Domination versus edge domination

TL;DR

This paper investigates the relationship between domination numbers and edge domination numbers in regular graphs, proposing a conjecture and providing bounds and verifications for specific graph classes.

Contribution

It introduces a conjecture relating domination and edge domination numbers in regular graphs and proves bounds supporting this conjecture, including for cubic claw-free graphs.

Findings

Proves bounds on the domination number relative to the edge domination number for regular graphs.

Verifies the conjecture for cubic claw-free graphs.

Provides specific bounds for graphs with degree 3.

Abstract

We propose the conjecture that the domination number $\gamma(G)$ of a $\Delta$-regular graph $G$ with $\Delta\geq 1$ is always at most its edge domination number $\gamma_e(G)$, which coincides with the domination number of its line graph. We prove that $\gamma(G)\leq \left(1+\frac{2(\Delta-1)}{\Delta 2^{\Delta}}\right)\gamma_e(G)$ for general $\Delta\geq 1$, and $\gamma(G)\leq \left(\frac{7}{6}-\frac{1}{204}\right)\gamma_e(G)$ for $\Delta=3$. Furthermore, we verify our conjecture for cubic claw-free graphs.

Equations

$$\Delta (n-2{\gamma }_e(G))\le 2(\Delta -1){\gamma }_e(G),$$

$$\mathbb{E}[|B|]=\sum _{u\in V(G)\setminus V(M)}\mathbb{P}[u\in B]\le \frac{|V(G)\setminus V(M)|}{2^{\Delta }}=\frac{n-2{\gamma }_e(G)}{2^{\Delta }}.$$

$$\gamma (G)\le |D|+\mathbb{E}[|B|]={\gamma }_e(G)+\frac{n-2{\gamma }_e(G)}{2^{\Delta }}\stackrel{(\text{ref}e1)}{\le }{\gamma }_e(G)+\frac{2(\Delta -1){\gamma }_e(G)}{\Delta 2^{\Delta }},$$

$$\gamma (G)\le {\gamma }_e(G)+\mathbb{E}[|B|]={\gamma }_e(G)+\sum _{u\in R}\mathbb{P}[u\in B].$$

$$\mathbb{E}[|B|]=\frac{|R|}{8}.$$

$$N_{G}(x)\cap\{u,v,u^{\prime},v^{\prime}\}\in\big{\{}\{u,u^{\prime}\},\{v,v^{\prime}\}\big{\}}$$

$$N_{G}(y)\cap\{u,v,u^{\prime},v^{\prime}\}\in\big{\{}\{u,v^{\prime}\},\{v,u^{\prime}\}\big{\}},$$

$$R^{(1)}=R\setminus (R_0\cup R_1\cup R_2\cup R_3),$$

$$r^{(1)}$$

$$\mathbb{E}[|B|]\le \frac{|R|}{8}-\frac{p}{8}=\frac{1}{8}\left(n-2{\gamma }_e(G)-r_0\right)-\frac{p}{8}.$$

$$\mathbb{P}[z_t\in B]=\mathbb{P}[z\in B]=\mathbb{P}[z^{\prime }\in B]=\mathbb{P}[z^{\prime \prime }\in B]=0.$$

$$|R^{(t+1)}|$$

$$|R^{(t+1)}|=|R^{(t)}\setminus \{v\in R:v\text{mbox}hasaneighborinV({\tau }_t)\cup S({\tau }_t)\}|\ge |R^{(t)}|-7-3\cdot 9=|R^{(t)}|-34.$$

$$t\ge \frac{r^{(1)}}{34}\stackrel{(\text{ref}er1)}{\ge }\frac{r-13r_0-28p}{34},$$

$$\frac{t}{8}\ge \frac{r-13r_0-28p}{272}.$$

$$\mathbb{E}[|B|]$$

$$\gamma (G)\le {\gamma }_e(G)+\mathbb{E}[|B|]\le \frac{79}{68}{\gamma }_e(G)=\left(\frac{7}{6}-\frac{1}{204}\right){\gamma }_e(G),$$

$$D^{\prime }=(D\setminus \{u_1,u_2,\dots ,u_{k+1}\})\cup \{v_1,v_2,\dots ,v_{k+1}\}$$

Full text

1 Institute of Optimization and Operations Research, Ulm University, Germany, {julien.baste,maximilian.fuerst,elena.mohr,dieter.rautenbach}@uni-ulm.de

