# HG changeset patch
# User Christian Urban <christian dot urban at kcl dot ac dot uk>
# Date 1462536952 -3600
# Node ID 55f097ab96c9c95bec4f065457df1c9168ac4d04
# Parent  5d85dc9779b13312793b9a447ef89ab3cb480751
updated

diff -r 5d85dc9779b1 -r 55f097ab96c9 handouts/scala-ho.tex
--- a/handouts/scala-ho.tex	Wed Apr 06 15:55:01 2016 +0100
+++ b/handouts/scala-ho.tex	Fri May 06 13:15:52 2016 +0100
@@ -3,6 +3,8 @@
 \usepackage{../langs}
 \usepackage{marvosym}
 
+%cheat sheet
+%http://worldline.github.io/scala-cheatsheet/
 
 \begin{document}
 
diff -r 5d85dc9779b1 -r 55f097ab96c9 hws/proof.pdf
Binary file hws/proof.pdf has changed
diff -r 5d85dc9779b1 -r 55f097ab96c9 hws/proof.tex
--- a/hws/proof.tex	Wed Apr 06 15:55:01 2016 +0100
+++ b/hws/proof.tex	Fri May 06 13:15:52 2016 +0100
@@ -1,10 +1,6 @@
 \documentclass{article}
-\usepackage{charter}
-\usepackage{hyperref}
-\usepackage{amssymb}
-\usepackage{amsmath}
+\usepackage{../style}
 
-\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions
 \begin{document}
 
 \section*{Proof}
@@ -14,16 +10,16 @@
 \begin{center}
 \begin{tabular}{c}
 \begin{tabular}[t]{rcl}
-  $r$ & $::=$  & $\varnothing$  \\
-         & $\mid$ & $\epsilon$       \\
+  $r$ & $::=$  & $\ZERO$  \\
+         & $\mid$ & $\ONE$       \\
          & $\mid$ & $c$                         \\
          & $\mid$ & $r_1 \cdot r_2$ \\
          & $\mid$ & $r_1 + r_2$  \\
          & $\mid$ & $r^*$                 \\
   \end{tabular}\hspace{10mm}
 \begin{tabular}[t]{r@{\hspace{1mm}}c@{\hspace{1mm}}l}
-$L(\varnothing)$ & $\dn$ & $\varnothing$ \\
-$L(\epsilon)$       & $\dn$ & $\{\texttt{""}\}$       \\
+$L(\ZERO)$ & $\dn$ & $\varnothing$ \\
+$L(\ONE)$       & $\dn$ & $\{\texttt{""}\}$       \\
 $L(c)$                   & $\dn$ & $\{\texttt{"}c\texttt{"}\}$                         \\
 $L(r_1 \cdot r_2)$   & $\dn$ & $L(r_1) \,@\, L(r_2)$ \\
 $L(r_1 + r_2)$         & $\dn$ & $L(r_1) \cup L(r_2)$  \\
@@ -37,9 +33,9 @@
 
 \begin{center}
 \begin{tabular}{lcl}
-  $der\, c\, (\varnothing)$    & $\dn$ & $\varnothing$  \\
-  $der\, c\, (\epsilon)$          & $\dn$ & $\varnothing$ \\
-  $der\, c\, (d)$                      & $\dn$ & if $c = d$ then $\epsilon$ else $\varnothing$\\
+  $der\, c\, (\ZERO)$    & $\dn$ & $\ZERO$  \\
+  $der\, c\, (\ONE)$          & $\dn$ & $\ZERO$ \\
+  $der\, c\, (d)$                      & $\dn$ & if $c = d$ then $\ONE$ else $\ZERO$\\
   $der\, c\, (r_1 + r_2)$        & $\dn$ & $(der\, c\, r_1) + (der\, c\, r_2)$ \\
   $der\, c\, (r_1 \cdot r_2)$ & $\dn$  & if $nullable(r_1)$\\
        & & then $((der\, c\, r_1) \cdot r_2) + (der\, c\, r_2)$\\ 
@@ -53,7 +49,7 @@
 regular expressions. We can establish that $\forall r.\; P(r)$ holds, provided we can show the following:
 
 \begin{enumerate}
-\item $P(\varnothing)$, $P(\epsilon)$ and $P(c)$ all hold,
+\item $P(\ZERO)$, $P(\ONE)$ and $P(c)$ all hold,
 \item $P(r_1 + r_2)$ holds under the induction hypotheses that 
 $P(r_1)$ and $P(r_2)$ hold,
 \item $P(r_1 \cdot r_2)$ holds under the induction hypotheses that 
@@ -86,25 +82,25 @@
 {\bf Proof}
 
