structured/0704.0309.structured.json

{"paper_meta":{"paper_id":"arxiv:0704.0309","title":"0704.0309","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0704.0309v3 [cs.CC] 13 Jul 2007\nThe Complexity of HCP in Digraps with Degree\nBound Two\nGuohun Zhu\nGuilin University of Electronic Technology,\nNo.1 Jinji Road,Guilin, Guangxi, 541004,P.R.China\nccghzhu@guet.edu.cn\nAbstract. The Hamiltonian cycle problem (HCP) in digraphs D with\ndegree bound two is solved by two mappings in this paper. The first\nbijection is between an incidence matrix Cnm of simple digraph and an\nincidence matrix F of balanced bipartite undirected graph G; The sec-\nond mapping is from a perfect matching of G to a cycle of D. It proves\nthat the complexity of HCP in D is polynomial, and finding a second\nnon-isomorphism Hamiltonian cycle from a given Hamiltonian digraph\nwith degree bound two is also polynomial. Lastly it deduces P = NP\nbase on the results.\n1\nIntroduction\nIt is well known that the Hamiltonian cycle problem(HCP) is one of the standard\nNP-complete problem [1]. As for digraphs, even when the digraphs on this case:\nplanar digraphs with indegree 1 or 2 and outdegree 2 or 1 respectively, it is still\nNP −Complete which is proved by J.Plesn ́ık [2].\nLet us named a simple strong connected digraphs with at most indegree 1 or\n2 and outdegree 2 or 1 as Γ digraphs. This paper solves the HCP of Γ digraphs\nwith following main results.\nTheorem 1. Given an incidence matrix Cnm of Γ digraph, building a mapping:F =\n C+\n−C−\n \n, then F is a incidence matrix of undirected balanced bipartite graph\nG(X, Y ; E), which obeys the following properties:\nc1. |X| = n,|Y | = n,|E| = m\nc2.\n∀xi ∈X ∧1 ≤d(xi) ≤2\n∀yi ∈Y ∧1 ≤d(yi) ≤2\nc3. G has at most n\n4 components which is length of 4.\nLet us named the undirected balanced bipartite graph G(X, Y : E) of Γ\ndigraph as projector graph.\n\nTheorem 2. Let G be the projector graph of a Γ graph D(V, A), determining\na Hamiltonian cycle in Γ digraph is equivalent to find a perfect match M in\nG and r(C′) = n −1, where C′ is the incidence matrix of D′(V, L) ⊆D and\nL = {ai|ai ∈D ∧ei ∈M}.\nLet the each component of G corresponding to a boolean variable, a mono-\ntonic function f(M) is build to represents the number of component in D. Based\non this function, the maximum number of non-isomorphism perfect matching is\nlinear, thus complexity of Γ digraphs has a answer.\nTheorem 3. Given the incidence matrix Cnm of a Γ digraph , the complexity\nof finding a Hamiltonian cycle existing or not is O(n4)\nThe concepts of cycle and rank of graph are given in section 2. Then theorems\n1,2,3 are proved in sections 3,4,5 respectively. The last section discusses the P\nversus NP in more detail.\n2\nDefinition and properties\nThroughout this paper we consider the finite simple (un)directed graph D =\n(V, A) (G(V, E), respectively), i.e. the graph has no multi-arcs and no self loops.\nLet n and m denote the number of vertices V and arcs A (edges E, respectively),\nrespectively.\nAs conventional, let |S| denote the number of a set S. The set of vertices V\nand set of arcs of A of a digraph D(V, A) are denoted by V = {vi|1 ≤i ≤n}\nand A = {aj|(1 ≤j ≤m) ∧aj =< vi, vk >, (vi ̸= vk ∈V )} respectively,\nwhere < vi, vk > is a arc from vi to vk. Let the out degree of vertex vi denoted\nby d+(vi), which has the in degree by denoted as d−(vi) and has the degree\nd(vi) which equals d+(vi) + d−(vi). Let the N +(vi) = {vj| < vi, vj >∈A}, and\nN −(vi) = {vj| < vj, vi >∈A}.\nLet us define a forward relation ⊲⊳between two arcs as following, ai ⊲⊳aj =\nvk iffai =< vi, vk > ∧aj =< vk, vj >. It is obvious that |ai ⊲⊳ai| = 0 .\nA cycle L is a set of arcs (a1, a2, . . . , al) in a digraph D, which obeys two\nconditions:\nc1. ∀ai ∈L, ∃aj, ak ∈L \\ {ai}, ai ⊲⊳aj ̸= aj ⊲⊳ak ∈V\nc2. |\nS\nai̸=aj∈L\nai ⊲⊳aj| = |L|\nIf a cycle L obeys the following conditions, it is a simple cycle.\nc3. ∀L′ ⊂L, L′ does not satisfy both conditions c1 and c2.\nA Hamiltonian cycle L is also a simple cycle of length n = |V | ≥2 in digraph.\nAs for simplify, this paper given a sufficient condition of Hamiltonian cycle in\ndigraph.\nLemma 1. If a digraph D(V, A) include a sub graph D′(V, L) with following\ntwo properties, the D is a Hamiltonian graph.\n2\n\nc1. ∀vi ∈D′ →d+(vi) = 1 ∧d−(vi) = 1,\nc2. |L| = |V | ≥2 and D′ is a strong connected digraph.\nA graph that has at least one Hamiltonian cycle is called a Hamiltonian\ngraph. A graph G=(V ; E) is bipartite if the vertex set V can be partitioned into\ntwo sets X and Y (the bipartition) such that ∃ei ∈E, xj ∈X, ∀xk ∈X \\ {xj},\n(ei ⊲⊳xj ̸= ∅→ei ⊲⊳xk = ∅) (ei, Y , respectively). if |X| = |Y |, We call that\nG is a balanced bipartite graph. A matching M ⊆E is a collection of edges\nsuch that every vertex of V is incident to at most one edge of M, a matching of\nbalanced bipartite graph is perfect if |M| = |X|. Hopcroft and Karp shows that\nconstructs a perfect matching of bipartite in O((m + n)√n) [3]. The matching\nof bipartite has a relation with neighborhood of X.\nTheorem 4. [4] A bipartite graph G = (X, Y ; E) has a matching from X into\nY if and only if |N(S)| ≥S, for any S ⊆X.\nLemma 2. A even length of simple cycle consist of two disjoin perfect matching.\nTwo matrices representation for graphs are defined as follows.\nDefinition 1. [5] The incidence matrix C of undirected graph G is a two di-\nmensional n × m table, each row represents one vertex, each column represents\none edge, the cij in C are given by\ncij =\n \n1, if vi ∈ej;\n0, otherwise.\n(1)\nIt is obvious that every column of an incidence matrix has exactly two 1\nentries.\nDefinition 2. [5] The incidence matrix C of directed graph D is a two dimen-\nsional n × m table, each row represents one vertex, each column represents one\narc the cij in C are given by\ncij =\n \n \n \n1,\nif < vi, vi >⊲⊳aj = vi;\n−1, if aj ⊲⊳< vi, vi >= vi;\n0,\notherwise.\n(2)\nIt is obvious to obtain a corollary of the incidence matrix as following.\nCorollary 1. Each column of an incidence matrix of digraph has exactly one 1\nand one −1 entries.\nTheorem 5. [5] The C is the incidence matrix of a directed graph with k com-\nponents the rank of C is given by\nr(C) = n −k\n(3)\nIn order to convince to describe the graph D properties, in this paper, we\ndenotes the r(D) = r(C).\n3\n\n3\nDivided incidence matrix and Projector incidence\nmatrix\nFirstly, let us divided the matrix of C into two groups.\nC+ = {cij|cij ≥0 otherwise 0 }\n(4)\nC−= {cij|cij ≤0 otherwise 0 }\n(5)\nIt is obvious that the matrix of C+ represents the forward arc of a digraph\nand C−matrix represents the backward arc respectively. A corollary is deduced\nas following.\nCorollary 2. A digraph D = (V, A) is strong connected if and only if the rank\nof divided incidence matrix satisfies r(C+) = r(C−) = |V |.\nSecondly, let us combined the the C+ and C−as following matrix.\nF =\n \nC+\n−C−\n \n(6)\nIn more additional, let F represents as an incidence matrix of undirected\ngraph G(X, Y ; E). The F is named as projector incidence matrix\nof C and\nG is named as projector graph , where X represents the vertices V + of D, Y\nrepresents the vertices of V −respectively. In another words we build a mapping\nF : D →G and denotes it as G = F(D). So the F(D) has 2n vertices and\nm edges if D has n vertices and m arcs. We also build up a reverse mapping:\nF −1 : G →D When G is a projector graph. To simplify, we also denotes the\narcs ai = F −1(ei), v+\ni = F −1(xi) and v−\ni = F −1(yi).\n3.1\nProof of Theorem 1\nFirstly, let us prove the theorem 1.\nProof. c1. Since Γ digraph is strong connected, then each vertices of Γ digraph\nhas at least one forward arcs, each row of C+ has at least one 1 entries, and\nthe U represents the C+ , so\n|U| = n\nthe same principle of C−, each row of C−has at least one −1 entries, and\nthe V represents the C−, so\n|V | = n\nSince the columns of F equal to the columns of C,\n|E| = m\n4\n\nc2. Since the degree of each vi of Γ digraph is 1 ≤d+(vi) ≤2,\n∀ui ∈U ∧1 ≤d(ui) ≤2\nSince the degree of each vi of Γ digraph is 1 ≤d−(vi) ≤2,\n∀vi ∈V ∧1 ≤d(vi) ≤2\nc3. Let us prove by contradiction, suppose there are k > n\n4 components with\nlength of 4 in G. Since D is strong connected, according to the corollary 2,\nr(F) = 3n\n2 −q ≥r(C+) = n, where q ≥k is number of components (in-\ncluding k components with length of 4). Thus q ≤n\n2 , then there are only x\ncomponents without length 4, where x is\nx = q −k < n\n4\n(7)\nSuppose the remind x components with length of t (at least t vertices con-\nnected by some edges), then 4k + xt = 3n\n2 . So tx = 3n\n2 −4k < n\n2 . According\nto the equation 7, the t < 2. It is contradict that the D is strong connected.