Hamiltonian chordal graphs are not cycle extendible
aa r X i v : . [ m a t h . C O ] D ec Hamiltonian chordal graphs are not cycle extendable
Manuel Lafond ∗ Ben Seamone †‡ June 9, 2018
Abstract
In 1990, Hendry conjectured that every Hamiltonian chordal graph is cycle extendable; thatis, the vertices of any non-Hamiltonian cycle are contained in a cycle of length one greater.We disprove this conjecture by constructing counterexamples on n vertices for any n ≥ P n and the bull. All graphs considered here are simple, finite, and undirected. A graph is
Hamiltonian if it has a cyclecontaining all vertices; such a cycle is a
Hamiltonian cycle . A graph G on n vertices is pancyclic if G contains a cycle of length m for every integer 3 ≤ m ≤ n . Let C and C ′ be cycles in G of length m and m + 1, respectively, such that V ( C ′ ) \ V ( C ) = { v } . We say that C ′ is an extension of C and that C is extendable (or, C extends through v to C ′ ). If every non-Hamiltonian cycle of G is extendablethen G is cycle extendable . If, in addition, every vertex of G is contained in a triangle, then G is fully cycle extendable . The study of pancyclic graphs was initiated by Bondy [3], who recognizedthat most of the sufficient conditions for Hamiltonicity known at the time in fact implied a morecomplex cycle structure. Hendry [12] introduced the concept of cycle extendability, and proved thatmany known sufficient conditions for a graph to be pancyclic in fact were sufficient for a graph tobe (fully) cycle extendable.Given a graph G and a set of vertices U ⊆ V ( G ), we denote by G [ U ] the subgraph obtained bydeleting from G all vertices except those in U ; G [ U ] is the subgraph induced by U , and a subgraphof G is an induced subgraph if it is induced by some U ⊆ V ( G ). A graph is chordal if it contains noinduced cycles of length 4 or greater. It is not hard to show that every Hamiltonian chordal graphis pancyclic (see Proposition 3.4), however the question of whether not every Hamiltonian chordalgraph is cycle extendable has remained open since 1990: Conjecture 1.1 (Hendry’s Conjecture) . [12] If G is a Hamiltonian chordal graph, then G is fullycycle extendable. ∗ Department d’Informatique et de recherche op´erationnelle, Universit´e de Montr´eal, Montreal, QC, Canada, [email protected] † Department d’Informatique et de recherche op´erationnelle, Universit´e de Montr´eal, Montreal, QC, Canada, [email protected] ‡ This work was supported by the Natural Sciences and Engineering Research Council of Canada. amiltonian chordal graphs are not cycle extendable n ≥
15 there exists a counterexample to Hendry’s Conjecture on n vertices and (b) forevery real number α > G with a non-extendable cycle C such that | V ( C ) | < α | V ( G ) | . The question then remains: for which subclasses of the class of chordal graphsis Hendry’s Conjecture true? In Section 3, we verify the conjecture for some particular chordalgraph classes based on forbidden induced subgraphs, and suggest some avenues for further researchin Section 4. We continue with some necessary definitions and properties of chordal graphs. A set of vertices X ⊆ V ( G ) which induces a complete subgraph of G is a clique . The neighbourhood of a vertex v ∈ V ( G ) is the set of vertices to which v is adjacent, which is denoted N G ( v ) (or N ( v ) if the graphin question is clear from context). A vertex v ∈ V ( G ) is called simplicial if N G ( v ) is a clique. A perfect elimination ordering of a graph G is an ordering of V ( G ), say v ≺ v ≺ v ≺ . . . ≺ v n , suchthat v i is simplicial in the graph G [ { v i , . . . , v n } ] for all i ∈ { , , . . . , n } . A vertex cut of a graph G is a set X ⊂ V ( G ) such that G − X is a disconnected graph. Let G and H be two graphs for which V ( G ) ∩ V ( H ) is a clique. We call the graph with vertex set V ( G ) ∪ V ( H ) and edge set E ( G ) ∪ E ( H )the clique sum of G and H ; this is also called a clique pasting of G and H .For a graph G , the following statements are well known to be equivalent: • G is chordal. • Every minimal vertex cut of every induced subgraph of G is a clique [8]. • G admits a perfect elimination ordering [8, 10].It easily follows that G is chordal if and only if G can obtained from two chordal graphs G and G ,with V ( G ) ( V ( G ) and V ( G ) ( V ( G ), via clique pasting.We build our counterexamples to Hendry’s Conjecture using the graph H given in Figure 1. a d eg hcb fz z Figure 1:
The base graph H Since f ≺ g ≺ z ≺ z ≺ b ≺ a ≺ c ≺ d ≺ e ≺ h is a perfect elimination ordering of H , H is achordal graph. Call the edges ab, de, ef, ch, and gh heavy ; these edges are highlighted in Figure 1. amiltonian chordal graphs are not cycle extendable H : C ∗ = abz z ghcdef aC = abchgf eda Note that C and C ∗ each contain every heavy edge of H . Furthermore, C ∗ is a Hamiltoniancycle of H and C spans every vertex of H except z and z . Lemma 2.1.
