Some question about finite automata and regular language I have started to studying formal languages, especialy finite automata and regular languages and I encoutered some dificulties, i.e. I have some questions:


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*For a finite automaton, the minimal automaton which accepts a language is isomorphic with given automaton? (I found this question in a test and i don't understand what really means)

*For a finite automaton, exists there a Turing Machine which accepts the language accepted by automaton?

*The languages accepted is finite? I think is false, but i'm not sure.

*Is a difference(as power compute) between Mealy and Moore automatas?

*For a finite automaton, accepted language intersect with rational languages is regular language?(From a test)(I read that rational language is same with regular language, and maybe the question is wrong). Any explainations are welcome here.
Thank you for any hints and answers!
 A: *

*An isomorphism between two finite automata $A_1 = (Q, \Sigma, \delta, q_0, F_1)$ and $A_2 = (Q', \Sigma, \delta', q_0', F_2)$ is a bijection $\varphi: Q \to Q'$ with certain conditions. This implies that $|Q| = |Q'|$ (same number of states in both automata) and this is false if your given automaton is not minimal.

*You can easily simulate a given finite automaton with a Turing machine since it is more powerful with its rewritable and erasable tape and the fact that it can traverse its tape in both directions (not only go to the end of the input like a DFA/NFA). To put it simply you can replace a transition $\delta(q,a) = q'$ from a finite automaton with $\delta(q,a) = (q',a,R)$ (keep $a$ on the tape and move right to next symbol in state $q'$) depending on your definition of a Turing machine.

*No, it doesn't have to be finite. Consider the language $L = \{a^n \mid n \geq 0\}$. It is clear that $L$ is infinite and that the automaton $A = (\{q_0\}, \{a\}, \delta, q_0, \{q_0\})$ with $\delta(q_0,a) = q_0$ accepts $L$.

*Mealy and Moore automata are equivalent, that means you can transform a Mealy automaton into a Moore one accepting the same language and vice versa. You can see examples of conversions here: https://www.tutorialspoint.com/automata_theory/moore_and_mealy_machines.htm 
Hence, Mealy and Moore automata are equal in their power of computation.

*I believe the term rational language is used synonymously with the term regular language. So if you have a (deterministic) finite automaton $A_1 = (Q, \Sigma, \delta, q_0, F_1)$ accepting the regular language $L$ and another regular language $L'$ then there exists a (deterministic) finite automaton $A_2 = (Q', \Gamma, \delta', q_0', F_2)$ (note that in general $\Gamma$ does not have to be $\Sigma$). You can construct a deterministic finite automaton accepting $L \cap L'$ with a tuple construction, that means $A = ((Q \times Q') \cup \{t\}, \Sigma \cup \Gamma, \hat{\delta}, (q_0,q_0'), F_1 \times F_2)$ with:
$\hat{\delta}((q,q'),a) = (\delta(q,a),\delta'(q',a))$ for all $a \in \Sigma \cap \Gamma$.
$\hat{\delta}((q,q'),b) = t$ for all $b \in (\Sigma\setminus\Gamma) \cup (\Gamma\setminus\Sigma)$  
$\hat{\delta}(t,c) = t$ for all $c \in \Sigma \cup \Gamma$. 
This automaton simulates both automata in its components with $t$ being a trap state if a letter of one alphabet is not in the other one. The automaton accepts if and only if both given automata would end in a final state, therefore it accepts the intersection. A necessary thing now would be to formally prove e.g. via induction that $A$ really accepts $L \cap L'$ but this is simple. It follows that $L \cap L'$ is a regular language.
