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,…,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 logloglogn−Θ(1),
for both types of oracles.
When advice is large, there are two natural time thresholds: Θ(n2)
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(n2) 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 nδ, for any δ<1/3.
For an instance oracle, advice of size O(nlogn) is enough to achieve time
O(n). We show that, with any advice of size o(nlogn), the time of
exploration must be at least nϵ, for any ϵ<2, and with any
advice of size O(n), the time must be Ω(n2).
We also investigate minimum advice sufficient for fast exploration of
hamiltonian graphs