2 Stage I. Stage II. Stage III (ii)

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1 Compettve Faclty Locato alog a Hghway Λ Hee-Kap Ah y Su-Wg Cheg z Otfred Cheog y Mordeca Gol z Reé va Oostrum y Aprl 10, 2001 Abstract We cosder a compettve faclty locato problem wth two players. Players alterate placg pots, oe at atme, to the playg area, utl each of them has placed pots. The area s the subdvded accordg to the earest-eghbor rule, ad the player whose pots cotrol the larger area ws. We preset a wg strategy for the secod player, where the area s a crcle or a le segmet. We also cosder a varato where players ca play more tha oe pot at a tme for the crcle area. 1 Itroducto The classcal faclty locato problem [5] asks for the optmum locato of a ew faclty (polce stato, super market, trasmtter, etc.) wth respect to a gve set of customers. Typcally, the fucto to be optmzed s the maxmum dstace from customers to the faclty ths results the mmum eclosg dsk problem studed by Megddo [8], Welzl [12] ad Aroov et al. [2]. Compettve faclty locato deals wth the placemet of stes by competg market players. Geometrc argumets are combed wth argumets from game theory to see how the behavor of these decso makers affect each other. Compettve locato models have bee studed may dfferet felds, such as spatal ecoomcs ad dustral orgazato [1, 9], mathematcs [6] ad operatos research [3, 7, 11]. Comprehesve overvews of compettve faclty locatos models are the surveys by Fresz et al. [11], Eselt ad Laporte [3] ad Eselt et al. [4]. We cosder a model where the behavor of the customers s determstc the sese that a faclty ca determe the set of customers more attracted to t tha to ay other faclty. Ths set s called the market area of the faclty. The collecto of market areas forms a tessellato of the uderlyg space. If customers choose the faclty o the bass of dstace some metrc, the tessellato s the Voroo Dagram of the set of facltes [10]. We address a compettve faclty locato problem that we call the Voroo Game. It s played by two players, Blue ad Red, who place a specfed umber,, of facltes a rego U. They alterate placg ther facltes oe at a tme, wth Blue gog frst. After all 2 facltes have bee placed, ther decsos are evaluated by cosderg the Voroo dagram of the 2 pots. The player whose facltes cotrol the larger area ws. Λ Part of the work was doe whle the frst, thrd ad ffth authors are at the Departmet of Computer Scece, HKUST, Hog Kog. y Departmet of Computer Scece, Utrecht Uversty, Netherlads. Emal: fheekap,otfred,reeg@cs.uu.l z Departmet of Computer Scece, HKUST, Hog Kog. Emal: fscheg,golg@cs.ust.hk 1

2 More formally, let fb g =1 ad fr g =1 be the respectve locatos of the blue ad red pots ad set B = jfu 2 U : m R = jfu 2 U : m d(u; b ) < m d(u; r )gj; d(u; r ) < m d(u; b )gj where d(u; v) s a uderlyg metrc ad j jdcates the area of a set. Blue ws f ad oly f B > R, Red ws f ad oly f R > B ad the game eds a te f B = R. The most atural Voroo Game s played a two-dmesoal area U usg the Eucldea metrc. Ufortuately obody kows how to w ths game, eve for very restrcted regos U. I ths ote we preset strateges for wg oe-dmesoal versos of the game, where the area s a crcle or a le segmet, ad varatos. I other words, we cosder compettve faclty locato alog a Australa hghway. The ext secto dscusses the smplest game, o the crcle. It s obvous that the secod player, Red, ca always acheve a te by playg o the atpode of Blue's move. Oe mght try to tweak ths strategy such that t results a w for Red. Ths does't seem to work, ad we preset stead a qute dfferet wg strategy for Red. Secto 3 descrbes how ths strategy remas a wg strategy eve f the rules of the game are drastcally relaxed. I Secto 4 we fally tur to the le segmet area. It would appear that Blue has a advatage here, because t ca play the mdpot ofthe segmet ts frst move. We show that ths does't help, ad prove that Red stll has a wg strategy. The strategy s qute smlar to the oe for the crcle case, but ts aalyss (because of a loss of symmetry) s more detaled. 