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如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决! \]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\[y\alpha+x\alpha_{j}+(x+y)\beta_{j}+x\gamma+y\gamma_{j}\leq\frac{1}{c}\sum_{i=1 }^{c}y\alpha+x\alpha_{i}+(x+y)\beta_{i}+x\gamma+y\gamma_{i}.\]Since\[\sum_{i=1}^{c}\alpha_{i}\leq(q-1)\alpha,\qquad\sum_{i=1}^{c}\beta_{i}\leq(p+q- 1)\beta\qquad\text{and}\qquad\sum_{i=1}^{c}\gamma_{i}\leq(p-1)\gamma,\]from (1) we obtain\[e(G_{j}) \leq\frac{1}{xy}(x\alpha_{j}+x\beta_{j}+x\gamma)(y\alpha+y\beta_ {j}+y\gamma_{j})\] \[\leq\frac{1}{4xy}\big{(}y\alpha+x\alpha_{j}+(x+y)\beta_{j}+x \gamma+y\gamma_{j}\big{)}^{2}\] \[\leq\frac{1}{4c^{2}xy}\big{(}(yc+x(q-1))\alpha+(x+y)(p+q-1) \beta+(xc+y(p-1))\gamma\big{)}^{2}. \tag{2}\]In particular, for \(x=1\) and \(y=1\), since \(p\leq q\) and \(\alpha+\gamma=n-\beta\), this gives\[e(G_{j}) \leq\frac{1}{4c^{2}}\big{(}(c+q-1)\alpha+2(p+q-1)\beta+(c+p-1) \gamma\big{)}^{2}\] \[\leq\frac{1}{4c^{2}}\big{(}(c+q-1)(\alpha+\gamma)+2(p+q-1)\beta \big{)}^{2}\] \[=\frac{1}{4c^{2}}\big{(}(c+q-1)n+(2p+q-1-c)\beta\big{)}^{2}.\]If \(c\leq t_{1}\), then the above expression is maximized at \(\beta=n\) and we obtain\[e(G_{j})\leq\frac{(p+q-1)^{2}}{c^{2}}n^{2},\]which gives the first bound in both cases of the theorem. This bound is achieved if \(A=\emptyset\), \(C=\emptyset\) and \(B\) is divided into \(\binom{c}{p+q-1}\) equal-sized sets, one for each subset of \(p+q-1\) colors assigned to the set, with edges in \(G_{i}\), for \(i\in[c]\), between any vertices from sets having color \(i\) assigned. This is depicted in Consider now \(c\geq t_{1}\) and take \(x=c-p-q+1\) and \(y=p\). This gives\[yc+x(q-1)=(x+y)(p+q-1)=(c-q+1)(p+q-1),\]so from (2) we obtain\[e(G_{j}) \leq\frac{1}{4c^{2}p(c\!-\!p\!-\!q\!+\!1)}\big{(}(c-q+1)(p+q-1)( \alpha+\beta)+((c-p-q+1)c+p(p-1))\gamma\big{)}^{2}\] \[=\frac{1}{4c^{2}p(c\!-\!p\!-\!q\!+\!1)}\big{(}(c-q+1)(p+q-1)n-(pq-( c-p-q+1)^{2})\gamma\big{)}^{2}.\]If \(c\leq t_{3}\), then the above expression is maximized at \(\gamma=0\) and gives\[e(G_{j})\leq\frac{(c-q+1)^{2}(p+q-1)^{2}}{4c^{2}p(c-p-q+1)}n^{2}.\] This bound is achieved when \(C=\emptyset\), the set \(B\) is of size \(\frac{(q-1)(p+q-1)-c(q-p-1)}{2p(c-p-q+1)}n\) and is divided into \(\binom{c}{p+q-1}\) equal-sized sets, one for each subset of \(p+q-1\) colors assigned to the set, while set \(A\) is of size \(\frac{(p+q-1)(c-2p-q+1)}{2p(c-p-q+1)}n\) and is divided into \(\binom{c}{q-1}\) equal-sized sets, one for each subset of \(q-1\) colors assigned to the set. We put edges in \(G_{i}\), for \(i\in[c]\), between any vertex from a set in \(A\cup B\) having color \(i\) assigned to any other vertex in \(A\) or a vertex in \(B\) having color \(i\) assigned. This