On the existence of minimally tough graphs having large minimum degrees

Let $m$ be an even number with $m\geq 4$. For each $i$ with $1\leq i\leq m$, let $v_{i,1},\ldots,v_{i,6}$ be six numbers. First, we add five edges $v_{i,1}v_{i,2},\ldots,v_{i,4}v_{i,5},v_{i,5}v_{i,1}$ (which forms a cycle). Next, we add four edges $v_{i,6}v_{i,j}$, where $j\in\{1,\ldots,4\}$. Finally, if $i$ is even, we add three edges $v_{i,1}v_{i+1,1},v_{i,2}v_{i+1,5},v_{i,5}v_{i+1,2}$, and if $i$ is odd, we add three edges $v_{i,4}v_{i+1,4},v_{i,3}v_{i+1,5},v_{i,5}v_{i+1,3}$ (if $i=m$, we compute $i+1$ modulo $m$ that means $i+1=1$). Call the resulting graph $G$. Obviously, this graph is planar and $4$-regular. For the smallest case $m$=4, the graph $G$ is illustrated in Figure 2.

We claim that $t(G)=3/2$. We first show that $t(G)\geq 3/2$. Let $S$ be an arbitrary cut-set of $G$. Define $G_{i}$ to be the induce subgraph of $G$ with the vertex set $V_{i}=\{v_{i,j}:1\leq j\leq 6\}$ and let $S_{i}=S\cap V_{i}$. Since $G_{i}$ has independence number at most two, we must have $\omega(G_{i}\setminus S_{i})\leq 2$. Let $\omega_{i}$ be the number of components of $G_{i}\setminus S_{i}$ which is not connected to a component of $G_{i+1}\setminus S_{i+1}$ in $G\setminus S$. Define $\theta_{i}=|S_{i}|-\frac{3}{2}\omega_{i}$. Let start with the following simple but useful subclaims.

Subclaim 1: If $G_{i}\setminus S_{i}$ is disconnected, then $|S_{i}|\geq 3$ or $S_{i}=\{v_{i,1},v_{i,4}\}$

Subclaim 2: If $\{v_{i,1},v_{i,5}\}\subseteq S_{i}$ or $\{v_{i,4},v_{i,5}\}\subseteq S_{i}$, then $\theta_{i}\geq 1/2$. In particular, in both cases, we have $\theta_{i}\geq 1$, if $|S_{i}|\geq 3$ or there is one component of $G_{i}\setminus S_{i}$ connecting to a component of $G_{i+1}\setminus S_{i+1}$.

If there is a component of $G$ having one vertex from every subgraph $G_{i}$, then $\omega(G\setminus S)\leq\sum_{1\leq i\leq m}(\omega(G_{i}\setminus S_{i})-1)+1$. Moreover, $\omega_{i}\leq 1$ and $\theta_{i}\geq 0$ for all indices $i$. Since $S$ is a cut-set, there is a subgraph $G_{i}$ in which $G_{i}\setminus S_{i}$ is disconnected. If $|S_{i}|\geq 3$, then $\theta_{i}\geq 3/2$. Otherwise, by Subclaim 1, $S_{i}=\{v_{i,1},v_{i,4}\}$. If there is a subgraph $G_{j}$ in which $G_{j}\setminus S$ is connected and $|S_{j}|>0$, then $\theta_{i}+\theta_{j}\geq 1/2+1=3/2$. Otherwise, there must be at least two other subgraphs $G_{j_{1}}$ and $G_{j_{2}}$ in which $G_{j_{t}}\setminus S_{j_{t}}$ is connected. Thus $\theta_{i}+\theta_{j_{1}}+\theta_{j_{2}}\geq 1/2+1/2+1/2=3/2$. Therefore, $\sum_{1\leq i\leq m}\theta_{i}\geq 3/2$ which implies that $|S|\geq\frac{3}{2}\omega(G\setminus S)$. Now, assume that there is no component of $G\setminus S$ having one vertex from every subgraph $G_{i}$. Therefore,

We say an index $i$ is good if $\theta_{i}\geq 0$, and bad otherwise. If all indices are good, then $\frac{|S|}{\omega(G\setminus S)}\geq 3/2$. Otherwise, by applying the following assertion, one can conclude that there is a partition $P$ of indices $\{1,\ldots,m\}$ such that for every $I\in P$, $\sum_{i\in I}\theta_{i}\geq 0$ and so $\sum_{1\leq i\leq m}\theta_{i}\geq 0$ and $\frac{|S|}{\omega(G\setminus S)}\geq 3/2$.

