2·1. General oriented graphs

In this section, we show how to deduce Proposition 1·2 and Proposition 1·3 from known results.

Let H be an arbitrary oriented graph. As before, let us write F for the underlying undirected graph of H. For a graph G, write D(G, H) for the number of H-free orientations of G and denote by N(G, F) the number of F-free spanning subgraphs of G. Moreover, write N(n, F) for the number of F-free graphs with vertex set [n]. Kozma and Moran [ Reference Kozma and Moran19 ] proved that N(G, F) is an upper bound for D(G, H).

Theorem 2·1 (Kozma--Moran [ Reference Kozma and Moran19 ]). Let F be an undirected graph and let H be an orientation of F. Then for any undirected graph G,

\begin{equation*}D(G,H)\leq N(G,F).\end{equation*}

In particular, for any $n\in \mathbb{N}$ ,

\begin{equation*}D(n,H)\leq N(n,F).\end{equation*}

Perhaps surprisingly, the proof uses an inequality about set systems. Given a set system $\mathcal{A}$ on ground set X, we say that $S\subset X$ is shattered by $\mathcal{A}$ if for every $T\subset S$ , there exists some $A\in \mathcal{A}$ with $A\cap S=T$ . Let us write $\mathrm{str}(\mathcal{A})$ for the collection of subsets of X which are shattered by $\mathcal{A}$ . A general version of the celebrated Sauer–Shelah lemma [ Reference Pajor22 ] states that $|\mathcal{A}|\leq |\mathrm{str}(\mathcal{A})|$ .

To see that this inequality implies Theorem 2·1, fix an orientation Q of G. Identify the power set of E(G) with the set of orientations of G by identifying $A\subset E(G)$ with the orientation of G which differs from Q precisely on the edge set A. Let $\mathcal{A}$ be the collection of subsets of E(G) which are identified with H-free orientations of G. Clearly, $|\mathcal{A}|=D(G,H)$ . On the other hand, assume that some $S\subset E(G)$ is shattered by $\mathcal{A}$ . Then the graph formed by the edges in S is F-free. Indeed, if it did contain F as a subgraph, then there would exist an orientation of the edges in S which contains a copy of H. This means that S could not be shattered by $\mathcal{A}$ . Thus, $|\mathrm{str}(\mathcal{A})|\leq N(G,F)$ and Theorem 2·1 follows.

The function N(n, F) has been extensively studied. Similarly to D(n, H), we have the trivial lower bound $N(n,F)\geq 2^{\mathrm{ex}(n,\,F)}$ since if G is an F-free graph, then any subgraph of G is also F-free. This has been shown to be almost tight first for complete graphs by Erdős, Kleitman and Rothschild [ Reference Erdős, Kleitman and Rothschild14 ] and then for general non-bipartite graphs by Erdős, Frankl and Rödl [ Reference Erdős, Frankl and Rödl13 ], who proved that $N(n,F)=2^{\mathrm{ex}(n,\,F)+o(n^2)}$ (see [ Reference Balogh, Bollobás and Simonovits6 ] for an improved error term). Combined with Theorem 2·1 and the trivial lower bound for D(n, H), this implies Proposition 1·2.

However, this gives an unsatisfactory answer for bipartite graphs F as in that case $\mathrm{ex}(n,\,F)=O(n^{2-\varepsilon_F})$ for some $\varepsilon_F>0$ . Since any F-free graph on vertex set [n] has at most $\mathrm{ex}(n,\,F)$ edges, we get a straightforward upper bound

\begin{equation*}N(n,F)\leq \sum_{k=0}^{\mathrm{ex}(n,\,F)} \binom{\binom{n}{2}}{k},\end{equation*}

which implies that $N(n,F)\leq 2^{C \mathrm{ex}(n,\,F)\log n}$ for some constant C. The logarithmic factor is necessary when F is acyclic with maximum degree at least 2. Indeed, in this case $\mathrm{ex}(n,\,F)=O(n)$ , but it is easy to see that there are $2^{\Omega(n\log n)}$ graphs on vertex set [n] with maximum degree 1. On the other hand, it is not known in general whether the logarithmic factor is necessary for graphs that contain a cycle. In this direction, settling a classical conjecture of Erdős it was shown by Morris and Saxton [ Reference Morris and Saxton21 ] that $N(n,C_{2k})=2^{O_k(n^{1+1/k})}$ , generalising a previous result of Kleitman and Winston [ Reference Kleitman and Winston17 ], and complementing the classical Bondy–Simonovits bound $\mathrm{ex}(n,C_{2k})=O_k(n^{1+1/k})$ [ Reference Bondy and Simonovits9 ]. Balogh and Samotij [ Reference Balogh and Samotij7, Reference Balogh and Samotij8 ] established a similar result for complete bipartite graphs in place of even cycles. These results were generalised greatly by Ferber, McKinley and Samotij [ Reference Ferber, McKinley and Samotij15 ]. They showed that if F is a graph containing a cycle and there are positive constants $\alpha$ and A such that $\mathrm{ex}(n,\,F)\leq An^{\alpha}$ , then there exists a constant C depending only on $\alpha$ , A and F such that for all n, $N(n,F)\leq 2^{Cn^{\alpha}}$ . This result, combined with Theorem 2·1 and $D(n,H)\geq 2^{\mathrm{ex}(n,\,F)}$ , implies Proposition 1·3.

