Deterministic rendezvous, treasure hunts and strongly universal exploration sequences

Transcription

1 Determnstc rendezvous, treasure hunts and strongly unversal exploraton sequences Amnon Ta-Shma Ur Zwck Abstract We obtan several mproved solutons for the determnstc rendezvous problem n general undrected graphs. Our solutons answer several problems left open by Dessmark et al. We also ntroduce an nterestng varant of the rendezvous problem whch we call the determnstc treasure hunt problem. Both the rendezvous and the treasure hunt problems motvate the study of unversal traversal sequences and unversal exploraton sequences wth some strengthened propertes. We call such sequences strongly unversal traversal (exploraton) sequences. We gve an explct constructon of strongly unversal exploraton sequences. The exstence of strongly unversal traversal sequences, as well as the soluton of the most dffcult varant of the determnstc treasure hunt problem, are left as ntrgung open problems. 1 Introducton In the rendezvous problem two robots are placed n an unknown envronment. The goal s to gve the two robots determnstc sequences of nstructons that ensure that they would eventually meet, no matter on whch graph they are placed, and no matter when they are actvated. Formally, we model the envronment by a fnte, connected, undrected graph G = (V, E). The edges ncdent on a vertex u V are numbered 1, 2,..., deg(u), n a predetermned manner, where deg(u) s the degree of u. The adversary chooses the envronment, and also when to actvate each robot, and where to place a robot when t s actvated. In addton, the adversary assgns each robot a unque label that s mportant for symmetry breakng. 1 An actve robot may traverse the graph. It does so by computng at each tme step a determnstc functon from the nformaton known to t, to an nstructon. The nformaton known to a robot ncludes: the robot s label, the number of tme unts that elapsed snce the robot was actvated and the degree of ts current vertex. 2 An nstructon for a robot currently n vertex u V, s a value from the set {0, 1,..., deg(u)}, wth the nterpretaton that 0 means that the robot stays n u, whle a value 0 < deg(u) means that the robot traverses the th edge leavng u. If a robot traverses the edge (u, v) E leavng u, then the robot fnds tself at v, the other endpont of ths edge. A soluton to the rendezvous problem s an algorthm such that for any adversary (.e., any envronment, any two dfferent labels, any actvaton tmes and any two ntal vertces) the two robots eventually meet. We assume the two robots move synchronously. 3 The two robots meet only when they School of Computer Scence, Tel Avv Unversty, Tel Avv 69978,Israel. E mal: {amnon,zwck}@tau.ac.l. Amnon thanks the Israel Scence Foundaton Grant No. 1090/10. Ur s research was supported by a GIF grant. A prelmnary verson of the paper appeared n SODA If the two robots are completely dentcal, the rendezvous problem cannot be solved determnstcally. Suppose, for example, that G s a rng on n vertces and that the edges are labeled so that out of every vertex, edge 1 goes clockwse, whle edge 2 goes ant-clockwse. If the two robots start the same tme at dfferent vertces and follow the same nstructons, they would never meet! 2 Notce that the robots do not know the sze of the graph. The knowledge of the graph sze makes the problem easy to solve. 3 For an asynchronous verson of the model, see De Marco et al. [DGK + 06].

2 are both actve and are at the same vertex at the same tme. In partcular, the two robots may traverse the same edge n opposte drectons and stll mss each other. As the robots can only meet when they are both actve, the cost of a soluton on a specfc nstance set by the adversary, s the number of tme unts that elapse from the actvaton of the second robot to the robots rendezvous. Dessmark et al. [DFKP06] (see also [DFP03] and [KP04]) presented a soluton of the rendezvous problem whch guarantees a meetng of the two robots after a number of steps whch s polynomal n n, the sze of the graph, l, the length of the shorter of the two labels, and τ, the dfference between ther actvaton tmes. More specfcally, the bound on the number of steps that they obtan s Õ(n 5 τl + n 10 l). Note that τ may be much larger then n 10 l, n whch case the frst term n the bound domnates the second. Dessmark et al. [DFKP06] ask whether t s possble to obtan a polynomal bound that s ndependent of τ. 1.1 Backtrackng. Kowalsk and Malnowsk [KM08] presented a determnstc soluton to the rendezvous problem that guarantees a meetng after at most Õ(n15 + l 3 ) steps. The number of steps performed by ther soluton s ndependent of τ. However, the soluton crucally uses backtrackng, as we explan now. As we sad before, the adversary chooses the envronment. I.e., t chooses the graph sze, n, the graph G = (V, E) tself, and also the edge labels. In general, the labelng s not assumed to be consstent,.e., an edge (u, v) E may be the -th edge of u but the j-th edge of v, where j, and n fact some graphs do not have any consstent labelng at all (e.g., the 3 cycle). The soluton Kowalsk and Malnowsk present, assumes that when a robot enters a node v after traversng an edge e = (u, v), t s told the ndex j of the edge e at v. In partcular, the robot may now choose the nstructon j, causng t to backtrack to v. Formally, the nformaton known to the robot now ncludes: The robot s label L, The number of tme unts T that elapsed snce the robot was actvated, The degree of the current vertex d, and, A sequence j 1,..., j T, such that the edge e t = (v t, u t ) traversed at tme t, has ndex j t at vertex u t. As before the robot s move s determned by a determnstc functon f(l; T ; d; j 1,..., j T ) {0, 1,..., d}. We call ths varant of the problem the rendezvous problem wth backtrackng. 1.2 Our results. A central result of our work s a determnstc soluton that guarantees a rendezvous wthn Õ(n5 l) tme unts after the actvaton of the second robot. When the unknown graph n whch the two robots are placed has maxmum degree d, our soluton guarantees a rendezvous after at most Õ(d 2 n 3 l) steps. If the graph s a smple d-regular graph, the number of steps s further reduced to Õ(dn 3 l). In addton to beng ndependent of τ, our soluton s more effcent than prevous solutons for all parameters and even f τ s small. More mportantly, our soluton (ncludng the fastest one) does not use backtrackng. In addton, we ntroduce the treasure hunt problem. In ths problem, a sngle robot s supposed to locate a treasure placed n an unknown locaton n an unknown envronment, modeled agan by a fnte, connected, undrected graph. The problem can be easly solved, f the treasure s present n the graph when the robot s actvated. A much more dffcult case s when the robot starts roamng the graph before the treasure s placed n the graph, and yet we would lke the number of steps that the

