Revolutionaries and spies on trees and unicyclic graphs

TL;DR

This paper analyzes a pursuit-evasion game involving revolutionaries and spies on trees and unicyclic graphs, determining the minimum number of spies needed to prevent revolutionaries from meeting unguarded.

Contribution

It provides exact spy count thresholds for trees and unicyclic graphs and characterizes unicyclic graphs requiring one more spy than the basic bound.

Findings

On trees, the minimum spies needed equals |V(G)|, (r/m)

On unicyclic graphs, the minimum spies needed equals |V(G)|, (\u222a r/m)

Characterization of unicyclic graphs where  r/m + 1 spies are necessary.

Abstract

A team of $r$ {\it revolutionaries} and a team of $s$ {\it spies} play a game on a graph $G$. Initially, revolutionaries and then spies take positions at vertices. In each subsequent round, each revolutionary may move to an adjacent vertex or not move, and then each spy has the same option. The revolutionaries want to hold an {\it unguarded meeting}, meaning $m$ revolutionaries at some vertex having no spy at the end of a round. To prevent this forever, trivially at least $\min\{|V(G)|,\FL{r/m}\}$ spies are needed. When $G$ is a tree, this many spies suffices. When $G$ is a unicyclic graph, $\min\{|V(G)|,\CL{r/m}\}$ spies suffice, and we characterize those unicyclic graphs where $\FL{r/m}+1$ spies are needed. \def\FL#1{\lfloor #1 \rfloor} \def\CL#1{\lceil #1 \rceil}