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1 Algorthms 2008, 1, 30-42; DOI: /a Artcle OPEN ACCESS algorthms ISSN andomzed Compettve Analyss for Two Server Problems Wolfgang Ben 1,,, Kazuo Iwama 2 and Jun Kawahara 2 1 Center for the Advanced Study of Algorthms, School of Computer Scence, Unversty of Nevada, Las Vegas, Nevada 89154, USA 2 School of Informatcs, Kyoto Unversty, Kyoto , Japan Emal addresses: ben@cs.unlv.edu; wama@kus.kyoto-u.ac.jp; jkawahara@kus.kyoto-u.ac.jp Author to whom correspondence should be addressed. esearch done whle vstng Kyoto Unversty as Kyoto Unversty Vstng Professor. eceved: 13 August 2008 / Accepted: 18 September 2008 / Publshed: 19 September 2008 Abstract: We prove that there exsts a randomzed onlne algorthm for the 2-server 3-pont problem whose expected compettve rato s at most Ths s the frst nontrval upper bound for randomzed k-server algorthms n a general metrc space whose compettve rato s well below the correspondng determnstc lower bound (= 2 n the 2-server case). Keywords: andomzed Algorthms; Onlne Algorthms; Server Problems. 1. Introducton The k-server problem, ntroduced by Manasse, McGeoch and Sleator [1], s one of the most fundamental onlne problems. In ths problem the nput s gven as k ntal server postons and a sequence p 1, p 2, of requests n the Eucldean space, or more generally n any metrc space. For each request p, the onlne player has to select, wthout any knowledge of future requests, one of the k servers and move t to p. The goal s to mnmze the total movng dstance of the servers. The k-server problem s wdely consdered nstructve to the understandng of onlne problems n general, yet, there are only scattered results. The most notable open problem s perhaps the k-server conjecture, whch states that the k-server problem s k-compettve. The conjecture remans open for k 3, despte years of effort by many researchers; t s solved for a very few specal cases, and remans open even for 3 servers when the metrc space has more than 6-ponts. The current best upper bound s 2k 1 gven by Koutsoupas and Papadmtrou n 1994 [2]. The conjecture s true for k = 2, for the

2 Algorthms 2008, a b c Fgure 1. 3 ponts on a lne C 1 L d 1 d 2 Fgure 2. Trangle CL lne [3], trees [4], and on fxed k + 1 or k + 2 ponts [5]. It s stll open for the 3-server problem on more than sx ponts and also on the crcle [6]. The lower bound s k whch s shown n the orgnal paper [1]. In the randomzed case, even less s known. Indeed, one of hardest problems n the area of onlne algorthms s to determne the exact randomzed compettveness of the k-server problem, that s, the mnmum compettveness of any randomzed onlne algorthm for the server problem. (As s customary, we mean by compettve rato of a randomzed algorthm ts expected compete rato.) Very lttle s known for general k. Bartal et al. [7] have an asymptotc lower bound, namely that the compettveness of any randomzed onlne algorthm for an arbtrary metrc space s Ω(log k/ log 2 log k). Even the case k = 2 s open for the randomzed 2-server problem, and, despte much effort, no randomzed algorthm for general metrc spaces wth compettveness strctly lower than 2 has been found. Surprsngly, the classc algorthm ANDOM SLACK [8], a very smple trackless algorthm, has been the algorthm wth best compettve rato for almost two decades now. Karln et al. [9] gave a lower e bound of, whch s the bound for the classcal sk rental problem. Ths bound s derved n a space e 1 wth three ponts, where two ponts are located closely together and the thrd pont s far from both of these. The best known lower bound s 1 + e , but ths lower bound requres a space wth at least four ponts, see [10]. (A lower bound very slghtly larger than 1 + e 1 2 s stated n [10], but wthout proof.) It s ndeed surprsng that no randomzed algorthm wth compettve rato better than 2 has been found, snce t seems ntutve that randomzaton should help. It should be noted