Consider the non-deterministic finite automaton (NFA) shown in the figure.
State $X$ is the starting state of the automaton. Let the language accepted by the NFA with $Y$ as the only accepting state be $L1$. Similarly, let the language accepted by the NFA with $Z$ as the only accepting state be $L2$. Which of the following statements about $L1$ and $L2$ is TRUE?
Both accepts it for L2 repeat state XZZXXZ
for L1 XZZXXY
L1 also accepts 101010.
Misprints : Edge $Y \rightarrow Z$ ( $0$ edge )
Edge $Z \rightarrow Y$ ( $1$ edge )
Writing $Y$ and $Z$ in terms of incoming arrows (Arden's method) :
$Y = X0 + Y0 + Z1$
$Z = X0 + Z1 + Y0$
Hence $Y=Z$. So, option (A).
@itsvkp1 yes the two DFA's will be exactly same except the final states, for L1 DFA the no of final states=4 whereas for L2 it is 1 only.. then how are two languages equal?
converting NFA to DFA,
both DFAs are same so,L1=L2
@Abhishek Tank how do you know after converting nfa to dfa that state yz represents y only and state xyz represents state z of nfa ???
Nice Explanation GFg
P.s Different ans due to different assumption.