Need Regular Expression for Finite Automata: Even number of 1s and Even number of 0s

How to write regular expression for a DFA using Arden theorem

Lets instead of language symbols 0,1 we take Σ = {a, b} and following is new DFA.

Notice start state is Q₀

You have not given but In my answer initial state is Q₀, Where final state is also Q₀.

Language accepted by is DFA is set of all strings consist of symbol a and b where number of symbol a and b are even (including Λ).

Some example strings are {Λ, aa, bb, abba, babbab }, there is no constraint of order and patter of appearance of symbol just both should be even number of time.
note: Λ is allowed because numberOf(a) and numberOf(b) is zero that is even.

As I said in my lined answer: How to write regular expression for a DFA every state stores some information. Below is a what information is stored in each state in above DFA.

Q₀: Even number of a and even number of b
Q₁: Odd number of a and even number of b
Q₂: Odd number of a and odd number of b
Q₃: Even number of a and odd number of b

(You can make DFAs for more interesting languages by changing set of final sates)
_{One should read the lined answer, because my approach to fined RE for DFA in both answer is different}

What is a Regular Expression?
The approach is explained below using Arden's Theorem, applicable on a transition diagram in which there is a single start state and no null move defined (our DFA is in this form). The technique is explained in a book: Formal Languages And Automata Theory

Remember 4.2 ARDEN THEOREM:

Let B and C be are two Regular Expressions over Σ. If C does not contain Λ, then for the equation A = B + AC has a unique (one and only one) solution A = BC*.

[Solution]:

Step-1: Write initial equation, one equation for corresponding to each state in DFA. This equation means how a state can be reach in a single step

So according to our DFA following 4-equations are possible:

1. Q₀ = Λ + Q₁a + Q₃b
2. Q₁ = Q₀a + Q₂b
3. Q₂ = Q₁b + Q₃a
4. Q₃ = Q₀b + Q₂a

In equation (1) extra Λ is because Q₀ is initial state, can be reached without any input (a point of start). Because Q₀ is also only a final state, a string consist of a, b is acceptable if it ends at Q₀. Value of Q₀ will give us required regular expression so our target is to simply equation-(1) in terms of a, b.

Step-2: Simplify equation using by putting value of states from other equations and using Arden's simplification equation.

Lets we first take equation-(4) and replace value of Q₂ from equation-(3).

Q₃ = Q₀b + Q₂a
Q₃ = Q₀b + (Q₁b + Q₃a) a
Q₃ = Q₀b + Q₁ba + Q₃aa

The last equation can be view in the form of Arden's equation A = B + AC. Where A is Q₃, B = Q₀b + Q₁ba and C = aa. So according to Arden's therm, equation Q₃ = Q₀b + Q₁ba + Q₃aa has a unique solution that is:

Q₃ = (Q₀b + Q₁ba)(aa)*

Or one can write this as follows:

5. Q₃ = Q₀b(aa)* + Q₁ba(aa)*

Logically you can check/understand eq-(5) means Q₃ can be reached in two ways (+) fist from by applying b on Q₀ then there is a loop with label aa on Q₃, second way is from Q₁ with application of ba.

In similar ways, we can simplify equation-(2)

Q₁ = Q₀a + Q₂b
Q₁ = Q₀a + (Q₁b + Q₃a)b
Q₁ = Q₀a + Q₁bb + Q₃ab

Use Arden's simplification rules here.

Q₁ = (Q₀a + Q₃ab)(bb)*

further simplify

6. Q₁ = Q₀a(bb)* + Q₃ab(bb)*

Now value of Q₃ from equation-(5) into equation-(6)

Q₁ = Q₀a(bb)* + (Q₀b(aa)* + Q₁ba(aa)* )ab(bb)*
Q₁ = Q₀a(bb)* + Q₀b(aa)* ab(bb)* + Q₁ba(aa)* ab(bb)*

Again improve this last equation using Arden law of simplification.

Q₁ = (Q₀a(bb)* + Q₀b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )*

take Q₀ conman:

7. Q₁ = Q₀(a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )*

Can you understand this equation, Its how you can go to Q₁ from state Q₀? We remember this solution as equation-(7)

As above we can evaluate value of Q₁ in terms of state Q₀ and a, b, Similarly we are to evaluate value for state Q₃. For this we can simple put value of state Q₁ from equation-(5) into equation-(7).

