\newcommand{\Xstirling}[2]{\genfrac{\{}{\}}{0pt}{}{#1}{#2}}
\newcommand*{\Xbar}[1]{\overline{#1\vphantom{\bar{#1}}}}
+\newcommand{\pnorm}[2]{\|{#1}\|^{Pos}_{#2}}
+\newcommand{\snorm}[2]{\|{#1}\|^{Sub}_{#2}}
\setlist{nosep}
%\setlength{\parskip}{0.5em}
\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} $p$ % $p \in Sub(t) \cup Sub(s)$
+ Given parse trees $t, s \in PT(e, w)$, we say that $t <_p s$ w.r.t. \emph{desision position} $p$ % $p \in Sub(t) \cup Sub(s)$
iff $\|t\|_p > \|s\|_p$ and $\|t\|_q = \|s\|_q \; \forall q < p$.
- We say that $t \lessdot s$ iff $t \lessdot_p s$ for some $p$.
+ We say that $t < s$ iff $t <_p s$ for some $p$.
\end{Xdef}
Note that in the definition \ref{order_on_PTs}
We extend their definition to non-canonical PTs.
\begin{XThe}
- Relation $\lessdot$ is strict total order on PTs.
+ Relation $<$ is a strict total order on PTs.
\end{XThe}
\end{cases}
\end{align*}
-IPTs have the same notion of \emph{positions} as PTs.
-Additionally, they have the notion of \emph{submatch positions}:
-for a given IPT $t$ the set of submatch positions is defined as
-$Sub(t) = \{ p \mid \exists t|_p = s^i \wedge i \neq 0 \}$.
+\emph{Positions} for IPTs are defined in the same way as for PTs, and
+$Pos(t)$ denotes the set of all positions of an IPT $t$.
+Additionally, we define a set of \emph{submatch positions} as
+$Sub(t) = \{ p \mid \exists t|_p = s^i : i \neq 0 \}$.
In other words, $Sub(t)$ is a subset of $Pos(t)$ that contains positions of subtrees with nonzero implicit submatch index.
Intuitively, this is the set of positions that are important for submatch extraction:
in the case of ambiguity we do not need to consider the full trees,
just the relevant parts of them.
%
-The definition of norm for IPTs is the same as for PTs, except that $Sub(t)$ is used instead of $Pos(t)$:
+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
-%\iffalse
-%\begin{figure}\label{fig_parse_trees}
-%\includegraphics[width=\linewidth]{img/trees.pdf}
-%\caption{
-%IRE and examples of IPTs for RE $(\epsilon|a^{0,\infty})(a|\epsilon)^{0,\infty}$ and string $a$.\\
-%Partial order:
-%$s \prec_1 t$,
-%$s \prec_1 u$,
-%$t \prec_{2.2} u$.
-%Total order:
-%$s <_1 t$,
-%$s <_1 u$,
-%$u <_{1.1} t$.
-%}
-%\end{figure}
-%\fi
-%
-% \begin{Xdef}
-% \emph{Lexicographic order on positions.}
-% Given positions $p = p_1. \dots .p_n$ and $q = q_1. \dots q_m$, we say that $p < q$ iff either
-% $\exists k < min(n,m) : p_k < q_k$,
-% or $n < m \wedge p_i = q_i \; \forall i = \overline{1, n}$.
-% \end{Xdef}
-
- \begin{Xdef}\label{norm_of_parse_tree}
- The \emph{norm} of IPT $t$ at position $p$ is:
- $$
- \|t\|_p =
- \begin{cases}
- -1 &\text{if } p \in Sub(t) \text{ and } t|_p = \varnothing^i \\
- |str(t|_p)| &\text{if } p \in Sub(t) \text{ and } t|_p = s^i \text{ where } s \neq \varnothing \\
- \infty &\text{if } p \not\in Sub(t)
- \end{cases}
- $$
+ \begin{Xdef}\label{tnorm_of_IPTs}
+ The \emph{p-norm} and \emph{s-norm} of an IPT $t$ at position $p$ are:
+ \begin{align*}
+ \pnorm{t}{p} =
+ \begin{cases}
+ -1 &\text{if } p \in Pos(t) \text{ and } t|_p = \varnothing^i \\
+ |str(t|_p)| &\text{if } p \in Pos(t) \text{ and } t|_p \neq \varnothing^i \\
+ \infty &\text{if } p \not\in Pos(t)
+ \end{cases}
+ \quad\quad\quad
+ \snorm{t}{p} =
+ \begin{cases}
+ -1 &\text{if } p \in Sub(t) \text{ and } t|_p = \varnothing^i \\
+ |str(t|_p)| &\text{if } p \in Sub(t) \text{ and } t|_p \neq \varnothing^i \\
+ \infty &\text{if } p \not\in Sub(t)
+ \end{cases}
+ \end{align*}
\end{Xdef}
%In other words, the norm is infinite for all positions not in $Sub(t)$,
%and for other subtrees it equals the number of symbols.
