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34 votes
34 votes

Which of the following are regular sets?

  1. $\left\{a^nb^{2m} \mid n \geq 0, m \geq 0 \right\}$

  2. $\left\{a^nb^m \mid n =2m  \right\}$

  3. $\left\{a^nb^m \mid n \neq m \right\}$

  4. $\left\{xcy \mid x, y, \in \left\{a, b\right\} ^* \right\}$

  1. I and IV only
  2. I and III only
  3. I only
  4. IV only
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2 Answers

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27 votes
27 votes

Answer is A.

Since in option $2$ and $3, n$ is dependent on $m$, therefore a comparison has to be done to evaluate those and hence are not regular.

I and IV are clearly regular sets.

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4 votes

I)Set of all strings containing any number of ‘a’ s followed by an even number of ‘b’ s. R.E=(a)$^{*}$(bb)$^{*}$.

IV) Strings containing a ‘c’. R.E= (a+b)$^{*}$c(a+b)$^{*}$.

Both these languages are regular as regular expressions exist.

By default a language is infinite. Eg : {a$^{n}$} it’s a infinite language.So both the languages II and III are infinite and comparison has to be done to evaluate these and hence are not regular.

 Answer: A


NOTE:

Every finite language is regular.

Infinite language + Comparison = Non-Regular.

Infinite language + No Comparison = Regular.


Edit: As nothing is mentioned about ‘c’ in option IV and there is a comma after y, So I think It’s a typo ‘c’ should also belongs to {a,b}$^{*}$. IV will be a complete language. Which is regular. R.E=(a+b)$^{*}$.

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