Search-and-Rescue Rendezvous

We consider a new type of asymmetric rendezvous search problem in which Agent II needs to give Agent I a `gift' which can be in the form of information or material. The gift can either be transfered upon meeting, as in traditional rendezvous, or it can be dropped o? by II at a location he passes, in the hope it will be found by I. The gift might be a water bottle for a traveller lost in the desert; a supply cache for Lieutenant Scott in the Antarctic; or important information (left as a gift). The common aim of the two agents is to minimize the time taken for I to either meet II or find the gift. We find optimal agent paths and droppo? times when the search region is a line, the initial distance between the players is known and one or both of the players can leave gifts. When there are no gifts this is the classical asymmetric rendezvous problem solved by Alpern and Gal in 1995 [10]. We exhibit strategies solving these various problems and use a `rendezvous algorithm' to establish their optimality.


Introduction
The rendezvous search problem asks how two (or more) non-communicating unit-speed agents, randomly placed in a known dark search region, can move about so as to minimize their expected meeting time. It was first proposed in [3] and formalized in [4]. Here we take the search region to be the real line. The form of the problem dealt with in this paper is called the 'playerasymmetric' (or 'indistinguishable player') version. This means that the players (agents) can agree before the start of the game which strategies each will adopt: for example if the search region were a circle, they could agree that Player I would travel clockwise and Player II counter-clockwise. In our problem they begin the game by being placed a known distance D apart on the line, but of course neither knows the direction to the other player. The solution to this problem given in (Alpern and Gal, 1995 [10]) achieves a rendezvous value (least expected meeting time) of 13D/8. Their solution is best presented as a modification of the simple 'wait for Mommy' strategy in which Player II (Baby) stays still and I (Mommy) searches optimally for an immobile target: Mommy first goes distance D in one direction (chosen randomly) and then 2D in the other. This gives an expected meeting time of (1/2) D + (1/2) (D + 2D) = 2D. If Baby optimizes against Mommy's strategy, he goes a distance D/2 in some direction (hoping to meet an oncoming Mommy) and then back to his starting location at time D (in case Mommy comes there from the other direction). If he has not met Mommy, he knows she is now at a distance 2D because she went in the wrong direction. So he goes a distance D is some direction and then back to his starting location again. The equally likely meeting times for this 'modified wait for Mommy strategy' are D/2, D, 2D and 3D with the stated average of 13D/8, which can be shown to be the best possible time. Two improved proofs of the optimality of this strategy pair are given in [1]. This paper introduces a new asymmetry into the rendezvous problem. We consider that one of the players (taken as I) is lost and needs a 'gift', say food or water. The other player (taken as II) is the rescuer, who has plenty of this resource to give. He can either give it to I by finding him, as in the original rendezvous problem, or by leaving a canteen of water or cache of food which is later found by the thirsty or hungry I. That is, the game ends when Player I either meets Player II or finds a gift he has left for him. This is the case where there is one gift and we denote this game G 1 .
Another version of this problem is where each player has a gift. First Each player has a gift, one must be found (8,8) [19.99968, 20] G and 2 Each player has a gift (0,0), (8,8), (4,x) [23.99968, 24]  consider the case where a boy and girl who like each other are trying to find each other at a rock concert. They can move about and hope to meet and they can also leave their phone number on some bulletin board. The game ends when they meet or when one of them finds the other's phone number and calls. More generally, the game ends when the players meet or when one finds a gift left by the other. Here the gift is interpreted as containing information rather than resources. We denote this game by G or 2 . A second case of two gifts is when two players are lost and one has food but needs water and the other has water and needs food. One can leave a cache of food along his path while the other can leave a canteen of water. (Each still has plenty left for himself or to give to the other upon meeting.) Here the game ends when either the two players meet or when both players have found the gift left by the other. We denote this game by G and 2 . We solve all of these rendezvous problems. For comparison it is easiest to take the initial distance as D = 16, so that the original rendezvous time (expected time for the game to end) of Alpern-Gal is 26 = 13(16)/8. This time goes down in all cases. Table 1 summarizes our main results, where τ denotes the optimal time(s) to drop off the gift(s) andR is the rendezvous value (least expected meeting time) When there is one gift, the rendezvous time is 21 (the gift is dropped at time D/4 = 4). When there are two gifts, but only one has to be found to end the game, the rendezvous time is 20 (the gifts are both dropped at time D/2 = 8). When there are two gifts and both must be found to end the game, the rendezvous time is 24, with the gifts being dropped at times (0, 0), (D/4 = 4, x) or (D/2 = 8, D/2 = 8). These results are summarized in Table  1.
It will be of interest to consider all of these problems in the two dimensional setting of a planar grid (Z 2 ), as initiated for asymmetric rendezvous [12] and studied in [18], and on arbitrary graphs as studied in [6]. The use of gifts could also be studied in the rendezvous contexts of Howard [22], Lim [26], Anderson and Essegaier [11], Han et al [21] and in other settings discussed in the survey Alpern [5]. Gifts might also be used in the discrete rendezvous problem solved by Weber [31].
The main tool that we use is to confine the rendezvous strategies to a finite set when the time (or times) of dropping the gifts is fixed, see section 5. Actually, in Proposition 5 we identify a finite set of particular events that can be turning points for the two players. In between turning points, players can only move in a fixed direction at maximal velocity. Then, we compute the game values for dropping times that vary over a grid with small diameter, and continuity properties of the rendezvous time provide bound for the optimal solution, see section 6.
The paper is divided into sections as follows. The next section is dedicated to the literature review. In section 3 we explain how the problem can be thought of in terms of a single Player I who wants to find all four 'agents' of Player II. This type of analysis goes back to [10]. In section 4 we give the solution to all of the versions of the problem discussed above. Simply presenting these strategies and evaluating the possible meeting times gives an upper bound on the rendezvous times. But the possibility of better strategies is not ruled out. Section 5 presents our algorithm for optimizing rendezvous strategies when the dropping-times are given, and thus proves the approximate or exact optimality of each of the strategies given in section 4. We show that they minimize the rendezvous times over the dropping time grid. In section 6, we show how the optimal values of the new games can be upper and lower bounded given exact values computed for dropping times in a regular mesh. Moreover, we show how bound the regions where the dropping-times leading to optimal strategy are. These results are summarized in Theorems 10 and 15. Sections 7 and 8 present numerical values for the lower and upper bounds. The lower bound are not computed as precisely as 'possible'. Improving the accuracy is only a matter of computing time.

