DDegree conditions forcing directed cycles ∗ Andrzej Grzesik † Jan Volec ‡ Abstract
Caccetta-H¨aggkvist conjecture is a longstanding open problem on degree conditionsthat force an oriented graph to contain a directed cycle of a bounded length. Motivatedby this conjecture, Kelly, K¨uhn and Osthus initiated a study of degree conditionsforcing the containment of a directed cycle of a given length. In particular, they foundthe optimal minimum semidegree, i.e., the smaller of the minimum indegree and theminimum outdegree, that forces a large oriented graph to contain a directed cycle ofa given length not divisible by 3, and conjectured the optimal minimum semidegreefor all the other cycles except the directed triangle.In this paper, we establish the best possible minimum semidegree that forces alarge oriented graph to contain a directed cycle of a given length divisible by 3 yet notequal to 3, hence fully resolve the conjecture of Kelly, K¨uhn and Osthus. We also findan asymptotically optimal semidegree threshold of any cycle with a given orientationof its edges with the sole exception of a directed triangle.
One of the most famous open problems in graph theory is the conjecture of Caccetta andH¨aggkvist from 1978.
Conjecture 1.1 (Caccetta-H¨aggkvist [6]) . Every n -vertex oriented graph G with mini-mum outdegree δ + ( G ) ≥ n(cid:96) contains a directed cycle of length at most (cid:96) . By an oriented graph, we understand a directed graph without loops and multipleedges. Since the time the conjecture was stated, it attracted a great amount of researchers.In particular, it was the topic of a 2006 workshop at the American Institute of Mathemat-ics. Partial results on Conjecture 1.1 include its proofs for large values of (cid:96) [6, 15, 17, 36],results with an additive error term in the bound for the cycle length [7, 28, 37] and a mul-tiplicative error term in the minimum outdegree assumption [5, 6, 13, 16, 25, 35], as well assolutions with an additional assumption on forbidden subgraphs [11, 30]. For more resultsand problems related to the Caccetta-H¨aggkvist conjecture, see a summary of Sullivan [33].A weaker conjecture, where n(cid:96) lower bounds the minimum semidegree, i.e., the smallerof the minimum outdegree and the minimum indegree, is also widely open, and is veryclosely related to a conjecture of Behzad, Chartrand and Wall [4] from 1970 on directedcages, i.e., directed graphs that are regular and have a prescribed girth.Motivated by these open problems, Kelly, K¨uhn and Osthus [21] conjectured the min-imum semidegree guaranteeing existence of a directed cycle of a given length. ∗ This work was partially supported by the National Science Centre grant 2016/21/D/ST1/00998 andby the ERC Consolidator Grant LaDIST 648509. † Faculty of Mathematics and Computer Science, Jagiellonian University, ul. Prof. St. (cid:32)Lojasiewicza 6,30-348 Krak´ow, Poland. E-mail:
[email protected] . ‡ Department of Mathematics, Faculty of Nuclear Sciences and Physical Engineering, Czech TechnicalUniversity in Prague, Trojanova 13, 120 00 Prague, Czech Republic. E-mail: [email protected] . a r X i v : . [ m a t h . C O ] F e b onjecture 1.2 (Kelly-K¨uhn-Osthus [21]) . For every (cid:96) ≥ there exists n := n ( (cid:96) ) suchthat every oriented graph G on n ≥ n vertices with δ ± ( G ) ≥ nk + k contains a directed cycleof length exactly (cid:96) , where k is the smallest integer greater than that does not divide (cid:96) . Observe that the conjectured semidegree threshold, if true, would be the best possible.Indeed, a so-called balanced blow-up of C k , i.e., an oriented graph on n = b · k verticessplit into k cyclically ordered parts of size b , where the edges goes from all the vertices ofeach part to all the vertices of the next part, has no (cid:96) -cycles and both the indegree andthe outdegree of every its vertex equals to b = nk .In their paper, Kelly, K¨uhn and Osthus proved Conjecture 1.2 for k = 3, i.e., when thecycle-length is not divisible by 3. Moreover, they showed that the corresponding bound( n + 1) / (cid:96) ≥ Theorem 1.3 ([21]) . For every (cid:96) ≥ there exists n := n ( (cid:96) ) such that every orientedgraph G on n ≥ n vertices with δ ± ( G ) ≥ n + contains a directed cycle of length exactly (cid:96) . Note that if Conjecture 1.1 is true for (cid:96) = 3, then the bound n +13 on the minimumsemidegree of a sufficiently large n -vertex oriented graph guarantees existence of any di-rected cycle of a fixed length. In 1978, Thomassen [38] actually conjectured that n/ n -vertex graphs with the minimum semidegree (cid:6) n − (cid:7) − n -vertex graph with the minimum semide-gree (cid:6) n − (cid:7) contains a Hamilton cycle. In fact, the results in [19, 21] lead to (cid:6) n − (cid:7) beingthe semidegree threshold for so-called pancyclicity, i.e., containing directed cycles of allpossible lengths.One of the natural ways to tackle Conjecture 1.2 is to consider its asymptotic relax-ation, i.e., adding an extra εn term to the minimum semidegree assumption, where ε > k ∈ { , } with an additional assumption on the cycle-lengths. Specifically, they as-sumed (cid:96) ≥
42 and (cid:96) ≥ k = 4 and k = 5, respectively. This was extended byK¨uhn, Osthus and Piguet [24] to an asymptotic version of Conjecture 1.2 for any k ≥ (cid:96) ≥ k . In this paper, we determine the exact value of the semidegree threshold of every directedcycle of length (cid:96) ≥ Theorem 1.4.
Fix an integer (cid:96) ≥ and let k be the smallest integer greater than thatdoes not divide (cid:96) . There exists n := n ( (cid:96) ) such that the following is true for every orientedgraph G on n ≥ n vertices: • If (cid:96) (cid:54)≡ mod and δ ± ( G ) ≥ nk + k − k , then G contains C (cid:96) , and • if (cid:96) ≡ mod and δ ± ( G ) ≥ n + , then G contains C (cid:96) .Moreover, for every (cid:96) ≥ the stated bound is the best possible. In particular, the threshold conjectured in [21] is correct only when either k = 3, or (cid:96) ≡ k ≥
5, i.e., when (cid:96) is divisible by 12, we also prove a stability-type result.2 heorem 1.5.
Fix an integer (cid:96) that is divisible by and let k be the smallest integergreater than that does not divide (cid:96) . There exists n := n ( (cid:96) ) such that every orientedgraph H on n ≥ n vertices with δ ± ( H ) ≥ nk (cid:0) − k (cid:1) that contains no closed walk oflength (cid:96) is homomorphic to C k . Observe now that Theorem 1.5 cannot hold in this form when (cid:96) = 3 a for odd integers a .Indeed, any balanced orientation of K n/ ,n/ yields an oriented graph G with semidegree δ ± ( G ) = n yet every closed walk in G has an even length. Nevertheless, when k = 4 and (cid:96) ≥
9, we prove the following stability-type result.
Theorem 1.6.
Fix an integer (cid:96) ≥ that is divisible by and not divisible by . Everyoriented graph H on n vertices with δ ± ( H ) ≥ n − o ( n ) that does not contain a closed walkof length (cid:96) can be made bipartite by removing o ( n ) vertices. The case of C is even more exceptional. Consider a balanced blow-up of C and placeinside one of the blobs an arbitrary one-way oriented bipartite graph B , i.e., a bipartitegraph with all its edges going from one part to the other one. Clearly, the obtained graphis C -free. However, if B is a one-way oriented K n , n then one must remove n vertices tomake the resulting graph bipartite.This paper is organized as follows. Section 2 introduces the notation and tools weuse. In Section 3, we construct oriented graphs with large minimum semidegree showingthat the threshold in Theorem 1.4 is the best possible. Most of the paper is devoted tothe proof of Theorem 1.4. Based on the discussion above, the presentation is divided intothree more or less independent parts: In Section 4, we focus on the cycle lengths (cid:96) with k ≥
5, Section 5 deals with the case (cid:96) ≥ k = 4, and we devote Section 6 to thesemidegree threshold of C . Finally, Section 7 is concerned with asymptotics of semidegreethresholds for cycles with a fixed non-cyclic orientation, and Section 8 concludes the paperby discussing possible future directions. Unless explicitly stated, all the graphs considered in this paper are oriented graphs , i.e.,directed graphs without loops and multiple edges (regardless of the direction). Given agraph G , we write V ( G ) and E ( G ) to denote its vertex-set and its edge-set, respectively.For two vertices u ∈ V ( G ) and w ∈ V ( G ), we write uw to denote the edge oriented from u to w . Note that if uw ∈ E ( G ) then wu / ∈ E ( G ), and vice versa.For a graph G and a vertex v ∈ V ( G ), we denote by N + G ( v ) and N − G ( v ) the outneighbor-hood and the inneighborhood of v , respectively. If the graph G is known from the context,we omit the subscript and simply write N + ( v ) and N − ( v ). More generally, for S ⊆ V ( G ),we define N + ( S ) := (cid:83) v ∈ S N + ( v ) and N − ( S ) := (cid:83) v ∈ S N − ( v ).The sizes of the outneighborhood and the inneighborhood of a vertex v are referredas the outdegree and the indegree of v , and we write δ + ( G ) and δ − ( G ) to denote theminimum outdegree and the minimum indegree over all the vertices of G , respectively. Bythe minimum semidegree δ ± ( G ) we mean the minimum value of δ + ( G ) and δ − ( G ).Whenever we write a path or a walk of length (cid:96) , we always mean a directed path ordirected walk with (cid:96) edges. For brevity, we write C (cid:96) or (cid:96) -cycle to denote a directed cycleof length (cid:96) , and P (cid:96) to denote a directed path of length (cid:96) . For a given integer s ≥
2, the s -shortcut is a path of length s with an additional edge from its first vertex to the last one,i.e., the graph obtained obtained from C s +1 after changing the orientation of one edge,see Figure 1. We call its unique vertex with outdegree 2 the source , and the unique vertexwith indegree 2 the sink . 3 P P P s Figure 1: The 2-shortcut, the 3-shortcut, the 4-shortcut and a general s -shortcut.Through the whole paper, we use the standard Bachmann-Landau notation: for twofunctions f and g from N to R , we write f ( n ) = o ( g ( n )) if lim f ( n ) g ( n ) = 0. With a slightabuse of notation, we also write f ( n ) ≥ o ( g ( n )) to denote that lim sup f ( n ) g ( n ) ≥ F is homomorphic to a graph H if there exists a map f : V ( F ) → V ( H ) that preserves the adjacencies, i.e., if uv ∈ E ( F ) then f ( u ) f ( v ) ∈ E ( H ). A blow-up of a graph H is a graph obtained from H by placing an independent set I v for every v ∈ V ( H ), and, for every uv ∈ E ( H ), connecting x to y for all x ∈ I u and y ∈ I v . In case | I v | = b for every v ∈ V ( G ), we call the obtained graph a b -blow-up of G , or a balancedblow-up of G . Note that a graph F is homomorphic to H if and only if F is a subgraphof some blow-up of H .For a fixed graph F , we say that a graph G is F -free if G has no (not necessarilyinduced) subgraph isomorphic to F . More generally, if F is a fixed family of graphs, wesay that G is F -free if G is F -free for every F ∈ F .One of the most important and powerful results in graph theory is the RegularityLemma introduced in 1976 by Szemer´edi [32]; see also [23, 31] for excellent surveys onthe topic. A well known application of the regularity method is the so-called
RemovalLemma [9]. Here we state its degree-form for oriented graphs, which can be derived, e.g.,using the regularity lemma for directed graphs of Alon and Shapira [2].
