Leader Election in Asynchronous Rings – Lower Bound

1. Leader Election in Asynchronous Rings – Lower Bound Based on slides provided by Prof. Jennifer Welch

2. Asynchronous Lower Bound on Messages • (n log n) lower bound for any leader election algorithm A that (1) works in an asynchronous ring (2) is uniform (doesn't use ring size) (3) elects maximum id (4) guarantees everyone learns id of winner necessary for result to hold necessary for this proof to work no loss of generality no loss of generality

3. Statement of Key Result • Theorem (3.5): For every n that is a power of 2 and every set of n ids, there is a ring using those ids on which any asynchronous leader election has a schedule in which at least M(n) messages are sent, where • M(2) = 1 and • M(n) = 2M(n/2) + (n/2 - 1)/2, n > 2. • Why does this give (n log n) result? • because M(n) = (n log n)

4. Discussion of Statement • power of 2: can be adapted for other case • "schedule": the sequence of events (and events only) extracted from an execution, i.e., discard the configurations • configuration gives away number of processors but we will want to use the same sequence of events in different size rings

5. Idea of Asynchronous Lower Bound • The number of messages, M(n), is described by a recurrence: • M(n) = 2 M(n/2) + (n/2 - 1)/2 • Prove the bound by induction • Double the ring size at each step • so induction is on the exponent of 2 • Show how to construct an expensive execution on a larger ring by starting with two expensive executions on smaller rings (2*M(n/2)) and then causing about n/4 extra messages to be sent

6. Open Schedules • To make the induction go through, the expensive executions must have schedules that are "open". • Definition of open schedule: There is an edge over which no message is delivered. • An edge over which no message is delivered is called an open edge.

7. Proof of Base Case • Suppose n = 2. • Suppose x > y. Then p0 wins and p1 must learn that the leader id is x. So p0 must send a message to p1. Number of messages >= 1. • Truncate immediately after the first message is delivered to get desired open schedule. • Edge on which first message is not delivered is open p0 p1 y x p1

8. Proof of Inductive Step • Suppose n ≥ 4. • By inductive hypothesis, there are two rings, R1 and R2 of size n/2 each:

9. Apply Inductive Hypothesis R1 has an open schedule 1 in which at least M(n/2) messages are delivered and e1 = (p1,q1) is an open edge. R2 has an open schedule 2 in which at least M(n/2) messages are delivered and e2 = (p2,q2) is an open edge.

10. Paste Together Two Rings • Paste R1 and R2 together over their open edges to make big ring R. • Next, build an execution of R with M(n) messages…

11. Paste Together Two Executions • Execute 1: procs. on left cannot tell difference between being in R1 and being in R. So they behave the same and deliver M(n/2) msgs in R. • Execute 2: procs. on right cannot tell difference between being in R2 and being in R. So they behave the same and deliver M(n/2) msgs in R. depends on uniform assumption

12. Generating Additional Messages • Now we have 2*M(n/2) msgs. • How to get the extra (n/2 - 1)/2 msgs? • Case 1: Without unblocking (delivering a msg on) ep or eq, there is an extension of 1 2 on R in which (n/2 - 1)/2 more msgs are sent. • Then this is the desired open schedule.

13. Generating Additional Messages • Case 2: Without unblocking (delivering a msg on) ep or eq, every extension of 1 2 on R leads to “quiescence”: • no proc. will send another msg. unless it receives one • no msgs are in transit except on ep and eq • Let 3 be any extension of 1 2 that leads to quiescence.

14. Getting n/2 More Messages • Let 4'' be an extension of 1 23 that leads to termination. • Claim: At least n/2 messages are sent in 4''. Why? • Each of the n/2 procs. in the half of R not containing the leader must receive a msg to learn the leader's id. • Before 4'' there has been no communication between the two halves of R (remember the open edges) depends on assumption that all learn leader's id

15. Getting an Open Schedule • Remember we want to use this ring R and this expensive execution as building blocks for the next larger power of 2, in which we will paste together two open executions • So we have to find an expensive open execution (with at least one edge over which no msg is delivered). • 1 23 4'' might not be open • A little more work is needed…

16. Getting an Open Schedule • As msgs in ep and eq are delivered in 4'', procs. "wake up" from quiescent state and send more msgs. • Sets of awakened procs.(P and Q) expand outward around ep and eq :

17. Getting an Open Schedule • Let 4'be the prefix of 4'' when n/2 - 1 msgs have been received • P and Q cannot meet in 4', since n/2 - 1 msgs are received in 4' (recall : R1 and R2 contain n/2 nodes each) • W.l.o.g., suppose that at least half of these msgs are received by processors in P (at least (n/2 - 1)/2 msgs) • Let 4 be the subsequence of 4' consisting of just the events involving procs. in P

18. Getting an Open Schedule • When executing 1234, processors in P behave the same as when executing 1234'. Why? • The only difference between 4and 4' is that 4is missing the events by procs. in Q. • But since there is no communication in 4 between procs. in P and procs. in Q, procs. in P cannot tell the difference. depends on asynchrony assumption

19. Case 2 – Wrap Up • Consider schedule 1234 . • During 1, M(n/2) msgs are received, none delivered over ep or eq • During 2, M(n/2) msgs are received, none delivered over ep or eq • During 3, all msgs are delivered except those over ep or eq • During 4, (n/2 - 1)/2 msgs are delivered, none delivered over eq • This is our desired schedule for the induction! • eq is the open edge for ring R