Literature Review
The rendezvous search problem was first proposed by Alpern [3] in a seminar given at the Institue for Advanced Study, Vienna. Many years passed before the problems presented there were properly modeled. The first model, where the players could only meet at a discrete set of locations, was analyzed by Anderson and Weber [19]. This difficult problem was later solved for three locations by Weber [31]. Rendezvous-evasion on discrete locations was studied by Lim [26] and solved for two locations (boxes) by Gal and Howard [30].
The rendezvous search problem for continuous space and time, including the infinite line, was introduced by Alpern [4]. The player-asymmetric form of the problem (used in this paper), where players can adopt distinct strategies, was introduced in Alpern and Gal [10]. Baston and Gal [15] allowed the players to leave markers at their starting points (e.g. the parachutes they used to arrive). The last two papers form the starting point of the present paper. We apply the same method as the one in this paper to the rendezvous problem on the line with markers. Our findings are that the solution in [15] seems optimal even if we allow the dropping times of markers to be chosen. Results are going to be published independently of this paper.
The corresponding player-symmetric problem on the line was developed by Anderson and Essegaier [11]. Their results have been successively improved by Baston [13], Gal [20], and Han et al [21]. These papers assumed that the initial distance between the players on the line was known. The version where the initial distance between the players is unknown was studied by Baston and Gal [14], Alpern and Beck [8,9] and Ozsoyeller et al [28].
The continuous rendezvous problem has also been extensively studied on finite networks: the unit interval and circle by Howard [22]; arbitrary networks by Alpern [6]; planar grids by Anderson and Fekete [12] and Chester and Tutuncu [18], the star graph by Kikuta and Ruckle [23].
The present paper is an application of rendezvous search to 'search-andrescue' operations. A different application of search theory to that area is in Alpern [7], where the Searcher must find the Hider (injured person) and then bring him back to a specified first aid location. An application of rendezvous to robotic exploration is given in Roy and Dudek [29]. An application of rendezvous to the communications problem of finding a common channel is given in Chang et al [17]. Using markers in communication networks to help matching publishers and consumers of information is suggested in [25,27,24]. These works have relevant applications to anonymous communication networks where the content of information is important (content based routing). It is observed that decentralized search strategies prove to be efficient in terms of congestion and the search times are well concentrated. A survey of the rendezvous search problem is given in Alpern [5].

