Leader Election in Synchronous Rings

1. Leader Election in Synchronous Rings Based on slides provided by Prof. Jennifer Welch

2. Leader Election in SynchronousRings • Here is a simple algorithm. • Group rounds into phases, each phase containing n rounds • In phase i, the processor with id i, if there is one, sends a message around the ring and is elected.

3. Example of Simple Synchronous Algorithm • n = 4, the smallest id is 7. • In phases 0 through 6 (corresponding to rounds 1 through 28), no message is ever sent. • At beginning of phase 7 (round 29), processor with id 7 sends message which is forwarded around the ring. Relies on synchrony and knowing n

4. Analysis of Simple Algorithm • Correctness: Easy to see. • Message complexity:O(n), which is optimal • Time complexity:O(n*m), where m is the smallest id in the ring. • not bounded by any function of n!

5. Another Synchronous Algorithm • Works in a slightly weaker model than the previous synchronous algorithm: • processors might not all start at same round; a processor either wakes up spontaneously or when it first gets a message • uniform (does not rely on knowing n)

6. Another Synchronous Algorithm • A processor that wakes up spontaneously is active; sends its id in a fast message (one edge per round) • A processor that wakes up when receiving a msg is relay; never in the competition • A fast message carrying id m becomes slow if it reaches an active processor; starts traveling at one edge per 2m rounds • A processor only forwards a msg whose id is smaller than any id is has previously sent • If a proc. gets its own id back, elects self

7. Analysis of Synchronous Algorithm • Correctness: convince yourself that the active processor with smallest id is elected. • Message complexity: Winner's msg is the fastest. While it traverses the ring, other msgs are slower, so they are overtaken and stopped before too many messages are sent.

8. Message Complexity • Divide msgs into three kinds: • fast msgs • slow msgs sent while the leader's msg is fast • slow msgs sent while the leader's msg is slow • Next, count the number of each type of msg.

9. Number of Type 1 Messages • Show that no processor forwards more than one fast msg: • Suppose pi forwards pj 's fast msg and pk 's fast msg. When pk 's fast msg arrives at pj : • either pjhas already sent its fast msg, so pk 's msg becomes slow, or • pj has not already sent its fast msg, so it never will • Number of type 1 msgs is n. pk pj pi

10. Number of Type 2 Messages (slow sent while leader's msg is fast) • Leader's msg is fast for at most n rounds • After n rounds it would have returned to leader • Slow msg i is forwarded n/2i times in n rounds • Max. number of msgs is when ids are small as possible (0 to n-1 and leader is 0) • Number of type 2 msgs is at most ∑n/2i ≤ n n-1 i=1

11. Number of Type 3 Messages (slow msgs sent while leader's is slow) • Maximum number of rounds during which leader's msg is slow is n*2L (L is leader's id). • No msgs are sent once leader's msg has returned to leader • Slow msg i is forwarded n*2L/2itimes during n*2Lrounds. • Worst case is when ids are L to L + n-1 • Number of type 3 msgs is at most ∑n*2L/2i ≤ 2n L+n-1 i=L

12. Total Number of Messages • We showed • number of type 1 msgs is at most n • number of type 2 msgs is at most n • number of type 3 msgs is at most 2n • Thus total number of msgs is at most 4n = O(n).

13. Time Complexity of Synchronous Algorithm • Running time is O(n 2x), where x is smallest id. • Even worse than previous algorithm, which was O(n x) • Both algorithms have two potentially undesirable properties: • rely on numeric values of ids to count • number of rounds is linear in n, but also depends on minimum id • Next result shows that to obtain linear msg complexity, an algorithm must rely on numeric values of the ids.

14. Synchronous Leader Election • Next result helps to show that to obtain linear msg complexity, an algorithm must rely on numeric values of the ids.

15. Comparison-Based LE Algorithms • We will show that if nodes cannot rely on the numeric values of the ids, then any synchronous LE algorithm has message complexity (n log n). • How do we formalize "not relying on numeric values of ids"? • Need a definition of "comparison-based", similar to that for sorting algorithms.

