Energy Consumption of Group Search on a Line ¹¹1This is the full version of the paper with the same title which will appear in the proceedings of the 46th International Colloquium on Automata, Languages and Programming 8-12 July 2019, Patras, Greece

We generalize the Two-Turn strategy for any values of $c,b$. We analyze the entire spectrum of values of $c,b$ for which the problem admits a solution, and for each of them we provide optimal (and in some cases nearly optimal) speed choices for our robots (Theorem 3.4). In particular, and somewhat surprisingly, our proof makes explicit how for any fixed $c$ the optimal speed choices do not simply "scale" with $b$; rather more delicate speed choices are necessary to comply with the speed and search time bounds. For the special case of $c\cdot b=9$, our results match with the specific Two-Turn strategy described in [15]. Our results further show that no Two-Turn strategy can achieve energy consumption less than $9d$ while keeping the search time at $9d$. In fact, we conjecture that this trade-off is impossible for any group search strategy that uses only constant memory robots.

In the unbounded-memory model, for the special case of $c=9$ and $b=1$, we give a novel search algorithm that achieves energy consumption of $8.42588d$, thus answering the second question above in the affirmative. This result shows that though two robots cannot search faster than one robot on the line [15], somewhat surprisingly, two robots can search using less total energy than one robot, in the same optimal time. Our algorithm uses robots that have unbounded memory, and a finite number of dynamically computed speeds. Note that our algorithm can be generalized for any $c,b$ with $cb=9$, and utilizes energy $8.42588b^{2}d$ (Theorem 4.7).

Several authors have investigated various aspects of mobile robot (agent) search, resulting in an extensive literature on the subject in theoretical computer science and mathematics (e.g., see [1, 29] for reviews). Search by constant-memory robots has been done mainly for finite-state automata (FSA) operating in discrete environments like infinite grids, their finite-size subsets (labyrinths) and other graphs. The main concern of this research was the feasibility of search, rather than time or energy efficiency. For example, [14] showed that no FSA can explore all labyrinths, while [8] proved that one FSA using two pebbles or two FSAs, communicating according to the F2F model can explore all labyrinths. However, no collection of FSAs may explore all finite graphs communicating in the F2F model [38] or wireless model [17]. On the other hand, all graphs of size $n$ may be explored using a robot having $O(\log n)$ memory [37].

Exploration of infinite grids is known as the ANTS problem [28], where it was shown that four collaborating FSAs in the semi-synchronous execution model and communicating according to the F2F scenario can explore an infinite grid. Recently, [13] showed that four FSAs are really needed to explore the grid (while three FSAs can explore an infinite band of the 2-dimensional grid).

Continuous environment cases have been investigated in several papers when the efficiency of the search is often represented by the time of reaching the target (e.g., see [2, 6, 7, 33]). Even in the case of continuous environment as simple as the infinite line, after the seminal papers [6, 7], various scenarios have been studied where the turn cost has been considered [27], the environment was composed of portions permitting different search speeds [25], some knowledge about the target distance was available [10] or where some other parameters are involved in the computation of the cost function [9] (e.g. when the target is moving).

The group search, sometimes interpreted as the evacuation problem has been studied first for the disc environment under the F2F [12, 18, 23, 26, 35] and wireless [18] communication scenarios and then also for other geometric environments (e.g., see [26]). Other variants of search/evacuation problems with a combinatorial flavour have been recently considered in [16, 19, 20, 30, 31]. Some papers investigated the line search problem in the presence of crash faulty [24] and Byzantine faulty agents [22]. The interested reader may also consult the recent survey [21] on selected search and evacuation topics.

The energy used by a mobile robot is usually considered as being spent solely for travelling. As a consequence, in the case of a single, constant speed robot the search time is proportional to the distance travelled and the energy used by a robot. Therefore the problems of minimization of time, distance or energy are usually equivalent for most robots' tasks. For teams of collaborating robots, the searchers often need to synchronize their walks in order to wait for information communicated by other searchers (e.g, see [12, 18, 35]), hence the time of the task and the distance travelled are different. However, the distance travelled by a robot and its energy used are still commensurable quantities.