Proposition 2.1.
Fix an (cid:96) -vertex graph F . If an n -vertex graph G contains o ( n (cid:96) ) copiesof F , then G contains an F -free subgraph H with | E ( G ) | − o ( n ) edges. Moreover, onecan remove o ( n ) vertices from H to get an F -free H (cid:48) ⊆ G with δ ± ( H (cid:48) ) = δ ± ( G ) − o ( n ) . In order to prove Theorem 1.4, we will make use of some results on Caccetta-H¨aggkvistconjecture. As we stated in the introduction, the conjecture has been extensively studiedand various relaxations has been considered. A particular interest is given to the so-calledtriangle case of the conjecture, i.e., when (cid:96) = 3.The triangle case of the conjecture states that every oriented graph G on n verticeswith δ + ( G ) ≥ n/ δ + ( G ) ≥ . n which forces an n -vertexoriented graph G to contain C . After a series of improvements due to Bondy [5], Shen [35]and Hamburger, Haxell and Kostochka [13], Hladk´y, Kr´al’ and Norin [16] applied the flagalgebra method of Razborov [29] and established the following: Theorem 2.2 ([16]) . Every n -vertex oriented graph G with δ + ( G ) ≥ . n contains C . Note that De Joannis de Verclos, Sereni and the second author [18] established animprovement to 0 . n , however, for our purposes the bound from [16] will be sufficient.Following an idea of Shen [37, Theorem 1], we use the bound from Theorem 2.2 toobtain not very strong yet for our purposes good enough bound for longer cycles.4 emma 2.3. Every n -vertex oriented graph G with δ + ( G ) ≥ (1 − . /m ) n contains adirected cycle of length at most m .Proof. Let c := 1 − . /m . Suppose the statement is not true, let G be a counterex-ample with the smallest number of vertices, and n := | V ( G ) | . We have that δ + ( G ) ≥ cn , G contains no directed cycle of length at most 3 m , but any graph G (cid:48) on n (cid:48) < n verticeswith δ + ( G (cid:48) ) ≥ cn (cid:48) does contain a cycle of length at most 3 m .Consider an arbitrary vertex x ∈ V ( G ), and let X := N + ( x ) and X i +1 := X i ∪ N + ( X i ).Clearly, for any 1 ≤ i < m , the induced graph G [ X i ] contain no directed cycle of length atmost 3 m . Since it has less vertices than G , there exists x i ∈ X i such that | N + ( x i ) ∩ X i |
There exists n such that there is no { C , C } -free oriented graph G on n ≥ n vertices with | N + ( v ) | = 0 . n for every v ∈ V ( G ) . Lemma 2.5.
There exists n such that there is no { C , C , C } -free oriented graph G on n ≥ n vertices with | N + ( v ) | = 0 . n for every v ∈ V ( G ) . The proofs of both lemmas are computer-aided, and the verification SAGE-scripts areavailable on http://honza.ucw.cz/proj/sd_cycle . The two lemmas in turn yield thefollowing.
Corollary 2.6.
There is n ∈ N such that the following is true for every oriented graph G on n ≥ n vertices.1. If δ + ( G ) ≥ . n then G contains C .2. If δ + ( G ) ≥ . n then G contains C .Proof. We prove only the first part as the other one is completely analogous. Supposethere exists an n -vertex oriented graph G with δ + ( G ) ≥ . n that has no C . By theminimum degree variant of Removal Lemma stated in the moreover part of Proposition 2.1,5e find H ⊆ G on n (cid:48) vertices with δ + ( H ) ≥ . n (cid:48) that is also C -free. Moreover, wemay assume that | N + ( v ) | = 0 . n (cid:48) for every v ∈ V ( H ) as otherwise we remove edgesgoing from vertices with larger outdegree.Let m := (cid:6) n n (cid:48) (cid:7) , where n is the constant in Lemma 2.4, and let H m be the m -blow-upof H . Clearly, H m is { C , C } -free and every vertex has outdegree 0 . n (cid:48) m . However,this contradicts Lemma 2.4. In order to concatenate different closed walks together, we use Kneser’s theorem, a gener-alization of the classical Cauchy-Davenport theorem.
Theorem 2.7 (Kneser’s theorem [22], see also [10]) . If A and B are nonempty subsets ofan additive group Z k , then | A + B | ≥ | A + Stab( A + B ) | + | B + Stab( A + B ) | − | Stab( A + B ) | , where Stab( A + B ) = { x ∈ Z k : x + A + B = A + B } . The main usage of Kneser’s theorem is the following lemma.
Lemma 2.8.
For an integer (cid:96) ≥ let k ≥ being the smallest integer greater than thatdoes not divide (cid:96) . Let A , . . . , A m be sets of non-zero remainders modulo k . By summingup some of the remainders from the sets A , . . . , A m , one can obtain the reminder (cid:96) mod k or at least | A | + . . . + | A m | different non-zero reminders modulo k .Proof. Let B i = A i ∪ { } . All the remainders that can be obtained by summing up someof the remainders from the sets A , . . . , A m are forming the set B + . . . + B m . Since forevery sets X any Y we have | X + Stab( X + Y ) | ≥ | X | and | Stab( X + Y ) | ≥ | Stab( Y ) | ,repeated application of Theorem 2.7 yields that | B + . . . + B m | ≥ | B | + | B + . . . + B m | − | Stab( B + . . . + B m ) |≥ | B | + | B | + | B + . . . + B m | − | Stab( B + . . . + B m ) |≥ . . . ≥ | B | + . . . + | B m | − ( m − | Stab( B + . . . + B m ) | . If Stab( B + . . . + B m ) is non-trivial, then k is not prime and we can write k = p q for some prime number p . Since (cid:96) is divisible by all integers smaller than k , in particularby p q − , we have (cid:96) ≡ cp q − mod k . In such case Stab( B + . . . + B m ) needs to contain (cid:96) mod k . Since 0 ∈ B + . . . + B m then also (cid:96) mod k is in B + . . . + B m .On the other hand, if Stab( B + . . . + B m ) is trivial, then we have | B + . . . + B m | ≥ | B | + . . . + | B m | − ( m −
1) = | A | + . . . + | A m | + 1 . Hence, by removing 0, we get what we wanted.An easy corollary of the above lemma is the following.
Corollary 2.9.
For an integer (cid:96) ≥ let k ≥ being the smallest integer greater than that does not divide (cid:96) . If S is a multiset of k − non-zero reminders modulo k , then thereexists S (cid:48) ⊆ S such that (cid:96) ≡ (cid:88) x ∈ S (cid:48) x mod k. We note that this corollary is also a consequence of [1, Lemma 4.2].6
Constructions of C (cid:96) -free graphs with the best possiblesemidegree We show that the minimum semidegree threshold in Theorem 1.4 is best possible separatelyfor odd and even values of k .For odd k , we need to construct an n -vertex graph G without C (cid:96) having minimumsemidegree δ ± ( G ) = nk + k − k . To construct such a graph, we start with a balanced blow-up of C k on n + ( k − / C (cid:96) . Now, we need to remove ( k − / C (cid:96) . In order to do so,we use ( k − / . . .. . . ...... ...... Figure 2: Maneuvers used for the constructions for odd k . In the first one, by adding themiddle vertex and incident edges, one can remove a vertex in the leftmost blob and in therightmost blob. In the remaining two maneuvers, by adding edges to arbitrarily chosentwo vertices in bottommost blobs, one can remove a vertex in the topmost blob.Without using any of the maneuvers one can obtain only cycles of lengths havingremainder 0 mod k . Each usage of the first maneuver gives a possibility to change theremainder mod k of the cycle possible to obtain by ( k + 1) / k − /
2. Similarly,each application of the second maneuver gives a possibility to change the remainder of theattainable cycle twice by ( k − /
2, and the third maneuver – twice by ( k + 1) /
2. Thus,we can combine them in the following way without creating C (cid:96) : • If (cid:96) mod k is in (0 , k/
4) then we take 2 (cid:96) − k − / − (cid:96) times the second maneuver. • If (cid:96) mod k is in ( k/ , k/
2) then we take k − − (cid:96) times the first maneuver and2 (cid:96) − ( k + 1) / • If (cid:96) mod k is in ( k/ , k/
4) then we take 2 (cid:96) − k − k − / − (cid:96) times the second maneuver. • Finally, if (cid:96) mod k is in (3 k/ , k ) then we take 2 k − − (cid:96) times the first maneuverand 2 (cid:96) − (3 k + 1) / k is even, i.e., for k being a power of 2, we need to construct a C (cid:96) -free graph on n vertices with minimum semidegree δ ± ( G ) = nk + k − k . Observe that for (cid:96) being even wehave (cid:96) ≡ k/ k . On the other hand, (cid:96) being odd yields k = 4. Analogously to thecase when k is odd, we start with a balanced blow-up of C k on n + ( k − / k − / C (cid:96) : 7 . . Figure 3: Maneuvers used for the constructions for even k . In the first one, by addingthe middle vertex and incident edges, one can remove a vertex in each topmost blob. Inthe second one, by adding the middle two vertices and incident edges, one can remove avertex in each blob except the bottomright blob. • If (cid:96) ≡ k/ k , then we use ( k − / k of the attainable cycle by 1 or −
1, so we will not obtain C (cid:96) . • If (cid:96) ≡ C (cid:96) . • Finally, if (cid:96) ≡ C . C (cid:96) with k ≥ Through the whole section, we assume that k ≥ k is defined from (cid:96) as in thestatement of Theorem 1.4. In particular, (cid:96) is divisible by all the numbers from 3 to k − k . Also, k needs to be a prime power.We start with proving Theorem 1.5, which is for closed walks instead of cycles, but onthe other hand, it gives an additional structural information. Proof of Theorem 1.5.