Subclaim 3: If $b$ is a bad index, then there exists an index $j\in\{1,2,3\}$ such that $\sum_{b\leq i\leq b+j}\theta_{i}\geq 0$ and all indices $b+1,\ldots,b+j$ are good.

Proof of Subclaim 3. By the isomorphic property, we may assume that $b$ is odd. Suppose the assertion false. This implies that $\theta_{b}+\max\{0,\theta_{b+1},\theta_{b+1}+\theta_{b+2},\theta_{b+1}+\theta_{b+2}+\theta_{b+3}\}\leq-1/2$. In addition, any component of $G_{b}\setminus S_{b}$ is not connected to a component of $G_{b+1}\setminus S_{b+1}$. Obviously, $\omega_{b}\in\{1,2\}$. To derive a contradiction, we shall consider the following two cases:

Case A: $\omega_{b}=1$.

In this case, $|S_{b}|\leq 1$ and $-3/2\leq\theta_{b}\leq-1/2$. If $S_{b}\cap\{v_{b,3},v_{b,4},v_{b,5}\}=\emptyset$, then $\{v_{b+1,3},v_{b+1,4},v_{b+1,5}\}\subseteq S_{b+1}$ and $\theta_{b+1}\geq 3/2$. Since $\theta_{b}+\theta_{b+1}\geq 0$, we derive a contradiction. Assume that $|S_{b}\cap\{v_{b,3},v_{b,4},v_{b,5}\}|=1$. In this case, $\theta_{b}=-1/2$. If $S_{b}=\{v_{b,5}\}$, then since the unique component of $G_{b}\setminus S_{b}$ is not connected to a component of $G_{b+1}\setminus S_{b+1}$, we must have $\{v_{b+1,4},v_{b+1,5}\}\subseteq S_{b+1}$ and $\theta_{b+1}\geq 1/2$ using Subclaim 2. Since $\theta_{b}+\theta_{b+1}\geq 0$, we derive a contradiction. If $S_{b}=\{v_{b,4}\}$, then it is not difficult to check that $S_{b+1}=\{v_{b+1,3},v_{b+1,5},v_{b+1,6}\}$ and $\theta_{b+1}=0$. Similarly, since any component of $G_{b+1}\setminus S_{b+1}$ is not connected to a component of $G_{b+2}\setminus S_{b+2}$, we must have $\{v_{b+2,1},v_{b+2,5}\}\subseteq S_{b+2}$ and $\theta_{b+2}\geq 1/2$ using Subclaim 2. Since $\theta_{b}+\theta_{b+1}+\theta_{b+2}\geq 0$, we derive a contradiction. If $S_{b}=\{v_{b,3}\}$, then it is not difficult to check that $S_{b+1}=\{v_{b+1,1},v_{b+1,3},v_{b+1,4}\}$ and $\theta_{b+1}=0$. Since any component of $G_{b+1}\setminus S_{b+1}$ is not connected to a component of $G_{b+2}\setminus S_{b+2}$, we must have $S_{b+2}=\{v_{b+2,2},v_{b+2,5},v_{b+2,6}\}$ and $\theta_{b+2}=0$. Therefore, we must have $S_{b+3}=\{v_{b+3,4},v_{b+3,5}\}$ and $\theta_{b+2}\geq 1/2$ using Subclaim 2. Since $\theta_{b}+\theta_{b+1}+\theta_{b+2}+\theta_{b+3}\geq 0$, we again derive a contradiction.

Case B: $\omega_{b}=2$.

In this case, by Subclaim 1, we must have $S_{b}=\{v_{b,1},v_{b,4}\}$ and so $\theta_{b}=-1$. Since there is not a component of $G_{b}\setminus S_{b}$ connecting to a component of $G_{b+1}\setminus S_{b+1}$, it is not difficult to check that $\{v_{b+1,3},v_{b+1,5}\}\subseteq S_{b+1}$, $\{v_{b+1,1},v_{b+1,2}\}\cap S_{b+1}=\emptyset$, and $\theta_{b+1}\geq 0$. Consequently, $S_{b+2}=\{v_{b+2,1},v_{b+2,2}\}$ and $\theta_{b+2}=1/2$ using Subclaim 2. This implies that $S_{b+3}=\{v_{b+3,4},v_{b+3,5}\}$ and $\theta_{b+3}=1/2$ using Subclaim 2 again. Since $\theta_{b}+\theta_{b+1}+\theta_{b+2}+\theta_{b+3}\geq 0$, we derive contradiction, as desired.

