Spanning trees of a claw-free graph whose reducible stems have few leaves

In this paper, we only consider finite graphs without loops or multiple edges. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. For any vertex $v\in V(G)$, we use $N_{G}(v)$ and $\deg_{G}(v)$ (or $N(v)$ and $\deg(v)$ if there is no ambiguity) to denote the set of neighbors of $v$ and the degree of $v$ in $G$, respectively. For any $X\subseteq V(G)$, we denote by $|X|$ the cardinality of $X$. Sometime, we denote by $|G|$ instead of $|V(G)|.$ We define $N_{G}(X)=\bigcup\limits_{x\in X}N_{G}(x)$ and $\deg_{G}(X)=\sum\limits_{x\in X}\deg_{G}(x)$. The subgraph of $G$ induced by $X$ is denoted by $G[X]$. We define $G-uv$ to be the graph obtained from $G$ by deleting the edge $uv\in E(G)$, and $G+uv$ to be the graph obtained from $G$ by adding a new edge $uv$ joining two non-adjacent vertices $u$ and $v$ of $G$. For two vertices $u$ and $v$ of $G$, the distance between $u$ and $v$ in $G$ is denoted by $d_{G}(u,v)$. We use $K_{n}$ to denote the complete graph on $n$ vertices. We write $A:=B$ to rename $B$ as $A$. We refer to [4] for terminology and notation not defined here.

For an integer $m\geq 2,$ let $\alpha^{m}(G)$ denote the number defined by

$\alpha^{m}(G)=\max\{|S|:S\subseteq V(G),d_{G}(x,y)\geq m\,$ for all distinct vertices $\,x,y\in S\}.$

For an integer $p\geq 2$, we define

$\sigma_{p}^{m}(G)=\min\{\deg_{G}(S):S\subseteq V(G),|S|=p,d_{G}(x,y)\geq m$ for all distinct vertices $\;x,y\in S\}.$

For convenience, we define $\sigma^{m}_{p}(G)=+\infty$ if $\alpha^{m}(G)<p$. We note that, $\alpha^{2}(G)$ is often written $\alpha(G)$, which is independence number of $G,$ and $\sigma_{p}^{2}(G)$ is often written $\sigma_{p}(G)$, which is minimum degree sum of $p$ independent vertices.

Let $T$ be a tree. A vertex of degree one is a leaf of $T$ and a vertex of degree at least three is a branch vertex of $T$. The set of leaves of $T$ is denoted by $L(T)$ and the set of branch vertices of $T$ is denoted by $B(T)$. The subtree $T-L(T)$ of $T$ is called the stem of $T$ and is denoted by $Stem(T)$.

For two distinct vertices $u,v$ of $T$, let $P_{T}[u,v]$ denote the unique path in $T$ connecting $u$ and $v.$ For a leaf $x$ of $T$, let $y_{x}$ denote the nearest branch vertex to $x$. For every leaf $x$ of $T$, we remove the path $P_{T}[x,y_{x})$ from $T$, where $P_{T}[x,y_{x})$ denotes the path connecting $x$ to $y_{x}$ in $T$ but not containing $y_{x}$. Moreover, the path $P_{T}[x,y_{x})$ is called the leaf-branch path of $T$ incident to $x$ and $y_{x}$ and denoted by $B_{x}$. The resulting subtree of $T$ is called the reducible stem of $T$ and denoted by $R\_Stem(T)$.

There are several sufficient conditions (such as the independence number conditions and the degree sum conditions) for a graph $G$ to have a spanning tree with a bounded number of leaves or branch vertices. Win [20] obtained the following theorem, which confirms a conjecture of Las Vergnas [16]. Beside that, recently, the author [6] also gave an improvement of Win by giving an independence number condition for a graph having a spanning tree which covers a certain subset of $V(G)$ and has at most $l$ leaves.

Later, Broersma and Tunistra gave the following degree sum condition for a graph to have a spanning tree with at most $l$ leaves.

