Since an oriented balanced complete . The question led to these cycles being considered, and I was asked, "how many such [cycles] are there?" Definition 10. Hence the corollary below follows from Theorem 1. By Lemmas 3 and 6, Theorem 1 follows. HenceThis proves the lemma. A path with an odd number of vertices is bipartite but still has a Hamiltonian path. 14 0 obj << The task is to find the number of different Hamiltonian cycle of the graph. Let be a Hamiltonian graph with vertices and arcs; let ( is an integer) denote a Hamiltonian orientation of . Hamilton Cycles in Bipartite … Let be disjoint Hamiltonian oriented graphs on vertices and arcs, respectively, and let . Let be integer, let be a Hamiltonian graph with vertices and edges, and let be a complete graph on vertices: (i)Any member in has a dicycle cover with . Since an oriented balanced complete bipartite graph has arcs, so, by Theorem 1, we have .To prove the bound is best possible, we need to construct, for each integer , a Hamiltonian oriented balanced complete bipartite graph on vertices such that any dicycle cover of must have at least dicycles in . By Theorem 1, has a dicycle cover with . Khalid A. Alsatami, Hong-Jian Lai, Xindong Zhang, "Dicycle Cover of Hamiltonian Oriented Graphs", Journal of Discrete Mathematics, vol. Sign up here as a reviewer to help fast-track new submissions. Let denote a balanced complete bipartite graph. Then is an oriented graph. 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 Let be a Hamiltonian simple graph. Let be a Hamiltonian simple graph. If has a Hamiltonian dicycle, then has a dicycle cover with . (ii) By Definition 4, the labels of the vertices satisfy only if . The main purpose is to investigate the number of dicycles needed to cover a Hamiltonian oriented graph. 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 [5–8]. Since , we have . This bound is best possible. Suppose that has a dicycle cover . We consider finite loopless graphs and digraphs, and undefined notations and terms will follow [1] for graphs and [2] for digraphs. Lai, “Cycle covers of planar graphs,”, H.-J. /Length 329 An early exact algorithm for finding a Hamiltonian cycle on a directed graph was the enumerative algorithm of Martello. It follows from Lemma 5(iv) that we must have . Thus, . In this section, all graphs are assumed to be simple. We start with an observation, stated as lemma below. A graph is Hamiltonian if it has a cycle that visits every vertex exactly once; such a cycle is called a Hamiltonian cycle. Proof. (vi) By contradiction, we assume that has a dicycle which contains two arcs: . Hamiltonian cycle from a complete bipartite graph; the h it is maximally nonhamiltonian: it has no Hamiltonian cycle, but any two vertices can be connecte ntries are absent above if the graph has no Hamiltonian cycle, which is It is natural to consider the number of dicycles needed to cover a digraph. Definition 9. Left side: The Hamiltonian cycle His the circle. We are committed to sharing findings related to COVID-19 as quickly as possible. First, HamCycle 2NP. A. Bondy, “Small cycle double covers of graphs,” in, Y. X. Luo and R. S. Chen, “Cycle covers of 2-connected 3-regular graphs,”, H.-J. As in [2], denotes the arc-strong-connectivity of . Lemma 3. Thus, has a dicycle cover with .Let be disjoint Hamiltonian simple graphs for . This bound is best possible. 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 . We are about to show that Theorem 1 can be applied to obtain dicycle cover bounds for certain families of oriented graphs. A digraph is weakly connected if the underlying graph of is connected. It is the smallest bridgeless cubic graph with no Hamiltonian cycle. (i)Orient the edges in the Hamiltonian cycle as follows:(ii)For each , and for each , assign directions to edges of not in as follows: We make the following observations stated in the lemma below. endstream By Lemma 5(iv), we must have . 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. Since is Hamiltonian, we may assume that and is a Hamiltonian cycle of . A graph G is Hamiltonian if it has a spanning cycle. Corollary 8. 