NP-Complete Problems. More reductions

Transcription

1 NP-Complete Problems More reductions

2 Definitions P: problems that can be solved in polynomial time (typically in n, size of input) on a deterministic Turing machine Any normal computer simulates a DTM NP: problems that can be solved in polynomial time on a non-deterministic Turing machine Informally, if we could guess the solution, we can verify the solution in P time (on a DTM) NP does NOT stand for non-polynomial, since there are problems harder than NP P is actually a subset of NP (we think)

3 Definitions, continued NP-hard At least as hard as any known NP problem (could be harder!) Set of interrelated problems that can be solved by reducing to another known problem NP-Complete A problem that is in NP and NP-hard Cook s Theorem SATISFIABILITY (SAT) is NP-Complete Other NP-Complete problems Reduce to SAT or previous reduced problem

4 Complexity Classes at-a-glance Image taken from Jeff Erickson's lecture notes, 1-nphard.pdf We will not discuss co-np

5 Correctness To prove unknown problem y is NP- Complete: Prove y is in NP Prove y is at least as hard as some x in NP- Complete Transform x into y in polynomial time such that x is yes if and only if y is yes Requires two proofs, one for each direction: x yes y yes, y yes x yes (or x no y no) Must also prove transformation is polynomial

6 Example: Independent Set Decision Problem: is there a set of k vertices such that none are adjacent? Proof of NP-Completeness: Is this in NP? Easy to check in k^2 (Adj. Matrix) or km (Adj. List) Is this NP-hard? Reduce known problem to unknown problem NOT the other way! We will use 3-SAT

7 Reductions Always solve a known problem by transforming it into an instance of the unknown problem Ex: 3SAT is NP-Complete. Independent Set is unknown. By transforming 3SAT into ISET and solving, we prove ISET is NP- Complete If you go the other way, you might just make a bad transformation: Sorting is an unknown. Transform sorting into 3SAT (somehow) and solve. This does NOT mean sorting is NP-Complete! Think of it this way: If I can solve x by transforming x into y, then y is at least as hard as x. (Corollary: if y can be solved quickly, so can x.)

8 3-SAT p ISET This connects each node directly to any node that is its complement. C = { v 1, v 2, v 3 }, {v 1, v 2, v 4 }, { v 2, v 4, v 5 }, {v 3, v 4, v 5 } v 2 v 1 v 2 v 3 v 1 v 3 v 2 v 4 v 4 v 5 v 4 v 5

9 3-SAT to ISET, continued The size of the transformation on the last slide is polynomial in the number of clauses (m) + number of literals (n) Adding opposing literals edges in the worst-case (full exploration of graph for each vertex) is still polynomial in m², so the time of the transformation is also polynomial Now we must prove 3SAT yes ISET yes 3SAT ISET Set target size for ISET = m Identify only one true literal in each clause of 3SAT, select that vertex in that triangle in ISET. Our set has no edges between nodes in the same triangle. Because this is a legal TA for 3SAT, we can t select a complementary version of a previously selected literal, so no edges will connect nodes in different triangles.

10 Visual demonstration of 3SAT ISET C = { v1, v2, v3}, {v1, v2, v4}, { v2, v4, v5}, {v3, v4, v5} Select nodes corresponding to literals in red. These form an independent set of size m. v 2 v 1 v 2 v 3 v 1 v 3 v 2 v 4 v 4 v 5 v 4 v 5

11 3SAT ISET So we have an ISET of size m, how does this give us a valid TA for 3SAT? At most one node comes from each triangle There are m triangles, so there must be one node in every triangle m clauses covered Because of the opposing literals edges, no conflicting truth assignments can be in ISET Assign the literals in the ISET to true; other literals don t matter

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13 NP-completeness proof structure 13

14 Traveling Salesman Problem TSP Optimization Problem A salesman must visit n cities. He must visit each city exactly one time and finish with the city at which he starts. Model possible routes as an undirected graph with weights on the edges. TSP Decision Problem c(i,j) is cost of traveling from city I to j Change it to a decision problem by making the problem to find a route of length at least k.