2·2. Directed path

In this subsection we will prove Theorem 1·4. Let G be an n-vertex graph. Our task is to show that there are at most $2^{3kn}$ orientations of G which do not contain $P_k$ . Let us fix a canonical ordering of the vertices of G. We will count the number of orientations with the help of the following algorithm. It takes as input an orientation of G. In each step, it processes a vertex and updates the current “state” for every vertex that has not been processed yet. We will show that, provided the orientation is $P_k$ -free, the potential states are severely restricted. We then use this to bound the number of possible orientations.

Algorithm: Initially, we assign to every vertex v a state $(a_v,b_v)=(0,0)$ . At step i we have a sequence of already processed vertices $v_1,\ldots, v_{i-1}$ and possibly the next vertex to be processed $v_i$ .

1. (i) If $v_i$ is not specified, it is chosen as a vertex in $V(G) \setminus \{v_1,\ldots, v_{i-1}\}$ with largest $a_{v_i}$ , breaking ties by choosing such a $v_i$ first in the canonical ordering.

2. (ii) We now proceed to process $v_i$ . We consider all edges between $v_i$ and not already processed vertices.

3. (i) If all these edges are oriented towards $v_i$ , we do not specify $v_{i+1}$ and continue to the next step.

4. (ii) Otherwise we choose $v_{i+1}$ to be the out-neighbour v of $v_i$ with largest value of $b_{v}$ (breaking ties by choosing such a v first in the canonical ordering) and

   1. (a) we increase $a_u$ by one for any (non-processed) out-neighbour u of $v_i$ and

   2. (b) we increase $b_u$ by one for any (non-processed) in-neighbour u of $v_i$ which had $b_u\le b_{v_{i+1}}$ .

Let us first look at what is going on here. The algorithm reveals the orientation of the edges of G bit by bit; specifically at step i it will reveal the orientation of all (not already revealed) edges incident to $v_i$ , the vertex we are currently processing. In particular, by the end of step i the algorithm has revealed the orientation of all edges incident to $v_1,\ldots, v_i$ . Note that the orientation of these edges determines uniquely the first i steps of the algorithm, regardless of how the remaining edges are oriented (in other words, the algorithm is by this point completely independent of the orientation of the remaining edges). Roughly speaking, we think of the states $a_v$ and $b_v$ as the length of a directed path ending and starting at v, respectively, only using the already revealed edges, or in other words using only already processed vertices (apart from v itself). This is captured more precisely in the following lemma.

Proof. Let $v_1,\ldots, v_n$ be the order in which the vertices are processed. Observe that both $a_v$ and $b_v$ are non-decreasing throughout the process and are never updated once we process v. Let $a_t,b_t$ denote the final value of $a_{v_{t}},b_{v_t}$ so in particular at the point when we process $v_t$ .

We are first going to show by induction on t that there is a path of length at least $a_t$ ending in $v_t$ which only uses previously processed vertices. For the base case $t=1$ , we know that $a_1=0$ so the claim trivially holds. Now given $v_t$ , let $i<t$ be the last step at which point $a_{v_t}$ was updated (if $a_{v_t}$ was never updated, then $a_t=0$ and our claim is trivial). Take the smallest j such that $v_{j},v_{j+1},\ldots,v_i$ form a directed path oriented towards $v_i$ . In particular, this means that $v_{j}$ was not an out-neighbour of $v_{j-1}$ and it was chosen as a vertex with maximum $a_{v}$ among all yet unprocessed vertices. As $v_t$ was not yet processed at this stage, it follows that we had $a_{v_t}\leq a_{v_j}$ at step j. Since after this point $a_{v_t}$ could only have been incremented when processing $v_j,\ldots, v_i$ , we know that $a_t \le a_j+i-j+1$ . In addition, we know by induction that there is a path of length $a_j$ ending at $v_j$ and using vertices only from $\{v_1,\dots,v_j\}$ to which we can append $v_{j}\ldots v_iv_t$ to obtain the desired path of length $a_j+i-j+1\ge a_t$ .

Turning now to the $b_v$ ’s, we are going to show by induction on t that there is a path of length at least $b_t-1$ starting with $v_t$ , which otherwise only uses vertices in $\{v_1,\ldots, v_{t-2}\}$ . The base case $t=1$ is trivial. Also, observe that if $b_{v_t}$ got incremented at most once, i.e. $b_t\leq 1$ , the claim also holds trivially. Let us now consider the penultimate (second to last) step i at which point $b_{v_t}$ was updated. Observe that by definition, our algorithm will only increment $b_{v_t}$ when processing $v_j$ if $v_tv_j$ is an edge and $v_jv_{j+1}$ is an edge. In particular, this implies that we did not update $b_{v_t}$ at step $t-1$ , and hence $i \le t-3$ , since we chose i to be the penultimate step which updated $b_{v_t}$ . The fact that we incremented $b_{v_t}$ when processing $v_i$ means that at that time $b_{v_t}$ was at most $b_{v_{i+1}}$ . Since this was the penultimate time $b_{v_t}$ was incremented, we know that $b_t \le b_{i+1}+2$ . By induction we can find a path of length $b_{i+1}-1$ starting with $v_{i+1}$ which otherwise only uses vertices from the set $\{v_1,\dots,v_{i-1}\}$ . This means that we can prepend $v_tv_iv_{i+1}$ to this path to obtain a new one of length $b_{i+1}-1+2 \ge b_t-1$ which only uses vertices from the set $\{v_1,\dots,v_{i+1}\}$ (apart from $v_t$ ). But $i\leq t-3$ , so $\{v_1,\dots,v_{i+1}\}$ is a subset of $\{v_1,\dots,v_{t-2}\}$ , as desired.

