Using $p$-row graphs to study $p$-competition graphs

TL;DR

This paper introduces the concept of p-row graphs and studies the set of positive integers p for which a given graph is a p-competition graph, extending previous results and characterizing competition-realizers for caterpillars.

Contribution

It generalizes row graphs to p-row graphs, introduces the condensation of graphs, and characterizes competition-realizers for various graphs including all caterpillars.

Findings

Characterization of competition-realizers for caterpillars.

Introduction of p-row graphs as a generalization.

Extension of previous results on p-competition graphs.

Abstract

For a positive integer $p$, the $p$-competition graph of a digraph $D$ is a graph which has the same vertex set as $D$ and an edge between distinct vertices $x$ and $y$ if and only if $x$ and $y$ have at least $p$ common out-neighbors in $D$. A graph is said to be a $p$-competition graph if it is the $p$-competition graph of a digraph. Given a graph $G$, we call the set of positive integers $p$ such that $G$ is a $p$-competition the competition-realizer of a graph $G$. In this paper, we introduce the notion of $p$-row graph of a matrix which generalizes the existing notion of row graph. We call the graph obtained from a graph $G$ by identifying each pair of adjacent vertices which share the same closed neighborhood the condensation of $G$. Using the notions of $p$-row graph and condensation of a graph, we study competition-realizers for various graphs to extend results given by Kim {\it et al.}~[$p$-competition graphs, {\it Linear Algebra Appl.} {\bf 217} (1995) 167--178]. Especially, we find all the elements in the competition-realizer for each caterpillar.

Equations

$$a_{ij}=\{\begin{array}{ll}1& \text{mbox}if{v}_i\in F_j;\\ 0& \text{mbox}otherwise.\end{array}$$

$$F_j=\{v_i\mid a_{ij}=1\}.$$

$$\{p\in[n]\mid G\mbox{ is a $p$-competition graph}\}$$

$${\Lambda }_M(u)=\{i\mid \text{the }i\text{th component of the row corresponding to }u\text{ in }M\text{ is }1\}.$$

$$p\le |{\Lambda }_M(u)\cap {\Lambda }_M(v)|\text{ and }p\le |{\Lambda }_M(u)\cap {\Lambda }_M(w)|.$$

$$|{\Lambda }_M(u)\cap {\Lambda }_M(v)|\le p\text{ and }|{\Lambda }_M(u)\cap {\Lambda }_M(w)|\le p.$$

$$\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& 0& 1& 1& 1& 1\\ 1& 1& 0& 0& 1& 1\end{array}\right].$$

$$\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 0& 0& 1& 1& 1& 0\\ 1& 0& 0& 0& 1& 1\end{array}\right].$$

$$V(G/\sim )=\{[u]\mid u\in V(G)\}\text{ and }E(G/\sim )=\{[u][v]\mid u,v\in V(G)\text{ and }uv\in E(G)\}.$$

$${\rm Y}(G)\supset \{p+i\mid p\in {\rm Y}(G/\sim ),i\in \{0,\dots ,n-m\}\}.$$

$$\left[\begin{array}{cccccc}1& 1& 0& 0& 1& 0\\ 0& 1& 0& 0& 1& 1\\ 0& 1& 0& 1& 0& 1\\ 1& 0& 1& 0& 1& 0\\ 0& 0& 1& 0& 1& 1\\ 0& 0& 1& 1& 0& 1\end{array}\right],$$

$$\{ST\mid S,T\subset [n],|S\cup T|\le k\}.$$

$$M=\left[\begin{array}{ccccccc}1& 1& 1& \cdots & 1& 1& 1\\ 0& 1& 1& \cdots & 1& 1& 1\\ 1& 0& 1& \cdots & 1& 1& 1\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 1& 1& 1& \cdots & 0& 1& 1\\ 1& 1& 1& \cdots & 1& 0& 1\end{array}\right].$$

$$m_{ij}=\{\begin{array}{ll}1& \text{mbox}if{v}_i\in C_j\\ 0& \text{mbox}if{v}_i\notin C_j.\end{array}$$

$$k(G)\ge {\theta }_e(G)-|V(G)|+2.$$

$$M_{p,n}:=\text{setcounter}MaxMatrixCols12\left[\begin{array}{cccccccccccc}1& 0& 0& 0& \cdots & 0& 0& 1& 1& 1& \cdots & 1\\ 1& 1& 0& 0& \cdots & 0& 0& 0& 1& 1& \cdots & 1\\ 1& 1& 1& 0& \cdots & 0& 0& 0& 0& 1& \cdots & 1\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& 1& 1& 1& \cdots & 1& 1& 0& 0& 0& \cdots & 0\\ 0& 1& 1& 1& \cdots & 1& 1& 1& 0& 0& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& 0& \cdots & 0& 1& 1& 1& 1& \cdots & 1\end{array}\right],$$

$${\rm Y}(P_n)=\{\begin{array}{ll}\{1,2\}& \text{mbox}ifn\in \{3,4\};\\ [n-3]& \text{mbox}ifn\ge 5.\end{array}$$

$$M_{2,4}^{\ast }:=\left[\begin{array}{cccc}1& 1& 0& 0\\ 1& 1& 1& 0\\ 0& 1& 1& 1\\ 0& 0& 1& 1\end{array}\right].$$

$${\rm Y}(G)\supset \{p+i\mid p\in [m-2]\text{ and }i\in [n-n^{\prime }]\cup \{0\}\}=[n-n^{\prime }+m-2].$$

$${\rm Y}(T)=\{\begin{array}{ll}[n-1]& \text{mbox}ifd(T)=2;\\ [n-2]& \text{mbox}ifd(T)=3;\\ [n-3]& \text{mbox}ifd(T)\ge 4\end{array}$$

$$p-d(T)+t-1\le k(T).$$

$$\alpha=\begin{cases}p-2&\mbox{if $d(T)=3$ and $p>2$,}\\ p-d(T)+2&\mbox{if $d(T)\geq 4$ and $p>d(T)-2$.}\end{cases}$$

$${\rm Y}(G)=\{\begin{array}{ll}[n-1]& \text{mbox}ifd(G)=2;\\ [n-2]& \text{mbox}ifd(G)=3;\\ [n-3]& \text{mbox}ifd(G)\ge 4\end{array}$$

$${\rm Y}(G/\sim )=\{\begin{array}{ll}[m-1]& \text{mbox}ifd(G/\sim )=2;\\ [m-2]& \text{mbox}ifd(G/\sim )=3;\\ [m-3]& \text{mbox}ifd(G/\sim )\ge 4,\end{array}$$

$${\rm Y}(G)\supset \{\begin{array}{ll}[n-1]& \text{mbox}ifd(G)=2;\\ [n-2]& \text{mbox}ifd(G)=3;\\ [n-3]& \text{mbox}ifd(G)\ge 4,\end{array}$$

$${\overline{\Lambda }}_M(v)\cap {\overline{\Lambda }}_M(x)={\overline{\Lambda }}_M(v)\cap {\overline{\Lambda }}_M(y)\subset {\overline{\Lambda }}_M(x)\cap {\overline{\Lambda }}_M(y).$$

$$|{\overline{\Lambda }}_M(v)|+|{\overline{\Lambda }}_M(x)|-|{\overline{\Lambda }}_M(v)\cap {\overline{\Lambda }}_M(x)|=|{\overline{\Lambda }}_M(v)\cup {\overline{\Lambda }}_M(x)|\le n-p$$

$$<|{\overline{\Lambda }}_M(x)\cup {\overline{\Lambda }}_M(y)|=|{\overline{\Lambda }}_M(x)|+|{\overline{\Lambda }}_M(y)|-|{\overline{\Lambda }}_M(x)\cap {\overline{\Lambda }}_M(y)|$$

Full text

Keywords: $p$-competition graph; $p$-edge clique cover; competition-realizer; $p$-row graph; condensation of a graph; caterpillar

2010 Mathematics Subject Classification: 05C05, 05C38, 05C62, 05C75

1 Introduction

Given a digraph $D=(V,A)$, the competition graph of $D$ is a simple graph having the same vertex set as $D$ and having an edge $uv$ if for some vertex $x\in V$, the arcs $(u,x)$ and $(v,x)$ are in $D$. The notion of competition graph is due to Cohen [1] and has arisen from ecology. Competition graphs also have applications in coding, radio transmission, and modelling of complex economic systems (see [13] and [14] for a summary of these applications and [4] for a sample paper on the modelling application). Since Cohen introduced the notion of competition graph, various variations have been defined and studied by many authors (see the survey articles by Kim [6] and Lundgren [11]). For recent work on this topic, see [3, 5, 9, 10, 15].

Kim et al. [7] introduced $p$-competition graphs as a variant of competition graph. For a positive integer $p$, the $p$-competition graph $C_{p}(D)$ of a digraph $D=(V,A)$ is defined to have the vertex set $V$ with an edge between two distinct vertices $x$ and $y$ if and only if, for some distinct vertices $a_{1},\ldots,a_{p}$ in $V$, the pairs $(x,a_{1})$, $(y,a_{1})$, $(x,a_{2})$, $(y,a_{2})$, $\ldots$, $(x,a_{p})$, $(y,a_{p})$ are arcs in $D$. Note that $C_{1}(D)$ is the ordinary competition graph, which implies that the notion of $p$-competition graph generalizes that of competition graph. A graph $G$ is called a $p$-competition graph if there exists a digraph $D$ such that $G=C_{p}(D)$. By definition, it is obvious that if a nonempty graph $G$ is a $p$-competition graph, then $p\leq|V(G)|$.

