A strengthening of the spectral color critical edge theorem: books and theta graphs
aa r X i v : . [ m a t h . C O ] F e b A strengthening of the spectral color critical edgetheorem: books and theta graphs * Mingqing Zhai a , Huiqiu Lin b † a School of Mathematics and Finance, Chuzhou University, Chuzhou, Anhui 239012, China b Department of Mathematics, East China University of Science and Technology, Shanghai 200237, China
Abstract
The Color Critical Edge Theorem of Simonovits states that for a given colorcritical graph H with chromatic number χ ( H ) = k +
1, there exists an n ( H ) such that theTur´an graph T n , k is the only extremal graph with respect to ex ( n , H ) provided n ≥ n ( H ).Nikiforov’s pioneer work on spectral graph theory implies that the color critical edgetheorem also holds if ex ( n , H ) is replaced by the maximum spectral radius and n ( H )is an exponential function on | H | . We conjecture that n ( H ) is a linear function of | H | .Previous evidences include complete graphs and odd cycles. In this paper, we confirmtwo new classes of graphs: books and theta graphs. Namely, we prove that every graphon n vertices and ρ ( G ) > ρ ( T n , ) contains a book of size greater than n . . This can be seenas a spectral version of a 1962 conjecture by Erd˝os, which states that every graph on n vertices and e ( G ) > e ( T n , ) contains a book of size greater than n . In addition, our resulton theta graphs implies that if G is a graph of order n and ρ ( G ) > ρ ( T n , ), then G containsa cycle of length t for every t ≤ n . This is related to an open question by Nikiforov whichasks to determine the maximum c such that if G is a graph of order n and ρ ( G ) > ρ ( T n , ),then G contains a cycle of length t for every t ≤ cn . Keywords:
Spectral extrema, Color critical edge theorem, Book, Theta graph,Consecutive cycles
AMS Classification:
The
Tur´an number ex ( n , H ) is the maximum number of edges in any graph of order n that contains no copy of H as a subgraph. A graph on n vertices is said to be extremal with respect to ex ( n , H ) if it does not contain a subgraph isomorphic to H and has ex-actly ex ( n , H ) edges. The Tur´an graph , denoted by T n , k , is the complete k -partite graph K n , n ,..., n k , where P ki = n i = n and ⌊ nk ⌋ ≤ n i ≤ ⌈ nk ⌉ . A main motivation for studying Tur´an * Supported by the National Natural Science Foundation of China (Nos. 11971445, 11771141 and12011530064). † Corresponding author. E-mail addresses: [email protected] (M. Zhai); [email protected] (H.Lin). ex ( n , H ) is known. By the well-known Tur´an Theorem [34], ex ( n , K k + ) = e ( T n , k ) for n ≥ k +
1, and T n , k is the only extremal graph. ex ( n , H ) is determined for some cases when H is acyclic, see ex ( n , P k ) [16,19], ex ( n , k · P ) [12], ex ( n , k · P ) [5,38], ex ( n , k · K , k ) [21].More extremal results on forests can be seen in [22,36]. F¨uredi and Gunderson [18] deter-mined ex ( n , C k + ) for all k and n and characterized all the extremal graphs. In particular, ex ( n , C k + ) = e ( T n , ) = ⌊ n ⌋ for n ≥ k − A ( G ) be the adjacency matrix of a graph G . The largest modulus of an eigenvalueof A ( G ) is called the spectral radius of G and denoted by ρ ( G ). Nikiforov [29] posed a Brualdi-Solheid-Tur´an type problem : what is the maximal spectral radius ρ of an H -freegraph of order n ? Let ex sp ( n , H ) = max { ρ ( G ) : | G | = n , H * G } . A graph G on n vertices is said to be extremal with respect to ex sp ( n , H ) if G is H -free and ρ ( G ) = ex sp ( n , H ). We use E x ( n , H ) to denote the set of extremal graphs with respect to ex sp ( n , H ).In the past decade, much attention has been paid to Brualdi-Solheid-Tur´an type prob-lem with extremal graphs, see ex sp ( n , K k + ) [24, 37], ex sp ( n , K s , t ) [1, 24, 28], ex sp ( n , P k )[29], ex sp ( n , S ki = P a i ) [6], ex sp ( n , S ki = S d i ) [7], ex sp ( n , C ) [24,40], ex sp ( n , C ) [41], ex sp ( n , C k + ) [29], ex sp ( n , F k ) [8] where F k is the friendship graph. For more results on the ex-tremal spectral graph theory, see [23, 30, 33].An edge e in a graph H is called a color critical edge , if χ ( H − e ) = χ ( H ) − χ ( H ) is the chromatic number of H . The following result is known as Simonovits’ colorcritical edge theorem. Theorem 1.1. (Simonovits, [32]) Let H be a given graph containing a color critical edgeand χ ( H ) = k + . Then there exists an n ( H ) such that if n ≥ n ( H ) then T n , k is the onlyextremal graph with respect to ex ( n , H ) . Having seen various spectral forms of Tur´an-type results, one can expect that moreresults that surround it can be cast in spectral form as well. This is indeed, for example,spectral Erd˝os-Stone-Bollob´as theorem [26] and spectral version of saturation problem[27]. By Nikiforov’s result (see Theorem 2, [27]) and a more careful calculation on theequality case, one can get the following spectral version of Simonovits’ color critical edgetheorem.
Theorem 1.2.
