Relationship between optimal k-distance dominating sets in a weighted graph and its spanning trees

Highlights

• It generalizes a result on optimal dominating sets of graphs to weighted graphs.

• Optimal dominating sets (ODS) have many important applications.

• The result may lead to efficient heuristics for an ODS in a weighted graph.

Abstract

We show that for each optimal k-distance dominating set D in connected graph G with link-weights $w(x,y)\ge 0$, there is a spanning tree $T(G,D,k)$ of G for which D is an optimal k-distance dominating set. This generalizes a similar result for the case of constant link-weights.

Introduction

We consider here a generalized form of the optimal k-hop dominating set problem. Let $G=(V,E)$ be a weighted graph with two or more nodes, where each link $(x,y)\in E$ has a weight $w(x,y)\ge 0$. A subset of nodes $D\subset V$ is called a k-distance dominating set, in short, a k-ddset if each node y in G has distance dist$(x,y)\le k$ from some node $x\in D$. Here, dist$(u,v)$ is the shortest length of a path in G connecting the nodes u and v (in short, an uv-path) based on the link lengths $w(x,y)$. We say D is an optimal k-ddset if $|D|$ is the smallest among all k-ddsets. If each $w(x,y)=1$, then the optimal k-ddset problem becomes the optimal k-hop dominating set problem. We refer the reader to [2] for the literature review and the many applications of k-ddset. We study here the relationship between optimal k-ddset of a general connected graph and those of its spanning trees.

Let $\mathrm{\Delta }(x)$ = $\{y:\text{dist}(x,y)\le k\}$; clearly, $x\in \mathrm{\Delta }(x)$. We say x dominates y or, equivalently, y is dominated by x if $y\in \mathrm{\Delta }(x)$. More generally, for $S\subset V$, we write $\mathrm{\Delta }(S)={\cup }_{x\in S}\mathrm{\Delta }(x)$ and for $y\in \mathrm{\Delta }(S)$ we say S dominates y. In particular, a set of nodes D is a k-ddset if and only if $\mathrm{\Delta }(D)=V$. For a disconnected graph G, a k-ddset of G is the union of a k-ddset for each of its connected components.

Let $D(G)$ denote an optimal k-ddset of a weighted graph G. If $G^{\prime }=(V,E^{\prime })$, where $E^{\prime }\subseteq E$, then every k-ddset of $G^{\prime }$ is a k-ddset of G. Thus, $|D(G^{\prime })|\ge |D(G)|$. In particular, for every spanning tree T in G, we have $|D(T)|\ge |D(G)|$. If the deletion of the links $E-E^{\prime }$ from G does not increase dist$(x^{\prime },y^{\prime })$ for any node pair $x^{\prime },y^{\prime }\in V$ and, more generally, if dist$(x,y)\le k$ in G implies dist$(x,y)\le k$ in $G^{\prime }$, then $\mathrm{\Delta }(x)$ in G is the same as $\mathrm{\Delta }(x)$ in $G^{\prime }$ for each node x and hence G and $G^{\prime }$ have the k-ddsets and thus $|D(G^{\prime })|=|D(G)|$. (Note that such a situation does not arise when $w(x,y)=c$ for each $(x,y)\in E$ for some $c>0$.) In particular, throwing away links $(x,y)\in E$ which have $w(x,y)>k$ does not affect the k-ddsets of G; more generally, if $w(x,y)>\text{dist}(x,y)$, then we can throw away the link $(x,y)$ without affecting the k-ddsets.

For an xy-path $\pi (x,y)=〈x,z_1,z_2,\cdots ,z_n,y〉$, where $n\ge 0$ and $z_i$ are the intermediate nodes, we denote its length by $|\pi (x,y)|={\sum }_{i=0}^{n}w(z_i,z_{i+1})$, where $z_0=x$ and $z_{n+1}=y$.

