The Idea. Leader Election. Outline. Why Rings? Network. We study leader election in rings. Specification of Leader Election YAIR. Historical reasons

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1 The Idea Leader Elecio Nework We sudy leader elecio i rigs Why Rigs? Oulie Specificaio of Leader Elecio Hisorical reasos origial moivaio: regeerae los oke i oke rig eworks Illusraes echiques ad priciples Good for lower bouds ad impossibiliy resuls YAIR Leader Elecio i Asychroous Rigs A O( 2 ) algorihm A O( log()) algorihm The Revege of he Lower Boud Leader elecio is sychroous rigs Breakig he Ω( log()) barrier

2 The Problem Los of variaios... Processes ca be i oe of wo fial saes eleced o-eleced I every execuio, exacly oe process (he leader) is eleced All oher processes are o-eleced Rig ca be uidirecioal or bidirecioal Processes ca be ideical or ca somehow be disiguishable from each oher The umber of processes may or may o be kow if o, uiform algorihms Commuicaios may be sychroous or asychroous Aoymous Neworks Call me Ishmael Processes have o uique IDs (ideical auomaa)...bu ca disiguish bewee lef ad righ Processes have uique IDs from some large oally ordered se (e.g. ) N + Operaios used o maipulae IDs ca be uresriced or limied (e.g. oly comparisos)

3 Commuicaio: Sychroous/Asychroous A Impossibiliy Resul I rouds Sychroous I each roud, a process delivers all pedig messages akes a execuio sep (possibly sedig oe or more messages)! Asychroous No upper boud o message delivery ime No ceralized clock No boud o relaive sped of processes Theorem There is o ouiform aoymous algorihm for leader elecio i sychroous rigs A Impossibiliy Resul A O( 2 ) Algorihm Le La ( 77), Chag & Robers ( 79) Theorem There is o ouiform aoymous algorihm for leader elecio i sychroous rigs Proof Suppose here exiss a aoymous ouiform algorihm A for R s.. R > 1 Lemma For every roud k of A i R, he saes of all he processes a he ed of roud k are he same Proof By iducio o k If some process eers he leader sae, hey all do upo receivig o message! sed uid i o lef (clockwise) upo receivig m from righ case! m.uid > uid i :!! sed m o lef! m.uid < uid i :!! discard m! m.uid = uid i :!! leader := i!! sed <ermiae, i > o lef!! ermiae edcase upo receivig <ermiae, i> from righ! leader := i! sed <ermiae, i> o lef! ermiae Asychroous ad Uiform Process wih highes uid is eleced leader - all oher uids are swallowed Time complexiy: O() Message complexiy: O( 2 ) Boud is igh:!!

4 A O( log ) Algorihm Hirscheberg & Siclair (1980) Bidirecioal rig I each phase k, p i : seds uid i oke lef ad righ oke ieded o ravel disace! 2 k ad ur back coiues ouboud oly if greaer ha okes o pah processes always forward iboud oke p i leader if i receives ow oke while goig ouboud A O( log ) Algorihm Hirscheberg & Siclair (1980) Bidirecioal rig I each phase k, proocol elecs oe wier (process wih highes uid) for each! k-eighborhood a k-eighborhood icludes! 2k+1! processes Afer O(log ) phases, here is oly oe wier! A O( log ) Algorihm Hirscheberg & Siclair (1980) Bidirecioal rig A O( log ) Algorihm Hirscheberg & Siclair (1980) Bidirecioal rig I each phase k, proocol elecs oe wier (process wih highes uid) for each! k-eighborhood a k-eighborhood icludes! 2k+1! processes Phase 0 I each phase k, proocol elecs oe wier (process wih highes uid) for each! k-eighborhood a k-eighborhood icludes! 2k+1! processes Phase 1 Afer O(log ) phases, here is oly oe wier! Afer O(log ) phases, here is oly oe wier!

