Clique Npc Proof

Proof that CLIQUE is NP-Complete In an undirected graph, a subgraph wherein each node is connected to each other node in the subgraph is a clique. Give an integer k  and a graph, the problem is to determine there is a clique of size k anywhere in the the graph. A clique of size k is called a k-clique.

Slide Credit: Mr. T. Widland

First, prove that CLIQUE is in NP It is simple to find a verifier for CLIQUE, using the nodes in the clique as the certificate: •1. Verify that the clique is a subgraph of the correct size •2. Check that the graph contains edges connecting the nodes in the clique. •If 1 and 2 are both true, accept, otherwise reject. Slide Credit: Mr. T. Widland

 Now, a polynomial time reduction to 3SAT • We will construct a graph G based on a formula φ . • Let k be the number of clauses in φ . • G will have a clique of size k iff φ is satisfiable.

Slide Credit: Mr. T. Widland

Construction of G • Each clause in φ will be represented as a set of three nodes, one for each literal. • Each literal in a clause will be represented as a node labeled as the literal. • The nodes will all be connected, except that no nodes from the same clause will be connected and no two nodes with contradictory labels will be connected(i.e. Slide Credit: Mr. T. Widland x1 and ~x1)

φ =(~x1∨x2∨ x3) ∧(x1∨∼ x2 ∨~x3) ∧ (x1 ∨x2 ∨x3)

x1

x2

x3

~x1

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An example

Proof  Theorem: φ is satisfiable iff G has a k-clique: Assume φ is satisfiable. In each clause, at least one literal is true. Select one node which has a true literal as its label from each clause. The nodes selected form a k-clique since the nodes are selected one from each clause. The labels are connected since since they cannot be contradictory. (cont) Slide Credit: Mr. T. Widland

Finishing the proof(the converse) Assume G has a k-clique. Assign each of the literals in the labels of the clique to be true. Because nodes from the same clause are not connected, a k-clique must visit k clauses. No contradictions will be encountered since contradictory labels are not connected. Thus each clause contains at least one true literal, which means that φ is satisfied. Slide Credit: Mr. T. Widland

Practice Problems optional -- turn in or bring to Office  Hours Minor:

Sipser 7.1, 7.2, 7.5, 7.10

Major: Sipser 7.6, 7.7, 7.12, 7.13, 7.14, 7.15, 7.19, 7.25, 7.26