--- a/handouts/ho03.tex	Thu Dec 11 21:13:48 2014 +0000
+++ b/handouts/ho03.tex	Thu Apr 09 07:42:23 2015 +0100
@@ -497,10 +497,10 @@
 left-hand side is $q_1$ and the right-hand side $q_0\,a$. The
 right-hand side is essentially all possible ways how to end up
 in $q_1$. There is only one incoming edge from $q_0$ consuming
-an $a$.  Therefore the right hand side is
+an $a$.  Therefore the right hand side is this
 state followed by character---in this case $q_0\,a$. Now lets
 have a look at the third equation: there are two incoming
-edges. Therefore we have two terms, namely $q_1\,a$ and
+edges for $q_2$. Therefore we have two terms, namely $q_1\,a$ and
 $q_2\,a$. These terms are separated by $+$. The first states
 that if in state $q_1$ consuming an $a$ will bring you to
 $q_2$, and the secont that being in $q_2$ and consuming an $a$
@@ -835,7 +835,7 @@
 for this is as follows: take a regular expression, translate
 it into a NFA and DFA that recognise the same language. Once
 you have the DFA it is very easy to construct the automaton
-for the language not recognised by an DFA. If the DAF is
+for the language not recognised by an DFA. If the DFA is
 completed (this is important!), then you just need to exchange
 the accepting and non-accepting states. You can then translate
 this DFA back into a regular expression.