2 The basc crcle game There are two players, Blue ad Red, each havg pots to play, where >1. They alterate placg these pots o crcle C, wth Blue placg the frst pot, Red the secod, Blue the thrd, etc., utl all 2 pots are played. We assume that pots caot le upo each other. Let fb g =1 be the locatos of the blue pots ad fr g =1 be those of the red oes. After all of the 2 pots have bee played each player receves a score equal to the total crcumferece of the crcle that s closer to that player tha to the other,.e., Blue ad Red have respectve scores B = jfx 2 C : m R = jfx 2 C : m d(x; b ) < m d(x; r )gj d(x; r ) < m d(x; b )gj The player wth the hghest score (the larger crcumferece) ws. The questo that we address here s, Does ether player have a wg strategy ad, f yes, what s t? We wll see below that the secod player, Red, always has a wg strategy. Before gvg the strategy we troduce some deftos. We parameterze the crcle usg the terval [0; 1], where the pots 0 ad 1 are detfed. Arcs o the crcle are wrtte as [x; y] mplyg the clockwse arc rug from x to y, as[:5;:6] or [:9;:1]: Defto 1 The pots u = ; =0; 1;:::; 1 are keypots. 2

3 0 3/4 1/4 1/2 Fgure 1: There are four keypots whe =4. Fgure 1 shows the keypots for =4. We call a arc betwee two clockwse cosecutve red/blue pots a terval. The teror of a terval s free of red/blue pots. At ay gve tme durg the game the crcle s parttoed to tervals. A terval s moochromatc f ts edpots have the same color, ad bchromatc f they have dfferet colors. A blue terval s a blue moochromatc oe, a red terval a red moochromatc oe. We deote the total legth of all red tervals by R m, ad the total legth of all the blue tervals by B m. A terval s called a key terval f both of ts edpots are keypots. The mportat thg to otce s that at the ed of the game the legth of each bchromatc terval s dvded equally amog the two players, so R B= R m B m ad Red ws f ad oly f R m >B m. We devse our strategy to force ths to happe. Sce we ca parameterze the crcle arbtrarly, we ca assume wthout loss of geeralty that Blue plays hs frst pot o 0 ad thus o a keypot. We ow descrbe Red's wg strategy. Fgure 2 shows a example. Red's Keypot Strategy Stage I: If there s a empty keypot the Red plays oto the keypot. Stage I eds after the last keypot splayed (by ether Red or Blue). Stage II: If there s o empty keypot ad t s ot Red's last move the Red plays her pot to a largest blue terval. We call ths breakg the blue terval. Stage II eds whe Blue plays hs last pot. Stage III: Red's last move. There are two possbltes: () f there exsts more tha oe blue terval the Red breaks a largest oe by placg her pot sde. () f there s oly oe blue terval defe `< 1 to be ts legth. Red's move s to go to a bchromatc key terval ad clam a red terval of legth larger tha ` by placg a red pot closer tha 1 ` to the blue edpot of the bchromatc key terval. The two followg lemmas wll be eeded. 3

4 Stage I 3 7 Stage II Stage III () Fgure 2: There are four pots to be played for both Blue ad Red. The whte dots represet Blue's pots ad the black dots represet Red's pots. We label the dots chroologcal order. Lemma 1 Let B be aset of b blue pots ad let R be a set of r red pots curretly o the crcle wth b r. Let (R) be the umber of red tervals they form ad (B) the umber of blue oes. The (B) (R) =b r. Proof: The proof wll be by ducto o r: If r =0the b blue pots form b blue tervals so (B) =b; (R) = 0 ad the codto (B) (R) =b r s satsfed. Now suppose that the lemma s true for all cofguratos of b blue pots ad r 1 red oes. Deletg ay red pot p from R leaves b blue ad r 1 red pots, so (B) (R fpg) =b r +1. We ow add the red pot p back to the cofgurato ad ask how the moochromatc tervals ca chage. There are three possble placemets of p : () sde a red terval, creasg (R) by oe