Energy Consumption of Group Search on a Line
Jurek Czyzowicz, Konstantinos Georgiou, Ryan Killick, Evangelos Kranakis, Danny Krizanc, Manuel Lafond, Lata Narayanan, Jaroslav Opatrny, Sunil Shende
EEnergy Consumption of Group Search on a Line ∗ Jurek Czyzowicz ∗ Konstantinos Georgiou † Ryan Killick § Evangelos Kranakis § Danny Krizanc ∗∗ Manuel Lafond ¶ Lata Narayanan †† Jaroslav Opatrny †† Sunil Shende ‡‡ April 23, 2019
Abstract
Consider two robots that start at the origin of the infinite line in search of an exit at an unknownlocation on the line. The robots can collaborate in the search, but can only communicate if they arrive atthe same location at exactly the same time, i.e. they use the so-called face-to-face communication model.The group search time is defined as the worst-case time as a function of d , the distance of the exit from theorigin, when both robots can reach the exit. It has long been known that for a single robot traveling at unitspeed, the search time is at least d − o ( d ) ; a simple doubling strategy achieves this time bound. It wasshown recently in [15] that k ≥ robots traveling at unit speed also require at least d group search time.We investigate energy-time trade-offs in group search by two robots, where the energy loss experiencedby a robot traveling a distance x at constant speed s is given by s x and is motivated by first principles inphysics and engineering. Specifically, we consider the problem of minimizing the total energy used by therobots, under the constraints that the search time is at most a multiple c of the distance d and the speed ofthe robots is bounded by b . Motivation for this study is that for the case when robots must complete thesearch in d time with maximum speed one ( b = 1; c = 9 ), a single robot requires at least d energy,while for two robots, all previously proposed algorithms consume at least d/ energy.When the robots have bounded memory and can use only a constant number of fixed speeds, wegeneralize an algorithm described in [3, 15] to obtain a family of algorithms parametrized by pairs of b, c values that can solve the problem for the entire spectrum of these pairs for which the problem is solvable.In particular, for each such pair, we determine optimal (and in some cases nearly optimal) algorithmsinducing the lowest possible energy consumption.We also propose a novel search algorithm that simultaneously achieves search time d and consumesenergy . d . Our result shows that two robots can search on the line in optimal time d whileconsuming less total energy than a single robot within the same search time. Our algorithm uses robotsthat have unbounded memory, and a finite number of dynamically computed speeds. It can be generalizedfor any c, b with cb = 9 , and consumes energy . b d . ∗ Universite du Qubec en Outaouais, Gatineau, Qu´ebec, Canada, [email protected] † Dept. of Mathematics, Ryerson University, Toronto, ON, Canada, [email protected] § School of Computer Science, Carleton University, Ottawa ON, Canada, ryankillick,[email protected] ∗∗ Department of Mathematics & Comp. Sci., Wesleyan University, Middletown, CT, USA, [email protected] †† Department of Comp. Sci. and Software Eng., Concordia University, Montreal, Qu´ebec, Canada, lata,[email protected] ‡‡ Department of Computer Science, Rutgers University, Camden, USA, [email protected] ¶ Department of Computer Science, Universit´e de Sherbrooke, Qu´ebec, Canada
[email protected] ∗ This is the full version of the paper with the same title which will appear in the proceedings of the 46th International Colloquiumon Automata, Languages and Programming 8-12 July 2019, Patras, Greece a r X i v : . [ c s . D M ] A p r Introduction
The problem of searching for a treasure at an unknown location in a specified continuous domain was initiatedover fifty years ago [6, 7]. Search domains that have been considered include the infinite line [2, 6, 7, 33],a set of rays [10, 11], the unit circle [12, 23, 36], and polygons [26, 32, 34]. Consider a robot (sometimescalled a mobile agent) starting at some known location in the domain and looking for an exit that is located atan unknown distance d away from the start. What algorithm should the robot use to find the exit as soon aspossible? The most common cost measure used for the search algorithm is the worst-case search time , as afunction of the distance d of the exit from the starting position. For a fixed-speed robot, the search time isproportional to the length of the trajectory of the robot. Other measures such as turn cost [27] and differentcosts for revisiting [9] have been sometimes considered.We consider for the first time the energy consumed by the robots while executing the search algorithm.The energy used by a robot to travel a distance x at speed s is computed as s x and is motivated from theconcept of viscous drag in fluid dynamics; see Section 2 for details on the energy model. For a single robotsearching on the line, the classic Spiral Search algorithm (also known as the doubling strategy) has searchtime d and is known to be optimal when the robot moves with unit speed. Since in the worst case, the robottravels distance d at unit speed, the energy consumption is d as well. Clearly, as the speed of the robotincreases, the time to find the exit decreases but at the same time, the energy used increases. Likewise, asthe speed of the robot decreases, the time to find the exit increases, while the energy consumption decreases.Thus there is a natural trade-off between the time taken by the robot to search for the exit and the energyconsumed by the robot. To investigate this trade-off, we consider the problem of minimizing the total energyused by the robots to perform the search when the speed of the robot is bounded by b , and the time for thesearch is at most a multiple c of the distance d from the starting point to the exit.Group search by a set of k ≥ collaborating robots has recently gained a lot of attention. In this case,the search time is the time when all k robots reach the exit. The problem has also been called evacuation ,in view of the application when it is desired that all robots reach and evacuate from the exit. Two modelsof communication between the robots have been considered. In the wireless communication model, therobots can instantly communicate with each other at any time and over any distance. In the face-to-facecommunication model (F2F), two robots can communicate only when in the same place at the same time. Inmany search domains, and for both communication models, group search by k ≥ agents has been shown totake less time than search by a single agent; see for example [23, 26].In this paper, we focus on group search on the line, by two robots using the F2F model. Chrobak et al [15] showed that group search in this setting cannot be performed in time less than d − o ( d ) , regardlessof the number of robots, assuming all robots use at most unit speed. They also describe several strategiesthat achieve search time d . In the first strategy, the two robots independently perform the Spiral Searchalgorithm, using unit speed during the entire search. Next, they consider a strategy first described in [3], thatwe call the Two-Turn strategy, whereby two robots head off independently in opposite directions at speed / ; when one of them finds the exit, it moves at unit speed to chase and catch the other robot, after whichthey both return at unit speed to the exit. Finally, they present a new strategy, called the Fast-Slow algorithmin which one robot moves at unit speed, while the other robot moves at speed 1/3, both performing a spiralsearch. The doubling strategy is very energy-inefficient, it uses energy d if the two robots always traveltogether, or D if the robots start by moving in opposite directions. The other two algorithms both useenergy d/ > d . Interestingly, the two strategies that achieve an energy consumption of d/ withsearch time d , both use two different and pre-computed speeds, but are quite different in terms of the robotcapacities needed. In the Two-Turn strategy, the robots are extremely simple and use constant memory; they2se only three states. In Fast-Slow and Spiral Search, the robots need unbounded memory, and performcomputations to determine how far to go before turning and moving in the opposite direction.Memory capability, time- and speed-bounded search, and energy consumption by a two-robot groupsearch algorithm on the line: these considerations motivate the following questions that we address in ourpaper:1. Is there a search strategy for constant-memory robots that has energy consumption < d ?2. Is there any search strategy that uses time d and energy < d ? 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 casesnearly optimal) speed choices for our robots (Theorem 3.4). In particular, and somewhat surprisingly, ourproof makes explicit how for any fixed c the optimal speed choices do not simply ”scale” with b ; rather moredelicate speed choices are necessary to comply with the speed and search time bounds. For the special caseof c · b = 9 , our results match with the specific Two-Turn strategy described in [15]. Our results further showthat no Two-Turn strategy can achieve energy consumption less than d while keeping the search time at d .In fact, we conjecture that this trade-off is impossible for any group search strategy that uses only constantmemory robots.In the unbounded-memory model, for the special case of c = 9 and b = 1 , we give a novel searchalgorithm that achieves energy consumption of . d , thus answering the second question above in theaffirmative. 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 optimaltime. Our algorithm uses robots that have unbounded memory, and a finite number of dynamically computedspeeds. Note that our algorithm can be generalized for any c, b with cb = 9 , and utilizes energy . b d (Theorem 4.7). Several authors have investigated various aspects of mobile robot (agent) search, resulting in an extensiveliterature 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 discreteenvironments like infinite grids, their finite-size subsets (labyrinths) and other graphs. The main concern ofthis research was the feasibility of search, rather than time or energy efficiency. For example, [14] showedthat 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 mayexplore all finite graphs communicating in the F2F model [38] or wireless model [17]. On the other hand, allgraphs 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 collabo-rating FSAs in the semi-synchronous execution model and communicating according to the F2F scenario canexplore an infinite grid. Recently, [13] showed that four FSAs are really needed to explore the grid (whilethree 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 searchis often represented by the time of reaching the target (e.g., see [2, 6, 7, 33]). Even in the case of continuousenvironment as simple as the infinite line, after the seminal papers [6, 7], various scenarios have been3tudied where the turn cost has been considered [27], the environment was composed of portions permittingdifferent search speeds [25], some knowledge about the target distance was available [10] or where someother 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 discenvironment under the F2F [12, 18, 23, 26, 35] and wireless [18] communication scenarios and then alsofor other geometric environments (e.g., see [26]). Other variants of search/evacuation problems with acombinatorial flavour have been recently considered in [16, 19, 20, 30, 31]. Some papers investigated the linesearch problem in the presence of crash faulty [24] and Byzantine faulty agents [22]. The interested readermay 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 aconsequence, in the case of a single, constant speed robot the search time is proportional to the distancetravelled and the energy used by a robot. Therefore the problems of minimization of time, distance orenergy are usually equivalent for most robots’ tasks. For teams of collaborating robots, the searchers oftenneed 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 distancetravelled 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 basedon 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 is tight for group searchtime with two mobile agents in the F2F model when both agents have unit maximal speeds. More exactly, itfollows from [15] that having more unit-speed robots cannot improve the group search time obtained by asingle robot. Nevertheless, our paper shows that using more robots can improve the energy spending, whilekeeping the group-search time still the best possible.[15] presents interesting examples of group search algorithms for two distinct speed robots communi-cating according to the F2F scenario. An interested reader may consult [4], where optimal group searchalgorithms for a pair of distinct maximal speed robots were proposed for both communication scenarios (F2Fand wireless) and for any pair of robots’ maximal speeds. It is interesting to note that, according to [4], forany distinct-speed robots with F2F communication, the optimal group search time is obtained only if one ofthe robots perform the search step not using its full speed. Paper Organization:
In Section 2 we formally define the evacuation problem EE bc , and proper notionsof efficiency. Our algorithms and their analysis for constant-memory robots is presented in Section 3, whilein Section 4 we introduce and analyze algorithms for unbounded-memory robots. All omitted proofs can befound in the Appendix. Also, due to space limitations, all figures appear in Appendix A. Two robots start walking from the origin of an infinite (bidirectional) line in search of a hidden exit at anunknown absolute distance d from the origin. The exit is considered found only when one of the robots walksover it. An algorithm for group search by two robots specifies trajectories for both robots and terminateswhen both robots reach the exit. The time by which the second robot reaches the exit is referred to as the search time or the evacuation time . Robot models:
The two robots operate under the F2F communication model in which two robots cancommunicate only when they are in the same place at the same time. Each robot can change its speed at anytime. We distinguish between constant-memory robots that can only travel at a constant number of hard-wiredspeeds, and unbounded-memory robots that can dynamically compute speeds and distances, and travel at any4ossible speed.
