The Lion and Man Game on Convex Terrains

Abstract

We study the lion-and-man game in which a lion (the pursuer) tries to capture a man (the evader). The players have equal speed and they can observe each other at all times. In this paper, we study the game on surfaces of convex terrains. We show that the lion can capture the man in finite number of steps determined by the terrain geometry.

Appendix

Remark 1

In the following proofs, we treat the disappearing vertex event as a no vertex event with edge length zero for the disappearing edges.

Lemma 8

Let $W_1$ and $W_2$ be two consecutive wavefronts, and $e$ be a point outside $W^i$ in the $XY$-plane. Suppose that $e$ is in the edge region of an edge $m_1$ in $W_1$. Then, the distance between $p_1$, the image of the projection of $e$ onto $W_1$, and $p_2$, the image of the projection of $e$ onto $W_2$, is less than $D$ where $D$ is the maximum distance between any two wavefronts in the $XY$-plane [12].

Lemma 9

Let $W_1$ and $W_2$ be two consecutive wavefronts, and $e$ be a point outside $W^i$ in the $XY$-plane. Suppose that $e$ is in the wedge region of a vertex $w_1$ in $W_1$. Then, the distance between $p_1 = w_1^i$, the image of the projection of $e$ onto $W_1$, and $p_2$, the image of the projection of $e$ onto $W_2$, is less than $D$ where $D$ is the maximum distance between any two wavefronts in the $XY$-plane [12].

Lemma 10

Consider the image of a wavefront $W$ in the $XY$-plane. Let $e$ be a point in the $XY$-plane which is outside $W^i$. Let $p$ be the projection of $e$ onto $W^i$. The line segment $ep$ makes two angles with $W^i$. Then both of these angles are larger than $\frac{\pi }{2}$ [12].

Lemma 11

Consider the image of a wavefront $W$ onto the $XY$-plane, i.e. $W^i$. Let $e_1$ be a point in the $XY$-plane which is outside the region enclosed by $W^i$. Moreover, let $e$ denote the projection of $e_1$ onto $W^i$ ( i.e. $\pi (e_1,W)$, see Definition 7). Then, for all points $q \in W^i$, we have [12]:

• In the $XY$-plane, $e$ is closer to $q$ than $e_1$. We show this by proving that $d_{W}(e, q) \le d_{XY}(e_1,q)$.

• In the $XY$-plane, $e$ is closer to $e_1$ than any other point $q$ on $W^i$. In other words, $d_{XY}(e,e_1) \le d_{XY}(q,e_1)$.