ad leavg (B) uchaged. () sde a blue terval, decreasg (B) by oe ad leavg (R) uchaged. () sde a bchromatc terval, creasg (R) by oe ad leavg (B) uchaged. After all three of these cases we fd that (B) (R) =(B) (R fpg) 1=b r. Lemma 2 Suppose that all keypots are covered ad Blue has just moved (possbly coverg the last keypot). If there s oly oe blue terval ad ths terval has legth < 1, the there exsts a bchromatc key terval. Proof: We apply the pgeo hole prcple: At most 2 1pots have bee played, of them o keypots. Cosder the crcle arcs of legth 1 formed by the keypots. Sce the blue terval has legth < 1, at least oe of ts edpots s sde a arc. That leaves oly 2 pots to have bee played sde the 1 remag arcs. Therefore, oe of the arcs must be free of pots, formg a key terval. Sce there s oly oe blue terval, there s o red terval by Lemma 1. Therefore, ths key terval s bchromatc. Theorem 1 The keypot strategy s a well-defed wg strategy for Red. Proof: We start wth a smple observato. Sce the crcle cotas oly keypots ad Blue's frst move covers the frst keypot, Red wll play oto at most 1 keypots. Thus Stage I always eds before Red plays her last pot. 4

5 Cosder Stage II. Lemma 1 mples that after each playby Blue (b = r + 1) there s always at least oe blue terval o the crcle, so Stage II of the strategy s deed well defed. We make two observatos cocerg the stuato after Stage II, whe Red has played her 1'st pot. The frst s that there s o blue key terval. Let k be the umber of keypots played by Blue durg the game. Red has covered the remag k keypots by the ed of Stage I. If k =1(the oly case whch Red skps Stage II), the there certaly s o blue key terval as there s oly oe blue keypot. Whe k > 1, Blue ca defe at most k 1 blue key tervals wth ts k keypots (sce Red has at least oe keypot). Note that sce all keypots are played by the ed of Stage I, all tervals Stages II ad III have legth at most 1. I partcular abluekey terval s loger tha ay other blue terval. Sce Red plays k 1pots Stage II, all blue key tervals are broke durg Stage II. The secod observato cocerg the stuato after stage II s that all red tervals are key tervals. Ths statemet s true at the ed of Stage I, as Red has so far oly played oto keypots, ad all keypots are covered. Durg Stage II, Red uses her pots to break blue tervals, ad therefore creates bchromatc tervals oly. Blue caot create red tervals, ad so, at the ed of Stage II, all red tervals are deed key tervals. We ow show why Stage III s well defed ad why Red ws. Suppose that Blue has just played hs last pot ad t s ow tme for Stage III, Red's last move. From Lemma 1 we kow that (B) 1. If (B) > 1 before Red's last move the the strategy s well defed: Red breaks a largest blue terval. Ths decreases (B) by 1 so the game eds wth (B) 1: By Lemma 1 we have (R) =(B) 1. But ow ote that from the observatos the precedg paragraphs all exstg red tervals are key tervals whle all exstg blue tervals have legth strctly less tha 1. Sce all red tervals are loger tha all blue tervals ad there are the same umber of red oes ad blue oes we fd that R m >B m ad Red ws. If (B) = 1 before Red's last move the strategy requres that the uque blue terval has legth `< 1, ad that there exsts a bchromatc key terval. The frst fact was already observed above, the secod fact follows from Lemma 2. After Red places her last pot Blue stll has oe blue terval of legth ` whle Red has oe red terval of legth >`. Thus R m >`= B m ad Red ws. 3 A modfed crcle game The basc game ca be modfed may dfferet ways. The smplest modfcato allows the players to play more tha oe pot at a tme. More complcated modfcatos permt the players (both or oe) to choose before each tur, how may pots they play. Suppose that there are k» rouds. Let f ad fl be the umbers of pots that Blue ad Red play respectvely roud. Suppose that the followg restrctos are placed. ffl 81»» k; f ;fl > 0. ffl 81» j» k; P j =1 f P j =1 fl. ffl P k =1 f = P k =1 fl =. ffl f1 <. 5

