# Regular languages and finite automata

Question 1 |

A | |

B | |

C | |

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 .
{} indicates an empty language. Concatenation of with any other language is . It works as 0 in multiplication.
L1* = * which is {}.
Union of and {} is {}

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.

1 and 3 only | |

2 and 4 only | |

2 and 3 only | |

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 |

Given the language L = {ab, aa, baa}, which of the following strings are in L*?

1) abaabaaabaa 2) aaaabaaaa 3) baaaaabaaaab 4) baaaaabaa

1, 2 and 3 | |

2, 3 and 4 | |

1, 2 and 4 | |

1, 3 and 4 |

**Regular languages and finite automata**

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Question 3 Explanation:

See question 2 of http://www.geeksforgeeks.org/automata-theory-set-2/

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

1, 2 and 3 | |

2, 3 and 4 | |

1, 2 and 4 | |

1, 3 and 4 |

**Regular languages and finite automata**

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Question 4 Explanation:

See question 2 of http://www.geeksforgeeks.org/automata-theory-set-2/

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 | |

B | |

C | |

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?

k+1 | |

n+1 | |

2^(n+1) | |

2^(k+1) |

**GATE CS 2011**

**Regular languages and finite automata**

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Question 6 Explanation:

See Question 3 of http://www.geeksforgeeks.org/automata-theory-set-4/

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 | |

B | |

C | |

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?

n-1 | |

n | |

n+1 | |

2n-1 |

**GATE CS 2010**

**Regular languages and finite automata**

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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)*?

The set of all strings containing the substring 00. | |

The set of all strings containing at most two 0’s. | |

The set of all strings containing at least two 0’s. | |

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?

There is unique minimal DFA for every regular language | |

Every NFA can be converted to an equivalent PDA. | |

Complement of every context-free language is recursive. | |

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:

Power of Deterministic PDA is not same as the power of Non-deterministic PDA. Deterministic PDA cannot handle languages or grammars with ambiguity, but NDPDA can handle languages with ambiguity and any context-free grammar. So every non-deterministic PDA can not be converted to an equivalent deterministic PDA.

There are 153 questions to complete.