Extremal problems of Erdős, Faudree, Schelp and Simonovits on paths and cycles
aa r X i v : . [ m a t h . C O ] F e b Extremal problems of Erd˝os, Faudree, Schelp and Simonovits on pathsand cycles
Binlong Li ∗ Jie Ma † Bo Ning ‡ Abstract
For positive integers n > d ≥ k , let φ ( n, d, k ) denote the least integer φ such that every n -vertexgraph with at least φ vertices of degree at least d contains a path on k + 1 vertices. Many years ago,Erd˝os, Faudree, Schelp and Simonovits proposed the study of the function φ ( n, d, k ), and conjecturedthat for any positive integers n > d ≥ k , it holds that φ ( n, d, k ) ≤ ⌊ k − ⌋⌊ nd +1 ⌋ + ǫ , where ǫ = 1 if k is odd and ǫ = 2 otherwise. In this paper we determine the value of the function φ ( n, d, k ) exactly.This confirms the above conjecture of Erd˝os et al. for all positive integers k = 4 and in a correctedform for the case k = 4. Our proof utilizes, among others, a lemma of Erd˝os et al. [3], a theorem ofJackson [6], and a (slight) extension of a very recent theorem of Kostochka, Luo and Zirlin [7], wherethe latter two results concern maximum cycles in bipartite graphs. Besides, we construct examplesto provide answers to two closely related questions raised by Erd˝os et al. We consider the following extremal problem asked by Erd˝os, Faudree, Schelp and Simonovits in [3]: forgiven positive integers n > d ≥ k , what is the minimum value ℓ such that every n -vertex graph withat least ℓ vertices of degree at least d contains a path P k +1 on k + 1 vertices? The goal of the presentpaper is to provide a complete solution for all positive integers n > d ≥ k .One of the best known results in extremal graph theory is the Erd˝os-Gallai Theorem [4], whichstates that any n -vertex graph with more than ( k − n/ k + 1 vertices. Sincethen, there has been many other extremal results on the existence of long paths in graphs with a largenumber of edges or vertices of high-degree. In this paper we investigate the following function, whosestudy was proposed by Erd˝os, Faudree, Schelp and Simonovits [3]. Definition 1.1.
For positive integers n > d ≥ k , define φ ( n, d, k ) to be the smallest integer φ such thatevery n -vertex graph with at least φ vertices of degree at least d contains a path P k +1 on k + 1 vertices. In this language, the well-known theorem of Dirac [2] asserts that φ ( n, d, d ) ≤ n and by that,we see that the function φ ( n, d, k ) is well-defined if and only if d ≥ k . A result of Bazgan, Li andWo´zniak [1] shows that φ ( n, d, d ) ≤ n . For the general case, Erd˝os, Faudree, Schelp and Simonovits [3]announced that for any k , there exists a constant c such that if n is large enough with respect to k ,then φ ( n, d, k ) ≤ ⌊ k − ⌋⌊ nd +1 ⌋ + c. They [3] further made the following conjecture (also see [5]). ∗ School of Mathematics and Statistics, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, China. Email:[email protected]. Partially supported by NSFC grant 12071370. † School of Mathematical Sciences, University of Science and Technology of China, Hefei, Anhui 230026, China. Email:[email protected]. Partially supported by NSFC grant 11622110, National Key Research and Development ProjectSQ2020YFA070080, and Anhui Initiative in Quantum Information Technologies grant AHY150200. ‡ College of Computer Science, Nankai University, Tianjin, 300071, China. Email: [email protected]. Partiallysupported by NSFC grant 11971346. onjecture 1.2 (Erd˝os, Faudree, Schelp and Simonovits, [3]) . Let n, d and k be any positive integerswith n > d ≥ k . Then φ ( n, d, k ) ≤ ⌊ k − ⌋⌊ nd +1 ⌋ + ǫ , where ǫ = 1 if k is odd and ǫ = 2 otherwise. Much attention in [3] was devoted to the special case when d + 1 ≤ n ≤ d + 1. In this case,the authors [3] showed that approximately k/ d are enough to ensure theexistence of P k +1 . They also commented that “unfortunately, even for this interval of values we arenot able to prove the exact statement of the conjecture.” However, as we shall see later, this case (i.e., d + 1 ≤ n ≤ d + 1) is a major difficulty that we face to and where several novel ideas take place in ourproof.Our main result determines the function φ ( n, d, k ) completely in the following statement. Theorem 1.3.
For any positive integers n, d and k with n > d ≥ k , the followings are true:(i) If k is odd, then φ ( n, d, k ) = k − q + 1 , where n = q ( d + 1) + r with ≤ r ≤ d .(ii) If k is even, then(a) for k = 2 , φ ( n, d,
2) = 1 ;(b) for k = 4 , φ ( n, d,
4) = ( q + 1 , ≤ r ≤ d ;2 q + 2 , d < r < d, where n = 2 qd + r with ≤ r < d ;(c) for k ≥ , φ ( n, d, k ) = ( k − q + 1 , ≤ r ≤ d − k ; k − q + 2 , d − k < r ≤ d, where n = q ( d + 1) + r with ≤ r ≤ d . We see immediately that φ ( n, d, k ) = ⌊ k − ⌋⌊ nd +1 ⌋ + 1 when k is odd, and φ ( n, d, k ) ≤ ⌊ k − ⌋⌊ nd +1 ⌋ + 2when k = 4 is even. However, the case k = 4 is different. In summary, we have the following. Corollary 1.4.
