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Diffstat (limited to 'mgr.tex')
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@@ -531,18 +531,24 @@ The question answering problem we will solve is: \queryproblem[% regular language $L$ over alphabet $\Sigma$, given by DFA $A$. -]{Branch Infix Regular Questions}{% +]{\label{branch-infix-problem} Branch Infix Regular Questions}{% 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 $L$? } -Let $n$ be the size of $V(T)$. We begin with a similar construction as in the -word case, i.e. we create a graph of $n+1$ copies of $A$'s states $Q$. The -vertex set of the graph will be $(V(T) \cup \{ \mathsf{root} \}) \times Q \}$ -and we can refer to a specific vertex as $x.q$ for $x \in V(T) \cup \{ -\mathsf{root} \}, q \in Q$. +Let $n$ be the size of $V(T)$. + +\begin{theorem}\label{branch-infix-theorem} + Problem \ref{branch-infix-problem} can be solved in time + \qptime{$O_{A}(n)$}{$O_{A}(1)$}. +\end{theorem} + +We begin with a similar construction as in the word case, i.e. we create a graph +of $n+1$ copies of $A$'s states $Q$. The vertex set of the graph will be $(V(T) +\cup \{ \mathsf{root} \}) \times Q \}$ and we can refer to a specific vertex as +$x.q$ for $x \in V(T) \cup \{ \mathsf{root} \}, q \in Q$. Again, each letter $a \in \Sigma$ defines a function $a : Q \to Q$, and these functions induce edges in our ``fattened'' tree: consider a vertex $x$ of $T$, @@ -607,7 +613,7 @@ subproblem. Consider the following question answering problem: -\queryproblem{Highest Marked Descendant on Path Questions}{% +\queryproblem{\label{highest-marked-problem} Highest Marked Descendant on Path Questions}{% 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 @@ -615,6 +621,11 @@ Consider the following question answering problem: exists. } +\begin{theorem}\label{highest-marked-theorem} + Problem \ref{highest-marked-problem} can be solved in time + \qptime{$O(n)$}{$O(1)$}. +\end{theorem} + We will build an index structure, constructible in linear time, that allows us to handle such questions in constant time. The structure is heavily inspired by \textcite{bender2000}, where a simple algorithm for answering LCA questions is @@ -710,6 +721,10 @@ check). See Figure \ref{marked-figure} for a visual example of the data structure described. +Constructing each of $\posttable$, $\prefixtable$, and $\indextable$ takes linear +time, RMQ is a \qptime{$O(n)$}{$O(1)$} problem, so we have proved Therom +\ref{highest-marked-theorem}. + \section{Answering branch infix regular questions} With the above problem solved, we are ready to finish our algorithm for branch @@ -757,6 +772,8 @@ another jump down. Handling each jump takes constant time, and the number of jumps is bounded by $|Q|$ since we can only jump to a lower color. Thus we can answer branch infix regular questions in time constant with respect to $|T|$. +This finishes the proof of Theorem \ref{branch-infix-theorem}. + \chapter{MSO Query Answering on Trees}\label{mso-query-answering} Now we have all the pieces necessary to present our main result, which we @@ -764,9 +781,7 @@ can formulate as Theorem \ref{mso-query-answering-theorem}: \begin{theorem}\label{mso-query-answering-theorem} Problem \ref{mso-query-answering-problem} can be solved in time - \qptime{$O_{\varphi}(n)$}{$O_{\varphi}(m \log m)$}. Furthermore, time - complexity with respect to the size of the tree automaton is exponential in - both the preprocessing and question answering phases. + \qptime{$O_{\varphi}(n)$}{$O_{\varphi}(m \log m)$}. \end{theorem} First, let's reduce Problem \ref{mso-query-answering-problem} to Problem @@ -824,7 +839,9 @@ This concludes the proof of the lemma, and Theorem \begin{theorem}\label{relabel-regular-queries-theorem} Problem \ref{relabel-regular-queries} can be solved in time - \qptime{$O_A(n)$}{$O_A(m \log m)$}. + \qptime{$O_A(n)$}{$O_A(m \log m)$}. Furthermore, time + complexity with respect to the size of the tree automaton is exponential in + both the preprocessing and question answering phases. \end{theorem} The rest of this chapter is dedicated to proving Theorem |