1.
Introduction
In this paper, we use G=(V(G),E(G)) to denote a graph with the vertex set V(G)={v1,v2,⋯,vn} and the edge set E(G). Let e(G)=m=|E(G)| be the number of edges of the graph G. For any vi∈ V(G), di=dvi=dG(vi) represents the degree of vi. Let (d1,d2,⋯,dn) be the degree sequence of G, where d1≤d2≤⋯≤dn. Let δ or δ(G) be the minimum degree of G, and NG(v) be the set of neighbors of a vertex v in G. The complete graph of order n is denoted by Kn, and the complete bipartite graph with the partite sets X and Y with |X|=m and |Y|=n is denoted by Km,n. The disjoint union of G and H, denoted by G+H, is the graph with vertex set V(G)∪V(H) and edge set E(G)∪(H). In particular, if G1=G2=⋯=Gk, then let kG1=G1+G2+⋯+Gk. Similarly, we use G∨H to denote the join of G and H. The complete graph on n−1 vertices together with an isolated vertex v is denoted by Kn−1+v.
Let A(G)=[aij] be the adjacency matrix of G, where aij=1 if vi is adjacent to vj, aij=0 otherwise. We use μ(G) to denote the maximum eigenvalue of A(G), which is called the spectral radius of G. Let D(G) be the diagonal degree matrix of G, and Q(G)=D(G)+A(G) be the signless Laplacian matrix of G. Let q(G) be the maximum eigenvalue of Q(G), which is called the signless Laplacian spectral radius.
We call the cycle (path) containing all vertices of G the Hamilton cycle (path) of G. If graph G contains a Hamiltonian cycle (path), then G is a Hamiltonian traceable graph. If for any positive integer k(3≤k≤n), G contains a cycle of length k, then G is called a pancyclic graph. Obviously, a bipartite graph is not a pancyclic graph. In addition, a Hamilton cycle graph is certainly Hamiltonian, but the converse is incorrect. Determining whether a given graph is a Hamilton graph or not is one of the most difficult classical problems in graph theory. In fact, it is an NP−complete problem.
In recent years, more and more worldwide researchers use the spectral theory of graphs to solve this problem. First, Fiedler and Nikiforov [1] established a sufficient condition on spectral radius for the existence of a graph with Hamilton paths and Hamilton cycles. Yu and Fan [2] used the spectral radius of the adjacency matrix or signless Laplacian matrix of the graph or its complement to give a sufficient condition for the graph to be Hamilton-connected. Lu et al. [3] gave the spectral sufficient conditions for a bipartite graph to be a Hamilton graph. Many scholars have studied similar problems under different spectral conditions since that time, see [4,5,6,7,8,9,10]. Recently, Yu et al. [11] has discussed for the first time the spectral sufficient conditions for a graph with a minimum degree δ(G)≥2 to be a pancyclic graph. In this paper, a new edge sufficient condition for a graph with a minimum degree δ(G)≥3 to be a pancyclic graph is established by using a similar method. Then, a sufficient condition on (signless Laplacian) spectral radius of a graph to be a pancyclic graph is given based on the edge sufficient condition.
2.
Preliminary
Given a graph G with order n, a vector X∈Rn, is called to be defined on G, if there is a 1-1 mapping φ from V(G) to vector, simply written Xu=φ(u).
X is defined on G if X is the eigenvector of A(G)(Q(G)). Let Xu denote the term of X corresponding to vertex u. One can find that when λ is an eigenvalue of G corresponding to the eigenvector, X if and only if X≠0,
The equation above is the eigen-equation of G. One can find when q is an signless Laplacian eigenvalue of G corresponding to the eigenvector X if and only if X≠0,
The equation above is the signless Laplacian eigen-equation of G.
Lemma 2.1 [12] Let G be a graph with n vertices, and its degree sequence is d1≤d2≤⋯≤dn. G is a pancyclic graph or bipartite graph if for all positive integers k there is always dn−k≥n−k and dk≤k<n/2.
Lemma 2.2 [13] Let G be a graph with n vertices and m edges, then
and the equality holds if and only if G=Kn or G=K1,n−1.
Lemma 2.3 [14] Let G be a graph with n vertices and m edges, then
and the equal sign is true if and only if G=Kn or G=K1,n−1 when G is connected. If G is not connected, the equal sign is true if and only if G=Kn−1+v.
Lemma 2.4 [15] Let G be a Hamiltonian graph and satisfy e(G)≥n2/4, then G is either a complete bipartite graph Kn2,n2 (where n is even), or a pancyclic graph.
Lemma 2.5 [16] If a Hamiltonian graph G of order n satisfies
then G is a pancyclic graph or G is a bipartite graph.
3.