The above lemma tells us that if our orientation was $P_k$ -free, then $a_v,b_v \le k$ throughout the process. Keeping this in mind, the following lemma considers the part of the orientation revealed before step i of our algorithm and gives a bound on the number of ways in which one can complete this partial orientation into a $P_k$ -free one.

Lemma 2·3. Assume that up to step i our algorithm processed vertices $v_1,\dots, v_{i-1}$ and assigned the state $(a_v,b_v)$ to any remaining vertex v. We fix the orientation of edges incident to vertices $v_1,\dots, v_{i-1}$ which leads to this state. There are at most

\begin{equation*}\prod_{v \in V(G) \setminus \{v_1,\dots, v_{i-1}\}} (k+2)\cdot 2^{2k-a_v-b_v}\end{equation*}

ways to orient the remaining edges to complete the orientation without creating a $P_k$ .

Before turning to the proof, note that for $i=1$ no part of the orientation was specified and $a_v,b_v=0$ for every v, so the lemma tells us there are at most $(k+2)^n2^{2kn}\le 2^{3kn}$ orientations of G without a $P_k$ , establishing Theorem 1·4.

Proof. Observe first that Lemma 2·2 guarantees that $a_v,b_v \le k$ as otherwise any orientation we produce would contain a $P_k$ .

We will prove this by reverse induction on i. For the base case of $i=n$ there are no remaining edges to orient so the claim holds trivially, since $a_{v_n},b_{v_n} \le k$ .

Let us assume that it holds if we start from step $i+1$ and any state. Given the orientation of the edges incident to $v_1,\ldots, v_{i-1}$ , we can determine which vertex will be $v_i$ . We will now consider all the ways in which we can extend our partial orientation to include the orientation of all edges incident to $v_i$ . That is, we consider all the ways we can orient the edges between $v_i$ and $W\;:\!=\;V(G) \setminus \{v_1\ldots, v_{i}\}$ . For each extended partial orientation, we will use the induction hypothesis to bound the number of ways it can be completed into a $P_k$ -free orientation. Summing over all choices for the orientation of the edges between $v_i$ and W, we will get the desired bound.

If $v_i$ has all its unspecified edges (i.e. edges to W) oriented towards $v_i$ then the state of every remaining vertex remains unchanged and the induction hypothesis tells us that there are $\prod_{v \in W} (k+2)\cdot 2^{2k-a_v-b_v}$ ways to complete it into a $P_k$ -free orientation.

Otherwise our algorithm will choose $v_{i+1}$ to maximise $b_v$ among out-neighbours $v \in W$ of $v_i$ . Assume this maximum is equal to b. Let us denote by d the number of neighbours $v\in W$ of $v_i$ (in the underlying undirected graph G) which have $b_v$ at most b. We know that any other neighbour of $v_i$ in W must be an in-neighbour of $v_i$ (by maximality of b). In particular, there are at most $2^{d}$ orientations of the edges incident to $v_i$ which result in this choice of b. For any such orientation, we claim that our algorithm increased the sum $\sum_{v \in W} (a_v+b_v)$ by d. Indeed, any out-neighbour of $v_i$ had its $a_v$ incremented and any in-neighbour among the d neighbours with $b_v \le b$ had their $b_v$ incremented. This means that regardless of how we orient, by the induction hypothesis the number of ways to complete any of these orientations is at most ${1}/{2^d}\prod_{v \in W} (k+2)\cdot 2^{2k-a_v-b_v}.$ Since there are $2^d$ possible orientations of the edges incident to $v_i$ (given this choice of b) and $k+1$ choices for $0\le b \le k$ , this gives at most $(k+1)\prod_{v \in W} (k+2)\cdot 2^{2k-a_v-b_v}$ different ways to complete our initial orientation. Adding this to the contribution of the case without any out-neighbours of $v_i$ and using $a_{v_i},b_{v_i} \le k$ we obtain the claimed bound.

2·3. General trees

In this section we will complete the proof of Theorem 1·5. As already discussed in Section 2.1, we always have $D(n,H)\le 2^{O(n\log n)}$ when H is an orientation of a forest. Let us now take $G=K_{\lfloor n/2 \rfloor,\lfloor n/2 \rfloor}$ (with an extra isolated vertex if n is odd), and orient all its edges from one part of the bipartition to the other (say from left to right), except for a single perfect matching which we orient in the other direction. Observe first that there are $\lfloor n/2 \rfloor !=2^{\Omega(n\log n)}$ such orientations, since this is the number of choices for the matching that we have. As any such orientation of G is 1-almost antidirected, this shows that if H is not 1-almost antidirected then $D(n,H)=2^{\Theta(n\log n)}$ , as claimed in the second part of Theorem 1·5.

To establish the first part, we need to show that for any 1-almost antidirected forest H there are at most $2^{O(n)}$ H-free orientations of any n-vertex graph. We will actually show that this holds for a certain universal oriented tree H which contains all k-vertex 1-almost antidirected oriented forests. We define this universal oriented tree recursively, layer by layer.

