COT 6410 Exam I Key
1) (a) is the Clique Problem, and it is known to be NP–Complete.
(b) is MaxClique, and it is not known to be in the set NP (let alone, NP–Complete) because
we know no way of checking a Yes answer in deterministic polynomial time.
MaxClique is NP–Hard: because it is possible to provide a Turing Reduction from
Clique. Let AMaxClique and AClique be the best algorithms for MaxClique and Clique,
respectively.
1) Accept an arbitrary instance of Clique (G, k).
2) i
k
2) While i
n and AMaxClique (G, i) = False, set i
≤
i+1
3) If i = n+1 then return (False)
else return (True)
Proof:
Suppose the given Clique instance is true. Then MaxClique must be
true for some i such that k
i
n. Thus, in Step 3, i < n+1 and the result is
≤ ≤
True.
Suppose the instance of Clique is false. Then, for every i such that k
i
n,
≤ ≤
then AMaxClique will return False, and results in i = n+1 for step 3.
MaxClique is in P if and only if P = NP (i.e., in NP–Easy) because we can also design a
Turing reduction from MaxClique to Clique.
1) Accept an arbitrary instance of MaxClique (G, k)
2) if AClique (G, k) = True and AClique (G, k+1) = False then return (True)
else return (False)
Proof:
If an instance of MaxClique is True, there is a clique of size k, but none
of size k+1. Therefore, on this graph, AClique will be true for k, but not for k+1.
If the instance of MaxClique is False,
then either the largest clique is less
than k, or greater than k. In the first case AClique(G, k) will be False and in
the second AClique(G, k+1) will be true. In
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 Fall '10
 Dutton
 Computational complexity theory, NPcomplete, Boolean satisfiability problem, arbitrary instance

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