**Problem Detail:**

I have this language:

And ive worked on a DFA where i can move on to create the RE. Here is my DFA:

What im trying to do now is using ardens theorem to get the Regular expression however I'm stuck. This is how I'm thinking:

a1 = 0a2 + e

a2 = 0a3 + 1a5

a3 = 0a3 + 1a4

a4 = 0a3 + 1a4

a5 = 0a6 + 1a5

a6 = 0a6 + 1a5

I dont describe the a7, "sink state" because it only results in empty set.

My problem here is that I will never be able to get describe the language because of how a3,a4,a5,a6 is ? Where did I go wrong?

After some work Ive came to this solution, is it correct?

0 + 00(11*0 + 0)* + 01(1 + 00*1)*00 *

###### Asked By : Lurr

#### Answered By : Rick Decker

Part of your confusion stems from the fact that there are two different formulations of Arden's Lemma, which I'll call LR and RL. To illustrate them, let's take a simple FA with three states, $P,Q,R$, where $P$ is the start state and $R$ is the only final state. The transitions are $$\begin{array}{c|cc} & 0 & 1\\ \hline P & Q & \varnothing\\ Q & R & Q\\ R & R & R \end{array}$$ It's clear with a moment's thought that the language accepted by this FA is denoted by the regular expression $01^*0(0+1)^*$, so let's see what the two versions of Arden's Lemma produce.

**LR**: In this version we generate the regular expression from left to right. If we have a state $X$ with a transition on $a$ to state $Y$ we will include the equation $X=aY$, along with any other terms that arise from transitions from $X$ to other states. Along with this, we add a "$+\epsilon$" to the expression for $X$ if $X$ happens to be a final state. Having made the collection of equations, we may apply the simplification rule that says that whenever we have $X=rX + s$ for regular expressions $r, s$ we can replace the equation by its solution $X=r^*s$. Note that here the goal is to product a regular expression for the start state.

In the example above, we'll have the equations $$\begin{align} P&=0Q\\ Q&=0R+1Q\\ R&=(0+1)R+\epsilon & \text{since R is final} \end{align}$$ Substituting, we see that $R=(0+1)^*\epsilon = (0+1)*$ and $Q=1^*(0R)=1^*0(0+1)^*$, and finally $P=0Q=01^*0(0+1)^*$, as we expected.

**RL**: In this version we construct the regular expression from right to left. The difference is that if we have a state $X$ with transition on $a$ to state $Y$ we include the equation $Y=Xa$, along with any similar terms. In this case, we add the "$+\epsilon$" only to the *start* state's equation and the substitution in this case says that if we have $X=Xr+s$ we have the solution $X=sr^*$.

Returning to our example, we'll have $$\begin{align} P&=\epsilon & \text{since $P$ is the start state and has no incoming transitions}\\ Q&=P0+Q1\\ R&=Q0+R(0+1) \end{align}$$ Now our goal is just opposite of what we had in the LR version: we want an expression for the *final* state, $R$. Working from the top down we have $Q=Q1+P0=Q1+0$ so $Q=01^*$ and $R=Q0(0+1)^*=01^*0(0+1)^*$: the same result generated in a different order.

Assuming you want to use the LR version, your equations should be $$\begin{align} a_1&=0a_2\\ a_2&=0a_3+1a_5+\epsilon\\ a_3&=0a_3+1a_4+\epsilon\\ a_4&=0a_3+1a_4\\ a_5&=0a_6+1a_5\\ a_6&=0a_6+1a_5+\epsilon \end{align}$$ As for your problem with $a_5, a_6$, for example, we could do this: $$\begin{align} a_5 &= 1^*(0a_6)=1^*0a_6 & \text{Arden}\\ a_6 &= 0^*(1a_5+\epsilon) &\text{Arden again, now substitute $a_6$}\\ a_5 &= 1^*0[0^*(1a_5+\epsilon)]\\ &= (1^*00^*1)a_5+1^*00^* &\text{so we have}\\ a_5 &= (1^*00^*1)^*1^*00^* \end{align}$$

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