5 A O( log ) Algorihm Hirscheberg & Siclair (1980) Bidirecioal rig A O( log ) Algorihm Hirscheberg & Siclair (1980) Bidirecioal rig I each phase k, proocol elecs oe wier (process wih highes uid) for each! k-eighborhood a k-eighborhood icludes! 2k+1! processes Phase 2 I each phase k, proocol elecs oe wier (process wih highes uid) for each! k-eighborhood a k-eighborhood icludes! 2k+1! processes Phase 3 Afer O(log ) phases, here is oly oe wier! Afer O(log ) phases, here is oly oe wier! Boudig message complexiy Lemma For every k 1 he umber of processes ha are phase k wiers are a mos Proof Two wiers cao have fewer ha processes bewee hem Message complexiy: 4 Phase 0 2 k +1 2 k Boudig message complexiy Lemma For every k 1 he umber of processes ha are phase k wiers are a mos Proof Two wiers cao have fewer ha processes bewee hem Message complexiy: log( 1) Phase 0 k=1 2 k +1 2 k # of phases before 1 wier

6 Boudig message complexiy Lemma For every k 1 he umber of processes ha are phase k wiers are a mos Proof Two wiers cao have fewer ha processes bewee hem Message complexiy: Phase 0 log( 1) {4 k k=1 # of phases before 1 wier # of messages per wier 2 k +1 2 k Boudig message complexiy Lemma For every k 1 he umber of processes ha are phase k wiers are a mos Proof Two wiers cao have fewer ha processes bewee hem Message complexiy: log( 1) Phase 0 k=1 # of phases before 1 wier 4 2 k { # of messages per wier 2 k 1 +1 # of wiers per phase 2 k +1 2 k Boudig message complexiy Lemma For every k 1 he umber of processes ha are phase k wiers are a mos Proof Two wiers cao have fewer ha processes bewee hem Message complexiy: log( 1) ermiaio Phase 0 k=1 # of phases before 1 wier 4 2 k { # of messages per wier 2 k 1 +1 # of wiers per phase 2 k +1 2 k Message complexiy Lemma For every k 1 he umber of processes ha are phase k wiers are a mos Proof Two wiers cao have fewer ha processes bewee hem Message complexiy: log( 1) ermiaio Phase 0 k=1 # of phases before 1 wier 4 2 k { # of messages per wier 2 k 2 k 1 < 8(log + 2) # of wiers per phase 2 k 1 +1

7 The Revege of he Lower Boud We have see: Facs a simple O( 2 ) algorihm a more clever O( log ) algorihm Ω( log ) lower boud i asychroous eworks Ω( log ) lower bouds i sychroous eworks whe usig oly comparisos Breakig hrough Ω( log ) Sychroous rigs UID are posiive iegers, maipulaed usig arbirary operaios No Uiform is kow o all!uidirecioal! commuicaio! O( ) messages! Uiform! is o kow!uidirecioal! commuicaio!o( ) messages! Wha abou ime complexiy? Ad ow, for somehig compleely differe RANDOMIZATION Wha is i good for? I geeral does o affec impossibiliy resuls leader elecio i aoymous eworks wors case bouds cosesus i fewer ha f +1 rouds Bu i makes a differece whe combied wih weakeig he problem saeme

8 Radomized leader elecio Trasiio fucio akes as ipu a radom umber from a bouded rage uder some fixed disribuio Theorem A secod look a aoymous rigs There is a radomized algorihm ha, wih probabiliy c>1/e elecs a leader i a sychroous rig sedig O( 2 ) messages Weaker problem defiiio for LE: Safey: I every global sae of every execuio, a mos oe process is i he eleced sae Liveess: A leas oe process is eleced wih some o-zero probabiliy The oe-sho algorihm The ieraed algorihm Iiially {!!! 1 wih probabiliy 1 1/ id i :=!!! 2 wih probabiliy 1/! sed id i o lef upo receivig S from righ! if S = he!! if id i is uique maximum of S he!!! eleced := rue!! else!!! eleced := false! else!! sed S id i o lef Oe execuio for each eleme of R = {1, 2} Algorihm ermiaes whe exacly oe process has id =2 Probabiliy of ermiaio c : ( ) ( ( = 1 1 ) 1 ( c> 1 1 ) 1 e Message complexiy: O( 2 ) ) 1 If oe execuio does o ermiae wih a leader, ry agai! How may imes? I he wors case, ifiiely may! Bu i he expeced case?

