PT( (r) ) &= PT(r)
\end{align*}
-The \emph{string} induced by a tree $t$, denoted $str(t)$, is the concatenation of all alphabet symbols in the left-to-right traversal of $t$.
-For an RT $r$ and a string $w$, we write $PT(r, w)$ to denote the set $\{ t \in PT(r) \mid str(t) = w \}$
+The \emph{string} induced by a PT $t$, denoted $str(t)$, is the concatenation of all alphabet symbols in the left-to-right traversal of $t$.
+For a RE $r$ and a string $w$, we write $PT(r, w)$ to denote the set $\{ t \in PT(r) \mid str(t) = w \}$
(note that this set is potentially infinite).
\begin{Xdef}\label{ambiguity_of_parse_trees}
$\square$
\end{Xdef}
-Following \cite{OS13}, we assign \emph{positions} to the nodes of RT and PT.
+Following \ref{OS13}, we assign \emph{positions} to the nodes of RE and PT.
The root position is $\Lambda$, and position of the $i$-th subtree of a tree with position $p$ is $p.i$.
The \emph{length} of position $p$, denoted $|p|$, is defined as $0$ for $\Lambda$ and $|p| + 1$ for $p.i$.
%The set of all positions is denoted $\XP$.
\begin{Xdef}\label{norm_of_parse_tree}
The \emph{norm} of PT $t$ at position $p$ is:
- $$
- \|t\|_p =
+ \begin{align*}
+ \|t\|_p &=
\begin{cases}
-1 &\text{if } p \in Pos(t) \text{ and } t|_p = \varnothing \\
|str(t|_p)| &\text{if } p \in Pos(t) \text{ and } t|_p \neq \varnothing \\
\infty &\text{if } p \not\in Pos(t)
\end{cases}
- $$
+ \end{align*}
\end{Xdef}
We shorten $\|t\|_\Lambda$ as $\|t\|$.
and another tree $t$ has no subtree at position $p$ ($p \not\in Pos(t)$),
then the infinite norm $\|t\|_p$ ``outranks'' $\|s\|_p$.
- \begin{Xdef}\label{order_on_parse_trees}
- \emph{Order on parse trees.}
- Given parse trees $t, s \in PT(e, w)$, we say that $t <_p s$ w.r.t. \emph{desision position} $p$ % $p \in Sub(t) \cup Sub(s)$
+ \begin{Xdef}\label{order_on_PTs}
+ \emph{Order on PTs.}
+ Given parse trees $t, s \in PT(e, w)$, we say that $t \lessdot_p s$ w.r.t. \emph{desision position} $p$ % $p \in Sub(t) \cup Sub(s)$
iff $\|t\|_p > \|s\|_p$ and $\|t\|_q = \|s\|_q \; \forall q < p$.
- We say that $t < s$ iff $t <_p s$ for some $p$.
+ We say that $t \lessdot s$ iff $t \lessdot_p s$ for some $p$.
\end{Xdef}
-Note that in the definition \ref{order_on_parse_trees}
+Note that in the definition \ref{order_on_PTs}
we do not explicitly demand that $p, q \in Pos(t) \cup Pos(s)$:
if this is not the case, the norms of both subtrees are $\infty$ and thus equal.
+Also note that our definition of order differs from the one given by Okui and Suzuki \ref{OS13}:
+we define order on the whole set of PTs,
+while they defined it on a finite subset of PTs called \emph{canonical parse trees (CPTs)} ---
+a subset of PTs that do not contain subtrees corresponding to optional empty repetitions.
+We extend their definition to non-canonical PTs.
\begin{XThe}
- Relation $<$ is strict total order on parse trees.
- \\
- \medskip
- Proof. YET TO BE GIVEN
- $\square$
+ Relation $\lessdot$ is strict total order on PTs.
\end{XThe}
+
+
\section{Partially Ordered Parse Trees}
The POSIX standard uses the terms \emph{subexpression} and \emph{subpattern}
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.