To the best of our knowledge, energy consumption as a function of mobile robot speed which is based on natural laws of physics (related to the drag force) has never been studied in the search literature before. Our present work is motivated by [15], which proves that the competitive ratio $9$ is tight for group search time with two mobile agents in the F2F model when both agents have unit maximal speeds. More exactly, it follows from [15] that having more unit-speed robots cannot improve the group search time obtained by a single robot. Nevertheless, our paper shows that using more robots can improve the energy spending, while keeping the group-search time still the best possible.

[15] presents interesting examples of group search algorithms for two distinct speed robots communicating according to the F2F scenario. An interested reader may consult  [4], where optimal group search algorithms for a pair of distinct maximal speed robots were proposed for both communication scenarios (F2F and wireless) and for any pair of robots' maximal speeds. It is interesting to note that, according to [4], for any distinct-speed robots with F2F communication, the optimal group search time is obtained only if one of the robots perform the search step not using its full speed.

Paper Organization: In Section 2 we formally define the evacuation problem $\textsc{EE}_{c}^{b}$, and proper notions of efficiency. Our algorithms and their analysis for constant-memory robots is presented in Section 3, while in Section 4 we introduce and analyze algorithms for unbounded-memory robots. All omitted proofs can be found in the Appendix. Also, due to space limitations, all figures appear in Appendix A.

In this section we propose solutions to the unbounded-speed problem $\textsc{EE}_{c}$. Since $\textsc{EE}_{c}$ is the same as $\textsc{EE}_{c}^{\infty}$, by Observation B.1, the problem is well-defined for every fixed $c>0$. Moreover, by the proof of Theorem 3.2 (see proof of Lemma C.1 in the Appendix) algorithm $\mathcal{N}_{s,r,k}$ induces a feasible solution for every $c>0$ as well, and the optimal speeds can be found by solving $\textsc{NLP}_{c}^{\infty}$. Indeed, in the remaining of the section we show how to choose optimal values for $s,r,k$ for solving $\textsc{EE}_{c}$ with $\mathcal{N}_{s,r,k}$. Let

whose exact values are the roots of an algebraic system and will be formally defined later. The main theorem of this section reads as follows.

A high level outline of the proof of Theorem 3.3 is as follows. First we show that any optimal choices of the speeds of $\mathcal{N}_{s,r,k}$ must satisfy the time constraint of $\textsc{NLP}_{c}^{\infty}$ tightly. Then, we show that finding optimal speeds $s,r,k$ of $\mathcal{N}_{s,r,k}$ for the general problem $\textsc{EE}_{c}$ reduces to problem $\textsc{EE}_{1}$. Finally, we obtain the optimal solution to $\textsc{NLP}_{1}^{\infty}$ by standard tools of nonlinear programming (KKT conditions).

In this section, we show how to choose optimal values for $s,r,k$ for solving $\textsc{EE}_{c}^{b}$ with $\mathcal{N}_{s,r,k}$, for the entire spectrum of $c,b$ values for which the problem is solvable by online algorithms.

The main result of this section is the following:

Theorem 3.4 is proven by solving $\textsc{NLP}_{c}^{b}$ of Theorem 3.2. In particular, the induced competitive ratio of $\mathcal{N}_{s,r,k}$ for the choices of Theorem 3.4 is summarized in Figure 1 (Appendix A). Speed values $s,r,k$, are chosen optimally when $cb$ is either at most $\gamma_{1}$ or at least $\gamma_{2}$ (i.e. optimizers to $\textsc{NLP}_{c}^{b}$ admit analytic description). The optimal speed parameters when $\gamma_{1}<cb<\gamma_{2}$ cannot be determined analytically (they are roots of high degree polynomials). The values that appear in Theorem 3.4 are heuristically chosen, but interestingly induce nearly optimal competitive ratio, see Figure 2 (Appendix A).