Therefore, $t(G)\geq 3/2$. From now on, for notational simplicity, for each odd index $i$, we set $S_{i}=\{v_{i,1},v_{i,4}\}$, and for even index $i$, we set $S_{i}=V_{i}\setminus\{v_{i,1},v_{i,4}\}$; recall that $V_{i}=\{v_{i,j}:1\leq j\leq 6\}$. If we set $S=\cup_{1\leq i\leq m}S_{i}$, then it is easy to check that $|S|=3m$ and $\omega(G\setminus S)=2m$. This implies that $t(G)\leq\frac{|S|}{\omega(G\setminus S)}=3/2$. Hence $t(G)=3/2$ and the claim is proved. It remains to show that $G$ is minimally $t(G)$-tough. Let $e$ be an arbitrary edge of $G$. We shall consider the following cases:

Case 1: Both ends of $e$ lie in the set $\{v_{i,2},v_{i,3},v_{i,6}\}$.

By the isomorphic property, we may assume that $i$ is odd. Let $v$ be the unique vertex in $\{v_{i,2},v_{i,3},v_{i,6}\}$ not incident with $e$. If we set $S=(\cup_{1\leq j\leq m}S_{j})\cup\{v\}$, then it is easy to check that $|S|=3m+1$ and $\omega((G-e)\setminus S)=2m+1$. This implies that $t(G-e)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{3m+1}{2m+1}<3/2$.

Case 2: $e\in\{v_{i,1}v_{i,6},v_{i,2}v_{i,1},v_{i,2}v_{i+1,5}\}$.

By the isomorphic property, we may assume that $i$ is even. If $e=v_{i,1}v_{i,6}$, then we set $S=(\cup_{1\leq j\leq m}S_{j})\setminus\{v_{i,6}\}$. Otherwise, we set $S=(\cup_{1\leq j\leq m}S_{j})\setminus\{v_{i,2}\}$. In both cases, it is easy to check that $|S|=3m-1$ and $\omega((G-e)\setminus S)=2m$. This implies that $t(G-e)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{3m-1}{2m}<3/2$.

Case 3: $e=v_{i,4}v_{i+1,4}$.

By the isomorphic property, we may assume that $i$ is odd. If we set $S=((\cup_{1\leq j\leq m}S_{j})\setminus\{v_{i-1,2},v_{i,4},v_{i+1,3}\})\cup\{v_{i,5}\}$, then it is not difficult to check that $|S|=3m-2$ and $\omega((G-e)\setminus S)=2m-1$. This implies that $t(G-e)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{3m-2}{2m-1}<3/2$.

Case 4: $e=v_{i,1}v_{i,5}$.

By the isomorphic property, we may assume that $i$ is even. If we set $S=((\cup_{1\leq j\leq m}S_{j})\setminus\{v_{i-2,2},v_{i-1,4},v_{i,3},v_{i,5}\})\cup\{v_{i-1,3},v_{i-1,6},v_{i-1,5},v_{i,4},v_{i+1,2}\},$ then it is not difficult to check that $|S|=3m+1$ and $\omega((G-e)\setminus S)=2m+1$. This implies that $t(G-e)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{3m+1}{2m+1}<3/2$.

The remaining edges are similar with respect to isomorphic property of the graph $G$. Hence the proof is completed. $\Box$

Let $H$ be the cubic graph illustrated in Figure 3 and let $G$ be the square of $H$. It is not difficult to show that $G$ contains only four maximum independent sets of size $4$ which are shown in Figure 3 by numbers $1,\ldots,4$.

Let $I$ be a maximum independent set of $G$. If we set $S=V(G)\setminus I$, then $t(G)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{|S|}{|I|}=\frac{9}{3}=3$. Let $S$ be a cut-set of $V(G)$. Since $G$ is $7$-connected, we must have $|S|\geq 7$. If $\omega(G\setminus S)=2$, then $\frac{|S|}{\omega(G\setminus S)}\geq\frac{7}{2}>3$. If $\omega(G\setminus S)\geq 3$, then according to the property of $G$, the subgraph $G\setminus S$ must consist of three isolated vertices and so $S=|V(G)|-3=9$ which implies that $\frac{|S|}{\omega(G\setminus S)}=3$. Consequently, $t(G)=3$.

