On Byzantine Agreement over (2,3)-Uniform ... - Cornell University

On Byzantine Agreement over (2, 3)-Uniform Hypergraphs 455
Fig. 1. Views: {p 1,p 2,p 3},{p ′ 1,p ′ 2,p ′ 3} & {p ′ 1,p 4,p 3}.
Observation 1 Byzantine agreement is not achievable on four node hypergraph
H tolerating the adversary A characterized by the adversary structure A adv =
{{p 2 }, {p 4 }} if {p 2 ,p 4 } disconnects the hypergraph.
We shall now continue with the proof of the necessity of conditions 1 and 2
of Theorem 1. Specifically we prove that Byzantine agreement is not achievable
among n processes tolerating t faults if the hypergraph is not (2,n)-hyper-(2t +
1)-connected.
On the contrary, assume a protocol Π exists for Byzantine agreement on
some hypergraph H(P, E) thatisnot(2,n)-hyper-(2t + 1)-connected. The main
idea of the proof is to construct a protocol Π ′ that achieves Byzantine agreement
on a four node hypergraph H ′ tolerating the adversary A characterized
by the adversary structure A adv = {{p 2 }, {p 4 }} where {p 2 ,p 4 } disconnects the
hypergraph H ′ . This leads to a contradiction since existence of such Π ′ violates
Observation 1.
Let n>(2t + 1), assume that H(P, E) isnot(2,n)-hyper-(2t + 1)-connected.
That is there exist a set of 2t processes that disconnect H. Partition this set
into two sets of t processes each, say P 2 and P 4 . On removal of the 2t processes,
the hypergraph disconnects into at least two components, let the processes in
one component be P 1 and the processes in the remaining components be P 3 .
Processes in P 1 and P 3 are disconnected from each other. Thus there does not
exist a hyperedge in E which has non empty intersection with both P 1 and P 3 .
Construct a hypergraph H ′ (P ′ ,E ′ ) on four processes where P ′ = {p 1 ,p 2 ,p 3 ,
p 4 }, {p i ,p j }∈E ′ if P i and P j are connected in H (there is a hyperedge e ∈ E
that has non empty intersection with each of P i ,P j )and{p i ,p j ,p k }∈E ′ if
P i ,P j and P k are connected in H (there is a hyperedge e ∈ E that has non
empty intersection with each of P i ,P j ,P k ). It follows from the earlier argument
that {p 1 ,p 3 } ∉ E ′ and hence {p 2 ,p 4 } disconnects H ′ .
Using the protocol Π, construct a protocol Π ′ for Byzantine agreement
among the processes in P ′ connected by the hypergraph H ′ by allowing p i simulate
the behavior of processes in P i for i =1, 2, 3, 4. Processes p i simulate the
behavior of processes in P i as follows: (a) Set the inputs of all players in P i as
the input of p i , (b) In each round, messages sent among p ∈ P i and q ∈ P j using
the hyperedge {p, q} ∈E are sent using the hyperedge {i, j} ∈E ′ ({i, j} ∈E ′