The proof of Theorem 3.4 is given by Lemma 3.6 (the case $cb\leq\gamma_{1}$), Lemma 3.7 (the case $cb\geq\gamma_{2}$), and Lemma 3.8 (the case $\gamma_{1}<cb<\gamma_{2}$). Next we state these Lemmata, and we sketch their proofs.

For proving Lemma 3.6, first we recall the known optimizer for the special case $cb=9$ (see Corollary C.6 within the Proof of Lemma 3.6 on page C.4), and we identify the tight constraints. Requiring that the exact same inequality constraints to $\textsc{NLP}_{c}^{b}$ remain tight, we ask how large can the product $cb$ be so as to have KKT condition hold true. From the corresponding algebraic system, we obtain the answer $cb\leq\gamma_{1}\approx 9.06609$.

Similarly, from Theorem 3.3 we know the optimizers to $\textsc{NLP}_{c}^{b}$ for large enough values of $cb$, and the corresponding tight constraints to the NLP. Again, using KKT conditions, we show that the same constraints remain tight for the optimizers as long as $cb\geq\gamma_{2}\approx 11.3414$. This way we obtain the following Lemma.

The case $\gamma_{1}<cb<\gamma_{2}$ can be solved optimally only numerically, since the best speed values are obtained by roots to a high degree polynomial. Nevertheless, the following lemma proposes a heuristic choice of speeds (that of Theorem 3.4) which is surprisingly close to the optimal (as suggested by Theorem 3.4, see also Figure 2).

The trick in order to find ``good enough'' optimizers to $\textsc{NLP}_{c}^{b}$ is to guess the subset of inequality constraints that remain tight when $\gamma_{1}<cb<\gamma_{2}$. First, we observe that constraint $r\leq b$ is tight for the provable optimizers for all $c,b$ when $cb\in[9,\gamma_{1}]\cup[\gamma_{2},\infty)$. As the only other constraint that switches from being tight to non-tight in the same interval is $k\leq b$, we are motivated to maintain tightness for constraints $r\leq b$ and the time constraint. Still the algebraic system associated with the corresponding KKT conditions cannot be solved analytically. To bypass this difficulty, and assuming we know (optimal) speed $s$, we use the tight time constraint to find speed $k$ as a function of $c,b,s$. From numerical calculations, we see that optimal speed $s$ is nearly optimal in $c$, and so we heuristically set $s=\alpha c+\beta$. We choose $\alpha,\beta$ so as to have $s$ satisfy optimality conditions for the boundary values $cb=\gamma_{1},\gamma_{2}$. After we identify all parameters to our solution, we compare the value of our solution to the optimal one (obtained numerically), and we verify (using numerical calculations) that our heuristic solution is only by at most 0.03 additively off. The advantage of our analysis is that we obtain closed formulas for the speed parameters for all values of $cb\geq 9$.

We adopt the language of [15] in order to discuss a structural property that any feasible evacuation algorithm for $\textsc{EE}_{9}^{1}$ satisfies. As a result, the purpose of this section is to provide high level intuition for our evacuation algorithm that is presented in subsequent sections.

We refer to the two robots (starting exploration from the origin) as $L$ and $R$, intended to explore to the left and to the right of the origin, respectively. The robot trajectories can be drawn on the Cartesian plane where point-location $(x,-t)$ will correspond to point $x$ on the line being visited by some robot at time $t$. The following Theorem is due to [15] and was originally phrased for the time-evacuation unit-speed robots' problem. We adopt the language of our problem.

Next we present some preliminaries toward describing our $k$-phase exploration algorithms. A phase is a pair $(s,r)$ where $s\in[0,1]$ is a speed and $r\in\mathbb{R}$ is a distance ratio, possibly negative. An $l$-phase algorithm is determined by a position $p_{0}$ on the line and a sequence $S=(s_{1},r_{1}),\ldots,(s_{k},r_{l})$ of $l$ phases (movement instructions). Whenever $r_{i}x<0$, movement will be to the left, whereas $r_{i}x>0$ will correspond to movement to the right.

