We investigate autonomous mobile robots in the Euclidean plane. A robot has a function called target function to decide the destination from the robots' positions. Robots may have different target functions. If the robots whose target functions are chosen from a set $\Phi$ of target functions always solve a problem $\Pi$, we say that $\Phi$ is compatible with respect to $\Pi$. If $\Phi$ is compatible with respect to $\Pi$, every target function $\phi \in \Phi$ is an algorithm for $\Pi$. Even if both $\phi$ and $\phi'$ are algorithms for $\Pi$, $\{ \phi, \phi' \}$ may not be compatible with respect to $\Pi$. From the view point of compatibility, we investigate the convergence, the fault tolerant ($n,f$)-convergence (FC($f$)), the fault tolerant ($n,f$)-convergence to $f$ points (FC($f$)-PO), the fault tolerant ($n,f$)-convergence to a convex $f$-gon (FC($f$)-CP), and the gathering problems, assuming crash failures. Obtained results classify these problems into three groups: The convergence, FC(1), FC(1)-PO, and FC($f$)-CP compose the first group: Every set of target functions which always shrink the convex hull of a configuration is compatible. The second group is composed of the gathering and FC($f$)-PO for $f \geq 2$: No set of target functions which always shrink the convex hull of a configuration is compatible. The third group, FC($f$) for $f \geq 2$, is placed in between. Thus, FC(1) and FC(2), FC(1)-PO and FC(2)-PO, and FC(2) and FC(2)-PO are respectively in different groups, despite that FC(1) and FC(1)-PO are in the first group.
翻译:我们研究了欧式平面中的自主移动机器人。机器人有一个名为目标函数的函数,用于确定机器人从其位置到达的目的地。机器人可能具有不同的目标函数。如果从目标函数集合Φ中选择的机器人总是解决问题Π,则我们称Φ与Π的兼容性。如果Φ与Π兼容,则每个目标函数Φ∈对Π都是算法。即使Φ和Φ'都是Π的算法,{Φ,Φ'}也可能与Π不兼容。从兼容性的角度出发,我们研究了自主移动机器人的收敛、容错(n,f)收敛(FC(f))、容错(n,f)收敛到f个点的(FC(f)-PO)、容错(n,f)收敛到凸f-边形的(FC(f)-CP)和聚集问题,假设有崩溃故障。所得结果将这些问题归类为三组:收敛、FC(1)、FC(1)-PO和FC(f)-CP组成第一组:每个总是缩小配置的凸包的目标函数集合都是兼容的。第二个组由聚集和$FC(f)-PO$组成。对于$f\ge2$,始终缩小配置的凸包的目标函数集合都不兼容。第三组,$FC(f)$对于$f\ge2$,被放置在第一组和第二组之间。因此,$FC(1)$和$FC(2)$,$FC(1)$ -PO和$FC(2)$ -PO,以及$FC(2)$和$FC(2)$ -PO分别位于不同的组中,尽管$FC(1)$和$FC(1)$ -PO位于第一组中。