# Minimal String

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Consider the following regular expression (RE)
RE = (aa + ab)+ (a + b)+ (a + b)*
How many minimal strings exist for the above RE?

I think answer should be 1 i.e ἑ but answer given is 4.
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Answer Matched if kleene closure replaced by Positive Closure

R.E. = (aa+ab)+ . (a+b) . (a+b)+

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@Shaik Masthan

How answer matched i.e 4 ?

I got 8. 1

yes, you are right, i mis-understood the question ( i read it as Minimal string length instead of no.of minimal strings, when i am thinking about My RE, it's my mistake )

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@Shaik Masthan

Here answer will be 4 only right ?

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didn't ger you
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what is the answer of the question ? if at last we have positive closure instead of kleene closure
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0
Question is

given $RE = (aa + ab)^+ (a + b)^+ (a + b)^+$

then what would be the number of minimal strings generated using the given $RE$ ?
1
aa a a

aa a b

aa b a

aa b b

ab a a

ab a b

ab b a

ab b b

total = 8

Four Minimal Strings are:-

1)aaa

2)aab

3)aba

4) abb

## Related questions

1
The number of strings present of length 10 in language $L=\left \{ a^{2n+1}.b^{2m+1}|n\geq 0,m\geq 0 \right \}$ are_________ My explanation: I think here minimum string ab , and $8$ gap is present. So, like counting we can solve it. That means these $8$ place we can fill with either $a$ or $b$ Where is wrong in it?