Formalization of the Problem(s)
We begin by presenting the formalization of the problem when there are no gifts, as given in [10]. Two players, I and II, are placed a distance D apart on the real line, and faced in random directions. They are restricted to moving at unit speed, so there position, relative to their starting point, is given by a function for some T sufficiently large so that rendezvous will have taken place. In fact optimal paths turn out to be much simpler. We will see that optimal paths are piecewise linear with slopes ±1 and so they can be specified by their turning points. Suppose I chooses path f ∈ F and i chooses path g ∈ F .
The meeting time depends on which way they are initially facing. If they are facing each other, the meeting time is given by If they are facing away from each other, the meeting time is given by If they are facing the same way, say both left, and I is on the left, the meeting time is given by If I is on the left and they are both facing right, the meeting time is given by To summarize, the four meeting times when strategies (paths) f and g are chosen are given by the four values, see Figure 1, The Rendezvous time for given strategies is their expected meeting time The Rendezvous ValueR is the optimum expected meeting time, There is a simple interpretation of the formula (2) as the average time for player I (whose position at time t is f (t)) to meet four agents of player II. We take as the origin of the line the starting point of Player I and we take his forward direction to be the positive direction on the line (up, if the line is depicted vertically). The four 'agents' of II start at +D and −D and face up or down, so their paths are ±D ± g (t) . The meeting times with these 'agents' are exactly the rendezvous times t i , i = 1, 2, 3, 4, see Figure 1.
It has been shown for the 'no gift' case, that optimal paths are of the form where the times f k are the turning points of the path f, namely If a player has a gift to drop off, we denote his strategy by where τ is the dropoff time and the f j are as above. We are now in a position to state and illustrate the initial result of the field, for the case of no gifts.
Theorem 1 (Alpern -Gal (1995b)) (G-Game ) An optimal solution pair for the asymmic rendezvous problem on the line, with initial distance D, is given by, see Figure 1,f The corresponding meeting times are with Rendezvous Valuē Note that in (6) we have introduced the subscripted times t j as the meeting times t i given in increasing order. The duration of the strategy pair is the final meeting time t 4 . We now illustrate the optimal strategiesf ,ḡ separately and then show how the solution can be seen by drawing the single path of I (f ) together with the paths of the four agents of player II (±D ± g (t)). We take D = 2 and draw the paths up to time t 4 = 3D = 6, see Figures 2 and 3.

Upper bounds for new games' solutions
In this section, we present the solutions of the new games that we introduce in this paper, see Table 1. These solutions embodies upper bounds on the rendezvous times.
Theorem 2 (G 1 -game) A solution for the asymmetric rendezvous problem on the line when one player has a gift, with initial distance D, is given by, see Figure 4f = The corresponding times are with Rendezvous Valuē The next theorem present a solution of the G or 2 game where both players have a gift. In this game, player I and agent i must rendezvous or at least one of player I or agent i must find the gift of the other.   The corresponding times are with Rendezvous Valuē The next theorem present a solution of the G and 2 game where both players have a gift. In this game, player I and agent i must rendezvous or (both) must find the gift of the other.

Theorem 4 (G and
2 -game) Solution for the asymmetric rendezvous problem on the line with two gifts (one for each player), with initial distance D is given by (the value x denotes any time, the gift is not used), see Figure 6 as well,f . The left solution is already known, see [16] that use marks, i.e. a mark is dropped off by a player in order to indicate that he was here. Although marks seem less powerful than gifts we havn't found better solution. In the middle, the gift of player I is not used, the immaterial dropping time is denoted by x.
the difference with the G 1 game is that in the G and 2 game both players are aware that the game is finished -they both get some information or gift from the other player. In the G 1 game the situation is asymmetric, it may happen that only the player that gets the gift is aware of the end of the game, the information flows in only one direction. Hence, although there is a solution for the G and 2 game that makes use of only one gift, the solutions to the G 1 and G and 2 are very different.