The starting point is the tree $H_{1,t}$ , which is defined as follows. It has a root v which has one in-neighbour $v_0$ and t out-neighbours $v_1,\ldots, v_t$ . Furthermore, $v_0$ has t in-neighbours and each of $v_1,\ldots, v_t$ has a single out-neighbour and t in-neighbours in addition to v. See Figure 2 for an illustration. In our recursive definition it will be convenient to have another building block which we call $H_{1,t}^{-}$ and which is obtained from $H_{1,t}$ by deleting the subtree rooted at $v_0$ . See Figure 3 for an illustration. $H_{s,t}$ is now defined by taking $H_{s-1,t}$ and appending a copy of $H_{1,t}$ to every out-leafFootnote ¹ and a copy of $H_{1,t}^{-}$ to every in-leaf. See Figure 4 for an illustration.

Fig. 2. $H_{1,2}$ .

Fig. 3. $H^-_{1,2}$ .

Fig. 4. Part of $H_{2,2}$ .

Observe first that $H_{k,k}$ contains any 1-almost antidirected tree on k vertices. This is due to the fact that in $H_{k,k}$ every non-leaf vertex at even depth has one in-neighbour and at least k out-neighbours and every vertex at odd depth has one out-neighbour and at least k in-neighbours, which allows one to simply greedily embed any 1-almost antidirected tree (starting from the root of $H_{k,k}$ to ensure that a leaf of $H_{k,k}$ is not encountered). Observe also that $H_{k+1,k}$ contains any 1-almost antidirected forest on k vertices. This follows since if we remove the root and its neighbours, the remaining oriented graph contains many (at least k) pairwise vertex-disjoint subtrees which are isomorphic to $H_{k,k}$ , and we can embed the 1-almost antidirected trees making up our 1-almost antidirected forest into separate ones. With this in mind, the following theorem is the key result we need to prove in order to establish the remaining case of Theorem 1·5.

Here and below, the notation $O_{k,s,t}$ means that the implied constant can depend on k, s and t. We will prove this theorem by induction on s. We will prove the case $s=1$ separately, as it will both serve as the base case and be useful in the induction step.

Proof. Set $T \;:\!=\;\max(|V(H_{1,t})|,kt)=O_{k,t}(1)$ .

Let D be an orientation of G which contains no $H_{1,\,t}$ with the root in X and in which there are at most k out-edges from each $y\in Y$ . Let $Y_1=Y$ and let $X_1$ be the set of vertices in X which have at least one in-edge in D. Note that there are at most $2^n$ possibilities for $X_1$ . Let $Y_2 \subseteq Y_1$ be the set of vertices in $Y_1$ which have at least $T+k$ neighbours in $X_1$ . There are at most $2^{(T+k)(|Y_1|-|Y_2|)}$ ways to orient the edges incident to $Y_1 \setminus Y_2$ . Consider the subset $X_2\subseteq X_1$ consisting of vertices which have an in-neighbour in $Y_2$ . Since every vertex $y \in Y_1 \setminus Y_2$ has at most $T+k$ neighbours in $X_1$ , there are at most $(T+k)(|Y_1|-|Y_2|)$ vertices in $X_1$ which have a neighbour in $Y_1 \setminus Y_2$ , and we can specify the subset of them which got removed from $X_1$ to obtain $X_2$ in at most $2^{(T+k)(|Y_1|-|Y_2|)}$ many ways. We repeat as long as we can, i.e. until we obtain subsets $X_\ell\subset X_1$ and $Y_\ell\subset Y_1$ which have the following properties: every vertex of $X_\ell$ has an in-neighbour in $Y_\ell$ and every vertex in $Y_\ell$ has at least $T+k$ neighbours inside $X_\ell$ . By the above counting, the number of possibilities for $X_\ell$ , $Y_\ell$ and the orientations of the edges incident to $(X\setminus X_\ell)\cup (Y\setminus Y_\ell)$ is at most $2^n\cdot 2^{2(T+k)|Y_1|}\leq 2^{n+2(T+k)n}$ .

If $Y_{\ell}=\emptyset$ , then we have already revealed the entire orientation of G, so there are at most $2^{n+2(T+k)n}$ such suitable orientations. Assume that $Y_{\ell}\neq \emptyset$ . We claim that this, together with the assumption that any vertex in $Y_\ell \subseteq Y$ has at most k out-neighbours in X, guarantees that we can find a copy of $H_{1,t}$ in $D[X_\ell\cup Y_\ell]$ with the root in $X_\ell$ , which is a contradiction. Indeed, fix one edge incoming from $Y_{\ell}$ at every vertex in $X_\ell$ . These edges span vertex disjoint out-directed stars of size at most k with centres in $Y_\ell$ . In particular, there are at least $|X_{\ell}|/k$ centres. Since each centre has at least $T+k$ neighbours in $X_\ell$ , at most k of which can be out-neighbours, there are at least T in-neighbours. Since $T \ge tk$ , this means that some vertex $v \in X_\ell$ is an in-neighbour of at least t distinct centres $v_1,\ldots, v_t$ of our out-stars. Picking one out-edge per star gives us an out-directed tree consisting of root v and t vertex-disjoint paths of length 2. Note also that it is guaranteed that there is an in-neighbour $v_0\in Y_{\ell}$ of v (which is distinct from $v_1,\ldots, v_t$ since they are its out-neighbours). What remains to be done is to find t in-neighbours for each of $v_0,\ldots, v_t$ which we can do greedily since each of them has at least $T \ge |V(H_{1,t})|$ in-neighbours in $X_\ell$ . So we found a copy of $H_{1,t}$ as claimed, and are done.

