On the Cycle Augmentation Problem: Hardness and Approximation Algorithms
In the k -Connectivity Augmentation Problem we are given a k -edge-connected graph and a set of additional edges called links . Our goal is to find a set of links of minimum size whose addition to the graph makes it ( k + 1)-edge-connected. There is an approximation preserving reduction from the men...
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| Published in: | Theory of computing systems Vol. 65; no. 6; pp. 985 - 1008 |
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| Main Authors: | , , , |
| Format: | Journal Article |
| Language: | English |
| Published: |
New York
Springer US
01.08.2021
Springer Nature B.V |
| Subjects: | |
| ISSN: | 1432-4350, 1433-0490 |
| Online Access: | Get full text |
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| Summary: | In the
k
-Connectivity Augmentation Problem we are given a
k
-edge-connected graph and a set of additional edges called
links
. Our goal is to find a set of links of minimum size whose addition to the graph makes it (
k
+ 1)-edge-connected. There is an approximation preserving reduction from the mentioned problem to the case
k
= 1 (a.k.a. the Tree Augmentation Problem or TAP) or
k
= 2 (a.k.a. the Cactus Augmentation Problem or CacAP). While several better-than-2 approximation algorithms are known for TAP, for CacAP only recently this barrier was breached (hence for
k
-Connectivity Augmentation in general). As a first step towards better approximation algorithms for CacAP, we consider the special case where the input cactus consists of a single cycle, the
Cycle Augmentation
Problem (CycAP). This apparently simple special case retains part of the hardness of the general case. In particular, we are able to show that it is APX-hard. In this paper we present a combinatorial
3
2
+
ε
-approximation for CycAP, for any constant
ε
> 0. We also present an LP formulation with a matching integrality gap: this might be useful to address the general case of the problem. |
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| Bibliography: | ObjectType-Article-1 SourceType-Scholarly Journals-1 ObjectType-Feature-2 content type line 14 |
| ISSN: | 1432-4350 1433-0490 |
| DOI: | 10.1007/s00224-020-10025-6 |