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CONVERSION FROM NFA TO DFA
In NFA, when a specific input is given to the current state, the machine goes to multiple states. It can
have zero, one or more than one move on a given input symbol. On the other hand, in DFA, when a
specific input is given to the current state, the machine goes to only one state. DFA has only one move
on a given input symbol.
Let, M = (Q, ∑, δ, q0, F) is an NFA which accepts the language L(M). There should be equivalent
DFA denoted by M' = (Q', ∑', q0', δ', F') such that L(M) = L(M').
Steps for converting NFA to DFA:
Step 1: Initially Q' = ϕ
Step 2: Add q0 of NFA to Q'. Then find the transitions from this start state.
Step 3: In Q', find the possible set of states for each input symbol. If this set of states is not in Q', then
add it to Q'.
Step 4: In DFA, the final state will be all the states which contain F(final states of NFA)
Example 1:
Convert the given NFA to DFA.
Solution: For the given transition diagram we will first construct the transition table.
State 0 1
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→q0 q0 q1
q1 {q1, q2} q1
*q2 q2 {q1, q2}
Now we will obtain δ' transition for state q0.
1. δ'([q0], 0) = [q0]
2. δ'([q0], 1) = [q1]
The δ' transition for state q1 is obtained as:
1. δ'([q1], 0) = [q1, q2] (new state generated)
2. δ'([q1], 1) = [q1]
The δ' transition for state q2 is obtained as:
1. δ'([q2], 0) = [q2]
2. δ'([q2], 1) = [q1, q2]
Now we will obtain δ' transition on [q1, q2].
1. δ'([q1, q2], 0) = δ(q1, 0) ∪ δ(q2, 0)
2. = {q1, q2} ∪ {q2}
3. = [q1, q2]
4. δ'([q1, q2], 1) = δ(q1, 1) ∪ δ(q2, 1)
5. = {q1} ∪ {q1, q2}
6. = {q1, q2}
7. = [q1, q2]
The state [q1, q2] is the final state as well because it contains a final state q2. The transition table for
the constructed DFA will be:
State 0 1
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→[q0] [q0] [q1]
[q1] [q1, q2] [q1]
*[q2] [q2] [q1, q2]
*[q1, q2] [q1, q2] [q1, q2]
The Transition diagram will be:
The state q2 can be eliminated because q2 is an unreachable state.
Example 2:
Convert the given NFA to DFA.
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Solution: For the given transition diagram we will first construct the transition table.
State 0 1
→q0 {q0, q1} {q1}
*q1 Φ {q0, q1}
Now we will obtain δ' transition for state q0.
1. δ'([q0], 0) = {q0, q1}
2. = [q0, q1] (new state generated)
3. δ'([q0], 1) = {q1} = [q1]
The δ' transition for state q1 is obtained as:
1. δ'([q1], 0) = ϕ
2. δ'([q1], 1) = [q0, q1]
Now we will obtain δ' transition on [q0, q1].
1. δ'([q0, q1], 0) = δ(q0, 0) ∪ δ(q1, 0)
2. = {q0, q1} ∪ ϕ
3. = {q0, q1}
4. = [q0, q1]
Similarly,
1. δ'([q0, q1], 1) = δ(q0, 1) ∪ δ(q1, 1)
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2. = {q1} ∪ {q0, q1}
3. = {q0, q1}
4. = [q0, q1]
As in the given NFA, q1 is a final state, then in DFA wherever, q1 exists that state becomes a final
state. Hence in the DFA, final states are [q1] and [q0, q1]. Therefore set of final states F = {[q1], [q0,
q1]}.
The transition table for the constructed DFA will be:
State 0 1
→[q0] [q0, q1] [q1]
*[q1] ϕ [q0, q1]
*[q0, q1] [q0, q1] [q0, q1]
The Transition diagram will be:
Even we can change the name of the states of DFA.
Suppose
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1. A = [q0]
2. B = [q1]
3. C = [q0, q1]
With these new names the DFA will be as follows:
