The stability method, eigenvalues and cycles of consecutive lengths
aa r X i v : . [ m a t h . C O ] F e b The stability method, eigenvalues and cycles ofconsecutive lengths
Binlong Li ∗ Bo Ning † Abstract:
Woodall proved that for a graph G of order n ≥ k + 3 where k ≥ e ( G ) ≥ (cid:0) n − k − (cid:1) + (cid:0) k +22 (cid:1) + 1 then G contains a C ℓ for each ℓ ∈ [3 , n − k ]. In this article, weprove a stability result of this theorem. As a byproduct, we give complete solutions to twoproblems in [12]. Our second part is devoted to an open problem by Nikiforov: what is themaximum C such that for all positive ε < C and sufficiently large n , every graph G of order n with spectral radius ρ ( G ) > q ⌊ n ⌋ contains a cycle of length ℓ for every ℓ ≤ ( C − ε ) n . Weprove that C ≥ by a method different from previous ones, improving the existing bounds.We also derive an Erd˝os-Gallai type edge number condition for even cycles, which may beof independent interest. Keywords: stability method; large cycles; spectral radius; signless Laplacianspectral radius; cycles of consecutive lengths; spectral inequalityAMS classification: 05C50; 05C35
In 1970s, Erd˝os [8] asked how many edges are needed in a graph on n vertices, in orderto ensure the existence of a cycle of length exactly n − r ? Woodall [26] determined theTur´an numbers of large cycles C ℓ for ℓ ∈ [ ⌊ n +32 ⌋ , n ] as follows. Theorem 1.1 (Woodall [26]) . Let G be a graph of order n ≥ k + 3 where k ≥ is aninteger. If e ( G ) ≥ (cid:0) n − k − (cid:1) + (cid:0) k +22 (cid:1) + 1 , then G contains a C ℓ for each l ∈ [3 , n − k ] . Define Γ as a graph which consists of a clique of n − k − k + 2vertices sharing one common vertex. The graph Γ shows Woodall’s theorem is sharp.In this paper, we shall first consider stability results of Woodall’s theorem following therecent trend. So it is natural to recall history of the related stability results of extremalresults on cycles.For non-hamiltonian graphs of order n with given minimum degree, Erd˝os [7] provedthe following result in 1962. Theorem 1.2 (Erd˝os [7]) . Let G be a graph on n vertices with δ ( G ) ≥ k where ≤ k ≤⌊ n − ⌋ . If G is non-hamiltonian then e ( G ) ≤ max ((cid:18) n − k (cid:19) + k , (cid:18) n − ⌊ n − ⌋ (cid:19) + (cid:22) n − (cid:23) ) . ∗ Department of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi 710072,P.R. China. Email: [email protected]. Partially supported by the NSFC grant (No. 11601429). † Corresponding author. College of Computer Science, Nankai University, Tianjin 300071, P.R. China.Email: [email protected]. Partially supported by the NSFC grant (No. 11601379).
1s a key lemma to attack the following problem: Among all non-hamiltonian graphsof order n which have minimum degree at least k , characterize the class of graphs whichattain the maximum spectral radius, the authors [17] proved a stability result of Erd˝os’theorem. This result was also proved by F¨uredi, Kostochka and Luo [9], independently. Theorem 1.3 (Li and Ning [17], F¨uredi, Kostochka and Luo [9]) . Let G be a graph oforder n ≥ k + 5 . If δ ( G ) ≥ k ≥ and e ( G ) > (cid:18) n − k − (cid:19) + ( k + 1) , then G is hamiltonian, unless G is a subgraph of K k ∨ ( kK + K n − k ) or a subgraph of K ∨ ( K n − k − + K k ) . In 1977, Kopylov [16] determined a sharp edge condition for the circumference of a2-connected graph. In 2016, F¨uredi, Kostochka, and Verstra¨ete [11] proved a stabilityversion of Erd˝os-Gallai theorem, and finally (together with Luo) [10] completed the sta-bility version of Kopylov’s theorem [16]. In fact, Kopylov’s theorem is a special case of aconjecture due to Woodall [26], which refers to the sharp edge condition for circumferenceof a 2-connected graph with given minimum degree. Recently, Ma and Ning [18] proved astability version of Woodall’s conjecture.In this paper, we shall prove a stability result of Theorem 1.1. Let us introduce somenotation.
Definition 1.
Let k and n ≥ k + 1 be integers. We define F n,k to be a family of graphs,such that a graph G ∈ F n,k if and only if G is a graph of order n in which there is a subgraph K ∼ = K n − k , and for each component H of G − V ( K ), V ( H ) is a clique and all vertices in H are adjacent to a same vertex in K . Specially, the graph L n,k ∼ = K ∨ ( K n − k − + K k )is the one in L n,k with maximum number of edges. Theorem 1.4.
