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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.

Leader election is one of the basic problems in distributed computing. 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 leader election is impossible in some networks, regardless of the allocated amount of time, even if nodes know the map of 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 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 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 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 algorithms with advice, this information is provided to all nodes at the start by an oracle knowing the entire network. The length of this string (its number of bits) is called the 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 and give tight (or almost tight) bounds on the minimum size of advice for these extremes. We also show that constant advice is not sufficient for leader election in all graphs, regardless of the allocated time.

We consider the problem of exploration of networks, some of whose edges are faulty. A mobile agent, situated at a starting node and unaware of which edges are faulty, has to explore the connected fault-free component of this node by visiting all of its nodes. The cost of the exploration is the number of edge traversals. For a given network and given starting node, the overhead of an exploration algorithm is the worst-case ratio (taken over all fault configurations) of its cost to the cost of an optimal algorithm which knows where faults are situated. An exploration algorithm, for a given network and given starting node, is called perfectly competitive if its overhead is the smallest among all exploration algorithms not knowing the location of faults. We design a perfectly competitive exploration algorithm for any ring, and show that, for networks modeled by hamiltonian graphs, the overhead of any DFS exploration is at most 10/9 times larger than that of a perfectly competitive algorithm. Moreover, for hamiltonian graphs of size at least 24, this overhead is less than 6% larger than that of a perfectly competitive algorithm., Comment: 17 pages, 1 figure

Topology recognition and leader election are fundamental tasks in distributed computing in networks. The first of them requires each node to find a labeled isomorphic copy of the network, while the result of the second one consists in a single node adopting the label 1 (leader), with all other nodes adopting the label 0 and learning a path to the leader. We consider both these problems in networks whose nodes are equipped with not necessarily distinct labels called colors, and ports at each node of degree $d$ are arbitrarily numbered $0,1,\dots, d-1$. Colored networks are generalizations both of labeled networks and anonymous networks. In colored networks, topology recognition and leader election are not always feasible. Hence we study two more general problems. The aim of the problem TOP (resp. LE), for a colored network and for input $I$ given to its nodes, is to solve topology recognition (resp. leader election) in this network, if this is possible under input $I$, and to have all nodes answer "unsolvable" otherwise. We show that nodes of a network can solve problems TOP and LE, if they are given, as input $I$, an upper bound $k$ on the number of nodes of a given color, called the size of this color. On the other hand we show that, if the nodes are given an input that does not bound the size of any color, then the answer to TOP and LE must be "unsolvable", even for the class of rings. Under the assumption that nodes are given an upper bound $k$ on the size of a given color, we study the time of solving problems TOP and LE in the $LOCAL$. We give an algorithm to solve each of these problems in arbitrary $n$-node networks of diameter $D$ in time $O(kD+D\log(n/D))$. We also show that this time is optimal, by exhibiting classes of networks in which every algorithm solving problems TOP or LE must use time $\Omega(kD+D\log(n/D))$.

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.

Two mobile agents, starting from different nodes of a network at possibly different times, have to meet at the same node. This problem is known as $\mathit{rendezvous}$. Agents move in synchronous rounds. Each agent has a distinct integer label from the set $\{1,\dots,L\}$. Two main efficiency measures of rendezvous are its $\mathit{time}$ (the number of rounds until the meeting) and its $\mathit{cost}$ (the total number of edge traversals). We investigate tradeoffs between these two measures. A natural benchmark for both time and cost of rendezvous in a network is the number of edge traversals needed for visiting all nodes of the network, called the exploration time. Hence we express the time and cost of rendezvous as functions of an upper bound $E$ on the time of exploration (where $E$ and a corresponding exploration procedure are known to both agents) and of the size $L$ of the label space. We present two natural rendezvous algorithms. Algorithm $\mathtt{Cheap}$ has cost $O(E)$ (and, in fact, a version of this algorithm for the model where the agents start simultaneously has cost exactly $E$) and time $O(EL)$. Algorithm $\mathtt{Fast}$ has both time and cost $O(E\log L)$. Our main contributions are lower bounds showing that, perhaps surprisingly, these two algorithms capture the tradeoffs between time and cost of rendezvous almost tightly. We show that any deterministic rendezvous algorithm of cost asymptotically $E$ (i.e., of cost $E+o(E)$) must have time $\Omega(EL)$. On the other hand, we show that any deterministic rendezvous algorithm with time complexity $O(E\log L)$ must have cost $\Omega (E\log L)$.

We study the time needed for deterministic leader election in the ${\cal LOCAL}$ model, where in every round a node can exchange any messages with its neighbors and perform any local computations. The topology of the network is unknown and nodes are unlabeled, but ports at each node have arbitrary fixed labelings which, together with the topology of the network, can create asymmetries to be exploited in leader election. We consider two versions of the leader election problem: strong LE in which exactly one leader has to be elected, if this is possible, while all nodes must terminate declaring that leader election is impossible otherwise, and weak LE, which differs from strong LE in that no requirement on the behavior of nodes is imposed, if leader election is impossible. We show that the time of leader election depends on three parameters of the network: its diameter $D$, its size $n$, and its level of symmetry $\lambda$, which, when leader election is feasible, is the smallest depth at which some node has a unique view of the network. It also depends on the knowledge by the nodes, or lack of it, of parameters $D$ and $n$.

