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Insert the $\langle \mathsf{key}, \mathsf{value} \...

Created on: March 12, 2025

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Insert the ⟨key,value⟩\langle \mathsf{key}, \mathsf{value} \rangle items
into an empty B-tree~\cite[\S 18]{Cormen2022} with minimum degree t=2t = 2:

\begin{enumerate}
\item ⟨32,I⟩,  ⟨17,I⟩,  ⟨9,X⟩,  ⟨21,C⟩,  ⟨11,E⟩\langle 32, I \rangle, \; \langle 17, I \rangle, \; \langle 9, X \rangle, \; \langle 21, C \rangle, \; \langle 11, E \rangle
\item ⟨2,A⟩,  ⟨28,E⟩,  ⟨36,G⟩,  ⟨3,E⟩,  ⟨4,T⟩\langle 2, A \rangle, \; \langle 28, E \rangle, \; \langle 36, G \rangle, \; \langle 3, E \rangle, \; \langle 4, T \rangle
\item ⟨18,B⟩,  ⟨13,E⟩,  ⟨6,T⟩,  ⟨7,R⟩,  ⟨37,I⟩\langle 18, B \rangle, \; \langle 13, E \rangle, \; \langle 6, T \rangle, \; \langle 7, R \rangle, \; \langle 37, I \rangle
\item ⟨26,L⟩,  ⟨33,N⟩,  ⟨20,U⟩,  ⟨24,I⟩,  ⟨30,D⟩\langle 26, L \rangle, \; \langle 33, N \rangle, \; \langle 20, U \rangle, \; \langle 24, I \rangle, \; \langle 30, D \rangle
\end{enumerate}

Show the state of the tree after every 5 insertions.
Depict each tree as a sequence of arrays for each layer.
For example, consider this B-tree:

\begin{center}
\footnotesize
\begin{tikzpicture}[level distance=30pt]
\tikzstyle{bplus}=[rectangle split, rectangle split horizontal,rectangle split ignore empty parts,draw]
\tikzstyle{every node}=[bplus]
\tikzstyle{level 1}=[sibling distance=80mm]
\tikzstyle{level 2}=[sibling distance=35mm]
\node {⟨9,I⟩\langle 9, I \rangle} [->]
child {node {⟨2,A⟩\langle 2, A \rangle \nodepart{two} ⟨6,T⟩\langle 6, T \rangle}
child[sibling distance=25mm] {node {⟨1,D⟩\langle 1, D \rangle}}
child {node {⟨3,S⟩\langle 3, S\rangle \nodepart{two} ⟨4,T⟩\langle 4, T\rangle \nodepart{three} ⟨5,A⟩\langle 5, A\rangle}}
child {node {⟨7,S⟩\langle 7, S\rangle \nodepart{two} ⟨8,T⟩\langle 8, T\rangle}}
}
child {node {⟨15,D⟩\langle 15, D \rangle}
child {node {⟨10,R⟩\langle 10, R \rangle \nodepart{two} ⟨12,U⟩\langle 12, U \rangle}}
child {node {⟨17,R⟩\langle 17, R\rangle \nodepart{two} ⟨19,E⟩\langle 19, E\rangle \nodepart{three} ⟨20,S⟩\langle 20, S\rangle}}
}
;\end{tikzpicture}
\end{center}

The tree above must be depicted as follows:

\begin{enumerate}
\item[(layer 1)]
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨9,I⟩\langle 9, I \rangle} \
\hline
\end{tabular}
\item[(layer 2)]
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨2,A⟩\langle 2, A \rangle}
& \makebox[\widthof{0000}][c]{⟨6,T⟩\langle 6, T \rangle} \
\hline
\end{tabular}\quad
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨15,D⟩\langle 15, D \rangle} \
\hline
\end{tabular}
\item[(layer 3)]
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨1,D⟩\langle 1, D \rangle} \
\hline
\end{tabular}\quad
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨3,S⟩\langle 3, S \rangle}
& \makebox[\widthof{0000}][c]{⟨4,T⟩\langle 4, T \rangle}
& \makebox[\widthof{0000}][c]{⟨5,A⟩\langle 5, A \rangle} \
\hline
\end{tabular}\quad
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨7,S⟩\langle 7, S \rangle}
& \makebox[\widthof{0000}][c]{⟨8,T⟩\langle 8, T \rangle} \
\hline
\end{tabular}\quad
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨10,R⟩\langle 10, R \rangle}
& \makebox[\widthof{0000}][c]{⟨12,U⟩\langle 12, U \rangle} \
\hline
\end{tabular}\quad
\begin{tabular}{|c|c|c|c|}
\hline \makebox[\widthof{0000}][c]{⟨17,R⟩\langle 17, R \rangle}
& \makebox[\widthof{0000}][c]{⟨19,E⟩\langle 19, E \rangle}
& \makebox[\widthof{0000}][c]{⟨20,S⟩\langle 20, S \rangle} \
\hline
\end{tabular}
\end{enumerate}

