Quote from: Come-from-Beyond on January 13, 2015, 09:15:07 PM

Quote from: bytemaster on January 13, 2015, 08:21:04 PM

Well for starters, there are 101 to make it odd, so it can never be evenly split...

I see that you chose 101 because it's 50 + 51. To be able to decide which group to join if the network splits...

Well, I don't get why you think that at least one group will include at least 51 delegates, what if we get 34 + 34 + 33?

Also, take a look at this (from http://research.microsoft.com/en-us/um/people/lamport/pubs/byz.pdf):

Quote

Reliable computer systems must handle malfunctioning components that give conflicting information
to different parts of the system. This situation can be expressed abstractly in terms of a group of
generals of the Byzantine army camped with their troops around an enemy city. Communicating only
by messenger, the generals must agree upon a common battle plan. However, one or more of them
may be traitors who will try to confuse the others. The problem is to find an algorithm to ensure that
the loyal generals will reach agreement. It is shown that, using only oral messages, this problem is
solvable if and only if more than two-thirds of the generals are loyal; so a single traitor can confound
two loyal generals. With unforgeable written messages, the problem is solvable for any number of
generals and possible traitors. Applications of the solutions to reliable computer systems are then
discussed.

If BitShares can be modelled in terms of Byzantine generals problem then 51 is an overkill (because even 1 would be enough) or 51 is not enough (you need at least 67). Or maybe you see a 3rd option?

Before this thread goes to total off-topic I'd like to remind about my question.