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Recently, there has been a growing interest in grid exploration by agents with limited capabilities. We show that the grid cannot be explored by three semi-synchronous finite automata, answering an open question by Emek et al. (Theor Comput Sci 608:255–267, 2015) in the negative. In the setting we consider, time is divided into discrete steps, where in each step, an adversarially selected subset of the agents executes one look–compute–move cycle. The agents operate according to a shared finite automaton, where every agent is allowed to have a distinct initial state. The only means of communication is to sense the states of the agents sharing the same grid cell. The agents are equipped with a global compass and whenever an agent moves, the destination cell of the movement is chosen by the agent’s automaton from the set of neighboring grid cells. In contrast to the four agent protocol by Emek et al., we show that three agents do not suffice for grid exploration. Keywords Finite automata · Graph exploration · Mobile robots 1 Introduction Semi-Synchrony Recently, there has been a growing interest in studying constant memory agents performing exploration Consider the problem of exploring an infinite grid with a set on an infinite grid. An infinite grid is a natural discrete version of mobile robots, ants, or agents. In practical applications, it is of a plane which disallows the bounded memory agents to often desirable to make use of inexpensive and simple devices make any use of the boundaries of the grid. Emek et al. [16] and therefore, a finite automaton is an attractive choice for introduced a model where the agents are able to communicate modeling these agents. Furthermore, neither reliable commu- by sensing each other’s states and showed a tight upper bound nication nor synchronous time is always available and thus, for the time needed for k agents to find a treasure at distance distributed and non-synchronous solutions are needed. Also D. As the first step into the model, let us introduce the way exploration models inspired by biology require these fea- that the semi-synchrony is defined. The time is divided into tures; for example models for ant foraging assume limited discrete time steps, and in each time step, an adversarially capabilities and distributed searching. In both settings men- chosen subset of the agents performs a look–compute–move tioned above, it is often reasonable to assume simple means cycle in parallel. In each cycle, the chosen agents first sense of communication of nearby agents. the states of all the other agents in the same cell and then, determined by their transition function, either stay still or Most of the research was done affiliated with ETH Zürich and move to an adjacent grid cell. We point out that in every University of Freiburg and supported by ERC Grant No. 336495 step, every agent performs the “look” action before any agent (ACDC). executes their “compute” step, i.e., agents sharing a cell and B Jara Uitto activated in the same time step see each other’s states before jara.uitto@aalto.fi any of them executes a state transition. This definition allows Sebastian Brandt an arbitrary discrepancy in the number of steps the agents brandts@ethz.ch perform but ensures that, whenever two agents meet, at least Roger Wattenhofer wattenhofer@ethz.ch In the deterministic case, exploring the grid and finding a treasure are equivalent. In the randomized case, considering a treasure is more con- ETH Zürich, Zürich, Switzerland venient as the exploration is equivalent to hitting every cell in expected finite time. Aalto University, Konemiehentie 2, 02150 Espoo, Finland 123 S. Brandt et al. one of them will be able to sense the presence of the other visit every node and/or edge of the graph by moving along agent. the edges. There is a wide selection of variants of graph All input parameters, such as D and k are unknown to exploration and one of the standard ways to classify these the agents and they are all initially located in the origin of variants is to divide them into directed and undirected vari- the grid. Motivated by the fact that ants are able to perform ants [1,11]. In the directed model, the edges of the graph only very precise path integration, it is assumed that the agents allow traversing into one direction, whereas in the undirected are endowed with a global compass. model, traversing both ways is allowed. Our work assumes Previous Results Following up on the above model, Emek the undirected graph exploration model. et al. [15] studied the minimum number of agents needed to Other typical parameters of the problem are the condi- explore the infinite grid, where exploring refers to reaching tions of a successful exploration and symmetry breaking any fixed cell within (expected) finite time. They showed that mechanisms. Some related works demand that the agents three randomized and four deterministic semi-synchronous are required to halt after a successful exploration [12]or agents are enough for the exploration task. We want to point that the agents must return to their starting point after the out that the asynchronous environment in their paper is exploration [3]. From the perspective of symmetry breaking, referred to as semi-synchronous in older literature [24,25]. one characterization is to break the problem into the case of The paper left two open questions: equipping nodes with unique identifiers [14,22] and into the case where nodes are anonymous [5,8,23]. Since the memory Can two agents controlled by a randomized FA solve of our agents is restricted to a constant amount of bits with the synchronous or asynchronous version of the ANTS respect to the size of the graph, the unique identifiers are not problem? helpful. Is there an effective FA-protocol for async-ANTS for The agents typically operate in look–compute–move cycles, three agents when no random bits are available? where they first gather the local information, then per- form local computations, and finally, decide to which node Very recently, Cohen et al. [10] solved the first question they move. This execution model can be divided into syn- by showing that two randomized agents do not suffice. The chronous [25], semi-synchronous [24,25] and asynchronous main result of this paper is a negative answer to the second variants [18,26], referred to as FSYN C, SSYN C, and question: ASYN C.Inthe FSYN C model, all agents execute their Theorem 1 Three semi-synchronous agents controlled by a cycles simultaneously in discrete rounds. In the SSYN C model only a subset (not necessarily proper) of the agents finite automaton are not sufficient to explore the infinite grid. is activated in every round and in the ASYN C model, the cycles are not assumed to be atomic. To avoid confusion, we Our result is obtained by solving two technical challenges. refer to the non-synchronous rounds as time steps. In this First, we carefully design an adversarial schedule for the paper, we consider the semi-synchronous model. Note that agents that, under the assumption that the agents actually since the ASYN C model is weaker than the SSYN C model, explore the entire grid, forces them to obey a movement pat- we directly obtain our lower bound result for the ASYN C tern with the following property: There is a fixed width w model as well. and fixed slope s such that at any point in time, all agents The standard efficiency measure of a graph exploration are contained in a band of width w and slope s. Second, algorithm executed in the FSYN C model is the number of we formally show that the agents cannot encode a super- synchronous rounds it takes until the graph is explored [22]. constant amount of information in their relative positions. In In the non-synchronous models, this measure is typically other words, while the relative distance can be unbounded generalized to the maximum delay between activation times and represent an unbounded amount of information, we can of any agent [9]. A widely-studied classic is the cow-path bound the amount of information the agents can infer from problem, where the goal of the cow is to find food or a trea- their relative positions. sure on a line as fast as possible. There is an algorithm with a constant competitive ratio for the case of a line and in the case of a grid, a simple spiral search is optimal and the prob- 2 Related work lem has been generalized to the case of many cows [4,21]. Some more recent work studied the time complexity of n Graph exploration is a widely studied problem in the com- distributed agents searching for a treasure at distance D on puter science literature. In the typical setting one or more agridand a Θ(D/n + D) bound was shown in the case of agents are placed on some node of a graph and the goal is to Turing machines without communication and in the case of communicating finite automata [16,17]. The ANTS problem in their context is the same as our grid exploration problem. 