Abstract

In this paper we consider the group search problem, or evacu- ation problem, in which k mobile entities (\({\cal M}{\cal E}\)s) located on the line perform search for a specific destination. The \({\cal M}{\cal E}\)s are initially placed at the same origin on the line L and the target is located at an unknown distance d, either to the left or to the right from the origin. All \({\cal M}{\cal E}\)s must simultaneously occupy the destination, and the goal is to minimize the time necessary for this to happen. The problem with k = 1 is known as the cow-path problem, and the time required for this problem is known to be 9d − o(d) in the worst case (when the cow moves at unit speed); it is also known that this is the case for k ≥ 1 unit-speed \({\cal M}{\cal E}\)s. In this paper we present a clear argument for this claim by showing a rather counter-intuitive result. Namely, independent of the number of \({\cal M}{\cal E}\)s, group search cannot be performed faster than in time 9d − o(d). We also examine the case of k = 2 \({\cal M}{\cal E}\)s with different speeds, showing a surprising result that the bound of 9d can be achieved when one \({\cal M}{\cal E}\) has unit speed, and the other \({\cal M}{\cal E}\) moves with speed at least 1/3.

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