Locally Irregular-Connected Graphs

Abstract

A graph G is locally irregular if every two adjacent vertices have distinct degrees and is locally irregular-connected if for every two vertices u and v of G, there is a locally irregular $u-v$ path in G. For a finite set S of two or more positive integers with maximum element k, it is known that there exists a graph of order $k+1$ with degree set S. This result is extended by showing that there is a locally irregular-connected graph of order $k+1$ with degree set S. Characterizations are established for all locally irregular-connected graphs of cycle rank at most 2. All sets S of positive integers are determined for which there is a locally irregular-connected graph of cycle rank at most 2 with degree set S. The minimum order of a locally irregular-connected graph with a prescribed degree set is determined as well. Other results and open questions are also presented.

1. Introduction

A graph G is irregular if every two vertices of G have distinct degrees. It is well known that no graph of order 2 or more is irregular.

Theorem 1 

([1]). For every integer $n\ge 2$, there is no irregular graph of order n.

The degree set of a graph G is $\mathcal{D}\left(G\right)=\{degv:v\in V(G\left)\right\}$. By Theorem 1, there is no graph G of order $n\ge 2$ for which $\left|\mathcal{D}\right(G\left)\right|=n$. For each integer $n\ge 2$, there are two graphs G of order n, such that $\left|\mathcal{D}\right(G\left)\right|=n-1$. If $G_1$ and $G_2$ are the two graphs of order $n\ge 2$ with $|\mathcal{D}\left(G_1\right)|=|\mathcal{D}\left(G_2\right)|=n-1$, then $G_2={\overline{G}}_1$ is the complement of $G_1$. Graphs with this property are sometimes referred to as antiregular graphs; that is, a nontrivial graph G is antiregular if exactly two vertices of G have the same degree (see [2,3,4], for example).

Theorem 2 

([2,3]). For every integer $n\ge 2$, there are exactly two antiregular graphs of order n, one of which is connected.

The connected antiregular graph of order n is denoted by $G_n$ and its complement ${\overline{G}}_n$ is the disconnected antiregular graph of order n. Then, $V\left(G_n\right)=\{v_1,v_2,\dots ,v_n\}$ where $E\left(G_n\right)=\{v_i{v}_j:i+j\ge n+1\}$ and $E\left({\overline{G}}_n\right)=\{v_i{v}_j:i+j\le n\}$. Thus, $\mathcal{D}\left(G_n\right)=\{1,2,\dots ,⌊{\displaystyle \frac{n}{2}}⌋,⌊{\displaystyle \frac{n}{2}}⌋+1,\dots ,n-1\}$ and $\mathcal{D}\left({\overline{G}}_n\right)=\{0,1,\dots ,⌊{\displaystyle \frac{n-1}{2}}⌋,⌊{\displaystyle \frac{n-1}{2}}⌋+1,\dots ,n-2\}$. If $n\ge 2$ is even, then the two vertices of the same degree ${\displaystyle \frac{n}{2}}$ in $G_n$ are adjacent; while if $n\ge 3$ is odd, then the two vertices of the same degree ${\displaystyle \frac{n-1}{2}}$ in $G_n$ are not adjacent. Figure 1 shows $G_n$ and ${\overline{G}}_n$ for $n=5,6$. More information on antiregular graphs can be found in [4,5,6,7,8,9,10]. Degree sets of other types of irregular graphs have also been studied (see [11,12], for example).

A graph G is locally irregular if every two adjacent vertices of G have distinct degrees (see [13], for example). An edge $e=uv$ in a graph with $degu\ne degv$ is an irregular edge, while e is a regular edge if $degu=degv$. Thus, a graph G is locally irregular if and only if every edge of G is irregular. The connected locally irregular graphs of order n with $3\le n\le 5$ are shown in Figure 2 where each vertex is labeled with its degree.

Since the two vertices of the same degree in the connected antiregular graph $G_n$ of odd order $n\ge 3$ are not adjacent, $G_n$ is locally irregular and ${\overline{G}}_n$ is not locally irregular for odd integers $n\ge 3$. Since the two vertices of the same degree in the graph $G_n$ of even order $n\ge 2$ are adjacent, $G_n$ is not locally irregular and ${\overline{G}}_n$ is locally irregular for even integers $n\ge 2$.

A graph G is locally irregular-connected if there is a locally irregular $u-v$ path for every two vertices u and v of G. While irregular graphs of order 2 or more do not exist, there are a number of properties possessed by the related class of locally irregular-connected graphs which we present. If a locally irregular graph G is connected, then every two vertices u and v of G are connected by a locally irregular $u-v$ path and so G is locally irregular-connected. The converse is not true, however. For example, the locally irregular-connected graph G of Figure 3 is not locally irregular since $uv$ is a regular edge. The edge $xy$ in the graph H of Figure 3 is regular. Since $xy$ is a bridge in H, it follows that $(x,y)$ is the only $x-y$ path in H and so there is no locally irregular $x-y$ path in H. Therefore, H is not locally irregular-connected.

The following two observations for locally irregular-connected graphs are often useful.

Observation 1.

If a graph G contains a regular edge that is a bridge, then G is not locally irregular-connected.

Observation 2.

A graph G is locally irregular-connected if G contains a connected spanning subgraph H that is locally irregular, that is, every edge of H is irregular. In particular, every connected locally irregular graph is locally irregular-connected.

Every connected antiregular graph of order at least 3 is locally irregular-connected, as we show next.

Proposition 1.

For each integer $n\ge 3$, the connected antiregular graph $G_n$ of order n is locally irregular-connected.

Proof. 