123 A tight lower bound for semi-synchronous collaborative grid exploration Our work does not focus on the time complexity of the In the beginning, all agents are placed in the same cell, problem, but rather on the computability, i.e, what is the min- called the origin. W.l.o.g., we will assume that the origin has imum number of agents that are required to find the treasure. the coordinates (0, 0). For the agents, all cells, including the The canonical algorithm in the case of little memory is the origin, are indistinguishable; in particular, they do not have random walk, where the classic result states that a random access to the coordinates of the cells. walk explores an n-node graph in polynomial time [2]. In Each agent is endowed with a compass, i.e., each agent the case of infinite grids, it was shown in a recent paper that, is able to distinguish between the four (globally consistent) even with a globally consistent orientation, two randomized cardinal directions in any cell and all agents have the same agents cannot locate the treasure in finite expected time [10]. notion of those directions. The behavior of each agent is By combining this result with previous work [7,15], it follows governed by a deterministic finite automaton. While we allow that this lower bound is tight. In the deterministic case, our the agents to use different finite automata, we will assume that lower bound of three deterministic semi-synchronous agents the agents use the same finite automaton but have different closes the remaining gap in the results of [15]. initial states. Since in all cases we consider, n is a constant, Another typical measure for efficiency is the number of the two formulations are equivalent. bits of memory needed per agent [12,19]. For example, it The only way in which communication takes place is the was shown by Fraigniaud et al., that Θ(D log Δ) bits are following: Each agent senses for any state q of the finite needed for a single agent to locate the treasure, where D automaton whether there is at least one other agent in the and Δ denote the diameter and the maximum degree of the same cell in state q. In each step of the execution, an agent graph, respectively. The memory of our agents is bounded by moves to an adjacent cell or stays in the current cell, solely a universal constant, independent of any graph parameters. based on its current state in the finite automaton and the subset Work that falls close to our work is the study of graph of states q for which another agent in state q is present in the exploration in labyrinths, i.e., graphs that can be seen as 2- current cell. dimensional grids, where some subset of the nodes cannot Given the above, we are set to describe our finite automa- be entered by the agents. The classic results state that all ton more formally. Let Q denote the set of states, with each co-finite (finite amount of cells not blocked) labyrinths can agent having its own initial state in Q. The set of input sym- be explored by two finite automata and an automaton with bolsis2 , the set of all subsets of Q, reflecting the fact two pebbles [6], and that finite labyrinths (finite amount of that for each state from Q an agent in this state might be cells are blocked) can be explored using one agent with four present or not in the considered cell. The transition func- pebbles [7], where a pebble is a movable marker. Further- tion δ : Q × 2 → Q ×{0, 1, 2, 3, 4} provides an agent more, it is known since long that there are finite and co-finite in state q ∈ Q (sensing a subset Q ⊆ Q of states present labyrinths where one pebble is not enough [20] and that no in the same cell) with a new state q ∈ Q and a movement, finite set of finite automata can explore all planar graphs [23]. where 1, 2, 3, 4 stand for the four cardinal directions while More recently, it was shown that Θ(log log n) pebbles for 0 indicates that the agent stays in the current cell. an agent with Θ(log log n) memory is the right answer for The SSYN C [24,25] environment in which the agents general graphs [13]. Notice that since we do not assume syn- perform their exploration is semi-synchronous. More specif- chronous communication between agents and a pebble can ically, we assume that the order of the steps of the agents always be simulated by a finite automaton, our result also is determined by an adversarial scheduler that knows the yields the same bound for the pebble model. finite automaton governing the agents’ behavior. Each step of an agent is a complete look–compute–move cycle, where first an agent senses for which states agents are present in the current cell, then it applies the transition function with the 3 Preliminaries sensed states and its own current state as input, and finally it moves as indicated by the result. Cycles of different agents The model may occur at the same time, in which case each of the agents completes the sensing before any of the agents starts to move. The model we use is the same as in [15]. We consider a group Cycles that do not occur at the same time have no overlap, of n agents whose task is to explore every cell of the infinite i.e., the movement performed in an earlier cycle is completed 2-dimensional grid where a cell is considered as explored before the sensing in a later cycle starts. Hence, we may con- sider the order of the individual components of the execution when it has been visited by at least one of the agents. We identify each cell of the grid with a pair of integers, i.e., the as given by a mapping of the agents’ steps to points in time. We call such a mapping a schedule. Since the look– grid can be considered as Z , with two cells being neighbors if and only if they differ in one coordinate by exactly 0 and compute–move cycles of the agents are atomic in nature, we in the other coordinate by exactly 1. can assume w.l.o.g. that the static configurations of the agents 123 S. Brandt et al. on the grid (including the information about the states they (from time t to time t ). Moreover, we call the time difference are currently in) occur at integer points in time t = 0, 1,... , t − t the travel period. and that the steps of the agents determining the transition Note that travel vector and travel period do not depend on from one configuration to a new one take place between these the choice of t and t (provided t and t satisfy the properties points in time. If an agent’s action is scheduled between time mentioned above). In the case of multiple agents, we use the t and t + 1, we say, for the sake of simplicity, that the action same definitions for any time segment where only a single takes place at time t. In order to prevent the adversary from agent is scheduled and does not encounter another agent. In delaying a single agent indefinitely, we adopt the common particular, we can only speak of a travel vector and a travel requirement that each agent is scheduled infinitely often. For period when there are two points in time (in the considered our lower bound we will only use adversarial schedules where time segment) where the scheduled agent repeats a state and no two agents are scheduled at the same time. at both times as well as in the time between, the agent is alone in its cell. Definitions and notation For the notion of distance between two cells we will use the 4 Techniques Manhattan distance. Let c = (x , y), c = (x , y ) be two cells of the infinite grid. Then, the distance between c and c In order to show our main result, we use a (large) proof by is defined as Dist(c, c ) =|x − x |+|y − y |. Moreover, we contradiction. In the following we give a (very informal and call the first coordinate of a cell the x-coordinate and the sec- possibly slightly inaccurate) high-level overview of how it ond coordinate the y-coordinate. We denote the cell an agent proceeds. Our assumption, that holds throughout the remain- a occupies at time t by c (a) = (x (a), y (a)). Similarly, we der of the paper, is that three agents actually suffice to explore t t t denote the state of the finite automaton in which agent a is at the grid. From this assumption, we derive a contradiction as time t by q (a).If a = a for some 1 ≤ i ≤ 3, then we also follows: t i i i i i write c , x , y , q instead of c (a ), x (a ), y (a ), q (a ), First, we fix an adversarial schedule for the three agents t i t i t i t i t t t t respectively. Moreover, we denote the number of states of the that has certain advantageous properties. (We will show that finite automaton governing the behavior of the three agents it is already possible to derive a contradiction for this specific by N . schedule.) Then, using the finiteness of the number of config- In our lower bound proof, we show for each finite automa- urations of agents in any bounded area, we show that for each ton that three agents governed by this automaton are not distance D there is a point in time such that from this time sufficient to explore the grid (or, more precisely, that there is