% $(\epsilon|a^{0,\infty})(a|\epsilon)^{0,\infty}$
-\begin{scope}[xshift=0.5in, yshift=0in]
+\small{
+
+\begin{scope}[xshift=0.7in, yshift=0in]
\node (a) {{
$\begin{aligned}
- & mark (((\epsilon|a^{0,\infty})(a|\epsilon)^{0,3} ))) = \big[ \\
- & \quad mark ( (\epsilon|a^{0,\infty})(a|\epsilon)^{0,3} ) ) = \big[ \\
+ & mark ((\epsilon|a^{0,\infty})(a|\epsilon)^{0,3} )) = \big[ \\
%
- & \quad\quad mark ( (\epsilon|a^{0,\infty}) ) = \big[ \\
+ & \quad mark ( (\epsilon|a^{0,\infty}) ) = \\
+ & \quad mark ( (\epsilon|a^{0,\infty})^{1,1} ) = \big[ \\
& \quad\quad mark ( \epsilon|a^{0,\infty} ) = \big[ \\
- & \quad\quad\quad\quad mark ( \epsilon ) = (0,0,\epsilon), \\
- & \quad\quad\quad\quad mark ( a^{0,\infty} ) = \big[ \\
- & \quad\quad\quad\quad\quad mark ( a ) = (0,0,a) \\
- & \quad\quad\quad\quad \big] = (0,0,(0,0,a)^{0,\infty}) \\
- & \quad\quad\quad \big] = (0,0,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
- & \quad\quad \big] = (1,1,(0,0,\epsilon)|(0,0,(0,0,a)^{0,\infty})) \\
+ & \quad\quad\quad mark ( \epsilon ) = (0,0,\epsilon), \\
+ & \quad\quad\quad mark ( a^{0,\infty} ) = \big[ \\
+ & \quad\quad\quad\quad mark ( a ) = (0,0,a) \\
+ & \quad\quad\quad \big] = (0,0,(0,0,a)^{0,\infty}) \\
+ & \quad\quad \big] = (0,0,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
+ & \quad \big] = (1,0,(1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1}) \\
%
- & \quad\quad mark ( (a|\epsilon)^{0,3} ) = \big[ \\
- & \quad\quad\quad mark ( (a|\epsilon) ) = \big[ \\
- & \quad\quad\quad\quad mark ( a|\epsilon ) = \big [ \\
- & \quad\quad\quad\quad\quad mark ( a ) = (0,0,a) \\
- & \quad\quad\quad\quad\quad mark ( \epsilon ) = (0,0,\epsilon) \\
- & \quad\quad\quad\quad \big] = (0,0,(0,0,a) \mid (0,0,\epsilon)) \\
- & \quad\quad\quad \big] = (1,1,(0,0,a) \mid (0,0,\epsilon)) \\
- & \quad\quad \big] = (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3}) \\
+ & \quad mark ( (a|\epsilon)^{0,3} ) = \big[ \\
+ & \quad\quad mark ( a|\epsilon ) = \big [ \\
+ & \quad\quad\quad mark ( a ) = (0,0,a) \\
+ & \quad\quad\quad mark ( \epsilon ) = (0,0,\epsilon) \\
+ & \quad\quad \big] = (0,0,(0,0,a) \mid (0,0,\epsilon)) \\
+ & \quad \big] = (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3}) \\
%
- & \quad\big] = (1,0,
- (1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
- & \quad\quad\quad\quad \cdot (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3})
- ) \\
- & \big] = (1,1,
- (1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
+ & \big] = (1,0,
+ (1,0,(1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1}) \\
& \quad\quad\quad \cdot (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3})
- ) \\
-%
-% & \IRE ((\epsilon|a^{0,\infty})(a|\epsilon)^{0,3} )) = \\
-% & \quad\quad\quad = (1,1, (2,2,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
-% & \quad\quad\quad\quad\quad\quad \cdot (3,0,(4,3,(0,0,a) \mid (0,0,\epsilon))^{0,3}))
+ )
\end{aligned}$
}};
\end{scope}
\begin{scope}[xshift=4in, yshift=0in]
\node (a) {{
$\begin{aligned}
- & enum (1,1,(1,1,
- (1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
- & \quad\quad\quad\quad\quad \cdot (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3}))) = \big[ \\
- & \quad enum (2,2,(1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))) = \big[ \\
- & \quad\quad enum (3,3,(0,0,\epsilon)) = (3,3,(0,0,\epsilon)) \\
- & \quad\quad enum (3,3,(0,0,(0,0,a)^{0,\infty})) = \big[ \\
