Post
Topic
Board Development & Technical Discussion
Re: New method of 51% attack?
by
StephenMorse
on 10/02/2015, 14:58:14 UTC
To mine two chains you need double the power, so in this scenario the “51%” would actually be 25.5% and 25.5%.

No, you have it backwards. To mine 2 chains you need to split your effort between the two chains, so you got 25.5% for each chain. I said it's “double” the power because the total power is double the power for each chain.

That's not the way it would work in the attack I'm describing. The malicious miner would still let others mine, it would just alternate which chain they are mining on to help the chain that fell behind catch back up. So instead of splitting 25.5% on two chains, it would be all 51% on chain A half the time and all 51% on chainB half the time. The disruptive part is that there would be switching back and forth all the time, and the two chains could contain very conflicting data.

If the miner were to try to create two 'empty' chains and to try to keep them both longer than the main chain, then you would need 2x as much hash power as the rest of the network, meaning you'd need >66.7% of the hashing power. That's not what I'm talking about, though.