2010 IEEE Intenational Confeence on Robotics and Automation Anchoage Convention Distict May 3-8, 2010, Anchoage, Alaska, USA Multi-Robot Flooding Algoithm fo the Exploation of Unknown Indoo Envionments Flavio Cabea-Moa, Jizhong Xiao and Pete Bass Abstact In this pape we study the poblem of multi-obot exploation of unknown indoo envionments that ae modeled as tees. Specifically, ou appoach conside that obots deploy and communicate with active landmaks in evey intesection they encounte. We pesent a novel algoithm that is guaanteed to completely exploe any tee with m edges and diamete D, by allowing k obots to be fed into the tee one at a time. We pove that the exploation time of the algoithm gows in linea popotion with the size of the tee and is not bigge than D + m. Simulation esults ae pesented that cooboate the theoetical analysis. I. INTRODUCTION The exploation of unknown indoo envionments is a poblem that emains open in the obotics community. Seveal appoaches have been poposed to deal with this poblem and can be mainly chaacteized by the way the envionment is modeled: as a geometic stuctue [1], [2], [3]; as a gid [4], [5]; o as a gaph [6], [7], [8], [9], [10]. In this pape we conside the exploation of an indoo envionment with multiple obots. These obots ae assisted by active landmaks (tokens) that ae deployed by the obots while pefoming the exploation. These tokens povide diections to the obots that find them along the way. Consideing these tokens as nodes of a gaph and the passages between them as edges, the exploation pocess can be conveniently epesented as the poblem of exploing a gaph: visiting all vetices and tavesing all edges. Exploation of gaphs has been a well-studied poblem, especially fo the single-obot case, fo both diected and undiected gaphs. Exploation of diected gaphs, whee obots can move only in one diection, has been studied fo example in [11] whee the minimum time of exploation is investigated. In [12], it is poven that by letting a single obot to mak the envionment with passive landmaks (pebbles) the exploation time is damatically impoved. In [13], it was shown that a single obot can lean its way though a diected gaph in time polynomial in n, using only one pebble, if the uppe bound on the total numbe of vetices n is known. If, on the othe hand, this uppe bound is not known, it is stated that θ(loglog(n)) pebbles ae both necessay and sufficient to pefom the exploation. Howeve, the authos in [13] do not mention how much the exploation time will be impoved when using moe than one pebble. Flavio Cabea-Moa is with the Depatment of Electical Engineeing, The Gaduate Cente of the City Univesity of New Yok, 365 5th Ave. New Yok, NY, 10016 USA, fcabea-moa@gc.cuny.edu Jizhong Xiao and Pete Bass ae with the Depatments of Electical Engineeing and Compute Science, The City College of the City Univesity of New Yok, 140th Steet & Convent Avenue, New Yok, NY 10031, USA jxiao@ccny.cuny.edu, pete@cs.ccny.cuny.edu Exploation using undiected gaphs is consideed in some othe woks. In [6] the goal is to achieve an efficient exploation unde the constaint that the obot must etun to the oot fo efueling peiodically. In [14] the exploation of all vetices and all edges is achieved while minimizing the total numbe of edge tavesals. In [15] and [16], undiected gaphs wee used to study the dynamic coveage poblem using a single obot and the least ecently visited (LRV) stategy. The obot deploys tokens that ae capable of communication and sensing. These tokens assist the obot by poviding instuctions, although this instuctions ae limited to only the fou cadinal diections. They poved that in a tee with m edges, the exploation time of LVR is 2m, whee thei definition of complete exploation is that evey edge has been tavesed at least once. Ideally, the exploation pocess must be accomplished as fast as possible and to this end, the use of multiple obots is expected to pefom faste than the single obot case. In [17], an exploation algoithm that exploes spase tees with k obots is pesented and it is shown that its competitive atio is influenced only by the density and the diamete of the tee. In this appoach, the communication exists exclusively among obots and it occus only when obots meet in the same vetex. In [18], k obots ae used to exploe a tee with m edges and diamete D, using a communication model that allows all obots to exchange infomation at evey step of the pocess though the use of tokens. The authos poposed an exploation algoithm with a unning time O( m logk +D), which is O( k logk ) lage than the optimal exploation time when the tee is known a-pioi. Finally, in [19], the authos pesented an algoithm fo the exploation of a tee with m edges and adius, using k obots, and