Conjecture 1.2 is true for any integer k = 4 and false for k = 4 . Our proof of Theorem 1.3 is inductive in its nature. For that it is crucial for us to manage thebase case when d + 1 ≤ n ≤ d + 1. It turns out in the proof of the base case that we make use ofresults on maximum cycles in bipartite graphs due to Jackson [6] and Kostochka, Luo and Zirlin [7](see Theorems 3.1 and 3.2, respectively). To be precise, we partition the vertex set into two parts X and Y , where X consists of vertices of degree at least d , and then apply the above results on maximumcycles to find few number of disjoint paths in the bipartite subgraph G ( X, Y ) to cover all vertices in X ; finally, an application of a lemma of Erd˝os et al. in [3] (see Lemma 3.6) will ensure the desiredlong path. We would like to point out that it seems to be an incredible coincidence that the boundswe need in this argument are exactly what the recent result of Kostochka, Luo and Zirlin [7] provided.The case when n ≥ d + 2 will be handled differently, which is essentially reduced to the base case.The remainder of this paper is organized as follows. In Section 2, we construct extremal graphs forthe function φ ( n, d, k ) and establish the lower bound of Theorem 1.3. In Section 3, we introduce thenotation, a lemma of Erd˝os et al., and results of Jackson and Kostochka, Luo and Zirlin on maximumcycles in bipartite graphs; we also provide some variance and extension of these cycle results for thecoming proof. In Section 4, we complete the proof of Theorem 1.3. In Section 5, we give betterconstructions to answer two questions in [3] which are closely related to Conjecture 1.2. In this section, we construct extremal graphs for the function φ ( n, d, k ). This will give the matchedlower bound of φ ( n, d, k ) in Theorem 1.3.We start with some notation. Let n > d ≥ k be positive integers and let G, H be two graphs. By G + H we mean the disjoint union of G and H , and we use k · G to denote the union of k disjoint copies2f the same graph G . Let K n be the n -vertex clique, I n be the graph induced by an independent set of n vertices, and K ,n be the star with n leaves. We define two special yet important graphs as following(see Figure 1): • The graph H d,k is obtained from the disjoint union of K ⌊ k − ⌋ and I d +1 −⌊ k − ⌋ by joining everyvertex of K ⌊ k − ⌋ to every vertex of I d +1 −⌊ k − ⌋ . • For even integers k ≥
4, let H ∗ d,k be the graph obtained from H d,k by adding a disjoint copy of I d +1 − k and joining every vertex in I d +1 − k to a fixed vertex of degree k − H d,k .Note that H d,k has d +1 vertices in total and ⌊ k − ⌋ vertices of degree at least d , while H ∗ d,k has 2 d +2 − k vertices in total and k vertices of degree at least d . In particular, H ∗ d, is the graph obtained from twodisjoint stars on d vertices by joining the two centers (we will also call it a double-star ).With the above notation, now we define the extremal graph G ( n, d, k ) for the function φ ( n, d, k ). Definition 2.1.
For positive integers n > d ≥ k , we define the graph G ( n, d, k ) as follows. • For k ∈ { , } , let G ( n, d, k ) = I n . • For k = 4 , write n = 2 qd + r with ≤ r < d and let G ( n, d, k ) = ( q · H ∗ d, + I r , if r ≤ d ; q · H ∗ d, + K ,d + I r − d − , otherwise. • For odd k ≥ , write n = q ( d + 1) + r with ≤ r ≤ d and let G ( n, d, k ) = q · H d,k + I r . • For even k ≥ , write n = q ( d + 1) + r with ≤ r ≤ d and let G ( n, d, k ) = ( q · H d,k + I r , if r ≤ d − k ;( q − · H d,k + H ∗ d,k + I r − d + k − , otherwise. It is straightforward to check the following fact on G ( n, d, k ). Lemma 2.2.
For any positive integers n > d ≥ k , the n -vertex graph G ( n, d, k ) contains no P k +1 andthus the lower bound of Theorem 1.3 holds. Let G be a graph. For disjoint subsets X, Y ⊆ V ( G ), we use G ( X, Y ) to denote the bipartite subgraphof G induced by two parts X and Y . Let P be a path or a cycle. By | P | , we mean the number ofvertices in P . For x, y ∈ V ( P ), let xP y be a subpath of P between x and y . When P is associatedwith an orientation, the successor and predecessor of x along the direction are denoted by x + and x − (if exist), respectively. We also denote by x ++ := ( x + ) + and x −− := ( x − ) − . For a subset S ⊆ V ( G ),we define N ( S ) to be the set of all vertices x ∈ V ( G ) \ S which is adjacent to some vertex in S . If S consists of a single vertex x , then we write N ( S ) as N ( x ).We now introduce two theorems on the existence of maximum cycles in bipartite graphs, whichprovide crucial tools for the proof of our main result. The first one is due to Jackson [6]. Throughout this paper, the word disjoint always mean for vertex-disjoint unless otherwise specified. ⌊ k − ⌋ H d,k d + 1 − ⌊ k − ⌋ K k − H ∗ d,k d + 1 − k Figure 1: H d,k (for all k ) and H ∗ d,k (for even k ) Theorem 3.1 (Jackson, [6, Theorem 1]) . Let G be a bipartite graph with two parts X and Y . If ≤ | X | ≤ d , | Y | ≤ d − , and every vertex in X has degree at least d , then G has a cycle containingall vertices in X . The following theorem, conjectured by Jackson [6] and proved by Kostochka, Luo and Zirlin [7]recently, extends the above theorem of Jackson to the setting of 2-connected graphs.
