The CFG
s---> as | bs | a | b
is equivalent to regular expression
A. | (a + b) |
B. | (a + b) (a + b)* |
C. | (a + b) (a + b) |
D. | None of these |
Option: B Explanation : Click on Discuss to view users comments. |
Consider the grammar :
S —> ABCc | Abc
BA —> AB
Bb —> bb
Ab —> ab
Aa —> aa
Which of the following sentences can be derived by this grammar
A. | abc |
B. | aab |
C. | abcc |
D. | abbb |
Option: A Explanation : Click on Discuss to view users comments. DIVYA said: (4:33pm on Monday 3rd October 2016)
S->Abc,s->abc so the answer is (A),S->ABCc,it is not non terminal terms in grammar
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Pumping lemma is generally used for proving that
A. | given grammar is regular |
B. | given grammar is not regular |
C. | whether two given regular expressions are equivalent or not |
D. | None of these |
Option: B Explanation : Click on Discuss to view users comments. |
The language of all words with at least 2 a's can be described by the regular expression
A. | (ab)*a and a (ba)* |
B. | (a + b)* ab* a (a + b)* |
C. | b* ab* a (a + b)* |
D. | all of these |
Option: D Explanation : Click on Discuss to view users comments. |
Any string of terminals that can be generated by the following CFG is
S-> XY
X--> aX | bX | a
Y-> Ya | Yb | a
A. | has atleast one 'b' |
B. | should end in a 'a' |
C. | has no consecutive a's or b's |
D. | has atleast two a's |
Option: D Explanation : Click on Discuss to view users comments. samra said: (11:33am on Saturday 9th June 2018)
what will be regular expression for this cfg?
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