16. Definition of Comparison-Based • First note that in the synchronous model, the behavior of an LE algorithm is totally determined by the distribution of the ids. Denote execution on ring R by exec(R). • An LE algorithm is comparison-based if, in any two "order-equivalent" rings R1 and R2, "matching" processors pi in R1 and pj in R2 have "similar" behaviors in exec(R1) and exec(R2).

17. Definition of Order-Equivalent • Rings R1 = (x1,x2,…,xn) and R2 = (y1,y2,…,yn) are order-equivalent if xi < xj if and only if yi < yj • Example: x3 y2 y1 x4 x2 y3 x1 x1 y4 x5 y5

18. Definition of Matching Processors • Processors pi in R1 and pj in R2 are matching if they are the same distance from the processor with minimum id. • Example: p0 (with id 18) in R1 is 2 hops from p3 ;p1 (with id 5) in R2 is 2 hops from p4

19. Definition of Similar Behaviors • Two processors have similar behaviors in two executions if, in every round • one processor sends a message to the left (right) if and only if the other one does • one processor is elected if and only if the other one is

20. Synchronous Lower Bound Theorem (3.18): For every n ≥ 8 that is a power of 2, there is a ring of size n on which any synchronous comparison-based algorithm sends (n log n) msgs. Proof: For each n, construct a highly symmetric ring Sn on which every comparison-based algorithm takes lots of messages. • Symmetry means many processors have order-equivalent neighborhoods and thus behave similarly (including sending msgs)

21. Proof Strategy • Define highly symmetric ring Sn • Show that the number of "active" rounds is at least n/8 • Show that in the k-th active round, at least n/(2(2k+1)) msgs are sent. • Do some arithmetic to show that ∑n/(2(2k+1)) = (n log n) n/8 k=1

22. Active Rounds • A round is active if at least one processor sends a msg • A proc. in a synchronous algorithm can potentially learn something even in an inactive round • cf. the simple synchronous alg • But in a comparison-based alg., a proc. can't learn about the order pattern of its ring in an inactive round • Note that the k-th active round might be much later than the k-th round

23. 2 3 0 1 A Highly Symmetric Ring Sn • Let pi 's id be rev(i) , the integer whose binary representation using log n bits is the reverse of i 's binary rep. Example: • For technical reasons, then multiply id by n+1 0 10 p0 p1 p2 15 p3 5

24. Why Lots of Active Rounds Show number of active rounds, T, in exec(Sn) is at least n/8. Proof: Suppose T < n/8. Let pi be the elected leader. • Number of T-neighborhoods order-equivalent to pi 's is at least n/(2(2T+1)) (L. 3.19, to be shown) • Since n ≥ 8 and T < n/8, algebra shows n/(2(2T+1)) ≥ 2. • So there is pj pi whose T-neighborhood is order-equivalent to pi's. • Then pj is also elected (L. 3.17, to be shown), contradiction.

25. Why Enough Msgs Sent in Each Active Round Show at least n/(2(2k + 1)) msgs are sent in kth active round (L. 3.21) Proof: Since the round is active, at least one proc., say pi, sends a msg. • There are at least n/(2(2k+1)) procs. whose k-neighborhood is order-equivalent to pi 's (L. 3.19, to be shown) • Each of those procs. sends a msg in kth active round, since pi does (L. 3.17, to be shown).

26. Why So Many O-E Neighborhoods L. 3.19: Show every k-nbrhood of Sn has at least n/(2(2k+1)) order-equivalent k-nbrhoods. Proof Sketch: Let N be a k-nbrhood (sequence of 2k+1 ids in the ring). • Let j be smallest power of 2 larger than 2k+1. • Break Sn into n/j segments of length j, with one segment encompassing N. • Claim: Each segment is order-equivalent to N by construction ofSn

27. Why Similar Behavior in Order-Equivalent Neighborhoods • Intuition is that two nodes in order-equivalent neighborhoods in the same ring will behave similarly, at least until differentiating information from beyond those neighborhoods has reached them. • Yet the definition of comparison-based algorithm only requires similar behavior for matching processors in different (albeit order-equivalent) rings.

