Regular languages and finite automata
Question 1 |
A | |
B | |
C | |
D |
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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 |
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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 |
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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 |
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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 |
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Question 5 Explanation:
Question 6 |
Definition of a language L with alphabet {a} is given as following.
L={What is the minimum number of states needed in DFA to recognize L?| k>0, and n is a positive integer constant}
k+1 | |
n+1 | |
2^(n+1) | |
2^(k+1) |
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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 |
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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 |
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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. |
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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.
The least possible string is ε 0 ε 0 ε = 00
The set of strings accepted is = { 00, 000, 100, 0010, 0000, 00100, 1001001,.....}
We can see from the set of accepted strings that all of the have at least two zeros which is the least possible string.
So option (C) is correct.
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. |
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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 155 questions to complete.