Counting graph orientations
with no directed triangles

Let $\vec{A}$ and $\vec{B}$ be arbitrary $K^{\circlearrowright}_{3}$-free orientations of $G[A]$ and $G[B]$ respectively. Suppose without loss of generality that $\vec{B}$ assigns the orientation of $\{u,v\}$ from $u$ to $v$ and consider the transitive ordering of $\vec{A}$. We estimate in how many ways one can include $u$ and $v$ in the ordering $(v_{1},\ldots,v_{r})$ while keeping it transitive. Since $\{u\}\cup N_{A}(u)$ and $\{v\}\cup N_{A}(v)$ are cliques, by Proposition 2.1 we have ${\rm{ext}}_{G}(u,A)\leq d_{A}(u)+1$ and ${\rm{ext}}_{G}(v,A)\leq d_{A}(v)+1$, which gives at most $(d_{A}(u)+1)(d_{A}(v)+1)$ positions to put the vertices $u$ and $v$ in the transitive ordering of $\vec{A}$. Note that there are $\binom{d_{A}(u,v)+1}{2}$ ways to place $\{u,v\}$ in the transitive ordering of $\vec{A}$ such that $u$ appears after $v$ and they have a common neighbor between them. But each such ordering induces a $K^{\circlearrowright}_{3}$. This finishes the proof. ∎

Let $d_{x}=d_{G}(x,A)$ and $d_{y}=d_{G}(y,A)$, and put $d=d_{x}+d_{y}$. If $d\leq r$, then by applying Proposition 2.1 twice, with $x$ and $y$, we have

Therefore, we assume that $d>r$. Note that since $G$ is $K_{r+1}$-free, we have $d_{x},d_{y}\leq r-1$. Applying Lemma 2.2 and using the fact that, for $d>r$, we have $d_{A}(x,y)\geq d-r$, we obtain

One can check that the right-hand side of (1) is a polynomial on $d$ of degree $2$ with negative leading coefficient and it is a growing function in the interval $(-\infty,2r+1)$. Since $d\leq 2(r-1)$, we have

∎

First, note that if $e_{G}(A,B)\leq 2$, then the trivial bound ${\rm{ext}}_{G}(A,B)\leq 2^{e(A,B)}$ implies ${\rm{ext}}_{G}(A,B)\leq 4$. Also, since $G$ is $K_{4}$-free, we have $e_{G}(A,B)\leq 3$. Thus, we may assume that $e_{G}(A,B)=3$, i.e., $G[A\cup B]$ is a $K_{4}^{-}$. Let $A=\{u_{1},u_{2}\}$ and $B=\{v_{1},v_{2}\}$ so that $u_{2}v_{2}$ is not an edge and consider an arbitrary orientation of the edges $\{u_{1},u_{2}\}$ and $\{v_{1},v_{2}\}$.

If the oriented edges are $u_{1}u_{2}$ and $v_{1}v_{2}$ (or, by symmetry, $u_{2}u_{1}$ and $v_{2}v_{1}$), then for the two possible orientations of $\{u_{1},v_{1}\}$, the orientation of one of the two remaining edges in $E_{G}(A,B)$ is forced. Thus, since there is only one edge left to orient in $E_{G}(A,B)$, which can be done in two ways, we have ${\rm{ext}}_{G}(A,B)\leq 4$.

It remains to consider the case where the oriented edges are $u_{1}u_{2}$ and $v_{2}v_{1}$ (or, by symmetry, $u_{2}u_{1}$ and $v_{1}v_{2}$). If $\{u_{1}v_{1}\}$ is oriented from $u_{1}$ to $v_{1}$, then the orientation of the two remaining edges in $E_{G}(A,B)$ are forced, which gives us one $K^{\circlearrowright}_{3}$-free orientation. On the other hand, if $\{u_{1}v_{1}\}$ is oriented from $v_{1}$ to $u_{1}$, then one can orient the both remaining edges in $E(A,B)$ in two ways, which in total gives that ${\rm{ext}}_{G}(A,B)\leq 5$. ∎

