We went over the procedure to find a minimal DFA equivalent to a given DFA, and then practiced this on two examples.
The initial partition is F = {0} and N = {1,2,3,4,5}. Looking at N, we find that 3 goes to F on 0 and to N on 1, while the other four go to N on either 0 or 1. So we must split N,
So our second partition is F = {0}, X = {3}, and Y = {1,2,4,5}. We look at N and find that states 1 and 4 go to Y on 0 and to X on 1, while states 2 and 5 go to Y on both 0 and 1. So we must split Y.
Now our third partition has F = {0}, X = {3}, U = {1,4}, and V = {2,5}. We must look at both U and V. For U, both states go to V on 0 and to X on 1. For V, both states go to U on 0 and to V on 1. So we have reached our final partition.
The final DFA has four states. F is the start state and only final state, and we have δ(F,0) = F and δ(F,1) = U. We also have
The initial partition has F = {0} and n = {1,2,3,4,5,6,7}. Looking at N, we find that 4 goes to F on 0 and to N on 1, while the other six states always go to N. So we must split N.
The second partition has F = {0}, X = {4}, and Y = {1,2,3,5,6,7}. Looking at Y, we find that states 2 and 6 go to X on 0 and to Y on 1, while the other four states always go to Y. So we must split Y.
The third partition has F = {0}, X = {4}, U = {2,6}, and V = {1,3,5,7}. We must look at U and V. We find that both states 2 and 6 go to X on 0 and to V on 1. We find that all four states in V go to U on 0 and to V on 1. So we have our final partition.
The DFA has four states {F,X,U,V}. F is the start state and only final state, and δ is as follows:
Last modified 15 May 2006