Theorem 3.2 (Kostochka et al., [7, Theorem 1.6]) . Let G be a 2-connected bipartite graph with twoparts X and Y . If ≤ | X | ≤ d , | Y | ≤ d − , and every vertex in X has degree at least d , then G hasa cycle containing all vertices in X . Our proof actually needs some intermediate statements from the proof of [7]. Let us give somenotation used in [7] first. Let G be bipartite with parts X and Y which is not a forest. For a cycle C and a vertex x in G , we say ( C, x ) is a tight pair if C is a longest cycle in G , x ∈ X \ V ( C ), and subjectto these, d C ( x ) := | N ( x ) ∩ V ( C ) | is maximum. Clearly G has a tight pair if and only if G has no cyclecontaining all vertices in X . The followings are collected from the proof of Theorem 1.6 in [7]. Lemma 3.3 (Kostochka et al., [7]) . Let G be a bipartite graph with two parts X and Y such that | X | ≤ d ≤ min { d ( x ) : x ∈ X } . Let ( C, x ) be a tight pair in G with c = | C | / . Then the followings hold:(i) If d C ( x ) ≤ and there is a path connecting two vertices in C and passing through x , then | Y | ≥ d − (see Case 1 in the proof of Theorem 1.6 in [7]);(ii) If ≤ d C ( x ) < c , then | Y | ≥ d − (i.e., Lemma 2.6 in [7]);(iii) If d C ( x ) = c , then for each x i ∈ X ∩ V ( C ) and each y ∈ N G − C ( x i ) , x i is a cut-vertex separating y from V ( C ) − x i (i.e., Lemma 2.7 in [7]). Let G be a connected graph which is not a forest. We say that G is essentially-2-connected , if G − V is 2-connected, where V denotes the set of vertices of degree one in G . We need a variance ofTheorem 3.2 for essentially-2-connected graphs. Lemma 3.4.
Let G be an essentially 2-connected bipartite graph with parts X and Y . Suppose that ≤ | X | ≤ d − , | Y | ≤ d − , and every vertex in X has degree at least d . Then G has a cyclecontaining all vertices in X . roof. Suppose for a contradiction that G has no cycle containing all vertices in X . Let ( C, x ) be atight pair of G and c = | C | /
2. So c < | X | . Since every vertex in X has degree at least d ≥
3, we see that X ∪ V ( C ) are contained in the 2-connected subgraph G − V and thus there is a path connecting twovertices in C and passing through x . If d C ( x ) < c , then by Lemma 3.3 (i) and (ii), we get | Y | ≥ d − d C ( x ) = c , i.e., x is adjacent to all vertices in Y ∩ V ( C ).We claim that every vertex y ∈ Y \ V ( C ) has degree one in G . Suppose otherwise that there existssome y ∈ Y \ V ( C ) with d ( y ) ≥
2. Since G is essentially 2-connected, there is a path P connecting twovertices in C and passing through y . By Lemma 3.3 (iii), the end-vertices of P are both in Y ∩ V ( C )and | P | ≥ G is bipartite. Let y , y be the end-vertices of P . If there exists a subpath Q between y and y in C of length two, then replacing Q with the path P in C , we can get a longer cycle than C , a contradiction. Fix an orientation of C . Then we have that y = y ++1 (and also y = y ++2 ) in C . If x ∈ V ( P ), then y ∈ V ( xP y i ) for some i ∈ { , } . Without loss of generality, suppose that y ∈ V ( xP y ). Replacing y y +1 y ++1 with the path y P x ∪ xy ++1 in C , again we have a longer cyclethan C , a contradiction. So x / ∈ V ( P ). Let C ′ be the cycle obtained from C by deleting the edgesin y y +1 y ++1 ∪ y y +2 y ++2 and adding the paths P and y ++1 xy ++2 . Then C ′ is a longer cycle than C , acontradiction. This proves the claim.Note that every vertex in X has at least d − c neighbors outside C . By the previous claim, theseneighbors all have degree one in G and thus are distinct for different vertices in X . This shows that | Y | ≥ c + | X | ( d − c ) ≥ c + ( c + 1)( d − c ) = c ( d − c ) + d . Since 2 ≤ c ≤ | X | − ≤ d −
2, now we can inferthat | Y | ≥ d −
4, a contradiction. This completes the proof of the lemma.We remark that the condition | X | ≤ d − | X | ≤ d by thefollowing examples. Let H = H ( X, Y ) be a complete bipartite graph with | X | = d and | Y | = d − G be the bipartite graph obtained from H by adding at least one new vertex x ′ for each vertex x ∈ X and then adding the edge xx ′ for every new vertex x ′ . Let Y be the part of G other than X .Then the size of Y can be any integer at least 2 d −
1, every vertex in X has degree at least d in G , and G is essentially 2-connected but has no cycle containing all vertices in X .The following lemma will be pivotal for the proof of our main result Theorem 1.3. Lemma 3.5.
