ace test

Question

Answer 1

See the language will accept

1. Episilon

2. occurance of a only

3.occurance of b only

And to sum it all up--->4. All strings over a and b.

These stings will be accepted by the language.

Thus states required will be 1.(initial and final states are same accepting all the strings)

Comments

Can u elaborate that regex plz, confused about that..

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Consider this:-

!)(a*b)^* Is a set which contains Epsilon and all strings ending in b.

2)Similarly (ab*)^* is a set which contains Epsilon and all strings starting with a.

Taking union of both the set plus the Kleen Closure will give us all the strngs possible.

you can also try deriving some patterns like 

{episilon, a,aa,aaa,b,bb,bbb,ab,abab,baba,abba,.....}

Deriving these may also give you some insight....

And please say if you need more explanation,I will be happy to provide that.....

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Pravin , 

Reply to your doubt :

if you would've removed kleen closure the on the entire set the language would be 

{strings starting with a or strings ending with b} (a*b)^* + (ab*)^* 

but it would not have been the set of all strings.

Try deriving (abbbba) .

For concept you may watch Ravindra Sir's lecture available on youtube or refer to any standard textbook.

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