Extremal Digraph Results for Topological Complete Subgraphs

Europ. J. Combinatorics (1998) 19, 687–694 Article No. ej980231 Extremal Digraph Results for Topological Complete Subgraphs C HRIS J AGGER We determine, to within a constant factor, the maximum size of a digraph that does not contain a topological complete digraph D K p of order p. Let t1 ( p) be defined for positive p by     |D| t1 ( p) = inf c; if e(D) ≥ + c|D| then D ⊂ T D K p , 2 where D denotes a digraph. We show that 2 1 2 16 p < t1 ( p) ≤ 44 p . We also obtain results for containing topological tournaments, and a Turán-type result for containing a topological transitive tournament and a transitive tournament. c 1998 Academic Press 1. I NTRODUCTION A topological graph (or subdivision) T G of a graph G consisting of vertices V (G) = {v1 , . . . , vr } is a graph with vertices u 1 , . . . , u r (the branch vertices), and independent paths (that is, vertex disjoint, except for the endvertices), Pi j joining u i to u j whenever vi v j ∈ E(G). A graph H with vertex set {v1 , . . . , vr } is a subcontraction (or minor) of graph G if there exist disjoint non-empty subsets V1 , . . . , Vr of V (G) with Vi being connected for each 1 ≤ i ≤ r , such that whenever vi v j ∈ E(H ), there exist vertices x ∈ Vi and y ∈ V j with x y ∈ E(G). In this case we write H ≺ G. (Throughout we shall use notation of Bollobás [2], thus E(G) is the edge set of the graph G, and later we shall use ŴG (x) for the neighbourhood of x, and δ(G) as the minimum degree of G.) There has been much interest in extremal results about the topological containment of complete subgraphs, and also about subcontracting complete subgraphs. In this paper we shall be looking at corresponding results for containing topological complete digraphs. For results about subcontractions of digraphs, see [7] and [8]. In the case of undirected graphs, extremal results for topological containment have generally been harder than those for subcontraction. Mader [13], and also Erdös and Hajnal [5], conjectured that there is a positive constant c such that any graph G of size at least cp 2 |G| contains a topological complete subgraph of order p. Jung [9] pointed out that complete bipartite graphs provide examples showing that c must be bigger than 1/16, and Ajtai, Komlós and Szemerédi [1] noticed that almost every graph is an example showing that c > 1/8. Furthermore, Luczak noticed that a random bipartite graph with edge probability 3/4 shows that c > 9/64. However, it is not even easy to show that a function t ( p) exists such that G ⊃ T K p provided e(G) ≥ t ( p)|G|. Mader [13] first found such a function t ( p), and he later showed [14] that t ( p) = 3.2 p−3 − p will do. Very recently there has been an enormous amount of progress in this area. In 1994 Komlós and Szemerédi [10] showed that for each η > 14, there exists a constant cη such that t ( p) = cη p 2 (log p)η is such a function. Independently, Alon and Seymour observed that a result of Robertson and Seymour [17] can be used to show that t ( p) = cp2 (log p)1/2 will do for some constant c. Finally, Bollobás and Thomason [3] improved this to t ( p) = 256 p 2 , which they improved again [4] to t ( p) = 22 p 2 , and now Komlós and Szemerédi [11] have improved this to t ( p) = p 2 /2. 