Equivalent formulation of Thomassen's conjecture using Tutte paths in claw-free graphs

TL;DR

This paper explores an equivalent formulation of Thomassen's conjecture, linking Hamilton cycles in line graphs to Tutte paths in claw-free graphs, and extends the concept to connect any two vertices with a maximal Tutte path.

Contribution

It establishes a new equivalent formulation of Thomassen's conjecture involving Tutte paths in claw-free graphs and extends the framework to connect any two vertices with such paths.

Findings

Proves the equivalence between Thomassen's and Jackson's conjectures.

Extends the formulation to connect any two vertices with a Tutte path.

Provides new insights into Hamiltonicity in claw-free graphs.

Abstract

We continue studying Thomassen's conjecture (every 4-connected line graph has a Hamilton cycle) in the direction of a recently shown equivalence with Jackson's conjecture (every 2-connected claw-free graph has a Tutte cycle), and we extend the equivalent formulation as follows: In each connected claw-free graph, every two vertices are connected by a maximal path which is a Tutte path.

Full text

1 Notation

We study Tutte paths in claw-free graphs, that is, in graphs that contain no copy of $K_{1,3}$ as an induced subgraph.

We recall that a Hamilton path (Hamilton cycle) is a path (cycle) containing all vertices of a graph. Considering a path $P$ in a graph $G$, we say that $P$ is a Tutte path if every component of $G-V(P)$ has at most $\min\{3,|V(P)|-1\}$ neighbours on $P$. In particular, a Hamilton path is a Tutte path. Similarly, a cycle $C$ is a Tutte cycle if $C$ is a Hamilton cycle or $|V(C)|\geq 4$ and every component of $G-V(C)$ has at most three neighbours on $C$.

We recall that an $(a,b)$-path is a path from vertex $a$ to vertex $b$, and we say that $P$ is a maximal $(a,b)$-path if there is no $(a,b)$-path $P^{\prime}$ such that $V(P)$ is a proper subset of $V(P^{\prime})$. We say that a graph is Tutte-connected (Hamilton-connected) if for every pair of its vertices $a,b$, there is a maximal $(a,b)$-path which is a Tutte path (an $(a,b)$-path which is a Hamilton path).

We let $N_{G}(x)$ denote the set of all vertices adjacent to vertex $x$ in a graph $G$, and $N_{G}[x]$ denote the set $N_{G}(x)\cup\{x\}$. Considering a graph $G$ and a set $A$ of its vertices, we let $\langle A\rangle_{G}$ denote the subgraph of $G$ induced by $A$. We let $G^{*}_{x}$ denote the so-called local completion of a graph $G$ at vertex $x$, that is, the graph obtained from $G$ by adding all possible edges among vertices of $N_{G}(x)$. In particular, the graph $\langle N_{G^{*}_{x}}(x)\rangle_{G^{*}_{x}}$ is a clique, meaning a complete subgraph (not necessarily maximal).

Although the present result is considering graphs with neither loops nor multiple edges, we shall use line graphs of multigraphs and hypergraphs for proving it. In accordance with [15], we define a hypergraph as a collection of subsets (edges) of a ground set. The rank of a hypergraph is the size of its maximum edge. (In particular, multiple edges are allowed and multigraphs are exactly hypergraphs of rank at most $2$.) The line graph of a graph (of a multigraph, of a hypergraph) $H$ is the graph whose vertex set is $E(H)$, and two of its vertices are adjacent if and only if the corresponding edges in $H$ have a vertex in common. (For short, line graph stands for the line graph of a graph.)

We shall recall some additional notation as needed in the text. For more definitions and concepts, we refer the reader to [4].

2 Introduction and results

Considering a cycle $C$ in a graph $G$ such that every component of $G-V(C)$ has at most three neighbours on $C$, we observe that if $G$ is $4$-connected, then $C$ is a Hamilton cycle. This fact was appreciated by Tutte [26] in proving that every $4$-connected planar graph has a Hamilton cycle. Since then, Tutte cycles and paths are used in the study of Hamilton properties of graphs. In particular, Thomassen [24] showed that all $4$-connected planar graphs are, in fact, Hamilton-connected. It was also shown that the result of Tutte extends to projective-planar graphs [22], and that every $4$-connected toroidal graph has a Hamilton path [23], and every $5$-connected such graph is Hamilton-connected [12].

Using the concept of Tutte-connectedness, we build on recent results of Čada et al. [7] and Li et al. [15], and we continue the study of the following conjecture formulated by Thomassen [25].

Conjecture A** ([25]).**

Every $4$-connected line graph has a Hamilton cycle.

Conjecture A remains open, even though partial results and numerous equivalent formulations of this problem are known (see survey [6]). For instance, Zhan [27] and independently Jackson (not published) showed that every $7$-connected line graph of a multigraph has a Hamilton cycle, and Kaiser and the second author [11] showed that every $5$-connected line graph of minimum degree at least $6$ is Hamilton-connected.

In [18], Ryjáček introduced a closure operation (a recurrent application of local completion operation on particular vertices), and showed that Conjecture A is equivalent to a seemingly stronger conjecture of Matthews and Sumner [16] saying that every $4$-connected claw-free graph has a Hamilton cycle.

We mention two additional equivalent formulations of Conjecture A which restate the problem in terms of Tutte cycles and Tutte paths (see Conjectures B and C). Conjecture B was stated by Jackson [9] and the equivalence was shown by Čada et al. [7] (using the closure technique of [18]). Recently, Li et al. [15] showed that the problem can be equivalently formulated as Conjecture C.

Conjecture B** ([9]).**

Every $2$-connected claw-free graph has a Tutte cycle.

Conjecture C** ([15]).**

For every pair of vertices $a,b$ of a connected line graph of a hypergraph of rank at most $3$, there is a maximal $(a,b)$-path which is a Tutte path.