Competition graphs are closely related to edge clique covers and the edge clique cover numbers of graphs. A clique of a graph $G$ is a subset of the vertex set of $G$ that induces a complete graph. We regard an empty set as a clique of $G$ for convenience. An edge clique cover of a graph $G$ is a family of cliques of $G$ such that the end vertices of each edge of $G$ are contained in a clique in the family. The minimum size of an edge clique cover of $G$ is called the edge clique cover number of $G$, and is denoted by $\theta_{e}(G)$. Dutton and Brigham [2] characterized a competition graph in terms of its edge clique cover number.

Theorem 1.1** (​​[2]).**

A graph $G$ with $n$ vertices is a competition graph if and only if $\theta_{e}(G)\leq n$.

A $p$-competition graph $G$ can be characterized in terms of the “$p$-edge clique cover number” of $G$. For a positive integer $p$, a $p$-edge clique cover ($p$-ECC for short) of a graph $G$ is defined to be a multifamily ${\mathcal{F}}=\{F_{1},\ldots,F_{r}\}$ of subsets of the vertex set of $G$ satisfying the following:

• •

For any $J\in\binom{[r]}{p}$, the set $\bigcap_{j\in J}F_{j}$ is a clique of $G$;

• •

The collection $\left\{\bigcap_{j\in J}F_{j}\mid J\in\binom{[r]}{p}\right\}$ covers all the edges of $G$,

where $\binom{[r]}{p}$ denotes the set of $p$-element subsets of the set $\{1,\ldots,r\}$ for a positive integer $r$. The minimum size $r$ of a $p$-edge clique cover of $G$ is called the $p$-edge clique cover number of $G$, and is denoted by $\theta_{e}^{p}(G)$. The following theorem characterizes $p$-competition graphs and so generalizes Theorem 1.1.

Theorem 1.2** (​​[7]).**

A graph $G$ with $n$ vertices is a $p$-competition graph if and only if $\theta_{e}^{p}(G)\leq n$.

In this paper, we introduce the notion of $p$-row graph of a matrix which generalizes the existing notion of row graph of a matrix. We also introduce the notion of the condensation of a graph that is obtained from a graph by identifying each pair of adjacent vertices which share the same closed neighborhood. Using these notions, we study competition-realizers for various graphs to extend results given by Kim et al. [$p$-competition graphs, Linear Algebra Appl. 217 (1995) 167–178]. Especially, we find all the elements in the competition-realizer for each caterpillar.

2 $p$-row graphs and competition-realizers

In this section, we introduce the notion of $p$-row graph of a matrix which generalize the notion of row graph of a matrix and the notion of competition-realizer for a graph. Then we study competition-realizers for various graphs in terms of $p$-row graph and the condensation of a graph. Particularly, we identify the graphs with $n$ vertices whose competition-realizers contain $n$ and $n-1$, respectively.

Definition 2.1**.**

Given a positive integer $p$ and a $(0,1)$-matrix $A$, a graph $G$ is called the $p$-row graph of $A$ if the vertices of $G$ are the rows of $A$, and two vertices are adjacent in $G$ if and only if their corresponding rows have common nonzero entries in at least $p$ columns of $A$.

If $p=1$, then $G$ is called the row graph of $A$, which was introduced by Greenberg et al. [4].

Suppose that a graph $G$ is a $p$-competition graph with the vertex set $\{v_{1},\ldots,v_{n}\}$. Then there exists a $p$-ECC $\mathcal{F}=\{F_{1},\ldots,F_{m}\}$ of $G$ for a nonnegative integer $m\leq n$ by Theorem 1.2. Now we define a square matrix $A=(a_{ij})$ of order $n$ by

[TABLE]

By the definition of $p$-ECC, it is easy to see that $G$ is isomorphic to the $p$-row graph of $A$. Conversely, suppose that a graph $G$ with $n$ vertices is isomorphic to the $p$-row graph of a square $(0,1)$-matrix $A$ of order $n$. Let

[TABLE]

and let $\mathcal{F}=\{F_{1},\ldots,F_{n}\}$. By the definition of $p$-row graph, a vertex $v_{s}$ and a vertex $v_{t}$ are adjacent if and only if the $s$th row and the $t$th row of $A$ have common nonzero entries in at least $p$ columns, which is equivalent to the statement that the pair of vertices $v_{s}$ and $v_{t}$ is contained in the sets in $\mathcal{F}$ corresponding to those columns. Then, by Theorem 1.2, $G$ is a $p$-competition graph.

Now we have shown the following statement:

Theorem 2.2**.**

A graph $G$ with $n$ vertices is a $p$-competition graph if and only if $G$ is isomorphic to the $p$-row graph of a square $(0,1)$-matrix of order $n$.

For simplicity’s sake, we denote $J_{m,n}$ for the $(0,1)$-matrix of size $m$ by $n$ such that every entry is 1, $I_{n}$ for the identity matrix of order $n$, and $O_{m,n}$ for the zero matrix of size $m$ by $n$.

For a graph $G$ with $n$ vertices, we denote the set

[TABLE]

by $\Upsilon(G)$ and call it the competition-realizer for $G$.

We make the following simple but useful observations.

Proposition 2.3**.**

Let $G$ be a graph with $n$ vertices. If $G$ is empty or complete, then $\Upsilon(G)=[n]$.

Proof.

If $G$ is empty, then $G$ is a $p$-row graph of $O_{n,n}$ and so, by Theorem 2.2, is a $p$-competition graph for any $p\in[n]$. If $G$ is complete, then $G$ is a $p$-row graph of $J_{n,n}$ and so, by the same theorem, is a $p$-competition graph for any $p\in[n]$. ∎

Proposition 2.4**.**

Given a graph $G$ with $n$ vertices, suppose that $G$ is a $p$-row graph of a matrix of size $n$ by $m$ for positive integers $p$ and $m\leq n$. Then $\Upsilon(G)\supset\{p+i\mid i\in[n-m]\cup\{0\}\}$.

Proof.

Let $M$ be an $n\times m$ matrix whose $p$-row graph is $G$. Take $i\in[n-m]\cup\{0\}$. Then we add $i$ all-one columns and $n-m-i$ all-zero columns to $M$ to obtain a square matrix of order $n$ whose $(p+i)$-row graph is $G$. By Theorem 2.2, $G$ is a $(p+i)$-competition graph and so $p+i\in\Upsilon(G)$. ∎

Proposition 2.5**.**

Let $G$ be a graph and $G^{\prime}$ be a graph obtained from $G$ by adding $k$ isolated vertices for a nonnegative integer $k$. Then $\Upsilon(G^{\prime})\supset\{p+i\mid i\in[k]\cup\{0\}\}$ for each $p\in\Upsilon(G)$.

Proof.

Let $|V(G)|=n$ and suppose $p\in\Upsilon(G)$. By Theorem 2.2, $G$ is a $p$-row graph of a square $(0,1)$-matrix $M$ of order $n$. Fix $i\in[k]\cup\{0\}$. We define a square $(0,1)$-matrix $M_{i}$ of order $n+k$ as follows:

\left[\vbox{\hrule height=18.36638pt,depth=18.36638pt,width=0.0pt}\right.$$\left.\vbox{\hrule height=18.36638pt,depth=18.36638pt,width=0.0pt}\right]$$M$$J_{n,i}$$O_{n,k-i}$$O_{k,n+k}$$M_{i}=.

Obviously, the $(p+i)$-row graph of $M_{i}$ is $G$ together with $k$ isolated vertices. By Theorem 2.2, $p+i\in\Upsilon(G^{\prime})$. ∎

For a $p$-row graph $G$ of a matrix $M$ and a vertex $u$ of $G$, we let

[TABLE]

A vertex $v$ of a graph $G$ is called a simplicial vertex if its neighbors form a clique in $G$.

Proposition 2.6**.**

Let $G$ be a $p$-row graph of a matrix $M$. Then, for a non-isolated non-simplicial vertex $u$, $|\Lambda_{M}(u)|\geq p+1$.

Proof.

By the condition on $u$, $u$ is adjacent to two nonadjacent vertices $v$ and $w$. Then

[TABLE]

Suppose that $|\Lambda_{M}(u)|\leq p$. Then

[TABLE]

Thus $|\Lambda_{M}(u)\cap\Lambda_{M}(v)|=|\Lambda_{M}(u)\cap\Lambda_{M}(w)|=p$ and so $\Lambda_{M}(u)\cap\Lambda_{M}(v)=\Lambda_{M}(u)\cap\Lambda_{M}(w)=\Lambda_{M}(u)$. Hence $\Lambda_{M}(u)\subset\Lambda_{M}(v)\cap\Lambda_{M}(w)$ and so $|\Lambda_{M}(v)\cap\Lambda_{M}(w)|\geq p$, which is a contradiction. ∎

The following proposition characterizes a graph $G$ with $n$ vertices and $n\in\Upsilon(G)$.

We denote a set of $m$ isolated vertices by $\mathcal{I}_{m}$. Technically, we let $\mathcal{I}_{0}=\emptyset$ and $K_{0}=\emptyset$.

Proposition 2.7**.**

Let $G$ be a graph with $n$ vertices. Then $G$ is an $n$-competition graph if and only if $G\cong K_{m}\cup\mathcal{I}_{n-m}$ for some integer $m\in\{0,\ldots,n\}$.

Proof.

