updated
authorChristian Urban <christian.urban@kcl.ac.uk>
Fri, 10 Apr 2020 16:30:52 +0100
changeset 720 ecbed0155f72
parent 719 0de3527e6ae3
child 721 e3c64f22dd31
updated
handouts/ho04.pdf
handouts/ho04.tex
Binary file handouts/ho04.pdf has changed
--- a/handouts/ho04.tex	Fri Apr 10 16:20:10 2020 +0100
+++ b/handouts/ho04.tex	Fri Apr 10 16:30:52 2020 +0100
@@ -310,20 +310,19 @@
 derivative functions. For this consider one of the ``squares'' from
 Figure~\ref{Sulz}:
 
-
-  \begin{center}
-  \begin{tikzpicture}[scale=2,node distance=1.2cm,
-                      every node/.style={minimum size=7mm}]
-  \node (r)  {$r$};
-  \node (rd) [right=of r]{$r_{der}$};
-  \draw[->,line width=1mm](r)--(rd) node[above,midway] {$\der\,c$};
-  \node (vd) [below=of r2]{$v_{der}$};
-  \draw[->,line width=1mm](rd) -- (vd);
-  \node (v) [left=of vd] {$v$};
-  \draw[->,line width=1mm](vd)--(v) node[below,midway] {$\inj\,c$};
-  \draw[->,line width=0.5mm,dotted](r) -- (v) node[left,midway,red] {\bf ?};
-  \end{tikzpicture}
-  \end{center}
+\begin{center}
+\begin{tikzpicture}[scale=2,node distance=1.2cm,
+                    every node/.style={minimum size=7mm}]
+\node (r)  {$r$};
+\node (rd) [right=of r]{$r_{der}$};
+\draw[->,line width=1mm](r)--(rd) node[above,midway] {$\der\,c$};
+\node (vd) [below=of r2]{$v_{der}$};
+\draw[->,line width=1mm](rd) -- (vd);
+\node (v) [left=of vd] {$v$};
+\draw[->,line width=1mm](vd)--(v) node[below,midway] {$\inj\,c$};
+\draw[->,line width=0.5mm,dotted](r) -- (v) node[left,midway,red] {\bf ?};
+\end{tikzpicture}
+\end{center}
 
 \noindent
 The input to the $\inj$-function is $r$ (on the top left), $c$ (the
@@ -336,7 +335,7 @@
 that $v_{der}$ is the value for how $r_{der}$ matches the corresponding string
 where $c$ has been chopped off.
 
-Let $r$ be $r_1 + r_2$. Then $r_{der}$
+For a concrete example, let $r$ be $r_1 + r_2$. Then $r_{der}$
 is of the form $(\der\,c\,r_1) + (\der\,c\,r_2)$. What are the possible
 values corresponding to $r_{der}$? Well, they can be only of the form
 $\Left(\ldots)$ and $\Right(\ldots)$. Therefore you have two
@@ -432,9 +431,9 @@
 
 Phew\ldots{}Sweat\ldots!\#@$\skull$\%\ldots Unfortunately, there is
 a gigantic problem with the described algorithm so far: it is very
-slow. We need to include in all this the simplification from Lecture
-2. And what rotten luck: simplification messes things up and we need
-to rectify the mess. This is what we shall do next.
+slow. To make it faster, we have to include in all this the simplification 
+from Lecture 2\ldots{}and what rotten luck: simplification messes things 
+up and we need to rectify the mess. This is what we shall do next.
 
 
 \subsubsection*{Simplification}