We will make sure that each time the loop is executed, position $x$ and corresponding time induce point-locations of the robots that lie in the boundary of the cone of Theorem 4.1. If a loop starts at location $x$, then it takes additional time $\sum_{i\in[l]}\frac{|r_{i}||x|}{s_{i}}$ to complete one iteration. We will be referring to quantity $1+\sum_{i\in[l]}\frac{|r_{i}|}{3s_{i}}$ as the expansion factor of Exploration $S$.

In this section we give a formal description of our evacuation algorithm. The most elaborate part of it is when robots are in Exploration States, in which they will perform $3$-phase exploration. It can be shown that $3$-phase exploration based evacuation algorithms that do not violate the constraints of problem $\textsc{EE}_{9}^{1}$ have expansion factor at most 4. Moreover, among those, the ones who minimize the induced energy consumption energy consumption make robots move at speed 1 in the first and third phase^‡‡‡The proof of these facts is lengthy and technical, and is not required for the correctness of our algorithm, rather it only justifies some parameter choices. Robot's speed in the second phase will be denoted by $s$.

We now present a specific $3$-phase exploration algorithm, that we denote by $\mathcal{A}\left(s\right)$, complying with the above conditions, with phases $(-1,1),(4s/(1-s),s)$ and $(4-4s/(1-s),1)$, where $s$ is an exploration speed to be determined later. Robot $L$ will execute the 3-phase exploration with starting position -1, while robot $R$ with starting position $2$. When subroutine $travel(v,p)$ is invoked, the robot sets its speed to $v$ and, from its current position, goes toward position $p$ on the line until it reaches it. We depict the trajectories of the robots while in the Exploration State in Figure 3.

Exploration State of $L$ $travel(1/3,-1)$ $k\leftarrow 0$ repeat     $travel(1,0)$     $travel(s,-4^{k+1}\cdot\frac{s}{1-s})$     $travel(1,-4^{k+1})$     $k\leftarrow k+1$ end Exploration State of $R$ $travel(1/3,2)$ $k\leftarrow 0$ repeat     $travel(1,0)$     $travel(s,2\cdot 4^{k+1}\cdot\frac{s}{1-s})$     $travel(1,2\cdot 4^{k+1})$     $k\leftarrow k+1$ end

A complete execution of one repeat loop within the Exploration State will be referred to as a round. Variable $k$ counts the number of completed rounds. Each robot stays in the Exploration State till the exit it found. When switching to the Chasing state (which happens only for the exit finder), robot remembers its current value of counter $k$, as well as the distance $d$ of the exit to the origin. Based on these values (as well as $s$) it calculates the most efficient trajectory in order to catch the other robot (predicting, when applicable, that the rendezvous can be realized while the other robot is approaching the exit finder). When the rendezvous is realized, robots store their current distance $p$ to the origin, as well as the time $t$ that has already passed. Then, robots need to travel distance $p+d$ to reach the exit. Knowing they have time $9d-t$ remaining, they go to the exit together as slow as possible to reach the exit in time exactly $9d$. Figure 4 provides an illustration of the behavior of the robots after finding the exit.

Chasing State $K\leftarrow 4^{k}$ if I am $R$ then     $K\leftarrow 2\cdot 4^{k}$ end if $s^{\prime}\leftarrow\min\left\{\dfrac{d}{4K-d/s},1\right\}$ Travel toward the other robot at speed $s^{\prime}$ until meeting it at distance $p$ from the origin, and at time $t$. Exit State $\bar{s}\leftarrow\frac{p+d}{9d-t}$ Go toward the exit with speed $\bar{s}$.

