The complexity of the proper orientation number
A proper orientation of a graph G=(V,E) is an orientation D of E(G) such that for every two adjacent vertices v and u, dD−(v)≠dD−(u) where dD−(v) is the number of edges with head v in D. The proper orientation number of G is defined as χ→(G)=minD∈Γmaxv∈V(G)dD−(v) where Γ is the set of proper orienta...
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Published in: | Information processing letters Vol. 113; no. 19-21; pp. 799 - 803 |
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Main Authors: | , |
Format: | Journal Article |
Language: | English |
Published: |
Amsterdam
Elsevier B.V
01-09-2013
Elsevier Sequoia S.A |
Subjects: | |
Online Access: | Get full text |
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Summary: | A proper orientation of a graph G=(V,E) is an orientation D of E(G) such that for every two adjacent vertices v and u, dD−(v)≠dD−(u) where dD−(v) is the number of edges with head v in D. The proper orientation number of G is defined as χ→(G)=minD∈Γmaxv∈V(G)dD−(v) where Γ is the set of proper orientations of G. We have χ(G)−1⩽χ→(G)⩽Δ(G), where χ(G) and Δ(G) denote the chromatic number and the maximum degree of G, respectively. We show that, it is NP-complete to decide whether χ→(G)=2, for a given planar graph G. Also, we prove that there is a polynomial time algorithm for determining the proper orientation number of 3-regular graphs. In sharp contrast, we will prove that this problem is NP-hard for 4-regular graphs.
•We study the computational complexity of the proper orientation number.•We show that, it is NP-complete to decide whether χ→(G)=2, for a given planar graph G.•Also, we prove that there is a polynomial time algorithm for determining the proper orientation number of 3-regular graphs. In sharp contrast, we will prove that this problem is NP-hard for 4-regular graphs. |
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Bibliography: | ObjectType-Article-2 SourceType-Scholarly Journals-1 ObjectType-Feature-1 content type line 23 |
ISSN: | 0020-0190 1872-6119 |
DOI: | 10.1016/j.ipl.2013.07.017 |