Non-Byzantine problem

Security: 2Q-N>=1 (equivalent to 2Q-N>0)
Activity: Q<=N-f
Judgment: There is no malicious node interference in Q, so it is possible to reach a consensus if Q is odd/even

N<2Q<=2(N-f)=>N>2f

N=2f+1, when Q is f+1, the inequality can be satisfied~

About Q
If we receive Q messages and a consensus is reached in the Q messages, we can consider that a consensus has been reached. According to the above, we know that the smaller Q is, the better:

Byzantium: Q=(N+f)/2+1
Non-Byzantine: Q=N/2+1
Because N=3f+1, we have to receive Q(2f+1) pieces of information every time, so we can judge

But if N=5f+1, we need to receive Q(3f+1) pieces of information before we can judge.

According to the situation of N=5f+1 (ie 6) and f=1, what is the problem if we receive 2f+1 (ie 3) messages?
The master sends 0 to a Q1=2, and sends 1 to a Q2=2

Q1 may receive 3 confirmation messages about 0 (including master->0+2 Q1 members), record 0
Q2 may receive 3 (including master->1+2 Q2 members) confirmation message about 1, record 1
Security cannot be guaranteed at this time.