In this section we are ready to provide the details for proving Theorem 4.7. Evacuation algorithm $\mathcal{A}\left(s\right)$ is not feasible to $\textsc{EE}_{c}^{b}$ for all values of speed parameter $s$ (of the Exploration States). We will show later that trajectories induce evacuation time at most $9d$ only if $s\in[1/3,1/2]$. In what follows, and even though we have not fixed the value of $s$ yet, we will assume that $s$ has some value between 1/3 and 1/2. The purpose of this section is to fix a value for parameter $s$, show that $\mathcal{A}\left(s\right)$ is feasible to $\textsc{EE}_{9}^{1}$, and subsequently compute the induced energy consumption and competitive ratio. As a reminder, each iteration of the repeat loop of the Exploration States is called a round, and $k$ is a counter for these rounds.

Let $X\in\{L,R\}$ be one of the robots. We define $K(X,k)=4^{k}$ if $X=L$, and $K(X,k)=2\cdot 4^{k}$ if $X=R$, i.e. the position of $X$ at the start of round $k$. We will often analyze 3 cases for the distance $d$ of the exit with respect to $K:=K(X,k)$ (as it also appears in the description of the Chasing State), associated with the following closed intervals

We may simply write $D_{1},D_{2}$ and $D_{3}$ if $K$ is clear from the context. Note that during the second phase of round $K$, robot $L$ explores $D_{1}$ and $D_{2}$, whereas $D_{3}$ is explored during the third phase. The same statement holds for $R$. The following lemma will be useful in analyzing the worst case evacuation time and energy consumption of our algorithm.

Using the lemma above, we can now prove that $\mathcal{A}\left(s\right)$ meets the speed bound and the evacuation time bound.

Lemma 4.3 allows us to derive the speed $s_{b1},s_{b2}$ and $s_{b3}$ at which both robots go toward the exit after meeting for the cases $d\in D_{1},d\in D_{2}$ and $d\in D_{3}$, respectively. We also know the speed $s_{c1}$ at which the exit-finder catches up to the other robot when $d\in D_{1}$. We define

The speed $s_{b2}$ is a simple rearrangement of the speed $\frac{d+qs}{9d-(8K+q)},\mbox{ where }q=\frac{d+d/s-4K}{1-s}$, and $s_{b3}$ is obtained by rearranging $\frac{d+2ds/(1-s)}{9d-(8K+2d+2ds/(1-s))}$.

Next compute the energy consumption. For given $K,d$ and $s$, denote by $E_{L}(K,d,s)$ the energy spent by robot $L$ from time $3$ to time $9d$ when it exits. Similarly, $E_{R}(K,d,s)$ is the energy spent by $R$ from time $6$ to time $9d$. Then, then energy consumption is $E(K,d,s):=\tfrac{1}{3}+E_{L}(K,d,s)+E_{R}(K,d,s)$. For any $K$ and $s$, we also define $F(K,s):=(K-1)(5-4s(s+1))$.

Denote by $E_{i}(k,d,s)$ the value of $E(K,d,s)$ when $d\in D_{i}$, $i=1,2,3$. Our intension now is to fix speed value $s$ that solves the following Nonlinear Program

For every $s\in[1/3,1/2]$ we show in Lemma 4.6 that $\frac{E_{1}(K,d,s)}{d}$ is decreasing in $d\in D_{1}$, that $\frac{E_{2}(K,d,s)}{d}$ is increasing in $d\in D_{2}$, and that $\frac{E_{3}(K,d,s)}{d}$ is decreasing in $d\in D_{3}$. Then, the best parameter $s$ can be chosen so as to make all worst case valued $\frac{E_{i}(K,d,s)}{d}$ equal (if possible) when $i=1,2,3$. The optimal $s$ can be found by numerically finding the roots of a high degree polynomial, and accordingly, we heuristically set $s=0.39403$, inducing the best possible energy consumption for algorithm $\mathcal{A}\left(s\right)$. All relevant formal arguments are within the proof of the next lemma.

By Lemma 4.6, we conclude that for the specific value of $s$, algorithm $\mathcal{A}\left(s\right)$ has competitive ratio $\frac{9^{2}}{2}8.42588\approx 341.24814,$ concluding the proof of Theorem 4.7.