By definition, $G$ is an $n$-competition graph if and only if $n\in\Upsilon(G)$. To show the “if” part, suppose that $G\cong K_{m}\cup\mathcal{I}_{n-m}$ for some integer $m$, $0\leq m\leq n$. By Proposition 2.3, $m\in\Upsilon(K_{m})$. By Proposition 2.5, $m+(n-m)\in\Upsilon(G)$. To show the “only if” part, suppose that $G$ is an $n$-competition graph. Then $G$ is isomorphic to the $n$-row graph of a matrix $M$ of order $n$ by Theorem 2.2. Suppose that $u$ is a non-isolated vertex in $G$. Then $u$ is adjacent to a vertex $v$ in $G$, so $|\Lambda_{M}(u)\cap\Lambda_{M}(v)|=n$. Thus we may conclude that each row of $M$ corresponding to a non-isolated vertex is the all-one vector in $\mathbb{R}^{n}$. Thus the subgraph of $G$ induced by non-isolated vertices is a clique. Hence $G=K_{m}\cup\mathcal{I}_{n-m}$ where $m$ is the number of non-isolated vertices in $G$. ∎

Corollary 2.8**.**

Suppose that a graph $G$ with $n$ vertices has no isolated vertices. Then $G$ is an $n$-competition graph if and only if $G=K_{n}$.

Corollary 2.9**.**

Let $G$ be a graph with $n$ vertices. Then $\Upsilon(G)=[n]$ if and only if $G\cong K_{m}\cup\mathcal{I}_{n-m}$ for some $m\in\{0,\ldots,n\}$.

Proof.

The “only if” part immediately follows by Proposition 2.7. To show the “if” part, suppose that $G\cong K_{m}\cup\mathcal{I}_{n-m}$ for some $m$, $0\leq m\leq n$. Let $M$ be a square $(0,1)$-matrix of order $n$ such that the first $m$ rows are all-one vector and the other $n-m$ rows are all-zero vector. Then it is easy to check that the $p$-row graph of $M$ is isomorphic to $G$ for each $p\in[n]$. ∎

Corollary 2.10**.**

If $G$ is a disjoint union of complete graphs with $n$ vertices, then $\Upsilon(G)\supset[n-1]$.

Proof.

Let $X_{1},\ldots,X_{k}$ be the complete components of $G$. Then $k\leq n$. Now fix $i\in[n-1]$. Then $k\leq{n\choose i}$. Let $S_{1},\ldots,S_{k}$ be distinct $i$-subsets of $[n]$. Let $M$ be a matrix of order $n$ satisfying $\Lambda_{M}(v)=[n]\setminus S_{j}$ for each vertex $v$ in $X_{j}$ for each $j\in[k]$. Then it is easy to check that $G$ is the $(n-i)$-row graph of $M$ and so $n-i\in\Upsilon(G)$ by Theorem 2.2. ∎

Example 2.11**.**

The competition-realizer may be empty for some graph. For example, for the complete bipartite graph $K_{3,3}$, $\Upsilon(K_{3,3})=\emptyset$. To see why, we note that the number of vertices of $K_{3,3}$ is $6$ and $\theta_{e}(K_{3,3})=9$. Therefore $1\notin\Upsilon(K_{3,3})$ by Theorem 1.2. By Proposition 2.6, $5\notin\Upsilon(K_{3,3})$ and $6\notin\Upsilon(K_{3,3})$. Suppose that $K_{3,3}$ is a $p$-competition graph for some $p\in\{2,3,4\}$. Then $G$ is isomorphic to the $p$-row graph of a square $(0,1)$-matrix $M_{p}$ of order $6$ by Theorem 2.2.

Consider the case $p=4$. Then each row of $M_{4}$ contains at least five $1$s by Proposition 2.6. This implies that any two rows of $M_{4}$ have at least four common $1$s and so $G$ is isomorphic to $K_{6}$, which is a contradiction. Thus $4\notin\Upsilon(K_{3,3})$.

Now consider the case $p=3$. Then each row of $M_{3}$ contains at least four $1$s by Proposition 2.6. This implies that any two rows of $M_{3}$ have at least two common $1$s. If there is a row containing at least five $1$s, then it shares at least three common $1$s with each of the other vertices, which is impossible. Thus each row of $M_{3}$ contains exactly four $1$s. Since $K_{3,3}$ has a partite set of size $3$, we may assume that $M_{3}$ contains the following submatrix by permuting columns, if necessary:

[TABLE]

Now we take a vertex $u$ in the other partite set. Then $u$ is adjacent to the vertex corresponding to each row of the above submatrix. To have $u$ and the vertex corresponding to the first row of the above submatrix be adjacent, $\Lambda_{M_{3}}(u)\cap\{1,2,3,4\}$ is one of $\{1,2,3\},\{2,3,4\},\{1,2,4\}$, $\{1,3,4\}$, and $\{1,2,3,4\}$. Then, in case of $\{1,2,3,4\}$, $u$ is not adjacent to the vertices corresponding to the second row and third row; in case of $\{1,2,3\}$ or $\{1,2,4\}$, $u$ is not adjacent to the vertex corresponding to the second row; in case of $\{1,3,4\}$ or $\{2,3,4\}$, $u$ is not adjacent to the vertex corresponding to the third row. Therefore we reach a contradiction. Thus $3\notin\Upsilon(K_{3,3})$.

Now consider the case $p=2$. Then each row of $M_{2}$ contains at least three $1$s by Proposition 2.6. Suppose that there is a row ${\bf r}_{1}$ containing at least four $1$s. Let $v_{1}$ be the vertex corresponding to ${\bf r}_{1}$ and $v_{2}$ and $v_{3}$ be the other vertices in the partite set to which $v_{1}$ belongs. We may assume that $\Lambda_{M_{2}}(v_{1})\supset\{1,2,3,4\}$. Since $v_{1}$ and $v_{2}$ are not adjacent, ${\bf r}_{1}$ has exactly four $1$s and we may assume that the row ${\bf r}_{2}$ corresponding to $v_{2}$ has $1$ in the fourth component through the sixth component. Then the row corresponding to $v_{3}$ must share at least two $1$s with ${\bf r}_{1}$ or ${\bf r}_{2}$ and we reach a contradiction. Thus each row of $M_{2}$ contains exactly three $1$s. If there are two vertices $w_{1}$ and $w_{2}$ in a partite set $W$ such that their corresponding rows do not share $1$s, then the row corresponding to the remaining vertex in $W$ must share at least two $1$s with one of the rows corresponding to $w_{1}$ and $w_{2}$, and we reach a contradiction. Therefore the rows corresponding to two vertices in the same partite set share exactly one $1$. Thus we may assume that $M_{2}$ contains the following submatrix by permuting columns, if necessary:

[TABLE]

Now we take a vertex $x$ in the other partite set $X$. Then $x$ is adjacent to each of the vertices corresponding to the rows of the above submatrix. Therefore $\Lambda_{M_{2}}(x)=\{1,3,5\}$. Since $x$ is arbitrarily chosen, the rows corresponding to the other two vertices in $X$ also have $1$ in the first, the third, and the fifth component, which is impossible. Hence we have shown that $\Upsilon({K_{3,3}})=\emptyset$.

Now we introduce the notion of condensation of a graph. Let $G$ be a graph with $n$ vertices. Two vertices $u$ and $v$ of $G$ are said to be homogeneous, denoted by $u\sim v$, if they have the same closed neighborhood. Clearly $\sim$ is an equivalence relation on $V(G)$. We denote the equivalence class containing a vertex $u$ of $G$ by $[u]$. Then we define a new simple graph $G/\!\!\sim$ for $G$ by

[TABLE]

See Figure 1 for an illustration.

We call $G/\!\!\sim$ the condensation of $G$ for a graph $G$.

We note the following:

$\Leftrightarrow$ Two vertices $u$ and $v$ are adjacent in $G$ $\Leftrightarrow$ $v\in N_{G}[u]$

$\Leftrightarrow$ $v\in N_{G}[u^{\prime}]$ for any $u^{\prime}\in[u]$ $\Leftrightarrow$ $u^{\prime}\in N_{G}[v]$ for any $u^{\prime}\in[u]$

$\Leftrightarrow$ $u^{\prime}\in N_{G}[v^{\prime}]$ for any $u^{\prime}\in[u]$ and $v^{\prime}\in[v]$

$\Leftrightarrow$ $u^{\prime}$ and $v^{\prime}$ are adjacent in $G$ for any $u^{\prime}\in[u]$ and $v^{\prime}\in[v]$.

Therefore $G/\!\!\sim$ is well-defined.

It is obvious that

• ($\star$)

for each isolated vertex in $G$, its equivalence class is isolated in $G/\!\!\sim$.

The notion of $p$-row graph provides a way of getting information on the competition-realizer for a graph $G$ from the competition-realizer for a simpler graph $G/\!\!\sim$ as seen in the following results.

Proposition 2.12**.**

The condensation of $G$ has the same diameter as $G$ for a connected non-complete graph $G$.

Proof.

Let $m$ be the diameter of $G$. Since $G$ is not complete, $m\geq 2$. Then there exists an induced $(u,v)$-path of length $m$ for some vertices $u$ and $v$ and there is no induced $(u,v)$-path of length $l$ for any $l<m$. Thus, by the definition of condensation of a graph, there exists an induced $([u],[v])$-path of length $m$ and there is no induced $([u],[v])$-path of length $l$ for any $2\leq l<m$ in $G/\!\!\sim$. If there exists a $([u],[v])$-path of length $1$, then $u$ and $v$ are adjacent, which contradicts the choice of $u$ and $v$ so that $d_{G}(u,v)=m\geq 2$. Therefore the diameter of $G/\!\!\sim$ is greater than or equal to $m$.

