# 线性代数网课代修|ENGO361代写|ENGO361英文辅导|ENGO361

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}