that generally randomzaton s qute powerful for onlne problems, snce t obvously reduces the power of the adversary. Such seems to be the case for the 2-server problem as well. To gve ntuton, consder a smple 2-server problem on the three equally spaced ponts a, b and c on a lne (See Fg. 1). It s easy to prove a lower bound of 2 for the compettve rato of any determnstc algorthm: The adversary always gves a request on the pont the server s mssng. Thus for any onlne algorthm, A, ts total cost s at least n the number of requests. But t turns out by a smple case analyss that the offlne cost s n/2. Suppose nstead that A s randomzed. Now f the request comes on b (wth mssng server), then A can decde by a con flp whch server (a or c) to move. An (oblvous) adversary knows A s algorthm completely but does not know the result of the con flp and hence cannot determne whch pont (a or c) has the server mssng n the next step. The adversary would make the next request on a but ths tme a has a server wth probablty 1/2 and A can reduce ts cost. Wthout gvng detals, t s not hard to show that ths algorthm A wth the randomzed acton for a request to b and a greedy acton one for others has a compettve rato of 1.5. Indeed, one would magne that t mght be qute straghtforward to desgn randomzed algorthms whch perform sgnfcantly better than determnstc ones for the 2-server problem. As mentoned above

3 Algorthms 2008, 1 32 ths has not been the case. Only few specal cases have yelded success. Bartal, Chrobak, and Larmore gave a randomzed algorthm for the 2-server problem on the lne, whose compettve rato s slghtly better than 2 ( ) [11]. One other result by Ben et. al. [12] uses a novel technque, the 78 knowledge state method, to derve a compettve randomzed algorthm for the specal case of Cross Polytope Spaces. Usng smlar technques a new result for pagng (the k-server problem n unform spaces) was recently obtaned. Ben et al. [13] gave an H k -compettve randomzed algorthm whch requres only O(k) memory for k-pagng. (Though the technques n the current paper are nspred by ths work, the knowledge state method s not used here.) Lund and engold showed that f specfc three postons are gven, then an optmal randomzed algorthm for the 2-server problem over those three ponts can be derved n prncple by usng lnear programmng [14]. However, they do not gve actual values of ts compettve rato and to ths date the problem s stll open even for the 2-server 3-ponts case. Fnally, we menton other somewhat related work n the realm of onlne computaton: see the work of Karln et al. [15] for sk-rental problems, engold et al. [16] for lst access problems, and Fat et al. [17] for pagng. Our Contrbuton. In ths paper, we prove that the randomzed compettve rato of the 2-server 3-pont problem n a general metrc space s at most and also we conjecture that t s at most e/(e 1) Thus we gve an upper bound that matches the lower bound wthn a small ε. The underlyng dea s to fnd a fnte set S of trangles (.e. three ponts) such that f the compettve rato for each trangle n S s at most c, then the compettve rato for all trangles n any metrc space s at most c δ(s) where δ(s) 1 s a value determned by S. To bound the compettve rato for each trangle n S, we apply lnear programmng. As we consder larger sets, the value of δ(s) becomes smaller and approaches 1. Thus the upper bound of the general compettve rato also approaches the maxmum compettve rato of trangles n S and we can obtan arbtrarly close upper bounds by ncreasng the sze of the computaton. Our result n ths paper strongly depends on computer smulatons smlar to earler work based on knowledge states. Indeed, there are several successful examples of such an approach, whch usually conssts of two stages; () reducng nfntely many cases of a mathematcal proof to fntely many cases (where ths number s stll too large for a standard proof ) and () usng computer programs to analyze the