Given the following two languages- L1-anbn-n- 0-n 100 L2-w-a-b-c-naw-nbw-ncwWhich of the following options is correct-

Option (c) Language nbsp; nbsp; nbsp;L_1={a^n b^n|n≥ 0,n 100} is context free pop all b^n element for each a^n element since we can stack. But language ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Intersection

Given the fol...

Question

Given the following two languages:
$L_{1} = {a^{n} b^{n} | n \geq 0, n \neq 100}$

$L_{2} = {w \in {a, b, c} * | n_{a} (w) = n_{b} (w) = n_{c} (w)}$

Which of the following options is correct?

L_{1}

is context free language,

L_{2}

is not context free language.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both

L_{1} a n d L_{2}

are not context free language.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

L_{1}

is context free language,

L_{2}

is not context free language.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Both

L_{1} a n d L_{2}

are context free language.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $L_{1}$ is context free language, $L_{2}$ is not context free language.
Option (c)
Language
$L_{1} = {a^{n} b^{n} | n \geq 0, n \neq 100}$ is context free pop all $b^{n}$ element for each $a^{n}$ element since we can stack.
But language
$L_{2} = {w \in {a, b, c} | n_{a} (w) = n b (w) = n_{c} (w)$ not compare all $a^{'} s, b^{'} s a n d c^{'} s$ simultaneously with more then one stack here then thus language $L_{2}$ bec.

Option (c) is correct.

Suggest Corrections

Similar questions

Q. Consider two languages,

L_{1} a n d L_{2}

Which of the following correctly represents

L_{1} \cdot L_{2}

, where

" \cdot "

is the concatenation operation?

L_{1} = {a^{n} | n \geq 0}

and

L_{2} = {b^{n} | n \geq 0}

Q. If

L_{1} = {a^{n} ∣ n \geq 0}

and

L_{2} = {b^{n} ∣ n \geq 0}

, consider

I.

L_{1} \cdot L_{2}

is a regular language

II.

L_{1} \cdot L_{2} = {a^{n} b^{n} ∣ n \geq 0}

Which one of the following is CORRECT?

Q. Conisder the following languages :

L_{1} : {w w | w ϵ {a}^{*}}

L_{2} : {w w | w ϵ {a b}^{*}}

Which of the above languages in non - regular but CFL ?

Q. Consider the following grammar G

S \to b S | a A | b

A \to b A | a B

B \to b B | a S | a

Let

N_{a} (w) a n d

N_{b} (w)

denote the number of a's and b's in a string w respectively. The language L(G)

\subseteq

{a,b}

^{+}

generated by G is

Q. Consider the following languages:

L_{1} = {a^{n} b^{m} c^{n + m} : m, n \geq 1}

L_{2} = {a^{n} b^{n} c^{2 n} : n \geq 1}

Which one of the following is TRUE?

Join BYJU'S Learning Program

Given the following two languages: L1={anbn|n≥0,n≠100} L2={w∈{a,b,c}∗|na(w)=nb(w)=nc(w)} Which of the following options is correct?

Given the following two languages:
$L_{1} = {a^{n} b^{n} | n \geq 0, n \neq 100}$

$L_{2} = {w \in {a, b, c} * | n_{a} (w) = n_{b} (w) = n_{c} (w)}$

Which of the following options is correct?