% dodaj opcję [licencjacka] dla pracy licencjackiej
% dodaj opcję [en] dla wersji angielskiej (mogą być obie: [licencjacka,en])
\documentclass[en]{pracamgr}

\usepackage{definitions}
\usepackage[backend=biber]{biblatex}
\addbibresource{mgr.bib}

\usepackage{amsmath}
\usepackage{amsfonts}

% Dane magistranta:
\autor{Marcin Chrzanowski}{370754}

\title{MSO Query Answering on Trees}
\titlepl{Odpowiadanie na zapytania MSO na drzewach}
\kierunek{Computer Science}

% Praca wykonana pod kierunkiem:
% (podać tytuł/stopień imię i nazwisko opiekuna
% Instytut
% ew. Wydział ew. Uczelnia (jeżeli nie MIM UW))
\opiekun{dr hab. Szymon Toruńczyk\\
  Institute of Informatics\\
  }

% miesiąc i~rok:
\date{August 2021}

%Podać dziedzinę wg klasyfikacji Socrates-Erasmus:
\dziedzina{
11.3 Informatyka\\
}

%Klasyfikacja tematyczna według ACM
\klasyfikacja{% TODO
  D. Software\\
  D.127. Blabalgorithms\\
  D.127.6. Numerical blabalysis}

\keywords{MSO, query answering, tree automata, RMQ}

\begin{document}
\maketitle

%tu idzie streszczenie na strone poczatkowa
\begin{abstract}
    We define relabel regular queries on trees, which, via the known equivalence
    between tree automata and MSO formulae on trees, happens to be a
    generalization of the MSO query answering problem on trees. We show these
    queries can be performed in time $O(m \log(m))$ with respect to query size
    (constant time in the case of MSO formulae with only first-order free
    variables) after preprocessing the input tree in linear time. Along the way,
    we show an algorithm for handling queries of the form ``Does the infix of a
    tree branch between nodes $x$ and $y$ belong to the regular language $L$''
    (for a previously fixed regular language $L$) in constant time after linear
    preprocessing.  Our approach is much simpler in presentation than a
    previously known solution due to \textcite{colcombet}.
\end{abstract}

\tableofcontents

\chapter{Introduction}
% \addcontentsline{toc}{chapter}{Introduction}

\queryproblem[%
    an MSO formula $\varphi(\vec{X})$ over trees with $k$ free
    second-order variables.
]{%
    MSO Query Answering on Trees
}{%
    a tree $T$.
}{%
    given a $k$-tuple of subsets of $T$'s vertices $\vec{W}
    \in \mathcal{P}(V(T))^{k}$, is $\vec{W}$ a satisfying assignment to
    $\vec{X}$? In other words, does $T \models \varphi(\vec{W})$?
}

Though we speak about MSO formulae with second-order variables only, first-order
variables are also supported simply by restricting to queries in which sets
assigned to single-order variables are singletons.

\queryproblem[%
    a deterministic bottom-up tree automaton $A$ over ranked alphabet $\Sigma$.
]{%
    Relabel Regular Queries on Trees
}{%
    a tree $T$ labeled with $\Sigma$.
}{%
    given $m$ relabelings $v_1 \mapsto a_1, \ldots, v_m \mapsto a_m$, where
    $v_i$ are vertices of $T$ and $a_i$ are elements of $\Sigma$, what state
    does $A$ arrive  at in the root of $T'$, where $T'$ is $T$ with each $v_i$'s
    label modified to the corresponding $a_i$.
}

\chapter{Preliminaries}\label{r:pojecia}

% \begin{quote}
  % Blaba, który jest blaba, nie jest prawdziwym blaba.

% \raggedleft\slshape tłum. z~chińskiego Robert Blarbarucki
% \end{quote}

\section{Definitions}
\subsection{Trees}

We work with finite trees whose vertices are labeled with letters from a finite
alphabet. More formally, given a finite alphabet $\Sigma$, for each $a \in
\Sigma$, $a$ is a tree, and if $t_1, \ldots, t_k$ are trees, then
$a(t_1, \ldots, t_k)$ is also a tree.