fntely many cases. See the work n [18 22] for desgn and analyss of such algorthms. In partcular, for onlne compettve analyss, Seden proved the currently best upper bound, , for onlne bnpackng [23]. Also wth ths approach, Horyama et al. [24] obtaned an optmal compettve rato for the onlne knapsack problem wth resource augmentaton by buffer bns. 2. Our Approach Snce we consder only three fxed ponts, we can assume wthout loss of generalty that they are gven n the two-dmensonal Eucldean space. The three ponts are denoted by L, C and, furthermore let d(c, L) = 1, d(c, ) = d 1, and d(l, ) = d 2 (see Fg. 2). Wthout loss of generalty we assume that 1 d 1 d 2 d 1 + 1, snce the actual lengths of the trangle s sdes are rrelevant and only ther ratos matter. The 2-server problem on L, C and s denoted by (1, d 1, d 2 ), where the two servers are

4 Algorthms 2008, 1 33 on L and ntally and the nput s gven as a sequence σ of ponts {L, C, }. (1, d 1, d 2 ) s also used to denote the trangle tself. The cost of an onlne algorthm A for the nput sequence σ s denoted by ALG A (σ) and the cost of the offlne algorthm by OP T (σ). Suppose that for some constant α 0, E[ALG A (σ)] r OP T (σ) + α, holds for any nput sequence σ. Then we say that the compettve rato of A s at most r. We frst consder the case that the three ponts are on a lne and both d 1 and d 2 are ntegers. In ths case, we can desgn a general onlne algorthm as follows. The proof s gven n the next secton. Lemma 1 Let n be a postve nteger. Then there exsts an onlne algorthm for (1, n, n + 1) whose compettve rato s at most C n = (1+ 1 n )n 1 n+1. (1+ 1 n )n 1 Note that f trangles 1 and 2 are dfferent, then good algorthms for 1 and 2 are also dfferent. However, the next lemma says that f 1 and 2 do not dffer too much, then one can use an algorthm for 1 as an algorthm for 2 wth small sacrfce on the compettve rato. Lemma 2 Suppose that there are two trangles 1 = (1, a 1, b 1 ) and 2 = (1, a 2, b 2 ) such that a 1 a 2 and b 1 b 2 and that the compettve rato of algorthm A for 1 s at most r. Let α = max( a 1 a 2, b 1 b2 ). Then the compettve rato of A for 2 s at most r α. Proof. For α let α = (1/α, a 1 /α, b 1 /α). Fx an arbtrary nput sequence σ and let the optmal offlne cost aganst σ be OP T 1, OP T 2 and OP T α for 1, 2 and α, respectvely. Snce α s smlar to 1 and the length of each sde s 1/α, OP T α s obvously (1/α)OP T 1. Snce every sde of 2 s at least as long as the correspondng sde of α, OP T 2 OP T α = (1/α)OP T 1. Let the expected cost of A aganst σ for 1 and 2 be ALG 1 and ALG 2, respectvely. Note that A moves the servers exactly n the same (randomzed) way for 1 and 2. Snce each sde of 2 s at most as long as the correspondng sde of 1, ALG 2 ALG 1. We have ALG 2 OP T 2 ALG 1 (1/α)OP T 1 = max( a 1 a 2, b 1 b2 ) ALG 1 OP T 1. Thus we can approxmate all trangles, whose α-value s at most wthn some constant, by a fnte set S of trangles. More precsely we ntroduce the noton of an approxmaton set as follows: Suppose that a target compettve rato,.e. a compettve rato one wshes to acheve, s r 0. Then we frst calculate the mnmum nteger n 0 such that r 0 n 0+2 n 0 C n0 +1, where C n0 +1 s the value gven n the statement of Lemma 1. Next we construct a set S such that for any two numbers a and b wth 1 a n 0 and b a + 1, there exst two trangles 1 = (1, a 1, b 1 ) and 2 = (1, a 2, b 2 ) n S such that the followng condtons are met: () a 2 < a a 1 and b 2 < b b 1, () there exsts an algorthm for 1 whose compettve rato s r 1, and () r 1 max( a 1 a 2, b 1 b2 ) r 0. We call such a set an r 0 -approx set. Lemma 3 If one can construct an r 0 -approx set S, then there s an onlne algorthm for the 2-server problem on three ponts, whose compettve rato s at most r 0. Proof. Consder the followng algorthm A(a, b) whch takes the values a and b of the trangle (1, a, b). Note that A(a, b) s an nfnte set of dfferent algorthms from whch we select one due