Let G be a graph of order n ≥ max { k + 17 , ( k +4)( k +5)2 } where k ≥ . If e ( G ) ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 32 (cid:19) , then G is weakly pancyclic with girth 3. Suppose that G contains no C n − k . Then one ofthe following holds:(a) G ⊆ L for some L ∈ L n,k +1 ;(b) G = L n,k +2 ∼ = K ∨ ( K n − k − + K k +2 ) ;(c) k = 0 and G ⊆ Γ n, := K ∨ ( K n − + 2 K ) ;(d) k = 1 and G = Γ n, := K ∨ ( K n − + 3 K ) . As a non-trivial byproduct, we give a solution to the following open problems proposedin [12]. By ρ ( G ) and q ( G ) we denote the spectral radius and signless Laplacian spectralradius of the graph G . Problem 1 ([12]) . Let G be a connected graph of order n and k ≥ n is sufficiently large compared to k .(a) Suppose that ρ ( G ) > ρ ( L n,k ). Does G contain a C n − k +1 ?(b) Suppose that q ( G ) > q ( L n,k ). Does G contain a C n − k +1 ?Our answer is the following. When k = 2, it implies all results in [12].2 heorem 1.5. Let k ≥ be an integer. Let G be a graph of order n . If either(a) ρ ( G ) ≥ ρ ( L n,k ) where n ≥ max { k + 11 , ( k +3)( k +4)2 } or,(b) q ( G ) ≥ q ( L n,k ) where n ≥ max { k + 11 , k + 2 k + 3 } ,then G contains a C ℓ for each ℓ ∈ [3 , n − k + 1] , unless G = L n,k . Our technique is to combine the stability methods in extremal graph theory withspectral technique. Compared with the original method in [17], we need to find such astability result of number of edges for Ω( √ n ) cycles of consecutive lengths, which is themain new point.Our second part is devoted to an open problem on cycles with consecutive lengths dueto Nikiforov [19].Bondy [2] proved that every hamiltonian graph G on n vertices contains cycles of alllengths ℓ ∈ [3 , n ] if e ( G ) ≥ n , unless n is even and G is isomorphic to K n , n . If one dropsthe condition that “ G is hamiltonian” in Bondy’s theorem, a theorem in Bollob´as’ textbook[1, Corrolary 5.4] states such a graph G contains all cycles C ℓ for each ℓ ∈ [3 , (cid:4) n +32 (cid:5) ].Nikiforov [19] considered cycles of consecutive lengths from a spectral perspective. Problem 2 (Nikiforov [19]) . What is the maximum C such that for all positive ε < C and sufficiently large n , every graph G of order n with ρ ( G ) ≥ q ⌊ n ⌋ contains a cycle oflength ℓ for every ℓ ≤ ( C − ε ) n .One may guess C = . However, the class of graphs G = K s ∨ ( n − s ) K where s = (3 −√ n (see [19]) shows C ≤ (3 −√ . Nikiforov [19] proved that C ≥ . Ning andPeng [21] slightly refined this as C ≥ . Only very recently, Mingqing Zhai and HuiqiuLin (private communication) improved these results to C ≥ .The second purpose of this article is to show that C ≥ by completely differentmethods. Theorem 1.6. Let ε be real with < ε < . Then there exists an integer N := N ( ε ) ,such that if G is a graph on n vertices with n ≥ N and ρ ( G ) > q ⌊ n ⌋ , then G containsall cycles C ℓ with ℓ ∈ [3 , ( − ε ) n ] . Let G be a graph. We use ω ( G ) to denote clique number of G . Let G and G be twovertex-disjoint graphs. The union of G and G , denoted by G + G , is defined to be agraph G ′ with V ( G ′ ) = V ( G ) ∪ V ( G ) and E ( G ′ ) = E ( G ) ∪ E ( G ). The join of G and G , denoted by G ∨ G , is a new graph obtained from G + G by adding all possibleedges from G to G .Let A ( G ) be the adjacency matrix of a graph G and D be the degree matrix of G .The spectral radius of G , denoted by ρ ( G ), is the largest eigenvalue of A ( G ). The signlessLaplacian spectral radius of G , denoted by q ( G ), is the largest eigenvalue of the signlessLaplacian matrix Q ( G ) := A ( G ) + D ( G ).The paper is organized as follows. In Section 2, we prove a sharp version of Woodall’stheorem and also a stability version of it. In Section 3, we answer Problem 1 completely.In Section 4, we consider Nikiforov’s open problem on cycles with consecutive lengths. Inthe last section, we mention some related problem for further study. If 0 < ε < − , then we can choose N = 2 . × ε − . Woodall’s theorem updated
We first refine Woodall’s Theorem on Tur´an number of large cycles as follows. We calla graph weakly pancyclic if it contains all cycles of lengths from the smallest one to thelargest one.
