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\documentclass[10pt,a4paper]{article}
\usepackage[margin=2cm]{geometry}
\usepackage{enumitem}
\setenumerate[1]{label=2.\arabic*.}
\setenumerate[2]{label=\textit{\alph*})}
% textcomp package is not available everywhere, and we only need the Copyright symbol
% taken from http://tex.stackexchange.com/a/1677/23992
\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}}
\usepackage{fancyhdr}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{0pt}
\fancyhead{}
\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps}
\pagestyle{fancy}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\DeclareMathAlphabet{\pazocal}{OMS}{zplm}{m}{n}
\usepackage{mathtools}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
\DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor}
\usepackage{minted}
\parindent0pt
\title{Algoritmen en Datastructuren - assignment 2}
\author{Camil Staps\\\small{s4498062}}
\begin{document}
\maketitle
\thispagestyle{fancy}
\begin{enumerate}
\item %todo
\item \begin{enumerate}
\item The algorithm is performs equally well on sorted and unsorted lists. However, it performs better on lists with size (close to) $2^n$ for some $n\in\mathbb N$, since it needs a minimum amount of \texttt{merge} calls in that case. The best case would be a list of size $2^n$ for some $n\in\mathbb N$, the worst case a list of size in between $2^n$ and $2^{n+1}$, that is, $1\frac12\cdot2^n$ for some $n\in\mathbb N$.
\item Dividing takes constant time, i.e. $\Theta(1)$. Conquering takes $T(\ceil{\frac n2})+T(\floor{\frac n2})$. Composing takes linear time, $\Theta(n)$ in the \texttt{merge} function and once again linear time in the final \texttt{for} loop in \texttt{mergesort}. That gives:
$$T(n) \leq T\left(\ceil*{\tfrac n2}\right) + T\left(\floor*{\tfrac n2}\right) + c, \qquad c\in\mathbb N$$
\item %todo
\item %todo
\end{enumerate}
\item \begin{enumerate}
\item \begin{proof}
Since $f\in\pazocal O(h)$, there exists a $c_f\in\mathbb N$ and $n_{0,f}\in\mathbb N$ s.t. for all $n>n_0$ we have $c_f\cdot h(n)\geq f(n)$. Similarly we have $c_g$ and $n_{0,g}$ with respect to $g$.
Then let $c=c_f + c_g, n_0 = \text{max}(n_{0,f},n_{0,g})$. By the above we know $(c_f + c_g)\cdot h(n) \geq f(n) + g(n)$ for all $n>n_0$. It follows from substitution that $c\cdot h(n)\geq f(n) + g(n)$, and therefore $f + g \in\pazocal O(h)$.
\end{proof}
\item \begin{proof}
For $n\in\mathbb N$ we know $n^m\geq n^k$ (since $k\leq m$). We can thus choose $c=1,n_0=0$ to find that for all $n>n_0$ it holds that $n^k \leq c\cdot n^m$. Therefore, $n^k\in\pazocal O\left(n^m\right)$.
\end{proof}
\item \begin{proof}
From (b) we know that all terms ($c_kn^k$, $c_{k-1}n^{k-1}$, etc.) are in $\pazocal O\left(n^k\right)$. From (a) it then follows that their sum is also in $\pazocal O\left(n^k\right)$.
\end{proof}
\end{enumerate}
\item \begin{enumerate}
\item \begin{minted}[tabsize=0,linenos,xleftmargin=20pt,fontsize=\footnotesize]{c}
int (float* a, float* b, unsigned int n, float* out) {
unsigned int ai, bi; // counters
bi = n - 1; // start b at the max. element
for (ai = 0; ai < n; ai++) { // start a at the min. element
while (a[ai] + b[bi] > 0) bi--; // as long as b > -a, choose lower b
if (a[ai] + b[bi] == 0) { // if a = -b, return a, b and success
out[0] = a[ai];
out[1] = b[bi];
return 0;
} // otherwise, choose higher a
}
return -1; // if all as checked but nothing found, return error
}
\end{minted}
The worst case is that we need the highest element from $a$ and the lowest element from $b$, meaning we need to iterate both lists fully. In that case we need $2n+c$ operations for some $n,c\in\mathbb N$: $2n$ for iterating and checking the elements, and $c$ for returning the correct elements.
This shows we have a worst-case complexity of $\pazocal O(n)$. The best case is to find the right elements directly; in this case we need constant time.
\item %todo
\end{enumerate}
\end{enumerate}
\end{document}
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