Let H be a given graph containing a color critical edge and χ ( H ) = k + .Then there exists an n ( H ) (where n ( H ) ≥ e | H | k (2 k + k + ) such that if n ≥ n ( H ) then T n , k isthe only extremal graph with respect to ex sp ( n , H ) . Set s = ⌊ nk ⌋ and suppose that T n , k is an extremal graph with respect to ex sp ( n , H ).Note that ρ ( T n , k ) = (cid:16) n − s − + p ( n − s − + s ( s + k − (cid:17) . Now let G ∗ be anextremal graph with respect to ex ( n , H ). By a well-known inequality ρ ( G ) ≥ e ( G ) n due toCollatz and Sinogowitz [9], we have e ( G ∗ ) ≤ n ρ ( G ∗ ) ≤ n ρ ( T n , k ) = (cid:16) n ( n − s − + n p ( n − s − + s ( s + k − (cid:17) . On the other hand, e ( T n , k ) = X d ( v ) =
12 ( n ( n − s − + s ( s + k ) . By a direct computation, one can find that e ( G ∗ ) ≤ n ρ ( T n , k ) < e ( T n , k ) +
1. It follows that e ( G ∗ ) ≤ e ( T n , k ) and T n , k is also an extremal graph with respect to ex ( n , H ). Thus we havethe following result. Fact 1.1.
Theorem 1.2 implies Theorem 1.1.
Our main interest of the paper is the following problem. We believe that this problemis very useful to study the existence of desired subgraphs.
Problem 1.1.
Can n ( H ) be a linear function on | H | in Theorem 1.2? Up to now, Problem 1.1 holds for complete graphs [24] and odd cycles [29]. In orderto find more support on Problem 1.1, we show the following two theorems. The proofsare given in Sections 3 and 4.Firstly, let B r + be the r + -book , that is, the graph obtained from K , r + by addingan edge within the partition set of two vertices. Obviously, B r + contains a color criticaledge and χ ( B r + ) = Theorem 1.3.
E x sp ( n , B r + ) = { T n , } for n ≥ r . The size of a book B r is the number r of triangles in it. The size of the largest book ina graph G is called the booksize of G . In 1962, Erd˝os [13] initiated the study of books ingraphs. Since then, books have attracted considerable attention in extremal graph theory(see, for example, [3, 14, 15, 20]). Erd˝os [13] proposed the following conjecture: Thebooksize of a graph G of order n and size e ( G ) > e ( T n , ) is greater than n . The con-jecture was proved by Edwards in an unpublished manuscript [11] and independently byKhadˇziivanov and Nikiforov in [20]. Moreover, they constructed a graph with n = r vertices, m > e ( T n , ) edges such that its booksize is n +
1. This implies that the booksize n is best possible. Since this paper is focused on the spectral extremal problems, it is naturalto ask what is the minimum booksize of a graph G with order n and the spectral radius ρ > ρ ( T n , )? Theorem 1.3 gives the following answer. Corollary 1.1.
For arbitrary positive integer n, the booksize of a graph G of order n and ρ ( G ) ≥ ρ ( T n , ) is greater than n . , unless G (cid:27) T n , k . Recall that n ρ ( T n , k ) < e ( T n , k ) +
1. If e ( G ) > e ( T n , ), then ρ ( G ) ≥ e ( G ) n ≥ e ( T n , ) + n > ρ ( T n , ) . Thus, we propose a stronger problem than Erd˝os’ conjecture for further research.
Problem 1.2.
For arbitrary positive integer n, whether the booksize of a graph G of ordern and ρ ( G ) > ρ ( T n , ) is great than n ? A generalized theta graph , denoted by θ ( l , l , . . . , l t ), is the graph obtained from t in-ternally disjoint paths of length respectively l , l , . . . , l t , with the same pair of endpoints,where l ≤ l . . . ≤ l t and l ≥
2. In particular, θ ( l , l ) (cid:27) C l . Thus, the problem of determin-ing ex ( n , θ ( l , l , . . . , l t )) generalizes the problem of determining ex ( n , C l ). When restrain l = l = · · · = l t = l , Faudree and Simonovits [17] showed that a θ ( l , l , . . . , l t )-freegraph on n vertices has O t , l ( n + / l ) edges. Recently, Bukh and Tait [4] improved their re-sult. When restrain t =
3, we obtain a theta graph θ ( p , q , r ). Verstra¨ete and Williford [35]gave a lower bound of order n / on the greatest size of n -vertex θ (4 , , θ (1 , , r +
1) which contain a color criticaledge. For convenience, we use θ r + to denote θ (1 , , r + χ ( θ r + ) = Theorem 1.4.
E x sp ( n , θ r + ) = { T n , } for n ≥ r if r is odd and n ≥ r if r is even. The study of consecutive cycles is an important topic in extremal graph theory. From[2], we known that if G is a graph of order n and e ( G ) > e ( T n , ), then G contains a cycleof length t for every t ≤ ⌊ n + ⌋ . In 2008, Nikiforov [25] studied the spectral conditionfor the existence of consecutive cycles. He proved that if G is a graph of order n and ρ ( G ) > ρ ( T n , ), then G contains a cycle of length t for every t ≤ n . Moreover, Nikiforovproposed the following problem. Problem 1.3. (Nikiforov, [25]) Determining the maximum constant c such that if n issu ffi ciently large and G is a graph of order n and ρ ( G ) > ρ ( T n , ) , then G contains a cycleof length t for every t ≤ cn. Nikiforov’s result implies that c ≥ . Moreover, Nikiforov [25] constructed a graph S n , k , which is the join of a complete graph of order k = ⌈ (3 − √ n ⌉ with an empty graph oforder n − k . One can see ρ ( S n , k ) > n ≥ ρ ( T n , ), but S n , k has no cycles of lengths longerthan 2 k . This implies that c ≤ − √ + n < . . Recently, Ning and Peng [31] improvedNikiforov’s result by showing c ≥ . As a direct corollary of Theorem 1.4, we get thefollowing result. Corollary 1.2.