We use the standard notions of root, child, sibling, ancestor, descendant, etc.

Binary trees are trees where each node has either no children (the node is a
leaf), or exactly two children (which, based on their order, can be called the
left and the right child).

\subsection{Tree automata}

Consider the case of binary trees labeled with $\Sigma$. A
\definedterm{deterministic, bottom-up tree automaton} (further called just a
\definedterm{tree automaton}) consists of

\begin{itemize}
    \item A finite set of \definedterm{states} $Q$.
    \item A set of \definedterm{accepting states} $F \subseteq Q$.
    \item A bottom-up \definedterm{transition function} $\delta : Q \times
        \Sigma \times Q \to Q$.
    \item An \definedterm{initializatoin function} $\iota : \Sigma \to Q$.
\end{itemize}

A \definedterm{run} of tree automaton $A$ over tree $T$ is a relabeling of $T$
with elements of $Q$ such that

\begin{itemize}
    \item Each leaf with label $a \in \Sigma$ is relabeled with $\iota(a)$.
    \item If an inner node $v$ has label $a \in \Sigma$, its left child got
        relabeled to $p \in Q$, and its right child got relabeled to $q \in Q$,
        then $v$ gets relabeled with $\delta(p, a, q)$.
\end{itemize}

A run is \definedterm{accepting} if $T$'s root gets relabeled to an accepting
state, that is a state $q \in F$. The set of all trees accepted by an automaton
$A$ is called the \definedterm{language recognized by $A$}, notated $L(A)$. We
call the class of all languages recognized by tree automata \definedterm{regular
tree languages}, analogously to regular languages recognized by finite state
automata.

We note that the expressive power of deterministic bottom-up tree automata is
the same as that of nondeterministic (either bottom-up or top-down) tree
automata.

\subsection{Monadic Second Order (MSO) Logic}
% def of MSO

From a logical point of view, the trees we work with can be seen as structures
over a signature with a single binary relation $E$ and unary relations $U_a$ for
each $a \in \Sigma$. $E(v, w)$ represents a (directed from parent to child) edge
from $v$ to $w$. $U_a(v)$ signifies that $v$'s label is $a$.

For convenience, we will also make use of the binary relation $\leq$, with $v
\leq w$ signifying that $v$ is an ancestor of $w$ (with every vertex being an
ancestor of itself; $<$ can be used to signify a strict ancestor). Note that in
the case of MSO on trees, $\leq$ could be defined using just the edge relation
$E$.

We make use of a fundamental theorem tying MSO logic on trees and tree automata:

\begin{theorem}\label{mso-to-automaton}
    For every MSO formula $\varphi$ over trees, there exists a tree automaton
    $A$ such that for every tree $T$, $T \models \varphi$ if, and only if $T \in
    L(A)$.
\end{theorem}

We note that the converse of this theorem is also true (i.e. that for every tree
automaton, there is a corresponding MSO formula), however we will use only the
MSO to automata direction in this work.

\section{Query answering problems}

Consider a computational problem whose inputs are of the form $(S, q) \in
\mathcal{S} \times Q$ for some set of \definedterm{structures} $\mathcal{S}$ and
some set of \definedterm{queries} $Q$. This induces a \definedterm{query
answering problem} which is divided into two phases:

\begin{description}
    \item[preprocessing] An input structure $S \in \mathcal{S}$ is given and a
        \definedterm{preprocessing algorithm} outputs an indexing structure
        $S'$.
    \item[queries] On-line queries $q_1, \ldots$ from $Q$ about $S$ need to be
        handled, with access to the preprocessing output $S'$.
\end{description}

We are interested in the time complexities of both phases. We use the following
notation for algorithms that have both a preprocessing and query phase: If
it takes $f(n)$ time to complete the preprocessing step for an input of size
$n$, and $g(n, m)$ time to then handle a query of size $m$, we say the algorithm
has time complexity \qptime{$f(n)$}{$g(n, m)$}.

