Prism-Hamiltonicity and Hamiltonicity of Graph Products
By
Pouria Salehi Nowbandegani
Dissertation
Submitted to the Faculty of the Graduate School of Vanderbilt University
in partial fulfillment of the requirements for the degree of
DOCTOR OF PHILOSOPHY in
Mathematics October 31st, 2020 Nashville, Tennessee
Approved:
Mark Ellingham, Ph.D.
Paul Edelman, Ph.D.
Michael Mihalik, Ph.D.
Denis Osin, Ph.D.
Jeremy Spinrad, Ph.D.
TABLE OF CONTENTS
Page
LIST OF FIGURES . . . iii
Chapter 1: Introduction . . . 1
1.1 Preliminaries . . . 1
1.2 Proof of Theorem 1.1.2 . . . 9
1.3 Outline . . . 11
Chapter 2: Hierarchy of Hamilton properties . . . 12
2.1 Hierarchy diagrams . . . 12
2.2 Repeated prisms . . . 21
Chapter 3: The Chv´atal-Erd˝os condition for prism-Hamiltonicity . . . 22
3.1 Preliminaries . . . 22
3.2 Proof of Theorem 3.1.4 . . . 25
3.3 Toughness conditions . . . 29
3.4 The Chv´atal-Erd˝os condition for Hamiltonicity ofGK3 . . . 30
Chapter 4: Toughness and prism-Hamiltonicity ofP4-free graphs . . . 36
4.1 Preliminaries . . . 36
4.2 Tools for proof of Theorem 4.1.3 . . . 39
4.3 Proof of Theorem 4.1.3 . . . 45
Chapter 5: Forbidden subgraphs and forbidden pairs for prism-Hamiltonicity . . . 49
5.1 Preliminaries . . . 49
5.2 Proof of Theorem 5.1.5 . . . 52
Chapter 6: Future work . . . 56
LIST OF FIGURES
Page
1.1 K2,4 . . . 2
1.2 N . . . 2
2.1 Nt . . . 19
2.2 Petersen Graph . . . 19
3.1 An even strong2-cactus . . . 23
4.1 An even2-cactus . . . 39
5.1 K1,t . . . 50
5.2 H1 . . . 50
5.3 N . . . 51
5.4 B . . . 52
5.5 Z1 . . . 52
5.6 H2 . . . 53
5.7 H3 . . . 54
5.8 Z2 . . . 54
5.9 H4 . . . 55
Chapter 1
Introduction
1.1 Preliminaries
In this dissertation, we consider only simple, finite, and undirected graphs unless otherwise stated. For any graph theory terms not explicitly defined in this dissertation please see [39]. LetG be a graph. Byκ(G)we mean the connectivity ofG, that is, the minimum number of vertices that need to be removed to disconnectG(orn−1, ifG =Kn). Byα(G)we mean the independence number ofG, that is, the maximum number of vertices with no edges between them. Thecartesian product GH of graphs G and H is a graph such that the vertex set of GH is the cartesian product V(G)×V(H) and two vertices (u, u0) and (v, v0) are adjacent in GH if and only if eitheru =v andu0 is adjacent tov0 inH, oru0 = v0 anduis adjacent tov inG. The prismover a graphGis the cartesian productGK2. Note that we can consider GK2 as two copies ofG, usually namedlevels, and a matching between corresponding vertices of these copies ofGwhose edges are called vertical edges. AHamilton cycle is a cycle containing all vertices of the graph.
A graph that contains a Hamilton cycle is called Hamiltonian. IfGK2 is Hamiltonian, we say thatGisprism-Hamiltonian. Ak-treeis a tree with maximum degree at mostk, and ak-walkis a closed walk with each vertex repeated at mostktimes.
Kaiser et al. [29] showed that the property of having a Hamiltonian prism is stronger than that of having a spanning2-walk and weaker than that of having a Hamilton path, i.e., for graphs with at least two vertices we have
Hamilton path⇒prism-Hamiltonian⇒spanning2-walk.
Both implications are sharp. Indeed ifGhas a Hamilton path, then it is prism-Hamiltonian, since we can find a Hamilton cycle inGK2by going along the Hamilton path in one level then switch- ing levels and going back. On the other hand, the complete bipartite graphK2,4(Figure 1.1 ) has no
Hamilton path but its prism is Hamiltonian. As for the second implication, any prism-Hamiltonian graph G has a spanning 2-walk that follows the edges of G corresponding to the edges of the Hamilton cycle in the prism. Anet,N (Figure 1.2), which is the graph constructed from a triangle
Figure 1.1:K2,4
together with one pendant edge attached to each vertex of the triangle, has a spanning2-walk but it is not prism-Hamiltonian.
Figure 1.2:N
Note that the net is not prism-Hamiltonian because of a parity issue. SupposeNK2contains
a Hamilton cycle. Whenever we visit a degree1vertex of a copy ofN along the Hamilton cycle of the prism over N, we must change the level, i.e., we must pass through a vertical edge of the prism. Also, we cannot change the level at triangle vertices along the Hamilton cycle otherwise we cannot visit at least one degree1vertex of each level. But since we must change the level an even number of times in order to build a Hamilton cycle, we should change the level at one or three of the triangle’s vertices, a contradiction. Also note that traveling along the outer face ofN generates a spanning2-walk ofN. Moreover, in [29], the authors construct graphs with arbitrarily large connectivity that have spanning2-walks, but are not prism-Hamiltonian. Thus, proving that a graph previously known to have a spanning 2-walk is prism-Hamiltonian strengthens what we know.
It is of interest to determine whether or not a graph fits in between the properties of having a Hamilton path and having a spanning2-walk. In particular, which graphs are prism-Hamiltonian even though they may not have a Hamilton path? Also, there are many interesting results on Hamiltonicity of graphs. It seems natural to ask similar questions about prism-Hamiltonicity. In the rest of this section we list some important results on prism-Hamiltonicity.
In 1884 Tait [37] conjectured that every3-connected cubic planar graph is Hamiltonian. Tutte in 1946 found a counterexample for this conjecture, but for the prism-Hamiltonicity version of the conjecture, in 1973 Rosenfeld and Barnette [35] showed a stronger result by proving the following theorem.
Theorem 1.1.1(Rosenfeld and Barnette [35]). Every 2-connected, planar, cubic graph is prism- Hamiltonian.
Their proof was based on the Four Color Theorem. In 1989, Fleischner [22] proved this re- sult independent of the Four Color Theorem. Moreover, in 1993 Paulraja [33] proved a modified version of Tait’s conjecture for prism-Hamiltonicity by showing the following theorem.
Theorem 1.1.2(Paulraja [33]). Every3-connected cubic graph is prism-Hamiltonian.
We will give a sketch of the proof of this theorem at the end of this section. Another direction
for this problem is keeping the planarity assumption and removing3-regularity.
Conjecture 1.1.3. [29, 35] Every3-connected planar graph is prism-Hamiltonian.
This conjecture was recently dissproved by ˇSpacapan [36]. However, the conjecture is true for some subclasses of3-connected planar graphs. The following result by Biebighauser and Elling- ham [4] proves the conjecture for triangulations.
Theorem 1.1.4. [4] LetGbe a triangulation of the plane, projective plane, torus, or Klein bottle.