We note that there are infinitely many claw-free graphs which are not line graphs of hypergraphs of rank at most $3$, and vice versa (it follows, for instance, from [10, Theorem 1], and the vice versa is trivial).

As the main result, we present the following reformulation of the problem (seemingly stronger than Conjecture B), and we show that they are, in fact, equivalent.

Conjecture 1**.**

For every pair of vertices $a,b$ of a connected claw-free graph, there is a maximal $(a,b)$-path which is a Tutte path.

Theorem 2**.**

Conjectures A and 1 are equivalent.

To prove Theorem 2, we introduce a closure operation preserving Tutte-connectedness (a similar concept can be found in [14]); that is, for a given graph $G$, we define a graph $G^{T}$ as follows. If $G$ is Tutte-connected, then we let $G^{T}$ be the complete graph obtained by adding all possible edges. Otherwise, we recurrently perform the local completion operation at vertices for which the resulting graph remains not Tutte-connected as long as there is at least one such vertex; and we let $G^{T}$ be a resulting graph. We say that $G^{T}$ is a Tutte-closure of $G$. (We note that if $G$ is not Tutte-connected, then there is a pair of vertices $a,b$ such that no maximal $(a,b)$-path in $G^{T}$ is a Tutte path; and for every vertex $x$ whose neighbourhood in $G^{T}$ is non-complete, $(G^{T})^{*}_{x}$ is Tutte-connected.)

We remark that alternatively the closure can be defined, for instance, with an additional condition that the local completion operation can only be applied to vertices whose neighbourhood is $2$-connected, and it should not affect the reasoning throughout the paper.

In order to prove Theorem 2, we show the following.

Theorem 3**.**

A Tutte-closure of a claw-free graph is a line graph of a hypergraph of rank at most $3$.

The proof of Theorem 3 is given in Section 6. We shall use two characterizations of line graphs of a multigraph (discussed in Section 3) in combination with properties of $2$-closed claw-free graphs (Section 4) and properties of maximal $(a,b)$-paths (Section 5).

We conclude this section by showing that Theorem 3 implies Theorem 2.

Proof of Theorem 2.

We suppose that Conjecture C is true. We consider a connected claw-free graph $G$ and its Tutte-closure $G^{T}$. Clearly, $G^{T}$ is connected. By Theorem 3, $G^{T}$ is a line graph of a hypergraph of rank at most $3$. By Conjecture C, $G^{T}$ is Tutte-connected, and so is $G$ (since the closure preserves the property), and thus Conjecture 1 is true.

Next, we suppose that Conjecture 1 is true. We consider a $4$-connected line graph $G$ and an adjacent pair of its vertices, say $a,b$. Clearly, $G$ is claw-free. We note that the $(a,b)$-path given by Conjecture 1 is, in fact, a Hamilton path. We extend this path by adding edge $ab$, and we conclude that Conjecture A is true. ∎

3 Line graphs of a multigraph (hypergraph)

An important ingredient of the argument is using two different characterizations of line graphs of multigraphs. The first characterization, due to Bermond and Meyer [2], describes the class in terms of forbidden induced subgraphs.

Theorem D** ([2]).**

For $i=1,2,\dots,7$, let $G_{i}$ be the graph depicted in Figure 1. A graph is a line graph of a multigraph if and only if it is $\{G_{1},G_{2},\dots,G_{7}\}$-free.

The second characterization, due to Krausz’s [13], describes it in terms of clique coverings. Viewing multigraphs as hypergraphs of rank at most $2$, we shall use the following generalization of Krausz’s characterization (see [1, 21]).

Theorem E** ([1, 21]).**

For every positive integer $r$, a graph $G$ is a line graph of a hypergraph of rank at most $r$ if and only if $E(G)$ can be covered by a system $\mathcal{K}$ of cliques so that every vertex of $G$ belongs to at most $r$ cliques of $\mathcal{K}$.

The basic idea of the proof of Theorem 3 is to find a suitable clique covering of a Tutte-closure and use the case $r=3$ of Theorem E (for hypergraphs of rank at most $3$). In finding such covering, we shall also use the case $r=2$ (for multigraphs).

Considering line graphs of a hypergraph of rank at most $r$, we remark that the terminology is not unitary; the graphs are also referred to as graphs of $\infty$-krausz dimension at most $r$ [8], as $r$-set representations [17], as graphs with clique cover number $r$ [10], or as edge intersection graphs of $r$-uniform hypergraphs [21]. For an overview of used terminology and for more details, we refer the reader to [10, 21].

4 On 2-closed claw-free graphs

In this section, we note that a Tutte-closure of a claw-free graph is, in fact, claw-free. In addition, we suppose that it is $2$-closed (the definition is to be found below), and we construct a desired clique covering. To this end, we shall use several lemmas on $2$-closed claw-free graphs (proven by Ryjáček and the second author in [20]).

We note the following (similar statements are shown, for instance, in [18, 5]).

Lemma 4**.**

Let $G$ be a $K_{1,k}$-free graph (where $k\geq 3$). For every vertex $x$, the graph $G^{*}_{x}$ is $K_{1,k}$-free.

Proof.

We consider a vertex $z$ and a maximum independent set $I$ of $\langle N_{G^{*}_{x}}(z)\rangle_{G^{*}_{x}}$, and we show that $\langle N_{G}(z)\rangle_{G}$ contains an independent set of size $|I|$ (and thus $|I|\leq k$). Clearly, $I$ is an independent set in $G$. In particular, we can assume that there is a vertex, say $v$, of $I\setminus N_{G}(z)$. We note that $x$ is adjacent to $v$ and $z$ and to no vertex of $I\setminus\{v\}$, and thus $(I\setminus\{v\})\cup\{x\}$ is an independent set in $\langle N_{G}(z)\rangle_{G}$. ∎

Following [3], we recall the concept of a $k$-closure. For a positive integer $k$ and a graph $G$, the $k$-closure of $G$ is the graph obtained from $G$ by recurrently performing the local completion operation at vertices whose neighbourhood induces a non-complete $k$-connected graph as long as this is possible. We say that $G$ is $k$-closed if $G$ is isomorphic to its $k$-closure.

