if NFA has $N$ states then DFA can have $M$ states where $1\leq M \leq 2^N$

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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

- $N^2$
- $2^N$
- $2N$
- $N!$

+31 votes

Best answer

+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

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

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