3 robot makes from the tme the treasure s placed untl the treasure s found to be polynomal n n, the number of vertces n the graph, no matter how long the robot has already been runnng. The treasure hunt problem corresponds to a verson of the rendezvous problem n whch one of the robots s completely passve. The second man result n the paper s an effcent algorthm for solvng the treasure hunt problem when backtrackng s allowed. We do not know f an effcent soluton s possble when backtrackng s not allowed. 1.3 Traversal sequences and non-backtrackng algorthms. Alelunas et al. [AKL + 79] showed a random walk, wth hgh probablty, covers all vertces of a graph wthn polynomal tme. Indeed, the rendezvous problem has a trval soluton f randomzaton s allowed: each robot smply performs a random walk on the graph. The two robots wll then meet, wth hgh probablty, after a polynomal (n the sze of the graph) number of steps, see, e.g., [Ald91, CTW93]. The random walk approach can be de-randomzed as follows. Let G = (V, E) be a d-regular graph. A sequence σ 1... σ k {1,..., d} k defnes a walk v 0, v 1,..., v k n G, that starts at v 0 V, where v s the σ -th neghbor of v 1. A walk v 0, v 1,..., v k s sad to cover a graph G f t vsts every vertex of G at least once. A unversal traversal sequence [AKL + 79] for n-vertex graphs s a predetermned sequence of nstructons that when executed on any n-vertex graph, from any startng vertex, defnes a walk that vsts all the vertces of the graph. Formally, Defnton 1.1. (Unversal Traversal Sequences (UTSs)) A sequence σ {1,..., d} l s a unversal traversal sequence (UTS) for connected d-regular graphs of sze at most n f for every such graph G, any numberng of ts edges, and any startng vertex v 0 n G, the walk defned by σ n G covers G. A neat probablstc argument was used by Alelunas et al. [AKL + 79] to show that there exst unversal traversal sequences for n-vertex graphs of length Õ(n5 ). In fact, almost all sequences of ths length are unversal traversal sequences! Indeed, the treasure hunt problem, wth the addtonal guarantee that the treasure s present n the graph when the robot s actvated, can be smply solved by lettng the robot follow the sequence U 1 U 2 U 4... U 2 k..., where U s a UTS for graphs of sze. A smlar strategy solves the rendezvous problem when the robots know the graph sze, or when the robots are smultaneously actvated. Now, consder the treasure hunt problem when the robot starts roamng the graph before the treasure s placed n the graph. Assume the robot employs the same strategy as before, namely, t follows the sequence U 1 U 2 U 4... U 2 k.... In ths case, the adversary may wat wth placng the treasure untl the robot s nsde some sequence U k, for k that s much larger than the actual graph sze n. The UTS property of U k s vald only when startng at the begnnng of the sequence, not when started from the mddle. Thus, t seems we have to wat untl the next sequence U 2k starts, whch ncurs a poly(k) loss, whch for a large k s super-polynomal n n. We are thus lead to the followng problem: Is there a fxed polynomal p( ), such that for every n 1 there s a sequence S n of nstructons of length p(n), such that for every 1 k n, every contguous subsequence of S n of length p(k) s a unversal traversal sequence for k-vertex graphs? Defnton 1.2. (Strongly Unversal Traversal Sequences (SUTSs)) Let p( ) be a polynomal. A possbly nfnte sequence σ = σ 1 σ 2..., where σ {1,..., d}, s a strongly unversal traversal sequence (SUTS) wth cover tme p( ) for connected d-regular graphs, f for any k 1, any contguous subsequence of σ of length p(k) s a UTS for connected d-regular graphs of sze k. Note that f for every n 1, there s a SUTS S n of length n wth cover tme p(k) k, then S 1 S 2 S 4 S 8... s an nfnte SUTS wth cover tme 2p( ).