Let n be an arbitrary positive integer and G be a graph of order n and ρ ( G ) > ρ ( T n , ) . Then G contains a cycle of length t for every t ≤ n . unless G (cid:27) T n , . Corollary 1.2 indicates that c ≥ without the condition “su ffi ciently large n ”. Inspiredby Nikiforov’s problem and current result, one may ask Problem 1.3 by replacing thecondition “su ffi ciently large n ” by “arbitrary positive integer n ”. Furthermore, we havethe following problem. Problem 1.4.
For arbitrary positive integer n, determining the maximum c ′ such that if Gis a graph of order n and ρ ( G ) > ρ ( T n , ) , then G contains a copy of θ r + for every r ≤ c ′ n. Theorem 1.4 implies that c ′ ≥ . Moreover, combining Theorems 1.3 and 1.4 withFact 1.1, we essentially determine the Tur´an numbers of B r + and θ r + . Corollary 1.3.
Let r be a positive integer. Then we have the following statements.(i) ex ( n , B r + ) = ⌊ n ⌋ for n ≥ r;(ii) ex ( n , θ r + ) = ⌊ n ⌋ for n ≥ r if r is odd and n ≥ r if r is even. All graphs considered here are undirected, loopless and simple. V ( G ) and E ( G ) arethe vertex set and edge set of a graph G , respectively. As usual, we use e ( G ) to denotethe size of G . For a vertex u ∈ V ( G ) and a subgraph H ⊆ G (possibly u < V ( H )), we use N H ( u ) to denote the set of neighbors of u in H and d H ( u ) to denote | N H ( u ) | . u ∼ v standsfor uv ∈ E ( G ). The subgraph of G induced by a vertex subset S is denoted by G [ S ]. Fortwo disjoint subsets A , B ⊆ V ( G ), let E ( A ) be the set of edges with both endpoints in A and E ( A , B ) be the set of edges with one endpoint in A and the other in B . Furthermore,we denote e ( A ) = | E ( A ) | and e ( A , B ) = | E ( A , B ) | .The following result is known as Erd˝os-Gallai theorem. Lemma 2.1 (Erd˝os, Gallai, [12]) . Let G be a P r + -free graph of order n. Then e ( G ) ≤ r n,and equality holds if and only if G is a union of K r + . Lemma 2.2 ( [39]) . Let G = < X , Y > be a bipartite graph, where | X | ≥ r , | Y | ≥ r − withr ≥ . If G does not contain a copy of P r + with both endpoints in X, thene ( G ) ≤ ( r − | X | + r | Y | − r ( r − . Equality holds if and only if G (cid:27) K | X | , | Y | , where | X | = r or | Y | = r − . Assume that H ∈ { B r + , θ r + } and G is an extremal graph with respect to ex sp ( n , H ).Let X = ( x , . . . , x n ) t be the Perron vector of G and u ⋆ ∈ V ( G ) with x u ⋆ = max { x i : i = , . . . , n } . Let A = N G ( u ⋆ ), B = V ( G ) \ ( A ∪ { u ⋆ } ) and γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) . Then we have the following lemma.
Lemma 2.3. | A | ≥ ⌈ n ⌉ , γ ( u ⋆ ) ≥ ⌊ n ⌋ , and G (cid:27) T n , if e ( A ) = . Proof.
Since T n , is H -free and G attains the maximum spectral radius, ρ ( G ) ≥ ρ ( T n , ) = ⌊ n ⌋⌈ n ⌉ = r ⌊ n ⌋ . Note that ρ ( G ) x u ⋆ = P u ∈ A x u ≤ | A | x u ⋆ . Then | A | ≥ ρ ( G ) ≥ q ⌊ n ⌋ > n − , which impliesthat | A | ≥ ⌈ n ⌉ . (1)Again by using eigen-equation, we have ρ ( G ) x u ⋆ = ρ ( G ) X u ∼ u ⋆ x u = X u ∼ u ⋆ X v ∼ u x v = | A | x u ⋆ + X u ∈ A d A ( u ) x u + X v ∈ B d A ( v ) x v ≤ | A | x u ⋆ + e ( A ) x u ⋆ + e ( A , B ) x u ⋆ = γ ( u ⋆ ) x u ⋆ . It follows that γ ( u ⋆ ) ≥ ρ ( G ) ≥ ⌊ n ⌋ . (2)Now we assume that e ( A ) =
0, then γ ( u ⋆ ) = | A | + e ( A , B ) ≤ | A | + | A || B | = | A | ( | B | + ≤ ⌊ n ⌋ , (3)since | A | + ( | B | + = n . Combining (2), we have γ ( u ⋆ ) = ⌊ n ⌋ . Consequently, all inequal-ities in (3) become equalities. Therefore, e ( A , B ) = | A || B | . Moreover, combining (1) wehave | A | = ⌈ n ⌉ and | B | + = ⌊ n ⌋ . This implies that G contains a copy of T n , as a spanningsubgraph. Furthermore, e ( B ) =
0, since G is B r + -free or θ r + -free. Thus, G (cid:27) T n , , asdesired. (cid:3) In this section, we give the proof of Theorem 1.3. Suppose that G is an extremal graphwith respect to ex sp ( n , B r + ). Let X = ( x , . . . , x n ) t be the Perron vector of G and u ⋆ be avertex with x u ⋆ = max { x i : i = , . . . , n } . Note that T n , has no B r + . Then ρ ( G ) ≥ ρ ( T n , ) = r ⌊ n ⌋ > r , (4)since n ≥ r . Moreover, we have the following claims. Claim 3.1. d A ( u ) ≤ r for each u ∈ A. Proof.
