2019-2-14-b-tree: +b-tree-predecessor solution

Author: hengxin

2019/2019-2/2019-2-14-b-tree/2019-2-14-b-tree.pdf

@@ -0,0 +1,32 @@
+% b-tree-predecessor.tex
+
+\begin{algorithm}[H]
+% \caption{Predecessor of a given key in B-tree.}
+% \label{b-tree-predecessor}
+\begin{algorithmic}[1]
+ \Procedure{B-Tree-Predecessor}{$T, x, i$} \Comment{\purple{$x.key_{i}$ in B-tree $T$}}
+ \If{$x.\leaf = 0$}
+ \State \Return the rightmost key in $x.c_{i}$
+ \ElsIf{$i \ge 2$} \Comment{$x.\leaf = 1$}
+ \State \Return $(x, i-1)$
+ \Else \Comment{\red{$x.\leaf = 1 \land i = 1$}}
+ \State $y \gets x.p$
+ \While{\red{$y \neq T.root \land x = y.c_{1}$}}
+ \Comment{\blue{exit: $y = T.root \lor x \neq y.c_{1}$}}
+ \State $x \gets y$
+ \State $y \gets y.p$
+ \EndWhile
+
+ \If{\red{$x = y.c_{1}$}} \Comment{\teal{$y = T.root \land x = y.c_{1}$}}
+ \State \Return ``no predecessor''
+ \Else \Comment{\teal{$x \neq y.c_{1}$}}
+ \State $j \gets 2$
+ \While{$y.c_{j} \neq x$}
+ \State $j \gets j + 1$
+ \EndWhile
+ \State \Return $(y, j-1)$ \Comment{\purple{$x = y.c_{j}$}}
+ \EndIf
+ \EndIf
+ \EndProcedure
+\end{algorithmic}
+\end{algorithm}

2019/2019-2/2019-2-14-b-tree/algs/b-tree-predecessor.tex

@@ -19,19 +19,67 @@
 %%%%%%%%%%%%%%%
 \begin{frame}{}
  \begin{exampleblock}{Predecessor (TC 18.2-3)}
- Explain how to find the \red{predecessor of a given key} stored in a B-tree.
+ Explain how to find the \red{predecessor of a given key $(x, i)$} stored in a B-tree.
  \end{exampleblock}
 
- \fig{width = 0.75\textwidth}{figs/b-tree-example}
+ \pause
+ \fig{width = 1.00\textwidth}{figs/b-tree-37-char}
+ \begin{center}
+ \teal{B-tree with 37 characters.}
+ \end{center}
+
+ \pause
+ \[
+ \red{x.\leaf = 0}
+ \]
+
+ \pause
+ \vspace{-0.30cm}
+ \[
+ \blue{H \qquad N} \qquad \teal{S \qquad V} \qquad \purple{Q}
+ \]
+
+ \pause
+ \vspace{0.20cm}
+ \begin{center}
+ find the \red{rightmost key} in $x.c_{i}$
+ \end{center}
 \end{frame}
 %%%%%%%%%%%%%%%
 
 %%%%%%%%%%%%%%%
 \begin{frame}{}
+ \[
+ (x, i)
+ \]
+ \fig{width = 1.00\textwidth}{figs/b-tree-37-char}
+
+ \[
+ \red{x.\leaf = 1}
+ \]
+
+ \vspace{-0.30cm}
+ \[
+ \blue{P \qquad U \qquad} \teal{T\qquad O \qquad} \purple{A}
+ \]
+
+ \pause
+ \[
+ i \ge 2 \implies (x, i-1)
+ \]
+ \[
+ i = 1 \implies \text{ find $(y, j)$ such that $x$ is the \red{leftmost key} in $y.c_{j + 1}$}
+ \]
+
+ \pause
+ \begin{center}
+ $A$ is the \purple{only} key which has no predecessor.
+ \end{center}
 \end{frame}
 %%%%%%%%%%%%%%%
 
 %%%%%%%%%%%%%%%
 \begin{frame}{}
+ \input{algs/b-tree-predecessor}
 \end{frame}
 %%%%%%%%%%%%%%%

2019/2019-2/2019-2-14-b-tree/figs/b-tree-27-char.png

@@ -97,9 +97,9 @@
  \Big( \left\lfloor \frac{n - T_{i}}{2^{i}} \right\rfloor + 2 \Big) \\}
  \onslide<5->{&= 1 + \sum_{i = 1}^{\infty} \teal{[T_{i} \le n]} 
  \Big( \left\lfloor \frac{n - 2^{i+1} - i + 1}{2^{i}} \right\rfloor \Big) \\}
- \onslide<6->{&= 1 + \sum_{i = 1}^{2^{i+1} + i - 1 \le n} 
+ \onslide<6->{&= 1 + \sum_{\substack{i \ge 1 \\ 2^{i+1} + i - 1 \le n}} 
  \left\lfloor \frac{n - i + 1}{2^{i}} \right\rfloor \\}
- \onslide<7->{&= 1 + \sum_{\red{i = 0}}^{2^{i+2} + i \le n} 
+ \onslide<7->{&= 1 + \sum_{\substack{\red{i \ge 0} \\ 2^{i+2} + i \le n}} 
  \left\lfloor \frac{n - i}{2^{i + 1}} \right\rfloor}
  \end{align*}
 \end{frame}

2019/2019-2/2019-2-14-b-tree/figs/b-tree-37-char.png

@@ -66,12 +66,11 @@
  \end{figure}
 }
 
-\usepackage{tikz}
+\usepackage{tikz, adjustbox}
 \usetikzlibrary{positioning, mindmap, shadows, calc}
 
-\newcommand{\titletext}{2-11 Heapsort}
-\usepackage{media9}
-\usepackage{multimedia}
+\newcommand{\titletext}{2-14 B-Trees}
+\newcommand{\leaf}{\textsl{leaf}}
 
 \newcommand{\thankyou}{
  \begin{frame}[noframenumbering]{}