6 The Red stll ws by followg exactly the same strategy as the prevous secto of frst fllg the keypots ad the breakg the largest blue tervals utl Red plays t's last pot whe t follows the Stage III rules. The proof that the strategy s well defed ad ws s almost exactly the same as the oe the prevous secto so we wll ot repeat t here. Note that ths geeralzato cludes both the orgal game ad the batched" verso whch each player plays the same umber (> 1) of pots at each tur. Note, too, that k; f ; ad fl eed ot be fxed advace. For example, Blue may decde at every move howmay pots he wll play ad the Red plays the same umber. We coclude ths secto by otg that the codto f1 <s essetal sce otherwse Blue would play the keypots, forcg a te. 4 The le segmet verso We ow move o to the verso of the game played o a le segmet. We cosder t to be horzotal ad parameterzed as [0; 1]. The scorg s the same as the basc crcle game except that the player wth the leftmost pot clams everythg betwee 0 ad the pot, ad the player wth the rghtmost pot clams everythg betwee the pot ad 1. We assume that > 1, ad pots caot le upo each other. Whe = 1, Blue ws by placg oto 1 2. We modfy some of the old deftos ad troduce ew oes: Defto 2 The pots u = 1 + ; =0; 1;:::; 1 are keypots. 2 The left segmet s the segmet from 0 to the leftmost red or blue pot. The rght segmet s the segmet from the rghtmost red or blue pot to1. The border terval s the uo of the left ad rght segmets. A terval s a secto of the le segmet wth red/blue edpots ad o red/blue pots ts teror. We cosder the border terval a terval. A terval, cludg the border terval, s moochromatc f ts edpots have the same color, ad bchromatc f they have dfferet colors. Wth ths defto of tervals, Lemmas 1 ad 2 are true for the le segmet as well. We deote the total legth of all of the blue tervals cludg, f approprate, the border terval by B m, the total legth of all of the red tervals aga cludg, f approprate, the border terval by R m. Whe the border terval s bchromatc, we useb b to deote the legth of the left/rght segmet wth a blue edpot ad R b to deote the legth of the left/rght segmet wth a red edpot. If the border terval s moochromatc, the B b = R b = 0: Sce all bchromatc o-border tervals are equally shared by both players R B =(R m +R b ) (B m +B b ) ad, as Secto 2, we desg our strategy so that Red fshes wth the rght had sde of the equato > 0: We ow troduce the le strategy, a modfed verso of the crcle strategy. Fgure 3 shows a example. Red's Le strategy Stage I: If there s a empty keypot the Red plays the keypot. If u0 or u 1 have ot yet bee played the Red should play oto oe of them frst. Stage I eds after the last keypot splayed by ether Red or Blue. Note that the game may fsh Stage I. 6

7 Stage I Stage II 8 Stage III Fgure 3: There are four pots to be played for both Blue ad Red. The whte dots represet Blue's pots ad the black dots represet Red's pots. We label the dots chroologcal order. Stage II: If there s o empty keypot ad t s ot Red's last move the () f there exsts at least oe blue o-border terval, the Red should break a largest blue o-border terval by placg her pot sde. () f the border terval s the oly blue terval, the there are two possble cases: (a) Oe of the blue edpots of the blue border terval s a keypot: Wthout loss of geeralty assume that t s u0 (the other case s symmetrc) ad the other edpot s1 `: From Stage I the other edpot caot be the keypot u 1 so `< 1 : Red ow places her ew pot atx where x s aywhere (`; u0): 2 (b) Nether of the edpots of the border terval are keypots: Let ` be the legth of the blue border terval; ` < 1. There must the exst a bchromatc key terval (Lemma 2). Red places her ew pot that terval to form a ew red terval of legth >`: Stage II eds after Blue plays hs last pot. Stage III: If Red s placg her last pot, we have two mutually exclusve