But what if the third bookkeeper is malicious? You need at least four bookkeepers to achieve byzantine fault tolerance with f=1.
But what if the third bookkeeper is malicious? You need at least four bookkeepers to achieve byzantine fault tolerance with f=1.
does this assume a maximum of one malicious bookkeeper?
Yes, you need 3f+1, so 4 for f=1, 7 for f=2.
https://lamport.azurewebsites.net/pubs/byz.pdf