Turán problems for k-geodetic digraphs
aa r X i v : . [ m a t h . C O ] F e b Tur´an problems for k -geodetic digraphs James Tuite a , Grahame Erskine a , Nika Salia b a Department of Mathematics and Statistics, Open University, Walton Hall, Milton Keynes, UK b Alfr´ed R´enyi Institute of Mathematics, Budapest, Hungary
Abstract
Tur´an-type problems have been widely investigated in the context of undirected simple graphs.Tur´an problems for paths and cycles in directed graphs have been treated by Bermond, Heydemannet al. Recently k -geodetic digraphs have received great attention in the study of extremal problems,particularly in a directed analogue of the degree/girth problem. Ustimenko et al studied the problemof the largest possible size of a diregular k -geodetic digraph with order n and gave asymptoticestimates. In this paper we consider the maximal size of a k -geodetic digraph with given order withno assumption of diregularity. We solve this problem for all k and classify the extremal digraphs.We then provide a partial solution for the more difficult problem of the largest number of arcs in astrongly-connected k -geodetic digraph with given order and provide lower bounds for these numbersthat we conjecture to be extremal for sufficiently large n . We close with some results on generalisedTur´an problems for k -geodetic digraphs. Keywords:
Digraph, Tur´an, Extremal, k -geodetic, Strong-connectivity
1. Introduction A graph here will refer to a simple undirected graph without loops. For a graph F we will saythat the graph H is F -free if H contains no subgraph isomorphic to F . A directed graph or digraph G is a set V ( G ) of vertices and a set A ( G ) of arcs, or ordered pairs of distinct vertices of G . If ( u, v )is an arc of G then we will write u → v . A directed path of length ℓ in a digraph G is a sequence x x . . . x ℓ − x ℓ of distinct vertices of G such that x i → x i +1 for 0 ≤ i ≤ ℓ −
1. A walk of length ℓ isdefined in the same way, but without the requirement that the vertices in the sequence be distinct.The distance d ( u, v ) between vertices u, v of a digraph G is the length of a shortest directed pathfrom u to v in G (or ∞ if no such path exists); note that we may have d ( u, v ) = d ( v, u ) in a digraph.For a vertex u of G the out-neighbourhood N + ( u ) of u is defined to be { v ∈ V ( G ) : u → v } ;similarly the in-neighbourhood of u is N − ( u ) = { v ∈ V ( G ) : v → u } . For a vertex u of G we define N + k ( u ) := { v ∈ V ( G ) | d ( u, v ) = k } , similarly N − k ( u ) := { v ∈ V ( G ) | d ( v, u ) = k } . The out-degree d + ( u ) of a vertex u of G is the number of out-neighbours of u , i.e. d + ( u ) = | N + ( u ) | , and the in-degree of u is d − ( u ) = | N − ( u ) | . A vertex with out-degree zero is a sink and a vertex with in-degreezero is a source . If there exists some integer d such that d + ( u ) = d for all u ∈ V ( G ) then G is d -out-regular or just out-regular . If further there exists some d such that d − ( u ) = d + ( u ) = d for allvertices u then G is diregular . The graph H formed by neglecting the orientations of the arcs of G is called the underlying graph of G . Email addresses: [email protected] (James Tuite), [email protected] (Grahame Erskine), [email protected] (Nika Salia)
Preprint submitted to Journal of Graph Theory February 10, 2021 directed cycle of length ℓ is a sequence of distinct vertices x , x . . . x ℓ − such that x i → x i +1 for 1 ≤ i ≤ ℓ − x ℓ − → x and the length of a shortest directed cycle in G is the girth of G . Adigraph G is strongly connected or strong if for any ordered pair of vertices ( u, v ) of G there exists adirected path from u to v in G ; trivially a digraph containing a source or a sink cannot be stronglyconnected. The converse of a digraph G is the digraph on the same vertex set as G obtained byreversing the direction of all arcs in G . For other graph-theoretical terminology we follow [4].A digraph G is k -geodetic if for any ordered pair of (not necessarily distinct) vertices ( u, v ) thereis at most one u, v -walk in G with length at most k . For any vertex u there is a trivial u, u -walk oflength zero, so k -geodecity in particular implies that there are no cycles of length at most k . Suchdigraphs have recently proven to be of interest in a directed analogue of the degree/girth problem;see for example [23]. The geodetic girth of a digraph G is the largest k such that G is k -geodetic.We say that an undirected graph H has a k -geodetic orientation if directions can be assigned toeach edge of H such that the resulting digraph G is k -geodetic. Observe that if a graph H has girthat least 2 k + 1, then every orientation of H will be k -geodetic.Tur´an type problems constitute one of the most investigated areas of extremal graph theory.Such a problem typically asks for the largest possible size of a graph G with a family F of forbiddensubgraphs. The first Tur´an problem to be solved was published in 1907 by Mantel. In [19], Mantelproved that the largest possible number of edges in a triangle-free graph with order n is given by ⌊ n ⌋ and this bound is achieved by the complete bipartite graph K ⌈ n ⌉ , ⌊ n ⌋ . This was later generalisedby Tur´an in 1941 as follows. Theorem 1. (Tur´an, [24]) The number of edges of a K r +1 -free graph H is at most (1 − r ) n . This theory was taken in an alternative direction by Erd˝os, who asked in 1975 for the largestpossible size of a graph with order n containing no C or C [11]. If this extremal size for givenorder n is denoted by f ( n ), then his conjecture can be written f ( n ) = ( + o (1)) / n / . It is knownthat 12 √ ≤ lim inf n →∞ f ( n ) n ≤ lim sup n →∞ f ( n ) n ≤ . This problem has been investigated more generally for graphs with girth ≥ g (see [18] for an example)but remains a famous open problem of extremal combinatorics. On the other hand, the problem ofthe largest possible size of a graph without a cycle of length four and a given odd