A. Pelc, R.N. Yadav, Using time to break symmetry: universal deterministic anonymous rendezvous, Proc. 31st ACM Symposium on Parallelism in Algorithms and Architectures (SPAA 2019), June 2019, Phoenix, AZ, USA, 85-92.
A. Pelc, R.N. Yadav, Using time to break symmetry: universal deterministic anonymous rendezvous, Proc. 31st ACM Symposium on Parallelism in Algorithms and Architectures (SPAA 2019), June 2019, Phoenix, AZ, USA, 85-92.
Abstract:
Two anonymous mobile agents navigate synchronously in an anonymous graph and have to meet at a node, using a deterministic algorithm. This is a symmetry breaking task called rendezvous, equivalent to the fundamental task of leader election between the agents. When is this feasible in a completely anonymous environment? It is known that agents can always meet if their initial positions are nonsymmetric, and that if they are symmetric and agents start simultaneously then rendezvous is impossible. What happens for symmetric initial positions with non-simultaneous start? Can symmetry between the agents be broken by the delay between their starting times? In order to answer these questions, we consider {\em space-time initial configurations} (abbreviated by STIC). A STIC is formalized as $[(u,v),\delta]$, where $u$ and $v$ are initial nodes of the agents in some graph and $\delta$ is a non-negative integer that represents the difference between their starting times. A STIC is {\em feasible} if there exists a deterministic algorithm, even dedicated to this particular STIC, which accomplishes rendezvous for it. Our main result is a characterization of all feasible STICs and the design of a universal deterministic algorithm that accomplishes rendezvous for all of them without {\em any } a priori knowledge of the agents. Thus, as far as feasibility is concerned, we completely solve the problem of symmetry breaking between two anonymous agents in anonymous graphs. Moreover, we show that such a universal algorithm cannot work for all feasible STICs in time polynomial in the initial distance between the agents.
B. Gorain, A. Pelc, Deterministic graph exploration with advice, ACM Transactions on Algorithms 15 (2019), 8:1-8:17
Preliminary version appeared in:
Proc. 44th International Colloquium on Automata, Languages and Programming (ICALP 2017), July 2017, Warsaw, Poland, 132:1-132:14.
Abstract:
We consider the fundamental 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, initially situated at some starting node $v$, has to visit all nodes and stop. The {\em time} of the exploration 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. Following the paradigm of {\em algorithms with advice}, this a priori information (advice) is provided to the agent by an {\em oracle}, in the form of a binary string, whose length is called the {\em size of advice}. We consider two types of oracles. The {\em 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 {\em map oracle} knows the port-numbered map of the graph but does not know the starting node of the agent. What is the minimum size of advice that must be given to the agent by each of these oracles, so that the agent explores the graph in a given time? We first determine the minimum size of advice to achieve exploration in polynomial time. We prove that some advice of size $\log\log\log n -c$, for any constant $c$, is sufficient for polynomial exploration, and that no advice of size $\log\log\log n -\phi(n)$, where $\phi$ is any function diverging to infinity, can help to do this. These results hold both for the instance and for the map oracles. On the other side of the spectrum, when advice is large, there are two natural time thresholds: $\Theta(n^2)$ for a map oracle, and $\Theta(n)$ for an instance oracle. This is because, in both cases, these time benchmarks can be achieved with sufficiently large advice (advice of size $O(n\log n)$ suffices). We show that, with a map oracle, time $\Theta(n^2)$ cannot be improved in general, regardless of the size of advice. What is then the smallest advice to achieve time $\Theta(n^2)$ with a map oracle? We show that this smallest size of advice is larger than $n^\delta$, for any $\delta <1/3$. For large advice, the situation changes significantly when we allow an instance oracle instead of a map oracle. In this case, advice of size $O(n\log n)$ is enough to achieve time $O(n)$. Is such a large advice needed to achieve linear time? We answer this question affirmatively. Indeed, we show more: 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 finally look at Hamiltonian graphs, as for them it is possible to achieve the absolutely optimal exploration time $n-1$, when sufficiently large advice (of size $O(n\log n)$) is given by an instance oracle. We show that a map oracle cannot achieve this: regardless of the size of advice, the time of exploration must be $\Omega(n^2)$, for some Hamiltonian graphs. On the other hand, even for the instance oracle, with advice of size $o(n\log n)$, optimal time $n-1$ cannot be achieved: indeed, we show that the time of exploration with such advice must sometimes exceed the optimal time $n-1$ by a summand $n^\epsilon$, for any $\epsilon <1$.
Y. Dieudonné, A. Pelc, Impact of knowledge on election time in anonymous networks, Algorithmica 81 (2019), 238-288.
Preliminary version appeared in:
Proc. 29th ACM Symposium on Parallelism in Algorithms and Architectures (SPAA 2017), July 2017, Washington, DC, USA, 207-215.
Abstract:
Leader election is one of the basic problems in distributed computing. This is a symmetry breaking problem: all nodes of a network must agree on a single node, called the leader. If the nodes of the network have distinct labels, then such an agreement means that all nodes have to output the label of the elected leader. For anonymous networks, the task of leader election is formulated as follows: every node $v$ of the network must output a simple path, which is coded as a sequence of port numbers, such that all these paths end at a common node, the leader. In this paper, we study deterministic leader election in arbitrary anonymous networks. It is well known that deterministic leader election is impossible in some networks, regardless of the allocated amount of time, even if nodes know the map of the network. This is due to possible symmetries in it. However, even in networks in which it is possible to elect a leader knowing the map, the task may be still impossible without any knowledge, regardless of the allocated time. On the other hand, for any network in which leader election is possible knowing the map, there is a minimum time, called the {\em election index}, in which this can be done. Informally, the election index of a network is the minimum depth at which views of all nodes are distinct. Our aim is to establish tradeoffs between the allocated time $\tau$ and the amount of information that has to be given {\em a priori} to the nodes to enable leader election in time $\tau$ in all networks for which leader election in this time is at all possible. Following the framework of {\em algorithms with advice}, this information (a single binary string) is provided to all nodes at the start by an oracle knowing the entire network. The length of this string is called the {\em size of advice}. For a given time $\tau$ allocated to leader election, we give upper and lower bounds on the minimum size of advice sufficient to perform leader election in time $\tau$. We focus on the two sides of the time spectrum. For the smallest possible time, which is the election index of the network, we show that the minimum size of advice is linear in the size $n$ of the network, up to polylogarithmic factors. On the other hand, we consider large values of time: larger than the diameter $D$ by a summand, respectively, linear, polynomial, and exponential in the election index; for these values, we prove tight bounds on the minimum size of advice, up to multiplicative constants. We also show that constant advice is not sufficient for leader election in all graphs, regardless of the allocated time.