“Tersine çevrilmiş” normal dilin düzenli olduğunu nasıl gösteririm?

Aşağıdaki soruya takılıp kaldım:

"Düzenli diller tam olarak sonlu otomatalar tarafından kabul edilen dillerdir. Bu gerçek göz önüne alındığında, eğer dili $L$bazı sonlu otomatlar tarafından kabul edilirse, $L^{R}$ de bir sonlu tarafından kabul edilir; ters $L^{R}$ kelimelerinin tümünden oluşur ."$L$

Bu yüzden düzenli bir dili verildiğinde (aslında tanım gereği) bazı sınırlı otomatalar tarafından kabul edildiğini biliyoruz, bu yüzden bizi başlangıç ​​durumundan kabul durumuna götüren uygun geçişlere sahip sınırlı bir durum kümesi var. , L cinsinden bir dizedir . İşleri basitleştirmek için yalnızca bir kabul eden devletin olduğu konusunda ısrar edebiliriz. Ters dili kabul etmek için tek yapmamız gereken geçişlerin yönünü tersine çevirmek, başlangıç ​​durumunu kabul etme durumuna ve kabul etme durumunu başlangıç ​​durumuna çevirmektir. Sonra orijinale kıyasla "geriye" dır ve dil kabul eden bir makine var L R .$L$$L$$L^{R}$

Sen hep dizeleri kabul eden FA kurulabileceğini göstermek zorunda $L^R$ dizeleri kabul eden bir sonlu Otomaton'u verilen $L$ . İşte bunu yapmak için bir prosedür.

1. Otomattaki tüm bağlantıları ters çevir
2. Yeni bir durum ekleyin ( $q_s$ )
3. İle etiketlenmiş bir bağlantı çizin $\epsilon$ devlet den $q_s$ her nihai duruma
4. Tüm nihai durumları normal durumlara dönüştürün
5. Başlangıç ​​durumunu son duruma getirme
6. Yap $q_s$ ilk durumu

Tüm bunları resmileştirelim; teoremi belirterek başlarız.

Teorem. Eğer $L$ düzenli bir dildir, o zaman böyledir $L^{R}$ .

Let $A = (Q_A, \Sigma_A, \delta_A, q_A, F_A)$ be a NFA and let $L = L(A)$. The $\epsilon$-NFA $A^{R}$ defined below accepts the language $L^{R}$.

1. $A^{R} = (Q_A \cup \{q_s\}, \Sigma_A, \delta_{A^R}, q_s, \{q_A\})$ and $q_s \notin Q_A$
2. $p \in \delta_A(q, a) \iff q \in \delta_{A^R}(p, a)$, where $a \in \Sigma_A$ and $q, p \in Q_A$
3. $\epsilon-closure(q_s) = F_A$

Proof. First, we prove the following statement: $\exists$ a path from $q$ to $p$ in $A$ labeled with $w$ if and only if $\exists$ a path from $p$ to $q$ in $A^R$ labeled with $w^R$ (the reverse of $w$) for $q, p \in Q_A$. The proof is by induction on the length of $w$.

1. Base case: $|w| = 1$
   Holds by definition of $\delta_{A^R}$
2. Induction: assume the statement holds for words of length $\lt n$ and let $|w| = n$ and $w = xa$
   Let $p \in \delta_A^*(q, w) = \delta_A^*(q, xa)$
   We know that $\delta_A^*(q, xa) = \cup_{p'}\delta_A(p', a)$ $\forall p' \in \delta_A^*(q, x)$
   $x$ and $a$ are words of fewer than $n$ symbols. By the induction hypothesis, $p' \in \delta_{A^R}(p, a)$ and $q \in \delta_{A^R}^*(p', x^R)$. This implies that $q \in \delta_{A^R}^*(p, ax^R) \iff p \in \delta_A^*(q, xa)$.

Letting $q = q_A$ and $p = s$ for some $s \in F_A$ and substituting $w^R$ for $ax^R$ guarantees that $q \in \delta_{A^R}^*(s, w^R)$ $\forall s \in F_A$. Since there is a path labeled with $\epsilon$ from $q_s$ to every state in $F_A$ (3. in the definition of $A^R$) and a path from every state in $F_A$ to the state $q_A$ labeled with $w^R$, then there is a path labeled with $\epsilon w^R = w^R$ from $q_s$ to $q_A$. This proves the theorem.

Notice that this proves that $(L^R)^R = L$ as well.

Please edit if there are any formatting errors or any flaws in my proof....