28. Order-Equivalent Means Similar L. 3.17: Show that if pi and pj have order-equivalent k-nbrhoods, then piand pj have similar behaviors through k-th active round. Proof Sketch: Construct another ring Sn' such that • piin Sn behaves same as pj in Sn' • pjin Sn' behaves similarly to pjin Sn

29. Order-Equivalent Means Similar • In more detail, construct Sn' such that • pi 's k-nbrhood in Snequalspi 's in Sn' • so behavior is same through k-th active round • Sn' and Sn are order-equivalent • pj in Sn' is matching to pj in Sn • so behavior is similar through k-th active round • Can be done since ids in Sn are spaced (multiplied by n+1) • So piand pj have similar behavior in Snthrough k-th active round

30. Proof Strategy • Define highly symmetric ring Sn • Show that the number of "active" rounds is at least n/8 • Show that in the k-th active round, at least n/(2(2k+1)) msgs are sent. • Do some arithmetic to show that ∑n/(2(2k+1)) = (n log n) n/8 k=1

31. This result can be further used to prove that any synchronous time-bounded algorithm (with running time not being dependent on identifiers) must have high message complexity (n log n) Recommended reading: Section 3.4.2.3 for further details

32. Anonymous Rings Revisited • Leader election is impossible in anonymous rings • No way to break symmetry • No deterministic algorithm which works in every execution • Another way to break symmetry, which works in some (but not all) executions is to use randomization.

33. Randomized Algorithm • In each computation step, the processor receives an additional input to its state transition function, a random number.

34. Revised LE Problem Definition • Weakened problem definition compared to original: • At most one leader is elected in every state of every admissible execution • same as previous definition • At least one leader is elected "with high probability". • weaker than previous definition • But what does "with high probability" mean?

35. Randomized LE Algorithm • Assume synchronous model • Initially: • set id to 1 with probability 1 - 1/n and to 2 with probability 1/n • send id to left • When msg M is received: • if M contains n ids then • if id is unique maximum in M then elect self • else not elected • else append id to M and send to left

36. Analysis of Randomized LE Alg. • Uses O(n2) msgs • There is never more than one leader • Sometimes there is no leader • leader is only elected if there is exactly one processor that sets its id to 2 • How often is there no leader, i.e., what is the probability? • Need some more definitions…

37. Random Choices and Probabilities • Since system is synchronous, an admissible execution of the algorithm is determined solely by the initial random choices. • Call this collection random choices RC = <r0, r1, …, rn-1> where each ri is either 1 or 2. • Let exec(RC) be the resulting execution. • Definition: For any predicate P on executions, Pr[P] is the probability of {RC : exec(RC) satisfies P} i.e., the proportion of random choices resulting in an execution that satisfies P.

38. Probability of Electing a Leader • Let P be the predicate "there is a leader". • Pr[P] = probability RC contains exactly one 2 ≥ 1/e for large n (converges from above)

39. Improving the Probability of Electing a Leader • If procs. notice there is no leader, they can try again. • Each iteration of the basic algorithm is a phase. • Keep trying until success. • Random choices that define an execution consist of an infinite sequence of 1's and 2's, one for each proc. • It is possible that algorithm doesn't terminate.

40. Probability of Not Terminating • Probability of terminating in a particular phase is (1 - 1/n)n-1 • Probability of not terminating in a particular phase is 1 -(1 - 1/n)n-1 • Probability of not terminating in k phases is (1 - (1 - 1/n)n-1)k since phases are independent. • Last expression goes to 0 as k increases.

41. Expected Number of Phases • Definition: The expected value of a random variable T is E[T] =  k Pr[T = k] • Let T be number of phases until termination. • Pr[T = k] = Pr[first k-1 phases fail & kth succeeds] = (1 - (1 - 1/n)n-1)k-1 (1 - 1/n)n-1 = (1 - p)k-1 p , where p = (1 - 1/n)n-1 ≥ 1/e • This is a geometric random variable with expected value p-1≤ e. • So expected number of phases is < 3. k