Comparing Theorem D to the following lemma, we note a similarity in the structure of line graphs of a multigraph and $2$-closed claw-free graphs.

Lemma F** ([20]).**

For $i=1,3,5,6,7$, let $G_{i}$ be the graph depicted in Figure 1. Every $2$-closed claw-free graph is $\{G_{1},G_{3},G_{5},G_{6},G_{7}\}$-free.

In particular, every $\{G_{2},G_{4}\}$-free induced subgraph of a $2$-closed claw-free graph is a line graph of a multigraph (and so it can be covered by a system of cliques given by the case $r=2$ of Theorem E). We view this subgraph as a starting point for the construction of a desired clique covering.

To manage the graphs $G_{2}$ and $G_{4}$, we use a concept of good walks (introduced in [20]). We recall that a walk is a sequence of vertices such that every two consecutive vertices are adjacent. Considering a walk $J=u_{0}u_{1}\dots u_{k+1}$ in a graph $G$, we say that $J$ is good if the following conditions are satisfied.

• •

$k\geq 4$,

• •

if $0\leq i\leq k-1$, then vertices $u_{i}$ and $u_{i+2}$ are adjacent in $G$,

• •

if $0\leq i\leq k-4$, then the graph $\langle\{u_{i},u_{i+1},\dots,u_{i+5}\}\rangle_{G}$ is a copy of either $G_{2}$ or $G_{4}$ (depicted in Figure 1).

We say that a good walk $J$ is maximal if there is no good walk $J^{\prime}$ in $G$ such that $J$ is a proper subwalk of $J^{\prime}$ (that is, $J^{\prime}$ being a subsequence of $J$ implies that $J$ is $J^{\prime}$).

We recall that the square of a graph $H$ is the graph on the same vertex set in which two vertices are adjacent if and only if their distance in $H$ is at most $2$. We shall use the following properties of good walks (proven in [20]).

Lemma G** ([20]).**

Let $G$ be a connected $2$-closed claw-free graph that is not the square of a cycle. If $J=u_{0}u_{1}\dots u_{k+1}$ is a good walk in $G$, then $u_{1}u_{2}\dots u_{k}$ is a path.

Lemma H** ([20]).**

Let $G$ be a connected $2$-closed claw-free graph that is not the square of a cycle. If $J=u_{0}u_{1}\dots u_{k+1}$ and $J=u^{\prime}_{0}u^{\prime}_{1}\dots u^{\prime}_{k^{\prime}+1}$ are maximal good walks in $G$ such that $u_{s}=u^{\prime}_{t}$, for some $1\leq s\leq k$ and $1\leq t\leq k^{\prime}$, then the following statements are satisfied.

1. (1)

$\{u_{1},u_{2},\dots,u_{k}\}=\{u^{\prime}_{1},u^{\prime}_{2},\dots,u^{\prime}_{k^{\prime}}\}$, 2. (2)

$k=k^{\prime}$, and $u_{i}=u^{\prime}_{i}$ or $u_{i}=u^{\prime}_{k-i+1}$ for every $i=1,2,\dots,k$.

Lemma I** ([20]).**

Let $G$ be a connected 2-closed claw-free graph that is not the square of a cycle. If $J=u_{0}u_{1}\dots u_{k+1}$ is a maximal good walk in $G$, then $N_{G}[u_{1}]\setminus\{u_{3}\}=N_{G}[u_{2}]\setminus\{u_{3},u_{4}\}$ and this set induces a clique in $G$.

Lemma J** ([20]).**

Let $G$ be a connected 2-closed claw-free graph. If $J=u_{0}u_{1}\dots u_{k+1}$ is a good walk in $G$ such that $k\geq 5$, then for every $i=3,4,\dots,k-2$, $u_{i}$ has degree $4$ in $G$.

In addition, we recall the following lemma (also proven in [20]); it shall be used in Section 6.

Lemma K** ([20]).**

Let $x$ be a vertex of a claw-free graph $G$. If there is a $2$-connected subgraph of $\langle N_{G}(x)\rangle_{G}$ containing two distinct pairs of non-adjacent vertices, then $\langle N_{G}(x)\rangle_{G}$ is $2$-connected.

We combine the results discussed in Sections 3 and 4, and we obtain a system of cliques covering a $2$-closed claw-free graph as follows.

Lemma 5**.**

If $G$ is a $2$-closed claw-free graph, then $E(G)$ can be covered by a system $\mathcal{K}$ of cliques so that every vertex of $G$ belongs to at most three cliques of $\mathcal{K}$. Furthermore, if a vertex belongs to three cliques of $\mathcal{K}$, then these are the only maximal cliques of $G$ containing this vertex.

Proof.

We can assume that $G$ is connected (otherwise, we apply the argument to each component of $G$). In addition, we observe that if $G$ is the square of a cycle, then it has the desired covering; and so we can assume that it is not.

For $i=1,2,\dots,7$, we let $G_{i}$ be the graph depicted in Figure 1, and we recall that $G$ is $\{G_{1},G_{3},G_{5},G_{6},G_{7}\}$-free by Lemma F. By definition, every induced copy of $G_{2}$ or $G_{4}$ gives a good walk (see Figure 2); and thus for each such induced subgraph, all its vertices are contained in a common maximal good walk.

By Lemma G, we have that if $u_{0}u_{1}\dots u_{k+1}$ is a maximal good walk in $G$, then $u_{1}u_{2}\dots u_{k}$ is a path; and we consider a maximal set $\mathcal{P}$ of these paths (meaning that the paths of $\mathcal{P}$ are pairwise distinct; in particular, if $u_{1}u_{2}\dots u_{k}$ is in $\mathcal{P}$, then $u_{k}u_{k-1}\dots u_{1}$ is not). By Lemma H, the paths of $\mathcal{P}$ are pairwise vertex-disjoint.

