Cyclic triangle factors in regular tournaments

Both Cuckler and Yuster independently conjectured that when $n$ is an odd positive multiple of $3$ every regular tournament on $n$ vertices contains a collection of $n/3$ vertex-disjoint copies of the cyclic triangle. Soon after, Keevash and Sudakov proved that if $G$ is an orientation of a graph on $n$ vertices in which every vertex has both indegree and outdegree at least $(1/2 - o(1))n$, then there exists a collection of vertex-disjoint cyclic triangles that covers all but at most $3$ vertices. In this paper, we resolve the conjecture of Cuckler and Yuster for sufficiently large $n$.


Introduction
Let H and G be graphs or directed graphs. An H-tiling of G is a collection of vertex-disjoint copies of H in G. An H-tiling C covers the set V (C) = C∈C V (C), and is called perfect or an H-factor if it covers V (G).
The celebrated Hajnal-Szemerédi Theorem [6] states that for every positive integer r, if n is a positive multiple of r and G is a graph on n vertices such that δ(G) ≥ (1 − 1/r)n, then G contains a K r -factor. The case when r = 3 is a corollary of an earlier result of Corrádi & Hajnal [3].
In this paper, we consider a similar problem in the context of oriented graphs, which are orientations of simple graphs, i.e., oriented graphs are directed graphs in which there is at most one directed edge between every pair of vertices and no loops. A tournament is an orientation of a complete graph. For an oriented graph G and v ∈ V (G), we denote the out-neighborhood of v and in-neighborhood of v by N + (v) and N − (v), respectively. We let N (v) = N + (v) ∪ N − (v) be the neighborhood of v, and we let d + (v) = |N + (v)| and d − (v) = |N − (v)| be the outdegree and indegree of v, respectively. The minimum semidegree of G is An oriented graph G on n vertices is a regular tournament if d + (v) = d − (v) = n−1 2 for every v ∈ G. A tournament is transitive if it contains no directed cycles, and the unique transitive tournament on r vertices is denoted T T r . Up to isomorphism, there are two different tournaments on three vertices: T T 3 and the three vertex cycle in which the edges are consistently oriented, which we denote by C 3 . We call C 3 and T T 3 the cyclic and transitive triangles, respectively There has been some prior work on minimum degree conditions that force an H-factor in directed graphs. See [17] and [5] for work on directed graphs, and [1] and [18] for oriented graphs. Also, [16] contains many additional interesting embedding problems for oriented graphs. This paper focuses on the following conjecture that Cuckler and Yuster made independently. Conjecture 1.1 (Cuckler 2008 [4], Yuster 2007 [19]). If n is an odd positive multiple of 3, then every regular tournament on n vertices has a cyclic triangle factor.
Keevash & Sudakov then proved the following approximate version of this conjecture. Theorem 1.2 (Keevash & Sudakov 2009 [11]). There exists c > 0 and n 0 such that for every n ≥ n 0 the following holds. If G is an oriented graph on n vertices and δ 0 (G) ≥ (1/2 − c)n, then there exists a cyclic triangle tiling that covers all but at most 3 vertices.
A corollary of our main result resolves Conjecture 1.1 for large tournaments. To see that the resolution of Conjecture 1.1 is a sharp result, consider the following construction from [11]. For a positive integer m, let G be a tournament on 3m vertices in which the edges are oriented so that there exists a partition {V 1 , V 2 , V 3 } of V (G) such that • |V 1 | = m − 1, |V 2 | = m, and |V 3 | = m + 1; • for i ∈ [3], the oriented graph induced by V i has minimum semidegree ⌊(|V i | − 1)/2⌋; and • no edges are directed from V 2 to V 1 , from V 3 to V 2 , and from V 1 to V 3 . We have that, To see why G has no cyclic triangle factor, let C be a cyclic triangle tiling of G and note that, for every C ∈ C, either C has one vertex in each of . Because |V 1 |, |V 2 | and |V 3 | are distinct modulo 3, we have that V (C) = V (G). Motivated by this example we make the following definitions. Definition 1.3 (Divisibility barrier and γ-extremal). Let G be an oriented graph. Call a partition P of V (G) a divisibility barrier if either P is the trivial partition {V (G)} and |V (G)| is not divisible by 3, or if P has exactly three parts, V 1 , V 2 , and V 3 , such that there are no edges directed from V 2 to V 1 , from V 3 to V 2 , and from V 1 to V 3 ; and |V 1 |, |V 2 |, and |V 3 | are not all equivalent modulo 3. For and the number of edges directed from V 2 to V 1 , from V 3 to V 2 , and from V 1 to V 3 are each at most γn 2 . An oriented graph is called γ-extremal if it contains a γ-extremal partition.
