stehau

W2/report/main.tex

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+\documentclass[a4paper]{article}
+\usepackage{listings}
+\usepackage{fontspec}
+\usepackage[margin=1in]{geometry}
+\usepackage{amsmath}
+\usepackage{multicol}
+
+\renewcommand{\thesection}{Task \arabic{section}}
+\renewcommand{\thesubsection}{\arabic{subsection})}
+\renewcommand{\thesubsubsection}{\alph{subsubsection})}
+
+\input{fasto.sty}
+
+\setmonofont[Scale=0.9]{Antikor Mono Medium}
+
+\setlength{\parskip}{5pt}
+\setlength{\parindent}{0pt}
+
+\title{W2 - IPS}
+\author{Nikolaj Gade (qhp695)}
+\date{May 2022}
+
+\newcommand{\reg}[1]{%
+\begin{center}
+  \large
+  \texttt{#1}
+\end{center}
+}
+
+\begin{document}
+\maketitle
+\section{}
+\subsection{}
+\begin{multicols}{2}
+\subsubsection{}
+The string must consist of any number of pairs of '\texttt{o}'s and '\texttt{g}'s. The pairs can be any combination of 2 characters, both of which are either '\texttt{o}' or '\texttt{g}'. Thus, each character can be written as, '\texttt{o|g}', and each pair as '\texttt{(o|g)(o|g)}'. The full string can be written as:
+
+\reg{((o|g)(o|g))*}
+
+\subsubsection{}
+The first character must be '\texttt{o}', and it must be followed by a character that is either '\texttt{o}' or '\texttt{g}'. Following that, the answer is the same as the previous problem:
+
+\reg{o(o|g)((o|g)(o|g))*}
+
+\columnbreak
+
+\subsubsection{}
+Keeping with the idea that an even number of characters can be split up in "pairs" of characters, an odd number of '\texttt{o}'s or '\texttt{g}'s \textit{must} result in an odd number of both. Thus, the string will either consist of an even number of '\texttt{o}'s, followed by an even number of '\texttt{g}'s, \textbf{or} an even number of '\texttt{o}'s, followed by '\texttt{og}', followed by an even number of '\texttt{g}'s. The full string can be written as:
+
+\reg{(oo)*(og)?(gg)*}
+
+\end{multicols}
+\subsection{}
+
+\begin{multicols}{2}
+\subsubsection{}
+'\textit{a}'s and '\textit{b}'s are placed at the same time, using the \textit{T} starting symbol. After all '\textit{a}'s and '\textit{b}'s are placed, the symbol turns to '\textit{S}', which places at least 1 '\textit{c}'.
+
+\[
+  T =
+  \begin{cases}
+    aTb \\
+    S
+  \end{cases}
+\]
+\[
+  S =
+  \begin{cases}
+    c \\
+    cS
+  \end{cases}
+\]
+
+\subsubsection{}
+Like with the previous problem, both the '\textit{a}'s and the '\textit{b}'s are placed at the same time, but the '\textit{b}'s are placed 2 at a time.
+
+\[
+  T =
+  \begin{cases}
+    aTbb \\
+    abb
+  \end{cases}
+\]
+
+\subsubsection{}
+Once again, '\textit{a}'s and '\textit{b}'s are placed at the same time with the starting symbol \textit{T}. Afterwards, any amount of '\textit{a}'s are placed before the '\textit{b}'s.
+
+\[
+  T =
+  \begin{cases}
+    aTb \\
+    S
+  \end{cases}
+\]
+\[
+  S =
+  \begin{cases}
+     \\
+    aS
+  \end{cases}
+\]
+\end{multicols}
+
+\newpage
+\subsection{}
+\subsubsection{}
+\texttt{\%nonassoc letprec} designates the \texttt{let} token as being non-associative, which means ambiguous implementations will lead to a syntax error.
+
+\subsubsection{}
+The order of the associativity declarations provide the precedence for the operators. So in the current way the code in written, the line \texttt{let x = 10 in x + 10 > 15} will be parsed as \texttt{let x = 10 in (x < 15)}, but if , it would be parsed as \texttt{(let x = 10 in x) < 15}.
+
+\subsubsection{}
+The code \texttt{\{ Let (Dec (fst \$2, \$4, \$3), \$6, \$1) \}} creates a \texttt{Let} instance, which containes the declared variable, the following expression, as well as the keyword.
+
+\section{}
+See code.
+
+\section{}
+\subsubsection{}
+\reg{filter (('a -> bool) * ['a]) -> ['a]}
+
+\subsubsection{}
+
+\begin{lstlisting}
+CheckExp(exp, vtable, ftable) = case exp of
+  filter(p, arr_exp) =>
+    let array_type = CheckExp(arr_exp, vtable, ftable)
+    let element_type = match array_type with
+      | Array(type) -> type
+      | _ -> Error()
+
+    let function_type = lookup(ftable, name(p))
+    match function_type with
+      | unbound -> Error()
+      | (input_type, output_type) ->
+        if input_type == element_type && output_type == bool then
+          Array(element_type)
+        else Error()
+      | _ -> Error()
+\end{lstlisting}
+
+\newpage
+\section{}
+
+
+\end{document}