For every path of $\mathcal{P}$ in sequence, we remove from $G$ all its interior vertices (that is, vertices $u_{2},\dots,u_{k-1}$); and we let $G^{\prime}$ denote the resulting graph. We note that $G^{\prime}$ is $\{G_{1},G_{2},\dots,G_{7}\}$-free, and thus it is a line graph of a multigraph by Theorem D; and we consider a system, say $\mathcal{K}_{m}$, of cliques given by Theorem E (covering $E(G^{\prime})$ so that every vertex belongs to at most two cliques).

In addition, we consider the system, say $\mathcal{K}_{r}$, of all distinct cliques given by Lemma I (that is, $\mathcal{K}_{r}$ contains $\langle N_{G}[u_{1}]\setminus\{u_{3}\}\rangle_{G}$ for every end-vertex $u_{1}$ of a path of $\mathcal{P}$, and $\langle N_{G}[u_{k}]\setminus\{u_{k-2}\}\rangle_{G}$ for every $u_{k}$). For every path of $\mathcal{P}$ in sequence, we remove from $\mathcal{K}_{m}$ all cliques containing its end-vertex (that is, $u_{1}$ or $u_{k}$), and then we add all cliques of $\mathcal{K}_{r}$; and we let $\mathcal{K}_{0}$ denote the resulting system.

We call an edge red if it is the first (last) edge of a path of $\mathcal{P}$. For every vertex $v$ of $G$, we note the following.

• •

If $v$ is in no path of $\mathcal{P}$, then it belongs to at most two cliques of $\mathcal{K}_{0}$.

• •

If $v$ is incident with a red edge, then it belongs to precisely one clique of $\mathcal{K}_{0}$.

• •

Otherwise, it belongs to none.

Furthermore, Lemma I implies that $N_{G}[u_{1}]\setminus\{u_{3}\}=N_{G}[u_{2}]\setminus\{u_{3},u_{4}\}$ and $N_{G}[u_{k}]\setminus\{u_{k-2}\}=N_{G}[u_{k-1}]\setminus\{u_{k-2},u_{k-3}\}$. By Lemma J, if $3\leq i\leq k-2$, then $u_{i}$ has degree $4$ in $G$, and thus $N_{G}(u_{i})=\{u_{i-2},u_{i-1},u_{i+1},u_{i+2}\}$ by the definition of a good walk. Consequently, we note that the only edges of $G$ which are not covered by $\mathcal{K}_{0}$ are the non-red edges whose both ends belong to a common path of $\mathcal{P}$ (namely, $u_{i}u_{i+1}$ for $i=2,\dots,k-2$ and $u_{i}u_{i+2}$ for $i=1,2,\dots,k-2$).

By definition, every three consecutive vertices of a path of $\mathcal{P}$ induce a clique (namely, a triangle); and we extend $\mathcal{K}_{0}$ by adding all these cliques, and we let $\mathcal{K}$ denote the resulting system. Clearly, every edge of $G$ is covered by a clique of $\mathcal{K}$. We note that every vertex of $G$ belongs to at most three cliques of $\mathcal{K}$, and furthermore a vertex belongs to three cliques if and only if it is an interior vertex of a path of $\mathcal{P}$. It remains to show that for each such vertex, these cliques are maximal and, in fact, the only maximal cliques containing this vertex.

We consider an interior vertex $u_{i}$ of a path of $\mathcal{P}$, and we discuss two cases. In case $3\leq i\leq k-2$, the considered cliques are $\langle\{u_{i-2},u_{i-1},u_{i}\}\rangle_{G}$, $\langle\{u_{i-1},u_{i},u_{i+1}\}\rangle_{G}$ and $\langle\{u_{i},u_{i+1},u_{i+2}\}\rangle_{G}$. We recall that $N_{G}(u_{i})=\{u_{i-2},u_{i-1},u_{i+1},u_{i+2}\}$ and $\langle N_{G}[u_{i}]\rangle_{G}$ is an induced subgraph of $G_{2}$ or $G_{4}$. The desired claim follows.

Otherwise, we suppose that $i=2$ (the argument for $i=k-1$ is similar). We recall that the considered cliques are $\langle N_{G}[u_{2}]\setminus\{u_{3},u_{4}\}\rangle_{G}$, $\langle\{u_{1},u_{2},u_{3}\}\rangle_{G}$ and $\langle\{u_{2},u_{3},u_{4}\}\rangle_{G}$. We observe that it is sufficient to show that $N_{G}(u_{2})\cap N_{G}(u_{3})=\{u_{1},u_{4}\}$ and $N_{G}(u_{2})\cap N_{G}(u_{4})=\{u_{3}\}$. By definition, $u_{3}$ is not adjacent to $u_{0}$, and $u_{4}$ is adjacent to neither $u_{0}$ nor $u_{1}$. To the contrary, we suppose that there is a vertex, say $v$, of $N_{G}(u_{2})\setminus\{u_{0},u_{1},u_{3},u_{4}\}$ which is adjacent to $u_{3}$ or $u_{4}$. Lemma I implies that $v$ is adjacent to $u_{0}$ and $u_{1}$, and also to both $u_{3}$ and $u_{4}$ (by considering the reversed walk). We note that $\langle\{u_{0},u_{1},u_{3},u_{4},v\}\rangle_{G}$ is a $2$-connected subgraph of $\langle N_{G}(u_{2})\rangle_{G}$ and $u_{0},u_{3}$ and $u_{1},u_{4}$ are two pairs of non-adjacent vertices. By Lemma K, $\langle N_{G}(x)\rangle_{G}$ is $2$-connected, contradicting the assumption that $G$ is $2$-closed. ∎

5 Local completions and maximal (a,b)-paths

With Lemma 5 on hand, we shall focus on vertices whose neighbourhood is non-complete and $2$-connected. We recall results of [3, 19] dealing with long paths in a local completion of a claw-free graph, and we combine and adapt them for maximal $(a,b)$-paths.