The proof contains two major claims about graph H : one boundsthe maximal possible length of a minimal path between any two vertices, and the otherone forbids certain subgraphs. Using the two, we will be able to understand the structureof H so that we can prove the theorem.Firstly, we use Lemma 2.3 to find a short path between any two vertices of H . Claim 4.1.
For any vertices x, y ∈ V ( H ) there exists a directed path from x to y of lengthat most (cid:4) k − (cid:5) . Additionally, for k = 7 , there exists such a path of length at most .Proof. We follow the main idea used in the proof of Lemma 2.3. Let m := (cid:4) k − (cid:5) , c :=1 − . /m , and fix an arbitrary vertex x ∈ V ( H ). We define X := N + ( x ), and X i +1 := X i ∪ N + ( X i ) for every i ∈ [5 m − i ∈ [5 m ], the induced graph H [ X i ] cannot contain a directed cycle of lengthat most 3 m ≤ k −
1, as otherwise H would contain a closed walk of length (cid:96) . Therefore,there exists x i ∈ X i such that | N + ( x i ) ∩ X i | < c | X i | by Lemma 2.3, and | X i +1 | > | X i | + | N + ( x i ) | − c | X i | ≥ nk (cid:18) − k (cid:19) + (1 − c ) | X i | . Combining this estimate with | X | ≥ nk (1 − k ) yields that | X i | > nk (cid:18) − k (cid:19) − (1 − c ) i c = nk (cid:18) − k (cid:19) − . i/m − . /m .
8n particular, | X m | > nk (cid:18) − k (cid:19) − . − . /m . (2)Recall that m = (cid:4) k − (cid:5) . For k ∈ { , , } we get | X m | > . n , | X m | > . n and | X m | > . n , respectively. Since (2) is ascending in terms of k when consideredseparately for each reminder modulo 3, we have that | X m | > n/ k . Note that we used the fact k (cid:54) = 6 since k must be a prime power. Therefore,there are paths of length at most 5 (cid:4) k − (cid:5) from x to more than half of the vertices of thegraph H .Now, fix y ∈ V ( H ) arbitrarily. Using the bound on the minimum indegree, we analo-gously find paths of length at most 5 (cid:4) k − (cid:5) to y from more than half of the vertices of H .Therefore, there is a path from x to y of length at most 10 (cid:4) k − (cid:5) .In order to show the additional part of the statement, we notice that in the case k = 7we have | X i | > . n > n/ i = 6 (note that 5 m = 10 in that case). Thisimprovement yields a path of length at most 12 from any x ∈ V ( H ) to any y ∈ V ( H ).Note that we established a bound valid also for small values of k , which is highlyoverestimating for big k . In particular, one can obtain better bound (below 1 . k ) assuming k is large enough. Nevertheless, for our purposes the bound proved in Claim 4.1 is goingto be sufficient.Next major step in the proof is forbidding shortcuts of a limited length. Recall thatan s -shortcut is a directed path of length s with an additional edge from its first vertexto the last one. Claim 4.2. H does not contain an s -shortcut for any s ≤ (cid:4) k (cid:5) + 1 .Proof. Since the cases k = 5 and k = 7 needs a more careful analysis, we split the proofinto three parts. Case k ≥ . Suppose for contradiction that the graph H contains an s -shortcut with s ≤ (cid:4) k (cid:5) + 1,and let u and v be its source and its sink vertex, respectively. From Claim 4.1 there is adirected path P of length at most 10 (cid:4) k − (cid:5) from v to u . Together with the vertices of the s -shortcut, we obtain two closed walks of length | P | + 1 and | P | + s .Sylvester’s solution [34] of the well-known Frobenius Coin Problem asserts that if anintegral multiple of gcd( a, b ) cannot be obtained as an integer-weighted sum of positiveintegers a and b , then it is at most ( a − b − −
1. Sincegcd( | P | + 1 , | P | + s ) ≤ s − < k − , we conclude that (cid:96) is divisible by gcd( | P | + 1 , | P | + s ). We also have | P | · ( | P | + s − ≤
103 ( k − · (cid:18)
103 ( k −
1) + k (cid:19) < k . Therefore, (cid:96) < k as otherwise H would contain a closed walk of length (cid:96) .We know that (cid:96) is greater than or equal to the least common multiple of all the integersless than k . It is known (see, e.g., [27]) that this is greater than 2 k − for k ≥
8. Since2 k − > k for k ≥
12, we need to address only the cases k = 8, k = 9, and k = 11. If k = 9 or k = 11, then the smallest possible value of (cid:96) is 840 > ·
24 or 2520 > · H contains a closed walk oflength (cid:96) ; a contradiction. 9t remains to consider the case k = 8, when (cid:96) is divisible by 420 and | P | ≤
20. However, | P | cannot be equal to 20 or 19, because it would create C or C , respectively, whichreadily yields to a closed walk of length (cid:96) . Therefore, | P | ( | P | + s − ≤ ·
22 = 396 < H contains no closed walk of length (cid:96) . Case k = 7 . This is the case of all (cid:96) divisible by 60 and not divisible by 7. It is enough to prove itfor (cid:96) = 60.Firstly, we will observe that for any x ∈ V ( H ) the sets N + ( x ) and N − ( x ) do notcontain directed cycles. By symmetry, it is enough to prove it for N + ( x ). Suppose on thecontrary that N + ( x ) contains a cycle, and so it contains a directed path P of length 6,say from u to v . From Claim 4.1 there is a directed path R of length at most 12 from v to x . Notice, that together with vertices from P it creates directed cycles of lengths | R | + 1 , | R | + 2 , . . . , | R | + 7. In each possible case of the length of R , one can find C , C or C , in particular a closed walk of length (cid:96) , which gives a contradiction. Since theset N + ( x ) does not contain directed cycles, it contains a vertex without outneighbors in N + ( x ). Similarly, the set N − ( x ) contains a vertex without inneighbors in N − ( x ).Knowing this, we can improve our bound in Claim 4.1 and prove that for any vertices x, y ∈ V ( H ) there is a directed path of length at most 10 from x to y . As previously,we define X = N + ( x ) and X i +1 = X i ∪ N + ( X i ) for i ≥
1. We have | X | ≥ n (1 − ).From the above observation, in X there is a vertex x such that N + ( x ) ∩ X = ∅ , so | X | ≥ n (1 − ). Now, for each i ≥ m = 2 for the induced graph H [ X i ] obtaining x i ∈ X i such that | N + ( x i ) ∩ X i | < c | X i | , where c = 1 − √ . < . | X | > . n , | X | > . n and | X | > n/
2. Analogously as in the end ofthe proof of Claim 4.1 it means that there is a directed path of length at most 10 from x to y .Using this observation for x = y we get that each vertex needs to appear in a cycle oflength at most 10. Since there is no closed walk of length 60, the only possible such shortcycles are of length 7, 8, and 9. Now notice that 60 can be expressed as a sum of multiplesof 11 and multiples of each of the numbers 7, 8 and 9, and so an existence of C , wouldcreate a closed walk of length 60. Thus, there is no C in H .We can now finish the proof as in the case k ≥
8. Consider an s -shortcut for s ≤ u being the source vertex and v being the sink vertex. From above observations thereis a directed path P of length at most 10 from v to u . Together with the vertices of the s -shortcut, we obtain two closed walks of length | P | + 1 and | P | + s . In each possible caseof the length of P and s ≤ C , which is forbidden. Case k = 5 . This is the case of all (cid:96) divisible by 12 and not divisible by 5. It is enough to prove itfor (cid:96) = 12. We follow the lines of the proof for the case k = 7.Firstly, we will prove that for any x ∈ V ( H ) the sets N + ( x ) and N − ( x ) do not containdirected cycles. By symmetry, it is enough to prove it for N + ( x ). Suppose on the contrarythat N + ( x ) contains a cycle, and so it contains a directed walk P of length 5, say from u to v (we allow that u = v ). From Claim 4.1 there is a directed path R of length atmost 10 from v to x . Notice, that together with vertices from P it creates closed walksof lengths | R | + 1 , | R | + 2 , . . . , | R | + 6. In each possible case of the length of R , one canfind a closed walk of length 12, which gives a contradiction. Since the set N + ( x ) does notcontain directed cycles, it contains a vertex without outneighbors in N + ( x ). Similarly,the set N − ( x ) contains a vertex without inneighbors in N − ( x ).Knowing this, we can improve our bound in Claim 4.1 and prove that for any vertices x, y ∈ V ( H ) there is a directed path of length at most 6 from x to y . As previously, we10efine X = N + ( x ) and X i +1 = X i ∪ N + ( X i ) for i ≥
1. We have | X | ≥ n (1 − ).From the above observation, in X there is a vertex x such that N + ( x ) ∩ X = ∅ , so | X | ≥ n (1 − ). Now, we use Corollary 2.6 for the induced graph H [ X ] obtaining x ∈ X such that | N + ( x ) ∩ X | < . | X | . Therefore, | X | > n/