Theorem 2.1.
Let G be a graph of order n ≥ max { k + 11 , ( k +3)( k +4)2 } , where k ≥ . If e ( G ) ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 22 (cid:19) , then G is weakly pancyclic with girth 3. Furthermore, one of the following is true:(a) G contains a C ℓ for each ℓ ∈ [3 , n − k ] ;(b) G = L n,k +1 ∼ = K ∨ ( K n − k − + K k +1 ) . The proof of Theorem 2.1 needs the following three lemmas. The circumference of G ,denoted by c ( G ), is the length of a longest cycle in G . The n -closure cl n ( G ), is defined tobe a graph of order n by recursively joining any pair of non-adjacent vertices with degreesum at least n till there is no such pair. Lemma 2.1 (Bondy and Chv´atal [4]) . Let G be a graph of order n and C ′ := cl n ( G ) .Then c ( G ) = c ( cl n ( G )) . Lemma 2.2 (Bondy [2]) . Let G be a graph of order n . If c ( G ) = c and e ( G ) > c (2 n − c )4 ,then G is weakly pancyclic with girth 3. For the last lemma, its original form in [17] needs the condition “ k ≥ k = 0. This lemma is the key tool for our proof. Lemma 2.3.
Let G be a graph of order n ≥ k + 5 , where k ≥ . If G = cl n ( G ) and e ( G ) > (cid:0) n − k − (cid:1) + ( k + 1) then ω ( G ) ≥ n − k .Proof. Recall that the case of k ≥ k = 0. Suppose thatthere exist two vertices x, y ∈ V ( G ) such that d ( x ) + d ( y ) ≤ n −
1. Let H := G − { x, y } .Then e ( G ) ≤ e ( H ) + d ( x ) + d ( y ) ≤ (cid:0) n − (cid:1) + n − (cid:0) n − (cid:1) + 1, a contradiction. Thus, forany two nonadjacent vertices, the degree sum of them is at least n . By the definition of n -closure, G = K n and so ω ( G ) = n .We are in stand for proving Theorem 2.1. Proof of Theorem 2.1.
Suppose that G is a graph satisfying the condition. We firstshow that G is weakly pancyclic with girth 3. Let c := c ( G ). By Lemma 2.2, we only needto show that (cid:0) n − k − (cid:1) + (cid:0) k +22 (cid:1) > c (2 n − c )4 . If not, then we have nc − c ≥ n − (2 k + 3) n k + 1)( k + 2) , which implies that c − nc + 2( n − (2 k + 3) n ) + 4( k + 2)( k + 1) ≤ . However, the discriminant of quadratic form ∆ = (2 n ) − (cid:0) n − (2 k + 3) n ) + 4( k + 1)( k + 2) (cid:1) < n ≥ k + 5, a contradiction. This proves the first part of the theorem.Now let G ′ = cl n ( G ). Since e ( G ′ ) ≥ e ( G ) ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 22 (cid:19) ≥ (cid:18) n − k − (cid:19) + ( k + 2) + 1for n ≥ max { k + 11 , ( k +3)( k +4)2 } , by Lemma 2.3, ω ( G ′ ) ≥ n − k −
1. This implies that c ( G ′ ) ≥ n − k −
1. If c ( G ′ ) ≥ n − k , then c ( G ) = c ( G ′ ) ≥ n − k by Lemma 2.1. Recall that G is weakly pancyclic, implying that (a) holds. So assume that c ( G ′ ) ≤ n − k −
1. Since c ( G ′ ) ≥ ω ( G ′ ), we have ω ( G ′ ) = n − k − S be a clique of G ′ with | S | = n − k −
1, let K = G ′ [ S ] and H = G − S . Thus K is complete. Let H be an arbitrary component of H . If | N G ′ ( H ) ∩ S | ≥
2, thenclearly c ( G ′ ) ≥ n − k , a contradiction. Thus we conclude that | N G ′ ( H ) ∩ S | ≤ H of H . Specially, every vertex v ∈ V ( H ) has | N G ′ ( v ) ∩ S | ≤
1. Now e ( G ′ − S ) = e ( G ′ ) − e ( K ) − e G ′ ( S, V ( H )) ≥ e ( G ) − e ( K ) − e G ′ ( S, V ( H )) ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 22 (cid:19) − (cid:18) n − k − (cid:19) − ( k + 1) = (cid:18) k + 12 (cid:19) . Since | V ( H ) | = k + 1, we infer that V ( H ) is a ( k + 1)-clique and equality holds in theabove formula. This implies that G = G ′ and | N G ( v ) ∩ S | = 1 for every v ∈ V ( H ). Recallthat | N ( H ) ∩ S | = 1. All vertices in H have a common neighbor in S . We obtain that G = L n,k +1 , and (b) holds. The proof is complete.We further prove a stability result of Theorem 1.1 as follows. Proof of Theorem 1.4.
