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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 that language $Double(L)$.

I shouldn't use equivalence between automatas and regular expressions

Proof attempt

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.
    1. $\Sigma$ is the alphabet. For the example it is : $\{a,b\}$.
    1. $\delta : Q × \Sigma → Q$ is a function, called the transition function,

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

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

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1 Answers 1

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I don't understand your proof attempt. I don't know what it means to "construct an input string", since your construction should hold for all strings. I also don't understand why you say "After having read the entire string w, it checks whether the following char is the same." The automaton should not wait until after it has read the entire string, but rather make sure that any symbol read in the original automaton is read twice in the constructed one. Lastly, your example automaton is merely that: an example.

Now, for a proof, the idea is the following. Let $A$ be a DFA that recognizes the language $L$. We are now going to construct an NFA $A'$ that recognizes the language $D(L)$. The alphabet, start, and accept states are the same as for $A$. However, for each transition in the original DFA $\delta(q,\alpha) = q'$, we add a new state $s_{q,q'}$ and make transitions $\delta(q,\alpha) = s_{q,q'}$ and $\delta(s_{q,q'},\alpha) = q'$. In this way, we ensure that whenever the automaton $A$ reads a symbol, $A'$ reads the same symbol twice.