\n3.2\nThe cycle in digraph corresponding matching in projector graph\nSecondly, let us given the properties after mapping Hamiltonian cycle L of D\ninto the sub graph M of projector graph G.\nLemma 3. If a Hamiltonian cycle L of D mapping into a forest M of projector\ngraph G, the forest M consist of |L| number of trees which has only two node\nand one edge, and M has a unique perfect matching.\nProof. Let the Γ digraph D(V, A) has a sub digraph D′(V, L) which exists one\nHamiltonian cycle and |L| = n, the incidence matrix C of L could be permutation\nas follows.\nC =\n \n \n \n \n \n \n \n \n1\n0\n0\n. . . 0\n−1\n−1 1\n0\n. . . 0\n0\n0\n−1 1\n. . . 0\n0\n0\n0\n−1 . . . 0\n0\n0\n0\n0\n. . . 0\n0\n0\n0\n0\n. . . −1 1\n \n \n \n \n \n \n \n \n.\n(8)\nLet\nF =\n \nC+\n−C−\n \nIt is obvious that each row of F has only one 1 entry and each column of F\nhas two 1 entries.\nAccording to theorem 1, F represents a balanced bipartite graph G(X, Y ; E)\nthat each vertex has one edge connected, and each edge ei connect on vertex\nxi ∈X , another in Y , in another words, ∃ei ∈E xj ∈X,∀xk ∈X \\ {xj}, ei ⊲⊳\nxj ̸= ∅→ei ⊲⊳xk = ∅(ei, Y ,respectively). According the matching definition,\nM is a matching, since |E| = |L|, E is a perfect matching. and pair of vertices\nbetween X and Y only has one edge, so M is a forest, and each tree has only\ntwo node with one edge.\n5\n\n4\nProof of Theorem 2\nProof. ⇒Let the Γ digraph D(V, A) has a sub digraph D′(V, L) which is a\nHamiltonian cycle and |L| = n, let matrix C′ represents the incidence matrix of\nD′, so r(C′) = n −1; According to lemma 3, the projector graph F(D′) has a\nperfect matching, thus F(D) also has a perfect matching.\n⇐Let G(X, Y ; E) be a projector graph of the Γ graph D(V, A),M is a perfect\nmatching in G. Let D′(V, L) be a sub graph of D(V, A) and L = {ai|ai ∈D∧ei ∈\nM}. Since r(L) = n −1, D′(V, L) is a strong connected digraph. it deduces that\n∀vi ∈D′,d+(vi) ≥1 ∧d−(vi) ≥1. Suppose ∃vi ∈D′, d+(vi) > 1 (d−(vi) > 1\nrespectively), Since |M| = n, it deduces that Pn\ni=1 d(vi) > 2n + 1, which imply\nthat |L| > n. this is contradiction with L = {ai|ai ∈D ∧ei ∈M} and |M| = n.\nSo ∀vi ∈D′, d+(vi) = d−(vi) = 1, According the lemma 1, D′ has a Hamiltonian\ncycle.\n5\nNumber of perfect matching in projector graph\nLet us considering the number of perfect matching in G . Firstly, let us consid-\nering a example as shown in figure 1.\nFigure 1. Original Digraph D\n✻\n♠\na8\n✲\n♠a1\n✛\n♠\na22\n❄\n✲\n♠a2\na9\n✛\n✻\n♠\na21\na10\n♠✲\na3\n✛\n♠\na20\n❄\n✲\n♠a4\na11\n✛\n✻\n♠\na12\na19\n♠✲\na5\n✛\n♠\na18\n❄\n✲\n♠a6\na13\n✛\n✻\n♠\na17\na14\n♠✲\na7\n✛\n♠\na16\n♠\n❄a15\nThen the projector graph is shown in figure 2.\nFigure 2. Projector graph G\n♠\n♠\ne1\n♠\n♠\ne8\n♠\n♠\ne22\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne9\ne2\nG1\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne10\ne21\n♠\n♠\ne3\n♠\n♠\ne20\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne11\ne4\nG2\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne19\ne12\n♠\n♠\ne5\n♠\n♠\ne18\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne6\ne13\nG3\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne17\ne14\n. . .\nGiven a perfect matching M, each component(cycle) in G has two partition\nedges belong to M. Let us code component Gi which |Gi| > 2 and matching M\nto a binary variable.\nGi =\n 1, if Gi ∩M = {ej, ek, . . .};\n0, if Gi ∩M = {el, eq, . . .}.\n(9)\n6\n\nNow there are two cases for the number of perfect matching.\nLabel edge. In that cases, the Code(M1) = {0, 0, 1} is different with Code(M2) = {0, 1, 0}.\nIf there are k number of components(cycles), then there are 2k perfect match-\ning.\nUnlabel edge. In that cases, the Code(M1) = {0, 0, 1} is isomorphic to Code(M2) = {0, 1, 0}.\nThe same principle that Code(M3) = {0, 1, 1} is isomorphic to Code(M4) =\n{1, 1, 0} but is not isomorphic to Code(M1).\nThen let us summary the maximal number of perfect matching in these two\ncases.\nLemma 4. The maximal number of labeled perfect matching in a projector graph\nG is 2\nn\n4 , but the maximal number of unlabeled perfect matching in a projector\ngraph G is n\n2 .\nProof. According to the theorem 1, there at most n\n4 components with a compo-\nnents which is length of k = 4. When k=2, there are only one perfect matching\nin G; When k = 4, there are n\n4 components which is C4, and so on when k = 6,\nthere are n\n6 components which is C6, etc, so on. According to the lemma 2, each\nsimple cycle has divided the perfect matching into two class. So maximal number\nperfect matching in the non isomorphism cycle which is 2\nn\n4 . Since in unlabeled\ncases, every C4 cycle is isomorphism, the maximal number of perfect matching\nis 2 ∗n\n4 = n\n2 .\nReview the example 1 again, it is easy find that follow proposition.\nProposition 1. Given two perfect matching M1 and M2 in projector graph G,\nif code(M1) = code(M2), then the r(F −1(M1)) = r(F −1(M2)).\n5.1\nProof of Theorem 3\nNow let us proof the theorem 3.\nProof. Let G be a project balanced bipartition of D. According theorem 1, the\nΓ graph is equivalent to find a perfect match M in a project G.\nAccording to the lemma 4, the maximal number non isomorphism perfect\nmatching in G is only n.\nThus it is only need exactly enumerate all of non isomorphism perfect match-\ning M, then obtain the value = r(F −1(M)),if value = n −1, then the ei ∈M\nis also ei ∈C, where C ⊂D is a Hamiltonian cycle.\nSince the complexity of rank of matrix is O(n3), finding a simple cycle in\na component with degree 2 is O(n2), and obtaining a perfect matching of a\nbipartite graph is O((m + n)√n) < O(n2) [3]. Then all exactly algorithms need\nto calculate the n time o(n3). Thus the complexity is O(n4).\nSince the non isomorphism perfect matching comes from the coding of edges\nin the component of G, it is not easy implementation.\n7\n\nLet us give two recursive equation to obtain a perfect matching M from G.\nSuppose there are k component G1, G2, . . . Gk in G where Gi is a component\nwith degree 2 and |Ei| ≥3.\nM ′ =\n \nM(t) ⊗Gt, Gt is a cycle ;\nM(t),\notherwise.\n(10)\nM(t + 1) =\n M ′,\nif r(F −1(M ′)) > r(F −1(M(t))) ;\nM(t), otherwise.\n(11)\nwhere t ≤k −1, when t = 0, M(0) is the initial perfect matching from G.\nWhen r(F −1(M(t))) = n −1, According the theorem 1, the A = F −1(M(t))\nis a Hamiltonian cycle solution. If all of r(F −1(M(t))) < n −1, then there has\nno Hamiltonian cycle in D.\nSince the non isomorphism perfect matching M in G is poset, the function\nr(F −1(M)) in G is monotonic, so this approach is exactly approach.\nLet us give a example to illustrate the approach in detail.\nExample 1. Considering the digraph D in figure 1, then the projector graph G\nin figure 2.\nLet M(0) = {e1, e8, e22, e9, e10, e3, e20, e11, e19, e5, e18, e6, e17, e7, e15, e16}.\nThus the r(F −1(M(0)) = n−3. Let M ′ = r(F −1(M(0)⊗G3),then r(F −1(M ′) =\nn−4, thus M(1) = M(0) and then turn to G2,G1. At last it obtain the solution.\nConsidering the equation 11, let it substituted by following equations when\nr(M ′) = n −1 and t < k −1.\nM(t + 1) = M ′ if r(F −1(M ′)) ≥r(F −1(M(t)))\n(12)\nIt is obvious that all non-isomorphism Hamiltonian cycle could obtain by the\nrepeat check the equation 12 and the equation r(M ′) = n −1.\nIn conversely, if a Hamiltonian cycle of Γ digraphs is given, it represents a\nperfect matching M in its projector graph G. Thus the equation 12 and Theo-\nrem 3 follows a corollary.\nCorollary 3. Given a Hamiltonian Γ digraph, the complexity of determining\nanother non-isomorphism Hamiltonian cycle is polynomial time.\n5.2\nThe HCP in digraph with bound two\nLet us extend the Theorem 3 to digraphs with d+(v) ≤2 and d−(v) ≤2 in this\nsection.\nTheorem 6. The complexity of finding a Hamiltonian cycle existing or not in\ndigraphs with degree d+(v) ≤2 and d−(v) ≤2 is polynomial time.\n8\n\nProof. Suppose a digraph D(V, A) having a vertex vi is shown as figure 3, which\nis d(vi) = 2 ∧d−(vi) = 2\nFigure 3. A vertex with degree than 2\n✟✟✟✟✟\n✟\n✯\n❍❍❍❍❍\n❍\n❥♠\na1\na2\n✟✟✟✟✟\n✟\n✯\n❍❍❍❍❍\n❍\n❥\na3\na4\nLet us spilt this vertex to two vertices that one of vertex has degree with\nin degree 2 or out degree 1 , another vertex has degree with in degree 1 or out\ndegree 2 as shown in figure 4. Then the D is derived to a new Γ graph S.