0195-6698/98/060687 + 08 $30.00/0 c 1998 Academic Press 688 C. Jagger In the directed case there are several analogous functions it is interesting to consider. We start by defining a topological digraph. For a digraph D, with vertex set {v1 , . . . , v p } and edge set E(D), a topological digraph T D is a digraph consisting of at least p vertices, including {x1 , . . . , x p }, such that there exists an independent directed path Pi j from xi to x j whenever vi v j ∈ E(D). (Thus Pi j and Pkl are vertex disjoint, with the possible exception of the endvertices.) Let D K p denote the complete digraph of order p (that is, the digraph with edges both ways between each pair of vertices). Note that, unlike the undirected case, the number of edges required to contain a T D K p is not linear in the order of the graph, as, for example, if we take D to be the complete transitive tournament (that is, a tournament T such that for all vertices u,v,w, edges uv, vw ∈ E(T ) imply uw ∈ E(T )) then D certainly does not contain T D K p for any p ≥ 2. Thus we define   |D| + c|D| then D ⊃ T D K p }. t1 ( p) = inf{c; if e(D) ≥ 2 1 2 p < t1 ( p) ≤ 44 p 2 . In Section 3 we shall move on to In Section 2 we shall show that 16 consider topological containment of tournaments, including the special case of containing a topological transitive tournament (which is rather different to that of all other tournaments), and uses a Turán type of proof. We also note that the same method will also obtain good bounds for containing a transitive tournament. Possibly the most interesting result is Theorem 6. 2. T OPOLOGICAL C ONTAINMENT OF C OMPLETE D IGRAPHS 1 2 p < t1 ( p) ≤ 44 p 2 (where t1 ( p) is as The aim of this section is to prove Theorem 2, that 16 defined in the introduction). To obtain the lower bound we simply need to construct a digraph
p2 n edges. To obtain the upper bound we need to introduce the D 6 ⊃ T D K p that has n2 + 16 idea of being k-linked. D EFINITION . A graph G is k-linked if, given any disjoint sets of vertices {s1 , . . . , sk } and {t1 , . . . , tk }, there are disjoint paths P1 , . . . , Pk such that Pi joins si to ti , for all 1 ≤ i ≤ k. Notice that if G is k-linked, then G − X is (k−⌈|X |/2⌉)-linked. For, if we let l = k−⌈|X |/2⌉, then, given {s1 , . . . , sl } and {t1 , . . . , tl } in G − X , extend to {s1 , . . . , sk } and {t1 , . . . , tk } in G, in such a way that X ⊂ {sl+1 , . . . , sk } ∪ {tl+1 , . . . , tk }, then k-linking of G gives the appropriate links in G − X . Also notice that δ(G) ≥ 2k − 1, for if there is a vertex v with d(v) < 2k − 1, let s1 be v, and pick {s2 , . . . , sk , t1 , . . . , tk } so that all neighbours of v are elements of {s2 , . . . , sk , t2 , . . . , tk }; t1 cannot be a neighbour of v, therefore there is not a path from s1 to t1 which is disjoint from {s2 , . . . , sk , t2 , . . . , tk }. Menger’s theorem states that if G has connectivity κ(G) then any two vertices have κ(G) vertex disjoint paths between them. A similar result for k-linking, say that G has connectivity ck implies that G is k-linked, would clearly be useful, although it is easy to find graphs with connectivity 2k − 1 which are not k-linked. Until recently no such result was known, although Larman and Mani [12] had noticed that if a graph has connectivity 2k and contains a T K 3k then it is k-linked. In 1994 Bollobás and Thomason [4] proved that if κ(G) ≥ 22k, then G is k-linked. The following lemma relies on techniques used in that paper. We let T 2 K p be a graph consisting of p vertices with two independent paths joining each pair of vertices. L EMMA 1. If G is p 2 -linked then G ⊃ T 2 K p . Topological complete subgraphs 689 P ROOF. Let X = {x1 , . . . , x p }. As G is p 2 -linked we know from our earlier observations that δ(G) ≥ 2 p 2 − 1 and thus |ŴG−X (xi )| ≥ (2 p − 1) p, for 1 ≤ i ≤ p. Therefore we can pick disjoint subsets Y1 , . . . , Y p of G − X each of order 2 p − 2, and with Yi ⊂ Ŵ(xi ), for 1 ≤ i ≤ p. Let Y = Y1 ∪ . . . ∪ Y p , so |Y | = 2 p( p − 1). Now G − X is certainly p( p − 1)-linked, so we can find vertex disjoint paths which link up vertices from Y . That is, if we partition Y into two equal sets Y ′ and Y ′′ then we can find p( p − 1) vertex disjoint paths linking the two sets in any way we desire. Thus we can use the set X together with the appropriate paths to form a T 2 K p as required. (For example, Y1 is in Ŵ(x1 ) and has order 2 p − 2, so two vertices can be used to form disjoint paths to two vertices ✷ of Y2 , and hence to x2 .) T HEOREM 2. 1 16 p 2 < t1 ( p) ≤ 44 p 2 . P ROOF. To obtain the upper bound we need to use a result of Mader [15] which states that if e(G) ≥