Let G be a bipartite graph with parts X and Y . Suppose every vertex in X has degree atleast d . Then the followings are true:(i) If | X | ≤ d + 1 and | Y | ≤ d − , then G has a path containing all vertices in X ;(ii) If G is connected, | X | ≤ d and | Y | ≤ d − , then G has a path containing all vertices in X ;(iii) Let t ≥ be any integer. If | X | ≤ d + t and | Y | ≤ d + 2 t − , then there exist at most t + 1 disjoint paths in G containing all vertices in X .Proof. (i). It is obvious when | X | = 1. So assume | X | ≥
2. Let G ′ be the graph obtained from G byadding a new vertex y and joining y to every vertex in X . Then every vertex in X has degree at least d + 1 in G ′ . Since | X | ≤ d + 1 and | Y ∪ { y }| ≤ d = 2( d + 1) −
2. By Theorem 3.1, G ′ has a cycle C containing all vertices in X . The vertex y may be contained in C or not. In either case, one can find apath in G containing all vertices in X (by considering C − y ). This proves (i).(ii). Similarly we may assume | X | ≥
2. Let G ′ be obtained from G by adding a new vertex y andjoining y to every vertex in X . Let Y ′ = Y ∪ { y } . Then we have that 2 ≤ | X | ≤ d , | Y ′ | ≤ d − X has degree at least d + 1 in G ′ . We claim that G ′ is essentially 2-connected. Tosee this, consider a spanning tree T in G (note that G is connected). Let G ′′ be obtained from T byadding the vertex y and joining y to every vertex in X . Clearly we have G ′′ ⊆ G ′ , and by definition,5 ′′ is essentially 2-connected. This implies that G ′ is essentially 2-connected. Now applying Lemma3.4, we can conclude that G ′ has a cycle C containing all vertices in X . The vertex y lies on C or not.In either case, C − y (and thus G ) contains a path containing all vertices in X . This proves (ii).(iii). Let G ′ be obtained from G by deleting all isolated vertices in Y , adding t new vertices y , ..., y t and then joining every y i for i ∈ [ t ] to all vertices in X . Let Y be the set of all isolated vertices of G in Y and let Y ′ = ( Y \ Y ) ∪ { y , ..., y t } . So G ′ is a connected bipartite graph with parts X and Y ′ suchthat | X | ≤ d + t , | Y ′ | ≤ | Y | + t ≤ d + t ) −
3, and every vertex in X has degree at least d + t in G ′ .By (ii), G ′ has a path P containing all vertices in X . By deleting the vertices y , ..., y t , we can obtainat most t + 1 disjoint paths (in G ) from P containing all vertices of X . This proves Lemma 3.5.Lastly, we need the following convenient lemma due to Erd˝os, Faudree, Schelp and Siminovits [3].An analog for cycles can be found in [8]. Lemma 3.6 (Erd˝os et al., [3, Lemma 1]) . Let G be a graph with at most d + 1 vertices and P be anyfamily of disjoint paths P i , where both end-vertices of each P i ∈ P have degree at least d in G . Then G contains a path Q such that both its end-vertices have degree at least d in G and S P i ∈P V ( P i ) ⊆ V ( Q ) . For given positive integers n > d ≥ k , let φ be the function φ ( n, d, k ) defined in Theorem 1.3 throughoutthis section. To complete the proof of Theorem 1.3, in view of Lemma 2.2, it suffices to prove thatany n -vertex graph with at least φ vertices of degree at least d contains a path P k +1 . (1)We will prove this by contradiction. Consider any positive integers d ≥ k for which (1) fails. Thenthere exists a counterexample G to the statement (1) as follows:(a). G is an n -vertex graph with at least φ = φ ( n, d, k ) vertices of degree at least d ,(b). G does not contain any path P k +1 (on k + 1 vertices),(c). subject to (a) and (b), n is minimum, and(d). subject to (a), (b) and (c), G has the minimum number of edges.We proceed the proof by proving a sequence of claims. Our first claim is the following. In the restof the proof, we say a vertex is high if it has degree at least d in G and low otherwise. A path is a high-end path if both its end-vertices are high vertices. Claim 4.1. G has no high-end path on at least k − vertices.Proof. Suppose that P is a high-end path on at least k − u, v be end-vertices of P . If | P | ≥ k + 1, then there is nothing to prove. If | P | = k , as d ( u ) ≥ d ≥ k , u has a neighbor u ′ outside V ( P ), and P ∪ uu ′ is a desired path P k +1 . So | P | = k −
1. Then u has a neighbor u ′ outside V ( P ),and v has a neighbor v ′ outside V ( P ) ∪ { u ′ } . Now u ′ u ∪ P ∪ vv ′ is a path P k +1 . In any case, we get acontradiction. Claim 4.2.
We may assume that k ≥ . roof. We first point out that the case k ∈ { , } is trivial: a graph G contains P if and only if G hasa vertex of degree at least 1, and G contains P if and only if G has a vertex of degree at least 2.Now consider the case k = 3. So G contains no P . Then each component H of G is a K , K , K or a star K ,s for some s ≥
2. Note that only the component K ,s with s ≥ d can have one high vertex.It follows that H has at most ⌊ | V ( H ) | d +1 ⌋ high vertices. So G has at most X each component H (cid:22) | V ( H ) | d + 1 (cid:23) ≤ (cid:22) nd + 1 (cid:23) = φ − k = 4. In this case, G contains no P . Let H be any component of G . ByClaim 4.1, H has no high-end path on 3 vertices. This shows that H has at most two high vertices,and if u, v are the two high vertices of H , then uv must be a cut-edge of H and thus N ( u ) ∩ N ( v ) = ∅ .It also follows that if H has only one high vertex, then | V ( H ) | ≥ d + 1 and if H has two high vertices,then | V ( H ) | ≥ d . So we can conclude that H has at most | V ( H ) | d high vertices (and if H has exactly | V ( H ) | d high vertices, then | V ( H ) | = 2 d ). Therefore G has at most X each component H | V ( H ) | d ≤ nd high vertices. This gives that the number of high vertices of G is at most φ −