2 Department of Mathematics and Applied Mathematics, University of Johannesburg, Auckland Park, 2006, South Africa, mahenning@uj.ac.za

[TABLE]

1 Introduction

We consider finite, simple, and undirected graphs, and use standard terminology. Let $G$ be a graph. A set $D$ of vertices of $G$ is a dominating set in $G$ if every vertex in $V(G)\setminus D$ has a neighbor in $D$, and the domination number $\gamma(G)$ of $G$ is the minimum cardinality of a dominating set in $G$. For a set $M$ of edges of $G$, let $V(M)$ denote the set of vertices of $G$ that are incident with an edge in $M$. The set $M$ is a matching in $G$ if the edges in $M$ are pairwise disjoint, that is, $|V(M)|=2|M|$. A matching $M$ in $G$ is maximal if it is maximal with respect to inclusion, that is, the set $V(G)\setminus V(M)$ is independent. Let the edge domination number $\gamma_{e}(G)$ of $G$ be the minimum size of a maximal matching in $G$. A maximal matching in $G$ of size $\gamma_{e}(G)$ is a minimum maximal matching.

A natural connection between the domination number and the edge domination number of a graph $G$ becomes apparent when considering the line graph $L(G)$ of $G$. Since a maximal matching $M$ in $G$ is a maximal independent set in $L(G)$, the edge domination number $\gamma_{e}(G)$ of $G$ equals the independent domination number $i(L(G))$ of $L(G)$. Since $L(G)$ is always claw-free, and since the independent domination number equals the domination number in claw-free graphs [1], $\gamma_{e}(G)$ actually equals the domination number $\gamma(L(G))$ of $L(G)$. While the domination number [7] and the edge domination number [11], especially with respect to computational hardness and algorithmic approximability [3, 4, 5, 6, 8, 10], have been studied extensively for a long time, little seems to be known about their relation. For regular graphs, we conjecture the following:

Conjecture 1**.**

If $G$ is a $\Delta$-regular graph with $\Delta\geq 1$, then $\gamma(G)\leq\gamma_{e}(G)$.

The conjecture is trivial for $\Delta\leq 2$, and fails for non-regular graphs, see Figure 1. As pointed out by Felix Joos [9], for $\Delta\geq 13$, Conjecture 1 follows by combining the known results $\gamma(G)\leq\frac{(1+\ln(\Delta+1))n}{\Delta+1}$ (cf. [2]) and $\gamma_{e}(G)\geq\frac{\Delta n}{4\Delta-2}$ (cf. (1) below), that is, it is interesting for small values of $\Delta$ only. Furthermore, he observed that the union of two triangles plus a perfect matching shows that Conjecture 1 is tight for $\Delta=3$.

Our contributions are three results related to Conjecture 1. A simple probabilistic argument implies a weak version of Conjecture 1, which, for $\Delta\leq 12$, is better than the above-mentioned consequence of [2] and (1).

Theorem 1**.**

If $G$ is a $\Delta$-regular graph with $\Delta\geq 1$, then $\gamma(G)\leq\left(1+\frac{2(\Delta-1)}{\Delta 2^{\Delta}}\right)\gamma_{e}(G)$.

For cubic graphs, Theorem 1 implies $\gamma(G)\leq\frac{7}{6}\gamma_{e}(G)$, which we improve with our next result. Even though the improvement is rather small, we believe that it is interesting especially because of the approach used in its proof.

Theorem 2**.**

If $G$ is a cubic graph, then $\gamma(G)\leq\left(\frac{7}{6}-\frac{1}{204}\right)\gamma_{e}(G)$.

Finally, we show Conjecture 1 for cubic claw-free graphs.

Theorem 3**.**

If $G$ is a cubic claw-free graph, then $\gamma(G)\leq\gamma_{e}(G)$.

All proofs are given in the following section.

2 Proofs

We begin with the simple probabilistic proof of Theorem 1, which is also the basis for the proof of Theorem 2.

Proof of Theorem 1.

Let $M$ be a minimum maximal matching in $G$. Since every vertex in $V(G)\setminus V(M)$ has $\Delta$ neighbors in $V(M)$, and every vertex in $V(M)$ has at most $\Delta-1$ neighbors in $V(G)\setminus V(M)$, we have

[TABLE]

where $n$ is the order of $G$.

