Question 4848 – UGC NET CS 2018-DEC Paper-2
November 30, 2023Software-Engineering
November 30, 2023UGC NET CS 2018-DEC Paper-2
| Question 7 |
Consider the following two languages:
L 1 = {x | for some y with | y| = 2 |x| ,xy∈ L and L is regular language}
L 2 = { x | for some y such that |x| = |y|, xy∈ L and L is regular language}
Which one of the following is correct?
L 1 = {x | for some y with | y| = 2 |x| ,xy∈ L and L is regular language}
L 2 = { x | for some y such that |x| = |y|, xy∈ L and L is regular language}
Which one of the following is correct?
| Both L 1 and L 2 are regular languages | |
| Both L 1 and L 2 are not regular languages | |
| Only L 1 is regular language | |
| Only L 2 is regular language |
Question 7 Explanation:
if L is a regular language then we always have a string “w” which can be broken into “xy” such that |y|=2|x|
Consider a language L= {a 3n | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, …….}
then L 1 = {a n | n ≥ 0} as every string in L can be broken into three equal parts .
DFA for L
DFA for L 1
By same logic L 2 is also regular, here we have to break each string of L into two equal parts and also every even length string from L will belongs to L 1 , since we can break only even length string into two equal parts such that |x| = |y| if “xy ε L”
Consider a language L= {a 3n | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, …….}
then L 1 = {a n | n ≥ 0} as every string in L can be broken into three equal parts .
DFA for L
DFA for L 1
By same logic L 2 is also regular, here we have to break each string of L into two equal parts and also every even length string from L will belongs to L 1 , since we can break only even length string into two equal parts such that |x| = |y| if “xy ε L”
Correct Answer: A
Question 7 Explanation:
if L is a regular language then we always have a string “w” which can be broken into “xy” such that |y|=2|x|
Consider a language L= {a 3n | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, …….}
then L 1 = {a n | n ≥ 0} as every string in L can be broken into three equal parts .
DFA for L
DFA for L 1
By same logic L 2 is also regular, here we have to break each string of L into two equal parts and also every even length string from L will belongs to L 1 , since we can break only even length string into two equal parts such that |x| = |y| if “xy ε L”
Consider a language L= {a 3n | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, …….}
then L 1 = {a n | n ≥ 0} as every string in L can be broken into three equal parts .
DFA for L
DFA for L 1
By same logic L 2 is also regular, here we have to break each string of L into two equal parts and also every even length string from L will belongs to L 1 , since we can break only even length string into two equal parts such that |x| = |y| if “xy ε L”