ThenGis prism-Hamiltonian
Moreover, Biebighauser and Ellingham [4] proved Conjecture 1.1.3 for bipartite graphs:
Theorem 1.1.5. Every3-connected planar bipartite graph is prism-Hamiltonian.
For a graphG, defineσk(G)to be the minimum degree sum of an independent set ofkvertices ofG. Ore’s theorem [31] provides the sharp lower bound onσ2(G)which implies Hamiltonicity.
It states that a graphGon at least3vertices withσ2(G)≥nis Hamiltonian. K. Ozeki answered a similar question for prism-Hamiltonicity by showing the following theorem.
Theorem 1.1.6(Ozeki [32]). Every connected graphGof ordern ≥2withσ3(G) ≥nis prism- Hamiltonian
The k-closure of a graph G, denoted Clk(G), is the unique graph obtained from G by re- cursively joining pairs of nonadjacent vertices whose degree sum is at leastk, until no such pair remains. The famous Bondy-Chv´atal theorem [6] states that a graphGon nvertices is Hamilto- nian if and only ifCln(G)is Hamiltonian. In 2007, Kr´al and Stacho [30] obtained a similar result on prism-Hamiltonicity.
Theorem 1.1.7(Kr´al and Stacho [30]). A graphGof ordernhas a Hamiltonian prism if and only if the graphCl4n
3 −43(G)has a Hamiltonian prism.
Moreover, in same paper it is shown that 4n3 −43 cannot be improved to less than 4n3 −5.
The middle-levels graphBk is the bipartite induced subgraph formed by the middle levels of the(2k+ 1)-cube. Thus, its vertices are thek-subsets and(k+ 1)-subsets of a(2k+ 1)-set, with two subsets adjacent if one is contained in the other. Whether Bk is Hamiltonian for all k is a well-known open question, Most likely it was first asked by Havel [25]. In 2005, Hor´ak et al. [26]
proved the following theorem to answer the prism-Hamiltonicity version of this question.
Theorem 1.1.8(Hor´ak et al. [26]). Fork ≥1,Bkis prism-Hamiltonian, andBkalso has a closed spanning trail using every vertex at most twice.
Every connected simple graph G has an acyclic orientation. Let AO(G) be a graph whose vertices are the acyclic orientations ofGand whose edges join orientations that differ by reversing the direction of a single edge. Hamiltonicity of the acyclic orientations of a given graph G is still an open problem. In 1995, Pruesse and Ruskey [34] proved the following theorem on prism- Hamiltonicity version of this problem.
Theorem 1.1.9(Pruesse and Ruskey [34]). IfGis a connected simple graph thenAO(G)is prism- Hamiltonian.
TheKneser graphK(n, k)has as its vertex set allk-element subsets of an n-element set, and two vertices are adjacent if the corresponding subsets are disjoint. Theodd graphOkis a special case of a Kneser graph, as Ok = K(2k + 1, k) (e.g., O1 is the triangle and O2 is the Petersen graph). In 1978 Biggs [5] conjectured thatOk is Hamiltonian for all k >2. In 2011, Bueno and Hor´ak [8] showed that ifk >2is even thenOk is prism-Hamiltonian. More recently, De Campos Mesquita et al. proved the following more general result.
Theorem 1.1.10(De Campos Mesquita et al. [13]). Okis prism-Hamiltonian for allk >2.
ForS ⊆V(G)the subgraph induced onV(G)−Sis denoted byG−S; we abbreviateG− {v}
toG−v. The number of components ofGis denoted byc(G). The graph is said to bet-toughfor a real numbert≥0if|S| ≥ t·c(G−S)for eachS ⊆V(G)withc(G−S)≥2. Thetoughness τ(G)is the largest real numbertfor whichGist-tough, or∞ifGis complete. Positive toughness implies thatGis connected. IfGhas a Hamilton cycle it is well known thatGis1-tough.
In 1973, Chv´atal [10] conjectured that for some constant t0, everyt0-tough graph is Hamilto- nian. Thomassen (see [3, p. 132]) showed that there are non-Hamiltonian graphs with toughness greater than 32. Enomoto, Jackson, Katerinis and Saito [20] showed that every2-tough graph has a 2-factor (2-regular spanning subgraph), but also for everyε >0constructed(2−ε)-tough graphs with no 2-factor, and hence no Hamilton cycle. Bauer, Broersma and Veldman [2] constructed (94 −ε)-tough non-Hamiltonian graphs for everyε >0. Thus, any sucht0is at least 94.
There have been a number of papers on toughness conditions that guarantee the existence of more general spanning structures in a graph. Recall that ak-tree is a tree with maximum degree at mostk, and ak-walk is a closed walk with each vertex repeated at mostk times. Ak-walk can be obtained from a k-tree by visiting each edge of the tree twice. Note that a spanning2-tree is a Hamilton path and if a graph has at least three vertices then a spanning 1-walk is a Hamilton cycle. Win [40] showed that fork ≥ 3, every k−21 -tough graph has a spanningk-tree, and hence a spanningk-walk. Ellingham and Zha [19] showed that every4-tough graph has a spanning2-walk.
In 1990, Jackson and Wormald made the following conjecture.
Conjecture 1.1.11 (Jackson and Wormald [27]). For each integer k ≥ 2, every connected k−11 - tough graph has a spanningk-walk.
Recently, M. Hasanvand [24] proved that every k−11 -tough graph for k ≥ 2 has a spanning k-walk, which means that the Jackson-Wormald Conjecture is true. Since Chv´atal’s conjecture seems far from reach, the prism-Hamiltonian version of the conjecture seems to be the next step.
Kaiser et al. [29, Conjecture 4] made the following conjecture, which is analogous to those of Chv´atal and of Jackson and Wormald.
Conjecture 1.1.12 (Kaiser et al. [29]). There exists a constant t1 such that the prism over any t1-tough graph is Hamiltonian.
Kaiser et al. also showed thatt1must be at least 98.
The main idea for proving that a given graphGis prism-Hamiltonian is to find the existence of a spanning subgraph ofG which provesGto be prism-Hamiltonian. Paulraja [33] characterized
the spanning subgraphs which force prism-Hamiltonicity by proving the following theorem. We will define an SEEP-subgraph below.
Theorem 1.1.13. A graphGis prism-Hamiltonian if and only ifGhas an SEEP-subgraphH.
A subgraphHof a connected graphGis called anEP-subgraphofGif the following conditions are satisfied:
(i) His a connected spanning subgraph ofG;
(ii) ∆(H)≤4; and
(iii) H =E∪P, whereE is an edge-disjoint union of cycles andP is a vertex-disjoint union of paths, such that no vertex of a path ofP is of degree4inH, andE andP are edge-disjoint.
It is easy to observe that in an EP-subgraph H ofG, if we duplicate the edges of the paths ofP, we get an Eulerian multigraphH0 of maximum degree4. For the rest of this chapter,H0 always represents the Eulerian multigraph that arises out of an EP-subgraphHof a graphG. IfH0admits an even cycle decomposition, then we call the EP-subgraphH ofGanEEP-subgraphofG. Now, we are going to use colors to indicate which levels of a given edge are used (in the prism over the graph). Suppose blue indicates one level and yellow indicates the other level. If H is an EEP- subgraph of a graph G, thenH0 admits an even cycle decomposition, and hence it is possible to 2-color, alternately with the colors blue and yellow, the edges of each of the even cycles (arising out of the decomposition). Clearly, this 2-coloring yields a 2-coloring of the edges of H0 such that two blue edges and two yellow edges are incident with each vertex of degree4. We call an EEP-subgraphHof a graphGanSEEP-subgraphifH0with some such bicoloring admits an Euler tour using the following traversal rules:
(i) Any vertex can be used as the starting vertex of the tour.