1, a contradiction.We now discuss several useful properties that the graph G has. Claim 4.3. If u is a low vertex in G , then every neighbor of u has degree exactly d and thus is high.Proof. Suppose, otherwise, that there is a vertex v ∈ N ( u ) with d ( v ) ≤ d − d ( v ) ≥ d + 1. Then G ′ = G − uv is a graph satisfying ( a ) , ( b ) and ( c ), but having less edges than G . This violates ( d ) andthe choice of G . Claim 4.4. G is connected.Proof. Suppose that G is not connected. Then G is a disjoint union of two subgraphs G and G . For i ∈ { , } , let | V ( G i ) | := n i = q i ( d + 1) + r i for 0 ≤ r i ≤ d , and let n = q ( d + 1) + r for 0 ≤ r ≤ d .As n = n + n , we have that either (1) q = q + q and r = r + r ≤ d , or (2) q = q + q + 1 and0 ≤ r = r + r − ( d + 1) ≤ d . Note that each G i contains no P k +1 .First consider that k is odd, or k ≥ r i ≤ d − k for both i ∈ { , } . In this case, since G i has no P k +1 , each G i has at most q i ⌊ k − ⌋ high vertices. So G has at most ( q + q ) ⌊ k − ⌋ ≤ q ⌊ k − ⌋ ≤ φ − k ≥ r and r is at most d − k . Without loss ofgenerality, assume that r ≤ d − k and r > d − k . Then G has at most q k − high vertices and G has at most q k − + 1 high vertices. Thus G has at most ( q + q ) k − + 1 high vertices. Note thateither q = q + q + 1, or q = q + q and d ≥ r = r + r > d − k . In both cases, we see that thenumber of high vertices in G is at most ( q + q ) k − + 1 ≤ φ −
1, a contradiction.Finally, k ≥ r i > d − k for both i ∈ { , } . In this case, each G i has at most q i k − + 1high vertices, and so G has at most ( q + q ) k − +2 high vertices. Since r + r > d − k ≥ d , we must havethat q = q + q + 1. Thus the number of high vertices in G is at most ( q + q ) k − + 2 ≤ q k − ≤ φ − k ≥
6. This contradiction completes the proof of Claim 4.4. We also see that in this case, if G has exactly nd high vertices, then every component of G forms a double-star H ∗ d, . laim 4.5. G has exactly φ high vertices.Proof. Suppose that G has at least φ + 1 high vertices. If every vertex is high, then δ ( G ) ≥ d ≥ k and by Erd˝os-Gallai Theorem [4], there is a path of length d ≥ k in G , a contradiction. So there issome low vertex, say u in G . By Claim 4.4, G is connected, so there exists some v ∈ N ( u ). By Claim4.3, v is high. Then G ′ = G − uv is a graph satisfying ( a ) , ( b ) and ( c ), but with less number of edges,contradicting the choice of G . This proves the claim.In the rest of the proof, we write n = q ( d + 1) + r for some integers q ≥ ≤ r ≤ d . Since k ≥
5, we have φ = ⌊ k − ⌋ q + 1 + ǫ , where ǫ = 1 if k is even and r > d − k , and ǫ = 0 otherwise. Claim 4.6.
We have n ≥ d + 2 and thus q ≥ .Proof. Suppose that d < n ≤ d + 1. So q = ⌊ nd +1 ⌋ = 1 and n = d + 1 + r . We will reach a contradictionto Claim 4.1 by finding a high-end path in G on at least k − X be the set of all highvertices in G and Y = V ( G ) \ X . Let G ′ = G ( X, Y ) be the spanning bipartite subgraph of G with parts X and Y . We have | X | = φ by Claim 4.5.Suppose that k is odd. Then | X | = φ = k +12 , and | Y | = n − | X | ≤ d + 1 − k +12 . For every x ∈ X , d G ′ ( x ) ≥ d G ( x ) − ( | X | − ≥ d − k − := d . Since d ≥ k , we can derive that | X | = k +12 ≤ d and | Y | ≤ d + 1 − k +12 ≤ d −
1. By Lemma 3.5 (iii) with t = 1, G ′ has at most two disjoint paths (say P , P ) containing all vertices in X . We may assume that all end-vertices of P , P are in X , so P , P are high-end paths of G . As n ≤ d + 1, by Lemma 3.6, there is a high-end path P in G satisfying V ( P ) ∪ V ( P ) ⊆ V ( P ) and thus | P | ≥ | P | + | P | = 2 | X | − k − k is even and r ≤ d − k . Then | X | = φ = k and | Y | = n −| X | = ( d +1+ r ) −| X | ≤ d + 1 + ( d − k ) − k = 2 d + 1 − k . Also for every x ∈ X , d G ′ ( x ) ≥ d − k + 1 := d . Since d ≥ k , we see | X | ≤ d and | Y | ≤ d −
1. By Lemma 3.5 (i), G ′ has a path P containing all vertices in X . We mayview P as a path with both end-vertices in X . So P is a high-end path of G with | P | = 2 | X | − k − k is even and d ≥ r > d − k . In this case, | X | = φ = k + 1, | Y | = n − | X | = ( d + 1 + r ) − | X | ≤ d − k , and for every x ∈ X , d G ′ ( x ) ≥ d − k := d . Since d ≥ k ,one can deduce that | X | ≤ d + 1 and | Y | ≤ d . By Lemma 3.5 (iii) with t = 2, G ′ has at mostthree disjoint paths P , P , P containing all vertices in X , all of which can be viewed as high-end pathsof G . Using Lemma 3.6, G has a high-end path P containing all vertices in P ∪ P ∪ P . Therefore | P | ≥ | P | + | P | + | P | = 2 | X | − (cid:0) k + 1 (cid:1) − k − Claim 4.7. If T is any set of at least d + 1 vertices, then T ∪ N ( T ) contains at least ⌊ k +12 ⌋ high vertices.Proof. Suppose that T ∪ N ( T ) contains at most ⌊ k − ⌋ high vertices. Let T ′ be any subset of T with | T ′ | = d + 1, and let G ′ be obtained from G by deleting all vertices in T ′ . Then n ′ := | V ( G ′ ) | = n − ( d + 1) > d and G ′ has at least φ − (cid:4) k − (cid:5) = φ ( n ′ , d, k ) high vertices. Thus G ′ is a counterexamplesmaller than G (which satisfies ( a ) and ( b ) but violates ( c )), a contradiction.Since G is connected and has φ ≥ G . Now we choose ahigh-end path P in G such that the number of high vertices in P is maximum, and subject to this, | P | is maximum. By Claim 4.1, we have | P | ≤ k −
2. Note that as q ≥
2, we have φ ≥ ⌊ k − ⌋ q + 1 ≥ k − V ( P ).Let u , u be the two end-vertices of P . We assign the orientation of P from u to u . Claim 4.8.