Proposition L** ([3]).**

Let $a,b$ and $x$ be vertices of a claw-free graph $G$ such that $N_{G}(x)$ induces a $3$-connected graph in $G$. If $G^{*}_{x}$ has an $(a,b)$-path of length $\ell$, then $G$ has an $(a,b$)-path of length at least $\ell$.

Following [19], we let $V^{x}_{i}(P)$ denote the set of all vertices $y$ of $V(P)\cap N_{G}(x)$ such that $|N_{P}(y)\cap N_{G}[x]|=i$ (for $i=0,1,2$ and given vertex $x$ and path $P$ in a graph $G$). If $V^{x}_{1}(P)$ is non-empty, then we let $a^{x}_{P}$ $(b^{x}_{P})$ denote the first (last) vertex on $P$ belonging to $V^{x}_{1}(P)$, respectively.

We recall that a cut is a set of vertices whose removal results in a disconnected graph.

Proposition M** ([19]).**

Let $a,b$ and $x$ be vertices of a claw-free graph $G$ such that the connectivity of $\langle N_{G}(x)\rangle_{G}$ is $2$, and let $R$ be a minimum cut in $\langle N_{G}(x)\rangle_{G}$. Let $P^{*}$ be a longest $(a,b)$-path in $G^{*}_{x}$. Then every longest $(a,b)$-path in $G$ is shorter than $P^{*}$ if and only if $\{a,b\}$ is a cut in $\langle N_{G}(x)\rangle_{G}$ and every longest $(a,b)$-path $P$ in $G^{*}_{x}$ has the following properties:

1. (1)

$x$* belongs to $V(P)$,* 2. (2)

$a,a^{x}_{P},b,b^{x}_{P}$* are distinct vertices,* 3. (3)

$a^{x}_{P}b^{x}_{P}$* is an edge of $G$,* 4. (4)

every component of ${\langle N_{G}(x)\rangle}_{G}-R$ contains a vertex not belonging to $V^{x}_{0}(P)$, 5. (5)

for every pair of vertices $u,v$ of $V^{x}_{1}(P)$ such that $u,v$ are in different components of ${\langle N_{G}(x)\rangle}_{G}-R$, if the $(u,v)$-subpath of $P$ contains an interior vertex, then it contains a vertex of $N_{G}[x]\setminus V^{x}_{0}(P)$ as an interior vertex.

Moreover, if the longest $(a,b)$-path in $G$ is shorter than $P^{*}$, then the following are satisfied:

1. (6)

for every longest $(a,b)$-path $Q$ in $G^{*}_{x}$, we have $a^{x}_{P}=a^{x}_{Q}$ and $b^{x}_{P}=b^{x}_{Q}$, 2. (7)

$a^{x}_{P}a$* and $b^{x}_{P}b$ are edges of $G$.*

As a corollary of Lemmas L and M, we prove the following statement (which is formally stronger since longest paths are maximal).

Corollary 6**.**

Let $a,b$ and $x$ be vertices of a claw-free graph $G$ such that $N_{G}(x)$ induces a $2$-connected graph in $G$, and let $R$ be a minimum cut in ${\langle N_{G}(x)\rangle}_{G}$. Let $P^{*}$ be a maximal $(a,b)$-path in $G^{*}_{x}$. Then there is no $(a,b)$-path $P^{\prime}$ in $G$ such that $V(P^{*})=V(P^{\prime})$ if and only if $\{a,b\}$ is a cut in $\langle N_{G}(x)\rangle_{G}$ and every $(a,b)$-path $P$ in $G^{*}_{x}$ such that $V(P^{*})=V(P)$ has the following properties:

1. (1)

$x$* belongs to $V(P)$,* 2. (2)

$a,a^{x}_{P},b,b^{x}_{P}$* are distinct vertices,* 3. (3)

$a^{x}_{P}b^{x}_{P}$* is an edge of $G$,* 4. (4)

every component of ${\langle N_{G}(x)\rangle}_{G}-R$ contains a vertex not belonging to $V^{x}_{0}(P)$, 5. (5)

for every pair of vertices $u,v$ of $V^{x}_{1}(P)$ such that $u,v$ are in different components of ${\langle N_{G}(x)\rangle}_{G}-R$, if the $(u,v)$-subpath of $P$ contains an interior vertex, then it contains a vertex of $N_{G}[x]\setminus V^{x}_{0}(P)$ as an interior vertex.

Moreover, if there is no $(a,b)$-path $P^{\prime}$ in $G$ such that $V(P^{*})=V(P^{\prime})$, then the following are satisfied:

1. (6)

for every $(a,b)$-path $Q$ in $G^{*}_{x}$ such that $V(P^{*})=V(Q)$, we have $a^{x}_{P}=a^{x}_{Q}$ and $b^{x}_{P}=b^{x}_{Q}$, 2. (7)

$a^{x}_{P}a$* and $b^{x}_{P}b$ are edges of $G$.*

Proof.

For simplicity, we let $S$ denote the graph $\langle V(P^{*})\rangle_{G}$ and $S^{*}_{x}$ denote $\langle V(P^{*})\rangle_{G^{*}_{x}}$.

We note several facts which we shall use in proving both implications. Clearly, $S$ is claw-free, and $S^{*}_{x}$ is, in fact, the local completion of $S$ at $x$, and $P^{*}$ is a longest path in $S^{*}_{x}$. Furthermore, we show that for both implications we can use that the graphs $\langle N_{S}(x)\rangle_{S}$ and $\langle N_{G}(x)\rangle_{G}$ are the same. We observe that in both cases $P^{*}$ contains an edge incident with two vertices of $N_{G}[x]$ (this follows either from property $(1)$ or from the assumption that there is no $(a,b)$-path $P^{\prime}$ in $G$ such that $V(P^{*})=V(P^{\prime})$). Since $P^{*}$ is maximal and $\langle N_{G}(x)\rangle_{G^{*}_{x}}$ is a clique, we get that $P^{*}$ contains all vertices of $N_{G}[x]$, which yields the claim.

