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

There exists a color $i$ which appears at most $\left\lfloor\frac{a}{c}\right\rfloor$ times in the minimum vertex covers, which gives that$e(G_{i})\leq\left\lfloor\frac{a}{c}\right\rfloor(n-1)+b\leq\left\lfloor\frac {n(q-1)-b}{c}\right\rfloor(n-1)+b.$Note that increasing $b$ by $1$ either increases the above bound by $1$ or decreases it by $n-2$, and the decrease happens after at most $c-1$ increases. Since $n>c$, the maximum value of the bound is achieved in the last moment before the first decrease, which occurs for $b$ equal to the remainder of $n(q-1)$ when divided by $c$, as desired.The sharpness of the bound follows from a modification of a construction described in the proof of . We enumerate the vertex set by consecutive integers from $1$ to $n$ and assign to every vertex $j\in[n]$ all colors from $(j-1)(q-1)$ to $j(q-1)-1$ considered modulo $c$. This way each vertex has $q-1$ colors assigned and in total we have $n(q-1)$ assignments. Now, remove the last $r$ assignments. Since $r$ is the remainder of $n(q-1)$ when divided by $c$, after the removal each color was assigned the same number of times. For every $i\in[c]$, add to $G_{i}$ an edge from each vertex with color $i$ assigned to any other vertex. This gives $\left\lfloor\frac{n(q-1)}{c}\right\rfloor(n-1)$ edges in each graph. Note that every vertex $v$ having $x$ assignments removed has positive outdegree in exactly $q-1-x$ graphs, so we may still add edges from $v$ to arbitrary $x$ vertices and avoid creating a rainbow $S_{0,q}$. This addition increases the number of edges in each graph by the total number of removed assignments, which is equal to $r$. Altogether we obtain the required number of edges in each graph