cycle has beensolved for all odd cycles except the triangle and in all cases the asymptotic behaviour of the Tur´annumber is the same as the largest possible size of a bipartite graph with no cycle of length four. Theorem 2. (Erd˝os, Simonovits, [12]) The maximum possible number of edges in a graph on n vertices containing no C or C as a subgraph is at most √ (1 + o (1)) n / . Tur´an problems in directed graphs is a relatively unexplored area. Bermond et al in [2] foundthe largest possible size of a strongly-connected digraph with girth g , and Heydemann discussedthe Tur´an problem for directed paths in [17]. The geodetic girth of a digraph has proven to be aninteresting ‘girth-like’ parameter. The largest size of a diregular k -geodetic digraph was discussedin [25] and asymptotic estimates obtained. In this paper we determine the solution of this problemwithout the assumption of diregularity.In Section 2 of this paper we determine the largest possible size of a k -geodetic digraph withorder n . It turns out that it is more difficult to determine the largest possible order of a strongly-connected k -geodetic digraph with order n ; we solve this problem in Section 3 for k = 2, classifythe extremal digraphs in Section 4 and in Section 5 for larger k we give constructions that providea lower bound for the size of such digraphs that we conjecture to be extremal. Finally in Section 6we study some generalised Tur´an problems for k -geodetic digraphs.2 wvx (a) This digraph is not 2-geodetic. u wv (b) This digraph is 2-geodetic but isnot 3-geodetic. Figure 1: Digraphs with low geodetic girth
2. The largest size of a k -geodetic digraph It is natural to ask for the largest possible size of a digraph with order n and no directed cyclesof length ≤ g . However, this question must be posed carefully, as there exists an acyclic tournamentfor every order n . Bermond et al in [2] proved the following theorem for directed cycles in stronglyconnected digraphs. Theorem 3. (Bermond et al in [2] ) Let D be a strong digraph of order n , size m and girth g . Let k ≥ . Then if m ≥
12 ( n + (3 − k ) n + k − k ) , we must have g ≤ k . This expression is best possible. Surprisingly, and in contrast to the undirected case, this result shows that, asymptoticallyspeaking, a graph can have ‘almost all’ arcs present and still have arbitrarily large girth, evenif the graph is strongly connected. We now solve the related Tur´an problem for another ‘girth-like’ parameter, namely the geodetic girth. This problem can be put into the form of a forbiddensubgraph problem, as every violation of k -geodecity in G can be identified with the occurrence ofa specific subdigraph of G ; in [25] these subdigraphs are referred to as ‘hooves’ or ‘commutativediagrams’. Two examples are shown in Figure 1.In the papers [21, 25] Ustimenko et al prove that, if f ( n, k ) is the largest size of a diregular k -geodetic digraph with order n then for fixed k we have f ( n, k ) ∼ n k +1 k and give a family of digraphs,now known as the permutation digraphs , that meet this asymptotic bound. These digraphs wereintroduced in [13] (some further properties of these digraphs are given in [6]). For d, k ≥ P ( d, k ) is defined as follows. The vertices of P ( d, k ) are all permutations x x . . . x k − of length k of symbols from the set [ d + k ] = { , , . . . , d + k − } . A vertex x x . . . x k − has an arc to all permutations of the form x x . . . x k − x k for any x k
6∈ { x , x , . . . , x k − } . Anexample, P (2 , P ( d, k ) is diregular with out-degree d and is k -geodetic. The order of P ( d, k ) is n = ( d + k )( d + k − · · · ( d + 1) and has size m = nd ∼ n k +1 k . We will see in the final section that these digraphs have other interesting extremalproperties.The result of Ustimenko also holds in the more general setting of out-regular digraphs as we cansee from the following short argument. Theorem 4.
For k ≥ the largest size ex out ( n, k ) of an out-regular k -geodetic digraph with order n satisfies ex out ( n ; k ) ∼ n k +1 k as n → ∞ .Proof. It is known that the order n of a k -geodetic digraph with minimum out-degree d is boundedbelow by the directed Moore bound M ( d, k ) = 1 + d + d + · · · + d k (see [20]). Hence n ≥ d k and,rearranging, d ≤ n /k . The size m of an out-regular k -geodetic digraph G with order n thus satisfies m = nd ≤ n k +1 k . 3 Figure 2: P (2 , What can we say about the maximum number of arcs in a k -geodetic digraph if we make noassumption of out-regularity? We make the following definition. Definition 5.
For n, k ≥ define ex ( n ; k ) to be the largest possible size of a k -geodetic digraph on n vertices. In contrast to the out-regular case it is easy to provide a quadratic lower bound for the numbers ex ( n ; k ). Lemma 6.
For n, k ≥ we have ex ( n ; k ) ≥ ⌊ n / ⌋ .Proof. Orienting all edges of the complete bipartite graph K ⌈ n/ ⌉ , ⌊ n/ ⌋ in the same direction yieldsa k -geodetic digraph.We now offer two separate proofs to show that this lower bound is optimal. The first proofuses a simple counting argument and induction. If a digraph G is k -geodetic, every subdigraphof G must also be k -geodetic. This approach yields a simple upper bound for ex ( n ; k ) in terms of ex ( m ; k ) for m < k . Lemma 7.
For any k ≤ m ≤ n − we have ex ( n ; k ) ≤ n ( n − m ( m − ex ( m ; k ) . Proof.
We count the pairs (
F, e ), where F is a subset of m vertices and e is an arc with both end-points in F . Let F be any subset of m vertices. In the induced subdigraph there can be at most ex ( m ; k ) arcs. Therefore there are at most (cid:0) nm (cid:1) ex ( m ; k ) such pairs. For each arc e there are exactly (cid:0) n − m − (cid:1) subsets containing the endpoints of e , so it follows that ex ( n ; k ) (cid:18) n − m − (cid:19) ≤ (cid:18) nm (cid:19) ex ( m ; k ) . Rearranging yields the result.
Theorem 8.
For n ≥ and k ≥ we have ex ( n ; k ) = ⌊ n ⌋ . yz z ′ Figure 3: There is no 2-geodetic orientation of K − Proof.