(ii) If an edge of one color is used to reach a vertex and there is another edge of the same color incident with the vertex, then the other edge of the same color must be used to leave the vertex.
To see how this Euler tour gives a Hamilton cycle in the prism, we consider the two levels ofG to be blue and yellow, at a vertex ofGwhere there are two edges of each color we go through the copies ofv without switching levels, and at a vertex of G where there is only one edge of each color we switch levels using a vertical edge.
Depending on the problem, we can work on a special class of SEEP-subgraphs. As an example of these kinds of ideas we will outline the proof of Theorem 1.1.2 from [33] in the next section.
1.2 Proof of Theorem 1.1.2
In this section we outline the proof of Theorem 1.1.2 from [33]. Recall that Theorem 1.1.2 states that every3-connected cubic graph is prism-Hamiltonian.
Proof of Theorem 1.1.2. We start the proof with the following claim.
Claim 1.2.1. LetBbe a2-connected bipartite subgraph of a3-connected graphG. ThenGhas a 2-connected spanning bipartite subgraphH containingB.
Proof. SupposeBis not spanning. Letv ∈G−B. By Menger’s theorem there are three internally disjoint paths fromv toB. Since we have three paths, a parity argument shows that we can choose two of them so that together with B they form a2-connected bipartite subgraph ofGcontainingB. Hence we can extendB to a2-connected spanning bipartite subgraph.
Claim 1.2.2. IfGis a2-connected bipartite graph of maximum degree at most3, thenGis prism- Hamiltonian.
Proof. Let Gbe a2-connected bipartite simple graph with maximum degree 3. It is not hard to prove that we can find a set of vertex-disjoint cyclesCand a set of vertex-disjoint pathsP such that P andCare edge-disjoint and H = C∪P is a connected spanning subgraph ofG. If necessary, delete some edges ofP fromHto get a spanning connected subgraphH1ofHsuch that the cycles ofH1 are precisely the cycles ofC. It is clear to see that if we contract the cycles to vertices in H1, the resulting graph would be a tree. Consider the setCof cycles ofH1and the vertex-disjoint union of nontrivial paths P00 = E(H1 − ∪Ci∈CE(Ci)). Clearly, C ∪P00 = H1 is a spanning connected subgraph of H and hence a spanning subgraph of G. If we duplicate the edges of the paths ofP00inH1, we get an eulerian graphH10 of maximum degree4. Clearly,C0 =C∪ {2-cycles that arise from the duplication of the edges ofP00}is an even cycle decomposition ofH10. One can easily check that any bicoloring of the edges of the cycles of C0 alternately with yellow and blue proves thatGhas an SEEP-subgraphH.
Let G be a 3-connected cubic graph. Pick two vertices of G. Similarly to the proof of out first claim, there are two internally disjoint paths of the same parity between these vertices, hence Gcontains an even cycle. Therefore using the first claimGhas a2-connected spanning bipartite subgraph H. By the second claim, Ghas a prism-Hamiltonian subgraph and hence Gis prism- Hamiltonian.
1.3 Outline
In Chapter 3, we will answer West’s question of finding a Chv´atal-Erd˝os condition, i.e., a bound on α(G) in terms of κ(G), guaranteeing prism-Hamiltonicity by proving that α(G) ≤ 2κ(G) implies prism-Hamiltonicity. This is joint work with M. Ellingham and has been published as [17]. Moreover, we will find the Chv´atal-Erd˝os condition for Hamiltonicity ofGK3.
In Chapter 4, we show that for the class ofP4-free graphs, the three properties of being prism- Hamiltonian, having a spanning2-walk, and being 12-tough are all equivalent. This is joint work with M. Ellingham and S. Shan and has been published as [18].
In Chapter 5, we will characterize the forbidden subgraphs and forbidden pairs that imply prism-Hamiltonicity. The last chapter includes some future problems and directions.
Chapter 2
Hierarchy of Hamilton properties
2.1 Hierarchy diagrams
In this chapter we investigate generalizations of prism-Hamiltonicity that form a hierarchy of properties. As we mentioned in Chapter 1, Kaiser et al. [29] showed that the property of having a Hamiltonian prism is stronger than that of having a spanning 2-walk and weaker than that of having a Hamilton path. Recall that Paulraja [33] proved that a graph Gis prism-Hamiltonian if and only ifGhas an SEEP-subgraphH.
It is a natural generalization to study the Hamiltonicity of other cartesian products of a graphG and graphH ∈ {Ct, Kt}fort≥2as weaker properties than Hamiltonicity ofG. It is obvious that Hamiltonicity ofGCtimplies Hamiltonicity ofGKt, as the first product is a spanning subgraph of the second product. The following theorem provides a sufficient condition for Hamiltonicity of GCt.
Theorem 2.1.1(Batagelj and Pisanski [1]). LetT be a tree with maximum degree∆(T)≥2. Then TCtis Hamiltonian if and only if∆(T)≤t.
Theorem 2.1.1 shows that if G contains a spanningt-tree then GCt is Hamiltonian. In the next lemma, we will find a sufficient condition for Hamiltonicity ofGKtfort ≥3.
Definition 2.1.2. At-cactusis a connected graph such that:
(i) Each block is either a cycle or an edge.
(ii) Each vertex lies in at mosttblocks.
A vertex of at-cactus is calledsaturatedif it lies in exactlytblocks. By astrongcactus we mean mean a cactus where each vertex belongs to at most one cycle. At-cactus is called anevent-cactus if all of its cycles have even length.
We usually do not consider a one-vertex graph to be a t-cactus. Note that a subgraph of a t-cactus is not necessarily at-cactus. For example, consider aC6 with two pendant edges attached to each of two nonadjacent vertices of the cycle. This graph is a3-cactus but if we delete one of the edges of theC6 the new graph is not a3-cactus.
We show later, in Theorem 4.2.2, that having a spanning even2-cactus implies prism-Hamilton- icity. However, here we will be concerned witht-cacti fort≥3.
Before stating the lemma let us define some terminology that we need for the proof. InGH the copiesGwof G,w∈V(H), will be calledlevels(similar to levels for prisms). If|V(H)|=t, we can assign an integer between1andtto each level. Moreover, the copiesvHofH,v ∈V(G), will be calledclusters.
Lemma 2.1.3. Ift≥3andGis at-cactus, thenGKtis Hamiltonian.
Proof. We prove this lemma by induction on the number of blocks ofG. Indeed, we prove that for everyt-cactusGthere is a Hamilton cycleZ inGKt such that wheneverv ∈ V(G)belongs to only one block ofGwe have thatZ ∩(vKt)is a single path.
