Regular Expression

Question

Which of the following language is accepted by the following finite automata?
(a) (110)∗01 (b) 0+(1(11)∗10)+0
(c) 0+(1(11)∗101)+0 (d) (11+10)∗01 

Comments

none of the above.

even the options given are wrong.

$a+b+a=a+b$ only ,

then why $b$ and $c$ are given as $0+(1(11)^{∗}10)+0$ and $0+(1(11)^{∗}101)+0$

we can write $b$ as $0+(1(11)^{∗}10)$

and $c$ as  $0+(1(11)^{∗}101) $

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anyways coming to your question,

option $a$ is incorrect

as $(110)^{∗}01$ should accept $01$ while it is not accepting.

option $b$ is incorrect

as $0+(1(11)^{∗}10)+0$ should accept $110$ while it is not accepting.

option $c$ is incorrect

as $0+(1(11)^{∗}101)+0$ should accept $1101$ while it is not accepting.

option $d$ is incorrect

as $(11+10)^{∗}01$ should accept $01$ while it is not accepting.

 

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RE for given FSM.

1((11)*101)*100+0

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your regular expression is accepting $10$ wwhich it should not!

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Thanks for pointing it out. I've edited. Please check now, although I've checked, but you can check it.

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aambazinga seems okk to me!

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Ok. Thanks.

Answer 1

• DFA accept string 1100 . But in option non of regular expression genrate 1100 . So all option are wrong

 

 

Comments

Thank you very much

Answer 2

Yes each and every options are wrong.

Answer 3

string generated from option (A)=11001 does not reach to final state,so it is incorrect.

string generated from option (B)=1111110 does not reach to any final state so it's incorrect.

string generated from from option (c)=111101 does not reach to final state it;s is also incorrect.

string generated from option (D)=1101 does not reach to any state so it is also incorrect,

so,no any option is correct.