Energy model:
A robot moving at constant speed s traversing an interval of length x is defined to use energy s · x . This model is well motivated from first principles in physics and engineering and corresponds to theenergy loss experienced by an object moving through a viscous fluid [5]. In particular, an object moving withconstant speed s will experience a drag force F D proportional ∗ to s . In order to maintain the speed s over adistance x the object must do work equal to the product of F D and x resulting in a continuous energy lossproportional to the product of the object’s squared speed and travel distance. For simplicity we have taken theproportionality constant to be one.The total energy that a robot uses traveling at speeds s , s , . . . , s t , traversing intervals x , x , . . . , x t ,respectively, is defined as P ti =1 s i · x i . For group search with two robots, the energy consumption is definedas the sum total of the two robots’ energies used until the search algorithm terminates.For each d > there are two possible locations for the exit to be at distance d from the origin: we willrefer to either of these as input instances d for the group search problem. Our goal is to solve the following optimized search problem parametrized by two values, b and c : (cid:73) Problem EE bc : Design a group search algorithm for two robots in the F2F model that minimizes theenergy consumption for d -instances under the constraints that the search time is no more than c · d and therobots use speeds that are at most b . When there are no speed limits on the robots (i.e. b = ∞ ), we abbreviateEE ∞ c by EE c . Note that b, c are inputs to the algorithm, but d and the exact location of the exit are not known.As it is standard in the literature on related problems, we assume that the exist is at least a known constantdistance away from the origin. In this work, we pick the constant equal to 2, although our arguments canbe adjusted to any other constant. It is not difficult to show that EE bc is well defined for each b, c > with bc ≥ , and the optimal offline solution, for instance d , is for both robots to move at speed c to the exit.This offline algorithm has energy consumption dc (see Observation B.1 in Appendix B). Consider an onlinealgorithm for EE bc , which on any instance d has energy consumption at most e ( c, b, d ) . The competitive ratio of the algorithm is defined as sup d> c d e ( c, b, d ) . Due to [15], and when b = 1 , no online algorithm (for two robots) can have evacuation time less than d − (cid:15) (for any (cid:15) > and for large enough d ). By scaling, using arbitrary speed limit b , we obtain thefollowing fact. Observation 2.1.
No online F2F algorithm can solve EE bc if cb < . bc with Constant-Memory Robots In this section we propose a family of algorithms for solving EE bc (including b = ∞ ). The family uses analgorithm that is parametrized by three discrete speeds: s , r and k . The robots use these speeds depending onfinite state control as follows: (cid:73) Algorithm N s,r,k : Robots start moving in opposite directions with speed s until the exit is found by oneof them. The finder changes direction and moves at speed r > s until it catches the other robot. Together thetwo robots return to the exit using speed k . ∗ The constant of proportionality has (SI) units kg/m and depends, among other things, on the shape of the object and the densityof the fluid through which it moves. emma 3.1 (Proof on page 16) . Let b, c be such that there exist s, r, k for which N s,r,k is feasible. Then, forinstance d of EE bc , the induced evacuation time of N s,r,k is d · T ( s, r, k ) and the induced energy consumptionis d · E ( s, r, k ) , where T ( s, r, k ) := 2( k + r ) k ( r − s ) + 1 s , E ( s, r, k ) := rr − s (cid:16) s + r + 2 k (cid:17) We propose a systematic way in order to find optimal values for s, r, k of algorithm N s,r,k for optimizationproblem EE bc (including b = ∞ ), whenever such values exist. Theorem 3.2 (Proof on page 16) . Algorithm N s,r,k gives rise to a feasible solution to problem EE bc if andonly if bc ≥ . For every such b, c > , the optimal choices of N s,r,k can be obtained by solving Non LinearProgram: min s,r,k ∈ R E ( s, r, k ) (NLP bc ) s.t. T ( s, r, k ) ≤ cr ≥ s ≤ s, r, k ≤ b where functions E ( · , · , · ) , T ( · , · , · ) are as in Lemma 3.1. Moreover, if s , r , k are the optimizers to NLP bc ,then the competitive ratio of N s ,r ,k equals c · E ( s , r , k ) . The following subsections are devoted to solving NLP bc , effectively proving Theorem 3.4. First inSection 3.1 we solve the case b = ∞ and we use our findings to solve the case of bounded speeds b in thefollow-up Section 3.2. N s,r,k for the Unbounded-Speed Problem In this section we propose solutions to the unbounded-speed problem EE c . Since EE c is the same as EE ∞ c , byObservation B.1, the problem is well-defined for every fixed c > . Moreover, by the proof of Theorem 3.2(see proof of Lemma C.1 in the Appendix) algorithm N s,r,k induces a feasible solution for every c > aswell, and the optimal speeds can be found by solving NLP ∞ c . Indeed, in the remaining of the section weshow how to choose optimal values for s, r, k for solving EE c with N s,r,k . Let σ ≈ . , ρ ≈ . , κ ≈ . , (1)whose exact values are the roots of an algebraic system and will be formally defined later. The main theoremof this section reads as follows. Theorem 3.3 (Proof on page 18) . Let σ, ρ, κ as in (1) . For every c > , the optimal speeds of N s,r,k forproblem EE c are s = σc , r = ρc , k = κc . Moreover, the competitive ratio of the corresponding solution isindependent of c and equals ρ ( κ + ρ + σ ) ρ − σ ≈ . . A high level outline of the proof of Theorem 3.3 is as follows. First we show that any optimal choices ofthe speeds of N s,r,k must satisfy the time constraint of NLP ∞ c tightly. Then, we show that finding optimalspeeds s, r, k of N s,r,k for the general problem EE c reduces to problem EE . Finally, we obtain the optimalsolution to NLP ∞ by standard tools of nonlinear programming (KKT conditions).6 .2 (Sub)Optimal Choices of N s,r,k for the Bounded-Speed Problem In this section, we show how to choose optimal values for s, r, k for solving EE bc with N s,r,k , for the entirespectrum 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.
Let γ ≈ . , γ = ρ ≈ . , and σ, ρ, κ as in (1) . For every c, b > with cb ≥ ,the following choices of speeds s, r, k are feasible for N s,r,k ≤ cb ≤ γ γ < cb < γ cb ≥ γ s − √ ( bc ) − bc +9+ bc − c . b − . b c σ/cr b b ρ/ck b bsbcs − b − cs − s κ/c The induced competitive ratio is given by: f ( x ) := x (cid:16) x (cid:16) x − p ( x − x − (cid:17) + p ( x − x −
1) + 3 (cid:17) , ≤ x ≤ γ x (cid:16) (0 . − . x ) + . . − .x )2( x ( x ( x − . − . . +1 (cid:17) . x +0 . , γ < x < γ . x ≥ γ and the induced energy, for instances d , is f ( cb ) dc . Moreover, the competitive ratio depends only on theproduct cb .In particular, the speeds’ choices are optimal when cb ≤ γ and when cb ≥ γ . When γ < cb < γ , thederived competitive ratio is no more than 0.03 additively off from that induced by optimal choices of s, r, k . Corollary 3.5.