Since G is B r + -free, the claim holds immediately. (cid:3) Claim 3.2. d B ( u ) + d B ( u ) ≤ | B | + r − for any u u ∈ E ( A ) . Proof.
By the way of contradiction, assume that d B ( u ) + d B ( u ) ≥ | B | + r for some u u ∈ E ( A ). Note that | B | + r ≤ d B ( u ) + d B ( u ) = | N B ( u ) ∪ N B ( u ) | + | N B ( u ) ∩ N B ( u ) |≤ | B | + | N B ( u ) ∩ N B ( u ) | . Then | N B ( u ) ∩ N B ( u ) | ≥ r , and then G contains a B r + , a contradiction. (cid:3) Claim 3.3. x u + x u ≤ | B | + r + ρ ( G ) − r x u ⋆ for any u u ∈ E ( A ) . Proof.
Suppose that x u + x v = max uv ∈ E ( A ) ( x u + x v ). Now we only need to show that x u + x v ≤ | B | + r + ρ ( G ) − r x u ⋆ . By Claim 3.1, we have ρ ( G ) x u = x u ⋆ + X w ∈ N A ( u ) x w + X w ∈ N B ( u ) x w ≤ x u ⋆ + rx v + d B ( u ) x u ⋆ and ρ ( G ) x v = x u ⋆ + X w ∈ N A ( v ) x w + X w ∈ N B ( v ) x w ≤ x u ⋆ + rx u + d B ( v ) x u ⋆ . Combining the above two inequalities and using Claim 3.2, we have( ρ ( G ) − r )( x u + x v ) ≤ x u ⋆ + ( d B ( u ) + d B ( v )) x u ⋆ ≤ ( | B | + r + x u ⋆ . Then the claim holds. (cid:3)
We use e ( A , B ) to denote the number of edges missing in E ( A , B ), that is, e ( A , B ) = | A || B | − e ( A , B ). For the sake of simplicity, we use d B ( u ) instead of e ( { u } , B ), where u ∈ A .Now we have the following claim. Claim 3.4. e ( A , B ) ≥ r e ( A )( | B | − r + . Proof.
By Claim 3.2, d B ( u ) + d B ( u ) ≤ | B | + r − u u ∈ E ( A ). Then d B ( u ) + d B ( u ) ≥ | B | − ( | B | + r − = | B | − r + . Since d A ( u ) ≤ r for each u ∈ A , d B ( u ) appears at most r times in P u u ∈ E ( A ) (cid:16) d B ( u ) + d B ( u ) (cid:17) .Thus we have e ( A , B ) = X u ∈ A d B ( u ) ≥ r X u u ∈ E ( A ) (cid:16) d B ( u ) + d B ( u ) (cid:17) ≥ r e ( A )( | B | − r + , as required. (cid:3) Now we give the final proof of Theorem 1.3. Since ρ ( G ) x v = P u ∼ v x u , we have ρ ( G ) x u ⋆ = ρ ( G ) X u ∼ u ⋆ x u = X u ∼ u ⋆ X v ∼ u x v = | A | x u ⋆ + X u ∈ A d A ( u ) x u + X v ∈ B d A ( v ) x v = | A | x u ⋆ + X u u ∈ E ( A ) ( x u + x u ) + X v ∈ B d A ( v ) x v ≤ | A | x u ⋆ + e ( A ) | B | + r + ρ ( G ) − r x u ⋆ + e ( A , B ) x u ⋆ by Claim 3.3 . Note that e ( A , B ) = | A || B | − e ( A , B ). Then by Claim 3.4, we have ρ ( G ) ≤ | A | + e ( A ) | B | + r + ρ ( G ) − r + | A || B | − r e ( A )( | B | − r + = | A | ( | B | + + ( | B | + r + ρ ( G ) − r − | B |− r + r ) e ( A ) . (5)If ρ ( G ) > r (1 + | B | + r + | B |− r + ), then | B | + r + ρ ( G ) − r < | B |− r + r and thus ρ ( G ) < | A | ( | B | + ≤ ⌊ n ⌋ , whichcontradicts (4).If ρ ( G ) = r (1 + | B | + r + | B |− r + ), then ρ ( G ) ≤ | A | ( | B | + ≤ ⌊ n ⌋ . Combining (4), we have ρ ( G ) = ⌊ n ⌋ . From Lemma 2.3, we have | A | = ⌈ n ⌉ and | B | + = ⌊ n ⌋ . Furthermore, by ρ ( G ) = r (1 + | B | + r + | B |− r + ) = r ( | B | + | B |− r + , we have ρ ( G ) = ( 2 r ⌊ n ⌋ − r ) ⌊ n ⌋ ≤ ( 2 r ⌊ n ⌋ − r ) ⌊ n ⌋ = ( 2 r ⌊ n ⌋ − r ) ρ ( G ) ≤ ρ ( G ) , (6)since n ≥ r . Therefore, ⌊ n ⌋ = ⌊ n ⌋ , which implies that n is even. Consequently,2 r ⌊ n ⌋ − r = r n − r ≤ . Thus, (6) becomes a strict inequality, also a contradiction.If ρ ( G ) < r (1 + | B | + r + | B |− r + ), then | B | + r + ρ ( G ) − r > | B |− r + r . By (5), we have ρ ( G ) ≤ | A | ( | B | + + ( | B | + r + ρ ( G ) − r − | B | − r + r ) r | A | by Claim 3.1 < | A | ( | B | + + ( | B | + r + r − | B | − r + r ) r | A | since ρ ( G ) > r by (4) = | A | ( | B | + + r ) ≤ ⌊ ( | A | + | B | + + r ) ⌋ = ⌊ ( n + r ) ⌋ since | A | + | B | + = n ≤ ⌊ n ⌋ as n ≥ r , a contradiction. This completes the proof. In this section, we give the proof of Theorem 1.4. Suppose that G is an extremal graphwith respect to ex sp ( n , θ r + ). X = ( x , . . . , x n ) t is the Perron vector of G and u ⋆ ∈ V ( G )with x u ⋆ = max { x i : i = , . . . , n } . A = N G ( u ⋆ ), B = V ( G ) \ ( A ∪ { u ⋆ } ) and γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) . Proof of Theorem 1.4 for odd r . When r = θ r + (cid:27) B r + . Therefore, by Theorem 1.3, E x sp ( θ , n ) = { T n , } for n ≥ . In the following, we may assume that r ≥