The argument used here is similar to Theorem 2.1. However,more details are needed. We first claim that G is weakly pancyclic with girth 3. ByLemma 2.2, we shall show that (cid:0) n − k − (cid:1) + (cid:0) k +32 (cid:1) > c (2 n − c )4 . Suppose to the contrary that c − nc + 2( n − (2 k + 5) n ) + 4( k + 2)( k + 3) ≤
0. However,(2 n ) − (cid:0) n − (2 k + 5) n ) + 4( k + 2)( k + 3) (cid:1) < n ≥ k + 7, a contradiction. This proves the first part of the theorem.Let G ′ := cl n ( G ). If c ( G ′ ) ≥ n − k , then by Lemma 2.1, c ( G ) = c ( G ′ ) ≥ n − k .Recall that G is weakly pancyclic, implying that G contains C n − k . So we assume that c ( G ′ ) ≤ n − k −
1. Since e ( G ′ ) ≥ e ( G ) ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 32 (cid:19) ≥ (cid:18) n − k − (cid:19) + ( k + 3) + 1for n ≥ ( k +4)( k +5)2 . By Lemma 2.3, ω ( G ′ ) ≥ n − k − n ≥ k + 17. If ω ( G ′ ) ≥ n − k .then c ( G ′ ) ≥ ω ( G ′ ) ≥ n − k , a contradiction. Now we assume that ω ( G ′ ) = n − k − ω ( G ′ ) = n − k −
1. Let S be a clique of G ′ with | S | = ω ( G ′ ), K = G ′ [ S ] and H = G ′ − S .Case A. ω ( G ′ ) = n − k −
1. Let H be an arbitrary component of H . If | N G ′ ( H ) ∩ S | ≥
2, then c ( G ′ ) ≥ n − k (recall that S is a clique of G ′ ), a contradiction. Thus, everycomponent H of G ′ − S satisfies | N G ′ ( H ) ∩ S | ≤
1. It follows G ⊆ G ′ ⊆ F ∈ F n,k +1 forsome F , and (a) holds.Case B. ω ( G ′ ) = n − k −
2. Set T = { v ∈ V ( H ) : | N G ′ ( v ) ∩ S | ≥ } . We distinguishthe following subcases. 5ase B.1. | T | = 0. In this case, every vertex v ∈ V ( H ) has | N G ′ ( v ) ∩ S | ≤
1. Now e ( G ′ − S ) = e ( G ′ ) − e ( K ) − e G ′ ( S, V ( H )) ≥ e ( G ) − e ( K ) − e G ′ ( S, V ( H )) ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 32 (cid:19) − (cid:18) n − k − (cid:19) − ( k + 2) = (cid:18) k + 22 (cid:19) . Since | V ( H ) | = k + 2, we infer that V ( H ) is a ( k + 2)-clique and equality holds in theabove formula. This implies that G = G ′ and | N G ( v ) ∩ S | = 1 for every v ∈ V ( H ). If | N ( H ) ∩ S | ≥
2, then clearly c ( G ) ≥ n − k , a contradiction. This implies that all verticesin H have a common neighbor in S . We obtain that G = L n,k +2 , and (b) holds.Case B.2. | T | = 1. Let v be the unique vertex in T . Let H be an arbitrary componentof H − v . If v ∈ N G ′ ( H ), then N G ′ ( H ) ∩ S = ∅ ; for otherwise c ( G ′ ) ≥ n − k . Furthermore,If | N G ′ ( H ) ∩ S | ≥
2, then there are two independent edges between S and V ( H ) (noticethat in G ′ , every vertex in H has at most 1 neighbors in S ), implying that c ( G ′ ) ≥ n − k , acontradiction. Thus, | N G ′ ( H ) ∩ ( S ∪ { v } ) | ≤ H of G ′ − ( S ∪ { v } ).This implies that G ⊆ G ′ ⊆ F ∈ F n,k +1 and (a) holds.Case B.3. | T | ≥
