Let L1 be the language represented by the regular expression b*ab*(ab*ab*)* and L2 = {w in (a+b)* |

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GATE-CS is a competitive examination conducted in India. This question is from the 2024 GATE-CS Theory of Computation paper.

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Question

Let L1 be the language represented by the regular expression b*ab*(ab*ab*)* and L2 = {w in (a+b)* | |w|

Accepted Answer

Answer: 15Concept -- Regular Languages and String Counting: L1 = b*ab*(ab*ab*)* is the language of strings with an odd number of 'a's (strings over {a,b} where the count of 'a' is odd). This follows because b*ab* matches strings with exactly 1 'a', and (ab*ab*)* appends pairs of 'a's (even number), so the total count of 'a's remains odd.L2 contains all strings of length 0 to 4. Total strings in L2: 2^0 + 2^1 + 2^2 + 2^3 + 2^4 = 1 + 2 + 4 + 8 + 16 = 31. Strings in L2 with even number of 'a's (including zero 'a's): by symmetry, half of all strings have even count and half have odd count of 'a's. But the empty string has 0 'a's (even). Total strings = 31 (odd number). Even 'a's count: 16. Odd 'a's count: 15. Therefore |L1 intersection L2| = 15.

Let L1 be the language represented by the regular expression bab(abab)* and L2 = {w in (a+b)* | |w|

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Let L1 be the language represented by the regular expression bab(abab)* and L2 = {w in (a+b)* | |w|

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