4 SUTSs, f exst, would solve both the rendezvous and the treasure hunt problems. We do not know, however, whether SUTSs exst, and ths s the man open problem rased by our work. Instead, we can show usng the standard probablstc method the exstence of slghtly weaker SUTSs whch are suffcent for solvng the rendezvous problem, but not the treasure hunt problem. Defnton 1.3. A sequence σ {1,..., d} l s a strongly unversal traversal sequence (SUTS) wth cover tme p( ) for connected d-regular graphs of sze [n, n ], f σ s of length at least p(n ) and for any nteger n, n n n, any contguous subsequence of σ of length p(n) s a UTS for connected d-regular graphs of sze n. For brevty, we say that such a sequence s an [n, n ]-SUTS. Lemma 1.1. For every n, there s an [log n, n]-suts wth cover tme p(k) = O(d 2 k 3 log k) for d-regular graphs. Proof: Alelunas et al. [AKL + 79] showed, by boundng the cover tme of random walks on graphs, that the probablty that a random sequence of length O(d 2 n 3 log n) over the alphabet {1, 2,..., d} s not a UTS for d-regular graphs of sze n, s at most 2 n2. By the unon bound, the probablty that such a random sequence s not a [log n, n]-suts, s at most O(n d 2 n 3 log n) 2 log2 n < 1, ths s because there are less than n graph szes, only O(d 2 n 3 log n) places to start a subsequence, and the probablty that a specfc subsequence s not a UTS for graphs of sze s s at most 2 s2. Thus, the probablty that such a random sequence s a [log n, n]-suts s postve, whch shows that sequences wth the requred propertes do exst. 4 Kowalsk and Malnowsk [KM08] defne unversal cover walks whch are essentally equvalent to our SUTSs, and almost unversal cover walks whch are essentally equvalent to our [n, n ]-SUTSs. They prove a lemma that s essentally equvalent to Lemma 1.1. It s worthwhle notng that the standard probablstc method cannot be used to obtan SUTSs that satsfy the condtons of Defnton 1.2. In partcular, note that a random sequence of length Ω(n) would contan, wth very hgh probablty, Ω(log d n) consecutve 1 s, and t s easy to see that such a subsequence s not a UTS for d-regular graphs of constant sze. It s temptng to try and prove the exstence of SUTSs usng the Lovász local lemma (see Erdös and Lovász [EL75] or Alon and Spencer [AS00]). Unfortunately, all our attempts to prove the exstence of SUTSs usng the local lemma faled. 1.4 Exploraton sequences and backtrackng algorthms. One of the motvatons for the ntroducton of unversal traversal sequences by Alelunas et al. [AKL + 79] was an attempt to obtan a determnstc log-space algorthm for the s-t connectvty problem n undrected graphs. Such an algorthm was recently obtaned by Rengold [Re08], thus resolvng ths major open problem. Rengold [Re08] log-space algorthm for the undrected s-t connectvty problem does not provde unversal traversal sequences for general graphs. It does provde, however, a log-space, and hence polynomal tme, constructon of unversal exploraton sequences, a closely related noton prevously ntroduced by Koucký [Kou02]. Roughly speakng, exploraton sequences can replace traversal sequences when backtrackng s allowed. We analogously defne strong unversal exploraton sequences (see Secton 3) and show the exstence of strong unversal exploraton sequences wth polynomal cover tmes. Furthermore, our constructon uses (non-strong) unversal exploraton sequences as black-boxes, and s effcent otherwse. In partcular, when usng Rengold s explct constructon of polynomally long unversal exploraton sequences [Re08], we get an explct constructon of strongly unversal exploraton sequences. Ths s used n our soluton of the treasure hunt problem when backtrackng s allowed. 4 The proof gven actually proves the exstence of [c log n, n]-suts wth cover tme p(k) = O(d 2 k 3 log k), for some c > 0. We do not need ths slghtly stronger result.

5 1.5 Further remarks. All graphs n ths paper are assumed to be connected. In the dscusson so far we often assumed that the graph G n whch the robots are placed s a d-regular graph, for some d 3. We now explan why ths smplfyng assumpton s mmateral. Our constructons compose unversal traversal and exploraton sequences. The only property of these sequences used n the analyss s that the n th sequence s good for all graphs of sze at most n and regular degree n. The nstructon set of these sequences nclude 0 for stayng at the current vertex, and {1,..., n} for traversng the th edge. We enforce a smple rule. If a robot at vertex u wth deg(u) < n gets an nstructon > deg(u), then t smply stays n place. We can thnk of ths as addng n deg(u) self loops to u, n partcular makng the new degree of u exactly n. To see why ths works, suppose G s the graph n whch the two robots walk, and that G has n vertces and varyng degrees deg(u) n. Let G be the graph G wth n deg(u) self loops added at each vertex u. G s a graph on n vertces wth regular degree n. In partcular, the unversal sequences are good for G and the constructon works. So far we were only nterested n the exstence of a determnstc functon that guarantees fast rendezvous, and we gnored the complexty of the functon f tself. Indeed, our non-backtrackng algorthms (as well as the solutons of Dessmark et al. [DFKP06] and [KM08]) rely on the exstence of unversal traversal sequences. As we sad before, the exstence of unversal traversal sequences s proved usng the probablstc method. Thus, even though the determnstc solutons mentoned above requre only a polynomal number of steps, t s not currently known how to compute these steps determnstcally n polynomal tme. Ths lead Dessmark et al. [DFKP06] to ask whether there s a soluton n whch the number of steps and the computaton requred to determne them are both polynomal. When backtrackng s allowed, we answer ths last queston n the affrmatve, usng our explct constructon of unversal exploraton sequences. We do not have such a constructon when backtrackng s not allowed. Fnally, we remark that our exploraton sequences were mproved and shortened by Xn [Xn07]. 2 Determnstc rendezvous solutons 2.1 An nformal dscusson. A natural dea s to have each robot run a sequence of UTSs for graphs of ncreasng szes, e.g., the nfnte sequence U 1 U 2 U 4... U 2 k..., where U n s a UTS for graphs of sze n. The sequence U n s guaranteed to cover every graph G of sze n, from any startng pont. A robot runnng U n s thus guaranteed of meetng the other robot, f the other robot s statonary (at least for long enough). The sequence U 1 U 2 U 4... U 2 k... provdes, therefore, a soluton for the (easy) verson of the treasure hunt problem n whch the treasure s placed n the graph before the robot s actvated. Note that f the treasure s placed n a graph of sze n after the robot s actvated, then the robot may have already started mplementng UTS for graphs that are much larger than n, and there s no bound that depends only on n, on the number of steps that the robot would make before t fnds the treasure. If both robots mplement the sequence U 1 U 2 U 4... U 2 k..., then we have no guarantee that the two robots would meet. Part of the problem, as explaned n the ntroducton, s the need to break the symmetry. Assume, therefore, that one of the two robots s assgned the label 0 and the other the label 1. The case of longer labels wll be dealt wth later. The next dea we explore s to have robot 0 follow the nfnte sequence and robot 1 follow the opposte sequence 0 U 1 U 1 0 U 2 U 2 0 U 4 U U 2 k U 2 k... U 1 0 U 1 U 2 0 U 2 U 4 0 U 4... U 2 k0 U 2 k.... Ths would ndeed provde a soluton to the rendezvous problem, provded that the two robots are actvated at the same tme. It s possble to modfy ths soluton so that t would work when the

