Show that if Automata $A$ recognize language $L$, there also exists an automata $A′$ that recognize $Double(L)$

Question

I don't know if this question should have better been on math.Stackexchange

Let be the operation $Double$ on the words on an $\Sigma$ alphabet which inserts after each character a copy of this character. Thus, $D(ab) = aabb$, $D(abaab) = aabbaaaabb$, etc... We had to prove that these regular expressions are closed by this operation and it was a sucess.

We now have to prove this properties for automata. In other words, we have to prove that if we have a language $L$ such that it exists an automata $A$ that recognize $(L=L(A))$, there also exists an automata $A'$ that recognize tha language $Double(L)$.

I shouldn't use equivalence between automatas and regular expressions

goal : I therefore deduce that I have to prove that autmomatas are closed by the Double operation.

the hypothese : is that it exists a language $L$ such taht it exists an automata $A$ which recognizes it.

proof attempt : $D(L)$ being a language, it must be a automata associated.

But:

1. has(have ?) every language(s?) an associated automata?
2. Isn't this proof too short or taking its goal has an hypothesis ?

Proof attempt n°2

Following Rick Decker's advises, here is the second attempt to prove that if we have a language $L$ such that it exists an automata $A$ that recognize $(L=L(A))$, there also exists an automata $A'$ that recognize tha language $Double(L)$ :

To prove it, we are goint to construct an automata $A'$ such that $A'=D(L)$.

The idea is to construct an input string of $w$ that we reads from left to right. After having read the entire string $w$, it checks whether the following char is the same. If it is the case we remain in the final state. Otherwise, we go to a transitional state and if the following char isn't exactly the same, we go to the bin state.

• 1. $Q=\{q_0, q_1, q_2, p\}$, $q_1,q_2$ are waiting states, $p$ is a bin state.
• 2. $\Sigma$ is the alphabet. For the example it is : $\{a,b\}$.
• 3. $\delta : Q × \Sigma → Q$ is a function, called the transition function,

$$\begin{array}{c|cc|c|c|} & a & b\\ \hline q_0 & q_2 & q_1\\ q_1 & p& q_0\\ q_2 & q_0&p\\ p & p & p\\ \hline \end{array}$$

• 4. $q=q_0$.
• 5. $F=q_0$ is the final state. It corresponds to the intial state because the empty set is accepted by the automata.

Answer 1

You said you showed this result using regular expressions. Take the idea you used then and apply it to automata. You assume you're given an automaton $A$ for the language $L$. You need to construct an automaton $A'$ which accepts all and only the strings in $D(L)$.

Here's a hint to get you started: if you have a rule in $A$ of the form $\delta(p, a) = q$, meaning that in state $p$, seeing input $a$, you go to state $q$, how would you modify that rule so that from state $p$, seeing $a$ and $a$ again, will go to state $q$? Perhaps you might introduce another state along the way from $p$ to $q$?

Answer 2

Unfortunately, your second attempt is not correct. Since your automaton recognized all the words of the type $aabbbbaabb$, etc. And not the $Double(L)$ for a given regular language $L$.

You should proceed as follows. If $L\subseteq\{a,b\}^*$ is regular, then there is a DFA which recognizes it. Let $\mathcal{A}=\langle\{a,b\},Q,F,q_0,\delta\rangle$ the description for such a DFA. Assume $Q=\{q_0,q_1,\ldots,q_k\}$ and $F=\{q_k\}$ (WLOG we assume only one final state).

Now we are supposed to describe the DFA for $Double(L)$ provided $\mathcal{A}$. Here is the description of $D(\mathcal{A})=\langle\{a,b\},Q',F',q'_i,\delta'\rangle$.

$q'_i=q_i$

$Q'=Q\cup\bar{Q^a}\cup\bar{Q^b}\cup\{q_P\}\;$ where $\;\bar{Q^a}=\{\bar{q^a_j},\,q_j\in Q\}$ and $\bar{Q^b}$ is defined similarly; $q_P$ is for the sink state.

$F'=\{\bar{q^a_k},\bar{q^b_k}\}$

where all the "barred" symbols are new symbols.

The idea is to use the "barred" states to check the "Double" property. Hence here is how we define the function $\delta'$ on the basis of the old $\delta$.

Step 1: $\delta'=\emptyset$

Step 2: if $(q_i,a,q_j)\in\delta$ then $\delta'=\delta'\cup(q_i,a,\bar{q^a_j})\cup(\bar{q^a_j},a,q_j)\cup(\bar{q^a_j},b,q_P)$ and $\delta=\delta\setminus\{(q_i,a,q_j)\}$

Step 3: if $(q_i,b,q_j)\in\delta$ then $\delta'=\delta'\cup(q_i,b,\bar{q^b_j})\cup(\bar{q^b_j},b,q_j)\cup(\bar{q^b_j},a,q_P)$ and $\delta=\delta\setminus\{(q_i,b,q_j)\}$

Step 4: if $\delta\ne\emptyset$ then goto Step 2.

The idea is that you store in the barred state the next symbol you expect; if the next symbol is the expected symbol then you go on, otherwise send everything to the sink.