\nFigrue 4 A vertex in D is mapping to a vertex in Γ digraph\n✟✟✟✟✟\n✟\n✯\n✲❤\na1\na2\n✲\na3\n❤\na4\n✟✟✟✟✟\n✟\n✯\n✲\nIt is obvious that each vertex in the Γ graph S has increase 1 vertices and 1\narcs of D. Suppose the worst cases is each vertex in D has in degree 2 and out\ndegree 2, the total vertices in S has 2n vertices.\nAccording to the theorem 3, obtain a Hailtonian cycle L′ in S is no more\nthen O(n4), then the D will has a Hamiltonian cycle L′ = L ∩A.\n6\nDiscussion P versus NP\nThe P versus NP is a famous open problem in computer science and math-\nematics, which means to determine whether very language accepted by some\nnondeterministic algorithm in polynomial time is also accepted by some deter-\nministic algorithm in polynomial time [6]. Cook give a proposition for the P\nversus NP.\nProposition 2. If L is NP-complete and L ∈P, then P = NP.\nAccording above proposition and the result above section, P versus NP\nproblem has a answer.\nTheorem 7. P = NP\n9\n\nProof. As the result of [2], the complexity of HCP in digraph with bound two\nis NP −complete. According the theorem 6, the complexity of HCP in digraph\nwith bound two is also P, thus according to proposition 2, P = NP.\nIn fact, the [2] proves that 3SAT ⪯p HCP of Γ digraph, since 3SAT is a\nNPC problem, which also implies that P = NP.\n7\nConclusion\nAccording to the theorem 6, the complexity of determining a Hamiltonian cycle\nexistence or not in digraph with bound degree two is in polynomial time. And\naccording to the theorem 7, P versus NP problem has closed, P = NP.\nAcknowledgements\nThe author would like to thank Prof. Kaoru Hirota for valuable suggestions,\nthank Prof. Jørgen Bang-Jensen who called mine attention to the paper [2], and\nthank Andrea Moro for useful discussions.\nReferences\n1. Papadimitriou, C. H. Computational complexity , in Lawler, E. L., J. K. Lenstra, A.\nH. G. Rinnooy Kan, and D. B. Shmoys, eds., The Traveling Salesman Problem: A\nGuided Tour of Combinatorial Optimization. Wiley, Chichester, UK. (1985), 37–85\n2. J.Plesn ́ık,The NP-Completeness of the Hamiltonian Cycle Problem in Planar di-\ngraphs with degree bound two, Journal Information Processing Letters, Vol.8(1978),\n199–201\n3. J.E. Hopcroft and R.M. Karp , An n5/2 Algorithm for Maximum Matchings in\nBipartite Graphs . SIAM J. Comput. Vol.2, (1973), 225–231\n4. P. Hall, On representative of subsets, J. London Math. Soc. 10, (1935), 26–30\n5. Pearl, M, Matrix Theory and Finite Mathematics,McGraw-Hill, New York,(1973),\n332–404.\n6. Stephen Cook. The P Versus NP Problem ,”http://citeseer.ist.psu.edu/302888.html”\n,2000.\n10","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0704.0309v3 [cs.CC] 13 Jul 2007\nThe Complexity of HCP in Digraps with Degree\nBound Two\nGuohun Zhu\nGuilin University of Electronic Technology,\nNo.1 Jinji Road,Guilin, Guangxi, 541004,P.R.China\nccghzhu@guet.edu.cn\nAbstract. The Hamiltonian cycle problem (HCP) in digraphs D with\ndegree bound two is solved by two mappings in this paper. The first\nbijection is between an incidence matrix Cnm of simple digraph and an\nincidence matrix F of balanced bipartite undirected graph G; The sec-\nond mapping is from a perfect matching of G to a cycle of D. It proves\nthat the complexity of HCP in D is polynomial, and finding a second\nnon-isomorphism Hamiltonian cycle from a given Hamiltonian digraph\nwith degree bound two is also polynomial. Lastly it deduces P = NP\nbase on the results.\n1\nIntroduction\nIt is well known that the Hamiltonian cycle problem(HCP) is one of the standard\nNP-complete problem [1]. As for digraphs, even when the digraphs on this case:\nplanar digraphs with indegree 1 or 2 and outdegree 2 or 1 respectively, it is still\nNP −Complete which is proved by J.Plesn ́ık [2].\nLet us named a simple strong connected digraphs with at most indegree 1 or\n2 and outdegree 2 or 1 as Γ digraphs. This paper solves the HCP of Γ digraphs\nwith following main results.\nTheorem 1. Given an incidence matrix Cnm of Γ digraph, building a mapping:F =\n C+\n−C−"},{"paragraph_id":"p2","order":2,"text":", then F is a incidence matrix of undirected balanced bipartite graph\nG(X, Y ; E), which obeys the following properties:\nc1. |X| = n,|Y | = n,|E| = m\nc2.\n∀xi ∈X ∧1 ≤d(xi) ≤2\n∀yi ∈Y ∧1 ≤d(yi) ≤2\nc3. G has at most n\n4 components which is length of 4.\nLet us named the undirected balanced bipartite graph G(X, Y : E) of Γ\ndigraph as projector graph."},{"paragraph_id":"p3","order":3,"text":"Theorem 2. Let G be the projector graph of a Γ graph D(V, A), determining\na Hamiltonian cycle in Γ digraph is equivalent to find a perfect match M in\nG and r(C′) = n −1, where C′ is the incidence matrix of D′(V, L) ⊆D and\nL = {ai|ai ∈D ∧ei ∈M}.\nLet the each component of G corresponding to a boolean variable, a mono-\ntonic function f(M) is build to represents the number of component in D. Based\non this function, the maximum number of non-isomorphism perfect matching is\nlinear, thus complexity of Γ digraphs has a answer.\nTheorem 3. Given the incidence matrix Cnm of a Γ digraph , the complexity\nof finding a Hamiltonian cycle existing or not is O(n4)\nThe concepts of cycle and rank of graph are given in section 2. Then theorems\n1,2,3 are proved in sections 3,4,5 respectively. The last section discusses the P\nversus NP in more detail.\n2\nDefinition and properties\nThroughout this paper we consider the finite simple (un)directed graph D =\n(V, A) (G(V, E), respectively), i.e. the graph has no multi-arcs and no self loops.\nLet n and m denote the number of vertices V and arcs A (edges E, respectively),\nrespectively.\nAs conventional, let |S| denote the number of a set S. The set of vertices V\nand set of arcs of A of a digraph D(V, A) are denoted by V = {vi|1 ≤i ≤n}\nand A = {aj|(1 ≤j ≤m) ∧aj =< vi, vk >, (vi ̸= vk ∈V )} respectively,\nwhere < vi, vk > is a arc from vi to vk. Let the out degree of vertex vi denoted\nby d+(vi), which has the in degree by denoted as d−(vi) and has the degree\nd(vi) which equals d+(vi) + d−(vi). Let the N +(vi) = {vj| < vi, vj >∈A}, and\nN −(vi) = {vj| < vj, vi >∈A}.\nLet us define a forward relation ⊲⊳between two arcs as following, ai ⊲⊳aj =\nvk iffai =< vi, vk > ∧aj =< vk, vj >. It is obvious that |ai ⊲⊳ai| = 0 .\nA cycle L is a set of arcs (a1, a2, . . . , al) in a digraph D, which obeys two\nconditions:\nc1. ∀ai ∈L, ∃aj, ak ∈L \\ {ai}, ai ⊲⊳aj ̸= aj ⊲⊳ak ∈V\nc2. |\nS\nai̸=aj∈L\nai ⊲⊳aj| = |L|\nIf a cycle L obeys the following conditions, it is a simple cycle.\nc3. ∀L′ ⊂L, L′ does not satisfy both conditions c1 and c2.\nA Hamiltonian cycle L is also a simple cycle of length n = |V | ≥2 in digraph.\nAs for simplify, this paper given a sufficient condition of Hamiltonian cycle in\ndigraph.\nLemma 1. If a digraph D(V, A) include a sub graph D′(V, L) with following\ntwo properties, the D is a Hamiltonian graph.\n2"},{"paragraph_id":"p4","order":4,"text":"c1. ∀vi ∈D′ →d+(vi) = 1 ∧d−(vi) = 1,\nc2. |L| = |V | ≥2 and D′ is a strong connected digraph.\nA graph that has at least one Hamiltonian cycle is called a Hamiltonian\ngraph. A graph G=(V ; E) is bipartite if the vertex set V can be partitioned into\ntwo sets X and Y (the bipartition) such that ∃ei ∈E, xj ∈X, ∀xk ∈X \\ {xj},\n(ei ⊲⊳xj ̸= ∅→ei ⊲⊳xk = ∅) (ei, Y , respectively). if |X| = |Y |, We call that\nG is a balanced bipartite graph. A matching M ⊆E is a collection of edges\nsuch that every vertex of V is incident to at most one edge of M, a matching of\nbalanced bipartite graph is perfect if |M| = |X|. Hopcroft and Karp shows that\nconstructs a perfect matching of bipartite in O((m + n)√n) [3]. The matching\nof bipartite has a relation with neighborhood of X.\nTheorem 4. [4] A bipartite graph G = (X, Y ; E) has a matching from X into\nY if and only if |N(S)| ≥S, for any S ⊆X.\nLemma 2. A even length of simple cycle consist of two disjoin perfect matching.\nTwo matrices representation for graphs are defined as follows.\nDefinition 1. [5] The incidence matrix C of undirected graph G is a two di-\nmensional n × m table, each row represents one vertex, each column represents\none edge, the cij in C are given by\ncij ="},{"paragraph_id":"p5","order":5,"text":"1, if vi ∈ej;\n0, otherwise.\n(1)\nIt is obvious that every column of an incidence matrix has exactly two 1\nentries.\nDefinition 2. [5] The incidence matrix C of directed graph D is a two dimen-\nsional n × m table, each row represents one vertex, each column represents one\narc the cij in C are given by\ncij ="},{"paragraph_id":"p6","order":6,"text":"1,\nif < vi, vi >⊲⊳aj = vi;\n−1, if aj ⊲⊳< vi, vi >= vi;\n0,\notherwise.