6 dfference between the actvaton tmes of the two robots s polynomal n the sze of the graph. We explan the soluton n Secton 2.2. We say, n ths case, that the two robots are roughly algned. We are thus left wth the case n whch the dfference between the actvaton tme of the two robots s large, relatve to the sze of the graph n whch the two robots are placed. To solve ths case we ntroduce dle perods not only before or after the completon of a full UTS U 2 k, but also between every two adjacent steps n such a sequence. More specfcally, let us say a robot s n block f t s walkng accordng to U. (We assume here that s a power of 2.) A robot n block rests for f tme unts after each nstructon of U that t makes. The total number of tme unts requred for mplementng block s therefore U (f + 1). Suppose the two robots run an n-vertex graph. The crucal parameter used n the analyss of our rendezvous solutons s the ndex κ = κ(n) of the block n whch the robot who started frst s at the tme the second robot starts mplementng ts block n. (For smplcty, we assumed here that n s a power of 2). If f κ > U n (f n + 1) then the second robot has enough tme to mplement the sequence U n, ncludng all the dle steps added to t, whle the frst robot sleeps and the two robots would meet. If f κ U n (f n + 1) then we have an upper bound on the dfference between the startng tmes of the two robots and a meetng would take place as the two robots are roughly algned. 2.2 A basc soluton of the rendezvous problem. We now descrbe n more detal our basc soluton to the rendezvous problem. We agan assume that the two robots are assgned the labels 0 and 1. The extenson of the soluton to the case of general labels s farly straghtforward, and wll be descrbed later on. We begn by ntroducng some notaton. For any sequence σ and a bt b {0, 1}, let { σ b σ f b = 1, = 0 σ f b = 0. In other words, σ 1 s just σ, whle σ 0 s a sequence of 0 s whose length s equal to the length of σ. Also let Let σ m 1...m k = σ m 1 σ m 2... σ m k. D k (σ 1... σ m ) = σ 1 0 k σ 2 0 k... 0 k σ m 0 k. In other words, the sequence D k (σ) s obtaned from the sequence σ by nsertng an dle perod of length k after each step of σ. Fnally, for l {0, 1} we let l = 1 l. Let U n be a UTS for graphs of sze at most n, and of length u n = U n = Θ(n c ), for some c 1. The sequence that a robot wth label l {0, 1} runs s: D u1 1((U 1 U 1 ) l l) D u2 1((U 2 U 2 ) l l) D u2 k 1((U 2 ku 2 k) l l) We let W = D (U ), B (0) = 0 W 0 W W W and B (1) = W W 0 W 0 W. The sequence used by robot 0 s B (0) 1 B(0) 2 B(0) 4... and the sequence used by robot 1 s B (1) 1 B(1) 2 B(1) We call B (l) the -th block of robot l (even though ths s a slght abuse of notaton: s always a power of two, and B s the log -th block). We call the subsequences W W and 0 W 0 W chunks. Each block s therefore composed of two chunks. We let w = W and b = B (0) = B (1). Note that w n = u 2 n = Θ(n 2c ) and b n = 4w n = 4u 2 n = Θ(n 2c ). We make the techncal assumpton that 4u n u 2n, for every n = 2. 5 It follows that 16w n w 2n 5 Ths essentally corresponds to the assumpton that u n = U n = Ω(n 2 ) that we can assume wthout loss of generalty.

7 and 16b n b 2n, for every n = 2. In partcular, we get that for every j 1, j 1 =0 b 2 < 1 15 b 2j. As a consequence, we get the followng useful property: If one of the robots s actvated when the other robot s at block B l, then by the tme the second robot starts executng block B 2l, the block that follows B l, the frst robot has fnshed much less than a fourth of ths block. We are now ready to prove: Theorem 2.1. If for every k 1, U k s a UTS for graphs of sze k, and U k = Θ(k c ), c 2, then the two robots meet after at most Θ(n 4c ) steps after the actvaton of the second robot, where n s the sze of the graph n whch the two robots are placed. Proof: Let n be the sze of the graph on whch the two robots are placed. We assume that n s a power of 2. (Otherwse, take the smallest power of 2 larger than n.) Let κ be the ndex of the block n whch the frst robot to be actvated s n at the tme the second robot to be actvated reaches block n. Clearly κ n. We consder two cases: Case 1: u κ b n. As the frst robot sleeps for u κ unts of tme after each step that t takes, the second robot mplements a full copy of W n whle the frst robot s statonary and the two robots are guaranteed to meet. Ths case s depcted n Fgure 2.2. (It s assumed n the fgure that the second robot has label 0. The other case s smlar.) Fgure 1: Case 1 n the proof of Theorem 2.1 (u κ b n ). Case 2: u κ < b n. As u κ = Θ(κ c ) and b n = Θ(n 2c ), t follows that κ O(n 2 ). The second robot to be actvated thus fnshes executng ts κ-th block, and starts executng hs 2κ-th block, after O(κ 2c ) = O(n 4c ) tme unts. When ths happens, the frst robot, as dscussed n the paragraph just before Theorem 2.1, s stll executng the frst forth of ts 2κ-th block. The stuaton s therefore as depcted n Fgures 2.2 and 2.2. In the frst fgure the frst robot has label 0 whle n the second fgure the frst robot has label 1. In both cases one player s asleep whle the other player executes a whole copy of W 2κ. The two robots are therefore guaranteed to meet before the 2κ-th block s over. 2.3 An Improved soluton. To mprove the tme bound we need to make several changes. Frst, we replace each UTS U n wth a [log n, n]-suts S n for d-regular graphs wth cover tme p(k) = O(k c ) and length S n = p(n) (see Lemma 1.1). Second, we reduce the delay tme performed after each step from f k = S k to f k = (log k) c (log log k) 2c and we change the delay scheme. Instead of watng f k tme unts after each step n block k, the robot wats j tme unts after every j steps of the block, for j beng a power of two up to f k. More specfcally, after the th step n block k the robot wats for r tme unts, where r s the largest power of two for whch r and r f k. We stll assume the labels are 0 and 1. Formally, let D k (σ 1... σ m ) = σ 1 0 r 1 σ 2 0 r 2... σ m 0 r m.