First, we show the above statement whenGis a single block. SupposeGis a single block, i.e., Gis an edge or a cycle. Note thatv ∈V(G)is either a degree1vertex ifGis an edge or a degree 2vertex ifGis a cycle. Ifvis a degree1vertex, i.e.,Gis an edge, thenZmust visit the vertices of vKtwith one entrance and one exit, sayZ enters in level1, visits levels2, . . . , t−1, and exits in levelt. Doing this at both vertices of the edge creates a Hamilton cycle inGKt. Now suppose v ∈V(G)is a vertex of degree2, i.e.,Gis a cycle. IfGis an even cycle thenZ can enter and exit in level 1andt alternatingly to visit the cluster of each vertex ofG. If Gis an odd cycle then Z enters and exits alternatingly in levels1andtwhen visiting the cluster of each vertex ofGexcept for two consecutive vertices where it enters the cluster of the first vertex at level1and exits at level 2, and enters the cluster of the second vertex at level2and exits at levelt. Therefore ifGis a single block thenGKthas a Hamilton cycleZsuch that for each vertexvofG,Z∩(vKt)is a single path.
Now suppose G is a t-cactus with more than one block. Pick a vertex v of G that belongs
to more than one block of G. Let F1, . . . Fr be the components of G −v and for 1 ≤ i ≤ r let Gi consist of Fi together with v and all edges from a vertex of Fi to v. Note that all Gi’s are t-cactus graphs. Moreover, the vertex v belongs to a single block of each t-cactus Gi for 1≤i≤ r. Therefore, by the induction hypothesis eachGiKthas a Hamilton cycleZi such that Zi∩(vKt) is a single path. After possibly applying an automorphism ofGithat permutes levels, we may assume that Zi is the union of the pathPi = Zi ∩(vKt)from (v, i)to (v, i+ 1) and another pathQi from(v, i)to(v, i+ 1)whereE(Qi)∩(vKt) = ∅. Then the cycleQ1∪ · · · ∪ Qr∪(v, r+ 1)(v, r+ 2). . .(v, t)(v,1)is a Hamilton cycle forGKt. Note that, since a vertexu that belongs to a single block ofGalso belong to a single block of one of theGi’s,Z∩(uKt)is a single path.
We may also consider more a general Dvoˇr´ak-Postle product, GDPH, which is a family of graphs that are constructed by blowing up each vertex ofG into a copy of H and for each edge ofGadding an arbitrary perfect matching between the appropriate copies ofH [16]. Dvoˇr´ak and Postle came up with this idea of a modified cartesian product in order to generalize a list-coloring problem. It should be noted that in their product, they put an arbitrary matching between clusters but here we consider only perfect matchings in order to have a natural generalization of prisms. A graphGisDP-prism-Hamiltonianif all members of the familyGDPK2are prism-Hamiltonian.
ReplacingK2 byH ∈ {Ct, Kt}introduces more generalization of prisms. The familyGDPHis called Hamiltonian if all members of the family are Hamiltonian.
Theorem 2.1.4. If a graphGhas a spanningt-tree thenGDPKtis Hamiltonian.
Proof. SupposeT is a spanningt-tree ofG. Suppose J is a member of the familyGDPKt. By the definition of the DP-product, TCtis isomorphic to a spanning subgraph of J. On the other hand, by Theorem 2.1.1TCtis Hamiltonian and henceJ is Hamiltonian. ThereforeGDPKtis Hamiltonian.
The following diagrams shows the implications that we already know, where ‘Ham’ means Hamiltonian. Recall thatG[H]denotes the lexicographic product of graphsGandH. Note that a
spanning2-tree is just a Hamilton path.
Ham. path=sp. 2-tree
⇓6⇑(a) ⇒
6⇐ (b)
GDPK2 Ham. sp. even2-cactus
⇓6⇑(c) ⇒
6⇐ (d) GK2 Ham.
⇓6⇑(e)
G[K2]Ham⇔sp. 2-walk
The forward implications are obvious because one family of graphs is a subset of another, or one graph is a subgraph of another (b, c, e), except (a) which follows from Theorem 2.1.4 and (d) which follows from Theorem 4.2.2 of Chapter 4.
For the backwards non-implications we have:
(a) In the next proposition, Proposition 2.1.5, we will show that the complete bipartite graph K2,4does not have a Hamilton path but it is DP-prism-Hamiltonian.
(b) Any graph with a spanning even2-cactus that does not have a Hamilton path is a counterex- ample to the backwards implication, such as a graph that consists of an even cycle with a pendant edge attached at each vertex. Another example is a graph consisting of a complete graph on an even number of vertices with a pendant edge attached at each vertex.
(c) Consider a graphGthat consists of an even cycle with a pendant edge at each vertex. Now in the class GDPK2 consider a graph J such that the corresponding perfect matching in J for each edge ofGconnects the same levels of clusters to each other except for exactly one cycle edge ofGwhose corresponding perfect matching is crossing (i.e., connects level1of one cluster to level 2of the other cluster and vice versa). The graphJ is not Hamiltonian, since if it has a Hamilton cycle, the Hamilton cycle must switch level between entering a cycle vertex cluster from another cycle vertex cluster and leaving to go to the next cycle vertex cluster, which is impossible due to the parity and the crossed matching.
(d) Consider four paths A = ua1a2a3v, B = ub1b2b3v, C = uc1c2c3c4v, D = ud1d2d3d4v. Now letGbe the graph that containsA∪B ∪C∪D, and a pendant edge at every vertex except u and v. The graph G is an SEEP. Note that G has only two even cycles A∪ B and C ∪D.
Color the edges of A ∪B alternately blue and yellow, and similarly for C ∪ D, and duplicate each pendant edge into one blue edge and one yellow edge. This satisfies the requirements of the proof of Theorem 1.1.13. Note thatG has no spanning even2-cactus. To see that first note that Ghas no Hamilton path. Moreover, the only two even cycles have two vertices in common and hence ifGhas a spanning even2-cactus it must use only one of the cycles. But we cannot build a spanning even2-cactus using one of those because any spanning subgraph containing just one of these cycles must have a vertex incident with three edge blocks. We can obviously extend this to a family of examples by lengthening each ofA,B,C,Dby an even number of edges.
(e) Recall that Kaiser et al. [29] construct graphs with arbitrarily large connectivity that have spanning2-walks, but are not prism-Hamiltonian. Moreover, in Chapter 1, we proved that a net has a spanning2-walk, but is not prism-Hamiltonian.
Proposition 2.1.5. For a graph G having a Hamilton path is strictly stronger than being DP- prism-Hamiltonian.
Proof. We will show that the complete bipartite graphK2,4 is DP-prism-Hamiltonian while it has no Hamilton path. Since the number of vertices in the two partite sets must differ by at most one for a Hamilton path,K2,4has no Hamilton path.
Let J be an element of K2,4DPK2. We assign a value of0 or1 to each edge of K2,4. The value of an edge is1if its corresponding matching in J flips the levels and otherwise it is0. An argument similar to the one that existence of a spanning even2-cactus implies prism-Hamiltonicity shows that if we can always find a spanning2-cactus with a single cycle having an even edge value sum then the graph is DP-prism-Hamiltonian.
Now suppose {x, y} is the partite set of cardinality 2in K2,4 and lets1, s2, s3, s4 denote the edge value sums of the fourxy-paths inK2,4. At least two of thesi’s have the same parity, which creates a C4 with even edge value sum. Joining one of the remaining vertices toxand the other
one toybuilds our desired structure.
The following diagram shows properties related to having a spanning3-tree or3-walk.
sp. 3-tree
⇓(f) ⇒
6⇐ (g)
GDPK3Ham. sp. 3-cactus
⇓(h) ⇐(i)
GK3 Ham.