- & \quad\quad\quad enum (3,3,(0,0,a)) = (3,3,(0,0,a)) \\
- & \quad\quad \big] = (3,3,(0,0,(0,0,a)^{0,\infty})) \\
- & \quad \big] = (3,3,(2,2,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))) \\
+ & enum (1,1, \; (1,0,
+ (1,0,(1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1}) \\
+ & \quad\quad\quad\quad\quad\quad \cdot (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3})
+ ) \;) = \big[ \\
+ %
+ & \quad enum (2,1, \; (1,0,(1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1}) \;) = \big[ \\
+ & \quad\quad enum (3,1, \; (1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \;) = \big[ \\
+ & \quad\quad\quad enum (4,2, \; (0,0,\epsilon)) = (4,2, \; (0,0,\epsilon) \;) \\
+ & \quad\quad\quad enum (4,2, \; (0,0,(0,0,a)^{0,\infty}) \;) = \big[ \\
+ & \quad\quad\quad\quad enum (4,2, \; (0,0,a)) = (4,2, \; (0,0,a) \;) \\
+ & \quad\quad\quad \big] = (4,2, \; (0,0,(0,0,a)^{0,\infty}) \;) \\
+ & \quad\quad \big] = (4,2, \; (3,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \;) \\
+ & \quad \big] = (4,2, \; (2,0,(3,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1}) \;) \\
%
- & \quad enum (3,3, (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3}) ) = \big[ \\
- & \quad\quad enum (4,3, (1,1,(0,0,a) \mid (0,0,\epsilon)) ) = \big[ \\
- & \quad\quad\quad enum (5,4, (0,0,a) ) = (0,0,a) \\
- & \quad\quad\quad enum (5,4, (0,0,\epsilon) ) = (0,0,\epsilon) \\
- & \quad\quad \big] = (5,4,(4,3,(0,0,a) \mid (0,0,\epsilon))) \\
- & \quad \big] = (5,4,(3,0,(4,3,(0,0,a) \mid (0,0,\epsilon))^{0,3})) \\
+ & \quad enum (4,2, \; (1,0,(1,1,(0,0,a) \mid (0,0,\epsilon))^{0,3}) \;) = \big[ \\
+ & \quad\quad enum (5,2, \; (1,1,(0,0,a) \mid (0,0,\epsilon)) \;) = \big[ \\
+ & \quad\quad\quad enum (6,3, \; (0,0,a) ) = (6, 3, \; (0,0,a) \;) \\
+ & \quad\quad\quad enum (6,3, \; (0,0,\epsilon) ) = (6, 3, \; (0,0,\epsilon) \;) \\
+ & \quad\quad \big] = (6,3, \; (5,2,(0,0,a) \mid (0,0,\epsilon)) \;) \\
+ & \quad \big] = (6,3, \; (4,0,(5,2,(0,0,a) \mid (0,0,\epsilon))^{0,3}) \;) \\
%
- & \big] = (5,4,(1,1,
- (2,2,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
- & \quad\quad\quad\quad\quad \cdot (3,0,(4,3,(0,0,a) \mid (0,0,\epsilon))^{0,3})
- )) \\
- \\ \\ \\ \\ \\ \\ \\
+ & \big] = (6,3, \; (1,0,
+ (2,0,(3,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1}) \\
+ & \quad\quad\quad\quad\quad \cdot (4,0,(5,2,(0,0,a) \mid (0,0,\epsilon))^{0,3})
+ ) \;) \\
\end{aligned}$
}};
\end{scope}
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- \node (c) {{
- $\begin{aligned}
- & \IRE ((\epsilon|a^{0,\infty})(a|\epsilon)^{0,3} ))
- = (1,1, (2,2,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))
- \cdot (3,0,(4,3,(0,0,a) \mid (0,0,\epsilon))^{0,3}))
- \end{aligned}$
- }};
-\end{scope}
+%\begin{scope}[xshift=2.5in, yshift=-2.4in]
+% \node (c) {{
+% $\begin{aligned}
+% & \IRE ((\epsilon|a^{0,\infty})(a|\epsilon)^{0,3} ))
+% = (1,0, (2,0,(3,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))^{1,1})
+% \cdot (4,0,(5,2,(0,0,a) \mid (0,0,\epsilon))^{0,3}))
+% \end{aligned}$
+% }};
+%\end{scope}
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\end{tikzpicture}
It is defined via a composition of two functions,
$mark$ that transforms REs into IREs with submatch indices in the boolean range $\{0, 1\}$,
and $enum$ that substitutes boolean indices with consecutive numbers:
-$\IRE(e) = r$ where $(\Xund, \Xund, r) = enum(1, 1, mark((e)))$.