%We shorten $\|t\|_\Lambda$ as $\|t\|$.
-The definition of order on IPTs is exactly the same as on PTs:
+Accordingly, IPTs have two definitions of order:
+\emph{p-order} based on p-norm
+and \emph{s-order} based on s-norm.
- \begin{Xdef}\label{order_on_parse_trees}
- \emph{Order on IPTs.}
- Given parse trees $t, s \in PT(e, w)$, we say that $t <_p s$ w.r.t. \emph{desision position} $p$ % $p \in Sub(t) \cup Sub(s)$
- iff $\|t\|_p > \|s\|_p$ and $\|t\|_q = \|s\|_q \; \forall q < p$.
+ \begin{Xdef}\label{total_order_on_IPTs}
+ \emph{P-order on IPTs.}
+ Given parse trees $t, s \in PT(e, w)$, we say that $t <_p s$ w.r.t. \emph{desision position} $p$
+ iff $\pnorm{t}{p} > \pnorm{s}{p}$ and $\pnorm{t}{q} = \pnorm{s}{q} \; \forall q < p$.
We say that $t < s$ iff $t <_p s$ for some $p$.
\end{Xdef}
-As in the case of PTs, we do not explicitly demand that $p, q \in Sub(t) \cup Sub(s)$:
-if this is not true, then the norms of both subtrees are $\infty$ and therefore equal.
-However, unlike PTs, the order on IPTs is not total:
+ \begin{Xdef}\label{partial_order_on_IPTs}
+ \emph{S-order on IPTs.}
+ Given parse trees $t, s \in PT(e, w)$, we say that $t \prec_p s$ w.r.t. \emph{desision position} $p$ % $p \in Sub(t) \cup Sub(s)$
+ iff $\snorm{t}{p} > \snorm{s}{p}$ and $\snorm{t}{q} = \snorm{s}{q} \; \forall q < p$.
+ We say that $t \prec s$ iff $t \prec_p s$ for some $p$.
+ \end{Xdef}
+
+P-order is defined in exactly the same way as the order on PTs.
+S-order, however, has an important difference: is not total, as
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}.
+For such trees s-order is not defined: none of them is less than the other one,
+and therefore they are called \emph{incomparable}.
+From submatch extraction perspective, incomparable IPTs are indistinguishable.
- \begin{Xdef}\label{order_on_parse_trees}
+ \begin{Xdef}\label{incomparable_IPTs}
IPTs $t$ and $s$ are \emph{incomparable}, denoted $t \sim s$,
- iff neither $t < s$, nor $s < t$.
+ iff neither $t \prec s$, nor $s \prec t$.
\end{Xdef}
Incomparability relation is an equivalence relation: it is
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 a strict weak order on IPTs.
+%as theorem \ref{theorem_porder_on_IPTs} states, p-order is a strict total order.
+%As the theorem \ref{theorem_sorder_on_IPTs} states, s-order is a strict weak order.
+
+ \begin{XThe}\label{theorem_porder_on_IPTs}
+ P-order $<$ is a strict total 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*}
+ \begin{XThe}\label{theorem_sorder_on_IPTs}
+ S-order $\prec$ is a strict weak order on IPTs.
+ \end{XThe}
-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.
+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.
+Consider the IPTs $t$ and $u$ on figure \ref{fig_mark_enum}.
+On one hand, we have $t \prec_{2.2} u$, because $\snorm{t}{2.2} = \infty > 0 = \snorm{u}{2.2}$ and s-norms at all preceding submatch positions agree.
+On the other hand, we have $u <_{1.1} t$, because $\pnorm{t}{1.1} = -1 < 0 = \pnorm{u}{1.1}$
+and p-norms at all preceding positions agree.
+So, p-order is not necessarily an extension of s-order.
%
-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}.
+However, the two orders are ``compatible'' in the following sense:
+they agree on the notion of the minimal tree (the exact meaning of this 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.