cases: () f there exsts more tha oe blue terval, the Red should break a largest o-border oe. () f there exsts oly oe blue terval, the let ts legth be `; we wll see below that `< 1 : Red should go to a bchromatc key terval (oe wll exst from Lemma 2) ad clam a red terval of legth >`as follows. If the bchromatc key terval s ot the border oe, Red ca do ths by creatg a ew red terval of legth >`: If the bchromatc key terval s the border oe, the Red already possesses 1 2 of t because t has all of ether [0;u0] or [u 1; 1]: Red ca therefore go to the other sde t does ot possess, ad grab eough legth to have a red border terval of legth >`. Theorem 2 The le strategy s a well-defed wg strategy for Red. 7

8 Proof: Note that ths strategy dffers at least oe major aspect from the crcle strategy: sce we have lost crcular symmetry t caot be guarateed that Blue plays oto at least oe keypot, ad so t s possble that the game wll ed Stage I, wth Red playg all keypots. I ths case, all red tervals (cludg, possbly, the border terval) are key tervals ad all blue tervals have legth < 1. By Lemma 1 Blue ad Red have the same umber of moochromatc tervals, so B m < R m. If the border terval s moochromatc, the B b = R b = 0 ad Red ws. If the border terval s bchromatc, the oe of ts edpots must be the red pot u0 or u 1. Ths mples that B b <R b = 1 2, ad Red ws. I what follows we may therefore assume that Blue plays oto at least oe keypot durg the game. We wll show that at the ed of the game t wll always be true that B m <R m ad, f the border terval s bchromatc, the B b» R b. The theorem wll follow. Frst ote that uder ths assumpto Stage I always eds wth all keypots covered, ad Stage III s reached. Note further that Red's frst move soto ether u0 or u 1. We cosder Stage II. After Blue's every move there exsts at least oe blue terval (possbly the border terval) by Lemma 1. If there s oly oe such blue terval, there s o red terval. Thus, oe of the two codtos () or () of Stage II holds. The strategy s clearly well defed cases () ad ()(a), the valdty of ()(b) follows from Lemma 2. We wll eed oe more observato. Lemma 3 After Blue's last move, there s o blue key terval. Proof: Let k be the umber of keypots played by Blue, where 1» k»d 2 e. There are therefore at most k 1 blue key tervals after Stage I. Red occupes k keypots Stage I, ad so Stage II lasts for k 1 rouds. Ths s suffcet for all blue key tervals to be broke, sce a blue key terval s loger tha ay other blue terval Stage II () ad Stage II ()(a). We owprove that Red ws. There are two cases. Lemma 4 Assume that case () of Stage II ever occurs. key tervals after Blue's last move. The all red tervals are Proof: After Stage I all red tervals are key tervals, sce Red has oly played keypots ad all keypots are covered. Durg Stage II Red uses all her pots to break blue tervals (sce case ()(b) does ot occur), ad so creates oly bchromatc tervals. As Blue caot create red tervals, all red tervals remag after Stage II are deed key tervals. We exame the result of Stage III, assumg that case () of Stage II dd ot occur. Assume frst that Red plays case () of Stage III. There are the equal umbers of red ad blue tervals left after the last move. Sce all blue tervals have legth < 1 by Lemma 3, ad all red tervals are key tervals by Lemma 4, we have B m <R m. If the border terval s moochromatc, the B b = R b = 0 ad Red ws. If the border terval s bchromatc, the ts red edpot must be u0 or u 1 (sce Red plays case () of Stage II oly), ad so B b» R b = 1 2 ad Red ws. O the other had, assume ow that Red plays case () of Stage III. The Blue has total legth `, Red has >`, ad so Red ws. We ow cosder the remag case, where case () of Stage II does occur. 8

