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Distributed Computing 2. Leader Election – ring network Shmuel Zaks zaks@cs.technion.ac.il ©

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message passing asynchronous 9 4 5 8 6 ? Leader election

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motivation who starts? Leader election, maximum finding, spanning tree (Leader election)

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Unidirectional ring Bidirectional rings Complete networks General networks (Leader election)

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Unidirectional ring phases, unique execution (Leader election) 5 2 8 4 1 6 3 7

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Bidirectional ring (Leader election) 5 2 8 4 1 6 3 7 sense of direction L R L L L L L L L R R R R R R R

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Bidirectional ring (Leader election) 5 2 8 4 1 6 3 7 no sense of direction R R L L L L L L L R R L R R R R

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Sense of Direction : For each process p in a bidirectional ring, its left and right neighbors are termed left(p) and right(p) respectively. If right(left(p)) = p for every p, then there is a sense of direction (otherwise – no sense of direction)

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LeLann’s algorithm state := candidate; send (my_id); receive (nid); while nid ≠ my_id do if nid > my_id then state:=no_leader; send (nid); receive (nid); od; if state=candidate then state:=leader.

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(LeLann’s algorithm) 5 2 8 4 1 6 3 7 messages: 64 time: 8 My_id nid 1 5

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Theorem: LeLann’s algorithm terminates, and exactly one processor is in state= leader. Message complexity: O(n 2 ) (worst and average) Time complexity: O(n) (LeLann’s algorithm)

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Chang and Roberts algorithm state := candidate; send (my_id); receive (nid); while nid ≠ my_id do if nid > my_id then state:=no_leader; send (nid); receive (nid); od; if state=candidate then state:=leader.

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or: state := candidate; send (my_id); while state ≠ leader do receive (nid); if nid > my_id then send (nid); od; if nid = my_id then state:=leader. (Chang and Roberts’ algorithm)

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5 2 8 4 1 6 3 7 messages: 2+1+8+2+1+1+4+1= 20 time: 8

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Theorem: Chang and Roberts’s algorithm terminates, and exactly one processor is in state= leader. Message complexity: O(n 2 ) (worst) Time complexity: O(n) (Chang and Roberts’ algorithm)

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Theorem: The average message complexity of Chang and Roberts’s algorithm is O(n log n). assume all rings equally probably (for the proof – assume ids are 1,2,…, n) (Chang and Roberts’ algorithm)

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K i P(i,k) – probability that id i makes exactly k steps (Chang and Roberts’ algorithm)

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or: Consider all rings Each id makes 1 step – times Identity of P i : makes 2 nd step iff it is largest among Pi, Pi+1, which happens times Identity of Pi: makes 3 rd step iff it is largest among Pi, Pi+1,Pi+2, which happens times, etc … (Chang and Roberts’ algorithm)

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Bidirectional rings 5 2 8 4 1 6 3 7 messages: ? time: ?

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Phases 1,2,… processors start phase k no. of phases messages time = O(n) Hirschberg and Sinclair’s algorithm

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5 2 8 4 1 6 3 7 messages: ? time: ? Franklin’s algorithm

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5 2 8 4 1 6 3 7 messages: ? (Franklin’s algorithm) messages: 16messages: 32messages: 48

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no. of phases messages time = O(n) (Franklin’s algorithm) Exercise: what is the expected number of active processors after the first phase?

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Peterson’s 1 st Algorithms This algorithm is a modification of Franklin’s algorithm for unidirectional ring. The basic idea is, during a phase, each active process receives the temporary identifier of its nearest active neighbor and that neighbor’s nearest active neighbor’s temporary identifier, then applies Franklin’s strategy. DKRP

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Each node maintains four variables: state { candidate, relay, leader} tid – temporary identity ntid – first id received nntid – second id received (Peterson’s 1 st Algorithms)

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state := candidate ; tid := id ; while state relay do begin [start phase] send( tid ); receive( ntid ); if ntid = id then state := leader ; if tid > ntid then send( tid ); else send( ntid ); receive( nntid ); if nntid = id then state := leader ; if ntid max( tid, nntid ) then tid := ntid else state := relay ; end; (now state = relay )

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while state leader do begin receive( tid ); if tid = id then state := leader ; send( tid ); end (now state = relay )

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 4 2 6 7 5 3 1 8 42 6 7 3 tid:=id; candidate [start phase] : send(tid); receive(ntid); phase 1a

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 5 5 8 8 4 6 7 7 5 5 8 84 6 7 7 candidate: if tid > ntid then send(tid); else send(ntid); receive(nntid); phase 1b

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 5 5 8 84 6 7 7 candidate: if ntid max(tid, nntid) then tid:=ntid else state := relay; 7 5 8 3 1 4 phase 1c

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 5 5 8 84 6 7 7 7 5 8 3 1 4 phase 2a candidate [start phase] : send(tid); receive(ntid); relay: … 5 7 5 5 5 88 8 8 8 7 7

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 5 5 8 84 6 7 7 7 5 8 3 1 4 phase 2b 5 5 88 8 8 7 7 candidate: if tid > ntid then send(tid); else send(ntid); receive(nntid); relay: … 8 7 8 8 8 7 7 88 8 8

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 5 5 8 84 6 7 7 7 5 8 3 1 4 phase 2c 5 5 88 8 8 7 7 8 8 7 7 88 8 8 candidate: if ntid max(tid, nntid) then tid:=ntid else state := relay; relay: … 8

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5 2 8 4 1 6 3 7 tid ntid nntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 5 5 8 84 6 7 7 7 5 8 3 1 4 phase 3a 5 5 88 8 8 7 7 8 8 7 7 88 8 8 8 candidate [start phase] : send(tid); receive(ntid); relay: … 8 8 8

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Exercises: 1. why send max{tid,ntid}? 2. what happens if n=2? 3. what happens if n=1?