onwards, there are always at least two agents that have dis- an adversarial schedule for this automaton under which the tance at least D. However, since we can prove that any two agents do not explore every cell of the grid). In this context, agents must meet infinitely often, there must be infinitely we consider the number N as a constant, which also implies many travels between the two far-away agents (which are that the result of applying any fixed function to N is a con- not always the same agents). We show that the vector along stant as well. For the proof of our lower bound we require which such a travel takes place must have a fixed slope that is another intuitive definition. Let be an infinite line in the the same for all such travel vectors (from a sufficiently large Euclidean plane and d some positive real number. Let B be point in time on). Otherwise, there would exist two subse- the set of all points in the plane with integer coordinates and quent travels forth and back of different slope, which would Euclidean distance at most d to .Let B be the set of all imply that the traveling agent on its way back would miss the grid cells that have the same coordinates as some point in B. agent it is supposed to meet (which is the agent from whose Then we call B a band. position the first of the two travels started, roughly speaking). This also holds if the traveling agent explores some area to A single agent the left and right of its travel direction (during its travel), since the distance D between the two endpoints can be made Consider a single agent a moving on the grid. Since the arbitrarily large. number of states of its finite automaton is finite, a must The crucial part of the proof is to show that the state of the repeat a state at some point, i.e., there must be points in traveling agent at the end of its travel does not depend on the time t , t such that q (a) = q (a) and q (a) = q (a) for exact vector between the start and the endpoint of its travel, t t t t all t < t < t .Asshown in [15], agent a will then, start- but only on this vector “modulo” some other vector v that ing at time t , repeat the exact behavior it showed starting at is obtained by combining all of the finitely many possible time t regarding both movement on the grid and updating of traveling vectors of the aforementioned fixed slope. Prov- its state. We call the 2-dimensional vector c (a) − c (a) = ing this statement enables us to show that, at the start of a t t (x (a) − x (a), y (a) − y (a)) the travel vector of agent a travel, the information (1) about the states and relative loca- t t t t 123 A tight lower bound for semi-synchronous collaborative grid exploration tions “modulo v” of the agents, and (2) about which agent would repeat the exact behavior between the first and the is scheduled next and which is the traveling agent, are suf- second assumption of q infinitely often afterwards, thus ficient to determine the same information at the start of the iterating through the exact same movement on and on. next travel. Since there are only finitely many of these infor- 3. If none of the two above cases occurs, i.e., a would move mation tuples (exactly because they contain only the modulo on indefinitely without meeting any other agent or being version of the relative locations), at some point a tuple has in the same state in the same cell as before, then we to occur again. Hence, in a sense, the whole configuration schedule as follows: Let (x , y) be the travel vector of a ’s consisting of the three agents repeats its previous movement movement, and p the travel period. Then the subschedule i i from this point on, at least if one ignores any movement in the S ends at the first time t (strictly after the start of S ) j j direction of the fixed slope. Thus, in each repetition between for which the following property is satisfied: two occurrences of the information tuple, the whole config- At time t, agent a is in a state that it already assumed uration moves by some fixed (and always the same) vector, before in subschedule S (i.e., a is guaranteed to repeat r r which implies that the agents explore “at most half” of the its behavior from now on), and for each cell (x , y ) t t i r grid. occupied by an agent a , r = i, we have that 1) x −x > t t i r i r p if x > 0, and x −x < −p if x < 0, and 2) y −y > p t t t t i r if y > 0, and y − y < −p if y < 0. The definition t t of the travel vector ensures that there is such a (finite) 5 The schedule point in time t. Note that Case 3 can only occur if x = 0 or y = 0. Moreover, if this case actually occurs, then From this section on, we assume that three semi-synchronous the complete subsequent schedule is adapted according agents whose behavior is governed by a finite automaton suf- to the following special rule (overriding all of the above): fice to explore the grid. Let a , a and a be these agents. We 1 2 3 After time t, the two agents a , r = i , are scheduled for start our proof by contradiction by specifying a schedule that we assume to be the adversarial schedule for the remainder one time step each (in arbitrary order), then agent a is scheduled for p time steps, i.e., exactly one travel period, of this paper: We first schedule agent a for some number of time steps, and then we iterate this new scheduling. then agent a , then a , and then we iterate, again starting with 2 3 a . The number of steps an agent is scheduled can vary. In 1 Observe that according to this schedule, the number of other words, we can describe our schedule as a sequence time steps a scheduled agent can stay put in a cell during one of its subschedules is upper bounded by N . Also note that in 1 2 3 1 2 3 1 S = S , S , S , S , S , S , S ,... each of the three cases, the number of steps in the subschedule 1 1 1 2 2 2 3 is positive (and finite). For an illustration of Cases 2 and 3, i see Fig. 1. We now collect a few lemmas that highlight certain of subschedules where in each subschedule S only agent a properties of the three cases. is scheduled. The number of time steps in a subschedule S is determined as follows: Lemma 1 Case 3 cannot occur. Proof Recall that we assume (globally) that the three agents 1. If there is a (finite) number u > 0 of time steps after explore the entire infinite grid. Assume that Case 3 occurs which agent a is in a cell occupied by another agent, and let a denote the agent that would move on indefinitely then the subschedule S ends after u time steps where min without meeting another agent. Then, at the beginning of the u denotes the smallest such u. min first iteration according to the special rule, the distance of 2. If Case 1 does not apply, but there is a (finite) number agent a to any of the other agents is more than p in at least u > 0 of time steps after which a is in the same state one (of x- and y-) direction and a moves away from the in the same cell as it was at some earlier point in time i agents according to the travel vector. After each of the other during S , then do the following: agents makes a step, this distance is still at least p. Hence, Fix a total order on the state space of a ’s finite automaton. agent a cannot encounter one of the other agents during its (This total order can be chosen arbitrarily, but in each next p steps, since in total it moves away from the other application of Case 2 for agent a the same order has to agents, according to the specification of Case 3. be used.) Let q be the smallest state according to this order which a assumes at least twice in the same cell (if we If x = y = 0, agent a stays within a constant distance from the scheduled a indefinitely). Then S ends after the smallest cell where the subschedule started. Hence, if Case 1 does not occur, positive number of steps after which a is in state q and in there must be a state/cell combination that is assumed twice (due to the a cell where a would assume q at least twice. Note that number of cells within this constant distance being finite), implying that the property that a would assume q twice implies that it Case 2 must occur. 