Main results
Theorem 3.1 Let G be a connected graph with n(n≥5) vertices and m edges, and its minimum degree δ(G)≥3. If
then G is a pancyclic graph or a bipartite graph or G∈NP1={(3K1+Kn−6)∨K3, (4K1+K3)∨K4, (6K1+K2)∨K6, 9K1∨K8, 8K1∨K7, (K1+K1,7)∨K6, K2,8∨K5, K6∨(2K1+K1,6), K4∨2K1∨(K2+K1,6), K5∨(K1+K1,6), (K2+K1,5)∨K5, (2K1∨K4)∨7K1, (K1+K2+K1,4)∨K5, (2K1+K1,5)∨K5, (K1+K2∨6K1)∨K4, (K1,5+K2)∨2K1∨K3, 7K1∨K6, K5∨(2K1+K1,3+K2), K4∨[K1+K1∨(K1,4+K2)], K4∨(K1+2K1∨6K1), K3∨K3,7, K5∨(3K1+K1,4), K3∨2K1∨(K1+K1,4+K2), K4∨[K1∨(K1+K1,5)+K1], K3∨2K1∨(2K1+K1,5), (5K1+K2)∨K5, K4∨(K1∨(4K1+K2)+K1), 7K1∨K5, (5K1+K2)∨2K1∨K3, 6K1∨K5, K4∨(K1,4+K2), K4∨(K1+K1,5), K3∨K2,6, K4∨(K1+K1,3+K2), K4∨(2K1+K1,4), K3∨[(K2∨5K1)+K1], K2∨2K1∨(K1,4+K2), K4∨(K2+K1,2+2K1), K3∨(K1+K2,5), K2∨K3,6, K4∨(K1,3+3K1), K2∨2K1∨(K1+K1,3+K2), [K1∨(K1,4+K1)+K1]∨K3, K2∨2K1∨(2K1+K1,4), (4K1+K2)∨K4, K3∨[K1+K1∨(3K1+K2)], 6K1∨K4, (4K1+K2)∨2K1∨K2, 5K1∨K4, (K1,3+K2)∨K3, (K1+K1,4)∨K3, 5K1∨2K1∨K2, (K1+K1,2+K2)∨K3, (2K1+K1,3)∨K3, (K1,3+K2)∨2K1∨K1}.
Proof: Suppose that G is neither a pancyclic graph nor a bipartite graph. According to Lemma 2.1, there is a positive integer k makes 3≤dk≤k<n/2 and dn−k≤n−k−1 hold at the same time. Then we have
thus
Since
thus (k−3)(2n−3k−10)≤0. Next, the following two cases are discussed.
Case 1 Assume that (k−3)(2n−3k−10)=0, i.e., k=3 or 2n−3k−10=0. All the above inequalities are equal and m=(n−32)+9.
Case 1.1 If k=3, then G is a graph with d1=d2=d3=3,d4=d5=⋯=dn−3=n−4,dn−2=dn−1=dn=n−1. Three vertices with degree n−1 must be connected to other vertices, then a K3 is obtained. The three vertices with degree 3 are not connected with other vertices, and a 3K1 is obtained. The remaining n−6 vertices with degree n−4 must be connected to each other to ensure that the degree is n−4, so they induce a Kn−6. According to the above analysis, the graph G is (3K1+Kn−6)∨K3.
Case 1.2 If 2n−3k−10=0, then we get n≤19 because 2n−103=k<n2, and hence n=11, k=4, or n=14, k=6, or n=17, k=8. The corresponding permissible graphic sequences are (4,4,4,4,6,6,6,10,10,10,10), (6,6,6,6,6,6,7,7,13,13,13,13,13,13), (8,8,8,8,8,8,8,8,8,16,16,16,16,16,16,16,16). Analysis shows that the graphs G are (4K1+K3)∨K4, (6K1+K2)∨K6 and 9K1∨K8 respectively.
Case 2 Assume that (k−3)(2n−3k−10)<0, i.e., k>3 and 2n−3k−10<0. Because 2n−10<3k<32n, then, n=15,k=7 or n=13,k=6 or n=12,k=5 or n=11,k=5 or n=10,k=4 or n=9,k=4.
Case 2.1 If n=15, k=7. According to equation (3.3), we have 150≤Σ15i=1di≤154. All possible degree sequences and their corresponding graphs are shown in Table 1.
Case 2.2 If n=13, k=6. According to equation (3.3), we have 108≤Σ13i=1di≤114. All possible degree sequences and their corresponding graphs are shown in Table 2.
Case 2.3 If n=12, k=5. According to equation (3.3), we have 90≤Σ12i=1di≤92. All possible degree sequences and their corresponding graphs are shown in Table 3.
Case 2.4 If n=11, k=5. According to equation (3.3), we have 74≤Σ11i=1di≤80. All possible degree sequences and their corresponding graphs are shown in Table 4.
Case 2.5 If n=10, k=4. According to equation (3.3), we have 60≤Σ10i=1di≤62. All possible degree sequences and their corresponding graphs are shown in Table 5.
Case 2.6 If n=9, k=4. According to equation (3.3), we have 48≤Σ9i=1di≤52. All possible degree sequences and their corresponding graphs are shown in Table 6.