First, suppose that $n\ge 3$ is odd. Since the two vertices of the same degree in $G_n$ are not adjacent, it follows that $G_n$ is locally irregular and so $G_n$ is locally irregular-connected by Observation 2. Next, suppose that $n\ge 4$ is even. Let u and v be two vertices of $G_n$. We show that there is a locally irregular $u-v$ path in $G_n$. The vertex w of degree $n-1$ in $G_n$ is adjacent to every vertex of $G_n$. If $w\notin \{u,v\}$, then $(u,w,v)$ is a locally irregular $u-v$ path in $G_n$; while if $w\in \{u,v\}$, then $(u,v)$ is a locally irregular $u-v$ path in $G_n$. □

2. Degree Sets of Locally Irregular-Connected Graphs

Since no regular graph is locally irregular-connected, every graph G with $\left|\mathcal{D}\right(G\left)\right|=1$ is not locally irregular-connected. On the other hand, for each integer $k\ge 2$, there is a locally irregular-connected graph G with $\left|\mathcal{D}\right(G\left)\right|=k$. For example, if $S=\{d_1,d_2,\dots ,d_k\}$ is a set of any k distinct positive integers, then the complete k-partite graph $G=K_{d_1,d_2,\dots ,d_k}$ is a locally irregular-connected graph with $\mathcal{D}\left(G\right)=\{n-d_i:1\le i\le k\}$. In fact, more can be said.

Theorem 3.

For every pair $k,n$ of integers with $2\le k\le n-1$, there exists a locally irregular-connected graph $G_{k,n}$ of order n with $\left|\mathcal{D}\right(G_{k,n}\left)\right|=k$.

Proof. 

For $k=n-1$, the connected antiregular graph $G_{n-1,n}=G_n$ of order n has $\mathcal{D}\left(G_{n-1,n}\right)=\{1,2,\dots ,n-1\}$ and is locally irregular-connected by Proposition 1. For $k=2$ and $n\ge 4$, the star $G_{2,n}=K_{1,n-1}$ of order n has $\mathcal{D}\left(G_{2,n}\right)=\{1,n-1\}$. For $k=3$ and $n\ge 5$, the double star $G_{3,n}=S_{2,n-2}$ of order n has central vertices of degrees 2 and $n-2\ge 3$ and so $\mathcal{D}\left(G_{3,n}\right)=\{1,2,n-2\}$. Since $K_{1,n-1}$ and $S_{2,n-2}$ are locally irregular, these graphs are locally irregular-connected.

For $4\le k\le n-2$, let $G_{k+1}$ be the connected antiregular graph of order $k+1$. Then $\mathcal{D}\left(G_{k+1}\right)=\{1,2,\dots ,k\}$ and so $\left|\mathcal{D}\right(G_{k+1}\left)\right|=k$. The vertex w of degree k in $G_{k+1}$ is adjacent to all other vertices of $G_{k+1}$. Let $G_{k,n}$ be the graph obtained from $G_{k+1}$ and ${\overline{K}}_{n-k-1}$ by joining each vertex of ${\overline{K}}_{n-k-1}$ to the vertex w in $G_{k+1}$. Thus, ${deg}_{G_{k,n}}w=k+(n-k-1)=n-1$. This implies that $\mathcal{D}\left(G_{k,n}\right)=(\mathcal{D}\left(G_{k+1}\right)\setminus \left\{k\right\})\cup \{n-1\}$ and so $\left|\mathcal{D}\right(G_{k,n}\left)\right|=k$. The graphs $G_{5,8}$ and $G_{6,8}$ are shown in Figure 4. We now show that $G_{k,n}$ is locally irregular-connected. That is, there is a locally irregular $u-v$ path in $G_{k,n}$ for every two vertices u and v of $G_{k,n}$. First, suppose that $u,v\in V\left(G_{k+1}\right)$. Since $G_{k+1}$ is locally irregular-connected by Proposition 1, there is a locally irregular $u-v$ path in $G_{k+1}$ and in $G_{k,n}$ as well. Next, suppose that at least one of u and v is not in $G_{k+1}$, say $v\notin V\left(G_{k+1}\right)$. If $u\notin V\left(G_{k+1}\right)$, then $(u,w,v)$ is a locally irregular $u-v$ path in $G_{k,n}$. If $u\in V\left(G_{k+1}\right)$, then for a locally irregular $u-w$ path P in $G_{k+1}$, the $u-v$ path $(P,v)$ in $G_{k,n}$ is locally irregular. Therefore, $G_{k,n}$ is locally irregular-connected.

It was shown in [14] that every finite set of positive integers is the degree set of some graph. □

Theorem 4 

([14]). For every set $S=\{a_1,a_2,\dots ,a_k\}$ of positive integers with $a_1<a_2<\cdots <a_k$, there exists a graph G with $\mathcal{D}\left(G\right)=S$. Furthermore, the minimum order of such a graph G is $a_k+1$.

A finite set S of distinct positive integers is realizable if there is a locally irregular-connected graph G such that $\mathcal{D}\left(G\right)=S$. Theorem 4 suggests the following question.

Question 1: Which finite sets S of distinct positive integers are realizable?

If S is a realizable set of positive integers for a locally irregular-connected graph G of order n and $max\left(S\right)$ is the largest integer in S, then $n\ge max\left(S\right)+1$. This observation suggests another question.

Question 2: For a realizable set S of positive integers, what is the minimum order of a locally irregular-connected graph G with $\mathcal{D}\left(G\right)=S$?

By applying the technique used in the proof of Theorem 4, an analogous result for locally irregular-connected graphs can be established which provides answers to Questions 1 and 2. For this purpose, a lemma is presented. For two vertex-disjoint graphs F and H, the graph $F\vee H$ is the join of F and H and $F+H$ is the union of F and H.

Lemma 1.

Let $G=F\vee H$ be the join of two vertex-disjoint graphs F and H. If ${deg}_{G}x\ne {deg}_{G}y$ for every two vertices x and y where $x\in V\left(F\right)$ and $y\in V\left(H\right)$, then G is locally irregular-connected.

Proof. 

Let u and v be two vertices of $G=F\vee H$. If $\{u,v\}\subseteq V\left(F\right)$ or $\{u,v\}\subseteq V\left(H\right)$, say the former, then $(u,w,v)$ is a locally irregular $u-v$ path in G for some vertex w of H. If one of u and v is in F and the other is in H, then $(u,v)$ is a locally irregular $u-v$ path in G. Hence, G is locally irregular-connected. □

Theorem 5.