show that the exploation time is 2m k + O(k 1 ), impoving the esult in [18] and almost eaching the lowe bound max( 2m k,2). Without exception all the afoementioned appoaches fo collective exploation assume that all k obots, located at the oigin, ente the tee all at once in the fist step. In pactical implementations howeve, it will neve be physically possible to make k obots ente an envionment o tavese a passage all at once if the numbe of obots is lage enough. Moeove, having k obots enteing one paticula edge all at once make them vulneable to the envionment. Futhemoe, in some paticula scenaios, such as a long path with no banches, the contibution of the k obots is null since all the obots will tavese the entie path, as if they wee pefoming the exploation individually. In this pape, we popose a novel algoithm in which 978-1-4244-5040-4/10/$26.00 2010 IEEE 5478
obots ente an undiected tee one at a time and only one obot is allowed to tavese an edge at each step. On evey vetex, obots deploy tokens that late on will instuct the obots towad unexploed/open edges, peventing obots to ente aleady exploed subtees. Consequently, the decision about the next move of a obot is mainly taken by the tokens. This ceates a decentalized system, feeing obots esouces fo othe tasks. We deive an uppe bound fo the exploation time, poving that it depends linealy on the size of the tee and that is neve bigge than D +m. We also pefom simulation to study the behavio of the algoithm in tees of diffeent sizes. The pape is oganized as follows: Section II descibes ou poposed flooding algoithm. Its analysis on tees is pefomed in Section III. Simulations showing the behavio of the algoithm in diffeent tees ae pesented in Section IV. Finally,concluding emaks ae offeed in Section V. A. Peliminaies II. FLOODING ALGORITHM k Robots ae initially located at the oot of an unknown tee T = (V,E), containing n vetices and m edges. Each obot is uniquely identified with an intege numbe fom 1 to k. Robots exploe the tee by moving along the edges that link a pai of vetices. All obots ae identical and exploe the tee following the same exploation algoithm. Define a paent vetex as one with at least one edge leading to anothe vetex in a downwad diection. The latte vetex is called a child vetex. A leaf is a vetex with no childen. Define a downwad edge ě v as an edge that leads fom a vetex v to one of its childen. That is, fo any vetex v, the numbe of downwad edges can be expessed as: ě v = deg(v) 1. Define a token (active landmak) as a bookkeeping device being deployed in each vetex. Evey token has communication capabilities and is able to stoe at least the following infomation: The numbe of edges conveging in a vetex The ID of the obots that have visited it A list of available edges fo obots to exploe The diection that a obot will take in the next step Each obot is capable of: Caying an unlimited numbe of tokens. Assessing whethe it has eached a vetex and deploy a token if none is pesent in it. Distinctively identifying evey edge on that vetex. At evey step, a obot located at a vetex can eithe tavese an edge o emain in its cuent position. Define an unused/unexploed edge as one that still has not been tavesed by any obot. Define an open edge as one though which a obot has left a vetex v, without etuning back to it no sending a message that it has eached a leaf (i.e., the obot is still exploing downwads the subtee ooted at v). If a vetex has some open edges, the token will send a obot to each open edge in inceasing ode. Fo instance, let assume vetex v has open edges labeled 2, 5, 8 and 10. The next obot aiving at v will be sent to edge 2 the next step. Any obot that aives at v late will be sent to edge 5, and so foth, until one obot has been sent to each open edge distibuting the obots as evenly as possible. Assume that at each step, each token is able to tansmit its own infomation about available edges to its paent token and that this tansmission is instantaneous. At the beginning, the fist obot deploys a token at. Each time a obot eaches a vetex whee a token has been deployed befoe, it will ead and wite its infomation. Define back-pointe as a list of all edges a obot has visited in ode to be in its cuent position. Each edge in the list must have associated the ID of the two tokens the edge is connected to. This list can be used late on by the obot to back tace its steps to. A obot aiving at a leaf will dop a token, making the place, and the token will tansmit to its paent that thee ae no banches available at this end. This will pevent obots to ente the edge again and will infom that the obot cuently located at the leaf is etuning back to its paent in the next step. B. One-by-one Flooding Algoithm The main