An Erd\H{o}s-Gallai-type theorem for keyrings

TL;DR

This paper proves a new Erdős-Gallai-type theorem establishing that graphs with sufficiently high average degree contain specific keyring subgraphs with leaves and a minimum number of edges.

Contribution

It introduces a novel theorem characterizing the existence of keyring subgraphs in graphs based on average degree constraints.

Findings

Graphs with average degree > k-1 contain keyrings with r leaves and at least k edges.

The theorem applies for all r ≤ (k-1)/2.

Provides a new extremal condition for keyring subgraphs.

Abstract

A keyring is a graph obtained by appending $r \geq 1$ leaves to one of the vertices of a cycle. We prove that for every $r \leq (k-1)/2$, any graph with average degree more than $k-1$ contains a keyring with $r$ leaves and at least $k$ edges.

Equations

$$|N_G(w)\backslash X|$$

$$\le$$

$$d_G(X)$$

$$\le$$

$$X_1=\{u_2,u_3,\dots ,u_{m-\lambda +1}\},$$

$$X_2=\{u_{\lambda -1},u_{\lambda },\dots ,u_{m-2}\},$$

$$X_3=\{u_1,u_2,\dots ,u_{m-1}\}\backslash (X_1\cup X_2).$$

$$|N_H(u_0)\cap (X_1\cup X_2)|$$

$$\sum _{i=0}^{m-1}(t_i+2r_i)=2d_G(X)>m\cdot (k-1).$$

$$d_H(u_i)=t_i$$

Full text

1 Introduction

The graphs considered in this paper are simple and undirected. The sets of vertices and edges of a graph $G$ are denoted by $V(G)$ and $E(G)$, respectively. Their sizes are $v(G)=|V(G)|$ and $e(G)=|E(G)|$. For $u\in V(G)$, we denote by $N_{G}(u)$ the set of vertices adjacent to $u$ in $G$, and $d_{G}(u)=|N_{G}(u)|$ is the degree of $u$. For a subset of vertices $X\subset V(G)$, the number of edges of $G$ with at least one end in $X$ is denoted by $d_{G}(X)$.

Let $\mathcal{D}_{k}$ be the class of all graphs whose average degree is strictly more than $k-1$. In other words, $G\in\mathcal{D}_{k}\;$ if $\;2e(G)>(k-1)v(G)$. Erdős and Gallai proved the following two statements.

Theorem 1.1 (Theorem 2.6 EG )

Any graph $G\in\mathcal{D}_{k}$ contains a path of $k$ edges.

Theorem 1.2 (Theorem 2.7 EG )

Any graph $G\in\mathcal{D}_{k}$ contains a cycle with at least $k$ edges.

In fact, the statement of Theorem 2.7 in EG is even stronger: an $n$-vertex graph without cycles of length $k$ (or more) has at most $(n-1)(k-1)/2$ edges, and less than that if $n-1$ is not a multiple of $k-2$. Faudree and Schelp Faudree and independently Kopylov Kopylov determined (for every $n$ and $k$) the largest number of edges in an $n$-vertex graph without a $k$-edge path. They also described the extremal graphs. Kopylov Kopylov did the same for 2-connected graphs without cycles of length $k$ and more.

Together, Theorem 1.1 and a simple observation that every graph $G\in\mathcal{D}_{k}$ contains a star with $k$ edges, led to the following conjecture formulated by Erdős and Sós (see Erd ).

Erdős–Sós Conjecture. *Any graph $G\in\mathcal{D}_{k}$ contains every tree with $k$ edges. *

Let $\mathcal{T}_{k}$ be the class of $k$-edge trees $T$ such that every $G\in\mathcal{D}_{k}$ contains $T$ as a subgraph. The Erdős–Sós conjecture states that all $k$-edge trees belong to $\mathcal{T}_{k}$. Ajtai, Komlós, Simonovits, and Szemerédi AKSS1 ; AKSS2 ; AKSS3 proved that there exists $k_{0}$ such that the conjecture holds for all $k>k_{0}$. Still, the general case has not been solved, and only partial results have been obtained.

