial time Goal: Transform arbitrar CLIQUE instance into VC instance such that CLIQUE answer is es iff VC answer is es

16 NP-Complete Problems #, #2 Definition: A verte cover of G =( V, E) is V V such that ever edge in E is incident to some v V. VC(G,k): Given undirected G = (V, E) and integer k, does G have a verte cover with k vertices? CLIQUE(G,k): Does G contain a clique of size k? Given that CLIQUE is NP-complete, prove that VC is NP-complete Reduce from to

17 Step : VC is in NP This is true since we can guess a cover of size k and check it in polnomial time Step #2 is Reduction: Transform arbitrar CLIQUE instance into VC instance such that CLIQUE answer is es iff VC answer is es

18 Step 2: Reduce CLIQUE to VC Transform arbitrar CLIQUE instance into VC instance such that CLIQUE answer is es iff VC answer is es G

19 CLIQUE(G,k) VC(G,n-k) Claim: CLIQUE(G, k) has same answer as VC ( G, n - k), where n = V. Observe: If U V is a clique in G, then V \ U is VC in G.

20 VC(G,n-k) CLIQUE(G,k) Observe: If D Vis a VC in G, then G has no edge between vertices in V \ D. So, V\D is a clique in G G Transform is polnomial time.

21 Given that X is NP-C, prove that Y is NP-C Remember: reduce from X to Y Step : Y is in NP Step 2: Transform arbitrar instance inst X of X into an instance inst Y of Y such that X(inst X ) is YES iff Y(inst Y ) is YES Practice problems are on HW and in slides; will consolidate a week from toda

22 Given that X is NP-C, prove that Y is NP-C X = Hamilton Path, Y = Hamilton Ccle (Rudrata = Hamilton) X = Hamilton Ccle, Y = Hamilton Path X = SAT, Y = 3-SAT X = CLIQUE, Y = DOM_SET

23 Techniques for Dealing with Hard Problems Implicit enumeration Backtrack Branch-and-Bound Ad hoc methods (heuristics) Provabl good (bounded performance ratio) Christofides method for metric Traveling Salesman Problem Engineer s method for Number Partitioning Not provabl good metaheuristics Genetic Optimization (evolutionar algorithms) Simulated Annealing Tabu Search

24 Backtrack This development is from T. C. Hu, Combinatorial Algorithms, Addison-Wesle, 982 Set up a - correspondence between configurations and possible solution sequences (or, partial solution vectors). Decision Tree: The root corresponds to the initial state of the problem (usuall is null, means no decision is made), and each branch corresponds to a decision concerning one parameter.

25 Backtrack Eample: 3-Coloring Three colors: R, G, B 4 Graph A graph is legall colored with k colors if each verte has a color (label) between and k (inclusive), and no adjacent vertices have the same color.

26 Backtrack Eample: 3-Coloring Three colors: R, G, B G B G Graph 5 R 4G 3R 4B B,2G B smmetr, can use an two colors for vertices and 2 4R 3B Tree of Colorings 4G B A graph is legall colored with k colors if each verte has a color (label) between and k (inclusive), and no adjacent vertices have the same color. Does the ordering of the vertices (, 2, 3, 4, 5) matter? 5R

27 Domino Effect Suppose a solution is represented b a vector. A partial solution is represented b a partial vector. If a partial vector does not satisf the solution requirements, there is no point in etending the partial vector into a more complete solution Domino Effect: P k false P k+ false Domino effect must hold for backtrack to work!

28 Backtrack Eample: Non-Attacking Queens Put N queens on an N N chessboard such that no queen attacks another queen A queen in chess can attack an square on the same row, column, or diagonal. Given an n n chessboard, we seek to place n queens onto squares of the chessboard, such that no queen attacks another queen. The eample shows a placement (red squares) of four mutuall nonattacking queens.

29 Backtrack Eample: Non-Attacking Queens

30 Branch-and-Bound (B&B) Variant of backtrack with costs Associate a cost with a partial solution, such that the cost of a parent is alwas less than or equal to the cost of its child in the decision tree do not branch from an internal node whose cost is higher than the current bound = cost of the minimumcost complete solution found so far the bound is updated if a better solution is found Ke points Used for optimization problems Cost-driven Bounding prunes the decision tree, saves time

31 What Does B&B Look Like for the TSP?

32 What Does B&B Look Like for the TSP?

33 Branch-and-Bound Eample: Game Tree In games (e.g., chess) can model the different stages of the game b a rooted tree Don t need to consider all possible situations of a game Can predict the outcome of the game using the concept of branch-and-bound Questions Who is the winner if both plaers pla optimall? How much is the paoff? E.g., we ma initiall onl know the paoff of the terminal nodes (endstates) of the game.

34 Two-Plaer Game Tree Paoff of the Game = amount the st plaer receives at the end of game st plaer wants to maimize paoff (Call her Ma) 2 nd plaer wants to minimize paoff (Call her Min)

35 C Game Tree Eample A B -3 I F J M D E G H K L N We can see that the value of the game is (node A) b eamining (bottom-up) the entire game tree. In general, (bottom-up) eamination of an entire game tree is not feasible. Appling DFS with branch-and-bound using - pruning improves efficienc.

36 NP-Completeness Proof: DS Definition: A dominating set D of G = (V, E) is D V such that ever v V is either in D or adjacent to at least one verte of D Problem: given G, k does G have a dominating set of size k? Theorem: DS is NP-complete Proof: (reduction from VC) DS NP Given instance (G,k) of VC, make ever edge of G into a triangle and answer DS.

37 NP-Completeness Proof: DS G: v w G : z u vz G has DS D of size k iff G has VC C of size k. Without loss of generalit, assume DV, D is a DS ever edge hits some verte in D D is a VC in G C is a VC new and old vertices dominated v z vu vw zu w u uw

38 NP-Completeness Proof: 3SAT 3SAT: Given a Boolean epression in CNF such that each clause has eactl 3 variables, determine satisfiabilit. Theorem: 3SAT is NP-Complete Proof: 3SATNP Let E be an arbitrar instance of SAT arbitrar clause of E For k 4 )... ( 2 k C ) (... ) ( ) ( ) ( ' k k k C

39 NP-Completeness Proof: 3SAT C is satisfiable iff C is satisfiable: C = at least one i = C = C = at least one i = can set i s so that C = ) ( ) ( ) ( ) ( ' ) ( ) ( ) ( ' ) ( C C C C

40 NP-Completeness Proof: CLIQUE Problem: Does G=(V,E) contain a clique of size k? Theorem: Clique is NP-Complete. (reduction from SAT) Idea: Make column of k vertices for each clause with k variables. No edge within a column. All other edges present ecept between and

41 NP-Completeness Proof: CLIQUE Eample: G = G has m-clique (m is the number of clauses in E), iff E is satisfiable. (Assign value to all variables in clique) ) ( ) ( ) ( z z z E z z z