For every set $S=\{a_1,a_2,\dots ,a_k\}$ of $k\ge 2$ positive integers where $a_1<a_2<\cdots <a_k$, there is a locally irregular-connected graph G of order $a_k+1$ such that $\mathcal{D}\left(G\right)=S$. Consequently, the minimum order of a locally irregular-connected graph G with $\mathcal{D}\left(G\right)=S$ is $a_k+1$.

Proof. 

We proceed by induction on $k\ge 2$.

⋆

For $k=2$, the order of the graph $G=K_{a_1}\vee {\overline{K}}_{a_2-a_1+1}$ is $a_2+1$. Since ${deg}_{G}v=a_2$ if $v\in V\left(K_{a_1}\right)$ and ${deg}_{G}v=a_1$ if $v\in V\left({\overline{K}}_{a_2-a_1+1}\right)$, it follows that $\mathcal{D}\left(G\right)=\{a_1,a_2\}$. By Lemma 1, G is locally irregular-connected.

⋆

For $k=3$, the order of the graph $G=K_{a_1}\vee ({\overline{K}}_{a_3-a_2}+K_{a_2-a_1+1})$ is $a_3+1$. Since ${deg}_{G}v=a_3$ if $v\in V\left(K_{a_1}\right)$, ${deg}_{G}v=a_1$ if $v\in V\left({\overline{K}}_{a_3-a_2}\right)$, and ${deg}_{G}v=a_2$ if $v\in V\left(K_{a_2-a_1+1}\right)$, it follows that $\mathcal{D}\left(G\right)=\{a_1,a_2,a_3\}$. By Lemma 1, G is locally irregular-connected.

Hence, the statement is true for $k=2,3$. Let $k\ge 2$. Suppose that for every set $S=\{a_1,a_2,\dots ,a_j\}$ of j positive integers where $2\le j\le k$ and $a_1<a_2<\cdots <a_j$, there is a locally irregular-connected graph H of order $a_j+1$ such that $\mathcal{D}\left(H\right)=S$.

Next, let $T=\{b_1,b_2,\dots ,b_{k+1}\}$ be a set of $k+1$ positive integers, where $b_1<b_2<\cdots <b_{k+1}$. We show that there is a locally irregular-connected graph G of order $b_{k+1}+1$ with $\mathcal{D}\left(G\right)=T$. Let $T^{\prime }=\{b_2-b_1,b_3-b_1,\dots ,b_k-b_1\}$. Then $|T^{\prime }|=k-1\ge 2$ and $1\le b_2-b_1<b_3-b_1<\cdots <b_k-b_1$. By the induction hypothesis, there is a locally irregular-connected graph $G^{\prime }$ of order $(b_k-b_1)+1$ with $\mathcal{D}\left(G^{\prime }\right)=T^{\prime }$. Let $G=K_{b_1}\vee ({\overline{K}}_{b_{k+1}-b_k}+G^{\prime })$. Then the order of G is $b_{k+1}+1$. Since ${deg}_{G}v=b_{k+1}$ if $v\in V\left(K_{b_1}\right)$, ${deg}_{G}v=b_1$ if $v\in V\left({\overline{K}}_{b_{k+1}-b_k}\right)$, and ${deg}_{G}v=b_i$ for some integer i with $2\le i\le k$ if $v\in V\left(G^{\prime }\right)$, it follows that $\mathcal{D}\left(G\right)=T$. By Lemma 1, G is locally irregular-connected. □

3. Trees and Unicyclic Graphs

In this section, we characterize locally irregular-connected trees and locally irregular-connected unicyclic graphs and determine all sets S of positive integers that are the degree sets of graphs in these two classes as well as determine the minimum order of each such graph with degree set S. We begin with trees.

Proposition 2.

A tree T is locally irregular-connected if and only if T contains no regular edges.

Proof. 

First, suppose that a tree T contains a regular edge. Since every edge of a tree is a bridge, it follows by Observation 1 that T is not locally irregular-connected. Next, suppose that T contains no regular edges. It then follows by Observation 2 that T is locally irregular-connected. □

A caterpillar is a tree of order 3 or more, the removal of whose end-vertices produces a path called the spine of the caterpillar.

Theorem 6.

Let $S=\{a_1,a_2,\dots ,a_k\}$ be a set of $k\ge 2$ positive integers. There is a locally irregular-connected tree T with $\mathcal{D}\left(T\right)=S$ if and only if $1\in S$. Furthermore, if $1\in S$, then the minimum order of a locally irregular-connected tree T with $\mathcal{D}\left(T\right)=S$ is $2+{\sum }_{i=1}^k(a_i-1)$.

Proof. 

First, suppose that T is a locally irregular-connected tree with $\mathcal{D}\left(T\right)=S$. Since every nontrivial tree T contains at least two end-vertices, $1\in S$. Next, let $S=\{a_1,a_2,\dots ,a_k\}$, where $a_1=1$. If $k=2$, then let $T=K_{1,a_2}$ be the star of order $a_2+1$ with $\mathcal{D}\left(T\right)=\{1,a_2\}$. If $k\ge 3$, let T be the caterpillar whose spine is the path $(v_1,v_2,\dots ,v_{k-1})$ where $deg{v}_i=a_{i+1}$ for $1\le i\le k-1$. The order of T is $2+{\sum }_{i=1}^k(a_i-1)$ and $\mathcal{D}\left(T\right)=S$. Since every edge of T is irregular, T is locally irregular-connected.

Next, let T be a locally irregular-connected tree of order n with $\mathcal{D}\left(T\right)=S=\{a_1,a_2,\dots ,a_k\}$. Then, the size of T is $n-1$. Since T contains at least one vertex of degree $a_i$ for $1\le i\le k$ and the remaining vertices have degree at least 1, it follows that

$$2(n-1)=\sum _{v\in V\left(T\right)}degv\ge \left({\displaystyle \sum _{i=1}^k}1·a_i\right)+(n-k)·1={\displaystyle \sum _{i=1}^k}(a_i-1)+n$$

and so $n\ge 2+{\sum }_{i=1}^k(a_i-1).$ We saw that there is a locally irregular-connected tree of order $2+{\sum }_{i=1}^k(a_i-1)$ having degree set S. Therefore, the minimum order of a locally irregular-connected tree T with $\mathcal{D}\left(T\right)=S$ is $2+{\sum }_{i=1}^k(a_i-1)$. □

We now turn to unicyclic graphs. A unicyclic graph is a connected graph with exactly one cycle. Thus, a unicyclic graph is obtained from a tree by adding an edge that joins two nonadjacent vertices of the tree. We now characterize of all locally irregular-connected unicyclic graphs and determine all realizable sets for locally irregular-connected unicyclic graphs as well as determine the minimum order of such graphs with a prescribed degree set.

