Problem

Source: Iran Team selection test 2024 - P1

Tags: combinatorics



Let $G$ be a simple graph with $11$ vertices labeled as $v_{1} , v_{2} , ... , v_{11}$ such that the degree of $v_1$ equals to $2$ and the degree of other vertices are equal to $3$.If for any set $A$ of these vertices which $|A| \le 4$ , the number of vertices which are adjacent to at least one verex in $A$ and are not in $A$ themselves is at least equal to $|A|$ , then find the maximum possible number for the diameter of $G$. (The distance between two vertices of graph is the number of edges of the shortest path between them and the diameter of a graph , is the largest distance between arbitrary pairs in $V(G)$. ) Proposed by Alireza Haqi