For c = 9 , b = 1 , the bounded-memory robot algorithm N s,r,k has energy consumption d/ and competitive ratio 378. Theorem 3.4 is proven by solving NLP bc of Theorem 3.2. In particular, the induced competitive ratioof 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 γ or at least γ (i.e. optimizers to NLP bc admit analyticdescription). The optimal speed parameters when γ < cb < γ cannot be determined analytically (theyare roots of high degree polynomials). The values that appear in Theorem 3.4 are heuristically chosen, butinterestingly 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 ≤ γ ), Lemma 3.7 (the case cb ≥ γ ), andLemma 3.8 (the case γ < cb < γ ). Next we state these Lemmata, and we sketch their proofs. Lemma 3.6 (Proof on page 20) . For every c ∈ (9 /b, γ /b ] , where γ ≈ . , the optimizers to NLP bc are k = r = b , and s b = − √ ( bc ) − bc +9+ bc − c . The induced competitive ratio is f ( cb ) , (see definition of f ( x ) for x ≤ γ in statement of Theorem 3.4), and the energy consumption, for instances d , is f ( cb ) dc . For proving Lemma 3.6, first we recall the known optimizer for the special case cb = 9 (see Corollary C.6within the Proof of Lemma 3.6 on page 20), and we identify the tight constraints. Requiring that the exactsame inequality constraints to NLP bc remain tight, we ask how large can the product cb be so as to have KKTcondition hold true. From the corresponding algebraic system, we obtain the answer cb ≤ γ ≈ . .Similarly, from Theorem 3.3 we know the optimizers to NLP bc for large enough values of cb , and thecorresponding tight constraints to the NLP. Again, using KKT conditions, we show that the same constraintsremain tight for the optimizers as long as cb ≥ γ ≈ . . This way we obtain the following Lemma.7 emma 3.7 (Proof on page 23) . For every c > ρ/b ≈ . /b , the optimal speeds of N s,r,k for EE bc are s = σ/c, r = ρ/c, k = κ/c, i.e. they are the same as for EE ∞ c . If the target is placed at distance d from theorigin, then the induced energy equals . dc . Moreover, the induced competitive ratio is . , andis independent of b, c . The case γ < cb < γ can be solved optimally only numerically, since the best speed values are obtainedby roots to a high degree polynomial. Nevertheless, the following lemma proposes a heuristic choice ofspeeds (that of Theorem 3.4) which is surprisingly close to the optimal (as suggested by Theorem 3.4, seealso Figure 2). Lemma 3.8 (Proof on page 23) . The choices of s, r, k of Theorem 3.4 when γ < cb < γ are feasible.Moreover, the induced competitive ratio is at most 0.03 additively off from the competitive ratio induced bythe optimal choices of speeds (evaluated numerically). The trick in order to find “good enough” optimizers to NLP bc is to guess the subset of inequalityconstraints that remain tight when γ < cb < γ . First, we observe that constraint r ≤ b is tight for theprovable optimizers for all c, b when cb ∈ [9 , γ ] ∪ [ γ , ∞ ) . As the only other constraint that switches frombeing tight to non-tight in the same interval is k ≤ b , we are motivated to maintain tightness for constraints r ≤ b and the time constraint. Still the algebraic system associated with the corresponding KKT conditionscannot be solved analytically. To bypass this difficulty, and assuming we know (optimal) speed s , we usethe tight time constraint to find speed k as a function of c, b, s . From numerical calculations, we see thatoptimal speed s is nearly optimal in c , and so we heuristically set s = αc + β . We choose α, β so as to have s satisfy optimality conditions for the boundary values cb = γ , γ . After we identify all parameters to oursolution, 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 advantageof our analysis is that we obtain closed formulas for the speed parameters for all values of cb ≥ . bc with Unbounded-Memory Robots In this section we prove Theorem 4.7, that is we solve EE bc by assuming that the two robots have unboundedmemory, and in particular that they can perform time and state dependent calculations and tasks. Note that,by scaling, our results hold for all b, c for which cb = 9 . For simplicity our exposition is for the natural case c = 9 and b = 1 . Also, as before, d will denote the unknown distance of the exit from the origin, still the exitis assumed, for the purposes of performance analysis, to be at least 2 away from the origin.Throughout the execution of our evacuation algorithm, robots can be in 3 different states (similar tothe case of constant-memory robots). First, both robots start with the Exploration State and they remainin this until the exit is located. While in the exploration state, robots execute an elaborate exploration thatrequires synchronous movements in which robots, at a high level, stay in good proximity, still they expandthe searched space relatively fast. Then, the exit finder enters the
Chasing State in which the robot, dependingon its distance from the origin, calculates a speed, at which to move in order to catch and notify the otherrobot. Lastly, when the two robots meet, they both enter the
Exit State in which both robots move toward theexit with the smallest possible speed while meeting the time constraint.Our algorithm takes as input the values of c = 9 , b = 1 , and use a speed value s ≤ b , that will be chosenlater. When the exit finder switches its state from Exploration to Chasing, it remembers the distance d of theexit to the origin, as well as the value k of a counter that was used while in the Exploration State. When theexit finder catches the other robot, they both switch to the Exit State, and they remember their distance p t that their rendezvous was realized. The speed of their Exit Statewill be determined as a function of p, d, t (and hence of s, c, b as well). l -Phase Explorations We adopt the language of [15] in order to discuss a structural property that any feasible evacuation algorithmfor EE satisfies. As a result, the purpose of this section is to provide high level intuition for our evacuationalgorithm 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 theleft and to the right of the origin, respectively. The robot trajectories can be drawn on the Cartesian planewhere 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. Theorem 4.1.
For any feasible solution to EE , the point-location of any robot lies within the cone spannedby vectors (cid:0) − − (cid:1) , (cid:0) − (cid:1) . Next we present some preliminaries toward describing our k -phase exploration algorithms. A phase is a pair ( s, r ) where s ∈ [0 , is a speed and r ∈ R is a distance ratio , possibly negative. An l -phasealgorithm is determined by a position p on the line and a sequence S = ( s , r ) , . . . , ( s k , r l ) of l phases(movement instructions). Whenever r i x < , movement will be to the left, whereas r i x > will correspondto movement to the right. We will make sure that each time the loop is executed, position x and corresponding l - PHASE E XPLORATION : GIVEN p AND S = ( s , r ) , . . . , ( s l , r l ) Go to p at speed / repeat x ← current position for i = 1 , . . . , l do Travel at speed s i for a distance of r i · x endend time induce point-locations of the robots that lie in the boundary of the cone of Theorem 4.1. If a loop startsat location x , then it takes additional time P i ∈ [ l ] | r i || x | s i to complete one iteration. We will be referring toquantity P i ∈ [ l ] | r i | s i as the expansion factor of Exploration S . A ( s ) : The Exploration, Chasing and Exit States In this section we give a formal description of our evacuation algorithm. The most elaborate part of it iswhen robots are in Exploration States, in which they will perform -phase exploration. It can be shown that -phase exploration based evacuation algorithms that do not violate the constraints of problem EE haveexpansion factor at most 4. Moreover, among those, the ones who minimize the induced energy consumptionenergy consumption make robots move at speed 1 in the first and third phase † . Robot’s speed in the secondphase will be denoted by s . † The proof of these facts is lengthy and technical, and is not required for the correctness of our algorithm, rather it only justifiessome parameter choices
9e now present a specific -phase exploration algorithm, that we denote by A ( s ) , complying with theabove conditions, with phases ( − , , (4 s/ (1 − s ) , s ) and (4 − s/ (1 − s ) , , where s is an explorationspeed to be determined later. Robot L will execute the 3-phase exploration with starting position -1, whilerobot R with starting position . 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 ofthe robots while in the Exploration State in Figure 3.E XPLORATION S TATE OF
Ltravel (1 / , − k ← repeat travel (1 , travel ( s, − k +1 · s − s ) travel (1 , − k +1 ) k ← k + 1 end E XPLORATION S TATE OF