No extension of C in H contains every heavy edge.Proof. Suppose, to the contrary, such an extension exists. We may remove from consideration anyedge incident to e or h that is not heavy, as well as the edge z z . The remaining available edges forour desired extension are shown in Figure 2. Since C cannot extend through z , any extension mustcontain the edges az and gz . We may now remove from consideration every other edge incidentto a or g . This leaves no remaining edges incident to f to include in an extension of C , and henceno such extension exists. a d eg hcb fz z Figure 2:
Available edges of H for an extension of C that contains all heavy edges Theorem 2.2.
For any n ≥ , there exists a counterexample to Hendry’s Conjecture on n vertices.Proof. Let G be a graph obtained from H by pasting a clique onto each heavy edge of H so that | V ( G ) | = n ≥
15. Since G is obtained from H and a disjoint set of complete graphs by clique pasting, G is chordal. Let D ∗ and D be cycles of G obtained from C ∗ and C , respectively, by replacing eachheavy edge xy with a Hamiltonian xy -path through the clique which was pasted onto xy to obtain G . We see that D ∗ is a Hamiltonian cycle of G and that D is a cycle that spans every vertex of G except z and z . Furthermore, D cannot be extended in G , otherwise C could have been extendedin H using every heavy edge, a contradiction of Lemma 2.1.For any fixed counterexample on n vertices constructed in the proof of Theorem 2.2, consider thegraph obtained by pasting a clique of size k onto the edge z z . Such a graph is still Hamiltonianand a cycle D as given in the proof of Theorem 2.2 cannot be extended. Since we have a cycle oflength n − n + k − Theorem 2.3.
For any real number α > , there exists a Hamiltonian chordal graph G with anon-extendable cycle C satisfying | V ( C ) | < α | V ( G ) | . amiltonian chordal graphs are not cycle extendable H – any set of 5 Hamiltonianchordal graphs { G , G , G , G , G } will suffice, where the edge of G i pasted onto a heavy edge of H can be chosen to be any edge from any Hamiltonian cycle in G i . Even though Hendry’s Conjecture is not true in general, it is still interesting to consider sufficientconditions for a chordal graph to be fully cycle extendable. A graph is H -free if it contains noinduced subgraph isomorphic to H , and it is H -free for a set of graphs H if it is H -free for every H ∈ H . The remainder of this paper is concerned with graphs characterized by forbidden inducedsubgraphs.Chordal graphs are one obvious example of a graph class characterized by forbidden inducedsubgraphs; they are by definition { C , C , C , . . . } -free. A strongly chordal graph is defined to be achordal graph in which even cycle of length at least 6 has a chord that connects vertices at an odddistance from one another along the cycle. Strongly chordal graphs can also be characterized byforbidden induced subgraphs. A k -sun is a chordal graph G whose vertices can be partitioned intotwo sets X = { x , x , . . . , x k } and Y = { y , y , . . . , y k } such that x i is adjacent only to y i and y i +1 in G (subscripts taken modulo k ). A graph is a sun if it is a k -sun for some k . Farber [9] showedthat a graph is strongly chordal if and only if it is chordal and sun-free.We now summarize the classes for which Hendry’s Conjecture is known to hold. A Hamiltonianchordal graph is fully cycle extendable if it is also • planar [13], • a spider intersection graph (the intersection graph of subtrees of a subdivided star) [1], • strongly chordal and ( K , + e )-free [2], or • strongly chordal and hourglass-free [2]. K , + e hourglass Figure 3: K , + e and the hourglassThe result on spider intersection graphs generalizes previous results on interval graphs [2, 6] andsplit graphs [2].One can obtain other classes of graphs for which Hendry’s Conjecture holds by looking at resultson locally connected graphs. A graph G is locally connected if N ( v ) induces a connected subgraphof G for every v ∈ V ( G ). Proposition 3.1.