3. We first give several claims.
Claim 4.1. e ( H ) ≤ r | H | for each component H of G [ A ] , and | B | ≥ r. Proof.
Since G is θ r + -free, one can see that G [ A ] is P r + -free. By Lemma 2.1, we have e ( H ) ≤ r | H | for each component H of G [ A ]. It follows that e ( A ) ≤ r | A | and then γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) ≤ (1 + r + | B | ) | A | = (1 + r + | B | )( n − − | B | ) . Suppose that | B | ≤ r −
1. Then γ ( u ⋆ ) ≤ r · r = r < ⌊ n ⌋ , since n ≥ r . This contradicts (2). So the claim holds. (cid:3) An A-path of G is a path consisting of edges in E ( A , B ) and with both endpoints in A .The length of a path P is denoted by l ( P ). Claim 4.2.
There exists an A-path of length r − in G. Proof.
By Lemma 2.3, we know that | A | ≥ ⌈ n ⌉ . By way of contradiction, suppose thatthere exists no A -path of length 3 r −
1. Then by Lemma 2.2, e ( A , B ) ≤ r − | A | + r − | B | − (3 r − r − . Combining with e ( A ) ≤ r | A | by Claim 4.1, we have γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) ≤ r − | A | + r − | B | − (3 r − r − = r −
12 ( | A | + | B | + − r | B | − r − r − ≤ (5 r − n − r − r − r −
54 since | B | ≥ r ≤ rn − r < ⌊ n ⌋ as n ≥ r , a contradiction. So the claim holds. (cid:3) The set of internal vertices of a path P is denoted by V in ( P ). For two vertices u , v ∈ V ( P ), the distance between u and v in P is denoted by d P ( u , v ). A vertex of a θ r + is saidto be a head , if it is of degree two and it belongs to a triangle of θ r + . The following twoclaims give some structural properties of G . Claim 4.3.
Let P be an A-path in G and u ∈ A \ V ( P ) . The following statements hold.(i) If l ( P ) ≥ r + and u ∼ u ′ for some u ′ ∈ A \ V in ( P ) , then e ( { u , u ′ } , B ) ≥ .(ii) If l ( P ) ≥ r + and u ∼ u ′ ∼ u ′′ for some u ′ , u ′′ ∈ A \ V ( P ) , then e ( { u } , B ) ≥ r + . Proof. (i) Let P = u v u v · · · u r + v r + u r + be an A -path of length r +
3, where u i ∈ A and v i ∈ B . Pick two pairs of vertices { v , v + r − } and { v , v + r − } . Clearly, d P ( v i , v i + r − ) = r − i ∈ { , } . If e ( { u , u ′ } , { v i , v i + r − } ) ≥
3, we have two independent edges, say uv i , u ′ v i + r − ,in E ( { u , u ′ } , { v i , v i + r − } ). Therefore, uv i u i + v i + · · · u i + r − v i + r − u ′ is an A -path of length r + uu ⋆ u ′ , we get a θ r + with its head u ⋆ , a contradiction. Thus e ( { u , u ′ } , { v i , v i + r − } ) ≤ e ( { u , u ′ } , { v i , v i + r − } ) ≥ i ∈ { , } . If r ≥
5, then { v , v + r − } ∩ { v , v + r − } = ∅ . It follows that e ( { u , u ′ } , B ) ≥ , asrequired. Now consider r =
3. It su ffi ces to show e ( { u , u ′ } , { v , v , v } ) ≥
3. Notethat e ( { u , u ′ } , { v , v } ) ≥ e ( { u , u ′ } , { v , v } ) ≥
2. If e ( { u , u ′ } , { v , v , v } ) ≤
2, then e ( { u , u ′ } , { v } ) = e ( { u , u ′ } , { v , v } ) =
0. This implies that e ( { u , u ′ } , { v , v } ) =
4. No-tice that u ′ < V in ( P ). Either u ′ < { u , u , u } or u ′ < { u , u , u } . Without loss of generality,assume that u ′ < { u , u , u } , then u ′ u ⋆ u v u is an A -path of length 4. Together with thetriangle u ′ v u , we get a θ with its head v , a contradiction. Therefore, e ( { u , u ′ } , B ) ≥ P = u v u v · · · u r + v r + u r + be an A -path of length 2 r +
2, where u i ∈ A and v i ∈ B . If u ∼ v i for some i ∈ { , , . . . , r + } , then we can select a vertex v j ∈ V ( P )such that d P ( v i , v j ) = r −
3. Let P v i → v j be a subpath of P with endpoints v i and v j . Then u ′ uP v i → v j u j + u ⋆ is a path of length r +
1. Combining with the the triangle u ′ u ′′ u ⋆ , we geta θ r + with its head u ′′ , a contradiction. It follows that N G ( u ) ∩ { v , . . . , v r + } = ∅ and thenthe result holds. (cid:3) Claim 4.4.