Now, we prove the statement. First, we suppose that there is no $(a,b)$-path $P^{\prime}$ in $G$ such that $V(P^{*})=V(P^{\prime})$. Clearly, there is no such path $P^{\prime}$ in $S$. In other words, every longest $(a,b)$-path in $S$ is shorter than $P^{*}$. By Lemma L, $\langle N_{S}(x)\rangle_{S}$ is not $3$-connected, which implies that its connectivity is $2$. By Lemma M, $\{a,b\}$ is a cut in $\langle N_{S}(x)\rangle_{S}$ (and in $\langle N_{G}(x)\rangle_{G}$). Furthermore, properties $(1),(2),\dots,(7)$ are satisfied for every longest $(a,b)$-path in $S^{*}_{x}$, that is, for every $(a,b)$-path $P$ such that $V(P^{*})=V(P)$. We observe that the properties are also satisfied in $G^{*}_{x}$.

Next, we suppose that $\langle N_{G}(x)\rangle_{G}$ is $2$-connected and $\{a,b\}$ is a cut in $\langle N_{G}(x)\rangle_{G}$ (and in $\langle N_{S}(x)\rangle_{S}$) and properties $(1),(2),\dots,(5)$ are satisfied for every $(a,b)$-path $P$ in $G^{*}_{x}$ such that $V(P^{*})=V(P)$. We note that the properties are satisfied for every longest $(a,b)$-path in $S^{*}_{x}$. By Lemma M, every longest $(a,b)$-path in $S$ is shorter than $P^{*}$, that is, there is no $(a,b)$-path in $S$ whose vertex set is $V(P^{*})$, and thus there is none in $G$. ∎

6 Proof of Theorem 3

In this section, we prove Theorem 3, that is, a Tutte-closure of a claw-free graph is a line graph of a hypergraph of rank at most $3$.

Let us give a brief outline of the proof. We use Corollary 6 and Lemma K and show that a Tutte-closure of a claw-free graph contains at most one vertex whose neighbourhood is non-complete and $2$-connected. We remove this vertex and consider the resulting $2$-closed graph and its clique covering given by Lemma 5. We extend this covering to the whole graph, and we use the case $r=3$ of Theorem E.

Proof of Theorem 3.

We let $G^{T}$ be a Tutte-closure of a claw-free graph. We note that $G^{T}$ is claw-free (by Lemma 4).

We show that if $G^{T}$ is not connected or if it is $2$-closed, then the statement is satisfied. In the former case, the components of $G^{T}$ are complete graphs (by definition). In the latter case, there is a system of cliques covering $E(G^{T})$ so that every vertex of $G^{T}$ belongs to at most three cliques (by Lemma 5). In both cases, $G^{T}$ is a line graph of a hypergraph of rank at most $3$ (by Theorem E).

We can assume that $G^{T}$ is connected and it is not $2$-closed. By the definition of a Tutte-closure, there is a pair of vertices $a,b$ such that no maximal $(a,b)$-path in $G^{T}$ is a Tutte path; and we fix one such pair $a,b$.

We show the following.

Claim 1**.**

Let $x$ be a vertex whose neighbourhood in $G^{T}$ is non-complete and $2$-connected. Then $\{a,b\}$ is a cut in $\langle N_{G^{T}}(x)\rangle_{G^{T}}$ (in particular, $x$ is adjacent to $a$ and $b$).

Proof of Claim 1.

By definition, the graph $(G^{T})^{*}_{x}$ has a maximal $(a,b)$-path, say $P$, which is a Tutte path, and there is no $(a,b)$-path $P^{\prime}$ in $G^{T}$ such that $V(P)=V(P^{\prime})$. By Corollary 6, $\{a,b\}$ is a cut in $\langle N_{G^{T}}(x)\rangle_{G^{T}}$. ∎

We consider a vertex $x$ whose neighbourhood in $G^{T}$ is non-complete and $2$-connected, and a maximal $(a,b)$-path $P$ in $(G^{T})^{*}_{x}$ which is a Tutte path. We note that property (2) of Corollary 6 implies that that the set $V^{x}_{1}(P)$ is non-empty. In particular, there is an edge, say $e$, of $P$ whose both ends belong to $N_{G^{T}}[x]$; and we use this fact and show that we can choose $P$ so that no interior vertex of $P$ belongs to $V^{x}_{0}(P)$. If there is such vertex, say $i$, then we consider the graph $\langle N_{P}(i)\cup\{i,x\}\rangle_{G^{T}}$, and we note that the vertices of $N_{P}(i)$ are joint by edge (since $G^{T}$ is claw-free). We modify $P$ by adding this edge and removing its edges incident with $i$, and by adding the edges joining $i$ to the vertices incident with $e$ and removing $e$. The choice of $P$ follows.

We fix this choice of $x$ and $P$. We consider vertices $a^{x}_{P}$ and $b^{x}_{P}$ given by Corollary 6 (we recall that these are the first and the last vertex on $P$ belonging to $V^{x}_{1}(P)$), and we simplify the notation by letting $a^{\prime}$ denote $a^{x}_{P}$, and $b^{\prime}$ denote $b^{x}_{P}$. We let $a^{+}$ denote the neighbour of $a$ on $P$.

We observe that the graph $\langle N_{G^{T}}(x)\rangle_{G^{T}}-\{a,b\}$ has precisely two components (since $G^{T}$ is claw-free, $\langle N_{G^{T}}(x)\rangle_{G^{T}}$ is $2$-connected, and $\{a,b\}$ is a cut in $\langle N_{G^{T}}(x)\rangle_{G^{T}}$ by Claim 1). Furthermore, properties (2) and (3) of Corollary 6 imply that $a^{\prime}$ and $b^{\prime}$ belong to the same component of $\langle N_{G^{T}}(x)\rangle_{G^{T}}-\{a,b\}$; and we let $C^{\prime}$ denote the set of all vertices of this component, and $C$ denote the set of all vertices of the other component.

