CSCE 668 DISTRIBUTED ALGORITHMS AND SYSTEMS Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668 DISTRIBUTED ALGORITHMS AND SYSTEMS CSCE 668 Spring 2014 Prof. Jennifer Welch

Asynchronous Lower Bound on Messages (n log n) lower bound for any leader election algorithm A that (1) works in an asynchronous ring necessary for result to hold (2) is uniform (doesn't use ring size) necessary for this proof to work (3) elects maximum id no loss of generality (4) guarantees everyone learns id of winner Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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 uniform asynchronous leader election algorithm 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) (cf. how to solve recurrences) Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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 relies on assumption of uniform algorithm (ring size unknown) Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

Idea of Asynchronous Lower Bound The lower bound on 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 Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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. Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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 at least one message to p1. Truncate execution immediately after the first message is sent (before it is delivered) to get desired open schedule with M(2) = 1 message. p0 p1 x p1 y Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

Proof of Inductive Step Suppose n ≥ 4. Split S (set of ids) into two halves, S1 and S2. By inductive hypothesis, there are two rings, R1 and R2: Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

Apply Inductive Hypothesis R1 has an open schedule 1 in which at least M(n/2) messages are sent and e1 = (p1,q1) is an open edge. R2 has an open schedule 2 in which at least M(n/2) messages are sent and e2 = (p2,q2) is an open edge. Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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… Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

Paste Together Two Executions depends on uniform assumption Execute 1: procs. on left cannot tell difference between being in R1 and being in R. So they behave the same and send M(n/2) msgs in R. Next… Execute 2: procs. on right cannot tell difference between being in R2 and being in R. So they behave the same and send M(n/2) msgs in R. no interference bcs α1 and α2 are open Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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. Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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 without unblocking ep or eq. Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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. Until 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 Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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… Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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 : Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

Getting an Open Schedule Let 4' be the prefix of 4'' when n/2 - 1 msgs have been sent P and Q cannot meet in 4', since less than n/2 msgs are sent in 4' W.l.o.g., suppose the majority of these msgs are sent by procs 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 Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

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 4 and 4' is that 4 is 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 Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668

Wrap Up Consider schedule 1234 . During 1, M(n/2) msgs are sent, none delivered over ep or eq During 2, M(n/2) msgs are sent, none delivered over ep or eq During 3, all msgs are delivered except those over ep or eq; possibly some more msgs are sent During 4, (n/2 - 1)/2 msgs are sent, none delivered over eq (why??) This is our desired schedule for the induction! Set 4: Asynchronous Lower Bound for LE in Rings CSCE 668