A counterexample to prism-hamiltonicity of 3-connected planar graphs

TL;DR

This paper constructs a specific counterexample disproving the conjecture that all 3-connected planar graphs are prism-hamiltonian, challenging a long-standing assumption in graph theory.

Contribution

The paper provides the first known counterexample to the conjecture that every 3-connected planar graph is prism-hamiltonian.

Findings

Counterexample to the conjecture is constructed

Disproves the universal property of prism-hamiltonicity in 3-connected planar graphs

Challenges previous assumptions in graph theory

Abstract

The prism over a graph $G$ is the Cartesian product of $G$ with the complete graph $K_2$. A graph $G$ is hamiltonian if there exists a spanning cycle in $G$, and $G$ is prism-hamiltonian if the prism over $G$ is hamiltonian. In [M.~Rosenfeld, D.~Barnette, Hamiltonian circuits in certain prisms, Discrete Math. 5 (1973), 389--394] the authors conjectured that every 3-connected planar graph is prism-hamiltonian. We construct a counterexample to the conjecture.

Figures

Equations

$$M=\{(a_{\ell },a),(a_{\ell },b),(d_{\ell +1},a),(d_{\ell +1},b)\}.$$

Full text

In [M. Rosenfeld, D. Barnette, Hamiltonian circuits in certain prisms, Discrete Math. 5 (1973), 389–394] the authors conjectured that every 3-connected planar graph is prism-hamiltonian. We construct a counterexample to the conjecture.

Key words: Hamiltonian cycle, planar graph

AMS subject classification (2010): 05C10, 05C45

1 Introduction

In 1956 Tutte proved in [16] that every 4-connected planar graph has a Hamilton cycle. On the other hand there exist 3-connected planar graphs that are not hamiltonian. An example of such cubic graph was first found by Tutte in [15], thereby disproving the Tait’s conjecture.

A $k$-tree is a tree with maximum degree $k$, and a $k$-walk in $G$ is a closed walk that visits every vertex of $G$ at most $k$ times. It is well known (see for example [9] and [2]), that the following implications hold.

$G$ is hamiltonian $\Rightarrow$ $G$ is traceable $\Rightarrow$ $G$ is prism-hamiltonian $\Rightarrow$ $G$ has a spanning 2-walk $\Rightarrow$ $G$ has a spanning 3-tree

In this hierarchy the property of being hamiltonian is the strongest and existence of a spanning 3-tree is the weakest. It was proved recently that in $P_{4}$-free graphs prism-hamiltonicity is equivalent to existence of a spanning 2-walk, see [5]. However for general graphs these properties are not equivalent, moreover they are also not equivalent in the class of 3-connected planar graphs.

In [1] it was proved that every 3-connected planar graph has a spanning 3-tree. This was strengthened in [6], where the authors prove that every 3-connected planar graph has a spanning 2-walk.

In this paper we address the following conjecture of Rosenfeld and Barnette, see [13], [7] and [9]222The conjecture is mentioned in the abstract of [13], and in [7] page 1145. Rosenfeld mentioned the conjecture during many of his talks [14], before it was formulated in [9] as Conjecture 1.1..

Conjecture 1.1

Any 3-connected planar graph is prism-hamiltonian.

In [13] and [7] the conjecture is given as a special case of a broader conjecture, claiming that every graph of a simple 4-polytope is hamiltonian. We construct a counterexample to Conjecture 1.1.

Many classes of graphs are prism-hamiltonian: chordal 3-connected planar graphs (also known as kleetopes) [9], planar near-triangulations [2], Halin graphs [9], line graphs of bridgeless graphs [9], 3-connected cubic graphs [12] and [4], and graphs that fulfil special degree conditions [11]. A characterization of prism-hamiltonian graphs is also given in [10]. Hamiltonicity of $k$-fold prisms is studied in [13], where the authors prove that for every $k\geq 2$ the $k$-fold prism $G\Box Q_{k}$ over a 3-connected planar graph $G$ is hamiltonian (here $Q_{k}$ denotes the $k$-cube).

