Constructing edge-disjoint spanning trees in augmented cubes

TL;DR

This paper presents a method for constructing the maximum possible number of edge-disjoint spanning trees in augmented cubes, which are a variation of hypercubes with superior properties, enhancing reliable communication protocols.

Contribution

It provides an optimal construction of n-1 edge-disjoint spanning trees in n-dimensional augmented cubes, improving upon previous results for hypercube variants.

Findings

Constructed n-1 edge-disjoint spanning trees in AQn for n > 2

Proved the construction is optimal with respect to the number of trees

Enhanced reliability in communication protocols using these trees

Abstract

Let T1, T2,.... Tk be spanning trees in a graph G. If for any pair of vertices u and v of G, the paths between u and v in every Ti( 0 < i < k+1) do not contain common edges then T1, T2,.... Tk are called edge-disjoint spanning trees in G. The design of multiple edge-disjoint spanning trees has applications to the reliable communication protocols. The n-dimensional augmented cube, denoted as AQn, a variation of the hypercube, possesses some properties superior to those of the hypercube. For AQn (n > 2), construction of n-1 edge-disjoint spanning trees is given the result is optimal with respect to the number of edge-disjoint spanning trees.

Equations

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Full text

S. A. Mane

Center for Advanced Studies in Mathematics, Department of Mathematics,

Savitribai Phule Pune University, Pune-411007, India.

Keywords: edge-disjoint spanning trees, Augmented cubes

1. Introduction

A graph $G$ is a triple consisting of a vertex set $V(G)$, an edge set $E(G)$, and a relation that associates with each edge two vertices called its endpoints[12]. The topology of the network is modeled in the form of a graph whose vertices correspond to nodes, while edges represent direct physical connections between nodes. This paper deals with the well-established problem of handling the maximum possible number of communication requests without using a single physical link more than once, known as the edge-disjoint spanning trees Problem. The hypercubes $Q_{n}$ are one of the most versatile and efficient interconnection networks discovered for parallel computation. Many variants of the hypercube have been proposed. The augmented cube, proposed by Choudam and Sunitha[2], is one of such variations. An $n-$dimensional augmented cube $AQ_{n}$ can be formed as an extension of $Q_{n}$ by adding some links. For any positive integer $n$, $AQ_{n}$ is a vertex transitive, $(2n-1)$-regular and $(2n-1)$-connected graph with $2^{n}$ vertices. $AQ_{n}$ retains all favorable properties of $Q_{n}$ since $Q_{n}\subset AQ_{n}$. Moreover, $AQ_{n}$ possesses some embedding properties that $Q_{n}$ does not. The main merit of augmented cubes is that their diameters are about half of those of the corresponding hypercubes.

A tree $T$ is called a spanning tree of a graph $G$ if $V(T)=V(G)$. Two spanning trees $T_{1}$ and $T_{2}$ in $G$ are edge-disjoint if $E(T_{1})\cap E(T_{2})=\phi$.

The edge-disjoint spanning trees(EDSTs for short) problem has received a great deal of attention in recent years because of its numerous applications on interconnection networks such as fault-tolerant broad casting and secure message distribution[3, 4, 5, 6, 7, 8, 9, 10, 13].

Barden, Hadas, Davis and Williams[1] proved there exist $n$ EDSTs in a hypercube of dimension $2n$ and provided a construction for obtaining the maximum number of EDSTs in a hyperube. Wang, Shen and Fan[11] proved existence of $n-1$ EDSTs in an augmented cube $AQ_{n}$ and they asked, “how to derive an effective algorithm that constructs edge-disjoint spanning trees based on our algorithm, in an augmented cube?” Motivated by this question, we provide the construction for obtaining the maximum number of EDSTs ($n-1$ EDSTs ) in augmented cube $AQ_{n}$ ( $n\geq 3$ ).

2. Preliminaries

The definition of the $n-$dimensional augmented cube is stated as the following. Let $n\geq 1$ be an integer. The $n$-dimensional augmented cube, denoted by $AQ_{n}$, is a graph with $2^{n}$ vertices, and each vertex $u$ can be distinctly labeled by an $n$-bit binary string, $u=u_{1}u_{2}....u_{n}$. $AQ_{1}$ is the graph $K_{2}$ with vertex set $\{0,1\}$. For $n\geq 2$, $AQ_{n}$ can be recursively constructed by two copies of $AQ_{n-1}$, denoted by $AQ^{0}_{n-1}$ and $AQ^{1}_{n-1}$, and by adding $2^{n}$ edges between $AQ^{0}_{n-1}$ and $AQ^{1}_{n-1}$ as follows:

Let $V(AQ^{0}_{n-1})=\{0u_{2}....u_{n}:u_{i}\in\{0,1\},2\leq i\leq n\}$ and $V(AQ^{1}_{n-1})=\{1v_{2}....v_{n}:v_{i}\in\{0,1\},2\leq i\leq n\}$. A vertex $u=0u_{2}....u_{n}$ of $AQ^{0}_{n-1}$ is joined to a vertex $v=1v_{2}....v_{n}$ of $AQ^{1}_{n-1}$ if and only if for every $i$, $2\leq i\leq n$ either

1. $u_{i}=v_{i}$; in this case an edge $\langle u,v\rangle$ is called a hypercube edge and we say $v=u^{h}$, or

2. $u_{i}=\overline{v_{i}}$; in this case an edge $\langle u,v\rangle$ is called a complement edge and we say $v=u^{c}$.

Let $E^{h}_{n}=\{\langle u,u^{h}\rangle:u\in V(AQ^{0}_{n-1})\}$ and $E^{c}_{n}=\{\langle u,u^{c}\rangle:u\in V(AQ^{0}_{n-1})\}$. See Fig.$1$.

\tiny{0}$$\tiny{1}$$\tiny{00}$$\tiny{10}$$\tiny{01}$$\tiny{11}$$\tiny{000}$$\tiny{010}$$\tiny{001}$$\tiny{011}$$\tiny{101}$$\tiny{100}$$\tiny{110}$$\tiny{111}$$\tiny AQ_{1}$$\tiny AQ_{2}$$\tiny AQ_{3}Fig.1$$\tiny AQ^{1}_{1}$$\tiny AQ^{0}_{1}$$\tiny AQ^{0}_{2}$$\tiny AQ^{1}_{2}

For undefined terminology and notations see [12].

3. Construction of edge-disjoint spanning trees in augmented cubes

Our proof is by induction.

As $AQ_{n}$ is $(2n-1)$-regular, $|E(AQ_{n})|=(2n-1)(2^{n-1})=n2^{n}-2^{n-1}$. When we construct any spanning tree on $2^{n}$ vertices of $AQ_{n}$ we need exactly $2^{n}-1$ edges hence we can construct at most $n-1$ EDSTs. Still, for $n\geq 3$, $2^{n-1}+n-1(<2^{n}-1)$ number of edges remain uncovered by these $n-1$ EDSTs, but by our method, we are able to construct the tree on $2^{n-1}+n$ vertices containing uncovered edges.

Theorem 3.1**.**

Let $n\geq 3$ be an integer. There exist $n-1$ edge-disjoint spanning trees in augmented cube $AQ_{n}$.

Proof.

First if $n=3$, we construct two EDSTs $T_{1}$ and $T_{2}$ as follows. See Fig.2.

\tiny{000}$$\tiny{001}$$\tiny{010}$$\tiny{011}$$\tiny{111}$$\tiny{101}$$\tiny{100}$$\tiny{110}Fig.2(a). Spanning tree $T_{1}$ in AQ_{3}$$\tiny{000}$$\tiny{001}$$\tiny{010}$$\tiny{011}$$\tiny{111}$$\tiny{101}$$\tiny{100}$$\tiny{110}Fig.2(b). Spanning tree $T_{2}$ in $AQ_{3}$

Edges uncovered by $T_{1}$ and $T_{2}$ again form a tree say $T_{3}$ on $7$ vertices. See Fig.3.

\tiny{000}$$\tiny{001}$$\tiny{010}$$\tiny{011}$$\tiny{111}$$\tiny{101}$$\tiny{100}$$\tiny{110}Fig.3. Tree $T_{3}$ in $AQ_{3}$

Let $AQ_{n+1}$ be decomposed into two augmented cubes say $AQ^{0}_{n}$ and $AQ^{1}_{n}$ with vertex set say $\{u^{0}_{i},v^{0}_{i}:1\leq i\leq 2^{n-1}\}$ and $\{u^{1}_{i},v^{1}_{i}:1\leq i\leq 2^{n-1}\}$ respectively. Denote by $T^{0}_{1},T^{0}_{2},.......,T^{0}_{(n-1)}$ the EDSTs in $AQ^{0}_{n}$. Let the identical corresponding EDSTs in $AQ^{1}_{n}$ be denoted by $T^{1}_{1},T^{1}_{2},.......,T^{1}_{(n-1)}$. Let the vertices of the tree say $T^{0}_{n}$ which is made up of uncovered edges of $n-1$ EDSTs in $AQ^{0}_{n}$ be denoted by $u^{0}_{1},u^{0}_{2},....u^{0}_{2^{n-1}},v^{0}_{1},v^{0}_{2},....v^{0}_{n}$ and the identical corresponding vertices of the corresponding tree say $T^{1}_{n}$ in $AQ^{1}_{n}$ be denoted by $u^{1}_{1},u^{1}_{2},....u^{1}_{2^{n-1}},v^{1}_{1},v^{1}_{2},....v^{1}_{n}$.

