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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图片描述

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