-Note the extra pair of parentheses around $e$:
-according to the POSIX standard the root sub-RE is always the submatch group number zero,
-even if it has no explicit parentheses.
+$\IRE(e) = r$ where $(\Xund, \Xund, r) = enum(1, 1, mark(e))$.
+Note that we consider $(e)$ as a special case of repetition $(e)^{1,1}$:
+this allows us to handle all parenthesized sub-RE uniformly.
+An example of constructing an IRE from a RE is given on figure \ref{fig_mark_enum}.
+The reverse transformation is also possible by erasing all indices
+and adding parentheses around subexpressions with nonzero explicit submatch index.
+Therefore RE and IRE are equivalent representations.
\begin{align*}
&\begin{aligned}
mark &: \XR_\Sigma \longrightarrow \XIR_\Sigma \\
- mark &(x) = (0, 0, x) \\
- &\text{where } x \in \{\epsilon, \alpha\}
- \\
+ mark &(x) \mid_{x \in \{\epsilon, \alpha\}} = (0, 0, x) \\
%
- mark &(e_1 \circ e_2) = (i, 0,
+ mark &(e_1 \circ e_2) \mid_{\circ \in \{|,\cdot\}} = (i, 0,
(i, j_1, r_1) \circ
(i, j_2, r_2)
) \\
- &\text{where } \circ \in \{|,\cdot\} \\
- &\space{\hphantom{where }}(i_1, j_1, r_1) = mark(e_1) \\
- &\space{\hphantom{where }}(i_2, j_2, r_1) = mark(e_2) \\
+ &\text{where } (i_1, j_1, r_1) = mark(e_1) \\
+ &\space{\hphantom{where }}(i_2, j_2, r_2) = mark(e_2) \\
&\space{\hphantom{where }}i = i_1 \vee i_2 \\
%
- mark &(e^{n, m}) = (i, 0, (i, j, r)^{n, m}) \\
+ mark &(e^{n, m}) \mid_{e = (e_1)} = (1, 0, (1, 1, r)) \\
+ &\text{where } (\Xund, \Xund, r) = mark(e_1) \\
+ %
+ mark &(e^{n, m}) \mid_{e \neq (e_1)} = (i, 0, (i, j, r)) \\
&\text{where } (i, j, r) = mark(e) \\
%
- mark &((e)) = (1, 1, r) \\
- &\text{where } (\Xund, \Xund, r) = mark(e)
+ mark &((e)) = mark((e)^{1, 1})
\end{aligned}
%
&&\begin{aligned}
enum &: \YZ \times \YZ \times \XIR_\Sigma \longrightarrow \YZ \times \YZ \times \XIR_\Sigma \\
- enum &(\bar{i}, \bar{j}, (i, j, x)) = (\bar{i} + i, \bar{j} + j, (\bar{i} \times i, \bar{j} \times j, x)) \\
- &\text{where } x \in \{\epsilon, \alpha\}
+ enum &(\bar{i}, \bar{j}, (i, j, x)) \mid_{x \in \{\epsilon, \alpha\}}
+ = (\bar{i} + i, \bar{j} + j, (\bar{i} \times i, \bar{j} \times j, x))
\\
- enum &(\bar{i}, \bar{j}, (i, j, r_1 \circ r_2)) = (i_2, j_2, (\bar{i} \times i, \bar{j} \times j, \bar{r}_1 \circ \bar{r}_2)) \\
- &\text{where } \circ \in \{|,\cdot\} \\
- &\space{\hphantom{where }}(i_1, j_1, \bar{r}_1) = enum(\bar{i} + i, \bar{j} + j, r_1) \\
+ enum &(\bar{i}, \bar{j}, (i, j, r_1 \circ r_2)) \mid_{\circ \in \{|,\cdot\}}
+ = (i_2, j_2, (\bar{i} \times i, \bar{j} \times j, \bar{r}_1 \circ \bar{r}_2)) \\
+ &\text{where } (i_1, j_1, \bar{r}_1) = enum(\bar{i} + i, \bar{j} + j, r_1) \\