Let $m^{\prime}$ be the diameter of $G/\!\!\sim$ for a nonnegative integer $m^{\prime}$. Since $G$ is non-complete, $m^{\prime}\geq 1$. Then there exists an induced $([x],[y])$-path of length $m^{\prime}$ in $G/\!\!\sim$ and there is no induced $([x],[y])$-path of length $l$ for any $l<m^{\prime}$ and some vertices $x$ and $y$ in $G$. Therefore, by the definition of condensation of a graph, there is an induced $(x,y)$-path of length $m^{\prime}$ and there is no induce $(x,y)$-path of length $l$ for any $l<m^{\prime}$ in $G$. Thus the diameter of $G/\!\!\sim$ is less than or equal to $m$ and this completes the proof. ∎

Proposition 2.13**.**

A graph $G$ is a $p$-competition graph if and only if there exists a square matrix $M$ such that $G$ is a $p$-row graph of $M$ and the rows corresponding to two homogeneous vertices are identical.

Proof.

The “if” part is obvious. To show the “only if” part, suppose that a graph $G$ is a $p$-competition graph for some positive integer $p$. Then, by Theorem 2.2, there exists a square matrix $M^{\prime}$ such that $G$ is a $p$-row graph of $M^{\prime}$. If there are at least two homogeneous vertices, then we fix one row among the rows corresponding to them and replace the remaining rows with the fixed row. We denote by $M$ the matrix obtained by repeatedly applying the above procedure. It is easy to see that $G$ is a $p$-row graph of $M$. ∎

Proposition 2.14**.**

Given a graph $G$ with $n$ vertices, suppose that $G/\!\!\sim$ is a $p$-row graph of a matrix $M$ satisfying the property that $M$ has $m$ columns for a positive integer $m\leq n$ and every row of $M$ has at least $p$ $1$s. Then $\Upsilon(G)\supset\{p+i\mid i\in\{0,\ldots,n-m\}\}$.

Proof.

Let $n_{j}$ be the size of equivalence class under $\sim$ corresponding to the $j$th row of $M$ for each $j$, $1\leq j\leq m$. We replace the $j$th row of $M$ with $n_{j}$ copies of it to obtain the matrix $M^{*}$ which contains $M$ as a submatrix. We note that the size of $M^{*}$ is $n\times m$. Suppose that $u$ and $v$ are two distinct vertices in $G$. Then let $\mathbf{r}_{u}$ and $\mathbf{r}_{v}$ be the rows of $M^{*}$ corresponding to $u$ and $v$, respectively. If $u$ and $v$ are not homogenous, then they belong to distinct equivalence classes under $\sim$ and the following are true:

• $\Leftrightarrow$

Two vertices $u$ and $v$ are adjacent in $G$

• $\Leftrightarrow$

$[u]$ and $[v]$ are adjacent in $G/\!\!\sim$

• $\Leftrightarrow$

the row corresponding to $[u]$ and the row corresponding to $[v]$ have at least $p$ common $1$s in $M$

• $\Leftrightarrow$

$u$ and $v$ are adjacent in the $p$-row graph of $M^{*}$.

Suppose that $u$ and $v$ are homogenous. Then the rows ${\bf r}_{u}$ and ${\bf r}_{v}$ are identical. By the hypothesis, every row of $M$ has at least $p$ $1$s. Thus ${\bf r}_{u}$ and ${\bf r}_{v}$ have at least $p$ common $1$s and so $u$ and $v$ are adjacent in the $p$-row graph of $M^{*}$. Hence $G$ is the $p$-row graph of $M^{*}$ and so, by Proposition 2.4, $G$ is a $(p+i)$-competition graph, that is, $p+i\in\Upsilon(G)$ for any $i\in\{0,\ldots,n-m\}$. ∎

For a positive integer $p$ and the $p$-row graph $G$ of a matrix $M$, each non-isolated vertex in $G$ has at least $p$ $1$s in the row of $M$ corresponding to it and so the following corollary is immediately true by the above proposition.

Corollary 2.15**.**

Given a graph $G$ with $n$ vertices, suppose that $G/\!\!\sim$ has no isolated vertices and is a $p$-row graph of a matrix $M$ having $m$ columns for a positive integer $m\leq n$. Then $\Upsilon(G)\supset\{p+i\mid i\in\{0,\ldots,n-m\}\}$.

Corollary 2.16**.**

Given a graph $G$ with $n$ vertices, suppose that the number of non-isolated vertices in $G/\!\!\sim$ is $m$ for a positive integer $m\leq n$. Then

[TABLE]

Proof.

Suppose $p\in\Upsilon(G)$. Then, by Theorem 2.2, $G/\!\!\sim$ is a $p$-row graph of a matrix $M$ having $m$ columns. Thus by Corollary 2.15, $p+i\in\Upsilon(G)$ for any $i\in\{0,\ldots,n-m\}$. ∎

Remark 2.17**.**

Even if $G$ is a $p$-competition graph, $G/\!\!\sim$ may not be a $(p-i)$-competition graph for some $i\in[n-m]\cup\{0\}$ where $n=|V(G)|$ and $m=|V(G/\!\!\sim)|$. For example, the graph $G_{2}$ in Figure 1 is a $2$-competition graph. To see why, we note that $G_{1}$ is the $2$-row graph of the matrix

[TABLE]

so $G_{1}$ is a $2$-competition graph by Theorem 2.2. Since $G_{1}$ is isomorphic to $G_{2}/\!\!\sim$ and has no isolated vertices, $G_{2}$ is a $2$-competition graph by Corollary 2.16. Yet, $G_{1}$, which is isomorphic to $G_{2}/\!\!\sim$, is not a $1$-competition graph by Theorem 1.1 since $|V(G_{1})|=6<7=|E(G_{1})|=\theta_{e}(G_{1})$.

A union $G\cup H$ of two graphs $G$ and $H$ is the graph having its vertex set $V(G)\cup V(H)$ and edge set $E(G)\cup E(H)$. In this paper, the union of $G$ and $H$ means their disjoint union which has an additional condition $V(G)\cap V(H)=\emptyset$. A join $G\vee H$ of two vertex-disjoint graphs $G$ and $H$ is the graph having its vertex set $V(G)\cup V(H)$ and edge set $E(G)\cup E(H)\cup\{uv\mid u\in V(G),v\in V(H)\}$.

For a positive integer $n$, a nonnegative integer $k\leq n$, and the power set $\mathcal{P}([n])$ of $[n]$, we denote by ${\Psi}_{n,k}$ the simple graph with the vertex set $\mathcal{P}([n])$ and the edge set

[TABLE]

Theorem 2.18**.**

Let $G$ be a connected graph with $n$ vertices and $k$ be a nonnegative integer less than $n$. Then $G$ is an $(n-k)$-competition graph if and only if $G/\!\!\sim$ is isomorphic to an induced subgraph of $\Psi_{n,k}$.

Proof.

To show the “only if” part, suppose that $G$ is an $(n-k)$-competition graph. Then, by Proposition 2.13, there exists a square matrix $M$ of order $n$ such that $G$ is an $(n-k)$-row graph of $M$ and the rows corresponding to two homogeneous vertices are identical. Let $M^{\prime}$ be a submatrix of $M$ obtained by taking all the distinct rows of $M$. Then obviously $G/\!\!\sim$ is an $(n-k)$-row graph of $M^{\prime}$. Therefore two distinct vertices $[x]$ and $[y]$ are adjacent in $G/\!\!\sim$ if and only if $|\Lambda_{M^{\prime}}([x])\cap\Lambda_{M^{\prime}}([y])|\geq n-k$ if and only if $|([n]\setminus\Lambda_{M^{\prime}}([x]))\cup([n]\setminus\Lambda_{M^{\prime}}([y]))|\leq k$ if and only if $[n]\setminus\Lambda_{M^{\prime}}([x])$ and $[n]\setminus\Lambda_{M^{\prime}}([y])$ are adjacent in $\Psi_{n,k}$. Hence we have shown that $G/\!\!\sim$ is isomorphic to an induced subgraph of $\Psi_{n,k}$.

To show the “if” part, suppose that $G/\!\!\sim$ is isomorphic to an induced subgraph of $\Psi_{n,k}$. Then each vertex $[v]$ of $G/\!\!\sim$ is assigned a subset $S_{v}$ of $[n]$ so that two distinct vertices $[v]$ and $[w]$ are adjacent in $G/\!\!\sim$ if and only if $|S_{v}\cup S_{w}|\leq k$. If $|V(G/\!\!\sim)|=1$, then $G$ is complete and so $\Upsilon(G)=[n]$ by Proposition 2.3. Thus, if $|V(G/\!\!\sim)|=1$, then $n-k\in\Upsilon(G)$. Now suppose that $|V(G/\!\!\sim)|\geq 2$. Since $G$ is connected by the hypothesis, $G/\!\!\sim$ is connected and so every vertex $[v]$ in $G/\!\!\sim$ is adjacent to a vertex, which implies $|S_{v}|\leq k$. We denote by $M^{\prime\prime}$ the matrix with $n$ columns and with each row corresponds to a vertex of $G/\!\!\sim$ in such a way that $[n]\setminus\Lambda_{M^{\prime\prime}}([v])=S_{v}$. Then it is easy to see that $G/\!\!\sim$ is an $(n-k)$-row graph of $M^{\prime\prime}$. By Corollary 2.16, we can conclude that $n-k\in\Upsilon(G)$. ∎

Lemma 2.19**.**

The star graph $K_{1,n}$ is an $n$-competition graph for a positive integer $n$.