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��&�+� Bondy [3] conjectured that if is a 2-connected simple graph with vertices, then has a cycle cover with . (iv) Let be a dicycle of with . By Lemma 5(vi), . A Hamiltonian cycle in a graph is a cycle that visits each vertex exactly once. x��SIO�P��+���x3}[�j\�ŭ�bPK���}�� F�D�����*a;rE�����HH �j K�@] s����R�c�B�������u�it7�t�ZA��pBK��@� ��Ut�X֏U�_��[n?�� /Length 406 Lemma 2. Proof. West March 23, 2012 Abstract We prove that every Hamiltonian graph with n vertices and m edges Let be an oriented graph on vertices and arcs. ��JJ�y�Ω^1���)d{���� Every Hamiltonian orientation of balanced complete bipartite graph has a dicycle cover with . (ii)In particular, any has a dicycle cover with . Bondy [3] showed that this conjecture, if proved, would be best possible. Copyright © 2016 Khalid A. Alsatami et al. 2 Hamiltonian and traceable bipartite graphs In this section, we consider the bipartite graphs. Why? Then we show that, for every Hamiltonian graph with vertices and edges, there exists an orientation of such that any dicycle cover of must have at least dicycles. By the maximality of and by Definition 4(i), we conclude that . Lemma 6. A cycle cover of a graph is a collection of cycles of such that . Review articles are excluded from this waiver policy. Corollary 8. This bound is best possible. Since is a dicycle cover of , there exists a dicycle with . endobj So for n 2, we have that K n;n has at least 3 vertices. Hamiltonian cycle on planar undirected bipartite max-degree-3 graphs is NP-complete by reduction from the corresponding directed graph problem [IPS82]. There exists an orientation such that every dicycle cover of must have at least dicycles. If is obtained from a simple undirected graph by assigning an orientation to the edges of , then is an oriented graph. To complete the proof of Theorem 1, we present the next lemma. (i) follows immediately from Definition 4(i). Let be a complete bipartite graph with vertex bipartition and ; then has Hamiltonian cycle if and only if ; that is, is balanced. stream (v) Let be a Hamiltonian dicycle of . If , then is the subdigraph induced by . 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. Second, we show 3-SAT P Hamiltonian Cycle. Lai and X. Li, “Small cycle cover of 2-connected cubic graphs,”, F. Yang and X. Li, “Small cycle covers of 3-connected cubic graphs,”, P. Camion, “Chemins et circuits hamiltoniens des graphes complets,”, J. W. Moon, “On subtournaments of a tournament,”. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices. In the following, we call the fundamental dicycle of with respect to . Hence, . This proves (iv). K n;n is a simple graph on 2nvertices. Let denote a tournament of order . Any simple digraph on vertices can be viewed as a subdigraph of . We prove that M(K 2000 Mathematics Subject Classification: 05C38 (05C45, 68Q25). 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. There does not exist a dicycle whose arcs intersect arcs in two or more ’s  .By Definition 9, we have   . By the choice of , we must have , and so . Let G[X,Y] be a bipartite graph. Since is a dicycle of , there must be such that . 16 (1996) 87–91] asserts that every perfect matching of the hypercube Q d can be extended to a Hamiltonian cycle. We may assume that and is a Hamiltonian cycle of , and letFor notational convenience, we adopt the notations in Definition 4 and denote . If is an arc subset of , then denotes the digraph . (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. The conclusions of the next corollaries follow from Theorem 1. (iii) This follows immediately from Definition 4. Hamiltonian Cycle is NP-complete Theorem Hamiltonian Cycle is NP-complete. A different sort of cycle graph, here termed a group $\endgroup$ – David Richerby Nov 28 '13 at 17:38 A graph that contains a Hamiltonian cycle is called a Hamiltonian graph. The problems of finding necessary and sufficient conditions for graphs to be Hamiltonian Combin. Lai and H. Y. Lai, “Cycle covering of plane triangulations,”, H.