What is the accepting state for this finite automaton ?

Question

What is the complement of given DFA accepting :?

What is the Regular expression for this FA ?

Comments

complement of this

RE - (10+11+0)(0+1)* +1

which is (0+1)⁺

Answer 1

Its accepting string is only { Eps }

Compliment of it is

(1+0+1 (0+1)) (0+1)*  =  (1+0+(10+11)) (0+1)*

Regular expression accepted by it

1(0+1)* +0(0+1)* + (10+11)(0+1)* =(0+1)+ 

So accepting everything other than {Eps}  over alphabet {0,1}

(1+0+1(0+1))(0+1)∗=(1+0+(10+11))(0+1)∗(1+0+1(0+1))(0+1)∗=(1+0+(10+11))(0+1)∗(1+0+1(0+1))(0+1)∗=(1+0+(10+11))(0+1)∗

(1+0+1(0+1))(0+1)∗=(1+0+(10+11))(0+1)∗

(1+0+1(0+1))(0+1)∗=(1+0+(10+11))(0+1)∗

Answer 2

Complement of the DFA is , it will accept  all string except the є(null string). i.e r.e = (0+1)^{+
B and C will be the final state} 
 

Regular expression of this DFA - r.e = є (i.e null string)

Answer 3

it is only accepting epsilon

let L₁={∊},then

complement of L₁     is  L₂ = (∑^*-L1) = (a + b)⁺  

complement of given DFA accepting   L₂  = (a + b)⁺