Question 17383 – NTA UGC NET Dec 2023 Paper-2
March 22, 2024NTA UGC NET Dec 2023 Paper-2
March 22, 2024NTA UGC NET Dec 2023 Paper-2
| Question 15 |
Let A={a, b} and L=A*. Let x={anbn, n>0). The languages L U X and X are respectively:
(1)
(2)
(3)
(4)
(1)
(2)
(3)
(4)
| Not regular, Regular | |
| Regular, Regular | |
| Regular, Not regular | |
| Not Regular, Not Regular |
Question 15 Explanation:
L = A*: As mentioned before, this is a regular language representing all strings over alphabet A (a and b), including the empty string.
X = {a^n b^n, n>0}:This language contains strings with an equal number of a’s and b’s, both counts being greater than 0. The key thing to notice is that the number of a’s and b’s must be the same, and both must be greater than 0.
L U X: Combining L (all strings over A) with X (strings with equal and positive a’s and b’s) using union will still result in a regular language. L essentially adds all other strings not included in X (e.g., strings with only a’s or b’s, or strings with unequal counts).
X:This is where the irregularity comes in. The condition of having an equal and positive number of a’s and b’s requires more than just a regular expression to recognize. Regular expressions can represent repetition, concatenation, and alternation, but checking for equality of two variable-length parts (a’s and b’s) within the string goes beyond what they can handle.
Therefore, we have:
L U X: Regular because it’s the union of two regular languages.
X: Not regular because it requires a condition that regular expressions cannot capture.
X = {a^n b^n, n>0}:This language contains strings with an equal number of a’s and b’s, both counts being greater than 0. The key thing to notice is that the number of a’s and b’s must be the same, and both must be greater than 0.
L U X: Combining L (all strings over A) with X (strings with equal and positive a’s and b’s) using union will still result in a regular language. L essentially adds all other strings not included in X (e.g., strings with only a’s or b’s, or strings with unequal counts).
X:This is where the irregularity comes in. The condition of having an equal and positive number of a’s and b’s requires more than just a regular expression to recognize. Regular expressions can represent repetition, concatenation, and alternation, but checking for equality of two variable-length parts (a’s and b’s) within the string goes beyond what they can handle.
Therefore, we have:
L U X: Regular because it’s the union of two regular languages.
X: Not regular because it requires a condition that regular expressions cannot capture.
Correct Answer: C
Question 15 Explanation:
L = A*: As mentioned before, this is a regular language representing all strings over alphabet A (a and b), including the empty string.
X = {a^n b^n, n>0}:This language contains strings with an equal number of a’s and b’s, both counts being greater than 0. The key thing to notice is that the number of a’s and b’s must be the same, and both must be greater than 0.
L U X: Combining L (all strings over A) with X (strings with equal and positive a’s and b’s) using union will still result in a regular language. L essentially adds all other strings not included in X (e.g., strings with only a’s or b’s, or strings with unequal counts).
X:This is where the irregularity comes in. The condition of having an equal and positive number of a’s and b’s requires more than just a regular expression to recognize. Regular expressions can represent repetition, concatenation, and alternation, but checking for equality of two variable-length parts (a’s and b’s) within the string goes beyond what they can handle.
Therefore, we have:
L U X: Regular because it’s the union of two regular languages.
X: Not regular because it requires a condition that regular expressions cannot capture.
X = {a^n b^n, n>0}:This language contains strings with an equal number of a’s and b’s, both counts being greater than 0. The key thing to notice is that the number of a’s and b’s must be the same, and both must be greater than 0.
L U X: Combining L (all strings over A) with X (strings with equal and positive a’s and b’s) using union will still result in a regular language. L essentially adds all other strings not included in X (e.g., strings with only a’s or b’s, or strings with unequal counts).
X:This is where the irregularity comes in. The condition of having an equal and positive number of a’s and b’s requires more than just a regular expression to recognize. Regular expressions can represent repetition, concatenation, and alternation, but checking for equality of two variable-length parts (a’s and b’s) within the string goes beyond what they can handle.
Therefore, we have:
L U X: Regular because it’s the union of two regular languages.
X: Not regular because it requires a condition that regular expressions cannot capture.
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