2. Analogously as inthe end of the proof of Claim 4.1 it means that there is a directed path of length at most 6from x to y .Using this observation for x = y we get that each vertex needs to appear in a cycle oflength at most 6. Since there is no closed walk of length 12, each vertex appears in C .Thus, there is no C in H .We can now finish the proof as in the previous cases. Consider an s -shortcut for s ≤ u being the source vertex and v being the sink vertex. From above observations thereis a directed path P of length at most 6 from v to u . Together with the vertices of the s -shortcut, we obtain two closed walks of length | P | + 1 and | P | + s . In each possiblecase of the length of P and s ≤ C , which isforbidden.We are now ready to analyze the structure of graph H and finish the proof of Theo-rem 1.5.Take any directed path x x . . . x k − of length k − H and consider the following k + 1 sets: for 1 ≤ i ≤ (cid:98) k/ (cid:99) + 1 the sets N − ( x i ), and for (cid:98) k/ (cid:99) ≤ j ≤ k − N + ( x j ). Sum of their sizes is above n , so they need to have a non-empty intersection.Notice that if for i and i (cid:48) any two sets N − ( x i ) and N − ( x i (cid:48) ) intersect, then we obtain an s -shortcut, where s is at most (cid:98) k/ (cid:99) + 1 contradicting Claim 4.2. From the same reason,the sets N + ( x j ) and N + ( x j (cid:48) ) cannot intersect for any j and j (cid:48) . If the sets N − ( x i ) and N + ( x j ) intersect, then we obtain a forbidden 2-shortcut or a directed cycle of length atmost k . Since in H there are no cycles of length less than k , there must be always C k .Take such a directed k -cycle y y . . . y k and define sets Y i = N + ( y i ). From now on,we denote them cyclically, in particular, by Y we understand Y k . Notice that they aredisjoint, because otherwise we obtain a shortcut forbidden by Claim 4.2. Let T be the setof the remaining vertices of H . We know that | T | ≤ n − k (cid:18) nk (cid:18) − k (cid:19)(cid:19) = n k . For any y i the set N − ( y i ) must be contained in Y i − ∪ T , because otherwise we obtaina cycle of length less than k or a forbidden 2-shortcut. Moreover, a vertex in T cannothave edges to two different vertices y i and y i (cid:48) , because it creates a forbidden shortcut.Therefore, we can move from (cid:83) Y i to T at most n/ k vertices to ensure that for all i , allvertices of Y i − have an edge to y i .Now, consider any vertex in Y i . It can have edges only to Y i +1 ∪ T and edges from Y i − ∪ T , otherwise it creates forbidden cycles or shortcuts. In particular, any vertex in Y i has edges to at least nk (cid:0) − k (cid:1) > nk vertices of Y i +1 . Similarly from Y i − .We will now describe a procedure that moves one-by-one every vertex t ∈ T to one ofthe sets Y i for some i ∈ [ k ], at each step maintaining the following conditions for every j ∈ [ k ] and every vertex v ∈ Y j : • v has edges only to Y j +1 ∪ T and hence N + ( v ) ∩ Y j +1 > nk , and • v has edges only from Y j − ∪ T and hence N − ( v ) ∩ Y j − > nk .At the end of this procedure, the set T will be empty so the vertex-partition ( Y i ) witnessesthat H is a subgraph of a blow-up of C k , i.e., H is homomorphic to C k .11ake any vertex v ∈ T . Since | T | ≤ n/ k , the vertex v needs to have more than nk outneighbors in (cid:83) Y i . If v has edges to two non-consecutive sets Y j and Y j (cid:48) then it creates aforbidden shortcut, so it can have edges only to Y j and Y j +1 for some j ∈ [ k ]. Now supposefor contradiction it has edges to both, and let w ∈ N + ( v ) ∩ Y j and u ∈ N + ( v ) ∩ Y j +1 . N + ( w ) ∩ N + ( v ) = ∅ because the graph H has no 2-shortcut. However, | N + ( w ) ∩ Y j +1 | > nk , thus | N + ( v ) ∩ Y j +1 | < nk . Similarly, N − ( u ) ∩ N + ( v ) = ∅ , but | N − ( u ) ∩ Y j | > nk ,thus | N + ( v ) ∩ Y j | < nk . In total, we have | N + ( v ) | < nk ; a contradiction.By the same reasoning applied to the inneighborhood of v , we conclude there areindices i, o ∈ [ k ] such that v has inneigbors and outneighbors only inside the sets Y i and Y o , respectively. In the case o = i , we can find a 3-shortcut contradicting Claim 4.2. If o = i + 1 or o = i − N + ( v ) and N − ( v ) which togetherwith v creates a 2-shortcut or a directed triangle. Finally, in the other cases with o (cid:54) = i + 2,we can find a directed cycle of length less than k . Therefore, o = i + 2 so the vertex v canbe moved to Y i +1 without violating the conditions of the procedure.We showed that the graph H is homomorphic to C k , which finishes the proof of The-orem 1.5.Using Theorem 1.5, we are ready to prove Theorem 1.4 for k ≥
5. Recall that (cid:96) is aninteger divisible by all positive integers less than k but not divisible by k , and supposethere exists a C (cid:96) -free graph G on n vertices with δ ± ( G ) ≥ nk + k − k for infinitely many n .By the moreover part of Proposition 2.1, one can remove o ( n ) vertices and o ( n ) edgesfrom G in order to find its subgraph H that satisfies the assumptions of Theorem 1.5.Thus, the vertices of the initial graph G can be partitioned into sets T and blobs X i for1 ≤ i ≤ k , where | T | = o ( n ), | X i | = nk + o ( n ) and each vertex in blob X i has edges to allbut o ( n ) vertices in X i +1 and from all but o ( n ) vertices in X i − .Let us refer to the edges inside one blob or between two blobs that do not fulfill thecycle order as extra edges. Notice that if we have a set of at least k − k blobs andusing one extra edge each time (making at most k + 1 steps between them). In particular,if (cid:96) ≥ ( k − k + 2) then G contains C (cid:96) . Thus, unless (cid:96) = 12 or (cid:96) = 24, there cannotbe such a set, and we can move at most k − T in order toremove all the extra edges. In the case (cid:96) = 24, note that 4 disjoint edges inside some blobcan also be connected to obtain C , so we can move at most 3 vertices from each blobto T in order to remove all the edges inside blobs. Now, any set of 3 disjoint extra edgesbetween the blobs contains extra edges that can be combined by going around the blobswhich would again yield C , hence we can move additional at most 2 vertices from theblobs to T to remove such edges. We conclude that in case (cid:96) = 24, there are at most 17vertices that are incident to all the extra edges. Finally, in the remaining case (cid:96) = 12,note that in any set of 4 disjoint extra edges from a blob X i to a blob X j for some indicies i, j ∈ [ k ] such that j (cid:54) = i + 1, one can find a subset of extra edges that can be connectedby going around the blobs to construct C . Thus, by moving at most 60 vertices fromblobs to T we can remove all the extra edges. We conclude that we may assume that H is an induced subgraph of G on n − o ( n ) vertices that is homomorphic to C k .The proof of Theorem 1.4 for k ≥ Lemma 4.3.
Fix an integer (cid:96) that is divisible by and let k be the smallest integergreater than that does not divide (cid:96) . Let G be an n -vertex graph with δ ± ( G ) ≥ nk + k − k .If G contains an induced subgraph on n − o ( n ) vertices that is homomorphic to C k , then G contains C (cid:96) . roof. Assume by contrary, that there exists an n -vertex C (cid:96) -free graph G with δ ± ( G ) ≥ nk + k − k for arbitrarily large n , and its vertices can be partitioned into set T := V ( G ) \ V ( H )and sets X i for 1 ≤ i ≤ k , where | T | = o ( n ), | X i | = nk + o ( n ), and each vertex in X i hasedges only to vertices in X i +1 ∪ T and only from vertices in X i − ∪ T .We refer to the sets X , X , . . . , X k as blobs and say that a blob X i is an in-pointing blob or an out-pointing blob of a vertex t ∈ T if | N − ( t ) ∩ X i | ≥ k or | N + ( t ) ∩ X i | ≥ k ,respectively. Using just edges in H one can obtain only cycles of lengths divisible by k .We will use the semidegree assumption for finding sufficiently many other edges in G thatallows us to change the remainder modulo k and construct a cycle of length (cid:96) .A sidewalk is a path of length 2 starting and finishing in blobs that has the middlevertex in T . The value of a sidewalk is the change in the reminder modulo k that isachieved by this sidewalk. Formally, if X i is an in-pointing blob and X j is an out-pointingblob of a vertex t ∈ T , then t with its corresponding neighbors create a sidewalk of value i − j +2 mod k . We say that two sidewalks are compatible if they are either vertex-disjoint,or, they share exactly one vertex which is then the sink of one of the sidewalks and thesource of the other one. A set of sidewalks is called compatible if and only if there exists acycle in G that contains all of them. In particular, the sidewalks are pairwise compatible.We say that a remainder r ∈ [ k −
1] can be combined using a set of sidewalks if r can be written as a sum of the sidewalk values modulo k , where the sum is taken over acompatible subset of sidewalks. Our aim is to show that the remainder (cid:96) mod k can becombined using some set of sidewalks. After we do so, then it only remains to connectthe chosen sidewalks to make C (cid:96) . Main tool we will use in combining the sidewalks isLemma 2.8 and its consequence Corollary 2.9. In our case it means that the remainder (cid:96) mod k can be combined using any compatible set of k − Claim 4.4. If t ∈ T has x in-pointing blobs and y out-pointing blobs, then we can combinethe reminder (cid:96) mod k or there are at least x + y − different non-zero values of sidewalkscontaining t .Proof. As a set A take all the x possible values of sidewalks using one chosen out-pointingblob. And as a set B take the set of distances in the cyclic order from the chosen out-pointing blob to all the out-pointing blobs (including himself at distance 0). The set ofpossible values of the sidewalks using t is exactly the set A + B . If Stab( A + B ) is non-trivial, then k is not prime and (cid:96) ∈ Stab( A + B ), thus one can combine the reminder (cid:96) modulo k . Otherwise, from Kneser’s theorem (Theorem 2.7) we get that | A + B | ≥ x + y − x + y − t . Claim 4.5.