Rtravel (1 / , k ← repeat travel (1 , travel ( s, · k +1 · s − s ) travel (1 , · k +1 ) k ← 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 currentvalue of counter k , as well as the distance d of the exit to the origin. Based on these values (as well as s ) itcalculates the most efficient trajectory in order to catch the other robot (predicting, when applicable, that therendezvous can be realized while the other robot is approaching the exit finder). When the rendezvous isrealized, 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 d − t remaining, they go tothe exit together as slow as possible to reach the exit in time exactly d . Figure 4 provides an illustration ofthe behavior of the robots after finding the exit.C HASING S TATE K ← k if I am R then K ← · k end s ← min (cid:26) d K − d/s , (cid:27) Travel toward the other robot at speed s until meeting it at distance p from the origin, and at time t . E XIT S TATE ¯ s ← p + d d − t Go toward the exit withspeed ¯ s . s In this section we are ready to provide the details for proving Theorem 4.7. Evacuation algorithm A ( s ) isnot feasible to EE bc for all values of speed parameter s (of the Exploration States). We will show later thattrajectories induce evacuation time at most d only if s ∈ [1 / , / . In what follows, and even though wehave not fixed the value of s yet, we will assume that s has some value between 1/3 and 1/2. The purpose ofthis section is to fix a value for parameter s , show that A ( s ) is feasible to EE , and subsequently compute10he induced energy consumption and competitive ratio. As a reminder, each iteration of the repeat loop of theExploration States is called a round, and k is a counter for these rounds. Proposition 4.2 (Proof on page 24) . For every k ≥ , and at the start of its k -th round, • robot L is at position − k at time · k , and • robot R , is at position · k at time · k . Let X ∈ { L, R } be one of the robots. We define K ( X, k ) = 4 k if X = L , and K ( X, k ) = 2 · 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 theexit with respect to K := K ( X, k ) (as it also appears in the description of the Chasing State), associatedwith the following closed intervals D ( K ) := [ K, Ks/ ( s + 1)] , D ( K ) := [4 Ks/ ( s + 1) , Ks/ (1 − s )] , D ( K ) := [4 Ks/ (1 − s ) , K ] . We may simply write D , D and D if K is clear from the context. Note that during the second phase ofround K , robot L explores D and D , whereas D is explored during the third phase. The same statementholds for R . The following lemma will be useful in analyzing the worst case evacuation time and energyconsumption of our algorithm. Lemma 4.3 (Proof on page 24) . Suppose that robot X ∈ { L, R } finds the exit at distance d when its roundcounter has value k . Let p and t be, respectively, the position and time at which X first meets with the otherrobot after having found the exit, and set K := K ( X, k ) . Then the following hold:1. If d ∈ D , then p = 0 and t = 8 K .2. If d ∈ D , then | p | = d + ds − Ks − s and t = 8 K + d + d/s − K − s .3. If d ∈ D , then | p | = 2 ds/ (1 − s ) and t = 8 K + 2 d + 2 ds/ (1 − s ) . Using the lemma above, we can now prove that A ( s ) meets the speed bound and the evacuation timebound. Lemma 4.4 (Proof on page 25) . For any s ∈ [1 / , / , evacuation algorithm A ( s ) is feasible to EE . Lemma 4.3 allows us to derive the speed s b , s b and s b at which both robots go toward the exit aftermeeting for the cases d ∈ D , d ∈ D and d ∈ D , respectively. We also know the speed s c at which theexit-finder catches up to the other robot when d ∈ D . We define s b := d d − K , s c = d K − d/s , s b := d − Ksd (8 − s − /s )+4 K (2 s − , s b := d (1+ s ) d (7 − s )+8 K ( s − The speed s b is a simple rearrangement of the speed d + qs d − (8 K + q ) , where q = d + d/s − K − s , and s b is obtainedby rearranging d +2 ds/ (1 − s )9 d − (8 K +2 d +2 ds/ (1 − s )) .Next compute the energy consumption. For given K, d and s , denote by E L ( K, d, s ) the energy spent byrobot L from time to time d when it exits. Similarly, E R ( K, d, s ) is the energy spent by R from time totime d . Then, then energy consumption is E ( K, d, s ) := + E L ( K, d, s ) + E R ( K, d, s ) . For any K and s ,we also define F ( K, s ) := ( K − − s ( s + 1)) .11 emma 4.5 (Proof on page 26) . Suppose that robot X ∈ { L, R } finds the exit at distance d when its roundcounter has value k , and let K := K ( X, k ) . Then E ( K, d, s ) = 13 + F ( K, s ) + 3 K + d ( s + s c + 2 s b ) if d ∈ D F ( K, s ) + 3 K + (cid:18) d − Ks − s (cid:19) (1 + s + 2 s b ) if d ∈ D F ( K, s ) + 3 K − Ks ( s + 1) + 2 d − s ( s + s b ( s + 1) + 1) if d ∈ D . Denote by E i ( k, d, s ) the value of E ( K, d, s ) when d ∈ D i , i = 1 , , . Our intension now is to fix speedvalue s that solves the following Nonlinear Program min s ∈ [1 / , / ( max ( sup d ∈ D ,k ≥ ,X E ( K, d, s ) d , sup d ∈ D ,k ≥ ,X E ( K, d, s ) d , sup d ∈ D ,k ≥ ,X E ( K, d, s ) d )) . For every s ∈ [1 / , / we show in Lemma 4.6 that E ( K,d,s ) d is decreasing in d ∈ D , that E ( K,d,s ) d isincreasing in d ∈ D , and that E ( K,d,s ) d is decreasing in d ∈ D . Then, the best parameter s can be chosen soas to make all worst case valued E i ( K,d,s ) d equal (if possible) when i = 1 , , . The optimal s can be found bynumerically finding the roots of a high degree polynomial, and accordingly, we heuristically set s = 0 . ,inducing the best possible energy consumption for algorithm A ( s ) . All relevant formal arguments are withinthe proof of the next lemma. Lemma 4.6 (Proof on Page 27) . On instance d of EE , algorithm A ( s ) induces energy consumption at most . d , when s = 0 . . By Lemma 4.6, we conclude that for the specific value of s , algorithm A ( s ) has competitive ratio . ≈ . , concluding the proof of Theorem 4.7. Theorem 4.7.
For every c, b > with cb = 9 , there is an evacuation algorithm for unbounded-memoryautonomous robots solving EE bc inducing energy consumption . b d for instances d , and competitiveratio 341.24814. Acknowledgements
Research supported by NSERC discovery grants, NSERC graduate scholarship, and NSF.
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Figures
Figure 1: The competitive ratio of algo-rithm N s,r,k (vertical axis) for the entirespectrum of cb ≥ (horizontal axis). Redcurve corresponds to the case cb ≤ γ ,blue curve to the case cb ∈ ( γ , γ ) andgreen curve to the case cb ≥ γ . Thecurve is continuous and differentiable forall cb ≥ . Figure 2: Comparison between the competitive ratioachieved by using optimal speed parameters to NLP bc ofTheorem 3.2 (calculated numerically using software) andthe competitive ratio achieved by the choices of Theo-rem 3.4. The vertical axis is the difference of the com-petitive ratios, and the horizontal axis corresponds to thevalues of cb ∈ ( γ , γ ) (for all other values of cb the differ-ence is provably 0). − k − k +1 · k · k +1 − · k s − s Figure 3: A representation ofposition (x-axis, vertical dashedline is ) and time (y-axis), andthe trajectory followed by thetwo robots (solid lines). The twodiagonal dashed lines form the“1/3 cone” of Theorem 4.1. − k − k +1 · k ≤ − · k ss +1 − k − k +1 · k ≥ − · k ss +1 Figure 4: The robots’ behavior when the exit is found by L isindicated by the bold line. In the first case (left), the catch-upspeed is slower than (and the rendezvous is realized at theturning point of the non-finder), whereas it is in the secondcase (right). 15 Observation B.1
Observation B.1. EE bc is well defined for each b, c > with bc ≥ , and the optimal solution, given thatinstance d is known, equals dc . Proof: [Proof of Observation B.1] Given that the location of the exit is known, and by symmetry, it isimmediate that both robots have the same optimal speed, call it s , and they move in the direction of theexit. The induced evacuation time is then d/s , and the induced evacuation energy is d · s . For a feasiblesolution we require that d/s ≤ c · d and that < s ≤ b , and hence, the optimal offline solution is obtained asthe solution to min s { s : 1 /c ≤ s ≤ b } . For a feasible solution to exist, we need bc ≥ . Moreover, it isimmediate that the optimal choice is s = 1 /c , inducing energy consumption d · s = 2 d/c . C Proofs Omitted from Section 3.
C.1 Lemma 3.1
Proof: [Proof of Lemma 3.1] Consider the moment that the exit is located, after time d/s time of searching.The robot that now chases the other speed- s robot at constant speed r > s will reach it after d/ ( r − s ) time.To see this note that the configuration is equivalent to that the speed- s robot is immobile and the other robotmoves at speed r − s , having to traverse a total distance of d . Moreover, the speed- s robot traverses anadditional length ds/ ( r − s ) segment till it is caught, being a total of ds/ ( r − s ) + 2 d away from the exit.Once robots meet, the walk to the exit at speed k , which takes additional time (2 ds/ ( r − s ) + 2 d ) /k . Overallthe evacuation time equals ds + 2 dr − s + 2 ds/ ( r − s ) + 2 dk = d (cid:18) k + r ) k ( r − s ) + 1 s (cid:19) . Similarly we compute the total energy till both robots reach the exit. The energy spent by the finder is d · s + (cid:18) dsr − s + 2 d (cid:19) · r + (cid:18) dsr − s + 2 d (cid:19) · k , while the energy spent by the non finder is (cid:18) d + 2 dsr − s (cid:19) · s + (cid:18) dsr − s + 2 d (cid:19) · k . Adding the two quantities and simplifying gives the promised formula.
C.2 Theorem 3.2
Proof: [Proof of Theorem 3.2] Note that in NLP ∞ c that aims to provide a solution to EE c , constraints s, r, k ≤ ∞ are simply omitted. In particular, the theorem above claims that when b = ∞ , i.e. when speedsare unbounded, algorithm N s,r,k always admits some feasible solution. In what follows, we prove all claimsof the theorem.By Lemma 3.1, the energy performance of N s,r,k equals d · E ( s, r, k ) , and the induced evacuationtime is d · T ( s, r, k ) . For the values of s, r, k to be feasible, we need that < s, r, k ≤ b , that r > k andthat d · T ( s, r, k ) ≤ cd . Clearly the latter time constraint simplifies to the time constraint of NLP bc , while16he objective value can be scaled by d > without affecting the optimizers to the NLP, if such optimizersexist. Finally note that even though the strict inequalities become non strict inequalities in the NLP, speedsevaluations for which any of s, r, k is 0 or r = k violates the time constraint (for any fixed c > ). Therefore,NLP bc correctly formulates the problem of choosing optimal values for N s,r,k for solving EE bc .The next two lemmas show that the Naive algorithm can solve problem EE bc for the entire spectrum of c, b values for which the problem admits solutions, as per Lemma 2.1. Lemma C.1.