Let P be an A-path in G starting from some vertex u ∈ A. If l ( P ) ≥ r − andu ∼ u ′ ∼ u ′′ for some u ′ , u ′′ ∈ A, then either u ′ ∈ V ( P ) or u ′′ ∈ V ( P ) . Proof.
Let P u → u ′′′ be an A -path of length r − u and u ′′′ . Then u ′ P u → u ′′′ u ⋆ is a path of length r +
1. Combining with the triangle u ′ u ′′ u ⋆ , we get a θ r + with its head u ′′ , a contradiction. So the claim holds. (cid:3) A component H of G [ A ] is said to be good , if 2 e ( H ) + e ( V ( H ) , B ) < | H || B | . Clearly, H is not good if H is an isolated vertex in G [ A ]. The following claim gives a characterizationof good nontrivial components in G [ A ]. Claim 4.5.
Let H be a component of G [ A ] and P be an A-path of length r − in G. Thefollowing statements hold.(i) If | H | = and V ( H ) * V ( P ) , then H is good.(ii) If | H | ≥ and V ⋆ ( H ) * V ( P ) , where V ⋆ ( H ) = { v ∈ V ( H ) : d H ( v ) ≥ } , then H is good. Proof. (i) Let V ( H ) = { u , u ′ } . Since V ( H ) * V ( P ), we may assume without loss ofgenerality that u < V ( P ). If r ≥
5, then l ( P ) = r − ≥ r +
4. Therefore, whether u ′ ∈ V ( P ) or not, there exists an A -subpath P of P with l ( P ) = r + u ′ < V in ( P ).By Claim 4.3 (i), we have e ( { u , u ′ } , B ) ≥
3. It follows that2 e ( H ) + e ( V ( H ) , B ) ≤ + | H || B | − < | H || B | , and then H is good.If r =
3, then l ( p ) =
8. Let P = u v · · · u v u , where u i ∈ A and v i ∈ B . If u ′ , u ( u is the central vertex of P ), then there exists an A -subpath P of P with l ( P ) = r + u ′ < V in ( P ). Similar as above, H is good. If u ′ = u , then e ( { u , u } , { v , v } ) ≥ uv , u v , in E ( { u , u } , { v , v } ). Thus uv u v u is an A -path of length 4 and then we get a θ with its head u ⋆ ). Similarly, e ( { u , u } , { v , v } ) ≥
2. It follows that2 e ( H ) + e ( V ( H ) , B ) ≤ + | H || B | − < | H || B | , as desired.(ii) We first show V ( H ) ∩ V ( P ) = ∅ . Suppose to the contrary, then there exists a path P ⋆ ⊆ H with one endpoint in V ( P ) and the other in V ⋆ ( H ) \ V ( P ), since V ⋆ ( H ) * V ( P ).Furthermore, we can find an edge uu ′ ∈ E ( P ⋆ ) with u ∈ V ( P ) and u ′ ∈ V ⋆ ( H ) \ V ( P ). Bythe definition of V ⋆ ( H ), we have d H ( u ′ ) ≥
2. Let u ′′ ∈ N H ( u ′ ) \{ u } . Note that l ( P ) ≥ r − A -subpath P of P , starting from u , such that l ( P ) ≥ r −
1. Since u ′ < V ( P ) and then u ′ < V ( P ), by Claim 4.4, we have u ′′ ∈ V ( P ) ⊆ V ( P ) . Note that l ( P ) ≥ r − u , u ′′ ∈ V ( P ). One can see that there exists either an A -subpath P of P starting from u with l ( P ) ≥ r − u ′′ < V ( P ), or an A -subpath P of P starting from u ′′ with l ( P ) ≥ r − u < V ( P ). By Claim 4.4, in either case we have u ′ ∈ V ( P ), acontradiction. Therefore, V ( H ) ∩ V ( P ) = ∅ .Note that l ( P ) = r − ≥ r +
2. If H (cid:27) K , s ( s ≥ u ∈ V ( H ) there exist u ′ , u ′′ ∈ V ( H ) such that u ∼ u ′ ∼ u ′′ . By Claim 4.3 (ii), we have e ( { u } , B ) ≥ r +
1. Hence, we have2 e ( H ) + e ( V ( H ) , B ) ≤ s − + | H || B | − ( r + s < | H || B | , H is not a star, then for each vertex u ∈ V ( H ) there exist u ′ , u ′′ ∈ V ( H )such that u ∼ u ′ ∼ u ′′ . Again by Claim 4.3 (ii), we have e ( { u } , B ) ≥ r +
1. Combiningwith Claim 4.1, we have2 e ( H ) + e ( V ( H ) , B ) ≤ r | H | + | H || B | − ( r + | H | < | H || B | , as required. (cid:3) For a subset A ′ ⊆ A , an A ′ -path of G is a path consisting of edges in E ( A ′ , B ) and withboth endpoints in A ′ . Claim 4.2 states that there exists an A -path of length 3 r − G .The following claim gives a stronger characterization. Claim 4.6.
Let P be an A-path of length r − in G and A ′ = A \ V ( P ) . Then there alsoexists an A ′ -path P ′ of length r − . Proof.