We also mention that some special subclasses of 3-connected planar graphs are hamiltonian. Every triangulation of the plane with at most 3 separating triangles is hamiltonian, see [8]. This result was extended recently, in [3], where it is proved that all 3-connected planar graphs with at most three 3-cuts are hamiltonian.

2 The counterexample

Let $G=(V(G),E(G))$ and $H=(V(H),E(H))$ be graphs. The graph $G\cap H$ is the graph with vertex set $V(G)\cap V(H)$ and edge set $E(G)\cap E(H)$, and $G\cup H$ is the graph with vertex set $V(G)\cup V(H)$ and edge set $E(G)\cup E(H)$. If $S\subseteq V(G)$, then $G-S$ is the graph obtained from $G$ by removing all vertices in $S$, and all edges incident to a vertex in $S$. If $S=\{x\}$ is a single vertex, we write $G-x$ instead of $G-\{x\}$. If $P$ is a path in $G$ with endvertices $u$ and $v$, then we say that $P$ starts in $u$ and ends in $v$, or vice versa. A block of a graph is a maximal connected subgraph without a cutvertex. Let $K_{2}$ be the complete graph on two vertices, and let us denote $V(K_{2})=\{a,b\}$. For $n\in\mathbb{N}$, we use $[n]$ to denote the set of positive integers not larger than $n$. This notation is used throughout the article.

The Cartesian product of graphs $G$ and $H$ is the graph with vertex set $V(G)\times V(H)$, where vertices $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are adjacent in $G\Box H$ if $x_{1}x_{2}\in E(G)$ and $y_{1}=y_{2}$, or $x_{1}=x_{2}$ and $y_{1}y_{2}\in E(H)$. The prism over $G$ is the Cartesian product $G\Box K_{2}$.

A Hamilton path (cycle) in a graph $G$ is a spanning path (cycle) in $G$. A graph is a hamiltonian graph if it has a Hamilton cycle, and it is traceable if it has a Hamilton path.

Let $H_{i}$ be the graph shown in Fig. 1. Observe that vertices $b_{i}$ and $c_{i}$ are cutvertices of $H_{i}$, and that $H_{i}$ has exactly three blocks. Let $J_{i}$ be the block of $H_{i}$ containing vertex $a_{i}$, $L_{i}$ the block containing vertices $b_{i}$ and $c_{i}$ and $R_{i}$ the block of $H_{i}$ containing vertex $d_{i}$.

Let $Z$ be the subgraph of $J_{i}$ induced by vertices $a_{i},x_{1},\ldots,x_{8}$ (see Fig. 1). Note that $Z$ is the union of two 5-cycles with a common vertex. Since $\{(x_{3},a),(x_{3},b)\}$ is a vertex separator in $Z\Box K_{2}$, and the cycle containing $a_{i}$ is odd, we have the following lemma.

Lemma 2.1

The prism $Z\Box K_{2}$ has no Hamilton path with endvertices $(a_{i},a)$ and $(a_{i},b)$.

**Proof. **Let $C_{1}$ be the 5-cycle of $Z$ containing vertex $a_{i}$, and $C_{2}$ the other 5-cycle. If $P$ is a path in $Z\Box K_{2}$ with endvertices $(a_{i},a)$ and $(a_{i},b)$ that contains all vertices of $C_{2}\Box K_{2}$, then one neighbor of $(x_{3},a)$ in $P$ is contained in $(C_{1}-x_{3})\Box K_{2}$ and the other is contained in $(C_{2}-x_{3})\Box K_{2}$ (the same is true for vertex $(x_{3},b)$). The lemma follows form the fact that $C_{1}$ is odd. $\square$

Lemma 2.2

Let $P$ be a path in $J_{i}\Box K_{2}$ with the following properties:

• (i)

One endvertex of $P$ is $(a_{i},a)$ or $(a_{i},b)$, and the other is $(b_{i},a)$ or $(b_{i},b)$

• (ii)

$P$* contains all vertices of $(J_{i}-\{a_{i},b_{i}\})\Box K_{2}$.*

Then $(b_{i},a)\in P$ and $(b_{i},b)\in P$.

