CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings1 CPSC 668 Distributed Algorithms and Systems Fall 2009 Prof. Jennifer Welch

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings2 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

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings3 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) (cf. how to solve recurrences)

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings4 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

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings5 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

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings6 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.

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings7 Suppose n = 2. Suppose x > y. Then p 0 wins and p 1 must learn that the leader id is x. So p 0 must send a message to p 1. Truncate immediately after the message is sent (before it is delivered) to get desired open schedule. Proof of Base Case p0p0 p1p1 p1p1 x y

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings8 Proof of Inductive Step Suppose n ≥ 4. Split S (set of ids) into two halves, S 1 and S 2. By inductive hypothesis, there are two rings, R 1 and R 2 :

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings9 Apply Inductive Hypothesis R 1 has an open schedule  1 in which at least M(n/2) messages are sent and e 1 = (p 1,q 1 ) is an open edge. R 2 has an open schedule  2 in which at least M(n/2) messages are sent and e 2 = (p 2,q 2 ) is an open edge.

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings10 Paste Together Two Rings Paste R 1 and R 2 together over their open edges to make big ring R. Next, build an execution of R with M(n) messages…

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings11 Paste Together Two Executions Execute  1 : procs. on left cannot tell difference between being in R 1 and being in R. So they behave the same and send M(n/2) msgs in R. Execute  2 : procs. on right cannot tell difference between being in R 2 and being in R. So they behave the same and send M(n/2) msgs in R.

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings12 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) e p or e q, 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.

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings13 Generating Additional Messages Case 2: Without unblocking (delivering a msg on) e p or e q, 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 e p and e q Let  3 be any extension of  1  2 that leads to quiescence.

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings14 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)

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings15 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…

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings16 Getting an Open Schedule As msgs in e p and e q are delivered in  4 '', procs. "wake up" from quiescent state and send more msgs. Sets of awakened procs.(P and Q) expand outward around e p and e q :

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings17 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

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings18 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.

CPSC 668Set 4: Asynchronous Lower Bound for LE in Rings19 Wrap Up Consider schedule  1  2  3  4. During  1, M(n/2) msgs are sent, none delivered over e p or e q During  2, M(n/2) msgs are sent, none delivered over e p or e q During  3, all msgs are delivered except those over e p or e q ; possibly some more msgs are sent During  4, (n/2 - 1)/2 msgs are sent, none delivered over e q (why??) This is our desired schedule for the induction!