8 Fgure 2: Case 2 n the proof of Theorem 2.1. The frst robot has label 0. Fgure 3: Case 2 n the proof of Theorem 2.1. The frst robot has label 1. Let S n be a [log n, n]-suts for graphs of sze at most n, and of length s n = S n = Θ(n c ), for some c 1. The sequence that a robot wth label l {0, 1} runs s: D 1 ((S 1 S 1 ) l l) D 2 ((S 2 S 2 ) l l) D 2 k((s 2 ks 2 k) l l) Before we proceed, we prove the followng useful clam: Clam 1. If a robot that s n block k and m s a power of two, 8 m f k, then every tme nterval of length t = 2m log m contans n t a contnuous watng perod of length at least m. Proof. Assume not. Fx t consecutve steps of the robot whle n block k that do not contan a watng perod of length m. The t steps nclude some orgnal steps σ q+1,..., σ q+l nterleaved n between dle steps. There are at most l watng perods of 2 tme n the length t sequence, for 2 = 1, 2, 4, 8,..., m 2 2, and n partcular the sequence length s at most l + l + 2 l l m l m/2 l(1 + log m) + m. Thus, t = 2m log m l(1 + log m) + m. Ths mples l m + 1. In partcular, there exsts at least one nstructon σ q+ (for < l) after whch there s a watng perod of length m. A contradcton. We are now ready to prove: Theorem 2.2. The two robots meet after Õ(nc ) steps, where n s the sze of the graph. Proof: Let κ be the block n whch the frst robot s n, when the second robot reaches block n. We also let T n be the tme needed to execute the n th block D n ((S n S n ) l l), where D s the modfed delay scheme. Notce that T n = O( S n f n ) = Õ(nc ). We now have several cases.

9 Assume f κ S n log 2 n. In ths case the robots are not algned and the frst robot starts much earler than the second one. Let us look at the tme when the second robot starts the actve part of block n log 2 n, and let us denote m = S n log 2 n = p(n log2 n) = O(n c log 2c n). Durng that perod m the frst robot has a contnuous watng perod that lasts at least 2 log m tme (because m f κ). m However, 2 log m = O(nc log 2c 1 n) O(n c ) f n log 2 n = p(n)f n log 2 n, and wthn that perod the second robot executes at least p(n) steps. As the sequence the second robot runs s a strong UTS, the robot covers the whole graph and the two robots meet. We can therefore assume that f κ = (log κ) c (log log κ) 2c n c (log n) 2c and n partcular log κ n and κ 2 n. Also, f κ 10n log 3 n then the robots are roughly algned and the same analyss apples as n case 2 of Theorem 2.1, and the two robots meet wthn tme Õ(nc ). Thus we can assume 10n log 3 n κ 2 n. Let t be the tme when the second robot starts executng block 2n log 3 n (notce that ths happens wthn O(T n log 3 n ) = Õ(nc ) tme from the tme the second robot s placed n the graph). As κ 10n log 3 n, the block lengths of the frst robot are much larger than those of the second robot. Thus, we ether get a long perod where the frst robot s actve, or a long perod where t s passve (and we mght need to wat for a short whle for such a perod, f the frst robot s close to a block end). As the second robot keeps changng between actve and passve states of length T n log 3 n, we get to a stuaton where one robot s actve and the other s passve for T n log 3 n tme. If the frst robot s passve durng the executon of S n log 3 n we are fne. Otherwse, the frst robot s actve and the second s passve for T n log 3 n tme. Now, wthn any T n log 3 n p(n log3 n) p(n log 2 n) log p(n log 2 n) tme unts, the frst robot does at least p(n log 2 n) p(n) steps (the condton κ 2 n s mportant here, because t guarantees that f κ n c log 2c n = p(n log 2 n)). Also, as n log κ and because S κ s a [log κ, κ] SUTS, t does not matter where the frst robot s n hs sequence, and t covers the whole graph. In partcular the two robots meet. 2.4 Dealng wth arbtrary labels. We modfy the labels. We start wth a label l = l 1... l c. Followng Dessmark et al. we let l = M(l) = l 1 l 1... l c l c 01 {0, 1} 2(c+1). Ths ensures that even f l s a prefx of l, M(l) s not a prefx of M(l ). We now descrbe the sequence each robot runs. For a bt b, let D n,b = D n ((S n S n ) bb ). For a sequence b = b 1,..., b k, let D n,b = D n,b1... D n,bk. For a robot wth label l, the n th block n the sequence (for n a power of two) s: Dn,M(l). Theorem 2.3. The two robots meet after Õ(l nc ) tme, where n s the sze of the graph, and l s the length of the shorter label. Proof: The proof follows the prevous ones. We call the robot who started executng block n frst, the frst robot (note that ths tme t may be that the robot who was actvated later becomes the frst robot). The other robot s called the second robot. We let κ be the block length of the frst robot, when the second robot starts block n. If κ > 2 n then the same argument as before apples. Ths s because that argument dd not use the labels, and also dd not rely on the number of chunks n a block. The same apples to the case where 10n log 3 n κ 2 n. Thus, we can focus on the case κ 10n log 3 n.

