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Noah L. Schrick 2022-04-21 00:26:09 -05:00
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Schrick-Noah_Project-Writeup.aux
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Schrick-Noah_Project-Writeup.tex
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\tableofcontents
\section{Problem Introduction}
Red-Black trees are binary search trees that contain an additional one-bit field that denotes the color of a node as either ``red" or ``black". These trees have additional structural requirements that must be followed, as well as the traditional binary tree requirements. Red-Black trees are also balanced, so a height is guaranteed as a function of \textit{n}. A full description of Red-Black trees were discussed in lecture, and the algorithms textbook by Cormen also gives a full work-through of the data structure. The main advantage of Red-Black trees is that the worst-case for searching, inserting, and deleting is $\mathcal{O}({log(n)})$. For this project, an implementation of Red-Black is provided in C++, with results shown for given problems.
\section{Program Platform and Submission Files}
This problem was solved using C++ on a Linux system.
@ -68,7 +69,7 @@ The cleanup function was implemented by following the pseudocode of the three ca
\subsubsection{Display}
\section{Results}
\subsection{Part 1.B: ``Tree 1"}
As a note, the ``.'' represents the right child, and the ` represents the left child. The terminal-based print does appear to merge together, so in some subtrees it may appear that there are two red nodes next to each other. However, despite this first-glance look, it is verifiable that there are no two red nodes next to each other.
\begin{figure}[htp]
\centering
\includegraphics[width=\linewidth]{"./images/b_init_tree.png"}
@ -101,7 +102,6 @@ The cleanup function was implemented by following the pseudocode of the three ca
\label{fig:b_41}
\end{figure}
\subsection{Part 1.C: ``Tree 2"}
\begin{figure}[htp]
\includegraphics[width=\linewidth]{"./images/c_init_tree.png"}
\vspace{.2truein} \centerline{}
@ -124,5 +124,18 @@ The cleanup function was implemented by following the pseudocode of the three ca
\end{figure}
\section{Part 2: Red-Black Discussion}
\subsection{Question 2.1}
When inserting a node into a red-black tree and then immediately deleting the same node, the red-black tree before the first insertion is not necessarily the same as the tree after the deletion. An example of this can be seen in Figure \ref{fig:2_1}. When inserting node 100 into the tree, it gets inserted as the right child of node 28. Since both node 28 and node 100 are red, it causes a rotation in the tree. After the node is deleted, the tree does not get rotated back, so the tree now has node 15 as the left child of node 28, instead of node 28 being the right child of node 15.
\begin{figure}[htp]
\centering
\includegraphics[width=\linewidth]{"./images/2_1.png"}
\vspace{.2truein} \centerline{}
\caption{Part 2.1: Example Tree}
\label{fig:2_1}
\end{figure}
\subsection{Question 2.2}
Maintaining the black height of nodes during the insertion process does not adversely affect the asymptotic performance of the process. Both Case 1 and Case 2 do not change the black height, so no work must be performed in these cases. In Case 1, black height is also not directly changed. In Case 1, the black node may shift down, but the height itself does not change. However, as a result of Case 1, we may color root as red, which then gets corrected to black. This part does change the black height. However, recoloring is constant time, and the black height only needs to be altered for the nearest neighbors. Similar to the rotation process, there is no need to walk down the entirety of the tree; only the nearest children may need an adjustment, which is constant time.
\end{document}

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\contentsline {section}{\numberline {3}Programming Approach}{3}{}%
\contentsline {subsection}{\numberline {3.1}Node Class}{3}{}%
\contentsline {subsection}{\numberline {3.2}Red-Black Tree Class}{3}{}%
\contentsline {subsubsection}{\numberline {3.2.1}Constructing the Problem}{3}{}%
\contentsline {subsubsection}{\numberline {3.2.2}Generating the Solution}{3}{}%
\contentsline {subsubsection}{\numberline {3.2.3}Printing the solution}{3}{}%
\contentsline {section}{\numberline {4}Results}{3}{}%
\contentsline {subsection}{\numberline {4.1}Part 1.B: ``Tree 1"}{3}{}%
\contentsline {subsection}{\numberline {4.2}Part 1.C: ``Tree 2"}{5}{}%
\contentsline {subsubsection}{\numberline {3.2.1}Insert}{4}{}%
\contentsline {subsubsection}{\numberline {3.2.2}Delete}{4}{}%
\contentsline {subsubsection}{\numberline {3.2.3}Tree Cleanup}{4}{}%
\contentsline {subsubsection}{\numberline {3.2.4}Display}{4}{}%
\contentsline {section}{\numberline {4}Results}{4}{}%
\contentsline {section}{\numberline {5}Part 2: Red-Black Discussion}{5}{}%
\contentsline {subsection}{\numberline {5.1}Question 2.1}{5}{}%
\contentsline {subsection}{\numberline {5.2}Question 2.2}{7}{}%

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src/main.cpp
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tree2.del(tree2.get_root(), 221);
tree2.display(tree2.get_root(), nullptr, false);
std::cout << "" << std::endl;
RB tree3 = RB();
std::vector<int> part_2 {9, 12, 15, 28, 10};
for (int k : part_2){
Node* newnode = new Node(k);
tree3.ins(tree3.get_root(), newnode);
}
std::cout << "Initial Part 2.1 Tree:" << std::endl;
tree3.display(tree3.get_root(), nullptr, false);
std::cout << "" << std::endl;
std::cout << "Inserting 100 into Tree 3" << std::endl;
Node* newnode1 = new Node(100);
tree3.ins(tree3.get_root(), newnode1);
tree3.display(tree3.get_root(), nullptr, false);
std::cout << "" << std::endl;
std::cout << "Removing 100" << std::endl;
tree3.del(tree3.get_root(), 100);
tree3.display(tree3.get_root(), nullptr, false);
}