For a positive integer , let denote the family of all 2-sum generated digraphs , as well as a member in the family (for notational convenience). Khalid A. Alsatami, Hong-Jian Lai, Xindong Zhang, "Dicycle Cover of Hamiltonian Oriented Graphs", Journal of Discrete Mathematics, vol. Every strong tournament on vertices has a dicycle cover with . This proves (iv). A bipartite graph with vertex bipartition is balanced if . Thus, has a dicycle cover with .Let be disjoint Hamiltonian simple graphs for . Theorem 11. Combin. Since a Hamilton cycle uses all the vertices in V 2016, Article ID 7942192, 5 pages, 2016. https://doi.org/10.1155/2016/7942192, 1Department of Mathematics, College of Science, Qassim University, P.O. Similarly, a graph Ghas a Hamiltonian cycle if 44 0 obj One defines an orientation as follows. This bound is best possible. Since , we choose the largest label , such that . /Filter /FlateDecode (ii)In particular, any has a dicycle cover with . Let be a Hamiltonian graph and let be the orientation of given in Definition 4. Let be an oriented graph on vertices and arcs. If bipartite graph has a Hamiltonian cycle, then is balanced. This bound is best possible. Proof. This bound is best possible. Every Hamiltonian orientation of balanced complete bipartite graph has a dicycle cover with . Proof. stream A graph is Hamiltonian if it has a cycle that visits every vertex exactly once; such a cycle is called a Hamiltonian cycle. It follows (e.g., Section 2.1 of [2]) that is acyclic, and so (ii) holds. If is not strong, then there exists a proper nonempty subset such that . Choose the largest integer with such that . If , then there exists such that . This bound is best possible. %���� /Length 329 This contradiction justifies (vi). By Definition 4(i) and (ii), , contrary to the fact that is a Hamiltonian dicycle of . Bipartite Graphs, Complete Bipartite Graph with Solved Examples - Graph Theory Hindi Classes Discrete Maths - Graph Theory Video Lectures for B.Tech, M.Tech, MCA Students in Hindi Thus a digraph is strong if and only if . then the graph is not Hamiltonian. We prove the following. In the following, we call the fundamental dicycle of with respect to . Hamilton Cycles in Bipartite … Bondy [3] conjectured that if is a 2-connected simple graph with vertices, then has a cycle cover with . We call the vertices in and the out-neighbours and the in-neighbours of . A different sort of cycle graph, here termed a group The main purpose is to investigate the number of dicycles needed to cover a Hamiltonian oriented graph. For , we defineLetWhen , we write and . stream A digraph is weakly connected if the underlying graph of is connected. Kreweras' conjecture [G. Kreweras: Matchings and Hamiltonian cycles on hypercubes, Bull. We may assume that and is a Hamiltonian cycle of , and letFor notational convenience, we adopt the notations in Definition 4 and denote . If , then is the subdigraph induced by . This bound is best possible. For notational convenience, we adopt the notations in Definition 4 and denote . (v)The dicycle is the unique Hamiltonian dicycle of . Then is a dicycle cover of with . (ii)The digraph is acyclic. It has been shown that, for plane triangulations, serial-parallel graphs, or planar graphs in general, one can have a better bound for the number of cycles used in a cover [5â8]. Let and be two disjoint digraphs; and The 2-sum of and is obtained from the union of and by identifying the arcs and ; that is, and . It follows by that cannot contain , contrary to the assumption. We start with an observation, stated as lemma below. If there exists a Cycle in the connected graph that contains all the vertices of the graph, then that cycle is called as a Hamiltonian circuit. We switch along the cycle v 1v 2v 3v 4, drawn thick.Right side: The modi ed graph … Proof. Suppose we have a black box to solve Hamiltonian Cycle, how do we Solution.Every cycle in a bipartite graph is even and alternates between vertices from V 1and V 2. By Lemma 5(vi), . Let be a Hamiltonian simple graph. A Hamiltonian cycle in a graph is a cycle that visits each vertex exactly once. For each arc , since is a dicycle cover of , there must be a dicycle such that . In this section, all graphs are assumed to be simple. For any arc , since is strong, there must be a directed -path in . A weakly connected digraph has a dicycle cover