The following is the main result of this paper.
Theorem 1.4. There exists c > 0 and n 0 such that for every n ≥ n 0 and for every oriented graph G on n vertices with δ 0 (G) ≥ (1/2 − c)n the following holds. G has a cyclic triangle factor if and only if G does not have a divisibility barrier.
The following corollary resolves Cuckler and Yuster's conjecture for sufficiently large regular tournaments. Corollary 1.5. There exists n 0 such that when n is a multiple of 3 and n ≥ n 0 the following holds. If G is an oriented graph on n vertices and δ 0 (G) ≥ n/2 − 1, then G has a cyclic triangle factor.
Proof. Assuming that Theorem 1.4 holds, we only need to show that G does not contain a divisibility barrier. For a contradiction, assume that {V 1 , V 2 , V 3 } is a divisibility barrier. Since n is divisible by 3 and |V 1 |, |V 2 |, and |V 3 | are not all equivalent modulo 3, we have that |V 1 |, |V 2 | and |V 3 | are distinct modulo 3. Therefore, there exists a labeling {i, j, k} = [3] a contradiction. A similar argument holds when i − 1 ≡ j (mod 3).
We are not sure how large the constant c can be in Theorem 1.4, and we do not compute the value of c that our proof implies. An example of Keevash & Sudakov [11], which we will present below, implies that the constant c cannot be larger than 1/18. This suggests the following problem. Problem 1.6. What is the smallest φ ≥ 0 such that there exists n 0 such that for every n ≥ n 0 every oriented graph G on n vertices with δ 0 (G) ≥ (4/9 + φ)n contains either a divisibility barrier or a cyclic triangle factor?
For m ≥ 1, let G be an oriented graph on n = 9(m + 1) vertices and let P = {V 1 , V 2 , V 3 } be a partition of V (G) such that |V 1 | = 3m + 1 and |V 2 | = |V 3 | = 3m + 4. Suppose that for every pair i, j ∈ [3] such that j ≡ i + 1 (mod 3), there is a directed edge from every vertex in V i to every vertex in V j . Further suppose that, for every i ∈ [3], the vertices of V i can be cyclically ordered so that, for every v ∈ V i , the intersection of the out-neighborhood of v and V i is exactly the (|V i |− 1)/3 vertices that succeed v in this ordering. Note that every cyclic triangle has at least one vertex in V 1 , so |C| ≤ |V 1 | = n/3 − 2 for every cyclic triangle tiling C, and Note that P is not a divisibility barrier, because the three parts of P all have the same size modulo 3. Additionally, G cannot have a divisibility barrier, because for every partition P ′ = P, such that |P ′ | ≤ 3, there exists a part U ∈ P ′ such that there exist x, y ∈ U such that x and y are in different parts, say V i and V j , of P. But then either P ′ is not a divisibility barrier or U contains all of the vertices of the third part, say V k , of P. Then, by similar logic, one can argue that either P ′ is not a divisibility barrier or U = V (G). Since 3 divides |V (G)|, P ′ is not a divisibility barrier.