15 TSP Suppose someone has proven that the problem of determining if a graph G has a Hamiltonian cycle has been shown to be NP-complete. And we want to show that the Traveling Salesman Problem (TSP) is NP-complete. Two steps show that TSP is in the class NP show that HAM reduces to TSP in polynomial time

16 Step 1 Show that TSP is in the class NP. If we are given a set of n+1 vertices that we claim constitute a tour of length at most k, then we can check to see that each vertex occurs exactly one time check to see that beginning and ending vertices are the same check to see that edges exist between adjacent vertices in the graph see if the sum of the edges is k

17 Step 1 continued We can show that each of these steps can be done in polynomial time Therefore TSP is an element of NP

18 Step 2 Now we need to show that any HAM problem can be reduced to a TSP problem. HAM Reduction TSP Known NPC Problem Problem we want to show to be NPC

19 HAM TSP The Two Problems A Hamiltonian cycle of an undirected graph G=(V,E) is a simple cycle that contains each vertex in V. The Hamiltonian-cycle problem asks if a given graph contains a Hamiltonian cycle. A TSP circuit in of a undirected graph with edge weights visits each vertex exactly one time and begins and ends at the same vertex. The TSP problem asks if a given graph contains a TSP circuit of length less than k.

20 The Reduction from A to B Consider an instance of problem A. Show how that instance can be changed to an instance of B. Prove that the answer to B is yes iff the answer to Q is yes Prove that the algorithm for the reduction runs in polynomial time.

21 The Reduction Creating an Instance Now we need to show that any HAM problem can be reduced to a TSP problem. Let G=(V,E) be an instance of HAM Construct an instance of TSP as follows: Form the complete graph G =(V,E ) Define the cost of the edges as c(i,j) = 0 if (i,j) is in E, 1 otherwise We look for a tour of length 0

22 G G HAM TSP

23 Step 2 continued Now we must show that graph G has a Hamiltonian cycle iff G has a tour of cost at most 0. a) G has Hamiltonian cycle implies G has tour of cost at most n. Suppose that graph G has a Hamiltonian cycle h. Each edge in h belongs to E and thus there is an edge from v i to v i+1 with a cost of 0 in G Thus, h is a tour in G with a cost of 0.

24 Step 2 continued b) G has a tour of cost at most 0 implies G has as Hamiltonian cycle. Suppose that G has a tour h of cost at most 0. Since the cost of the edges is 0 or 1, none of the edges can have a cost of 1if the cost of the tour is at most 0 Therefore all edges are in E. So h visits each vertex exactly once, contains only edges in E, and is therefore a Hamiltonian cycle in G.

25 NPC proof -- CLIQUE Definition: a clique in an undirected graph G=(V,E) is a subset V' V of vertices, each pair of which is connected by an edge in E, i.e., a clique is a complete subgraph of G. Size of a clique is the number of vertices in the clique. Optimization problem: find the maximum clique. Decision problem: whether a clique of given size k exists in the graph? CLIQUE={<G,k>: G is a graph with a clique of size k.} Intuitive solution:??? 27

26 CLIQUE is NP-complete Theorem :CLIQUE problem is NP-complete. Proof: CLIQUE NP: given G=(V,E) and a set V' V as a certificate for G. The verifying algorithm checks for each pair of u,v V', whether <u,v> E. time: O( V' 2 E ). CLIQUE is NP-hard: show 3-CNF-SAT p CLIQUE. The result is surprising, since from boolean formula to graph. 28

27 CLIQUE is NP-complete Reduction from 3-CNF-SAT to CLIQUE. Suppose =C 1 C 2 C k be a boolean formula in 3-CNF with k clauses. We construct a graph G=(V,E) as follows: For each clause C r =(l 1r l 2r l 3r ), place a triple of v 1r, v 2r, v 3 r into V Put the edge between two vertices v i r and v j s when: r s, that is v i r and v j s are in different triples, and Their corresponding literals are consistent, i.e, l i r is not negation of l j s. Then is satisfiable if and only if G has a clique of size k. 29

28 =(x 1 x 2 x 3 ) ( x 1 x 2 x 3 ) (x 1 x 2 x 3 ) and its reduced graph G C 1 =x 1 x 2 x 3 30