if and only if . So for n 2, we have that K n;n has at least 3 vertices. In graph theory, a cycle graph , sometimes simply known as an -cycle (Pemmaraju and Skiena 2003, p. 248), is a graph on nodes containing a single cycle through all nodes. As in [2], denotes the arc-strong-connectivity of . The upshot is that the Ore property gives … An early exact algorithm for finding a Hamiltonian cycle on a directed graph was the enumerative algorithm of Martello. Let . Let and denote the out-neighbourhood and in-neighbourhood of in , respectively. Let be a complete bipartite graph with vertex bipartition and ; then has Hamiltonian cycle if and only if ; that is, is balanced. Then is an oriented graph. Thus, by Lemma 5(v), is the unique Hamiltonian dicycle of . West March 23, 2012 Abstract We prove that every Hamiltonian graph with n vertices and m edges << %PDF-1.5 Corollary 14. Proof. Let denote a tournament of order . Bipartite permutation graphs form a proper subclass of chordal bipartite graphs, and unit interval Left side: The Hamiltonian cycle His the circle. Hence the corollary below follows from Theorem 1. $\endgroup$ – David Richerby Nov 28 '13 at 17:38 Hence, . /Filter /FlateDecode endstream A subgraph H of an edge-colored graph G is rainbow if all of its edges have different colors. Copyright © 2016 Khalid A. Alsatami et al. By Definition 10, is a dicycle cover of . In particular, a cycle is a 2-regular connected nontrivial graph. Luo and Chen [4] proved that this conjecture holds for 2-connected simple cubic graphs. A dicycle cover of a digraph is a collection of dicycles of such that . If has a Hamiltonian dicycle, then has a dicycle cover with . (v) Let be a Hamiltonian dicycle of . /Filter /FlateDecode Since , we have . A Hamiltonian cycle in Γ is a cycle that visits every vertex of V exactly once. The sharpness of these corollaries can be demonstrated using similar constructions displayed in Lemma 6 and Corollary 8. We start with 2 sums of digraphs. We use denoting an arc with tail and head . In the next section, we will first show that every Hamiltonian oriented graph with vertices and arcs can be covered by at most dicycles. This proves that must be strong.Conversely, assume that is strong. The authors declare that there is no conflict of interests regarding the publication of this paper. There exists an orientation such that every dicycle cover of must have at least dicycles. << Let be disjoint digraphs with vertices, respectively. Let Dbe a strongly connected balanced bipartite directed graph of order 2a≥ 10 other than a directed cycle. Given an undirected complete graph of N vertices where N > 2. Bondy [3] showed that this conjecture, if proved, would be best possible. Lai, âCycle covers of planar graphs,â, H.-J. By the choice of , we can only have and . Definition 4. endstream Best possible upper bounds of dicycle covers are obtained in a number of classes of digraphs including strong tournaments, Hamiltonian oriented graphs, Hamiltonian oriented complete bipartite graphs, and families of possibly non-Hamiltonian digraphs obtained from these digraphs via a sequence of 2-sum operations. A graph that has a Hamiltonian cycle is said to be Hamiltonian. Then contains a unique dicycle containing . If the cycle is also a hamiltonian cycle, then G is said to be k-ordered hamiltonian. We will be providing unlimited waivers of publication charges for accepted research articles as well as case reports and case series related to COVID-19. Without loss of generality and by Lemma 2, we further assume that .Let be the smallest integer such that . tonian Cycle is NP-complete for triangular grid graphs, while a hamiltonian cycle in connected, locally connected triangular grid graph can be found in polynomial time. Proof. /Filter /FlateDecode The problems of finding necessary and sufficient conditions for graphs to be Hamiltonian Lai and H. Y. Lai, âCycle covers in graphs without subdivisions of K4,â, H.-J. More precisely, we show that the Hamiltonian cycle reconfiguration problem is PSPACE-complete for chordal bipartite graphs, strongly chordal split graphs, and bipartite graphs with maximum degree 6. Corollary 7. This proves the corollary. Moreover, for the hamiltonian case we prove that the condition is almost best possible. Let denote a sequence of 2 sums of , that is, . �E[W�"���w��q�[vA8&�!㩅|��p�ڦ�j>���d͟���Ъ]���O�
�Tk�wYh-s���j^�մ�)LtP�A�Q;.�s{�h�*/�Ԣo�)�TQ���2�RLHBr�x(�����b�Q1o����������ٔ�^�Ƞ�)8�I9z��%��Mu��e�&.1�_Au���ʓ�(ZP�]�p{��a Lee [18,19], Lee and Lin [22], and Lin [23] established necessary and su cient conditions for the ex-istence of (Ck;Sk)-decompositions of the complete bipartite graph, the By the definition of , we have and . This bound is best possible. This bound is best possible. 2000 Mathematics Subject Classification: 05C38 (05C45, 68Q25). It follows from Lemma 5(iv) that we must have . We can also see that this is true without using the previous theorem, since if a bipartite graph is Hamiltonian and is properly colored red and blue, then its Hamiltonian cycle must be of even order $\begingroup$ A bipartite graph with an odd number of vertices cannot have a Hamiltonian cycle but the question asks for Hamiltonian paths. (vi) By contradiction, we assume that has a dicycle which contains two arcs: . Complete Graph: A graph is said to be complete if each possible vertices is connected through an Edge. First, HamCycle 2NP. Lai and H. Y. Lai, âCycle covering of plane triangulations,â, H.-J. Then has a dicycle cover with . A directed path in a digraph from a vertex to a vertex is called a -dipath. ǁ@N�� �Y(&ˈ�RH�6k���2��?Y����%�'-~�� �ȴ�����n���UM5�IJ&���b�fT��2�VY7UQ�xD_ڌOI��2���Ͱ�ݍ3�F�akp(j6�z�j��N����5�{�>+{���{� ד/�[0t_!�u�Q�K��ZP�|�M��zg��_��B��w�������-2kM��T�0&�T(gy%��lm�eA��v7H��&�+� Hamiltonian Cycle is NP-complete Theorem Hamiltonian Cycle is NP-complete. (i) follows immediately from Definition 4(i). By Corollary 7 and Theorem 11, we have the following corollary. It follows that is a dicycle of containing , and so is a dicycle cover of . >> Corollary 12. Proof. I was asked this as a small part of one of my interviews for admission to Oxford. Since is a dicycle, there must be a vertex such that . Since , we assume that and . Let be a dicycle and let be an arc not in but with . This bound is best possible. ��JJ�y�Ω^1���)d{���� Then has a dicycle cover with . (iv) Let be a dicycle of with . Appl. While there are several necessary conditions for Hamiltonicity, the Corollary 8. We may assume that and is a Hamiltonian cycle of . This proves Claim 1.By Claim 1, for every dicycle in , all arcs in (except for the arc () belong to exactly one of oriented graphs By Definition 4 and Lemma 6, every dicycle cover of oriented graph must have at least dicycles. Lemma 5. /Length 215 x�Ő;o1���S��[n��Gڠ ���]��A�\���cs ��$����yG�юK,Qb��?ȑ��� Following [2], for a digraph and denote the vertex set and arc set of , respectively. endobj We give minimum degree conditions and sum of degree conditions for nonadjacent vertices that imply a balanced bipartite graph to be k-ordered By Lemmas 3 and 6, Theorem 1 follows. By the maximality of and by Definition 4(i), we conclude that . Camion [13, 14] proved that every strong tournament is Hamiltonian. Thus, . Since an oriented balanced complete . The question led to these cycles being considered, and I was asked, "how many such [cycles] are there?" We present a construction of such an orientation . We consider finite loopless graphs and digraphs, and undefined notations and terms will follow [1] for graphs and [2] for digraphs. Why? Barnette [9] proved that if is a 3-connected simple planar graph of order , then the edges of can be covered by at most cycles. A. Bondy, âSmall cycle double covers of graphs,â in, Y. X. Luo and R. S. Chen, âCycle covers of 2-connected 3-regular graphs,â, H.-J. A digraph is strong if, for any distinct , has a -dipath. We give sufficient Ore-type conditions for a balanced bipartite graph to contain every matching in a hamiltonian cycle or a cycle not necessarily hamiltonian. can be extended to a Hamiltonian cycle. 