A spider $S(k_{1},k_{2},\ldots,k_{r})$ is a tree obtained from $r$ disjoint paths of lengths $k_{1},k_{2},\ldots,k_{r}$ by combining their starting vertices into one. The combined vertex has degree $r$ and is called the center. Obviously, $S(k_{1},k_{2})$ is just a path of length $k_{1}+k_{2}$. Woźniak Woz proved that $S(k_{1},k_{2},\ldots,k_{r})\in\mathcal{T}_{k}$ if $k_{1},k_{2},\ldots,k_{r}\leq 2$. Fan and Sun FS proved that $S(k_{1},k_{2},\ldots,k_{r})\in\mathcal{T}_{k}$ when $r=3$ or $k_{1},k_{2},\ldots,k_{r}\leq 4$. Very recently, Fan, Hong and Liu FHL proved that all spiders belong to $\mathcal{T}_{k}$. McLennan McL proved that $T\in\mathcal{T}_{k}$ when $T$ is a tree of diameter 4.

It was mentioned in Section 3 of Mos that Perles proved the Erdős–Sós conjecture for caterpillars (those are trees which do not contain $S(2,2,2)$ as a subgraph). The proof was published in Kalai only recently.

A vertex which is adjacent to a leaf is called a preleaf. It was proved in Sid that if a tree $T$ has a preleaf which is adjacent to at least $(k-1)/2$ leaves then $T\in\mathcal{T}_{k}$. We will prove in Section 2 a stronger statement:

Theorem 1.3

If a tree $T$ with $k$ edges has $p$ preleaves and one of them is adjacent to at least $(k-p-1)/2$ leaves then $T\in\mathcal{T}_{k}$.

Trees and cycles are not the only subgraphs whose existence can be deduced from the graph’s average degree. As a referee of this paper pointed out, it was Turán who formulated the very problems for which the Erdős–Gallai theorems provide the answers. He asked the maximal number of edges in an $n$-vertex graph which does not contain a lasso, that is a cycle and a path having one common vertex. Fan and Sun FS solved the lasso problem (but did not formulate their result explicitly) within the proof of their Theorem 3.1. In this paper, we consider a similar forbidden pattern. A keyring $C_{r}(l)$ is a $(l+r)$-edge graph obtained from a cycle of length $l$ by appending $r$ leaves to one of its vertices. This vertex has degree $r+2$ and is called the center of the keyring. In Section 3, we prove an analog of Theorem 1.2 for keyrings:

Theorem 1.4

For any positive integer $r\leq(k-1)/2$, every graph $G\in\mathcal{D}_{k}$ contains a keyring with $r$ leaves and at least $k$ edges.

A graph is called $k$-minimal if it belongs to $\mathcal{D}_{k}$ but none of its proper subgraphs belongs to $\mathcal{D}_{k}$. Obviously, any graph from $\mathcal{D}_{k}$ contains a $k$-minimal subgraph. Thus, in proofs of Theorems 1.4, 1.3 and similar statements, instead of considering all graphs $G\in\mathcal{D}_{k}$, it is sufficient to consider only those which are $k$-minimal. The main help comes from the simple observation:

Remark 1

If a graph $G$ is $k$-minimal then for any subset of vertices $X\subset V(G)$ the number of edges of $G$ with at least one end in $X$ is strictly more than $\frac{k-1}{2}|X|$.

2 Proof of Theorem 1.3

Let $m(T)$ denote the largest number of leaves connected to a single preleaf in $T$, $L(T)$ denote the set of leaves of $T$, and $P(T)$ denote the set of preleaves. In this proof, we will keep $k$ fixed and use induction in $m(T)$. The basis of induction is the case $m(T)=k$. In this case, $T$ is a star and belongs to $\mathcal{T}_{k}$. Now we are going to prove the inductive step. Suppose that $m(T)=m<k$ and the statement of the theorem holds for all trees $T^{\prime}$ where $m(T^{\prime})>m$. (We will make use of the assumption $m\geq(k-p-1)/2$ later.)

