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.  contextfree language 
C.  contextsensitive language 
D.  recursively enumerable language 
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.  contextsensitive language 
D.  contextfree language 
Option: B Explanation : Click on Discuss to view users comments. 
The logic of pumping lemma is a good example of
A.  pigeonhole principle 
B.  divideandconquer technique 
C.  recursion 
D.  iteration 
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 
Option: B Explanation : Click on Discuss to view users comments. Jaya said: (3:01am on Saturday 2nd February 2013)
I have a doubt that the correct answer should be B is always Context free
ads said: (12:06pm on Friday 18th October 2013)
answer is D because if {fae}intersection{a*b*}so answer is regular
ads said: (12:07pm on Friday 18th October 2013)
answer is D because if {fae}intersection{a*b*}so answer is regular

For two regular languages
L_{1} = (a + b)* a and L_{2} = b (a + b ) *
,
the intersection of L_{1} and L_{2} is given by
A.  (a + b ) * ab 
B.  ab (a + b ) * 
C.  a ( a + b ) * b 
D.  b (a + b ) * a 
Option: D Explanation : Click on Discuss to view users comments. ritesh kumar said: (2:17am on Friday 8th February 2013)
how it can be done