We show the following.

Claim 2**.**

No vertex of $N_{G^{T}}[x]\setminus\{a,b,a^{\prime}\}$ is adjacent to $a^{+}$. Furthermore, $N_{G^{T}}(a)\cap C^{\prime}=\{a^{\prime}\}$ and $N_{G^{T}}(b)\cap C^{\prime}=\{b^{\prime}\}$ (in particular, $ab^{\prime}$ and $a^{\prime}b$ are non-edges in $G^{T}$).

Proof of Claim 2.

To the contrary, we suppose that there is such vertex, say $w$. We recall that $P$ contains all vertices of $N_{G^{T}}[x]$ and no interior vertex of $P$ belongs to $V^{x}_{0}(P)$ (by the choice of $P$). In particular, we can choose an edge of $P$ incident with two vertices of $N_{G^{T}}[x]$ one of which is $a^{\prime}$, and analogously choose an edge incident with $w$ (it might be the same edge). We consider the subpaths of $P$ obtained by removing these edges (this edge) and removing edge $aa^{+}$, and we connect two of these subpaths using edge $a^{+}w$. We observe that each of the obtained subpaths has both ends in $N_{G^{T}}[x]$, and so we can join these subpaths into an $(a,b)$-path (since $N_{(G^{T})^{*}_{x}}[x]$ induces a clique); and we let $P^{\prime}$ denote the resulting path. Clearly, $P^{\prime}$ is a maximal $(a,b)$-path (since $V(P)=V(P^{\prime})$). We note that $a=a^{x}_{P^{\prime}}$, a contradiction with property (2) of Corollary 6.

We show that $N_{G^{T}}(a)\cap C^{\prime}=\{a^{\prime}\}$ (the argument for showing $N_{G^{T}}(b)\cap C^{\prime}=\{b^{\prime}\}$ is similar). By item (7) of Corollary 6, $a^{\prime}$ is adjacent to $a$, and thus $a^{\prime}$ belongs to $N_{G^{T}}(a)\cap C^{\prime}$. For every vertex $u$ of $C^{\prime}\setminus\{a^{\prime}\}$, we show that it is not adjacent to $a$. To this end, we consider a vertex, say $v$, of $N_{G^{T}}(a)\cap C$ and the graph $\langle\{a,a^{+},u,v\}\rangle_{G^{T}}$. First, we observe that vertices $a,a^{+},u,v$ are distinct (in particular, $a^{+}$ is distinct from $u$ and $v$ since $a^{+}$ is not adjacent to $x$ by the first part of Claim 2). Next, we discuss the edges. By definition, $a$ is adjacent to $a^{+}$ and $v$, and $u$ is not adjacent to $v$ (since they are in different components of $\langle N_{G^{T}}(x)\rangle_{G^{T}}-\{a,b\}$). By the first part of Claim 2, $a^{+}$ is adjacent to neither $u$ nor $v$. We conclude that $u$ is not adjacent to $a$ (since $G^{T}$ is claw-free). ∎

We let $A$ denote the set $N_{G^{T}}(a)\setminus N_{G^{T}}[x]$, and $B$ denote $N_{G^{T}}(b)\setminus N_{G^{T}}[x]$. We study sets $A$, $B$, $C$ and $C^{\prime}$ (see Figure 3) in the following claims.

Claim 3**.**

Each of sets $A$, $B$, $C\cup\{a\}$, $C\cup\{b\}$ and $C^{\prime}$ induces a clique in $G^{T}$.

Proof of Claim 3.

We suppose that some of these sets contains a pair of non-adjacent vertices. To obtain a contradiction, we assign a pair of vertices to each of the sets, namely we assign $\{a,x\}$ to $A$, $\{b,x\}$ to $B$, $\{x,b^{\prime}\}$ to $C\cup\{a\}$, $\{x,a^{\prime}\}$ to $C\cup\{b\}$, and $\{x,v\}$ to $C^{\prime}$ where $v$ is a vertex of $C$. We consider the $4$-tuple consisting of the assigned pair of vertices and the pair of non-adjacent vertices, and we observe that it induces a claw in $G^{T}$, a contradiction. ∎

Claim 4**.**

No vertex of $A\cup B$ is adjacent to a vertex of $C$. Furthermore, both $A\cup\{a,a^{\prime}\}$ and $B\cup\{b,b^{\prime}\}$ induces a clique in $G^{T}$.

Proof of Claim 4.

We shall use Claims 2 and 3 and show the statements for every vertex $u$ of $A$ (the arguments for $B$ are similar).

To the contrary, we suppose that there is a vertex, say $v$, of $C$ which is adjacent to $u$. We note that $a^{+}$ is not adjacent to $v$ (in particular, $a^{+}$ is distinct from $u$). Considering the graph $\langle\{a,a^{+},a^{\prime},v\}\rangle_{G^{T}}$, we note that $a^{+}$ is adjacent to $a^{\prime}$ (since $G^{T}$ is claw-free). We observe that $\langle\{v,u,a^{+},a^{\prime},x\}\rangle_{G^{T}}$ is a $2$-connected subgraph of $\langle N_{G^{T}}(a)\rangle_{G^{T}}$ and $u,x$ and $a^{\prime},v$ are two pairs of non-adjacent vertices. By Lemma K, $\langle N_{G^{T}}(a)\rangle_{G^{T}}$ is $2$-connected, and we obtain a contradiction with Claim 1.

Furthermore, we recall that $A$ induces a clique and $a$ is adjacent to $a^{\prime}$ and $u$. Considering $\langle\{a,a^{\prime},u,v\}\rangle_{G^{T}}$ (where $v$ is a vertex of $C$), we note that $u$ is adjacent to $a^{\prime}$ (since $G^{T}$ is claw-free). Thus, $A\cup\{a,a^{\prime}\}$ induces a clique. ∎

Claim 5**.**

$C\cup\{x\}=N_{G^{T}}(a)\cap N_{G^{T}}(b)$.

