A Note on Distance-Fall Colorings

TL;DR

This paper introduces a new type of graph coloring called distance-$k$ fall coloring, proves its existence for certain classes of graphs, and confirms a conjecture about the size of distance-$d$-dominating sets in trees.

Contribution

It establishes the existence of distance-$k$ fall colorings for connected 3-colorable graphs and trees, confirming an old conjecture about dominating sets in trees.

Findings

Connected 3-colorable graphs have a distance-2 fall 3-coloring.

Trees of order at least $k$ have a $k$-coloring with vertices within distance $k-1$ of every color.

Proves the conjecture that trees have small independent distance-$d$-dominating sets.

Abstract

We say a proper coloring of a graph is distance-$k$ fall if every vertex is within distance $k$ of at least one vertex of every color. We show that if $G$ is a connected graph of order at least $3$ that is $3$-colorable, thenit has a distance-2 fall 3-coloring. Further, for every integer $k\ge 2$, if $T$ is a tree of order at least $k$, then $T$ has a $k$-coloring such that every vertex is within distance $k-1$ of every color. This proves an old conjecture of Beineke and Henning that every tree of order $n$ has an independent distance-$d$-dominating set of size at most $n/(d + 1)$.

Full text

Keywords: Fall coloring; distance-$k$ domination; chromatic number

MSC-class: 05C12; 05C15; 05C69

1 Introduction

A fall coloring is a proper coloring of the vertices of a graph such that every vertex sees every color. That is, for each vertex $v$ its closed neighborhood $N[v]$ contains all colors. Not all graphs have a fall coloring. The simplest example is the $5$-cycle. In this paper we consider proper colorings where every vertex is “near” to every color. We say a coloring is distance-$k$ fall if every vertex is within distance $k$ of at least one vertex of every color. For example, for a graph of diameter $2$, every proper coloring is distance-$2$ fall; and every odd cycle has a $3$-coloring that is distance-$2$ fall.

The terminology “fall coloring” was introduced in the 2000 paper by Dunbar et al. [5]. But the concept is older, having been studied before as partitioning a graph into independent dominating sets; see the references of [5]. Recall that a set $S$ of vertices is independent if no two elements of $S$ are joined by an edge. A set $S$ is dominating if every vertex is either in $S$ or adjacent to at least one vertex of $S$. An independent dominating set is one that is both independent and dominating. More generally, a set $S$ is distance-$k$ dominating if every vertex is within distance $k$ of at least one vertex of $S$. Thus a distance-$k$ fall coloring is a partition of the vertex set into independent distance-$k$ dominating sets. The parameter independent distance-$k$ domination, that is, the minimum size of an independent distance-$k$ dominating set, was originally studied by Beineke and Henning [1].

2 Distance-Fall Colorings and Chromatic Number

It is immediate that if a (connected nontrivial) graph is bipartite, then the bipartite coloring is a fall coloring. So the next question to consider is $3$-colorable graphs. For a coloring, we say a vertex is $d$-good if every color appears within distance $d$ of the vertex; otherwise the vertex is $d$-bad. The following theorem shows that if the graph is $3$-colorable then there is a proper coloring with $3$ colors such that every vertex is $2$-good.

Theorem 2.1**.**

If $G$ is a connected graph of order at least $3$ that is $3$-colorable, then $G$ has a distance-$2$ fall $3$-coloring.

Proof: Suppose to the contrary, that $G$ does not admit a distance-$2$ fall $3$-coloring. Take a proper $3$-coloring of $G$ that minimizes the number of $2$-bad vertices. Without loss of generality, choose a $2$-bad vertex $v\in V(G)$. Then $N(v)$ is monochromatic.

If there exists a vertex $w_{1}\in N(v)$ with $\deg(w_{1})=1$, then by recoloring $w_{1}$ with the missing color in $N[v]$, we obtain a proper $3$-coloring of $G$ with strictly fewer $2$-bad vertices, contradicting the minimality of the original coloring.

Therefore, assume $\deg(w)\geq 2$ for all $w\in N(v)$. Since $v$ is $2$-bad, we have $N(v)\cap N(w)=\emptyset$ and $N(w)$ is monochromatic for every $w\in N(v)$. That is, $v$ and all vertices in its second neighborhood share the same color.

Now recolor $v$ with the missing color in $N[v]$. Vertex $v$ then becomes $2$-good, and every neighbor of $v$ remains $2$-good. Moreover, no previously $2$-good vertex becomes $2$-bad under this recoloring. Thus, the total number of $2$-bad vertices decreases, contradicting the minimality of the original coloring.

Therefore, there exists a proper $3$-coloring of $G$ with zero $2$-bad vertices.

$\Box$

It is perhaps interesting to note that an independent distance-$2$ dominating set is related to an independent isolating set. An isolating (or vertex-edge dominating) set $S$ can be viewed as a $2$-dominating set $S$ with the added condition that if vertex $v$ is at distance $2$ from $S$, then all its neighbors are at distance $1$ from $S$. An independent isolating set is an isolating set that is also independent. See for example [2, 3, 6]. It is easy to see that Theorem 2.1 does not extend to a partition into three independent isolating sets (for example $C_{5}$). However, in [3] it was shown that it almost does: specifically, if $G$ is $3$-colorable there exist three sets $A$, $B$, and $C$ of vertices, such that each is an independent isolating set, their union is all vertices, and such that $A$ and $B$ overlap in at most one vertex, while $C$ is disjoint from $A\cup B$.