⇓(j)
G[K3]Ham⇔sp.3-walk
The forward implications follow because one family of graphs is a subset of another, or one graph is a subgraph of another (g, h, j), except (i) which follows from Lemma 2.1.3 and (f) which follows from Theorem 2.1.4.
For the backward non-implication (g), consider three cyclesC1, C2 andC3 of lengths at least 3 sharing a vertex v and two pendant edges attached to every vertex except v. This graph is a 3-cactus. But it does not have a spanning3-tree, since we can use at most three edges incident to v, which does not allow us to cover the pendant edges.
Our final diagram shows properties related to having a spanning t-tree ort-walk whent ≥ 4.
In that case,KtandCtare not isomorphic, so we have more properties than whent = 3.
sp. t-tree
⇓(k) ⇒
6⇐ (`) GDPCtHam. (m)
⇒ GDPKtHam. sp. t-cactus
⇓(n) ⇓(o) ⇐(p)
sp. t-tree (q)
⇒ GCtHam. (r)
⇒6⇐ GKtHam.
⇓(s) ⇓(t)
G[Ct]Ham (u)
⇒ G[Kt]Ham⇔sp. t-walk
Note that ‘sp. t-tree’ appears in two places in the diagram. The forward implications follow because one family of graphs is a subset of another, or one graph is a subgraph of another (`, m, n, o, r, s, t, u), except (p) which follows from Lemma 2.1.3 and (q) which follows from Theorem 2.1.4.
For the backwards non-implications:
(`) We can constructt-cacti with no spanningt-tree in a way similar to (g).
(r) This follows by the next proposition, Proposition 2.1.6.
We will also show that having a spanning t-tree does not imply GDPCt is hamiltonian, in Example 2.1.7. It seems thatGDPCtbeing Hamiltonian is a very strong property. It does not fit in between having a spanningt-tree and a having a spanningt-walk like the other properties, and examples where this property fails involve structures like the Petersen, graph so it is likely to be very difficult to prove any sort of results about this propoerty.
Proposition 2.1.6. Supposet ≥ 4. For a graphG, having a spanningt-cactus is not a sufficient condition for Hamiltonicity of GCt. Hence, Hamiltonicity of GCt is strictly stronger as a condition onGthan Hamiltonicity ofGKt.
Proof. In this proof the levels, or equivalently the vertices they correspond to inCt, are numbered in order around the cycle and level numbers are interpreted modulot. LetGbe the graph Ntthat contains a triangle and t − 1 pendant edges attached to each vertex of the triangle; see Figure 2.1. Note thatGis a t-cactus and henceGKt is Hamiltonian. We will show that GCt is not Hamiltonian, which shows that having a spanningt-cactus does not imply Hamiltonicity ofGCt. Supposea,bandcare the vertices of the triangle inGand supposeN(a) ={a1, . . . at−1, b, c}, N(b) = {b1, . . . bt−1, a, c}andN(c) = {c1, . . . ct−1, a, b}. SupposeGCthas a Hamilton cycleZ. The Hamilton cycleZ projects to a spanningt-walk inG. Theorefore, for a degree1vertex vi of G, adjacent tov ∈ {a, b, c}, the Hamilton cycle enters and exitsviCtexactly one time, otherwise it violates the degree condition of thet-walk atv. Hence, at each visit of a degree1vertex ofGwe need to change the level fromitoi±1. Therefore ifZ entersaCtat levelithen after visiting all
Figure 2.1:Nt
t−1pendant edges ata,Z must exit in levelaCtat leveli±1and similarly forbandc. In other words, in each visit of aCt, bCt and cCt we must switch level from i to i±1. Therefore, following the Hamilton cycle fromaCttobCtand thencCt, if it uses levelj along the edge abit must use levelj ±1alongbc, and then levelj+ 2, j orj−2alongca, but sincet >3this contradicts the fact that it enters and leavesaCtat levels that differ by1. ThereforeGCtis not Hamiltonian.
Figure 2.2: Petersen Graph
Example 2.1.7. K2DPC5 is not Hamiltonian and hence for a graph Ghaving a spanningt-tree does not in general imply Hamiltonicity ofGDPCt.
Proof. The graphK2 has a spanning5-tree, but the Petersen graph (Figure 2.2), which belongs to K2DPC5, is not Hamiltonian.
We already know that some of the implications in our diagrams cannot be reversed. We would like to determine which of the remaining implications can be reversed (if any), and which cannot.
Also, we are interested in finding the Chv´atal-Erd˝os conditions that imply any of these generaliza- tions of Hamiltonicity. Some results of this kind will be given in Chapter 3.
2.2 Repeated prisms
Since taking a cartesian product withK2creates a graph that is more likely to be Hamiltonian, it makes sense to consider repeated cartesian products withK2of a connected graphG, i.e., cartesian products ofGwith powers ofK2. This way we can provide a measure of how close a connected graph is to being Hamiltonian. Hence a natural question is to ask for a given connected graphG whether there is a positive integer n such that GK2n is Hamiltonian. If the answer is yes, we would like to know the smallest suchn. The theorem below shows that thisn exists and gives an upper bound for it.
Theorem 2.2.1. If G is a connected graph with ∆ ≥ 2 (i.e., with at least three vertices) then GK2nis Hamiltonian forn≥ dlog2(∆(G)e.
Proof. If ∆(G) = 2 thenG is a path or cycle, and the result holds for n = 1and hence for all suitablen. So we may assume that ∆(G) ≥ 3. Note that forn ≥ 2, K2n contains aC2n. Setting n=dlog2(∆(G))eimplies thatK2nhas a cycle of lengthm= 2n≥∆(G). SinceGhas a spanning
∆(G)-tree, by theorem 2.1.1GCm is Hamiltonian and hence GK2nis Hamiltonian asCm is a spanning subgraph ofK2n.
Chapter 3
The Chv´atal-Erd˝os condition for prism-Hamiltonicity
3.1 Preliminaries
In this chapter we study the Chv´atal-Erd˝os condition, that is an upper bound for independence number in terms of connectivity, that implies prism-Hamiltonicity and Hamiltonicity of other graph products. Chv´atal and Erd˝os proved the following theorem.
Theorem 3.1.1(Chv´atal and Erd˝os [11]). LetGbe a graph with at least three vertices. Ifα(G)≤ κ(G), thenGis Hamiltonian.
Suppose G is a graph with |V(G)| ≥ 2 and α(G) ≤ κ(G) + 1. By adding a new vertex v adjacent to all vertices of G, we construct G0 which satisfies the hypothesis of Theorem 3.1.1.
HenceG0 is Hamiltonian, so thatG=G0−v has a Hamilton path. This also holds if|V(G)|= 1, giving the following corollary.
Corollary 3.1.2. LetGbe a graph. Ifα(G)≤κ(G) + 1, thenGhas a Hamilton path.
Moreover, it is known thatα(G)≤2κ(G)implies existence of a spanning2-walk forG[27].
Problem 3.1.3(West [38]). Givenk, what is the largest value of asuch that ifGis a graph with κ(G) =kandα(G) =a, then the prism overGis Hamiltonian?
Fora > k, the complete bipartite graphKk,a isk-connected and has independence numbera.
When a > 2k, the prism over Kk,a is not Hamiltonian, since deleting the 2k vertices of degree a+ 1leavesacomponents. Hence the answer to this problem is at most2k.
