Theory of Computing

Lecture 18MAS 714

Hartmut Klauck

Before

• We have shown NP-completeness of 3-SAT:– a k-CNF is a CNF in which all clauses have 3 literals

(or less)– k-SAT={set of all k-CNF that are satisfiable}

• Notes: Not hard to give a reduction from CNF-SAT to 3-SAT

• 2-SAT is in P (and hence not NP-complete unless P=NP)

• 3-SAT is the basis of our reductions from now on

Reductions

• We will show some reductions from 3-SAT to a few problems

• Due to transitivity we can use reductions from any of the problems for which we already have a reduction from 3-SAT– Often this is easier

Vertex Cover

• For an undirected Graph G a vertex cover is a subset S of the vertices such that every edge has at least one endpoint in S

• VC={G,k such that G has a vertex cover of size at most k}

• Theorem: VC is NP-complete• Part 1: VC in NP is easy (guess S and check all

edges)

Reduction VC

• Part 2: we reduce from 3-SAT• Let F be a formula with variables x1,…,xn and 3-

clauses z1,…,zm

• For each xi we use two vertices vi and wi connected by an edge (variable gadget)

• For each zi we use three vertices ai, bi , ci

connected to form a triangle (clause gadget)

Reduction VC• vi corresponds to the literal xi

• wi to the literal :xi

• We connect ai, bi, ci with the corresponding literal from clause zi

• We set k to n+2m

• Observation: For any vertex cover S one of vi and wi needs to be in S, and at least 2 of ai, bi, ci

• 1) Assume x satisfies F. We show that a small S exists.• If xi=0 we choose wi else vi for S

• For each clause gadget at least one of ai, bi, ci is connected to a vertex in S. Choose other 2 vertices per clause gadget to form a VC of size n+2m

Reduction VC

• 2) Now assume there is a VC S of size at most k=n+2m• S must contain either vi or wi

• S must contain two vertices for each clause gadget (triangle)

• Hence it cannot contain vi and wi

• Set x accordingly (xi=1 iff vi2 S)• Claim: this is a satisfying assignment• Proof: every clause gadget must have one vertex not in S,

connected to a literal node in S• Hence x satisfies F

Clique

• We want to show that MaxClique is NP-complete– Surely it is in NP

• Reduction from VC• Observation/reduction:• (G,k)2 VC iff (Gc,n-k)2 Max Clique– Gc: complement of G

Observation

• If S is a vertex cover in G then all edges in G have one endpoint in S, there are no edges going from V-S to V-S

• Hence in the complement of G, V-S is a clique• Gc,n-k in MaxClique

• Note: we have shown that VC has a 2-approximation algorithm

• Max Clique does not have a n1-² approximation unless P=NP– Notice how a VC of size · 2k implies only a Clique of size ¸ n-2k,

which can be much smaller than n-k, e.g., k=n/2

SetCover

• Input: set of m subsets s1,…,sn of S, |S|=m

• Task: find the smallest set of si such thatS= [ i2I si

We want |I| minimal• SetCover:{s1,…,sn,k: there is I µ {1,…,n}, |I|=k, [ i2I

si= [i=1…n si }

• Theorem: SetCover is NP-complete– Part 1: 2 NP: guess I

SetCover

• Part 2: Reduction from VC• G,k is mapped to a SetCover instance as

follows: G=(V,E)• Universe: E• Subsets: s1,…,sn

• si: {edges adjacent to vertex i}• k unchanged• Then: (G,k)2 VC iff (s1,…,sn,k) 2 SetCover

Hamiltonian Cycle

• A Hamiltonian cycle in a directed G=(V,E) is a cycle that visits each vertex once

• HC={G: G has a Hamiltonian cycle}• Theorem: HC is NP-complete

• In NP: guess the cycle• Hardness: reduction from VC

HC

• G=(V,E) and k given (input to VC)• Assume edges in E are ordered (e1,…,em)

• New vertices: a1,…,ak

and vertices (u,e,b) where u2 V, e2 E is incident to u and b2{0,1}

• Edges: (u,e,0) to (u,e,1) type 1(u,e,b) to (v,e,b) where e={u,v} type 2(u,e,1) to (u,e’,0) where e,e’ inc. to u and there is no e’’ with e<e’’<e’ inc. to u type 3(u,e,1) to all ai where e is max. edge at u type 4all ai to (u,e,0) where e is the min. edge at u type 5

HC

• Take vertex cover S of size k• Find the cycle: for v2 S consider all the vertices (v,e,b) and

all (u,e,b) where e={u,v}• We traverse these from (v,e,0) in ascending order of e

– (v,e,0) to (v’,e,0) to (v’,e,1) to (v,e,1) to (v,e’,0) etc. until (v,e,1) for max e at v reached. Then to a1 and to the next (u,e,0) with u the next vertex in S and e min. at u

• A Hamitonian cycle exists• Assume a Hamiltonian cycle ex. It must traverse all ai.

Group all vertices visited between ai and ai+1

• Claim: vertex cover of size k exists

Traveling Salesman

• Input: matrix of distances (edge weights),k• Output: is there a Hamiltonian cycle of total

edge weight at most k?• Reduction from HC:– Use adjacency matrix as weights, k=n

Subgraph Isomorphism

• Given G,H, is H isomorphic to a subgraph of G?– subgraph: subset of vertices and edges– isomorphic: same up to renaming vertices

• Theorem: Subgraph isomorphism is NP-complete

• Reduction from Clique: Map G,k to G and H, where H is a k-clique (undirected case)

• Reduction from HC: Map G to G and H, where H is an n-cycle (directed case)

Note

• Graph Isomorphism (are G,H isomorphic?)is not known to be in P or NPC

Subset Sum

• Input: set S of integers, target t• Is there a subset S’µ S such that the elements

in S’ sum to t?

• Theorem: Subset Sum is NP-complete• Reduction from 3-SAT