There exists a compatible set of at most k − sidewalks using which one cancombine the reminder (cid:96) mod k .Proof. Each vertex in T has less than 4 k neighbors, which are not in out-pointing orin in-pointing blobs, so the total set X of such vertices has size at most 4 k | T | = o ( n ).Let T (cid:48) ⊆ T be the maximal set of vertices in T creating a compatible set of sidewalks ofnon-zero values with vertices not from X . If | T (cid:48) | ≥ k −
1, then from Corollary 2.9 we havethe wanted set of sidewalks, thus assume | T (cid:48) | < k − z be the sum of the numbers of in-pointing and out-pointing blobs of the verticesin T (cid:48) . Using Claim 4.4 for each vertex in T (cid:48) and Lemma 2.8 we get that using sidewalkscontaining T (cid:48) one can combine the remainder (cid:96) mod k or at least z − | T (cid:48) | non-zeroremainders. If z − | T (cid:48) | ≥ k −
1, then we have the remainder (cid:96) mod k , so assume that k − > z − | T (cid:48) | . 13ake a set of vertices that contains exactly one vertex from each blob, avoiding theset X . By summing up their indegrees and outdegrees we get at least 2 kδ ± ( G ) ≥ n + k − z neighbors in T (cid:48) , so to or from T \ T (cid:48) they have atleast 2 n + k − − z − n − | T | ) = k − − z + 2 | T | > | T | − | T (cid:48) | )edges. This means that there exists a vertex t ∈ T \ T (cid:48) with at least 3 neighbors among thechosen vertices. It creates a sidewalk of a non-zero value. Since each in-pointing or out-pointing blob of t contains at least 2 k neighbors of t and | T (cid:48) | < k −
1, then one can chooseto the sought set sidewalks that are using different vertices in blobs for different verticesin T (cid:48) ∪ { t } . This yields a compatible set of sidewalks, which contradicts the maximalityof T (cid:48) .In order to finish the proof of Lemma 4.3 it remains to prove that the found set of atmost k − C (cid:96) . We can connect the sidewalksby going around the k blobs and using one sidewalk each time. Between the sidewalks wemake at most k + 1 steps, so in total we make at most ( k − k + 3). It means that if (cid:96) ≥ ( k − k + 3), which is true for any considered here (cid:96) except 12 and 24, then we canconnect the sidewalks and obtain C (cid:96) .In the case (cid:96) = 24, which means k = 5, if we have at most 3 sidewalks then we canconnect them to obtain C , because 24 ≥ ·
8. The only set of 4 sidewalks that do notcontain a smaller subset of sidewalk that can be combined to obtain the remainder 4, isthe set containing 4 sidewalks o value 1. One can easily check that independently howsuch sidewalks will be distributed, one can always connect them in a specific order tocreate C .In the remaining case (cid:96) = 12, a more detailed analysis is needed, since one can have4 compatible sidewalks of value 3 that cannot be connected to create C . Thus, we focusnow on the case (cid:96) = 12, and so k = 5 and δ ± ( G ) ≥ n +25 .We will reduce the set T to a smaller set T (cid:48) and extend the blobs to X (cid:48) i for i = 1 , . . . , T to the appropriate sets based on the adjacenciesto the blobs they have. Notice that there are no sidewalks of value 2, since such a sidewalkcan be easily extended to obtain C . Take a vertex t ∈ T that is not creating a sidewalk ofvalue 1 and that creates a sidewalk of value 0. It cannot have inneighbors and outneighborsin the same blob and it has an inneighbor in X i − and an outneighbor in X i +1 for some i .Thus, extend blob X i by taking X (cid:48) i := X i ∪ { t } . Notice that there are no edges insideextended blobs and there are no edges between the consecutive blobs, that do not fulfillthe cyclic order. Since t ∈ X (cid:48) i has appropriate neighbors in X i − and X i +1 , every vertex in T \ { t } still cannot create a sidewalk of value 2 with the extended blobs X (cid:48) i . Thus, we cancontinue this procedure and assign all vertices from T that creates sidewalks of value 0and do not create sidewalks of value 1 to the respective blobs. We end up with a graphwhich is a blow-up of C , possibly with some edges between the non-consecutive blobs andwith a set T (cid:48) (possibly empty) of vertices that do not create sidewalks of value 0 with theextended blobs, or that create sidewalks of value 1.Notice that if there exist compatible sidewalks of values 3 and 4, or two compatiblesidewalks of value 1, then one can easily obtain C .Firstly assume that there is a blob of size smaller than n +25 −
1, say it is X (cid:48) . It impliesthat every vertex in X (cid:48) is a source of at least two sidewalks of non-zero values and everyvertex in X (cid:48) is a sink of at least two sidewalks of non-zero values. Moreover only one vertexin T (cid:48) (the one possible vertex creating sidewalks of value 1) can be counted simultaneouslyas outneighbor of X (cid:48) and inneighbor of X (cid:48) . If we obtained this way compatible sidewalks14f values 3 and 4, three compatible sidewalks of value 4, or four compatible sidewalksof value 3, then we can find C . The only remaining option is to have two compatiblesidewalks of value 3 and one sidewalk of value 1. But now, if any vertex in X (cid:48) or X (cid:48) is creating more sidewalks of non-zero value, then it needs to be of value 3 and we canobtain C . So X (cid:48) needs to be of size at least n +25 −
2, be fully connected to X (cid:48) and X (cid:48) and | T (cid:48) | ≥
1. It implies that there is a different blob smaller than n +25 , which givesadditional sidewalk of value 3 that can be combined with the previously proven sidewalksin order to obtain C .Assume now that every blob is of size at least n +25 −
1. If there is no sidewalk ofvalue 1, then there are at least two blobs of size smaller than n +25 , say X (cid:48) i and X (cid:48) j . Itimplies that every vertex in X (cid:48) i − and X (cid:48) j − is the source of a sidewalk of a non-zero valueand every vertex in X (cid:48) i +1 and X (cid:48) j +1 is the sink of a sidewalk of a non-zero value. This givesa set of 4 compatible sidewalks of value 3 or value 4. In both cases we can obtain C .Thus, assume there is a sidewalk of value 1. It needs to use a vertex in T (cid:48) and twoconsecutive blobs, say X (cid:48) and X (cid:48) . It also implies that there are at least 3 blobs of sizesmaller than n +25 . If two of them are neither X (cid:48) nor X (cid:48) , then as before we have a set of4 compatible sidewalks of values 3 or 4, which leads to C . In the remaining case we getthat | X (cid:48) | = | X (cid:48) | = | X (cid:48) i | = n +25 − i (cid:54) = 1 ,
2. It implies that every vertex in X (cid:48) and X (cid:48) i − is the source of a sidewalk of a non-zero value and every vertex in X (cid:48) and X (cid:48) i +1 is the sink of a sidewalk of a non-zero value. This leads to a set of 4 compatible sidewalksof value 3 or value 4, and consequently to C . C (cid:96) with k = 4 and (cid:96) ≥ In this section we assume that (cid:96) ≥ k = 4, which means that (cid:96) is divisible by 3 andnot divisible by 4. We start with proving Theorem 1.6, which is a similar stability-typetheorem as in the previous section. Proof of Theorem 1.6.
Let H be an n -vertex graph with δ ± ( H ) ≥ n − o ( n ) that doesnot contain a closed walk of length (cid:96) . In particular, H is C -free. If H does not containtransitive triangles, then the underlying undirected graph is triangle-free and has minimumdegree n − o ( n ), so Andr´asfai-Erd˝os-S´os Theorem [3] yields it is bipartite. Therefore, it isenough to find o ( n ) vertices in H that covers all its transitive triangles. Analogously tothe previous section, we start with bounding the diameter of the graph H . Claim 5.1.
For any vertices x, y ∈ V ( H ) there exists a directed path of length at least 2and at most 6 from x to y .Proof. The proof is similar to the proof of Claim 4.1. Denote X = N + ( x ) and X i +1 = X i ∪ N + ( X i ) for i ≥
1. For each i ≥ H [ X i ] cannot contain a directedtriangle, so from Lemma 2.3 there exists x i ∈ X i such that | N + ( x i ) ∩ X i | < . | X i | .Thus, | X i +1 | > | X i | + | N + ( x i ) | − . | X i | . In particular, since | X | ≥ . n − o ( n ), then | X | ≥ . n − o ( n ) and | X | ≥ . n − o ( n ).It means that there is a directed path of length at most 3 starting in x to more than halfof the vertices of the graph H .Now, take the vertex y ∈ V ( H ) and do analogously for a graph obtained from H byreversing the edges. This way we find a directed path of length at most 3 to y from morethan half of the vertices of H . It means that there is a directed path from x to y of lengthat least 2 and at most 6. 15nowing this, we can control the structure of outneighborhood and inneighborhood ofevery vertex in H . Claim 5.2.
For every vertex x ∈ V ( H ) , there exists a vertex v in N + ( x ) with N + ( v ) ∩ N + ( x ) = ∅ and a vertex w in N − ( x ) with N − ( w ) ∩ N − ( x ) = ∅ .Proof. By symmetry, it is enough to prove it for N + ( x ), where x is an arbitrary vertex of H . Assuming the contrary, we get that in N + ( x ) there must exists a directed walk oflength 4 from some vertex a to some vertex b (it may happen that a = b ). Consider anydirected path of length (cid:96) − b terminating in some vertex c . This way, weobtained walks from x to c of lengths (cid:96) − (cid:96) − (cid:96) − (cid:96) − (cid:96) − P from c to x of length at least 2 and atmost 6. In each possible case of the length of P , we can obtain a closed walk of length (cid:96) .With this, we can improve our bound in Claim 5.1. Claim 5.3.
For any vertices x, y ∈ V ( H ) , there exists a directed path of length at least 2and at most 5 from x to y .Proof. From the proof of Claim 5.1 we know that there exists a directed path of lengthat most 3 to at least 0 . n − o ( n ) vertices of the graph H . From Claim 5.2 there exists w ∈ N − ( y ) without inneighbors in N − ( y ), so there exists a directed path of length atmost 2 to y from at least 0 . n − o ( n ) vertices of H . This means that there is a directedpath from x to y of length at least 2 and at most 5. Claim 5.4.
