Set 4: Asynchronous Lower Bound for LE in Rings
CSCE 668
DISTRIBUTED ALGORITHMS AND
SYSTEMS
CSCE 668
Fall 2011
Prof. Jennifer Welch
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Asynchronous Lower Bound on Messages
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
(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
no loss of generality
Set 4: Asynchronous Lower Bound for LE in Rings
CSCE 668
Statement of Key Result
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
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
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

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
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
The lower bound on the number of messages, M(n), is
described by a recurrence:



Prove the bound by induction
Double the ring size at each step


M(n) = 2 M(n/2) + (n/2 - 1)/2
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
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


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
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
Suppose n = 2.
x p0




pp11 y
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 immediately after the first message is
sent (before it is delivered) to get desired open
schedule.
Set 4: Asynchronous Lower Bound for LE in Rings
CSCE 668
Proof of Inductive Step
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


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
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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
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

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
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

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.
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.
Set 4: Asynchronous Lower Bound for LE in Rings
CSCE 668
Generating Additional Messages
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


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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
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
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
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

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)

Set 4: Asynchronous Lower Bound for LE in Rings
CSCE 668
Getting an Open Schedule
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



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
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

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
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 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
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


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.
Set 4: Asynchronous Lower Bound for LE in Rings
CSCE 668
Wrap Up
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





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