**Proof. **Let $P$ be a path as declared in the lemma. Suppose that $(a_{i},a)\notin P$. Then $P$ starts in $(a_{i},b)$. Let $X=\{x_{1},\ldots,x_{8}\}$ and $Y=J_{i}\setminus(X\cup\{a_{i},b_{i}\})$. Now there are only two possibilities, either the second vertex of $P$ is contained in $Y\times V(K_{2})$ or it is contained in $X\times V(K_{2})$. Since $J_{i}$ is symmetric we may assume, without loss of generality, that the latter is true (hence the second vertex of $P$ is $(x_{1},b)$ or $(x_{4},b)$). Since $(a_{i},a)\notin P$, $P$ will eventually go through $(b_{i},a)$ or $(b_{i},b)$ and enter $Y\times V(K_{2})$ to cover vertices of $Y\times V(K_{2})$. Since $P$ ends with $(b_{i},a)$ or $(b_{i},b)$ we conclude that $(b_{i},a)\in P$ and $(b_{i},b)\in P$.

If $(a_{i},b)\notin P$ the proof is analogous. Assume therefore that $(a_{i},a),(a_{i},b)\in P$. Now, if $(a_{i},a)(a_{i},b)\in E(P)$, then the argument is basically the same as above. Either the third vertex of $P$ is contained in $Y\times V(K_{2})$ or it is contained in $X\times V(K_{2})$, with the same reasoning as before. So assume that $(a_{i},a)(a_{i},b)\notin E(P)$, and assume also that $P$ starts in $(a_{i},b)$. Now the only difference of this case is the fact that $P$ might enter $X\times V(K_{2})$, cover all vertices of $X\times V(K_{2})$, and then go through $(a_{i},a)$ to enter $Y\times V(K_{2})$. Since the graph induced by $X\cup\{a_{i}\}$ is $Z$, this is not possible by Lemma 2.1. $\square$

Note that Lemma 2.2 can be applied to blocks $J_{i}$ and $L_{i}$, since they are isomorphic (the isomorphism takes $a_{i}$ to $c_{i}$, so these two vertices have the same role when applying Lemma 2.2.)

Lemma 2.3

There is no path $P$ in $(J_{i}\cup L_{i})\Box K_{2}$ with the following properties:

• (i)

One endvertex of $P$ is $(a_{i},a)$ or $(a_{i},b)$, and the other is $(c_{i},a)$ or $(c_{i},b)$

• (ii)

$P$* contains all vertices of $((J_{i}\cup L_{i})-\{a_{i},c_{i}\})\Box K_{2}$.*

**Proof. **Suppose on the contrary, that there is such a path $P$. Let $P_{1}=P\cap(J_{i}\Box K_{2})$ and $P_{2}=P\cap(L_{i}\Box K_{2})$. Then $P_{1}$ and $P_{2}$ fulfil the properties (i) and (ii) of Lemma 2.2. We apply Lemma 2.2 to $J_{i}$ and $L_{i}$ to find that $(b_{i},a),(b_{i},b)\in P_{1}\cap P_{2}$. It follows that $(b_{i},a)(b_{i},b)\in E(P)$. We may assume, without loss of generality that $(b_{i},a)(u,a)\in E(P)$ and $(b_{i},b)(v,b)\in E(P)$ for some $u\in J_{i}$ and $v\in L_{i}$. By symmetry of $J_{i}$ we may assume (without loss of generality) that $u\in Y$ (where $X$ and $Y$ are defined as in Lemma 2.2). Since $P_{1}$ contains all vertices of $X\times V(K_{2})$, we find that $P_{1}$ will eventually go through vertex $(a_{i},a)$ or $(a_{i},b)$ and enter $X\times V(K_{2})$. Then $P_{1}\cap(Z\Box K_{2})$ is a Hamiltonian path in $Z\Box K_{2}$, with endvertices $(a_{i},a)$ and $(a_{i},b)$. This contradicts Lemma 2.1. $\square$

It’s straightforward to check the following lemma (which is due to the fact that $R_{i}$ is an odd cycle).