We turn to a discussion of several query problems with known solutions, both to
serve as examples and because we will be using them in our algorithm.

\subsection{Lowest Common Ancestor}

\queryproblem{%
    Lowest Common Ancestor (LCA) Queries
}{%
    a tree $T$.
}{%
    given vertices $x$ and $y$, find the vertex $z$ that's an ancestor of both
    $x$ and $y$, and is their lowest (i.e. furthest from the root) common
    ancestor.
}

\textcite{tarjan1984} were the first to show an optimal \qpoptimal{}
algorithm for LCA. \textcite{schieber1988} used a similar approach but
simplified the indexing structure, keeping the same time complexities.
\textcite{berkman1993} showed a completely new approach to the problem, which
relies on answering range minimum queries (see below) about an array of properly
ordered tree vertices. \textcite{bender2000} offer a simpler presentation of the
algorithm in \cite{berkman1993} and note the equivalence between the LCA and RMQ
problems.

\subsection{Range Minimum Query (RMQ)}

\queryproblem{Range Minimum Queries}{%
    an array $A$ of integers.
}{%
    given indices $i$ and $j$, return the index of the smallest
    element in the subarray $A[i, j]$.
}

As mentioned above, \textcite{bender2000} show an \qpoptimal{}
algorithm for the RMQ problem.

Their method is as follows. First they show that a special case of the RMQ
problem, $\pm1$ RMQ, can be solved in \qpoptimal. This restriction of the problem
is enough to handle LCA queries. Then, for the general RMQ case, a Cartesian
tree\footnote{A Cartesian tree of a list is a binary tree with the list's
minimum element in its root, the root's children being Cartesian trees of the
left and right sublists around the minimal element. It can be constructed in
linear time.} of the list is built, and LCA queries on this tree correspond to
range minimum queries on the list.

\subsection{Word infix regular queries}\label{wordinfix}

\queryproblem[%
    regular language $L$ over alphabet $\Sigma$, given by DFA $A$.
]{Word Infix Regular Queries}{%
    a word $w \in \Sigma^*$.
}{%
    given indices $i$ and $j$ with $1 \leq i < j \leq |w|$, does the infix $w[i,
    j]$ belong to $L$?
}

This problem has a very elegant \qpoptimal{} solution. We present the full
construction as we will be generalizing its internals for the tree case in
Chapter \ref{branchinfix}.

We begin by replacing each letter of $w$ with the set of states of $A$, $Q$.

Since $A$ is deterministic, each letter $a \in \Sigma$ can be seen as a function
$a : Q \to Q$, and we draw these functions as directed edges between successive
copies of $Q$. For example, if $A$ in state $q$, reading $a$, would move to
state $q'$, then, for any copy of $Q$ corresponding to an instance of $a$ in the
original word, there will be an edge from $q$ there to $q'$ in the successive
copy of $Q$.

Now we will color the vertices of the graph we just constructed with the colors
$1, 2, \ldots, |Q|$ in such a way that

\begin{enumerate}
    \item every copy of $Q$ has one vertex of each of the $|Q|$ colors;
    \item when a vertex of color $i$ has an edge to a vertex of color $j$ in a
        successive copy of $Q$, then $i \geq j$.
\end{enumerate}

The second point basically means that we will be trying to draw single-color
paths, but when paths need to join, it is the higher-colored path that gets cut
off.

The construction is as follows:

\begin{enumerate}
    \item Color an arbitrary vertex of the first copy of $Q$ with the color $1$.
    \item Follow the deterministic edges to the end of the word, coloring all
        vertices along this path with $1$.
    \item Now color another uncolored vertex of the first copy of $Q$ with the
        color $2$.
    \item Try following edges as far as possible, coloring all vertices with
        $2$.
    \item If your run into an already colored vertex, pick an arbitrary
        uncolored vertex in this copy of $Q$ to color with $2$ and continue from
        here.
    \item Repeat steps 3.-5. for each successive color up to $|Q|$.
\end{enumerate}

Additionally, for each vertex $v$, we store the index of the next copy of $Q$ in
which the path of this vertex's color is broken by a lower color. Put this
information in table \breaktable.