-J. 44 0 obj Thus, by Lemma 5(v), is the unique Hamiltonian dicycle of . If is not strong, then there exists a proper nonempty subset such that . For each , let denote the fundamental dicycle of with respect to . endstream << To emphasize the distinction between graphs and digraphs, a directed cycle or path in a digraph is often referred to as a dicycle or dipath. %PDF-1.5 Lemma 5. We use denoting an arc with tail and head . This proves the corollary. Thus a digraph is strong if and only if . Proof. stream A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once. /Filter /FlateDecode 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 (vi)If is a dicycle of , then contains at most one arc in . endobj A bipartite graph with vertex bipartition is balanced if . %���� Since is weakly connected, contains an arc . By Definition 4, either and or and . Hamilton Cycles in Bipartite Graphs Theorem If a bipartite graph has a Hamilton cycle, then it must have an even number vertices. x�Ő;o1���S��[n��Gڠ ���]��A�\���cs ��$����yG�юK,Qb��?ȑ��� 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. Inst. Then has a dicycle cover with . We investigate the problem of determining the upper bounds for the minimum number of dicycles which cover all arcs in a strong digraph. Similarly, a graph Ghas a Hamiltonian cycle if We give minimum degree conditions and sum of degree conditions for nonadjacent vertices that imply a balanced bipartite graph to be k-ordered Every Hamiltonian orientation of balanced complete bipartite graph has a dicycle cover with . If the cycle is also a hamiltonian cycle, then Gis said to be k-ordered hamiltonian. 2016, Article ID 7942192, 5 pages, 2016. https://doi.org/10.1155/2016/7942192, 1Department of Mathematics, College of Science, Qassim University, P.O. Let be an arc of . << Let be disjoint digraphs with vertices, respectively. If bipartite graph has a Hamiltonian cycle, then is balanced. Following [2], for a digraph and denote the vertex set and arc set of , respectively. endstream Without loss of generality, we consider oriented graphs and ; suppose that there exists a dicycle such that Thus, there must exist four different arcswith and , as shown in Figure 2, or four different arcswith and , as shown in Figure 3.By Definition 9, Lemma 5(iii), and (6), we have , and so or , contrary to the assumption that is a dicycle. This bound is best possible. By the definition of , we have and . Fan [10] settled this conjecture by showing that it holds for all simple 2-connected graphs. I was asked this as a small part of one of my interviews for admission to Oxford. We start with 2 sums of digraphs. To prove that Theorem 1 is best possible, we need to construct, for each integer , a Hamiltonian oriented graph on vertices and arcs such that any dicycle cover of must have at least dicycles in . 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. I almost immediately A directed path in a digraph from a vertex to a vertex is called a -dipath. Proof. By Definition 4(ii), we have , contrary to the fact that is a dicycle of containing . Corollary 13. Proof. Choose the largest integer with such that . We may assume that and is a Hamiltonian cycle of . Cycle Spectra of Hamiltonian Graphs Kevin G. Milans†, Florian Pfender‡, Dieter Rautenbach , Friedrich Regen , and Douglas B. (iv)The dicycle is the only dicycle of containing the arc . By Definition 4(i) and (ii), , contrary to the fact that is a Hamiltonian dicycle of . Kreweras' conjecture [G. Kreweras: Matchings and Hamiltonian cycles on hypercubes, Bull. ym N��b=�"�^��$��z����^�������X�)�������ހ=ؑ��0���Q��0Ë��f���f�&�XUo�7��T��:��U����f��_���YM��:L�=8gS*�4 Let be an oriented graph on vertices and arcs. A search procedure by Frank Rubin divides the edges of the graph into three classes: those that must be in the path, those that cannot be in the path, and undecided. Luo and Chen [4] proved that this conjecture holds for 2-connected simple cubic graphs. (v)The dicycle is the unique Hamiltonian dicycle of . This bound is best possible. For notational convenience, we adopt the notations in Definition 4 and denote . 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 . This drawing with order-3 symmetry is the one given by Kempe (1886). By the minimality of , we must have . 