10 Assume κ n = 10n log 3 n,.e., the robots are almost algned. We now redefne frst and second. We call the robot who started executng block n frst, the frst robot, and the other the second robot. Now, f the frst robot gets to run a whole chunk of block n before the second robot starts block n, then the frst robot chunks are much larger than the second robot chunks, and so one robot s asleep and the other s awake for the duraton of a whole chunk of the second robot, and as before the two robots meet. Otherwse, the offset between the two robots, when the second robot starts block n, s less than one chunk. Now, a robot wth label l, runs (S m S m ) M(l). As M(l 1 ),M(l 2 ) are not prefxes of each other, there s an ndex 1 2(l + 1), such that they dffer at ndex. As the chunk lengths are now equal (because they run the same block) the offset between the robots s kept, and s less than a chunk length. In partcular, one robot s asleep whle the other s awake for a whole executon of S m. The two robots meet! 3 Unversal and strongly unversal exploraton sequences Let G = (V, E) be a d-regular graph. As we saw n Secton 1.3, a sequence σ 1 σ 2 σ k {0, 1, 2,..., d} k and a startng vertex v 0 V defne a walk v 0, v 1,..., v k n G. In a smlar way, a sequence τ 1 τ 2 τ k {0, 1,..., d 1} k and a startng edge e 0 = (v 1, v 0 ) E defne a walk v 1, v 0,..., v k as follows: For 1 k, f (v 1, v ) s the s-th edge of v, let e = (v, v +1 ) be the (s + τ )-th edge of v, where we assume here that the edges of v are numbered 0, 1,..., d 1, and that s+τ s computed modulo d. Defnton 3.1. (Unversal Exploraton Sequences (UXSs)) A sequence τ 1 τ 2 τ l {0, 1,..., d 1} l s a unversal exploraton sequence for d-regular graphs of sze at most n f for every connected d-regular graph G = (V, E) on at most n vertces, any numberng of ts edges, and any startng edge (v 1, v 0 ) E, the walk obtaned vsts all the vertces of the graph. The exstence of UXS of length O(d 2 n 3 log n) for d-regular graphs of sze at most n can be shown usng the same probablstc argument used to show the exstence of such UTS. However, whle we do not have explct polynomal-sze UTS, Rengold [Re08] obtans an explct constructon of polynomal-sze UXS: Theorem 3.1. ([Re08]) There exsts a constant c 1 such that for every d 3 and n 1, a UXS of length O(n c ) for d-regular graphs of sze at most n can be constructed, determnstcally, n polynomal tme. Rengold s explct UXSs can be easly used to turn our basc determnstc soluton for the rendezvous problem presented n the prevous secton nto an explct soluton, n the varant of the model n whch a robot s told whch edge t used to enter a vertex. Note that the knowledge of ths edge s needed to trace the walk defned by the UXS. We note, however, that the explct soluton obtaned s much less effcent than the non-explct soluton, as the constant c n Rengold s constructon s large. As a natural analog of SUTS, we can defne: Defnton 3.2. (Strongly Unversal Exploraton Sequences (SUXSs)) A possbly nfnte sequence τ = τ 1 τ 2..., where τ {0, 1,..., d 1}, s a strongly unversal exploraton sequence (SUXS) for d-regular graphs wth cover tme p( ), f for any n 1, any contguous subsequence of τ of length p(n) s a UXS for d-regular graphs of sze n.

11 Whle we cannot show the exstence of strongly unversal traversal sequences (SUTSs), even nonexplctly, the man Theorem of ths secton shows that strongly unversal exploraton sequences (SUXSs) do exst and they can be constructed determnstcally n polynomal tme. Theorem 3.2. If for every n 1 there are UXS of length O(n c ) for d-regular graphs of sze at most n, then there s an nfnte SUXS for d-regular graphs wth cover tme O(n 2c ). Furthermore, f the UXSs can be constructed determnstcally n polynomal tme, then so can the SUXS. 3.1 The sequence. The crucal property of exploraton sequences used n the proof of Theorem 3.2 s that walks defned by exploraton sequences can be reversed. For τ = τ 1 τ 2 τ k {0, 1,..., d 1} k, we let τ 1 = τ 1 k τ 1 k 1 τ 1 1 1, where τ = d τ. It s not dffcult to check that a walk defned by an exploraton sequence τ can be backtracked by executng the sequence 0τ 1 0. Note that f e 0, e 1,..., e k s the sequence of edges defned by τ, startng wth e 0, then executng 0τ 1 0, startng wth e k defnes the sequence e k, e k, e k 1,..., e 0, e 0, where e s the reverse of edge e,.e. f e = (u, v), then e = (v, u). Also, t s not dffcult to see that f τ s a unversal exploraton sequence for graphs of sze at most n, then so s τ 1. In ths secton t s more convenent to let U n be a unversal sequence of length n, rather than a unversal sequence for graphs of sze n. Let U n be a sequence of length n whch s a unversal exploraton sequence for d-regular graphs of sze at most γn α, for some γ > 0 and 0 < α < 1. We are nterested n sequences U n only when n s a power of 2. We may assume that for every k = 2 and n = 2 j, where < j, U k s a prefx of U n. If ths condton does not hold, we can replace the sequence U n by the sequence Ūn = U 1 U 1 U 2 U 4... U n/2. (Recall that n s assumed to be a power of 2. The sequence U 1 s used twce to ensure that Ūn = n.) We now defne recursvely a sequence S n. The recurson base s S 1 = U 1. Assume that U n = u 1 u 2... u n and that n 2. Defne, (3.1) S n = u 1 S r1 0Sr u 2 S r2 0Sr u 3... u S r 0S 1 0 u +1 u n 1 S rn 1 0Sr 1 n 1 0 u n where for every 1 < n, we let r = 2 j be the largest power of 2 where 2 2j but 2 2(j+1) (and so r s at most but s much smaller f has only few 2 factors). Before we begn the proof of Theorem 3.2 we show the sequence S n s short. Lemma 3.1. For every n = 2 j, where j 1, we have S n < 20n. r Proof: Let s n = S n. It s not dffcult to see that S 2 = U 2 + ( U 2 1) 2( S 1 + 1) + We remnd the reader that U l = l. It follows that s /2 1 j=1 /2 1 j=1 U2 1 s 2 j s 2 j 1 2 2j 2 2j. 2( S 2 j S 2 j 1 ) The values s 1,..., s 8 can be computed drectly from the defnton. Dong that gves s 1 = 1, s 2 = 5, s 4 = 15, s 8 = 31 and s 16 = 87. The clam that s n < 20n for every n 16 then follows usng nducton. (In fact, lm n s n /n = )