Proposition 3.

A unicyclic graph G is locally irregular-connected if and only if G has at most one regular edge and the regular edge (if it exists) must lie on the unique cycle in G.

Proof. 

If G has no regular edge, then G is locally irregular-connected by Observation 2. Thus, let G be a graph with exactly one regular edge e that lies on the cycle. Then $G-e$ is a spanning tree of G and has no regular edge. Again, it follows by Observation 2 that G is locally irregular-connected.

Next, we verify the converse. Suppose that G is a unicyclic graph that contains two or more regular edges. Let $C=(v_1,v_2,\dots ,v_{\ell },v_1)$ be the unique cycle of order $\ell \ge 3$ in G and let e and f be two regular edges of G. It follows by Observation 1 that e and f are not bridges and so they lie on the cycle C. We may assume that $e=v_1{v}_2$ and $f=v_i{v}_{i+1}$ where $2\le i\le \ell -1$. There are exactly two $v_1-v_2$ paths in G (both on C), namely $P_1=(v_1,v_2)$ and $P_2=(v_1,v_{\ell },v_{\ell -1}\dots ,v_2)$. Since $P_1$ contains e and $P_2$ contains f, neither $P_1$ nor $P_2$ is locally irregular. Thus, there is no locally irregular $v_1-v_2$ path in G and so G is not locally irregular-connected. □

We now determine all sets S of positive integers that are the degree sets of locally irregular-connected unicyclic graphs and determine the minimum order of such a unicyclic graph G with $\mathcal{D}\left(G\right)=S$. First, we show that the degree set of a locally irregular-connected unicyclic graph contains at least three elements.

Proposition 4.

There is no locally irregular-connected unicyclic graph G with $\left|\mathcal{D}\right(G\left)\right|=2$.

Proof. 

Assume, to the contrary, that there is a locally irregular-connected unicyclic graph G with $\mathcal{D}\left(G\right)=\{a_1,a_2\}$, where $1\le a_1<a_2$. Let $C=(v_1,v_2,\dots ,v_{\ell },v_1)$ be the unique cycle of G for some integer $\ell \ge 3$. Since C is not locally irregular-connected, $G\ne C$. Hence, G contains at least one end-vertex and so $1\in S$. This implies that $deg{v}_i=a_2$ for $1\le i\le \ell$ and so every edge on C is regular, which contradicts Proposition 3. □

Theorem 7.

Let $S=\{a_1,a_2,\dots ,a_k\}$ be a set of $k\ge 3$ positive integers. There is a locally irregular-connected unicyclic graph G with $\mathcal{D}\left(G\right)=S$ if and only if $1\in S$. Furthermore, if $1\in S$, then the minimum order of a locally irregular-connected unicyclic graph G with $\mathcal{D}\left(G\right)=S$ is $a_2+{\sum }_{i=1}^3(a_i-1)$ if $k=3$ and is ${\sum }_{i=1}^k(a_i-1)$ if $k\ge 4$.

Proof. 

First, suppose that G is a locally irregular-connected unicyclic graph with $\mathcal{D}\left(G\right)=S$. Then G is not a cycle. Thus, G contains at least one end-vertex and so $1\in S$. Next, let $S=\{a_1,a_2,\dots ,a_k\}$, where $1=a_1<a_2<\cdots <a_k$.

⋆

For $k=3$, let G be the unicyclic graph with the unique cycle $(v_1,v_2,v_3,v_1)$ such that $deg{v}_1=deg{v}_2=a_2$ and $deg{v}_3=a_3$. Then, the order of G is $2(a_2-2)+(a_3-2)+3=a_2+{\sum }_{i=1}^3(a_i-1)$ and $\mathcal{D}\left(G\right)=\{1,a_2,a_3\}$. Since G has exactly one regular edge that lies on the cycle, it follows by Proposition 3 that G is locally irregular-connected.

⋆

For $k\ge 4$, let G be the unicyclic graph with the unique cycle $(v_1,v_2,\dots ,v_{k-1},v_1)$ such that $deg{v}_i=a_{i+1}$ for $1\le i\le k-1$. Then, the order of G is ${\sum }_{i=2}^k(a_i-2)+(k-1)={\sum }_{i=2}^k(a_i-1)={\sum }_{i=1}^k(a_i-1)$, where $a_1-1=0$ and $\mathcal{D}\left(G\right)=S$. Since every edge of G is irregular, G is locally irregular-connected.

For $S=\{1,4,5\}$ or $S=\{1,2,3,4,5\}$, some locally irregular-connected unicyclic graphs G are shown in Figure 5 with $\mathcal{D}\left(G\right)=S$.

Among all locally irregular-connected unicyclic graphs whose degree set is S = {$a_1$, $a_2$, $\dots ,$ $a_k$}, where $1=a_1<a_2<\cdots <a_k$, let G be one of minimum order n. Then, the size of G is n as well.

⋆

First, suppose that $k=3$. Let $C=(v_1,v_2,\dots ,v_{\ell },v_1)$ be the unique cycle of G for some integer $\ell \ge 3$. Since (1) $deg{v}_i\in \{a_2,a_3\}$ for $1\le i\le \ell$ and (2) C contains at most one regular edge by Proposition 3, it follows that $\ell =3$ and $C=(v_1,v_2,v_3,v_1)$ contains exactly one regular edge. Thus, exactly two vertices on C have the same degree. Therefore, $deg{v}_1+deg{v}_2+deg{v}_3\ge 2a_2+a_3$ and the remaining $n-3$ vertices of G are end-vertices. Thus,

$$2n=\sum _{v\in V\left(G\right)}degv\ge 2a_2+a_3+(n-3)=a_2+{\displaystyle \sum _{i=1}^3}(a_i-1)+n$$

and so $n\ge a_2+{\sum }_{i=1}^3(a_i-1)$. Since there is a locally irregular-connected unicyclic graph of order $a_2+{\sum }_{i=1}^3(a_i-1)$ with degree set S, it follows that $n=a_2+{\sum }_{i=1}^3(a_i-1)$.

