L = (an bn an | n = 1,2,3) is an example of a language that is
A. | context free |
B. | not context free |
C. | not context free but whose complement is CF |
D. | both (b) and (c) |
Option: D Explanation : Click on Discuss to view users comments. nirosa1234@gmail.com said: (10:02pm on Saturday 24th August 2013)
it is not regular so it is context free only
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If Σ = (0, 1), L = Σ* and R = (0n 1nsuch that n > 0 )
then languages L ∪ R and R respectively are
A. | Regular, Regular |
B. | Regular, Not regular |
C. | Not regular, Not regular |
D. | None of these |
Option: B Explanation : Click on Discuss to view users comments. Nancy said: (3:48am on Thursday 15th August 2013)
L is regular and R is Context free which is not regular. Regular U CFG => Regular
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FSM can recognize
A. | any grammar |
B. | only CG |
C. | Both (a) and ( b ) |
D. | only regular grammar |
Option: D Explanation : Click on Discuss to view users comments. |
Set of regular languages over a given alphabet set is closed under
A. | union |
B. | complementation |
C. | intersection |
D. | All of these |
Option: D Explanation : Click on Discuss to view users comments. gunjan said: (2:51pm on Tuesday 22nd October 2013)
regular languages are closed under union,intersection.
kl said: (2:50pm on Tuesday 12th November 2013)
i have read closer properties of regular language.these are union,intersection complement ,homomorphic,kleen closer
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Which of the following statement is correct?
A. | All languages can not be generated by CFG |
B. | Any regular language has an equivalent CFG |
C. | Some non regular languages can't be generated by CFG |
D. | both (b) and (c) |
Option: D Explanation : CFG is a higher than regular language. So we can draw a regular equivalent to CFG. And some non regular like context sensitive can't be generated by cfg. So, option 3 and 2 are correct. So answer is 'D'. Click on Discuss to view users comments. anita said: (2:09pm on Tuesday 2nd May 2017)
option a should be also correct because all language can not be come into cfg
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