Let the set $D$ arise by selecting, for every edge in $M$, one of the two incident vertices independently at random with probability $1/2$. Clearly, $|D|=\gamma_{e}(G)$. If $u$ is a vertex in $V(G)\setminus V(M)$, then $u$ has no neighbor in $D$ with probability at most $1/2^{\Delta}$. Note that $u$ might be adjacent to both endpoints of some edge in $M$ in which case it always has a neighbor in $D$. If $B$ is the set of vertices in $V(G)\setminus V(M)$ with no neighbor in $D$, then linearity of expectation implies

[TABLE]

Since $D\cup B$ is a dominating set in $G$, the first moment method implies

[TABLE]

which completes the proof. ∎

The next proof arises by modifying the previous proof.

Proof of Theorem 2.

Clearly, we may assume that $G$ is connected. Let $M$ be a minimum maximal matching in $G$. Let $R_{0}$ be the set of vertices from $V(G)\setminus V(M)$ that are adjacent to both endpoints of some edge in $M$, and let $R$ be $(V(G)\setminus V(M))\setminus R_{0}$. Also in this proof, we construct a random set $D$ containing exactly one vertex from every edge in $M$. Note that every vertex from $R_{0}$ will always have a neighbor in $D$. Again, let $B$ be the set of vertices in $R$ with no neighbor in $D$. As before, we will use the estimate

[TABLE]

Initially, we choose $D$ exactly as in the proof of Theorem 1, which implies

[TABLE]

In order to obtain an improvement, we iteratively modify the random choice of $D$ in such a way that $\mathbb{E}[|B|]$ becomes smaller. We do this using two operations. Each individual operation leads to some reduction of $\mathbb{E}[|B|]$, and we ensure that all these reductions combine additively. While the first operation leads to a reduction of $\mathbb{E}[|B|]$ regardless of additional structural properties of $G$, our argument that the second operation leads to a reduction is based on the assumption that the first operation has been applied as often as possible.

The first operation is as follows.

• •

If there are two edges $uv$ and $u^{\prime}v^{\prime}$ in $M$ such that the set $X$ of vertices $x$ in $R$ with

[TABLE]

is larger than the set $Y$ of vertices $y$ in $R$ with

[TABLE]

see Figure 2, then we couple the random choices for the pair $\{uv,u^{\prime}v^{\prime}\}$ in such a way that $D$ contains $\{u,v^{\prime}\}$ with probability $1/2$ and $\{u^{\prime},v\}$ with probability $1/2$.

The choice for the coupled pair $\{uv,u^{\prime}v^{\prime}\}$ will remain independent of all other random choices involved in the construction of $D$. Furthermore, the two edges in a coupled pair will not be involved in any other operation modifying the choice of $D$.

Let $\pi$ be a coupled pair $\{uv,u^{\prime}v^{\prime}\}$. By construction, we obtain $\mathbb{P}[x\in B]=0$ for every vertex $x$ in $X$. Now, consider a vertex $y$ in $Y$. The two neighbors of $y$ in the two coupled edges are either both in $D$ or both outside of $D$, each with probability exactly $1/2$. We will ensure that the third neighbor of $y$, which is necessarily in a third edge from $M$, will belong to $D$ still with probability exactly $1/2$. By the independence mentioned above, we have $\mathbb{P}[y\in B]=1/4$. Recall that, for the choice of $D$ as in the proof of Theorem 1, each vertex from $X\cup Y$ belongs to $B$ with probability exactly $1/8$. Hence, by coupling the pair $\pi$, the expected cardinality $\mathbb{E}[|B|]$ of $B$ is reduced by $(|X|-|Y|)/8$, which is at least $1/8$.

The second operation is as follows.

• •

We select a suitable vertex $z$ from $R$ such that it has no neighbor in any of the coupled edges. If the edges $u_{1}v_{1}$, $u_{2}v_{2}$, and $u_{3}v_{3}$ from $M$ are such that $u_{1}$, $u_{2}$, and $u_{3}$ are the three neighbors of $z$, then we derandomize the selection for these three edges, and $D$ will always contain $u_{1}$, $u_{2}$, and $u_{3}$. We call $\{u_{1}v_{1},u_{2}v_{2},u_{3}v_{3}\}$ a derandomized triple with center $z$.

We will first couple a maximal number of pairs, and then derandomize triples one after the other as long as possible.

Let $\tau=\{u_{1}v_{1},u_{2}v_{2},u_{3}v_{3}\}$ be the next triple to be derandomized at some point. Let $S(\tau)$ be the set of all vertices that are incident with an edge $e$ from $M\setminus\tau$ such that some vertex in $R$ has a neighbor in $V(\tau)$ as well as in $e$, see Figure 3.

