Let L be a language recognizable by a finite automaton. The language
REVERSE (L) = {w such that w is the reverse of v where v ∈ L } is a
| A. | regular language |
| B. | context-free language |
| C. | context-sensitive language |
| D. | recursively enumerable language |
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Option: A Explanation : Click on Discuss to view users comments. |
The grammars G = ( { s }, { 0, 1 }, p , s)
where p = (s —> 0S1, S —> OS, S —> S1, S —>0} is a
| A. | recursively enumerable language |
| B. | regular language |
| C. | context-sensitive language |
| D. | context-free language |
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Option: B Explanation : Click on Discuss to view users comments. |
The logic of pumping lemma is a good example of
| A. | pigeon-hole principle |
| B. | divide-and-conquer technique |
| C. | recursion |
| D. | iteration |
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Option: A Explanation :
The pigeon hole principle is nothing more than the obvious remark: if you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon. It is surprising how useful this can be as a proof strategy.
In the theory of formal languages in computability theory, a pumping lemma or pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language. The proofs of these lemmas typically require counting arguments such as the pigeonhole principle. So the answer is 'A'
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The intersection of CFL and regular language
| A. | is always regular |
| B. | is always context free |
| C. | both (a) and (b) |
| D. | need not be regular |
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Option: B Explanation : Click on Discuss to view users comments. Jaya said: (6:01pm on Friday 1st February 2013)
I have a doubt that the correct answer should be B- is always Context free
ads said: (2:06am on Friday 18th October 2013)
answer is D because if {fae}intersection{a*b*}so answer is regular
ads said: (2:07am on Friday 18th October 2013)
answer is D because if {fae}intersection{a*b*}so answer is regular
imran said: (8:36pm on Sunday 24th December 2017)
if we have intersection of set of A ( being regular) with B ( being a CF) then result will always be empty or finite or at maximum A.. in every case the result will be a regular language. so the correct option seems D. for example intersection of (a b)* and EQUAL will be language Equal, hence a CF language. also by definition every language which is Regular is also CF by definition. hence option A, C can never occur. so correct option is B only. Plz give some example that result can be a non Context Free language to advocate that correct option as D
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