Theorem 1

For each optimal k-ddset D of a connected graph G, there is a spanning tree $T(G,D,k)$ of G such that D is an optimal k-ddset of $T(G,D,k)$.

Proof

For the moment, let $D=\{x_1,x_2,\dots ,x_m\}$ be an arbitrary k-ddset of G. First, we show the construction of the tree $T(G,D,k)$. We use the optimality of D for G at the very end to show that D is also an optimal k-ddset of $T(G,D,k)$.

We partition V into m disjoint subsets $V_i\supseteq \{x_i\},1\le i\le m$, such that the subgraph $G_i$ of G on $V_i$ is connected and $D_i=\{x_i\}$ is an optimal k-ddset of $G_i$ for each i. Let $d_j(y)$ = dist$(x_j,y)$ for $1\le j\le m$ and $d(y)$ = min {$d_j(y):1\le j\le m$}. For each $y\in V-D$, $0\le d(y)\le k$ because D is a k-ddset of G. Fix a node $y\in V-D$, let i be the smallest index such that $d_i(y)=d(y)$. Let $\pi (y)=〈x_i,y_1,y_2,\dots ,y_{p-1},y〉$ be a shortest $x_{i}y$-path of length $d(y)={\sum }_1^{p}w(y_{n-1},y_n)$, where $y_0=x_i$ and $y_p=y$. Clearly, each $y_q\in V-D$ for $1\le q\le p-1$. We first show that each $d(y_q)=d_i(y_q)={\sum }_1^{q}w(y_{n-1},y_n)$. Clearly, $d(y_q)\le d_i(y_q)$. If possible, suppose for some q, we have $d(y_q)<d_i(y_q)={\sum }_1^{q}w(y_{n-1},y_n)$ and $i^{\prime }$ is the smallest index such that $d_{i^{\prime }}(y_q)=d(y_q)$. This means $d(y)\le d_{i^{\prime }}(y)\le d_{i^{\prime }}(y_q)$ + dist$(y_q,y)<d_i(y_q)$ + dist$(y_q,y)<{\sum }_1^{q}w(y_{n-1},y_n)+{\sum }_{q+1}^{p}w(y_{n-1},y_n)=d_i(y)$, a contradiction, and therefore we have $d(y_q)=d_i(y_q)$. Now, if $i^{\prime }<i$, then as before we get $d_{i^{\prime }}(y)\le d_{i^{\prime }}(y_q)+\text{ dist}(y_q,y)=d_i(y_q)+\text{ dist}(y_q,y)=d_i(y)$ and hence $d_{i^{\prime }}(y)=d_i(y)$. This contradicts the choice of i for y and thus $i^{\prime }=i$ for each node $y_q\in \pi (y)$. We now let each $V_i$ = {$y:y=x_i$ or $y\in V-D$ and i is the smallest index such that $d_i(y)=d(y)$}. The argument above shows that if $y\in V_i$ and $y\ne x_i$ then there is an $x_{i}y$-path $\pi (y)$ in G such that all intermediate nodes in $\pi (y)$ are also in $V_i$. Thus, the subgraph $G_i$ of G on $V_i$ is connected.

Now we build the spanning tree $T(G,D,k)$. First, choose a spanning tree $T_i$ of the subgraph $G_i$ on $V_i$, rooted at $x_i$, such that the $x_{i}y$-path in $T_i$ for each $y\in G_i$ is a shortest path of length $d(y)\le k$; let $L_i$ be the links in $T_i$. Because the sets $V_i$ form a partition of V, the links $L_1\cup L_2\cup \dots \cup L_m$ do not contain any cycle. We now add $m-1$ other links $L_0$ of G to $L_1\cup L_2\cup \dots \cup L_m$ to form a spanning tree $T(G,D,k)$ of G. Clearly, D is a k-ddset of $T(G,D,k)$. Finally, if D is not an optimal k-ddset of $T(G,D,k)$, then it has a k-ddset $D^{\prime }$ such that $|D^{\prime }|<|D|$. This implies $D^{\prime }$ is a k-ddset of G, a contradiction if D is an optimal k-ddset of G. This completes the proof. □

