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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?

1. $L1 = L2$
2. $L1 \subset L2$
3. $L2 \subset L1$
4. None of the above

edited by
L2 accepts 101010 but L1 doesn't then how L1=L2.

PS : L1 also accepts.

Both accepts it for L2 repeat state XZZXXZ

for L1  XZZXXY

edited

L1 also accepts 101010.

Misprints : Edge $Y \rightarrow Z$ ( $0$ edge )
Edge $Z \rightarrow Y$ ( $1$ edge )

Explanation:

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).

by

I want to know how to write state diagram into this kind of expressions ...

Y=X0+Y0+Z1
Z=X0+Z1+Y0

Any resource to learn that properly...
Study ardens theorem etc
convert this nfa to dfa , you will get the dfa with same final states  for both the cases , so option A is correct
by

Time Consuming ...
edited

@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?

Wrong DFA
puja mishra, it is not at all time consuming, u just need to be thorough

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.

Can we apply only in NFA ?