Lemma 2.4

If $P$ and $P^{\prime}$ are disjoint paths in $R_{i}\Box K_{2}$ such that both paths have one endvertex in $\{(c_{i},a),(c_{i},b)\}$ and the other in $\{(d_{i},a),(d_{i},b)\}$, then $V(P)\cup V(P^{\prime})\neq V(R_{i}\Box K_{2})$.

Now we define counterexamples to the conjecture. Let $P_{i}$ and $Q_{i}$ be paths on the boundary walk of $H_{i}$ from vertex $a_{i}$ to $d_{i}$, and note that there are exactly two such paths, the “upper” and the “lower”.

Let $G_{n}$ be the graph obtained from graphs $H_{i}$, where $i\in[n]$, and two additional vertices $x$ and $y$. To construct the graph $G_{n}$ we first identify vertex $d_{i}$ with $a_{i+1}$ for $i\in[n-1]$. We may embed the obtained graph in the plane so that $P_{i}$ is always the “upper” path and $Q_{i}$ the “lower” path. Then we draw edges from $x$ to every vertex in $P_{i}$ and from $y$ to every vertex in $Q_{i}$, for $i\in[n]$. Clearly, this can be done so that the obtained graph $G_{n}$ is a plane graph.

Lemma 2.5

For every $n\in\mathbb{N}$, the graph $G_{n}$ is 3-connected.

**Proof. **Suppose that the claim is false and that $S\subseteq V(G_{n})$ is a vertex separator of size 2. Note that $\{x,y\}$ is not a vertex separator in $G_{n}$, therefore we may assume (by symmetry) that $x\notin S$.

If $y\in S$, then $S=\{y,u\}$, where $u\neq x$. Observe that every block of $G_{n}-\{x,y\}$ contains at most one vertex of $S$ (the vertex $u$). Therefore, if $B$ is a block of $G_{n}-\{x,y\}$, then $B-S$ is connected. Since $x$ is adjacent to more than one vertex of $B$ (in fact $x$ is adjacent to at least three vertices of $B$) we find that $(V(B)\cup\{x\})\setminus S$ induces a connected graph. Since this is true for every block $B$, the graph $G_{n}-S$ is connected.

If also $y\notin S$, then $S=\{u,v\}$, where $u,v\notin\{x,y\}$. If $B-S$ is a connected graph, for every block $B$ of $G_{n}-\{x,y\}$, the argument is the same as above. Assume therefore that there is a block $B$ of $G_{n}-\{x,y\}$ such that $B-S$ is disconnected. This is possible only if $u,v\in V(B)$. It follows that there is at most one block $B$ such that $B-S$ is not connected. Note that every connected component of $B-S$ containes an external vertex of $B$ (a vertex incident to the unbounded face of $B$). It follows that every connected component of $B-S$ is adjacent to $x$ or $y$ in $G_{n}$. The argument is completed by noting that $x$ and $y$ are adjacent to all other blocks of $G_{n}-\{x,y\}$ (by more than one edge), and all these blocks remain connected in $G_{n}-S$. $\square$

Theorem 2.6

If $n>25$, then $G_{n}$ is not prism-hamiltonian.

**Proof. **Suppose that $n>25$, and that $C$ is a Hamilton cycle in $G_{n}\Box K_{2}$. There are at most 8 edges of $C$ incident to a vertex in $T=\{(x,a),(x,b)),(y,a),(y,b)\}$. Since $n>25$ there exist consecutive graphs $H_{\ell}$ and $H_{\ell+1}$, such that $C$ has no edge with one endvertex in $(H_{\ell}\cup H_{\ell+1})\Box K_{2}$ and the other in $T$.

Let $D_{j},j\in[k]$ be connected components of $((H_{\ell}\cup H_{\ell+1})\Box K_{2})\cap C$, and let

[TABLE]

Each $D_{j}$ is a path or an isolated vertex. If it is a path, it has both endvertices in $M$. If $D_{j}$ is an isolated vertex, then it is a vertex of $M$ (these properties follow from the assumption that no vertex in $(H_{\ell}\cup H_{\ell+1})\Box K_{2}$ is adjacent to a vertex in $T$ by an edge of $C$). It follows that $k\leq 3$. Moreover there is either one component $D_{j}$ which is a path, or there are two such components $D_{j}$.