Note that if the famous Caccetta-Häggkvist Conjecture [2] is false and, for some ψ > 1/3 and all n, there exists a C 3 -free oriented graph on n vertices with minimum semidegree ψn, then a similar example would imply that φ must be strictly greater than 0 in Problem 1.6. We define E + (A, B) = {uv ∈ E(G) : u ∈ A and v ∈ B} and E − (A, B) = E + (B, A). We will often write cyclic and transitive triangles C as abc when V (C) = {a, b, c}. For V 1 , V 2 , V 3 ⊆ V (G), cyc(V 1 , V 2 , V 3 ) and trn(V 1 , V 2 , V 3 ) count, respectively, the number of cyclic and transitive triangles with vertex set {v a , v b , v c } such that {a, b, c} = [3] and v i ∈ V i for i ∈ [3]. We abbreviate cyc(A, A, A) and trn(A, A, A) as cyc(A) and trn(A), respectively. We will often replace {v} with v in this notation.
We define the strong β-out-neighborhood of A to be the set of vertices x ∈ A such that d − (x, A) ≥ |A| − βn and we denote this set by SN + β (A). We define the strong β-in-neighborhood similarly. Throughout the paper, we write 0 < α ≪ β ≪ γ to mean that we can choose the constants α, β, γ from right to left. More precisely, there are increasing functions f and g such that, given γ, whenever we choose β ≤ f (γ) and α ≤ g(β), all calculations needed in our proof are valid. Hierarchies of other lengths are defined in the obvious way. For real numbers x and y, we write x = y ± c to mean that y − c ≤ x ≤ y + c.
In our proof, we use a very small part of the theory and notation developed in [10], [9], [7], [12], and [14]. It is based on the absorbing method of Rödl, Rucińksi and Szemerédi [15]. This theory was developed for hypergraphs, so we define, for every oriented graph G, the hypergraph H(G) to be the 3-uniform hypergraph in which xyz is in an edge if and only if xyz is a cyclic triangle in G. Clearly a cyclic triangle factor in G is equivalent to a perfect matching in H(G).
Let V be a set of order n and let P = {V 1 , . . . , V d } be a partition of V . We say that P is trivial if |P| = 1, and, for η > 0, we call P an η-partition if |V i | ≥ ηn for every i ∈ [d]. Let H be a k-uniform hypergraph with vertex set V . We let For every subset U of V the index vector with respect to P, denoted i P (U ), is the vector defined by Let I P (H) = {i P (e) : e ∈ E(H)} be the set of edge-vectors and, for µ > 0, let We call an additive subgroup of Z d a lattice, and we let L P (H) and L µ P (H) be the lattices generated by I P (H) and I µ P (H), respectively. Clearly if M is a collection of vertex-disjoint edges in H, then i P (V (M )) ∈ L P (H). Therefore, H does not have a perfect matching if i P (V (H)) / ∈ L P (H). For example, suppose that G is an oriented graph with a divisibility barrier.
such that the entries of i P (V (G)) are each different modulo 3, and, for every e ∈ E(H(G)), the entries of i P (e) are each the same modulo 3, so i P (V (G)) / ∈ L P (H(G)). We let u i ∈ Z d be the ith unit vector, i.e., u i is the vector in which the ith component is 1 and 2.1. Overview. Our proof follows the stability method, i.e., we divide the proof into two cases depending on whether G is γ-extremal. The case when G is γ-extremal is handled in the following lemma, the proof of which we defer until Section 2.5.
Lemma 2.1 (Extremal case). Suppose that 0 < 1/n ≪ c, γ ≪ 1, n is divisible by 3, and G is an oriented graph on n vertices. If δ 0 (G) ≥ (1/2 − c)n and G is γ-extremal, then G has a triangle factor if and only if G does not have a divisibility barrier.
The proof in the case when G is not γ-extremal follows the absorbing method, and the following lemma of Lo & Markström [14] serves as our absorbing lemma.  3. Suppose that 0 < 1/n ≪ c ≪ β ≪ 1/ℓ < 1, n is divisible by 3, and that G is an oriented graph on n vertices. If δ 0 (G) ≥ (1/2 − c)n and V (G) is (H(G), β, ℓ)-closed, then G has a triangle factor.