A connected chordal graph is -connected if and only if it is locally connected.Proof. Chartrand and Pippert [5] proved that every connected and locally connected chordal graphis 2-connected, so we need only consider the “only if” portion of the statement. Suppose G is a2-connected chordal graph and let x, y ∈ N ( v ) for some v . Since G is 2-connected, there exists a amiltonian chordal graphs are not cycle extendable xv that contains y ; equivalently, there is an xy -path that avoids v . Suppose that everysuch path has vertices not in N ( v ). Let P be a shortest such path, and let Q be a segment of P with ends a, b lying in N ( v ) and all internal vertices not in N ( v ). By minimality of P , we have that Q + avb is an induced cycle of length at least 4, a contradiction.A connected and locally connected graph G is known to be cycle extendable if it also satisfiesone of the following conditions: • ∆( G ) = 5 and δ ( G ) ≥ • G is almost claw-free [14], and hence claw-free (originally shown in [7, 12]), • G is { K , , K , + e } -free with δ ( G ) ≥ • P -free; • { bull , K , } -free; • { bull , K ∨ P } -free;Recall that the join of two graphs G and H , denoted G ∨ H , is the graph with vertex set V ( G ) ∪ V ( H ) and edge set E ( G ) ∪ E ( H ) ∪ { gh : g ∈ V ( G ) , h ∈ V ( H ) } . P bull K , K ∨ P Figure 4:
Forbidden induced subgraphs considered in Sections 3.1 and 3.2 P -free chordal graphs We begin with some technical results on cycles in chordal graphs, particularly as they relate to vertexcuts or cutsets. For two vertices u, v ∈ V ( G ), a uv -separator is a set of vertices X ⊂ V ( G ) such that u and v lie in different connected components of G − X . A minimal uv -separator is a uv -separatorwhich has no proper subset which is also a uv -separator. A separator is a uv -separator for some u, v ∈ V ( G ), and a minimal separator is a minimal uv -separator for some u, v ∈ V ( G ). A vertex v is complete to a set X ⊆ V ( G ) \ { v } if v is adjacent to every vertex in X , and v is anticomplete to X if it is non-adjacent to every vertex in X . For two disjoint subsets of V ( G ), say X and Y , we say X is complete (anticomplete) to Y if every vertex in X is complete (anticomplete) to Y .As stated earlier, it was shown by Dirac [8] that if G is a chordal graph and X ⊆ V ( G ) is aminimal separator (i.e. a minimal vertex cut), then X is a clique. Furthermore, since every chordalgraph has a perfect elimination ordering, it follows that every chordal graph has a simplicial vertex,and if G is not a clique then G contains at least two nonadjacent simplicial vertices. The following,more general statement, easily follows: amiltonian chordal graphs are not cycle extendable Proposition 3.2. If G is a chordal graph and X is a minimal separator, then every connectedcomponent of G − X contains a simplicial vertex of G . As a corollary, we obtain a more general result on (not necessarily minimal) clique separators:
Proposition 3.3. If X is a clique separator of a chordal graph G , then each connected componentof G − X contains a simplicial vertex of G .Proof. Suppose that Q is a connected component of G − X . Let H be the graph obtained from G [ X ∪ V ( Q )] by adding a vertex v and all edges from v to X . For any u ∈ V ( Q ), X is a uv -separatorin H , and hence X contains a minimal uv -separator, Y . Since v is complete to X , it is complete to X \ Y , and hence H − Y has Q as a connected component (and H [ { v } ∪ ( X \ Y )] as the other). ByProposition 3.2, Q contains a simplicial vertex in H , say x . However, H [ X ∪ V ( Q )] = G [ X ∪ V ( Q )]and all neighbours of x in G lie in V ( Q ) or X , and thus x is a simplicial vertex in G as well.The following simple proposition implies that every Hamiltonian chordal graph is pancyclic andthus “cycle reducible”; it is this fact that originally inspired Hendry’s Conjecture. Proposition 3.4.
