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Consider the following languages.
L1 = {w xy x ⏐w, x , y ∈ (0 + 1)+}
L2 = {xy ⏐x , y ∈ (a + b)*, ⏐x ⏐=⏐y ⏐, x ≠ y}
Which one of the following is TRUE?
  • a)
    L1 is regular and L2 is context-free.
  • b)
    L1 is context-free hut L2 is not context-free.
  • c)
    Neither L1 nor L2 is context-free.
  • d)
    L1 is context-free but not regular and L1 is context-free.
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider the following languages.L1 = {w xy x w, x , y ∈ (0 + 1)+...
L1 = {w xyz ⏐w, x , y ∈ (0 + 1)+} is regular because by putting x as “0” and then “1” we can create a subset w 0 y 0 + w 1y1
r = (0 + 1)+ 0 (0 + 1)+ 0 + (0 + 1)+ 1(0 + 1)+ 1
Which can be shown to cover all other strings obtained by putting x as “00”, “01”, “10” and “11” etc.
Hence we can write regular expression r to represent given language “L1” and hence it is regular.
L2 has strings which can be broken into two parts x and y with not only equal length but also x ≠ y . The x ≠ y involves, string matching which FA cannot do and hence L2 is not regular.
It is however, context free because even length strings always have ⏐x⏐=⏐y⏐. So only one strings matching required and hence CFL.
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Most Upvoted Answer
Consider the following languages.L1 = {w xy x w, x , y ∈ (0 + 1)+...
Explanation:

L1:
- Language L1 consists of strings of the form "wxyxw" where w, x, y belong to the set (0+1)+, meaning they can be any combination of 0s and 1s.
- This language can be represented by the regular expression (0+1)*(0+1)(0+1)(0+1)*(0+1).
- Since L1 can be represented by a regular expression, it is a regular language.

L2:
- Language L2 consists of strings of the form "xyx" where x and y belong to the set (a+b)*, meaning they can be any combination of 'a's and 'b's.
- Additionally, the condition x=y is imposed, which means that x and y must be the same in L2.
- The condition x≠y is also given, which contradicts the previous condition, making L2 impossible to generate.
- Since L2 cannot be generated by any context-free grammar, it is not a context-free language.
Therefore, the correct answer is option 'A': L1 is regular and L2 is not context-free.
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Consider the following languages.L1 = {w xy x w, x , y ∈ (0 + 1)+}L2 = {xy x , y ∈ (a + b)*, x =y , x ≠ y}Which one of the following is TRUE?a)L1 is regular and L2 is context-free.b)L1 is context-free hut L2 is not context-free.c)Neither L1 nor L2 is context-free.d)L1 is context-free but not regular and L1 is context-free.Correct answer is option 'A'. Can you explain this answer?
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Consider the following languages.L1 = {w xy x w, x , y ∈ (0 + 1)+}L2 = {xy x , y ∈ (a + b)*, x =y , x ≠ y}Which one of the following is TRUE?a)L1 is regular and L2 is context-free.b)L1 is context-free hut L2 is not context-free.c)Neither L1 nor L2 is context-free.d)L1 is context-free but not regular and L1 is context-free.Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider the following languages.L1 = {w xy x w, x , y ∈ (0 + 1)+}L2 = {xy x , y ∈ (a + b)*, x =y , x ≠ y}Which one of the following is TRUE?a)L1 is regular and L2 is context-free.b)L1 is context-free hut L2 is not context-free.c)Neither L1 nor L2 is context-free.d)L1 is context-free but not regular and L1 is context-free.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following languages.L1 = {w xy x w, x , y ∈ (0 + 1)+}L2 = {xy x , y ∈ (a + b)*, x =y , x ≠ y}Which one of the following is TRUE?a)L1 is regular and L2 is context-free.b)L1 is context-free hut L2 is not context-free.c)Neither L1 nor L2 is context-free.d)L1 is context-free but not regular and L1 is context-free.Correct answer is option 'A'. Can you explain this answer?.
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