2. Let v be a vertex in T and u , u be two vertices in N G ′ ( v ) ∩ S . Forany other vertex v ∈ T , we have that N G ′ ( v ) ∩ S = { u , u } , for otherwise c ( G ′ ) ≥ n − k .Furthermore, N G ′ ( v ) = { u , u } . In brief, we have N G ′ ( T ) ∩ S = { u , u } . If there aretwo vertices in T which are adjacent in G ′ , then c ( G ′ ) ≥ n − k , a contradiction. So T isindependent in G ′ . For any vertex v ∈ V ( G ) \ ( S ∪ T ), we claim that | N G ′ ( v ) ∩ ( S ∪ T ) | ≤ v / ∈ T , v cannot have two neighbors in S . If N G ′ ( v ) contains two vertices in T orcontains one vertex in T and one vertex in S , then we have c ( G ′ ) ≥ n − k , a contradiction.Set t = | T | . Notice that 2 ≤ t ≤ k + 2. Now e ( G ′ ) = e ( K ) + e G ′ ( S, T ) + e G ′ ( S ∪ T, V ( G ) \ ( S ∪ T )) + e ( H − T ) ≤ (cid:18) n − k − (cid:19) + 2 t + ( k + 2 − t ) + (cid:18) k + 2 − t (cid:19) = (cid:18) n − k − (cid:19) + (cid:18) k + 32 (cid:19) + t − (2 k + 1) t ≤ e ( G ) + t ( t − k − . This implies that t ≥ k + 1. Combining with 2 ≤ t ≤ k + 2, it can only be that k = 0and t = 2, or k = 1 and t = 3. In each case V ( G ) = S ∪ T . For the first case, we have G ⊆ G ′ = Γ n, , and (c) holds. For k = 1 and t = 3, G ′ = Γ n, . Moreover, equality holdsin the above inequalities, implying that G = G ′ and (d) holds. Let G be a graph and u, v ∈ V ( G ). We use G [ u → v ] to denote a new graph obtainedfrom G , by replacing all edges uw by vw , where w ∈ N G ( u ) \ ( N G ( v ) ∪ { v } ). FollowingBrouwers’ book, we call this as “Kelmans operation”.In this article, we need some results on spectral properties of graphs under Kelmansoperation. These theorems will play important roles in our answers to Problem 1. Theorem 3.1 (Csikv´ari [5]) . Let G be a graph and u, v ∈ V ( G ) . Let G ′ := G [ u → v ] .Then ρ ( G ′ ) ≥ ρ ( G ) . heorem 3.2 (Li and Ning [17]) . Let G be a graph and u, v ∈ V ( G ) . Let G ′ := G [ u → v ] .Then q ( G ′ ) ≥ q ( G ) . The following spectral inequalities help us to invert our problems into ones in extremalstyle.
Theorem 3.3 (Hong [15]) . Let G be a graph on n vertices and m edges. If δ ( G ) ≥ then ρ ( G ) ≤ √ m − n + 1 . Theorem 3.4 (Das [6]) . Let G be a graph on n vertices and m edges. Then q ( G ) ≤ mn − + n − . The following two lemmas will be used to determine the extremal graphs.
Lemma 3.1.
Let G be a graph. Suppose that G is a subgraph of a member in F n,k , where n ≥ k + 1 .(a) If ρ ( G ) ≥ ρ ( L n,k ) , then G = L n,k .(b) If q ( G ) ≥ q ( L n,k ) , then G = L n,k .Proof. (a) Let F ∈ F n,k with G ⊆ F . Since G ⊆ F , ρ ( G ) ≤ ρ ( F ), with equality if andonly if G = F (recall that F is connected). Let K be the complete subgraph of F with | K | = n − k . Let H , H , . . . , H t be the components of F − K , and let v i , i ∈ [1 , t ], be theunique vertex in N ( H i ) ∩ V ( K ). By a series of Kelmans operation from v i to v for all v i = v , we get a graph F ′ which is a subgraph of L n,k +1 . By Theorem 3.1, ρ ( G ) ≤ ρ ( F ) ≤ ρ ( F ′ ) ≤ ρ ( L n,k ) , equality holds if and only if G = F = F ′ ∼ = L n,k . This proves the statement (a).(b) The proof is almost the same as the one of (a). We just use Theorem 3.2 insteadof Theorem 3.1 in the whole proof. We omit the details. Lemma 3.2.
