updated
authorChristian Urban <christian.urban@kcl.ac.uk>
Tue, 29 Sep 2020 21:52:52 +0100
changeset 768 34f77b976b88
parent 767 bdd12391d345
child 769 f9686b22db7e
updated
hws/hw01.pdf
hws/hw02.pdf
hws/hw02.tex
hws/hw03.pdf
hws/hw04.pdf
hws/hw04.tex
hws/hw05.pdf
hws/hw06.pdf
hws/hw07.pdf
hws/hw08.pdf
hws/hw09.pdf
hws/proof.pdf
progs/pow.scala
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Binary file hws/hw02.pdf has changed
--- a/hws/hw02.tex	Tue Sep 29 19:35:11 2020 +0100
+++ b/hws/hw02.tex	Tue Sep 29 21:52:52 2020 +0100
@@ -62,7 +62,7 @@
       expressions with respect to a character. (Hint: The
       derivative is defined recursively.)
 
-\item Assume the set $Der$ is defined as
+\item  Assume the set $Der$ is defined as
 
   \begin{center}
     $Der\,c\,A \dn \{ s \;|\;  c\!::\!s \in A\}$
Binary file hws/hw03.pdf has changed
Binary file hws/hw04.pdf has changed
--- a/hws/hw04.tex	Tue Sep 29 19:35:11 2020 +0100
+++ b/hws/hw04.tex	Tue Sep 29 21:52:52 2020 +0100
@@ -41,7 +41,13 @@
   In case they can, can you give the corresponding token
   sequences.
 
-\item Assume that $s^{-1}$ stands for the operation of reversing a
+\item Assume $r$ is nullable. Show that
+  \[ 1 + r + r\cdot r \;\equiv\; r\cdot r
+  \]
+
+  holds.
+
+\item \textbf{(Deleted)} Assume that $s^{-1}$ stands for the operation of reversing a
   string $s$. Given the following \emph{reversing} function on regular
   expressions
 
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--- a/progs/pow.scala	Tue Sep 29 19:35:11 2020 +0100
+++ b/progs/pow.scala	Tue Sep 29 21:52:52 2020 +0100
@@ -63,7 +63,9 @@
 pow(B, 4)
 pow(B, 4).size
 
-
+val C = Set("a", "")
+pow(C, 4)
+pow(C, 4).size
 
 val SA = Set("a", "b", "c", "d")
 val SB = Set("0", "1", "2", "3", "4", "5", "6")