5 Algorthms 2008, 1 34 to the values of a and b. If a n 0, then we select the maxmum nteger n such that a n. Then A(a, b) uses the algorthm for (1, n + 1, n + 2) of Lemma 1. Clearly we have a n + 1 and b n + 2. Therefore, by Lemma 2, the compettve rato of ths algorthm for (1, a, b) s at most (recall that C n+1 s the compettve rato of ths algorthm for (1, n + 1, n + 2) gven n Lemma 1) ( n + 1 max a, n + 2 ) C n+1 n + 2 b n C n+1 n C n0 +1 r 0. n 0 By a smple calculaton we have that n+2 C n n+1 = n+2 (1+ n) 1 n 1 n+1 monotoncally decreases, whch n (1+ n) 1 n 1 mples the nequalty second to last. If a < n 0, then we have the two trangles 1 and 2 satsfyng the condtons () to () above. Then we use the algorthm for 1 guaranteed by condton (). Its compettve rato for (1, a, b) s obvously at most r 0 by Lemma Three Ponts on a Lne We now prove Lemma 1. To ths end we make use of a state dagram, called the offset graph, whch ndcates the value of the work functon W (s, σ) [25]. ecall that W (s, σ) gves the optmal offlne cost under the assumpton that all requests gven by σ have been served and the fnal state after σ must be s, where s s one of (L, C), (L, ) and (C, ) n our stuaton. Fg. 3 shows the offset graph, G OP n T for (1, n, n + 1). Each state contans a trple (x, y, z), whch represents three values for the work functon; the frst for when (C, ) are covered (leavng L blank), the next for when (L, ) are covered, and the last for when (L, C) are covered. For nstance, n the fgure the top mddle state s the ntal state, whch s denoted by V L. ecall that the ntal server confguraton s (L, ). Ths state contans (n, 0, 1), whch means that W ((L, C), φ) = n, W ((L, ), φ) = 0, and W ((C, ), φ) = 1. For example, to see that W ((L, C), φ) = n, consder that ntally the request sequence s empty denoted by φ. The ntal server confguraton s (L, ); thus n order to change ths confguraton nto (L, C), one can optmally move a server from to C at cost n. Therefore, W ((L, C), φ) = n. Consder now state V 3 the fourth state from the top. Its trple gves the value of the work functon for the request sequence CLC, namely, W ((L, C), CLC), W ((L, ), CLC) and W ((C, ), CLC). Note that ths request sequence CLC s obtaned by concatenatng the labels of arrows from the ntal state V L to V 3. For the work functon value of W ((L, ), CLC) we have W ((L, ), CLC) = 4. Ths s calculated from the prevous state V 2 n the followng way: Server poston (L, ) can be reached from prevous confguraton (L, ) (= 2) plus 2 (= the cost of movng a server on L to C and back to L) or from prevous confguraton (C, ) (= 3) plus 1 (= the cost of movng a server on C to L),.e. n both cases a value of 4. From state V 3, there s an arrow to V C as a result of request. Carryng out a smlar calculaton, one can see that the trple should change from (n, 4, 3) to (n + 4, 4, 3) n ths transton. However, the trple n V C s (n + 1, 1, 0). Ths s because of an offset value of 3 on the arrow from V 3 to V C. Namely, (n + 1, 1, 0) n V C s obtaned from (n + 4, 4, 3) by reducng each value by 3. Because of the use of offset values a fnte graph can be used to represent the potentally nfntely many values of the work functon. Thus one can conclude that (n, 0, 1) n the ntal state V L also means

6 Algorthms 2008, 1 35 v LC v L v C (0, n, n+1), n (n, 0, 1) L, 1 (n+1, 1, 0) L, n, 2n-2 v 1, 2 v 2 v 3 v 2n-3 v 2n-2 (n, 2, 1) v 2n-1 C, 0, 1 L, 0 (n, 2, 3) C, 0 (n, 4, 3) L, 0 C, 0 (n, 2n-2, 2n-3) L, 0 (n, 2n-2, 2n-1) C, 0 (n, 2n, 2n-1), 3, 2n-3, 2n-1 S LC S L S C L L S 1 p 1-p 1 1 S 2 p 1-p 1 1 S 3 p 1-p 2 2 S 2n-3 p 1-p n-1 n-1 S 2n-2 p 1-p n-1 n-1 S 2n-1 p 1-p n n C L C L C L C Fgure 3. Offset graph Fgure 4. State dagram of the algorthm (n+4, 4, 5), (n+8, 8, 9), by traversng the cycle V L V 1 V 2 V 3 V C repeatedly. We leave t to the reader to formally verfy that Fg. 3 s a vald offset graph for (1, n, n + 1). We ntroduce another state graph, called the