We are going to show that $G$ is minimally $t(G)$-tough. Let $e=uv$ be an arbitrary edge of $G$. If $e\in E(H)$, then it is not difficult to check that there is a unique edge $e^{\prime}=u^{\prime}v^{\prime}$ of $H$ which is not connected to $e$ by an edge in $G$. If we set $S=V(G)\setminus\{u,v,u^{\prime},v^{\prime}\}$, then $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{8}{3}<3=t(G)$. If $e\not\in E(H)$, then we may assume that $u$ and $v$ have a common neighbor $w$ in $H$ for which $w$ and $v$ lie in a triangle in $H$. In this case, we again have $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{8}{3}<3=t(G)$, where $S=V(G)\setminus(I\cup\{u\})$ and $I$ is the maximum independent set of $G$ including $v$. Hence the proof is completed. $\Box$

Let $G_{i}$ be a copy of the complete graph $K_{n}$ with vertices $v_{i,1},\ldots,v_{i,n}$, where $i=1,2$. For each $j$ with $1\leq j\leq m$, we add an edge $v_{1,j}v_{2,j}$ and call the resulting graph $G$. The special case $n=7$ and $m=4$ is illustrated in Figure 4 (the right graph).

If we set $S=\{v_{1,p}:1\leq p\leq m\}$, then $\omega(G\setminus S)=2$ and $|S|=m$ which implies that $t(G)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{m}{2}$. On the other hand, if $S$ is a cut-set of $G$, then $\omega(G\setminus S)=2$ and $|S|\geq m$, because $\alpha(G)=2$ and $G$ is $m$-connected. This implies that $\frac{|S|}{\omega(G\setminus S)}\geq\frac{m}{2}$ and so $t(G)=\frac{m}{2}$. We are going to show that $G$ is minimally $t(G)$-tough. Let $e$ be an arbitrary edge of $G$. If $e=v_{1,j}v_{2,j}$, then $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{m-1}{2}<\frac{m}{2}=t(G)$, where $S=\{v_{1,p}:1\leq p\leq m\text{ and }p\neq j\}$. If $e=v_{i,j_{1}}v_{i,j_{2}}$, then $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{n}{3}<\frac{m}{2}=t(G)$, where $S=(V(G_{i})\setminus\{v_{i,j_{1}},v_{i,j_{2}}\})\cup\{v_{3-i,j_{1}},v_{3-i,j_{2}}\}$. This completes the proof. $\Box$

Let $G_{i}$ be a copy of the complete graph $K_{n}$ with vertices $v_{i,1},\ldots,v_{i,n}$, where $i=1,2,3$. For each $p$ with $1\leq p\leq n$, we insert two edges $v_{1,p}v_{2,p}$ and $v_{2,p}v_{3,p}$ into these graphs and call the resulting graph $G$. This graph is isomorphic to the Cartesian product $K_{n}\square P_{3}$. The special case $n=5$ is illustrated in Figure 5 (the left graph).

We claim that $t(G)=\frac{n+1}{3}$. If we set $S=\{v_{2,p}:1\leq p<n\}\cup\{v_{1,n},v_{3,n}\}$, then $|S|=n+1$ and $\omega(G\setminus S)=3$. Hence $t(G)\leq\frac{|S|}{\omega(G\setminus S)}=\frac{n+1}{3}$. It remains to show that $t(G)\geq\frac{n+1}{3}$. Let $S$ be an arbitrary cut-set of $V(G)$. If $|S|\geq n+1$, then $\frac{|S|}{\omega(G\setminus S)}\geq\frac{|S|}{\alpha(G\setminus S)}\geq\frac{n+1}{3}$. Assume that $|S|\leq n$. According to the construction of $G$, we must have $\omega(G\setminus S)=2$. More precisely, $|S\cap\{v_{1,p},v_{2,p},v_{3,p}\}|=1$ for all integers $p$ with $1\leq p\leq n$. Therefore, $\frac{|S|}{\omega(G\setminus S)}=\frac{n}{2}\geq\frac{n+1}{3}$. Hence the claim is proved. We are going to show that $G$ is minimally $t(G)$-tough. Let $e$ be an arbitrary edge of $G$. If $e=v_{2,j}v_{i,j}$, then $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{n}{3}<\frac{n+1}{3}=t(G)$, where $S=\{v_{2,p}:1\leq p\leq n\text{ and }p\neq j\}\cup\{v_{4-i,j}\}$. If $e=v_{i,j_{1}}v_{i,j_{2}}$ and $i\neq 2$, then $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{n}{3}<\frac{n+1}{3}=t(G)$, where $S=\{v_{i,p}:1\leq p\leq n\text{ and }p\neq j_{1},j_{2}\}\cup\{v_{2,j_{1}},v_{2,j_{2}}\}$. If $e=v_{2,j_{1}}v_{2,j_{2}}$, then $t(G-e)\leq\frac{|S|}{\omega((G-e)\setminus S)}=\frac{n+2}{4}<\frac{n+1}{3}=t(G)$, where $S=\{v_{2,p}:1\leq p\leq n\text{ and }p\neq j_{1},j_{2}\}\cup\{v_{1,j_{1}},v_{1,j_{2}},v_{3,j_{1}},v_{3,j_{2}}\}$. This completes the proof of the first part.