construction is possible only if the mentioned sizes of sets \(A\) and \(B\) are non-negative, which occurs when \(c\geq t_{1}\) and \(c\leq t_{2}\), and so it proves the second bound in both cases of the theorem. Note that if \(q\leq p+1\), then the size of \(B\) is always positive, which explains why \(t_{2}\) is defined in this way. This construction is illustrated in In order to prove the third bound in the first case of the theorem, consider \(c\) satisfying \(t_{2}\leq c\leq t_{4}\) and take \(x=c\) and \(y=q-1\). From (2) we get\[e(G_{j})\leq\frac{1}{4c^{3}(q-1)}\big{(}2c(q-1)\alpha+(c+q-1)(p+q-1)\beta+(c^{ 2}+(p-1)(q-1))\gamma\big{)}^{2}.\]The assumption \(c\leq t_{4}\) implies that \(c^{2}+(p-1)(q-1)\leq 2c(q-1)\), so\[e(G_{j}) \leq\frac{1}{4c^{3}(q-1)}\big{(}2c(q-1)(\alpha+\gamma)+(c+q-1)(p+ q-1)\beta\big{)}^{2}\] \[=\frac{1}{4c^{3}(q-1)}\big{(}2c(q-1)n-(c(q-p-1)-(q-1)(p+q-1)) \beta\big{)}^{2}.\]Since \(c\geq t_{2}\), the expression above is maximized at \(\beta=0\) and gives\[e(G_{j})\leq\frac{q-1}{c}n^{2},\] which proves the third bound in the first case of the theorem. This is the same bound as when forbidding a rainbow \(S_{0,q}\), so it is achieved when \(B=C=\emptyset\), while set \(A\) is divided into \(\binom{c}{q-1}\) equal-sized sets, one for each subset of \(q-1\) colors assigned to the set, and we put edges in \(G_{i}\), for \(i\in[c]\), between any vertex from a set having color \(i\) assigned to any other vertex. This is depicted for \(q=3\) and \(c=3\) in Finally, consider the last remaining bound, where we have \(c\geq t_{3}\) and \(c\geq t_{4}\). From (2) for \(x=c-p+1\) and \(y=c-q+1\) we obtain\[e(G_{j}) \leq\frac{1}{4c^{2}(c\!-\!p\!+\!1)(c\!-\!q\!+\!1)}\big{(}(c^{2}-(p \!-\!1)(q\!-\!1))(\alpha+\gamma)+(2c-p-q+2)(p+q-1)\beta\big{)}^{2}\] \[=\frac{1}{4c^{2}(c\!-\!p\!+\!1)(c\!-\!q\!+\!1)}\big{(}(c^{2}-(p\! -\!1)(q\!-\!1))n-((c-p-q+1)^{2}-pq)\beta\big{)}^{2}.\]Since \(c\geq t_{3}\), the above expression is maximized at \(\beta=0\) and gives\[e(G_{j})\leq\frac{(c^{2}-(p-1)(q-1))^{2}}{4c^{2}(c-p+1)(c-q+1)}.\]This is achieved when \(B=\emptyset\), the set \(A\) is of size \(\frac{(c-p+1)^{2}+(p-1)(q-p)}{2(c-p+1)(c-q+1)}n\) and is divided into \(\binom{c}{q-1}\) equal-sized sets, one for each subset of \(q-1\) colors assigned to the set, similarly, the set \(C\) is of size \(\frac{(c-q+1)^{2}-(q-1)(q-p)}{2(c-p+1)(c-q+1)}n\) and is divided into \(\binom{c}{p-1}\) equal-sized sets, one for each subset of \(p-1\) colors assigned to the set. We put edges in \(G_{i}\), for \(i\in[c]\), from any vertex in \(A\) having color \(i\) assigned and any vertex in \(C\) to any vertex in \(A\) and any vertex in \(C\) having color \(i\) assigned图片描述

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