Let S i = N G − P ( u i ) for i ∈ { , } and S = S ∪ S . Then any vertex in S is a low vertex, S ∩ S = ∅ , and N P ( u ) ∪ { u } ⊆ V ( P ) \ ( N P ( u )) + . In particular, we have d P ( u ) + d P ( u ) ≤ | P | − . roof. First any vertex in S is a low vertex (as, otherwise, say s ∈ N G − P ( u ) is high, then P ∪ u s contradicts the choice of P ). If u , u has a common neighbor say v outside V ( P ), then v is a low vertexand let C = P ∪ u vu ; if there exists a vertex v ∈ N P ( u ) ∩ ( N P ( u )) + such that u v, u v − ∈ E ( G ),then let C = ( P − v − v ) ∪ { u v, u v − } . In any case C is a cycle containing all vertices in P . As wejust discussed before this claim, there exists some high vertex x outside V ( P ) (and thus outside V ( C )).Now let P ′ be any path in G from x to C and internally disjoint from C (recall Claim 4.4 that G isconnected). Then C ∪ P ′ contains a high-end path containing more high vertices than P , a contradictionto the choice of P .In the following claims, we investigate more properties on the sets S and S . Claim 4.9.
Every vertex in N ( S ) ∪ N ( S ) is high and lies in P , every vertex in ( N ( S ) \{ u } ) − ∪ ( N ( S ) \{ u } ) + is low, and moreover, | S | ≥ d + 3 .Proof. By Claim 4.3, any neighbor of a low vertex is a high vertex. So for i ∈ { , } , every vertexin N ( S i ) is high and must lie on P (otherwise, P can be extended to a longer high-end path, whichcontradicts the choice of P ). Consider any v ∈ S and x ∈ N ( v ) \{ u } . So x ∈ V ( P ) \{ u } . If x − ishigh, then P ′ = x − P u ∪ u vx ∪ xP u is a high-end path such that V ( P ) ( V ( P ′ ), a contradiction tothe choice of P . Thus we conclude that all vertices in ( N ( S ) \{ u } ) − are low, and similarly all verticesin ( N ( S ) \{ u } ) + are low. By Claim 4.8, one can get that | S | = | S | + | S | ≥ ( d ( u ) − d P ( u )) +( d ( u ) − d P ( u )) ≥ d − ( k − ≥ d + 3. Claim 4.10.
There is at most one vertex v ∈ V ( P ) such that v, v + ∈ N ( S ) . In particular, we have v ∈ N ( S ) \ N ( S ) and v + ∈ N ( S ) \ N ( S ) .Proof. First, both v, v + ∈ N ( S ) are high. If v ∈ N ( S ), then by Claim 4.9, v + is low, a contradiction.Thus v ∈ N ( S ) \ N ( S ), and similarly, v + ∈ N ( S ) \ N ( S ). Note that possibly v = u or v + = u .Suppose for a contradiction that there are two such vertices, say v and v . Assume that v ∈ V ( u P v ). Recall that v , v ∈ N ( S ) and v +1 , v +2 ∈ N ( S ). We remark that v +1 = v (because v − islow by Claim 4.9). So v , v +1 , v , v +2 appear on P in order. For i ∈ { , } , let x i ∈ S with x i v i ∈ E ( G )and y i ∈ S with y i v + i ∈ E ( G ) such that if v = u then choose x = x , and if v +2 = u then choose y = y . Now we define a new path P ′ as follows (see Figure 4): P ′ = u P v ∪ v x v ∪ v P v +1 ∪ v +1 y v +2 ∪ v +2 P u , if x = x and y = y ; u +1 P v ∪ v x u x v ∪ v P v +1 ∪ v +1 y v +2 ∪ v +2 P u , if x = x and y = y ; u P v ∪ v x v ∪ v P v +1 ∪ v +1 y u y v +2 ∪ v +2 P u − , if x = x and y = y ; u +1 P v ∪ v x u x v ∪ v P v +1 ∪ v +1 y u y v +2 ∪ v +2 P u − , if x = x and y = y . Then V ( P ′ ) = V ( P ) ∪ { x , x , y , y } , and the possible end-vertices u +1 and u − of P ′ can be low vertices.Let P ′′ be the path obtained from P ′ by removing its low end-vertices (which only can be u +1 and/or u − ). Note that u +1 (respectively, u − ) is an end-vertex of P ′ if and only if x = x (respectively, y = y ).By Claim 4.3, P ′′ is a high-end path and contains exactly the same high vertices as in P . However, P ′′ is longer than P , as | P ′′ | ≥ | P ′ | − x = x − y = y ≥ | P | + 2. This contradicts the choice of P .Now we can derive from Claims 4.9 and 4.10 that | N ( S ) | ≤ | P | +22 ≤ k . On the other hand, as | S | ≥ d + 3 (by Claim 4.9), S ∪ N ( S ) contains at least ⌊ k +12 ⌋ high vertices (by Claim 4.7). All verticesin S are low (by Claim 4.8), so | N ( S ) | ≥ ⌊ k +12 ⌋ . Combining the above, we have (cid:22) k + 12 (cid:23) ≤ | N ( S ) | ≤ | P | + 22 ≤ k . (2)9 u +1 v v +1 v v +2 u − u x = x y = y u u +1 v v +1 v v +2 u − u x x y = y u u +1 v v +1 v v +2 u − u x = x y y u u +1 v v +1 v v +2 u − u x x y y Figure 2: Key steps in the proof of Claim 4.10.This indicates that k is even, | P | = k − | N ( S ) | = k , and there is exactly one vertex v ∈ V ( P )satisfying Claim 4.10. Furthermore, by letting k = 2 s , we can express P = a b a · · · b j a j a j +1 b j +1 · · · b s − a s − for some 0 ≤ j ≤ s − a = u , a j = v, a j +1 = v + , a s − = u and N ( S ) = { a , a , ..., a s − } . Claim 4.11.