Let k = 2. The theorem is easily shown to be true for n = 4. Let n ≥ n −
1. Suppose that n = 2 r is even. Putting m = n − ex (2 r ; 2) ≤ r (2 r − r − r − r ( r −
1) = r as required. Now let n = 2 r + 1. Lemma 7 with m = 2 r gives ex (2 r + 1; 2) ≤ r (2 r + 1)2 r (2 r − r = (2 r + 1) r r − < r + r + 1 , so again the necessary inequality follows. As a k -geodetic digraph is also 2-geodetic for k ≥
2, no k -geodetic digraph can have more than ex ( n ; 2) arcs; at the same time, the digraph in Lemma 6 istrivially k -geodetic for any k ≥ k -geodetic digraph. We denote the complete graph on n vertices with oneedge deleted by K − n . K − is also often called the diamond graph . Lemma 9.
Every graph with a 2-geodetic orientation is K − -free.Proof. Suppose for a contradiction that a graph G contains two triangles x, y, z and x, y, z ′ , where z = z ′ . The only way to orient a triangle is to make it into a directed 3-cycle, so we can assumethat x → y → z → x and x → y → z ′ → x . This situation is shown in Figure 3. However there arenow two distinct directed paths from y to x of length two, violating 2-geodecity. Theorem 10.
For n ≥ and k ≥ we have ex ( n ; k ) = ⌊ n ⌋ and for n ≥ all extremal -geodeticdigraphs are orientations of complete balanced bipartite graphs K ⌈ n ⌉⌊ n ⌋ .Proof. As noted previously, it is sufficient to prove the upper bound for k = 2. A simple inductiveargument shows the well-known result [7, 8] that for n ≥ n and size > ⌊ n ⌋ contains a copy of K − and the unique K − -free graph with size ⌊ n ⌋ is K ⌈ n ⌉ , ⌊ n ⌋ . The result thereforefollows by Lemma 9. The graph in Figure 4 shows that n ≥ n ≥ ex ( n ; 2) arcs is an orientation of acomplete bipartite graph K ⌈ n ⌉⌊ n ⌋ . However, there are a large number of non-isomorphic orientationsof K ⌈ n ⌉⌊ n ⌋ , not all of which are 2-geodetic. We now completely classify all extremal 2-geodetic5 v v u u u Figure 4: A 2-geodetic digraph with order 6 and size 9 containing a triangle digraphs. We will label one partite set of our bipartite graph X and the other Y . If X contains asource then Y contains no sources and vice versa, so we can assume that any source of G lies in X .If X contains only sources then Y consists only of sinks, in which case we recover the constructionin Lemma 6. We can therefore assume that X contains a vertex that is neither a source nor a sink. Lemma 11.
Let K be a 2-geodetic orientation of a complete bipartite graph K s,t with partite sets X and Y , where s ≥ t ≥ . If x is any vertex of K that is neither a source nor a sink, then either d + ( x ) = 1 or d − ( x ) = 1 .Proof. Let x ∈ X be a vertex of K that is neither a source nor a sink. Suppose that d + ( x ) ≥ d − ( x ) ≥
2. Let y ∈ Y be an out-neighbour of x such that y is not a sink. Hence y → x ′ for some x ′ ∈ X − x . If any other out-neighbour y ′ of x has an arc to x ′ , then we would have two 2-paths x → y → x ′ and x → y ′ → x ′ , violating 2-geodecity, so it follows that x ′ has arcs to every vertex of N + ( x ) − { y } . Any in-neighbour y − of x can already reach every vertex of N + ( x ) by a 2-path via x . As x ′ has arcs to every vertex of N + ( x ) − { y } , it follows that x ′ → y − for every in-neighbour y − of x . However, there are at least two such in-neighbours y − and y − by assumption, so there existpaths x ′ → y − → x and x ′ → y − → x , a contradiction.It follows that every out-neighbour of x in Y is a sink and similarly every in-neighbour of x is asource. Let x ∗ ∈ X − { x } . Then if y + ∈ N + ( x ) , y − ∈ N − ( x ) we have two paths y − → x → y + and y − → x ∗ → y + , which is impossible. Hence we must have either d + ( x ) = 1 or d − ( x ) = 1. Theorem 12.
Let K be a 2-geodetic orientation of a complete bipartite graph K s,t , where s ≥ t ≥ .Then all of the arcs of K are oriented in the same direction, except for a matching (possibly of sizezero) in the opposite direction.Proof. As K is 2-geodetic X cannot contain both sources and sinks; for example if x ∈ X is asource and x ∈ X is a sink, then if y , y ∈ Y we have paths x → y → x and x → y → x ,which is impossible.Note that if all vertices are neither a source nor a sink then both partitions contain a vertexwhich is neither a source nor a sink. If s = t = 2 then by Lemma 11 we are done, so we mayassume that | Y | > X contains a vertex which is neither a source nor a sink. By Lemma 11any such vertex has either in-degree or out-degree one; without loss of generality, we assume that N + ( x ) = { y } . Then X contains no vertex x ′ ∈ X , d + ( x ′ ) >
1, otherwise there would be paths6 igure 5: A strongly connected digraph with n = 2 r and m = r (for r = 4) x ′ → y → x and x ′ → y → x if y , y = y a contradiction or there is a vertex y ′ in Y such thatwe have y ′ → x and y ′ → x ′ which leads us also to a contradiction since there are two paths from y ′ to y . Applying Lemma 11 we have the desired result.Combining Theorems 10 and 12 immediately yields a classification of all 2-geodetic digraphs oforder n and maximal size. Theorem 13.
Let G be a 2-geodetic digraph with order n and maximal size. For n ≥ , G isisomorphic to an orientation of K ⌈ n ⌉ , ⌊ n ⌋ with all arcs oriented in the same direction, except for amatching that is oriented in the opposite direction. The number of isomorphism classes of extremaldigraphs is n + 1 for odd n ≥ and n + 1 for even n ≥ .
