Distinguishing Views in Symmetric Networks: A Tight Lower Bound

The view of a node in a port-labeled network is an infinite tree encoding all walks in the network originating from this node. We prove that for any integers $n\geq D\geq 1$, there exists a port-labeled network with at most $n$ nodes and diameter at most $D$ which contains a pair of nodes whose (infinite) views are different, but whose views truncated to depth $\Omega(D\log (n/D))$ are identical

Deterministic Graph Exploration with Advice

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 $\log\log\log n -\Theta(1)$, for both types of oracles. When advice is large, there are two natural time thresholds: $\Theta(n^2)$ for a map oracle, and $\Theta(n)$ for an instance oracle, that can be achieved with sufficiently large advice. We show that, with a map oracle, time $\Theta(n^2)$ 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 $n^\delta$, for any $\delta <1/3$. 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 $n^\epsilon$, for any $\epsilon <2$, and with any advice of size $O(n)$, the time must be $\Omega(n^2)$. We also investigate minimum advice sufficient for fast exploration of hamiltonian graphs

Topology recognition with advice

In topology recognition, each node of an anonymous network has to deterministically produce an isomorphic copy of the underlying graph, with all ports correctly marked. This task is usually unfeasible without any a priori information. Such information can be provided to nodes as advice. An oracle knowing the network can give a (possibly different) string of bits to each node, and all nodes must reconstruct the network using this advice, after a given number of rounds of communication. During each round each node can exchange arbitrary messages with all its neighbors and perform arbitrary local computations. The time of completing topology recognition is the number of rounds it takes, and the size of advice is the maximum length of a string given to nodes. We investigate tradeoffs between the time in which topology recognition is accomplished and the minimum size of advice that has to be given to nodes. We provide upper and lower bounds on the minimum size of advice that is sufficient to perform topology recognition in a given time, in the class of all graphs of size $n$ and diameter $D\le \alpha n$, for any constant $\alpha< 1$. In most cases, our bounds are asymptotically tight