The following theorem is our answer to this question.
Theorem 3.1.4. Let G be a graph with at least two vertices. If α(G) ≤ 2κ(G), then G has a spanning even 2-cactusH in which each vertex belongs to at most one cycle, so that∆(H) ≤ 3.
Moreover, ifκ(G)≥3thenH has at most one cycle. HenceG0 is prism-Hamiltonian.
Recall that by a strong cactus we mean mean a cactus where each vertex belongs to at most one cycle. Therefore, Theorem 3.1.4 shows the existence of a spanning even strong2-cactus. This theorem shows that the Chv´atal-Erd˝os condition sufficient for being prism-Hamiltonian is the same as for the weaker property of having a spanning2-walk.
Here we list the results that we need in our proofs.
Theorem 3.1.5(Dirac [14]). LetSbe a set ofkvertices in ak-connected graphG, wherek ≥ 2.
Then there exists a cycle inGthrough every vertex ofS.
Theorem 3.1.6(Bondy and Lov´asz [7]). Let S be a set of k vertices in a k-connected graph G, wherek≥3. Then there exists an even cycle inGthrough every vertex ofS.
Theorem 3.1.7 (Jackson and Wormald [27]). The existence of a spanningt-tree implies the ex- istence of a spanning t-walk, and the existence of a spanning t-walk implies the existence of a spanning(t+ 1)-tree.
Figure 3.1: An even strong2-cactus
Lemma 3.1.8 ( ˇCada et al. [9]). IfG contains a spanning even strong2-cactus, thenG is prism- Hamiltonian.
Proof. Same as the proof of Claim 1.2.2 in the proof of Theorem 1.1.2.
Let us finish this section with some terminology. LetP,T, andCbe a path, a tree, and a cycle, respectively. Fora, b∈ V(P)byP[a, b]we mean the path froma tob alongP. Fora, b ∈V(T) byT[a, b]we mean the unique path fromatobinT. ForC we assume that one direction alongC has been designated as ‘forward’ or ‘clockwise’. Then fora, b ∈ V(C)byC+[a, b](orC−[a, b]) we mean the unique path froma tobalongC clockwise (or counterclockwise). If any of the end vertices of these paths is excluded we replace[by(or]by)or both, as appropriate. For example, P[a, b)meansP[a, b]−b.
3.2 Proof of Theorem 3.1.4
Recall that Theorem 3.1.4 states that if G is a connected graph then α(G) ≤ 2κ(G)implies thatGhas a spanning even strong 2-cactus, with at most one cycle ifκ(G) ≥ 3, and hence Gis prism-Hamiltonian.
Proof of Theorem 3.1.4. Ifα(G)≤κ(G) + 1then, by Corollary 3.1.2,Ghas a Hamilton path, and hence is prism-Hamiltonian by Lemma 3.1.8. So we may assume thatκ(G) + 2≤α(G)≤2κ(G).
Thus,κ(G)≥2.
We break the proof into two cases,κ(G) = 2andκ(G)≥ 3. Somewhat surprisingly, we have to work harder in the first case; in the second case Bondy and Lov´asz’s Theorem 3.1.6 does a significant amount of the work.
Case 1. Suppose thatκ(G) = 2. Sinceκ(G) + 2 = 4≤ α(G)≤2κ(G) = 4, we haveα(G) = 4.
By adding two adjacent vertices (a complete graph on two vertices,K2) toGthat are adjacent to all vertices ofG, we obtain a new graph, sayG0. Thenκ(G0) =α(G0) = 4. Therefore by Theorem 3.1.1G0 is Hamiltonian. Removing these two new vertices implies thatGhas a Hamilton path or two vertex-disjoint paths P1 and P2 that cover all vertices of G. In the former case Gis prism- Hamiltonian, so we assume the latter case. Letu1 andu2 be the end vertices ofP1 andv1 and let v2 be the end vertices ofP2.
Claim 3.2.1. Each ofP1andP2contains more than one vertex; otherwise,Gis prism-Hamiltonian.
Proof. Suppose u1 = u2 = u. Since G is 2-connected, there are two edges from u to P2, say ub1 and ub2. If b1 or b2 belongs to {v1, v2}, then G has a Hamilton path, and hence is prism- Hamiltonian. Now supposeb1 is the neighbor ofu closest to v1 in P2. Since Gis 2-connected, there exists an edgexy ∈ E(G)such that x ∈ V(P2[v1, b1))andy ∈ V(P2(b1, v2]). One of the cyclesP1[y, b2]∪b2ub1∪P2[b1, x]∪xy,P2[b1, b2]∪b2ub1orP2[x, y]∪xyis an even cycle and the even cycle together with remaining two path segments ofP2form a spanning even strong2-cactus, and henceGis prism-Hamiltonian.
Suppose u1 6= u2 and v1 6= v2. Since G is 2-connected, there are distinct vertices a1, a2 ∈ V(P1)andb1, b2 ∈ V(P2)such thata1a2, b1b2 ∈ E(G). We may assume that u1, a1, a2, u2 occur in that order onP1, andv1, b1, b2, v2 occur in that order onP2.
Claim 3.2.2. The orders of the paths P1[a1, a2]andP2[b1, b2]have different parity; otherwise, G is prism-Hamiltonian.
Proof. Suppose the orders of the pathsP1[a1, a2]andP2[b1, b2]have same parity. ThenP1[a1, a2]∪ a2b2∪P2[b2, b1]∪b1a1 is an even cycle. This cycle together with remaining path segments ofP1 andP2form a spanning even strong2-cactus, i.e., the even cycle together withP1−P1[a1, a2]and P2−P2[b1, b2]. ThereforeGis prism-Hamiltonian.
Claim 3.2.3. IfP2[x, y]∩P2[b1, b2]has at least one edge andxy∈E(G)\E(P2)forx, y ∈V(P2), thenP2[x, y]∪yxis an even cycle; otherwise,Gis prism-Hamiltonian.
Proof. SupposeP2[x, y]∪yxis an odd cycle. By Claim 3.2.2,P1[a1, a2]∪a2b2∪P2[b2, b1]∪b1a1 is an odd cycle. Then combining these two odd cycles form an even cycle which yields a spanning even strong2-cactus. (The same statement holds for edges between two vertices ofP1.)
Claim 3.2.4. The set{u1, u2, v1, v2}is an independent set; otherwise,Gis prism-Hamiltonian.
Proof. By contradiction suppose {u1, u2, v1, v2} is not an independent set. If uivj ∈ E(G) for i, j ∈ {1,2}thenGcontains a Hamilton path and hence it is prism-Hamiltonian.
Thus, we may assume thatu1u2 ∈E(G), i.e.,P1∪u1u2 is a cycle. By Claim 3.2.3,P1∪u1u2 is an even cycle. We can assume thatb1is the closest neighbor of a vertex ofP1onP2 tov1. Then, by2-connectedness and since b1 is the closest vertex tov1 adjacent to a vertex ofP1, there is an edgexysuch thatx ∈V(P2[v1, b1))andy ∈V(P2(b1, v2]). Then by Claim 3.2.3,P2[y, x]∪xyis an even cycle. ThereforeP1∪u1u2 andP2[x, y]∪yxare even cycles and together with the edge a1b1and remaining path segments ofP2 form a spanning even strong2-cactus.