We have N ( S ) = { a , ..., a j } and N ( S ) = { a j +1 , ..., a s − } .Proof. It suffices to show that N ( S ) ⊆ { a , ..., a j } and N ( S ) ⊆ { a j +1 , ..., a s − } . By symmetry, wewill only prove that N ( S ) ⊆ { a , ..., a j } . Suppose not. Then there exists some a ℓ ∈ N ( S ) with ℓ ≥ j + 1, and we may assume that subject to the condition ℓ ≥ j + 1, ℓ is minimal. By Claim 4.9,we see that ( a ℓ ) − is a low vertex. This implies that ℓ ≥ j + 2 and b ℓ − = ( a ℓ ) − is low. Thenby the minimality of ℓ , we have a ℓ − ∈ N ( S ). Let z ∈ S , z ∈ S be two vertices such that z a ℓ , z a ℓ − ∈ E ( G ). Then P ′ := a j P u ∪ u z a ℓ ∪ a ℓ P u ∪ u z a ℓ − ∪ a ℓ − P a j +1 is a high-end pathsuch that V ( P ′ ) = ( V ( P ) \{ b ℓ − } ) ∪ { z , z } and | P ′ | = | P | + 1. So P ′ contains all high vertices of P and is longer than P , a contradiction to the choice of P . This proves Claim 4.11.Hence by Claim 4.9, every a i for 0 ≤ i ≤ s − b i for 1 ≤ i ≤ s − Claim 4.12.
For any a i / ∈ { v, v + } , every vertex z ∈ N G − P ( a i ) is a low vertex such that N ( z ) ⊆ V ( P ) .On the other hand, every b i satisfies N ( b i ) ⊆ V ( P ) .Proof. First we consider b i for any 1 ≤ i ≤ s −
2. Without loss of generality, we may assume 1 ≤ i ≤ j .Suppose that b i has a neighbor z outside V ( P ). Then by Claim 4.3, z is high. Let w ∈ S be such that wa i ∈ E ( G ). Now zb i ∪ b i P u ∪ u wa i ∪ a i P u is a high-end path containing more high vertices than P ,a contradiction. This proves N ( b i ) ⊆ V ( P ). Now consider any a i / ∈ { v, v + } . We are done by Claims 4.8and 4.9 if a i ∈ { u , u } . Hence, without loss of generality assume that 1 ≤ i ≤ j −
1. Consider z ∈ N G − P ( a i ). Let w ′ ∈ S be such that w ′ a i +1 ∈ E ( G ) and let P ′ := za i ∪ a i P u ∪ u w ′ a i +1 ∪ a i +1 P u .If z is high, then P ′ is a high-end path containing more high vertices than P , a contradiction. So anysuch z must be low. Suppose for a contradiction that z has a neighbor z ′ outside V ( P ). Then byClaim 4.3, z ′ is high. So z ′ z ∪ P ′ is a high-end path containing more high vertices than P , again acontradiction. Thus we can conclude that N ( z ) ⊆ V ( P ).We are ready to complete the proof of Theorem 1.3. Recall that G has a high vertex (say x ) outside V ( P ). Let U = V ( P ) \{ v, v + } . We can derive from Claim 4.12 that { v, v + } is a 2-cut of G separatingthe vertex x from the set S ∪ U . Let D be the union of (at most two) components in G − { v, v + } S ∪ U . Claim 4.12 also shows that all high vertices in D ∪ N ( D ) are those in V ( P ), i.e., vertices in N ( S ) = { a , a , ..., a s − } . By Claim 4.8, we have | S ∪ V ( P ) | = | S | + | S | + | P | ≥ d G − P ( u )+ d G − P ( u )+( d P ( u )+ d P ( u ))+1 = d ( u )+ d ( u )+1 ≥ d +1 . Either v or v + has some neighbor not in S ∪ V ( P ). Without loss of generality we assume that v hassome neighbor not in S ∪ V ( P ). By Claim 4.4, v has degree d . Also note that vv + ∈ E ( G ). So v hasat most d − S ∪ U . Set S ′ = ( S ∪ U ) \ N ( v ) . Then using | S ∪ V ( P ) | ≥ d + 1, we have | S ′ | ≥ | S ∪ U | − ( d −
2) = ( | S ∪ V ( P ) | − − ( d − ≥ d + 1 . By Claim 4.7, S ′ ∪ N ( S ′ ) contains at least s = k high vertices. However, as S ′ is a subset in D , bydefinition we see S ′ ∪ N ( S ′ ) ⊆ ( D ∪ N ( D )) \{ v } . We have pointed out that all high vertices in D ∪ N ( D )are a , a , ..., a s − . So S ′ ∪ N ( S ′ ) contains at most s − In this section, we consider two questions related to Conjecture 1.2 that are raised by Erd˝os, Faudree,Schelp and Simonovits in [3]. We will provide better constructions than the ones given in [3], whichgive (negative and positive) answers to their questions.It is natural to consider the analog of Definition 1.1 for long cycles. For integers n > d ≥ k ≥ θ ( n, d, k ) to be the smallest integer θ such that every n -vertex graph with at least θ verticesof degree at least d contains a cycle on at least k + 1 vertices. In this language, the well-knownDirac’s theorem [2] states that θ ( n, d, d ) ≤ n . Improving Dirac’s theorem, Woodall [9] proved that θ ( n, d, d ) ≤ ( d +2)( n − d if d is even and θ ( n, d, d ) ≤ d ( n − d − otherwise; while for the general case, he [9]showed that θ ( n, d, k ) ≤ ( k +3)( n − d . Recall the graph H d,k +1 that it contains no cycles on at least k + 1vertices and has d + 1 vertices in total, where ⌊ k ⌋ vertices have degree d (call them high vertices) and allother vertices have degree ⌊ k ⌋ (call them low vertices). By considering the graphs consisting of blocks H d,k +1 , the authors of [3] raised the following question: Is it possible that θ ( n, d, k ) ≤ ⌊ k ⌋⌊ n − d ⌋ + 2?In the following proposition, we give a negative answer to the above question. Proposition 5.1.