5. Q₃ = Q₀b(aa)* + Q₁ba(aa)*
. Q₃ = Q₀b(aa)* + Q₀(a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* ba(aa)*
8. Q₃ = Q₀ ( b(aa)* + (a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* ba(aa)* )

Now, in equation number (1) put value of state Q₃ and Q₁ from equation number (8) and (7) receptively.

Q₀ = Λ + Q₁a + Q₃b
Q₀ = Λ + Q₀(a(bb)* + (aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* a + Q₀ ( b(aa)* + (a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* ba(aa)* ) b

Now, Last time apply Arden solution to find value of state Q₀ in terms of symbols a and b.

Q₀ = Λ + ( (a(bb)* + (aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* a + ( b(aa)* + (a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* ba(aa)* ) b )*

that is same as (we can discard Λ here) RE:

( (a(bb)* + (aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* a + ( b(aa)* + (a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* ba(aa)* ) b )*

This is the RE you where looking for.

I am not sure that it can be further simplified. I am leaving it as an exercise for you.

In linked question I suggested a non-formal and analytical method but it was hard to apply and find RE for this DFA and this question demonstrate power of Arden's theorem and step by step solution.

Edit:

My previous regular expression is correct but hard to grapes because unsymmetrical form. Below I am writing new form of RE that is more symmetrical.

We have equation-(5), (6) as follows:

5. Q₃ = Q₀b(aa)* + Q₁ba(aa)*
6. Q₁ = Q₀a(bb)* + Q₃ab(bb)*

Both are symmetrical in construction and easy to learn. (read my comment after eq-(5) above)

To evaluate value of state Q₁ in terms of Q₀, I putted value of Q₃ from equation-(5) into equation-(6) that gives me equation-(7) as follows:

7. Q₁ = Q₀(a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )*

Similarly, to evaluate value of state Q₃ in terms of Q₀, we can put value of Q₁ from equation-(6) into equation-(5) that will give us new form of equation-(8) as follows:

Q₃ = Q₀b(aa)* + Q₁ba(aa)*
Q₃ = Q₀b(aa)* + (Q₀a(bb)* + Q₃ab(bb)* ) ba(aa)*
Q₃ = Q₀b(aa)* + Q₀a(bb)* ba(aa)* + Q₃ab(bb)* ba(aa)*

Now, we can have equation-(8) in our desired form:

8. Q₃ = Q₀(b(aa)* + a(bb)* ba(aa)* )(ab(bb)* ba(aa)* )*

Now, we have equation-(1), (7), (8):

1. Q₀ = Λ + Q₁a + Q₃b
7. Q₁ = Q₀(a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )*
8. Q₃ = Q₀(b(aa)* + a(bb)* ba(aa)* ) (ab(bb)* ba(aa)* )*

Now, we can have equation-(8) in our desired form:

8. Q₃ = Q₀(b(aa)* + a(bb)* ba(aa)* )(ab(bb)* ba(aa)* )*

Now, we have equation-(1), (7), (8):

1. Q₀ = Λ + Q₁a + Q₃b
7. Q₁ = Q₀(a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )*
8. Q₃ = Q₀(b(aa)* + a(bb)* ba(aa)* ) (ab(bb)* ba(aa)* )*

Now put value of state Q₁ and Q₃ into equation-(1):

Q₀ = Λ + Q₀(a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* a + Q₀(b(aa)* + a(bb)* ba(aa)* ) (ab(bb)* ba(aa)* )* b

can also be written as:

Q₀ = Λ + Q₀ ( (a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* a + (b(aa)* + a(bb)* ba(aa)* ) (ab(bb)* ba(aa)* )* b )

Next, apply Arden's theorem on this equation, and we get the final RE:

Regular Expression for even numbers of 'a' and even numbers of 'b':

( (a(bb)* + b(aa)* ab(bb)* ) (ba(aa)* ab(bb)* )* a + (b(aa)* + a(bb)* ba(aa)* ) (ab(bb)* ba(aa)* )* b )*

Can one step further simplified as below:

((a + b(aa)*ab)(bb)*(ba(aa)*ab(bb)*)*a + (b + a(bb)*ba)(aa)*(ab(bb)*ba(aa)*)*b)*