3. The largest size of a strongly-connected 2-geodetic digraph
For even n , if the matching mentioned in Theorem 13 is chosen to be a perfect matching, then theresulting digraph is strongly connected. Therefore for even n there is always a strongly-connected2-geodetic digraph with ex ( n ; 2) arcs. Such a digraph is shown in Figure 5. However, it is easilyseen that if n is odd, then all of the 2-geodetic digraphs given in Theorem 13 contain either a sourceor a sink and so are not strongly connected. This leads us to make the following definition. Definition 14.
For n ≥ k +1 and k ≥ , ex ∗ ( n ; k ) is the largest possible size of a strongly-connected k -geodetic digraph with order n . From the former observation we know that for r ≥ ex ∗ (2 r ; 2) = r . We turn to thequestion of determining ex ∗ (2 r + 1; 2). Taking a strongly connected 2-geodetic digraph with order2 r and size r and expanding one arc into a directed triangle shows that ex ∗ (2 r + 1; 2) ≥ r + 2(this construction is shown in Figure 6). We now show that this lower bound is optimal; in fact weprove slightly more: that any 2-geodetic digraph with larger size contains either a source or a sink. Theorem 15.
For all r ≥ we have ex ∗ (2 r + 1; k ) = r + 2 and any 2-geodetic digraph with largersize contains a source or a sink. For r ≥ the underlying graph of any 2-geodetic digraph G withorder n = 2 r + 1 and size m = r + 2 that has no sources or sinks is isomorphic to a graph formedfrom a triangle T and a copy of K r − ,r − by joining every vertex in K r − ,r − to exactly one vertexof T . igure 6: A strongly connected digraph with n = 2 r + 1 and m = r + 2 for r = 4 with the triangle in bold Proof.
The result for r ≤ r ≥
3. Let G bea 2-geodetic digraph with size m ≥ r + 2 and H be the underlying graph of G . Then H is K − -freeby Lemma 9. We will proceed to show that H contains a triangle with a special substructure.Suppose that H is triangle-free. As the size of H is at least r + 2, it follows by the stabilityresults of [10, 22] that H is bipartite. We will name the larger partite set X and the smaller Y . Weclaim that there are at least three vertices of X that are connected to every vertex of Y . Otherwise,setting t = | X | , the size of H is bounded above by f ( t ) = ( t − r − t ) + 2(2 r + 1 − t ) = − t + 2 rt + 2 , where r + 1 ≤ t ≤ r + 1 as | X | > | Y | . The function f ( t ) has its maximum at t = r + 1, where f ( r + 1) = r + 1 < m .Let X ′ be the set of vertices in X that are adjacent to every vertex of Y and let H ′ be thecomplete bipartite subgraph of H with partite sets X ′ and Y . By Theorem 12 all edges of H between X and Y are directed in the same direction with the exception of a matching M of size ≤ r in the opposite direction. Taking the converse of G if necessary we can assume that the edges of M are directed from Y to X , with all other edges in the other direction. By assumption G containsno sources or sinks, so every vertex of X ′ must be incident with an edge of the matching M .Therefore if two vertices y, y ′ of Y have a common out-neighbour x , then there will be a vertex x ′ ∈ X ′ that has arcs to both y and y ′ and hence has two paths of length two to x , violating2-geodecity. Hence the out-neighbourhoods of the vertices of Y are pairwise disjoint. No vertexof X is a source and so each vertex of X has at least one in-neighbour in Y . As | X | > | Y | theremust be a vertex y ∈ Y that has (at least) two out-neighbours x and x in X . If x and x hada common out-neighbour y ′ ∈ Y then there would be two paths of length two from y to y ′ , so wehave N + ( x ) ∩ N + ( x ) = ∅ . Hence there are at most two arcs incident to { x , x } and at most r − { x , x } , so there are at most r + 1 arcs incident with x or x . If we delete x and x we would thus obtain a 2-geodetic digraph with order n = 2( r − ≥ m ′ , where m ′ ≥ m − r − ≥ r − > ( r − , contradicting Theorem 8. Therefore H contains a triangle T .8et us label the vertices of T by x, y, z . As G is K − -free every vertex of V ( G ) − T is adjacentto at most one vertex of T , so deleting T from G removes at most n = 2 r + 1 arcs. By Theorem 8the size of G − T is at most ( r − .Thus m ≤ (2 r + 1) + ( r − = r + 2, with equality if andonly if G − T is an extremal digraph given in Theorem 13 and every vertex of H − T is adjacentwith exactly one vertex of T .
4. Classification of extremal 2-geodetic digraphs without sources and sinks
In the previous section it was shown that for r ≥ n = 2 r + 1 has at most ex ∗ (2 r + 1; 2) = r + 2 arcs. In this section we will classify thestrongly-connected 2-geodetic digraphs that achieve this bound. Our analysis will focus on the case r ≥
5, i.e. odd n ≥
11. Computer search shows that there is a unique extremal strongly-connected2-geodetic digraph with size r + 2 for r = 1, 3 extremal digraphs for r = 2, 29 solutions for r = 3and 19 solutions for r = 4; and any 2-geodetic digraphs with larger size contain either a source ora sink.Let G be a 2-geodetic digraph with order n = 2 r + 1 ≥
11, size r + 2 and no sources or sinksand let H be the underlying undirected graph of G . By Theorem 15 H contains a triangle T withvertices x, y, z , which is oriented in G as x → y → z → x and each vertex in H − T is adjacent toexactly one of x, y or z . Furthermore, G − T must be one of the r orientations of K r − ,r − given inTheorem 13. Let the bipartition of K r − ,r − be X, Y , where X = { x , . . . , x r − } , Y = { y , . . . , y r − } .By Theorem 13 we can assume that x i → y i for 1 ≤ i ≤ r − − s for some 0 ≤ s ≤ r −
1, with allother edges oriented in the other direction. Obviously there are s sources and s sinks in G − T .We will say that a partite set is covered by a subset T ′ of T if all of its neighbours in T belongto T ′ ; in particular, if all of the neighbours of a partite set, say X , are the same vertex of T , say x ,then X is covered by x . We call a vertex in T bad if it has neighbours in both partite sets of H − T . Lemma 16.