Motivating by Theorem 1.1, a natural question is whether we can find sharp sufficient conditions of $\sigma_{l+1}(G)$ for a connected graph $G$ has a few leaves. This question is still open. But, in certain graph classes, the answers have been determined.

For a positive integer $t\geq 3,$ a graph $G$ is said to be $K_{1,t}-$ free graph if it contains no $K_{1,t}$ as an induced subgraph. If $t=3,$ the $K_{1,3}-$ free graph is also called the claw-free graph. About this graph class, Kano, Kyaw, Matsuda, Ozeki, Saito and Yamashita proved the following theorem.

For other graph classes, we refer the readers to see [2], [3], [5], [14], [15] and [17] for examples.

The first main purpose of this paper is to give a new short proof for Theorem 1.3 base on the new technique of Gould and Shull in [5].

Moreover, many researchers studied spanning trees in connected graphs whose stems have a bounded number of leaves or branch vertices (see [7], [12], [13], [18] and [19] for more details). We introduce here some results on spanning trees whose stems have a few leaves or branch vertices.

Recently, Ha, Hanh and Loan gave a sufficient condition for a graph to have a spanning tree whose reducible stem has few leaves. In particular, they proved the following theorem.

After that, Ha, Hanh, Loan and Pham also gave a sufficient condition for a graph to have a spanning tree whose reducible stem has few branch vertices.

Very recently, Hanh stated the following theorem.

The open question is whether we may give a sharp condition of $\sigma_{3k+3}(G)$ to show that $G$ has a spanning tree whose reducible stem has at most $k$ leaves.

For the last purpose of this paper, we will give an affirmative answer to this question. In particular, we prove the following theorem.

To show that our result is sharp, we will give the following example. Let $k\geq 2$ and $m\geq 1$ be integers, and let $R_{0},R_{1},...,R_{k}$ and $H_{0},H_{1},...,H_{k}$ be $2k+2$ disjoint copies of the complete graph $K_{m}$ of order $m$. Let $D$ be a complete graph with $V(D)=\{w_{i}\}_{i=0}^{k}.$ Let $\{x_{i},x_{iy},x_{iz}\}_{i=0}^{k}$ be $3k+3$ vertices not contained in $\cup_{i=0}^{k}\bigg{(}V(R_{i})\cup V(H_{i})\cup\{w_{i}\}\bigg{)}$. Join $x_{iy}$ to all the vertices of $V(R_{i})$ and $x_{iz}$ to all the vertices of $V(H_{i})$ for every $0\leq i\leq k$. Adding $3k+3$ edges $x_{i}x_{iy},x_{i}x_{iz},x_{iy}x_{iz}$ and joining $x_{i}$ to $w_{i}$ for every $0\leq i\leq k$. Let $G$ denote the resulting graph (see Figure 1). Then, we have $|G|=(k+1)(2m+4).$ Moreover, take a vertex $u_{i}\in V(R_{i})$ and a vertex $v_{i}\in V(H_{i})$ for for every $0\leq i\leq k.$ We obtain

	$\displaystyle\sigma_{3k+3}(G)$	$\displaystyle=$	$\displaystyle\sum\limits_{i=0}^{k}\bigg{(}\deg_{G}(u_{i})+\deg_{G}(v_{i})+\deg_{G}(x_{i})\bigg{)}$
		$\displaystyle=$	$\displaystyle(k+1)m+(k+1)m+3(k+1)=(k+1)(2m+3)$
		$\displaystyle=$	$\displaystyle|G|-k-1.$

But $G$ has no spanning tree whose reducible stem has $k$ leaves. Hence the condition of Theorem 1.10 is sharp.

Before begining to prove Theorem 1.3 we recall some definitions in [5].

We note here that pseudoadjacency (with respect to any tree) is a weaker condition than adjacency, while pseudoindependence is a stronger condition than independence.

Now we are ready to prove Theorem 1.3.
Proof of Theorem 1.3. Suppose that $G$ has no spanning tree with at most $l$ leaves. Choose some spanning tree $T$ of $G$ such that:
(T1) $|L(T)|$ is as small as possible.

By the assumption, $T$ must have at least $l+1$ leaves.

