Leader election is one of the fundamental problems in distributed computing: a single node, called the leader, must be specified. This task can be formulated either in a weak way, where one node outputs 'leader' and all other nodes output 'non-leader', or in a strong way, where all nodes must also learn which node is the leader. If the nodes have distinct identifiers, then such an agreement means that all nodes have to output the identifier of the elected leader. For anonymous networks, the strong version of leader election requires that all nodes must be able to find a path to the leader, as this is the only way to identify it. In this paper, we study variants of deterministic leader election in arbitrary anonymous networks. Leader election is impossible in some anonymous networks, regardless of the allocated amount of time, even if nodes know the entire map of the network. This is due to possible symmetries in the network. However, even in networks in which it is possible to elect a leader knowing the map, the task may be still impossible without any initial knowledge, regardless of the allocated time. On the other hand, for any network in which leader election (weak or strong) is possible knowing the map, there is a minimum time, called the 'election index', in which this can be done. We consider four formulations of leader election discussed in the literature in the context of anonymous networks : one is the weak formulation, and the three others specify three different ways of finding the path to the leader in the strong formulation. Our aim is to compare the amount of initial information needed to accomplish each of these "four shades" of leader election in minimum time. Following the framework of 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 size of advice. We show that the size of advice required to accomplish leader election in the weak formulation in minimum time is exponentially smaller than that needed for any of the strong formulations. Thus, if the required amount of advice is used as a measure of the difficulty of the task, the weakest version of leader election in minimum time is drastically easier than any version of the strong formulation in minimum time.

Broadcast is one of the fundamental network communication primitives. One node of a network, called the source, has a message that has to be learned by all other nodes. We consider broadcast in radio networks, modeled as simple undirected connected graphs with a distinguished source. Nodes communicate in synchronous rounds. In each round, a node can either transmit a message to all its neighbours, or stay silent and listen. At the receiving end, a node v hears a message from a neighbour w in a given round if v listens in this round and if w is its only neighbour that transmits in this round. If more than one neighbour of a node v transmits in a given round, we say that a collision occurs at v. We do not assume collision detection: in case of a collision, node v does not hear anything (except the background noise that it also hears when no neighbour transmits). We are interested in the feasibility of deterministic broadcast in radio networks. If nodes of the network do not have any labels, deterministic broadcast is impossible even in the four-cycle. On the other hand, if all nodes have distinct labels, then broadcast can be carried out, e.g., in a round-robin fashion, and hence O(łog n)-bit labels are sufficient for this task in n-node networks. In fact, O(łog Δ)-bit labels, where Δ is the maximum degree, are enough to broadcast successfully. Hence, it is natural to ask if very short labels are sufficient for broadcast. Our main result is a positive answer to this question. We show that every radio network can be labeled using 2 bits in such a way that broadcast can be accomplished by some universal deterministic algorithm that does not know the network topology nor any bound on its size. Moreover, at the expense of an extra bit in the labels, we can get the following additional strong property of our algorithm: there exists a common round in which all nodes know that broadcast has been completed.

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,…, d −1. A mobile agent, initially situated at some starting node v , has to visit all nodes and stop. The 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 , to explore the graph in a given time, using a deterministic algorithm. Following the paradigm of algorithms with advice , 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. 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 −ϕ ( n ), where ϕ 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: Θ ( n 2 ) for a map oracle, and Θ ( 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 Θ ( n 2 ) cannot be improved in general, regardless of the size of advice. What is then the smallest advice to achieve time Θ ( n 2 ) with a map oracle? We show that this smallest size of advice is larger than n δ , for any δ < 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 ϵ , for any ϵ < 2, and with any advice of size O ( n ), the time must be Ω( 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 Ω( n 2 ), for some Hamiltonian graphs. However, 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 ϵ , for any ϵ < 1.