Question

Given language L = (0+1)*0^n, define a DFA with as few states as possible with n > 0.

Comments

where you stucked while solving it?

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minimum 2 states needed 

 

Answer 1

State = 1

(0+1)*0* = (0+1)* so minimizedDFA has 1 state

Comments

The state can be 2.
Because the given language is (0+1)*0^n, here n>0
So, n can be 1,2,3,4
let take n =1;

the first smaller string can be 0^1 which is equal to 0.

Please correct me if I am wrong.

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 thedrift611, your regular expression accept $\epsilon$ also but in given regular expression $n>0$ 

r0h17, $L=(0+1)^*0^n/n>0$. this can be written as $(0+1)^*0^+$

minimum length of string is $0$.

strings ending with 0’s. $L=(0,00,000…,10,110,1110..010,00110,001100...\infty)$

invalid strings are: $(\epsilon,1,11,111,….,01,011,0011,...\infty)$