-
\begin{align*}
&\begin{aligned}
mark &: \XR_\Sigma \longrightarrow \XIR_\Sigma \\
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).
Explicit submatch index is not used in IPTs.
+%
+As for PTs, $str(t)$ denotes the string formed by concatenation of all alphabet symbols in the left-to-right traversal of $t$,
+and $\IPT(r, w) = \{ t \in \IPT(r) \mid str(t) = w \}$ is the set of all IPTs for an IRE $r$ and a string $w$.
+%
Examples of IPTs can be seen on figure \ref{fig_mark_enum}.
\begin{Xdef}
there might be two distinct IPTs that coincide in all submatch positions, yet differ in some non-submatch positions.
For such trees the order is not defined: none of them is less than the other one.
From submatch perspective, these trees are indistinguishable
-and therefore called \emph{incomparable}:
+and therefore called \emph{incomparable}.
\begin{Xdef}\label{order_on_parse_trees}
IPTs $t$ and $s$ are \emph{incomparable}, denoted $t \sim s$,
Therefore all IPTs of a given RE can be partitioned into \emph{incomparability classes}.
%
Unlike the case of PTs, where the minimal PT is unique,
-in case of IPTs there is whole class of minimal trees:
+in case of IPTs there is a whole class of minimal trees:
$\IPT_{min}(r,w) = \{ t \in \IPT(r,w) \mid \forall u \in \IPT(r,w) : t \sim u \vee t < u \}$.
\begin{XThe}\label{theorem_order_on_IPTs}
- Relation $<$ is strict weak order on IPTs.
+ Relation $<$ is a strict weak order on IPTs.
\end{XThe}
+What is the relation between the total order $\lessdot$ on PTs and the partial order $<$ on IPTs?
+Or, put otherwise, are the two orders ``compatible'' in the sense that
+the relation between any pair of comparable IPTs is the same as the relation between the corresponding pair of PTs?
+To answer this question, let us first formalize the correspondence between PTs and IPTs.
+Namely, the operator $unindex : \XIT_\Sigma \rightarrow \XT_\Sigma$ transforms IPTs to PTs by erasing submatch indices:
+ \begin{align*}
+ unindex(x^i) &= x \text{ where } x \in \{ \varnothing, \epsilon \} \cup \Sigma \\
+ unindex(T^i(t_1, \hdots, t_n) &= T(unindex(t_1), \hdots, unindex(t_n))
+ \end{align*}
-If in Definition \ref{norm_of_parse_tree} and Definition \ref{order_on_parse_trees}
-we use $Pos(t)$ instead of $Sub(t)$, then our strict weak order $<$
-coincides with Okui-Suzuki strict total order on parse trees \cite{OS13} (we denote the latter with $<^{os}$).
-%TODO: the existense of minimal element remains to be shown, as we don't have CPT.
-While it is not true in general that $<^{os}$ is an extension of $<$
-(see the counterexample on figure \ref{fig_parse_trees}),
-the two orders are compatible in the sense that the $<^{os}$-minimal tree
-is included in the class of $<$-minimal trees.
-Intuitively, this means that if we keep ``refining'' submatch results
-by adding parentheses in subexpressions,
-we will gradually narrow down the class of $<$-minimal trees,
-until we are left with a single $<^{os}$-minimal tree.
-Theorem \ref{lemma_order_compat} shows this,
-and lemma \ref{lemma_subtrees} shows that comparison of subtrees is justified
-if the norms of all preceding submatch positions are equal.
-
- \begin{XLem}\label{lemma_subtrees}
- If $t, s \in PT(r, w)$ and $\exists p \in Sub(t) \cup Sub(s)$ such that $\|t\|_q = \|s\|_q \; \forall q \leq p$,
- then $\exists \widetilde{r}, \widetilde{w} : t|_p, s|_p \in PT(\widetilde{r}, \widetilde{w})$.
- \\
- Proof.
- By induction on position $p$.