9 Lemma 5 Assume that case () of Stage II occurs at least oce. After Red's last move Stage II, there s o blue o-border terval ad B < R. Proof: We prove that the statemet s true after the last occurrece of case () Stage II, ad after each subsequet move by Red. Cosder the last occurrece of case (). Before Red's move, the border terval s the oly blue terval. If Red plays case ()(a), there s o blue terval at all after Red's move. We have the B m = R m = 0 ad B b <R b, whch mples B < R. If Red plays case ()(b), Red clams a red terval loger tha the blue border terval. So after Red's move we have B < R, asb m <R m ad B b = R b =0. Cosder ow the remag moves of Stage II. I all these moves Red plays case (), adsowe ca deduce that Blue uses hs move to create a ew blue o-border terval. Red mmedately breaks ths blue terval. Ths leaves B m, R m, B b, ad R b uchaged ad destroys the oly blue o-border terval. The clam therefore remas true after each subsequet move by Red, ad partcular after Red's last move Stage II. We cosder the stuato rght before Blue's last move. By Lemma 5, there s o blue oborder terval, ad B < R. If the border terval s blue before Blue's last move ad Blue uses that move to create a blue o-border terval move, Red plays case () of Stage III. Ths breaks the ew blue o-border terval, returg B ad R to ther state before Blue's last move, ad so Red ws. I all other cases there s a sgle blue terval after Blue's last move. It caot be a key terval by Lemma 3, ad so, by Lemma 2, there s a bchromatc key terval. Red clams a red terval loger tha the blue terval, ad ws. 5 Coclusos We have gve strateges for oe-dmesoal compettve faclty locato, allowg the secod player, Red, to w. The marg by whch Red ws s very small, however, ad fact Blue ca make t as small as he wats. Is there a strategy that would allow Red to w by a larger marg? The aswer s o our argumets ca easly be trasformed to a strategy that allows the frst player, Blue, to lose by a arbtrarly small marg. Do our fdgs have ay bearg o the two-dmesoal Voroo Game? The cocept of keypots tured out to be essetal to our strateges. We have see that a player goverg all keypots caot possbly lose the game. Is there a smlar cocept two dmesos? A atural attempt would be to defe a uform square grd of keypots. Perhaps surprsgly, a player goverg ths grd ca stll lose the game by a cosderable marg. 6 Ackowledgmets The authors would lke to thak Dr. Jacob Ecco for tally troducg them to ths problem. We also thak Sul Arya, Mark de Berg, ad Joachm Gudmudsso for helpful dscussos. 9

10 Refereces [1] S.P. Aderso. Equlbrum exstece a lear model of spatal competto. Ecoomca, 55: , [2] B. Aroov, M. va Kreveld, R. va Oostrum, ad K. Varadaraja. Faclty locato o terras. I K.Y. Chwa ad Oscar H. Ibarra, edtors, Proc. 9th Iteratoal Symposum of Algorthms ad Computato, volume 1533 of Lecture Notes Comput. Sc., pages Sprger-Verlag, [3] H.A. Eselt ad G. Laporte. Compettve spatal models. Europea Joural of Operatoal Research, 39: , [4] H.A. Eselt, G. Laporte, ad J.-F. Thsse. Compettve locato models: A framework ad bblography. Trasportato Scece, 27:44 54, [5] R.L. Fracs ad J.A. Whte. Faclty Layout ad Locato. Pretce-Hall, Ic., [6] S.L. Hakm. Locato wth spatal teractos: Compettve locato ad games. I R.L. Fracs ad P.B. Mrchada, edtors, Dscrete Locato Theory, pages Wley, New York, [7] M. Labbé ad S.L. Hakm. Market ad locatoal equlbrum for two compettors. Operatos Research, 39: , [8] N. Megddo. Lear-tme algorthms for lear programmg R 3 ad related problems. SIAM J. Comput., 12: , [9] A. Okabe ad M. Aoyagy. Exstece of equlbrum cofguratos of compettve frms o a fte two-dmesoal space. Joural of Urba Ecoomcs, 29: , [10] F. P. Preparata ad M. I. Shamos. Computatoal Geometry: A Itroducto. Sprger-Verlag, New York, NY, [11] R.L. Tob T.L. Fresz ad T. Mller. Exstece theory for spatally compettve etwork faclty locato models. Aals of Operatos Research, 18: , [12] Emo Welzl. Smallest eclosg dsks (balls ad ellpsods). I H. Maurer, edtor, New Results ad New Treds Computer Scece, volume 555 of Lecture Notes Comput. Sc., pages Sprger-Verlag,