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SLIDES BY OSAMA ASKOURA
EECS 6117
Distributed Computing
Set3
Consensus Agreement
Slide
2
Imagine a Byzantine agreement problem
where processes cannot be malicious (lie or
cheat or behave inconsistently)
Instead, processes can only behave
consistently or die (fail.)
This is what we define as a consensus
agreement with halting failures
Consensus in
Model
Synchronous
Non Anonymous Processes
Message-Passing system
Slide
2
Consensus algorithm 1:
• Pref : v (your input)
• Round 1: send your input to all
• Round 2:
for i =1 to f+1
• Keep sending input to all
• Pref =max value received in round
• Output pref
This means that we can tolerate any number
of failures (unlike Byzantine), but we need f+1
rounds!
Model
Synchronous
Non Anonymous Processes
Message-Passing system
Slide
2
CLAIM:
We cannot solve consensus in less than f+1 rounds
PROOF (Contradiction):
Suppose, exists algorithm that uses < f+ 1 rounds
Ifα1 and α2 are 2 executions of the algorithm,
α1 ~ α2 means some correct processes get same sequence of
messages in both executions
Ifα1 ~ α2 then some processes will output same result in both
executions
α1 ~~α2 if α1 ~ β1 ~ β2 ~ βn ~α2
Ifα1 ~~ α2 then they have same output.
Model
Synchronous
Non Anonymous Processes
Message-Passing system
Proof (continued):
Let α1 be failure-free execution where p1..pi have input 0 and
pi+1….. Pn have input 1
Inputs
P1 P2 P3 P4 …. Pn
α0 1 1 1 1 ….. 1 Outputs 1
α1 0 1 1 1 ….. 1
α2 0 0 1 1 ….. 1
α3 0 0 0 1 ….. 1
α4 0 0 0 0 ….. 1
αn 0 0 0 0 ….. 0 Outputs 0
Slide
2
PROOF (continued):
To contradict, prove:
α0 (no failure) ~β1 ~ α1  no failure
We know that α0 outputs 0 and αn outputs 1.
We will show that for each I αi ~~ αi+1 up to αn, so proc will
Output same result in α0 and αn;but we know they don’t!
(Contradiction)
Model
Synchronous
Non Anonymous Processes
Message-Passing system
Proc 1 fails
immediately
Slide
2
PROOF (continued):
Lemma:
Suppose ψ is an execution with ≤ 1failure per round (failure-sparse)
And is same as execution γ up to end of first r rounds and ψ has
No failures in round r+1 or later. ( 0 ≤r≤f )
Then ψ ~~ γ
Model
Synchronous
Non Anonymous Processes
Message-Passing system
Slide
2
PROOF (continued):
Proof of Lemma (By reverse induction on r):
Base step: r = f
Then ψ &γ are identical in first f rounds,
Since the algorithm uses f rounds (ψ ~~γ)
Inductive step
Assume claim is true for r+1; Prove claim for r
If γhas no failures in round r+1, then (ψ ~~γ) by inductive
hypothesis
We use fact that r ≥ f+1 when we kill q.
Model
Synchronous
Non Anonymous Processes
Message-Passing system
Summary for Synchronous
Conensus & Byzantine
Slide
2
Byzantine agreement is solvable iff n > 3f and
connectivity > 2f in an arbitrary graph
Halting failures (Consensus) is solvable iff connectivity >
f in an arbitrary graph and must take f + 1 rounds
How About
Byzantine/Consensus in
Asynchronous systems?
Slide
2
In Asynchronous systems,
1 halting failure makes the problem of agreement
insolvable. This is remains true even if you solve the
Halting Problem.
Impossibility for halting failures implies impossibility
for Byzantine failures as well in asynchronous systems
Slide
18
EECS6117notesFW15-16,Instructor:EricRuppert
HagitAttiyaandJenniferWelch.DistributedComputing:Fundamentals,SimulationsandAdvanced
Topics,2ndedition. Wiley,2004.
NancyA.Lynch.DistributedAlgorithms. MorganKaufmann,1996.
References

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Distributed Computing Set 3 - Topics of non-Byzantine Consensus

  • 1. SLIDES BY OSAMA ASKOURA EECS 6117 Distributed Computing Set3
  • 3. Slide 2 Imagine a Byzantine agreement problem where processes cannot be malicious (lie or cheat or behave inconsistently) Instead, processes can only behave consistently or die (fail.) This is what we define as a consensus agreement with halting failures Consensus in Model Synchronous Non Anonymous Processes Message-Passing system
  • 4. Slide 2 Consensus algorithm 1: • Pref : v (your input) • Round 1: send your input to all • Round 2: for i =1 to f+1 • Keep sending input to all • Pref =max value received in round • Output pref This means that we can tolerate any number of failures (unlike Byzantine), but we need f+1 rounds! Model Synchronous Non Anonymous Processes Message-Passing system
  • 5. Slide 2 CLAIM: We cannot solve consensus in less than f+1 rounds PROOF (Contradiction): Suppose, exists algorithm that uses < f+ 1 rounds Ifα1 and α2 are 2 executions of the algorithm, α1 ~ α2 means some correct processes get same sequence of messages in both executions Ifα1 ~ α2 then some processes will output same result in both executions α1 ~~α2 if α1 ~ β1 ~ β2 ~ βn ~α2 Ifα1 ~~ α2 then they have same output. Model Synchronous Non Anonymous Processes Message-Passing system
  • 6. Proof (continued): Let α1 be failure-free execution where p1..pi have input 0 and pi+1….. Pn have input 1 Inputs P1 P2 P3 P4 …. Pn α0 1 1 1 1 ….. 1 Outputs 1 α1 0 1 1 1 ….. 1 α2 0 0 1 1 ….. 1 α3 0 0 0 1 ….. 1 α4 0 0 0 0 ….. 1 αn 0 0 0 0 ….. 0 Outputs 0
  • 7. Slide 2 PROOF (continued): To contradict, prove: α0 (no failure) ~β1 ~ α1  no failure We know that α0 outputs 0 and αn outputs 1. We will show that for each I αi ~~ αi+1 up to αn, so proc will Output same result in α0 and αn;but we know they don’t! (Contradiction) Model Synchronous Non Anonymous Processes Message-Passing system Proc 1 fails immediately
  • 8. Slide 2 PROOF (continued): Lemma: Suppose ψ is an execution with ≤ 1failure per round (failure-sparse) And is same as execution γ up to end of first r rounds and ψ has No failures in round r+1 or later. ( 0 ≤r≤f ) Then ψ ~~ γ Model Synchronous Non Anonymous Processes Message-Passing system
  • 9. Slide 2 PROOF (continued): Proof of Lemma (By reverse induction on r): Base step: r = f Then ψ &γ are identical in first f rounds, Since the algorithm uses f rounds (ψ ~~γ) Inductive step Assume claim is true for r+1; Prove claim for r If γhas no failures in round r+1, then (ψ ~~γ) by inductive hypothesis We use fact that r ≥ f+1 when we kill q. Model Synchronous Non Anonymous Processes Message-Passing system
  • 11. Slide 2 Byzantine agreement is solvable iff n > 3f and connectivity > 2f in an arbitrary graph Halting failures (Consensus) is solvable iff connectivity > f in an arbitrary graph and must take f + 1 rounds
  • 13. Slide 2 In Asynchronous systems, 1 halting failure makes the problem of agreement insolvable. This is remains true even if you solve the Halting Problem. Impossibility for halting failures implies impossibility for Byzantine failures as well in asynchronous systems