Regular languages and finite automata

 Question 1
Consider the languages L1 = and L2 = {a}. Which one of the following represents L1 L2* U L1*
 A A B B C C D D
GATE CS 2013    Regular languages and finite automata
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Question 1 Explanation:
L1 L2* U L1* Result of L1 L2* is [Tex]\phi[/Tex]. {[Tex]\phi[/Tex]} indicates an empty language. Concatenation of [Tex]\phi[/Tex] with any other language is [Tex]\phi[/Tex]. It works as 0 in multiplication. L1* = [Tex]\phi[/Tex]* which is {[Tex]\epsilon[/Tex]}. Union of [Tex]\phi[/Tex] and {[Tex]\epsilon[/Tex]} is {[Tex]\epsilon[/Tex]}
 Question 2
Consider the DFA given. Which of the following are FALSE?
1. Complement of L(A) is context-free.
2. L(A) = L((11*0+0)(0 + 1)*0*1*)
3. For the language accepted by A, A is the minimal DFA.
4. A accepts all strings over {0, 1} of length at least 2. 
 A 1 and 3 only B 2 and 4 only C 2 and 3 only D 3 and 4 only
GATE CS 2013    Regular languages and finite automata
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Question 2 Explanation:
1 is true. L(A) is regular, its complement would also be regular. A regular language is also context free. 2 is true. 3 is false, the DFA can be minimized to two states. Where the second state is final state and we reach second state after a 0. 4 is clearly false as the DFA accepts a single 0.
 Question 3
W hat is the complement of the language accepted by the NFA shown below?
 A A B B C C D D
GATE CS 2012    Regular languages and finite automata
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Question 3 Explanation:
The given alphabet contains only one symbol {a} and the given NFA accepts all strings with any number of occurrences of ‘a’. In other words, the NFA accepts a+. Therefore complement of the language accepted by automata is empty string.
 Question 4
Given the language L = {ab, aa, baa}, which of the following strings are in L*?
1) abaabaaabaa
2) aaaabaaaa
3) baaaaabaaaab
4) baaaaabaa 
 A 1, 2 and 3 B 2, 3 and 4 C 1, 2 and 4 D 1, 3 and 4
GATE CS 2012    Regular languages and finite automata
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Question 4 Explanation:
 Question 5
Consider the set of strings on {0,1} in which, every substring of 3 symbols has at most two zeros. For example, 001110 and 011001 are in the language, but 100010 is not. All strings of length less than 3 are also in the language. A partially completed DFA that accepts this language is shown below. The missing arcs in the DFA are
 A A B B C C D D
GATE CS 2012    Regular languages and finite automata
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Question 5 Explanation:
 Question 6
Definition of a language L with alphabet {a} is given as following.
             L={| k>0, and n is a positive integer constant}
What is the minimum number of states needed in DFA to recognize L?
 A k+1 B n+1 C 2^(n+1) D 2^(k+1)
GATE CS 2011    Regular languages and finite automata
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Question 6 Explanation:
 Question 7
A deterministic finite automation (DFA)D with alphabet {a,b} is given below Which of the following finite state machines is a valid minimal DFA which accepts the same language as D?
 A A B B C C D D
GATE CS 2011    Regular languages and finite automata
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Question 7 Explanation:
Options (B) and (C) are invalid because they both accept ‘b’ as a string which is not accepted by give DFA. (D) is invalid because it accepts "bba" which are not accepted by given DFA.
 Question 8
Let w be any string of length n is {0,1}*. Let L be the set of all substrings of w. What is the minimum number of states in a non-deterministic finite automaton that accepts L?
 A n-1 B n C n+1 D 2n-1
GATE CS 2010    Regular languages and finite automata
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Question 8 Explanation:
We need minimum n+1 states to build NFA that accepts all substrings of a binary string. For example, following NFA accepts all substrings of “010″ and it has 4 states.
 Question 9
Which one of the following languages over the alphabet {0,1} is described by the regular expression: (0+1)*0(0+1)*0(0+1)*?
 A The set of all strings containing the substring 00. B The set of all strings containing at most two 0’s. C The set of all strings containing at least two 0’s. D The set of all strings that begin and end with either 0 or 1.
GATE-CS-2009    Regular languages and finite automata
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Question 9 Explanation:
The regular expression has two 0′s surrounded by (0+1)* which means accepted strings must have at least 2 0′s.
 Question 10
Which one of the following is FALSE?
 A There is unique minimal DFA for every regular language B Every NFA can be converted to an equivalent PDA. C Complement of every context-free language is recursive. D Every nondeterministic PDA can be converted to an equivalent deterministic PDA.
GATE-CS-2009    Regular languages and finite automata
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Question 10 Explanation:
Deterministic PDA cannot handle languages or grammars with ambiguity, but NDPDA can handle languages with ambiguity and any context-free grammar. So every nondeterministic PDA can not be converted to an equivalent deterministic PDA.
There are 80 questions to complete.

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