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authorKarl Berry <karl@freefriends.org>2013-04-23 22:16:26 +0000
committerKarl Berry <karl@freefriends.org>2013-04-23 22:16:26 +0000
commit2f29eaac5f1a5540c3e514e0f6b828ec29d56caa (patch)
tree8b9218f8d4d1d13ef978d9857ee4a7bc4176939a /Master/texmf-dist/doc/latex/vhistory/en_sets.tex
parent14c6ba96f2f119d83f5bee918b4dbfb3b5327ae1 (diff)
vhistory (21apr13)
git-svn-id: svn://tug.org/texlive/trunk@30080 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/vhistory/en_sets.tex')
-rw-r--r--Master/texmf-dist/doc/latex/vhistory/en_sets.tex17
1 files changed, 8 insertions, 9 deletions
diff --git a/Master/texmf-dist/doc/latex/vhistory/en_sets.tex b/Master/texmf-dist/doc/latex/vhistory/en_sets.tex
index d3f6ff745cb..21f17115878 100644
--- a/Master/texmf-dist/doc/latex/vhistory/en_sets.tex
+++ b/Master/texmf-dist/doc/latex/vhistory/en_sets.tex
@@ -35,7 +35,7 @@ The items of a set are separated with \texttt{|}. The set $A$ could therefore be
The set $L$ is defined with:\\
\mbox{}\hspace{2em}\verb$\newset{\sL}{}$\\
-\verb$\newset$ creates a new set. This set will be sorted in alphabetical order and duplicates will be removed. So it would be no matter, if in the definition of $A$ after ``Charly'' a second ``Alice'' was inserted.
+\verb$\newset$ creates a new set. This set will be sorted in alphabetical order and duplicates will be removed. So it would be no matter if in the definition of $A$ after ``Charly'' a second ``Alice'' was inserted.
The effort for sorting and duplicate deleting is unnecessary at this point. If you want to skip these expensive steps, you can create a set with the command
\verb$\newsetsimple$, too.
@@ -52,9 +52,8 @@ Inspectores help you to retrieve informations about sets and to print sets.
\paragraph{Printing:}A set can be printed using the command\\
\mbox{}\hspace{2em}\verb$\listset$.\\
-The elements will be put in the sequence they are in the set. A comma is used as separator.
-
-\verb$\listset{\sA}$ therefore leads to the following output:\\
+The elements will be put in the sequence they are in the set. A comma is used as separator.\\
+\mbox{}\hspace{2em}\verb$\listset{\sA}$ therefore leads to the following output:\\
\centerline{\listset{\sA}}
Sometimes you might want to separate the items in a different way, for example with a \texttt{\&} to put them in a table. In this case a (temporary) redefinition of\\
@@ -68,7 +67,7 @@ where \texttt{<counter>} is the name of a \LaTeX\ counter, which afterwards will
\mbox{}\hspace{2em}\verb$\sizeofset{\sB}\is{mycounter}$\sizeofset{\sB}\is{mycounter}\\
\mbox{}\hspace{2em}\verb$\arabic{mycounter}$\\
leads to the output: ``\arabic{mycounter}''
-If you determine the size of set $L$, the result\sizeofset{\sL}\is{mycounter}---as you might have expected---is ``\arabic{mycounter}''.
+If you determine the size of set $L$, the result\sizeofset{\sL}\is{mycounter} (as you might have expected) is ``\arabic{mycounter}''.
\paragraph{Testing for membership:}By using the command\\
\mbox{}\hspace{2em}\verb$\iselementofset{$$e$\verb$}{$$S$\verb$}$\\
@@ -87,14 +86,14 @@ Table \ref{tab:ops} contains some examples. The result of the operation is a sor
\mbox{}\hspace{2em}\verb|\minussets{|$S_1$\verb|}\minus{|$S_2$\verb|}\to{|$R$\verb|}|.\\
If $S_1$ is a sorted set, $R$ will be sorted, too. If $S_1$ contains duplicates, $R$ might also contains these duplicate elements. Table \ref{tab:ops} contains several examples for the usage of this command.
-Colloquially you can formulate the operation as follows: Check for every element $e$ in $S_1$, if $e \in S_2$ is true. If not, insert $e$ into $R$. And that's exactly the way it has been implemented!
+Colloquially you can formulate the operation as follows: Check for every element $e$ in $S_1$ if $e \in S_2$ is true. If not, insert $e$ into $R$. And that's exactly the way it has been implemented!
\paragraph{Intersection of sets:} The operation $R := S_1 \cap S_2$ is made possible with the command\\
\mbox{}\hspace{2em}\verb|\intersectsets{|$S_1$\verb|}{|$S_2$\verb|}\to{|$R$\verb|}|.\\
As above: If $S_1$ is a sorted set, $R$ will be sorted, too. If $S_1$ contains duplicates, $R$ might also contains these duplicate elements. Table \ref{tab:ops} contains several examples for the usage of this command, too.
-This operation can colloquially be written down as: Check for every element $e$ in $S_1$, if $e \in S_2$ is true. If yes, insert $e$ into $R$. If you compare this with the formulation above, one can recognize that the only difference is the small word ``yes''.
-In the source code, this expresses in a missing \verb|\else|. Acutally amazingly simple, if you remember the formal relation $S_1 \cap S_2 \equiv S_1 \backslash (S_1\backslash S_2)$, which lets one expect a much higher complexity.
+This operation can colloquially be written down as: Check for every element $e$ in $S_1$ if $e \in S_2$ is true. If yes, insert $e$ into $R$. If you compare this with the formulation above, one can recognize that the only difference is the small word ``yes''.
+In the source code, this expresses in a missing \verb|\else|. Acutally amazingly simple if you remember the formal relation $S_1 \cap S_2 \equiv S_1 \backslash (S_1\backslash S_2)$, which lets one expect a much higher complexity.
\begin{table}%[htb]
\begin{center}
@@ -173,4 +172,4 @@ The given complexity classes can help you to arrange a set of operations in the
\caption{Complexity classes of set operations}%
\label{tab:complexity}%
\end{center}
-\end{table} \ No newline at end of file
+\end{table}