Consider an arbitrary orientation of the edges of $G[B]$. We may assume that $d_{B}(u)\geq 3$, as otherwise we have ${\rm{ext}}_{G}(u,B)\leq 4$. Since $G$ is $K_{4}$-free and $G[B]$ induces a copy of $K_{4}^{-}$, the vertex $u$ must have exactly three neighbors in $B$, which span an induced path $v_{1}v_{2}v_{3}$. By symmetry, we assume that $\{v_{1},v_{2}\}$ is oriented from $v_{1}$ to $v_{2}$. If we orient $uv_{1}$ from $u$ to $v_{1}$, then the orientation of $\{u,v_{2}\}$ is forced, which leaves two possible orientations for the edge $\{u,v_{3}\}$. On the other hand, if we orient $uv_{1}$ from $u$ to $v_{1}$, we just apply Proposition 2.1 to conclude that ${\rm{ext}}_{G}(u,\{v_{2},v_{3}\})\leq 3$. Combining the possible orientations, we obtain ${\rm{ext}}_{G}(u,B)\leq 5$. ∎

By Proposition 2.1, there are three orientations of $T_{1}=G[\{a,b,c\}]$ (resp. $T_{2}=G[\{c,d,e\}]$) compatible with $\vec{T}$, and hence there are at most nine orientations of $E(P)$ compatible with $\vec{T}$. In these orientations, each direction of $\{b,c\}$ and $\{c,d\}$ appears at least once. If $\{b,d\}$ is oriented towards $d$ (resp. towards $b$), then the orientations in which $\{b,c\}$ and $\{c,d\}$ are oriented, respectively, towards $b$ and $c$ (resp. $c$ and $d$) are not compatible with $\vec{T}$. Therefore, there are at most eight orientations of $E(P)$ compatible with $\vec{T}$. Now, suppose that $\{a,c\}$ and $\{b,d\}$ are oriented towards $a$ and $d$. If we orient $\{b,c\}$ towards $c$ (resp. $b$), then $\{a,b\}$ must be oriented towards $a$ (resp. $\{c,d\}$ must be oriented towards $d$), and there are three orientations of $E(T_{2})$ (resp. four orientations of $\{\{a,b\},\{d,e\}\}$) from which we can complete a compatible orientation of $E(P)$. Therefore, there are at most seven orientations of $E(P)$. ∎

Let $A=\{x_{1},x_{2},x_{3}\}$ and $B=\{y_{1},y_{2},y_{3}\}$. Since $G$ is $K_{4}$-free, $y_{i}$ cannot be adjacent to every vertex of $A$, for $i=1,2,3$. This implies that $d_{A}(y_{i})\leq 2$, for $i=1,2,3$. Analogously, we have $d_{B}(x_{i})\leq 2$, for $i=1,2,3$. Thus the set $E$ of edges in $G$ joining $A$ and $B$ induces a set of paths and cycles. Since $G$ is $K_{4}$-free, $E$ does contain a cycle of length $4$. If $|E|\leq 3$, then ${\rm{ext}}_{G}(A,B)\leq 2^{|E|}\leq 8$, as desired. If $|E|=4$, then some vertex, say $x_{1}\in A$, is incident to two edges of $E$, say $\{x_{1},y_{1}\}$ and $\{x_{1},y_{2}\}$, which implies that ${\rm{ext}}_{G}(x_{1},B)\leq 3$, and hence ${\rm{ext}}_{G}(A,B)\leq{\rm{ext}}_{G}(x_{1},B)\cdot 2^{|E\setminus\{\{x_{1},y_{1}\},\{x_{1},y_{2}\}\}|}\leq 12$, as desired. If $|E|=5$, then $|E|$ induces a path of length $5$, say $x_{1}y_{1}x_{2}y_{2}x_{3}y_{3}$. In this case, note that $\{x_{1},x_{2}\}$ and $\{y_{1},y_{2}\}$ are disjoint cliques of size $2$, and hence, by Lemma 2.4, we have ${\rm{ext}}_{G}(\{x_{1},x_{2}\},\{y_{1},y_{2}\})\leq 5$. Since each edge in $E$ either joins $\{x_{1},x_{2}\}$ to $\{y_{1},y_{2}\}$, or is adjacent to $x_{3}$, we have ${\rm{ext}}_{G}(A,B)\leq{\rm{ext}}_{G}(\{x_{1},x_{2}\},\{y_{1},y_{2}\})\cdot{\rm{ext}}_{G}(x_{3},B)\leq 15$, as desired.