Proof. Introduce constants η and α so that Therefore, by Theorem 1.2, with 2c and G ′ playing the roles of c and G, respectively, there exists a cyclic triangle tiling Note that, with Lemmas 2.1 and 2.3, the following lemma implies Theorem 1.4.
The proof of Lemma 2.4 relies on the following four lemmas.
We prove Lemmas 2.5 and 2.6 in Section 2.2, Lemma 2.7 in Section 2.3, and Lemma 2.8 in Section 2.4.
Proof of Lemma 2.4. Introduce additional constants, β ′ , µ and α so that By Lemma 2.6, with β ′ and ℓ ′ playing the roles of β and ℓ, respectively, there exists , then merge the parts that correspond to the 2-transferral, i.e., consider the new partition . Continue to merge the parts that correspond to 2-transferrals until we have a partition P such that L µ P (H(G)) is 2-transferral-free. By Lemma 2.5, we can assume that P is an (H(G), β, ℓ)-closed 0.1-partition of V (G) for some ℓ ≤ 5 3 · 8 = 1000. If |P| = 1, then V (G) is (H(G), β, ℓ)-closed which contradicts our assumptions, so we can assume that P is non-trivial. By Lemma 2.7, there exists A ∈ P such that cyc(A, A, A) ≤ αn 3 and by Lemma 2.8 we have that G is γ-extremal. 5 2.2. Proofs of Lemmas 2.5 and 2.6. We start this section with a proof of Lemma 2.5.
We will first show that if u, v ∈ V (G) are (H(G), β ′ , ℓ)-reachable, then u and v are also (H(G), β, 4ℓ + 1)-reachable. In particular, this will imply that, because P is (H(G), β ′ , ℓ)-closed, P is also (H(G), β, 4ℓ + 1)-closed. Let T be the set of (12ℓ + 2)-tuples (x 1 , x 2 , . . . , x 12ℓ+2 ) such that Since u and v are (H(G), β ′ , ℓ)-reachable, there are at least β ′ n 3ℓ−1 ways to select the first 3ℓ − 1 entries of a tuple in T . For every such selection, there are exactly (6 cyc(V (G))) 3ℓ+1 ways to select the remaining 9ℓ + 3 entries of a tuple in T . To see that 6 cyc(V (G)) ≥ 2(β ′ ) 2 n 3 , let V q be the part in P of largest cardinality. We have that is not empty, which implies that v is in at least β ′ n 2 cyclic triangles. Therefore, because 6 cyc(V (G)) is the number of ordered triples (x, y, z) such that G[{x, y, z}] is a cyclic triangle in G, we have that 6 cyc(V (G)) ≥ 2(β ′ ) 2 n 3 . Hence, so u and v are (H(G), β, 4ℓ + 1)-reachable. Now we will complete the proof by showing that if u 0 ∈ V i and v 3 ∈ V j , then u 0 and v 3 are (H(G), β, 4ℓ + 1)-reachable. By assumption, there are A, B ∈ P such that cyc(V i , A, B) and cyc(A, B, V j ) are both at least µn 3 . Let T be the set of (12ℓ + 2)-tuples that satisfy the following: • u 1 , u 2 , u 3 is a cyclic triangle with u 1 ∈ A and u 2 ∈ B, and u 3 ∈ V j ; and Since cyc(V i , A, B) and cyc(A, B, V j ) are both at least µn 3 and V i , V j , A, B are all (H(G), β ′ , ℓ)closed, we have that so u 0 and v 3 are (H(G), β, 4ℓ + 1)-reachable.
Since we can assume c < 1/2, This further implies that, for every pair of disjoint subsets A and B of V , Proof. To prove the first part of the lemma, note that To prove the second part of the lemma, note that if u ∈ A and v ∈ N + (u, B), then N + (u, B)).

