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CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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CLIQUE is NP-complete

Statement

CLIQUE is NP-complete.

Facts & Assumptions

Given: An arbitrary CLIQUE instance (G,k).

[L1]

The previous theorem gives a polynomial-time many-one reduction from 3-SAT to CLIQUE, by 3SAT polynomial-time many-one reduces to CLIQUE.

[L2]

The language 3-SAT is NP-complete, by 3-SAT is NP-complete.

[L3]

A language is NP-complete exactly when it is NP-hard and belongs to NP, by NP-hard and NP-complete languages.

[F1]

A yes-instance of CLIQUE consists of a graph with k pairwise adjacent chosen vertices, by Clique, independent set, and vertex cover decision problems.

Proof

technique · direct
1.1F1givenconstruct

First reject malformed instances and any k>∣V(G)∣; such an instance cannot contain a k-clique. Otherwise k≤∣V(G)∣, so a certificate is a list of at most ∣V(G)∣ vertices. Check that exactly k listed vertices are distinct and every pair is adjacent. The certificate length and verification time are polynomial in the graph encoding length, even when k is written in binary. Thus CLIQUE lies in NP.

1.2L1L2L3

Because [L2] says that 3-SAT is NP-complete, every language in NP reduces to 3-SAT. Composing any such reduction with the fixed reduction [L1] shows that every language in NP reduces to CLIQUE. Hence CLIQUE is NP-hard.

2.1L3step 1.1step 1.2∎

Step 1.1 gives CLIQUE ∈NP, and step 1.2 gives NP-hardness. Therefore CLIQUE is NP-complete by [L3].

Depends on

Used by

Dependency tree · two levels

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Sources