In rendezvous, two agents traverse network edges in synchronous rounds and have to meet at some node. In treasure hunt, a single agent has to find a stationary target situated at an unknown node of the network. We study tradeoffs between the amount of information ($\mathit{advice}$) available $\mathit{a\ priori}$ to the agents and the cost (number of edge traversals) of rendezvous and treasure hunt. Our goal is to find the smallest size of advice which enables the agents to solve these tasks at some cost $C$ in a network with $e$ edges. This size turns out to depend on the initial distance $D$ and on the ratio $\frac{e}{C}$, which is the $\mathit{relative\ cost\ gain}$ due to advice. For arbitrary graphs, we give upper and lower bounds of $O(D\log(D\cdot \frac{e}{C}) +\log\log e)$ and $\Omega(D\log \frac{e}{C})$, respectively, on the optimal size of advice. For the class of trees, we give nearly tight upper and lower bounds of $O(D\log \frac{e}{C} + \log\log e)$ and $\Omega (D\log \frac{e}{C})$, respectively.

The leader election task calls for all nodes of a network to agree on a single node. If the nodes of the network are anonymous, the task of leader election is formulated as follows: every node $v$ of the network must output a simple path, 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 anonymous trees. Our aim is to establish tradeoffs between the allocated time $\tau$ and the amount of information that has to be given $\textit{a priori}$ to the nodes to enable leader election in time $\tau$ in all trees for which leader election in this time is at all possible. Following the framework of $\textit{algorithms with advice}$, this information (a single binary string) is provided to all nodes at the start by an oracle knowing the entire tree. The length of this string is called the $\textit{size of advice}$. For an allocated time $\tau$, we give upper and lower bounds on the minimum size of advice sufficient to perform leader election in time $\tau$. We consider $n$-node trees of diameter $diam \leq D$. While leader election in time $diam$ can be performed without any advice, for time $diam-1$ we give tight upper and lower bounds of $\Theta (\log D)$. For time $diam-2$ we give tight upper and lower bounds of $\Theta (\log D)$ for even values of $diam$, and tight upper and lower bounds of $\Theta (\log n)$ for odd values of $diam$. For the time interval $[\beta \cdot diam, diam-3]$ for constant $\beta >1/2$, we prove an upper bound of $O(\frac{n\log n}{D})$ and a lower bound of $\Omega(\frac{n}{D})$, the latter being valid whenever $diam$ is odd or when the time is at most $diam-4$. Finally, for time $\alpha \cdot diam$ for any constant $\alpha <1/2$ (except for the case of very small diameters), we give tight upper and lower bounds of $\Theta (n)$.

Two mobile agents, starting from different nodes of an unknown network, have to meet at the same node. Agents move in synchronous rounds using a deterministic algorithm. Each agent has a different label, which it can use in the execution of the algorithm, but it does not know the label of the other agent. Agents do not know any bound on the size of the network. In each round an agent decides if it remains idle or if it wants to move to one of the adjacent nodes. Agents are subject to delay faults: if an agent incurs a fault in a given round, it remains in the current node, regardless of its decision. If it planned to move and the fault happened, the agent is aware of it. We consider three scenarios of fault distribution: random (independently in each round and for each agent with constant probability 0 < p < 1), unbounded adver- sarial (the adversary can delay an agent for an arbitrary finite number of consecutive rounds) and bounded adversarial (the adversary can delay an agent for at most c consecutive rounds, where c is unknown to the agents). The quality measure of a rendezvous algorithm is its cost, which is the total number of edge traversals. For random faults, we show an algorithm with cost polynomial in the size n of the network and polylogarithmic in the larger label L, which achieves rendezvous with very high probability in arbitrary networks. By contrast, for unbounded adversarial faults we show that rendezvous is not feasible, even in the class of rings. Under this scenario we give a rendezvous algorithm with cost O(nl), where l is the smaller label, working in arbitrary trees, and we show that \Omega(l) is the lower bound on rendezvous cost, even for the two-node tree. For bounded adversarial faults, we give a rendezvous algorithm working for arbitrary networks, with cost polynomial in n, and logarithmic in the bound c and in the larger label L.

Two mobile agents, starting from different nodes of an $n$-node network at possibly different times, have to meet at the same node. This problem is known as rendezvous. Agents move in synchronous rounds using a deterministic algorithm. In each round, an agent decides to either remain idle or to move to one of the adjacent nodes. Each agent has a distinct integer label from the set $\{1,...,L\}$, which it can use in the execution of the algorithm, but it does not know the label of the other agent. If $D$ is the distance between the initial positions of the agents, then $\Omega(D)$ is an obvious lower bound on the time of rendezvous. However, if each agent has no initial knowledge other than its label, time $O(D)$ is usually impossible to achieve. We study the minimum amount of information that has to be available a priori to the agents to achieve rendezvous in optimal time $\Theta(D)$. This information is provided to the agents at the start by an oracle knowing the entire instance of the problem, i.e., the network, the starting positions of the agents, their wake-up rounds, and both of their labels. The oracle helps the agents by providing them with the same binary string called advice, which can be used by the agents during their navigation. The length of this string is called the size of advice. Our goal is to find the smallest size of advice which enables the agents to meet in time $\Theta(D)$. We show that this optimal size of advice is $\Theta(D\log(n/D)+\log\log L)$. The upper bound is proved by constructing an advice string of this size, and providing a natural rendezvous algorithm using this advice that works in time $\Theta(D)$ for all networks. The matching lower bound, which is the main contribution of this paper, is proved by exhibiting classes of networks for which it is impossible to achieve rendezvous in time $\Theta(D)$ with smaller advice.