We have the following claims.

Case 1: Suppose $g(e,s)\not=g(e,t)$. Then $e_{s}=g(e,t)$ and $e_{t}=g(e,s)$, so $se_{t},te_{s}\in E(G)$. We consider the spanning tree

$$T^{\prime}:=\left\{\begin{array}[]{ll}T-e+se_{t},&\;\mbox{ if }e=uu_{t},\\ T-\{e,uu_{t}\}+\{se_{t},te_{s}\},&\;\mbox{ if }e\not=uu_{t}.\end{array}\right.$$

Hence, $|L(T^{\prime})|<|L(T)|.$ This violates (T1). So case 1 does not happen.

Case 2: Suppose $g(e,s)=g(e,t)$. Define $a:=g(e,s)=g(e,t)$. Then $e_{s}=e_{t}\not\in\{s,t\}$ and denoted by vertex $z$. Since $G[a,z,s,t]$ is not $K_{1,3}$-free , so we have either $st\in E(G)$ or $zt\in E(G)$ or $zs\in E(G)$. Consider the tree

$$T^{\prime}:=\left\{\begin{array}[]{ll}T-uu_{t}+st,&\;\mbox{ if }st\in E(G),\\ T-\{e,uu_{t}\}+\{zt,sa\},&\;\mbox{ if }zt\in E(G),\\ T-\{e,bb_{s}\}+\{zs,ta\},&\;\mbox{ if }zs\in E(G).\end{array}\right.$$

Hence, $|L(T^{\prime})|<|L(T)|.$ This violates the condition (T1). So case 2 does not happen.
Therefore, the claim 2.4 is proved. $\Box$

Let $s_{t}\in N_{T}(b)\setminus\{s\}.$ Then there exists some vertex $s_{i}\in N_{T}(b)\setminus\{s_{t},s\}$ such that $s_{t}s_{i}\in E(G).$ Set $e:=bs_{t}.$ To complete Claim 2.5, we will need only to prove that $x$ is not an oblique neighbor of $e$ with respect to $T$ for every $x\in L(T).$ Indeed, to the contrary, assume that there exists some vertex $x\in L(T)$ such that $x$ is an oblique neighbor of $e$ with respect to $T.$ Consider the tree

$$T^{\prime}:=\left\{\begin{array}[]{ll}T-\{e,bs_{i}\}+\{bx,s_{t}s_{i}\},&\;\mbox{ if }g(e,x)=b,x\not=s_{t},\\ T-bs_{i}+s_{t}s_{i},&\;\mbox{ if }g(e,x)=b,x=s_{t},\\ T-e+xs_{t},&\;\mbox{ if }g(e,x)=s_{t}.\end{array}\right.$$

Hence, $|L(T^{\prime})|<|L(T)|.$ This violates with (T1).
Claim 2.5 holds. $\Box$

For any $v,x\in V(T)$, we now have $vx\in E(G)$ if and only if $v$ is an oblique neighbor of $xx_{v}$ with respect $T$. Therefore, the number of edges of $T$ with $v$ as an oblique neighbor equals the degree of $v$ in $G$. Combining with Claims 2.4 and 2.6, we obtain that

$$\sigma_{l+1}(G)\leq|E(T)|-l=|V(T)|-1-l=|G|-1-l,$$

which contradicts the assumption of Theorem 1.3. The proof of Theorem 1.3 is completed.  

For two distinct vertices $u,v$ of $T$, let $P_{T}[u,v]$ denote the unique path in $T$ connecting $u$ and $v.$ We define the orientation of $P_{T}[u,v]$ is from $u$ to $v$. For each vertex $x\in V(P_{T}[u,v])$, we denote by $x^{+}$ and $x^{-}$ the successor and predecessor of $x$ in $P_{T}[u,v]$, respectively, if they exist. For any $X\subseteq V(G),$ set $(N(X)\cap P_{T}[u,v])^{-}=\{x^{-}|x\in V(P_{T}[u,v])\setminus\{u\}$ and $x\in N(X)\}$ and $(N(X)\cap P_{T}[u,v])^{+}=\{x^{+}|x\in V(P_{T}[u,v])\setminus\{v\}$ and $x\in N(X)\}.$

Proof of Theorem 1.10. Suppose to the contrary that there does not exist a spanning tree $T$ of $G$ such that $|L(R\_Stem(T))|\leq k.$ Then every spanning tree $T$ of $G$ satisfies $|L(R\_Stem(T))|\geq k+1$.