Any bad vertex of T has degree four in H . If n ≥ then there is at most one badvertex.Proof. It is easily seen that if a bad vertex has degree ≥ H , and hence ≥ K r − ,r − , then H contains a copy of K − , which is impossible by Lemma 9. As any bad vertex of T is adjacent to one vertex of X and one vertex of Y for r ≥ T can bebad. Furthermore if two vertices of T are bad then the third vertex of T would have also have tobe bad. Lemma 17.
If there is no bad vertex in T , then either X or Y is covered by a single vertex of T and s ≤ .Proof. If there is no bad vertex, then the neighbours of any vertex of T in K r − ,r − must be entirelycontained in one partite set, so one partite set is covered by one vertex of T and the other partiteset is covered by the other two vertices of T .Concerning the value of s , suppose that s ≥ X is covered by x (the argument for Y is similar). There must be an arc from every sink in X to x . But any source in Y has arcs to allthe sinks in X and hence will have multiple 2-paths to x , contradicting 2-geodecity. Lemma 18.
Any vertex of T with neighbours in X has at most one in-neighbour in X . Any vertexof T that is joined to ≥ non-sink vertices of X has no in-neighbour among the non-sink verticesof X . Substituting ‘source’ for ‘sink’ and ‘out-neighbour’ for ‘in-neighbour’, the analogous resultshold for Y . z yx x x x x y y y y y (a) A xy zx x x x x y y y y y (b) B , Figure 7: The graphs A and B , Proof.
Suppose that a vertex of T , say x , has ≥ x i and x j in X . For any l ∈{ , , . . . , r − } − { i, j } we have y l → x i → x and y l → x j → x , a contradiction.Now let x be adjacent to vertices x i and x j in X , where we now assume that x i and x j are notsinks in G − T . If x i → x , then as x has at most one in-neighbour in X we must have x → x j .Hence there are paths x i → x → x j and x i → y i → x j , a contradiction. The results for Y follow ina similar manner.First we shall deal with the case that T has no bad vertices. Assume firstly that X is coveredby x . Suppose that s = 1 (Figure 7a). Then x r − and y r − are the sink and the source of G − T respectively. Now x must have an arc from the sink so that it does not remain a sink in G ; henceby Lemma 18 we have x → x i for 1 ≤ i ≤ r − Y is covered by y and z and either y or z has anarc to the source y r − .If z has an arc to Y then there would be multiple 2-paths from z to a non-sink vertex in X and similarly if z has an in-neighbour y i in Y , then there would be 2-paths y i → z → x and y i → x r − → x . Therefore z has no neighbours in Y , so y must have an arc to y r − and by Lemma18 y i → y for 1 ≤ i ≤ r −
2. This yields the 2-geodetic digraph A r , an example of which is shownin Figure 7a. This digraph is isomorphic to its converse.Now let X be covered by x and s = 0. By Lemma 18 x → x i for 1 ≤ i ≤ r −
1. By reasoningsimilar to the previous case, y and z can have no out-neighbours in Y . Let the resulting digraph inwhich y has t in-neighbours in Y be denoted by B r,t for 0 ≤ t ≤ r − B r,t isa 2-geodetic extremal digraph.The case of Y being covered by one vertex of T is symmetric. In particular we shall denote theconverse of B r,t by B ′ r,t . We have B ′ r, ∼ = B r, and B ′ r,r − ∼ = B r,r − , but otherwise these digraphs are10 xz x x x x x y y y y y (a) C yxz x x x x x y y y y y (b) D Figure 8: The digraphs C and D pairwise non-isomorphic.We now turn to the case that there is a bad vertex; say z is bad. Hence d ( z ) = 4 in H . It followsby Lemma 16 that x and y each have r − K r − ,r − and each is connected to justone partite set. Lemma 19. If z is bad, then s ≤ . If z is joined to a source in Y , then X is covered by { y, z } and Y is covered by { x, z } . Likewise, if z is joined to a sink in X , then X is covered by { x, z } and Y is covered by { y, z } . If s = 2 , then z is connected to a source in Y and a sink in X .Proof. Suppose that s ≥
3. The bad vertex z is adjacent to one vertex of Y in H − T , so the vertexof T that also has edges to Y must have arcs to two or more sources in Y , violating Lemma 18.This reasoning also demonstrates that if s = 2, then z is connected to a source in Y and a sink in X . For any s ≤
2, suppose that z is joined to a source in Y . Suppose that X is covered by { x, z } .Then z has a 2-path to every vertex of X via the source, but by Lemma 18 x has an out-neighbour x i ∈ X , so there will also be a 2-path from z to x i via x , violating 2-geodecity. Hence X must becovered by { y, z } and hence Y is covered by { x, z } . The other statement is symmetric to this one.Let s = 2. The sources in G − T are y r − and y r − and the sinks are x r − and x r − . ByLemma 19 we can assume that z → y r − and x r − → z . Also by Lemma 19 X is covered by { y, z } .There must be an arc from x r − to y so that x r − is not a sink in G and y → x i for 1 ≤ i ≤ r − x to y r − . However, we now have two 2-paths from x to the vertices in { x , . . . , x r − } , one via y and the other via y r − , a contradiction. It follows that s ≤ s = 1. The sink and source of G − T are x r − and y r − respectively (Figure 8a). Supposethat z is joined to x r − and y r − . By Lemma 19 X is covered by { y, z } and Y is covered by { x, z } .By Lemma 18 y → x i for 1 ≤ i ≤ r − y i → x for 1 ≤ i ≤ r −
2. This gives the singlesolution C r , an example of which is shown in Figure 8a. Note that the digraph C r is isomorphic toits converse.Suppose that z is joined to the source y r − but is not joined to the sink x r − of G − T ; say z has an edge to x r − in H − T . By Lemma 19 X is covered by { y, z } and Y is covered by { x, z } .11ence there is an arc x r − → y and by Lemma 18 x has at most one out-neighbour in Y − y r − , sothat there is a vertex y i with y i → x . Hence there would be paths y i → x → y and y i → x r − → y in G , a contradiction. We will get a similar contradiction if z is joined to the sink x r − in X , butnot to the source y r − in Y .Finally let z be joined to x i and y j where 1 ≤ i, j ≤ r −