Let G be a Hamiltonian chordal graph. If v is a simplicial vertex of G , then G − v is Hamiltonian. We may now show that Hendry’s Conjecture holds for P -free graphs. Theorem 3.5.
Every P -free Hamiltonian chordal graph is fully cycle extendable.Proof. Let G be a P -free Hamiltonian chordal graph, and let C be a non-Hamiltonian cycle in G .We prove the statement by induction on | V ( G ) | . The statement can be easily checked for sufficientlysmall graphs, say for | V ( G ) | ≤
5. Assume that | V ( G ) | ≥ P -free Hamiltonianchordal graph G ′ with | V ( G ′ ) | < | V ( G ) | , any non-Hamiltonian cycle C ′ extends in G ′ . Let Q be aconnected component of G − V ( C ) and let X ⊆ V ( C ) be those vertices of C with a neighbour in Q (note that X necessarily contains at least 2 vertices).If X is a clique, then Q contains a simplicial vertex by Proposition 3.3, say v . By Proposition3.4, G − v is Hamiltonian. Clearly C is a cycle in G − v . If C is Hamiltonian in G − v , then aHamiltonian cycle in G is an extension of C in G . If C is not Hamiltonian in G − v , then C extendsin G − v by induction and hence also in G .Suppose, then, that there exist x, y ∈ X which are nonadjacent; let a and b denote the neighboursof x on C and let c and d denote the neighbours of y on C . We will show that C contains an edgewhose ends share a neighbour outside of C , and hence C easily extends.First, suppose that each of x and y have a cycle neighbour which does not lie in X , say a and c (we do not assume that these vertices are distinct). Since xy / ∈ E ( G ), a shortest xy -path in G [ V ( Q ) ∪ { x, y } ] has length at least 2. Let P be such a path, and consider the subgraph of G induced by V ( P ) ∪ { a, c } . If a = c , then ax + P + ya is a cycle of length at least 4 in G . However,since a has no neighbours in Q and P is minimal, this cycle is chordless, a contradiction. Thus, weassume that a = c . The only possible edges induced by the vertices of the path ax + P + yc (otherthan those in the path themselves) are ay, ac, and xc . Since G is P free, at least one such edge mustbe present, however any combination creates a chordless cycle of length at least 4, a contradiction.We now may assume that one of x or y has both of its cycle neighbors in X . Without loss ofgenerality, we may assume that a ∈ X . Let Y a = V ( Q ) ∩ N G ( a ) and Y x = V ( Q ) ∩ N G ( x ). Supposethat Y a ∩ Y x = ∅ . Out of all paths connecting some vertex in Y a to some vertex in Y x , let P beone of minimum length; say that P joins s ∈ Y a and t ∈ Y x . By construction, no internal vertex of amiltonian chordal graphs are not cycle extendable P is adjacent to either a or x in G , and hence P + txas is an induced cycle of length at least 4, acontradiction. It follows that Y a ∩ Y x is nonempty. If t ∈ Y a ∩ Y x , then C ′ = C − ax + atx is ourdesired extension of C in G . Bull-free chordal graphs are of interest to us for two reasons. The first is that bull-free graphs arehistorically tied to the study of perfect graphs, of which chordal graphs form a well-known subclass.The second is that there is some evidence that Hendry’s Conjecture may hold for strongly chordalgraphs (see [1, 2, 6]), and bull-free chordal graphs are strongly chordal. To see this, recall that agraph is strongly chordal if and only if it is chordal and sun-free. Since every sun contains a bull,any bull-free chordal graph is sun-free, and hence strongly chordal.While we cannot yet show that bull-free Hamiltonian chordal graphs are fully cycle extendable,we can show that { bull , X } -free Hamiltonian chordal graphs are fully cycle extendable for tworeasonably large subgraphs X . To do this, we need the following simple