带建议的确定性图探索
Deterministic Graph Exploration with Advice
摘要：
我们考虑图探索的任务。一个$n$ -节点图有未标记的节点，并且在度为$d$的任何节点上的所有端口都是任意编号的$0,\dots, d-1$ .移动代理必须访问所有节点并停止。探索时间是边缘遍历的次数。我们考虑的问题有多少知识的代理人有先验，为了在给定的时间内探索图，使用确定性算法。这种先验信息（建议）由一个oracle以二进制字符串的形式提供给代理，其长度称为建议的大小。我们考虑两种类型的神谕。实例oracle知道探索问题的整个实例，即，图的端口编号映射和此映射中代理的起始节点。映射预言机知道图的端口编号映射，但不知道代理的起始节点。我们首先考虑在多项式时间内进行探索，并确定实现它的建议的确切最小大小。对于两种类型的预言机，这个大小都是$\log\log\log n -Θ(1)$。当通知很大时，有两个自然的时间阈值：映射oracle的$Θ(n^2)$和实例oracle的$Θ(n)$，这可以通过足够大的通知来实现。我们表明，与地图的甲骨文，时间$Θ(n^2)$不能提高一般，无论大小的意见。我们还表明，实现这一时间的最小尺寸的建议是大于$n^δ$，任何$δ<1/3$。对于实例oracle，大小为$O(n\log n)$的通知足以实现时间$O(n)$。我们表明，与任何建议的大小$o(n\log n)$，探索的时间必须至少$n^ε$，对于任何$ε<2$，和任何建议的大小$O(n)$，时间必须是$Ω(n^2)$。我们还调查了最小的意见，足以快速探索汉密尔顿图。
英文摘要：
We consider the task of graph exploration. An $n$-node graph has unlabeled nodes, and all ports at any node of degree $d$ are arbitrarily numbered $0,\dots, d-1$. A mobile agent has to visit all nodes and stop. The exploration time is the number of edge traversals. We consider the problem of how much knowledge the agent has to have a priori, in order to explore the graph in a given time, using a deterministic algorithm. This a priori information (advice) is provided to the agent by an oracle, in the form of a binary string, whose length is called the size of advice. We consider two types of oracles. The instance oracle knows the entire instance of the exploration problem, i.e., the port-numbered map of the graph and the starting node of the agent in this map. The map oracle knows the port-numbered map of the graph but does not know the starting node of the agent. We first consider exploration in polynomial time, and determine the exact minimum size of advice to achieve it. This size is $\log\log\log n -Θ(1)$, for both types of oracles. When advice is large, there are two natural time thresholds: $Θ(n^2)$ for a map oracle, and $Θ(n)$ for an instance oracle, that can be achieved with sufficiently large advice. We show that, with a map oracle, time $Θ(n^2)$ cannot be improved in general, regardless of the size of advice. We also show that the smallest size of advice to achieve this time is larger than $n^δ$, for any $δ<1/3$. For an instance oracle, advice of size $O(n\log n)$ is enough to achieve time $O(n)$. We show that, with any advice of size $o(n\log n)$, the time of exploration must be at least $n^ε$, for any $ε<2$, and with any advice of size $O(n)$, the time must be $Ω(n^2)$. We also investigate minimum advice sufficient for fast exploration of hamiltonian graphs.
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