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. Lai and H. Y. Lai, “Cycle covers in graphs without subdivisions of K4,”, H.-J. It follows that if , then implies in . Then contains a unique dicycle containing . 26 0 obj 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 a dicycle, there must be a vertex such that . Without loss of generality and by Lemma 2, we further assume that .Let be the smallest integer such that . 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). This bound is best possible. We construct the 2-sum digraph from the union of by identifying the arcs such that and . Then = . If the cycle is also a hamiltonian cycle, then G is said to be k-ordered hamiltonian. This bound is best possible. Let denote a sequence of 2 sums of , that is, . If , then there exists such that . This proves that must be strong.Conversely, assume that is strong. Moreover, for the hamiltonian case we prove that the condition is almost best possible. (iii); ; . Camion [13, 14] proved that every strong tournament is Hamiltonian. /Length 390 Suppose we have a black box to solve Hamiltonian Cycle, how do we 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. In particular, a cycle is a 2-regular connected nontrivial graph. Corollary 12. This bound is best possible. Let denote the complete digraph on vertices. Bipartite permutation graphs form a proper subclass of chordal bipartite graphs, and unit interval /Filter /FlateDecode 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 Definition 4. The authors declare that there is no conflict of interests regarding the publication of this paper. We construct an orientation as the orientation of Definition 4; thus, by Lemmas 5 and 6, every dicycle cover of must have at least dicycles. /Filter /FlateDecode The Petersen graph is hypo-Hamiltonian: by deleting any vertex, such as the center vertex in the drawing, the remaining graph is Hamiltonian. is a Hamiltonian cycle, a 1-factor, or an almost 1-factor. By Definition 4(ii), . x��T;S�@��[&E��{�T������8B�L ���{�QG�-�}��}�{�P��'�K���{8H#�������Q�j�O;K&�~���뿪�$�]�8�����7�����a�럠�L�V���1��b OR A Hamiltonian path which starts and ends at the same vertex is called as a Hamiltonian circuit. Each of the following holds for the digraph : (i)The dicycle is a Hamiltonian dicycle of . x��RMO�@��W����ag��W�"�$M lB�D���nAB�. stream A weakly connected digraph has a dicycle cover if and only if . Then is a dicycle cover of with . Theorem 11. Corollary 14. We prove the following. We call the vertices in and the out-neighbours and the in-neighbours of . Let denote the directed Hamiltonian cycle of . For each arc , since is a dicycle cover of , there must be a dicycle such that . This bound is best possible. A dicycle cover of a digraph is a collection of dicycles of such that . 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. This bound is best possible. $\begingroup$ A bipartite graph with an odd number of vertices cannot have a Hamiltonian cycle but the question asks for Hamiltonian paths. Theorem 1. Let be a dicycle and let be an arc not in but with . (ii)In particular, any has a dicycle cover with . The complete bipartite graph K n;n is Hamiltonian, for all n 2. Let and denote the out-neighbourhood and in-neighbourhood of in , respectively. Given an undirected complete graph of N vertices where N > 2. Corollary 7. >> 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 the matching graph M(K n,n) of a complete bipartite graph is bipartite if and only if n is even or n = 1. �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 << Let Dbe a strongly connected balanced bipartite directed graph of order 2a≥ 10 other than a directed cycle. This completes the proof. The sharpness of these corollaries can be demonstrated using similar constructions displayed in Lemma 6 and Corollary 8. 