For every c , problem NLP ∞ c admits an optimal solution. Proof: [Proof of Lemma C.1] Consider the redundant constraints s, r, k ≥ /c that can be derived by theexisting constraints of NLP ∞ c (note that if all speeds are not at least /c then clearly the time constraint isviolated). For the same reason, it is also easy to see that r − s ≥ /c , since again we would have a violationof the time constraint.Next, it is easy to check that s = 7 /c, r = 14 /c, k = 7 /c is a feasible solution, hence the NLP is notinfeasible. The value of the objective for this evaluation is /c . But then, notice that the objective isbounded from below by s + r + 2 k . Hence, if an optimal solution exists, constraints s, r, k ≤ √ /c are valid for the optimizers. We may add these constraints to NLP ∞ c , resulting into a compact (closed andbounded) feasible region. But then, note that the objective is continuous over the new compact feasible region,hence from the extreme value theorem it attains a minimum. Lemma C.2.
There exist s, r, k for which N s,r,k induces a feasible solution to EE bc if and only if c ≥ /b . Proof: [Proof of Lemma C.2] Consider the problem of minimizing completion time of the Naive Algorithm,given that the speeds are all bounded above by b . The corresponding NLP that solves the problem reads as. min 2( k + r ) k ( r − s ) + 1 s (2) s.t. r ≥ s ≤ s, r, k ≤ b Note that it is enough to prove that the optimal value to (2) is /b . Indeed, that would imply that no speedsexist that induce completion time less than /b , making the corresponding feasible region of NLP bc empty if c < /b .Now we show that the optimal value to (2) is /b , by showing that the unique optimizers to the NLP are r = k = b and s = b/ . Indeed, note that ∂∂r (cid:18) k + r ) k ( r − s ) + 1 s (cid:19) = − k + s ) k ( r − s ) which is strictly negative for all feasible s, r, k with r > s . Hence, there is no optimal solution for which r < b , as otherwise by increasing r one could improve the value of the objective. Similarly we observe that ∂∂k (cid:18) k + r ) k ( r − s ) + 1 s (cid:19) = − rk ( r − s ) which is again strictly negative for all feasible s, r, k with r > s . Hence, there is no optimal solution forwhich k < b , as otherwise by increasing k one could improve the value of the objective.17o conclude, in an optimal solution to (2) we have that r = k = b , and hence one needs to find s minimizing g ( s, b, b ) = b − s + s . For this we compute ∂∂s g ( s, b, b ) = 4( b − s ) − s and it is easy to see that g ( s, b, b ) = 0 if and only if s = b/ or s = − b (and the latter is infeasible). Atthe same time, g ( s, b, b ) is convex when s ≤ b because ∂ ∂s g ( s, b, b ) = b − s ) + s > , hence s = b/ corresponds to the unique minimizer.The last component of Theorem 3.2 that requires justification pertains to the competitive ratio. Nowfix b, c > for which bc ≥ , and let E ( s , r , k ) be the optimal solution to NLP bc (corresponding to theoptimal choices of algorithm N s,r,k ). By Lemma 3.1 the induced energy consumption is d · E ( s , r , k ) .Then, the competitive ratio of the algorithm is sup d> c d d · E ( s , r , k ) = c · E ( s , r , k ) . C.3 Theorem 3.3
Proof: [Proof of Theorem 3.3] First we observe that s = σc , r = ρc , k = κc are indeed feasible to NLP ∞ c (forevery c > ), since T (cid:18) σc , ρc , κc (cid:19) = c (cid:18) κ + ρ ) κ ( ρ − σ ) + 1 σ (cid:19) and in particular, for the values of σ, ρ, κ described above we have (cid:16) κ + ρ ) κ ( ρ − σ ) + σ (cid:17) ≈ (from the formaldefinition of σ, ρ, κ that appears later, it will be clear that expression will be exactly equal to 1). Moreover, byTheorem 3.2, the competitive ratio of N σc , ρc , κc is c · E (cid:18) σc , ρc , κc (cid:19) = ρ (cid:0) κ + ρ + σ (cid:1) ρ − σ , as claimed.In the remaining of the section we prove that the choices for s, r, k of Theorem 3.3 are indeed optimal for N s,r,k . First we establish a structural property of optimal speeds choices for N s,r,k . Lemma C.3.
For any c > , optimal solutions to NLP ∞ c satisfy constraint T ( s, r, k ) ≤ c tightly. Proof:
Consider an optimal solution ¯ s, ¯ r, ¯ k . As noted before, we must have ¯ s, ¯ r, ¯ k > and ¯ r > ¯ s , asotherwise the values would be infeasible.Next note that the time constraint can be rewritten as k ≥ rscrs − cs − r − s For the sake of contradiction, assume that the time constraint is not tight for ¯ s, ¯ r, ¯ k . Then, there is (cid:15) > sothat ¯ s, ¯ r, k is a feasible solution, where k = ¯ k − (cid:15) > . But then, the objective value strictly decreases, acontradiction to optimality.We will soon derive the optimizers to NLP ∞ c using Karush-Kuhn-Tucker (KKT) conditions. Before that,we observe that solutions are scalable with respect to c , which will also allow us to simplify our calculations.18 emma C.4. Let s , r , k be the optimizers to NLP ∞ inducing optimal energy E . Then, for any c , theoptimizers to NLP ∞ c are ¯ s = s /c, ¯ r = r /c, ¯ k = k /c , and the induced optimal energy is c E . Proof:
Note that the triplet ( s, r, k ) is feasible to NLP ∞ c (for a specific c ) if and only if the triplet ( c · s, c · r, c · k ) is feasible to NLP ∞ . Moreover, it is straightforward that when speeds are scaled by c , theinduced energy is scaled by c . Hence, for every c > there is a bijection between feasible (and optimal)solutions to NLP ∞ c and NLP ∞ .We are therefore motivated to solve NLP ∞ , and that will allow us to derive the optimizers for NLP ∞ c ,for any c > . Lemma C.5.
The optimal solution to
NLP ∞ is obtained for s = σ ≈ . , r = ρ ≈ . , k = κ ≈ . and the optimal NLP value is ρ ( κ + ρ + σ ) ρ − σ ≈ . . Proof:
By KKT conditions, we know that, necessarily, all minimizers of E ( s, r, k ) satisfy the conditionthat −∇ E ( s, r, k ) is a conical combination of tight constraints (for the optimizers). Lemma C.3 assertsthat T ( s, r, k ) = 1 has to be satisfied for all optimizers s, r, k . At the same time, recall that, by the proofof Lemma C.1, none of the constraints r ≥ s and s, r, k ≥ can be tight for an optimizer. Hence, KKTconditions imply that any optimizer s, r, k satisfies, necessarily, the following system of nonlinear constraints −∇ E ( s, r, k ) = λ ∇ T ( s, r, k ) T ( s, r, k ) = 1 λ ≥ More explicitly, the first equality constraints is r − r ( k + r ) ( r − s ) k s − r +3 r s + s ( r − s ) krs − r = λ k + r ) k ( r − s ) − s − k + s ) k ( r − s ) rk ( s − r ) From the 3rd coordinates of the gradients, we obtain that λ = 2 k , which directly implies that the dualmultiplier λ preserves the correct sign for the necessary optimality conditions.Hence, the original system of nonlinear constraints is equivalent to that r − r (cid:0) k + r (cid:1) ( r − s ) = 2 k (cid:18) k + r ) k ( r − s ) − s (cid:19) − k s + 2 r − r s − s = 4 k ( k + s )2( k + r ) k ( r − s ) + 1 s = 1 Using software numerical methods, we see that the above algebraic system admits the following 3 real rootsfor ( s, r, k ) : (2 . , . , . multiplicity 1 )( − . , . , . multiplicity 2 ) ( σ, ρ, κ ) = (2 . , . , . . To verify that indeed ( σ, ρ, κ ) is a minimizer, we compute ∇ E ( s, r, k ) = r ( k + r ) ( r − s ) ( r + s ) ( − k + r + s − rs ) ( r − s ) kr ( r − s ) ( r + s ) ( − k + r + s − rs ) ( r − s ) ( r − sr +3 s r + s +2 k s ) ( r − s ) − ks ( r − s ) kr ( r − s ) − ks ( r − s ) rr − s . Moreover, ∇ E ( σ, ρ, κ ) = . − . . − . . − . . − . . which has eigenvalues . , . , . , hence it is PSD. As a result, f ( s, r, k ) is locally convex at ( σ, ρ, κ ) , and therefore ( σ, ρ, κ ) is a local minimizer to NLP ∞ . As we showed earlier, ( σ, ρ, κ ) is the onlycandidate optimizer, hence a global minimizer as well.Lemma C.5 together with Lemma C.4 imply that for any c > the optimal solution to NLP ∞ c is exactlyfor s = σc , r = ρc , k = κc and hence, the proof of Theorem 3.3 follows. C.4 Lemma 3.6
Proof: [Proof of Lemma 3.6]An immediate corollary from the proof of Lemma C.2 (within the proof of Theorem 3.2) is the following
Corollary C.6.
