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The regular expression for the language recognized by the finite state automaton of figure is ______

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0* 1 1* 0 (0+1)* this is the language in which every 1 is must followed by 0

@Bikram sir my doubt is how to complement this ?
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0* 1 1* 0 (0+1)* why this is wrong
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i also make the sm mistake :P

Using Arden's theoram to find the regular expression :

A = ε + A0 = ε0* = $0^*$

$B = A1 + B1 = 0^*1 +B1 = 0^*11^*$

As there are two final states, we should union both RE.

$=0^∗+0^∗11^∗=0∗(e+11^∗)=0^∗(e+1+)=0^∗1^∗=0∗+0^∗11^∗=0^∗(e+11^∗)=0^∗(e+1^+)=0^∗1^∗$

Note that $e+1^+=1^∗$

PS : Ardens theoram -  Let P and Q be two regular expressions. If P does not contain null string, then R = Q + RP has a unique solution that is R = QP*

$L=0^*1^*$
$L$ contains all binary strings where any $1$ is not followed by any $0$.

edited
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It can also be 0*11*+0*
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@gatemate  u right R.E for each final state and apply in b/w '+' mans OR then u got 0*+0*11* then take common 0*(epsilion+11*) ans as we know (epsilon+rr*)=r*  so here we get 0*1*
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Yes, I am also getting this answer.
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Tell me if there is a mistake in my logic as C is dead state we don't need to consider expression going into that state.

Hence expression will be 0*.1.1* is a.a*=a+

hence it reduces to 0*.1+?

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@sripo your regular expression is not accepting epsilon it should be 0*1*

A grammer not containing 10 as substring

Clearly , looking into DFA we can say that C is Dead state.

Using Arden's theoram to find the regular expression :

A = ε + A0ε0* = 0*

B = A1 + B1 = 0*1 +B1 = 0*11* = 0*1*

PS : Ardens theoram -

Let P and Q be two regular expressions.

If P does not contain null string, then R = Q + RP has a unique solution that is R = QP*

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How did you simplified B = A1 + B1 = 0*1 +B1 = 0*11*.

according to Arden's theorem shouldn't it be: B=B.1+0*11* =1*0*1 ?

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B= 0*1 + B1

R=B, Q=0*1 and P=1

then by arden's theorem,

0*11*

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u have to union both the A and B values becoz there are two final states n given FA

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yes, @once_2019, you're correct. We should union both.

$= 0^* + 0^*11^* = 0*(e + 11^*) = 0^*(e+1^+) = 0^*1^*$

Note that $e + 1^+ = 1^*$

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