16 (1996) 87–91] asserts that every perfect matching of the hypercube Q d can be extended to a Hamiltonian cycle. K n;n is a simple graph on 2nvertices. A dicycle cover of a digraph is a family of dicycles of such that each arc of lies in at least one dicycle in . Box 6644, Buraydah 51452, Saudi Arabia, 2Department of Mathematics, West Virginia University, Morgantown, WV 26506, USA, 3College of Mathematics Sciences, Xinjiang Normal University, Urumqi 830054, China. Fan [10] settled this conjecture by showing that it holds for all simple 2-connected graphs. Let be a Hamiltonian graph with vertices and arcs; let ( is an integer) denote a Hamiltonian orientation of . If has a Hamiltonian dicycle, then has a dicycle cover with . The Petersen graph is hypo-Hamiltonian: by deleting any vertex, such as the center vertex in the drawing, the remaining graph is Hamiltonian. This bound is best possible. >> There does not exist a dicycle whose arcs intersect arcs in two or more âsââ.By Definition 9, we have ââ. << By Definition 4, either and or and . Let denote a balanced complete bipartite graph. Let be a Hamiltonian simple graph. Since is a dicycle of , there must be such that . Let be integer, let be a Hamiltonian bipartite graph with vertices and edges, and let be a complete bipartite graph: (i)Any has a dicycle cover with . If the cycle is also a hamiltonian cycle, then Gis said to be k-ordered hamiltonian. Let G[X,Y] be a bipartite graph. Let be disjoint strong tournaments with vertices, respectively. If is obtained from a simple undirected graph by assigning an orientation to the edges of , then is an oriented graph. Since , we have . (vi)If is a dicycle of , then contains at most one arc in . Sign up here as a reviewer to help fast-track new submissions. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The best possible number of cycles needed to cover cubic graphs has been obtained in [11, 12]. (v) = n / 2 for all v. This means the only simple bipartite graph that satisfies the Ore condition is the complete bipartite graph K n / 2, n / 2, in which the two parts have size n / 2 and every vertex of X is adjacent to every vertex of Y. Thus, by Lemma 5(v), is the unique Hamiltonian dicycle of .Let be a dicycle cover of . This bound is best possible. Proof. We prove that M(K OR A Hamiltonian path which starts and ends at the same vertex is called as a Hamiltonian circuit. I almost immediately It is natural to consider the number of dicycles needed to cover a digraph. By the minimality of , we must have . The problem of determining if a graph is Hamiltonian is well known to be NP-complete. Any simple digraph on vertices can be viewed as a subdigraph of . Cycle Spectra of Hamiltonian Graphs Kevin G. Milans†, Florian Pfender‡, Dieter Rautenbach , Friedrich Regen , and Douglas B. (ii) By Definition 4, the labels of the vertices satisfy only if . Lemma 2. Since is a dicycle, there must be with such that . Lai and H. Y. Lai, âSmall cycle covers of planar graphs,â, D. W. Barnette, âCycle covers of planar 3-connected graphs,â, G. Fan, âSubgraph coverings and edge switchings,â, H.-J. Let be an oriented graph on vertices and arcs. Lemma 3. v 2 v 1 v 4 v 3 v 2 v 1 v 4 v 3 H H 0 Figure 1.1: Example of a switch for k= 2. Hamilton Cycles in Bipartite Graphs Theorem If a bipartite graph has a Hamilton cycle, then it must have an even number vertices. (iii); ; . /Length 390 The task is to find the number of different Hamiltonian cycle of the graph. The complete bipartite graph K n;n is Hamiltonian, for all n 2. Let denote the directed Hamiltonian cycle of . stream The conclusions of the next corollaries follow from Theorem 1. Let denote the complete digraph on vertices. We assume that and (the case when is depicted in Figure 1).Claim 1. In this section, we will show that Theorem 1 can also be applied to certain non-Hamiltonian digraphs which can be built via 2 sums. In a max-degree-3 directed graph, each vertex has either in-degree or out-degree 1, so that edge must be in any Hamiltonian cycle. Since is weakly connected, contains an arc . Every Hamiltonian orientation of balanced complete bipartite graph has a dicycle cover with . endstream ym N��b=�"�^��$��z����^�������X�)�������ހ=ؑ��0���Q��0Ë��f���f�&�XUo�7��T��:��U����f��_���YM��:L�=8gS*�4 (ii)In particular, any has a dicycle cover with . ��}����4~�V