Choose $T$ to be a spanning tree of $G$ such that

• $($C0$)$

  $|L(R\_Stem(T))|$ is as small as possible,

• $($C1$)$

  $|R\_Stem(T)|$ is as small as possible, subject to (C0),

• $($C2$)$

  $|L(T)|$ is as small as possible, subject to (C0) and (C1).

Set $L(R\_Stem(T))=\{x_{1},x_{2},...,x_{l}\},l\geq k+1.$ By the definition of reducible stem of $T$, we have the following claim.

We obtain the following claim as a corollary of Claim 3.3.

Set $U_{1}=\{y_{i},z_{i}\}_{i=1}^{l}$. For each $i\in\{1,...,l\}$ we also set $x_{iy}\in N_{T}(x_{i})\cap V(B_{y_{i}})$ and $x_{iz}\in N_{T}(x_{i})\cap V(B_{z_{i}}).$

Now, we choose $T$ to be a spanning tree of $G$ satisfying

• $($C3$)$

  $\sum_{i=1}^{l}\deg_{T}(x_{i})$ is as small as possible, subject to (C0)-(C2),

• $($C4$)$

  $\sum_{i=1}^{l}\bigg{(}|B_{y_{i}}|+|B_{z_{i}}|\bigg{)}$ is as large as possible, subject to (C0)-(C3).

Set $U=U_{1}\cup L(R\_Stem(T)).$

If $y_{i}x_{iz}\in E(G).$ Consider the spanning tree $T^{\prime}:=T+y_{i}x_{iz}-x_{iz}x_{i}.$ Then $|L(R\_Stem(T^{\prime}))|\leq|L(R\_Stem(T))|.$ If $\deg_{T}(x_{i})=3$ then $x_{i}$ is not a branch vertex of $T^{\prime}.$ Hence $|R\_Stem(T^{\prime})|<|R\_Stem(T)|,$ this contradicts either the conditions (C0) or (C1). Otherwise, we have $|L(R\_Stem(T^{\prime}))|=|L(R\_Stem(T))|,$ $|R\_Stem(T^{\prime})|=|R\_Stem(T)|$ and $|L(T^{\prime})|<|L(T)|,$ this contradicts the condition (C2). Now, since $G[x_{i},t,x_{iz},y_{i}]$ is not $K_{1,3}$-free, we obtain that $tx_{iz}\in E(G)$ or $ty_{i}\in E(G).$ We consider the spanning tree

$$T^{\prime}:=\left\{\begin{array}[]{ll}T+tx_{iz}-x_{iz}x_{i},&\;\mbox{ if }tx_{iz}\in E(G),\\ T+ty_{i}-x_{iy}x_{i},&\;\mbox{ if }ty_{i}\in E(G).\end{array}\right.$$

If $\deg_{T}(x_{i})=3$ then we obtain $|L(R\_Stem(T^{\prime}))|\leq|L(R\_Stem(T))|$ and $|R\_Stem(T^{\prime})|<|R\_Stem(T)|,$ a contradiction with (C0) or (C1). Otherwise, we have $L(R\_Stem(T^{\prime}))=L(R\_Stem(T))=\{x_{i}\}_{i=1}^{k+1},$ $|R\_Stem(T^{\prime})|=|R\_Stem(T)|,$ $|L(T^{\prime})|=|L(T)|$ and $\sum_{i=1}^{l}\deg_{T^{\prime}}(x_{i})<\sum_{i=1}^{l}\deg_{T}(x_{i}).$ This also violates the condition (C3).
Therefore, the proof of Claim 3.6 is completed. $\Box$