Regular languages and Finite automata

Question 1
The length of the shortest string NOT in the language (over Σ = {a b,} of the following regular is expression is ______________. a*b*(ba)*a*
A
3
B
4
C
5
D
6
       Theory-of-Computation       Regular Languages and Finite Automata       Gate 2014 Set -03
Question 1 Explanation: 
The regular expression generate all the strings of length 0 , 1 and 2
{ϵ, a, b, aa, ab, ba, bb}
Let’s check all the string of length 3.
The given regular expression generates {aaa, aab, aba, abb, baa, bba, bbb}
But it doesn’t generate the string “bab”, hence the shortest string not generated by regular expression has length 3 (string “bab”).
Question 2
Consider the languages L1 = ϕ and L= {a}. Which one of the following represents L1L2* ∪ L1*?
A
{є}
B
ϕ
C
a*
D
{є,a}
       Algorithms       Regular languages and Finite automata       Gate 2013
Question 2 Explanation: 
As we know, for any regular expression R,
Rϕ = ϕR=ϕ
So L1 L2 * = ϕ
and L1 * = {ϕ}* = {ϵ}
So L1L2* ∪ L1* = {ϵ}
Question 3
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
       Theory-of-Computation       Regular Languages and Finite Automata       Gate 2013
Question 3 Explanation: 
L(A) is regular and its complement is also regular (by closure property) and every regular is CFL also. So Complement of LA is context-free.
The regular expression corresponding to the given FA is

Hence we have regular expression: (11*0 +0) (0+1)*
Since we have (0+1)* at the end so if we write 0*1* after this it will not have any effect, the reason is whenever string ends with the terminals other than 1*0* there we can assume 1*0* as epsilon.
So it is equivalent to (11*0 +0) (0+1)*0*1*
The given DFA can be minimised, since the non-final states are equivalent and can be merged and the min DFA will have two states which is given below:

Hence statement 3 is false.
Since DFA accept string “0” whose length is one, so the statement “A accepts all strings over {0, 1} of length at least 2” is false statement.
Question 4
Let L be a regular language and M be a context-free language, both over the alphabet Σ. Let Lc and Mc denote the complements of L and M respectively. Which of the following statements about the language if Lc ∪ Mc is TRUE?
A
It is necessarily regular but not necessarily context-free.
B
It is necessarily context-free.
C
It is necessarily non-regular.
D
None of the above.
       Theory-of-Computation       Regular Languages and finite automata       Gate 2005-IT
Question 4 Explanation: 
Context-free languages not closed under complementation. So, Lc ∪ Mc is neither regular nor context-free. It might be context sensitive language.
Question 5
Which of the following statements is TRUE about the regular expression 01*0?
A
It represents a finite set of finite strings.
B
It represents an infinite set of finite strings.
C
It represents a finite set of infinite strings.
D
It represents an infinite set of infinite strings.
       Theory-of-Computation       Regular languages and finite automata       Gate 2005-IT
Question 5 Explanation: 
The given expression01*0 is regular. So this is a finite string. So options C and D are false and * is placed. So this is infinite set.
So, given regular expression represents an infinite set of finite strings.
Question 6
The language {0n 1n 2n | 1 ≤ n ≤ 106} is
A
regular
B
context-free but not regular
C
context-free but its complement is not context-free
D
not context-free
       Theory-of-Computation       Regular languages and finite automata       Gate 2005-IT
Question 6 Explanation: 
In this the value of n is finite then we can be able to construct a finite state automata for this language.
So, given language is regular.
Question 7

Consider the non-deterministic finite automaton (NFA) shown in the figure.

State X is the starting state of the automaton. Let the language accepted by the NFA with Y as the only accepting state be L1. Similarly, let the language accepted by the NFA with Z as the only accepting state be L2. Which of the following statements about L1 and L2 is TRUE? Correction in Question: There is an edge from Z->Y labeled 0 and another edge from Y->Z labeled 1 - in place of double arrowed and no arrowed edges.

A
L1 = L2
B
L1 ⊂ L2
C
L2 ⊂ L1
D
None of the above
       Theory-of-Computation       Regular Languages and Finite Automata       Gate 2005-IT
Question 7 Explanation: 
Based on Arden's theorem write the Y and Z in terms of incoming arrows,
Y = X0 + Y0 + Z1
Z = X0 + Y0 + Z;
⇒ X = Z;
⇒ L1 = L2
Question 8
Consider the context-free grammar E → E + E E → (E * E) E → id
where E is the starting symbol, the set of terminals is {id, (,+,),*}, and the set of nonterminals is {E}.
Which of the following terminal strings has more than one parse tree when parsed according to the above grammar?
A
id + id + id + id
B
id + (id* (id * id))
C
(id* (id * id)) + id
D
((id * id + id) * id)
       Theory-of-Computation       Regular languages and finite automata       Gate 2005-IT
Question 8 Explanation: 
Let's draw more than one possible tree for id + id + id + id.
There are 8 questions to complete.
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