We want to construct $n$ EDSTs in $AQ_{n+1}$, of which first $n-2$ EDSTs are constructed from $T^{0}_{i}$ and $T^{1}_{i}$ by adding a hypercube edge $\langle v^{0}_{i},v^{1}_{i}\rangle\in E(AQ_{n+1})$ to connect two internal vertices $v^{0}_{i}\in V(T^{0}_{i})$ and $v^{1}_{i}\in V(T^{1}_{i})$, for $1\leq i\leq n-2$. See Fig.4.

T^{0}_{1}$$T^{0}_{2}$$T^{0}_{n-2}$$T^{1}_{1}$$T^{1}_{2}$$T^{1}_{n-2}$$v^{1}_{1}$$v^{1}_{2}$$v^{1}_{n-2}$$v^{0}_{1}$$v^{0}_{2}$$v^{0}_{n-2}Fig.4. AQ^{0}_{n}$$AQ^{1}_{n}

The $(n-1)^{th}$ EDST is constructed from $T^{0}_{(n-1)}$ by adding all complement edges $E^{c}_{n+1}=\{\langle u,u^{c}\rangle:u\in V(AQ^{0}_{n})\}$.See Fig.5.

AQ^{0}_{n}$$AQ^{1}_{n}$$T^{0}_{(n-1)}$$u^{0}_{1}$$(u^{0}_{1})^{c}$$u^{0}_{2^{n-1}}$$v^{0}_{1}$$(v^{0}_{1})^{c}$$(u^{0}_{2^{n-1}})^{c}$$(v^{0}_{2^{n-1}})^{c}$$v^{0}_{2^{n-1}}Fig.5

The $n^{th}$ EDST is constructed from $T^{1}_{(n-1)}$ by adding to it hypercube edges $\langle v^{0}_{i},v^{1}_{i}\rangle\in E(AQ_{n+1})$ to connect internal vertices $v^{0}_{i}\in V(AQ^{0}_{n})$ and $v^{1}_{i}\in V(T^{1}_{(n-1)})$ $n\leq i\leq 2^{n-1}$. The $2^{n-1}+(n-1)$ vertices in $AQ^{0}_{n}$ which are not connected to $T^{1}_{(n-1)}$ via hypercube edges are $u^{0}_{i}$ ($1\leq i\leq 2^{n-1}$) and $v^{0}_{1},v^{0}_{2},....v^{0}_{n-1}$, but the edge $\langle v^{0}_{n},v^{1}_{n}\rangle$ is included in $T^{1}_{(n-1)}$, by adding to $T^{1}_{(n-1)}$ the edges of the tree $T^{0}_{n}$ through the vertex $v^{0}_{n}$ connect all remaining vertices $v^{0}_{1},v^{0}_{2},....v^{0}_{n-1}$ and $u^{0}_{i}$ ($1\leq i\leq 2^{n-1}$) to $T^{1}_{(n-1)}$. See Fig.6.

AQ^{0}_{n}$$AQ^{1}_{n}$$T^{1}_{(n-1)}$$T^{0}_{n}$$v^{0}_{2^{n-1}}$$v^{1}_{2^{n-1}}Fig.6v^{0}_{n}$$v^{1}_{n}$$v^{0}_{n+1}$$v^{1}_{n+1}

Still, we have uncovered edges of $AQ_{n+1}$ namely edges of tree $T^{1}_{n}$, the hypercube edge $\langle v^{0}_{n-1},v^{1}_{n-1}\rangle$ and hypercube edges $\langle u^{0}_{i},u^{1}_{i}\rangle$ ($1\leq i\leq 2^{n-1}$). We can easily observe that these uncovered hypercube edges along with the tree $T^{1}_{n}$ again form a new tree on $(2^{n-1}+n)+(2^{n-1}+1)=2^{n}+n+1$ vertices. See Fig.7.

AQ^{0}_{n}$$AQ^{1}_{n}$$T^{1}_{n}$$u^{0}_{1}$$u^{1}_{1}$$u^{0}_{2^{n-1}}$$u^{1}_{2^{n-1}}Fig.7v^{0}_{n-1}$$v^{1}_{n-1}

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Acknowledgment: The author gratefully acknowledges the Department of Science and Technology, New Delhi, India for the award of Women Scientist Scheme for research in Basic/Applied Sciences.