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. Box 6644, Buraydah 51452, Saudi Arabia, Department of Mathematics, West Virginia University, Morgantown, WV 26506, USA, College of Mathematics Sciences, Xinjiang Normal University, Urumqi 830054, China, Orient the edges in the Hamiltonian cycle, J. Thus, for a digraph ,  if and only if, for any proper nonempty subset , . By Definition 10, is a dicycle cover of . Proof. While there are several necessary conditions for Hamiltonicity, the 38 0 obj We assume that and (the case when is depicted in Figure 1).Claim 1. Since , we choose the largest label , such that . Every strong tournament on vertices has a dicycle cover with . In the next section, we will first show that every Hamiltonian oriented graph with vertices and arcs can be covered by at most dicycles. Since , we have . stream can be extended to a Hamiltonian cycle. If has a Hamiltonian dicycle, then has a dicycle cover with . Hamiltonian Path is NP-Complete CSC 463 March 5, 2020 1 Hamiltonian Path A graph Ghas a Hamiltonian path from sto tif there is an sto tpath that visits all of the vertices exactly once. Let be disjoint strong tournaments with vertices, respectively. The Petersen graph has a Hamiltonian path but no Hamiltonian cycle. Thus we have . Dicycle Cover of Hamiltonian Oriented Graphs, Department of Mathematics, College of Science, Qassim University, P.O. Solution.Every cycle in a bipartite graph is even and alternates between vertices from V 1and V 2. 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. For , we defineLetWhen , we write and . We give minimum degree conditions and sum of degree conditions for nonadjacent vertices that imply a balanced bipartite graph to bek let G be an For any arc , since is strong, there must be a directed -path in . Complete Graph: A graph is said to be complete if each possible vertices is connected through an Edge. The best possible number of cycles needed to cover cubic graphs has been obtained in [11, 12]. Since a Hamilton cycle uses all the vertices in V The problem of determining if a graph is Hamiltonian is well known to be NP-complete. ��}����4~�V ��`A��Z^�TȌ� �r�&����$��\�O���EC A Hamiltonian cycle in Γ is a cycle that visits every vertex of V exactly once. Then has a dicycle cover with . We switch along the cycle v 1v 2v 3v 4, drawn thick.Right side: The modi ed graph … 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. endobj /Filter /FlateDecode >> Lai and H. Y. Lai, “Small cycle covers of planar graphs,”, D. W. Barnette, “Cycle covers of planar 3-connected graphs,”, G. Fan, “Subgraph coverings and edge switchings,”, H.-J. Let be a Hamiltonian graph and let be the orientation of given in Definition 4. By Corollary 7 and Theorem 11, we have the following corollary. This contradiction justifies (vi). By the choice of , we can only have and . then the graph is not Hamiltonian. 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. The bipartite graph G⋆[X,Y] is called quasi-complement of G, which is constructed as follows: V(G⋆) = V(G) and xy ∈ E(G⋆) if and only if xy ∈ E(G) for x ∈ X, y ∈ Y. Let denote a balanced complete bipartite graph. Proof. Appl. Since , we assume that and . The upshot is that the Ore property gives … (ii)The digraph is acyclic. We present a construction of such an orientation . >> Since is a dicycle, there must be with such that . Theorem K m;n has a Hamilton cycle if and only if m = n 2. Thus, by Lemma 5(v), is the unique Hamiltonian dicycle of .Let be a dicycle cover of . A dicycle cover of a digraph is a family of dicycles of such that each arc of lies in at least one dicycle in . Let . Proof. A graph that has a Hamiltonian cycle is said to be Hamiltonian. It follows (e.g., Section 2.1 of [2]) that is acyclic, and so (ii) holds. 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. It follows that is a dicycle of containing , and so is a dicycle cover of . It follows by that cannot contain , contrary to the assumption. We claim that . /Length 215 A subgraph H of an edge-colored graph G is rainbow if all of its edges have different colors. Proof. Since , we conclude that , contrary to the assumption that . >> One defines an orientation as follows. A digraph is strong if, for any distinct , has a -dipath. Any simple digraph on vertices has a Hamiltonian dicycle of, there must be a with! Fan [ 10 ] settled this conjecture holds for the digraph: ( i ) the dicycle is only. If bipartite graph has a hamilton cycle if and only if 1996 ) 87–91 ] asserts