Proof.

It is obvious that $K_{1,n}$ is the $n$-row graph of the following square matrix of order $n+1$:

[TABLE]

∎

The following proposition characterizes a graph $G$ with $n$ vertices and $n-1\in\Upsilon(G)$.

Proposition 2.20**.**

Let $G$ be a graph with $n$ vertices. Then $G$ is an $(n-1)$-competition graph if and only if $G\cong(K_{n_{0}}\vee(K_{n_{1}}\cup\cdots\cup K_{n_{k}}))\cup\mathcal{I}_{m}$ for some nonnegative integers $k,n_{0},n_{1},\ldots,n_{k}$, and $m$ satisfying $m+\sum_{i=0}^{k}n_{i}=n$.

Proof.

We show the “if” part. If $G$ is empty, then, by Proposition 2.3, $G$ is an $(n-1)$-competition graph. Now suppose that $G$ is a nonempty graph. Let $G^{\prime}$ be the subgraph of $G$ resulting from deleting all the isolated vertices in $G$. It is easy to check that $G^{\prime}/\!\!\sim$ is an empty graph or a star graph by the hypothesis. Suppose that $|V(G^{\prime}/\!\!\sim)|=1$. Then $G^{\prime}$ is a complete graph and so $\Upsilon(G^{\prime})=\{1,\ldots,|V(G^{\prime})|\}$. By Proposition 2.5, $\Upsilon(G)=[n]$ and so $G$ is an $(n-1)$-competition graph. Now suppose that $|V(G^{\prime}/\!\!\sim)|\geq 2$. If $G^{\prime}/\!\!\sim$ is an empty graph, then $G^{\prime}$ is a disjoint union of complete graphs and so, by Corollary 2.10 and Proposition 2.5, $G$ is an $(n-1)$-competition graph. Suppose that $G^{\prime}/\!\!\sim$ is not an empty graph. Then, by Proposition 2.3 and Lemma 2.19, $|V(G^{\prime}/\!\!\sim)|-1\in\Upsilon(G^{\prime}/\!\!\sim)$. Then $|V(G^{\prime})|-1\in\Upsilon(G^{\prime})$ by Corollary 2.16. By Proposition 2.5, $G$ is an $(n-1)$-competition graph.

Now we show the “only if” part. Suppose that $G$ is an $(n-1)$-competition graph. Then $G$ is isomorphic to the $(n-1)$-row graph of a matrix $M$ of order $n$ by Theorem 2.2. Suppose that there exists a row, say $\mathbf{r}$, of $M$ such that $\mathbf{r}$ contains at most $n-2$ $1$s. Then the vertex in $G$ corresponding to $\mathbf{r}$ is an isolated vertex, so $G$ is still the $(n-1)$-row graph of the matrix resulting from replacing $\mathbf{r}$ with all-zero row. Therefore we may assume that $M$ contains the rows of exactly three types:

the row with $n$ $1$s;

• 2.

the row with $n-1$ $1$s;

• 3.

the row with [math] $1$s.

Let $V_{1}$, $V_{2}$, and $V_{3}$ be the sets of vertices corresponding to rows of Type i for each $i=1,2,3$. Then the vertex set of $G$ is partitioned into $V_{1}$, $V_{2}$, and $V_{3}$. Obviously, each vertex in $V_{1}$ is adjacent to each vertex of $G$ not in $V_{3}$ and each vertex in $V_{3}$ is isolated. We note that two rows of Type 2 are identical if and only if their corresponding vertices in $V_{2}$ are adjacent in $G$. Therefore each vertex in $V_{2}$ is a simplicial vertex, that is, a vertex whose neighbors form a clique. Let $n_{0}=|V_{1}|$ and $m=|V_{3}|$. If $V_{2}=\emptyset$, then $G\cong K_{n_{0}}\cup\mathcal{I}_{m}$ by the above observation. Now suppose that $V_{2}\neq\emptyset$. Then the subgraph of $G$ induced by $V_{2}$ is a disjoint union of cliques. Let $W_{1},W_{2},\ldots,W_{k}$ be the vertex sets of those cliques and let $|W_{i}|=n_{i}$ for $1\leq i\leq k$. Then $m+\sum_{i=0}^{k}n_{i}=n$. By the above observation again, $G\cong(K_{n_{0}}\vee(K_{n_{1}}\cup\cdots\cup K_{n_{k}}))\cup\mathcal{I}_{m}$. ∎

Theorem 2.21**.**

Let $G$ be a graph with $n$ vertices. Then $\Upsilon(G)=[n-1]$ if and only if $G\cong H\cup\mathcal{I}_{m}$ for some integer $m$, $0\leq m\leq n-3$ and some graph $H$ for which $H/\!\!\sim$ is an induced subgraph of a star graph and has more than one vertex.

Proof.

To show the “if” part, suppose that $G\cong H\cup\mathcal{I}_{m}$ for some integer $m$, $0\leq m\leq n-3$ and some graph $H$ for which $H/\!\!\sim$ is an induced subgraph of a star graph $Q$ with more than one vertex. We denote the number of vertices in $H/\!\!\sim$ by $t$. Take $p\in[t-1]$. We construct a square $(0,1)$-matrix $M$ of order $t$ in the following way. If $H/\!\!\sim$ contains a center of $Q$, then the row of $M$ corresponding to it is the all-one vector. The rows of $M$ corresponding to the vertices in $H/\!\!\sim$ which are not a center of $Q$ are mutually distinct, and the number of $1$s in each of them is $p$. Such a matrix $M$ exists since ${t\choose p}\geq t$. It is easy to check that $H/\!\!\sim$ is isomorphic to the $p$-row graph of $M$. Thus $[t-1]\subset\Upsilon(H/\!\!\sim)$ by Theorem 2.2. By Proposition 2.14, $[n-m-1]\subset\Upsilon(H)$. Now, by Proposition 2.5, $[n-1]\subset\Upsilon(G)$. By Proposition 2.7, $n\notin\Upsilon(G)$ and so $\Upsilon(G)=[n-1]$.

To show the “only if” part, suppose that $\Upsilon(G)=[n-1]$. Then, by Proposition 2.20, $G\cong(K_{n_{0}}\vee(K_{n_{1}}\cup\cdots\cup K_{n_{k}}))\cup\mathcal{I}_{m}$ for some nonnegative integers $k,n_{0},n_{1},\ldots,n_{k}$, and $m$ satisfying $m+\sum_{i=0}^{k}n_{i}=n$. If there is at most one nonzero integer among $n_{1},n_{2},\ldots,n_{k}$, then $G\cong K_{n-l_{1}}\cup\mathcal{I}_{l_{1}}$ for some integer $l_{1}$, $0\leq l_{1}\leq n$ and so, by Corollary 2.9, $\Upsilon(G)=[n]$, which is a contradiction. Therefore there are at least two nonzero integers among $n_{1},n_{2},\ldots,n_{k}$ and $H:=K_{n_{0}}\vee(K_{n_{1}}\cup\cdots\cup K_{n_{k}})$ is an induced subgraph of $G$. It is easy to check that $H/\!\!\sim$ is an induced subgraph of a star graph and has more than one vertex. Finally we show that $m\leq n-3$. Suppose that there are exactly two nonzero integers among $n_{0},n_{1},n_{2},\ldots,n_{k}$. If $n_{0}$ and $n_{i}$ are nonzero integers for some $i$, $1\leq i\leq k$, then $G\cong K_{n_{0}+n_{i}}\cup\mathcal{I}_{l_{2}}$ for some integer $l_{2}$, $0\leq l_{2}\leq n$ and we reach a contradiction as before. If $n_{i}$ and $n_{j}$ are nonzero integers for some $i$ and $j$, $1\leq i<j\leq k$, then $G\cong K_{n_{i}}\cup K_{n_{j}}\cup\mathcal{I}_{l_{3}}$ for some integer $l_{3}$, $0\leq l_{3}\leq n$, which implies $n_{i}\geq 2$ and $n_{j}\geq 2$. Therefore $m\leq n-4$ if there are exactly two nonzero integers among $n_{0},n_{1},n_{2},\ldots,n_{k}$. If there are at least three nonzero integers among $n_{0},n_{1},n_{2},\ldots,n_{k}$, then it is obvious that $m\leq n-3$. ∎

A hole of a graph is a cycle of length greater than or equal to $4$ which is an induced subgraph of the graph. A graph without holes is said to be chordal.

For $p=n$ or $n-1$, a $p$-competition graph is a chordal graph by Propositions 2.7 and 2.20. As a matter of fact, an $(n-2)$-competition graph is also chordal by Corollary 2.24.

An induced path of a graph means a path as an induced subgraph of the graph.

Theorem 2.22**.**

If a graph $G$ with $n$ vertices contains two internally disjoint induced paths of length $2$ (whose internal vertices are nonadjacent), then $\Upsilon(G)\subset[n-3]$.

Proof.

