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+15 votes
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Let $S$ and $T$ be languages over $\Sigma=\{a.b\}$ represented by the regular expressions $(a+b^*)^*$ and $(a+b)^*$, respectively. Which of the following is true?

  1. $S \subset T$
  2. $T \subset S$
  3. $S = T$
  4. $S \cap T = \phi$
asked in Theory of Computation by Veteran (59.5k points)
edited by | 1.1k views
+2
C is correct

S=(a+b*)*=(a*(b*)*)*=(a*b*)*

T=(a+b)*=(a*b*)*     so S=T
+1
can anyone explain option D?
0
If two sets are equal then how intersection can be null??

A={1,2,3} B ={1,2,3} so when A=B then their intersection can't be null.

3 Answers

+24 votes
Best answer

(c)  $S=T$. Both generates all strings over $\Sigma$.

answered by Veteran (358k points)
edited by
+1
Can anyone explain in more detail? I am confused that regular expression S will generate more number of b's as compared to expression T. If they need to be equal they both should always generate same set of symbol but number of b's generated by expression S can be more.

Is this solution is an identity of regular expression? or we need to evaluate set of string and make the comparison.
0

vupadhayay

see the above comment both are equal

+4 votes
C is correct

S=(a+b*)*=(a*(b*)*)*=(a*b*)*

T=(a+b)*=(a*b*)*     so S=T
answered by Active (3.4k points)
0
Bt string abbbabbb is not accepted by T which is accepted by S...then how S=T?Please correct me if I m wrong...
+1
T = (a+b)*  this language is set of all possible strings it will have all strings belongs to alphabet  Σ={a.b}

(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)

now take a from first(a+b) then b from second(a+b)and so on... in this way ur string will be accepted by T..
0
Ohh...now I understand Thank You
0 votes
answer is C
answered by (191 points)


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