9 The ieraed algorihm Summary If oe execuio does o ermiae wih a leader, ry agai! How may imes? I he wors case, ifiiely may! Bu i he expeced case? Expeced value of T: E[T ]= x T x Pr[T = x] Probabiliy of success i ieraio i : c (1 c) i 1 Expeced umber of ieraios: i c (1 c) i 1 d = c (1 c) i d 1 = c dc dc 1 (1 c) = 1/c < e i=0 i=0 No deermiisic soluio for aoymous rigs No soluio for uiform aoymous rigs (eve whe usig radomizaio) Proocols wih O( 2 ) ad O( log ) messages for uiform rigs Ω( log ) lower boud o message complexiy for pracical proocols O() message complexiy for uiform sychroous rigs

10 Clock sychroizaio Clock sychroizaio??? Give me all files ha have chaged sice 12:00pm There you go... Wai... wha??? 13:00 13:00 12:01 Wha is he ime? Hard ruh: clocks drif apar Clock drif Exeral vs ieral sychroizaio Boud o drif: ρ (1 ρ)( ) H() H( ) (1 + ρ)( ) H() (clock ime) (1 + ρ) Exeral Clock Sychroizaio: keeps clock wihi some maximum deviaio from a exeral ime source. Ieral Clock Sychroizaio: keeps clocks wihi some maximum deviaio from each oher. ρ is ypically small (10-6 ) ρ ρ =1+ρ 1 1+ρ =1 ρ (1 ρ) (real ime) exchage of ifo abou imig eves of differe sysems ca ake acios a real-ime deadlies ca measure duraio of disribued aciviies ha sar o oe process ad ermiae o aoher ca oally order eves ha occur i a disribued sysem

11 Probabilisic Clock Sychroizaio (Crisia) Maser-Slave archiecure Maser ca be coeced o exeral ime source Slaves read maser s clock ad! adjus heir ow How accuraely ca a slave read he maser s clock? The Idea Clock accuracy depeds o message roudrip ime if roudrip is small, maser ad slave cao have drifed by much! No upper boud o message delivery, so o ceraiy of accurae eough readig... bu very accurae readig ca be achieved by repeaed aemps Seup ad assumpios The proocol Goal: Sychroize he slave s clock wih he maser (real ime) = x (real ime) mi P () slave P () ime=? ime= T Q() T Q(x) maser Q() Assume ha miimum delay is kow Assume ha clock drifs are kow ( ρ for boh) Quesio: wha is Q(x)?

12 Ideal sceario Problem #1: message delay (real ime) P () mi mi = x Q(x) =T + mi Perfec sychroizaio! P () Q() P () mi mi + α 2d mi + β T 2d mi β =2d 2mi Q() T Q(x) =T +2d mi Q() T T + mi Assume o clock drif P () Q() 2d mi mi T Q(x) =T + mi β =0 Problem #2: slave drif Problem #3: maser drif P () mi + α 2d 2D mi + β P () mi + α 2d 2D mi + β = x Q() T Q() T 2d(1 ρ) 2D 2d(1 + ρ) Durig he maser s clock drifs Eve if you kow β, here is sill some uceraiy!

13 Crisia s algorihm Crisia s algorihm mi + α 2d mi + β = x α, β 0 Naive esimaio: Q(x) =T +(mi + β) (ake maser s drif io accou) P () 2D Q(x) [T +(mi + β)(1 ρ),t +(mi + β) (1 + ρ) ] 0 β 2d 2mi (ake delay io accou) ime=? ime= T Q(x) [T +(mi+0)(1 ρ),t +(mi+2d 2mi)(1 + ρ)] =[T +(mi)(1 ρ),t +(2d mi)(1 + ρ)] Q() T Q(x) =? 2d 2D(1 + ρ) (ake slaveʼs drif io accou) Q(x) [T +(mi)(1 ρ),t +(2D(1 + ρ) mi)(1 + ρ)] =[T +(mi)(1 ρ),t +2D(1 + 2ρ) mi(1 + ρ)] Slave s esimaio ad precisio Slave s bes guess: Maximum error: Q(x) =T + D(1 + 2ρ) mi ρ e = D(1 + 2ρ) mi You ca keep ryig, uil you achieve he required precisio H() (clock ime) Adjusig he clock Afer sychroizig: If slave simply ses creae ime discoiuiies. P (x) =Q(x) H() (clock ime), i could (real ime) (real ime)

