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The Complexity of Approximation

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1 The Complexity of Approximation
- Lalitha Pragada

2 Theorem 8.26 Approximating the optimal solution to Clique within any constant ratio is NP-hard.

3 Proof: Develop a Gap-amplifying procedure, show that it turns any constant-ratio approximation into an approximation scheme, then appeal to theorem 8.23 to conclude that no constant-ratio approximation can exist.

4 Let G be any graph on n vertices
Let G be any graph on n vertices. Consider the new graph G2 on n2 vertices, where each vertex of G has been replaced by a copy of G itself and vertices in two copies corresponding to two vertices joined in the original are connected with all possible n2 edges connecting a vertex in one copy to a vertex in the other.

5 We can claim that G has a clique of size k iff G2 has a clique of size k2. The k copies of the clique of G corresponding to the k clique vertices in G form a clique of size k2 in G2.

6 The “If” part is harder since we have no a priori constraint on the composition of the clique in G2. However, two copies of G in the larger graph are either fully connected to each other or not at all.

7 So, if two vertices in different copies belong to the large clique, then the two copies must be fully connected and an edge exists in G between the vertices corresponding to the copies. If two vertices in the same copy belong to the large clique, then these two vertices are connected by an edge in G.

8 Thus, every edge used in large clique corresponds to an edge in G
Thus, every edge used in large clique corresponds to an edge in G. So, if the large clique has vertices in k or more distinct copies, then G has a clique of size k or more . If large clique has vertices in at most k distinct copies then it must include at least k vertices from same copy. So, G has a clique of size at least k.

9 Assume we have an approximation algorithm A For clique with absolute ratio E. Given some graph G with largest clique of size k we compute G2 ;run A on G2. We get a clique of size at least E k2 and then recover from this clique of size at least squareroot(E k2) = k*squareroot(E ). This new procedure , call it A ‘ runs in polynomial time if A does and has ratio RA ‘ = Squareroot(RA )

10 So more generally, j applications of this scheme yield procedure A j with absolute ratio j RA Given any approximation ratio E we can apply the scheme [log E ] times to obtain a log RA procedure with desired ratio.

11 Since [log E ] is a constant and since each
log RA application can run in a polynomial time we can say that we derived a polynomial-time approximation scheme for clique. But clique is OPTNP-hard and thus, according to theorem 8.23 it cannot be in PTAS . That’s the desired contradiction.


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