The unique solution to
NLP bc when c = 9 /b is given by r = k = b, s = b , inducing energy b d , and competitive ratio cb ) = 378 . Next we find solutions for c > /b so that r, k ≤ b remain tight. Since, when c = 9 /b , there is only oneoptimizer s = b/ , r = b, k = b , two inequality constraints are tight. The next calculations investigate thespectrum of c for which the same constraints remain tight for the optimizer.We write 1st order necessary optimality conditions for NLP bc , given that the candidate optimizer satisfiesthe time constraint, and the two r, k ≤ b speed bound constraints tightly −∇ E ( s, r, k ) = λ ∇ T ( s, r, k ) + λ + λ T ( s, r, k ) = cr = bk = bλ , λ , λ ≥ s we obtain that s , = ±√ b c − bc + 9 + bc − c For each s ∈ { s , s } , the first gradient equality defines a linear system over λ , λ , λ whose solutionsare λ = ( s − s s − , λ = − − s − s + 2( s − s − , λ = − (cid:0) s − s − s + 2 (cid:1) ( s − s − .λ = bs (3 b − s ) b − s , λ = − b − b s − s ( b − s )( b − s ) , λ = − (cid:0) b − b s − bs + s (cid:1) ( b − s )( b − s ) respectively. As long as all dual multiplies λ i = λ i ( s ) are positive, corresponding solution ( s, b, b ) is optimalto R c , provided that ∇ f ( s, b, b ) (cid:31) .First we claim that s cannot be part of an optimizer. Indeed, λ ( s ) = − b (cid:16) bc − p ( bc − bc −
1) + 3 (cid:17) (cid:16) bc + p ( bc − bc − − (cid:17) c (cid:16) bc + 3 p ( bc − bc − − (cid:17) Recall that bc > , and hence the denominator of λ ( s ) as well as bc + p ( bc − bc − − are strictlypositive. But then, the sign of λ ( s ) is exactly the opposite of bc − p ( bc − bc −
1) + 3 . Define function h ( x ) := 5 x − p ( x − x −
1) + 3 over the domain x > . It is easy to verify that h ( x ) preserves positivesign (in fact min x ≥ h ( x ) = h (cid:16) (cid:16) √ (cid:17)(cid:17) = 8 √ > Hence, λ ( s ) < that concludes ourclaim.Next we investigate the spectrum of c for which all λ i ( s ) remain non-negative.Our next claim is that for all bc > we have that s ( c ) < b/ . Indeed, consider function d ( x ) := 3 p x − x + 9 − bx + 9 . It is easy to see that d ( bc ) = 6 c (cid:16) b − s ( c ) (cid:17) . But then, elementary calculations show that min x ≥ d ( x ) = d (9) = 0 , proving that s ( c ) < b/ as claimed.Next we investigate the sign of λ ( s ) , λ ( s ) , λ ( s ) . For this, introduce function t ( x ) = p ( x − x − ,and note that λ ( s ) = − b ( − bc + t ( bc ) + 3) (5 bc + t ( bc ) + 3)4 c ( bc − t ( bc ) + 3)) ,λ ( s ) = 30 ( t ( bc ) + 3) + bc ( bc ( − bc + 3 t ( bc ) + 32) − t ( bc ) + 11))4 c ( bc − ,λ ( s ) = bc ( − t ( bc ) + bc ( − bc + 3 t ( bc ) + 22) + 49) −
12 ( t ( bc ) + 3)4 c ( bc − . Claim 1: λ ( s ) > for all c > /b .Define d ( x ) = x − t ( x )) and d ( x ) = 3 + 5 x + t ( x ) . Note that sign ( λ ( s )) = − sign ( d ( bc )) · sign ( d ( bc )) . Simple calculus shows that d ( x ) is strictly decreasing in x ≥ , and d (9) = 0 , and therefore d ( bc ) < for all c > /b . Similarly, it is easy to see that d ( x ) is strictly increasing in x > , and d (9) = 45 . Therefore d ( bc ) > for all c > /b . Overall this implies that λ ( s ) is positive for all c > /b .21laim 2: λ ( s ) > for all c ∈ (9 /b, . /b ) .First we observe that the denominator of λ ( s ) preserves positive sign for c > /b . So we focus on thesign of the numerator we we abbreviate by d ( x ) = 30(3 + t ( x )) + x ( x (32 − x + 3 t ( x )) − t ( x ))) .Note that d ( x ) = 0 is equivalent to that (cid:16) x − x + 30 (cid:17) t ( x ) − (cid:16) x − x + 55 x − (cid:17) = 0 ⇔ (cid:16) x − x + 30 (cid:17) t ( x ) − (cid:16) x − x + 55 x − (cid:17) = 0 ⇔ − x − x (3 x ( x (2 x −
25) + 60) − Degree-3 polynomial x ( x (2 x −
25) + 60) − has only one real root, which is r (cid:16) √
10 + 1055 (cid:17) + q − √
10 + 75 ! ≈ . Hence, λ ( s ) > for all c ∈ (9 /b, . /b ) Claim 3: λ ( s ) > for all c ∈ (9 /b, . /b ) .First we observe that the denominator of λ ( s ) preserves positive sign for c > /b . So we focus on the signof the numerator we we abbreviate by d ( x ) = x ( x (3 t ( x ) − x + 22) − t ( x ) + 49) − t ( x ) + 3) . Notethat d ( x ) = 0 is equivalent to that (cid:16) x − x − (cid:17) t ( x ) − (cid:16) x − x − x + 36 (cid:17) = 0 ⇔ (cid:16) x − x − (cid:17) t ( x ) − (cid:16) x − x − x + 36 (cid:17) = 0 ⇔ − x − x (3 x (2( x − x −
3) + 49) = 0
The roots of degree-3 polynomial x (2( x − x −
3) + 49 are γ = 3 + √
38 cos (cid:18)
13 tan − (cid:18) √ (cid:19)(cid:19) ≈ . γ = 3 + r
572 sin (cid:18)
13 tan − (cid:18) √ (cid:19)(cid:19) + r (cid:18) − cos (cid:18)
13 tan − (cid:18) √ (cid:19)(cid:19)(cid:19) ≈ . γ = 3 − r
572 sin (cid:18)
13 tan − (cid:18) √ (cid:19)(cid:19) + r (cid:18) − cos (cid:18)
13 tan − (cid:18) √ (cid:19)(cid:19)(cid:19) ≈ − . We conclude that λ ( s ) preserves positive sign for all c ∈ (9 /b, γ /b ) .Overall, we have shown that feasible solution s = −√ b c − bc +9+ bc − c , r = k = b satisfies necessary1st order optimality conditions. We proceed by checking that s , r , k satisfy 2nd order sufficient conditions,which amounts to showing that ∇ f ( s , b, b ) (cid:31) . Indeed, ∇ f ( s , b, b ) = b ( b − s ) − ( b + s ) ( b +4 s b − s ) b − s b − ( b + s ) ( b +4 s b − s ) b ( b − s b +3 s b + s ) b s ( s − b ) b − s b s ( s − b ) b b − s ) b
22y setting q := s /b = −√ b c − bc +9+ bc − bc , we obtain the simpler form ∇ f ( s , b, b ) = b ( b − s ) − q − (cid:0) − q + 4 q + 1 (cid:1) − q ( − q − (cid:0) − q + 4 q + 1 (cid:1) (cid:0) q + 3 q − q + 1 (cid:1) q − q − q q − q − q ) (3)When bc > we have that q < / , q is decreasing in the product of bc > , and it remains positive. Theeigenvalues of the matrix that depends only on q and for any q ∈ (0 , / can be obtained using a closedformula (they are real roots of a degree-3 polynomial). In Figure 5 we depict their behavior. Since alleigenvalues are all positive, the candidate optimizer is indeed a minimizer. Figure 5: The eigenvalues of matrix (3) as a function of q ∈ (0 , / (and scaled by b / ( b − s ) ). C.5 Lemma 3.7
Proof: [Proof of Lemma 3.7] By Theorem 3.3, we know the optimizers to NLP ∞ c ; s = σ/c, r = ρ/c, k = κ/c, . These optimizers satisfy the speed bound constraints s, r, k ≤ b as long as max { s, r, k } ≤ b , i.e. ρ/c ≤ b . Hence, when c ≥ ρ/b , Non Linear Programs NLP ∞ c , NLP bc have the same optimizers. C.6 Lemma 3.8
Proof: [Proof of Lemma 3.8] First, we observe that constraint r ≤ b is tight for the provable optimizers forall c, b when cb ∈ [9 , γ ] ∪ { γ } . As the only other constraint that switches from being tight to non-tight in thesame interval is k ≤ b , we are motivated to maintain tightness for constraints r ≤ b and the time constraint.Given that speed s is chosen (to be determined later), we fix r = b , and we set k = k ( c, b ) where k ( c, b ) := 2 bsbcs − b − cs − s so as to satisfy the time constraint tightly (by solving T ( s, r, b ) = c for k ). It remains to determine values forspeed s . To this end, we heuristically require that s = s ( c, b ) where s ( c, b ) := α · c + β α, β that we allow to depend on b . In what follows we abbreviate s ( c, b ) by s ( c ) . Let s ( c ) , s be the chosen values for speed s as summarized by the statement of Theorem 3.4 when cb ≤ γ and cb ≥ γ , respectively. We require that s ( γ /b ) = s ( γ /b ) , s ( γ /b ) = s ( γ /b ) inducing a linear system on α, β . By solving the linear system, we obtain α = b (cid:18) − γ σ + γ γ − q γ − γ + 9 γ − γ (cid:19) γ ( γ − γ ) γ β = b (cid:18) − γ σ + γ γ − q γ − γ + 9 γ − γ (cid:19) γ γ ( γ − γ ) Using the known values for γ , γ , σ , we obtain s ( c ) = 0 . b − . b c , as promised. It remainsto argue that s ( c, b ) , together with r = b , and k ( c, b ) are feasible when γ < cb < γ .The fact that s ( c ) complies with bounds ≤ s ≤ b follows immediately, since s ( c ) is a linear strictlydecreasing function in c , and both s ( γ /b ) , s ( γ /b ) satisfy the bounds by construction. We are thereforeleft with checking that ≤ k ( c, b ) ≤ b which is equivalent to that bcs − b − cs − s ≥ ⇔ b ( bc ( bc (0 . − . bc ) + 0 . − . ≥ Define degree-2 polynomial function g ( x ) = x ( x (0 . − . x ) + 0 . − . and observe that it sufficies to prove that g ( x ) ≥ for all x ∈ ( γ , γ ) . The roots of g can be numericallycomputed as − . , , . , . , proving that g preserves positive sign in ( γ , γ ) as wanted.Finally, the claims regarding the induced energy and competitive ratio is implied by Theorem 3.2 andobtained by evaluating the given choices of s, r, k in E ( s, r, k ) . D Proofs Omitted from Section 4.