To handle the query ``does $w[i, j] \in L$'', we look at vertex $v$, which is
the vertex of $A$'s initial state in the $i$th copy of $Q$ and note its color
$c$. Now we want to answer the following question: if we follow the edges of the
graph until the $j$th copy of $Q$, what color will we end in? First, look at $k
:= \breaktable[v]$. If $k \geq j$, then we know that in the $j$th copy of $Q$,
the path we're interested in still has color $c$. Look at the state in this copy
of $Q$ that's colored with $c$, if it's an accepting state answer YES, if not,
answer NO.

If $k < j$, then jump to the $k$th copy of $Q$ and take the edge from the vertex
colored $c$ here to the next copy of $Q$. The vertex $v'$ we end up in will be
colored with color $c' < c$. Continue as we did before, by looking at $k' :=
\breaktable[v']$, comparing it to $j$, and either halting if $k' \geq j$, or
jumping again otherwise. Because with each jump we move to a color strictly
smaller than before, the number of jumps is bounded by the number of colors,
$|Q|$. Thus the query is answered in time constant with respect to $|w|$.

\chapter{Branch Infix Regular Queries}\label{branchinfix}

Before solving our main problem, that of MSO queries on trees, we generalize
word infix regular queries (section \ref{wordinfix}) to trees. This will be a
vital step in the MSO queries algorithm, but is an interesting result on its
own.

The query problem we will solve is:

\queryproblem[%
    regular language $L$ over alphabet $\Sigma$.
]{Branch Infix Regular Queries}{%
    a $\Sigma$-labeled tree $T$.
}{%
    given a vertex $x$ and its descendant $y$, does the word given by labels on
    the path from $x$ to $y$ belong to $L$?
}

We begin with a similar construction as in the word case, i.e. we replace each
vertex of the tree with a copy of the states of $A$, $Q$. Again, each letter
defines a function $a : Q \to Q$, and these functions induce edges in our
``fattened'' tree: if $A$ in state $q$, reading letter $a$ goes to $q'$, then a
copy of $Q$ corresponding to an internal node of $T$ will have edges going to
$q'$ in all of the copies of $Q$ corresponding to the children of this node in
$T$.

We also color all vertices with $1, \ldots, |Q|$ in this graph, analogously to
how we did so in the word case: begin by coloring an arbitrary vertex in the
root with $1$, then follow edges downwards, coloring all visited vertices with
$1$. Then begin the same process with the next color, restarting with a
different vertex in a given copy of $Q$ if we run into an already colored
vertex.

This process will again lead to a coloring with the desired properties that

\begin{enumerate}
    \item every copy of $Q$ has one vertex of each of the $|Q|$ colors;
    \item when a vertex of color $i$ has an edge to a vertex of color $j$ in a
        child copy of $Q$, then $i \geq j$.
\end{enumerate}

Now let's consider how we could answer a query ``is the branch infix from $x$
down to $y$ in $L$?''. Here we can't proceed exactly as in the word case. We
don't have a \breaktable\ table since a vertex in an internal node can have
arbitrarily many points below it where its color is broken by a lower one.
Somehow we need to be able to find such a break point that is ``in the direction
of $y$ from $x$''.

In the next section we formulate this question formally and solve it as a
subproblem.

\section{Highest Marked Descendant on Path}

We can now reduce the problem to the following:

\queryproblem{Highest Marked Descendant on Path Queries}{%
    a tree $T$ with set $M \subseteq V(T)$ of marked vertices.
}{%
    given a vertex $x$, its descendant $y$, find the node $z \in M$ that
    is the highest marked node on the path between $x$ and $y$, if such $z$
    exists.
}

We will build an index structure, constructible in linear time, that allows us
to handle such queries in constant time. The structure is heavily inspired by
\textcite{bender2000}, where a simple algorithm for computing LCA queries is
presented.