Let n, k be integers where k ≥ . Then(a) ρ ( L n,k ) > ρ ( L n,k +1 ) for n ≥ k + 4 ; ρ ( F n, ) > ρ ( Γ n, ) for n ≥ ; ρ ( F n, ) > ρ ( Γ n, ) for n ≥ .(b) q ( L n,k ) > q ( L n,k +1 ) for n ≥ k + 4 ; q ( F n, ) > q ( Γ n, ) for n ≥ ; q ( F n, ) > q ( Γ n, ) for n ≥ .Proof. (a) Let V ( L n,k +1 ) = X ∪ Y ∪ { z } , where X ∪ { z } is the ( k + 2)-clique in L n,k +1 and Y ∪ { z } is the ( n − k − L n,k +1 . Choose x ∈ X . L n,k can be obtained from L n,k +1 by deleting all edges xx ′ for x ′ ∈ X and adding all edges xy ′ for y ′ ∈ Y .Let M be the Perron vector with respect to ρ ( L n,k +1 ), where x, y, w correspond to theeigencomponent of vertices in X , the vertices in Y and the vertex z . Let ρ := ρ ( L n,k +1 ).By eigenequation, we have ρ x = kx + z and ρ y = ( n − k − y + z . It follows that( ρ − k ) x = ( ρ − ( n − k − y . Since n ≥ k + 4, we have y > x . Then by Rayleigh quoit,we have ρ ( L n,k ) − ρ ( L n,k +1 ) ≥ n − k − xy − kx = 2 x (( n − k − y − kx > . This proves ρ ( L n,k ) > ρ ( L n,k +1 ) for n ≥ k + 4.Let M ′ be the Perron vector with respect to q ( L n,k +1 ), where x, y, w correspond to theeigencomponent of vertices in X , the vertices in Y and the vertex z . Let q := q ( L n,k +1 ).7y eigenequation, we have q x = (2 k + 1) x + z and ρ y = (2 n − k − y + z . It followsthat ( q − (2 k + 1)) x = ( q − (2 n − k − y . If n ≥ k + 4, then y > x . Then by Rayleighquoit, we have q ( L n,k ) − q ( L n,k +1 ) ≥ ( n − k − x + y ) − k ( x + x ) > . This proves q ( L n,k ) > q ( L n,k +1 ) for n ≥ k + 4.(b) ρ ( Γ n, ) ≤ p e ( Γ n, ) − n + 1 = √ n − n + 15 < n − ρ ( K n − ) < ρ ( F n, ) for n ≥ q ( Γ n, ) ≤ e ( Γ n, ) n − + n − ≤ n −
2) = q ( K n − ) < q ( F n, ) for n ≥ ρ ( Γ n, ) ≤ p e ( Γ n, ) − n + 1 = √ n − n + 25 < n − ρ ( K n − ) < ρ ( F n, ) for n ≥ q ( Γ n, ) ≤ e ( Γ n, ) n − + n − ≤ n −
3) = q ( K n − ) < q ( F n, ). Proof of Theorem 1.5. If G is disconnected, then we can add some edges betweendifferent components recursively, and get a connected graph G ′ with ρ ( G ′ ) > ρ ( G ) and q ( G ′ ) > q ( G ). Since the added edges are not contained in any cycle, if G ′ contains somecycles, then so does G . Thus we only deal with the case that G is connected.Suppose that (a) holds. Furthermore, suppose that G does not contains a C ℓ for every ℓ ∈ [3 , n − k + 1]. We shall show that G = L n,k .By Theorem 3.3, we have p e ( G ) − n + 1 ≥ ρ ( G ) ≥ ρ ( L n,k +1 ) ≥ n − k − . It follows that 2 e ( G ) ≥ ( n − k − + n −
1. Note that( n − k − + n − ≥ (cid:18) n − k − (cid:19) + (cid:18) k + 22 (cid:19) for n ≥ ( k +2) . By Theorem 1.4, G is weakly pancyclic with girth 3 for n ≥ max { k +11 , ( k +3)( k +4)2 } . Furthermore, if G does not contain a C n − k +1 , then one of the following istrue: (1) G ⊆ F for some F ∈ F n,k ; (2) G = L n,k +1 ; (3) k = 1 and G ⊆ Γ n, , or k = 2 and G ⊆ Γ n, . By Lemma 3.1 and Lemma 3.2, G = L n,k .Suppose that (b) holds. By Theorem 3.4, we obtain2 e ( G ) n − n − ≥ q ( G ) ≥ q ( F k +1 ) ≥ n − k − , which implies that e ( G ) ≥ n − (2 k +1) n +2 k . Note that n − (2 k +1) n +2 k ≥ (cid:0) n − k − (cid:1) + (cid:0) k +22 (cid:1) for n ≥ k + 2 k + 2. By Theorem 1.4, G is weakly pancyclic with girth 3. Furthermore, if G does not contain a C n − k +1 , then one of the following is true: (1) G ⊆ F for some F ∈ F n,k ;(2) G = L n,k +1 ; (3) k = 1 and G ⊆ Γ n, , or k = 2 and G ⊆ Γ n, . By Lemma 3.1 andLemma 3.2, G = L n,k .The proof is complete. This section is devoted to an open problem by Nikiforov [19]. Before the proof, we collectvarious results that will be used in our arguments.We first prove one edge condition for even cycles.
Theorem 4.1.
Let G be a graph on n vertices and e ( G ) edges. If G contains no evencycle of length more than k , where k ≥ is an integer, then e ( G ) ≤ (2 k +1)( n − . heorem 4.2 (Voss and Zuluaga [25]) . (1) Every 2-connected graph G with δ ( G ) ≥ r ≥ having at least r + 1 vertices contains an even cycle of length at least r . (2) Every 2-connected non-bipartite graph G with δ ( G ) ≥ r ≥ having at least r + 1 vertices containsan odd cycle of length at least r − . Theorem 4.3 (Ore [23]) . Let G be a graph on n vertices. If G contains no Hamiltoncycle, then e ( G ) ≤ (cid:0) n − (cid:1) + 1 . A graph is called a theta graph if it consists of three paths starting and ending withtwo same vertices and are internal-disjoint. The following lemma is very basic.