&\space{\hphantom{where }}(i_2, j_2, \bar{r}_2) = enum(i_1, j_1, r_2)
\\
enum &(\bar{i}, \bar{j}, (i, j, r^{n,m})) = (i_1, j_1, (\bar{i} \times i, \bar{j} \times j, \bar{r}^{n,m})) \\
\end{aligned}
\end{align*}
-An example of constructing an IRE from a RE is given on figure \ref{fig_mark_enum}.
-The reverse transformation is also possible:
-we can transform IRE back to RE
-by erasing all indices
-and adding parentheses around subexpressions with nonzero explicit submatch index.
-Therefore RE and IRE are equivalent representations.
-\\
-
-%\iffalse
-%Below is an example of IRE construction for RE $(\epsilon|a^{0,\infty})(\epsilon|a)^{0,\infty} )$
-%(the resulting IRE can be seen on figure \ref{fig_parse_trees}):
-%
-% \begin{align*}
-% &\begin{aligned}
-% & mark ( (\epsilon|a^{0,\infty})(\epsilon|a)^{0,\infty} ) ) = \big[ \\
-% %
-% & \quad mark ( (\epsilon|a^{0,\infty}) ) = \big[ \\
-% & \quad mark ( \epsilon|a^{0,\infty} ) = \big[ \\
-% & \quad\quad\quad mark ( \epsilon ) = (0,0,\epsilon), \\
-% & \quad\quad\quad mark ( a^{0,\infty} ) = \big[ \\
-% & \quad\quad\quad\quad mark ( a ) = (0,0,a) \\
-% & \quad\quad\quad \big] = (0,0,(0,0,a)^{0,\infty}) \\
-% & \quad\quad \big] = (0,0,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
-% & \quad \big] = (1,1,(0,0,\epsilon)|(0,0,(0,0,a)^{0,\infty})) \\
-% %
-% & \quad mark ( (\epsilon|a)^{0,\infty} ) = \big[ \\
-% & \quad\quad mark ( (\epsilon|a) ) = \big[ \\
-% & \quad\quad\quad mark ( \epsilon|a ) = \big [ \\
-% & \quad\quad\quad\quad mark ( \epsilon ) = (0,0,\epsilon) \\
-% & \quad\quad\quad\quad mark ( a ) = (0,0,a) \\
-% & \quad\quad\quad \big] = (0,0,(0,0,\epsilon)|(0,0,a)) \\
-% & \quad\quad \big] = (1,1,(0,0,\epsilon) \mid (0,0,a)) \\
-% & \quad \big] = (1,0,(1,1,(0,0,\epsilon) \mid (0,0,a))^{0,\infty}) \\
-% %
-% & \big] = (1,1,
-% (1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
-% & \quad\quad\quad \cdot (1,0,(1,1,(0,0,\epsilon) \mid (0,0,a))^{0,\infty})
-% )
-% \end{aligned}
-% %
-% &&\begin{aligned}
-% & enum (1,1,(1,1,
-% (1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
-% & \quad\quad\quad\quad\quad \cdot (1,0,(1,1,(0,0,\epsilon) \mid (0,0,a))^{0,\infty}))) = \big[ \\
-% & \quad enum (2,2,(1,1,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))) = \big[ \\
-% & \quad\quad enum (3,3,(0,0,\epsilon)) = (3,3,(0,0,\epsilon)) \\
-% & \quad\quad enum (3,3,(0,0,(0,0,a)^{0,\infty})) = \big[ \\
-% & \quad\quad\quad enum (3,3,(0,0,a)) = (3,3,(0,0,a)) \\
-% & \quad\quad \big] = (3,3,(0,0,(0,0,a)^{0,\infty})) \\
-% & \quad \big] = (3,3,(2,2,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty}))) \\
