The Structure of Graphs Whose Vertices of Degree at Least Three Are at a Distance of at Least Three

Abstract

Let G be a connected graph in which the distance between any two distinct vertices of degree of at least three is at least three. We find the structure of G. It turns out that G decomposes into a tree and a matching.

1. Introduction

In this paper, the considered graphs are finite and simple; they have neither loops nor parallel edges. The vertex set of a graph H will be denoted by $V\left(H\right)$, while its edge set will be denoted by $E\left(H\right)$. Let $G=(V,E)$ be a graph. If $e=uv\in E$, then we say u and v are neighbors, u and v are endpoints of e, v is a neighbor of u, and e is incident to u (also to v), and we say that u is incident to e. Two edges, $e=uv$ and $e^{\prime }=u^{\prime }{v}^{\prime }$, are independent if $\{u,v\}\cap \{u^{\prime },v^{\prime }\}=\varphi$. A matching M in G is a set of independent edges in G. The matching M is said to be a perfect match between the vertices of $S\subseteq V$ if the set of endpoints of edges in M is equal to S. The set of all neighbors of a vertex u is denoted by $N_G\left(u\right)$. The degree of a vertex u, denoted by $d_G\left(u\right)$, is the number of its neighbors. A leaf in G is a vertex of degree one. The length of a path or a cycle is the number of its edges. The distance between two vertices, x and y, in G, denoted by $dis{t}_G(x,y)$, is the length of the smallest path in G between x and y. The graph G is connected if there is a path between any two of its vertices. A graph is acyclic (also called a forest) if it has no cycle. A star T is a tree where one of the vertices, say x, is adjacent in T to all the other vertices of T; that is, the vertex set of a star T is $\left\{x\right\}\cup N_T\left(x\right)$. The subscript G in all of the previous notations is sometimes omitted if the graph is sufficiently clear from the context.

In addressing specific problems in graph theory, uncovering the structural properties of certain graphs often proves instrumental and informative, providing key insights that aid in tackling such challenges (for example, see [1,2,3]).

The distribution of the degree of the vertices significantly influences the characterization and detection of structural patterns in a graph. For example, if the degree of every vertex is two, then G is a disjointed union of cycles. In 1736, Euler proved that a connected graph G has an Eulerian circuit if and only if the degree of every vertex is even [4]; that is, we can traverse every edge exactly once starting from any vertex x and returning back to x. According to Brooks’ theorem [5], if G is a graph of maximum degree k that is neither a clique nor an odd cycle, then its vertex set can be partitioned into, at most, k stable sets. Moreover, according to Vizing’s theorem [6], its edge set can be partitioned into k or $k+1$ sets of independent edges. Mader [7,8] proved that high average degree forces a given graph as a minor. Also, if the average degree of a graph is at least $4k$, then it has a $(k+1)$-connected subgraph [9]. It is well known that if T is a tree on n vertices and the minimum degree of a graph G is at least $n-1$, then G contains (a copy of) T.

We investigate the structure of graphs in which any two distinct vertices of degree of at least three are separated by a distance of at least three. Leveraging this structural constraint, we derive an upper bound on the number of edges such graphs can possess. It will be shown that these graphs are sparse, meaning their edge count is significantly lower than the theoretical maximum ${\displaystyle \frac{n(n-1)}{2}}$, where n is the number of vertices (see Theorem 2). Such graphs are well-suited as underlying structures in sparse arc routing problems [10,11,12]. In fact, in the literature on Capacitated Arc Routing Problems (CARP), the structure of the underlying graph plays a crucial role in modeling and algorithm design. For example, the work by Tfaili et al. [10] studies CARP under sparse underlying graphs and shows that algorithmic solutions such as greedy heuristics and tabu-search can effectively exploit the small number of edges to solve min–max and robust variants of CARP under conditions of uncertainty. Similarly, Tfaili et al. [11,12] develop approaches for solving CARP on sparse graphs, including model transformations to sparse capacitated vehicle routing and specialized heuristics that take advantage of sparseness to construct feasible solutions and improve them efficiently with tabu search under dynamic graph settings. These studies focus specifically on routes defined over sparse graphs. Therefore, understanding and characterizing the structure of sparse underlying graphs can be exploited to further improve algorithmic performance.

2. Theoretical Results

We begin with the following simple lemma.

Lemma 1.

Let G be a connected graph such that for any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$. Then, the sets $\left\{x\right\}\cup N_G\left(x\right),x\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ are disjoint.

Proof. 

Let $x\ne y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$. Since $dis{t}_G(x,y)\ge 3$, $xy\notin E\left(G\right)$; hence, $x\notin N\left(y\right)$ and $y\notin N\left(x\right)$.