Proof of Claim 5.

We note that every vertex of $C\cup\{x\}$ belongs to $N_{G^{T}}(a)\cap N_{G^{T}}(b)$ (by Claim 3).

To the contrary, we suppose that there is a vertex, say $z$, of $N_{G^{T}}(a)\cap N_{G^{T}}(b)$ not belonging to $C\cup\{x\}$. By Claim 2, $z$ does not belong to $C^{\prime}$. Consequently, $z$ does not belong to $N_{G^{T}}[x]$, and so it belongs to $A\cap B$, and thus it is adjacent to $a^{\prime}$ and $b^{\prime}$ (by Claim 4). Using Claims 2, 3 and 4, we observe that the graph $\langle\{a,z,b^{\prime},x\}\rangle_{G^{T}}$ is a $2$-connected subgraph of $\langle N_{G^{T}}(a^{\prime})\rangle_{G^{T}}$ and $a,b^{\prime}$ and $x,z$ are two pairs of non-adjacent vertices. By Lemma K, $\langle N_{G^{T}}(a^{\prime})\rangle_{G^{T}}$ is $2$-connected. By Claim 1, $a^{\prime}$ is adjacent to $b$, which contradicts Claim 2. ∎

Now, we show that $x$ is the only vertex whose neighbourhood in $G^{T}$ is non-complete and $2$-connected. To the contrary, we suppose that there is such vertex $y$ distinct from $x$. By Claim 1, $y$ is adjacent to $a$ and $b$, and thus it belongs to $C$ by Claim 5. In particular, $x$ and $y$ are adjacent. Furthermore, $\{a,b\}$ is a cut in $\langle N_{G^{T}}(y)\rangle_{G^{T}}$. We consider a component of $\langle N_{G^{T}}(y)\rangle_{G^{T}}-\{a,b\}$ not containing $x$ (clearly, this component contains no vertex of $N_{G^{T}}[x]$); and we choose a vertex, say $z$, of this component such that $z$ is adjacent to $a$. Thus, $z$ belongs to $A$. By Claim 4, we get that $z$ is not adjacent to $y$, a contradiction.

Since the neighbourhood of every vertex except of $x$ is complete or not $2$-connected, the graph $G^{T}-x$ is $2$-closed, and so Lemma 5 gives a suitable clique covering of $G^{T}-x$.

In addition, we shall define a system $\mathcal{K}_{x}$ of cliques covering all edges incident with $x$. We discuss two cases. In case $a$ and $b$ are not adjacent, we recall that each of sets $C\cup\{a\}$, $C\cup\{b\}$ and $C^{\prime}$ induces a clique in $G^{T}$ (by Claim 3), and we let $\mathcal{K}_{x}$ be the system consisting of these three cliques. Otherwise, we note that $C\cup\{a,b\}$ induces a clique and we let $\mathcal{K}_{x}=\{\langle C\cup\{a,b\}\rangle_{G^{T}},\langle C^{\prime}\rangle_{G^{T}}\}$.

We show that every clique of $\mathcal{K}_{x}$ is a maximal clique in $G^{T}-x$. We consider a set $S$ of vertices inducing a clique of $\mathcal{K}_{x}$. To the contrary, we suppose that there is a vertex, say $u$, such that $S\cup\{u\}$ induces a clique in $G^{T}-x$. We discuss two cases. In case $S$ is $C\cup\{a\}$ or $C\cup\{b\}$ or $C\cup\{a,b\}$, we note that $u$ does not belong to $C^{\prime}$, and so it belongs to $A\cup B$. By Claim 4, $u$ is not adjacent to a vertex of $C$, a contradiction. Otherwise, we have that $S=C^{\prime}$. By definition, $u$ is adjacent to $a^{\prime}$ and $b^{\prime}$. In particular, $a^{+}$ is distinct from $u$ (by Claim 2). Considering $\langle\{a^{\prime},x,a^{+},u\}\rangle_{G^{T}}$, we note that $a^{+}$ is adjacent to $u$ (since $G^{T}$ is claw-free). We observe that $\langle\{a,x,b^{\prime},u,a^{+}\}\rangle_{G^{T}}$ is a $2$-connected subgraph of $\langle N_{G^{T}}(a^{\prime})\rangle_{G^{T}}$ and $a,b^{\prime}$ and $a^{+},x$ are two pairs of non-adjacent vertices. By Lemma K, $\langle N_{G^{T}}(a^{\prime})\rangle_{G^{T}}$ is $2$-connected, a contradiction.

We consider a system, say $\mathcal{K}_{0}$, of cliques covering $G^{T}-x$ by Lemma 5; and we shall extend the system to $G^{T}$. For every clique $K$ of $\mathcal{K}_{x}$, we discuss two cases. In case $K$ belongs to $\mathcal{K}_{0}$, we replace $K$ by the clique $\langle V(K)\cup\{x\}\rangle_{G^{T}}$ in the system. Otherwise, we note that no vertex of $K$ belongs to three cliques of $\mathcal{K}_{0}$ (by Lemma 5 since $K$ is a maximal clique in $G^{T}-x$). If $K$ is induced by $C^{\prime}$ or $C\cup\{a\}$ or $C\cup\{a,b\}$, then we add $\langle V(K)\cup\{x\}\rangle_{G^{T}}$ to the system. If $K$ is induced by $C\cup\{b\}$, then we add $\langle\{b,x\}\rangle_{G^{T}}$.

We note that the resulting system of cliques covers $E(G^{T})$ so that every vertex of $G^{T}$ belongs to at most three cliques. We conclude that $G^{T}$ is a line graph of a hypergraph of rank at most $3$ (by Theorem E). ∎