Let $G_{0}$ be the graph obtained from $G$ by deleting the vertex $v_{2,n}$ and adding the new edge $v_{1,n}v_{3,n}$. Obviously, $G_{0}$ is $n$-regular. The special case $n=5$ is illustrated in Figure 5 (the right graph). For notional simplicity, let us use $H$ instead of $G_{0}$. We claim that $t(H)=\frac{n+1}{3}$. If we set $S=\{v_{2,p}:1\leq p\leq n-2\}\cup\{v_{1,n-1},v_{3,n-1},v_{3,n}\}$, then $|S|=n+1$ and $\omega(H\setminus S)=3$. Hence $t(H)\leq\frac{|S|}{\omega(H\setminus S)}=\frac{n+1}{3}$. It remains to show that $t(H)\geq\frac{n+1}{3}$. Let $S$ be an arbitrary cut-set of $V(H)$. If $|S|\geq n+1$, then $\frac{|S|}{\omega(H\setminus S)}\geq\frac{|S|}{\alpha(H\setminus S)}\geq\frac{n+1}{3}$. Assume that $|S|\leq n$. According to the construction of $H$, we must have $\omega(H\setminus S)=2$. More precisely, $|S\cap\{v_{1,n},v_{3,n}\}|=1$ and $|S\cap\{v_{1,p},v_{2,p},v_{3,p}\}|=1$ for all integers $p$ with $1\leq p\leq n-1$. Therefore, $\frac{|S|}{\omega(H\setminus S)}=\frac{n}{2}\geq\frac{n+1}{3}$. Hence the claim is proved. We are going to show that $H$ is minimally $t(H)$-tough. Let $e$ be an arbitrary edge of $H$. If $e=v_{1,n}v_{3,n}$, then $t(H-e)\leq\frac{|S|}{\omega((H-e)\setminus S)}=\frac{n}{3}<\frac{n+1}{3}=t(H)$, where $S=\{v_{2,p}:1\leq p\leq n-2\}\cup\{v_{1,n-1},v_{3,n-1}\}$. If $e=v_{2,j}v_{i,j}$, then $t(H-e)\leq\frac{|S|}{\omega((H-e)\setminus S)}=\frac{n}{3}<\frac{n+1}{3}=t(H)$, where $S=\{v_{2,p}:1\leq p\leq n-1\text{ and }p\neq j\}\cup\{v_{4-i,j},v_{1,n}\}$. If $e=v_{i,j_{1}}v_{i,j_{2}}$ and $i\neq 2$, then $t(H-e)\leq\frac{|S|}{\omega((H-e)\setminus S)}=\frac{n}{3}<\frac{n+1}{3}=t(H)$, where $S=\{v_{i,p}:1\leq p\leq n\text{ and }p\neq j_{1},j_{2}\}\cup\{v_{2,j_{1}},v\}$ in which $v=v_{2,j_{2}}$ when $n\not\in\{j_{1},j_{2}\}$, and $v=v_{4-i,n}$ when $j_{1}\neq j_{2}=n$. If $e=v_{2,j_{1}}v_{2,j_{2}}$, then $t(H-e)\leq\frac{|S|}{\omega((H-e)\setminus S)}=\frac{n+2}{4}<\frac{n+1}{3}=t(H)$, where $S=\{v_{2,p}:1\leq p\leq n-1\text{ and }p\neq j_{1},j_{2}\}\cup\{v_{1,j_{1}},v_{1,j_{2}},v_{3,j_{1}},v_{3,j_{2}},v_{1,n}\}$. Hence the proof is completed. $\Box$