Assume that there exists $z\in N_G\left(x\right)\cap N_G\left(y\right)$. Then, $xzy$ is a path in G between x and y. Hence, $dis{t}_G(x,y)\le 2$, which is a contradiction. Thus, $N_G\left(x\right)\cap N_G\left(y\right)=Ø$. Therefore, $(\left\{x\right\}\cup N_G\left(x\right))\cap (\left\{y\right\}\cup N_G\left(y\right))=Ø$. □

The existence of spanning trees with specified structural properties, such as bounded degree, bounded numbers of leaves, or branch vertices, has been extensively studied in graph theory (see the survey by Ozeki and Yamashita [13]). In this context, the below lemma guarantees a spanning tree that preserves the degrees of high-degree vertices and other non-leaf vertices under a specific structural condition. This lemma serves as the cornerstone of our findings.

Lemma 2.

Let G be a connected graph such that for any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$. Then, there is a spanning tree T of G such that

• $\forall x\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$; we have $d_T\left(x\right)=d_G\left(x\right)$.
• $\forall x\in V\left(G\right)$ such that x is not a leaf of T; we have $d_T\left(x\right)=d_G\left(x\right)$.

Proof. 

Let H be the graph defined by $V\left(H\right)=V\left(G\right)$, and $E\left(H\right)$ consists only of all the edges incident to vertices of a degree of at least 3. Note that H consists of disjoint stars only due to Lemma 1. So, H is an acyclic spanning graph of G. Let T be a maximal acyclic spanning subgraph of G containing H. Assume that T is not connected. Since $T\subseteq G$ and G are connected, we can find $e\in E\left(G\right)$, which joins two connected components of T. Hence, $T+e$ is also the acyclic spanning graph of G containing H. This contradicts the maximality of T. Thus, T is connected. Since $H\subseteq T$ and T are connected and acyclic, T is a tree containing H. Now, suppose that $x\in V\left(G\right)$ such that $d_G\left(x\right)\ge 3$. Then, $d_G\left(x\right)\ge d_T\left(x\right)\ge d_H\left(x\right)=d_G\left(x\right)$. So, $d_T\left(x\right)=d_G\left(x\right)$, and T is the required tree. Now, suppose that $x\in V\left(G\right)$ such that x is not a leaf of T. If $d_G\left(x\right)\ge 3$, then we have already proved that $d_T\left(x\right)=d_G\left(x\right)$. So, we may suppose that $d_G\left(x\right)\le 2$. Since x is not a leaf of T, $d_T\left(x\right)\ge 2$. But $d_G\left(x\right)\ge d_T\left(x\right)$. Then, $d_G\left(x\right)=d_T\left(x\right)=2$. □

We observe that in the previous lemma, the condition “For any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$” may be relaxed without compromising the validity of the result. Specifically, it suffices to impose the following weaker structural constraint: “The graph H with $V\left(H\right)=V\left(G\right)$ and $E\left(H\right)=\{xy\in E\left(G\right);d_G\left(x\right)\ge 3\}$ is a forest.”

Theorem 1.

Let G be a connected graph without leaves such that for any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$. Then, there is a spanning tree T of G, and there is a matching M in G such that

• $\forall x\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$; we have $d_T\left(x\right)=d_G\left(x\right)$.
• $\forall x\in V\left(G\right)$ such that x is not a leaf of T; we have $d_T\left(x\right)=d_G\left(x\right)$.
• $E\left(G\right)=E\left(T\right)\cup M$ and $E\left(T\right)\cap M=Ø$.
• M is a perfect matching between the leaves of T.

Proof. 

By Lemma 2, there exists a spanning tree T of G such that

• $\forall x\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$; we have $d_T\left(x\right)=d_G\left(x\right)$.
• $\forall x\in V\left(G\right)$ such that x is not a leaf of T; we have $d_T\left(x\right)=d_G\left(x\right)$.

Now, we will show the existence of the perfect matching M between the leaves of T. Let y be any leaf of T. Since $d_G\left(y\right)\ge 2$ and $d_T\left(y\right)=1$, there is an edge $e=y{y}^{\prime }\in E\left(G\right)\backslash E\left(T\right)$ for some vertex $y^{\prime }$. Since $d_G\left(y\right)\ne d_T\left(y\right)$, $d_G\left(y\right)\le 2$; thus, $d_G\left(y\right)=2$. Hence, $e=y{y}^{\prime }$ is the unique edge in $E\left(G\right)\backslash E\left(T\right)$ incident to y. Again, since $e=y{y}^{\prime }\in E\left(G\right)\backslash E\left(T\right)$, $d_G\left(y^{\prime }\right)\ne d_T\left(y^{\prime }\right)$. Hence, $y^{\prime }$ is a leaf of T. Therefore, $e=y{y}^{\prime }$ is an edge between two leaves of T, and it is the unique edge in $\in E\left(G\right)\backslash E\left(T\right)$ incident to y. Therefore, $M=\{y{y}^{\prime };y$ is a leaf of T, and $y{y}^{\prime }\in E\left(G\right)\backslash E\left(T\right)\}$ is a perfect matching between the leaves of T.