observation: Lemma 3.6. If C is a cycle in a chordal graph and uv ∈ E ( C ) , then u and v share a commonneighbour on C .Proof. There is a shortest cycle through uv in G [ V ( C )]. Since G [ V ( C )] is chordal, this cycle musthave length 3; the vertex in this 3-cycle distinct from u and v is the desired common neighbour. Theorem 3.7. If G is a { bull , K , } -free or { bull , K ∨ P } -free Hamiltonian chordal graph, then G is fully cycle extendable.Proof. Let X be either K , or K ∨ P . Let G be a minimal { bull , X } -free Hamiltonian chordalgraph which is not cycle extendable, and let C be a non-Hamiltonian cycle which is not extendable.By only making use of the fact that G is Hamiltonian, bull-free, and chordal, we will show that G necessarily contains both K , and K ∨ P , a contradiction.Consider the vertices of C in some cyclic order. For a vertex a ∈ V ( C ), we denote by a − and a + the vertices immediately preceding and succeeding a along C , respectively. For two vertices a, b ∈ V ( C ), let C [ a, b ] denote the segment of C from a to b with respect to the cyclic ordering (thatis, containing a + and b − ).Let C ∗ be a Hamiltonian cycle of G . There must be a segment of C ∗ with at least 3 verticeswhose ends lie on C and whose internal vertices are disjoint from V ( C ). Choose Z = uz · · · z k v tobe a shortest such segment; for notation purposes let z = u and z k +1 = v , and let ˆ Z denote theinternal vertices of Z . We now argue the presence or absence of edges in the induced subgraph of G having vertex set V ( C ) ∪ V ( Z ). Figure 5 displays the edges which we argue are present.We first argue that uv / ∈ E ( G ). Suppose, to the contrary, that uv ∈ E ( G ). This implies that uv + Z is a cycle in G and so, by Lemma 3.6, u and v share a common neighbour on Z , say z . Ifˆ Z = { z } , then G − z is Hamiltonian (we replace the segment uzv in C ∗ with uv ) and we have that C isextendable in G − z by the minimality of G . If | ˆ Z | ≥
2, then consider the graph G ′ = G − (cid:16) ˆ Z \ { z } (cid:17) .This graph is still Hamiltonian (we replace Z in C ∗ with uzv ), so C extends in G ′ by the minimalityof G , and hence also extends in G . Thus, uv / ∈ E ( G ).Now, we note that u − and u + must be non-adjacent to z and that v − and v + must be non-adjacent to z k , otherwise C is extendable. We also may deduce that u is not adjacent to any of z , . . . , z k , v is not adjacent to any of z , . . . , z k − , and no edge connects two vertices of ˆ Z except amiltonian chordal graphs are not cycle extendable Z itself. If any of these were not the case, then there are vertices of Z that may bedeleted that maintain Hamiltonicity, and C then extends by the minimality argument.Among all vertices on C [ u, v ], let x be the neighbour of u that is closest to v . Similarly, let y denote the vertex on C [ v, u ] that is the neighbour of u closest to v . By Lemma 3.6, u and x mustshare a common neighbour on C [ x, y ] ∪ yux . Since u is adjacent only to x and y on this cycle, thiscommon neighbour must by y and so xy ∈ E ( G ).We now show that z i x, z i y ∈ E ( G ) and z i x − , z i x + , z i y − , z i y + / ∈ E ( G ) for every i = 1 , . . . , k .Consider the cycle Z ∪ ux ∪ C [ x, v ]. Note that the only neighbours of u on this cycle are x and z . By Lemma 3.6, u and x must share a common neighbour on this cycle, and hence z x ∈ E ( G ).The non-extendability of C implies that x − z , x + z / ∈ E ( G ). Note that we may now deduce that x is distinct from both u + and v − . Now consider the subgraph induced by { u, x, x + , z , z } . Thetriangle uxz together with the edges z z and xx + form a bull. We have already deduced that ux + , uz , x + z / ∈ E ( G ); since