\n(2)\nIt is obvious to obtain a corollary of the incidence matrix as following.\nCorollary 1. Each column of an incidence matrix of digraph has exactly one 1\nand one −1 entries.\nTheorem 5. [5] The C is the incidence matrix of a directed graph with k com-\nponents the rank of C is given by\nr(C) = n −k\n(3)\nIn order to convince to describe the graph D properties, in this paper, we\ndenotes the r(D) = r(C).\n3"},{"paragraph_id":"p7","order":7,"text":"3\nDivided incidence matrix and Projector incidence\nmatrix\nFirstly, let us divided the matrix of C into two groups.\nC+ = {cij|cij ≥0 otherwise 0 }\n(4)\nC−= {cij|cij ≤0 otherwise 0 }\n(5)\nIt is obvious that the matrix of C+ represents the forward arc of a digraph\nand C−matrix represents the backward arc respectively. A corollary is deduced\nas following.\nCorollary 2. A digraph D = (V, A) is strong connected if and only if the rank\nof divided incidence matrix satisfies r(C+) = r(C−) = |V |.\nSecondly, let us combined the the C+ and C−as following matrix.\nF ="},{"paragraph_id":"p8","order":8,"text":"C+\n−C−"},{"paragraph_id":"p9","order":9,"text":"(6)\nIn more additional, let F represents as an incidence matrix of undirected\ngraph G(X, Y ; E). The F is named as projector incidence matrix\nof C and\nG is named as projector graph , where X represents the vertices V + of D, Y\nrepresents the vertices of V −respectively. In another words we build a mapping\nF : D →G and denotes it as G = F(D). So the F(D) has 2n vertices and\nm edges if D has n vertices and m arcs. We also build up a reverse mapping:\nF −1 : G →D When G is a projector graph. To simplify, we also denotes the\narcs ai = F −1(ei), v+\ni = F −1(xi) and v−\ni = F −1(yi).\n3.1\nProof of Theorem 1\nFirstly, let us prove the theorem 1.\nProof. c1. Since Γ digraph is strong connected, then each vertices of Γ digraph\nhas at least one forward arcs, each row of C+ has at least one 1 entries, and\nthe U represents the C+ , so\n|U| = n\nthe same principle of C−, each row of C−has at least one −1 entries, and\nthe V represents the C−, so\n|V | = n\nSince the columns of F equal to the columns of C,\n|E| = m\n4"},{"paragraph_id":"p10","order":10,"text":"c2. Since the degree of each vi of Γ digraph is 1 ≤d+(vi) ≤2,\n∀ui ∈U ∧1 ≤d(ui) ≤2\nSince the degree of each vi of Γ digraph is 1 ≤d−(vi) ≤2,\n∀vi ∈V ∧1 ≤d(vi) ≤2\nc3. Let us prove by contradiction, suppose there are k > n\n4 components with\nlength of 4 in G. Since D is strong connected, according to the corollary 2,\nr(F) = 3n\n2 −q ≥r(C+) = n, where q ≥k is number of components (in-\ncluding k components with length of 4). Thus q ≤n\n2 , then there are only x\ncomponents without length 4, where x is\nx = q −k < n\n4\n(7)\nSuppose the remind x components with length of t (at least t vertices con-\nnected by some edges), then 4k + xt = 3n\n2 . So tx = 3n\n2 −4k < n\n2 . According\nto the equation 7, the t < 2. It is contradict that the D is strong connected.\n3.2\nThe cycle in digraph corresponding matching in projector graph\nSecondly, let us given the properties after mapping Hamiltonian cycle L of D\ninto the sub graph M of projector graph G.\nLemma 3. If a Hamiltonian cycle L of D mapping into a forest M of projector\ngraph G, the forest M consist of |L| number of trees which has only two node\nand one edge, and M has a unique perfect matching.\nProof. Let the Γ digraph D(V, A) has a sub digraph D′(V, L) which exists one\nHamiltonian cycle and |L| = n, the incidence matrix C of L could be permutation\nas follows.\nC ="},{"paragraph_id":"p11","order":11,"text":"1\n0\n0\n. . . 0\n−1\n−1 1\n0\n. . . 0\n0\n0\n−1 1\n. . . 0\n0\n0\n0\n−1 . . . 0\n0\n0\n0\n0\n. . . 0\n0\n0\n0\n0\n. . . −1 1"},{"paragraph_id":"p12","order":12,"text":".\n(8)\nLet\nF ="},{"paragraph_id":"p13","order":13,"text":"C+\n−C−"},{"paragraph_id":"p14","order":14,"text":"It is obvious that each row of F has only one 1 entry and each column of F\nhas two 1 entries.\nAccording to theorem 1, F represents a balanced bipartite graph G(X, Y ; E)\nthat each vertex has one edge connected, and each edge ei connect on vertex\nxi ∈X , another in Y , in another words, ∃ei ∈E xj ∈X,∀xk ∈X \\ {xj}, ei ⊲⊳\nxj ̸= ∅→ei ⊲⊳xk = ∅(ei, Y ,respectively). According the matching definition,\nM is a matching, since |E| = |L|, E is a perfect matching. and pair of vertices\nbetween X and Y only has one edge, so M is a forest, and each tree has only\ntwo node with one edge.\n5"},{"paragraph_id":"p15","order":15,"text":"4\nProof of Theorem 2\nProof. ⇒Let the Γ digraph D(V, A) has a sub digraph D′(V, L) which is a\nHamiltonian cycle and |L| = n, let matrix C′ represents the incidence matrix of\nD′, so r(C′) = n −1; According to lemma 3, the projector graph F(D′) has a\nperfect matching, thus F(D) also has a perfect matching.\n⇐Let G(X, Y ; E) be a projector graph of the Γ graph D(V, A),M is a perfect\nmatching in G. Let D′(V, L) be a sub graph of D(V, A) and L = {ai|ai ∈D∧ei ∈\nM}. Since r(L) = n −1, D′(V, L) is a strong connected digraph. it deduces that\n∀vi ∈D′,d+(vi) ≥1 ∧d−(vi) ≥1. Suppose ∃vi ∈D′, d+(vi) > 1 (d−(vi) > 1\nrespectively), Since |M| = n, it deduces that Pn\ni=1 d(vi) > 2n + 1, which imply\nthat |L| > n. this is contradiction with L = {ai|ai ∈D ∧ei ∈M} and |M| = n.\nSo ∀vi ∈D′, d+(vi) = d−(vi) = 1, According the lemma 1, D′ has a Hamiltonian\ncycle.\n5\nNumber of perfect matching in projector graph\nLet us considering the number of perfect matching in G . Firstly, let us consid-\nering a example as shown in figure 1.\nFigure 1. Original Digraph D\n✻\n♠\na8\n✲\n♠a1\n✛\n♠\na22\n❄\n✲\n♠a2\na9\n✛\n✻\n♠\na21\na10\n♠✲\na3\n✛\n♠\na20\n❄\n✲\n♠a4\na11\n✛\n✻\n♠\na12\na19\n♠✲\na5\n✛\n♠\na18\n❄\n✲\n♠a6\na13\n✛\n✻\n♠\na17\na14\n♠✲\na7\n✛\n♠\na16\n♠\n❄a15\nThen the projector graph is shown in figure 2.\nFigure 2. Projector graph G\n♠\n♠\ne1\n♠\n♠\ne8\n♠\n♠\ne22\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne9\ne2\nG1\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne10\ne21\n♠\n♠\ne3\n♠\n♠\ne20\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne11\ne4\nG2\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne19\ne12\n♠\n♠\ne5\n♠\n♠\ne18\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne6\ne13\nG3\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne17\ne14\n. . .\nGiven a perfect matching M, each component(cycle) in G has two partition\nedges belong to M. Let us code component Gi which |Gi| > 2 and matching M\nto a binary variable.\nGi =\n 1, if Gi ∩M = {ej, ek, . . .};\n0, if Gi ∩M = {el, eq, . . .}.\n(9)\n6"},{"paragraph_id":"p16","order":16,"text":"Now there are two cases for the number of perfect matching.\nLabel edge. In that cases, the Code(M1) = {0, 0, 1} is different with Code(M2) = {0, 1, 0}.\nIf there are k number of components(cycles), then there are 2k perfect match-\ning.\nUnlabel edge. In that cases, the Code(M1) = {0, 0, 1} is isomorphic to Code(M2) = {0, 1, 0}.\nThe same principle that Code(M3) = {0, 1, 1} is isomorphic to Code(M4) =\n{1, 1, 0} but is not isomorphic to Code(M1).\nThen let us summary the maximal number of perfect matching in these two\ncases.\nLemma 4. The maximal number of labeled perfect matching in a projector graph\nG is 2\nn\n4 , but the maximal number of unlabeled perfect matching in a projector\ngraph G is n\n2 .\nProof. According to the theorem 1, there at most n\n4 components with a compo-\nnents which is length of k = 4. When k=2, there are only one perfect matching\nin G; When k = 4, there are n\n4 components which is C4, and so on when k = 6,\nthere are n\n6 components which is C6, etc, so on. According to the lemma 2, each\nsimple cycle has divided the perfect matching into two class. So maximal number\nperfect matching in the non isomorphism cycle which is 2\nn\n4 . Since in unlabeled\ncases, every C4 cycle is isomorphism, the maximal number of perfect matching\nis 2 ∗n\n4 = n\n2 .\nReview the example 1 again, it is easy find that follow proposition.\nProposition 1. Given two perfect matching M1 and M2 in projector graph G,\nif code(M1) = code(M2), then the r(F −1(M1)) = r(F −1(M2)).\n5.1\nProof of Theorem 3\nNow let us proof the theorem 3.\nProof. Let G be a project balanced bipartition of D. According theorem 1, the\nΓ graph is equivalent to find a perfect match M in a project G.\nAccording to the lemma 4, the maximal number non isomorphism perfect\nmatching in G is only n.\nThus it is only need exactly enumerate all of non isomorphism perfect match-\ning M, then obtain the value = r(F −1(M)),if value = n −1, then the ei ∈M\nis also ei ∈C, where C ⊂D is a Hamiltonian cycle.