According to the definition of M, we have $E\left(T\right)\cap M=Ø$. Now, we will prove that $E\left(G\right)=E\left(T\right)\cup M$. Assume that there exists $e=yz\in E\left(G\right)\backslash \left(E\right(T)\cup M)$. If y is not a leaf of T, then $d_T\left(y\right)=d_G\left(y\right)$ according to the definition of T. Hence, $e\in E\left(T\right)$, which is a contradiction. Suppose that y is a leaf of T. Then, there exists a unique edge $yx\in E\left(T\right)$ (according to the definition of a leaf), and there exists a unique edge $y{y}^{\prime }\in M$ (according to the definition of M). Hence, $x,y^{\prime }$ and z are three distinct neighbors of y. Thus, $d_G\left(y\right)\ge 3$. According to the definition of T, we get $d_T\left(y\right)=d_G\left(y\right)$. Hence, $e=yz\in E\left(T\right)$, a contradiction. □

Corollary 1.

Let G be a connected graph on n vertices without leaves such that for any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$. Then, $n\le \left|E\left(G\right)\right|\le {\displaystyle \frac{3}{2}}(n-1)$.

Proof. 

Let G be a graph, as in the statement. By Theorem 1, there is a spanning tree T of G, and there is a matching M in G such that $E\left(G\right)=E\left(T\right)\cup M$. Let l be the number of leaves of T. Note that $l\le n-1$. The result follows because $\left|E\right(T\left)\right|=n-1$ and $\left|M\right|\le {\displaystyle \frac{l}{2}}\le {\displaystyle \frac{n-1}{2}}$. □

Note that we consider the empty set to be a matching, and it is called the empty matching.

Theorem 2.

Let G be a connected graph such that for any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$. Then, there is a spanning tree T of G, and there is a matching M (possibly empty) in G such that $E\left(G\right)=E\left(T\right)\cup M$.

Proof. 

The proof proceeds by induction on $n=\left|V\right(G\left)\right|$. Clearly, the statement is true for small values of n. Let G be a graph, and suppose that $\left|V\right(G\left)\right|=n$. If G does not have a leaf, then the result follows according to Theorem 1. Otherwise, let u be a leaf of G and let $vu\in E\left(G\right)$. It is clear that removing a leaf does not create shorter distances between vertices of G that are of a degree at least 3. Let $G^{\prime }=G-u$. According to the induction hypothesis, there exists a spanning tree $T^{\prime }$ of $G^{\prime }$ and a matching $M^{\prime }$ in $G^{\prime }$ such that $E\left(G^{\prime }\right)=E\left(T^{\prime }\right)\cup M^{\prime }$. Take $T=T^{\prime }+vu$ and $M=M^{\prime }$. Then, $T^{\prime }$ is a spanning tree of G and $E\left(G\right)=E\left(T\right)\cup M$. □

Corollary 2.

Let G be a connected graph on n vertices such that for any two distinct vertices $x,y\in V\left(G\right)$ with $d_G\left(x\right)\ge 3$ and $d_G\left(y\right)\ge 3$, we have $dis{t}_G(x,y)\ge 3$. Then, $n-1\le \left|E\left(G\right)\right|\le {\displaystyle \frac{3}{2}}n-1$.

Proof. 

Let G be a graph as in the statement. According to Theorem 2, there is a spanning tree T of G, and there is a matching M in G such that $E\left(G\right)=E\left(T\right)\cup M$. The result follows because $\left|E\right(T\left)\right|=n-1$ and $\left|M\right|\le {\displaystyle \frac{n}{2}}$. □

3. Conclusions

We have identified the structure of the graphs in which every pair of distinct vertices of degree of at least three is separated by a distance of at least three. In particular, we demonstrated that such graphs can be decomposed into the union of a tree and a matching. This naturally leads to a broader question: Can a similar structural characterization be established for graphs in which vertices of degree of at least four are mutually at a distance of at least three? More generally, what structural properties emerge when the degree threshold is replaced by an arbitrary integer $k\ge 3$?