D.1 Proposition 4.2
Proof: [Proof of Proposition 4.2] The exploration algorithm explicitly ensures that round k − ends (andhence that round k begins) at the claimed position, so only the time needs to be proved. Note that thestatement is true for both robots when k = 0 . Suppose that when L starts its ( k − -th round, it is at position − k − at time · k − . The round ends at time · k − + 4 k − + s · k · s − s + (4 k − k · s − s ) = 3 · k ,and the k -th round will start at the claimed time. The proof is identical for R . D.2 Lemma 4.3
Proof: [Proof of Lemma 4.3]
Case 1: d ∈ D . Let Y = { L, R } \ { X } be the robot other than X . Observethat by Proposition 4.2, X finds the exit at time K + K + d/s = 4 K + d/s . Then X goes towards Y atspeed s = d K − d/s , and so it will reach position at time K + d/s + K − d/sd d = 8 K . We know that attime K , robot Y is starting a round at position K or − K , then goes towards at full speed. Hence Y isat position at time K , where it meets X . 24 ase 2: d ∈ D = [4 Ks/ ( s + 1) , Ks/ (1 − s )] . As before, the exit is found at time K + d/s . Assume forsimplicity that X = L (the case X = R is identical by symmetry). After finding the exit at position − d , L goes full speed to the right. Thus at time t = 4 K + d/s + d + d + ds − Ks − s , it arrives at position d + ds − Ks − s .We show that at this time, R is in its second phase and is at this position. Notice that t = 4 K + d/s + d + d + ds − Ks − s = 8 K + d/s + d − K + d + ds − Ks − s = 8 K + d/s + d − K − s ≤ K (1 + s/ (1 − s ) ) the latter inequality being obtained from d ≤ Ks/ (1 − s ) . Now, R enters its second travel phasewhen at position at time K , and the phase ends a time K + 1 /s · Ks/ (1 − s ) = 8 K (1 + 1 / (1 − s )) .Since s ≤ / , we get K (1 + 1 / (1 − s )) ≥ K (1 + s/ (1 − s ) ) ≥ t . Therefore R is still in its secondphase at time t , and it follows that its position at this time is s ( t − K ) = d + ds − Ks − s . Hence L and R meet. It is straightforward to see that L and R could not have met before time t , and thus L and R meet atthe claimed time and position. Case 3: d ∈ D = [4 Ks/ (1 − s ) , K ] . Again assume that X = L . Thistime L finds the exit at position − d at time K + K + s · Ks − s + d − Ks − s = 8 K + d . Going full speedto the right, at time t = 8 K + 2 d + 2 ds/ (1 − s ) , it reaches position ds/ (1 − s ) . Since d ≤ K , we have t ≤ K (2 + s/ (1 − s )) = 8 K (1 + 1 / (1 − s )) . As in the previous case, the second phase of R ends at time K (1 + 1 / (1 − s )) ≥ t . Thus at time t , R is at position s ( t − K ) = 2 ds + 2 ds / (1 − s ) = 2 ds/ (1 − s ) .Again, one can check that L and R could not have met before, which concludes the proof. D.3 Lemma 4.4
Proof: [Proof of Lemma 4.4] Let X be the robot that finds the exit at distance d , and let K := K ( X, k ) . Weshow that all speeds are at most 1, as well as that the evacuation time is at most d . There are three cases toconsider. Case 1: d ∈ D . By Lemma 4.3, after meeting, both robots need to travel distance d and have d − K timeremaining. By the last line of the exit phase algorithm, they go back at speed s b := d d − K and make it intime, provided that speed s b is achievable, i.e. < s b ≤ . Clearly s b > since d − K > . If weassume d d − K > , we get d > d − K , leading to d < K , a contradiction. Case 2: d ∈ D . By Lemma 4.3, the robots meet at position p such that | p | = d + ds − Ks − s at time t =8 K + d + d/s − K − s . The robots use the smallest speed s b := p + dt − d that allows the them to reach the exit in time d . We must check that < s b ≤ . We argue that if the two robots used speed to get to the exit aftermeeting, they would make it before time d . Since s b allows the robots to reach the exit in time exactly d ,it follows that < s b ≤ .First note that since d ≤ Ks/ (1 − s ) , we have K ≥ d (1 − s ) /s . Using speed from the point theymeet, the robots would reach the exit at time 25 K + d + d/s − K − s + d + d + ds − Ks − s = 4 K (cid:18) − s − s (cid:19) + d (cid:18) s + 1 /s − s (cid:19) = 4 K · − s − s + d · s + 1 s (1 − s ) ≤ d · − ss · − s − s + d · s + 1 s (1 − s )= d (cid:18) − ss + 3 s + 1 s (1 − s ) (cid:19) where we have used the fact that − s ≤ in the inequality. It is straighforward to show that d (cid:16) − s − s + s +1 s (1 − s ) (cid:17) ≤ d when / ≤ s ≤ / , proving our claim. Case 3: d ∈ D . Again according to Lemma 4.3, the robots meet at position p satisfying | p | = 2 ds/ (1 − s ) at time t = 8 K + 2 d + 2 ds/ (1 − s ) . The robots go towards the exit at speed s c := p + d d − t . As in the previouscase, we show that s c is a valid speed by arguing that the robots have enough time if they used their fullspeed. If they do use speed after they meet, they reach the exit at time t = 8 K + 3 d + 4 ds/ (1 − s ) . Since d ≥ Ks/ (1 − s ) , we have K ≤ d (1 − s ) / (4 s ) . Therefore t ≤ · ( d (1 − s ) / (4 s )) + 3 d + 4 ds/ (1 − s ) = d · s − s +2 s (1 − s ) . One can check that this is d or less whenever / ≤ s ≤ / . D.4 Lemma 4.5
Proof: [Proof of Lemma 4.5] For any K := 4 i power of with i ≥ , define B L ( K , s ) as the energy spentby L after reaching position − K for the first time without having found the exit, ignoring the initial / energy spent to get to position . The quantity B L ( K , s ) is the sum of energy spent in each of the first i − rounds, and so B L ( K , s ) = i − X j =0 (cid:16) j + s (4 j +1 s/ (1 − s )) + 4 j +1 − j +1 s/ (1 − s ) (cid:17) = i − X j =0 (cid:16) j +1 (5 / s − s ) / (1 − s )) (cid:17) = 4 / i − / − s ( s + 1))= 1 / K − − s ( s + 1)) We define B R (2 K , s ) similarly for R , i.e. B R (2 K , s ) is the energy spent by R when its ( i − -th roundis finished and it reached position K for the first time, ignoring the initial / energy to get at position .We get B R (2 K , s ) = i − X j =0 (cid:16) · j + s (2 · j +1 s/ (1 − s )) + 2 · j +1 − · j +1 s/ (1 − s ) (cid:17) = 2 B L ( K , s ) = 2 / K − − s ( s + 1))
26e may now calculate the three possible cases of energy. Assume that X ∈ { L, R } finds the exit and Y = { L, R } \ { X } . Observe that B X ( K, s ) + B Y (2 K, s ) = ( K − − s ( s + 1)) = F ( K, s ) We implictly use Lemma 4.3 for the distance traveled by X to catch up to Y after finding the exit, andthe distance traveled back by both robots. In the E ( K, d, s ) expressions that follow, for clarity we partitionthe terms into 3 brackets, which respectively represent the energy spent by X to find the exit and catch up to Y , the energy spent by Y before being caught, and the energy spent by both robots to go to the exit. Case 1: d ∈ D . The total energy spent is E ( K, d, s ) = h B X ( K, s ) + K + s d + s c d i + [ B Y (2 K, s ) + 2 K ] + h s b d i = F ( K, s ) + 3 K + d ( s + s c + 2 s b ) Case 2: d ∈ D . In this case, the energy spent is E ( K, d, s ) = (cid:20) B X ( K, s ) + K + s d + d + d + ds − Ks − s (cid:21) + (cid:20) B Y (2 K, s ) + 2 K + s (cid:18) d + ds − Ks − s (cid:19)(cid:21) + (cid:20) s b (cid:18) d + d + ds − Ks − s (cid:19)(cid:21) = F ( K, s ) + 3 K + (cid:18) d − Ks − s (cid:19) (1 + s + 2 s b ) Case 3: d ∈ D . The energy spent is E ( K, d, s ) = h B X ( K, s ) + K + s Ks/ (1 − s ) + d − Ks/ (1 − s ) + d + 2 ds/ (1 − s ) i + h B Y (2 K, s ) + 2 K + s (2 ds/ (1 − s )) i + h s b ( d + 2 ds/ (1 − s )) i = F ( K, s ) + 3 K + 4 Ks/ (1 − s )( s −
1) + d (cid:18) s − s (2 + 2 s + 2 s b ) + 2 + 2 s b (cid:19) = F ( K, s ) + 3 K − Ks ( s + 1) + 2 d − s ( s + s b ( s + 1) + 1) D.5 Lemma 4.6
Proof: [Proof of Lemma 4.6] We analyze each case separately.