\nSince the complexity of rank of matrix is O(n3), finding a simple cycle in\na component with degree 2 is O(n2), and obtaining a perfect matching of a\nbipartite graph is O((m + n)√n) < O(n2) [3]. Then all exactly algorithms need\nto calculate the n time o(n3). Thus the complexity is O(n4).\nSince the non isomorphism perfect matching comes from the coding of edges\nin the component of G, it is not easy implementation.\n7"},{"paragraph_id":"p17","order":17,"text":"Let us give two recursive equation to obtain a perfect matching M from G.\nSuppose there are k component G1, G2, . . . Gk in G where Gi is a component\nwith degree 2 and |Ei| ≥3.\nM ′ ="},{"paragraph_id":"p18","order":18,"text":"M(t) ⊗Gt, Gt is a cycle ;\nM(t),\notherwise.\n(10)\nM(t + 1) =\n M ′,\nif r(F −1(M ′)) > r(F −1(M(t))) ;\nM(t), otherwise.\n(11)\nwhere t ≤k −1, when t = 0, M(0) is the initial perfect matching from G.\nWhen r(F −1(M(t))) = n −1, According the theorem 1, the A = F −1(M(t))\nis a Hamiltonian cycle solution. If all of r(F −1(M(t))) < n −1, then there has\nno Hamiltonian cycle in D.\nSince the non isomorphism perfect matching M in G is poset, the function\nr(F −1(M)) in G is monotonic, so this approach is exactly approach.\nLet us give a example to illustrate the approach in detail.\nExample 1. Considering the digraph D in figure 1, then the projector graph G\nin figure 2.\nLet M(0) = {e1, e8, e22, e9, e10, e3, e20, e11, e19, e5, e18, e6, e17, e7, e15, e16}.\nThus the r(F −1(M(0)) = n−3. Let M ′ = r(F −1(M(0)⊗G3),then r(F −1(M ′) =\nn−4, thus M(1) = M(0) and then turn to G2,G1. At last it obtain the solution.\nConsidering the equation 11, let it substituted by following equations when\nr(M ′) = n −1 and t < k −1.\nM(t + 1) = M ′ if r(F −1(M ′)) ≥r(F −1(M(t)))\n(12)\nIt is obvious that all non-isomorphism Hamiltonian cycle could obtain by the\nrepeat check the equation 12 and the equation r(M ′) = n −1.\nIn conversely, if a Hamiltonian cycle of Γ digraphs is given, it represents a\nperfect matching M in its projector graph G. Thus the equation 12 and Theo-\nrem 3 follows a corollary.\nCorollary 3. Given a Hamiltonian Γ digraph, the complexity of determining\nanother non-isomorphism Hamiltonian cycle is polynomial time.\n5.2\nThe HCP in digraph with bound two\nLet us extend the Theorem 3 to digraphs with d+(v) ≤2 and d−(v) ≤2 in this\nsection.\nTheorem 6. The complexity of finding a Hamiltonian cycle existing or not in\ndigraphs with degree d+(v) ≤2 and d−(v) ≤2 is polynomial time.\n8"},{"paragraph_id":"p19","order":19,"text":"Proof. Suppose a digraph D(V, A) having a vertex vi is shown as figure 3, which\nis d(vi) = 2 ∧d−(vi) = 2\nFigure 3. A vertex with degree than 2\n✟✟✟✟✟\n✟\n✯\n❍❍❍❍❍\n❍\n❥♠\na1\na2\n✟✟✟✟✟\n✟\n✯\n❍❍❍❍❍\n❍\n❥\na3\na4\nLet us spilt this vertex to two vertices that one of vertex has degree with\nin degree 2 or out degree 1 , another vertex has degree with in degree 1 or out\ndegree 2 as shown in figure 4. Then the D is derived to a new Γ graph S.\nFigrue 4 A vertex in D is mapping to a vertex in Γ digraph\n✟✟✟✟✟\n✟\n✯\n✲❤\na1\na2\n✲\na3\n❤\na4\n✟✟✟✟✟\n✟\n✯\n✲\nIt is obvious that each vertex in the Γ graph S has increase 1 vertices and 1\narcs of D. Suppose the worst cases is each vertex in D has in degree 2 and out\ndegree 2, the total vertices in S has 2n vertices.\nAccording to the theorem 3, obtain a Hailtonian cycle L′ in S is no more\nthen O(n4), then the D will has a Hamiltonian cycle L′ = L ∩A.\n6\nDiscussion P versus NP\nThe P versus NP is a famous open problem in computer science and math-\nematics, which means to determine whether very language accepted by some\nnondeterministic algorithm in polynomial time is also accepted by some deter-\nministic algorithm in polynomial time [6]. Cook give a proposition for the P\nversus NP.\nProposition 2. If L is NP-complete and L ∈P, then P = NP.\nAccording above proposition and the result above section, P versus NP\nproblem has a answer.\nTheorem 7. P = NP\n9"},{"paragraph_id":"p20","order":20,"text":"Proof. As the result of [2], the complexity of HCP in digraph with bound two\nis NP −complete. According the theorem 6, the complexity of HCP in digraph\nwith bound two is also P, thus according to proposition 2, P = NP.\nIn fact, the [2] proves that 3SAT ⪯p HCP of Γ digraph, since 3SAT is a\nNPC problem, which also implies that P = NP.\n7\nConclusion\nAccording to the theorem 6, the complexity of determining a Hamiltonian cycle\nexistence or not in digraph with bound degree two is in polynomial time. And\naccording to the theorem 7, P versus NP problem has closed, P = NP.\nAcknowledgements\nThe author would like to thank Prof. Kaoru Hirota for valuable suggestions,\nthank Prof. Jørgen Bang-Jensen who called mine attention to the paper [2], and\nthank Andrea Moro for useful discussions.\nReferences\n1. Papadimitriou, C. H. Computational complexity , in Lawler, E. L., J. K. Lenstra, A.\nH. G. Rinnooy Kan, and D. B. Shmoys, eds., The Traveling Salesman Problem: A\nGuided Tour of Combinatorial Optimization. Wiley, Chichester, UK. (1985), 37–85\n2. J.Plesn ́ık,The NP-Completeness of the Hamiltonian Cycle Problem in Planar di-\ngraphs with degree bound two, Journal Information Processing Letters, Vol.8(1978),\n199–201\n3. J.E. Hopcroft and R.M. Karp , An n5/2 Algorithm for Maximum Matchings in\nBipartite Graphs . SIAM J. Comput. Vol.2, (1973), 225–231\n4. P. Hall, On representative of subsets, J. London Math. Soc. 10, (1935), 26–30\n5. Pearl, M, Matrix Theory and Finite Mathematics,McGraw-Hill, New York,(1973),\n332–404.\n6. Stephen Cook. The P Versus NP Problem ,”http://citeseer.ist.psu.edu/302888.html”\n,2000.\n10"}],"pages":[{"page":1,"text":"arXiv:0704.0309v3 [cs.CC] 13 Jul 2007\nThe Complexity of HCP in Digraps with Degree\nBound Two\nGuohun Zhu\nGuilin University of Electronic Technology,\nNo.1 Jinji Road,Guilin, Guangxi, 541004,P.R.China\nccghzhu@guet.edu.cn\nAbstract. The Hamiltonian cycle problem (HCP) in digraphs D with\ndegree bound two is solved by two mappings in this paper. The first\nbijection is between an incidence matrix Cnm of simple digraph and an\nincidence matrix F of balanced bipartite undirected graph G; The sec-\nond mapping is from a perfect matching of G to a cycle of D. It proves\nthat the complexity of HCP in D is polynomial, and finding a second\nnon-isomorphism Hamiltonian cycle from a given Hamiltonian digraph\nwith degree bound two is also polynomial. Lastly it deduces P = NP\nbase on the results.\n1\nIntroduction\nIt is well known that the Hamiltonian cycle problem(HCP) is one of the standard\nNP-complete problem [1]. As for digraphs, even when the digraphs on this case:\nplanar digraphs with indegree 1 or 2 and outdegree 2 or 1 respectively, it is still\nNP −Complete which is proved by J.Plesn ́ık [2].\nLet us named a simple strong connected digraphs with at most indegree 1 or\n2 and outdegree 2 or 1 as Γ digraphs. This paper solves the HCP of Γ digraphs\nwith following main results.\nTheorem 1. Given an incidence matrix Cnm of Γ digraph, building a mapping:F =\n C+\n−C−\n \n, then F is a incidence matrix of undirected balanced bipartite graph\nG(X, Y ; E), which obeys the following properties:\nc1. |X| = n,|Y | = n,|E| = m\nc2.\n∀xi ∈X ∧1 ≤d(xi) ≤2\n∀yi ∈Y ∧1 ≤d(yi) ≤2\nc3. G has at most n\n4 components which is length of 4.\nLet us named the undirected balanced bipartite graph G(X, Y : E) of Γ\ndigraph as projector graph."},{"page":2,"text":"Theorem 2. Let G be the projector graph of a Γ graph D(V, A), determining\na Hamiltonian cycle in Γ digraph is equivalent to find a perfect match M in\nG and r(C′) = n −1, where C′ is the incidence matrix of D′(V, L) ⊆D and\nL = {ai|ai ∈D ∧ei ∈M}.\nLet the each component of G corresponding to a boolean variable, a mono-\ntonic function f(M) is build to represents the number of component in D. Based\non this function, the maximum number of non-isomorphism perfect matching is\nlinear, thus complexity of Γ digraphs has a answer.\nTheorem 3. Given the incidence matrix Cnm of a Γ digraph , the complexity\nof finding a Hamiltonian cycle existing or not is O(n4)\nThe concepts of cycle and rank of graph are given in section 2. Then theorems\n1,2,3 are proved in sections 3,4,5 respectively. The last section discusses the P\nversus NP in more detail.\n2\nDefinition and properties\nThroughout this paper we consider the finite simple (un)directed graph D =\n(V, A) (G(V, E), respectively), i.e. the graph has no multi-arcs and no self loops.\nLet n and m denote the number of vertices V and arcs A (edges E, respectively),\nrespectively.