Let us now define the oriented tree $H^*_{s,t,t'}$ by modifying $H_{s,t}$ so that every vertex in the penultimate layer has t ^′ instead of t out-leaves attached to it. Note that if $t'\geq t$ , then $H_{s,t'}$ contains $H^*_{s,t,t'}$ as a subgraph.

In the induction step, we will use the following key lemma.

Proof. Throughout the proof, we assume that t ^′ is sufficiently large.

Take a subgraph K in D which is isomorphic to $H^*_{s,t,t'}$ with the root in X and in which every leaf is the root of an $H_{1,t'}$ . Choose also a subgraph L(w) isomorphic to $H_{1,t'}$ with root w for every leaf w in K. Observe that for every leaf w of K, w has a unique in-neighbour in L(w), call this vertex f(w) and note that $f(w)\in Y$ .

Claim. K has a subgraph K ^′ isomorphic to $H_{s,t}$ with the same root as K such that for every out-leaf w in K ^′, the vertex f(w) is not in V(K ^′), and all these f(w)’s are distinct.

Proof of Claim. To get a subgraph of $H^*_{s,t,t'}$ isomorphic to $H_{s,t}$ , we need to keep t of the t ^′ out-leaves for every vertex of the penultimate layer. We can do this one by one in an arbitrary order. We just need to pay attention that for each out-leaf w that we keep, the vertex f(w) should be different from every vertex that is already in K ^′, and moreover all the f(w)’s for different w’s should be different. This can be done; indeed, every f(w) is in Y and every $y\in Y$ has at most k out-edges in D, so at any point the number of forbidden choices for w is at most $2|V(H_{s,t})|\cdot k$ , hence $t'>2|V(H_{s,t})|\cdot k$ suffices.

It remains to extend K ^′ to a copy of $H_{s+1,t}$ . For this, we need to “attach” a copy of $H_{1,t}$ to each out-leaf in K ^′, and we need to attach a copy of $H_{1,t}^-$ to each in-leaf in K ^′ in a way that all new vertices are distinct from each other and from the vertices of K ^′.

We begin by extending K ^′ by joining f(w) to w for each out-leaf w of K ^′. By the above claim, all of these f(w)’s are distinct and disjoint from K ^′. Next for every leaf w of K ^′ we want to append t vertex disjoint (apart from sharing the start vertex w) paths of length 2 directed away from w. This can be done since L(w) gives us $t'>|V(H_{s+1,t})|$ vertex disjoint paths of length 2 starting at w, so no matter how many vertices we already embedded we still have one of these paths available.

It remains to attach t in-neighbours to each vertex in the penultimate layer of our partial $H_{s+1,t}$ . Since the vertices in the penultimate layer are in the middle layer of some L(w), we know that each of these vertices has at least t ^′ in-neighbours in D so once again we will always have enough of them available.

Proof of Theorem 2·4. We will use induction on s. The case $s=1$ is Lemma 2·5. Assume now that we have verified the statement for $s-1$ . Let k, s, t be positive integers and let $t'=t'(k,s-1,t)$ from Lemma 2·6.

Let D be an $H_{s,t}$ -free orientation of G with the property that there are at most k out-edges from every $y\in Y$ . Let $X_1$ be the set of vertices in X which are roots of a copy of $H_{1,t'}$ in D and let $X_2=X\setminus X_1$ . Clearly there are at most $2^n$ possibilities for $X_1$ . The oriented graph $D\lbrack X_2\cup Y\rbrack$ contains no $H_{1,t'}$ with the root in $X_2$ , so the number of possibilities for the orientation of $G\lbrack X_2\cup Y\rbrack$ is at most $2^{O_{k,t'}(n)}$ by Lemma 2·5. Moreover, since D is $H_{s,t}$ -free, Lemma 2·6 implies that $D\lbrack X_1\cup Y\rbrack$ contains no copy of $H^*_{s-1,t,t'}$ , and hence also no copy of $H_{s-1,t'}$ , with the root in $X_1$ . Then by induction there are at most $2^{O_{k,s-1,t'}(n)}$ possibilities for the orientation of $G\lbrack X_1\cup Y\rbrack$ . Combining all three upper bounds, the result follows.

Finally let us deduce the first part of Theorem 1·5.

Proof. The proof is by induction on k. For the base case note that the $k\le 2$ case is trivial. Let us now assume that the statement holds for forests with $k-1$ vertices. We may w.l.o.g. assume that H has an in-leaf. Let H ^′ denote the oriented forest obtained from H with this leaf removed. Given an orientation D of G, let X be the subset of V(G) consisting of vertices with out-degree at least k. There are $2^n$ different options for X. Let $Y=V(G) \setminus X$ . Observe first that if we could find H ^′ inside X then we could extend it to a copy of H in D since every vertex of X has out-degree at least k. This means that by induction there are at most $2^{O_{k-1}(n)}$ many ways to orient the edges inside X. There are at most $|Y|k \le kn$ edges inside Y, so the edges inside Y can be oriented in at most $2^{kn}$ many ways. Finally, the number of ways to orient the edges between X, Y in a way that any vertex in Y has at most k out-edges and without creating a copy of $H_{k+1,k}$ is at most $2^{O_k(n)}$ by Theorem 2·4. Since $H \subseteq H_{k+1,k}$ , this completes the proof.