algorthm graph. Fg. 4 shows the algorthm graph, G ALG n, for (1, n, n + 1). Notce that G ALG n s smlar to G OP n T. Each state ncludes a trple (q 1, q 2, q 3 ) such that q 1 0, q 2 0, q 3 0 and q 1 +q 2 +q 3 = 1, whch means that the probabltes of confguratons (C, L), (L, ) and (C, ) are q 1, q 2 and q 3, respectvely. (Snce the most recent request must be served, one of the three values s zero. In the fgure, therefore, only two probabltes are gven. For example, n S 1, the probabltes for (L, C)(= p 1 ) and for (C, )(= 1 p 1 ) are gven. In our specfc algorthm G ALG n, we defne those values as follows: S LC = (1, 0, 0), S L = (0, 1, 0), S C = (0, 0, 1), S 2 1 = (p, 0, 1 p ) ( = 1,..., n), S 2 = (p, 1 p, 0) ( = 1,..., n 1) n where p s (1+ 1 n ) 1. n+1 (1+ 1 n )n 1 We descrbe how to transform an algorthm graph nto the actual algorthm. Suppose for example that the request sequence s CL. Assume the algorthm s n state S 2, and suppose that the next request s C. The state transton from S 2 to S 3 occurs. Suppose that S 2 has confguraton-probablty pars (C 1, q 1 ), (C 2, q 2 ), and (C 3, q 3 ) (C 1 = (L, C), C 2 = (L, ) and C 3 = (C, )) and S 3 has (C 1, r 1 ), (C 2, r 2 ), and (C 3, r 3 ). We ntroduce varables x j (, j = 1, 2, 3) such that x j s equal to the probablty that the

7 Algorthms 2008, 1 36 confguraton before the transton s C and the confguraton after the transton s C j. Indeed, by a slght abuse of notaton the x j values can be consdered to be the algorthm tself. The x j values also allow us to calculate the cost of the algorthm as descrbed next. The average cost for a transton s gven by cost = 3 3j=1 =1 x j d(c, C j ), where d(c, C j ) s the cost to change the confguraton from C to C j. We can select the values of x j n such a way that they mnmze the above cost under the condton that 3 j=1 x j = q, 3=1 x j = r j. In the case of three ponts on the lne, t s straghtforward to solve ths lnear program n general. If the servers are on L and C and the request s, then a greedy move (C ) s optmal. If the servers are on L and and the request s C, then the optmal probablty s just a proportonal dstrbuton due to d(l, C) and d(c, ). The x j values also show the actual moves of the servers. For example, f the servers are on L and n S 2, we move a server n L to C wth probablty x 23 /q 2 and to C wth probablty x 21 /q 2. From the values x j, one can also obtan the expected cost of an algorthm for each transton, as follows: cost(s LC, S L ) = n, cost(s C, S L ) = 1, cost(s L, S 1 ) = np p 1, cost(s 2 1, S 2 ) = 1 p ( = 1,..., n 1), cost(s 2, S 2+1 ) = n(p +1 p ) + 1 p +1 ( = 1,..., n 1), cost(s 2 1, S C ) = (n + 1)p ( = 1,..., n), cost(s 2, S L ) = np ( = 1,..., n 1), cost(s 2n 1, S LC ) = (n + 1)(1 p n ). We are now ready to prove Lemma 1. ecall that G OP T n and G ALG n sequence σ, we can thus assocate the same sequence, λ(σ), of transtons n G OP T n offlne cost for λ(σ) can be calculated from G OP T n and the onlne cost from G ALG n are the same graph. Wth a request and G ALG n. The. Consderng these two costs, we obtan the compettve rato for σ. We need only consder the cycles of the type consdered below, as other cycles can be obtaned as dsjont unons. (1) S 1, S 2,..., S 2h 1, S C, S L (h = 1,..., n 1) (2) S 1, S 2,..., S 2h, S L (h = 1,..., n 1) (3) S 1, S 2,..., S 2n 1, S LC, S L. For sequence (1), the OPT cost s 2h and ALG cost s 2np h + 2h 2 h 1 j=1 p j = 2hC n. Smlarly, for sequence (2), OP T = 2h and ALG < 2hC n and for sequence (3) OP T = 2n and ALG = 4n 2 n j=1 p j = 2nC n. Thus the compettve rato s at most C n for any of these sequences, whch proves the lemma. 4. Constructon of a Fnte Set of Trangles For trangle 1 = (1, a, b) and d > 0, let 2 = (1, a, b ) be any trangle such that a d a a and b d b b. Then as shown n Sec. 2 the compettve rato for 2, denoted by f( 2 ), can be wrtten as f( 2 ) max ( a a, b ) ( a f( b 1 ) max a d, ) b f( 1 ) b d a a d f( 1).