-% %
-% & \quad enum (3,3, (1,0,(1,1,(0,0,\epsilon) \mid (0,0,a))^{0,\infty}) ) = \big[ \\
-% & \quad\quad enum (4,3, (1,1,(0,0,\epsilon) \mid (0,0,a)) ) = \big[ \\
-% & \quad\quad\quad enum (5,4, (0,0,\epsilon) ) = (0,0,\epsilon) \\
-% & \quad\quad\quad enum (5,4, (0,0,a) ) = (0,0,a) \\
-% & \quad\quad \big] = (5,4,(4,3,(0,0,\epsilon) \mid (0,0,a))) \\
-% & \quad \big] = (5,4,(3,0,(4,3,(0,0,\epsilon) \mid (0,0,a))^{0,\infty})) \\
-% %
-% & \big] = (5,4,(1,1,
-% (2,2,(0,0,\epsilon) \mid (0,0,(0,0,a)^{0,\infty})) \\
-% & \quad\quad\quad\quad\quad \cdot (3,0,(4,3,(0,0,\epsilon) \mid (0,0,a))^{0,\infty})
-% ))
-% \end{aligned}
-% \end{align*}
-%\fi
-
-\begin{figure}\label{fig_mark_enum}
-\includegraphics[width=\linewidth]{img/mark_enum.pdf}
-\caption{
-An example of constructing IRE for RE $(\epsilon|a^{0,\infty})(a|\epsilon)^{0,3}$ using $mark$ and $enum$\\
-and some examples of IPT for the resulting IRE and string ``$a$''.
-S-norm of each subtree is marked with $\#$.
-}
-\end{figure}
-
Just like REs denote sets of PTs, IREs denote sets of \emph{IPTs} --- \emph{indexed parse trees}.
The only difference between PTs and IPTs is that each node of an IPT has an associated number ---
the implicit submatch index inherited from the corresponding IRE node (denoted with a superscript).
%
Examples of IPTs can be seen on figure \ref{fig_mark_enum}.
-\FloatBarrier
-
\begin{Xdef}
\emph{Indexed parse trees (IPT)} over finite alphabet $\Sigma$, denoted $\XIT_\Sigma$, are:
\begin{enumerate}
IPTs have two definitions of norm, one for $Pos$ and one for $Sub$,
which we call \emph{p-norm} and \emph{s-norm} respectively:
+\FloatBarrier
+
\begin{Xdef}\label{tnorm_of_IPTs}
The \emph{p-norm} and \emph{s-norm} of an IPT $t$ at position $p$ are:
\begin{align*}
S-order $\prec$ is a strict weak order on IPTs.
\end{XThe}
+\begin{figure}\label{fig_mark_enum}
+\includegraphics[width=\linewidth]{img/mark_enum.pdf}
+\caption{
+An example of constructing IRE for RE $(\epsilon|a^{0,\infty})(a|\epsilon)^{0,3}$ using $mark$ and $enum$\\
+and some examples of IPT for the resulting IRE and string ``$a$''.
+S-norm of each subtree is marked with $\#$.
+}
+\end{figure}
+
What is the relation between the total p-order $<$ and the partial s-order $\prec$ on IPTs?
Can we say that p-order is an extension of s-order on non-comparable IPTs?
We show by the means of a counterexample that the answer to the above question is negative.
In the rest of the paper the words ``order'' and ``norm`` refer to s-order and s-norm.
+\FloatBarrier
\section{Parenthesized expressions}
projects / re2c / commitdiff