Recall that RMQ queries can be answered in constant time after linear preprocessing of
the input array.

We turn to describing the index structure for our tree problem.

First, we create the array \posttable\ of length $n$, which is the post-order of
the nodes.

Next, we label each node of the tree with its pre-order number. We create the
array \prefixtable\ with the corresponding pre-order labels of the nodes in
\posttable, i.e.  if $\posttable[i] = v$, then $\prefixtable[i]$ is $v$'s
pre-order number.

Finally, for each node of the tree, we record its index in \posttable\ in the
array \indextable.

\begin{observation}
    given a node $x$ and its descendant $y$, looking at the range
    $\prefixtable[\indextable[y], \indextable[x] - 1]$, all the values in this
    range are descendants of $x$, and the values smaller than
    $\prefixtable[\indextable[y]]$ correspond to ancestors of $y$.  In
    particular, the minimum of that range corresponds to the highest ancestor of
    $y$ that's a descendant of $x$.
\end{observation}

In our problem, we only care about ancestors of $y$ that are colored black. So
we perform one final modification of our data structure: for all non-black
vertices $v$, we change $\prefixtable[\indextable[v]]$ to $\infty$ (which can be
represented by $n+1$, an integer greater than any node's pre-order label). With
\prefixtable\ modified like this, we observe that now the minimum value of
$\prefixtable[\indextable[y], \indextable[x] - 1]$ corresponds exactly to the
answer of our queries -- the highest black node between $x$ and $y$.

We preprocess \prefixtable\ for RMQ queries in linear time.

Now when given a query $x$, $y$, we:

\begin{enumerate}
    \item Lookup $i := \indextable[y]$ and $j := \indextable[x] - 1$.
    \item Perform an RMQ query on $\prefixtable[i, j]$, giving us index $k$ of
        the minimal value in that range.
    \item Lookup the corresponding vertex as $z := \posttable[k]$. This is the
        answer to our query.
\end{enumerate}

\section{Solving branch infix regular queries}

With the above problem solved, we are ready to finish our solution to the branch
infix regular query problem. Recall that we have replaced the input tree $T$'s
vertices with copies of $Q$, then connected and colored them in the same way as
in the word case. We additionally remember in each vertex a unique identifier of
the connected component of vertices of the same color that it's in (call this
data $\componenttable[v]$).

For each color $c$, we will preprocess the above graph for queries that answer
the question ``if we start in a vertex $v$ colored with $c$, which is the first
copy of $Q$ on the path towards the copy of $Q$ that contains $y$?'' by creating
a copy of $T$ in which we mark the vertices where an edge from a $c$ colored
vertex points to a vertex with a lower color. This tree we preprocess for the
queries from the previous section.

Now query answering can proceed similar to the word case. Given a query ``does
the word on vertices from $x$ down to $y$ belong to $L$?'' find the copy of $Q$
corresponding to $x$ and consider the color $c$ corresponding to the initial
state here. If in the copy of $Q$ corresponding to $y$, the vertex of color $c$
has the same \componenttable\ information as the initial state we were
considering, we can immediately answer whether or not $A$ will accept the word
on this path based on whether this state is accepting.

Otherwise, we need to jump down to a lower copy of $Q$. We do this by issuing a
highest marked descendant on path query on the marked tree we created for color
$c$. The answer to this query exactly corresponds to the point where the path of
color $c$ from our initial state merges into a lower color $c'$ on the path
towards $y$. From here we continue as at the beginning, checking
\componenttable\ to see if we can answer immediately, or taking another jump
down. Handling each jump takes constant time, and the number of jumps is bounded
$|Q|$ since we can only jump to a lower color. Thus we can answer branch infix
regular queries in constant time.

