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QUESTION: 1

According to the given transitions, which among the following are the epsilon closures of q1 for the given NFA?

Δ (q1, ε) = {q2, q3, q4}

Δ (q4, 1) =q1

Δ (q1, ε) =q1

Solution:

The set of states which can be reached from q using ε-transitions, is called the ε-closure over state q.

QUESTION: 2

State true or false?Statement: An NFA can be modified to allow transition without input alphabets, along with one or more transitions on input symbols.

Solution:

It is possible to construct an NFA with ε-transitions, presence of no input symbols, and that is called NFA with ε-moves.

QUESTION: 3

State true or false?Statement: ε (Input) does not appears on Input tape.

Solution:

ε does not appears on Input tape, ε transition means a transition without scanning a symbol i.e. without moving the read head.

QUESTION: 4

Statement 1: ε- transition can be called as hidden non-determinism.

Statement 2: δ (q, ε) = p means from q it can jump to p with a shift in read head.

Which among the following options is correct?

Solution:

The transition with ε leads to a jump but without any shift in read head. Further, the method can be called one to introduce hidden non-determinism.

QUESTION: 5

Δ (q1, ε) = {q2, q3, q4}

Δ (q4, 1) =q1

Δ (q1, ε) =q1

Solution:

The set of states which can be reached from q using ε-transitions, is called the ε-closure over state q.

QUESTION: 6

Predict the total number of final states after removing the ε-moves from the given NFA?

Solution:

The NFA which would result after eliminating ε-moves can be shown diagramatically.

QUESTION: 7

For NFA with ε-moves, which among the following is correct?

Solution:

Due to the presence of ε symbol, or rather an epsilon-move, the input alphabets unites with it to form a set including ε.

QUESTION: 8

Which among the following is false?

ε-closure of a subset S of Q is:

Solution:

All the mentioned are the closure properties of ε and encircles all the elements if it satisfies the following options:

a) Every element of S ϵ Q

b) For any q ϵ ε(S), every element of δ (q, ε) is in ε(S)

c) No other element is in ε(S)

QUESTION: 9

The automaton which allows transformation to a new state without consuming any input symbols:

Solution:

NFA-l or e-NFA is an extension of Non deterministic Finite Automata which are usually called NFA with epsilon moves or lambda transitions.

QUESTION: 10

e-transitions are

Solution:

An epsilon move is a transition from one state to another that doesnt require any specific condition.

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