12 3.2 S n s strong. The key lemma we are usng n the analyss s: Lemma 3.2. Let k and n 2k 2 be powers of 2. Then, every subsequence T of S n or Sn 1 s 2k k 2 contans, as a contguous subsequence, a full copy of S k or S 1 k. of length The proof of ths lemma uses some nce combnatoral propertes of the recurson. Before we gve a proof to the lemma, we show that t suffces for provng that S n s strong (Theorem 3.2). Proof: (of Theorem 3.2) Suppose there are UXS of length p(n) = O(n c ) for d regular graphs of length at most n. Let E n = S 2p 2 (n). The nfnte sequence s E = E 1 E 2 E 4 E 8... E 2 k.... Then, by Lemma 3.2, any subsequence of E of length O(p 2 (n)) contans n t a full copy of S p(n) or S 1 p(n). W.l.o.g., let us assume t contans S p(n). We now look at the recursve defnton of S p(n) and gnore all the recursve calls of S j (for j < p(n)) and ther nverses. We can gnore these parts because the sequence Sj 1 reverses the acton of S j. We are left wth U p(n) = u 1,..., u p(n). However, U p(n) s a UXS for graphs of sze n. In partcular, any subsequence of E of sze O(p 2 (n)) = O(n 2c ) covers all graphs of sze n. Ths, n partcular, gves (an explct or non-explct) soluton to the treasure hunt problem: all the robot has to do s to run the SUXS. The adversary decdes when to put the treasure, but then the subsequence of length p(n) startng at ths pont s a UXS and the robot fnds the treasure. 3.3 The combnatoral propertes of S n. We now turn to provng Lemma 3.2 whch essentally clams that n any sub-sequence of length s 2k 2 there s a full copy of S k or S 1 k. For the proof of ths lemma we use only the recursve defnton of S n. I.e., the clam s true for any sequence u 1,..., u n, and we do not use the fact that U n s a UXS. We frst determne what s Sn 1. We have: (3.2) Sn 1 = n (3.3) 0S rn 1 0S 1 r n 1 n 1 0S r n 2 0S 1 r n 2 n 2 u S r 0Sr 1 = un 1 0S r1 0Sr 1 1 n 1 0S r 2 0Sr 1 2 n 2 u S r n 0Sr 1 n 2 0S r1 0Sr S rn 1 0Sr 1 n 1 1. where we used the clam that r = r n that we soon state and prove. The sequence Sn 1 thus dffers from S n only n elements that orgnate from U n and n the algnment of the 0 s. The fact r = r n s one of several combnatoral propertes of the constructon that we now state. Clam 2. Let n be a power of If k n then the sequence S k s a prefx of S n. 2. For every 1 < n, we have r = r n. 3. For any k and, f r k then r = r mod k Denote x = max{2 2 x, Z + },.e., x s the largest power of 2 less than or equal to x. Then, the frst half of S n s equal to S n/2 S n/2 0. Smlarly, the second half of S n starts wth a copy of S 1 n/2. We defer the proof of the clam to later on, and we turn to the proof of Lemma 3.2. Proof: (of Lemma 3.2) Fx k. We prove the clam by nducton on n. If n = 2k 2 then the clam s vacuously satsfed as S n and Sn 1 are too short to contan a subsequence of length s 2k = s n + 1.