8 Algorthms 2008, 1 37 b fnte set of squares (trangles) X 0 g(x ) = 0 n0 n C n0 b [2, 3; 1] 2 X d (a,b ); C = r g(x ) = a a - d r [3, 4; 2] [2, 2; 1] 0 a 0 1 a Fgure 6. Dvson of a square Fgure 5. Coverng Ω (The last nequalty comes from the fact that a b.) ecall that trangle (1, a, b) always satsfes 1 a b a + 1, whch means that (a, b) s n the shaded area of Fg. 5, whch we denote by Ω. Consder pont (a, b ) n ths area and the square X of sze d, whose rght upper corner s (a, b ) (Fg. 5). Such a square s also denoted by [a, b ; d ]. Then for any trangle whose (a, b )-values are wthn ths square a (some porton of t may be outsde Ω), ts compettve rato can be bounded by a d f( (1, a, b )), whch we call the compettve rato of the square X and denote by g(x ) or g([a, b ; d ]). Addtonally, for = 0 the value g(x 0 ) s bounded by n 0 C n 0 2 n 0. Consder a fnte set of squares X 0, X 1,..., X k = [a k, b k ; d k ],..., X m wth the followng propertes (see also Fg. 5): (1) The rght-upper corners of all the squares are n Ω. (2) X 0 s the rghtmost square, whch must be [, + 1; 2] for some. (3) The area of Ω between a = 1 and must be covered by those squares, or any pont (a, b) n Ω such that 1 a must be n some square. Suppose that all the values of g(x ) for 0 m are at most r 0. Then one can easly see that the set S = =0,m { (1, a, b ), (1, a d, b d )} of trangles satsfes condtons () to () gven n Sec. 2,.e., we have obtaned the algorthm whose compettve rato s at most r 0. The ssue s how to generate those squares effcently. Note that g(x) decreases f the sze d of the square X decreases. Specfcally, we can subdvde each square nto squares of smaller sze to obtan an mproved compettve rato. However, t s not the best polcy to subdvde all squares evenly snce g(x) for a square X of the same sze substantally dffers n dfferent postons wthn Ω. Note ths phenomenon especally between postons close to the orgn (.e., both a and b are small) and those far from the orgn (the former s larger). Thus our approach s to subdvde squares X dynamcally, or to dvde the one wth the largest g(x) value n each step. Also observe that as squares move to the rght n Fg. 5 then the g value decreases due to Lemma 1.

9 Algorthms 2008, states g-value = , , states g-value = Fgure 7. Approxmaton of a square We gve now an nformal descrpton of the procedure for generatng the squares, the formal descrpton s gven n Procedures 1 and 2. Broadly speakng one starts wth square near the orgn, then subdvdes ths square, and contnues by creatng a new square to the rght. Due to Lemma 1 ths process s bounded. The procedure begns wth a sngle square [2, 3; 2]. Note that ts g-value s poor (n fact, not bounded). Next, square [2, 3; 2] s splt nto four squares of sze 1 as shown n Fg. 6: [1, 3; 1], [1, 2; 1], [2, 2; 1] and [2, 3; 1]. For each of these squares, the g-value s then calculated n the followng way: If the square s of the form [, + 1; l] for some, l, the g-value can be calculated mmedately by Lemma 1. Otherwse, we use the lnear program descrbed n [14] to determne the compettve rato of trangle (1, a, b). Note that ths lnear program makes use of state graphs smlar to those n Fg. 3 and Fg. 4. Next, square [3, 4; 2] of sze 2 s added. In general, f the procedure dvdes [, + 1; 2] of sze 2 and [ + 1, + 2; 2] of sze 2 does not exst, then square [ + 1, + 2; 2] s added. Thus at ths stage there are four squares of sze 1 (two of these are, n fact, outsde Ω) and one square of sze 2. The procedure further dvdes that square (nsde Ω) whose g value s the worst. One contnues n ths way and takes the worst g-value as an upper bound of the compettve rato. An ssue regardng the effcency of the procedure s that the number of states of the state dagram used by the algorthm for a small square (or for the correspondng trangle) becomes large. Ths leads to potentally excessve computaton tmes for the LP nvolved. Consder, for example, the trangle (1, 170, ) (or the square [, 213; 1 ]). It turns out that