2k|G| then G contains a k-connected subgraph. For any digraph D of order n with e(D) ≥ n2 + 44 p 2 n the corresponding simple graph G that has an undirected edge x y only if E(D) contains both x y and yx, has at least 44 p 2 n edges. Hence, by Mader’s result there is a 22 p 2 -connected subgraph H of G. Thus, by Bollobás and Thomason’s result H is p2 -linked, and by Lemma 1, H ⊃ T 2 K p . But then certainly D ⊃ T D K p . Now to obtain the lower bound take a graph of order n, where n is an even multiple of p 2 /8 (which we shall assume for simplicity is an integer). Fix an ordering on the vertices and form a transitive tournament from this ordering by putting in edges from each vertex to all larger ones in the ordering. Let the first p 2 /8 vertices be the set V1 , the next p 2 /8 vertices be V2 , and so on, up to V2r , where 2r = 8n/ p 2 . For all 1 ≤ i ≤ r add in edges from V2i to V2i−1 . It is easy to see that D[V2i ∪ V2i−1 ] 6 ⊃ T D K p , for 1 ≤ i ≤ r . (For example, suppose l of
the p branch vertices of the T D K p are in V2i , and p − l in V2i−1 , then there must be 2l vertices in V2i−1 to form paths between the l branch vertices in V2i , in addition to the p − l branch vertices there. When |V2i | = p 2 /8 there are not enough vertices to form a T D K p .) Furthermore, since T D K p is strongly connected, any T D K p in D cannot consist of vertices from V2i ∪ V2i−1 and V2 j ∪ V2 j−1 , (i 6= j), hence D 6⊃ T D K p . It just remains to note that D has    2 2   n n n p2 p + 2 = + n 2 2 16 p /4 8 edges. Hence t1 ( p) > p 2 /16. 3. ✷ D IGRAPHS C ONTAINING T OPOLOGICAL T OURNAMENTS In this section we look at how many edges are required to ensure that a digraph contains a subdivision of every tournament T p of order p. Similarly to our earlier definition of t1 ( p), we define     |D| t2 ( p) = inf c; if e(D) ≥ + c|D| then D ⊃ T T p for every tournament T p . 2 Clearly t1 ( p) ≥ t2 ( p), and in fact it turns out that t2 ( p) is also of order p 2 . The following theorem obtains tight bounds for t2 ( p) (up to a constant factor). Note that the lower bound corresponds to digraphs which not only do not contain every tournament T p , but in fact do not contain any strongly connected tournaments at all. 690 C. Jagger T HEOREM 3. 1 32 p 2 < t2 ( p) ≤ t ( p) ≤ p2 /2.
P ROOF. For the upper bound we notice that if D is a digraph with order n and n2 + t ( p)n edges then let G be the corresponding graph consisting of an undirected edge x y whenever both x y ∈ E(D) and yx ∈ E(D). G has at least t ( p)n edges, thus G ⊃ T K p , and so D ⊃ T T p , for any tournament T p . For the lower bound we form a digraph in the following way: assume for simplicity that n is a multiple of p 2 /8, and that p 2 /32 is an integer (although trivial modifications ensure that the same method works generally). Now order the vertices and form a transitive tournament from this ordering as in the previous theorem. Let V1 be the set of the first p 2 /32 vertices in the ordering, V2 be the next p 2 /32 vertices, and so on up to V4r , where 4r = 32n/ p2 . Remove all edges x y such that x, y ∈ Vi , 1 ≤ i ≤ 4r . Add in all edges x y such that x ∈ V4i−1 ∪ V4i and y ∈ V4i−3 ∪ V4i−2 , 1 ≤ i ≤ r , and also all those with x ∈ V4i , y ∈ V4i−1 , or x ∈ V4i−2 , y ∈ V4i−3 , 1 ≤ i ≤ r , to form a digraph D. The number of edges in D is      2  2 2  2 2    2 n n n 1 n p /32 p p n p + n. + + = + − 2 16 32 p 2 /16 32 2 2 2 p 2 /32 p 2 /8 Now let T p be a strongly connected tournament of order p. If D ⊃ T T p then the T T p must lie in D[V4i−3 ∪ . . . ∪ V4i ] for some 1 ≤ i ≤ r . Now we use a similar argument to that used in the