Optimal strategy when the dropping time is known
In this section we show how computing the optimal strategy of the games G 1 , G or 2 , G and 2 if the dropping times are known. First observe that in all solutions presented in section 4 the players move at maximal speed and in direct way toward the location where he finds/drops off a gift or meets the other player. We prove in Proposition 5 that strategies that departs from this principles cannot be optimal. Hence, if we know the dropping times there are only a finite set of strategies that are candidate for being optimal (compare with the original strategy space (1)). By testing all elements of the finite set (with a program) we identify optimal strategies.
The following result is a generalization of , Lemma 5.1 of [10] and Theorem 16.10 of [2].
Proposition 5 Let G be any asymmetric rendezvous game on the line where each player has at most one gift. Then in any Nash equilibrium (NE) for G (and in particular at any optimal strategy pair) each player moves at unit speed in a fixed direction (no turns) on each of the time intervals J determined by times 0, and the following times c: 1. The meeting times c = t i when he meets the agent i of the other player. This contradicts that f was a best response to g.

Corollary 6
Optimal strategy pair (f, g) in the games G, G 1 , G or 2 , G and 2 admit a representation as in (4), (5). In particular, there are only a finite set of strategies candidate for optimality.
Proof. By proposition 5 players move at full speed and turning points must coincide with location where the player drops/finds a gift or meets the other player. The time interval between any two such events is deterministic, depending on the configuration and can be computed. If such an interval corresponds to a turning point then it leads to the definition of the f i in (4), (5). The total set of strategies in the form (4), (5) can be listed by selecting at each event either to change the direction of continue in the same direction. Because there is a finite set of such events and a finite set of directions there is a finite set of strategies in the form (4), (5).
In the following we provide two examples of exact solutions whose computation is possible because of Proposition 5 (Corollary 6).
In Figure 7 we show a procedure that enumerates all strategies in the form (4), solving the game G. The state of the system is described by the distance vector (d 1 , d 2 , d 3 , d 4 , t) where d i is the distance for player I to agent i and t the current time.
Initially (d 1 , d 2 , d 3 , d 4 , t) = (1, 1, 1, 1, 0) and player I has met with agent i if d i = 0, in which case the value of d i is frozen in the subsequent computation (this is why we introduced the operator +). At the time where turning is an option by Proposition 5, the algorithm checks whether a direction makes sense and, if yes, try it. Indeed, there are situations where going in a direction makes no sense, for instance player I cannot go forward if agents 1 and 4 have met already with player I. Figure 8 summarizes all meaningful direction choices given the current state of the system (d 1 , d 2 , d 3 , d 4 , t) as well as the new state reached after the motion. A systematic exploration of all trategies that that are eligible for optimality can be implemented by a recursive procedure, see Figure 7. Execution of this program provides a new proof of the optimal strategy of G in [10].
The programs for solving the other games are very similar. Few more cases are allowed. For instance, for a player moving in any direction always makes sense provided there is a player in the direction and independently of the moving direction of this player. For, player I going in the forward direction makes always sense provided agent 1 or 2 is not yet found. Actually, even if the agent goes the wrong direction, player I can still go for the gift (symmetric situations admit the same argument).
To conclude this section we can illustrate how to enumerate all solution by considering the game G 1 when the dropping time is z < 1/4, see Figure  9. At the onset player I and II are always going in the forward direction by symmetry. Because z < 1/4, player II drops off the gift before any other if (d 1 > 0) print("player I : F player II : B time : t") print("player I : F player II : F time : t") print("player I : B player II : F time : t") print("player I : B player II : B time : t") print("final time t") } Figure 7: Pseudo-code of the program that enumerates candidate for optimal strategies of G. The recursive procedure prints out all the checked strategies and the total time. We use an operator denoted + , i.e. x +y which semantics is: if x > 0 then x + y else 0. event. Hence after time z there are two strategies for player II, either it is a turning point either not. By direct inspection with a program we checked that not turning at time z is not optimal. Player I meets agent 4 at time z + d/2 and now both player I and II are allowed to change direction. We again checked by direct inspection that the best strategy is that player I turns and player II not. Continuing in this way, considering turning points only when this is allowed by Proposition 5 we are able to generate all strategies and select the optimal one. In this case the best strategy for player I is [D−z] while for player II it is [z; z].

