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The string $1101$ does not belong to the set represented by

1. $110^*(0 + 1)$
2. $1(0 + 1)^*101$
3. $(10)^*(01)^*(00 + 11)^*$
4. $(00 + (11)^*0)^*$
edited | 4k views
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option d generate 110110, here 1101 is as substring so option d is valid or not???
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No. 1101 as a string must belong to the set.

Only (a) and (b) can generate $1101$.

In (c) after $11$, we can not have $01$ and so $1101$ cannot be generated.

In (d) Every $11$ followed by $0$ and no single occurrence of $1$ is possible. So it cannot generate $1101$ or $11011$.

edited
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does really order matters in every case ??
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concatenation is not commutative that's why order matters.
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@Arjun Sir (d) can generate 1101 through (00+(11)*0)*. As firstly it will choose 11 from (00+11) then it will got 0, so at this time string will become 110 and similarly again it will repeat the same, so string will become 11011 or 110110 which contains 1101 as a substring. Correct me if I am wrong.
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generate means it must be a string- not substring.
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(00 + (11)*0)*
Let us write this as (00 + (11)*0) (00 + (11)*0) by taking * = 2
Now, let me choose the second part of each of this i.e. (11)*0(11)*0 = 110110
110110 is a string and 1101 is a substring of 110110. Therefore D is incorrect.

Thank you Arjun Sir
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sir  in optn D "1101" as a substring is generated.. b'cz if for whole* I'm putting 2 and both time I'm selecting( (11)*0) * then 110110 is coming  as a string and 1101 is a substring over here..
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sir can you please expend the option 4 ..i am not getting how to expend

(11)* = ????
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Sir, I think option a can't be true .. Even though it generates 1101 but in worst case option a can generate 110 or 111 (by taking * as 0 ).

Sir, in exam should I assume worst case or just solve normally ? plz help.
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sir will  you give finite automata's for this regular expressions?
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@Arjun Sir which option to mark if such question comes in exam
c and d both are right

I think option (c) is more appropriate to choose from, because even option (d) can generate 11011 or 110110 in which 1101 is a substring.

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yeahh.  I think D  is not allowed because let  (00+(11)*0)*  can be written as (a+b)*
and we know 'bb'  is allowed in
(a+b)* so in  (00+(11)*0)*  110110110...   is allowed. now see bold substring is 1101.

option d is quite inviting to choose it as we can have only even number of 1's with this regular set where as our original string contains 3 1's

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That's right. But what about (c)?

from option c and d we can't generate string 1101.

therefore both c and d are correct

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