diffeence between ou algoithm and those in the liteatue is that ous allows obots to ente the tee one at a time and only one obot is allowed to tavese an edge in each step. In addition, each token exchanges infomation with its paent about the next destination of the obot in the vetex. As a esult, no two obots will ente the same vetex on the same step. Algoithm 1: One-by-one Flooding Algoithm 1 Let R be a obot aiving at a vetex v though edge e; 2 Let e j be the ID of each edge at v; 3 Let j be the total numbe of edges at v; 4 if thee is no token pesent at v then 5 R dops one token at v; 6 R follows one of the j available edges in next step; 7 else if thee ae unused edges at v then 8 R follows one of the unused edges in next step; 9 else if thee ae open edges at v then 10 if open edges to be used by othe obots in next step then 11 R will stay put at v in next step; 12 else 13 R follows one open edge in next step; 14 end 15 else 16 R etuns to the paent vetex of v using its back-pointe in next step; 17 end The algoithm exhibits fou popeties: 1) No edge is used moe than twice by the same obot. 2) Some obots might emain static in some vetices (they will not ente an edge unless an instuction to do so is given by the token at the vetex). 3) The exploation is pefomed by the numbe of obots that is moe appopiate fo the size of the tee. 5479
4) No obot etuns to befoe evey edge has been tavese and evey vetex has been visited at least once. III. RESULT AND ANALYSIS ON TREES Define finished tee as a tee in which all edges have been tavesed and all vetices have been visited at least once by any obot, and the finished exploation time (t e ) as the time needed fo the tee to be finished. Define completed tee as a tee that is finished and in which all obots ae back at. The complete exploation time (t c ) is the time equied fo the tee to be completed. Conside a tee T with m edges, ooted at vetex and define d as the degee of. Assume that thee ae k obots stationed at and that it is guaanteed that thee will be obots available at all times in to be fed into the tee until the exploation is finished. Claim 1: Any tee can be consideed as being fomed by d individual subtees, all of them ooted at, whee the degee of each oot is equal to 1. Poof: By definition, the algoithm will feed one obot into each available unused/open edge at evey step. At the stat of the exploation, thee ae d unused edges in, along with k obots. The fist d obots will ente one of the d unused edges in the fist step since thee is nothing that pevent them to do so (ecall that only a token can instuct the obot to emain static in a vetex). The feeding pocess stats at the same time on all the subtees of. Wheneve a subtee is finished, the obots in that subtee will stat etuning to, but this will not affect the movement of obots in the othe subtees of. Thus, the exploation of a subtee of is independent of the exploation of any othe subtee, and as such evey subtee can be thought of as an individual tee whee its oot has degee 1. Define a path S = [a 0,a 1,a 2,,a l 1,a l,a l+1 ] as the path fom the oot to the vetex that will be exploed last, whee a 0 = and the last element is a leaf. Define T s as the subtee of whee S is specified. Since we ae inteested in the wost pefomance, at evey vetex the decision of what edge a obot should take next step belongs to the advesay. That is, if vetex a i has j unexploed edges, the edge that leads to a i+1 will be taken afte all othe j 1 edges has been taken befoe. Claim 2: The complete exploation time of tee T equals the time needed to completely exploe the subtee T s. Poof: Poof of claim 2 follows diectly fom claim 1: since each subtee of is independent of each othe, and because the subtee T s is the subtee that will be completely exploed last, any othe subtee will be completed befoe T s. Consequently, T s will detemine the time in which tee T is completely exploed. Due to claim 2, fom now on the analysis of tee T will be efeed as the analysis of T s. Claim 3: The total time to complete the exploation t c equals the total time to finish the exploation t e plus the numbe of obots k m inside the tee at time t e. That is, t c = t e +k m (1) Poof: One of the popeties of the algoithm states that all obots that ente the tee will emain in it until the exploation is finished eithe by moving along open/unexploed edges, o by emaining static in a vetex. If at time t e thee ae k m obots inside the tee, and since only one obot is allowed to tavese an edge in each step, then it will take k m steps to evacuate all obots to. Claim 4: With algoithm one-by-one, the maximum numbe of obots enteing a tee equals t e. That is, k mmax = t e (2) Poof: Accoding to the algoithm, at each step one obot entes a tee. If