⋆

Next, suppose that $k\ge 4$. Since G contains at least one vertex of degree $a_i$ for $1\le i\le k$ and the remaining $n-k$ vertices have degree at least 1, it follows that

$$2n=\sum _{v\in V\left(G\right)}degv\ge \left({\displaystyle \sum _{i=1}^k}1·a_i\right)+(n-k)·1={\displaystyle \sum _{i=1}^k}(a_i-1)+n$$

and so $n\ge {\sum }_{i=1}^k(a_i-1).$ Since there is a locally irregular-connected unicyclic graph of order ${\sum }_{i=1}^k(a_i-1)$ with degree set S, it follows that $n={\sum }_{i=1}^k(a_i-1)$.

□

4. Graphs of Cycle Rank 2

The number of edges that must be deleted from a connected graph G of order n and size m to obtain a spanning tree of G is its cycle rank $m-n+1$. Thus, the cycle rank of a tree is 0 and the cycle rank of a unicyclic graph is 1. By Propositions 2 and 3, a graph G has cycle rank 0 if and only if G has no regular edge and a graph G has cycle rank 1 if and only if G has at most one regular edge and this regular edge is not a bridge. We now turn our attention to locally irregular-connected graphs of cycle rank 2 and determine all realizable sets for these graphs and determine the minimum order with a prescribed degree set. For this purpose, we first introduce some additional definitions and describe the structure of graphs of cycle rank 2.

A graph H is a subdivision of a graph G if $H=G$ or H is obtained from G by inserting vertices of degree 2 into one or more edges of G. If G is a graph of order n and size m having cycle rank 2, then $m=n+1$ and G contains at least two cycles. Therefore, G has a subgraph that is isomorphic to one of the following three graphs:

(1)

a graph obtained from two edge-disjoint cycles $C_1$ and $C_2$ by identifying a vertex in $C_1$ and a vertex in $C_2$, as shown in Figure 6a,

(2)

a graph obtained from two vertex-disjoint cycles $C_1$ and $C_2$ and a nontrivial path P by identifying an end-vertex of P with a vertex of $C_1$ and identifying the other end-vertex of P with a vertex of $C_2$, as shown in Figure 6b,

(3)

a subdivision of $K_4-e$ consisting of three internally disjoint paths $P_i$$(1\le i\le 3$), as shown in Figure 6c, where at least two paths $P_i$$(1\le i\le 3$) have length 2 or more.

A graph G of cycle rank 2 is of type I if G contains exactly two edge-disjoint cycles and is of type II if G contains a subdivision of $K_4-e$. The minimum degree $\delta \left(G\right)$ of a graph G of cycle rank 2 is either 1 or 2.

Theorem 8.

A graph G of cycle rank 2 is locally irregular-connected if and only if G has at most two regular edges and satisfies one of the following two conditions, depending on whether G is of type I or type II.

(1)

If G is of type I, then each regular edge of G lies on a cycle and no cycle contains more than one regular edge.

(2)

If G is of type $II$, where $\mathcal{P}=\{P_1,P_2,P_3\}$ is the set of the three paths forming the cycles in G, then each regular edge lies on some path in $\mathcal{P}$ and no path in $\mathcal{P}$ contains more than one regular edge.

Proof. 

We consider two cases, according to whether G is of type I or type II.

Case 1. G is of type I. We show that G is locally irregular-connected if and only if G has at most two regular edges such that each regular edge lies on a cycle and no cycle contains more than one regular edge.

First, suppose that G has at most two regular edges where each regular edge lies on a cycle and no cycle contains more than one regular edge. If G has no regular edge, then G is locally irregular-connected by Observation 2. If G contains exactly one regular edge e lying on a cycle, then $G-e$ is a spanning subgraph of G and has no regular edge. By Observation 2 again, G is locally irregular-connected. If G contains exactly two regular edges e and f lying on different cycles in G. Then $G-\{e,f\}$ is a spanning tree of G and has no regular edge. Once again, by Observation 2, G is locally irregular-connected.

For the converse, suppose that G contains three or more regular edges. We show that G is not locally irregular-connected. Let $e_1,e_2,e_3$ be three regular edges of G. It follows by Observation 1 that $e_i$ ($i=1,2,3$) is not a bridge and so $e_i$ lies on a cycle. This implies that at least one of the two cycles in G contains two or more regular edges. We may assume that $C=(v_1,v_2,\dots ,v_{\ell },v_1)$ is a cycle in G for some integer $\ell \ge 3$ and $e_1=v_1{v}_2$ and $e_2=v_i{v}_{i+1}$ are regular edges for some integer i with $2\le i\le \ell -1$. There are exactly two $v_1-v_2$ paths in G (both on C), namely $Q_1=(v_1,v_2)$ and $Q_2=(v_1,v_{\ell },v_{\ell -1}\dots ,v_2)$. Since $Q_1$ contains $e_1$ and $Q_2$ contains $e_2$, neither $Q_1$ nor $Q_2$ is locally irregular. Thus, there is no locally irregular $v_1-v_2$ path in G and so G is not locally irregular-connected.

Case 2. G is of type II. Let $\mathcal{P}=\{P_1,P_2,P_3\}$ be the set consisting of the three paths forming the cycles in G. We show that G is locally irregular-connected if and only if G has at most two regular edges such that each regular edge lies on some path in $\mathcal{P}$ and no path in $\mathcal{P}$ contains more than one regular edge.