2. Suppose that X is covered by { x, z } and Y by { y, z } . If i = j , then the triangle is oriented as x i → y i → z → x i ; however this yieldspaths y i → z → x and y i → x r − x , so we must have i = j . Without loss of generality we can set i = r − j = r −
3. The triangle is now oriented as y r − → x r − → z → y r − . There is an arc x r − → x so by Lemma 18 there are arcs x → x l for 1 ≤ l ≤ r −
3. In this case we would havepaths z → y r − → x and z → x → x .Hence we can assume that X is covered by { y, z } and Y by { x, z } . By Lemma 18 y has at leasttwo out-neighbours in X , so if x has any out-neighbour in Y then there would be more than one2-path from x to an out-neighbour of y in Y . In particular we must have z → y r − , a case that wehave already considered.Now we can set s = 0. Suppose that z is joined to x and y . As y → x , we must orient thetriangle z, x , y as z → y → x → z . If X is covered by { x, z } and Y is covered by { y, z } , thenby Lemma 18 x → x i for 2 ≤ i ≤ r − z → y → x and z → x → x .Hence X must be covered by { y, z } and Y must be covered by { x, z } . By Lemma 18 we have y → x . Hence there are paths x → y → x and x → z → x .Therefore we can assume that z is joined to x and y . We must have z → x and y → z . If X is covered by { y, z } and Y by { x, z } then by Lemma 18 y → x i and y i → x for 2 ≤ i ≤ r − D r shown in Figure 8b. D r is isomorphic to its converse. If X is coveredby { x, z } and Y by { y, z } then by a suitable redrawing of the digraph it can be seen that we obtaina solution isomorphic to C r in Figure 8a.This completes our classification of the strongly-connected 2-geodetic digraphs with order n =2 r + 1 and size r + 2. We therefore have the following theorem. Theorem 20. If G is a 2-geodetic digraph with order n = 2 r + 1 ≥ , size m = r + 2 and nosources or sinks, then G is either isomorphic to one of A r , B r, , B r,r − , C r or D r or is isomorphicto a member of the family B r,t , B ′ r,t for some ≤ t ≤ r − . The digraphs in this list are pairwisenon-isomorphic and so there are r + 1 distinct solutions up to isomorphism.
5. Extremal strongly-connected k -geodetic digraphs for k ≥ The analysis of Section 3 naturally raises the question: what is the largest possible size of astrongly-connected k -geodetic digraph with order n for k ≥
3? In this section we provide upperand lower bounds for ex ∗ ( n ; k ) for k ≥ n .As any k -geodetic digraph with k ≥ k ≥ ex ∗ ( n ; k ) < ex ∗ ( n ; 2), where strict inequality follows from the fact that the extremal digraphs for k = 2 are not 3-geodetic. However, it is easy to provide a better upper bound for larger k . Lemma 21.
For k ≥ any k -geodetic digraph without sources or sinks has strictly fewer than n k arcs.Proof. Let G be a k -geodetic digraph without sinks. Suppose that G contains a vertex u with out-degree d + ( u ) ≥ nk . As every vertex has out-degree at least one, it follows that | N + t ( u ) | ≥ d + ( u ) = nk for 1 ≤ t ≤ k , where N + t ( u ) denotes the set of vertices at distance t from u . As G is k -geodetic,all of the vertices in these sets are distinct, so it follows that n ≥ k nk , a contradiction. Hence12he maximum out-degree of G is ∆ + < nk and, summing over all vertices of G , the size of G is m < n nk = n k .It follows that lim sup n →∞ ex ( n ; k ) n ≤ k . We now provide constructions that show that k ≤ lim inf n →∞ ex ( n ; k ) n .Let the quotient and remainder when n is divided by k be r and s respectively, i.e. n = kr + s .We assume that s ≤ r . Form the digraph G ( n, k ) as follows (Figure 9). The vertex set of G ( n, k )consists of vertices u i,j for 1 ≤ i ≤ r and 1 ≤ j ≤ k , as well as s further vertices v , v , . . . v s .We define the adjacencies of G ( n, k ) as follows.i) u i,j → u i,j +1 for 1 ≤ i ≤ r and 1 ≤ j ≤ k − u i,k → v i for 1 ≤ i ≤ s iii) u i,k → u j, for s + 1 ≤ i ≤ r and 1 ≤ j ≤ s iv) u i,k → u i ′ , for s + 1 ≤ i, i ′ ≤ r and i = i ′ v) v t → u i, for 1 ≤ t ≤ s and all i in the range 1 ≤ i ≤ r .This digraph is k -geodetic and has size m = rs + ( k − r + s + ( r − s )( r −
1) = r + ( k − r + 2 s. If r + 1 ≤ s ≤ k −
1, then we have ⌊ nk ⌋ ≤ k −
2, which is equivalent to n ≤ k − k −
1. Thereforethese digraphs will certainly exist for n ≥ k − k . The arcs in part iii) can also be directed to u j, ; combined with taking the converse of the resulting digraphs, this generates several differentisomorphism classes.These digraphs admit a particularly simple description when k | n . Let n = kr for some r ≥ G ( kr, k ) is k -geodetic and has order kr and size r ( r − r ( k −
1) = r +( k − r = n k + ( k − nk .It has vertices u i,j , where 1 ≤ i ≤ r and 1 ≤ j ≤ k . We define the adjacencies as follows.i) u i,j → u i,j +1 for 1 ≤ i ≤ r and 2 ≤ j ≤ k − u i, → u i ′ , for 1 ≤ i, i ′ ≤ r and i = i ′ ,iii) u i,k → u i, for 1 ≤ i ≤ r .It can also be observed that G ( kr, k ) can be derived from the extremal strongly-connected 2-geodeticdigraphs of order 2 r (i.e. the orientation of K r,r with a perfect matching pointing in one directionand all other arcs directed in the opposite direction) by extending the perfect matching into pathsof length k −
1. The digraph G (24; 6) is shown in Figure 10.Table 1 displays the results of computational work on the values of ex ∗ ( n ; k ) for some smallvalues of n and k ≥
3. It can be seen that the digraph G ( n, k ) has largest possible size whenever n = kr + s , where s ≤ min { r, k − } . In fact for n and k in the above range such that k | n we can sayfurther that the underlying undirected graph of G ( n, k ) is the unique graph with size n k + ( k − nk that has a strongly-connected k -geodetic orientation. This leads us to make the following conjecture.13 u u u u u u u u u u u u u u u u u u u u u u u u u u u u u v v v Figure 9: G (33 , G (24 , /k Table 1: ex ∗ ( n ; k ) for some small values of n and k Conjecture 22. If n ≥ k + 1 and n ≤ ( k + 1) ⌊ nk ⌋ (in particular for n ≥ k − k ), ex ∗ ( n ; k ) = ⌊ nk ⌋ − ( k + 2) ⌊ nk ⌋ + 2 n. Also for r ≥ the underlying graph of G ( kr, k ) is the unique graph with largest size to have astrongly-connected k -geodetic orientation. Theorem 15 proves this conjecture for k = 2.
