\documentclass[10pt,a4paper]{article} \usepackage[margin=2cm]{geometry} \usepackage{minted} \usepackage{enumitem} \setenumerate[1]{label=\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{amsfonts} \usepackage{enumitem} \setenumerate[1]{label=\alph*.} \parindent0pt \title{Operating Systems - assignment 4} \author{Camil Staps\\\small{s4498062}} \begin{document} \maketitle \thispagestyle{fancy} \section*{6.11} %todo \section*{6.18} \begin{enumerate} \item If there are two types of philosophers, the fourth condition of deadlock (circular wait) is not fulfilled. Lefties and righties can never be in a chain waiting for resources. Therefore, deadlock cannot occur. \item On a table with two philosophers, of which one is a lefty and one a righty, it is obvious to see that one philosopher will start eating and when he is done, the other one can grab the forks. No starvation occurs. If on a table with $k$ philosophers no starvation occurs, and we add another philosopher $p$, this new philosopher is placed between two philosophers who, at some point, will eat. Then at some point the first fork $p$ wants to take will be free, and also at some point the other fork $p$ needs will be free. Hence, $p$ will also eat. From the principle of induction it now follows that starvation cannot occur on a table with $n\ge2$ philosophers, as long as there is at least one lefty and one righty. \end{enumerate} \end{document}