previous theorem. Assume there are x j branch vertices in V4i− j , 0 ≤ j ≤ 3, then x3
x2
x1 x0
+ x1 + x2 + x3 ≤ p 2 /8, with x0 + x1 + x2 + x3 = p. This is + x + + + 0 2 2 2 2 not possible, hence we have the result. ✷ This theorem indicates a way to obtain bounds for specific tournaments as long as they have a largest strongly connected component of order k ≥ 2 (or equivalently,
as long as the longest directed cycle has length k ≥ 2). Thus, if a digraph of order n has n2 + t ( p)n edges then it will certainly contain a topological tournament, T p . But furthermore, there are digraphs not containing a topologicaltournament  with largest strongly connected subgraph of order k ≥ 2,
k 2 k−2 , 2 n edges, as given by the following theorem. with at least n2 + max 32 T HEOREM 4. If T p is a tournament with largest strongly connected subgraph having order k > 1, then there exist graphs G of order n not containing a T T p with     2 n k k−2 , n + max 32 2 2 edges. P ROOF. Assume for simplicity that n is a multiple of k − 1. Fix an ordering on the vertices and form a transitive tournament. Then we take the first k − 1 vertices and add in all remaining edges between these vertices. We then take the next k −1 vertices and add in all their edges, and so on. The strongly connected component of order k in the tournament cannot have vertices from different (k − 1)-blocks, but equally clearly
cannot have all k vertices from one block of order k − 1. This produces a digraph with n2 + (k − 2)n/2 edges. To obtain the other bound we assume for simplicity that k is a multiple of eight (so that k 2 /32 is an integer), then take a digraph D of order n a multiple of k 2 /32 and form D using the just same construction as in the previous theorem, letting Vi have k 2 /32 vertices, for each n
k2 2 1 ≤ i ≤ 32n/k . Now D has 2 + ( 32 + 12 )n edges. Let Tk′ ⊂ T p be a strongly connected Topological complete subgraphs 691 component of order k ≥ 2. Then, as before, if D ⊃ T T p then D ⊃ T Tk′ , so T Tk′ must lie D[V4i−3 ∪ . . . ∪ V4i ] for some 1 ≤ i ≤ r . As before this is impossible and hence we have the result. ✷ So far we have only considered how many edges are required to ensure the topological containment of a non-transitive tournament. Now we look at the case when we wish to ensure the containment of a topological transitive tournament, denoted T T pt (that is, the case k = 1). In this case there are no strongly connected components, thus the digraph does not necessarily need to have any strongly connected components in order to contain a T T pt . This problem in fact seems to be a bit like some extremal problems of Turán-type. If D is a digraph of order n, we define t3 ( p, n) = min{e : if |D| = n and e(D) ≥ e then D ⊃ T T pt }.

It is clear that t3 ( p, n) ≤ t2 ( p)n + n2 , and thus t3 ( p, n) ≤ n2 + p 2 n/2. However, it is not immediately clear how we can do significantly better than this. At this point, in order to be consistent with standard notation, we shall slightly abuse notation and use T p (n) to denote the p-partite Turán graph of order n, and T p to denote a tournament of order p. A lower bound for t3 ( p, n) is obtained by letting D be the Turán graph T p−1 (n) with all the edges oriented transitively, which we shall call DT p−1 (n). (Thus DT p−1 (n) consists of p − 1 independent sets D1 , . . . , D p−1 of order ⌊n/( p − 1)⌋ or ⌈n/( p − 1)⌉, such that for all 1 ≤ i < j ≤ p − 1, if x ∈ V (Di ) and y ∈ V (D j ), then the edge x y is in E(D).) The next theorem shows that the bound given by DT p−1 (n) is essentially the right one, but first we require the following result, due to Erdös and Stone [6]. L EMMA 5. Given any integers p and m, and ǫ > 0, then there exists n 0 such that for any n ≥ n 0 , any graph G of order n, with e(G) ≥ e(T p−1 (n)) + ǫn 2 contains a complete p-partite graph with vertex classes of size at least m.