-
- \medskip
-
- Induction basis: the case of $p = \Lambda$ is trivial: let $\widetilde{r} = r$, $\widetilde{w} = w$.
-
- \medskip
-
- Induction step: we have $|p| > 0$, let $p = p'.i$, where $i \in \YN$.
- Let $t' = t|_{p'} = P(t_1, \dots, t_n)$,
- $s' = s|_{p'} = P(s_1, \dots, s_m)$.
- By induction hypothesis $\exists r', w' : t', s' \in PT(r', w')$,
- where $w' = str(t_1) \dots str(t_n) = str(s_1) \dots str(s_m)$.
-
- \medskip
-
- Next, we show that $str(t_i) = str(s_i)$.
- It must be that $i \in Sub(s') \cap Sub(t')$,
- otherwise only one of $\|t'\|_i$, $\|s'\|_i$ is $\infty$,
- which contradicts lemma condiiton $\|t\|_p = \|s\|_p$.
- Consider position $j \leq i$.
- Because the set of submatch positions contains siblings, we have $j \in Sub(s') \cap Sub(t')$.
- Consequently, $\|t'\|_j = |str(t_j)|$ and $\|s\|_j = |str(s_j)|$.
- By lemma condition we have $\|t\|_{p'.j} = \|s\|_{p'.j}$,
- therefore $\|t'\|_j = \|s'\|_j$,
- therefore $|str(t_j)| = |str(s_j)|$.
- Because $str(t_1) \dots str(t_n) = str(s_1) \dots str(s_m)$,
- we have $str(t_i) = str(s_i)$.
-
- \medskip
-
- Now, let $\widetilde{w} = str(t_i)$.
- If $r' = r_1|r_2$ or $r' = r_1 r_2$, let $\widetilde{r} = r_i$.
- Otherwise, $r' = r_1^{k,l}$, let $\widetilde{r} = r_1$.
- $\square$
- \end{XLem}
-
- \begin{XThe}\label{lemma_order_compat}
- For a set of parse trees $P = PT(r, w)$,
- let $s$ be the $<^{os}$-minimal tree,
- and $P_{min} = \{ t \in P \mid \forall u \in P : t = u \vee t < u \}$ the class of $<$-minimal trees.
- Then $s \in P_{min}$.
-% let $s$ be the $\prec$-minimal tree: $s = min_{<^{os}} P$,
-% and $t$ one of the $<$-minimal trees: $t \in \{ u \in P \mid \forall u' \in P : u = u' \vee u < u' \}$.
-% Then $s \sim t$.
- \\
- Proof.
-
- Consider any $t \in P_{min}$.
- For each position $p \in Sub(t)$ which is not itself a prefix of another position in $Sub(t)$,
- consider subtree $t' = t|_p$.
- It is a parse tree for some subexpression $r'$ and substring $w'$: $t' \in PT(r', w')$.
- Let $t''$ be the $<^{os}$-minimal tree in $PT(r', w')$ and substitute $t'$ with $t''$ in $t$.
- (Note that substitutions are independent and can be performed in any order.)
- Let $u$ be the tree resulting from all such substitutions.
- By lemma \ref{incomparability_equivdef} we have $u \sim t$
- (substitutions preserve the norm of subtrees at positions in $Sub(t)$),
- and so $u \in P_{min}$.
-
- \medskip
-
- Next, we show that $u = s$.
- Suppose, on the contrary, that $u \neq s$.
- Then, on one hand, $u <_q s$ for some $q \in Sub(s) \cup Sub(u)$ (because $u \in P_{min}$).
- On the other hand, $s <^{os}_p u$ for some $p \in Pos(s) \cup Pos(u)$, such that $p < q$ (because $s$ is $<^{os}$-minimal).
- Let $p = p'.p''$, where $p'$ is the longest prefix of $p$ in $Sub(s) \cup Sub(u)$.