conditions ae adequate, the algoithm will keep feeding obots until time t e is eached. Since t e is the time when the last unexploed edge is tavesed, afte t e all obots in T will stat etuning to. Theoem 1: Algoithm one-by-one will completely exploe any tee in time at most D +m. That is, t c D +m (3) Poof: In the fist pat of the poof we will conside how the algoithm pefom in smalle tees. Define minimum tee as a 3-edge tee withd = 2 andm = 3 that will always be exploed by two obots in timet e = 3 (Fig. 1). As a esult, the time fo complete exploation will be: t c = t e +k m = 5. Note that t c can also be defined, fo this configuation, as: t c = D + m = 5. Define extended minimum tee as a tee with D = 2 and m e = j + 1 edges, whee j is the total numbe of leaves. The extended minimum tee will also be exploed by at most two obots (Fig. 2). In geneal, the exploation time of any extended minimum tee with two obots can be expessed as: step 1 Fig. 1. t e = m e (4) t c = t e +k m = m e +2 = m e +D (5) step 2 step 3 ( te = 3) Two obots exploing the minimum tee path of obot 1 path of obot 2 Now, conside a tee with diamete D 3. Accoding to the definition of intenal vetex, the tee with the minimum numbe of edges and diamete D 3 has the configuation shown in Fig. 3a. Fig. 3b shows that this tee can be decomposed into D 1 minimum tees. Since each minimum tee will be exploed in t e = m e with two obots, then the total 5480
= edundant obots 1 2 3 j j-1 step 1 step 2 step 3 edundant edges path of obot 1 path of obot 2 step 4 step j+1 ( t e = j+1) Fig. 2. Two obots exploing an extended minimum tee a) b) Fig. 3. a) Tee with minimum numbe of intenal vetices and D 3 (dots epesent obots located at vetices at time t e) b) Decomposition into D 1 minimum tees a0 exploation time will be equal to t e = (D 1)m e. Howeve, this esult is inaccuate: thee ae some edundant edges, as shown in Fig. 3b. Fo each D 1 minimum tees foming T thee will be D 2 edundant tems. Consequently, the total exploation time can be expessed as, t e = (D 1)m e (D 2) (6) The total numbe of edges in T can be obtained fom the following analysis: each minimum tee has m e edges, and thee ae D 1 minimum tees and D 2 edundant edges that belong to moe than one minimum tee. Thus, the total numbe of edges in T can be expessed as, m = (D 1)m e (D 2) (7) 1 2 3 a1 1 2 3 4 N h-1 N a2 a3 Fig. 4. N-ay tee ooted at a 1 a l-2 a l-1 a l a l+1 h Compaison of (6) and (7) leads to: t e = m. The numbe of obots inside the tee in Fig. 3a at time t e can be obtained following a simila analysis: each minimum tees is exploed by two obots. Thee ae some obots that ae consideed in moe than one minimum tee (cf. Fig. 3b). That is, fo each pai of minimum tees, thee is one edundant obot. Thus, the total numbe of obots at time t e can be expessed as: k m = 2(D 1) (D 2) = D (8) Using claim 3, we can expess t c as, t c = t e +k m = m+d (9) Thus, theoem 1 holds fo any small tee of diamete D and with the minimum numbe of edges in it. In the second pat of the poof, we study the behavio of the algoithm in any lage tee. To this end, we will use a full N-ay tee. A full N-ay tee is one in which all leaves have the same depth (diamete) and all intenal vetices have degee N, whee by definition N 2 and D N. Conside thee exists a N-ay tee ooted at a 1 with diamete h. The path S is defined in the N th subtee as shown in Fig. 4. The total numbe of leaves in a N-ay tee equals N h, and the numbe of intenal vetices can be calculated as follows, 1+N +N 2 + +N h 1 = h 1 N i i=0 = Nh 1 N 1 (10) Since it is assumed that the N-ay tee is ooted at a 1, the total numbe of vetices n in T equals, n = Nh 1 N 1 +Nh +1 (11) Whee the one accounts fo vetex a 0. The total numbe of edges in T is, m = n 1 = Nh 1 N 1 +Nh = Nh+1 1 N 1 Since h = D 1, then (12) can be ewitten as, (12) m = ND 1 N 1 = ND 1 +N D 2 + +N +1(13) 5481
If the one-by-one algoithm is used in the N-ay tee, the fist subtee of a 1 will be finished fist, then the second subtee and so foth. Once finished, a subtee will be blocked fo othe obots to ente. By time t e, it is likely that some obots have etuned to a 1 and enteed any othe open subtee. This will block obots at to ente the tee at evey step, and as a esult the total numbe of obots in the tee at time t e will be less than the maximum possible numbe of obots (ecall claim 2). That is, As a esult, k m < t e (14) t c = t e +k m < 2t e (15) Now we can look fo t e : if the N-ay tee has in total N h leaves, thee ae N h 1 extended minimum tees in T. Assume that all extended minimum tees can be labeled with numbes fom 1 to N h 1, whee the latte coesponds to the one ooted at a l 1 that leads to the last vetex in the path S. Assume that all the N h 1 extended minimum tees will be