\chapter{MSO Query Answering on Trees}

We now have all the pieces necessary to present our \qptime{$n$}{$m \log m$}
algorithm for MSO query answering on trees. We fix an MSO formula
$\varphi(\vec{X})$ and take a tree $T$ of size $n$. We proceed with several
reductions so that we end up with needing to solve the relabel regular queries
problem, which we solve in the remainder of this chapter.

First of all, if $T$ is not binary, we can use a standard encoding to encode it
inside of a binary tree $T'$, modifying $\varphi$ to $\varphi'$, such that $T
\models \varphi(\vec{W})$ if and only if $T' \models \varphi'(\vec{W})$.

Now, to use Theorem \ref{mso-to-automaton}, we turn the formula into an MSO
sentence without free variables by adding a unary relation $U_X$ for each
variable $X \in \vec{X}$ and instead of considering the formula $\varphi'$, we
now consider the sentence

Selecting a model in which the vertices in set $W$ are exactly those colored
with unary relation $U_X$ is equivalent to a valuation of $\vec{X}$ in which the
variable $X$ is set to $W$.

\begin{equation*}
    \varphi'' = \exists_{X_1} \ldots \exists_{X_k}
        (\forall{}_x x \in X_i \implies U_{X_i}(x)) \land
        \varphi'(X_1, \ldots, X_k)
\end{equation*}

\section{Relabel Regular Queries on Trees}

Fix a tree automaton $A$ and take a $\Sigma$-labeled binary tree $T$.

We'll preprocess the tree in such a way that when given a relabel query, we'll
be able to partition the tree into linearly (with respect to the query's size)
many parts, then, in a bottom-up fashion, compute the state in the root of each
part in the relabeled tree. The computation for each part will take constant
time.

\subsection{LCA partition}

Consider a set of tree vertices $W$. We'll say that $W$ is \definedterm{LCA
closed} if for every pair of vertices $v, w \in W$, it is the case that
$\mathlca(v, w) \in W$ (note that one of $v$ or $w$ might be their LCA). We also
require that the tree's root is in $W$ (just to ensure the partition we define
next covers the whole tree).

The \definedterm{subtree of $v$} is $v$ along with all its descendants.
We also consider subtrees with holes: the \definedterm{subtree of $v$ with hole
$w$} is the subtree of $v$ minus the subtree of $w$ (for $w$ a descendant of
$v$).

For an LCA closed set $W$ in tree $T$, we define the \definedterm{LCA partition
with respect to $W$} to be a partition of $T$'s vertices into subtrees, subtrees
with holes, and individual vertices created according to the following rules:

\begin{enumerate}
    \item If $v \in W$ is a maximal element with respect to the ancestor
        relation $<$ (i.e. there are no other elements of $W$ in the subtree of
        $v$), then the subtree of $v$ is a part of the partition.
    \item If $v \in W$ has descendants that are in $W$ only in its left subtree,
        let $w$ be the highest such descendant (there is a unique highest
        descendant because $W$ is LCA closed). The subtree of $v$ with hole $w$
        is a part of the partition.
    \item Symmetrically if $v \in W$ only has descendants that are in $W$ in its
        right subtree.
    \item If $v \in W$ has descendants that are in $W$ in both its subtrees, let
        $v_l$ and $v_r$ be $v$'s left and right child, respectively. Treat $v_l$
        and $v_r$ as if they were in $W$ for the purposes of defining the
        partition. Put $v$ into a singleton part on its own.
\end{enumerate}

Note that in the last rule, both $v_l$ and $v_r$ will either themselves be
elements of $W$ or will have descendants in $W$ in only one of their subtrees.
Indeed, if for example $v_l$ wasn't in $W$ but it had descendants from $W$ in
both its subtrees, then $W$ wouldn't have been LCA closed.

Thus the partition covers the entire tree with $O(|W|)$ parts.