\color{blue}
\textbf{Answer:} \emph{(replace with your answer)}
\color{black} solve it in latex in english. you must refer to Cormen, T.H., Leiserson, C.E., Rivest, R.L. and Stein, C., 2022. Introduction to algorithms, Fourth Edition. MIT press.
[GTG] M. T. Goodrich, R. Tamassia, and M. H. Goldwasser. Data Structures and Algorithms
in Java. WILEY 2014.
you need to write answer and solution in (replace with your answer) in document

you can also refer to this solution, but you have to write the solution yourself, solution is in photos.

Thought

Reasoned about B-tree insertion process for 1m 59s

Answer

latex
\documentclass[11pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{array}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}

\noindent
\textbf{Solution (following the standard B-tree insertion algorithm of 
\cite[\S 18]{Cormen2022} and \cite{GTG}):}

\vspace*{1em}

We are to insert the following $ \langle \mathsf{key}, \mathsf{value} \rangle $ pairs 
into an initially empty B-tree of minimum degree $t=2$ (so each node 
may contain up to $2t - 1 = 3$ keys). 
They are inserted in batches of five:
$$
\begin{aligned}
&\text{1st batch: } 
  \langle 32,I\rangle,\;\langle 17,I\rangle,\;\langle 9,X\rangle,\;
  \langle 21,C\rangle,\;\langle 11,E\rangle,\\
&\text{2nd batch: } 
  \langle 2,A\rangle,\;\langle 28,E\rangle,\;\langle 36,G\rangle,\;
  \langle 3,E\rangle,\;\langle 4,T\rangle,\\
&\text{3rd batch: } 
  \langle 18,B\rangle,\;\langle 13,E\rangle,\;\langle 6,T\rangle,\;
  \langle 7,R\rangle,\;\langle 37,I\rangle,\\
&\text{4th batch: }
  \langle 26,L\rangle,\;\langle 33,N\rangle,\;\langle 20,U\rangle,\;
  \langle 24,I\rangle,\;\langle 30,D\rangle.
\end{aligned}
$$

We show below the state of the B-tree \emph{after each group of five insertions}. 
All nodes are depicted (layer by layer) as small arrays of up to three keys.

\medskip
\hrule
\medskip

\paragraph{After the first five insertions}
$$
\bigl\langle (32,I),\, (17,I),\, (9,X),\, (21,C),\, (11,E)\bigr\rangle
$$
Insertion of these five causes one split, leaving $\langle 17,I\rangle$ 
in the root and two children.  

\noindent
\textbf{Layer 1:}
$$
\begin{tabular}{|c|}
\hline
$\langle 17,I\rangle$ \\
\hline
\end{tabular}
$$

\noindent
\textbf{Layer 2:}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 9,X\rangle$ & $\langle 11,E\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 21,C\rangle$ & $\langle 32,I\rangle$\\
\hline
\end{tabular}

\medskip
\hrule
\medskip

\paragraph{After the second five insertions}
$$
\bigl\langle (2,A),\, (28,E),\, (36,G),\, (3,E),\, (4,T)\bigr\rangle
$$
Now the root ends with three keys.  
(The child that reached four keys $\{21,28,32,36\}$ split, promoting 
$\langle 28,E\rangle$ upward.)

\noindent
\textbf{Layer 1:}
$$
\begin{tabular}{|c|c|c|}
\hline
$\langle 9,X\rangle$ & $\langle 17,I\rangle$ & $\langle 28,E\rangle$\\
\hline
\end{tabular}
$$

\noindent
\textbf{Layer 2:}
\quad
\begin{tabular}{|c|c|c|}
\hline
$\langle 2,A\rangle$ & $\langle 3,E\rangle$ & $\langle 4,T\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|}
\hline
$\langle 11,E\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|}
\hline
$\langle 21,C\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 32,I\rangle$ & $\langle 36,G\rangle$\\
\hline
\end{tabular}

\medskip
\hrule
\medskip

\paragraph{After the third five insertions}
$$
\bigl\langle (18,B),\, (13,E),\, (6,T),\, (7,R),\, (37,I)\bigr\rangle
$$
When inserting $\langle 6,T\rangle$, the leftmost child of the root was full, 
causing a cascade of splits.  The new root becomes $\langle 9,X\rangle$ 
alone, and we now have \emph{three} levels.