\nAs conventional, let |S| denote the number of a set S. The set of vertices V\nand set of arcs of A of a digraph D(V, A) are denoted by V = {vi|1 ≤i ≤n}\nand A = {aj|(1 ≤j ≤m) ∧aj =< vi, vk >, (vi ̸= vk ∈V )} respectively,\nwhere < vi, vk > is a arc from vi to vk. Let the out degree of vertex vi denoted\nby d+(vi), which has the in degree by denoted as d−(vi) and has the degree\nd(vi) which equals d+(vi) + d−(vi). Let the N +(vi) = {vj| < vi, vj >∈A}, and\nN −(vi) = {vj| < vj, vi >∈A}.\nLet us define a forward relation ⊲⊳between two arcs as following, ai ⊲⊳aj =\nvk iffai =< vi, vk > ∧aj =< vk, vj >. It is obvious that |ai ⊲⊳ai| = 0 .\nA cycle L is a set of arcs (a1, a2, . . . , al) in a digraph D, which obeys two\nconditions:\nc1. ∀ai ∈L, ∃aj, ak ∈L \\ {ai}, ai ⊲⊳aj ̸= aj ⊲⊳ak ∈V\nc2. |\nS\nai̸=aj∈L\nai ⊲⊳aj| = |L|\nIf a cycle L obeys the following conditions, it is a simple cycle.\nc3. ∀L′ ⊂L, L′ does not satisfy both conditions c1 and c2.\nA Hamiltonian cycle L is also a simple cycle of length n = |V | ≥2 in digraph.\nAs for simplify, this paper given a sufficient condition of Hamiltonian cycle in\ndigraph.\nLemma 1. If a digraph D(V, A) include a sub graph D′(V, L) with following\ntwo properties, the D is a Hamiltonian graph.\n2"},{"page":3,"text":"c1. ∀vi ∈D′ →d+(vi) = 1 ∧d−(vi) = 1,\nc2. |L| = |V | ≥2 and D′ is a strong connected digraph.\nA graph that has at least one Hamiltonian cycle is called a Hamiltonian\ngraph. A graph G=(V ; E) is bipartite if the vertex set V can be partitioned into\ntwo sets X and Y (the bipartition) such that ∃ei ∈E, xj ∈X, ∀xk ∈X \\ {xj},\n(ei ⊲⊳xj ̸= ∅→ei ⊲⊳xk = ∅) (ei, Y , respectively). if |X| = |Y |, We call that\nG is a balanced bipartite graph. A matching M ⊆E is a collection of edges\nsuch that every vertex of V is incident to at most one edge of M, a matching of\nbalanced bipartite graph is perfect if |M| = |X|. Hopcroft and Karp shows that\nconstructs a perfect matching of bipartite in O((m + n)√n) [3]. The matching\nof bipartite has a relation with neighborhood of X.\nTheorem 4. [4] A bipartite graph G = (X, Y ; E) has a matching from X into\nY if and only if |N(S)| ≥S, for any S ⊆X.\nLemma 2. A even length of simple cycle consist of two disjoin perfect matching.\nTwo matrices representation for graphs are defined as follows.\nDefinition 1. [5] The incidence matrix C of undirected graph G is a two di-\nmensional n × m table, each row represents one vertex, each column represents\none edge, the cij in C are given by\ncij =\n \n1, if vi ∈ej;\n0, otherwise.\n(1)\nIt is obvious that every column of an incidence matrix has exactly two 1\nentries.\nDefinition 2. [5] The incidence matrix C of directed graph D is a two dimen-\nsional n × m table, each row represents one vertex, each column represents one\narc the cij in C are given by\ncij =\n \n \n \n1,\nif < vi, vi >⊲⊳aj = vi;\n−1, if aj ⊲⊳< vi, vi >= vi;\n0,\notherwise.\n(2)\nIt is obvious to obtain a corollary of the incidence matrix as following.\nCorollary 1. Each column of an incidence matrix of digraph has exactly one 1\nand one −1 entries.\nTheorem 5. [5] The C is the incidence matrix of a directed graph with k com-\nponents the rank of C is given by\nr(C) = n −k\n(3)\nIn order to convince to describe the graph D properties, in this paper, we\ndenotes the r(D) = r(C).\n3"},{"page":4,"text":"3\nDivided incidence matrix and Projector incidence\nmatrix\nFirstly, let us divided the matrix of C into two groups.\nC+ = {cij|cij ≥0 otherwise 0 }\n(4)\nC−= {cij|cij ≤0 otherwise 0 }\n(5)\nIt is obvious that the matrix of C+ represents the forward arc of a digraph\nand C−matrix represents the backward arc respectively. A corollary is deduced\nas following.\nCorollary 2. A digraph D = (V, A) is strong connected if and only if the rank\nof divided incidence matrix satisfies r(C+) = r(C−) = |V |.\nSecondly, let us combined the the C+ and C−as following matrix.\nF =\n \nC+\n−C−\n \n(6)\nIn more additional, let F represents as an incidence matrix of undirected\ngraph G(X, Y ; E). The F is named as projector incidence matrix\nof C and\nG is named as projector graph , where X represents the vertices V + of D, Y\nrepresents the vertices of V −respectively. In another words we build a mapping\nF : D →G and denotes it as G = F(D). So the F(D) has 2n vertices and\nm edges if D has n vertices and m arcs. We also build up a reverse mapping:\nF −1 : G →D When G is a projector graph. To simplify, we also denotes the\narcs ai = F −1(ei), v+\ni = F −1(xi) and v−\ni = F −1(yi).\n3.1\nProof of Theorem 1\nFirstly, let us prove the theorem 1.\nProof. c1. Since Γ digraph is strong connected, then each vertices of Γ digraph\nhas at least one forward arcs, each row of C+ has at least one 1 entries, and\nthe U represents the C+ , so\n|U| = n\nthe same principle of C−, each row of C−has at least one −1 entries, and\nthe V represents the C−, so\n|V | = n\nSince the columns of F equal to the columns of C,\n|E| = m\n4"},{"page":5,"text":"c2. Since the degree of each vi of Γ digraph is 1 ≤d+(vi) ≤2,\n∀ui ∈U ∧1 ≤d(ui) ≤2\nSince the degree of each vi of Γ digraph is 1 ≤d−(vi) ≤2,\n∀vi ∈V ∧1 ≤d(vi) ≤2\nc3. Let us prove by contradiction, suppose there are k > n\n4 components with\nlength of 4 in G. Since D is strong connected, according to the corollary 2,\nr(F) = 3n\n2 −q ≥r(C+) = n, where q ≥k is number of components (in-\ncluding k components with length of 4). Thus q ≤n\n2 , then there are only x\ncomponents without length 4, where x is\nx = q −k < n\n4\n(7)\nSuppose the remind x components with length of t (at least t vertices con-\nnected by some edges), then 4k + xt = 3n\n2 . So tx = 3n\n2 −4k < n\n2 . According\nto the equation 7, the t < 2. It is contradict that the D is strong connected.\n3.2\nThe cycle in digraph corresponding matching in projector graph\nSecondly, let us given the properties after mapping Hamiltonian cycle L of D\ninto the sub graph M of projector graph G.\nLemma 3. If a Hamiltonian cycle L of D mapping into a forest M of projector\ngraph G, the forest M consist of |L| number of trees which has only two node\nand one edge, and M has a unique perfect matching.\nProof. Let the Γ digraph D(V, A) has a sub digraph D′(V, L) which exists one\nHamiltonian cycle and |L| = n, the incidence matrix C of L could be permutation\nas follows.\nC =\n \n \n \n \n \n \n \n \n1\n0\n0\n. . . 0\n−1\n−1 1\n0\n. . . 0\n0\n0\n−1 1\n. . . 0\n0\n0\n0\n−1 . . . 0\n0\n0\n0\n0\n. . . 0\n0\n0\n0\n0\n. . . −1 1\n \n \n \n \n \n \n \n \n.\n(8)\nLet\nF =\n \nC+\n−C−\n \nIt is obvious that each row of F has only one 1 entry and each column of F\nhas two 1 entries.\nAccording to theorem 1, F represents a balanced bipartite graph G(X, Y ; E)\nthat each vertex has one edge connected, and each edge ei connect on vertex\nxi ∈X , another in Y , in another words, ∃ei ∈E xj ∈X,∀xk ∈X \\ {xj}, ei ⊲⊳\nxj ̸= ∅→ei ⊲⊳xk = ∅(ei, Y ,respectively). According the matching definition,\nM is a matching, since |E| = |L|, E is a perfect matching. and pair of vertices\nbetween X and Y only has one edge, so M is a forest, and each tree has only\ntwo node with one edge.\n5"},{"page":6,"text":"4\nProof of Theorem 2\nProof. ⇒Let the Γ digraph D(V, A) has a sub digraph D′(V, L) which is a\nHamiltonian cycle and |L| = n, let matrix C′ represents the incidence matrix of\nD′, so r(C′) = n −1; According to lemma 3, the projector graph F(D′) has a\nperfect matching, thus F(D) also has a perfect matching.\n⇐Let G(X, Y ; E) be a projector graph of the Γ graph D(V, A),M is a perfect\nmatching in G. Let D′(V, L) be a sub graph of D(V, A) and L = {ai|ai ∈D∧ei ∈\nM}. Since r(L) = n −1, D′(V, L) is a strong connected digraph. it deduces that\n∀vi ∈D′,d+(vi) ≥1 ∧d−(vi) ≥1. Suppose ∃vi ∈D′, d+(vi) > 1 (d−(vi) > 1\nrespectively), Since |M| = n, it deduces that Pn\ni=1 d(vi) > 2n + 1, which imply\nthat |L| > n. this is contradiction with L = {ai|ai ∈D ∧ei ∈M} and |M| = n.\nSo ∀vi ∈D′, d+(vi) = d−(vi) = 1, According the lemma 1, D′ has a Hamiltonian\ncycle.\n5\nNumber of perfect matching in projector graph\nLet us considering the number of perfect matching in G . Firstly, let us consid-\nering a example as shown in figure 1.\nFigure 1. Original Digraph D\n✻\n♠\na8\n✲\n♠a1\n✛\n♠\na22\n❄\n✲\n♠a2\na9\n✛\n✻\n♠\na21\na10\n♠✲\na3\n✛\n♠\na20\n❄\n✲\n♠a4\na11\n✛\n✻\n♠\na12\na19\n♠✲\na5\n✛\n♠\na18\n❄\n✲\n♠a6\na13\n✛\n✻\n♠\na17\na14\n♠✲\na7\n✛\n♠\na16\n♠\n❄a15\nThen the projector graph is shown in figure 2.