6. Generalised Tur´an problems for k -geodetic digraphs Recently the following extension of Tur´an’s problem has received a great deal of attention: givengraphs T and H what is the largest possible number of copies of T in an H -free graph with order n ?Erd˝os considered this problem in 1962 [9] when T and H are complete graphs. The largest numberof 5-cycles in a triangle-free graph was treated in [14, 16] and the converse problem of the largestnumber of triangles in a graph without a given odd cycle C k +1 is discussed in [5, 15]. The problemwas considered in greater generality in [1]. To investigate this problem in digraphs we define thefollowing notation. Definition 23.
For any digraph Z and k ≥ we denote the largest number of copies of Z in a k -geodetic digraph by ex ( n ; Z ; k ) . Observe that if Z is a directed arc then ex ( n ; Z ; k ) = ex ( n ; k ). We will study the asymptoticsof the function ex ( n ; Z ; k ) in the cases that Z is a directed ( k + 1)-cycle or a directed path. Webegin with the function ex ( n ; C k +1 ; k ), where k ≥ C k +1 is a directed ( k + 1)-cycle. Earlierwe made use of the fact that any arc in a 2-geodetic digraph is contained in at most one triangle;a similar principle applies for larger k . Lemma 24.
Every arc in a k -geodetic digraph is contained in at most one directed ( k + 1) -cycle.Proof. Suppose that an arc xy is contained in two distinct ( k + 1)-cycles. Then y has distinct pathsof length k to x , violating k -geodecity. 15e now utilise an inductive approach to give an upper bound on the number of directed ( k + 1)-cycles in a k -geodetic digraph. Lemma 25.
Every k -geodetic digraph with order n contains a vertex with out-degree ≤ n /k .Proof. Assume the contrary. Then for any vertex x the set N + k ( x ) will contain at least n vertices,a contradiction. Theorem 26. ex ( n ; C k +1 ; k ) ≤ Σ ni =1 i /k = kk + 1 n k +1 k + O ( n k ) . Proof.
We claim that ex ( n ; C k +1 ; k ) ≤ Σ ni =1 i /k . This is easily verified for small n , giving a basis for induction. Assume that the result is truefor digraphs with order n − k -geodetic digraph G with order n and ex ( n ; C k +1 ; k )directed ( k +1)-cycles and, subject to this, the smallest possible size m . In particular, every arc of G is contained in a unique C k +1 , for otherwise deleting this arc would yield a k -geodetic digraph withthe same number of ( k + 1)-cycles but smaller size. It follows that we can pair off the in- and out-neighbours of every vertex according to the corresponding ( k + 1)-cycles. Hence d − ( x ) = d + ( x ) forevery vertex x of G and every vertex x is contained in exactly d − ( x ) = d + ( x ) directed ( k + 1)-cycles.By Lemma 25, G contains a vertex x with out-degree ≤ n /k . Deleting this vertex, we obtain a k -geodetic digraph with order n − n − i =1 i /k copies of C k +1 .Deleting x destroyed at most n /k ( k + 1)-cycles, so the result follows by induction.In fact the upper bound is tight up to a multiplicative constant. We can show this using thepermutation digraphs P ( d, k ) that were discussed in Section 2. The permutation digraph P ( d, k )has order n = ( d + k )( d + k − . . . ( d + 1) and size dn . It is easily seen that each arc of P ( d, k ) iscontained in a unique ( k + 1)-cycle; for example 0123 . . . ( k − → . . . ( k − k is contained inthe unique ( k + 1)-cycle0123 . . . ( k − → . . . ( k − k → . . . ( k − k → · · · → k . . . ( k − → . . . ( k − . Hence P ( d, k ) contains ndk +1 copies of C k +1 . Therefore asymptotically ex ( n ; C k +1 ; k ) is at least k +1 n k +1 k . In particular, ex ( n ; C ; 2) must lie somewhere between n / and n / . We show thatthe lower bound is correct. Theorem 27. ex ( n ; C ; 2) = 13 n / + O ( n ) . Proof.
