The Degree/Diameter Problem in Maximal Planar Bipartite graphs

The $(\Delta,D)$ (degree/diameter) problem consists of finding the largest possible number of vertices $n$ among all the graphs with maximum degree $\Delta$ and diameter $D$. We consider the $(\Delta,D)$ problem for maximal planar bipartite graphs, that is, simple planar graphs in which every face i...

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Bibliographic Details
Published inThe Electronic journal of combinatorics Vol. 23; no. 1
Main Authors Dalfó, Cristina, Huemer, Clemens, Salas, Julián
Format Journal Article
LanguageEnglish
Published 18.03.2016
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Summary:The $(\Delta,D)$ (degree/diameter) problem consists of finding the largest possible number of vertices $n$ among all the graphs with maximum degree $\Delta$ and diameter $D$. We consider the $(\Delta,D)$ problem for maximal planar bipartite graphs, that is, simple planar graphs in which every face is a quadrangle. We obtain that for the $(\Delta,2)$ problem, the number of vertices is $n=\Delta+2$; and for the $(\Delta,3)$ problem, $n= 3\Delta-1$ if $\Delta$ is odd and $n= 3\Delta-2$ if $\Delta$ is even. Then, we prove that, for the general case of the $(\Delta,D)$ problem, an upper bound on $n$ is approximately $3(2D+1)(\Delta-2)^{\lfloor D/2\rfloor}$, and another one is $C(\Delta-2)^{\lfloor D/2\rfloor}$ if $\Delta\geq D$ and $C$ is a sufficiently large constant. Our upper bounds improve for our kind of graphs the one given by Fellows, Hell and Seyffarth for general planar graphs. We also give a lower bound on $n$ for maximal planar bipartite graphs, which is approximately $(\Delta-2)^{k}$ if $D=2k$, and $3(\Delta-3)^k$ if $D=2k+1$, for $\Delta$ and $D$ sufficiently large in both cases.
ISSN:1077-8926
1077-8926
DOI:10.37236/4468