During all changes of the initial random choice of $D$ performed so far, we ensure that the following property holds just before we derandomize the triple $\tau$:

[TABLE]

All coupled pairs and derandomized triples will be disjoint.

For every edge in $M$ that does not belong to any coupled pair or derandomized triple, we select the endpoint that is added to $D$ exactly as in the proof of Theorem 1, that is, with probability $1/2$ independently of all other random choices involved in the construction of $D$.

We fix a maximal collection ${\cal P}$ of pairwise disjoint coupled pairs $\pi_{1},\ldots,\pi_{p}$.

Let $S_{\rm paired}$ be the set of the $4p$ vertices from $V(M)$ that are incident with some of the $2p$ paired edges. Let $R_{1}$ be the set of vertices in $R$ with exactly one neighbor in $S_{\rm paired}$, and let $R_{2}$ be the set of vertices in $R$ with at least two neighbors in $S_{\rm paired}$. Note that the sets $R_{0}$, $R_{1}$, and $R_{2}$ are disjoint by definition. Let $S_{\rm paired}^{\prime}$ be the set of vertices from $V(M)\setminus S_{\rm paired}$ that are incident with an edge in $M$ that contains a neighbor of some vertex in $R_{0}\cup R_{2}$. Let $R_{3}$ be the set of vertices from $R\setminus(R_{0}\cup R_{1}\cup R_{2})$ that have a neighbor in $S_{\rm paired}^{\prime}$. All sets are illustrated in Figure 4. Let

[TABLE]

$r=|R|$, $r^{(1)}=|R^{(1)}|$, and $r_{i}=|R_{i}|$ for $i\in\{0,1,2,3\}$.

Since $G$ has at most $8p$ edges leaving $S_{\rm paired}$, we have $2r_{2}+r_{1}\leq 8p$, which implies $r_{1}+7r_{2}\leq 28p$. By definition, we obtain $|S_{\rm paired}^{\prime}|\leq 4r_{0}+2r_{2}$. Considering the number of edges leaving $S_{\rm paired}^{\prime}$, we obtain $r_{3}\leq 3|S_{\rm paired}^{\prime}|\leq 12r_{0}+6r_{2}$. Therefore,

[TABLE]

Note that, only coupling the pairs $\pi_{1},\ldots,\pi_{p}$ and not derandomizing any triple, we have

[TABLE]

If $r_{0}+p$ is large enough, then this already yields the desired improvement. Since we cannot guarantee this, we now form derandomized triples one by one with centers from $R^{(1)}$. For every selected triple to be derandomized, we remove suitable vertices from $R^{(1)}$ in order to ensure (2). Suppose that we have already formed $t-1$ such derandomized triples with centers $z_{1},\ldots,z_{t-1}$, then the center $z_{t}$ for the triple $\tau_{t}$ will be selected from $R^{(t)}$, where $t$ is initially $1$, and $R^{(t+1)}$ is obtained from $R^{(t)}$ by removing every vertex from $R^{(t)}$ that has a neighbor in $V(\tau_{t})\cup S(\tau_{t})$. This ensures that all coupled pairs and derandomized triples are disjoint as well as (2).

Now, we analyze the reduction of $\mathbb{E}[|B|]$, or rather the reduction of the upper bound on $\mathbb{E}[|B|]$ given in (4), incurred by some derandomized triple $\tau_{t}$ with center $z_{t}$. Let $e_{1}$, $e_{2}$, and $e_{3}$ in $M$ be such that $e_{i}=u_{i}v_{i}$ for $i\in[3]$ and $z_{t}$ is adjacent to $u_{1}$, $u_{2}$, and $u_{3}$, that is, $\tau_{t}=\{u_{1}v_{1},u_{2}v_{2},u_{3}v_{3}\}$.

We consider two cases.

Case 1 Some vertex $z$ in $R$ distinct from $z_{t}$ has three neighbors in $V(\tau_{t})$.