For any ≤ a ≤ (cid:96) − there are no vertices x and y in H for which thereexist directed paths of lengths a , a + 1 , a + 2 and a + 3 from x to y . In particular, forevery vertex v ∈ V ( H ) , there is no path of length in N + ( v ) or in N − ( v ) .Proof. Consider any directed path of length (cid:96) − − a starting in y going to some vertex z (if the path has length 0 then z = y ). This way, we obtained walks from x to z of lengths (cid:96) − (cid:96) − (cid:96) − (cid:96) −
2. From Claim 5.3 there is a path from z to x of length between2 and 5. In each case we can find a closed walk of length (cid:96) .Using the proved claims, we can finish the proof of the theorem.As noted at the beginning of the proof, H contains a transitive triangle. Denote itssource by u and sink by v . There is no path of length 2 or 3 from v to u , as otherwise itcreates a closed walk of length (cid:96) in H . If there is a path of length 4 from v to u , then itcreates a closed walk of length (cid:96) for all considered here (cid:96) except (cid:96) = 9. But in this case,we can use Claim 5.3 for two edges on the constructed C to obtain a contradiction withClaim 5.4 or a closed walk of length 9. Thus, there is no path of length smaller than 5from v to u .Let A := N + H ( v ) and D := N − H ( u ). From Claim 5.4 there exists a vertex a ∈ A without outneighbors in A , let B := N + H ( a ). Similarly, there is a vertex d ∈ D withoutinneighbors in D , let C := N − H ( d ). The sets A , B , C and D are disjoint and each is ofsize at least δ ± ( H ) = n − o ( n ), so by removing the remaining vertices from H we keep thesemidegree assumption of H . Notice that an edge from B to A (or from D to C ) is givinga contradiction with Claim 5.4, so there are no such edges. Let b ∈ B be a vertex withoutoutneighbors in B . Vertex b can have outneighbors only in C , so it is connected with | C | − o ( n ) vertices in C . Similarly, there is a vertex c ∈ C having | B | − o ( n ) inneigboursin B . Remove those o ( n ) vertices from H (as before, it keeps the semidegree assumptionof H ). 16ow notice that if at least two of the sets A , B , C and D contain any edges, then weobtain a contradiction with Claim 5.4 for (cid:96) ≥
15 and there is C forbidden in the case (cid:96) = 9. Thus, at least three of the sets A , B , C and D contain no edges. If B (or bysymmetry C ) contains no edges, all the vertices in B have n − o ( n ) outneighbors in C ,thus all but o ( n ) vertices in C have n − o ( n ) inneighbors in B . Remove those o ( n ) verticesfrom H . If A (or by symmetry D ) contains no edges, then each vertex in A has n − o ( n )outneighbors in B , and so by removing o ( n ) vertices from H we can obtain that all verticesin B have n − o ( n ) inneighbors in A . There cannot be any edges from C to A , as otherwisewe can find a forbidden directed triangle in H , which implies that each vertex in A has atleast n − o ( n ) inneighbors in D . Thus, by removing o ( n ) vertices from H , we can obtainthat all vertices in D have n − o ( n ) outneighbors in A . Finally, since C or D contains noedges, we can remove o ( n ) vertices from H to obtain that all vertices in C have n − o ( n )outneighbors in D and all vertices in D have n − o ( n ) inneighbors in C . This implies thatthere are no edges from D to B and inside any of the sets A , B , C and D . Hence, byremoving the starting transitive triangle, we obtain the sought bipartite graph.Analogously to Theorem 1.5 being crucial for our proof of Theorem 1.4 when k ≥ (cid:96) ≥ (cid:96) ,and suppose there exists an n -vertex graph G contradicting Theorem 1.4 for an arbitrarylarge n . As in the previous section, a combination of Proposition 2.1 and Theorem 1.6yields that we can remove o ( n ) vertices and o ( n ) edges from G in order to find a bipartite H ⊆ G with δ ± ( H ) ≥ n − o ( n ). Now we show that there are o ( n ) vertices and o ( n ) edgesin H such that their removal yields a subgraph of a blow-up of C with the minimumsemidegree at least n − o ( n ). Claim 5.5. G contains a subgraph H (cid:48) homomorphic to C with δ ± ( H (cid:48) ) ≥ n − o ( n ) .Proof. Note that it is enough to find such a subgraph H (cid:48) inside H . Let ( L, R ) be anybipartition of H . Firstly, we show that there exist disjoint sets A ⊆ L and D ⊆ R of atleast n − o ( n ) vertices, without edges in G from A to D .If (cid:96) is even, then take an arbitrary path P in H of length (cid:96) − v L ∈ L and v R ∈ R be its source and sink. We set A := N + H ( v R ) \ P and D := N − H ( v L ) \ P . Clearly,both A and D have sizes at least n − o ( n ), and no edge in G goes from A to D since suchan edge would create a copy of C (cid:96) .If (cid:96) is odd and there exists v ∈ V ( G ) \ V ( H ), then without loss off generality, v hasan inneighbor in one part (say L ), and an outneighbor in the other one. In the case V ( G ) = V ( H ) there must be an edge in G inside the larger part, which also yields anexistence of a vertex v ∈ V ( G ) that has inneighbor in L and outneighbor in R . Let v L bean inneighbor of v in L , and P be an arbitrary path in H of length (cid:96) − v in R and sink in some v R ∈ R , that is not going through v L . Nowwe set A := N + H ( v R ) \ ( P ∪ { v L , v } ) and D := N − H ( v L ) \ ( P ∪ { v R , v } ). As before, both A and D have sizes at least n − o ( n ), and no edge in G goes from A to D since such an edgewould create a copy of C (cid:96) .Now, let C := L \ A and B := R \ D . Clearly N + H ( A ) ⊆ B and N − H ( D ) ⊆ C and both B and C have sizes at most n + o ( n ). Hence, the lower bound on δ ± ( H ) yields that only o ( n ) edges can go from B to A so there are at least n − o ( n ) edges of H that goes from B to C . Analogously, at least n − o ( n ) edges of H goes from C to D . Therefore, wehave found a subgraph of G with n − o ( n ) edges that is homomorphic to C . Since only o ( n ) vertices of any such subgraph can have the indegree or the outdegree below n − o ( n ),removing all the low-degree vertices from it yields the sought subgraph H (cid:48) ⊆ G .17he proof of Theorem 1.4 for k = 4 and (cid:96) ≥ Lemma 5.6.
Fix an integer (cid:96) ≥ divisible by but not divisible by , and let G be an n -vertex graph with the minimum semidegree δ ± ( G ) ≥ (cid:40) ( n + 1) / when (cid:96) ≡ mod , ( n + 2) / otherwise.If G contains a subgraph H homomorphic to C with δ ± ( H ) ≥ n − o ( n ) , then G contains C (cid:96) .Proof. Suppose for contradiction that G is C (cid:96) -free. Since H is a subgraph of a blow-upof C , we refer to its parts as blobs. Let us now look what are the edges in G thatthe vertices of a blob might induce. Clearly, no blob can contain a path of length 3.Also, if there are 3 disjoint edges in G inside the blobs of H , then we find C (cid:96) for anyconsidered here (cid:96) . Indeed, when (cid:96) = 9, any edge inside a blob creates C , and for (cid:96) ≥ H to get rid of all edges insideblobs. Similarly, any 3 disjoint edges in G between a given pair of the non-consecutiveblobs of H creates C (cid:96) , hence removing at most 4 vertices in each blob destroys all suchedges.We refer to the parts of H in their cyclic order as A , B , C and D , and call themblobs. By the discussion in the previous paragraph, we may assume that V ( H ) inducesa bipartite graph in G , and the only edges induced by V ( H ) that are not in E ( H ) goesbetween consecutive blobs in the reverse order. However, there can be only o ( n ) suchedges as δ ± ( H ) ≥ n − o ( n ).Let T := V ( G ) \ V ( H ). Recall a sidewalk is a path of length 2 starting and finishingin V ( H ) with the middle vertex in T , and the value of a sidewalk is the change in theremainder modulo 4 of cycle lengths that can be achieved by using this sidewalk. Alsorecall that two sidewalks are compatible if they are either vertex-disjoint, or, they shareexactly one vertex which is then the sink of one of the sidewalks and the source of theother one. As in the proof of Theorem 1.4 for k ≥
5, our aim is to find a compatible setof sidewalks that can be combined to a copy of C (cid:96) .We call a blob-transversal any set of 4 vertices of G that contains exactly one vertexfrom each blob. For a blob-transversal S , let e ( S, T ) be the number of edges between S and T (regardless of their direction). Since each vertex of H have neighbors only inthe neighboring blobs or in T , summing up the sizes of the neighborhoods of a fixedblob-transversal S yields the following estimate: e ( S, T ) + 2( n − | T | ) ≥ · δ ± ( G ) . (3)If a vertex in T creates a sidewalk of value (cid:96) mod 4, then we easily find C (cid:96) . Therefore,there are no sidewalks of such a value. Let us now restrict adjacencies between the verticesin T and blob-transversals. Claim 5.7.
For every t ∈ T and blob-transversal S there are at most neighbors of t inside S .Proof. Suppose for contrary that there is such a vertex t ∈ T , a blob-transversal S ,and 4 edges between them. If t has 4 outneighbors (or inneighbors) in S , then pick aninneighbor (or outneighbor) of t outside S ∪ T . Such a neighbor together with t and S creates a sidewalk of any value modulo 4 contradicting that there is no sidewalk of value (cid:96) mod 4. 18herefore, not all the edges between t and S have the same direction and S ∪ { t } contains sidewalks of both values 1 and 3. This readily yields a contradiction unless (cid:96) ≡ S ∪ { t } .Fix a blob-transversal S (cid:48) that is disjoint from S , and let t (cid:48) (cid:54) = t be a vertex in T that hasat least three neighbors inside S (cid:48) . If S (cid:48) ∪{ t (cid:48) } contains no sidewalk of a non-zero value, thenall the edges between t (cid:48) and S (cid:48) must have the same orientation. However, the vertex t (cid:48) must be adjacent to n/ − o ( n ) vertices of H using edges with the other orientation, whichyields a sidewalk of a non-zero value disjoint from S ∪ { t } . Since its value cannot be 2,it must be either 1 or 3. In both cases we combine the new sidewalk with the one of thesame value in S ∪ { t } to find a copy of C (cid:96) .This claim is already sufficient for ruling out the possibility that (cid:96) is even. Claim 5.8. If (cid:96) ≥ and (cid:96) ≡ mod , then G contains C (cid:96) .Proof. By (3), we have e ( S, T ) ≥ | T | + 4 for every blob-transversal S . Together withClaim 5.7, it yields that for every blob-transversal S there are at least 4 vertices in T withexactly 3 neighbors in S . In particular, every blob-transversal is in 4 different sidewalksof a non-zero value. Since these sidewalk values can be only 1 or 3, we find two disjointsidewalks of the same value (simply consider three vertex-disjoint blob-transversals). As (cid:96) ≥
18, any two disjoint sidewalks of the same value can be combined and create C (cid:96) .For the rest of the proof, we assume (cid:96) is an odd multiple of 3. Claim 5.9.
