A Look at Byzantine Generals Problem

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A Look at Byzantine Generals Problem

• R J Walters

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Outline

• Introduce BGP, and OM(x)
• Some discussion about which algorithm to use
• Some results
• Conclusion

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Byzantine Generals Problem

• A classic distributed agreement problem
• Imagine a city under siege
• How do the leaders of the forces outside decide
  whether to attack or wait?
• What is the effect if some of these leaders are
  traitors?

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What to do?

• Question If all the loyal forces attack, can
  they overwhelm the city?
• Suppose they can, but only just
• If the traitors can get some forces to attack and
  others no to, the city will stand
• The Loyal Generals need to vote on what to do and
  then act on the outcome

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Solving BGP

• A method for a single loyal General to
  communicate their decision is all you need
• Solution depends on the type of message
• Written signed by the sender
• Oral can be altered by anyone who handles it
• Im only looking at oral messages

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Example of the problem
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Example of the problem
What is C to think of A? Is A loyal, if so
whats his vote?
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The Solution

• A scheme whereby messages are forwarded
• Get this right and your fellow loyal players will
  be able to calculate the correct message, no
  matter what a few traitors do
• (Remember we are now concerned with one loyal
  General passing their opinion to the other loyal
  Generals)

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More on OM solution

• The OM(x) algorithm
• There must be at least 2x 1 Loyal Generals
• x is the maximum number of traitors

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Introduction to OM(x)
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First round I say x
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What 2 sends (round 2)
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What 2 hears about 1s view
(I say) he said y
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This is OM(1)

• Considering J, they get
• The message from H direct
• Five reports of that message
• There is nothing one traitor acting alone can do
  to interfere with the meaning received by J
• (Loyal Generals report what they hear accurately)
• And the same applies for the others (K-O)
• (Obvious?)

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Next round

• For OM(2) messages are forwarded again
• This is necessary if you are to guarantee the
  right result when there are two traitors

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(I say) he says he said z
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What 3 receives
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What about these messages?

• Now each player has a bunch of third hand
  reports of each of the second hand messages
• They use these to arrive at the correct versions
  of the second hand messages
• And then deduce the original message
• (As before)

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Whats happening?

• My diagrams have reached their limit
• In each round, each player passes on what they
  hear to all of the others
• They dont send messages back to people who have
  already seen them
• So, first round 6 messages
• Then 5 more for each of those 6
• And 4 more for each of the 30

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Shortcomings of the solution

• The number of messages is exploding
• Round 1 6,
• Round 2 6 x 5 30,
• Round 3 6 X 5 x 4 120.
• Remember we are just looking at one. There are 6
  other Generals who have to pass on their view

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Total number of messages

• 7 Generals
• OM(0) 7 x 6 42
• OM(1) 7 x (6 6x5) 252
• OM(2) 7 x (6 6x5 6x5x4) 1092
• OM(3)

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Shortcomings of the solution

• In order to apply it you need to know how many
  traitors are present
• What if there are more?
• Or less?
• How can you possibly know?

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So what do you do?

• Anything works if there are no traitors
• Use OM(1) and you are safe from the actions of a
  single traitor
• Use OM(2) and you guarantee safety from the
  actions of as many as two traitors
• OM(3) ? Does it work?

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So, how many traitors?

• We dont know who they are
• We could guess (estimate)
• How?
• Probability of failure?
• Whatever, we cannot guarantee the result.

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So what?

• Amongst 7 Generals, if the probability of bad
  behaviour is 5
• The chance of zero traitors is quite good
• The chance of no more than 1 is even better
• The chance of exactly 2 is tiny
• 3 or more is very unlikely

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The Numbers for failure at 5
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So what?

• OM(2) requires more than four times the effort to
  perform
• In effect, the extra work and messages of OM(2)
  guarantee the result if there are exactly two
  traitors
• Is it worth it?

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Implementation

• I couldnt understand the Lamport paper so I
  built an implementation of the OM algorithm using
• VB and MSMQ
• It didnt behave quite as I expected

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Observation

• You get agreement where the guarantees apply
• But you also get agreement pretty often when it
  is not guaranteed
• Two types of traitor. Both ignore what they
  receive
• A always sends the same value
• B sends a random value

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General Behaviour

• Decide if you are loyal or not
• If loyal, pick your value
• Otherwise
• A picks value to go in every message
• B reselects value each time it sends a message
• Run the algorithm and report your conclusion
• (Uses the random number generator)

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Result of 1000 runs
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Result of 1000 runs
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My conclusion

• Not surprisingly, the B behaviour is more
  disruptive
• In the situation I looked at
• Unless it is vital you agree in the presence of
  two or less traitors, or you expect traitors to
  be disruptive or to collude
• Its not worth the effort of doing OM(2)
• BGP fully described by Lamport, Shostak and Pease
  in their 1982 paper

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