First, suppose that $z$ is adjacent to $u_{1}$ and $u_{2}$. In this case, the pair $e_{1}$ and $e_{2}$ could be coupled and added to ${\cal P}$, contradicting the choice of ${\cal P}$. Next, suppose that $z$ is adjacent to $v_{1}$, $v_{2}$, and $v_{3}$. Since the pair $e_{1}$ and $e_{2}$ cannot be coupled and added to ${\cal P}$, there are two vertices $z^{\prime}$ and $z^{\prime\prime}$ in $R$ such that $z^{\prime}$ is adjacent to $u_{1}$ and $v_{2}$, and $z^{\prime\prime}$ is adjacent to $u_{2}$ and $v_{1}$. Since the pair $e_{2}$ and $e_{3}$ cannot be coupled and added to ${\cal P}$, the vertex $z^{\prime}$ is adjacent to $u_{3}$, which implies the contradiction that the pair $e_{1}$ and $e_{3}$ could be coupled and added to ${\cal P}$.

Hence, by symmetry, we may assume that $z$ is adjacent to $u_{1}$, $v_{2}$, and $v_{3}$. Since the pair $e_{2}$ and $e_{3}$ cannot be coupled and added to ${\cal P}$, there are two vertices $z^{\prime}$ and $z^{\prime\prime}$ in $R$ such that $z^{\prime}$ is adjacent to $u_{2}$ and $v_{3}$, and $z^{\prime\prime}$ is adjacent to $u_{3}$ and $v_{2}$. If $z^{\prime\prime}$ is adjacent to $v_{1}$, then, considering the pair $e_{1}$ and $e_{2}$, it follows that $z^{\prime}$ must be adjacent to $v_{1}$. In this case, the connected graph $G$ has order $10$, and $\{u_{1},u_{2},u_{3}\}$ is a dominating set, which implies the statement. Hence, we may assume that $z^{\prime\prime}$ is not adjacent to $v_{1}$. A symmetric argument implies that $z^{\prime}$ is not adjacent to $v_{1}$. See Figure 5 for an illustration.

Our derandomized choice of adding always $u_{1}$, $u_{2}$, and $u_{3}$ to $D$ yields

[TABLE]

Furthermore, property (2) implies $\mathbb{P}[w\in B]=1/4$ for every neighbor $w$ of $v_{1}$ in $R$. Since $v_{1}$ has at most two such neighbors, derandomizing the triple $\tau_{t}$ additionally reduces the upper bound on $\mathbb{E}[|B|]$ given in (4) by at least $\frac{4}{8}-\frac{2}{8}=\frac{1}{4}$. Since $z^{\prime}$ and $z^{\prime\prime}$ both have at most one neighbor not in $V(\tau_{t})$, and at most two neighbors of $v_{1}$ in $R$ both have at most two neighbors not in $V(\tau_{t})$, we obtain $|S(\tau_{t})|\leq 12$, and

[TABLE]

Case 2 $z_{t}$* is the only vertex in $R$ that has three neighbors in $V(\tau_{t})$.*

Since the pair $e_{1}$ and $e_{2}$ cannot be coupled and added to ${\cal P}$, we may assume, by symmetry, that there is a vertex $z$ in $R$ that is adjacent to $u_{1}$ and $v_{2}$. Since the pair $e_{2}$ and $e_{3}$ cannot be coupled and added to ${\cal P}$, we may assume that there is a vertex $z^{\prime}$ in $R$ such that either $z^{\prime}$ is adjacent to $u_{3}$ and $v_{2}$ or $z^{\prime}$ is adjacent to $u_{2}$ and $v_{3}$.

If $z^{\prime}$ is adjacent to $u_{3}$ and $v_{2}$, then the assumption of Case 2 implies the contradiction that the pair $e_{1}$ and $e_{3}$ can be coupled and added to ${\cal P}$. Hence, we may assume that $z^{\prime}$ is adjacent to $u_{2}$ and $v_{3}$. Since the pair $e_{1}$ and $e_{3}$ cannot be coupled and added to ${\cal P}$, there is a vertex $z^{\prime\prime}$ in $R$ adjacent to $u_{3}$ and $v_{1}$. See Figure 6 for an illustration.

The choice of ${\cal P}$ implies that no vertex from $R^{(t)}$ distinct from $z_{t}$, $z$, $z^{\prime}$, and $z^{\prime\prime}$ has two neighbors in $V(\tau_{t})$. Arguing as above, we obtain that derandomizing the triple $\tau_{t}$ additionally reduces the upper bound on $\mathbb{E}[|B|]$ given in (4) by at least $\frac{4}{8}-\frac{3}{8}=\frac{1}{8}$. Similarly as in Case 1, it follows that $|S(\tau_{t})|\leq 18$, and that

[TABLE]

Since we derandomize as many triples as possible, it follows that the number $t$ of derandomized triples satisfies

[TABLE]

and that the joint reduction of the upper bound on $\mathbb{E}[|B|]$ given in (4) is at least

[TABLE]

Altogether, coupling all $p$ pairs in ${\cal P}$, and derandomizing the $t$ triples, we obtain

[TABLE]

Therefore,

[TABLE]

which completes the proof. ∎

We proceed to the final proof.