Next, we consider the pancyclicity of these graphs. For convenience, we put the maximum length cycle L(G) of G in the appendix. In the following description, when it comes to the description of the graphs, we will use the sequence number of the graphs in the appendix.
As can be seen from the appendix, according to Lemma 2.5, all the graphs of No.5-18 in the appendix are pancyclic. Others belong to NP1 and they are neither pancyclic nor bipartite.
The proof is complete.
Theorem 3.2 Let G be a connected graph with n(n≥5) vertices and m edges, and its minimum degree δ(G)≥3. If
then G is a pancyclic graph or a bipartite graph.
Proof: Suppose that G is neither a pancyclic graph nor a bipartite graph. Because a complete graph is a pancyclic graph and δ(K1,n−1)=1, then G cannot be a complete graph or K1,n−1. In such a situation, the equation in Lemma 2.2 is invalid. By Lemma 2.2, √n2−8n+31≤μ(G)<√2m−n+1, i.e., m>(n−32)+9.
By Theorem 3.1, G is a pancyclic graph or a bipartite graph or G∈NP1. It can be calculated that (3K1+Kn−6)∨K3, (4K1+K3)∨K4, (6K1+K2)∨K6, 9K1∨K8 satisfy m=(n−32)+9, which is contradiction. Next, we will study whether the graphs of No.19-69 satisfy the spectral radius condition in the Theorem 3.2. Note that G∈NP2 and NP2 is the set of graphs of No.19-69 in the appendix.
Take (2K1∨K4)∨7K1 as an example:
Let X=(X1,X2,………,X13)T be the eigenvector corresponding to μ(G), where Xi(1≤i≤2) corresponds to the vertex of degree 11, Xi(3≤i≤6) corresponds to the vertex of degree 12 and Xi(7≤i≤13) corresponds to the vertex of degree 6. By eigen-equation (2.1), then we have
Transform the above equations into the matrix equation (A′(G)−μ(G)I)X′=0, where X′=(X1,X3,X7)T and
Let f(x)=|xI−A′(G)|=x3−3x2−50x−70, then the maximum root of f(x)=0 is μ(G)=9.235. It can be calculated that μ(G)<√132−8×13+31, which is contradiction. The remaining graphs are studied in the same way, and the results are shown in Table 7.
From Table 7, all graphs in NP2 satisfy μ(G)<√n2−8n+31, a contradiction. The proof is complete.
Theorem 3.3 Let G be a connected graph with n(n≥5) vertices and minimum degree δ(G)≥3. If
then G is a pancyclic graph or a bipartite graph or 7K1∨K6 or 6K1∨K5 or 5K1∨K4.
Proof: Suppose that G is neither a pancyclic graph nor a bipartite graph. Because a complete graph is a pancyclic graph and δ(K1,n−1)=1, then G cannot be a complete graph or K1,n−1. In such a situation, the equation in Lemma 2.3 is invalid. By Lemma 2.3, 24n−1+2n−8≤q(G)<2mn−1+n−2, i.e., m>(n−32)+9.
By Theorem 3.1, G is a pancyclic graph or a bipartite graph or G∈NP1. It can be calculated that the graphs of (3K1+Kn−6)∨K3, (4K1+K3)∨K4, (6K1+K2)∨K6, 9K1∨K8 satisfy m=(n−32)+9, which is contradiction. Next, we will study whether the graphs of No.19-69 satisfy the signless Laplacian spectral radius condition in the Theorem 3.3. Note that G∈NP2.
Take (2K1∨K4)∨7K1 as an example:
Let X=(X1,X2,………,X13)T be the eigenvector corresponding to μ(G), where Xi(1≤i≤2) corresponds to the vertex of degree 11, Xi(3≤i≤6) corresponds to the vertex of degree 12 and Xi(7≤i≤13) corresponds to the vertex of degree 6. By eigen-equation (2.2), then we have
Transform the above equations into the matrix equation (Q′(G)−q(G)I)X′=0, where X′=(X1,X3,X7)T and
Let f(x)=|xI−Q′(G)|=x3−32x2+271x−536, then the maximum root of g(x)=0 is q(G)=19.528. It can be calculated that q(G)<2413−1+2×13−8, which is contradiction. The remaining graphs are studied in the same way, and the results are shown in Table 7.
From Table7, all graphs in NP2 except G=7K1∨K6 and G=6K1∨K5 and G=5K1∨K4 satisfy q(G)<24n−1+2n−8, a contradiction.
The proof is complete.
Acknowledgments
This work is supported by the Natural Science Foundation of China(11871077), the NSF of Anhui Province(1808085MA04), the NSF of Department of Education of Anhui Province(KJ2017A362), and the NSF of Anhui Province(KJ2018A0964), the Deanship of Scientific Research (DSR) at King Abdulaziz University under grant no (RG-19-135-38), the 2019 Teaching and research project of Hefei Preschool Education College (hyzzd2019003).
Conflict of interest
The author declare that he has no competing interest.
Appendix