It is, however, unclear if the Theorem 2.1 generalizes to larger number of colors. The next step would be to determine whether every connected $4$-colorable graph of order at least $4$ has a $4$-coloring such that every vertex is within distance three of all colors. This is best possible because of path–complete graph, meaning the graph that consists of a path and a complete graph joined by an edge. However, in the case of trees, the above theorem does generalize:

Theorem 2.2**.**

For integer $k\geq 2$, if $T$ is a tree of order at least $k$, then $T$ has a $k$-coloring such that every vertex is within distance $k-1$ of every color.

Proof: Let $T$ be a tree of diameter $d$. If $d<k$, then any coloring that uses each color at least once automatically has the desired property.

So, assume $d\geq k$. Let $P:=v_{1},v_{2},\ldots,v_{d}$ be a diametral path in $T$. Define a coloring $c:V(P)\to\{0,1,2,\ldots,k-1\}$ on the vertices of $P$ such that $c(v_{i})=i\pmod{k}$.

Now, let $e=uv$ be a central edge of $P$, and let $T_{u}$ and $T_{v}$ be the components of $T-e$ containing $u$ and $v$, respectively. If $T\setminus P$ denotes the forest induced by the vertices of $T$ not contained in $P$, then color every vertex of $T\setminus P$ such that, in each of the components $T_{u}$ and $T_{v}$, any two vertices at the same distance from $u$ or $v$ receive the same color.

We claim that this coloring has the desired property. If a vertex $w$ lies within distance $(k-1)/2$ of the edge $e$, then $w$ is within distance $k-1$ of each color on $P$. Otherwise, the unique path from $w$ to $P$ enters $P$ at some vertex and continues along $P$ via $e$; the first $k$ vertices on this path all have distinct colors. Hence the theorem follows. $\Box$

Theorem 2.1 is equivalent to saying that the vertices of a $3$-colorable graph can be partitioned into three (disjoint) independent distance-$2$ dominating sets. And thus the independent distance-$2$ domination number of a $3$-colorable graph is at most $n/3$, where $n$ is the order. This generalizes Theorem 2 of [1], which proved the bound for trees. Furthermore, Theorem 2.2 shows that the independent distance-$k$ domination number of a tree is at most $n/(k+1)$, which establishes the Conjecture at the end of [1].

We note that Theorem 2.2 and thus the conjecture in [1] was recently also proved by Bujtás et al. [4]. Indeed, they showed that Theorem 2.2 generalizes to bipartite graphs.

There is a version of Theorem 2.1 for general graphs with a slightly weaker distance condition. A partial coloring means a proper coloring where only some of the vertices are colored.

Theorem 2.3**.**

If $G$ is a connected graph of order at least $3$, then $G$ has a partial $3$-coloring such that every vertex is within distance $3$ of every color.

Proof: Consider any partial $3$-coloring that maximizes the number of colored vertices. Note that any uncolored vertex $v$ has neighbors of every color, since it is not possible to color $v$. Thus, out of all partial $3$-colorings with the maximum number of colored vertices, take the one with the minimum number of $2$-bad vertices and let $v$ be such a vertex.

We know $v$ is colored. Furthermore, all of its neighbors are colored. If some vertex $w$ at distance $2$ from $v$ is uncolored, then $v$ is $3$-good since $N(w)$ contains all colors. So assume every vertex within distance two of $v$ is colored. Then, as in the proof of Theorem 2.1, one can re-color either $v$ or a neighbor of $v$ so that $v$ and its neighbors are $2$-good. Note that if this recoloring makes an uncolored vertex bad, then we have a contradiction of the original requirement that the maximum number of vertices were colored. So every vertex that is $2$-bad is $3$-good; or in other words, every vertex is $3$-good. $\Box$

This strengthens the first result in [1] for the case that $k=3$.

We conclude with a brief comment about graph operations. Kaul and Mitillos [7] showed that if a graph $G$ has a fall $k$-coloring and a graph $H$ has a $k$-coloring, then the cartesian product $G\Box H$ has a fall $k$-coloring. The same idea works here: if $G$ has a distance-$d$ fall $k$-coloring $f_{G}$ and $H$ has a $k$-coloring $f_{H}$, then the cartesian product $G\Box H$ has a distance-$d$ fall $k$-coloring. As per usual in the cartesian product, consider both colorings as assigning integers in the range $1$ to $k$ and take the coloring of vertex $(g,h)$ in the product to be the sum $f_{G}(g)+f_{H}(h)$ modulo $k$.

We note further that if $G$ has a distance-$d_{G}$ fall $k_{G}$-coloring and $H$ has a distance-$d_{H}$ fall $k_{H}$-coloring, then the cartesian product $G\Box H$ has a distance-$(d_{G}+d_{H})$ fall $k_{G}k_{H}$-coloring. Simply take the coloring of vertex $(g,h)$ in the product to be the ordered pair $(f_{G}(g),f_{H}(h))$.

Similar results can be shown for other products.

Acknowledgment

The authors would like to thank Ben Gobler for suggesting the concept.