This implies that
where the last inequality follows from the fact that n − 2δ 0 (G) ≤ 2cn. This observation, with the first part of the lemma gives us that Letting m = e + (A, B) and |A| = a, we have that, by the convexity of f (x) = x 2 , where the last inequality follows because, by Lemma 2.10, m ≤ n 2 /5. Therefore, to show that |N | ≥ (1/8 − 10α)n and complete the proof, it suffices to prove that |T (V )| ≥ (1/64 − 2c)n 3 . Let m = e + (N + (v), N − (v)). By Lemma 2.12, we have that This completes the proof, because Lemma 2.11 implies that m ≥ (1/8 − 2c)n 2 , so Proof of Lemma 2.6. Let α = 1/1000, δ = 2/9 and δ ′ = 1/8 − 1/100. By Lemmas 2.11 and 2.13, we have that δ 1 (H(G)) ≥ (1/8 − 2c)n 2 > δ n−1 2 and that Ñ H(G) (α, 1, v) ≥ (1/8 − 10α)n = δ ′ n for every v ∈ V (G). Therefore, by Lemma 2.9, there exists P = {V 1 , . . . , V d } a (H(G), β, 8)-closed partition of V (G) such that such that d ≤ min{⌊1/δ⌋ , ⌊1/δ ′ ⌋} = 4 and |V i | ≥ (δ ′ − α)n > n/9 for every i ∈ [d]. 8 2.3. Proof of Lemma 2.7. In this section, we prove Lemma 2.7. In an effort to explain the structure of the proof at a high level, we first informally discuss how to derive a contradiction in the following situation. Suppose that G is a regular tournament on n vertices that has an η-partition P of V (G) with three distinct parts A, B, D ∈ P such that • A is the largest part in P, • every cyclic triangle with two vertices in A has one vertex in B, and • every cyclic triangle with two vertices in B has one vertex in D.
These conditions are an idealized version of the conditions we will use to produce a contradiction to prove Lemma 2.7. Let and a similar lower bound holds for the cardinality of C + = N + (x − , C). Let v ∈ C − and note that, because no cyclic triangle contains both v and x + and has its third vertex in A, we have that N + (v) = A. Similarly N − (v) = A for every v ∈ C + . Therefore, C + and C − are disjoint, so, with the fact that |A| ≥ |B|, we have that |C|/2 ≥ |C + |, |C − | ≥ |C|/2 − 1/2 and |B| ≥ |A| − 1.
Since |B| is close to |A| and A was the largest part in P, we can apply similar logic to G[B] to find y + , y − ∈ B such that d + . Therefore, the sets A − and A + almost partition A and then |A − | cannot be much smaller than |A|/2. Because G[A − ] is a transitive tournament, there exists x ∈ A − such that d + (x, A − ) = |A − | − 1. Now consider the out-neighborhood of x. Recall that N − (v) = A for every v ∈ C + , so N + (x) contains (A − − x) ∪ C + . Moreover, since xy + is an edge and there are no triangles with one vertex in A and two vertices in B, the out-neighborhood of x also contains all of B. Note that |A − ∪ B ∪ C + | is roughly n/2 + |B|/2, so, since |B| ≥ ηn, we have contradicted the fact that G is a regular tournament. Lemma 2.14. Suppose that 0 < 1/n ≪ α, c ≪ ξ, β ≪ 1 and that G is an oriented graph on n vertices such that δ 0 (G) ≥ (1/2 − c)n. Lemma 2.15. Suppose that 0 < 1/n ≪ c, α ≪ β < 1, and that G = (V, E) is an n-vertex oriented graph such that δ 0 (G) ≥ (1/2 − c)n. If A ⊆ V and cyc(A) ≤ αn 3 , then, for σ ∈ {+, −}, there exists Proof. Pick ξ so that 0 < 1/n ≪ c, α ≪ ξ ≪ β ≪ 1.