2·4. Odd cycles

We will assume some familiarity with the basic directed regularity lemma, the specific details needed are given in Section 2 of [ Reference Alon and Yuster3 ]. Since the use of regularity in our argument is essentially the same as in both [ Reference Alon, Balogh, Keevash and Sudakov2, Reference Alon and Yuster3 ], we will not go into more technical details of these parts and will refer the reader to either of these papers, with the goal of making the key part of the argument easier to follow.

The following lemma says that if there are many orientations of G which are $C_{2k+1}$ -free then G is not far from being bipartite. It is analogous to [ Reference Alon and Yuster3 , Lemma 2·1] which replaces $C_{2k+1}$ with an arbitrary tournament.

Proof. Let us fix $0<\varepsilon \ll \eta \ll \beta \ll \alpha \ll \delta$ as needed for various points of the upcoming argument.

Let $\overrightarrow{G}$ be a $C_{2k+1}$ -free orientation of G. We apply the directed regularity lemma to $\overrightarrow{G}$ to obtain an $\varepsilon$ -regular partition $V(\overrightarrow{G})=V_1 \cup \cdots \cup V_m$ (all $V_i$ ’s should have sizes as equal as possible, and all but $\varepsilon m^2$ pairs $(V_i, V_j)$ should satisfy that linear sized subsets have about the same density of edges in each direction as the density of edges between $V_i,V_j$ in the same direction). We then consider a cluster (di)graph C of density $\eta$ (its vertices are the parts of our partition and two parts are joined by a directed edge if they are $\varepsilon$ -regular and the density of edges in the corresponding direction is at least $\eta$ ). Note that C can have bidirected edges.

We first want to show that there exists some orientation $\overrightarrow{G}$ for which the resulting cluster graph has at least $m^2/4-\beta m^2$ edges directed both ways. We claim that if this is not the case, then there would be too few (less than $2^{\lfloor n^2/4\rfloor}$ ) orientations possible. Since the regularity lemma guarantees that $m \le M=M(\varepsilon)$ , there are at most $M^n$ choices for the $\varepsilon$ -regular partition $\mathcal{P},$ at most $2^{\binom{M}{2}}$ choices for which pairs are $\varepsilon$ -regular and $4^{\binom{M}{2}}$ choices for C. In total there are at most $M^n2^{3M^2/2}$ choices for $\mathcal{P},$ regular pairs and C. Let us now bound how many orientations could give rise to a fixed choice. There are few edges inside parts of our fixed $\mathcal{P}$ and between non- $\varepsilon$ -regular pairs (at most $\varepsilon n^2$ in both cases) and each edge may be oriented in two ways, so the total contribution of these edges to the number of orientations is at most a factor of $2^{2\varepsilon n^2}$ . For any $\varepsilon$ -regular pair $(V_i,V_j)$ which is not an edge of C in one of the directions, there are at most about $\eta n^2/m^2$ directed edges in that direction. An easy estimate tells us that the edges between $V_i$ and $V_j$ can be oriented like this in at most $2^{c_{\eta}n^2/m^2}$ many ways where $c_{\eta} \to 0$ as $\eta \to 0$ and $c_{\eta}$ only depends on $\eta$ . Since there are at most $m^2$ such pairs $(V_i,V_j)$ , orienting edges between them contributes at most a factor of $2^{c_{\eta}n^2}$ to the total number of orientations. Finally, for any edge of C directed both ways, there are at most $2^{(n/m)^2}$ orientations of the edges between the corresponding pair of parts, but since we are assuming that C has at most $m^2/4-\beta m^2$ such edges, they contribute at most a factor of $2^{n^2/4-\beta n^2}$ to the total number of orientations. Putting it all together we get at most

\begin{equation*}M^n2^{3M^2/2}\cdot 2^{2\varepsilon n^2} \cdot 2^{c_{\eta}n^2} \cdot 2^{n^2/4-\beta n^2}\end{equation*}

orientations. Choosing $\eta$ to be small enough compared to $\beta$ gives us a contradiction to having at least $2^{\lfloor {n^2/4} \rfloor}$ orientations.

Let now $\overrightarrow{G}$ be an orientation for which the resulting cluster graph C has at least $m^2/4-\beta m^2$ edges directed both ways. We claim that C can not contain a bidirected triangle missing only a single directed edgeFootnote ² as otherwise $\overrightarrow{G}$ would contain a $C_{2k+1}$ . This is a consequence of a standard embedding lemma. One can easily deduce it from the classical (undirected) embedding lemma (see e.g. [ Reference Komlós and Simonovits18 , Lemma 2·1]) by first refining the partition (splitting each part into k parts of size as equal as possible, which preserves the regularity while density drops to at worst $\eta-\varepsilon\ge \eta/2$ ) then only keeping edges in the desired direction, and applying the usual embedding lemma; see Figure 5 for an illustration.

Fig. 5. Illustration of how we find $C_5$ .

