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Give context-free grammars generating the following languages.

  1. The set of strings over the alphabet $\{a,b\}$ with more $a's$ than $b's$
  2. The complement of the language $\{a^{n}b^{n}\mid n\geq 0\}$
  3. $\{w\#x\mid w^{R}$ $\text{is a substring of $x$ for }$ $w,x \in\{0,1\}^{*}\}$
  4. $\{x_{1}\#x_{2}\#...\#x_{k}\mid k\geq 1,$ $\text{each}$  $ x_{i}\in\{a,b\}^{*},$ $\text{and for some i and }$ $ j,x_{i}=x_{j}^{R}\}$
in Theory of Computation 48 views
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question 3 is a regular language I think?

RE=(a+b)*#(a+b)*
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How come?
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not sure... just guessed... I could be wrong
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You should never guess in these topics :)

1 Answer

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a.  S->AaA

     A->AA | aAB | bAa | a |  ε

     B->b

b.  $\overline{L}$={$a^nb^m | n>m>1$}  $\cup$  {$a^nb^m | m>n>1$} $\cup$ {(a+b)*b(a+b)*a(a+b)*}

     Grammar for $\overline{L}$:

     S->A | B | DbDaD

     A->aaAb | aaCb

     B->aBbb | aCbb

     C->aCb | ε

     D->aD | bD | a | b | ε

edited by
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Can it generate $aab?$
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@Arjun sir

I've changed it

pls check is it correct now?

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For b part, you havent given bounds for $m,n$. Also $bab$ won't be generated rt?
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thanks for pointing out that @Arjun sir

corrected :P

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