123 S. Brandt et al. vector (from time t onwards) which would imply that S is not of type 2. This implies that if a ’s subschedule would also continue at and after time t + N + 1 on an empty grid, then a would cycle through the same movement on and on, starting from time t . Hence, if there is a cell c that is visited by a in some q p steps state q in the (continued) movement after time t , then there must also be a point in time before t (during S ) at which q p steps a visits c in state q. It follows from the definition of our schedule that S ends before time t , yielding a contradiction p steps to our assumption. Lemma 3 Any subschedule S of type 1, where agent a ends Fig. 1 In the upper figure, Case 2 of our schedule is shown. Note that in the same cell from which it started, consists of at most the agent already stops when it visits the cell on the right (in state q) N (2N + 1) time steps. More generally, any subschedule S for the first time (unless this happens after 0 time steps). As the agent chooses the repeated state/cell pair in which it ends its subschedule of type 1, where a ends in a cell of distance at most D from the according to some arbitrary ordering of the repeated pairs, it will not cell from which it started, consists of at most N (2N + 1 + D) necessarily stop at the start of the loop. In the lower figure, we see Case 3 time steps. of our schedule. One agent would move arbitrarily far away if scheduled sufficiently long. By letting this agent move away far enough and then Proof We start by proving the special case where a ends scheduling it sufficiently often for a long enough period of time, we in the same cell from which it started. Suppose for a con- make sure that it will not interact anymore with any of the other two tradiction that there is a subschedule S as described in the agents lemma that consists of more than N (2N + 1) time steps. Let t and u denote the points in time when S starts and ends, The direction of the travel vector also ensures that the respectively. Since a does not encounter any other agent distance to the other agents is again increased to more than between time t and time u, it behaves like a single agent on p (in at least one direction). Thus, the same arguments hold an empty grid between t and u. In particular, there is a travel for the next iteration, and we obtain by induction that agent vector (x , y) of agent a from time t + 1 to time u − 1 since a will never encounter another agent after the occurrence of N (2N + 1) − 1 > N . Case 3. It follows that, if three agents suffice to explore the For reasons of symmetry, we can assume w.l.o.g. that x > grid, then also a team of two agents and a separate single agent 0 and y ≥ 0. Note that x = 0 = y is not possible since in can explore the grid without any communication between the that case a would cycle through the same (cyclic) movement team and the single agent. From [15], we know that this is over and over without meeting any other agent, which would not possible since a team of two agents (hence, also a single i imply that S is not of type 1. Let p be the travel period agent) can only explore a band of constant width. which, according to its definition, is at most N.Let q be the state whose second occurrence during S (excluding the Following Lemma 1, we will assume in the following occurrence of the state at the beginning of S ) comes earliest. that Case 3 does not occur, i.e., each agent’s subschedule j Let t be the time when q occurs for the first time. Since ends because it encounters another agent or because it enters i i t ≤ t + N , we know that x ≥ x − N . a pair state/cell that it would assume more than once (if t Now, as in each travel period a increases the x-coordinate scheduled indefinitely). This allows us to group the possi- of the cell it occupies by at least 1, it follows that at time ble subschedules of an agent into two categories: We say that i i t + 2N · p the x-coordinate of the cell a occupies is at least a subschedule S is of type 1 if S ends because of the con- j j x + N . Furthermore, since in each further travel period agent dition given in Case 1, and of type 2 if S ends because of a would advance by at least one cell in (positive) x-direction the condition given in Case 2. in total and p ≤ N , after time t + 2N · p agent a will never have an x-coordinate of less than x +1, i.e., it will never reach Lemma 2 Any subschedule of type 2 consists of at most N t i i i c then. But a also cannot have visited c (= c ) between time time steps. t t u t + 1 and t + 2N · p since t + 2N · p ≤ t + N (2N + 1) and Proof Assume for a contradiction that there is a subschedule we assumed that S consists of more than N (2N + 1) time S of type 2 that consists of at least N + 1 time steps and steps. Thus, we obtain a contradiction, which proves the first starts at some time t. Then, by the pigeonhole principle, there lemma statement. must be two points in time t < t < t ≤ t + N + 1 such that For the more general second statement, by an analogous i i i i q = q . Moreover, it must also hold that c = c since proof we obtain that after time t + 2N · p + D · p agent a t t t t otherwise a would move according to some non-zero travel will never have an x-coordinate of less than x + 1 + D, i.e., 123 A tight lower bound for semi-synchronous collaborative grid exploration it will never reach c then. But, since t + 2N · p + D · p ≤ t + N (2N + 1 + D), a also cannot have visited c between time t +1 and t +2N · p + D · p, under the assumption that S consists of more than N (2N + 1 + D) time steps. Hence, this assumption must be false, and the lemma statement follows. Finally, we show that the number of consecutive steps an agent is scheduled for (and hence also which steps the agent takes) from a given snapshot of the system onwards does not depend on the execution history. Lemma 4 Let t < t < t , and assume that some subsched- Fig. 2 An example showing a possible movement (red) of an agent ule S of agent a begins at time t and ends at time t . Then i whose travel vector is given by the black arrows. The agent performs the total movement given by the travel vector in at most N time steps, the number t −t of steps agent a is scheduled for from time or more precisely, during one travel period t onwards is exactly the number of steps agent a would be scheduled for if its subschedule started at time t (given the same location of all three agents as in the former scenario at they showed starting at time t, by Lemma 4. Hence, at time time t ). In particular, the definition of our schedule ensures t +(t −t ) the agents will again be, with (possibly) translated that the location of the agents combined with the information absolute coordinates, in the exact same configuration and so which agent is scheduled next uniquely determines all further on. moves taken by the agents. i i i i Define (x , y) = (x − x , y − y ), where i = 1(which t t t t implies that this equation also holds for i = 2, 3). Vector Proof This follows directly from Lemma 1, the stopping con- (x , y) describes the total movement of each of the agents ditions of Cases 1 and 2, and the fact that, at any point in time, during each of the (repeating) time periods of length t − t. the decision an agent takes (if it is scheduled) does not depend It follows that each cell that has not been explored by time on the execution history. t must be at distance at most t − t from some cell that is obtained by adding a multiple of the vector (x , y) to one cell 1 2 3 from {c , c , c }; otherwise it will never be explored. Since t t t each such cell at distance at most t − t (which is constant) 6 Traveling and meeting must lie in a band of constant width and “direction” (x , y) that 1 2 3 contains c , c or c , there are infinitely many cells that must t t t Having defined and studied the schedule, we now proceed have been explored before time t. This yields a contradiction. with our lower bound proof as described in Sect. 4. The next lemma shows that for each distance there is a point in time For any distance D, we denote by T the smallest time T after which the farthest two agents are never closer than this for which it holds that at any time t ≥ T the largest pairwise distance. distance of the three agents is at least D. In the following Lemma 5 For each distance D there is a time T such that we collect a number of useful definitions regarding the meet- at any time t ≥ T the largest pairwise distance of the three ings of different agents. In particular, we distinguish between agents is at least D. three different types of agents at times when one agent is traveling from another agent to the far-away agent whose Proof Suppose that the lemma statement is not true. Then existence is certified by Lemma 5. For an illustration of a there is an infinite sequence T of points in time such that travel over a large distance (i.e., allowing for multiple travel at each of these points in time the largest pairwise distance periods), see Fig. 2. of the three agents is less than D. Since the distances of the agents are less than D at all points in time from T and the Definition 1 For any t ≥ 0, we define the meeting set M as number of states the three agents can be in is finite, it follows the set of agents that are not alone in the cell they occupy, at that there must be points in time t , t ∈ T such that (1) each time t. We call the infinite sequence (M , M ,...) the meet- 0 1 j j agent is in the same state at t and t ,(2) x −x = x −x and t t ing sequence. If for a subsequence (M , M ,..., M ) of t t t t +1 t +i j j i i y − y = y − y for all i , j ∈{1, 2, 3}, i = j, and (3) the the meeting sequence it holds that i > 0, M =∅, M =∅ t t t +i t t same agent is scheduled to move next. Since the agents are and M =∅ for all 0 < j < i, then we call the pair t + j oblivious of the absolute coordinates of the grid, this