Bounding the optimal solutions
Given the algorithm described in section 5 a way a obtaning reasonnable knowledge about the optimal solution of one of the games is to define a mesh of values that corresponds to the dropping times. For each dropping time we compute the exact optimal solution of the game. Computing the minimal values leads to an upper bound. In the following we show how the game values that are not computed (the dropping time does not belongs to the mesh) can be bounded. For the game G 1 , we denote x(l) the optimal game value computed by direct inspection when the dropping time is l. Proposition 7 shows that x(l) and x(l + α) and related by x(l) ≥ x(l + α) − α, α ≥ 0. Hence, if the value x(l + α) is computed it can be used to bound the exact solution for x(l) and, if this bound is larger that upper bound computed (see Section 5 ) the optimal solution cannot be with dropping time in the interval [l, l + α]. This result is stated in Theorem 10. Proposition 9 shows that the function x(l) is indeed continuous, see Figure 10.
For the games G or 2 , G and 2 the derivation of the bound is similar but requires few extra work. The result is stated in Theorem 15. The analogue of Proposition 7 is Proposition 11. In this case there are two dropping times and delaying only one dropping time without modifying the other is easy only if the delayed dropping time is the latest. This is the reason why in Proposition 11 we distinguish l 1 ≥ l 2 or l 1 ≤ l 2 . Corollaries 12, 13 and 14 specialize the results of Proposition 11 to identify the region of dropping times that cannot lead to optimal solution. A direct application of these corollaries leads to Theorem 15, this is illustrated on Figure 11.
Proof. Let be given an optimal strategy with value x(l). Considering the same strategy but instead of starting at time 0, players are still until time α.
The resulting strategy leads to the estimate (10).
Proposition 8 (Bounds in G 1 ) We denote x(l) the value of G 1 if the gift is dropped off at time l. For α > 0 we have that Proof. Let x(l) the optimal strategy if the gift is dropped off at time l. The modified strategy follows strategy x(l) until time l − α at which it drops thegift. If the gift (or absence) is never used by the strategy this does not change the remaining rendezvous time and hence the modified strategy statisfies (11). Else, write t the smallest time at which an agent of player II finds the gift or notice the absence. The modified strategy follows x(l) until time t. If the agent of player II finds the gift before time t the modified strategy continues as x(l). Indeed, smaller rendezvous time are consistent with (11). If the agent of player II does not find the gift before time t because of the new position of the gift then, player I strategy is stopped at time t until time t + 2α. In the meantime, the agent of player II continues for α unit of time and then back to the same position by time t + 2α. At this time, the positions of the players are the same as the one at time t but now the agent of player II knows whether the gift was dropped off or not. We are then in the same situation as with the strategy x(l) but delayed by 2α. Since the reasoning can be applied at most four times, we obtain the bound (11).
Proposition 9 For G 1 the function x(l) is continuous.
Theorem 10 (Bounding optimal solutions for G 1 ) Let be given a regular mesh {d i } i=0,...,N of dropping time and the corresponding optimal solutions of G 1 when the dropping times are in the mesh, we denote these values {G 1 (d i )} i=0,...,N . Then, the optimal value of G 1 given that the dropping is constrained to belong to . Moreover, the optimal dropping times are contained in the set Proposition 11 (Bounds for G or 2 , G and 2 ) We denote x(l 1 , l 2 ) the value of the game if the gifts are dropped off at times l 1 and l 2 . For α > 0 we have that Proof. The idea of the proof of this proposition is analogous to the one of propositions 7 and 8. Consider the first line of (12) and the strategy that leads to x(l 1 , l 2 ). If l 1 ≥ l 2 we keep this strategy unchanged until time l 1 , then we freeze the motions of both players for a time span of α and, finally resume the motion. In the second strategy the second player drops off the gift at time l 1 + α and all the subsequent meeting time are delayed at most by α units of time. This leads to the first line of (12). The following lines are proved in a similar way.
Proof. We get using the two first lines of (12) The first inequality comes by adding γ to l 1 and l 2 and the middle inequality of (12). By assumption we have −α ≤ −β + γ, hence l 1 − β + γ ≥ l 2 + α − β + γ ≥ l 2 , and the first line of (12) applies leading to the second inequality. To conclude we use the hypothesis and direct computations.
Theorem 15 (Bounds for G or 2 , G and 2 ) Let be given a regular mesh {(d i , d j )} for i, j = 0, . . . , N of dropping times and the corresponding optimal solutions of the game {X(d i , d j )} i,j=0,...,N , where X = G or 2 or G and 2 . Then, the optimal value of X given that the dropping times belong to    computed around the minimal solution found for marker dropping times (0, 0). for various combination of marker dropping times. The minimal value computed is 24 for dropping times (0, 0), (8,8), (4, x) and (x, 4). Figure 14 nation.On the right of Figure 12 we see the excluded region by Theorem 15 that shows the optimal dropping times are in the region [