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Chapter 8: Q19E (page 281)

Akiteis a graph on an even number of vertices, say 2n, in which of the vertices form a clique and the remaining vertices are connected in a 鈥渢ail鈥� that consists of a path joined to one of the vertices of the clique. Given a graph and a goal , the KITE problem asks for a subgraph which is a kite and which contains 2g nodes. Prove that KITE is NP-complete.

Short Answer

Expert verified

The kite problem is reduced to clique and is NP-complete.

Step by step solution

01

Define a Clique

A clique of a graph is the subgraph of the graph. The clique problem is an undirected graph with a goal .Then the clique problem is the subset of vertices that are adjacent to each other and which form a complete graph.

A clique is defined by the following equation Cx=鈭�i-0wGcixi.

02

Prove KITE is an NP-Complete

Consider the kite graph with 2n vertices, n vertices form a clique, and the other n vertices are connected by a tail path, and the tail ends are connected to a vertex in the clique. Given a graph G and target g , the kite subgraph with 2g vertices need to be calculated.

By adding Vvertices to the graph GV,E,then connecting the Vvertices with the original vertices in G has a one-to-one correspondence to obtain G'. Then the maximum clique problem can be reduced to the Kite problem.

If G has a kite subgraph with 2g vertices if and only if there is a clique of size g in G when the following conditions satisfy.

There is a clique of size g in G鈬�G'forms a kite subgraph with 2g vertices.

If G' has 2g kite subgraph, then there must have a group of size g.

Thus, the kite is an NP-complete problem.

Therefore, the kite problem is reduced to clique and is NP-complete.

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Most popular questions from this chapter

Search versus decision. Suppose you have a procedure which runs in polynomial time and tells you whether or not a graph has a Rudrata path. Show that you can use it to develop a polynomial-time algorithm for RUDRATA PATH (which returns the actual path, if it exists).

Show that for any problem 鈭� in NP, there is an algorithm which solves n in time O 2pnwhere is the size of the input instance and p(n)is a polynomial (which may depend on鈭� ).

In task scheduling, it is common to use a graph representation with a node for each task and a directed edge from task i to j task if i is a precondition for j. This directed graph depicts the precedence constraints in the scheduling problem. Clearly, a schedule is possibe if and only if the graph is acyclic; if it isn鈥檛, we鈥檇 like to identify the smallest number of constraints that must be dropped so as to make it acyclic.

Given a directed graph G=(V,E), a subset E'鈯�Eis called a feedback arc set if the removal of edges E' renders G acyclic.

FEEDBACK ARC SET (FAS): Given a directed graph G=(V,E)and a budget , find a feedback arc set of role="math" localid="1658907144825" 鈮�bedges, if one exists.

(a)Show that FAS is in NP.

FAS can be shown to be NP-complete by a reduction from VERTEX COVER. Given an instance (G,b)of VERTEX COVER, where G is an undirected graph and we want a vertex cover of size 鈮�b, we construct a instance (G',b)of FAS as follows. If G=(V,E)has vertices v1,K,vnthen make G'=(V',E')a directed graph with 2n verticesw1,w1',k,wn,wn',andn+2|E|(directed) edges:

  • (wi,wi')foralli=1,2,k,n
  • (wi',wj)and(wj',wi)forevery(vi,vj)鈭�E.
  • Show that if G contains a vertex cover of size b, then G' contains a feedback arc set of size b .
  • Show that if G' contains a feedback arc set of size b, then G contains a vertex cover of size (at most) b. (Hint: Given a feedback arc set of size b in G', you may need to first modify it slightly to obtain another one which is of a more convenient form, but is of the same size or smaller. Then, argue that G must contain a vertex cover of the same size as the modified feedback arc set.)

Consider the CLIQUE problem restricted to graphs in which every vertex has degree at most v. Call this problem CLIQUE-3 .

(a) Prove that CLIQUE-3 is in NP .

(b) What is wrong with the following proof of NP-completeness for CLIQUE-3 ? We know that the CLIQUE problem in general graphs is NP-complete, so it is enough to present a reduction from CLIQUE-3 to CLIQUE . Given a graph G with vertices of degree 鈮�3, and a parameter g, the reduction leaves the graph and the parameter unchanged: clearly the output of the reduction is a possible input for the CLIQUE problem. Furthermore, the answer to both problems is identical. This proves the correctness of the reduction and, therefore, the NP-completeness of CLIQUE-3 .

(c) It is true that the VERTEX COVER problem remains NP-complete even when restricted to graphs in which every vertex has degree at most 3 . Call this problem VC-3 . What is wrong with the following proof of NP-completeness for CLIQUE ? We present a reduction from VC-3 to CLIQUE-3 . Given a graph G=(V,E) with node degrees bounded by 3 , and a parameter b , we create an instance of CLIQUE-3 by leaving the graph unchanged and switching the parameter to |V|-b. Now, a subset C鈯�Vis a vertex cover in G if and only if the complementary set V-C is a clique in G. Therefore G has a vertex cover of size鈮�bif and only if it has a clique of size 鈮�|V|-b. This proves the correctness of the reduction and, consequently, the NP-completeness of CLIQUE-3 .

(4)Describe an O(V)algorithm for CLIQUE-3 .

Show that if P=NP then the RSA cryptosystem (Section 1.4.2) can be broken in polynomial time.
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