Note that | B | ≥ r , | A | ≥ ⌈ n ⌉ ≥ r and | A \ A ′ | = | V ( P ) ∩ A | = r + . Hence, | B | > r − and | A ′ | = | A | − r + > r − . If G does not contain an A ′ -path of length 3 r − e ( A ′ , B ) ≤ r − | A ′ | + r − | B | − (3 r − r − . Moreover, e ( A \ A ′ , B ) ≤ | A \ A ′ || B | = r + | B | . By Claim 4.1, we have 2 e ( A ) ≤ r | A | . Combining these inequalities, we have γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) ≤ (1 + r ) | A | + r − | A ′ | + r | B | − (3 r − r − = (1 + r ) | A | + r −
32 ( | A | − r +
12 ) + r | B | − (3 r − r − = r − | A | + r | B | − r (3 r − = r ( | A | + | B | + − r + | A | − r (3 r − ≤ rn − r + n − r since | A | ≥ n and 3 r − ≥ r < rn − r < ⌊ n ⌋ as n ≥ r , a contradiction. So the claim holds. (cid:3) In the rest, we only need to show that e ( A ) =
0, and then by Lemma 2.3, we completethe proof. By the way of contradiction, suppose that e ( A ) , . Then G [ A ] containsnontrivial components. Let A = { u ∈ A : d A ( u ) = } and A = A \ A . If all nontrivialcomponents of G [ A ] are good, then we have2 e ( A ) + e ( A , B ) < | A || B | . γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) + e ( A , B ) < | A | + | A || B | + | A || B | = | A | ( | B | + ≤ ⌊ n ⌋ , a contradiction. Thus, G [ A ] has a nontrivial component H which is not good. Thenby Claim 4.6, there exist two A -paths, P and P ′ , such that l ( P ) = l ( P ′ ) = r − V ( P ) ∩ V ( P ′ ) ∩ A = ∅ . Furthermore, by Claim 4.5, if | H | =
2, then V ( H ) ⊆ V ( P ) and V ( H ) ⊆ V ( P ′ ); if | H | ≥
3, then V ⋆ ( H ) ⊆ V ( P ) and V ⋆ ( H ) ⊆ V ( P ′ ). Both cases contradictthe fact V ( P ) ∩ V ( P ′ ) ∩ A = ∅ . This completes the proof of Theorem 1.4 for odd r . Proof of Theorem 1.4 for even r . We first give four claims.
Claim 4.7.
If P is an A-path of length r starting from u , then N A ( u ) ⊆ V ( P ) . Proof.
Let P = u v · · · u r v r u r + , where u i ∈ A and v i ∈ B . Then P ⋆ = u v · · · u r v r u r + u ⋆ is a path of length r +
1. If there exists a vertex u ∈ N A ( u ) \ V ( P ), then P ⋆ together withthe triangle u uu ⋆ consists a θ r + with its head u , a contradiction. (cid:3) Claim 4.8. | B | ≥ r and G contains an A-path of length r. Proof.
We first show that | B | ≥ r . If not, suppose that | B | ≤ r −
1. Since G is θ r + -free, G [ A ] is P r + -free. Then by Lemma 2.1, 2 e ( A ) ≤ r | A | . It follows that γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) ≤ (1 + r ) | A | + | A || B | = (1 + r + | B | )( n − − | B | ) ≤ r ( n − r ) since | B | ≤ r − < ⌊ n ⌋ as n ≥ r and r ≥ , a contradiction. It follows that | B | ≥ r , as required.Now, suppose that G does not contain an A -path of length 2 r . Then by Lemma 2.2, e ( A , B ) ≤ ( r − | A | + r | B | − r ( r − . Furthermore, γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) ≤ (1 + r ) | A | + e ( A , B ) ≤ r | A | + r | B | − r ( r − = r ( | A | + | B | + − r ( | B | + r + ≤ rn − r (2 r +
1) since | B | ≥ r < ⌊ n ⌋ as n ≥ r , a contradiction. So the claim holds. (cid:3) Claim 4.9.
Let P be an A-path of length r in G and A ′ = A \ V ( P ) . Then there also existsan A ′ -path P ′ of length r. Proof.