13 Assume, therefore, that the clam holds for every m = 2 j that satsfes 2k 2 m < n = 2 j. We show that t also holds for n. Let T be a subsequence of S n of length s 2k Essentally the same argument works f T s a subsequence of such length of Sn 1 (because Equaton (3.2) shows Sn 1 s smlar to S n ). We consder the followng cases: Case 1: T s completely contaned n a subsequence S m or S 1 m of S n, for some m < n. The clam then follows mmedately from the nducton hypothess. Case 2: T s completely contaned n a subsequence S m 0Sm 1 of S n, for some m < n. In ths case, T = T 0T, where T s a suffx of S m and T s a prefx of Sm 1. Ether T 1 2 s 2k 2 or T 1 2 s 2k 2. As a suffx of S m s the same as a prefx of Sm 1 we focus on just one case. We assume for concreteness that T 1 2 s 2k 2. As T s a prefx of Sm 1, and T 1 2 s 2k 2, t follows that m 2k2. Now, S 1 k s almost a prefx of Sm 1, n the sense that they dffer only n symbols that orgnate drectly from S m (ths follows from Equaton (3.2) and Clam 2(1)). In partcular, a prefx of Sm 1 of length 1 2 s 2k 2, half the length of S 2k 2, ends wth a full copy of S k, followed by 0 (by Clam 2(4)). Case 3: T contans a symbol u l of S n that orgnates from U n. In ths case, T = T u l T. Agan, we ether have T 1 2 s 2k 2 concreteness, that T 1 2 s 2k2. The other case s analogous. Let or T 1 2 s 2k2. Assume agan, for S n,l = u l S rl 0Sr 1 l 0 u l+1 u n 1 S rn 1 0Sr 1 n 1 0 u n be the suffx of S n that starts wth the symbol u l that orgnates from the l-th symbol of U n. We clam that the prefx of S n,l of length 1 2 s 2k 2 contans a copy of S k. Let l = l/k 2 k 2 be the frst ndex after l whch s dvsble by k 2. Clearly r l k and r k for every l < l. Also, S k s a prefx of S rl (by Clam 2(1)) and so S = u l S rl 0Sr 1 l 0 u l S k s a prefx of S n,l whch ends wth a full copy of S k. For every l < l we have r = r mod k 2 (by Clam 2(3)), and so up to changng the symbols u the sequence S s contaned n the frst of half of S 2k 2 (whch also ends wth S k ), and hence the length of S s at most 1 2 s 2k2, as requred. Fnally, we turn to provng the clam: Proof. (of Clam 2) 1. By nspectng Equaton (3.1) and because U k s a prefx of U n. 2. n s a power of 2. Now, a number that s a power of 2 that dvdes 1 n must also dvde n. Hence ff n. It follows that the largest power of two that dvdes s also the largest power of 2 that dvdes n and so r = r n. 3. Frst, clearly for any k a power of 2 and any, r mod k r. We know, however, that k s large enough,.e., r k, and so r 2 and r2 k2 and t follows that r 2 mod k2. Thus, r r mod k 2 and therefore r = r mod k We already saw that S n/2 s a prefx of S n. By nspectng Equaton (3.1) we see that n fact the frst half of S n s S n/2 S rn/2 0. If n/2 s a power of 2, n/2 = 2, then r n/2 = 2 /2 = 2 /2 = n/2, whch completes the frst asserton n tem (4). The second asserton s smlarly proved.

14 4 Concludng remarks and open problems We obtaned mproved determnstc solutons for the rendezvous problem that are ndependent of τ, the dfference between the actvaton tmes of the two robots. Furthermore, we get close to the length of a UTS. Wth backtrackng, we obtan a polynomal tme, explct soluton. The technque used n the paper rases the queston whether there exst strongly UTSs. We defne the treasure hunt problem whch s the varant of the problem where one robot s statc and always stays n the place where t s put. Strong UTS exst ff the treasure hunt problem has a soluton that s ndependent of τ. Standard probablstc arguments used to show the exstence of (non-strong) unversal traversal and exploraton sequences cannot be used to prove the exstence of strongly unversal sequences, wth any cover tme. We can, however, show an explct constructon of strong unversal exploraton sequences. We do not know wether strong unversal traversal sequences exst. We beleve ths last queston deserves further study. Acknowledgements We thank Oded Goldrech for a dscusson that led to the defnton of the treasure hunt problem. References [AKL + 79] R. Alelunas, R.M. Karp, R.J. Lpton, L. Lovász, and C. Rackoff. Random walks, unversal traversal sequences, and the complexty of maze problems. In FOCS, pages , [Ald91] D.J. Aldous. Meetng tmes for ndependent Markov chans. Stochastc Processes and ther applcatons, 38(2): , [AS00] N. Alon and J. Spencer. The Probablstc Method. John Wley, [CTW93] D. Coppersmth, P. Tetal, and P. Wnkler. Collsons among random walks on a graph. SIAM Journal on Dscrete Mathematcs, 6(3): , [DFKP06] A. Dessmark, P. Fragnaud, D. Kowalsk, and A. Pelc. Determnstc rendezvous n graphs. Algorthmca, 46(1):69 96, [DFP03] A. Dessmark, P. Fragnaud, and A. Pelc. Determnstc rendezvous n graphs. In ESA, pages , [DGK + 06] G. De Marco, L. Gargano, E. Kranaks, D. Krzanc, A. Pelc, and U. Vaccaro. Asynchronous determnstc rendezvous n graphs. Theoretcal Computer Scence, 355(3): , [EL75] P. Erdös and L. Lovasz. Problems and results on 3-chromatc hypergraphs and some related questons. In A. Hajnal et al., edtors, Infnte and Fnte Sets, volume 11 of Colloq. Math. Soc. Janos Bolya, pages North-Holland, [KM08] D. R. Kowalsk and A. Malnowsk. How to meet n anonymous network. Theoretcal Computer Scence, 399(1-2): , [Kou02] M. Koucký. Unversal traversal sequences wth backtrackng. Journal of Computer and System Scences, 65(4): , [KP04] D.R. Kowalsk and A. Pelc. Polynomal determnstc rendezvous n arbtrary graphs. In ISAAC, pages , [Re08] Omer Rengold. Undrected connectvty n log-space. J. ACM, 55(4), [Xn07] Qn Xn. Faster treasure hunt and better strongly unversal exploraton sequences. In Proceedngs of the 18th nternatonal conference on Algorthms and computaton, pages Sprnger-Verlag, 2007.