Case 1: d ∈ D = [ K, Ks/ ( s + 1)] . According to Lemma 4.5, we have E ( K, d, s ) /d = 1 /d · (( K − − s ( s + 1)) + 3 K + d ( s + s c + 2 s b ))= ( K − − s ( s + 1)) + 3 Kd + s + (cid:18) d K − d/s (cid:19) + 2 (cid:18) d d − K (cid:19) d = K . Plugging in s = 0 . , the above evaluates to . − . /K . We claim that E ( K, d, s ) /d is a decreasing function over interval D , and thereforeattains its maximum when d = K . Assuming this is true, adding the initialization energy of / omitted sofar and given that d ≥ K , the energy ratio is at most . − . /K + 1 / (3 K ) ≤ . We now prove that E ( K, d, s ) /d is decreasing over the interval D . Let f ( K, d, s ) := d (9 d − K ) f ( K, d, s ) := d ( d − Ks ) f ( K, d, s ) := − K (cid:0) s + s − (cid:1) + 4 s ( s + 1) − d , and observe that E ( K, d, s ) d = 2 f ( K, d, s ) + s f ( K, d, s ) + f ( K, d, s ) + s . The plan is to prove that ∂∂d E ( K, d, s ) /d < . For this we calculate ∂∂d f ( K, d, s ) := − dK (9 d − K ) ∂∂d f ( K, d, s ) := − dKs ( d − Ks ) ∂∂d f ( K, d, s ) := 4 K (cid:0) s + s − (cid:1) − s ( s + 1) + 5 d , Now we claim that all ∂∂d f i ( K, d, s ) are increasing functions in d , for i = 1 , , . Indeed, first, ∂ ∂d f ( K, d, s ) = 32 K (9 d + 4 K )(9 d − K ) > since d ≥ K . Hence ∂∂d f ( K, d, s ) is increasing in d .Second, ∂ ∂d f ( K, d, s ) = 16 Ks ( d + 2 Ks )( d − Ks ) is positive (and well defined), since d ≤ ks/ (1 + s ) . Hence ∂∂d f ( K, d, s ) is increasing in d .Third, we show that ∂∂d f ( K, d, s ) is increasing in d . For this it is enough to prove that K (cid:0) s + s − (cid:1) − s ( s + 1) + 5 < . For s = 0 . (and in fact for all s ∈ ( − , ) the strict inequality can be written as K ≥ s +4 s − s +4 s − , which we show next it is satisfied. Indeed, it is easy to see that s +4 s − s +4 s − ≤ / (which isattained for s = 0 ), while K ≥ , hence the claim follows.28o resume, we showed that ∂∂d f i ( K, d, s ) are increasing functions in d , for i = 1 , , . Recalling that s = 0 . , and since d ≤ ks/ (1 + s ) , we obtain that ∂∂d E ( K, d, s ) /d ≤ ∂∂d f ( K, Ks/ (1 + s ) , s ) + s ∂∂d f ( K, Ks/ (1 + s ) , s ) + ∂∂d f ( K, Ks/ (1 + s ) , s )= ( s + 1) ( s (4 K ( s ( s + 1)(49 s (7 s −
6) + 76) − − s (7 s (28 s (7 s + 1) − − K s (7 s − = 2 . − . KK . Since K ≥ , the latter quantity is clearly negative. This shows that ∂∂d E ( K, d, s ) /d is negative (in the givendomain), hence E ( K, d, s ) /d is decreasing in d . Case 2: d ∈ D = [4 Ks/ ( s + 1) , Ks/ (1 − s )] . In this case, the energy ratio E ( K, d, s ) /d is /d · (cid:18) / K − − s ( s + 1)) + 3 K + (cid:18) d − Ks − s (cid:19) (1 + s + 2 s b ) (cid:19) = ( K − − s ( s + 1)) + 3 Kd + (cid:18) d − Ksd (1 − s ) (cid:19) s + 2 (cid:18) d − Ksd (8 − s − /s ) + 4 K (2 s − (cid:19) ! We will show that this expression achieves its maximum at d = 4 Ks/ (1 − s ) . When s = 0 . , thenabove yields . − . /K . Given that / (3 d ) ≤ / (3 . K ) , this implies that theenergy ratio is at most . − . /K + 1 / (3 . K ) ≤ . We prove that E ( K, d, s ) /d is an increasing function over interval D . First we compute ∂∂d E ( K,d,s ) d andwe substitute s = 0 . to find d ( d − . K ) g ( K, d ) , where g ( K, d ) := d (7 . K + 2 . d K ( − . K − . dK (8 . K + 1 . K ( − . K − . . Note that d ≥ Ks/ (1 + s ) ≈ . K , and hence d ( d − . K ) > for all values of d underconsideration. Therefore the lemma will follow if we show that g ( K, d ) ≥ as well. g ( K, d ) is a degree-3 polynomial with positive leading coefficient. It attains a local minimum at thelargest real root of ∂∂d g ( K, d ) = 3 d (7 . K + 2 . dK ( − . K − . K (8 . K + 1 . which is d ( K ) := K (cid:16) . K + 0 . p K (129 . K + 8 . . (cid:17) K + 0 . K > , we have d ( K ) < Ks/ (1 + s ) ≈ . K .From the above, it follows that g ( K, d ) is monotonically increasing for d ≥ Ks/ (1 + s ) , and therefore g ( K, d ) ≥ g ( K, Ks/ (1 + s )) = (0 . K + 0 . K ≥ as wanted. Case 3: d ∈ D = [4 Ks/ (1 − s ) , K ] . The energy ratio E ( K, d, s ) /d is /d · (cid:18) / K − − s ( s + 1)) + 3 K − Ks ( s + 1) + 2 d − s ( s + s b ( s + 1) + 1) (cid:19) = 1 / K − − s ( s + 1)) + 3 K − Ks ( s + 1) d + 2 s + 21 − s + 2 s + 21 − s (cid:18) d (1 + s ) d (7 − s ) + 8 K ( s − (cid:19) In this case, we claim that this expression is decreasing over D and achieves its maximum at d =4 Ks/ (1 − s ) . When s = 0 . , the above gives . − . /K (which is the same asin case 2, as one should expect). Given that / (3 d ) ≤ / (7 . K ) , we get that the energy ratio is at most . − . /K + 1 / (7 . K ) ≤ . Let us prove that E ( K, d, s ) /d is indeed decreasing. Note that E ( K, d, s ) /d equals s + 1) − s g ( K, d, s ) + h ( K, d, s ) + 2 (cid:0) s + 1 (cid:1) − s where g ( K, d, s ) := d ( d (7 − s ) + 8 K ( s − , h ( K, d, s ) := 8 K (cid:0) − s − s + 1 (cid:1) + 4 s ( s + 1) − d . In what follows we prove that both g ( K, d, s ) , h ( K, d, s ) are strictly decreasing when d ≥ Ks/ (1 − s ) ,implying the claim of the lemma.First we show that h ( K, d, s ) is decreasing. For that note that, using the fixed value of s = 0 . , wehave h ( K, d, s ) = (3 . K − . /d , and the latter expression (in d ) is clearly strictly decreasing forall constants K > .Now we show that g ( K, d, s ) is strictly decreasing for all d ≥ Ks/ (1 − s ) . For that observe that for thespecific constant s , and since d ≥ Ks/ (1 − s ) ≈ . K , we have that | d (7 − s ) + 8 K ( s − | = d (7 − s ) + 8 K ( s − > . Hence, to show that g ( K, d, s ) is strictly decreasing, it is enough to prove that q ( K, d, s ) := d ( d (7 − s ) + 8 K ( s − is strictly decreasing in d ≥ Ks/ (1 − s ) . First observe that the rational function is well defined for thesevalues of d , since the denominator becomes 0 only when d = K ( s − s − < Ks/ (1 − s ) (the last inequality iseasy to verify). To that end, we compute ∂∂d q ( K, d, s ) = 8 K ( s − d (7 − s ) + 8 K ( s − which is of course negative for the given value of s <1