14 Adjusig he clock Adjusig he clock Logical clock C() =H()+A() C(x) =L : eed o adjus so ha C(x + α) =M + α Hardware clock Adjusme fucio Use liear adjusme fucio A() =mh()+n m = M L,N = L (1 + m)h α H() (clock ime) H() (clock ime) C() (logical ime) M α C() (logical ime) α L L M (real ime) (real ime) x (real ime) x (real ime) Nework Time Proocol Hierarchical srucure Each level is called a sraum The oldes disribued proocol sill ruig o he Iere Hierarchical archiecure Laecy-olera, jier-olera, faulolera.. very olera! Sraum 0: aomic clocks Sraum 1: ime servers wih direc coecios o sraum 0 Sraum 2: Use sraum 1 as ime sources ad work as server o sraum 3 ec... Accuracy is loosely coupled wih sraum level 3 2 1

15 Very olera. How? Marzullo s algorihm Tolerace o jier, laecy, fauls: redudacy Each machie seds NTP requess o may oher servers o he same or he previous sraum The sychroizaio proocol bewee wo machies is similar o Crisia s algorihm For each respose, we geerae a uple <T,δ> which defies a ierval [T-δ,T+δ] How o combie hose iervals? [8,12] [11,13] [10,12] [11,12] Give M source iervals, fid he larges ierval ha is coaied i he larges umber of source iervals 10±2 11±1 12±1 11.5± Marzullo s algorihm Give M source iervals, fid he larges ierval ha is coaied i he larges umber of source iervals The iuiio Visi he edpois lef-o-righ Cou how may source iervals are acive a each ime [8,12] [11,13] [14,15] 10±2 12±1 14.5±0.5 Icrease cou a sarig pois, decrease a edig pois 10±2 12±1 14.5± ± ±

16 Preprocessig For each source ierval [T 1,T 2], creae 2 uples of he form <ime, ype>: <T 1,-1> (sar of ierval) <T 2,+1> (ed of ierval) Sor all uples accordig o ime Example: Source iervals: [8,12], [11,13], [14,15] Tuples: <8,-1> <12,+1> <11,-1> <13,+1> <14, -1> <15, +1> Sored: <8,-1> <11,-1> <12,+1> <13,+1> <14, -1> <15, +1> 10±2 11.5±0.5 12±1 14.5± The algorihm bes=0, cou=0 for all uples<ime[i],ype[i]> { cou = cou - ype[i] if(cou>bes) { bes=cou bessar=ime[i] besed=ime[i+1] } } reur [bessar, besed] Noes: cou: umbers of acive iervals bes: bes umbers of acive iervals we have see cou=cou-ype[i] : if i s a sarpoi (ype=-1), icrease cou, else decrease i if(cou>bes) : if his is he highes umber of acive iervals we have see, le he bes ierval be [ ime[i], ime[i+1] ] If he ex poi is a sarpoi, i will replace his bes ierval If he ex poi is a edpoi, i will ed his bes ierval The algorihm a work Sored: <8,-1> <11,-1> <12,+1> <13,+1> <14, -1> <15, +1> Ii: bes=0, cou=0 <8,-1> : cou = cou - (-1) = 1 Is cou>bes? Yes bes=1, bessar=8, besed=11 <11,-1> : cou = cou - (-1) = 2 Is cou>bes? Yes bes=2, bessar=11, besed=12 <12,+1> : cou = cou - (+1) = 1 Is cou>bes? No <13,+1> : cou = cou - (+1) = 0 Is cou>bes? No <14, -1> : cou = cou - (-1) = 1 Is cou>bes? No <15, +1 : cou = cou - (+1) = 0 Is cou>bes? No 10±2 12± reur [11,12] 14.5±0.5 NTP imesamps How o represe ime? Tuesday April 19h 2011, 17:55:00? CDT? NTP: 64-bi UTC imesamp 32 bis 32 bis offse i secods sub-secod precisio offse = #secods sice Jauary 1, 1900 Wraps aroud every 2 32 secods = 136 years Firs wrap-aroud: 2036 Soluio: 128-bi imesamp. Eough o provide uambiguous ime represeaio uil he uiverse goes dim