First, suppose that G has at most two regular edges such that each regular edge lies on some path in $\mathcal{P}$ and no path in $\mathcal{P}$ contains more than one regular edge. If G has no regular edge, then G is locally irregular-connected by Observation 2. If G contains exactly one regular edge lying on a path in $\mathcal{P}$, then $G-e$ is a connected spanning subgraph of G and has no regular edge. By Observation 2 again, G is locally irregular-connected. If G contains exactly two regular edges e and f lying on different paths in $\mathcal{P}$, then $G-\{e,f\}$ is a spanning tree of G and has no regular edge. Once again, by Observation 2, G is locally irregular-connected.

For the converse, suppose that G contains three or more regular edges. We show that G is not locally irregular-connected. Let $e_1,e_2,e_3$ be three regular edges of G. It follows by Observation 1 that $e_i$ ($i=1,2,3$) is not a bridge and so $e_i$ lies on a path in $\mathcal{P}$.

⋆

First, suppose that there is path in $\mathcal{P}$ containing two or more regular edges, say $P_1$ contains the regular edges $e_1$ and $e_2$. Let $P_1=(u_1,u_2,\dots ,u_{\ell })$ for some integer $\ell \ge 3$ and $e_1=u_i{u}_{i+1}$ and $e_2=u_j{u}_{j+1}$ where $1\le i<j\le \ell -1$. There are exactly three $u_i-u_{i+1}$ paths in G, namely $P_1=(u_i,u_{i+1})$, $P_2=(u_i,\dots ,u_1,P_2,u_{\ell },\dots ,u_{i+1})$, and $P_3=(u_i,\dots ,u_1,P_3,u_{\ell },\dots ,u_{i+1})$. Since $P_1$ contains $e_1$ and both $P_2$ and $P_3$ contain $e_2$, there is no locally irregular $u_i-u_{i+1}$ path in G and so G is not locally irregular-connected.

⋆

Next, suppose that each path $P_i$ ($i=1,2,3$) in $\mathcal{P}$ contains exactly one regular edge, say $e_i\in E\left(P_i\right)$ for $i=1,2,3$. Let $P_1=(u_1,u_2,\dots ,u_{\ell })$ for some integer $\ell \ge 2$ and $e_1=u_i{u}_{i+1}$, where $1\le i\le \ell -1$. There are exactly three $u_i-u_{i+1}$ paths in G, namely $P_1=(u_i,u_{i+1})$, $P_2=(u_i,\dots ,u_1,P_2,u_{\ell },\dots ,u_{i+1})$, and $P_3=(u_i,\dots ,u_1,P_3,u_{\ell },\dots ,u_{i+1})$. Since $e_i\in E\left(P_i\right)$ for $i=1,2,3$, there is no locally irregular $u_i-u_{i+1}$ path in G and so G is not locally irregular-connected.

Next, we determine all sets S of positive integers that are the degree sets of locally irregular-connected graphs of cycle rank at most 2 as well as the minimum order of such a graph G with $\mathcal{D}\left(G\right)=S$. We begin with 2-element sets S. □

Proposition 5.

Let $S=\{a_1,a_2\}$ be a set of two positive integers where $a_1<a_2$. There is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ if and only if $S=\{2,3\}$ or $S=\{2,4\}$. In each case, the minimum order of a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ is $a_2+1$.

Proof. 

If $S=\{a_1,a_2\}$ where $a_1=2$ and $a_2\in \{3,4\}$, there is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ (see Figure 7, for example).

Next, suppose that G is a locally irregular-connected graph of cycle rank 2 with $\mathcal{D}\left(G\right)=\{a_1,a_2\}$ where $1\le a_1<a_2$. Then, $a_1$ is the minimum degree $\delta \left(G\right)$ of G and $a_2$ is the maximum degree $\mathrm{\Delta }\left(G\right)$ of G. Since the minimum degree of every graph of cycle rank 2 is either 1 or 2, it follows that $a_1\in \{1,2\}$. Let C be one of the cycles of G. By Theorem 8, there is at most one regular edge on C and so at least two vertices on C have different degrees. Hence, $\{degv:v\in V\left(C\right)\}=\{a_1,a_2\}$. Thus, $a_1\ge 2$. So $a_1=2$ and $a_2\ge 3$. Let $n_1$ and $n_2$ denote the number of vertices of degree $a_1$ and $a_2$ in G, respectively. Since G is a graph of cycle rank 2, its size is $n_1+n_2+1$ and so

$$2(n_1+n_2+1)=a_1{n}_1+a_2{n}_2.$$

Since $a_1=2$, it follows that $(a_2-2)n_2=2$. Because $a_2-2$ and $n_2$ are positive integer, either $a_3=3$ and $n_2=2$ or $a_2=4$ and $n_1=1$. Therefore, $a_2\in \{3,4\}$.

If G is a locally irregular-connected graph of cycle rank 2 with $\mathcal{D}\left(G\right)=S$, then the order of G is at least $a_2+1$ and so the minimum order of a locally irregular-connected graph with degree set S is at least $a_2+1$. Since the graph $K_4-e$ of order 4 is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=\{2,3\}$ and the graph $2K_2\vee K_1$ of order 5 is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=\{2,4\}$, the minimum order of a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=\{a_1,a_2\}$ is $a_2+1$. □

Proposition 6.

Let $S=\{a_1,a_2,a_3\}$ be a set of three positive integers with $1=a_1<a_2<a_3$. There is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ if and only if $1\in S$. Furthermore, if $1\in S$, then the minimum order of a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ is

$$\begin{array}{cc}6& if S=\{1,2,3\}\hfill \\ a_3+1& if S=\{1,2,a_3\}where a_3\ge 4\hfill \\ a_3+3a_2-6& if S=\{1,a_2,a_3\}where a_2\ge 3.\hfill \end{array}$$

Proof. 

First, suppose that G is a locally irregular-connected graph of cycle rank 2 with $\mathcal{D}\left(G\right)=S$. Since $\left|\mathcal{D}\right(G\left)\right|=3$, it follows that G is not one of the graphs in Figure 6. Hence, G contains an end-vertex and so $1\in S$. Next, let $S=\{a_1,a_2,a_3\}$ where $1=a_1<a_2<a_3$.

