CPSC 668 Distributed Algorithms and Systems

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CPSC 668Distributed Algorithms and Systems

• Fall 2006
• Prof. Jennifer Welch

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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
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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 gt 2.
• Why does this give ?(n log n) result?
• because M(n) ?(n log n)
• (cf. how to solve recurrences from CPSC 311)

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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

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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 (2M(n/2)) and then
  causing about n/4 extra messages to be sent

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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.

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Proof of Base Case

• Suppose n 2.
• Suppose x gt y. Then p0 wins and p1 must learn
  that the leader id is x. So p0 must send a
  message to p1. Truncate immediately after the
  message is sent (before it is delivered) to get
  desired open schedule.

p0
p1
y
p1
x
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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

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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.
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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

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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.
• 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.

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Generating Additional Messages

• Now we have 2M(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.

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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.

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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
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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

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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

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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

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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
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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!