There are no sidewalks of value and no edges between consecutive blobs thatdo not agree with the cyclic order of the blobs.Proof. Assuming the contrary, there is a directed path uvw with u and w from the sameblob. Consider a blob-transversal S using an outneighbor of u and an inneighbor of w .From (3) there is a vertex t (cid:54) = v in T connected with 3 vertices from S . If it creates asidewalk of a non-zero value then it can be combined with the assumed sidewalk of value 2to find C (cid:96) for every (cid:96) ≥ t . But then, with any inneighbor (or outneighbor) of t , we have a sidewalk of value 1or 3. Combining it with the assumed path uvw we find C (cid:96) .We will now reduce the set T to a smaller set T (cid:48) and extend the blobs to A (cid:48) , B (cid:48) , C (cid:48) and D (cid:48) by consecutively assigning vertices from T to the appropriate sets based onthe adjacencies to the blobs they have. From Claim 5.9 each vertex t ∈ T cannot haveinneighbors and outneighbors in the same blob. If t creates a sidewalk of value 0, bysymmetry we assume t has an inneighbor in A and an outneighbor in C , then extend theblob B by taking B (cid:48) := B ∪ { t } . Notice that since t ∈ B (cid:48) has appropriate neighbors in A and C , Claim 5.9 holds for the extended blobs. Thus, we can continue this procedureand assign all vertices from T that creates sidewalks of value 0 to the respective blobs A (cid:48) , B (cid:48) , C (cid:48) or D (cid:48) . We end up with a graph which is a blow-up of C , possibly with someedges inside the blobs or between the diagonal blobs and with a set T (cid:48) (possibly empty)of vertices that do not create sidewalks of value 0 with the extended blobs. Claim 5.10. If (cid:96) ≡ mod , then G contains C (cid:96) .Proof. In this case (cid:96) ≥ δ ± ( G ) ≥ n +14 and there are no sidewalks of value 3, so thereare no edges between the diagonal blobs. In each blob A (cid:48) , B (cid:48) , C (cid:48) and D (cid:48) consider avertex that does not have out-neighbors in its blob. It exists, because otherwise we have19hree disjoint edges inside blobs, which leads to an (cid:96) -cycle. Notice that such vertices canhave outneighbors only in the next blobs or in T (cid:48) , but each vertex in T (cid:48) can have onlyone inneighbor among the chosen vertices. Summing up their outdegrees we get at least4 δ ± ( G ) ≥ n + 1 > n , which gives a contradiction. Claim 5.11. If (cid:96) ≡ mod , then G contains C (cid:96) .Proof. In this case (cid:96) ≥ δ ± ( G ) ≥ n +24 and there are no sidewalks of value 1, so thereare no edges inside blobs. By symmetry, let B (cid:48) be the smallest blob. In particular it issmaller than n +24 . This means that every vertex in A (cid:48) has an outneighbor in C (cid:48) or T (cid:48) , i.e.,it is the source of a sidewalk of value 3. Similarly, every vertex in C (cid:48) has an inneighbor in A (cid:48) or T (cid:48) and so is the sink of a sidewalk of value 3 (each vertex in T (cid:48) cannot be in bothtypes of mentioned sidewalks). If any vertex in A (cid:48) or C (cid:48) has more such neighbors, thenwe have another sidewalk of value 3 that can be combined with those starting in A (cid:48) andterminating in C (cid:48) to obtain an (cid:96) -cycle. So, | B (cid:48) | ≥ n − . This implies that there must beanother blob that is smaller than n +24 , which similarly creates a sidewalk of value 3. Itcan be easily combined with the sidewalks starting in A (cid:48) and terminating in C (cid:48) to obtainan (cid:96) -cycle.Since Claims 5.8, 5.10 and 5.11 cover all the cycle-lengths (cid:96) ≥ C Through the whole section, let G be an oriented graph on n vertices with δ ± ( G ) ≥ n + .Assume by contradiction, that G is C -free.We follow similar lines as in Section 4 and Section 5. Let us start with an analogueof Claim 5.3. Claim 6.1.
For any two vertices x, y ∈ V ( G ) and a set Z ⊆ V ( G ) of size at most ,there is a path of length at most from x to y that is internally vertex-disjoint from Z .Proof. Let X := N + ( x ) \ Z and for i ∈ { , } , let X i +1 := X i ∪ N + ( X i ) \ Z . Clearly, | X | ≥ . n − G [ X ] is C -free, thus applying Corollary 2.6 to G [ X ] yields avertex x ∈ X such that | N + ( x ) ∩ X | ≤ . | X | . Therefore, | X | ≥ | X | + | N + ( x ) | − . | X | − | Z | > . n − G [ X ] yields a vertex x ∈ X satisfying | N + ( x ) ∩ X | ≤ . | X | , and hence | X | ≥ | X | + | N + ( x ) | − . | X | − | Z | > . n − . Similarly for the inneighborhood of y , we have | Y | > . n −
10, where Y := N − ( y ) \ Z .Applying Corollary 2.6 to G with all the edges being reversed, we find a set Y ⊆ V ( G ) \ Z of size | Y | > . n −
20 containing only vertices from which there exist a path to y oflength at most 2. Since | X | + | Y | > n , the two sets must intersect. Therefore, G containsa path of length at most 5 from x to y that avoids the set Z .20n particular, the previous claim yields that there cannot be a path of length 3 in theoutneighborhood and in the inneighborhood of a given vertex. Using this, we can actuallyimprove the previous claim. Claim 6.2.
For any two vertices x ∈ V ( G ) and y ∈ V ( G ) : • if xy ∈ E ( G ) then there exists a path of length , or from x to y , which avoidsany fixed vertex, and • if xy / ∈ E ( G ) then there exists a path of length , or from x to y , which avoidsany fixed vertices.Proof. For brevity, let Z be the set of vertices we want to avoid. We define A := N + ( x ) \ Z , • B := N + ( A \ { y } ) \ Z , • D := N − ( y ) \ Z , • C := N − ( D \ { x } ) \ Z . • Since there is no path of length 3 in N + ( x ), there exists a vertex v ∈ A \ { y } ⊆ N + ( x )such that | N + ( v ) ∩ N + ( x ) | ≤
1. Thus, the set A ∪ B has size at least 2 δ ± ( G ) − | Z | − C ∪ D has size at least 2 δ ± ( G ) − | Z | −
1. If those two sets overlap thenthere exists the sought path.Firstly, suppose that | A ∪ B | + | C ∪ D | ≥ δ ± ( G ) − | Z | ≥ n + 2 − | Z | yet the two setsdo not overlap. If xy ∈ E ( G ) (hence | Z | = 1), then G has at least n + 2 − | Z | > n vertices,which is not possible. On the other hand, if xy / ∈ E ( G ) (hence | Z | = 3 and vertices x and y are not in A ∪ B ∪ C ∪ D ), then G has n + 4 − | Z | > n vertices; a contradiction.We conclude that if A ∪ B is disjoint from C ∪ D , then at least one of the considered setscontains exactly 2 δ ± ( G ) − | Z | − δ ± ( G ) − | Z | .By symmetry, we may assume | A ∪ B | = 2 δ ± ( G ) − | Z | −
1. Since there is no pathof length 3 inside N + ( x ) yet | A ∪ B | < δ ± ( G ) − | Z | , we conclude that every vertex in A \ { y } has an outneighbor in A \ { y } , so G [ A \ { y } ] must be a so-called tiling of C , i.e., | A \ { y }| / A \ { y } must havean edge to almost all the vertices in B . If there is no path of length at least 2 and atmost 4 from x to y , then every vertex in B needs to be adjacent to almost every vertexin C . Since any two vertices in the set A \ { y } with two inneighbors in C create C , allbut one of the vertices in A \ { y } must have an edge from almost every vertex in D \ { x } .Finally, almost every vertex in C is adjacent to almost every vertex in D . Therefore, wecan find C in G by taking two edges of a directed triangle contained inside A \ { y } andthen walking along the sets B , C and D . Claim 6.3.
Graph G contain no C with a chord.Proof. Firstly, we show that G has no 4 vertices spanning C with exactly one chord.Suppose for contradiction there are 4 vertices a , b , c and d with edges ab , bc , cd , da and ac . Claim 6.2 yields a path P of length 2, 3 or 4 from d to a avoiding c . If P has length 4we get C with vertex c . If P has length 3, then it needs to avoid vertex b , so it alsocreates C . Thus, there is a path of length 2 from d to a . Similarly, the path from c to d avoiding a needs to be of length 2. However, those two paths of length 2 yield to C .Now, suppose there is C , again with vertices a, b, c and d , that has two chords ac and bd . As before, there needs to be a vertex e and edges de and ea . The only way not tocreate C by considering a path from Claim 6.2 from c to d avoiding a is that the onlypath of length 3 from c to d uses both vertices e and b , and so there are edges ce and eb .Now, consider a path from Claim 6.2 from c to d avoiding e . There are no paths of length21 and 3, and a path of length 4 avoiding the vertex a creates C . Therefore, a needs tobe in N + ( c ) ∪ N + ( N + ( c ) \ { d } ). This means that there exists a vertex f , different from d and e , such that cf and f a are both edges in G . Since the vertices a , b , c and f create C with a chord, the previous paragraph yields there needs to be an edge bf or f b . However,in both cases we find C , which finishes the proof. Claim 6.4.
Graph G cannot contain a transitive triangle.Proof. Suppose for contrary there is a transitive triangle with edges ab , bc and ac . FromClaim 6.2 there is a path of length 2, 3 or 4 from c to a avoiding b . Since length 4creates C and length 2 creates C with a chord, which is forbidden by Claim 6.3, it musthave length 3. Say its edges are cd , de and ea . Claim 6.3 gives also that there are no otheredges between vertices a , c , d and e .We assert that the sets N − ( a ), N + ( c ), N − ( d ) \{ b } and N + ( e ) \{ b } are pairwise-disjoint: • N − ( a ) ∩ N + ( c ) = ∅ because it yields C with a chord contradicting Claim 6.3. • If there is x ∈ N − ( a ) ∩ N − ( d ) \ { b } then from Claim 6.3 there is no edge cx andfrom Claim 6.2 there is a path of length 2, 3 or 4 from c to x avoiding a , b and e .Path of length 2 needs to avoid also vertex d , so in each case we have C . • N − ( a ) ∩ N + ( e ) \ { b } = ∅ because it creates C . • N + ( c ) ∩ N − ( d ) \ { b } = ∅ because it creates C . • If there is x ∈ N + ( c ) ∩ N + ( e ) \ { b } then from Claim 6.3 there is no edge xa andfrom Claim 6.2 there is a path of length 2, 3 or 4 from x to a avoiding b , c and d .Path of length 2 needs to avoid also vertex d , so in each case we have C . • Having x ∈ N − ( d ) \ { b } ∩ N + ( e ) \ { b } from Claim 6.2 we get a path of length 2, 3or 4 from d to e avoiding vertex x . Any path of length 2 together with x creates C with a chord contradicting Claim 6.3. Path of length 3 avoids also vertices a and c ,and so creates C . Path of length 4 creates C with vertex x .Observe that the vertex b cannot be in both sets N − ( d ) and N + ( e ), because it createsa forbidden C with a chord. Thus | N − ( a ) ∪ N + ( c ) ∪ N − ( d ) ∪ N + ( e ) | ≥ δ ± ( G ) > n ,which finishes the proof of the claim.Now, we are ready finish the proof of the case (cid:96) = 6. For any edge ab ∈ E ( G ) considerthe sets N + ( a ), N − ( a ), N + ( b ) and N − ( b ). Since | N + ( a ) | + | N − ( a ) | + | N + ( b ) | + | N − ( b ) | ≥ δ ± ( G ) ≥ n + 2 , Claim 6.4 yields that there are at least two directed triangles containing the edge ab .Pick an arbitrary directed triangle xyz in G , and consider the directed triangles T xy , T yz and T zx different from xyz for the edges xy , yz and zx , respectively. Clearly, thesethree directed triangles must be edge-disjoint which readily yields a copy of C in G . As noted in [21], Conjecture 1.2 (hence our Theorem 1.4) yields the asymptotic semidegreethreshold for any given cycle with a fixed orientation apart from the directed triangle.Through this whole section, we write oriented cycle to denote a cycle with an arbitraryorientation of its edges, not necessarily the cyclic orientation as in the directed cycle C (cid:96) .22n order to state the main result of this section, we need to define the so-called cycle-type. For an oriented cycle C , we define the cycle-type t ( C ) to be the absolute value of thedifference between the number of edges oriented forwards in C and the number of edgesoriented backwards in C with respect to the cyclic orientation of C . In particular, the (cid:96) -cycle has a cycle-type (cid:96) , while cycles of cycle-type 0 are precisely those orientations forwhich there is a homomorphism into a directed path of an appropriate length. Observethat if t ( C ) ≥
3, then the cycle-type is equal to the maximum length of a directed cycleinto which there is a homomorphism from C .For an arbitrary oriented cycle C with t ( C ) ≥ k ( C ) be defined as the smallestinteger greater than 2 that does not divide t ( C ). Notice that if C is an (cid:96) -cycle for (cid:96) ≥ k ( C ) coincides with the definition of k in the previous sections.We are now ready to state the main result of this section. Theorem 7.1.
