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Give an example problem of converting NFA to DFA.
Problem
Consider a Non-deterministic finite automata (NFA) and convert that NFA into equivalent Deterministic Finite Automata (DFA).
Solution
Let’s construct NFA transition table for the given diagram −
States\inputs | a | b |
---|---|---|
->q0 | {q0,q1} | q0 |
q1 | q2 | q1 |
q2 | q3 | q3 |
q3 | - | q2 |
DFA cannot have multiple states. So, consider {q0,q1} as a single state while constructing DFA.
Let’s convert the above table into equivalent DFA
States\inputs | a | b |
---|---|---|
->q1 | [q0,q1] | q0 |
[q0,q1] | [q0q1q2] | [q0q1] |
*[q0q1q2] | [q0q1q2q3] | [q0q1q3] |
*[q0q1q2q3] | [q0q1q2q3] | [q0q1q2q3] |
*[q0q1q3] | [q0q1q2] | [q0q1q2] |
In DFA the final states are q2 and q3, wherever q2 and q3 are present that state becomes a final state.
Now the transition diagram for DFA is as follows −
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