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Find the regular expression

No 2 a's and 2 b's should come together?
in Theory of Computation by Junior (517 points) | 841 views
Is d answer (ab)^*+(ba)^*???

​​​​​​(epsilon+b) (ab) *(a+epsilon) 

epsilon symbol would hold for NULL. Isn't it?
And why not ba??. Coz it says No 2a's and no 2b's should be together. The minimum possible symbols acceptable are a,b,ab,ba. Isn't it?

Why don't u try to cover d basic?? Unlike start with the fundamentals of R.E. For dat ur Engineering Textbook on TOC is more den enough. Always grasp d concept den hit at d question. Yeah!!!. Lectures of IIT Profs especially dat of Mam Kamla Krithivasan must refer. K. Hope dat it helps u. :)


In your case devshree,  there can be many cases with different symbol...and the question here is no 2 consecutive a and no 2 consecutive b 's  can comes together..

L={a, b, ab, ba, aba,bab,  abab, baba.....,}  isn't it
Yes. :)
Thnq for ur suggestion.:) . Really nptel course are so boring I have watched her video.but.for toc its better to watch shai simonson video rather than nptel.. :)
it means select epsilon or b from the first part

From the second part u can get any number of ab

And from the third part u can select a or epsilon.
No probs. Work out d way u feel it's better. Yeah. :)

4 Answers

+1 vote

No two a's and b's should come language is L=(Epsilon+b)(ab)* (a+Epsilon).

L={Epsilon, b, a, ab,abab,....,bab,babab,....aba,ababa,....}

by Junior (787 points)
0 votes
(a+€) (ba) * (b+€) + a+b+ab+€

Explanation :

The only possible strings is a followed by b and b followed by a tht is  bababa.....  

So we get (ba) *

But strings could also start with a and could also end with b

So we add (€+a) and (b+€) at the start and the end

But the minimum possible string in our regular expression is ba which means we have left some strings

So we add ab+a+b+€ in the end to cover all possible strings
by Junior (575 points)
0 votes
alphabets= { a , b }

L={ epsilon , a , b , ab , ba , aba , bab , abab ,...........................} it's infinite language
It may have these two answers
1)         (epsilon+b)(ab)*(epsilon+a)


2)         (epsilon+a)(ba)*(epsilon+b)

Both of them are correct
by Junior (987 points)
0 votes

RE : (a+ba)(ba)* + (b+ab)(ab)* + ϵ

by (207 points)
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