G is chordal and bull-free, we must have that xz ∈ E ( G ). It followsthat x − z , x + z / ∈ E ( G ). We then iterate this argument with the triangle z i z i +1 x and the edges xx + and z i +1 z i +2 to argue that x is complete to ˆ Z and that x − and x + are anticomplete to ˆ Z .Now, if x + = v − , then xv ∈ E ( G ), since { x, v − , v, z k } cannot induce a 4-cycle and v − z k / ∈ E ( G ).If x + = v − , then since { x, x + , z k − , z k , v } cannot induce a bull, we must also have xv ∈ E ( G ). Anidentical argument gives that y is complete to ˆ Z ∪ { v } , that y is distinct from u − and v + , and that y − and y + are anticomplete to ˆ Z . Since G is chordal and u − z , u + z , v − z k , v + z k / ∈ E ( G ), it is alsoeasy to deduce that { u − , u + , v − , v + } is anticomplete to ˆ Z .We now show that { u − , u, u + } is anticomplete to { v − , v, v + } , that the vertices in { u + , x + , v + , y + } are pairwise non-adjacent, and that the vertices in { u − , x − , v − , y − } , are pairwise non-adjacent.If uv − ∈ E ( G ), then Z ∪ vv − u would be a chordless cycle of length at least 4, a contradic-tion. An identical argument holds for uv + , vu − , and vu + . Now, if any of u + x + , x + v + , v + y + or y + u + are edges of G , then C easily extends (for instance, u + x + ∈ E ( G ) gives the extension C − uu + − xx + + u + x + + uz x ). Hence, u + x + , x + v + , v + y + , y + u + / ∈ E ( G ). If x + y + ∈ E ( G ), then C extends to C − xx + − yy + + x + y + + xz y , a contradiction. Also, if any edge of G has one endin { u − , u + } and the other in { v − , v + } , then G contains an induced cycle of length at least 5, acontradiction. Thus, { u − , u, u + } is anticomplete to { v − , v, v + } and the vertices in { u + , x + , v + , y + } are pairwise non-adjacent. An identical argument gives that the vertices of { u − , x − , v − , y − } arepairwise non-adjacent.We claim that x is adjacent to u − , u + , v + and that y is adjacent to u − , u + , v − . We begin with x and u + . If u + = x − , then { u, u + , x, z } is a 4-cycle and u + x is the only possible chord. Otherwise,the triangle uxz and the edges uu + and xx + form a bull, and ux + , z x + , u + z , u + x + are non-edges.Hence, we must have u + x ∈ E ( G ). An identical argument for the set of vertices { u, u − , x, x + , z } shows that xu − ∈ E ( G ). To show that xv + ∈ E ( G ) requires a little more work. We first note thatwe have a bull consisting of the triangle xz k v and the edges xx + and vv + . We have shown that x + , v + , and z k are mutually non-adjacent, and so one of x + v or xv + must be an edge. If xv + ∈ E ( G ),then we are done. Suppose that x + v ∈ E ( G ). We now consider the bull with triangle xx + v andpendant edges ux and vv + . The only possible edge which has not yet been ruled out is xv + , and soit must be an edge in this case as well. Symmetric arguments gives that y is adjacent to u − , u + , v − .Consider the bull with triangle z xy and pendant edges xv + and yv − . The edges v − z and v + z are forbidden, and so one of the remaining three edges ( xv − , yv + , v − v + ) must be present. Regardlessof the presence of v − v + , one of xv − , yv + must be an edge since G is chordal. We may say withoutloss of generality (by symmetry) that xv − ∈ E ( G ). Finally, we consider the bull with triangle z xy and pendant edges xx − and yy − . Since x − , y − , and z are pairwise non-adjacent, we must have amiltonian chordal graphs are not cycle extendable u u + u − vxyx − x + v − v + y − y + z z z k Figure 5:
Edges of a subgraph in G as described in the proof of Theorem 3.7that one of xy − or yx − is an edge of G .We now see that the following subgraphs are induced in G : • K , : { x, u − , x − , v − , y − , z } if xy − ∈ E ( G ) and { y, u − , x − , v − , y − , z } if yx − ∈ E ( G ); • K ∨ P : { x, y, u − , u, z , v, v − } if Z has one internal vertex and { x, y, u − , u, z , z , z } otherwise(recall that v = z k +1 ).Hence we have arrived at our desired contradiction. As we have mentioned, many of the classes for which Hendry’s Conjecture is known to hold arestrongly chordal—interval graphs (shown to be strongly chordal in [4]), strongly chordal graphswhich are also either ( K , + e )-free or hourglass-free (stated on page 4), and { bull , K , } -free and { bull , K ∨ P } -free graphs (Corollary 3.7). Furthermore, no counterexample to Hendry’s Conjecturewhich was constructed in Section 2 is strongly chordal. To see this, let H + be the graph obtainedfrom H (Figure 1) by joining a vertex x to the vertices g and h (see Figure 6). The vertices { a, b, f, g, h, x } induce a 3-sun in H + . Since H + is an induced subgraph of every one of the graphsconstructed in Theorems 2.2 and 2.3, each such counterexample contains a sun and is thus not strongly chordal. As such, we pose the following question: Question 4.1.
Is every Hamiltonian strongly chordal graph fully cycle extendable?
In Theorem 3.5, we showed that being P -free is a sufficient condition for a Hamiltonian chordalgraph to be fully cycle extendable. However, if we take a closer look at the construction methodgiven in Section 2, we see that there exists a counterexample to Hendry’s Conjecture that is P -free. amiltonian chordal graphs are not cycle extendable a x d eg hcb fz z Figure 6:
A 3-sun in H + Consider again the base graph H given in Figure 1. Note that a and e are universal vertices, so anyinduced path of H of length 4 or greater contains neither vertex; Figure 7 shows H with the edgesincident to a and e removed. a d eg hcb fz z Figure 7: H without the edges incident to a or e We can now see that H − { a, e, h } is a path, and hence an induced path of H on 7 vertices. Callthis path P . Let H ∗ be the graph obtained from H by pasting a triangle onto each heavy edge of H .Clearly H ∗ contains P as an induced path, and each end of P will be incident to a degree 2 vertexin H ∗ . We can thus extend P to an induced path on 9 vertices, and this is the longest such pathin H ∗ . Hence, not every P -free Hamiltonian chordal graph is fully cycle extendable. We thus askthe following: Question 4.2.
What is the largest integer r for which every P r -free Hamiltonian chordal graph isfully cycle extendable? Based on the discussion above and Theorem 3.5, the answer to Question 4.2 is at least 5 and atmost 9.It is also worth noting that every counterexample given in Section 2 has a vertex cut of size 2;this prompts the following question:
Question 4.3.
Does there exist a value k > such that every k -connected Hamiltonian chordalgraph is fully cycle extendable? amiltonian chordal graphs are not cycle extendable τ ( G ), is the minimum value of | X | c ( G − X ) taken over all vertex cuts X , where c ( G − X ) denotes the number of connected components of G − X . A graph is t -tough if τ ( G ) ≥ t . Note that every t -tough graph is 2 t -connected and that every Hamiltonian graph is1-tough. As such, we immediately see that every counterexample given in Section 2 is 1-tough,which prompts the following, more restrictive, question: Question 4.4.
Does there exist a value t > such that every t -tough Hamiltonian chordal graph isfully cycle extendable? The authors thank Geˇna Hahn for the stimulating discussions. We are also grateful for the carefulreading of this paper by the referees and for their helpful comments.
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