implies (t , t + i ) a meeting pair.Now,let (t , u) be a meeting pair that from time t on, the agents will repeat the exact behavior such that |M |= 2 =|M | and M = M . Then we call t u t u 123 S. Brandt et al. (t , u) a travel meeting pair. Moreover, we call the (uniquely agent is not contained in the meeting set M . Hence, if a u i defined) agent a contained in M ∩ M a traveling agent (for is not scheduled at all between time t and time u, then the t u (t , u)), the agent contained in M \{a} a source agent and source agent must be scheduled at most once (because of the the agent contained in M \{a} a destination agent. specification of our schedule) which implies that its distance from c at time u is at most N , by Lemma 2. But since in In order to continue according to our high-level proof idea this case a and the destination agent meet at c at time u,we from Sect. 4, we need a few helping lemmas that highlight obtain a contradiction to the fact that T ≥ T . Thus, we 2N +1 properties of the previous definitions. We start with a lemma know that a is scheduled at least once between time t and that shows an important property of the meeting sequence: time u. Now, assume for a contradiction that the first subschedule Lemma 6 Each of the three agents is contained in infinitely of a between time t and time u is of type 2. This implies that many of the M from the meeting sequence. if one would schedule a on and on, it would repeat a state in the same (empty) cell after at most N + 1 time steps and then Proof Suppose that there is an agent a that is not contained cycle through (a part of) the same movement it performed in infinitely many of the M , i.e., there is a point in time u before. Hence, even if there are more subschedules for a such that a ∈ / M for all t ≥ u. Then, starting from time i t than one (between time t and time u), it will never reach a cell u, the exploration by the two agents a , r = i is entirely that has a distance of more than N from c . Since analogous independent of the exploration by agent a since they never statements hold for the source agent, we know that at time meet again. Thus, we get a contradiction analogously to the u the distance between the source agent and the cell where argumentation in the proof of Lemma 1. a and the destination agent meet is at most 2N which again contradicts our specification of T . Thus, we know that the Next, we study travel meeting pairs more closely. In first subschedule of a is of type 1. Lemma 7, we present bounds on the number of subsched- i It follows that a ’s subschedule ends exactly at time u since ules of the different types of agents in the time frame given i the subschedule must end with a meeting the destination by a travel meeting pair, and examine the types of the sub- i agent, which also implies that a is scheduled exactly once schedules. Afterwards, in Lemma 8, we bound the number of i between time t and time u. Moreover, the subschedules of time steps between two subsequent travel meeting pairs from the source and the destination agent (if they are scheduled above. In both cases, the results only hold from a large enough at all between time t and time u) must be of type 2 since point in time onwards, but this is sufficient for our purposes (t , u) is a (travel) meeting pair. Furthermore, by the nature since before that point in time only a constant number of of our schedule, the source and the destination agent must be cells were explored. Note that, in general, we do not attempt scheduled at most once between time t and time u. to minimize the dependence on N in our bounds as showing the finiteness of certain parameters is, again, sufficient for Lemma 8 There is a point in time T such that the following our purposes. Instead we prefer to choose the simplest argu- holds: If (t , u) and (t , u ) are travel meeting pairs such that ments that lead to the desired finiteness results, even if they T ≤ t < t and there exists no travel meeting pair (t , u ) augment the actual bound by a few factors of N . with t < t < t , then t − u ≤ 8(N + 1) . Lemma 7 There is a point in time T such that, for each travel Proof Observe that from the definition of a travel meeting meeting pair (t , u) with t ≥ T , the following properties hold: pair it follows that t ≥ u. Set T := T and let N (4N +1)+1 t , u, t , u be as described in the lemma. W.l.o.g., let a and 1. The traveling agent for (t , u) is scheduled exactly once a be the agents contained in M .Let t < t < ··· < t 2 u 1 2 k (for a number of time steps) between time t and time u. be exactly the points in time t between u and t for which 2. The subschedule of the traveling agent is of type 1 and M =∅ holds. It follows that all M are identical to M = t t u j j ends exactly at time u. M . 3. The source and the destination agent for (t , u) are sched- We claim that k <(2N + 1)(N + 1). Suppose for a uled at most once (for a number of time steps). contradiction that k ≥ (2N + 1)(N + 1). Then, there must 4. If the source or the destination agent is scheduled, then be at least 2N + 1 indices g ∈{1,..., k} such that a or its subschedule is of type 2. a are scheduled to move at time t since each subschedule 2 g of a between time u and t is of type 2 and hence consists Proof Recall the definition of T for any distance D.Let of at most N time steps, by Lemma 2. It follows that there T ≥ T , and consider an arbitrary travel meeting pair must be some 1 ≤ g < h ≤ k such that a and a are in the 2N +1 1 2 (t , u) with t ≥ T and traveling agent a . Observe that if exact same (pair of) states at time t and at time t , and the i g h the source agent is scheduled between time t and time u, same agent is scheduled next. By Lemma 4, this implies that then its subschedules must be of type 2, because the source a and a go through the same movement that they executed 1 2 123 A tight lower bound for semi-synchronous collaborative grid exploration between time t and t , over and over again, starting from Using Lemma 8, we show in the following that for any g h time t , until at least one of them encounters agent a . travel meeting pair (t , u), the information about the states of h 3 During this movement, i.e., anytime between t and u , our the agents, which two agents are in the same cell, and who is agents a and a cannot move too far from each other as we scheduled next, all at time u, already uniquely determines a 1 2 show in the following: If an agent executes a subschedule of lot of information about the agents at the starting time of the type 2 and then its next subschedule is again of type 2 (and no next travel meeting pair. Again, this result only holds from a other agent is in the cell from which this second subschedule sufficiently large point in time onwards. This concludes our starts), then the agent ends both subschedules in the same cell collection of helping lemmas. and the same state, due to the specification of type 2 subsched- ules. Hence, if a and a together perform three consecutive Lemma 9 There is a point in time T such that the following 1 2 subschedules of type 2 (disregarding any subschedules of holds: a ), then in each subsequent subschedule they will repeat the For any two subsequent travel meeting pairs (t , u), (t , u ) 1 2 3 next same movement as in the last subschedule, until one of them with T ≤ t < t , the tuple (q , q , q , a , M ) uniquely u u u u 4 1 2 3 1 1 2 2 3 encounters a . Since we already established that a and a determines the tuple (q , q , q , c − c , c − c , c − 3 1 2 u u t t t t t t next next 3 next would repeat the movement they executed between time t c , a , M ), where a , resp. a , denotes the agent g t u u t t and t (which includes subschedules of type 1) over and over scheduled at time u, resp. t . again if none of them met a , it cannot be the case that there are three consecutive subschedules of type 2 (of a and a ). Proof Let T be sufficiently large so that the condition T ≥ 1 2 Hence, between any two subschedules (of the agents a and T 5 holds and Lemmas 7 and 8 apply. Let t, u, t , 8(N +1) +4N +1 a ) of type 1 between time t and time u , there are at most u be as described in the lemma. Observe that the subschedule 2 g two subschedules of type 2 (of those agents). Now by Lem- of the traveling agent for (t , u) ends exactly at time u,by next mas 2 and 3, the fact that after each subschedule of type 1 Lemma 7, and thus the subschedule of a actually starts at the agents a and a are in the same cell, and our established u. Moreover, due to the choice of T , the agent not contained 1 2 observation about the cyclic movement following t ,itfol- in M will not be in the same cell as another agent until at h u lows that the maximum distance of the two agents between least (and including) time u + 8(N + 1) + 4N , while the time t and time u is at most N (2N +1+ D) where D = 2N . two agents contained in M are in the same cell at time u. g u Hence, when one of the two agents encounters agent a , then Hence, if we knew of