Lemma 4.1.
Let G be a graph containing no theta graphs. Then each component of G isan edge or a cycle. Proof of Theorem 4.1. If n ≤ k + 1, then e ( G ) ≤ (cid:0) n (cid:1) ≤ (2 k +1)( n − . If n = 2 k + 2, thenby Theorem 4.3, we have e ( G ) ≤ (cid:0) k +12 (cid:1) + 1 ≤ (2 k +1)( n − . Next, we assume n ≥ k + 3.Let k = 1. We shall prove that if a graph on n vertices contains no even cycles then e ( G ) ≤ n − . By Lemma 4.1, every component of G is an edge or an odd cycle. Let c be the number of components which are odd cycles. We use induction to prove that e ( G ) ≤ n + c − ≤ n − n − = n − . In the following, we suppose k ≥ v ∈ V ( G ) with d G ( v ) = δ ( G ), and G ′ := G − v . Note that G ′ satisfies that v ( G ′ ) ≥ k + 2 and G ′ contains no even cycle of length more than 2 k . By inductionhypothesis, if d ( v ) ≤ k , then we have e ( G ) = e ( G ′ ) + δ ≤ (2 k +1)( n − + k < (2 k +1)( n − , asrequired. Thus, δ ( G ) ≥ k + 1 ≥
3. If G is 2-connected, then by Theorem 4.2, G containsan even cycle of length at least 2 k + 2, a contradiction. Thus, G contains a cut-vertexor is disconnected. For each case, we use induction to each component and compute thenumber of edges. The proof is complete.The following spectral inequality was originally proposed by Guo, Wang and Li [14] asa conjecture and proved by Sun and Das [24]. Theorem 4.4 (Sun and Das [24]) . Let G be a graph with minimum degree δ ( G ) ≥ . Forany v ∈ V ( G ) , we have ρ ( G − v ) ≥ ρ ( G ) − d ( v ) + 1 . By Theorem 4.4, we deduce a result for graphs with isolated vertices.
Lemma 4.2.
Let G be a graph. For any v ∈ V ( G ) , we have ρ ( G ) ≤ ρ ( G − v ) + 2 d ( v ) . For a graph G , denote by ec ( G ) the length of a longest even cycle of G and oc ( G ) thelength of a longest odd cycle of G . Theorem 4.5 (Gould, Haxell and Scott [13]) . For every real number c > , there exists aconstant K := K ( c ) = . × c depending only on c such that the following holds. Let G bea graph with n ≥ Kc vertices and minimum degree at least cn . Then G contains a cycleof length t for every even t ∈ [4 , ec ( G ) − K ] and every odd t ∈ [ K, oc ( G ) − K ] . Now we give the proof of Theorem 1.6.
Proof of Theorem 1.6. If G is disconnected, for example, G contains t components,then we can add t − G ′ . Note that ρ ( G ′ ) ≥ ρ ( G ). For any integer k ≥ contains a cycle of length k if and only if G ′ contains a cycle of length k . Thus, we canassume that G is connected.By Theorem 3.3, we have n − ≤ ρ ( T n, ) < ρ ( G ) ≤ m − n + 1 . One can compute that 2 m ≥ n +4 n +34 . Thus, the average degree d ( G ) := mn > n .Let H be a subgraph of G defined by a sequence of graphs G , G , . . . , G k such that:(1) G = G , H = G k ;(2) for every i ∈ [0 , k − v i ∈ V ( G i ) such that d G i ( v i ) ≤ n and G i +1 = G i − v i ;(3) for every v ∈ V ( G k ), d G k ( v ) > n .We claim that d ( H ) > n . Suppose not the case. Then there is a smallest i ∈ [1 , k ] with d ( G i ) ≤ n . This implies that d ( G i − ) = 2 d ( v i − ) + | G i | d ( G i ) | G i | + 1 ≤ n , a contradiction. Thus, we conclude that d ( H ) > n and δ ( H ) > n .Case A: Even cycle. Note that e ( H ) = d ( H ) | H | > n ( | H |− . By Theorem 4.1, ec ( H ) > n . Recall that δ ( H ) > n . By Theorem 4.5, H contains all even cycles C ℓ with ℓ ∈ [4 , ec ( G ) − K ] if | H | ≥ · · K , where K = K ( ) = . × ( ) be the constant in Theorem4.5. Clearly | H | > n . Let n be an integer satisfying( i ) n ≥ · · K ; ( ii ) εn ≥ K. Now if n ≥ max { . × , . × ε } , then G contains