Automata Theory | Set 6

4

Following questions have been asked in GATE CS 2010 exam.

1) Let L={w ∈ (0 + 1)*|w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expression below represents L?
(A) (0*10*1)*
(B) 0*(10*10*)*
(C) 0*(10*1*)*0*
(D) 0*1(10*1)*10*

Answer (B)
Option (A) is incorrect because it cannot accept “110”
Option (C) is incorrect because it accept a string with single 1.
Option (D) is incorrect because it cannot accept 11101



2) Let L1 be a recursive language. Let L2 and L3 be languages that are recursively enumerable but not recursive. Which of the following statements is not necessarily true?
(A) L2 – L1 is recursively enumerable.
(B) L1 – L3 is recursively enumberable
(C) L2 ∩ L1 is recursively enumberable
(D) L2 ∪ L1 is recursively enumberable

Answer (B)



3) Consider the languages L1={0i1j | i != j}, L2={0i1j | i = j}, L3 = {0i1j | i = 2j+1}, L4 = {0i1j | i != 2j}. Which one of the following statements is true?
(A) Only L2 is context free
(B) Only L2 and L3 are context free
(C) Only L1 and L2 are context free
(D) All are context free

Answer (D)
A Pushdown Automata can be built for all four languages.



4) 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

Answer (C)
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.

NFA that accepts all substrings of 010

NFA that accepts all substrings of 010, the substrings are: 0, 01, 10 and 010

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