diff -r b794db0b79db -r b1258b7d2789 Attic/Myhill.thy --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/Attic/Myhill.thy Mon Jul 25 13:33:38 2011 +0000 @@ -0,0 +1,331 @@ +theory Myhill + imports Myhill_2 +begin + +section {* Preliminaries \label{sec_prelim}*} + +subsection {* Finite automata and \mht \label{sec_fa_mh} *} + +text {* + +A {\em determinisitc finite automata (DFA)} $M$ is a 5-tuple +$(Q, \Sigma, \delta, s, F)$, where: +\begin{enumerate} + \item $Q$ is a finite set of {\em states}, also denoted $Q_M$. + \item $\Sigma$ is a finite set of {\em alphabets}, also denoted $\Sigma_M$. + \item $\delta$ is a {\em transition function} of type @{text "Q \ \ \ Q"} (a total function), + also denoted $\delta_M$. + \item @{text "s \ Q"} is a state called {\em initial state}, also denoted $s_M$. + \item @{text "F \ Q"} is a set of states named {\em accepting states}, also denoted $F_M$. +\end{enumerate} +Therefore, we have $M = (Q_M, \Sigma_M, \delta_M, s_M, F_M)$. Every DFA $M$ can be interpreted as +a function assigning states to strings, denoted $\dfa{M}$, the definition of which is as the following: +\begin{equation} +\begin{aligned} + \dfa{M}([]) &\equiv s_M \\ + \dfa{M}(xa) &\equiv \delta_M(\dfa{M}(x), a) +\end{aligned} +\end{equation} +A string @{text "x"} is said to be {\em accepted} (or {\em recognized}) by a DFA $M$ if +$\dfa{M}(x) \in F_M$. The language recoginzed by DFA $M$, denoted +$L(M)$, is defined as: +\begin{equation} + L(M) \equiv \{x~|~\dfa{M}(x) \in F_M\} +\end{equation} +The standard way of specifying a laugage $\Lang$ as {\em regular} is by stipulating that: +$\Lang = L(M)$ for some DFA $M$. + +For any DFA $M$, the DFA obtained by changing initial state to another $p \in Q_M$ is denoted $M_p$, +which is defined as: +\begin{equation} + M_p \ \equiv\ (Q_M, \Sigma_M, \delta_M, p, F_M) +\end{equation} +Two states $p, q \in Q_M$ are said to be {\em equivalent}, denoted $p \approx_M q$, iff. +\begin{equation}\label{m_eq_def} + L(M_p) = L(M_q) +\end{equation} +It is obvious that $\approx_M$ is an equivalent relation over $Q_M$. and +the partition induced by $\approx_M$ has $|Q_M|$ equivalent classes. +By overloading $\approx_M$, an equivalent relation over strings can be defined: +\begin{equation} + x \approx_M y ~ ~ \equiv ~ ~ \dfa{M}(x) \approx_M \dfa{M}(y) +\end{equation} +It can be proved that the the partition induced by $\approx_M$ also has $|Q_M|$ equivalent classes. +It is also easy to show that: if $x \approx_M y$, then $x \approx_{L(M)} y$, and this means +$\approx_M$ is a more refined equivalent relation than $\approx_{L(M)}$. Since partition induced by +$\approx_M$ is finite, the one induced by $\approx_{L(M)}$ must also be finite, and this is +one of the two directions of \mht: +\begin{Lem}[\mht , Direction two] + If a language $\Lang$ is regular (i.e. $\Lang = L(M)$ for some DFA $M$), then + the partition induced by $\approx_\Lang$ is finite. +\end{Lem} + +The other direction is: +\begin{Lem}[\mht , Direction one]\label{auto_mh_d1} + If the partition induced by $\approx_\Lang$ is finite, then + $\Lang$ is regular (i.e. $\Lang = L(M)$ for some DFA $M$). +\end{Lem} +The $M$ we are seeking when prove lemma \ref{auto_mh_d2} can be constructed out of $\approx_\Lang$, +denoted $M_\Lang$ and defined as the following: +\begin{subequations} +\begin{eqnarray} + Q_{M_\Lang} ~ & \equiv & ~ \{ \cls{x}{\approx_\Lang}~|~ x \in \Sigma^* \}\\ + \Sigma_{M_\Lang} ~ & \equiv & ~ \Sigma_M \\ + \delta_{M_\Lang} ~ & \equiv & ~ \left (\lambda (\cls{x}{\approx_\Lang}, a). \cls{xa}{\approx_\Lang} \right) \\ + s_{M_\Lang} ~ & \equiv & ~ \cls{[]}{\approx_\Lang} \\ + F_{M_\Lang} ~ & \equiv & ~ \{ \cls{x}{\approx_\Lang}~|~ x \in \Lang \} +\end{eqnarray} +\end{subequations} +It can be proved that $Q_{M_\Lang}$ is indeed finite and $\Lang = L(M_\Lang)$, so lemma \ref{auto_mh_d1} holds. +It can also be proved that $M_\Lang$ is the minimal DFA (therefore unique) which recoginzes $\Lang$. + + + +*} + +subsection {* The objective and the underlying intuition *} + +text {* + It is now obvious from section \ref{sec_fa_mh} that \mht\ can be established easily when + {\em reglar languages} are defined as ones recognized by finite automata. + Under the context where the use of finite automata is forbiden, the situation is quite different. + The theorem now has to be expressed as: + \begin{Thm}[\mht , Regular expression version] + A language $\Lang$ is regular (i.e. $\Lang = L(\re)$ for some {\em regular expression} $\re$) + iff. the partition induced by $\approx_\Lang$ is finite. + \end{Thm} + The proof of this version consists of two directions (if the use of automata are not allowed): + \begin{description} + \item [Direction one:] + generating a regular expression $\re$ out of the finite partition induced by $\approx_\Lang$, + such that $\Lang = L(\re)$. + \item [Direction two:] + showing the finiteness of the partition induced by $\approx_\Lang$, under the assmption + that $\Lang$ is recognized by some regular expression $\re$ (i.e. $\Lang = L(\re)$). + \end{description} + The development of these two directions consititutes the body of this paper. + +*} + +section {* Direction @{text "regular language \finite partition"} *} + +text {* + Although not used explicitly, the notion of finite autotmata and its relationship with + language partition, as outlined in section \ref{sec_fa_mh}, still servers as important intuitive + guides in the development of this paper. + For example, {\em Direction one} follows the {\em Brzozowski algebraic method} + used to convert finite autotmata to regular expressions, under the intuition that every + partition member $\cls{x}{\approx_\Lang}$ is a state in the DFA $M_\Lang$ constructed to prove + lemma \ref{auto_mh_d1} of section \ref{sec_fa_mh}. + + The basic idea of Brzozowski method is to extract an equational system out of the + transition relationship of the automaton in question. In the equational system, every + automaton state is represented by an unknown, the solution of which is expected to be + a regular expresion characterizing the state in a certain sense. There are two choices of + how a automaton state can be characterized. The first is to characterize by the set of + strings leading from the state in question into accepting states. + The other choice is to characterize by the set of strings leading from initial state + into the state in question. For the second choice, the language recognized the automaton + can be characterized by the solution of initial state, while for the second choice, + the language recoginzed by the automaton can be characterized by + combining solutions of all accepting states by @{text "+"}. Because of the automaton + used as our intuitive guide, the $M_\Lang$, the states of which are + sets of strings leading from initial state, the second choice is used in this paper. + + Supposing the automaton in Fig \ref{fig_auto_part_rel} is the $M_\Lang$ for some language $\Lang$, + and suppose $\Sigma = \{a, b, c, d, e\}$. Under the second choice, the equational system extracted is: + \begin{subequations} + \begin{eqnarray} + X_0 & = & X_1 \cdot c + X_2 \cdot d + \lambda \label{x_0_def_o} \\ + X_1 & = & X_0 \cdot a + X_1 \cdot b + X_2 \cdot d \label{x_1_def_o} \\ + X_2 & = & X_0 \cdot b + X_1 \cdot d + X_2 \cdot a \\ + X_3 & = & \begin{aligned} + & X_0 \cdot (c + d + e) + X_1 \cdot (a + e) + X_2 \cdot (b + e) + \\ + & X_3 \cdot (a + b + c + d + e) + \end{aligned} + \end{eqnarray} + \end{subequations} + +\begin{figure}[h!] +\centering +\begin{minipage}{0.5\textwidth} +\scalebox{0.8}{ +\begin{tikzpicture}[ultra thick,>=latex] + \node[draw,circle,initial] (start) {$X_0$}; + \node[draw,circle,accepting] at ($(start) + (3.5cm,2cm)$) (ac1) {$X_1$}; + \node[draw,circle,accepting] at ($(start) + (3.5cm,-2cm)$) (ac2) {$X_2$}; + \node[draw,circle] at ($(start) + (6.5cm,0cm)$) (ab) {$X_3$}; + + \path[->] (start) edge [bend left] node [midway, above] {$a$} (ac1); + \path[->] (start) edge [bend right] node [midway, below] {$b$} (ac2); + \path[->] (ac1) edge [loop above] node [midway, above] {$b$} (ac1); + \path[->] (ac2) edge [loop below] node [midway, below] {$a$} (ac2); + \path[->] (ac1) edge [bend right] node [midway, left] {$c$} (ac2); + \path[->] (ac2) edge [bend right] node [midway, right] {$c$} (ac1); + \path[->] (ac1) edge node [midway, sloped, above] {$d$} (start); + \path[->] (ac2) edge node [midway, sloped, above] {$d$} (start); + + \path [draw, rounded corners,->,dashed] (start) -- ($(start) + (0cm, 3.7cm)$) + -- ($(ab) + (0cm, 3.7cm)$) node[midway,above,sloped]{$\Sigma - \{a, b\}$} -- (ab); + \path[->,dashed] (ac1) edge node [midway, above, sloped] {$\Sigma - \{b,c,d\}$} (ab); + \path[->,dashed] (ac2) edge node [midway, below, sloped] {$\Sigma - \{a,c,d\}$} (ab); + \path[->,dashed] (ab) edge [loop right] node [midway, right] {$\Sigma$} (ab); +\end{tikzpicture}} +\end{minipage} +\caption{An example automaton}\label{fig_auto_part_rel} +\end{figure} + + Every $\cdot$-item on the right side of equations describes some state transtions, except + the $\lambda$ in \eqref{x_0_def_o}, which represents empty string @{text "[]"}. + The reason is that: every state is characterized by the + set of incoming strings leading from initial state. For non-initial state, every such + string can be splitted into a prefix leading into a preceding state and a single character suffix + transiting into from the preceding state. The exception happens at + initial state, where the empty string is a incoming string which can not be splitted. The $\lambda$ + in \eqref{x_0_def_o} is introduce to repsent this indivisible string. There is one and only one + $\lambda$ in every equational system such obtained, becasue $[]$ can only be contaied in one + equivalent class (the intial state in $M_\Lang$) and equivalent classes are disjoint. + + Suppose all unknowns ($X_0, X_1, X_2, X_3$) are solvable, the regular expression charactering + laugnage $\Lang$ is $X_1 + X_2$. This paper gives a procedure + by which arbitrarily picked unknown can be solved. The basic idea to solve $X_i$ is by + eliminating all variables other than $X_i$ from the equational system. If + $X_0$ is the one picked to be solved, variables $X_1, X_2, X_3$ have to be removed one by + one. The order to remove does not matter as long as the remaing equations are kept valid. + Suppose $X_1$ is the first one to remove, the action is to replace all occurences of $X_1$ + in remaining equations by the right hand side of its characterizing equation, i.e. + the $ X_0 \cdot a + X_1 \cdot b + X_2 \cdot d$ in \eqref{x_1_def_o}. However, because + of the recursive occurence of $X_1$, this replacement does not really removed $X_1$. Arden's + lemma is invoked to transform recursive equations like \eqref{x_1_def_o} into non-recursive + ones. For example, the recursive equation \eqref{x_1_def_o} is transformed into the follwing + non-recursive one: + \begin{equation} + X_1 = (X_0 \cdot a + X_2 \cdot d) \cdot b^* = X_0 \cdot (a \cdot b^*) + X_2 \cdot (d \cdot b^*) + \end{equation} + which, by Arden's lemma, still characterizes $X_1$ correctly. By subsituting + $(X_0 \cdot a + X_2 \cdot d) \cdot b^*$ for all $X_1$ and removing \eqref{x_1_def_o}, + we get: + \begin{subequations} + \begin{eqnarray} + X_0 & = & \begin{aligned} + & (X_0 \cdot (a \cdot b^*) + X_2 \cdot (d \cdot b^*)) \cdot c + + X_2 \cdot d + \lambda = \\ + & X_0 \cdot (a \cdot b^* \cdot c) + X_2 \cdot (d \cdot b^* \cdot c) + + X_2 \cdot d + \lambda = \\ + & X_0 \cdot (a \cdot b^* \cdot c) + X_2 \cdot (d \cdot b^* \cdot c + d) + \lambda + \end{aligned} \label{x_0_def_1} \\ + X_2 & = & \begin{aligned} + & X_0 \cdot b + (X_0 \cdot (a \cdot b^*) + X_2 \cdot (d \cdot b^*)) \cdot d + X_2 \cdot a = \\ + & X_0 \cdot b + X_0 \cdot (a \cdot b^* \cdot d) + X_2 \cdot (d \cdot b^* \cdot d) + X_2 \cdot a = \\ + & X_0 \cdot (b + a \cdot b^* \cdot d) + X_2 \cdot (d \cdot b^* \cdot d + a) + \end{aligned} \\ + X_3 & = & \begin{aligned} + & X_0 \cdot (c + d + e) + ((X_0 \cdot a + X_2 \cdot d) \cdot b^*) \cdot (a + e)\\ + & + X_2 \cdot (b + e) + X_3 \cdot (a + b + c + d + e) \label{x_3_def_1} + \end{aligned} + \end{eqnarray} + \end{subequations} +Suppose $X_3$ is the one to remove next, since $X_3$ dose not appear in either $X_0$ or $X_2$, +the removal of equation \eqref{x_3_def_1} changes nothing in the rest equations. Therefore, we get: + \begin{subequations} + \begin{eqnarray} + X_0 & = & X_0 \cdot (a \cdot b^* \cdot c) + X_2 \cdot (d \cdot b^* \cdot c + d) + \lambda \label{x_0_def_2} \\ + X_2 & = & X_0 \cdot (b + a \cdot b^* \cdot d) + X_2 \cdot (d \cdot b^* \cdot d + a) \label{x_2_def_2} + \end{eqnarray} + \end{subequations} +Actually, since absorbing state like $X_3$ contributes nothing to the language $\Lang$, it could have been removed +at the very beginning of this precedure without affecting the final result. Now, the last unknown to remove +is $X_2$ and the Arden's transformaton of \eqref{x_2_def_2} is: +\begin{equation} \label{x_2_ardened} + X_2 ~ = ~ (X_0 \cdot (b + a \cdot b^* \cdot d)) \cdot (d \cdot b^* \cdot d + a)^* = + X_0 \cdot ((b + a \cdot b^* \cdot d) \cdot (d \cdot b^* \cdot d + a)^*) +\end{equation} +By substituting the right hand side of \eqref{x_2_ardened} into \eqref{x_0_def_2}, we get: +\begin{equation} +\begin{aligned} + X_0 & = && X_0 \cdot (a \cdot b^* \cdot c) + \\ + & && X_0 \cdot ((b + a \cdot b^* \cdot d) \cdot (d \cdot b^* \cdot d + a)^*) \cdot + (d \cdot b^* \cdot c + d) + \lambda \\ + & = && X_0 \cdot ((a \cdot b^* \cdot c) + \\ + & && \hspace{3em} ((b + a \cdot b^* \cdot d) \cdot (d \cdot b^* \cdot d + a)^*) \cdot + (d \cdot b^* \cdot c + d)) + \lambda +\end{aligned} +\end{equation} +By applying Arden's transformation to this, we get the solution of $X_0$ as: +\begin{equation} +\begin{aligned} + X_0 = ((a \cdot b^* \cdot c) + + ((b + a \cdot b^* \cdot d) \cdot (d \cdot b^* \cdot d + a)^*) \cdot + (d \cdot b^* \cdot c + d))^* +\end{aligned} +\end{equation} +Using the same method, solutions for $X_1$ and $X_2$ can be obtained as well and the +regular expressoin for $\Lang$ is just $X_1 + X_2$. The formalization of this procedure +consititues the first direction of the {\em regular expression} verion of +\mht. Detailed explaination are given in {\bf paper.pdf} and more details and comments +can be found in the formal scripts. +*} + +section {* Direction @{text "finite partition \ regular language"} *} + +text {* + It is well known in the theory of regular languages that + the existence of finite language partition amounts to the existence of + a minimal automaton, i.e. the $M_\Lang$ constructed in section \ref{sec_prelim}, which recoginzes the + same language $\Lang$. The standard way to prove the existence of finite language partition + is to construct a automaton out of the regular expression which recoginzes the same language, from + which the existence of finite language partition follows immediately. As discussed in the introducton of + {\bf paper.pdf} as well as in [5], the problem for this approach happens when automata + of sub regular expressions are combined to form the automaton of the mother regular expression, + no matter what kind of representation is used, the formalization is cubersome, as said + by Nipkow in [5]: `{\em a more abstract mathod is clearly desirable}'. In this section, + an {\em intrinsically abstract} method is given, which completely avoid the mentioning of automata, + let along any particular representations. + *} + +text {* + The main proof structure is a structural induction on regular expressions, + where base cases (cases for @{const "NULL"}, @{const "EMPTY"}, @{const "CHAR"}) are quite straightforward to + proof. Real difficulty lies in inductive cases. By inductive hypothesis, languages defined by + sub-expressions induce finite partitiions. Under such hypothsis, we need to prove that the language + defined by the composite regular expression gives rise to finite partion. + The basic idea is to attach a tag @{text "tag(x)"} to every string + @{text "x"}. The tagging fuction @{text "tag"} is carefully devised, which returns tags + made of equivalent classes of the partitions induced by subexpressoins, and therefore has a finite + range. Let @{text "Lang"} be the composite language, it is proved that: + \begin{quote} + If strings with the same tag are equivalent with respect to @{text "Lang"}, expressed as: + \[ + @{text "tag(x) = tag(y) \ x \Lang y"} + \] + then the partition induced by @{text "Lang"} must be finite. + \end{quote} + There are two arguments for this. The first goes as the following: + \begin{enumerate} + \item First, the tagging function @{text "tag"} induces an equivalent relation @{text "(=tag=)"} + (defiintion of @{text "f_eq_rel"} and lemma @{text "equiv_f_eq_rel"}). + \item It is shown that: if the range of @{text "tag"} (denoted @{text "range(tag)"}) is finite, + the partition given rise by @{text "(=tag=)"} is finite (lemma @{text "finite_eq_f_rel"}). + Since tags are made from equivalent classes from component partitions, and the inductive + hypothesis ensures the finiteness of these partitions, it is not difficult to prove + the finiteness of @{text "range(tag)"}. + \item It is proved that if equivalent relation @{text "R1"} is more refined than @{text "R2"} + (expressed as @{text "R1 \ R2"}), + and the partition induced by @{text "R1"} is finite, then the partition induced by @{text "R2"} + is finite as well (lemma @{text "refined_partition_finite"}). + \item The injectivity assumption @{text "tag(x) = tag(y) \ x \Lang y"} implies that + @{text "(=tag=)"} is more refined than @{text "(\Lang)"}. + \item Combining the points above, we have: the partition induced by language @{text "Lang"} + is finite (lemma @{text "tag_finite_imageD"}). + \end{enumerate} + +We could have followed another approach given in appendix II of Brzozowski's paper [?], where +the set of derivatives of any regular expression can be proved to be finite. +Since it is easy to show that strings with same derivative are equivalent with respect to the +language, then the second direction follows. We believe that our +apporoach is easy to formalize, with no need to do complicated derivation calculations +and countings as in [???]. +*} + + +end