Example 1

Fig. 1(i) shows a weighted connected graph G. For $k=4$, no $\mathrm{\Delta }(x)$ contains all the vertices and thus an optimal k-ddset has size ≥2. One of the many optimal 4-ddsets here is $D=\{a,f\}$; the node b with any of $\{e,f,g,h,k\}$ also gives an optimal 4-ddset. For $D=\{a,f\}$, Fig. 1(ii) shows the tree $T_a$ of shortest-paths from a to the nodes $\mathrm{\Delta }(a)=\{a,b,c,i,j\}$ and the tree $T_f$ of shortest-paths from f to the nodes $\mathrm{\Delta }(f)=\{d,e,f,g,h,k\}$, which is disjoint from $\mathrm{\Delta }(a)$. We can obtain a spanning tree $T(G,D,k)$ for which D is an optimal 4-ddset as shown in Fig. 1(ii), which consists of $T_a$, $T_f$ and the link $(c,d)$.

Corollary 1

If $T(G,D,k)$ is a spanning tree of G as in Theorem 1, then every optimal k-ddset of $T(G,D,k)$ is an optimal k-ddset of G.

Proof

Follows immediately from the argument in the last part of the proof of Theorem 1. □

We briefly mention the notion of an optimal k-ddset for a given exclusion-set of nodes $X\subset V$, where we want to find an optimal k-ddset $D_X(G)$ in G which is disjoint from X. Here, although we do not need to dominate the nodes in X we may need them to form paths from $D_X(G)$ to nodes in $V-X$. Thus, we cannot find $D_X(G)$ by simply considering the subgraph of G on the nodes $V-X$. Typically, the optimal $|D_X(G)|$ will be larger than the optimal $|D(G)|$. We show below how to convert an exclusion-set version of the optimal k-ddset problem to the standard (unconstrained) form. If $y,z\in V-X$ and there is at least one yz-path ${\pi }_X(x,y)$ = $〈y,x_1,x_2,\cdots ,x_n,z〉,n\ge 1$, such that all its intermediate nodes are in X, then let $d_X(y,z)$ = min$\{w(y,z),|〈y,x_1,x_2,\cdots ,x_n,z〉|:x_i\in X\text{ and }n\ge 1\}$, where we take $w(y,z)=\infty$ if $(y,z)\notin E$. In particular, $d_x(y,z)$ is undefined if and only if $(y,z)\notin E$ and ${\pi }_X(y,z)$ does not exist. Let $G_X(V-X,E_X)$ be the reduced graph with the nodes $V-X$ and the links $E_X=\{(y,z):y,z\in V-X\text{ and either }(y,z)\in E\text{ or }{d}_X(y,z)\text{is defined}\}$. We take the link-weight $w_X(y,z)=d_X(y,z)$ in $G_X(V-X,E_X)$ if the latter is defined and, otherwise, $w_X(y,z)=w(y,z)$. It is clear that the distance between any two nodes in $V-X$ is the same in G and in $G_X$. Thus, the k-ddsets of G with the exclusion-set X are exactly the same as the k-ddsets of $G_X$, and hence the same is true from the optimal k-ddsets.

Example 2

For the graph G in Fig. 1(i) and $X=\{a\}$, $G_X$ is the same as $G-\{x\}$ and $\{b,f\}$ s an optimal 4-ddset of $G_X$ and is also an optimal 4-ddset of G which is disjoint from X. If we choose $X=\{a,b\}$, then $G_X$ equals $G-X$ plus the link $(c,j)$ with weight $w_X(c,j)=3$ and $\{f,j\}$ is an optimal 4-ddset of $G_X$ and is also an optimal 4-ddset of G which is disjoint from X.