In particular, this tells us that the graph consisting only of the bidirected edges of C is both triangle-free and has at least $m^2/4-\beta m^2$ edges. The stability theorem of Simonovits [ Reference Simonovits25 ] shows that there is a bipartition $V(C)=W_1 \cup W_2$ with at most $\alpha m^2$ bidirected edges within a part (using that $\alpha \gg \beta$ ). Hence, the bipartite subgraph consisting of the bidirected edges of C between $W_1$ and $W_2$ has at least $m^2/4-(\beta+\alpha) m^2$ edges. If C had in addition more than $8(\alpha+\beta)m^2$ directed edges inside parts, we would find a bidirected triangle with one directed edge removed in C. Indeed, more than $4(\alpha+\beta)m^2$ of these additional edges must be inside a single part, say $W_1$ , and we can pass to a bipartite subgraph of size more than $2(\alpha+\beta)m^2$ within $W_1$ . Taking into account these edges might also be bidirected there are more than $(\alpha+\beta)m^2$ distinct pairs spanning a directed edge. These edges together with the bidirected edges between $W_1$ and $W_2$ make a subgraph of C with more than $m^2/4$ edges, so by Mantel’s theorem they give a triangle. This triangle has at most one edge inside $W_1$ (since the edges we used inside $W_1$ form a bipartite graph), so it has at least two bidirected edges, as desired.

It follows from the above that there are at most $\alpha m^2+8(\alpha+\beta)m^2$ edges of C inside $W_1$ and $W_2$ . Remove all edges of G which correspond to such edges of C. Moreover, remove all edges within $V_i$ ’s and between pairs $(V_i,V_j)$ corresponding to non-edges in C. The remaining subgraph of G is bipartite (with the parts being the union of $V_i$ ’s corresponding to $W_1$ and to $W_2$ ). Since there are at most $\varepsilon n^2$ edges within $V_i$ ’s, at most $\varepsilon n^2$ edges between non- $\varepsilon$ -regular pairs and at most $2\eta n^2$ edges between $\varepsilon$ -regular pairs which are non-edges in C, we have in total removed at most $(\alpha+8(\alpha+\beta))n^2+\varepsilon n^2+\varepsilon n^2+2\eta n^2\leq \delta n^2$ edges, as desired.

The following lemma replaces the embedding [ Reference Alon and Yuster3 , Lemma 3·1]. Let us introduce some notation for convenience. Given an oriented graph D and an integer k, we say that a pair of disjoint subsets $W_1,W_2\subseteq V(D)$ with $|W_i|\ge 2k$ is k-rich if for any $X_i \subseteq W_i, |X_i|\ge |W_i|/20$ , D has at least $\frac{1}{10}|X_1||X_2|$ edges from $X_1$ to $X_2$ , as well as at least ${1}/{10}|X_1||X_2|$ edges from $X_2$ to $X_1$ .

Proof. We iteratively find our directed path. Assume that for some $i\leq k$ , we have found a path $v_1 v_2 \cdots v_{2i-1}$ and a subset $V_{2i-1}\subseteq W_2 \setminus \{v_2,v_4,\ldots,v_{2i-2}\}$ of at least $|W_2|/20$ out-neighbours of $v_{2i-1}$ . Then since D has at least ${1}/{10}|V_{2i-1}||W_1\setminus \{v_1,v_3,\ldots,v_{2i-1}\}|$ edges oriented from $V_{2i-1}$ to $W_1 \setminus \{v_1,v_3,\ldots,v_{2i-1}\}$ , there must be a vertex $v_{2i}$ in $V_{2i-1}$ with a set $V_{2i}$ of at least $(|W_1|-k)/10\ge |W_1|/20$ out-neighbours in $W_1 \setminus\{v_1,v_3,\ldots,v_{2i-1}\}$ . Repeating from the other side completes the iteration. After k iterations, we find the desired path.

We now turn to the proof of the main result in this section.

Proof of Theorem 1·1. Let $n_0=n_0(\delta^2, k)$ be given by Lemma 2·8 applied with $\delta^2$ in place of $\delta$ , for some sufficiently small $\delta$ .

Let us take a graph G on $n>n_0^2+n_0$ vertices which has at least $2^{\lfloor {n^2/4} \rfloor+m}$ $C_{2k+1}$ -free orientations for some $m \ge 0$ . We will show that if G is not the Turán graph, then either we can find a vertex v such that $G\setminus v$ has at least $2^{\lfloor {(n-1)^2/4} \rfloor+m+1}$ $C_{2k+1}$ -free orientations or we can find distinct vertices u and v such that $G\setminus \{u,v\}$ has at least $2^{\lfloor {(n-2)^2/4} \rfloor+m+2}$ $C_{2k+1}$ -free orientations. We then iterate (note that no subgraph we consider can any longer be a Turán graph since it has too many orientations, so also edges) as long as our graph has at least $n_0$ vertices. When we stop, we obtain a graph with less than $n_0$ vertices which has at least $2^{n_0^2}$ orientations, which is impossible, since it has at most $n_0^2/2$ edges.

Let us assume that G is not the Turán graph on $n \ge n_0$ vertices and proceed to find a suitable vertex v.

Every vertex needs to have degree at least $\lfloor {n/2} \rfloor$ as otherwise its edges contribute at most a factor of $2^{\lfloor {n/2} \rfloor-1}$ to the number of orientations so it would immediately work as our vertex v above.

Let $V_1,V_2$ form a bipartition of V(G) which minimises the number of edges within parts. Since $n \ge n_0$ , by Lemma 2·8 we have at most $\delta^2 n^2$ edges within parts. This implies $|V_1|,|V_2| \le (1/2+\delta)n$ as otherwise G would have less than $n^2/4$ edges, so too few orientations. Similarly, there can be at most $\delta^2 n^2$ edges missing between parts.

