NFA to DFA Revision as of 22:53, 26 July 2014 by Arjun Suresh (talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Conversion

• Q: I know that $2^n$ are the maximum states in DFA for an n$$ state NFA. But what about the minimum no of states in a DFA for an $n$ state NFA?

A: I would say 1 because the $n$ states can be connected through $ε$ moves and hence all of these will be combines to a single start state in DFA.

• Q: After converting from NFA to DFA and we get $2^n$ states. Now we apply minimization- how many states can we get after this?

In the worst case this can remain $2^n$ - that is no minimization can be possible. We can see this in the following example

<graphviz caption='NFA'> digraph NFA { rankdir=LR; node [shape = none] ""; node [shape = doublecircle] q1; node [shape = circle];

""-> q0; q0->q1 [label="a"]; q1->q0 [label="a"]; q1->q1 [label="a"]; q1 -> q0 [label="b"]; }

<graphviz caption='DFA'> digraph DFA { rankdir=LR; node [shape = none] ""; node [shape = doublecircle] q1, q12; node [shape = circle];

""-> q0; q0->q1 [label="a"]; q1->q0 [label="a"]; q1->q1 [label="a"]; q1 -> q0 [label="b"]; }