1 T HEOREM 6. t3 ( p, n) ≤ n2 (1 − p−1 + o(1)), the o(1) denoting a term which tends to zero as n → ∞. P ROOF. Consider a digraph D of order n that does not contain T T pt . If D has p 2 n/2 double edges then the corresponding simple graph G, consisting of an edge wherever there was a double edge in D, contains T K p , and so certainly D ⊃ T T pt . Hence we may assume D has fewer than p 2 n/2 double edges. Delete one edge in each double edge to obtain a subgraph D1 of D, so that D1 is in fact an oriented graph. By the previous lemma, given ǫ > 0 and t, there exists n 0 such that if n ≥ n 0 , and e(D1 ) ≥ e(T p−1 (n)) + ǫn 2 , then D1 contains a complete oriented p-partite graph with vertex classes of size t. By a Ramsey-like argument, provided t is large enough there is a p-partite subgraph D2 with p( p − 1) vertices in each class such that between any two classes any two edges go the same way. (For example, by the usual Zarankiewicz bound, between any two classes there must be a (2-partite) subgraph with all edges from one side to the other with order approximately log2 t. Now, noting that there is a p-colouring of the edges of K p , we can pair off classes and find these 2-partite subgraphs for each pair of order log2 t. Next we pair off in a new way and find new 2-partite subgraphs of order log2 log2 t. We carry on in this way and provided that p iterations of the log function on t is at least p( p − 1), we have D2 as claimed.) If D2 is transitive then it is clear that D2 ⊃ T T pt . If not, it contains three classes which form a triangle, that is, V1 , V2 , V3 such that if x ∈ Vi and y ∈ V j , (i 6= j), then x y ∈ E(D2 ) if i ≡ j − 1 (mod 3). To form T T pt we take p vertices from V1 . Now there are p( p − 1) connections to be made. For any pair a, b, take x ∈ V2 and y ∈ V3 , then ax yb is the required path. Hence, provided n ≥ n 0 , then D ⊃ T T pt , and we have the desired contradiction. ✷ 692 C. Jagger It is clear that when n is small there are digraphs with more edges than DT p−1 (n) which do not contain a T T pt . For n < 2 p all sorts of digraphs have more edges without containing a T T pt , and for 2 p ≤ n ≤ p 2 /2 we can do better by taking a complete bipartite digraph. However, for n large enough (maybe n ≥ p2 ), it could be that t3 ( p, n) = e(T p−1 (n)) + 1, and that DT p−1 (n) is the unique extremal graph. The following Turán-type proof might be helpful to prove this. This would be particularly interesting in that unlike the other analogous results this one would have an easily obtainable unique extremal example. P ROPOSITION 7. If there is an n 0 ≥ 3( p − 1) such that DT p−1 (n 0 ) is the unique extremal digraph which does not contain T T pt , then DT p−1 (n) is the unique extremal digraph for all n
n ≥ n 0 , and t3 ( p, n) = 2 (1 − 1/( p − 1)). P ROOF. We proceed by induction on n. Given a particular value of n we assume that the theorem is true for all smaller values. For any digraph D 6⊃ T T pt of order n and with e(D) ≥ e(T p−1 (n)), remove edges until e(D) = e(T p−1 (n)). Thus δ(D) ≤ δ(T p−1 (n)). Pick a vertex x ∈ V (D) of minimum degree, then e(D − x) ≥ e(T p−1 (n)) − δ(T p−1 (n)) = e(T p−1 (n − 1)). By the inductive hypothesis, because D − x 6⊃ T T pt , we have D − x = DT p−1 (n − 1), and δ(D) = δ(T p−1 (n)). Now try adding x back into D − x in such a way that D 6⊃ T T pt , and x has degree δ(T p−1 (n)). Let D − x consist of the classes D1 , . . . , D p−1 described above. As n ≥ n 0 + 1 ≥ 3( p − 1) + 1 we may assume that each class has at least three vertices. There are three cases: (1) x has an edge connecting it to every class D1 , . . . , D p−1 . In this case if there are vertices y ∈ Di and z ∈ Di+1 for some i, such that x y ∈ E(D) and zx ∈ E(D), then pick vertices a ∈ Di \{y} and b ∈ Di+1 \{z}, together with a vertex from each of the other classes and also y. This forms a T T pt , since we have the path azx y. If there are not the vertices y and z as described then there must be a value of k, 0 ≤ k ≤ p − 1 such that if a vertex w ∈ Di is connected to x, then it is connected from w to x if and only if i ≤ k. In this case, x together with a vertex from every class (connected to x) will form a T pt , and hence D certainly contains T T pt . (2) x has a double edge to a vertex y ∈ Di . If x also has an edge connecting it to a vertex z ∈ Di , then either y has a path to z (via x) or z has a path to y. Hence y and z together with a