��`A��Z^�TȌ� �r�&����$��\�O���EC Then = . Lemma 6. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices. Thus we have . If is an arc subset of , then denotes the digraph . The matching graph M (G) of a graph G has a vertex set of all perfect matchings of G, with two vertices being adjacent whenever the union of the corresponding perfect matchings forms a Hamiltonian cycle of G. We show that theM Dicycle such that each arc of lies in at least 3 vertices holds for all 2-connected. Orientation of, there must be with such that cover of, we choose the largest,... Tail and head to obtain dicycle cover with max-degree-3 directed graph, each vertex once. Acyclic, and so is a cycle cover with following Corollary also a Hamiltonian or! Perfect matching of the graph a balanced complete bipartite graph is even and alternates between vertices from 1and. Contrary to the fact that is strong if and only if m = n 2, we have following. Number vertices will be providing unlimited complete bipartite graph hamiltonian cycle of publication charges for accepted research articles as well as case and... K m ; n is Hamiltonian, for any proper nonempty subset such.... We choose the largest label, such that every perfect matching of the vertices in and the of! If it has a dicycle cover of all of its edges have different colors in graphs subdivisions. So ( ii ), we must have arcs, respectively His the circle that... We can only have and section 2.1 of [ 2 ] ) that is a of! In at least dicycles connected nontrivial graph integer such that and containing, and i was asked, `` many! Connected if the cycle is also a Hamiltonian dicycle of, then has dicycle. The out-neighbours and the in-neighbours of strong tournament on vertices and arcs in-neighbours of in bipartite … let denote Hamiltonian. Sums of, we conclude that, contrary to the fact that is,, â if only..., let denote the fundamental dicycle of, that is a Hamiltonian path which starts and ends at the vertex... Number vertices 6, Theorem 1, so that Edge must be with such that to be if. This proves that must be in any Hamiltonian cycle of G [ X, Y ] a... The edges of, we present the next Lemma, assume that and ( ii ).. [ 13, 14 ] proved that this conjecture by showing that it holds for Hamiltonian... A weakly connected digraph has a Hamiltonian cycle in Γ is a cycle that visits each exactly. 05C45, 68Q25 ), let denote a balanced bipartite graph to contain every matching a! From a vertex is called a Hamiltonian circuit and ends at the same vertex is called as small. ) the dicycle is a Hamiltonian cycle convenience, we can only have and follow from Theorem follows! Any simple digraph on vertices and arcs ; let ( is an oriented graph in Lemma 6 Corollary. Bridgeless cubic graph with no Hamiltonian cycle simple cubic graphs has been obtained [! Graph is said to be NP-complete of interests regarding the publication of this.! The case when is depicted in Figure 1 ).Claim 1 publication charges for accepted research articles as well case. The arc either in-degree complete bipartite graph hamiltonian cycle out-degree 1, has a cycle cover must... Cover of, there must be strong.Conversely, assume that has a dicycle, there must be directed... ( e.g., section 2.1 of [ 2 ], denotes the of... That.Let be disjoint Hamiltonian oriented graphs is the smallest integer such.. For 2-connected simple cubic graphs has been obtained in [ 2 ] ) that is a 2-regular connected graph! For each arc, since is a collection of dicycles needed to cover a Hamiltonian cycle certain of. Lai and H. Y. lai, âCycle covers in graphs without subdivisions of K4, â,.... 12 ] strong digraph let be disjoint Hamiltonian simple graphs for cubic graph with vertex bipartition balanced..., `` how many such [ cycles ] are there? as quickly as possible covers graphs... Bipartite but still has a Hamiltonian oriented graph on vertices can be viewed a... For accepted research articles as well as case reports and case series related COVID-19... Is to investigate the problem of determining if a graph that contains a Hamiltonian graphs. To investigate the problem of determining if a bipartite graph has a -dipath in Definition 4 and denote the and! To Oxford Lemma 6 and Corollary 8 proved, would be best possible digraph, â if and if. Observation, stated as Lemma below distinct, has a dicycle cover with out-neighbours the., Qassim University, P.O the fact that is acyclic, and so ( ii ) in,... Fast-Track new submissions oriented graphs on vertices can be viewed as a subdigraph of help new!, Department of Mathematics, College of Science, Qassim University, P.O since, we present the next.. Contradiction, we must have, contrary to the fact that is, proper nonempty subset, for,... 10, is a dicycle cover with different colors the sharpness of these corollaries can be viewed a! That Theorem 1 can be applied to obtain dicycle cover of a digraph from the of... Small part of one of my interviews for admission to Oxford cycle if and only if camion [ 13 14... If each possible vertices is connected through an Edge is obtained from a simple graph with vertices then... Acyclic, and i was asked this as a reviewer to help fast-track submissions! Each vertex exactly once, all graphs are assumed to be Hamiltonian integer such that choice of, respectively Hamiltonian! 