Consider a $k$-minimal graph $G$. We need to show that $G$ contains $T$ as a subgraph. Let $u$ be a preleaf of $T$ with $m$ leaves. Let $v$ be another preleaf of $T$ (since $m<k$, $\;T$ is not a star and has at least two preleaves). Now we disconnect in $T$ one of the leaves attached to $v$ and reconnect it to $u$ instead. The resulting tree $T^{\prime}$ has $k$ edges and $m(T^{\prime})=m+1$. Since $|P(T^{\prime})|\geq|P(T)|-1$, we have $m(T^{\prime})=m+1\geq(k-|P(T^{\prime})|-1)/2$, and by the induction hypothesis, $G$ must contain a copy of $T^{\prime}$. We are going to transform this copy of $T^{\prime}$ into a copy of $T$ by changing the assignment of leaves to the preleaves. From now on, we will assume that $V(T^{\prime})$ is a subset of $V(G)$, and $E(T^{\prime})$ is a subset of $E(G)$. We define three non-overlapping sets of vertices: $A=P(T^{\prime})\cup\{v\}$, $B=L(T^{\prime})\;\backslash\;\{v\}$, $\;C=V(G)\;\backslash\;V(T^{\prime})$. Notice that $v$ in $T^{\prime}$ might remain a preleaf or become a leaf or neither. In any case, we want to ensure that $v$ belongs to $A$ and not to $B$. Thus, $|A|=|P(T)|=p$, $|B|=|L(T)|$. Consider a bipartite directed graph $F$ whose vertex set is $A\cup B\cup C$ with directed edges of two types:

$(a,b)$ where $a\in A$, $\;b\in(B\cup C)$ and $\{a,b\}\in E(G)$; and 2. 2.

$(b,a)$ where $a\in A$, $\;b\in B$, $\;a$ and $b$ are adjacent in $T^{\prime}$ (which means that leaf $a$ is connected to preleaf $b$).

If there exists a path in $F$ from $v$ either to $u$ or to $C$, we will be able to find a copy of $T$ in $G$. Indeed, let $(a_{0},b_{0},a_{1},b_{1},\ldots,a_{q},b_{q})$ be a simple path where $a_{0}=v$, $b_{q}\in C$, $\{a_{i},b_{i}\}\in E(G)$ $(i=0,1,\ldots,q)$, and $b_{i}\in B$, $\;a_{i+1}\in A$ are adjacent in $T^{\prime}$ $(i=0,1,\ldots,q-1)$. Then we can add to $T^{\prime}$ vertex $b_{q}$ as well as edges $\{a_{i},b_{i}\}$ for $i=0,1,\ldots,q$, remove edges $\{a_{i+1},b_{i}\}$ for $i=0,1,\ldots,q-1$ and remove one of the leaves connected to $u$. The resulting subgraph of $G$ is a copy of $T$.

Similarly, let $(a_{0},b_{0},a_{1},b_{1},\ldots,a_{q})$ be a simple path where $a_{0}=v$, $a_{q}=u$, $\{a_{i},b_{i}\}\in E(G)$ $(i=0,1,\ldots,q-1)$, and $b_{i}\in B$, $\;a_{i+1}\in A$ are adjacent in $T^{\prime}$ $(i=0,1,\ldots,q-1)$. Then we can add to $T^{\prime}$ edges $\{a_{i},b_{i}\}$ for $i=0,1,\ldots,q-1$ and remove edges $\{a_{i+1},b_{i}\}$ for $i=0,1,\ldots,q-1$. The resulting subgraph of $G$ is a copy of $T$.

Finally, consider the case when $C\cup\{u\}$ is unreachable from $v$ in $F$. We split A into two subsets: $X$ consists of the vertices that are reachable from $v$ in $F$, and $Y$ consists of the rest. Obviously, $v\in X$ and $u\in Y$. Let $Z$ be the set of $m+1$ leaves that are attached to $u$. There are no edges in $G$ between $X$ and $C\cup Y\cup Z$. Since $|X|+|Y|=p$ and $|Z|=m+1\geq(k-p+1)/2$, then for any $w\in X$ we can estimate:

[TABLE]

Using $|X|\leq p-1$, we get

[TABLE]

which, according to Remark 1, contradicts the $k$-minimality of $G$. ∎

3 Proof of Theorem 1.4

Lemma 1

Fix integers $\lambda\geq 2$ and $t\geq 1$. Let $H$ be a Hamiltonian graph with $m\geq\lambda$ vertices, and $u_{0}$ be one of them. If $d_{H}(u_{0})\geq 2t-1+\max\{2\lambda-m-1,2\}$ then there exists $l\geq\lambda$ such that $H$ contains a copy of $C_{t}(l)$ with the center at $u_{0}$.