Claim 3.2.5. There is no edgexywithx∈V(P1)− {a1, a2}andy∈V(P2)− {b1, b2}; otherwise, Gis prism-Hamiltonian.
Proof. Supposexy ∈ E(G)forx ∈ V(P1)andy ∈ V(P2). By Claim 3.2.2,P1[a1, a2]∪a2b2 ∪ P2[b2, b1]∪b1a1 is an odd cycle. Then one of the cyclesZ1 =P1[x, a1]∪a1b1∪P2[b1, y]∪yxor Z2 =P1[x, a2]∪a2b2∪P2[b2, y]∪yxis even and together with remaining path segments ofP1and P2 forms a spanning even strong2-cactus.
Claim 3.2.6. There is no edge xysuch that either (i) x ∈ {u1, u2} andy ∈ V(P2)− {b1, b2}or (ii)x∈V(P1)− {a1, a2}andy∈ {v1, v2}; otherwise,Gis prism-Hamiltonian.
Proof. Without loss of generality suppose that (i) holds with x = u1. Ifx 6= a1 then the result follows by Claim 3.2.5, so suppose thatx=u1 =a1. The proof of Claim 3.2.5 fails whenx=a1 because if we need to construct a spanning even strong2-cactus from the cycleZ1 then we would have to attach two path segments ofP1 atx=a1, creating a degree4vertex, which is not allowed.
However, since x = u1 = a1 here one of these path segments is trivial (just the single vertexu1) so this does not create a problem now, and we may proceed as in the proof of Claim 3.2.5.
Now we may suppose that {u1, u2, v1, v2} is an independent set. By Claim 3.2.2, the paths P1[a1, a2] and P2[b1, b2] have different parity. Without loss of generality we can assume that P2[b1, b2]has an odd number of vertices, and therefore there is a vertexx ∈ V(P2(b1, b2)). Since α(G) = 4, andS = {u1, u2, v1, v2}is an independent set,xis adjacent to some vertex in S. By Claim 3.2.6, we may assume thatxis adjacent to neitheru1 noru2. Without loss of generality we may assume that x is adjacent to v1. Then by Claim 3.2.3, the cycleP2[x, v1]∪v1x is even. If a1 = u1 then we have a spanning even strong2-cactus using the cycle P2[x, v1]∪v1x and paths P2[x, v2]andb1u1∪P1, so we may assume thata1 6=u1. By2-connectedness there is an edgeyz such thaty∈V(P1[u1, a1))andz ∈V(P2)∪V(P1(a1, u2]). So we have the following cases.
Case 1.1. Ifz ∈V(P1(a1, v2]), by Claim 3.2.3 we may assume thatyz∪P1[z, y]is an even cycle.
Then the cyclesv1x∪P2[x, v1]andyz ∪P1[z, y]together with the edge a1b1 and remaining path segments ofP1andP2 form a spanning even strong2-cactus.
Case 1.2. Supposez ∈ V(P2). By Claim 3.2.5 we can assume thatz =b1 orz = b2 which lead us to the following cases.
Case 1.2.1. Supposez =b2. Then we can assume that the cycleyb2a2∪P1[a2, y]is even; otherwise, the cycle P2[b1, b2]∪b2y ∪P1[y, a1]∪ a1b1 is even and yields a spanning even strong 2-cactus.
Therefore the even cycles yb2a2 ∪P1[a2, y] and v1x∪P2[x, v1] together with the edgea1b1 and remaining path segments ofP1 andP2form a spanning even strong2-cactus.
Case 1.2.2. Suppose z = b1. Then for the same reason as above we can assume that the cycle yb1a1 ∪P1[a1, y] is even. Therefore there is a vertex c ∈ V(P1(y, a1)). We can assume that ca1 ∈ E(P1). Since α(G) = 4, and S = {u1, u2, v1, v2} is an independent set, c is adjacent to some vertex in S. By Claim 3.2.6 we may assume that c is adjacent to neitherv1 nor v2. If u2c∈E(G)then by Claim 3.2.3,P1[c, u2]∪u2cis an even cycle and together withP2[v1, x]∪xv1 it yields a spanning even strong2-cactus. Hence we may assume thatu1c∈E(G).
Ifu1c /∈ E(P1), then we may assume thatP1[c, u1]∪u1cis an odd cycle; otherwise, together with P2[v1, x]∪ xv1 it yields a spanning even strong 2-cactus. If P1[c, u1] ∪ u1c is odd, then P1[c, a2]∪a2b2∪P2[b2, b1]∪b1y∪P1[y, u1]∪u1cis an even cycle and together with remaining path segments of P1 and P2 forms a spanning even strong 2-cactus. Therefore we may assume thatu1c∈ E(P1), which impliesy =u1. ThenP2[v1, x]∪xv1 together with pathsb1u1 ∪P1 and P2[x, v2]forms a spanning even strong2-cactus.
Case 2. Suppose thatk =κ(G)≥ 3. Letα =α(G)and lett =α−k ≥ 2. LetG0 be the graph G together with a Kt and all edges from these new t vertices to V(G). Then α(G0)=α(G) ≤ κ(G0)=κ(G) +t, hence by Theorem 3.1.1G0 is Hamiltonian. By removing thesetnew vertices, we can cover all the vertices of Gby r ≤ t vertex-disjoint paths, P1, P2, . . . , Pr. Let v1, . . . , vr be one of the end vertex of each of these r paths. By Theorem 3.1.6 there is an even cycle, say C, passing throughv1, . . . , vr. Now we put a direction on each of these rpaths starting from vi, 1≤i≤r. Our goal is attaching some paths toCto form a spanning even strong2-cactus.
SupposeC intersectsPi atwi1 =vi,wi2,. . .,wik
i, in that order alongPi. Letxik
i be the end of Pi other thanvi, and for1≤j ≤ki−1letxij be the vertex immediately beforewij+1onPi. Then we add the pathsPi[wji, xij],1 ≤i≤ r,1≤ j ≤ ki, toC. This process will form a spanning even strong2-cactus withCas its only cycle.
3.3 Toughness conditions
Recall that Chv´atal [10] conjectured that for some fixed teveryt-tough graph is Hamiltonian.
Kaiser et al. [29, Conjecture 4] make the natural related conjecture that for some fixedtallt-tough graphs are prism-Hamiltonian, and show thattmust be at least9/8.
While it appears very difficult to show that some constant toughness implies Hamiltonicity or even prism-Hamiltonicity, Chv´atal-Erd˝os conditions combined with some simple observations suffice to show that Ω(√
n)-tough graphs have these properties. As far as we can tell, no one has noted this before. Suppose G is a non-complete n-vertex t-tough graph; letα = α(G) and κ=κ(G). By [10, Propositions 1.3 and 1.4],κ ≥2tandt≤(n−α)/α, orn/(t+ 1)≥α. Using these, we obtain the following.
Proposition 3.3.1. Supposet >0,n≥3, andGis at-toughn-vertex graph.
(i) If2t(t+ 1) ≥n(e.g., ift ≥p
n/2), thenGis Hamiltonian.
(ii) If4t(t+ 1) ≥n(e.g., ift ≥√
n/2), thenGis prism-Hamiltonian.