Assume the latter, and suppose that $D_{1}$ and $D_{2}$ are paths. Since $C$ is a Hamilton cycle $D_{1}$ and $D_{2}$ contain all vertices of $(H_{\ell}\cup H_{\ell+1})\Box K_{2}$.

Case 1. Suppose that both, $D_{1}$ and $D_{2}$, have one endvertex in $\{(a_{\ell},a),(a_{\ell},b)\}$ and the other in $\{(d_{\ell+1},a),(d_{\ell+1},b)\}$. For every $u\in U=\{b_{\ell},c_{\ell},d_{\ell},b_{\ell+1},c_{\ell+1}\}$ the set $\{(u,a),(u,b)\}$ is a vertex separator in $(H_{\ell}\cup H_{\ell+1})\Box K_{2}$. It follows that for every $u\in U$ either $(u,a)\in D_{1}$ and $(u,b)\in D_{2}$, or $(u,a)\in D_{2}$ and $(u,b)\in D_{1}$. Therefore $P=D_{1}\cap(R_{\ell}\Box K_{2})$ and $P^{\prime}=D_{2}\cap(R_{\ell}\Box K_{2})$ are disjoint paths that partition the vertex set of the prism $R_{\ell}\Box K_{2}$. Moreover $P$ and $P^{\prime}$ have one endvertex in $\{(c_{\ell},a),(c_{\ell},b)\}$ and the other in $\{(d_{\ell},a),(d_{\ell},b)\}$. This contradicts Lemma 2.4.

Case 2. Suppose that $(a_{\ell},a),(a_{\ell},b)\in D_{1}$. Then $(a_{\ell},a),(a_{\ell},b)$ are endvertices of $D_{1}$, and $(d_{\ell+1},a),(d_{\ell+1},b)$ are endvertices of $D_{2}$. If $D_{1}$ contains a vertex of $(J_{\ell+1}-\{a_{\ell+1},b_{\ell+1}\})\Box K_{2}$, then $D_{1}$ contains vertices $(c_{\ell},a),(c_{\ell},b),(d_{\ell},a)$ and $(d_{\ell},b)$. Therefore $D_{1}\cap(R_{\ell}\Box K_{2})$ is a disjoint union of two paths $P$ and $P^{\prime}$ that partition the vertex set of the prism $R_{\ell}\Box K_{2}$. This contradicts Lemma 2.4. Otherwise $D_{2}$ contains a vertex of $(J_{\ell+1}-\{a_{\ell+1},b_{\ell+1}\})\Box K_{2}$, and therefore $D_{2}\cap(R_{\ell+1}\Box K_{2})$ is a disjoint union of two paths that partition the vertex set of the prism $R_{\ell+1}\Box K_{2}$, a contradiction.

Assume now that there is one component $D_{j}$ which is a path. Let $D_{1}$ be the only path, and note that $D_{1}$ contains all vertices of $((H_{\ell}\cup H_{\ell+1})\Box K_{2})-M$. If $(a_{\ell},a)$ and $(a_{\ell},b)$ are endvertices of $D_{1}$, we have Case 2 above. Otherwise one endvertex of $D_{1}$ is contained in $\{(a_{\ell},a),(a_{\ell},b)\}$ and the other in $\{(d_{\ell+1},a),(d_{\ell+1},b)\}$. It follows that $P=D_{1}\cap((J_{\ell}\cup L_{\ell})\Box K_{2})$ is a path that contains all vertices of $((J_{\ell}\cup L_{\ell})-\{a_{\ell},c_{\ell}\})\Box K_{2}$. Moreover, one endvertex of $P$ is $(a_{\ell},a)$ or $(a_{\ell},b)$, and the other is $(c_{\ell},a)$ or $(c_{\ell},b)$. This contradicts Lemma 2.3. $\square$

We note that a slight modification of the above proof is needed to prove that for sufficiently large $n$, $G_{n}\Box K_{2}$ has no Hamilton path.