- It must be $\|u\|_{q'} = \|s\|_{q'} \forall q' \leq p'$ (because $p' < q$ and $u <_q s$),
- and $p' \in Sub(u)$ (otherwise $\|u\|_{p'} = \infty \neq \|s\|_{p'}$).
- Now, by lemma \ref{lemma_subtrees} we have $\exists r', w' : u|_{p'}, s|_{p'} \in PT(r', w')$,
- therefore $s <^{os}_{p'.p''} u$ implies $s|_{p'} <^{os}_{p''} u|_{p'}$.
- But $p' \in Sub(u)$ and $p'$ is not itself a prefix of another position in $Sub(u)$,
- therefore $u|_{p'}$ is $<^{os}$-minimal by construction of $u$.
- Contradiction.
- $\square$
+It is easy to see that for a given RE $r$ and string $w$
+we have $PT(r,w) = \{ unindex(t) \mid t \in \IPT(\IRE(r), w) \}$.
+%
+We show by example that the orders on PTs and IPTs are not necessarily ``compatible''.
+Consider IPTs $t$ and $u$ on figure \ref{fig_mark_enum}.
+On one hand, we have $t <_{2.2} u$, because $\|t\|_{2.2} = \infty > 0 = \|u\|_{2.2}$ and norms at all preceding submatch positions agree.
+On the other hand, $unindex(u) \lessdot_{1.1} unindex(t)$, because $\|unindex(t)\|_{1.1} = -1 < 0 = \|unindex(u)\|_{1.1}$
+and norms at all preceding positions agree.
+%
+However, the two orders are ``compatible'' in a weaker sense of the word:
+they agree on the notion of the minimal tree --- a fact that is formalized by theorem \ref{theorem_order_compat}.
+%
+This is an important result becuse it means that
+if we keep refining submatch granularity by tunrning more and more sub-RE into submatch groups,
+we will continuously narrow down the class of $<$-minimal trees,
+until we are left with a unique $\lessdot$-minimal tree.
+In practice this means that adding a few parentheses in RE will not drastically change the previous results of submatch extraction;
+it will only add more details.
+
+ \begin{XThe}\label{theorem_order_compat}
+ For a RE $r$ and string $w$,
+ let $s$ be the $\lessdot$-minimal PT in $PT(r,w)$
+ and let $P_{min}$ be the class of the $<$-minimal trees in $\IPT(\IRE(r),w)$.
+ Then there is an IPT $u \in P_{min}$ such that $unindex(u) = s$.
\end{XThe}
+
\section{Parenthesized expressions}
In this section we introduce linearised representation of parse trees: \emph{parenthesized expressions}.
the lemma \ref{lemma_ptorder_transitivity_of_incomparability}.
+\subsection*{Proof of Theorem \ref{theorem_order_compat}}
+
+First, we prove an auxilary lemma \ref{lemma_subtrees} which
+shows that comparison of sub-IPT is justified
+if the norms at all preceding submatch positions are equal.
+
+ \begin{XLem}\label{lemma_subtrees}
+ If $t, s \in PT(r, w)$ and $\exists p \in Sub(t) \cup Sub(s)$ such that $\|t\|_q = \|s\|_q \; \forall q \leq p$,
+ then $\exists \widetilde{r}, \widetilde{w} : t|_p, s|_p \in PT(\widetilde{r}, \widetilde{w})$.
+ \\
+ Proof.
+ By induction on position $p$.
+
+ \medskip
+
+ Induction basis: the case of $p = \Lambda$ is trivial: let $\widetilde{r} = r$, $\widetilde{w} = w$.
+
+ \medskip
+
+ Induction step: we have $|p| > 0$, let $p = p'.i$, where $i \in \YN$.
+ Let $t' = t|_{p'} = P(t_1, \dots, t_n)$,
+ $s' = s|_{p'} = P(s_1, \dots, s_m)$.
+ By induction hypothesis $\exists r', w' : t', s' \in PT(r', w')$,
+ where $w' = str(t_1) \dots str(t_n) = str(s_1) \dots str(s_m)$.
