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Eliminating ε TransitionsNFA with ε can be converted to NFA without ε, and this NFA without ε can be converted to DFA. To do this, we will use a method, which can remove all the ε transition from given NFA. The method will be:
Example:Convert the following NFA with ε to NFA without ε. ![]() Solutions: We will first obtain ε-closures of q0, q1 and q2 as follows:
ε-closure(q0) = {q0}
ε-closure(q1) = {q1, q2}
ε-closure(q2) = {q2}
Now the δ' transition on each input symbol is obtained as:
δ'(q0, a) = ε-closure(δ(δ^(q0, ε),a))
= ε-closure(δ(ε-closure(q0),a))
= ε-closure(δ(q0, a))
= ε-closure(q1)
= {q1, q2}
δ'(q0, b) = ε-closure(δ(δ^(q0, ε),b))
= ε-closure(δ(ε-closure(q0),b))
= ε-closure(δ(q0, b))
= Ф
Now the δ' transition on q1 is obtained as:
δ'(q1, a) = ε-closure(δ(δ^(q1, ε),a))
= ε-closure(δ(ε-closure(q1),a))
= ε-closure(δ(q1, q2), a)
= ε-closure(δ(q1, a) ∪ δ(q2, a))
= ε-closure(Ф ∪ Ф)
= Ф
δ'(q1, b) = ε-closure(δ(δ^(q1, ε),b))
= ε-closure(δ(ε-closure(q1),b))
= ε-closure(δ(q1, q2), b)
= ε-closure(δ(q1, b) ∪ δ(q2, b))
= ε-closure(Ф ∪ q2)
= {q2}
The δ' transition on q2 is obtained as:
δ'(q2, a) = ε-closure(δ(δ^(q2, ε),a))
= ε-closure(δ(ε-closure(q2),a))
= ε-closure(δ(q2, a))
= ε-closure(Ф)
= Ф
δ'(q2, b) = ε-closure(δ(δ^(q2, ε),b))
= ε-closure(δ(ε-closure(q2),b))
= ε-closure(δ(q2, b))
= ε-closure(q2)
= {q2}
Now we will summarize all the computed δ' transitions:
δ'(q0, a) = {q0, q1}
δ'(q0, b) = Ф
δ'(q1, a) = Ф
δ'(q1, b) = {q2}
δ'(q2, a) = Ф
δ'(q2, b) = {q2}
The transition table can be:
State q1 and q2 become the final state as ε-closure of q1 and q2 contain the final state q2. The NFA can be shown by the following transition diagram:
Next TopicConversion from NFA to DFA
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