Let X = (Qx, ∑, δx, q0, Fx) be an NDFA which accepts the language L(X). We have to design an equivalent DFA Y = (Qy, ∑, δy, q0, Fy) such that L(Y) = L(X). The following procedure converts the NDFA to its equivalent DFA −
Input − An NDFA
Output − An equivalent DFA
Step 1 − Create state table from the given NDFA.
Step 2 − Create a blank state table under possible input alphabets for the equivalent DFA.
Step 3 − Mark the start state of the DFA by q0 (Same as the NDFA).
Step 4 − Find out the combination of States {Q0, Q1,... , Qn} for each possible input alphabet.
Step 5 − Each time we generate a new DFA state under the input alphabet columns, we have to apply step 4 again, otherwise go to step 6.
Step 6 − The states which contain any of the final states of the NDFA are the final states of the equivalent DFA.
Let us consider the NDFA shown in the figure below.
q | δ(q,0) | δ(q,1) |
---|---|---|
a | {a,b,c,d,e} | {d,e} |
b | {c} | {e} |
c | ∅ | {b} |
d | {e} | ∅ |
e | ∅ | ∅ |
Using the above algorithm, we find its equivalent DFA. The state table of the DFA is shown in below.
q | δ(q,0) | δ(q,1) |
---|---|---|
[a] | [a,b,c,d,e] | [d,e] |
[a,b,c,d,e] | [a,b,c,d,e] | [b,d,e] |
[d,e] | [e] | ∅ |
[b,d,e] | [c,e] | [e] |
[e] | ∅ | ∅ |
[c, e] | ∅ | [b] |
[b] | [c] | [e] |
[c] | ∅ | [b] |
The state diagram of the DFA is as follows −