one needs 514 states for the dagram and substantal computaton tme n solvng the LP s requred. However, note that there s a slghtly larger trangle, (1, 4, 5) (or the square [ 4, 5; 5 ]), whch needs only 12 states to solve the LP (Fg. 7). Thus one can shorten computaton tme by usng [ 4 3, 5 3 ; ] nstead of [, 213; 1 ], thereby tradng off the g-value of the former (= ), for that of the latter (= ). Although we do not have an exact relaton between the trangle and the number of states, we have observed that f the rato of the three sdes of the trangle can be represented by three small ntegers then the number of states s also small. In the procedure, therefore, we do not smply calculate g(x) for a square X, but we attempt to fnd X whch contans X and has such desrable propertes. Procedure 1 and 2 gve a formal descrpton of our method. Each square X = [a, b; d] s represented by p = (a, b, d, r), where r s an upper bound of g(x). The man procedure SQUAEGENEATION dvdes the square, whose g value s the worst, nto four half-szed squares and, f necessary, also creates a new rghtmost square of sze 2. Then the g-values of those new squares are calculated by procedure

10 Algorthms 2008, 1 39 d (a 2, b 2 ) (a 4, b 4 ) d d = 2 (a, b) = (a 1, b 1 ) (a 3, b 3 ) d + e 0 e 0 d + e 0 (a, b) d (a 0, b 0 ) y - b (a, b) x - a x, y Fgure 8. Lnes 9-13 Fgure 9. Lne 27 Fgure 10. Lnes CALCULATEC. As dscussed above, our method attempts to fnd a more sutable square, such that the number of states can be kept manageable. More formally, let the current square be X = [a, b; d]. Then one seeks to fnd X = [ã, b; d] whch contans X and where ã can be represented by β, such that α both α and β are ntegers and α s at most 31. (Smlarly for b.) We have confrmed that the number of states and LP computaton tme are reasonably small f α s at most ths sze; for detals, see also FINDAPPOXPOINT n Procedure 2. We note that we scan the value of α only n the range from 17 to 31. Ths s suffcent; for example, α = 10 can be covered by α = 20. The value α = 16 s not needed ether snce t should have been calculated prevously n the subdvson process. If g( X) s smaller than the g-value of the orgnal square (of double sze), then we use that value as the g-value of X. Otherwse we abandon such an approxmaton and calculate g(x) drectly. Now suppose that SQUAEGENEATION has termnated. Then for any p = (a, b, d, r) n P, t s guaranteed that r 0. Ths means that the set of squares whch satsfy the condtons (1) to (3) have been created. As mentoned there, we have also created the set of trangles satsfyng the condtons of Sec. 2. Thus by Lemma 3, we can conclude: Theorem 1 There s an onlne algorthm for the 2-server 3-pont problem whose compettve rato s at most 0. We now gve results of our computer experments: For the entre area Ω, the current upper bound s (recall that the conjecture s ). The number N of squares generated s 13285, n whch the sze m of smallest squares s 1/256 and the sze M of largest squares s 2. We also conducted experments for small subareas of Ω: (1) For [5/4, 7/4, 1/16]: The upper bound s (better than the conjecture but ths s not a contradcton snce our trangles are restrcted). (N, M, m) = (69, 1/64, 1/128). (2) For [7/4, 9/4, 1/4]: The upper bound s (N, M, m) = (555, 1/64, 1/2048). (3) For [10, 11, 1]: The upper bound s (N, M, m) = (135, 1/16, 1/32). We note that for our computatons the LP solver whch s part of Mathematca was used for our computatons and accuracy to wthn 5 dgts s guaranteed. 5. Concludng emarks There are at least two drectons for future research: Frst one mght prove that the compettve rato of the 2-server 3-pont problem s analytcally at most e/(e 1) + ε. Secondly we wsh to extend our current approach (.e., approxmaton of nfnte pont locatons by fnte ones) to four (or more) ponts. We have a partal result for the 4-pont case where two of the four ponts are close (obvously t s smlar to the 3-pont case), but the generalzaton does not appear easy.

11 Algorthms 2008, 1 40 Procedure 1 Procedure SquareGeneraton 1: procedure SQUAEGENEATION( 0 ) 2: p (2, 3, 2, C 2 2/(2 2) = ) 3: Mark p. 4: P {p} 5: whle p = (a, b, d, r) such that r > 0 do 6: p the pont n P whose r s maxmum. 7: P P \{p} 8: Let p = (a, b, d, r) 9: d d/2 10: a 1 a, b 1 b 11: a 2 a d, b 2 b 12: a 3 a, b 3 b d 13: a 4 a d, b 4 b d See Fg : for 1 to 4 do 15: f (a, b ) Ω then 16: r CALCULATEC(a, b, d, r) 17: P P {(a, b, d, r )} 18: end f 19: end for 20: f p s marked then 21: p (a + 1, b + 1, 2, C a+1 a/(a 2)). 22: Mark p. Unmark p. 23: P P {p } 24: end f 25: end whle 26: end procedure 27: procedure CALCULATEC(a, b, d, r) 28: (a 0, b 0 ) FINDAPPOXPOINT(a, b) See Fg : r 0 GETC FOMLP(a 0, b 0 ) 30: e 0 max(a a 0, b b 0 ) 31: r 0 r 0 a 0 /(a 0 d e 0 ) 32: f r 0 < r 0 then 33: return r 0 34: else 35: r 0 GETC FOMLP(a, b) 36: r 0 r 0 a 0 /(a 0 d) 37: return r 0 38: end f 39: end procedure

12 Algorthms 2008, 1 41 Procedure 2 Procedure FndApproxPont 1: procedure FINDAPPOXPOINT(a, b) 2: e mn 3: for 31 to 17 do 4: x a, y y, e max(x/ a, y/ b) 5: f e < e mn then 6: e mn e, mn, x mn x, y mn y 7: end f 8: end for 9: return (x mn / mn, y mn / mn ) 10: end procedure eferences 1. Manasse, M.; McGeoch, L. A.; Sleator, D. Compettve algorthms for server problems. J. Algorthms 1990, 11, Koutsoupas, E.; Papadmtrou, C. On the k-server conjecture. J. ACM 1995, 42, Chrobak, M.; Karloff, H.; Payne, T. H.; Vshwanathan, S. New results on server problems. SIAM J. Dscrete Math. 1991, 4, Chrobak, M.; Larmore, L. L. An optmal onlne algorthm for k servers on trees. SIAM J. Comput. 1991, 20, Koutsoupas, E.; Papadmtrou, C. Beyond compettve analyss. In Proc. 35th FOCS, pages IEEE, Ben, W.; Chrobak, M.; Larmore, L. L. The 3-server problem n the plane. In Proc. 7th European Symp. on Algorthms (ESA), volume 1643 of Lecture Notes n Comput. Sc., pages Sprnger, Bartal, Y.; Bollobas, B.; Mendel, M. A amsey-type theorem for metrc spaces and ts applcatons for metrcal task systems and related problems. In Proc. 42nd FOCS, pages IEEE, Coppersmth, D.; Doyle, P. G.; aghavan, P.; Snr, M. andom walks on weghted graphs and applcatons to on-lne algorthms. J. ACM 1993, 40, Karln, A.; Manasse, M.; McGeoch, L.; Owck, S. Compettve randomzed algorthms for nonunform problems. Algorthmca 1994, 11, Chrobak, M.; Larmore, L. L.; Lund, C.; engold, N. A better lower bound on the compettve rato of the randomzed 2-server problem. Inform. Process. Lett. 1997, 63, Bartal, Y.; Chrobak, M.; Larmore, L. L. A randomzed algorthm for two servers on the lne. In Proc. 6th ESA, Lecture Notes n Comput. Sc., pages Sprnger, Ben, W.; Iwama, K.; Kawahara, J.; Larmore, L. L.; Oravec, J. A. A randomzed algorthm for two servers n cross polytope spaces. In Proc. 5th WAOA, volume 4927 of Lecture Notes n Computer Scence, pages Sprnger, Ben, W.; Larmore, L. L.; Noga, J. Equtable revsted. In Proc. 15th ESA, volume 4698 of Lecture Notes n Computer Scence, pages Sprnger, 2007.

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