Let X be the set of vertices x in A such that cyc(x, A, A) ≤ α 1/2 n 2 . Since (|A| − |X|) · α 1/2 n 2 ≤ 3 cyc(A) ≤ 3αn 3 9 we have that Pick x + ∈ X so as to maximize d + (x + , X), and let X + = N + (x + , X) and X − = N − (x + , X). Note that, because of the minimum semidegree condition, |X − | ≥ |X| − |X + | − 2cn, and that α 1/2 n 2 ≥ cyc(x + , A, A) ≥ e + (X + , X − ). Therefore, with Lemma 2.14, we have that Therefore, by the selection of x + we have that, Proof. We will prove the lemma by showing that, with ξ/2 playing the role of ξ, there exist subsets C + and C − of C that meet the conditions of the lemma except that C + ∩ C − might not be empty. This will prove the lemma, because then, by the minimum semidegree condition, and this, with the fact that |A| ≥ ηn, implies that |C + ∩ C − | ≤ ξn/2, which means that the sets C + \ C − and C − \ C + are disjoint sets that meet the conditions of the lemma. Furthermore, we will only prove that such a C + exists, because the existence of the desired set C − follows by a similar argument. Let X be the set of vertices x in A such that cyc(x, A, C) ≤ α 1/2 n 2 . Because we have that |X| ≥ |A| − 3α 1/2 n. Since cyc(X, X, X) ≤ cyc(A, A, A) ≤ αn 3 , Lemma 2.15 implies that there exists x ∈ X such that . Because every edge directed from C + to N − (x, A) corresponds to a triangle in cyc(x, A, C), we have that We also have that This completes the proof of the lemma.
Proof of Lemma 2.7. Pick β, α ′ and ξ so that For a contradiction, assume that Because P is an η-partition we have that 1/d ≥ η which implies that Let A = V x be the largest part in P. By (4) and (5) (4) and (5) Note that v 1 = 2u x +u y ∈ I µ P (H(G)) and v 2 = 2u y +u z ∈ I µ P (H(G)).
By Lemma 2.16, with B and F playing the roles of A and C, respectively, there are disjoint subsets F + and F − of F such that, for σ ∈ {−, +}, (11) e −σ (B, F σ ) ≤ ξn 2 .
Note that A ⊆ F and fix σ ∈ {−, +} so that |F −σ ∩ A| ≥ |F σ ∩ A|. By (12) and the selection of σ, we have that Note that Lemma 2.14, with (7), implies that |A \ SN σ β (C σ )| ≤ βn/2, 11 and Lemma 2.14, with (11) and (13), implies that and, with (9), we have that Therefore, with the fact that |B| ≥ ηn, we have From this, we then argue that |A| cannot be much larger than n/3. At a high-level, we use Lemma 2.17 to prove Lemma 2.8 in the following way (the actual proof of Lemma 2.8 appears after the proof of Lemma 2.17). We start with a non-trivial η-partition of G such that L µ P (H(G)) is 2-transferral-free and such that there exists A ∈ P such that cyc(A, A, A) is o(n 3 ). We then use Lemma 2.17 to get a partition {S + , S − } of A such that S + and S − have roughly the same size and such that e + (A, S − ) and e + (S + , A) are both o(n 2 ). We can then finish the proof by showing that A, S + and S − each have roughly the same size (which is implied if |S + | is not much more than n/3) and that e + (S − , S + ) is o(n 2 ). To this end, we use the fact that L µ P (H(G)) is 2-transferral-free to show that cyc(S + , S + , S + ) is o(n 3 ), and then apply Lemma 2.17 again with S + now playing the role of A. |S| ≤ βn.
By assumption, there exists P a non-trivial η-partition of V (G) such that L µ P (H(G)) is 2-transferralfree and A ∈ P such that cyc(A, A, A) ≤ αn 3 . Since P is an η-partition, we have that (18) |A| ≥ ηn.
Applying Lemma 2.17 with S σ , 4β and ω playing the roles of A, α and ξ, respectively, implies that