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Paper:

TR97-004 | 19th February 1997 00:00

Approximating Dense Cases of Covering Problems

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Abstract:

We study dense instances of several covering problems. An instance of
the set cover problem with m sets is dense if there is \epsilon>0
such that any element belongs to at least \epsilon m sets. We show
that the dense set cover problem can be approximated with the
performance ratio c\log n for any c>0 and it is unlikely to be
NP-hard. We construct a polynomial-time approximation scheme for the
dense Steiner tree problem in n-vertex graphs, i.e. for the case
when each terminal is adjacent to at least \epsilon n vertices. We
also study the vertex cover problem in \epsilon-dense graphs.
Though this problem is shown to be still MAX-SNP-hard as in general
graphs, we find a better approximation algorithm with the performance
ratio 2\over{1+\epsilon}. The {\em superdense} cases of all these
problems are shown to be solvable in polynomial time.


Comment(s):

Comment #1 to TR97-004 | 30th April 1998 15:05

Evaluation of an Approximate Algorithm for the Everywhere Dense Vertex Cover Problem Comment on: TR97-004





Comment #1
Authors: Anton Eremeev
Accepted on: 30th April 1998 15:05
Downloads: 2331
Keywords: 


Abstract:

We generalize the DVC algorithm for the weighted case of vertex cover
problem (VCP) and study the performance of this algorithm. An extension
of result from TR97-004 for the weighted case is proposed in terms of a
new density parameter. Given a graph G(V,E) let there be \rho >0
such that w(O(v)) \geq \rho w(V) for any vertex v \in V. Here
w(O(v)) is the total weight of the neighbours of v, and w(V) is
the total weight of V. Then \frac{2}{1+\rho} performance guarantee
holds for an algorithm similar to DVC. The value \rho can be easily
estimated for everywhere \varepsilon-dense VCP, if there is such d
that w_u \leq dw_v for any pair of vertices u and v. The
generalization of DVC allows us to propose another polinomial-time
algorithm for the weighted VCP with the performance ratio as a function
on |V| which is less than \frac{2}{1+\rho}.




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