P, Q, R are three languages, if P and R are regular and if PQ = R, then
A.  Q has to be regular 
B.  Q cannot be regular 
C.  Q need not be regular 
D.  Q cannot be a CFL 
Option: C Explanation : Click on Discuss to view users comments. KALPANA SINGH said: (10:57pm on Sunday 4th June 2017)
As per decision rule only union, intersection and difference with other language and regular language are regular but no rule for concatenation it must be regular Q
AKASH MAHAJAN said: (11:07pm on Monday 18th September 2017)
by defination A is correct i.e Q has to be regular.Regular Grammer is closed under concatenation property, ie if A

A class of language that is closed under
A.  union and complementation has to be closed under intersection 
B.  intersection and complement has to be closed under union 
C.  union and intersection has to be closed under complementation 
D.  both (A) and (B) 
Option: D Explanation : Click on Discuss to view users comments. 
The productions
E—>E+E
E—>E—E
E>E*E
E —> E / E
E —> id
A.  generate an inherently ambiguous language 
B.  generate an ambiguous language but not inherently so 
C.  are unambiguous 
D.  can generate all possible fixed length valid computation for carrying out addition, subtraction, multipication and division, which can be expressed in one expression 
Option: B Explanation : Click on Discuss to view users comments. PENCHAL said: (7:22am on Monday 20th November 2017)
A CFL language L is inherently ambiguous if every CFG for L is ambiguous.in the problem every CFG L is ambiguous so the correct answer is A

Which of the folowing definitions below generates the same language as L, where
L = {xn yn such that n > = 1} ?
I. E —> xEy  xy
II. xy  (x+ xyy+)
III .x+y+
A.  I only 
B.  I and II 
C.  II and III 
D.  II only 
Option: A Explanation : Click on Discuss to view users comments. 
Following context free grammar
S —> aB  bA
A —>b  aS  bAA
B —> b  bS  aBB
generates strings of terminals that have
A.  equal number of a's and b's 
B.  odd number of a's and odd number b's 
C.  even number of a's and even number of b's 
D.  odd number of a's and even number of a's 
Option: A Explanation : Click on Discuss to view users comments. manjureka said: (2:12pm on Friday 12th June 2015)
Option A also wrong..why because the string aB>abs>abbA>abbbAA>abbbbb it may confuse please give me the clear solution for me.....
TJ said: (5:48pm on Wednesday 7th October 2015)
the given answer is incorrect. and for the above answer A, grammer must be A>aaSbAA and for S and B same as above. then it will work.
anubhav said: (1:37am on Thursday 30th June 2016)
option A is wrong as this grammar is accepting "bb" in which no. of a's and b's are not equal...
mozhi said: (4:33pm on Thursday 4th May 2017)
Suppose we derive llike S â€”> bA

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