If $W$ is not LCA closed, we will call the LCA partition with respect to $W$
simply the partition with respect to $W$'s LCA closure.

\subsection{Computing the LCA closure}

The LCA closure of a set of $m$ vertices $W$ can be computed in time $O(m \log
m)$ after linear preprocessing of the tree $T$. In the preprocessing step, in
addition to preprocessing for LCA queries, we will assign each vertex $v$ its
in-order number, $\infixtable[v]$.

Now to compute the closure of $W$, we first sort the vertices in $W$ with
respect to their in-order numbers, so we end up with a list $v_1, \ldots, v_m$
of vertices such that $\infixtable[v_1] < \ldots < \infixtable[v_m]$.

\begin{lemma}
    if $\infixtable[u] < \infixtable[v] < \infixtable[w]$, then $\mathlca(u, w)$
    is equal to either $\mathlca(u, v)$ or $\mathlca(v, w)$.
\end{lemma}

This may be proved by a simple case analysis.

By the above lemma, once $W$ has been sorted by in-order numbers, we can
complete our computation in time $O(m)$. It is enough to walk through the list
and issue LCA queries about successive pairs.

\section{Computing root states of parts}

Let's now discuss how we will use LCA paritions to solve the relabel query
problem. Our approach is the following: given a query $v_1 \mapsto a_1, \ldots,
v_m \mapsto a_m$, let $W = \{ v_1, \ldots, v_m \}$. Consider the partition of
$T$ with respect to $W$. We will show how to compute $A$'s state in the root of
each part of the partition, after the tree had been relabeled. Per our
definition of the partition, we have three types of parts to consider:

\begin{enumerate}
    \item A subtree rooted at vertex $v$.
    \item A singleton vertex $v$.
    \item A subtree rooted at vertex $v$ with hole $w$.
\end{enumerate}

The first case is trivial to compute: no descendants of $v$ had been relabeled,
so the states of the children of $v$ are the same as in $A$'s run on the
original tree $T$. We look them up in the precomputed run, then apply $A$'s
transition to those states and $v$'s new label.

We will compute the states of part roots in a bottom-up fashion, so the second
case is also simple: once we've computed the states in the children of $v$, we
again simply apply $A$'s transition function to those states and $v$'s new
label.

The third is the interesting case, that of a subtree with a hole. Consider the
path from the hole $w$ up to the root $v$. What we claim is that the question
``if $A$'s state in $w$ is $q$, is the state in $v$ going to be $q'$?'' can be
answered with a branch infix regular query.

This is the case because a finite automaton can compute $A$'s states along the
path from $w$ to $v$ when given access to each of the node's label, state of its
children in the run of $A$ on the original labeling of $T$, and the information
whether this node is a left or right child of its parent. Thus the language
$L_{q, q'}$ over alphabet $Q \times \Sigma \times Q \times \{ left, right \}$ of
proper semgents of a run of $A$ that start in state $q$ and end in state $q'$ is
a regular language. Our algorithm for branch infix regular queries dealt with
top-down queries, but regular languages are reversible. So we preprocess $T$ for
branch infix regular queries for each language $L_{q, q'}^R$.

Now to compute $v$'s state after the relabeling of $T$, given we've already
computed $w$'s new state $q$, take $v'$ -- $v$'s child in the direction of $w$,
find $q' \in Q$ such that the path from $v'$ to $w$ belongs to $L_{q, q'}$
(using branch infix regular queries; since $A$ is deterministic, there will be
exactly one such $q'$), then from $q'$, the state of $v$'s other child, and
$v$'s new label, use $A$'s transition function to compute $v$'s new state.

Since $T$'s root is the root of one of the parts of our LCA partition, in the
end we will know whether or not $A$ accepts $T$ after relabeling. Handling each
part takes constant time, and there are $O(m)$ parts to handle. We did require
$O(m \log m)$ time to sort the query's vertices when computing their LCA
closure, thus query handling takes time $O(m \log m)$.

\chapter{Conclusions}

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