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Given an arbitrary non-deterministic finite automaton (NFA) with $N$ states, the maximum number of states in an equivalent minimized DFA at least

1. $N^2$
2. $2^N$
3. $2N$
4. $N!$
edited | 2.9k views

Answer is (B) $2^N$.

In DFA any subset of the $N$ states (for $N$ element set $2^N$ subsets possible)  can become a new state and they can remain even when the DFA is minimized. So, maximum we can get $2^N$ states for the minimized DFA. (at least in question must be a typo for at most).

answered by Loyal (8.2k points)
edited

Example DFA-> 3rs symbol from right is x.

https://gateoverflow.in/544/gate1991_17-b

Here you have to use 2^3 symbols.

nth symbol from right is x , this DFA will have 2^n state.s
answered by Boss (42.5k points)

Answer is B. i.e., 2N

answered by Active (2.6k points)
–1 vote
Ans: B
answered by Loyal (7.1k points)