Let $G$ be a graph with $n$ vertices containing two internally disjoint induced paths $uvw$ and $xyz$ of length $2$ with $v$ and $y$ nonadjacent (see Figure 2). Then, by Propositions 2.7 and 2.20, $\Upsilon(G)\subset[n-2]$. Suppose, to the contrary, that $n-2\in\Upsilon(G)$. By Theorem 2.2, $G$ is isomorphic to the $(n-2)$-row graph of a square matrix $M$ of order $n$. If $|\Lambda_{M}(v)|\geq n-1$ and $|\Lambda_{M}(y)|\geq n-1$, then $|\Lambda_{M}(v)\cap\Lambda_{M}(y)|\geq n-2$ and so $v$ and $y$ are adjacent, which is impossible. Thus $|\Lambda_{M}(v)|\leq n-2$ or $|\Lambda_{M}(y)|\leq n-2$. Without loss of generality, we may assume that $|\Lambda_{M}(v)|\leq n-2$. Since $v$ is non-isolated, $|\Lambda_{M}(v)|=n-2$. Since $u$ and $v$ (resp. $w$ and $v$) are adjacent, $|\Lambda_{M}(u)\cap\Lambda_{M}(v)|\geq n-2$ (resp. $|\Lambda_{M}(w)\cap\Lambda_{M}(v)|\geq n-2$). Since $|\Lambda_{M}(v)|=n-2$, $\Lambda_{M}(v)\subset\Lambda_{M}(u)$ and $\Lambda_{M}(v)\subset\Lambda_{M}(w)$. Therefore $\Lambda_{M}(v)\subset\Lambda_{M}(u)\cap\Lambda_{M}(w)$ and so $|\Lambda_{M}(u)\cap\Lambda_{M}(w)|\geq n-2$. Then $u$ and $w$ are adjacent in $G$ and we reach a contradiction. ∎

Since a graph with the diameter at least four has two internally disjoint induced paths of length $2$, the following corollary is immediately true.

Corollary 2.23**.**

If the diameter of a graph $G$ is at least four, then $\Upsilon(G)\subset[n-3]$.

Corollary 2.24**.**

If $\Upsilon(G)\not\subset[n-3]$ for a graph $G$ with $n\geq 4$ vertices, then $G$ is chordal and has no induced path of length $4$.

Proof.

We will show that the contrapositive of the statement is true. If a $p$-competition graph with $n$ vertices is non-chordal or has an induced path of length $4$ for integers $n\geq 4$ and $p\in[n]$, then it contains two internally disjoint induced paths of length $2$ whose internal vertices are nonadjacent and the statement is true by Theorem 2.22. ∎

By Corollary 2.8, it is trivially true that if a connected graph $G$ with $n$ vertices is an $n$-competition graph, then the diameter of $G$ is $1$. By Proposition 2.20, the diameter of a connected $(n-1)$-competition graph which has $n$ vertices is at most $2$. The diameter of a connected $(n-2)$-competition graph which has $n$ vertices is at most $3$ by Corollary 2.24. However, interestingly, the diameter of a connected $(n-3)$-competition graph with $n$ vertices can be arbitrarily large, which will be shown by Proposition 3.4.

Proposition 2.25**.**

For a graph $G$ with $\theta_{e}(G)\leq|V(G)|$, $[|V(G)|-\theta_{e}(G)+1]\subset\Upsilon(G)$.

Proof.

Let $|V(G)|=n$ and $V(G)=\{v_{1},v_{2},\ldots,v_{n}\}$. There is an edge clique cover $\mathcal{C}:=\{C_{1},C_{2},\ldots,C_{\theta_{e}(G)}\}$ of $G$ as $\theta_{e}(G)$ is the edge clique number of $G$. We define an $n\times\theta_{e}(G)$ matrix $M=(m_{ij})$ as follows:

[TABLE]

Then $G$ is isomorphic to the $1$-row graph of $M$. Therefore the statement is true by Proposition 2.4. ∎

It is a well-known fact that any graph $G$ can be made into the competition graph of an acyclic digraph as long as it is allowed to add new isolated vertices to $G$. The smallest among such numbers is called the competition number of $G$ and denoted by $k(G)$. Opsut [12] showed that

[TABLE]

Corollary 2.26**.**

Let $G$ be a graph with $\omega$ components. If each component of $G$ has competition number one, then $[\omega+1]\subset\Upsilon(G)$.

Proof.

Let $G_{1}$, $G_{2}$, $\ldots$, $G_{\omega}$ be the components of $G$. Then, by (2), $|V(G_{i})|-\theta_{e}(G_{i})\geq 2-k(G_{i})=1$ for any $1\leq i\leq\omega$. Since $|V(G)|=\sum_{i=1}^{\omega}|V(G_{i})|$ and $\theta_{e}(G)=\sum_{i=1}^{\omega}\theta_{e}(G_{i})$, $|V(G)|-\theta_{e}(G)+1\geq\omega+1$. Thus, by Proposition 2.25, the corollary statement is true. ∎

It is known that a chordal graph and a forest both have the competition number at most $1$. Since any graph without isolated vertex has competition number at least $1$, the following corollaries immediately follow from Corollary 2.26.

Corollary 2.27**.**

For a chordal graph $G$ having $\omega$ components none of which is trivial, $[\omega+1]\subset\Upsilon(G)$.

Corollary 2.28**.**

For a forest $G$ having $\omega$ components none of which is trivial, $[\omega+1]\subset\Upsilon(G)$.

3 Competition-realizers for trees

In this section, we study $p$-competition trees. Especially, we completely characterize the competition-realizers for caterpillars.

Let $G$ be a $p$-competition graph. Then $G$ is isomorphic to the $p$-row graph of a matrix $M=(m_{ij})$. If $|\Lambda_{M}(v)|\leq p-1$, then $v$ is an isolated vertex in $G$, and so $G$ is still the $p$-row graph of the resulting matrix even if the row corresponding to $v$ is replaced by the row with $p-1$ $1$s. Thus we may conclude as follows:

• $(\S)$

If a $p$-competition graph $G$ is isomorphic to the $p$-row graph of a matrix $M$, then we may assume that $|\Lambda_{M}(v)|\geq p-1$ for each vertex $v$ in $G$.

Adding a pendant vertex $v$ to a graph $G$ means obtaining a graph $G^{\prime}$ such that $v\notin V(G)$, $V(G^{\prime})=V(G)\cup\{v\}$, and $E(G^{\prime})=E(G)\cup\{vu\}$ for a vertex $u$ in $G$.

Theorem 3.1**.**

If $G$ is a graph obtained from a $p$-competition graph by adding a pendant vertex, then $p\in\Upsilon(G)$.

Proof.

Let $G$ be a graph obtained from a $p$-competition graph $G^{\prime}$ with $n$ vertices by adding a pendant vertex $u$ at vertex $v$ in $G^{\prime}$. Since $G^{\prime}$ is a $p$-competition graph, $G^{\prime}$ is isomorphic to the $p$-row graph of a square matrix $M^{\prime}=(m^{\prime}_{ij})$ of order $n$. By ($\S$), we may assume that $|\Lambda_{M^{\prime}}(v)|\geq p-1$. Without loss of generality, we may assume that the row corresponding to $v$ is located at the bottom of $M^{\prime}$ and $\Lambda_{M^{\prime}}(v)=\{1,2,\ldots,|\Lambda_{M^{\prime}}(v)|\}$.

Now we define a matrix $M=(m_{ij})$ of order $n+1$ by

\left[\vbox{\hrule height=31.36612pt,depth=31.36612pt,width=0.0pt}\right.$$\left.\vbox{\hrule height=31.36612pt,depth=31.36612pt,width=0.0pt}\right]$$M^{\prime}[math] $\vdots$ [math] 1$$1\cdots 1\;0\cdots 0$$1$$u$$v$$M=$$p-1$$n-p+1$$1

It is easy to check that $G$ is the $p$-row graph of $M$. By Theorem 2.2, $G$ is a $p$-competition graph. ∎

Kim et al. [8] specified the length of a cycle which is a $p$-competition graph in terms of $p$.

Theorem 3.2** (​​[8]).**

Let $C_{n}$ be a cycle with $n$ vertices for a positive integer $n\geq 4$. Then $\Upsilon(C_{n})=[n-3]$.

In the proof of Theorem 3.2, $\mathcal{F}:=\{S_{0},\ldots,S_{n-1}\}$, where, for each $i=0,\ldots,n-1$, $S_{i}:=\{v_{i},v_{i+1},\ldots,v_{i+p}\}$, is given as a $p$-edge clique cover of $C_{n}=v_{0}v_{1}\ldots v_{n-1}v_{0}$ for which all the subscripts are reduced modulo $n$. The following square matrix of order $n$ is obtained from $\mathcal{F}$ by (1):

[TABLE]

where $i$th row (resp. column) is corresponding to $v_{i-1}$ (resp. $S_{i-1}$) and the $(i+1)$st row (we identify the $(n+1)$st row with the first row) is obtained by cyclically shifting the $i$th row by $1$ to the right for each $i$, $1\leq i\leq n$. Therefore the $p$-row graph of $M_{p,n}$ is isomorphic to $C_{n}$ and the following proposition is immediately true.

Lemma 3.3**.**

Let $n$ be an integer greater than or equal to $4$ and $p$ be a positive integer less than or equal to $n-3$. Then, for the matrix $M_{p,n}$ given in the equation (3), the following are true:

• (1)

*the *$k$*th row contains exactly $p+1$ *$1$s for each integer $k$, $1\leq k\leq n$;

• (2)

*the *$k$*th row and the *$(k+1)$*st row have common *$1$*s in exactly $p$ columns for each integer $k$, $1\leq k\leq n$ (we identify the *$(n+1)$st row with the first row);

• (3)

*the *$k$*th row and the *$l$*th row have common *$1$s in at most $p-1$ columns for integers $k$ and $l$ satisfying $1\leq k,l\leq n$ and $2\leq|k-l|$.

We denote the path graph with $n$ vertices by $P_{n}$.

Proposition 3.4**.**

For an integer $n\geq 3$,

[TABLE]

Proof.