all even cycles C ℓ with ℓ ∈ [4 , ( − ε ) n ].Case B: Odd cycle. Set h = | H | . By Lemma 4.2, we have ρ ( G ) ≤ ρ ( H ) + 2 k − X i =0 d G i ( v i ) ≤ ρ ( H ) + 2 k · n ρ ( H ) + kn , where G i , v i are those in the definition of H , and k = n − h . This implies that ρ ( H ) ≥ √ nh − . Since ρ ( G ) > q ⌊ n ⌋ , by Nosal’s theorem [22] and Mantel’s theorem, G containsa triangle, and so is non-bipartite. If h < n , then ρ ( H ) > √ nh − ≥ q ⌊ h ⌋ , and H isnon-bipartite as well. In any case we infer H is non-bipartite.Let F be a subgraph of H defined by a sequence of graphs H , H , . . . , H k such that:(1) H = H , F = H k ;(2) for every i ∈ [0 , k − v i of H i and H i +1 = H i − v i ;(3) H k has no cut-vertex.Note that the component number w ( H i +1 ) ≥ w ( H i )+1. Clearly w ( H ) ≤
8, for otherwise H will have a vertex of degree less than n . We claim that w ( F ) ≤
8. Suppose to the contrarythat there is a smallest i with w ( H i ) ≥
9. Notice that i ≤
8, implying that δ ( H i ) > n − w ( H i ) ≥ H i has a vertex with degree less that | H i | < n , a contradiction when n ≥ w ( F ) ≤
8, and specially, v ( F ) ≥ h − ρ ( H ) ≤ ρ ( F ) + 2 k − X i =0 d H i ( v i ) ≤ ρ ( F ) + 2 k ( h − ≤ ρ ( F ) + 14( h − . d ( H ) > n , we obtain h > n + 1. Thus, ρ ( F ) ≥ p ρ ( H ) − h − ≥ r nh − − h − r(cid:16) n − (cid:17) h + 554 > r(cid:16) n − (cid:17) (cid:16) n (cid:17) + 554 ≥ n − n ≥ F has no cut-vertex, i.e., every component of F is 2-connected. Let F bea component of F with ρ ( F ) = ρ ( F ). Thus we have δ ( F ) ≥ n − ρ ( F ) > n − | F | > n − δ ( F ) ≥ | F | . Recall that δ ( H ) ≥ n ≥ | H | , we assume that F = H . Thisimplies that F has a second component F . Since δ ( F ) ≥ n − k , we have | F | ≥ n − k + 1(here k is that in definition of F ). This implies that | F | ≤ h − k − ( n − k + 1) < h . Thus δ ( F ) ≥ n − ≥ n/ ≥ | F | when n ≥ F is non-bipartite. Recall that H is non-bipartite. So we assumethat F = H . By the analysis above we have | F | < h . Thus ρ ( F ) = ρ ( F ) ≥ (cid:16) n − (cid:17) h + 554 ≥ (cid:18) h − (cid:19) h + 554 > (7 h/ > | F | h ≥ h > n + 1, we have that F is non-bipartite when n ≥ oc ( F ) ≥ δ ( F ) − ≥ n −
15. By Theorem 4.5, F contains all oddcycles C ℓ for ℓ ∈ [ K, n − − K ], where K = K ( ) is the constant as in Theorem 4.5. Atheorem of Nikiforov [19, Theorem 1] states that there exists a sufficiently large N suchthat any graph of order n ≥ N has a cycle of length ℓ for every ℓ ∈ [3 , n ]. Let n be aninteger such that( i ) n ≥ max { , N } ; ( ii ) n ≥ K ; ( iii ) εn ≥ K + 15 . We only need n ≥ max { N, . × , . × ε } . Now if n ≥ max { n , n } , then G containsall cycles C ℓ with ℓ ∈ [3 , ( − ε ) n ].The proof is complete. Nikiforov [20] proposed two nice conjectures on cycles of small lengths. He conjecturedthat: (a) every graph on sufficiently large order n contains a C k +1 or a C k +2 if ρ ( G ) ≥ ρ ( S n,k ), unless G = S n,k where S n,k := K k ∨ ( n − k ) K ; and (b) every graph on sufficientlylarge order n contains a C k +2 if ρ ( G ) ≥ ρ ( S + n,k ), unless G = S + n,k where S + n,k is obtainedfrom S n,k by adding an edge in the n − k isolated vertices. One can easily compute that ρ ( S n,k ) = Ω( √ n ) and ρ ( S + n,k ) = Ω( √ n ). If these conjectures will be confirmed, then wemaybe obtain tight spectral conditions for C ℓ where ℓ ∈ [3 , Ω( √ n )] ∪ [ n − Ω( √ n ) , n ]. It ismysterious to determine tight spectral conditions for C ℓ , where 0 < lim n →∞ ℓn = c <
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