1. A Look at Byzantine Generals Problem R J Walters

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

3. 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”?

4. 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

5. 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 I’m only looking at oral messages

6. Example of the problem

7. What is C to think of A? Is A “loyal”, if so what’s his vote?

8. 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)

9. More on OM solution The OM(x) algorithm There must be at least 2x + 1 Loyal Generals x is the maximum number of traitors

10. Introduction to OM(x)

11. First “round” – I say x

12. What 2 sends (round 2)

13. “(I say) he said y” What 2 hears about 1’s view

14. 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?)

15. 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

16. (I say) he says he said z

17. What 3 receives

18. 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)

19. What’s happening? My diagrams have reached their limit In each round, each player passes on what they hear to all of the others They don’t 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…

20. 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

21. 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) …

22. 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?

23. 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?

24. So, how many traitors? We don’t know who they are We could guess (estimate) How? Probability of failure? Whatever, we cannot guarantee the result.

25. 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

26. The Numbers for failure at 5% Maximum no. traitors ProbabilityCum Probability 00.735091891 10.2321342810.967226172 20.0305439840.997770156 More than 20.0022298441

27. 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?

28. Implementation I couldn’t understand the Lamport paper so I built an implementation of the OM algorithm using VB and MSMQ It didn’t behave quite as I expected

29. 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

30. “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)

31. Result of 1000 runs Prob. Traitor A Mean No. Traitors Correct Results OM1OM2 001000 1/71.031000993 2/72.01992954 3/72.96996914

32. Result of 1000 runs Prob. Traitor B Mean No. Traitors Correct Results OM1OM2 001000 1/71.00854978 2/71.96628924 3/72.94508872

33. 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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