test-series

Question

Comments

no its not true

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Is it L1 and L3 ?

Answer 1

a) { b¹⁰⁰⁰⁺ⁿ | n>=0 } ∪ {ab¹⁰⁰⁰⁺ⁿ | n>=0 } ∪ {a²b¹⁰⁰⁰⁺ⁿ | n>=0 }  ∪ .............{a¹⁰⁰⁰b¹⁰⁰⁰⁺ⁿ | n>=0 } 

Reg ∪ Reg ∪ Reg .....∪ Reg = Reg     // iff it's not infinite union

b) Comparison between a and b , certainly CFL    // Remember n_a(w)=n_b(w) or n_a(w)-n_b(w)=0  well known CFL

c) It's (a+b)^*  so Reg

So in that way L₁ and L₃ are Regular languages.

Comments

how is $L_{3}$ regular ?how the regular expression came $(a+b)^{*}$ .can you ellaborate?

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sourav : $wcww^r$

case 1: $w\in (a+b)^*$ Put $\epsilon$ in place of w, you will get $(a+b)^*$

case 2: Put anything else from $w$

$(a+b)^*$ + more strings =$(a+b)^*$

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Praveen Saini can u explain L1 again???

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let $L_1=\{a^pb^q | 0\leq p\leq 4\leq q\}$

then regular expression of $L_1$ is $(b^4b^*+ab^4b^*+a^2b^4b^*+a^3b^4b^*+a^4b^4b^*)$

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thank u sir