Proof

Let $(u_{0},u_{1},\ldots,u_{m-1},u_{m}=u_{0})$ be a Hamiltonian cycle in $H$. Suppose first that $m=\lambda$. Since $m\geq 3$, we get $2\lambda-m-1=m-1\geq 2$ and $d_{H}(u_{0})\geq 2t+m-2\geq m=v(H)$ which is impossible. Therefore, we may assume $m\geq\lambda+1$. Denote

[TABLE]

[TABLE]

[TABLE]

Clearly, $|X_{1}\cup X_{2}|=|X_{1}|+|X_{2}|=2(m-\lambda)$ when $m-\lambda+1<\lambda-1$, and $|X_{1}\cup X_{2}|=m-3$ when $m-\lambda+1\geq\lambda-1$. Thus, $|X_{1}\cup X_{2}|=\min\{2m-2\lambda,m-3\}$ and $|X_{3}|=(m-1)-|X_{1}\cup X_{2}|=\max\{2\lambda-m-1,2\}$. Note that

[TABLE]

Thus, there exist $u_{i_{1}},u_{i_{2}},\ldots,u_{i_{2t-1}}\in N_{H}(u_{0})\cap(X_{1}\cup X_{2})$ where $i_{1}<i_{2}<\ldots<i_{2t-1}$. If $u_{i_{t}}\in X_{1}$ then $i_{t}\leq m-\lambda+1$. In this case, $u_{0}$ is adjacent to $u_{1},u_{i_{1}},u_{i_{2}},\ldots,\allowbreak u_{i_{t-1}}$ and belongs to the cycle $(u_{0},u_{i_{t}},u_{i_{t}+1},\ldots,u_{m}=u_{0})$ whose length is $m-i_{t}+1$. This produces a copy of $C_{t}(l)$ with the center at $u_{0}$ where $l=m-i_{t}+1\geq m-(m-\lambda+1)+1=\lambda$. Alternatively, if $u_{i_{t}}\in X_{2}$ then $i_{t}\geq\lambda-1$. In this case, $u_{0}$ is adjacent to $u_{i_{t+1}},u_{i_{t+2}},\ldots,u_{i_{2t-1}},u_{m-1}$ and belongs to the cycle $(u_{0},u_{1},\ldots,u_{i_{t}},u_{m}=u_{0})$ whose length is $i_{t}+1$. This produces a copy of $C_{t}(l)$ with the center at $u_{0}$ where $l=i_{t}+1\geq\lambda$. ∎

Proof of Theorem 1.4 We are going to show that any $k$-minimal graph $G$ contains a copy of $C_{r}(l)$. By Theorem 1.2, $G$ contains a cycle $(u_{0},u_{1},\ldots,u_{m-1},u_{m}=u_{0})$ of length $m\geq k$. Let $X=\{u_{0},u_{1},\ldots,u_{m-1}\}$. Then $d_{G}(u_{i})=t_{i}+r_{i}$ where $t_{i}=|N_{G}(u_{i})\cap X|$ and $r_{i}=|N_{G}(u_{i})\backslash X|$ for $i=0,1,\ldots,m-1$. According to Remark 1,

[TABLE]

Thus, there is such an index $i$ that $t_{i}+2r_{i}\geq k$. If $r_{i}\geq r$ then $G$ contains a copy of $C_{r}(m)$ with the center at $u_{i}$. Suppose $r_{i}<r$. Let $H$ be the subgraph of $G$ induced by $X$. We are going to apply Lemma 1 with parameters $t=r-r_{i}$ and $\lambda=k-r+1$. To be able to invoke it, we need to demonstrate that $\lambda\geq 2$ and

[TABLE]

Indeed, on one hand, since $t_{i}+2r_{i}\geq k$, we get $t_{i}\geq k-2r_{i}=2k-2r_{i}-k\geq 2k-2r_{i}-m$. On the other hand, since $r\leq(k-1)/2$, we get $t_{i}\geq k-2r_{i}\geq 2r-2r_{i}+1$. Also, $r\leq(k-1)/2$ implies $\lambda=k-r+1\geq 2$. By Lemma 1, $H$ contains a copy of $C_{t}(l)$ with the center at $u_{i}$ where $t=r-r_{i}$ and $l\geq\lambda=k-r+1$. Now we use the $r_{i}$ vertices from $N_{G}(u_{i})\backslash X$ to extend it to a copy of $C_{r}(l)$ where $l+r\geq k+1$. ∎

Acknowledgements.

The author would like to thank the referees for their suggestions and comments.