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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. $N!$
in Theory of Computation by Veteran (52.1k points)
edited by | 4.3k views
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if NFA has $N$ states then DFA can have  $M$ states where $1\leq M \leq 2^N$

4 Answers

+27 votes
Best answer

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).

by Loyal (8.1k points)
edited by
0

In DFA any subset of the N states

 This should be "In NFA any subset of the N states..."

+9 votes
Answer is 2^N.

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
by Boss (41.4k points)
0 votes

Answer is B. i.e., 2N

by Active (2.4k points)
0 votes
Ans: B
by Loyal (7.2k points)
Answer:

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