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Note that the bound in is not realizable for \(p\geq 2\) in any collection of directed graphs each having the same number of edges. Thus, for \(p\geq 2\) the optimal bound for \(\min_{1\leq i\leq c}e(G_{i})\) is different. On the other hand, when \(p=1\) the second bound in can be obtained in such a collection of directed graphs, so this theorem implies the optimal bound for \(\min_{1\leq i\leq c}e(G_{i})\) for \(c\geq 2q+2\sqrt{q}\). It occurs that the same construction gives the optimal bound already for \(c\geq\max\{q+\sqrt{q},2q-2\}\), but for smaller values of \(c\) there are different optimal constructions.The theorem below gives the optimal bound for \(\min_{1\leq i\leq c}e(G_{i})\) for any integers \(p\) and \(q\).**Theorem**.: _For integers \(q\geq p\geq 1\), \(c\geq p+q\) and \(n\), every collection of directed graphs \(G_{1},\ldots,G_{c}\) on a common set of \(n\) vertices containing no rainbow \(S_{p,q}\) satisfies the following. The values_\[t_{1}=2p+q-1,\qquad t_{2}=\begin{cases}\frac{(q-1)(p+q-1)}{q-p-1}&\text{ if } \quad q\geq p+2, \infty&\text{ if }\quad q\leq p+1,\end{cases}\]\[t_{3}=p+q-1+\sqrt{pq},\qquad t_{4}=q-1+\sqrt{(q-1)(q-p)}\]_satisfy \(t_{1}\leq t_{2}\), \(t_{1}\leq t_{3}\), and either \(t_{2}\leq t_{3}\leq t_{4}\) or \(t_{4}\leq t_{3}\leq t_{2}\). If \(t_{2}\leq t_{3}\leq t_{4}\), then_\[\min_{1\leq i\leq c}e(G_{i})\leq\begin{cases}\frac{(p+q-1)^{2}}{c^{2}}n^{2}+o( n^{2})&\text{ if }\quad c\leq t_{1}, \frac{(c-q+1)^{2}(p+q-1)^{2}}{4c^{2}p(c-p-q+1)}n^{2}+o(n^{2})&\text{ if }\quad t_{1}\leq c \leq t_{2}, \frac{q-1}{c}n^{2}+o(n^{2})&\text{ if }\quad t_{2}\leq c\leq t_{4}, \frac{(c^{2}-(p-1)(q-1))^{2}}{4c^{2}(c-p+1)(c-q+1)}n^{2}+o(n^{2})&\text{ if }\quad c\geq t_{4}.\end{cases}\]_While if \(t_{4}\leq t_{3}\leq t_{2}\), then_\[\min_{1\leq i\leq c}e(G_{i})\leq\begin{cases}\frac{(p+q-1)^{2}}{c^{2}}n^{2}+o( n^{2})&\text{ if }\quad c\leq t_{1}, \frac{(c-q+1)^{2}(p+q-1)^{2}}{4c^{2}p(c-p-q+1)}n^{2}+o(n^{2})&\text{ if }\quad t_{1}\leq c \leq t_{3}, \frac{(c^{2}-(p-1)(q-1))^{2}}{4c^{2}(c-p+1)(c-q+1)}n^{2}+o(n^{2})&\text{ if }\quad c\geq t_{3}.\end{cases}\]_Moreover, the above bounds are tight up to a lower order error term._Proof.: First, we prove the inequalities between thresholds \(t_{i}\) for \(i\in[4]\). Note that \(t_{3}\geq t_{1}\) since \(q\geq p\). If \(q\leq p+1\), then clearly \(t_{4}\leq t_{3}\leq t_{2}\) and \(t_{1}\leq t_{2}\). While for \(q\geq p+2\) we have\[t_{2}=\frac{(q-1)(p+q-1)}{q-p-1}=\frac{(q-p-1+p)(q-p-1+2p)}{q-p-1}=q+2p-1+\frac{ 2p^{2}}{q-p-1}\geq t_{1}.\]One can also verify that for \(q\geq p+2\) each of the inequalities \(t_{2}\leq t_{3}\) and \(t_{3}\leq t_{4}\) is equivalent to the inequality \(q(q-p-1)^{2}\geq p(p+q-1)^{2}\), so either \(t_{2}\leq t_{3}\leq t_{4}\) or \(t_{4}\leq t_{3}\leq t_{2}\), as desired.Consider a collection of directed graphs \(G_{1},\ldots,G_{c}\) on a common set \(V\) of \(n\) vertices containing no rainbow \(S_{p,q}\). Similarly as in the proof of we use a colored graph removal lemma to remove all homomorphic images of a rainbow \(S_{p,q}\) by deleting \(o(n^{2})\) total edges. Thus, we may assume that no vertex in \(V\) has nonzero indegree in \(p\) graphs and nonzero outdegree in \(q\) different graphs.We split the vertex set \(V\) into disjoint sets. Let \(B\) be the set of vertices incident to edges in at most \(p+q-1\) graphs, \(A\) be the set of vertices in \(V\setminus B\) that have nonzero outdegree in at most \(q-1\) graphs, and \(C\) be the set of vertices in \(V\setminus B\) having nonzero indegree in at most \(p-1\) graphs. Additionally, for each \(i\in[c]\), let \(A_{i}\subset A\) be the set of vertices in \(A\) that have nonzero outdegree in \(G_{i}\), similarly \(C_{i}\subset C\) be the set of vertices in \(C\) that have nonzero indegree in \(G_{i}\), while \(B_{i}\subset B\) be the set of vertices in \(B\) incident to edges in \(G_{i}\). For every \(i\in[c]\), we denote \(\alpha_{i}=|A_{i}|\), \(\beta_{i}=|B_{i}|\), \(\gamma_{i}=|C_{i}|\), \(\alpha=|A|\), \(\beta=|B|\) and \(\gamma=|C|\).Observe that for every \(i\in[c]\),\[e(G_{i})\leq(\alpha_{i}+\beta_{i}+\gamma)(\alpha+\beta_{i}+\gamma_{i}), \tag{1}\]because edges of \(G_{i}\) can go only from vertices in \(A_{i}\cup B_{i}\cup C\) to vertices in \(A\cup B_{i}\cup C_{i}\).For integers \(x,y\geq 1\), by averaging over all colors, there is \(j\in[c]\) such that\[y\alpha+x\alpha_{j}+(x+y)\beta_{j}+x\gamma+y\gamma_{j}\leq\frac{1}{c}\sum_{i=1 }^{c}y\alpha+x\alpha_{i}+(x+y)\beta_{i}+x\gamma+y\gamma_{i}图片描述

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