We first claim that there can be only few orientations for which there exists a pair of disjoint subsets $X_1\subseteq V_1, X_2\subseteq V_2$ , both of size at least $2\delta n$ , which have at most $|X_1||X_2|/10$ edges directed from $X_1$ to $X_2$ . The number of such orientations of edges between $X_1,X_2$ is at most

\begin{equation*}\sum_{i=0}^{|X_1||X_2|/10}\binom{e(X_1,X_2)}{i} \le \sum_{i=0}^{|X_1||X_2|/10}\binom{|X_1||X_2|}{i}\le 2^{|X_1||X_2|/2},\end{equation*}

where $e(X_1,X_2)$ stands for the number of edges between $X_1$ and $X_2$ . Since the total number of edges is at most $n^2/4+\delta^2 n^2$ , there are at most $2^{n^2/4+\delta^2 n^2-|X_1||X_2|/2} \le 2^{n^2/4-\delta^2 n^2}$ such orientations of the whole graph. Since we can choose $X_1$ and $X_2$ in at most $2^{2n}$ many ways, there can be at most $2^{2n}\cdot 2^{n^2/4-\delta^2 n^2}\le 2^{\lfloor {n^2/4} \rfloor-1}$ orientations for which such a pair $X_1,X_2$ exists.

Let us now consider only $C_{2k+1}$ -free orientations such that for any pair of disjoint subsets $X_1,X_2$ of size at least $2\delta n$ , there are at least $|X_1||X_2|/10$ edges oriented from $X_1$ to $X_2$ and also from $X_2$ to $X_1$ . Then any pair of subsets, both of size at least $40\delta n$ , is k-rich. We call such an orientation relevant and by the above counting and our assumption on the number of $C_{2k+1}$ -free orientations of G, there are at least $2^{\lfloor {n^2/4} \rfloor+m}-2^{\lfloor {n^2/4} \rfloor-1}\ge 2^{\lfloor {n^2/4} \rfloor+m-1}$ relevant orientations.

Case 1. Some vertex v has at least $800\delta n$ neighbours in its own part, say $V_1$ .

Note that v must have at least $800 \delta n$ neighbours in $V_2$ as well, by maximality of the number of edges between $V_1$ and $V_2$ . If in a relevant orientation v has at least $40\delta n$ out-neighbours and at least $40\delta n$ in-neighbours belonging to different parts, then since these sets make a k-rich pair we can use Lemma 2·9 to find a path of length $2k-1$ and join it with v to give a $C_{2k+1}$ , a contradiction. This means that at least 2 out of the 4 sets: out-neighbours of v in $V_1$ , in-neighbours of v in $V_1$ , out-neighbours of v in $V_2$ and in-neighbours of v in $V_2$ need to have size at most $40\delta n.$ These two parts can not belong to the same $V_i$ or be of different types in different parts. The only remaining option is for v to have at most $80\delta n$ in-neighbours or at most $80\delta n$ out-neighbours. In particular, its edges may be oriented in such a way in at most

\begin{equation*}2 \sum_{i=0}^{80 \delta n} \binom{d(v)}{ i}\le 2 \sum_{i=0}^{d(v)/10} \binom{d(v)}{ i} \le 2^{0.49d(v)} \le 2^{0.49n}\end{equation*}

many ways, where $d(v)\le n$ denotes the degree of v in G. In other words $G \setminus \{v\}$ must have at least

\begin{equation*}2^{\lfloor {n^2/4} \rfloor+m-1-0.49n}\ge 2^{\lfloor {(n-1)^2/4} \rfloor+m+1}\end{equation*}

$C_{2k+1}$ -free orientations, as desired.

Case 2. Every vertex of G has at most $800\delta n$ neighbours in its own part.

Since G is not the Turán graph there must exist an edge uv inside a part, say in $V_2$ . Both u and v have at least $\lfloor {n/2} \rfloor-800\delta n \ge n/3$ neighbours in $V_1$ , in particular they have $d(u,v) \ge n/8$ common neighbours in $V_1$ since parts have size at most $n/2+\delta n$ . If in a relevant orientation uv is an edge, then the set $W_1$ of out-neighbours of v in $V_1$ which are also in-neighbours of u has size at most $40 \delta n$ . This is due to Lemma 2·9 (applied with $W_1$ and $W_2=V_2\setminus \{u,v\}$ ) which allows us to find a path of length $2k-2$ starting and ending in $W_1$ , which in turn can be completed into a $C_{2k+1}$ using uv. This will severely reduce the number of possible orientations of edges incident to u or v. More precisely, the edges from u and v to their common neighbours can be oriented in at most

\begin{equation*}\sum_{i=0}^{40\delta n} \binom{d(u,v)}{i} \cdot 4^{i} \cdot 3^{d(u,v)-i}\le 41\delta n \cdot \binom{d(u,v)}{40\delta n} \cdot 4^{40\delta n} \cdot 3^{d(u,v)-40\delta n}\le 4^{0.99d(u,v)}.\end{equation*}

The same bound analogously holds if vu is the edge instead. In particular, there are at most

\begin{equation*}2^{d(u)+d(v)-0.02d(u,v)}\le 2^{n-n/1000}\end{equation*}

possible orientations of edges incident to u or v (we are using that both u and v have degree at most $n/2+ \delta n+800\delta n$ and that $\delta$ is small). In particular, the total number of orientations of $G \setminus \{u,v\}$ is at least

\begin{equation*}2^{\lfloor {n^2/4} \rfloor+m-1}/2^{n-n/1000} \ge 2^{\lfloor {(n-2)^2/4} \rfloor+m+2}\end{equation*}

and we are done.