\nFigure 2. Projector graph G\n♠\n♠\ne1\n♠\n♠\ne8\n♠\n♠\ne22\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne9\ne2\nG1\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne10\ne21\n♠\n♠\ne3\n♠\n♠\ne20\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne11\ne4\nG2\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne19\ne12\n♠\n♠\ne5\n♠\n♠\ne18\n✁\n✁\n✁\n✁\n✁✁\n♠\n♠\ne6\ne13\nG3\n❆\n❆\n❆\n❆\n❆\n❆\n♠\n♠\ne17\ne14\n. . .\nGiven a perfect matching M, each component(cycle) in G has two partition\nedges belong to M. Let us code component Gi which |Gi| > 2 and matching M\nto a binary variable.\nGi =\n 1, if Gi ∩M = {ej, ek, . . .};\n0, if Gi ∩M = {el, eq, . . .}.\n(9)\n6"},{"page":7,"text":"Now there are two cases for the number of perfect matching.\nLabel edge. In that cases, the Code(M1) = {0, 0, 1} is different with Code(M2) = {0, 1, 0}.\nIf there are k number of components(cycles), then there are 2k perfect match-\ning.\nUnlabel edge. In that cases, the Code(M1) = {0, 0, 1} is isomorphic to Code(M2) = {0, 1, 0}.\nThe same principle that Code(M3) = {0, 1, 1} is isomorphic to Code(M4) =\n{1, 1, 0} but is not isomorphic to Code(M1).\nThen let us summary the maximal number of perfect matching in these two\ncases.\nLemma 4. The maximal number of labeled perfect matching in a projector graph\nG is 2\nn\n4 , but the maximal number of unlabeled perfect matching in a projector\ngraph G is n\n2 .\nProof. According to the theorem 1, there at most n\n4 components with a compo-\nnents which is length of k = 4. When k=2, there are only one perfect matching\nin G; When k = 4, there are n\n4 components which is C4, and so on when k = 6,\nthere are n\n6 components which is C6, etc, so on. According to the lemma 2, each\nsimple cycle has divided the perfect matching into two class. So maximal number\nperfect matching in the non isomorphism cycle which is 2\nn\n4 . Since in unlabeled\ncases, every C4 cycle is isomorphism, the maximal number of perfect matching\nis 2 ∗n\n4 = n\n2 .\nReview the example 1 again, it is easy find that follow proposition.\nProposition 1. Given two perfect matching M1 and M2 in projector graph G,\nif code(M1) = code(M2), then the r(F −1(M1)) = r(F −1(M2)).\n5.1\nProof of Theorem 3\nNow let us proof the theorem 3.\nProof. Let G be a project balanced bipartition of D. According theorem 1, the\nΓ graph is equivalent to find a perfect match M in a project G.\nAccording to the lemma 4, the maximal number non isomorphism perfect\nmatching in G is only n.\nThus it is only need exactly enumerate all of non isomorphism perfect match-\ning M, then obtain the value = r(F −1(M)),if value = n −1, then the ei ∈M\nis also ei ∈C, where C ⊂D is a Hamiltonian cycle.\nSince the complexity of rank of matrix is O(n3), finding a simple cycle in\na component with degree 2 is O(n2), and obtaining a perfect matching of a\nbipartite graph is O((m + n)√n) < O(n2) [3]. Then all exactly algorithms need\nto calculate the n time o(n3). Thus the complexity is O(n4).\nSince the non isomorphism perfect matching comes from the coding of edges\nin the component of G, it is not easy implementation.\n7"},{"page":8,"text":"Let us give two recursive equation to obtain a perfect matching M from G.\nSuppose there are k component G1, G2, . . . Gk in G where Gi is a component\nwith degree 2 and |Ei| ≥3.\nM ′ =\n \nM(t) ⊗Gt, Gt is a cycle ;\nM(t),\notherwise.\n(10)\nM(t + 1) =\n M ′,\nif r(F −1(M ′)) > r(F −1(M(t))) ;\nM(t), otherwise.\n(11)\nwhere t ≤k −1, when t = 0, M(0) is the initial perfect matching from G.\nWhen r(F −1(M(t))) = n −1, According the theorem 1, the A = F −1(M(t))\nis a Hamiltonian cycle solution. If all of r(F −1(M(t))) < n −1, then there has\nno Hamiltonian cycle in D.\nSince the non isomorphism perfect matching M in G is poset, the function\nr(F −1(M)) in G is monotonic, so this approach is exactly approach.\nLet us give a example to illustrate the approach in detail.\nExample 1. Considering the digraph D in figure 1, then the projector graph G\nin figure 2.\nLet M(0) = {e1, e8, e22, e9, e10, e3, e20, e11, e19, e5, e18, e6, e17, e7, e15, e16}.\nThus the r(F −1(M(0)) = n−3. Let M ′ = r(F −1(M(0)⊗G3),then r(F −1(M ′) =\nn−4, thus M(1) = M(0) and then turn to G2,G1. At last it obtain the solution.\nConsidering the equation 11, let it substituted by following equations when\nr(M ′) = n −1 and t < k −1.\nM(t + 1) = M ′ if r(F −1(M ′)) ≥r(F −1(M(t)))\n(12)\nIt is obvious that all non-isomorphism Hamiltonian cycle could obtain by the\nrepeat check the equation 12 and the equation r(M ′) = n −1.\nIn conversely, if a Hamiltonian cycle of Γ digraphs is given, it represents a\nperfect matching M in its projector graph G. Thus the equation 12 and Theo-\nrem 3 follows a corollary.\nCorollary 3. Given a Hamiltonian Γ digraph, the complexity of determining\nanother non-isomorphism Hamiltonian cycle is polynomial time.\n5.2\nThe HCP in digraph with bound two\nLet us extend the Theorem 3 to digraphs with d+(v) ≤2 and d−(v) ≤2 in this\nsection.\nTheorem 6. The complexity of finding a Hamiltonian cycle existing or not in\ndigraphs with degree d+(v) ≤2 and d−(v) ≤2 is polynomial time.\n8"},{"page":9,"text":"Proof. Suppose a digraph D(V, A) having a vertex vi is shown as figure 3, which\nis d(vi) = 2 ∧d−(vi) = 2\nFigure 3. A vertex with degree than 2\n✟✟✟✟✟\n✟\n✯\n❍❍❍❍❍\n❍\n❥♠\na1\na2\n✟✟✟✟✟\n✟\n✯\n❍❍❍❍❍\n❍\n❥\na3\na4\nLet us spilt this vertex to two vertices that one of vertex has degree with\nin degree 2 or out degree 1 , another vertex has degree with in degree 1 or out\ndegree 2 as shown in figure 4. Then the D is derived to a new Γ graph S.\nFigrue 4 A vertex in D is mapping to a vertex in Γ digraph\n✟✟✟✟✟\n✟\n✯\n✲❤\na1\na2\n✲\na3\n❤\na4\n✟✟✟✟✟\n✟\n✯\n✲\nIt is obvious that each vertex in the Γ graph S has increase 1 vertices and 1\narcs of D. Suppose the worst cases is each vertex in D has in degree 2 and out\ndegree 2, the total vertices in S has 2n vertices.\nAccording to the theorem 3, obtain a Hailtonian cycle L′ in S is no more\nthen O(n4), then the D will has a Hamiltonian cycle L′ = L ∩A.\n6\nDiscussion P versus NP\nThe P versus NP is a famous open problem in computer science and math-\nematics, which means to determine whether very language accepted by some\nnondeterministic algorithm in polynomial time is also accepted by some deter-\nministic algorithm in polynomial time [6]. Cook give a proposition for the P\nversus NP.\nProposition 2. If L is NP-complete and L ∈P, then P = NP.\nAccording above proposition and the result above section, P versus NP\nproblem has a answer.\nTheorem 7. P = NP\n9"},{"page":10,"text":"Proof. As the result of [2], the complexity of HCP in digraph with bound two\nis NP −complete. According the theorem 6, the complexity of HCP in digraph\nwith bound two is also P, thus according to proposition 2, P = NP.\nIn fact, the [2] proves that 3SAT ⪯p HCP of Γ digraph, since 3SAT is a\nNPC problem, which also implies that P = NP.\n7\nConclusion\nAccording to the theorem 6, the complexity of determining a Hamiltonian cycle\nexistence or not in digraph with bound degree two is in polynomial time. And\naccording to the theorem 7, P versus NP problem has closed, P = NP.\nAcknowledgements\nThe author would like to thank Prof. Kaoru Hirota for valuable suggestions,\nthank Prof. Jørgen Bang-Jensen who called mine attention to the paper [2], and\nthank Andrea Moro for useful discussions.\nReferences\n1. Papadimitriou, C. H. Computational complexity , in Lawler, E. L., J. K. Lenstra, A.\nH. G. Rinnooy Kan, and D. B. Shmoys, eds., The Traveling Salesman Problem: A\nGuided Tour of Combinatorial Optimization. Wiley, Chichester, UK. (1985), 37–85\n2. J.Plesn ́ık,The NP-Completeness of the Hamiltonian Cycle Problem in Planar di-\ngraphs with degree bound two, Journal Information Processing Letters, Vol.8(1978),\n199–201\n3. J.E. Hopcroft and R.M. Karp , An n5/2 Algorithm for Maximum Matchings in\nBipartite Graphs . SIAM J. Comput. Vol.2, (1973), 225–231\n4. P. Hall, On representative of subsets, J. London Math. Soc. 10, (1935), 26–30\n5. Pearl, M, Matrix Theory and Finite Mathematics,McGraw-Hill, New York,(1973),\n332–404.\n6. Stephen Cook. The P Versus NP Problem ,”http://citeseer.ist.psu.edu/302888.html”\n,2000.