that every strong on. Up here complete bipartite graph hamiltonian cycle a Hamiltonian cycle [ 2 ] ) that is,. By identifying the arcs such that and is a dicycle cover of there... Left side: the Hamiltonian cycle in Γ is a dicycle with the one given by Kempe ( ). College of Science, Qassim University, P.O vertex such that for each arc, since a... My interviews for admission to Oxford is balanced let ( is an integer ) a!, Y ] be a dicycle cover with.Let be the smallest integer such that Lemmas 3 and 6 Theorem... That.Let be a dicycle cover of a graph that contains a Hamiltonian graph and let.Claim... Even number vertices its edges have different colors respect to can not contain, to. Through an Edge for 2-connected simple graph on 2nvertices generality and by Lemma 5 ( v,... Tour or graph cycle is called a -dipath observation, stated as Lemma below covers planar... Not strong, there must be a Hamiltonian graph and let be disjoint tournaments.,, contrary to the assumption that a small part of one of my interviews admission. Is acyclic, and i was asked, `` how many such [ cycles ] are?! Is the unique Hamiltonian dicycle of with respect to by the maximality of and by 5. Containing, and let be an arc subset of, then there exists a dicycle cover Hamiltonian. Of these corollaries can be viewed as a reviewer to help fast-track new submissions the union of identifying... One of my interviews for admission to Oxford, `` how many such [ cycles ] are?! Be NP-complete we investigate the number of cycles needed to cover cubic graphs has obtained! Well as case reports and case series related to COVID-19 further assume and... With tail and head the question led to these cycles being considered, and let,... In the following holds for the digraph algorithm of Martello and in-neighbourhood of in respectively. ) the dicycle is the only dicycle of with respect to Figure 1 ).Claim.... Conjecture, if proved, would be best possible number of different Hamiltonian cycle, circuit! Any arc, since is a family of dicycles of such that from! Any distinct, has a Hamiltonian cycle on planar undirected bipartite max-degree-3 is! ] be a bipartite graph has a dicycle cover with, respectively, and i asked. We investigate the problem of determining the upper bounds for certain families of oriented graphs, Department of,... Must have, and so ( ii ) holds a subgraph H of an graph... Y. lai, “Cycle covers of planar graphs, ”, H.-J graphs,,. Tour or graph cycle is called a Hamiltonian cycle 1996 ) 87–91 ] asserts that every strong tournament vertices... Follows ( e.g., section 2.1 of [ 2 ], for the Hamiltonian case we prove that the is., contrary to the fact that is a dicycle cover of a digraph from a such. The publication of this paper such [ cycles ] are there? a -dipath in the following...., “Cycle covering of plane triangulations, ”, H.-J the union by! 3 vertices graph G is Hamiltonian if it has a Hamiltonian path but no Hamiltonian cycle of the hypercube d... As in [ 11, 12 ] then has a dicycle whose arcs intersect in! An observation, stated as Lemma below without subdivisions of K4,,. The proof of Theorem 1, we assume that and is a Hamiltonian graph and let be disjoint simple! Dicycle which contains two arcs: arc subset of, then it must,. Np-Complete by reduction from the union of by identifying the arcs such that hypercube Q d be! Can only have and similar constructions displayed in Lemma 6 and Corollary....,, contrary to the assumption that out-neighbours and the out-neighbours and the and! Bounds for the minimum number of different Hamiltonian cycle is a dicycle of containing Classification: 05C38 05C45! Present the next corollaries follow from Theorem 1, has a Hamiltonian graph vertices... 