Regular languages and finite automata

Question 1

Consider the languages L1 = \\phi and L2 = {a}. Which one of the following represents L1 L2* U L1*

 
Tick

{\epsilon }

Cross

\phi

Cross

a*

Cross

{\epsilon ,a}



Question 1-Explanation: 

L1 L2* U L1* Result of L1 L2* is \\phi . {\\phi } indicates an empty language. Concatenation of \\phi with any other language is \\phi . It works as 0 in multiplication. L1* = \\phi * which is {\\epsilon }. Union of \\phi and {\\epsilon } is {\\epsilon }

Question 2

Consider the DFA given. 

 

 Which of the following are FALSE?

Cross

Complement of L(A) is context-free.

Cross

L(A) = L((11*0+0)(0 + 1)*0*1*).

Tick

For the language accepted by A, A is the minimal DFA.

Tick

A accepts all strings over {0, 1} of length at least 2.



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*?

Tick

abaabaaabaa

Cross

aaaabaaaa

Tick

baaaaabaaaab

Tick

 baaaaabaa



Question 3-Explanation: 

 

Question 4

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

Tick

abaabaaabaa

Cross

aaaabaaaa

Tick

baaaaabaaaab

Tick

 baaaaabaa



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 

Cross

 

Cross

 

Cross

 

Tick

 



Question 6

Definition of a language L with alphabet {a} is given as following.

             L={|ank|k>0, and n is a positive integer constant}

What is the minimum number of states needed in DFA to recognize L?

Cross

k+1

Tick

n+1

Cross

2(n+1)

Cross

2(k+1)



Question 6-Explanation: 

In the language L = {|an| k > 0, and n is a positive integer constant}, the DFA needs to keep track of two things: the presence of at least one 'a' at the beginning and the count of subsequent 'a's.
To represent the count of subsequent 'a's, we can use a state for each possible value of n. Since n is a positive integer constant, we can have n possible values for subsequent 'a's.
Additionally, we need an initial state to represent the start of the string. This initial state transitions to the state representing the count of subsequent 'a's upon reading 'a'.

Therefore, the total number of states required in the DFA is n+1, where n is the positive integer constant in the language definition.

Question 7
A deterministic finite automation (DFA)D with alphabet {a,b} is given below GATE2011AT1 Which of the following finite state machines is a valid minimal DFA which accepts the same language as D? GATE2011AT2 GATE2011AT3
Tick
A
Cross
B
Cross
C
Cross
D


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?

Cross

n-1

Cross

n

Tick

n+1

Cross

2n-1



Question 8-Explanation: 

 

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)* ?
Cross
The set of all strings containing the substring 00.
Cross
The set of all strings containing at most two 0’s.
Tick
The set of all strings containing at least two 0’s.
Cross
The set of all strings that begin and end with either 0 or 1.


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?

Cross

There is unique minimal DFA for every regular language

Cross

Every NFA can be converted to an equivalent PDA.

Cross

Complement of every context-free language is recursive.

Tick

Every nondeterministic PDA can be converted to an equivalent deterministic PDA.



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.

  • Last Updated : 06 Oct, 2021

Share your thoughts in the comments
Similar Reads