 \noindent
-According to 1.~above we need to prove $P(\varnothing)$, $P(\epsilon)$ and $P(d)$. Lets do this in turn.
+According to 1.~above we need to prove $P(\ZERO)$, $P(\ONE)$ and $P(d)$. Lets do this in turn.
 
 \begin{itemize}
-\item First Case: $P(\varnothing)$ is $L(der\,c\,\varnothing) = Der\,c\,(L(\varnothing))$ (a). We have $der\,c\,\varnothing = \varnothing$ 
-and $L(\varnothing) = \varnothing$. We also have $Der\,c\,\varnothing = \varnothing$. Hence we have $\varnothing = \varnothing$ in (a). 
+\item First Case: $P(\ZERO)$ is $L(der\,c\,\ZERO) = Der\,c\,(L(\ZERO))$ (a). We have $der\,c\,\ZERO = \ZERO$ 
+and $L(\ZERO) = \ZERO$. We also have $Der\,c\,\ZERO = \ZERO$. Hence we have $\ZERO = \ZERO$ in (a). 
 
-\item Second  Case: $P(\epsilon)$ is $L(der\,c\,\epsilon) = Der\,c\,(L(\epsilon))$ (b). We have $der\,c\,\epsilon = \varnothing$,
-$L(\varnothing) = \varnothing$ and $L(\epsilon) = \{\texttt{""}\}$. We also have $Der\,c\,\{\texttt{""}\} = \varnothing$. Hence we have 
-$\varnothing = \varnothing$ in (b). 
+\item Second  Case: $P(\ONE)$ is $L(der\,c\,\ONE) = Der\,c\,(L(\ONE))$ (b). We have $der\,c\,\ONE = \ZERO$,
+$L(\ZERO) = \ZERO$ and $L(\ONE) = \{\texttt{""}\}$. We also have $Der\,c\,\{\texttt{""}\} = \ZERO$. Hence we have 
+$\ZERO = \ZERO$ in (b). 
 
 \item Third  Case: $P(d)$ is $L(der\,c\,d) = Der\,c\,(L(d))$ (c). We need to treat the cases $d = c$ and $d \not= c$. 
 
-$d = c$: We have $der\,c\,c = \epsilon$ and $L(\epsilon) = \{\texttt{""}\}$. 
+$d = c$: We have $der\,c\,c = \ONE$ and $L(\ONE) = \{\texttt{""}\}$. 
 We also have $L(c) = \{\texttt{"}c\texttt{"}\}$ and $Der\,c\,\{\texttt{"}c\texttt{"}\} = \{\texttt{""}\}$. Hence we have 
 $\{\texttt{""}\} = \{\texttt{""}\}$ in (c). 
 
-$d \not=c$: We have $der\,c\,d = \varnothing$.
-We also have $Der\,c\,\{\texttt{"}d\texttt{"}\} = \varnothing$. Hence we have 
-$\varnothing = \varnothing$  in (c). 
+$d \not=c$: We have $der\,c\,d = \ZERO$.
+We also have $Der\,c\,\{\texttt{"}d\texttt{"}\} = \ZERO$. Hence we have 
+$\ZERO = \ZERO$  in (c). 
 \end{itemize}
 
 \noindent
@@ -196,7 +192,7 @@
 $Der\,c\,(L(r^*)) = Der\,c\,(L(r)^0 \cup \bigcup_{n \ge 1} L(r)^n) = (Der\,c\,L(r)^0) \cup Der\,c\,(\bigcup_{n \ge 1} L(r)^n)$
 \end{center}
 
-The first union ``disappears'' since $Der\,c\,(L(r)^0) = \varnothing$.
+The first union ``disappears'' since $Der\,c\,(L(r)^0) = \ZERO$.
 
 
 \end{document}