Path cover problems on bipartite interconnection networks

Abstract

A \emph{paired many-to-many $k$-disjoint path cover} (\emph{$k$-DPC} for short) of a graph is a set of $k$ disjoint paths joining $k$ disjoint source-sink pairs that cover all the vertices of the graph. Extending the notion of DPC, we define a {\em paired many-to-many bipartite $k$-DPC} (\emph{$k$-BiDPC}) of a bipartite graph $G$ to be a set of $k$ disjoint paths joining $k$ distinct source-sink pairs that altogether cover the same number of vertices as the maximum number of vertices covered when the source-sink pairs are given in the complete bipartite, spanning supergraph of $G$. We consider the problem of finding paired many-to-many $2$-DPC and paired many-to-many $1$-BiDPC in an $m$-dimensional bipartite HL-graph and the problem of finding paired many-to-many $k$-BiDPC in an $m$-dimensional hypercube.

It is proved that every $m$-dimensional bipartite HL-graph, under the condition that at most $m-3$ faulty edges removed, has a paired many-to-many $2$-DPC if the set of sources and sinks is balanced in the sense that it contains the same number of vertices from each part of the bipartition, where $m\geq 4$. Furthermore, every $m$-dimensional bipartite HL-graph, where $m\geq 4$, has a paired many-to-many 2-DPC in which the two paths have the same length if each source-sink pair is balanced in that source and sink do not have the same color. Using the $2$-DPC properties, it is proved that every $m$-dimensional bipartite HL-graph, under the condition that either at most $m-2$ faulty edges, or one faulty vertex and at most $m-3$ faulty edges removed has a paired many-to-many $1$-BiDPC, where $m\geq 3$. Using this result, we show that every $m$-dimensional hypercube, $Q_m$, under the condition that $f$ or less faulty elements (vertices and/or edges) are removed, has a paired many-to-many $k$-BiDPC joining $k$ distinct source-sink pairs for any $f$ and $k\geq 1$ subject to $f+2k\leq m$. This implies that $Q_m$ with $m-2$ or less faulty elements is strongly Hamilto...