\n10"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"with degree bound two is also polynomial. Lastly it deduces P = NP","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"Theorem 1. Given an incidence matrix Cnm of Γ digraph, building a mapping:F =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"c1. |X| = n,|Y | = n,|E| = m","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"G and r(C′) = n −1, where C′ is the incidence matrix of D′(V, L) ⊆D and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"L = {ai|ai ∈D ∧ei ∈M}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"Throughout this paper we consider the finite simple (un)directed graph D =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"and set of arcs of A of a digraph D(V, A) are denoted by V = {vi|1 ≤i ≤n}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"and A = {aj|(1 ≤j ≤m) ∧aj =< vi, vk >, (vi ̸= vk ∈V )} respectively,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"d(vi) which equals d+(vi) + d−(vi). Let the N +(vi) = {vj| < vi, vj >∈A}, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"N −(vi) = {vj| < vj, vi >∈A}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"Let us define a forward relation ⊲⊳between two arcs as following, ai ⊲⊳aj =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"vk iffai =< vi, vk > ∧aj =< vk, vj >. It is obvious that |ai ⊲⊳ai| = 0 .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"c1. ∀ai ∈L, ∃aj, ak ∈L \\ {ai}, ai ⊲⊳aj ̸= aj ⊲⊳ak ∈V","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"ai̸=aj∈L","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"ai ⊲⊳aj| = |L|","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"A Hamiltonian cycle L is also a simple cycle of length n = |V | ≥2 in digraph.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"c1. ∀vi ∈D′ →d+(vi) = 1 ∧d−(vi) = 1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"c2. |L| = |V | ≥2 and D′ is a strong connected digraph.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"graph. A graph G=(V ; E) is bipartite if the vertex set V can be partitioned into","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"(ei ⊲⊳xj ̸= ∅→ei ⊲⊳xk = ∅) (ei, Y , respectively). if |X| = |Y |, We call that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"balanced bipartite graph is perfect if |M| = |X|. Hopcroft and Karp shows that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"Theorem 4. [4] A bipartite graph G = (X, Y ; E) has a matching from X into","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"cij =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"cij =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"if < vi, vi >⊲⊳aj = vi;","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"−1, if aj ⊲⊳< vi, vi >= vi;","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"r(C) = n −k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"denotes the r(D) = r(C).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"C+ = {cij|cij ≥0 otherwise 0 }","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"C−= {cij|cij ≤0 otherwise 0 }","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"Corollary 2. A digraph D = (V, A) is strong connected if and only if the rank","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"of divided incidence matrix satisfies r(C+) = r(C−) = |V |.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"F =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"F : D →G and denotes it as G = F(D). So the F(D) has 2n vertices and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"arcs ai = F −1(ei), v+","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"i = F −1(xi) and v−","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"i = F −1(yi).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"|U| = n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"|V | = n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"|E| = m","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"r(F) = 3n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"2 −q ≥r(C+) = n, where q ≥k is number of components (in-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"x = q −k < n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"nected by some edges), then 4k + xt = 3n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"2 . So tx = 3n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"Hamiltonian cycle and |L| = n, the incidence matrix C of L could be permutation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"C =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"F =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"xj ̸= ∅→ei ⊲⊳xk = ∅(ei, Y ,respectively). According the matching definition,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"M is a matching, since |E| = |L|, E is a perfect matching. and pair of vertices","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"Hamiltonian cycle and |L| = n, let matrix C′ represents the incidence matrix of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"D′, so r(C′) = n −1; According to lemma 3, the projector graph F(D′) has a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"matching in G. Let D′(V, L) be a sub graph of D(V, A) and L = {ai|ai ∈D∧ei ∈","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"M}. Since r(L) = n −1, D′(V, L) is a strong connected digraph. it deduces that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"respectively), Since |M| = n, it deduces that Pn","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"i=1 d(vi) > 2n + 1, which imply","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"that |L| > n. this is contradiction with L = {ai|ai ∈D ∧ei ∈M} and |M| = n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"So ∀vi ∈D′, d+(vi) = d−(vi) = 1, According the lemma 1, D′ has a Hamiltonian","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"Gi =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"1, if Gi ∩M = {ej, ek, . . .};","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"0, if Gi ∩M = {el, eq, . . .}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"Label edge. In that cases, the Code(M1) = {0, 0, 1} is different with Code(M2) = {0, 1, 0}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"Unlabel edge. In that cases, the Code(M1) = {0, 0, 1} is isomorphic to Code(M2) = {0, 1, 0}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"The same principle that Code(M3) = {0, 1, 1} is isomorphic to Code(M4) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"nents which is length of k = 4. When k=2, there are only one perfect matching","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"in G; When k = 4, there are n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"4 components which is C4, and so on when k = 6,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"4 = n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"if code(M1) = code(M2), then the r(F −1(M1)) = r(F −1(M2)).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"ing M, then obtain the value = r(F −1(M)),if value = n −1, then the ei ∈M","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"M ′ =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"M(t + 1) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"where t ≤k −1, when t = 0, M(0) is the initial perfect matching from G.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"When r(F −1(M(t))) = n −1, According the theorem 1, the A = F −1(M(t))","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"Let M(0) = {e1, e8, e22, e9, e10, e3, e20, e11, e19, e5, e18, e6, e17, e7, e15, e16}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"Thus the r(F −1(M(0)) = n−3. Let M ′ = r(F −1(M(0)⊗G3),then r(F −1(M ′) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"n−4, thus M(1) = M(0) and then turn to G2,G1. At last it obtain the solution.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"r(M ′) = n −1 and t < k −1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"M(t + 1) = M ′ if r(F −1(M ′)) ≥r(F −1(M(t)))","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"repeat check the equation 12 and the equation r(M ′) = n −1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"is d(vi) = 2 ∧d−(vi) = 2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"then O(n4), then the D will has a Hamiltonian cycle L′ = L ∩A.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"Proposition 2. If L is NP-complete and L ∈P, then P = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"Theorem 7. P = NP","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"with bound two is also P, thus according to proposition 2, P = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"NPC problem, which also implies that P = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"according to the theorem 7, P versus NP problem has closed, P = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":18742,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}