Proof of Theorem 3.

Let $M$ be a minimum maximal matching in $G$. Let the set $D$ of $|M|$ vertices intersecting each edge in $M$ be chosen such that the set $B=\{u\in V(G)\setminus V(M):|N_{G}(u)\cap D|=0\}$ is smallest possible. For a contradiction, we may suppose that $B$ is non-empty. Let $C=\{u\in V(G)\setminus V(M):|N_{G}(u)\cap D|=1\}.$ Let $b$ be a vertex in $B$. Let $u_{-1}v_{-1}$, $u_{0}v_{0}$, and $u_{1}v_{1}$ in $M$ be such that $N_{G}(b)=\{v_{-1},v_{0},v_{1}\}$. Since $D$ intersects each edge in $M$, we have $u_{-1},u_{0},u_{1}\in D$. Since $G$ is claw-free, we may assume, by symmetry, that $v_{0}$ and $v_{1}$ are adjacent, which implies that $v_{-1}$ is not adjacent to $v_{0}$ or $v_{1}$. Let $x$ be the neighbor of $v_{-1}$ distinct from $u_{-1}$ and $b$. Since $G$ is claw-free, the vertex $x$ is adjacent to $u_{-1}$. If $x=u_{0}$, then $u_{0}$ has no neighbor in $C$, and exchanging $u_{0}$ and $v_{0}$ within $D$ reduces $|B|$, which is a contradiction. Hence, by symmetry between $u_{0}$ and $u_{1}$, the vertex $x$ is distinct from $u_{0}$ and $u_{1}$. Since exchanging $u_{1}$ and $v_{1}$ within $D$ does not reduce $|B|$, the vertex $u_{1}$ has a neighbor $c_{1}$ in $C$, which is necessarily distinct from $x$.

Now, let $\sigma:v_{1},u_{1},c_{1},v_{2},u_{2},c_{2},\ldots,v_{k},u_{k},c_{k}$ be a maximal sequence of distinct vertices from $V(G)\setminus\{u_{-1},u_{0},v_{-1},v_{0},b,x\}$ such that $u_{i}v_{i}\in M$, $u_{i}\in D$, $c_{i}\in C$, $u_{i}$ is adjacent to $c_{i}$ for every $i\in[k]$, and $v_{i+1}$ is adjacent to $u_{i}$ for every $i\in[k-1]$. Let $X=\{u_{-1},u_{0},v_{-1},v_{0},b,x\}\cup\{v_{1},u_{1},c_{1},v_{2},u_{2},c_{2},\ldots,v_{k},u_{k},c_{k}\}$, and see Figure 7 for an illustration.

Let $v_{k+1}$ be the neighbor of $u_{k}$ distinct from $v_{k}$ and $c_{k}$. Since $G$ is claw-free, the vertex $v_{k+1}$ is adjacent to $c_{k}$. Since $V(G)\setminus V(M)$ is independent, we have $u_{k+1}v_{k+1}\in M$ for some vertex $u_{k+1}$. Since $c_{k}\in C$ and $u_{k}\in D$, we obtain $v_{k+1}\not\in D$ and $u_{k+1}\in D$, which implies that the vertex $v_{k+1}$ does not belong to $X$.

If $u_{k+1}$ belongs to $X$, then $u_{k+1}=x$, and replacing $D$ with

[TABLE]

reduces $|B|$, which is a contradiction. Hence, the vertex $u_{k+1}$ does not belong to $X$. If $u_{k+1}$ has a neighbor $c_{k+1}$ in $C$, then, by the structural conditions, the vertex $c_{k+1}$ does not belong to $X$, and the sequence $\sigma$ can be extended by appending $v_{k+1},u_{k+1},c_{k+1}$, contradicting its choice. Hence, the vertex $u_{k+1}$ has no neighbor in $C$, and replacing $D$ with the set $D^{\prime}$ as above again reduces $|B|$. This final contradiction completes the proof. ∎

Acknowledgement We thank Felix Joos for pointing out that Conjecture 1 holds for large values of $\Delta$.