3 Concluding remarks

A cactus is a connected graph $G$ such that every block of $G$ is either a $K_{2}$ or a cycle. An even cactus is a cactus with no odd cycles. A good cactus is a cactus such that every vertex is contained in at most two blocks. A good even cactus is a cactus that is good and even simultaneously. A good vertex of a cactus $G$ is a vertex contained in exactly one block of $G$.

The authors of [4] considered good even cactuses with the additional property that no vertex is contained in two distinct cycles. They prove the following theorem (which is formulated in terms of our terminology).

Theorem 3.1

Every 3-connected cubic graph $G$ has a spanning subgraph $H$, such that $H$ is a good even cactus, and every vertex of $H$ is contained in at most one cycle of $H$.

It follows from the above theorem, and Proposition 3.2, that 3-connected cubic graphs are prism-hamiltonian. The restriction that no vertex of a cactus $G$ is contained in two cycles of $G$ is redundant, when questions of prism-hamiltonicity are addressed. This is justified by Proposition 3.2, and its corollary.

The partitioning result given in [6] implies that every 3-connected planar graph has a spanning good cactus. The existence of a spanning good catctus in $G$ implies the existence of a 2-walk in $G$, while the existence of a spanning good even cactus implies prism-hamiltonicity of $G$ - as mentioned in the introduction this is a stronger property. The method applied to prove the following proposition is inspired by [2].

Proposition 3.2

Every good even cactus is prism-hamiltonian.

**Proof. **We use induction to prove the following stronger statement. Every prism over a good even cactus $G$ has a Hamilton cycle $C$ such that for every good vertex $x$ of $G$, we have $(x,a)(x,b)\in E(C)$. This is clearly true for all even cycles and $K_{2}$. Let $G$ be a good even cactus and assume that the statement is true for all good even cactuses with fewer vertices than $|V(G)|$. If all vertices of $G$ are good, then $G$ is an even cycle or $K_{2}$. Otherwise, there is a vertex $u$, which is not a good vertex of $G$. Hence, $u$ is contained in exactly two blocks of $G$.

Let $G_{1}^{\prime}$ and $G_{2}^{\prime}$ be connected components of $G-x$, and let $G_{1}=G-G_{2}^{\prime}$ and $G_{2}=G-G_{1}^{\prime}$. Both, $G_{1}$ and $G_{2}$, are good even cactuses. Moreover, $x$ is a good vertex in $G_{i}$, for $i=1,2$. By induction hypothesis there is a Hamilton cycle $C_{i}$ n $G_{i}$ such that $C_{i}$ uses the edge $e=(x,a)(x,b)$ in $G_{i}$. The desired Hamilton cycle in $G$ is $(C_{1}\cup C_{2})-e$. Observe that every good vertex of $G$ is a good vertex of $G_{1}$ or $G_{2}$. It follows that for every good vertex $x$ of $G$, we have $(x,a)(x,b)\in E(C)$. $\square$

Corollary 3.3

Every graph $G$, that has a good even cactus $H$ as a spanning subgraph, is prism-hamiltonian.

The counterexample to prism-hamiltonicity of 3-connected planar graphs, given in Theorem 2.6, was constructed via a construction of a graph with no spanning good even cactus. We conclude this article with open problems.

Problem 3.4

Prove or disprove the following statement. If a 3-connected planar graph $G$ is prism-hamiltonian, then $G$ has a good even cactus $H$ as a spanning subgraph.

The following problem is due to Rosenfeld, in fact it’s conjectured that question (1) has a positive answer [14].

Problem 3.5

(1) Is every 4-connected 4-regular graph prism-hamiltonian ? (2) Is also every 3-connected 4-regular graph prism-hamiltonian ?

Acknowledgement: The author thanks M. Rosenfeld for helpful comments on the origin of Conjecture 1.1. This work was supported by the Ministry of Education of Slovenia [grant numbers P1-0297, J1-9109].