finite-automata
I'm having trouble trying to find a language expressed by this automation. I have constructed a next-state table to help me understand better. From my understanding, any string that starts with 0 takes my automaton A to an accepting state but what if it doesn't start with 0 i.e a 1? The automation would need 14n to take us back to S0 
I want to
- Find the language accepted by the automaton
- Find a regular expression that defines the same language
Indeed, there is an elegant way to compute this.
The process of finding the language accepted by an automaton $A = (Q,\Sigma, \delta, q_0, F)$ involves solving a system of equations over the monoid $(\Sigma, \circ, \epsilon)$ with $\epsilon$ denoting the empty word of the alphabet.
Denote as $\Xi_i$ the language recognized by the automaton $(Q, \Sigma, \delta, q_i, F)$ and $Q= \left\{q_0,..,q_n \right\}$. That is, create for each state of the initial automaton a new automaton with an initial state $q_i$, for $i=0,1,..,n$. Now denote $L_{ij}$ the language consisting of all the letters on the diagram of $A$ that start from $q_i$ and end at $q_j$. So from there there is the system
$$\Xi_i = \bigcup L_{ij} \Xi_j \cup Mi$$
where $M_i = \emptyset$ if state $i$ is not final, ${\epsilon}$ otherwise.
So to find the language of the automaton $A$ we got to solve the above system of equations and find $\Xi_0$ under the operations defined by the monoid.
To solve this system we can use the following lemma:
Lemma: Let $L.M \subset \Sigma^*$ languages. The smallest language satisfying the equation $$Z = LZ \cup M$$ is $L^* M$.
For example consider the automaton:
$A = (\left\{q_0,q_1,q_2\right\},\left\{a,b \right\},\delta, q_0,\left\{q_2 \right\})$ with the state diagram of $\delta$ as follows
The system of equations is
$$\begin{cases} \Xi_0 = a \Xi_0 \cup b \Xi_1 \\ \Xi_1 = b\Xi_0 \cup a \Xi_2 \\ \Xi_2 = (a \cup b) \Xi_0 \cup \left\{\epsilon \right\} \end{cases} $$
With substitution we obtain
$$\begin{cases} \Xi_0 = a \Xi_0 \cup b \Xi_1 \\ \Xi_1 = b \Xi_0 \cup a ((a \cup b) \Xi_0 \cup \left\{\epsilon \right\}) \end{cases} $$
$$\begin{cases} \Xi_0 = a \Xi_0 \cup b \Xi_1 \\ \Xi_1 = b \cup a ((a \cup b)) \Xi_0 \cup a) \end{cases} $$
Substituting to the first equation we get
$$ \Xi_0 = a \Xi_0 \cup b ((b \cup a ((a \cup b)) \Xi_0 \cup a) $$
$$ \Xi_0 = a \cup (b^2 \cup b a^2 \cup b a b) \Xi_0 \cup ba $$
Finally, applying our lemma
$$ \Xi_0 = (a \cup b^2 \cup b a^2 \cup b a b)^* ba $$
Thus the language accepted by the automaton is
$$L(A) = (a \cup b^2 \cup b a^2 \cup b a b)^* ba $$ as one can validate easily.
The second interpretation seems adequate here like the comment mentionned it. It's not very difficult though if you consider a modulo 2 calculus for number of a's or b's. There are only four possibilities so a four state automata is perfect to do the job:
Best Answer
Converting NFAs to regular expressions
The theory of this method is very well described on the CS Stack Exchange. I am directly going to apply it since it is so simple in this case that you might follow without knowing the algorithm yet.
I am going to introduce a system of equations to describe the language $S_n$ that is being accepted starting from state $s_n$ with the following equations:
$$\begin{align*}S_0 &\equiv 0S_0~\mid~1S_1~\mid~\epsilon\\ S_1 &\equiv 0S_1~\mid~1S_2\\ S_2 &\equiv 0S_2~\mid~1S_3\\ S_3 &\equiv 0S_3~\mid~1S_0\end{align*}$$
An expression of the form $S_m\equiv aS_n$ represents a transition from $s_m$ to $s_n$ by reading $a$ and the $\epsilon$ marks a state as a final state. We are looking for a final expression for $S_0$ (the language accepted starting from state $s_0$) and need some tricks to get to that point.
Now I am going to use Ardens Lemma which states (written using regular expressions and not sets of languages) that $$X\equiv \alpha X~\mid~\beta\implies X\equiv \alpha^*\beta$$ which helps to solve equations of the the form that is described on the left side. We can apply this to all four rules which yields the following (I am doing this backwards):
$$\begin{align*} S_3 &\equiv 0S_3~\mid~1S_0\color{red}{\equiv 0^*1S_0}\\ S_2 &\equiv 0S_2~\mid~1S_3\color{red}{\equiv 0^*1S_3}\\ S_1 &\equiv 0S_1~\mid~1S_2\color{red}{\equiv 0^*1S_2}\\ S_0 &\equiv 0S_0~\mid~1S_1~\mid~\epsilon\\ &\equiv 0S_0~\mid~(1S_1~\mid~\epsilon)\\ &\;\color{red}{\equiv 0^*(1S_1~\mid~\epsilon)} \end{align*}$$
Now we have to plug in $S_3$ into $S_2$ into $S_1$ into $S_0$:
$$\begin{align*} S_0 &\equiv 0^*(1S_1~\mid~\epsilon) \\ &\equiv 0^*(10^*10^*10^*1S_0~\mid~\epsilon) \\ &\equiv 0^*10^*10^*10^*1S_0~\mid~0^* \\ &\;\color{red}{\equiv(0^*10^*10^*10^*1)^*0^*}. \end{align*}$$
The very last equivalence in red follows from Ardens Lemma once again - just be careful how to handle the expression preceeding it.
My result supports J.-E. Pins solution and disproves Globe Theatres solution as well.