exploed by only one obot. This will guaantee that t c is maximized. Additionally, any extended minimum tee will be enteed by one obot afte the obot has tavesed at least h 2 steps (i.e., the time it takes to go fom a 1 to the oot of the last extended minimum tee of the subtee). Finally, the fist obot to ente the N (h 1)th extended minimum tee will do so afte a obot has been sent to any othe extended minimum tee. That is, the fist obot will ente N (h 1)th extended minimum tee afte N (h 1) 1 steps have passed. All these times ae with efeence to the N-ay tee which is ooted at a 1. In ode to efe them to a 0 we need to add one additional step. As a esult, the edge connecting a l to a l+1 will be tavesed at time, t e = (h 2)+2N +(N h 1 1)+1 Using (16) into (15), = N h 1 +2N +h 2 (16) IV. SIMULATION RESULTS We simulate the behavio of ou one-by-one flooding algoithm in diffeent tees of inceasing size (inceasing numbe of vetices n), stating with the minimum tee (n=4). Simulation esults show the uppe bound of exploation fo evey tee and the complete exploation time given by the algoithm. A typical un fo tees of up to 200 vetices is show in Fig. 5. The algoithm always exploes the tee in its entiety. Fig. 6 shows that fo a given numbe of Steps 300 250 200 150 100 50 Uppe bound ( D+m) Time fo complete exploation ( tc) 0 0 20 40 60 80 100 120 140 160 180 200 Numbe of Vetices Fig. 5. Simulation esults fo a typical un of the algoithm in tees of diffeent sizes fom n=4 to n=200 vetices diffeent configuation of tees ae possible, whee evey configuation will have diffeent exploation times.fig. 7 shows the simulation esults fo 50 andomly geneated tee configuations pe each n anging fom 4 to 100. The complete exploation time coesponds to the mean value of the 50 uns along with its standad deviation pe each value of n. The esult of the simulations (Figs. 5 and 7) show that t c < 2(N h 1 +2N +h 2) < 2N h 1 +4N +2h 4 (17) Expessing t c as a function of D yields, t c < 2N D 2 +4N +2D 6 (18) On the othe hand, the sum D +m esults on, D +m = D + ND 1 N 1 = N D 1 +N D 2 + +N +1+D (19) Compaison of (18) and (19) eveals that t c < D+m. Thus, theoem 1 holds fo any lage tee. Finally, combining both pats of the poof yields that fo any tee t c D +m. D = 2 D = 3 D = 4 Fig. 6. Example of diffeent tee configuations fo the same numbe of vetices (n=8) t c is always less o equal than the poposed uppe bound. In fact, as the size of the tee gows the gap between the uppe bound and t c inceases, guaanteeing that the uppe bound will neve be violated even if the size of the tee goes to infinite. Anothe inteesting esult of the simulations is that the gow of t c is almost linea with the size of the tee. This is a vey desiable chaacteistic since it shows its pedictable natue. Futhemoe, Fig. 7 shows that the 5482
Steps 150 100 50 Uppe bound ( D+m) Time fo complete exploation ( tc) obot in each step. We pove that the complete exploation time of the algoithm t c is neve lage than D + m. This poposed uppe bound is independent of the numbe of obots k needed to pefom the exploation, anothe diffeence with espect to othe appoaches in the liteatue. We simulate the behavio of the algoithm on tees of diffeent sizes and show that the analysis of section III poduces a tight uppe bound with espect to the complete exploation time. The simulations also show that both, the gowth of t c and the numbe of obots k needed to pefom the exploation ae linealy elated to the size of the tee. 0 0 10 20 30 40 50 60 70 80 90 100 Numbe of Vetices Fig. 7. Simulation esults fo 50 uns of the algoithm in tees of diffeent sizes fom n=4 to n=100 poposed uppe bound is tight with espect to the complete exploation time of the algoithm. Finally, Fig. 8 shows that the elationship between the numbe of obots k needed to exploe an unknown tee and the size of the tee is almost linea. This is an impotant fact about the algoithm since it glimpses the existence of a linea uppe bound in the numbe of obots needed to pefom the exploation of any unknown tee with ou algoithm. That is, thee exists a limit in the numbe of obots beyond which the exploation time will not be impoved. Numbe of Robots 40 35 30 25 20 15 10 5 0 0 10 20 30 40 50 60 70 80 90 100 Numbe of Vetices Fig. 8. Numbe of obots used fo completely exploe tees of diffeent sizes fom n=4 to n=100 using one-by-one algoithm V. CONCLUSIONS In this pape, we pesent a novel algoithm that is guaanteed to completely exploe any undiected tee with m edges and diamete D. The algoithm diffes fom othe woks in the liteatue by the fact that obots ae allowed to ente only one at a time and that edges can be tavesed by only one REFERENCES [1] A. Blum, P. Raghavan, and B. 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