Baltic Way 2019

The numbers $62$ and $2019$ have the following secret property: $$64+63+\cdots +3=\frac{64\cdot 65}{2}-3=2077>2019.$$ Let before the cops move the robber be in the vertex $A$. If $\mathrm{deg}A\le 62$, the cops can catch the robber in one, two or three moves (due to diameter $2$ property): denote $B_1,B_2,\dots$ the vertices adjacent to $A$, let in the first move the $i$-th cop go “towards” the $B_i$ (i.e. he either goes directly to $B_i$ if the corresponding edge exists or goes to the vertex adjacent to $B_i$ and in the next move to $B_i$); if the robber will leave the vertex $A$ and will go to some vertex $B_k$ he will be caught immediately or at the next move, otherwise the cops will occupy all vertices of $B_i$ in their second move and after that will catch the robber. If $\mathrm{deg}A\ge 63$, let one of the cops go to $A$ and after that always stay there guarding the vertex $A$ and its neighbouring vertices. It means that for the robber the safe part of the graph is decreased by at least $64$ vertices. Now, if before the cops move the “safe degree” of the robber’s vertex is at most $61$, the remaining “active” cops again can catch the robber in one-two-three moves. Otherwise, let one of the active cops go to the vertex of degree at least $62$ and after that always guard this vertex and its neighbours. The safe part of the graph for the robber is decreased once again by at least $63$ vertices. Continuing this process we will find after the $62^{\text{nd}}$ step that the safe area is empty.