29 CLIQUE is NP-complete Prove the above reduction is correct: If is satisfiable, then there exists a satisfying assignment, which makes at least one literal in each clause to evaluate to 1. Pick one this kind of literal in each clause. Then consider the subgraph V' consisting of the corresponding vertex of each such literal. For each pair v ir,v j s V', where r s. Since l ir,l j s are both evaluated to 1, so l i r is not negation of l js, thus there is an edge between v i r and v js. So V' is a clique of size k. If G has a clique V' of size k, then V' contains exact one vertex from each triple. Assign all the literals corresponding to the vertices in V' to 1, and other literals to 1 or 0, then each clause will be evaluated to 1. So is satisfiable. It is easy to see the reduction is in poly time. The reduction of an instance of one problem to a specific instance of the other problem. 31

30 Hamiltonian Path & H. Cycle HAMCYCLE is NP-Complete How can we show that HAMPATH is also NP- Complete? Take graph G and modify to G' Show transformation is polynomial Show that G' has HAMPATH if G has HAMCYCLE Show that G has HAMCYCLE if G' has HAMPATH

31 HAMPATH/HAMCYCLE, cont'd G V Split any vertex V into two identical vertices V1 and V2 (do not create E(V1,V2)). G' Run HAMPATH on G' If G had a HAMCYCLE, G' has a path through all the vertices except V1' and V2' starting at V1 and ending at V2 that corresponds to the cycle. Add V1' to the start of the path, V2' to the end, and it's a HAMPATH. If G' has a HAMPATH, it must start/end at V1'/V2' since they have degree 1, so G has a HAMCYCLE starting from V V1' V1 V2 V2' Then create V1' and V2' adjacent only to V1 and V2, respectively. (The transformation is trivially polynomial.)

32 Subset Sum is NPC SUNSET-SUM={<S,t>: S is a set of integers and there exists a S' S such that t= s S' s.} Theorem : SUBSET-SUM is NP-complete. 34

33 SUBSET-SUM is NPC SUBSET-SUM belongs to NP. Given a certificate S', check whether t is sum of S' can be finished in poly time. SUBSET-SUM is NP-hard (show 3-CNF- SAT p SUBSET-SUM). 35

34 SUBSET-SUM is NPC Given a 3-CNF formula =C 1 C 2 C k with literals x 1, x 2,, x n. Construct a SUBSET- SUM instance as follows: Two assumptions: no clause contains both a literal and its negation, and either a literal or its negation appears in at least one clause. The numbers in S are based on 10 and have n+k digits, each digit corresponds to (or is labeled by) a literal or a clause. Target t= (n 1 s and k 4 s) For each literal x i, create two integers: v i =0 01 (i) (l) 0 01 (w) 0 0, where x i appears in C l,,c w. v i '=0 01 (i) (m) 0 01 (p) 0 0, where x i appears in C m,,c p.. Clearly, v i and v i ' can not be equal in right k digits, moreover all v i and v i ' in S are distinct. For each clause C j, create two integers: s j = (j) 0 0, s j '= (j) 0 0. all s j and s j ' are called slack number. Clearly, all s j and s j ' in S are distinct. Note: the sum of digits in any one digit position is 2 or 6, so when there is no carries when adding any subset of the above integers. 36

35 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 37

36 SUBSET-SUM is NPC The above reduction is done in poly time. The 3-CNF formula is satisfiable if and only if there is a subset S' whose sum is t. suppose has a satisfying assignment. Then for i=1,,n, if x i =1 in the assignment, then v i is put in S', otherwise, then v i ' is put in S'. The digits labeled by literals will sum to 1. Moreover, for each digit labeled by a clause C j and in its three literals, there may be 1, 2, or 3 assignments to be 1. correspondingly, both s j and s j ' or s j ', or s j is added to S' to make the sum of the digit to 4. So S' will sum to Suppose there is a S' which sums to then S' contains exact one of v i and v i ' for i=1,,n. if v i S', then set x i =1, otherwise, v i ' S', then set x i =0. It can be seen that this assignment makes each clause of to evaluate to 1. so is satisfiable. 38