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P max – processor holding max_id Phase 1,2,… t p - number of non-relay processors starting phase p. Lemma: During the execution of the algorithm, a candidate processor that becomes relay will never be in a candidate state. Lemma: For every p, if t p ≥ 3 then t p+1 ≤ t p / 2.

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Lemma: if t p ≥ 3 then at the start of phase p: for p>1: The tid of a candidate is equal to the tid of the preceeding candidate in the previous phase. (corollary: for every p: each identity resides as the tid of at most one candidate processor.) If the id of P i resides in P k, then all processors P i,P i+1, …, P k-1 are relays. max_id resides as a tid of some processor. Lemma: if t p = 2 or t p = 1 then the algorithm terminates, with the processor holding max_id as a leader.

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Theorem: Peterson’s 1st algorithm always determines a unique processor – the one holding the largest identity - as a leader. Message complexity ≤ 2n log n Time complexity ≤ 2n-1 Exercise: show examples for worst cases and for best cases in terms of time and in terms of messages.

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Peterson’s 2 nd Algorithm improvement of Peterson’s 1 st algorithm Instead of comparing its id with both neighbors in the same time, a process first compares itself with its left neighbor, then its right neighbor.

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Each node maintains four variables: state { candidate, relay, leader} tid – temporary identity ntid – id received (Peterson’s 2 nd Algorithms)

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state := candidate ; tid := id ; while state relay do begin [compare to left, odd phase] send( tid ); receive( ntid ); if ntid = id then state := leader ; if tid < ntid then state := relay ; end; begin [compare to right, even phase] send( tid ); receive( ntid ); if ntid = id then state := leader ; if tid > ntid then state := relay else tid := ntid ; end; (now state = relay )

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while state leader do begin receive( tid ); if tid = id then state := leader ; send( tid ); end (now state = relay )

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5 2 8 4 1 6 3 7 tid ntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 begin [compare left] send(tid); receive(ntid); if ntid = id then state :=leader; if tid < ntid then state := relay; end; phase 1a

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5 2 8 4 1 6 3 7 tid ntid 5 3 1 8 7 6 2 4 5 1 8 42 6 7 3 phase 1b begin [compare right] send(tid); receive(ntid); if ntid = id then state :=leader; if tid > ntid then state := relay else tid := ntid ; end; relay: … 5 5 8 8 8 3 7 7 7 8

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5 2 8 4 1 6 3 7 tid ntid 7 3 1 8 7 8 2 4 5 1 8 42 6 7 3 phase 2a 5 5 8 8 8 3 7 7 begin [compare left] send(tid); receive(ntid); if ntid = id then state :=leader; if tid < ntid then state := relay; end; relay: …

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Theorem: Peterson’s 2nd algorithm always determines a unique processor as a leader. Message complexity ≤ 1.44… n log n Exercise: show an example where a processor whose id is not the largest is elected as a leader.

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phases p, p-1, … 1 (last phase) t k – no. of processors that remain candidates after phase k = no. of processors that start phase k-1. t p+1 = n t 1 = 1 t 2 ≥ 2

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Lemma: t k ≤ no. of processors that became relay during phase k+1 Proof: We show that for each processor that remained active after phase k there is a processor that became relay during phase k+1 (the previous phase)

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beginning of phase k+1 end of phase k end of phase k+1 = beginning of phase k p p p Case a: k is odd

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P survived phase k: end of phase k end of phase k+1 = beginning of phase k p p q q ? Hence p > q Case a: k is odd

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If in the beginning of phase k+1 all of these were already relays … end of phase k+1 = beginning of phase k p q then p would have become relay, contradiction. Case a: k is odd

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Lemma: t k ≤ no. of processors that became relay during phase k+1 Corollary: t k ≤ t k+2 - t k+1 t k + t k+1 ≤ t k+2 t k ≥ Fibonacci k+1 =

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n = t p+1 ≥ Fibonacci p+2 = message complexity ≤ np ≤ 1.44… n log n no. of phases = p ≤ 1.44… log n

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References E. Chang and R. Roberts, An improved algorithm for decentralized extrema-finding in circular configurations of processes, Communications of the ACM}, 22, 5, 1979, pp. 281-283.

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References D. Dolev, M. Klawe and M. Rodeh, An O(n log n) unidirectional distributed algorithm for extrema finding in a circle, Journal of Algorithms, 3, 1982, pp. 245-260.

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References W. R. Franklin, On an improved algorithm for decentralized extrema finding in circular configurations of processors, Communication of the ACM, 25, 1982, pp. 336-337.

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References D. S. Hirschberg and J. B. Sinclair, Decentralized extrema-finding in circular configuration of processors, Communications of the ACM, 23, 1980, pp. 627-628.

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References G. LeLann, Distributed systems - towards a formal approach, Information Processing Letters, 1977, pp. 155-160.

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References G. L. Peterson An O(nlogn) unidirectional algorithm for the circular extrema problem. ACM Trans. Program. Lang. Syst. 4,4 (Oct. 1982), 758-762.

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References N. Santoro, Sense of direction, topological awareness and communication complexity, SIGACT News, 16, 2, Summer 1984, pp. 50-56.

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