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 + ea2 = 0a3 + 1a5a3 = 0a3 + 1a4a4 = 0a3 + 1a4a5 = 0a6 + 1a5a6 = 0a6 + 1a5I 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 *
Write a regular expression - Contains an equal number of 01 and 10 subtrings
regular languages;automata;finite automata;regular expressions
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$$\begin{array}{c|cc} & 0 & 1\\ \hlineP & Q & \varnothing\\Q & R & Q\\R & R & R\end{array}$$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:$$\begin{align}a_5 &= 1^*(0a_6)=1^*0a_6 & \text{Arden}\\a_6 &= 0^*(1a_5+\epsilon) &\text{Arden again, now substitute $a_6$}\\a_5 &= 1^*0[0^*(1a_5+\epsilon)]\\    &= (1^*00^*1)a_5+1^*00^* &\text{so we have}\\a_5 &= (1^*00^*1)^*1^*00^*\end{align}$$