+we will continuously narrow down the class of $\prec$-minimal trees,
+until we are left with a unique $<$-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 $unindex(u) = s$.
+ For an IRE $r$ and string $w$,
+ let $t_{min}$ be the $<$-minimal tree in $\IPT(r,w)$
+ and let $T_{min}$ be the class of the $\prec$-minimal trees in $\IPT(r,w)$.
+ Then $t_{min} \in T_{min}$.
\end{XThe}
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.
+if the s-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 $\exists \widetilde{r}, \widetilde{w} : t|_p, s|_p \in PT(\widetilde{r}, \widetilde{w})$.
+ If $t, s \in \IPT(r, w)$ and $\exists p \in Sub(t) \cup Sub(s)$ such that $\snorm{t}{q} = \snorm{s}{q} \; \forall q \leq p$,
+ then $\exists \widetilde{r}, \widetilde{w} : t|_p, s|_p \in \IPT(\widetilde{r}, \widetilde{w})$.
\\
Proof.
By induction on position $p$.
-
- \medskip
+ \\[-1em]
Induction basis: the case of $p = \Lambda$ is trivial: let $\widetilde{r} = r$, $\widetilde{w} = w$.
-
- \medskip
+ \\[-1em]
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 $\exists r', w' : t', s' \in PT(r', w')$,
+ Let $t' = t|_{p'} = T(t_1, \dots, t_n)$,
+ $s' = s|_{p'} = T(s_1, \dots, s_m)$.
+ By induction hypothesis $\exists r', w' : t', s' \in \IPT(r', w')$,
where $w' = str(t_1) \dots str(t_n) = str(s_1) \dots str(s_m)$.
-
- \medskip
+ \\[-1em]
Next, we show that $str(t_i) = str(s_i)$.
It must be that $i \in Sub(s') \cap Sub(t')$,
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
+ \\[-1em]
Now, let $\widetilde{w} = str(t_i)$.
If $r' = r_1|r_2$ or $r' = r_1 r_2$, let $\widetilde{r} = r_i$.
$\square$
\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 $Sub(t)$,
+ Proof of theorem \ref{theorem_order_compat}.
+ \\[-1em]
+
+ Consider any $t \in T_{min}$.
+ 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 $t$.
+ It is an IPT for some sub-IRE $r'$ and substring $w'$: $t' \in \IPT(r', w')$.
+ Let $t''$ be the $<$-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 $\|u\|_{p'} = \infty \neq \|\widetilde{s}\|_{p'}$).
+ (because substitutions preserve the s-norm of subtrees at positions in $Sub(t)$),
+ and so $u \in T_{min}$.
+ We will show that $u = t_{min}$.
+ \\[-1em]
+
+ Suppose, on the contrary, that $u \neq t_{min}$.
%
- 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 $u$.
+ Then we have $t_{min} <_p u$ for some non-submatch decision position $p$.
+ Let $p = p'.p''$, where $p'$ is the longest prefix of $p$ that is a submatch position: $p' \in Sub(u) \cup Sub(t_{min})$.
+ \\[-1em]
+
+ It must be that for all submatch positions $q' < p'$ we have $\snorm{u}{q'} = \snorm{t_{min}}{q'}$,
+ because $u \in T_{min}$ and thus either $u \sim t_{min}$,
+ or $u \prec_q t_{min}$ for some $q \in Sub(u) \cup Sub(t_{min})$
+ (in which case it must be $q > p$, because $u \prec_q t_{min}$ implies $\snorm{u}{q} > \snorm{t_{min}}{q}$,
+ which in turn implies $\pnorm{u}{q} > \pnorm{t_{min}}{q}$,
+ which contradicts $t_{min} <_p u$ if $q \leq p$).
+ \\[-1em]
+
+ Therefore by lemma \ref{lemma_subtrees} we have $\exists r', w' : u|_{p'}, t_{min}|_{p'} \in \IPT(r', w')$.
+ On one hand, $t_{min} <_{p'.p''} u$ implies $t_{min}|_{p'} <_{p''} u|_{p'}$.
+ But on the other hand, $p' \in Sub(u)$ and $p'$ is not itself a prefix of another position in $Sub(u)$,
+ therefore $u|_{p'}$ is $<$-minimal by construction of $u$.
Contradiction.
+%
+% %
+% 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.
$\square$
projects / re2c / commitdiff