⋆

If $a_2=2$ and $a_3=3$, then let G be one of the two graphs of maximum degree 3 in Figure 8. By Theorem 8, these two graphs are locally irregular-connected. The graph of type I has order 8, while the graph of type II has order 6.

⋆

If $a_2=2$ and $a_3\ge 4$, then let G be the graph of cycle rank 2 obtained from the graph $K_4-e$ by adding $a_3-3$ pendant edges at each of the two vertices of degree 3 in $K_4-e$. This is illustrated in Figure 8 for $a_3=4$. Then, the order of G is $2a_3-2$ and $\mathcal{D}\left(G\right)=\{1,2,a_3\}$. By Theorem 8, G is locally irregular-connected.

⋆

If $a_2\ge 3$ and $a_3\ge 4$, then let G be the graph of cycle rank 2 obtained from the graph $K_4-e$ by adding $a_2-2$ pendant edges at each of the two vertices of degree 2 in $K_4-e$ and adding $a_3-3$ pendant edges at each of the two vertices of degree 3 in $K_4-e$. Then, the order of G is $2(a_2+a_3-3)$ and $\mathcal{D}\left(G\right)=\{1,a_2,a_3\}$. Since G has exactly one regular edge that lies on a cycle, it follows by Theorem 8 that G is locally irregular-connected.

Among all locally irregular-connected graphs of cycle rank 2 with degree set $S=\{a_1,a_2,a_3\}$, where $1=a_1<a_2<a_3$, let G be one of minimum order n. Then, the size of G is $n+1$. We consider two cases, according to whether $a_2=2$ or $a_2\ge 3$.

Case 1. $a_2=2$. There are two possibilities here depending on whether $a_3=3$ or $a_3\ge 4$.

Subcase $1.1$. $a_3=3$. First, suppose that G is of type I. Since $\mathrm{\Delta }\left(G\right)=3$, the two cycles $C_1$ and $C_2$ of order at least 3 are vertex-disjoint. Since G is locally irregular-connected, G contains no regular bridge. Thus, the path joining $C_1$ and $C_2$ has order at least 3. Since $1\in S$, there is at least one end-vertex in G. Therefore, $n\ge 8$. (In fact, $n=8$ is the minimum order of such a graph.)

Next, suppose that G is of type II. Since $1\in S$, it follows that $G\ne K_4-e$ and so $n\ge 5$. We show in fact that $n\ge 6$. Assume, to the contrary, that $n=5$. Since $S=\{1,2,3\}$, it follows that G is the graph obtained from $K_4-e$ by adding an end-vertex that is joined to a vertex of degree 2 in $K_4-e$. This implies that all three vertices on one cycle of G have degree 3. Hence, G is not locally irregular-connected by Theorem 8 and so $n\ge 6$. Since there is a locally irregular-connected graph of cycle rank 2 that has order 6 and degree set $\{1,2,3\}$, it follows that $n=6$.

Subcase $1.2$. $a_3\ge 4$. Let x be a vertex of degree $a_3$. First, suppose that G is of type I. Since there are at least 5 vertices lying on a cycle of G, it follows that G has at least 4 vertices (distinct from x) having degree at least 2 and the remaining $n-5$ vertices have degree at least 1. Thus,

$$2(n+1)\ge a_3+4·2+(n-5)$$

and so $n\ge a_3+1$. Next, suppose that G is of type II. Since there are at least 4 vertices lying on a cycle of G, it follows that G has at least 3 vertices (distinct from x) such that one has degree at least 3 and the other two have degree at least 2. The remaining $n-4$ vertices of G have degree at least 1. Thus,

$$2(n+1)\ge a_3+3+2·2+(n-4)$$

and so $n\ge a_3+1$. There are locally irregular-connected graphs of cycle rank 2 having order $a_3+1$ and degree set $\{1,2,a_3\}$ where $a_3\ge 4$. One such graph is obtained from two triangles with exactly one common vertex v by attaching $a_3-4$ pendant edges at v. Thus, $n=a_3+1$.

Case 2. $a_2\ge 3$. Let x be a vertex of degree $a_3\ge 4$. First, suppose that G is of type I. Since there are at least 5 vertices lying on a cycle of G, it follows that G has at least 4 vertices (distinct from x) having degree at least $a_2$ and the remaining $n-5$ vertices have degree at least 1. Thus,

$$2(n+1)\ge a_3+4·a_2+(n-5)$$

and so $n\ge a_3+4a_2-7$. Next, suppose that G is of type II. Since there are at least 4 vertices lying on a cycle of G, it follows that G has at least 3 vertices (distinct from x) having degree at least $a_2$. The remaining $n-4$ vertices of G have degree at least 1. Thus,

$$2(n+1)=\ge a_3+3a_2+(n-4)$$

and so $n\ge a_3+3a_2-6$. Since $a_3+4a_2-7>a_3+3a_2-6$, and it follows that $n\ge a_3+3a_2-6$ when $a_2\ge 3$. There are locally irregular-connected graphs of cycle rank 2 having order $a_3+3a_2-6$ and degree set $\{1,a_2,a_3\}$, where $a_2\ge 3$. One such graph is obtained from the graph $K_4-e$ by (1) adding $a_2-2$ pendant edges at each of the two vertices of degree 2 in $K_4-e$, (2) adding $a_2-3$ pendant edges at one vertex of degree 3 in $K_4-e$, and (3) adding $a_3-3$ pendant edges at the other vertex of degree 3 in $K_4-e$. Therefore, $n=a_3+3a_2-6$. □

Proposition 7.

Let $S=\{a_1,a_2,\dots ,a_k\}$ be a set of $k\ge 4$ positive integers with $a_1<a_2<\cdots <a_k$. There is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ if and only if $1\in S$. Furthermore, if $1\in S$, then the minimum order of a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ is

$$\begin{array}{cc}{\displaystyle {\sum }_{i=1}^4}(a_i-1)+a_2-3& if k=4\hfill \\ {\displaystyle {\sum }_{i=1}^k}(a_i-1)-2& if k\ge 5.\hfill \end{array}$$

Proof. 