Let G be a 2-geodetic digraph with order n and N := ex ( n ; C ; 2) directed triangles. Asbefore we can assume that every arc is contained in a unique triangle. Thus N = Σ v ∈ G d + ( v ). Forany vertex v we have Σ u ∈ N + ( v ) d + ( u ) = | N +2 ( v ) | ≤ n − − d + ( v ). By H¨older’s inequality N = 13 Σ v ∈ G d + ( v ) ≤ √ n p Σ v ∈ G ( d + ( v )) . In the sum Σ v ∈ G Σ u ∈ N + ( v ) d + ( u ) the term d + ( u ) appears d − ( u ) = d + ( u ) times, so N ≤ √ n q Σ v ∈ G Σ u ∈ N + ( v ) d + ( u ) ≤ √ n p Σ v ∈ G ( n − − d + ( v )) = √ n p n − n − N . N ≤ n n − n − N ) . Rearranging, N + n N − n n − ≤ . Solving the associated quadratic equation it follows that N ≤ ⌊ n √ n − − ⌋ . Remark 28.
For infinitely many n , the upper bound in Theorem 27 is at most n off from the lowerbound of the permutation digraph P ( d, . Based on this example, we make the following conjecture.
Conjecture 29.
For all k ≥ we have ex ( n ; C k +1 ; k ) = 1 k + 1 n k +1 k + O ( n k ) . We turn now to the problem of the largest number of directed paths of given length in a 2-geodetic digraph. Let P l be the path of length l (i.e. order l + 1). Surprisingly there are somedifferences between odd and even length paths; in the following theorem we show different lowerbounds. Theorem 30. If k ≥ and k divides l , then we have ex ( n ; P l ; k ) ≥ n ( l/k )+1 + O ( n l − k ) . In particular, if l is even then ex ( n ; P l ; 2) ≥ n ( l/ + O ( n ( l +1) / ) . If l is odd, we have ex ( n ; P l ; 2) ≥ ( n/ ( l +3) / . Proof.
Let l be even and let P ( d, k ) be a permutation digraph with degree d . P ( d, k ) has order( d + k )( d + k − . . . ( d + 1). From each vertex x there are at least d k ( d − d − . . . ( d − l + k ) = d l + O ( d l − ) distinct l -paths with initial vertex x , so there are d l + k + O ( d l + k − ) distinct l -paths in P ( d, k ). Thus there are n ( l/k )+1 + O ( n l − k ) distinct l -paths in P ( d, k ).Now let l be odd and consider an orientation of the complete bipartite graph K r,r where n = 2 r ,in which a perfect matching is oriented in one direction and all other arcs are oriented in the oppositedirection. We have already seen that this digraph is 2-geodetic. n of the vertices are the initialvertices of ( n ) ( l +1) / + O ( n ( l − / ) distinct l -paths, whereas the vertices in the other partite set arethe initial vertices of only O ( n ( l − / ) l -paths each. Multiplying by n yields the result. Theorem 31. If k ≥ and k divides l , then ex ( n ; P l ; k ) ≥ n ( l/k )+1 + O ( n l − k ) . In particular, for every even l ex ( n ; P l ; 2) = n ( l/ + O ( n l/ ) . roof. Let k ≥ l . We have a lower bound from Theorem 30. For an upper bound, consider apath of length l with vertices 0 , , . . . , l . By k -geodecity, given the two endpoints of a path of length k all of the intermediate vertices are determined. Hence we can only choose vertices 0 , k, k, . . . , l independently. Hence ex ( n ; P l ; k ) is at most n ( l/k )+1 .As for paths of odd length we have an asymptotically sharp result only for P . Theorem 32. ex ( n ; P ; 2) = ( n/ + O ( n ) Proof.
We have a lower bound from Theorem 30. For an upper bound we denote the max-imum out-degree by ∆ := max v ∈ V ( G ) { d + ( v ) } and we assume without loss of generality that∆ ≥ max v ∈ V ( G ) { d − ( v ) } . Let v be a vertex with d + ( v ) = ∆ and denote the out-neighbourhoodof v by N := N + G ( v ). Let us assume that v is a vertex for which N := N + ( v ) − N is largestpossible.Note that by 2-geodecity, for each fixed first vertex and last arc we have at most one path oflength three; similarly we have at most one path of length three for each fixed first arc and lastvertex. We will use make of this several times in the following argument.There are at most 2 n arcs starting from N ∪ N since G is 2-geodetic. Hence, by the observationof the previous paragraph, the number of 3-paths starting from N or N has quadratic order.Similarly there are at most a quadratic number of paths of length three with third vertex lying in N or N . Therefore since the desired upper bound for the number of 3-paths is cubic in order, wemay ignore paths of length three which contain a vertex from N or N as the first or third vertex.Let us denote the number of vertices in N by x . Since there are at most ( n − ∆ − x ) choices forthe first or the third vertex and at most ∆ choices for the last vertex after fixing the third vertex,there are at most ( n − ∆ − x ) ∆ + O ( n ) directed paths of length three. Using elementary calculusit is simple to check that we have ( n − ∆ − x ) ∆ ≤ n x ≥ n or ∆ ≤ −√ n . Hence if ∆ + x ≥ n we have the desired inequality and we are done.If ∆ + x < n and ∆ > −√ n , then we bound the number of paths of length three by a differentfunction. The number of arcs in G is at most 2 n + ∆ + ( n − ∆ − x ) x , where the 2 n term boundsthe number of arcs starting at N ∪ N , the second term bounds the number of arcs entering N and the third term bounds the number of arcs that are not incident from N ∪ N or incidentto N . So by choosing the first vertex and the last arc and neglecting terms of order O ( n ) thenumber of paths of length three is at most f ( x ) := ( n − ∆ − x )(∆ + ( n − ∆ − x ) x ). We have f ′ ( x ) = 4∆ x − n ∆ + 3 x − nx + n which is positive when x = 0 and is negative when x = n − ∆.Therefore the maximum of the function f ( x ) in the interval [0 , n − ∆] is attained at the smallestzero of f ′ ( x ), x = n − −√ n − n ∆+4∆ . Expanding and setting ∆ = zn shows that the number of3-paths minus n / n (cid:18) − − z + 59 z − z + 227 (1 − z + 4 z ) / (cid:19) . This function is negative in the interval [ −√ , ] , completing the proof of the result. Acknowledgements
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