which types the subschedules of the the other is at distance at most N (4N + 1). This contradicts three agents are until time u + 8(N + 1) + 2N , then we the fact that t ≥ T and proves the claim. could (deterministically) compute the exact behavior of the N (4N +1)+1 From the above, we obtain the following picture: There three agents up to time u + 8(N + 1) + 2N . are at most k <(2N + 1)(N + 1) subschedules of type 1 Fortunately, the types of the subschedules are uniquely 1 2 3 next between time u and t (since, when a subschedule of type 1 determined by the tuple (q , q , q , a , M ) inasimple u u u u next ends, the corresponding element from the meeting sequence way. Consider the first subschedule, i.e., the one of agent a next is non-empty). Between any two subschedules of type 1 (and starting at time u. Now consider the state of a whose sec- next possibly before the first/after the last) there are at most two ond occurrence would come earliest if we scheduled a subschedules of type 2 of agents a and a , which gives us a indefinitely (not counting the occurrence of a state at time u 1 2 total of at most 2 · (2N + 1)(N + 1) subschedules of type 2 and ignoring the existence of any other agents on the grid). If of agents a and a together between time u and t . For agent this second occurrence happens in the same cell as the previ- 1 2 next a , we obtain an upper bound of 1/2 · 3 · (2N + 1)(N + ous occurrence of the same state and a did not encounter 1) + 1 for the number of subschedules between time u and t any other agent after time u until the time of the second next (which are all of type 2), due to the facts that there are atmost occurrence, then the subschedule of a must be of type 2 3 · (2N + 1)(N + 1) subschedules of the other two agents according to the specification of our schedule. Otherwise, it in that time frame, and that the three agents are scheduled in must be of type 1. next a cyclic fashion. Now, by Lemmas 2 and 3, we obtain that Note that a could only have encountered another agent 2 2 t − u ≤ (7/2 · (2N + 1)(N + 1) + 1)N + (2N + 1)(N + until the second occurrence if these two agents are contained 1)N (2N +1+D) where D = 2N . Hence, t −u ≤ 8(N +1) . in M , due to our choice of T . More generally, any meeting during such a “simulation” (for determining if the respective subschedule is of type 2) must be between the two agents from M if the agent whose potential subschedule is simu- lated starts its subschedule at time u + 8(N + 1) + 2N at the This statement does not necessarily hold for only two consecutive sub- latest, again due to our choice of T . (Here, we use that such schedules of type 2 since the agent a who moved first may encounter the a simulation contains at most 2N time steps until the second second agent (who moved during the subsequent subschedule) during a’s next subschedule. occurrence.) 123 S. Brandt et al. Now, the specification of the type of the subschedule Lemma 10 There is a point in time T and a (possibly nega- in combination with the information contained in the tuple tive) ratio r such that each travel starting at time T or later 1 2 3 next (q , q , q , a , M ) uniquely determines the states of the has travel vector (x , y) with y/x = r. For the sake of sim- u u u u next agents at the time the subschedule of a ends, their relative plicity, assume that r is set to ∞ if x = 0. locations compared to time u and which agent is to move next. Then, we can iterate this argument for the second, Proof Let T be sufficiently large so that T ≥ T holds 2N +2 third, ... , subschedule (from time u on) and obtain that for and Lemmas 7 and 8 apply. Then we know that any travel each of these subschedules the exact movement of the sched- starting at time T or later actually has a travel vector (and 1 2 3 next uled agent is uniquely determined by (q , q , q , a , M ). period), by Lemmas 2 and 7. Now, consider two travel meet- u u u u Again, this argumentation holds up to (and including) time ing pairs (t , u) and (t , u ) with T ≤ t < t such that u + 8(N + 1) + 2N . By Lemma 8, we know that t ≤ there is no travel meeting pair (t , u ) with t < t < t . u + 8(N + 1) . On the other hand, we have that u > Let (x , y), (x , y ) be the travel vectors for the travels cor- u + 8(N + 1) + 2N , due to our choice of T . Hence, the responding to (t , u) and (t , u ), respectively. Assume that agent scheduled at time u + 8(N + 1) + 2N must be the y /x = y/x, where, again, we set the ratio to ∞ if the traveling agent for (t , u ), by Lemmas 2 and 7. denominator is 0. Note that not both of x and y (or x and y ) Moreover, the above considerations ensure that the exact can be 0. Let c and c be the cells at which the travel with 0 1 behavior of the agents up to (and including) time u + travel vector (x , y) starts and ends, respectively, and c and 8(N + 1) + 2N is uniquely determined by the tuple c analogously for the travel with travel vector (x , y ). 1 2 3 next (q , q , q , a , M ). Thus, also the traveling agent for By the characterization of the travel of a single agent u u u u 1 2 3 next (t , u ) is uniquely determined by (q , q , q , a , M ), and and the fact that the travel period is always at most N , u u u u also the parameter t − u. Since t ≤ u + 8(N + 1) ,it we know that there are positive integers b and b such that 1 2 3 next follows that (q , q , q , a , M ) uniquely determines the Dist(c , c + b · (x , y)) ≤ N and Dist(c , c + b · (x , y )) u 1 0 u u u u 1 0 1 2 3 1 1 2 2 3 3 next tuple (q , q , q , c − c , c − c , c − c , a , M ). ≤ N . Moreover, by Lemmas 2 and 7, the source agent u u u t t t t t t t for (t , u) travels at most a distance of N between time t and u since its subschedule is of type 2 if the agent 7 The travel vector and a modulo operation is scheduled at all. The same holds for the destination agent for (t , u ) between time t and u . By Lemma 8,it After collecting the above helping lemmas, we are now all set follows that Dist(c , c ) ≤ 8(N + 1) + 2N (since the to formally prove the (remaining) statements from our proof source agent for the first of the two travels is the destina- sketch. Before going through the statements one by one, let tion agent for the second) and Dist(c , c ) ≤ 8(N + 1) . us for convenience define the notion of a travel: Let (t , u) be a Combining our above distance observations, we also obtain travel meeting pair. By Lemma 7, we know that the traveling Dist(c , c + b · (x , y) + b · (x , y )) ≤ N +8(N +1) +N , agent for (t , u) is scheduled exactly once between t and u. which together with Dist(c , c ) ≤ 8(N + 1) + 2N implies We call the corresponding subschedule (or the movement Dist(c , c + b · (x , y) + b · (x , y )) ≤ 16(N + 1) + 4N . 0 0 during that subschedule) a travel. Recall the definition of Let D ≥ N be some positive integer. We now require, travel vector and travel period. Note that a travel only has a additionally to the above requirements regarding T , that travel vector (and period) if the traveling agent repeats a state T ≥ T . Also fix some arbitrary x , y, x , y such that (x , y) (in empty cells) during the travel. Furthermore, observe that and (x , y ) are possible travel vectors of a single agent. For if a travel has a travel vector, then at least one entry of the a contradiction, assume that x , y, x , y have the properties travel vector is non-zero, due to the choice of our schedule. specified at the beginning of the proof (which implies that We now prove the first of the remaining statements, namely, also all of the above conclusions hold). that after a certain point in time, any travel vector has the At the time when the first of the two considered trav- same slope. els starts there are two agents at c and c while the last 0 1 agent is at distance at most N from c . Hence, the dis- tance between c and c is at least D − N . This implies that 0 1 b · (|x|+|y|) ≥ Dist(c , c ) − N ≥ D − 2N . Analogously, 1 0 we obtain b · (|x |+|y |) ≥ D − 2N . Since x , y, x , y are To be a bit more precise, we can simply execute each (relative) step fixed, we can therefore make b and b arbitrarily large by of any agent between time u and t one after the other, as at each point in time, we have all the information necessary to compute the step increasing D. By increasing b and b , we can in turn make (in particular, which agent is scheduled, due to the above simulation Dist(c , c + b · (x , y) + b · (x , y )) arbitrarily large, since 0 0 argument). Note that for this, we don’t need the exact location of the y /x = y/x (which implies that there is an angle between agent not contained in M relative to the location of the agents contained ◦ ◦ the two vectors (x , y) and (x , y ) that is not 0 or 180 ). in M as we know that the former agent will not meet any other agent until time u + 8(N + 1) + 4N ≥ t + 2N . Hence, if D is sufficiently large, then the above inequality 123 A tight lower bound for semi-synchronous collaborative grid exploration Dist(c , c + b · (x , y) + b · (x , y )) ≤ 16(N + 