Thus, we may assume $|E|=6$, and hence $E$ induces the cycle $x_{1}y_{1}x_{2}y_{2}x_{3}y_{3}x_{1}$. By symmetry, we may assume that $\{x_{1},x_{2}\}$, $\{x_{1},x_{3}\}$ are both oriented towards $x_{1}$, and $\{y_{1},y_{3}\}$ is oriented towards $y_{1}$. Suppose $\{y_{2},y_{3}\}$ is oriented towards $y_{2}$. If we orient $\{x_{1},y_{1}\}$ towards $y_{1}$, then $x_{2}y_{1}$ must be oriented towards $y_{1}$, and, since $\{x_{1},x_{3}\}$ and $\{y_{2},y_{3}\}$ are oriented towards $x_{1}$ and $y_{2}$, by Proposition 2.6, there are $7$ compatible orientations of the edges in the path $x_{2}y_{2}x_{3}y_{3}x_{1}$ (see Figure 1(a)). If we orient $\{x_{1},y_{1}\}$ towards $x_{1}$, then $\{y_{3},x_{1}\}$ must be oriented towards $x_{1}$, and by Proposition 2.6, there are $8$ compatible orientations of the edges in the path $y_{1}x_{2}y_{2}x_{3}y_{3}$ (see Figure 1(b)). Thus, there are $15$ compatible orientations of $E$, as desired. Thus, we may assume that $\{y_{2},y_{3}\}$ is oriented towards $y_{3}$, and hence $\{y_{1},y_{2}\}$ must be oriented towards $y_{1}$. If we orient $\{x_{1},y_{1}\}$ towards $y_{1}$, then $\{x_{2},y_{1}\}$ must be oriented towards $y_{1}$, and by Proposition 2.6, there are $8$ compatible orientations of the edges in the path $x_{2}y_{2}x_{3}y_{3}x_{1}$ (see Figure 1(c)). If we orient $\{x_{1},y_{1}\}$ towards $x_{1}$, then $\{y_{3},x_{1}\}$ must be oriented towards $x_{1}$, and, since $\{x_{1},x_{3}\}$ and $\{x_{1},x_{2}\}$ are oriented towards $x_{1}$, regardless of the orientation of $\{x_{2},x_{3}\}$, by Proposition 2.6, there are $7$ compatible orientations of the edges in the path $y_{1}x_{2}y_{2}x_{3}y_{3}$ (see Figure 1(d)) Thus, there are $15$ compatible orientations of $E$, as desired. ∎

Let $K_{1,\ell,\ell}$ be a complete tripartite graph with vertex partition $\{v\}\cup A\cup B$. For $1\leq i,j\leq\ell$ there are $\binom{\ell}{i}\binom{\ell}{j}$ orientations of the edges incident to $v$ with exactly $i$ out-neighbors of $v$ in $A$, and $j$ in-neighbors of $v$ in $B$, sets which we denote by $A^{+}$ and $B^{-}$ respectively. For each of those orientations, the edges between $A^{+}$ and $B^{-}$ and between $A\setminus A^{+}$ and $B\setminus B^{-}$ are forced in any $K^{\circlearrowright}_{3}$-free orientation. Since any of the other $(\ell-i)j+(\ell-j)i$ edges can be oriented in two ways, we sum over $i$ and $j$ to get

∎

Let $H$ be as in the statement. Recall that $D(T,K^{\circlearrowright}_{3})=6$ and $D(e_{1},K^{\circlearrowright}_{3})=D(e_{2},K^{\circlearrowright}_{3})=2$. Moreover, ${\rm{ext}}_{H}(T,e_{i})\leq 8$ for $i=1,2$, by Corollary 2.3. Also, since $H[e_{1}\cup e_{2}]$ is triangle-free, $E_{H}(e_{1},e_{2})\leq 2$ and hence ${\rm{ext}}_{H}(e_{1},e_{2})\leq 4$. Throughout the proof, we use each of these bounds unless the structure of $H$ allows us to obtain a better bound.

First note that if there is at most one edge between $e_{1}$ and $e_{2}$, then ${\rm{ext}}_{H}(e_{1},e_{2})\leq 2$. In this case we use the bound

We count the number of orientations by considering different values of $E_{H}(e_{1}\cup e_{2},V(T))$. In particular, since $H$ is $K_{4}$-free, we have that $E_{H}(e_{i},V(T))\leq 4$ for $i=1,2$. First note that if $E(e_{i},V(T))=3$ for $i=1,2$, then ${\rm{ext}}_{H}(e_{i},T)\leq 6$. Therefore, if there are at most six edges between $H[e_{1}\cup e_{2}]$ and $T$, then either there are at most three edges between each $e_{i}$ and $T$, which implies ${\rm{ext}}_{G}(T,e_{1}),{\rm{ext}}_{G}(T,e_{2})\leq 6$; or, without loss of generality, there are at most two edges between $e_{1}$ and $T$, which implies ${\rm{ext}}_{G}(T,e_{1})\leq 4$. In both cases we have that ${\rm{ext}}_{G}(T,e_{1})\cdot{\rm{ext}}_{G}(T,e_{2})\leq 36$ and consequently that $D(H,K^{\circlearrowright}_{3})\leq 6\cdot 2\cdot 2\cdot 36\cdot 4<2^{12}$.

Thus, we assume that $7\leq E_{H}(e_{1}\cup e_{2},V(T))\leq 8$. Then, without loss of generality, we have $E_{H}(e_{1},V(T))=4$ and, by Turán’s Theorem, $H_{1}=H[e_{1}\cup V(T)]$ is isomorphic to $K_{1,2,2}$. If $E_{H}(e_{2},V(T))=3$, the aforementioned bounds and Lemma 3.1 yields

Finally, if $E_{H}(e_{i},V(T))=4$ for both $i=1,2$, then the graphs $H_{i}=H[e_{i}\cup V(T)]$ are isomorphic to $K_{1,2,2}$ with $v_{i}\in V(T)$ being the vertex of degree $4$ in $H_{i}$. Since $H$ does not contain two disjoint triangles, then $v_{1}=v_{2}$ and since $H[e_{1}\cup e_{2}]\simeq K_{2,2}$, we have in fact $H\simeq K_{1,3,3}$. Finally, Lemma 3.1 yields $D(H,K^{\circlearrowright}_{3})=2754<2^{12}$. ∎

Let $r=\omega(G)$. The proof follows by induction on $n$. By Proposition 1.1, the statement holds for $n=8$. If $9\leq n\leq 10$, then the result follows from Mantel’s Theorem (see [Ma1907]) for $r=2$, and from Proposition 3.3 for $r\geq 3$, as $n\leq 7+\min\{r,8\}$. Thus, assume $n\geq 11$ and suppose that the statement holds for any graph with less than $n$ vertices (but at least 8 vertices).

Let $K$ be a clique of $G$ of size $s=\min\{r,8\}$. If $n\leq 7+s$, then the result follows from Proposition 3.3, so we may assume that $n-s\geq 8$. Thus, we can apply the induction hypothesis for any subgraph of $G$ with at least $n-s$ vertices.