Let P = u v · · · u r v r u r + , where u i ∈ A and v i ∈ B . Then | A \ A ′ | = | V ( P ) ∩ A | = r + | A ′ | = | A | − ( r + > r , since | A | ≥ ⌈ n ⌉ and n ≥ r . If G does not contain an A ′ -path of length 2 r , then by Lemma 2.2, we have e ( A ′ , B ) ≤ ( r − | A ′ | + r | B | − r ( r − . (7)We first show that e ( A ′ , A \ A ′ ) =
0. Note that A \ A ′ = { u , . . . , u r + } . Suppose that uu i ∈ E ( A ) for some u ∈ A ′ and u i ∈ A \ A ′ . Without loss of generality, we may assumethat i ≤ r + . Then u i v i · · · u i + r − v i + r − u i + r u ⋆ is a path of length r +
1. Together with thetriangle u i uu ⋆ , we get a θ r + with its head u , a contradiction. Therefore, e ( A ′ , A \ A ′ ) = e ( A \ A ′ ) ≤ r . Let V = { u i : 1 ≤ i ≤ r + } and V = { u i : r + ≤ i ≤ r + } . If u i u j ∈ E ( G ) for some i , j with 1 ≤ i < j ≤ r + P ⋆ = u j v j · · · u j + r − v j + r − u j + r is an A -path of length r starting from u j . By Claim4.7, N A ( u j ) ⊆ V ( P ⋆ ) and then u i ∈ V ( P ⋆ ), a contradiction. This implies that V isan independent set. Similarly, V is also an independent set. Hence, G [ A \ A ′ ] is abipartite graph. Furthermore, for each i ≤ r , u i v i · · · u i + r − v i + r − u i + r is an A -path of length r starting from u i . Again by Claim 4.7, if u i u j ∈ E ( G ) for some j ≥ i + r +
1, we have acontradiction. Now one can see that N A \ A ′ ( u i ) ⊆ { u j : r + ≤ j ≤ r + i } for each i ≤ r (where N A \ A ′ ( u ) = ∅ ). Thus, d A \ A ′ ( u i ) ≤ i − i ≤ r . It follows that e ( A \ A ′ ) = r X i = d A \ A ′ ( u i ) ≤ r X i = ( i − = r ( r − < r . Note that G [ A ′ ] is P r + -free. Then by Lemma 2.1, 2 e ( A ′ ) ≤ r | A ′ | . Thus we have2 e ( A ) = e ( A ′ ) + e ( A \ A ′ ) < r | A ′ | + r . (8)Note that e ( A \ A ′ , B ) ≤ | A \ A ′ || B | = ( r + | B | . (9)Combining (7), (8) and (9), we have γ ( u ⋆ ) = | A | + e ( A ) + e ( A ′ , B ) + e ( A \ A ′ , B ) ≤ | A | + (2 r − | A ′ | + (2 r + | B | − ( 34 r − r ) = | A | + (2 r − | A | − ( r + + (2 r + | B | − ( 34 r − r ) = r ( | A | + | B | + + ( | B | + − ( 114 r + r ) ≤ (2 r +
12 ) n − ( 114 r + r ) since | B | + = n − | A | ≤ n < ⌊ n ⌋ as n ≥ r and r ≥ , a contradiction. So the claim holds. (cid:3) Claim 4.10.
Let H be a nontrivial component of G [ A ] . Then e ( H ) + e ( V ( H ) , B ) ≤ | H || B | ,and if equality holds then H (cid:27) K r + . Proof.
By Claim 4.8, there exists an A -path P with l ( P ) = r . We first show that either V ( H ) ⊆ V ( P ) or V ( H ) ∩ V ( P ) = ∅ . By the way of contradiction, suppose that we can pick u ∈ V ( H ) ∩ V ( P ) and u ′ ∈ V ( H ) \ V ( P ) such that u ∼ u ′ . Since l ( P ) = r , we can get an A -subpath P ⋆ of P of length r starting from u . By Claim 4.7, we have u ′ ∈ V ( P ⋆ ) ⊆ V ( P ),a contradiction. Hence V ( H ) ⊆ V ( P ) or V ( H ) ∩ V ( P ) = ∅ .Now by Claim 4.9, there exists another A -path P ′ of length 2 r with V ( P ) ∩ V ( P ′ ) ∩ A = ∅ . Similarly, we have either V ( H ) ⊆ V ( P ′ ) or V ( H ) ∩ V ( P ′ ) = ∅ . It follows thateither V ( H ) ∩ V ( P ) = ∅ or V ( H ) ∩ V ( P ′ ) = ∅ . Without loss of generality, assume that V ( H ) ∩ V ( P ) = ∅ .Let P = u v · · · u r v r u r + , where u i ∈ A and v i ∈ B . We next show that N G ( u ) ∩{ v , v , . . . , v r } = ∅ for each u ∈ V ( H ). Suppose to the contrary that u ∼ v i for some u ∈ V ( H ) and i ∈ { , , . . . , r } . Combining the edge uv i with the subpath of P of length r − v i , we can get an A -path of length r starting from u . By Claim 4.7, onecan see that N A ( u ) ⊆ V ( P ), which contradicts V ( H ) ∩ V ( P ) = ∅ .Now, we have d B ( u ) ≤ | B | − r for each u ∈ V ( H ). Thus e ( V ( H ) , B ) ≤ | H | ( | B | − r ) . Since H is P r + -free, by Lemma 2.1, we have 2 e ( H ) ≤ r | H | , with equality if and onlyif H (cid:27) K r + . It follows that 2 e ( H ) + e ( V ( H ) , B ) ≤ | H || B | , and if equality holds then H (cid:27) K r + . (cid:3) In the rest, we only need to show that e ( A ) =
0, then by Lemma 2.3, the proof iscompleted. Claim 4.10 implies that 2 e ( H ) + e ( V ( H ) , B ) ≤ | H || B | for any component H of G [ A ]. Furthermore, if G [ A ] contains a good component, then we can easily have2 e ( A ) + e ( A , B ) < | A || B | . It follows that γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) < | A | ( | B | + ≤ ⌊ n ⌋ , a contradiction. So may assume that all components of G [ A ] are not good, that is,2 e ( H ) + e ( V ( H ) , B ) = | H || B | (10)for any component H of G [ A ]. This implies that γ ( u ⋆ ) = | A | + e ( A ) + e ( A , B ) = | A | + | A || B | = | A | ( | B | + ≤ ⌊ n ⌋ , Combining with Lemma 2.3, we have γ ( u ⋆ ) = ⌊ n ⌋ , | A | = ⌈ n ⌉ and | B | = n − − | A | > r + . Now suppose that e ( A ) , H be a nontrivial component of G [ A ]. Then by (10)and Claim 4.10, we have H (cid:27) K r + . Note that G is θ r + -free. One can see that d H ( w ) ≤ w ∈ B . Therefore,2 e ( H ) + e ( V ( H ) , B ) ≤ e ( K r + ) + | B | = r ( r + + | B | < r | B | + | B | = | H || B | , since r + < | B | . This contradicts (10). So e ( A ) =
0. This completes the proof.5
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