To add to the automata-based transformations described above, you can also prove that regular languages are closed under reversal by showing how to convert a regular expression for $L$ into a regular expression for $L^R$. To do so, we'll define a function $REV$ on regular expressions that accepts as input a regular expression $R$ for some language $L$, then produces a regular expression $R'$ for the language $L^R$. This is defined inductively on the structure of regular expressions:

1. $REV(\epsilon) = \epsilon$
2. $REV(\emptyset) = \emptyset$
3. $REV(a) = a$ for any $a \in \Sigma$
4. $REV(R_1 R_2) = REV(R_2) REV(R_1)$
5. $REV(R_1 | R_2) = REV(R_1) | REV(R_2)$
6. $REV(R*) = REV(R)*$
7. $REV((R)) = (REV(R))$

You can formally prove this construction correct as an exercise.

Hope this helps!

Using regular expressions, prove that if $L$ is a regular language then the \emph{reversal} of $L$, $L^R=\{w^R: w\in L\}$, is also regular. In particular, given a regular expression that describes $L$, show by induction how to convert it into a regular expression that describes $L^R$. Your proof should not make recourse to NFAs.

We will assume that we are given a regular expression that describes $L$. Let us first look at the concatination operator ($\circ$), and then we can move onto more advanced operators. So our cases of concatenation deal with the length of what is being concatenated. So first we will break all concatenations from $ab$ to $a\circ b$. When dealing with these break the components up as much as possible: $(ab \cup a)b \rightarrow (ab \cup a) \circ b$, but you cannot break associative order between different comprehensions of course.

When $\emptyset^R \rightarrow \emptyset$

When $s = \epsilon$, we have the empty string which is already reversed thus the mechanism does not change

When $s$ is just a letter, as in $s \in \Sigma$, the reversal is just that letter, $s$

When $s = \sigma$, we have a single constituent so we just reverse that constituent and thus $\sigma^R$

When $s = (\sigma_0\sigma_1...\sigma_{k-1}\sigma_k)$ where k is odd, we have a regular expression which can be written as $(\sigma_0\circ\sigma_1...\sigma_{k-1}\circ\sigma_k)$. The reversal of these even length strings is simple. Merely switch the 0 index with the k index. Then Switch the 1 index with k-1 index. Continue till the each element was switched once. Thus the last element is now the first in the reg ex, and the first is the last. The second to last is the second and the second is the second to last. Thus we have a reversed reg ex which will accept the reversed string (first letter is the last etc.) And of course we reverse each constituent. Thus we would get $(\sigma_k^R\sigma_{k-1}^R...\sigma_{1}^R\sigma_0^R)$

When $s = (\sigma_0\sigma_1...\sigma_{k/2}...\sigma_{k-1}\sigma_k)$ where k is even, we have a regular expression generally which can be written as $(\sigma_0\circ\sigma_1...\sigma_{k-1}\circ\sigma_k)$. The reversal of these even length strings is simple. Merely switch the 0 index with the k index. Then Switch the 1 index with k-1 index. Continue till the each element was switched once, but the k/2 element (an integer because k is even). Thus the last element is now the first in the reg ex, and the first is the last. The second to last is the second and the second is the second to last. Thus we have a reversed reg ex which will accept the reversed string (first letter is the last etc.). And that middle letter. And of course we reverse each constituent. Thus we would get $(\sigma_k^R\sigma_{k-1}^R...\sigma_{k/2}^R...\sigma_{1}^R\sigma_0^R)$

Okay the hard part is done. Let us look to the $\cup$ operator. This is merely a union of sets. So given two strings, $s_1, s_2$, the reverse of $s_1 \cup s_2$ is only $s_1^R \cup s_2^R$. The union will not change. And this makes sense. This will only add strings to a set. It does not matter which order they are added to the set, all that matters is that they are.

The kleene star operator is the same. It is merely adding strings to a set, not telling us how we should construt the string persay. So to reverse a kleene star of a string $s$, is only $((s^R)^*)$. Reversal can just move through them.

Thus to reverse this $(((a\cup b) \circ (a))^* \cup ((a\cup b) \circ (b))^*)^R$ we simply follow the rules. To reverse the outer union we simply reverse its two components. To reverse this: $((a\cup b) \circ (a))^*$ kleene star, we simply reverse what is inside it $\rightarrow (((a\cup b) \circ (a))^R)^*$. Then to reverse a concatenation, we index and then switch greatest with least. So we start with $((a\cup b) \circ (a))^R$ and get $((a)^R \circ (a\cup b)^R)$. To reverse that single letter, we reach our base case and get $(a)^R \rightarrow (a)$. This process outlined above describes an inductive description of this change.