First, suppose that G is a locally irregular-connected graph of cycle rank 2 with $\mathcal{D}\left(G\right)=S$. Since $\left|\mathcal{D}\right(G\left)\right|\ge 4$, it follows that G is not one of the graphs in Figure 6. Hence, G contains an end-vertex and so $1\in S$. Next, let $S=\{a_1,a_2,\dots ,a_k\}$ where $1=a_1<a_2<\cdots <a_k$.

⋆

If $k=4$, then let G be the graph obtained from $K_4-e$ by (1) adding $a_2-2$ pendant edges at each of the two vertices of degree 2 in $K_4-e$, (2) adding $a_3-3$ pendant edges at one vertex of degree 3 in $K_4-e$, and (3) adding $a_4-3$ pendant edges at the other vertex of degree 3 in $K_4-e$. Then, the order of G is $2a_2+a_3+a_4-6$ and $\mathcal{D}\left(G\right)=S$.

⋆

If $k\ge 5$, then let F be the graph of cycle rank 2 obtained from the $(k-1)$-cycle $C=(v_1,v_2,\dots ,v_{k-1},v_1)$ by adding the edge $v_2{v}_{k-1}$. The graph G is obtained from F by (1) adding $a_{i+1}-2$ pendant edges at $v_i$ for $i=1$ and $3\le i\le k-2$ and (2) adding $a_{i+1}-3$ pendant edges at $v_i$ for $i=2$ and $i=k-1$. The order of G is

$$\begin{array}{ccc}\hfill (a_2-2)+(a_3-3)+{\displaystyle \sum _{i=4}^{k-1}}(a_i-2)+(a_k-3)+(k-1)& =& {\displaystyle \sum _{i=2}^k}(a_i-2)+k-3\hfill \\ & =& {\displaystyle \sum _{i=1}^k}(a_i-1)-2\hfill \end{array}$$

and $\mathcal{D}\left(G\right)=S$.

Among all locally irregular-connected graphs of cycle rank 2 with degree set S$=${$a_1$, $a_2$, …, $a_k$}, where $1=a_1<a_2<\cdots <a_k$, let G be one of minimum order n. Then, the size of G is $n+1$. We consider two cases, according to whether $k=4$ or $k\ge 5$.

Case 1. $k=4$. The graph G contains at least one vertex of degree $a_i$ for $1\le i\le 4$. Since there are at least four vertices that are not end-vertices, there are at least two non-end vertices of the same degree $a_i$ for some integer i with $2\le i\le 4$. The remaining $n-5$ vertices have degree 1 or more. Hence,

$$2(n+1)\ge \left({\displaystyle \sum _{i=1}^4}{a}_i\right)+a_2+(n-5)=\left({\displaystyle \sum _{i=1}^4}(a_i-1)\right)+a_2+(n-1)$$

so

$$n\ge \left({\displaystyle \sum _{i=1}^4}(a_i-1)\right)+a_2-3=2a_2+a_3+a_4-6.$$

Since there is a locally irregular-connected graph of cycle rank 2 that has order ${\sum }_{i=1}^4(a_i-1)+a_2-3$ and degree set S, it follows that $n=\left({\sum }_{i=1}^4(a_i-1)\right)+a_2-3$.

Case 2. $k\ge 5$. The graph G contains at least one vertex of degree $a_i$ for $1\le i\le k$ and the remaining $n-k$ vertices have degree 1 or more. Hence,

$$2(n+1)\ge \left({\displaystyle \sum _{i=1}^k}{a}_i\right)+(n-k)=\left({\displaystyle \sum _{i=1}^k}(a_i-1)\right)+n$$

and so $n\ge {\sum }_{i=1}^k(a_i-1)-2.$ Since there is a locally irregular-connected graph of cycle rank 2 that has order ${\sum }_{i=1}^k(a_i-1)-2$ and degree set S, it follows that $n={\sum }_{i=1}^k(a_i-1)-2$. □

Propositions 5–7 give rise to the following result.

Theorem 9.

For an integer $k\ge 2$, let $S=\{a_1,a_2,\dots ,a_k\}$ be a set of k positive integers with $a_1<a_2<\cdots <a_k$.

(1)

If $k=2$, then there is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ if and only if $S=\{2,3\}$ or $S=\{2,4\}$. In either case, the minimum order of a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ is $a_2+1$.

(2)

If $k\ge 3$, then there is a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ if and only if $1\in S$. Furthermore, if $1\in S$, then the minimum order of a locally irregular-connected graph G of cycle rank 2 with $\mathcal{D}\left(G\right)=S$ is

$$\begin{array}{cc}6& if S=\{1,2,3\}\hfill \\ a_3+1& if S=\{1,2,a_3\} where a_3\ge 4\hfill \\ a_3+3a_2-6& if S=\{1,a_2,a_3\} where a_2\ge 3\hfill \\ {\displaystyle {\sum }_{i=1}^4}(a_i-1)+a_2-3& if k=4\hfill \\ {\displaystyle {\sum }_{i=1}^k}(a_i-1)-2& if k\ge 5.\hfill \end{array}$$

5. In Closing

While a set $S=\{a_1,a_2,\dots ,a_k\}$ of $k\ge 2$ positive integers with $a_1<a_2<\cdots <a_k$. can be realized as the degree set of a locally irregular-connected graph of order $a_k+1$ by Theorem 5, the degree set of a locally irregular-connected graph of cycle rank at most 2 has a specific structure. Furthermore, the minimum order of a locally irregular-connected graph of cycle rank at most 2 with degree set S may not be $a_k+1$. Therefore, we conclude with the following problems.

Problem 1.

Let $S=\{a_1,a_2,\dots ,a_k\}$ be a set of $k\ge 2$ positive integers with $a_1<a_2<\cdots <a_k$ and let $r\ge 3$ be an integer.

(1)

Under what conditions is S the degree set $\mathcal{D}\left(G\right)$ of a locally irregular-connected graph G of cycle rank r?

(2)

What is the minimum order of a locally irregular-connected graph G of cycle rank r with $\mathcal{D}\left(G\right)=S$?