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The length of the shortest string NOT in the language (over $\Sigma = \{a, b\})$ of the following regular expression is _______.
$$a^*b^*(ba)^*a^*$$
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can anyone give nfa for this regular expression?

$R = a^*b^*(ba)^*a^*$

for finding shortest string that is not in language it is better to look strings of length $0$, then of length $1$ and so on

$\text{length}0 \{ \epsilon \}$  is in $L$

$\text{length}1 \{a, b\}$      all belong to $L$

$\text{length}2 \{aa, ab, ba, bb\}$      all belong to $L$

$\text{length}3 \{aaa, aab, aba, abb, baa, bab, bba, bbb\}$  bab does not belong to L

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Praveen Saini  how did u get this?

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0 length strings over {a,b} = $(a+b)^0= \epsilon$

1 length strings over {a,b} = $(a+b)^1=a,b$

2 length strings over {a,b} = $(a+b)^2= (a+b)(a+b)= aa,ab,ba,bb$.

so on
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@Praveen Saini why r  u not finding the language derived from R = a*b*(ba)*a* and complement it! can this be an approach ?

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yes that is same thing, you can find language of R, L(R), then complement, L'(R), then first element of L(R') will be bab.

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Good Explanation

Thqq
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@praveen Saini .i didn't get that approach compliment one .can u please elaborate!!!!!!!!!

although i understand the given approach.....help
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Please draw once then explain m not getting how to solve this???
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can u draw finite automata for the regular expression?
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@suneetha @ Ankit @ Phalkey

In the DFA of the language (for given regular expression), shortest string that is NOT belong to the Language is,the shortest string that reach to a non-final state from the start state $q_0$.

$q_0\xrightarrow{\text{bab}}q_3$

$q_0\xrightarrow{\text{baab}}q_5$

I hope now it is clear that $bab$ is the answer

The big give away is the concatenation (ba)*  followed by a*

the shortest length is 3.

To be completely sure of the right answer, construct an ɛ-NFA for the given language, convert it into an equivalent DFA and then complement it. The answer comes out to be bab.
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@Vishal please explain with Diagram ..

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