Proof. We may assume G is non-complete. If p ≥ 0 and 2pt(t + 1) ≥ n then pκ ≥ 2pt ≥ n/(t+ 1) ≥ α. Applying Theorem 3.1.1 when p = 1 and Theorem 3.1.4 when p = 2 gives the result.
3.4 The Chv´atal-Erd˝os condition for Hamiltonicity ofGK3
It is known [27, Theorem 5.3] thatα(G)≤ tκ(G)implies Hamiltonicity ofG[Kt](the lexico- graphic product ofGandKt), which is equivalent to the existence of a spanningt-walk inG. As an extension of Theorems 3.1.1 and 3.1.4 we can ask whetherα(G) ≤ tκ(G)implies Hamiltonicity of GKt whent ≥ 3. We can prove the following slightly weaker result. Recall that a simple consequence of Batagelj and Pisanski’s result, Theorem 2.1.1, is that if G has a spanning t-tree thenGCtis Hamiltonian.
Proposition 3.4.1. LetGbe a graph, andt ≥ 3an integer. Ifα(G) ≤ (t−1)κ(G)thenGCt, and henceGKt, is Hamiltonian.
Proof. We know that α(G) ≤ (t− 1)κ implies existence of a spanning (t−1)-walk in G. By Theorem 3.1.7 existence of a spanning(t−1)-walk implies the existence of a spanningt-tree and hence, by Theorem 2.1.1, Hamiltonicity ofGCt.
Recall that by at-cactus we mean a connected graph such that each block is either a cycle or an edge, and each vertex lies in at mosttblocks. Lemma 2.1.3 states that ift≥3andGis at-cactus, thenGKtis Hamiltonian.
Lemma 3.4.2. SupposeHis a2-cactus with exactly one cycleZ. IfR⊆V(H)with|R|=r ≥2, thenH−R has at most2rcomponents. Moreover,H−Rhas exactly2rcomponents if and only ifR ⊆V(Z),Ris an independent set inZ, anddegH(v) = 3for allv ∈R.
Proof. Observe that all vertices of H have degree at most3, and vertices of degree3must lie on Z. IfRcontains no vertices ofZ, then all of its vertices have degree at most2. Thus, deleting each vertex ofRadds at most one component, so thatc(H−R)≤r+ 1<2r. So we may assume that R∩V(Z)6=∅.
By deleting the first vertex of R ∩V(Z) we add exactly 1component. After that, since the remaining components are3-trees, deleting each vertex may add at most2components. Note that deleting a vertex will add exactly2components if and only if it is a degree3vertex of a3-tree and
hence a vertex ofZ, and the vertex is not adjacent to another vertex that has been already deleted.
Therefore,H−Rhas at most2rcomponents, andH−Rhas exactly2rcomponents if and only ifR ⊆V(Z),Ris an independent set inZ, anddegH(v) = 3for allv ∈R.
The following lemma generalizes the construction used in Case 2 of the proof of Theorem 3.1.4.
Lemma 3.4.3. Suppose G is a graph with κ(G) ≥ 2, and t ≥ 2. Suppose also that G has a subgraphH with componentsH1, H2, . . . , Hk, wherek ≤ κ(G)and∆(H)≤t. ThenGcontains at-cactusJ with exactly one cycleC, such thatV(H)⊆V(J), vertices inV(J)−V(H)belong toC but no other block ofJ, and|V(C)| ≥ k. In particular, ifH is a spanning subgraph of G thenJ is a spanningt-cactus inG.
Proof. Since∆(H)≤t,Hi has a spanningt-treeTi fori∈ {1, . . . , k}. In eachTi select a degree 1(or0) vertex and direct all edges ofTiaway from the selected vertex. Since the connectivity ofG is at least the number of selected vertices, by Theorem 3.1.5 there is a cycleC that passes through thekselected vertices. ConsiderC∪T1∪ · · · ∪Tk, and for each vertex inV(C)∩V(T1∪ · · · ∪Tk) that is not a selected vertex, delete its incoming edge. The result is a subgraphJ in whichC is the only cycle. InJ, vertices inV(J)−V(H) =V(C)−V(H)are incident only with (edges of)C.
Since each unselected vertex inV(C)∩V(T1∪ · · · ∪Tk)has at mostt−1outgoing edges, it is incident withC and at mostt−1other edge blocks. Selected vertices are incident withCand at most one other edge block. Vertices not onC are incident with at mostt edge blocks. Thus, the result is at-cactus that includesV(H). In particular ifH is a spanning subgraph ofGthenJ is a spanningt-cactus inG.
Theorem 3.4.4. Ifα(G)≤3κ(G), thenGK3is Hamiltonian.
Proof. To prove this theorem we break the problem into four cases and in each case we show the existence of a spanning2-cactus or a spanning3-cactus. Ifα(G)≤2κ(G)then by Theorem 3.1.4 Ghas a spanning2-cactus. Hence we may assume that2κ(G)< α(G)≤3κ(G).
Case 1. Supposeκ(G) = 1and hence2< α(G)≤3. Build a graphG0by adding a vertexvto the graphGthat is adjacent to all vertices ofG. By the construction ofG0, we haveα(G0) = α(G) = 3 andκ(G0) =κ(G) + 1 = 2. Thereforeα(G0) ≤κ(G0) + 1and hence by Corollary 3.1.2G0has a Hamilton pathP.
InG=G0−vthe subgraphP−vis a spanning subgraph that is either one path or two disjoint paths. IfGhas a Hamilton path then that is a spanning2-cactus ofG0. Hence we may assume that P −v =P1∪P2whereP1 andP2 are two disjoint paths covering all vertices ofG. Connectivity ofGimplies existence of an edgeeconnectingP1 andP2. The subgraphP1∪e∪P2 is a spanning 3-cactus ofG.
Case 2. Supposeκ(G) = 2and hence4< α(G)≤6. Build a graphG0by adding a vertexvto the graphGthat is adjacent to all vertices ofG. By the construction ofG0, we haveα(G0) = α(G)≤6 and κ(G0) = κ(G) + 1 = 3. Therefore α(G0) ≤ 2κ(G0) and hence by Theorem 3.1.4 G0 has a spanning2-cactusH with∆(H)≤3.
We have the following cases:
Case 2.1. SupposedegH(v) = 1, thenH−v is a spanning2-cactus ofG.
Case 2.2. Suppose degH(v) ≥ 2. If v is not a cutvertex of H then v belongs to a cycle of H.
Removing a vertex of a cycle ofH means that the cycle block gets replaced by one or two edge blocks at each remaining vertex of the cycle and hence we add at most one block containing each vertex after removingv. Theerefore,H−vis a spanning3-cactus ofG. Hence we may assume that vis a cutvertex ofH. In this caseH−vhas two components, sayH1andH2, sincevlies in exactly 2blocks by definition of a2-cactus. Since∆(H−v)≤∆(H)≤3andc(H−v)≤2 = κ(G),G has a spanning3-cactus by Lemma 3.4.3.
Case 3. Supposeκ(G) ≥ 3and κ(G) is even. Build a graphG0 by adding a complete graph on r = 12κ(G)vertices to the graphGsuch that every vertex of theKris adjacent to every vertex of G. By the construction ofG0, we haveα(G0) =α(G)andκ(G0) =κ(G) +r= 32κ(G). Therefore α(G0) ≤ 2κ(G0)and hence by Theorem 3.1.4G0 has a spanning2-cactusH with∆(H) ≤3that