vertex from each of the other classes forms a T T pt . Thus if there is a double edge there can be no other edges between x and vertices of Di . Thus there must be a vertex in every other class which is connected to x, but this case has already been done. (3) x has no edges to a particular set Dk , and edges connecting it (singly) to every other vertex. Suppose there is an i such that y, z ∈ Di and x y, zx ∈ E(D). Then y and z together with a vertex from all the other classes forms a T T pt . Hence for all y, z ∈ Di , for all 1 ≤ i ≤ p − 1, we have either x y, x z ∈ E(D) or yx, zx ∈ E(D). Now suppose there is an i < j, and y ∈ Di , z ∈ D j , with x y, zx ∈ E(D). Then we can find a path P from z to a vertex w ∈ D j , and so w, z together with a vertex not in P from each class forms a T T pt . Thus there exists 0 ≤ i ≤ p − 1 such that yx ∈ E(D) for all y ∈ D1 ∪ . . . ∪ Di \Dk , and x y ∈ E(D) for all y ∈ Di+1 ∪ . . . ∪ D p−1 \Dk . If i ≤ k − 2, then there is a path from x to a vertex w ∈ Dk , so we can take x, w together with appropriate vertices from the other classes to form a T T pt . The case i ≥ k + 1 is similar. Thus, as we are assuming there is no T T pt , we have that i = k ✷ or k − 1, but these both correspond to D = DT p−1 (n), as required. Whilst considering topological transitive tournaments, it is interesting to note what happens if we want enough edges to ensure containing a transitive tournament, T pt . A similar argument 693 Topological complete subgraphs n
1 + o(1)) to that used in Theorem 6 shows that any oriented graph of order n and 2 (1 − f ( p)−1 t edges contains a T p , where f ( p) is the smallest value such that any tournament of order f ( p) has a transitive subtournament of order p. Note that 2 p/2 ≤ f ( p) ≤ 2 p−1 (a simple induction gives the upper bound, a random argument gives the lower bound). For a digraph the most edges with no transitive tournament is 2e(DT p−1 (n)), using the usual Turán proof as in the previous proposition. The problem with considering the number of edges required to ensure that a digraph contains a T D K p , or indeed a non-transitive tournament, is that there are digraphs with a large number of edges that do not have any non-trivial (strongly) connected components. In particular, the transitive tournament is one, and so we require a large number of edges just to ensure that there are suitably large strongly connected components. Therefore it is worth considering whether there are any better results if we consider the minimum outdegree, or connectedness, instead of the number of edges. For example, is there a function g( p) such that a digraph being strongly g( p)-connected must contain a T D K p ? Unfortunately this is not known even for p = 3. Note that in the case of undirected graphs, the problem of determining bounds for the containment of T K p is similar no matter whether one is interested in obtaining bounds in terms of the number of edges, the minimum degree or the connectivity. However, in the case of digraphs these problems are rather different. Very few results are known about graphs with large minimum outdegree. Mader [16] showed that every digraph with minimum outdegree three contains a subdivision of the transitive tournament on four vertices (T T4t ), but it is not known whether large minimum outdegree implies the existence of a T T pt for higher values of p. A related problem is that of the existence of even directed cycles, or dicycles (note that a T D K 3 must contain an even dicycle). Thomassen [19] showed that a large minimum outdegree does not imply the existence of an even dicycle, and hence certainly does not imply the existence of a T D K 3 . He also showed [20] that a strongly 3-connected digraph contains an even dicycle. Earlier we used the result that κ(G) ≥ 22k implies that G is k-linked. However, Thomassen [18] showed that for all k, there are digraphs which are strongly k-connected but which are not 2-dilinked. (By 2-dilinking we mean that given any vertices a, b, c, d, there are directed paths from a to b and from c to d.) His example also shows that for any k and l, there are graphs that are strongly k-connected and contain a T D K l but which are not 2-dilinked, thus there is no analogous result to the one of Larman and Mani that we mentioned earlier. Thus it seems that dilinking will not be very useful for this question. 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