For every oriented cycle C (cid:54) = C and ε > there exists n = n ( C, ε ) suchthat the following is true for every oriented graph G on n ≥ n vertices: • If t ( C ) ≥ and δ ± ( G ) ≥ nk ( C ) + εn , then G contains C , and • if t ( C ) = 0 and δ ± ( G ) ≥ εn , then G contains C .Moreover, the bounds on the semidegree are asymptotically the best possible. In order to prove Theorem 7.1 for the case t ( C ) = 3, we need an auxiliary lemma. Let D denote the cycle on 5 vertices with t ( D ) = 3, i.e., the 4-shortcut depicted in Figure 1. Lemma 7.2.
Every n -vertex oriented graph G with δ ± ( G ) ≥ n +14 contains C or D .Proof. Suppose for contradiction there is an oriented graph G on n vertices with δ ± ( G ) ≥ ( n + 1) /
4, not containing C and D . For any vertex v ∈ V ( G ) both N + ( v ) and N − ( v )do not contain a path of length 3, since otherwise there is D in G . In particular, thereis s ∈ N + ( v ) such that N + ( s ) is disjoint from N + ( v ). Analogously to the reasoning inClaim 6.2, we find C in G that contains v .Let v , v , v and v be consecutive vertices of some copy of C in G . Note that thecopy must be induced since G is C -free. Also, as G is D -free, for i ∈ [4] the sets N + ( v i )and N − ( v i ) are disjoint from N + ( v i +1 ) and N − ( v i +1 ), respectively, where the indices aretaken modulo 4. Similarly, the C -freeness of G yields that the sets N − ( v i ) and N + ( v i +1 )are disjoint.By the bound on the outdegree of G , we have (cid:80) i ∈ [4] | N + ( v i ) | > n . Therefore, theremust be a common outneighbor of two non-consecutive vertices of the C . Without lossof generality, let x ∈ N + ( v ) ∩ N + ( v ). However, δ ± ( G ) ≥ ( n + 1) / y ∈ N + ( v ) ∩ N − ( v ) and v , v , y, v , x is a copy of D ; a contradiction.With this lemma we can prove Theorem 7.1. Proof of Theorem 7.1.
Since the case t ( C ) ≤ t ( C ) ≥
3. Let (cid:96) := | V ( C ) | and G be a graph contradictingthe statement of the theorem. If t ( C ) ≥ G contains an (cid:96) -blow-up of C t ( C ) . In particular, G contains C .Now suppose t ( C ) = 3 and C (cid:54) = C . If G contains C then we find C in the same way asin the case t ( C ) ≥
4. Therefore, by Lemma 7.2, G contains D . Since δ ± ( G ) ≥ (cid:96) +5, we cangreedily extend a copy of D to the graph D (cid:96) obtained by adding a path of length (cid:96) starting23n the sink of D and a path of length (cid:96) ending in the source of D . By supersaturation, G contains not only D (cid:96) , but also an (cid:96) -blow-up of D (cid:96) .In order to find a copy of C in G , it remains to prove that C is homomorphic to D (cid:96) .Without loss of generality, assume C has more forward edges than backward edges. Since t ( C ) = 3 and C is not a directed triangle, C contains two different vertices x and y suchthat there are exactly 4 more forward edges than backward edges in C between x and y .Let u and v be the respective source and sink of D in D (cid:96) , and P u and P v the pathsending in u and starting and v , respectively. Observe that we can homomorphically mapthe segment of C between x and y by mapping x to u , y to v and the other vertices of thesegment to V ( D (cid:96) ) in such a way that no edge of this segment is mapped to uv . On the otherhand, the segment of C between y and x can be mapped to the vertices of P u and P v so that x is mapped u , y to v and every edge of this segment is mapped to E ( P u ) ∪ E ( P v ) ∪ { uv } .Combining the mappings of the two segments yields a homomorphism from C to D (cid:96) .The moreover part of the theorem follows from the fact that balanced blow-ups of acycle of length k ( C ) are C -free. In this paper, for any given (cid:96) ≥ C (cid:96) in large enough oriented graphs. The threshold is essentially given by thedegrees in a balanced blow-up of C k , where k is the smallest integer greater than 2 thatdoes not divide (cid:96) . However, unless k = 3 or (cid:96) ≡ nk coming from the blow-ups of C k an extra term k − k to overcome certain number-theoreticflavored perturbations. What remains is to determine the semidegree threshold of C ,which is a long-standing open problem. Conjecture 8.1 ([14]) . Any n -vertex oriented graph G with the minimum semidegree δ ± ( G ) ≥ n contains C . Since Conjecture 8.1 has the same conjectured set of extremal constructions as thetriangle case of the Caccetta-H¨aggkvist conjecture (see, e.g., [30, Section 2] for its precisedescription), it is commonly believed that the difficulty of the two problems is very similar.Despite the extremal constructions are not exactly the balanced blow-ups of directedcycles, we have established in Theorem 1.5 a stability-type result for the cycle lengthsdivisible by 12. Specifically, every extremal graph in this case is close in the so-called edit-distance to a balanced blow-up of C k . The edit-distance of two graphs with the samenumber of vertices is the number of adjacencies one must alter in one of them to obtaina graph isomorphic to the other one. Using the terminology of graph limits theory (werefer the reader to the monograph by Lov´asz [26] for an excellent exposition to the topic),this translates to the fact that the unique extremal so-called oriented graphon is the onerepresenting the limit of balanced blow-ups of C k .As we have mentioned in the introduction, when the length of the forbidden cycleis an odd multiple of 3, there are infinitely many extremal limits. Indeed, any Eulerianorientation of a balanced complete bipartite graph is asymptotically extremal in this case.The fact that these are the only extremal limits follows from Theorem 1.6. However, whenforbidding C (cid:96) for (cid:96) ≡ (cid:96) (cid:54) = 6, a careful inspection of the arguments presentedin Section 5 yields the following analogue of Theorem 1.5 for these cycle-lengths: Proposition 8.2.
Fix an integer (cid:96) ≥ that is divisible by and not divisible by .Every oriented graph H on n vertices with δ ± ( H ) ≥ n − o ( n ) that contains no closed walkof length (cid:96) is homomorphic to C .
24n the case when (cid:96) = 6, we expect that the extremal graphs are close in the edit-distance to a blow-up of C where exactly one of the independent sets might be replacedby a one-way oriented bipartite graph. Similar to Proposition 8.2, maybe even a strongerstatement can be made if one is concerned with closed walks. Question 8.3.
Is it true that every oriented graph H on n vertices with δ ± ( H ) ≥ n − o ( n ) that contains no closed walk of length is homomorphic to the graph depicted in Figure 4? Figure 4: The conjectured homomorphic image of all the graphs with no closed walk oflength 6 and asymptotically the largest possible minimum semidegree.Recently, Theorem 1.3 was strengthen by Czygrinow, Molla, Nagle and Oursler [8],who proved that it is enough to assume the vertices of G have large outdegree. Theorem 8.4 ([8]) . For every (cid:96) ≥ there exists n := n ( (cid:96) ) such that every orientedgraph G on n ≥ n vertices with δ + ( G ) ≥ n + contains a directed cycle of lengthexactly (cid:96) . Motivated by the relations between Theorem 1.3 and Theorem 8.4 as well as Conjec-ture 8.1 and the triangle-case of the Caccetta-H¨aggkvist conjecture, it might be temptingto conjecture that the outdegree threshold of C (cid:96) is the same as the semidegree thresh-old of C (cid:96) for every length (cid:96) . However, this is not true and for most cycle-lengths thecorresponding outdegree threshold is strictly larger than the semidegree one. ...... ...... Figure 5: Maneuvers used for the constructions assuming only the minimum outdegree.In both maneuvers, by adding presented edges to an arbitrarily chosen vertex in one blob,one can remove a vertex in the topmost blob.For an example, fix an integer (cid:96) divisible by 3 such that the smallest positive integer k that does not divide (cid:96) is odd, and consider a balanced blow-up of a k -cycle on n + k − nk + k − k . If (cid:96) < k/
2, then using2 (cid:96) − k − (cid:96) − k − k +12 the remainders modulo k of the cycle lengths that can beobtained, while each usage of the second maneuver is changing those remainders by k − .There is no (cid:96) -cycle, since it is impossible to obtain (cid:96) mod k in such a way. Similarly, forthe case (cid:96) > k/
2, we can use 2 (cid:96) − k − k − (cid:96) − k − k ≥ k of any cycle length that can beobtained by 3 and k − Conjecture 8.5.
Fix an integer (cid:96) ≥ and let k be the smallest integer greater than thatdoes not divide (cid:96) . There exists n := n ( (cid:96) ) such that every oriented graph G on n ≥ n vertices with δ + ( G ) ≥ nk + 1 contains C (cid:96) . Acknowledgements.
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