\noindent
\textbf{Layer 1 (root):}
$$
\begin{tabular}{|c|}
\hline
$\langle 9,X\rangle$\\
\hline
\end{tabular}
$$

\noindent
\textbf{Layer 2:}
\quad
\begin{tabular}{|c|}
\hline
$\langle 3,E\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 17,I\rangle$ & $\langle 28,E\rangle$ \\
\hline
\end{tabular}

\noindent
\textbf{Layer 3:} 
Below $\langle 3,E\rangle$, we have two children:
$$
\begin{tabular}{|c|}
\hline
$\langle 2,A\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}
\hline
$\langle 4,T\rangle$ & $\langle 6,T\rangle$ & $\langle 7,R\rangle$\\
\hline
\end{tabular}
$$

Below $\langle 17,I\rangle \;\langle 28,E\rangle$, we have three children:
$$
\begin{tabular}{|c|c|}
\hline
$\langle 11,E\rangle$ & $\langle 13,E\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 18,B\rangle$ & $\langle 21,C\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}
\hline
$\langle 32,I\rangle$ & $\langle 36,G\rangle$ & $\langle 37,I\rangle$\\
\hline
\end{tabular}
$$

\medskip
\hrule
\medskip

\paragraph{After the fourth (final) five insertions}
$$
\bigl\langle (26,L),\, (33,N),\, (20,U),\, (24,I),\, (30,D)\bigr\rangle
$$
Several splits occur in the right subtree when we insert 
$\langle 33,N\rangle$ and $\langle 20,U\rangle$.
The final structure has height 3, with a two-key root:

\noindent
\textbf{Layer 1 (root):}
$$
\begin{tabular}{|c|c|}
\hline
$\langle 9,X\rangle$ & $\langle 21,I\rangle$ \\
\hline
\end{tabular}
$$

\noindent
\textbf{Layer 2:}
\quad
\begin{tabular}{|c|}
\hline
$\langle 3,E\rangle$ \\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|}
\hline
$\langle 17,I\rangle$ \\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 28,E\rangle$ & $\langle 36,G\rangle$\\
\hline
\end{tabular}

\noindent
\textbf{Layer 3:} 
Below $\langle 3,E\rangle$:
$$
\begin{tabular}{|c|}
\hline
$\langle 2,A\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}
\hline
$\langle 4,T\rangle$ & $\langle 6,T\rangle$ & $\langle 7,R\rangle$\\
\hline
\end{tabular}
$$

Below $\langle 17,I\rangle$:
$$
\begin{tabular}{|c|c|}
\hline
$\langle 11,E\rangle$ & $\langle 13,E\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|}
\hline
$\langle 18,B\rangle$ & $\langle 20,U\rangle$\\
\hline
\end{tabular}
$$

Below $\langle 28,E\rangle\;\langle 36,G\rangle$:
$$
\begin{tabular}{|c|c|}
\hline
$\langle 24,I\rangle$ & $\langle 26,L\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|c|c|}
\hline
$\langle 30,D\rangle$ & $\langle 32,I\rangle$ & $\langle 33,N\rangle$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|c|}
\hline
$\langle 37,I\rangle$\\
\hline
\end{tabular}
$$

\medskip

\noindent
Hence, this is the final B-tree of minimum degree $t=2$ after all 20 insertions.

\begin{thebibliography}{9}

\bibitem{Cormen2022}
T.\,H.~Cormen, C.\,E.~Leiserson, R.\,L.~Rivest, and C.~Stein.
\textit{Introduction to Algorithms}, Fourth Edition. MIT Press, 2022.

\bibitem{GTG}
M.\,T.~Goodrich, R.~Tamassia, and M.\,H.~Goldwasser.
\textit{Data Structures and Algorithms in Java}. Wiley, 2014.

\end{thebibliography}

\end{document}

Explanation Sketch (Informal):

  1. \textbf{Start empty.} Insert the first 5 keys ⟨32,I⟩,⟨17,I⟩,…\langle 32,I\rangle,\langle 17,I\rangle,\ldots one at a time, splitting when a node exceeds 3 keys.

  2. \textbf{Second 5 keys} ⟨2,A⟩,⟨28,E⟩,…\langle 2,A\rangle,\langle 28,E\rangle,\ldots again cause a split when we insert 3636, which pushes ⟨28,E⟩\langle 28,E\rangle up.
    We end with the root containing {9,17,28}\{9,17,28\}.

  3. \textbf{Third 5 keys} ⟨18,B⟩,⟨13,E⟩,…\langle 18,B\rangle,\langle 13,E\rangle,\ldots force a \emph{root} split when inserting 66.
    The new root is ⟨9,X⟩\langle 9,X\rangle.

  4. \textbf{Fourth 5 keys} ⟨26,L⟩,⟨33,N⟩,…\langle 26,L\rangle,\langle 33,N\rangle,\ldots cause splitting in the right subtree.
    Eventually 2121 is promoted up to the root, yielding the final tree with root keys ⟨9,X⟩,⟨21,I⟩\langle 9,X\rangle,\langle 21,I\rangle.

All intermediate and final states are shown above in “layered” array form.

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