Finished assignment 1 of 'compiler design'.

compiler/.gitignore

+*.aux
+*.pdf
+*.log
+*.toc

compiler/ass1.tex

+\documentclass[10pt,a4paper]{article}
+\usepackage{float}
+
+% aliases
+\newcommand{\tab}{\hspace*{1cm}}
+
+\title{Compiler Optimalisation Assignment 1: Loops}
+\author{Sander van Veen (6167969)}
+
+\begin{document}
+
+\maketitle
+
+\section{Natural loops} % (fold)
+\label{sec:Natural loops}
+
+Given flow graph $G = (V,E,v_0)$, where $V$ is a collection of all vertices
+(typically basic blocks), $E$ is a collection of edges (which represent the
+relation between the basic blocks) and $v_0$ is the entry node. A natural loop
+has the following definition:
+
+\begin{itemize}
+    \item A loop has a single entry point, which dominates the loop.
+    \item There must be a path back to the entry point of the loop.
+    \item Loops can be found by searching for edges of which their heads
+          dominate their tails, also know as backedges.
+    \item Given an backedge $n \rightarrow d$, the natural loop is $d$ plus the
+          nodes that can reach $n$ without going through $d$.
+\end{itemize}
+
+\noindent \textbf{Assignment: Give the natural loop(s) of the assignment's flow graph.}
+\\
+To answer this assignment, I'll go through the definition of a natural loop and
+discard the basic blocks which do not met the criteria.
+
+\begin{enumerate}
+    \item A loop has a single entry point, which dominates the loop.\\
+          Blocks $B_1$ and $B_2$ met this criteria. Block $B_5$
+          does not, since $B_3$ and $B_4$ both have an edge to $B_5$, which
+          makes $B_5$ a multiple entrance point. $B_3$ and $B_4$ do not
+          dominate the loop, since they are both a successor of $B_2$.
+    \item There must be a path back to the entry point of the loop.\\
+          Block $B_1$ does not have a path back to itself, so only block $B_2$
+          remains.
+    \item Loops can be found by searching for edges of which their heads
+          dominate their tails, also know as backedges. \\
+          Block $B_2$ has a backedge $B_5 \rightarrow B_2$.
+    \item Given an backedge $n \rightarrow d$, the natural loop is $d$ plus the
+          nodes that can reach $n$ without going through $d$.\\ Since $B_2$ is
+          the only remaining header, the assignment's flow graph has one
+          natural loop: $\{B_2, B_3, B_4, B_5\}$. Note: $B_6$ is not part of
+          the loop, because it is not stated if $B_6$ has a path back to $B_2$.
+\end{enumerate}
+
+% section Natural loops (end)
+
+\section{Reaching definition} % (fold)
+\label{sec:Reaching definition}
+
+\textbf{Assignment: Give the \texttt{gen} and \texttt{kill} sets of the basic blocks.}
+
+%A typical dataflow equation has the form:
+%\[\texttt{out}[S] = \texttt{gen}[S] \cup ( \texttt{in}[S] - \texttt{kill}[S])\]
+
+\begin{table}[H]
+\begin{tabular}{|l|l|l|} \hline
+$B_i$ & \texttt{gen} $[B_i]$  & \texttt{kill} $[B_i]$ \\ \hline
+1     & $\{d_1,d_2,d_3,d_4\}$ & $\{d_5,d_6,d_7,d_8,d_9\}$ \\ \hline
+2     & $\{d_5\}$             & $\{d_4,d_7,d_8\}$ \\ \hline
+3     & $\{d_6,7\}$           & $\{d_1,d_3,d_4,d_5,d_8\}$ \\ \hline
+4     & $\{d_8\}$             & $\{d_4,d_5\}$ \\ \hline
+5     & $\{d_9\}$             & $\{\}$ \\ \hline
+6     & undefined             & undefined \\ \hline
+\end{tabular}
+\end{table}
+
+\noindent Note: I'm not sure about the \texttt{gen} and \texttt{kill} set of
+$B_1$. I noticed $d_3$ will overwrite $d_1$, but I thought that doesn't mean
+that $d_1$ should be added to this \texttt{kill} set.
+% section Reaching definition (end)
+
+\section{Iterative algorithm for reaching definitions} % (fold)
+\label{sec:Iterative algorithm for reaching definitions}
+
+\textbf{algorithm} \\
+for each block B: \\
+\tab out[B] = gen[B] \\
+\\
+changed = true \\
+while changed: \\
+\tab changed = false \\
+\tab for each block B: \\
+\tab \tab in[B] = $\bigcup_{p \in pred(B)}$ out[p] \\
+\tab \tab oldout = out[B] \\
+\tab \tab out[B] = gen[B] $\cup$ (in[B] $-$ kill[B]) \\
+\tab \tab if out[B] $\neq$ oldout: \\
+\tab \tab \tab changed = true \\
+
+\noindent Using this algorithm, the following result is generated.
+
+\begin{table}[H]
+\begin{tabular}{|l|l|l|l|l|l|} \hline
+      & start   & first     & first       & second        & second      \\ \hline
+$B_i$ & \texttt{out}[$B_i$] & \texttt{in}[$B_i$] & \texttt{out}[$B_i$] &
+    \texttt{in}[$B_i$] & \texttt{out}[$B_i$] \\ \hline
+1     & 1,2,3,4 & $\emptyset$ & 1,2,3,4     & $\emptyset$   & 1,2,3,4     \\ \hline
+2     & 5       & 1,2,3,4,9   & 1,2,3,5,9   & 1,2,3,4,5,8,9 & 1,2,3,5,9   \\ \hline
+3     & 6,7     & 1,2,3,5,9   & 2,6,7,9     & 1,2,3,5,9     & 2,6,7,9     \\ \hline
+4     & 8       & 1,2,3,5,9   & 1,2,3,8,9   & 1,2,3,5,9     & 1,2,3,8,9   \\ \hline
+5     & 9       & 1,2,3,5,8,9 & 1,2,3,5,8,9 & 1,2,3,5,8,9   & 1,2,3,5,8,9 \\ \hline
+6     & undef.  & 1,2,3,8,9   & undef.      & 1,2,3,8,9     & undef.      \\ \hline
+\end{tabular}
+\end{table}
+
+% section Iterative algorithm for reaching definitions (end)
+
+\end{document}