For any integer k ≥ , there exist infinitely many integers d such that the followingholds. There exists some constant c = c ( d, k ) > such that θ ( n, d, k ) > ( ⌊ k ⌋ + c ) · n − d for infiniteintegers n .Proof. We show a slightly stronger statement: Let α, β be positive integers such that d = (1 + α ) ⌊ k ⌋ and n = 1 + d + αβd satisfy n > d ≥ k ≥
2. Then θ ( n, d, k ) ≥ n − d ⌊ k ⌋ + n − ( d +1) αd +1.We construct an n -vertex graph G as follows. Let H be a copy of H d,k +1 with a low vertex v .For each 1 ≤ i ≤ β , let H i be obtained from α copies of H d,k +1 by identifying one low vertex fromeach copy of H d,k +1 (call the resulting vertex u i ); let v i be a low vertex in H i other than u i . Finally,let G be obtained from H , H , . . . , H β by identifying v i − and u i for all 1 ≤ i ≤ β . Since each blockof G is a copy of H d,k +1 , we see that G contains no cycles of at least k + 1 vertices. However, G has(1 + αβ ) ⌊ k ⌋ + β = n − d ⌊ k ⌋ + n − ( d +1) αd vertices of degree at least d . This proves the proposition.Note that in the above proof, one can take α ≥ k and α ≥ k . For the cases d > k in Lemma 5.1, the constant c can be taken up to (i.e., when it corresponds to α = 2).Another question concerned in [3] is the restricted version of Conjecture 1.2 when the graph G isassumed to be connected. For positive integers n > d ≥ k , define ψ ( n, d, k ) to be the smallest integer ψ n -vertex connected graph with at least ψ vertices of degree at least d contains a path P k +1 on k + 1 vertices. Erd˝os et al. [3] observed that the graph, obtained from ⌊ n − d ⌋ copies of H d, ⌈ k ⌉ +1 by identifying at a fixed high vertex of each H d, ⌈ k ⌉ +1 , contains no P k +1 . This gives that ψ ( n, d, k ) ≥⌊ k − ⌋⌊ n − d ⌋ + 2 , which is approximately a half of the number of high vertices in Conjecture 1.2 (as k → ∞ ). They [3] asked if there is a better construction. We show that it is possible to improve theleading coefficient of n by a positive constant factor in the above lower bound of ψ ( n, d, k ). Proposition 5.2.
For any integer k ≥ , there exist infinitely many integers d such that the followingholds. There exists some constant c ′ = c ′ ( d, k ) > such that ψ ( n, d, k ) > ( ⌊ k − ⌋ + c ′ ) · n − d for infiniteintegers n .Proof. Let α ≥ β ≥ d be any integers. Let d = 1 + α ⌊ k − ⌋ and n = 1 + β (1 + αd ). Note thatthe graph H ∗ = H d, ⌈ k ⌉− contains ⌊ k − ⌋ vertices of degree d , and all other vertices have degree ⌊ k − ⌋ (call them low vertices). We will construct an n -vertex connected graph G to show that ψ ( n, d, k ) ≥⌊ k − ⌋ n − d + (cid:0) − d ⌊ k − ⌋ (cid:1) n − αd + 2 as follows. Let H be a star K ,β with leaves v i for 1 ≤ i ≤ β . Foreach 1 ≤ i ≤ β , let H i be obtained from α copies of H ∗ by identifying one low vertex from each H ∗ (call the resulting vertex u i ). Now let G be obtained from H , H , ..., H β by identifying v i and u i foreach 1 ≤ i ≤ β . Then G contains no path P k +1 and has αβ ⌊ k − ⌋ + β + 1 = (cid:16) n − d − βd (cid:17) ⌊ k − ⌋ + β + 1 = ⌊ k − ⌋ n − d + (cid:0) − d ⌊ k − ⌋ (cid:1) n − αd + 1 vertices of degree at least d , as desired.It would be very interesting to determine the functions θ ( n, d, k ) and ψ ( n, d, k ) exactly. Acknowledgements.
The authors are very grateful to Professor Douglas R. Woodall for sending acopy of [9].
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