OR Two Finite Automata
Consider the following two FAs:
FA1
accepts all words with a double a in them.
FA2
accepts all words ending with b.
Let’s
follow the algorithm to build FA3 that accepts the union of the two languages.
Transition
Table
|
Input |
a |
b |
Step 1: State z1 |
z1=x1+y1 |
z2=x2+y1 |
z3=x1+y2 |
Step 2: State z2 |
z2=x2+y1 |
Z4=x3+y1 |
Z3=x1+y2 |
Step 3: State z3 |
z3=x1+y2 |
Z2=x2+y1 |
Z3=x1+y2 |
Step 4: State z4 |
z4=x3+y1 |
Z4=x3+y1 |
Z5=x3+y2 |
Step 5: State z5 |
z5=x3+y2 |
Z4=x3+y1 |
Z5=x3+y2 |
Step
1:
Step 1: State z1 |
z1=x1+y1 |
z2=x2+y1 |
z3=x1+y2 |
Step
2:
Step 2: State z2 |
Z=x2+y1 |
Z4=x3+y1 |
Z3=x1+y2 |
Step
Final:
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