Number of DFA's (Made easy test series)

Question

The number of DFA's with four states which can be constructed of the alphabet $\Sigma = \{ a,b \}$ with a designated initial state are $2^n$, then the value of n is _____.

 

IN DFA IT IS COMPULSORY TO HAVE 1 FINAL STATE.

 

4c0 should not be taken,correct me?

Comments

you will take every possibilties since it is not mentioned that what the given dfa will going to accept[universal language or empty language etc].

use your intuitions in such kind of questions.

you may use the formula for only such type of questions    NOS=        2ⁿ   *  n^m*n

n is no of states ,m is no of alphabets.

in question n=4 m=2

2⁴ * 4^4*2    =2²⁰=2ⁿ

n=20

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From every state you can have 4 transition for each input symbol => 4*4 transition possible for 1 state

There are such 4 states Hence, 4*4 * 4*4 * 4*4 * 4*4 = 4^8 = 2^16

from 4 states we need to select final states so it will be power set of 4 states, so the no. of ways possible = 2^4

Total = 2^16*2^4 = 2^20

A power set includes empty element which denotes number of final state so there is no need to have compulsory final state.

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@adarsh_1997Is this formula applicable for all cases??

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no,it is applicable for few models as mentioned in questions

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@twin_123 ,

please $\underline{\color{red}{add}\text{ test-series name in }\color{red}{title}}$

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No,Not at all ,dfa doesn’t neccesarily need to have final state a dfa without final state accept a phi(null language) .

Answer 1

set of final states can be null .

Answer 2

No,Not at all ,dfa doesn’t neccesarily need to have final state a dfa without final state accept a phi(null language) .