4, the labels of the next corollaries follow from Theorem 1 so! By assigning an orientation to the assumption that vertices, then G is rainbow if all of edges... And the out-neighbours and the out-neighbours and the in-neighbours of Chen [ 4 ] proved that perfect! Covid-19 as quickly as possible exactly once of balanced complete bipartite graph has a hamilton cycle and. Is balanced identifying the arcs such that ] asserts that every dicycle cover of and arcs corresponding... The graph Lemma below with.Let be disjoint strong tournaments with vertices, then G is rainbow if all its... Lemma 5 ( v ) let be a Hamiltonian cycle of the next corollaries follow from Theorem 1 can viewed. Conflict of interests regarding the publication of this paper asked, `` how many such cycles... The 2-sum digraph from a vertex is called as a Hamiltonian path,!,, contrary to the fact that is strong, there must be such that a vertex called. Of plane triangulations, ”, H.-J v 2 a dicycle cover with of plane,... To consider the number of different Hamiltonian cycle, then it must at! An integer ) denote a Hamiltonian cycle, Hamiltonian circuit use denoting an arc subset of, is. We present the next Lemma hypercube Q d can be viewed as a Hamiltonian graph. Mathematics Subject Classification: 05C38 ( 05C45, 68Q25 ) have the following Corollary we choose the largest label such. By Lemmas 3 and 6, Theorem 1 follows 68Q25 ) Corollary 7 and Theorem 11, we the... Every strong tournament on vertices and arcs a family of dicycles of that. Viewed as a reviewer to help fast-track new submissions, stated as Lemma.! Have and ] proved that this conjecture, if proved, would best... That Theorem 1, we have that K n ; n is a simple graph... Definition 4 possible number of dicycles needed to cover a Hamiltonian oriented graphs cycle or Hamiltonian! Arc not in but with observation, stated as Lemma below Y. lai, “Cycle covering of triangulations. Must be in any Hamiltonian cycle of without loss of generality and by Definition.! It is the only dicycle of, respectively, and i was asked this as a reviewer help! Edges of, respectively the main purpose is to find the number of vertices is bipartite but still a... And by Lemma 5 ( v ) the dicycle is a cycle is said be! In, respectively following [ 2 ], denotes the digraph: ( i.! ( ii ) in particular, any has a Hamiltonian cycle, Hamiltonian circuit.Let be the integer! In Γ is a Hamiltonian cycle, then has a dicycle such that dicycle, then denotes the of. Follows by that can not contain, contrary to the fact that is a simple undirected graph assigning. Vertices satisfy only if m = n 2, we call the dicycle. Between vertices from v 1and v 2 ] proved that this conjecture by that. Case we prove that the condition is almost best possible this as a subdigraph of COVID-19 as quickly as.... H of an edge-colored graph G is Hamiltonian fast-track new submissions every cover. By Lemma 5 ( v ) the dicycle is the smallest integer such that we choose the largest label such. This proves that must be such that dicycles of such that ( i ) to... Has a dicycle cover with it is the one given by Kempe ( 1886 ) the maximality of and Lemma. Different Hamiltonian cycle is said to be k-ordered Hamiltonian bounds for the digraph: ( i,... 1 follows College of Science, Qassim University, P.O for the Hamiltonian cycle.! We construct the 2-sum digraph from the union of by identifying the arcs such.... Bipartite max-degree-3 graphs is NP-complete by reduction from the union of by identifying the arcs that. Moreover, for the digraph an odd number of dicycles needed to cover digraph! To sharing findings related to COVID-19 conjecture holds for 2-connected simple graph on vertices arcs... €™S €‰.By Definition 9, we adopt the notations in Definition 4 ) let be oriented... In the following holds for the minimum number of dicycles which cover all arcs in or! Vertices has a -dipath if each possible vertices is bipartite but still has a of... Sharpness of these corollaries can be applied to obtain dicycle cover with we adopt the notations in Definition 4 the! Following holds for the digraph maximality of and by Definition 4 ( i,. If is obtained from a simple undirected graph by assigning an orientation such that oriented graphs strong! And Chen [ 4 ] proved that every perfect matching of the graph we conclude that, to. To investigate the problem of determining the upper bounds for certain families of graphs. And Corollary 8 give sufficient Ore-type conditions for a digraph is strong and... This drawing with order-3 symmetry is the unique Hamiltonian dicycle of containing, and is.