1) + 4N vectors from R. Note that R cannot be empty since otherwise 0 0 is not satisfied anymore, which shows that y /x = y/x. it is not possible that the agents explore the entire grid, due Note that the magnitude D has to reach for this (in our to Lemmas 6 and 10. proof by contradiction) depends on x , y, x , y . However, Now, let w, z be integers and let b be the smallest integer since the number of possible travel vectors of a single agent such that w + bx ≥ 0. (This is well-defined since x > 0, due is bounded by the number of states in its finite automaton, to r =∞.) We define (w, z)(mod (x , y)) := (w + bx , z + we can simply derive a sufficiently large D for each of the by). For two cells (w , z ), (w , z ), we define (w , z ) finitely many possible combinations for x , y, x , y and then (w , z ) := (w − w , z − z )(mod (x , y)). choose a T that is larger than all of the T . Note that Definition 2 ensures that for any (w, z), (w , z ) Note that the exact value of r depends only on the finite where (w − w, z − z) is a multiple of (x , y), we have that automaton governing the behavior of the three agents. From (w, z)(mod (x , y)) = (w , z )(mod (x , y)). now on, we denote the ratio whose existence is certified by The following lemma shows that if an agent in state q Lemma 10 by r. W.l.o.g., we can (and will) assume that r ≥ 0 travels from some cell c to another (sufficienly distant) cell (and that r =∞), for reasons of symmetry. Recall that any c , it is not required to know the exact location of c relative travel vector has at least one non-zero entry. The next step on to c to determine the state in which the agent arrives at c : our agenda is essentially to show that the state of an agent at instead, it is sufficient to know the relative location of c and the end of a travel does not depend on (the full information c modulo the respective travel vector. Formally, we prove about) the vector between start and endpoint of that travel this by showing that any two choices for c that differ by a (and other parameters), but only on a reduced amount of multiple of the travel vector result in the agent having the information regarding this vector (and the other parameters). same state at c . More specifically, the required information about this vector is the result of applying a certain modulo operation to the Lemma 11 Let a be an agent, q a state from a’s finite automa- vector. ton and c, c , c cells of the grid such that the following We then proceed by showing that the information about (1) properties are satisfied: the states of the agents, (2) their relative locations after apply- ing the modulo operation, (3) which agents shared a cell most 1. Dist(c, c ) ≥ 2N and Dist(c , c ) ≥ 2N recently, and (4) which agent is scheduled next, at the start of 2. There is an integer b such that c − c = b · (w, z), where a travel, is enough to determine the exact same information (w, z) is agent a’s travel vector if it starts in state q. at the end of the travel. Now, we benefit from the previous 3. If agent a starts in cell c in state q on an otherwise empty reduction of information due to our modulo operation in the grid, then it arrives at c after finite time. sense that we can show that there are only constantly many 4. If agent a starts in cell c in state q on an otherwise empty combinations of relative locations of the three agents (that can grid, then it arrives at c after finite time. actually occur) after applying the modulo operation. This, in turn, implies that there are only constantly many possibili- Let q denote the state in which a arrives at c (for the first ties for the whole aforementioned information tuple at the time) when starting from c (in state q), and q the state in start and end of a travel, which will enable us to prove our which a arrives at c (for the first time) when starting from main theorem. We start by defining our modulo operation c (in state q). Then it holds that q = q . in Definition 2. Then we show a technical helping lemma, Lemma 11, which finally enables us to prove the aforemen- Proof If q is a state that agent a assumes only at the very tioned relation between the information tuple at the start and beginning (if scheduled on an otherwise empty grid with ini- end of a travel in Lemma 12. Note that for technical reasons, tial state q), then after at most N steps a assumes a state q Lemma 12 gives a slightly different statement than indicated that it assumes more than once (i.e., after one (and hence, above, dealing with travel meeting pairs instead of travels. each) travel period a will be again in state q and have trav- Definition 2 Let {(x , y ), (x , y ), . . . , (x , y )} be the set eled exactly “along” the travel vector). Let c be the vector 1 1 2 2 k k of travel vectors that the agents can have if you let one of them describing a s movement from the initial cell to the cell where explore the grid starting in an arbitrary state (which clearly a first assumes state q . From the properties given in the is a superset of the actually occurring travel vectors in our lemma, it follows that these properties are still satisfied if we ∗ ∗ ∗ multi-agent case). Let R be the subset of the above set that replace q by q , and c, c by c + c , c + c , respectively, contains exactly the vectors (x , y ) that satisfy y /x = r. except that we have to replace each occurrence of “2N”by j j j j From now on, denote by x the least common multiple of the “N ” in the first property. Hence, it is sufficient to prove the |x | from the vectors in R and set y := rx.Itfollows that lemma for those states q that a will assume again after exactly (x , y) is a (possibly negative) integer multiple of any of the one travel period (i.e., those states q that a assumes during its 123 S. Brandt et al. next periodic behavior), under the modification to the first prop- M , then a cannot have been scheduled at time t − 1as erty. Thus, in the following assume that q is of this kind, and otherwise its subschedule would not continue at time t due next that the first property only guarantees Dist(c, c ) ≥ N and to the specification of our schedule, and if a ∈ / M , then Dist(c , c ) ≥ N . (t , u) would not be a travel meeting pair. If c = c , then the lemma holds trivially, thus assume that Combining the argumentation, about the uniquely deter- c = c . W.l.o.g., we can assume that b > 0, which implies mined subschedules, from the proof of Lemma 9 with that, if agent a starts in cell c in state q (say, at time t), then Lemmas 2 and 7, we see that Q uniquely determines which a arrives at some point in time u > t in cell c in state q agent is the traveling agent for (t , u), which in turn uniquely (possibly a visited c before in some other state). Hence, if determines M . By Lemma 7, the last agent that is scheduled a does not visit cell c between time t and time u, then the before time u is exactly the traveling agent, which uniquely next lemma also holds since after arriving at c in state q, a will determines a . perform the exact same movement as if it started in c in state Moreover, the information which agent is the traveling q. agent together with the information which agent is scheduled next Thus, consider the last remaining case, i.e., assume that a at time t (i.e., a ) uniquely determines which agents are visits c for the first time at some time t < t < u. W.l.o.g., scheduled between time t and time u (and all of them are we can assume that w and z are non-negative and w ≥ z. scheduled only once, possibly for multiple subsequent time (Also recall that at least one of w and z is non-zero.) Let steps, if they are scheduled at all). Since no two agents meet c , c ,... be the cells that a visits in state q at and after between time t and time u, it follows that the states of the 0 1 time t, where c and c ,for some k > 0, are the cells that a source agent and the destination agent at time u are uniquely 0 k visits at time t and u, respectively, i.e., c = c and c = c . determined by the states of the three agents at time t (and 0 k Observe that c = c + (w, z) holds for each j. Denote the the information which two agents are contained in M ). Here j +1 j t x-coordinates of c and c = c by x and x , respectively. we use that the subschedules of those agents (if they are Since w ≥ z,itfollows that Dist(c , c ) ≥ Dist(c , c ) ≥ N scheduled at all) are of type 2, according to Lemma 7. for all j ≥ k if x ≤ x , and Dist(c , c ) ≥ Dist(c , c ) ≥ N Similarly, the exact vectors (possibly the vector (0, 0)) for all 0 ≤ j ≤ k if x ≥ x .Let h be the largest index such by which the source and the destination agent move are that a visits c in state q at or before time t . Then h < k, and uniquely determined by Q . By Lemma 7, the subsched- h t Dist(c , c ) ≤ N − 1 since traveling from c (in state q)to ule of the traveling agent is of type 1 and ends in the cell h h c (in state q) takes a at most one travel period, so at most that is occupied by the agent not contained in M . Since h+1 t N time steps. If x ≥ x , then we obtain a contradiction to addition and our modulo operation behave nicely (more our above observation, thus it follows that x < x . But this specifically, because (w, z) + (w , z ) (mod (x , y)) = implies Dist(c , c ) ≥ N for all j ≥ k which in turn implies (w, z)(mod (x , y)) + (w , z ) (mod (x , y)) for all inte- 1 2 1 3 2 3 for all j ≥ k that c cannot be visited by a between visiting gers w, z,w , z ), we get that c c , c c , and c c u u u u u u c (in state q) and c (in state q). Hence, a does not visit c are uniquely determined by the vectors by which source and j j +1 at or after time u. Since a performs the exact same movement destination agent move (combined with the information con- from time u onwards as if it would have initially started in c tained in Q ), and thus also by Q . t t in state q, it follows that agent a starting in c in state q never It remains to show that the state of the traveling agent at visits c , which is a contradiction to our assumptions. Thus, time u is uniquely determined by Q . Denote the traveling this last remaining case cannot occur, which completes the agent by a . Since t ≥ T , the distance between a and i 3N +1 i proof. the destination agent at time t is at least 3N + 1. When the subschedule of a starts at some time t ≤ t ≤ u,the Lemma 12 Let (t , u) be a travel meeting pair. Consider the destination agent may have moved from its location at time tuple t, but since the subschedule of the destination agent is of type 2 (as observed above), it has moved a distance of at most N , 1 2 3 1 2 1 3 2 3 next Q := (q , q , q , c c , c c , c c , a , M ), t t t t t t t t t t t t by Lemma 2. Hence, at time t the distance between a and the destination agent is at least 2N + 1. next where a again denotes the agent that is scheduled at Now, let c, c be cells with a distance of at least 2N +1to time t. There is a point in time T such that the following the location c of the destination agent at time t and assume holds: If t ≥ T , then Q uniquely determines the tuple that c c = c c . Then, according to the definition of 1 2 3 1 2 1 3 2 3 next Q = (q , q , q , c c , c c , c c , a , M ). u u u u u u u u u u u u our modulo operation, c − c is a (possibly negative) integer Proof Let T be sufficiently large so that T ≥ T holds multiple of (x , y), and thus also of the travel vector of a ,by 3N +1 Lemma 10.Thus,byLemma 11, it follows from the above and Lemmas 7 and 10 apply. Let (t , u) be a travel meeting pair with t ≥ T . We start by observing that the subschedule that q is uniquely determined by Q . Note that although a t i next next of a starts at time t. The reason for this is that if a ∈ may not be alone in its cell at the time its subschedule starts, t t 123 A tight lower bound for semi-synchronous collaborative grid exploration we can still apply Lemma 11 since after the first step of a Consider the sequence it is alone in its cell while all other requirements for Lemma 11 are still satisfied. This completes the proof. ((t , u ), (t , u ), . . . , (t , u )) k k k+1 k+1 h h of travel meeting pairs, where h is the smallest index such that h > k holds, h − k is even, and Q = Q . We examine 8 Three semi-synchronous agents do not t t k h the cells that are explored by the source agent for (t , u ) suffice k k between time t and t and by the destination agent for k k+1 (t , u ) (which is the same as the aforementioned source We now conclude our lower bound proof with Theorem 1. k+1 k+1 agent) between time t and t . Then we iterate this exam- Roughly speaking, Lemma 12 certifies that the behavior of k+1 k+2 ination, in each iteration increasing the indices by 2, and stop the agents between any two subsequent occurrences of the at time t . We say that the cells explored in the described way same fixed information tuple Q is reasonably similar. Since h are explored during even explorations. there are only finitely many different Q that actually occur, In the first iteration, we obtain the following picture, where it follows that the behavior of the agents loops, in a very we denote the source agent for (t , u ) (i.e., the destination informal sense. From this, we can derive a contradiction to k k agent for (t , u ))by a: The exact vector by which a the assumption that all cells are explored. k+1 k+1 moves between time t and u is uniquely determined by k k Theorem (Theorem 1 restated) Three semi-synchronous Q , as observed in the proof of Lemma 12. The exact vector agents controlled by a finite automaton are not sufficient to by which a moves between time u and t is uniquely k k+1 explore the infinite grid. determined by Q , by Lemma 9. Similarly, the exact vectors by which a moves between time t and u and between k+1 k+1 Proof Suppose for a contradiction that three agents suffice time u and t are uniquely determined by Q and k+1 k+2 t k+1 to explore the grid. From the definition of a travel meeting Q , respectively. k+1 pair and Lemma 6, it follows that there are points in time Moreover, by combining Lemmas 9 and 12, we see that t < u ≤ t < u ≤ t < ... such that (t , u ) is a travel 1 1 2 2 3 j j Q , Q , Q , and Q are all uniquely determined by u t u t k k+1 k+1 k+2 meeting pair for any j ≥ 1 and for every travel meeting pair Q . Thus, the exact vector by which a moves between time t t k (t , u ) there is a j ≥ 1 with t = t and u = u . j j and time t is uniquely determined by Q . Furthermore, by k+2 t Recall the definition of Q in Lemma 12 (in particular, Lemmas 2, 7 and 8, the number of cells a visits between time recall that t is assumedtobeapoint in timefor whichthereisa t and time t is bounded by a constant. Note that each Q k k+2 t travel meeting pair (t , u) “starting” in t). Let T be sufficiently also uniquely determines which agent is the traveling agent large so that T ≥ T holds (where T is just T for D = 1) 1 1 D (and hence which agent is the source/destination agent) for and Lemmas 7, 8, 9, 10 and 12 apply, and let k be an index (t , u ), as observed in the proof of Lemma 12. j j such that t ≥ T and there is a h > k with h − k even and k For the second, third, ... , iteration we obtain an analogous Q = Q . Such a k must exist since there is only a finite t t k h picture. Hence, the tuples Q , Q ,... are all uniquely t t k+2 k+4 number of tuples of the general form Q (after time T ) and the determined by Q , and the locations of the respective source number of travel meeting pairs is infinite, by Lemma 6.Note agents at times t , t ,... are all uniquely determined by k+2 k+4 that the finiteness of the number of tuples, in particular the Q and the location of the source agent for (t , u ) at time t k k finiteness of the (combinations of the) relative locations of the t . agents modulo (x , y), relies on the fact that the possible travel We obtain the following bigger picture: The location of vectors after time T are restricted by Lemma 10, together the source agent for (t , u ) at time t together with Q k k k t with the fact that in the time span given by a travel meeting uniquely determines both Q and the location of the source pair source and destination agent are scheduled for at most agent for (t , u ) at time t , which, in turn, uniquely deter- h h h N steps, by Lemmas 2 and 7. mine Q and the location of the source agent for h+(h−k) (t , u ) at time t , and so on. Hence, h+(h−k) h+(h−k) h+(h−k) More precisely, for any travel meeting pair (t , u) with t ≥ T,the there is a vector (w, z) such that the locations of the respec- vector between the locations of any two agents at time t can be written tive source agents at times t , t , t , t ,... are k h h+(h−k) h+2(h−k) in the form g·(x , y)+(v, w) where g is some integer and |v|≤|x|+2N , c, c + (w, z), c + 2(w, z), . . . , where c denotes the cell occu- |w|≤|y|+ 2N , by Lemmas 2, 7, 10, and the fact that any travel period pied by the respective source agent at time t . Moreover, is at most N . The finiteness of the number of tuples of the form Q now follows from the observation that the first coordinate of (g · (x , y) + since the number of cells explored during an even explo- (v, w)) (mod (x , y)) = (v, w) (mod (x , y)) is at least 0 and at most ration between time t and t (and similarly between time k h x (by Definition 2) and the absolute value of the second coordinate t and t for each j ≥ 0) is bounded by a h+ j (h−k) h+( j +1)(h−k) therefore is at most |y|+ 2N + (|x|+ 2N ) ·|y| (since the smallest constant (which follows from a similar observation above), integer b such that v + bx ≥ 0 holds must satisfy |b|≤|v|≤|x|+ 2N due to x > 0). we get that there is a constant L such that each cell explored 123 S. Brandt et al. during an even exploration has a distance of at most L to some 6. Blum, M., Kozen, D.: On the power of the compass (or, why mazes are easier to search than graphs). 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Distributed Computing – Springer Journals
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