For $n\geq 5$, then $\Upsilon(P_{n})\subset[n-3]$ by Corollary 2.24. By Corollary 2.8, $\Upsilon(P_{3})\subset\{1,2\}$. By Corollary 2.8 and Proposition 2.20, $\Upsilon(P_{4})\subset\{1,2\}$.

Now we show the other direction containment. If $n=3$ and $p\leq 2$, then $\{1,2\}\subset\Upsilon(P_{n})$ by Corollary 2.28. It is easy to check that $P_{4}$ is isomorphic to the $2$-row graph of the following matrix:

[TABLE]

Thus $2\in\Upsilon(P_{4})$. Now suppose that $n\geq 4$ and $p\leq n-3$. In $M_{p,n}$ given in (3), we replace $1$ in the $(1,n-p+1)$-entry with [math] to obtain a square matrix $M^{*}_{p,n}$ of order $n$. Let $G^{\prime}$ be the $p$-row graph of $M^{*}_{p,n}$. Then the first row and the second row of $M^{*}_{p,n}$ still share $p$ $1$s. Yet the first row and the $n$th row of $M^{*}_{p,n}$ share only $p-1$ $1$s. Thus, by (2) and (3) of Lemma 3.3, $G^{\prime}$ is isomorphic to a path graph with $n$ vertices. Hence $[n-3]\subset\Upsilon(P_{n})$ and this completes the proof. ∎

Proposition 3.5**.**

Let $T$ be a tree with the diameter $m$ for some integer $m$, $m\geq 3$. Then $\Upsilon(T)\supset[m-2]$.

Proof.

Take $p\in[m-2]$. Since the diameter of $T$ is $m$, there exists an induced path of length $m$. Since $m\geq p+2$, we may take a section $P$ of this path which has length $p+2$. Then $P$ is a $p$-competition graph by Proposition 3.4. Since $T$ can be obtained from $P$ by adding pendant vertices sequentially, $G$ is a $p$-competition graph by Theorem 3.1. ∎

Corollary 3.6**.**

Given a graph $G$ with $n$ vertices and diameter $m$, if $G/\!\!\sim$ is a tree with $n^{\prime}$ vertices, then $\Upsilon(G)\supset[n-n^{\prime}+m-2]$.

Proof.

Suppose that $G/\!\!\sim$ is a tree with $n^{\prime}$ vertices. Then, by Proposition 2.12, $G/\!\!\sim$ has diameter $m$. Thus, by Proposition 3.5, $\Upsilon(G/\!\!\sim)\supset[m-2]$. By Corollary 2.16,

[TABLE]

∎

A caterpillar is a tree with at least $3$ vertices the removal of whose pendant vertices produces a path. A spine of a caterpillar is the longest path of the caterpillar. In the following, for a caterpillar $T$ with $n$ vertices, we shall find all the positive integers $p$ such that $T$ is a $p$-competition graph in terms of $n$. To do so, we need the following lemmas.

Lemma 3.7**.**

Let $n$ and $t$ be positive integers such that either $(n,t)=(4,2)$ or $n\geq 4$ and $t\leq n-3$. Then, for any nonnegative integer $k$ and a path graph $P$ of length $n-1$, a caterpillar $T$ obtained by adding $k$ new vertices to $P$ in such a way that the added vertices are pendent vertices of $T$ adjacent to interior vertices of $P$ is a $(t+k)$-competition graph.

Proof.

Since either $(n,t)=(4,2)$ or $n\geq 4$ and $t\leq n-3$, $P$ is a $t$-competition graph by Proposition 3.4, which is actually the $t$-row graph of $M^{*}_{t,n}$ where $M^{*}_{t,n}$ is the matrix defined in the proof of the same lemma. Thus the statement is true for $k=0$. Now we assume that $k$ is a positive integer. Let $P=x_{1}x_{2}\cdots x_{n}$ and $y_{1},\ldots,y_{k}$ be the vertices added to $P$ as described in the theorem statement. There exists a map $\phi:[k]\to[n]$ such that $x_{\phi(i)}$ is a vertex on $P$ adjacent to $y_{i}$ for each $i$, $1\leq i\leq k$. By the way that $y_{1},\ldots,y_{k}$ were added, $\phi$ is well-defined. We define a $k\times n$ $(0,1)$-matrix $A$ so that the $i$th row of $A$ is the same as the row of $M^{*}_{t,n}$ corresponding to $x_{\phi(i)}$ for each $i$, $1\leq i\leq k$.

Now we consider the matrix $M$ defined as follows:

\left[\vbox{\hrule height=31.28305pt,depth=31.28305pt,width=0.0pt}\right.$$\left.\vbox{\hrule height=31.28305pt,depth=31.28305pt,width=0.0pt}\right]$$M_{t,n}^{*}$$J_{n,k}$$A$$J_{k,k}-I_{k}$$n$$k$$M:=.

For an example, see Figure 3. Then, for each $i$ and $j$, $1\leq i<j\leq k$,

• (M-1)

if $y_{i}$ and $y_{j}$ are adjacent to the same vertex on $P$, then the $i$th row and the $j$th row of the $(2,1)$-block of $M$ are identical and have exactly $t+1$ common $1$s; otherwise, the $i$th row and the $j$th row of the $(2,1)$-block of $M$ have at most $t$ common $1$s;

• (M-2)

if $y_{i}$ and $x_{l}$ are adjacent in $T$ for some $l\in[n]$, then the $l$th row of the $(1,1)$-block of $M$ and the $i$th row of the $(2,1)$-block of $M$ have exactly $t+1$ common $1$s; otherwise, the $l$th row of the $(1,1)$-block of $M$ and the $i$th row of the $(2,1)$-block of $M$ have at most $t$ common $1$s.

Let $G$ be the $(t+k)$-row graph of $M$. We denote the row of $M$ containing the row of $M^{*}_{t,n}$ corresponding to $x_{i}$ by $\mathbf{x}_{i}$ for each $i$, $1\leq i\leq n$, the row of $M$ containing the $i$th row of the $(2,1)$-block of $M$ by $\mathbf{y}_{i}$ for each $i$, $1\leq i\leq k$.

By the definition of $M^{*}_{t,n}$, the row of $M^{*}_{t,n}$ corresponding to $x_{i}$ and the row of $M^{*}_{t,n}$ corresponding to $x_{j}$ have at most $t-1$ common $1$s if and only if $|j-i|\geq 2$. Thus $\mathbf{x}_{i}$ and $\mathbf{x}_{j}$ have at most $t+k-1$ common $1$s if and only if $|j-i|\geq 2$ and therefore $P$ is an induced subgraph of $G$.

We note that the $i$th row and the $j$th row of the $(2,2)$-block of $M$ have exactly $k-2$ common $1$s for each $i$ and $j$, $1\leq i<j\leq k$. Thus, by (M-1), ${\bf y}_{i}$ and ${\bf y}_{j}$ have at most $t+k-1$ common $1$s for each $i$ and $j$, $1\leq i<j\leq k$. Thus $y_{i}$ and $y_{j}$ are not adjacent in $G$ for each $i$ and $j$, $1\leq i<j\leq k$.

We note that the $l$th row of the $(1,2)$-block of $M$ and the $i$th row of the $(2,2)$-block of $M$ have exactly $k-1$ common $1$s for each $i$, $1\leq i\leq k$ and each $l$, $1\leq l\leq n$. Thus, by (M-2), $y_{i}$ and $x_{l}$ are adjacent in $T$ if and only if ${\bf y}_{i}$ and ${\bf x}_{l}$ have at least $t+k$ common $1$s. Thus we have shown that $T$ is isomorphic to $G$. Hence $T$ is a $(t+k)$-competition graph by Theorem 2.2. ∎

Lemma 3.8**.**

Given an integer $n\geq 2$ and the star graph $K_{1,n}$, $\Upsilon(K_{1,n})=[n]$.

Proof.

By Corollary 2.8, $n+1\not\in\Upsilon(K_{1,n})$, so $\Upsilon(K_{1,n})\subset[n]$.

Now we show the converse containment. By Corollary 2.28, $\{1,2\}\subset\Upsilon(K_{1,n})$. By Lemma 2.19, $n\in\Upsilon(K_{1,n})$. Therefore $[n]\subset\Upsilon(K_{1,n})$ for $n=2,3$. Now suppose $n\geq 4$. Then $\{1,2,n\}\subset\Upsilon(K_{1,n})$ by the above argument. Thus it is sufficient to show that $p\in\Upsilon(K_{1,n})$ for $3\leq p\leq n-1$. Now we consider the following matrix $M$ of order $n+1$:

\left[\vbox{\hrule height=28.69969pt,depth=28.69969pt,width=0.0pt}\right.$$\left.\vbox{\hrule height=28.69969pt,depth=28.69969pt,width=0.0pt}\right]$$M_{p-2,n}$$1$$\vdots$$1$$1$$\cdots$$1$$1$$M=

where $M_{p-2,n}$ is the matrix defined in (3). Let $G$ be the $p$-row graph of $M$. In addition, let $\mathbf{r}_{i}$ denote the $i$th row of $M$ and $v_{i}$ be the vertex of $G$ corresponding to $\mathbf{r}_{i}$ for each $i$, $1\leq i\leq n+1$.

For each $i=1$, $\ldots$, $n$, the $i$th row of $M_{p-2,n}$ contains exactly $p-1$ $1$s by Lemma 3.3(1). Thus $\mathbf{r}_{i}$ contains exactly $p$ $1$s in $M$ and so $v_{i}$ and $v_{n+1}$ are adjacent in $G$ for each $i=1,\ldots,n$.