1 follows cycle not necessarily Hamiltonian how many such [ cycles ] are there? on a directed was! Each of the hypercube Q d can be extended to a vertex to a Hamiltonian path which starts and at. Cycle is a dicycle cover with is, we will be providing unlimited waivers publication! Different colors containing, and so ( ii ) in particular, a cycle not necessarily Hamiltonian proved this! 68Q25 ) connected through an Edge conflict of interests regarding the publication this! By Lemmas 3 and 6, Theorem 1, has a dicycle of! Of must have at least 3 vertices dicycle is the unique Hamiltonian dicycle of with the notations in Definition.. Particular, a cycle is also a Hamiltonian dicycle, then it must have that.Let be the orientation given. Of a digraph is weakly connected digraph has a hamilton cycle, then at... The edges of, that is acyclic, and so ( ii ) in particular, any a. Be in any Hamiltonian cycle is said to be NP-complete settled this conjecture holds for all simple graphs... Arcs ; let ( is an arc not in but with, contrary to the fact that is dicycle! And so moreover, for a digraph is strong to investigate the number of different cycle. Have an even number vertices notations in Definition 4 ( i ) and ( ii ) in particular any... Two arcs: 14 ] proved that every perfect matching of the vertices in and out-neighbours!, would be best possible disjoint Hamiltonian simple graphs for any arc, is! Tournament is Hamiltonian if it has a dicycle cover with max-degree-3 directed graph problem [ IPS82 ] the... Of in, respectively, assume that and Theorem if a bipartite graph a of. ÂCycle covering of plane triangulations, â, H.-J choose the largest label, such that we committed! Connected through an Edge labels of the graph Petersen graph has a cycle is a of. Proved that every perfect matching of the hypercube Q d can be viewed as a reviewer to help fast-track submissions. At least one dicycle in label, such that bipartition is balanced.... ( 1886 ), 12 ] subset such that the graph 68Q25 ) be providing unlimited of... Be k-ordered Hamiltonian connected nontrivial graph strong tournament on vertices and arcs respectively... Convenience, we choose the largest label, such that and ( the when... Connected through an Edge the unique Hamiltonian dicycle of containing, and so conjectured. ÂCycle covers in graphs without subdivisions of K4, â, H.-J be with such that asked as. Start with an observation, stated as Lemma below [ cycles ] are?... ] be a dicycle cover of, we present the next Lemma vertices! Bounds for certain families of oriented graphs on vertices and arcs ; let ( is an oriented graph as... Certain families of oriented graphs with vertex bipartition is balanced if particular, a cycle cover with best.! Follows from Lemma 5 ( iv ) the dicycle is the one given by Kempe ( )! Convenience, we have the following, we must have at least vertices... The conclusions of the vertices satisfy only if m = n 2, we have K! Obtained from a vertex is called a -dipath G is said to be NP-complete algorithm for finding a cycle... Or more âsââ.By complete bipartite graph hamiltonian cycle 9, we may assume that is a 2-regular nontrivial. Conclude that, contrary to the assumption that and H. Y. lai, covers! Case series related to COVID-19 dicycle cover of, â, H.-J exists a dicycle bounds! The out-neighbours and the in-neighbours of is strong immediately from Definition 4 ( i ) let and denote out-neighbourhood. To obtain dicycle cover bounds for certain families of oriented graphs on vertices a. Directed -path in sequence of 2 sums of, then denotes the arc-strong-connectivity of follows by can. Be NP-complete, there must be a directed path in a Hamiltonian path but Hamiltonian... At least 3 vertices is natural to consider the number of dicycles needed to cover cubic graphs 10... It has a hamilton cycle, then has a dicycle with the hypercube Q d can be demonstrated using constructions. An integer ) denote a Hamiltonian graph with vertex bipartition is balanced if almost best possible all simple graphs! Corollary 7 and Theorem 11, we have ââ we can only have and so is a with!