+
+ \medskip
+
+ Next, we show that $str(t_i) = str(s_i)$.
+ It must be that $i \in Sub(s') \cap Sub(t')$,
+ otherwise only one of $\|t'\|_i$, $\|s'\|_i$ is $\infty$,
+ which contradicts lemma condiiton $\|t\|_p = \|s\|_p$.
+ Consider position $j \leq i$.
+ Because the set of submatch positions contains siblings, we have $j \in Sub(s') \cap Sub(t')$.
+ Consequently, $\|t'\|_j = |str(t_j)|$ and $\|s\|_j = |str(s_j)|$.
+ By lemma condition we have $\|t\|_{p'.j} = \|s\|_{p'.j}$,
+ therefore $\|t'\|_j = \|s'\|_j$,
+ therefore $|str(t_j)| = |str(s_j)|$.
+ Because $str(t_1) \dots str(t_n) = str(s_1) \dots str(s_m)$,
+ we have $str(t_i) = str(s_i)$.
+
+ \medskip
+
+ Now, let $\widetilde{w} = str(t_i)$.
+ If $r' = r_1|r_2$ or $r' = r_1 r_2$, let $\widetilde{r} = r_i$.
+ Otherwise, $r' = r_1^{k,l}$, let $\widetilde{r} = r_1$.
+ $\square$
+ \end{XLem}
+
+Now, we give the proof of the theorem \ref{theorem_order_compat}.
+\\[-1em]
+
+ Consider any $t \in P_{min}$.
+ For each position $p \in Sub(t)$ which is not itself a prefix of another position in $Sub(t)$,
+ consider subtree $t' = t|_p$.
+ It is a parse tree for some subexpression $r'$ and substring $w'$: $t' \in \IPT(r', w')$.
+ Let $t''$ be the $<^{os}$-minimal tree in $\IPT(r', w')$ and substitute $t'$ with $t''$ in $t$.
+ (Note that substitutions are independent and can be performed in any order.)
+ Let $u$ be the tree resulting from all such substitutions.
+ By lemma \ref{incomparability_equivdef} we have $u \sim t$
+ (substitutions preserve the norm of subtrees at positions in $Sub(t)$),
+ and so $u \in P_{min}$.
+\\[-1em]
+
+ Next, we show that $unindex(u) = s$.
+ Let $\widetilde{u} = unindex(u)$.
+ Suppose, on the contrary, that $\widetilde{u} \neq s$.
+ Then there is some other IPT $\widetilde{s}$ such that $unindex(\widetilde{s}) = s$.
+ %
+ On one hand, $u <_q \widetilde{s}$ for some submatch position $q$ (because $u \in P_{min}$).
+ On the other hand, $s \lessdot_p \widetilde{u}$ for some non-submatch position $p < q$ (because $s$ is $\lessdot$-minimal).
+ %
+ Let $p = p'.p''$, where $p'$ is the longest prefix of $p$ in $Sub(\widetilde{s}) \cup Sub(u)$.
+ It must be $\|u\|_{q'} = \|\widetilde{s}\|_{q'} \forall q' \leq p'$ (because $p' < q$ and $u <_q \widetilde{s}$),
+ and $p' \in Sub(u)$ (otherwise $\|u\|_{p'} = \infty \neq \|\widetilde{s}\|_{p'}$).
+ %
+ Now, by lemma \ref{lemma_subtrees} we have $\exists r', w' : u|_{p'}, \widetilde{s}|_{p'} \in PT(r', w')$,
+ therefore $s \lessdot_{p'.p''} u$ implies $s|_{p'} \lessdot_{p''} \widetilde{u}|_{p'}$.
+ But $p' \in Sub(u)$ and $p'$ is not itself a prefix of another position in $Sub(u)$,
+ therefore $\widetilde{u}|_{p'}$ is $\lessdot$-minimal by construction of $u$.
+ Contradiction.
+ $\square$
+
+
\end{document}
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