By (2) and (3) of Lemma 3.3, the $i$th row and the $j$th row of $M_{p-2,n}$ have at most $p-2$ common $1$s for distinct $i$ and $j$ in $[n]$. Therefore $\mathbf{r}_{i}$ and $\mathbf{r}_{j}$ have at most $p-1$ common $1$s and so $v_{i}$ and $v_{j}$ are not adjacent in $G$ for distinct $i$ and $j$ in $[n]$. Thus $G$ is isomorphic to $K_{1,n}$. Hence $p\in\Upsilon(K_{1,n})$ and so $[n]\subset\Upsilon(K_{1,n})$. ∎

Theorem 3.9**.**

For a caterpillar $T$ with $n$ vertices,

[TABLE]

*where $d(T)$ denotes the diameter of $T$. *

Proof.

If $d(T)=2$, then $T\cong K_{1,n-1}$ and, by Lemma 3.8, $\Upsilon(T)=[n-1]$.

Suppose $d(T)\geq 3$. If $d(T)=3$, $\Upsilon(T)\subset[n-2]$ by Corollary 2.8 and Proposition 2.20. If $d(T)\geq 4$, $\Upsilon(T)\subset[n-3]$ by Corollary 2.23.

To show the converse containment, let $k(T)$ denote the number of vertices which are attached to the spine of $T$. Now take a positive integer $p\in[n-t]$ where $t=2$ if $d(T)=3$ and $t=3$ if $d(T)\geq 4$.

Since $d(T)$ is the length of the spine of $T$, $n=d(T)+1+k(T)$. Thus $p\leq(d(T)+1+k(T))-t$ or

[TABLE]

If either $d(T)=3$ and $p\leq 2$ or $d(T)\geq 4$ and $p\leq d(T)-2$, then the spine of $T$ is a $p$-competition graph by Proposition 3.4 and so $T$ is a $p$-competition graph by Theorem 3.1.

Now assume that one of the following: $d(T)=3$ and $p>2$; $d(T)\geq 4$ and $p>d(T)-2$. Let

[TABLE]

Then $\alpha\geq 1$ and, by (4), we have $\alpha\leq k(T)$. Let $T^{\prime}$ be a caterpillar obtained from $T$ by deleting some pendent vertices of $T$ so that $d(T^{\prime})=d(T)$ and $k(T^{\prime})=\alpha$. Then, by Lemma 3.7, $T^{\prime}$ is a $p$-competition graph and so, by Theorem 3.1, $T$ is a $p$-competition graph. ∎

Corollary 3.10**.**

Let $G$ be a graph with $n$ vertices such that $G/\!\!\sim$ is a caterpillar. Then

[TABLE]

where $d(G)$ denotes the diameter of $G$.

Proof.

By Theorem 3.9,

[TABLE]

where $m=|V(G/\!\!\sim)|$ and $d(G/\!\!\sim)$ denotes the diameter of $G/\!\!\sim$. Since $G/\!\!\sim$ has no isolated vertices,

[TABLE]

by Proposition 2.12 and Corollary 2.16. By Corollary 2.8, $n\not\in\Upsilon(G)$. Therefore $\Upsilon(G)=[n-1]$ if $d(G)=2$. By Proposition 2.20, $n-1\not\in\Upsilon(G)$ if $d(G)\geq 3$. Thus $\Upsilon(G)=[n-2]$ if $d(G)=3$. By Corollary 2.23, $\Upsilon(G)\subset[n-3]$ if $d(G)\geq 4$. Hence $\Upsilon(G)=[n-3]$ if $d(G)\geq 4$. ∎

Lemma 3.11**.**

Given a $p$-competition graph $G$ with $n$ vertices, suppose that $2^{r}+1$ neighbors of a vertex $v$ of $G$ form an independent set for some positive integer $r$. Then $p\geq n-r$ implies that there are two nonadjacent neighbors $x$ and $y$ of $v$ with $|\Lambda_{M}(x)|<|\Lambda_{M}(v)|$ and $|\Lambda_{M}(y)|<|\Lambda_{M}(v)|$ for any square matrix $M$ of order $n$ whose $p$-row graph is isomorphic to $G$.

Proof.

Let $M$ be a square matrix $M$ of order $n$ whose $p$-row graph is isomorphic to $G$. Since $v$ is not isolated, $p\leq|\Lambda_{M}(v)|\leq n$. For notational convenience, we let $\overline{\Lambda}_{M}(v)=[n]\setminus\Lambda_{M}(v)$. Now suppose $p\geq n-r$. Then $0\leq|\overline{\Lambda}_{M}(v)|\leq n-p\leq r$. Thus the number of subsets of $\overline{\Lambda}_{M}(v)$ is less than $2^{r}+1$. For each neighbor $w$ of $v$, $\overline{\Lambda}_{M}(v)\cap\overline{\Lambda}_{M}(w)$ is a subset of $\overline{\Lambda}_{M}(v)$. Since $v$ has $2^{r}+1$ neighbors which form an independent set by the hypothesis, there are two nonadjacent neighbors $x$ and $y$ of $v$ such that $\overline{\Lambda}_{M}(v)\cap\overline{\Lambda}_{M}(x)=\overline{\Lambda}_{M}(v)\cap\overline{\Lambda}_{M}(y)$ by the Pigeonhole principle. Since $\overline{\Lambda}_{M}(v)\cap\overline{\Lambda}_{M}(x)$ and $\overline{\Lambda}_{M}(v)\cap\overline{\Lambda}_{M}(y)$ are subsets of $\overline{\Lambda}_{M}(x)$ and $\overline{\Lambda}_{M}(y)$, respectively, we have

[TABLE]

Since $v$ is adjacent to $x$ and $y$, $|\overline{\Lambda}_{M}(v)\cup\overline{\Lambda}_{M}(x)|\leq n-p$ and $|\overline{\Lambda}_{M}(v)\cup\overline{\Lambda}_{M}(y)|\leq n-p$. Since $x$ and $y$ are not adjacent, $|\overline{\Lambda}_{M}(x)\cup\overline{\Lambda}_{M}(y)|>n-p$. Thus

[TABLE]

Then, by (5), $|\overline{\Lambda}_{M}(y)|>|\overline{\Lambda}_{M}(v)|$. By the same argument, one can show that $|\overline{\Lambda}_{M}(x)|>|\overline{\Lambda}_{M}(v)|$ and we complete the proof. ∎

By Proposition 3.5, $\Upsilon(T)\neq\emptyset$ for a tree $T$ with the diameter at least $3$. It is easy to see that $\Upsilon(T)\neq\emptyset$ for a tree $T$ with the diameter at most two. Thus $\max\Upsilon(T)$ exists for any tree $T$. We have shown that $|V(T)|-\max\Upsilon(T)\leq 3$ for a caterpillar $T$. One might think by this result that there exists a positive integer $t$ such that $|V(T)|-\max\Upsilon(T)\leq t$ for any tree $T$, yet it is not true by the following theorem.

Let $k$ be a positive integer. A $k$-ary tree is a rooted tree in which each vertex has no more than $k$ children. A full $k$-ary tree is a rooted tree exactly $k$ children or no children. A perfect $k$-ary tree is a full $k$-ary tree in which all pendant vertices are at the same depth. The depth of a vertex in a rooted tree is the distance between the vertex and the root. The height of a rooted tree is the number of edges on the longest path between its root and a pendant vertex.

Theorem 3.12**.**

For any positive integer $r$, there is a tree $T$ with $|V(T)|-\max\Upsilon(T)>r$.

Proof.

Let $T$ be a perfect $(2^{r}+1)$-ary tree with height $r+1$ and a root $x_{0}$. Then by the definition of $\Upsilon(G)$ for a graph $G$, $T$ is a $(\max\Upsilon(T))$-competition graph. Then $T$ is a $(\max\Upsilon(T))$-row graph of a matrix $M$. Suppose that $|V(T)|-\max\Upsilon(T)\leq r$. Then, by Lemma 3.11, $|\Lambda_{M}(x_{1})|<|\Lambda_{M}(x_{0})|\leq|V(T)|$ for some children $x_{1}$ of $x_{0}$. Then, by the same lemma again, $|\Lambda_{M}(x_{2})|<|\Lambda_{M}(x_{1})|\leq|V(T)|-1$ for some children $x_{2}$ of $x_{1}$. We apply the lemma repeatedly to have $|\Lambda_{M}(x_{r+1})|<|V(T)|-r$. Since $x_{r+1}$ is non-isolated, $|\Lambda_{M}(x_{r+1})|\geq\max\Upsilon(T)$. Thus $|V(T)|-\max\Upsilon(T)>r$ and we reach a contradiction. Hence $|V(T)|-\max\Upsilon(T)>r$. ∎

4 Closing Remarks

We have shown that $\Upsilon(K_{3,3})=\emptyset$. We would like to know whether or not $\Upsilon(K_{n,n})=\emptyset$ for any $n\geq 4$. We have characterized the graphs with $n$ vertices and the competition-realizer $[n]$ and $[n-1]$, respectively. It would be interesting to characterize the graphs with $n$ vertices and competition-realizer $[n-2]$. Finally we suggest to find the competition-realizer for a Lobster to extend our result which gives every element in the competition-realizer for a caterpillar.

5 Acknowledgement

This research was supported by the National Research Foundation of Korea(NRF) funded by the Korea government(MSIT) (No. NRF-2017R1E1A1A03070489) and by the Korea government(MSIP) (No. 2016R1A5A1008055).