HackMD/public/vendor/codemirror/mode/stex/index.html

110 lines
4 KiB
HTML

<!doctype html>
<title>CodeMirror: sTeX mode</title>
<meta charset="utf-8"/>
<link rel=stylesheet href="../../doc/docs.css">
<link rel="stylesheet" href="../../lib/codemirror.css">
<script src="../../lib/codemirror.js"></script>
<script src="stex.js"></script>
<style>.CodeMirror {background: #f8f8f8;}</style>
<div id=nav>
<a href="http://codemirror.net"><h1>CodeMirror</h1><img id=logo src="../../doc/logo.png"></a>
<ul>
<li><a href="../../index.html">Home</a>
<li><a href="../../doc/manual.html">Manual</a>
<li><a href="https://github.com/codemirror/codemirror">Code</a>
</ul>
<ul>
<li><a href="../index.html">Language modes</a>
<li><a class=active href="#">sTeX</a>
</ul>
</div>
<article>
<h2>sTeX mode</h2>
<form><textarea id="code" name="code">
\begin{module}[id=bbt-size]
\importmodule[balanced-binary-trees]{balanced-binary-trees}
\importmodule[\KWARCslides{dmath/en/cardinality}]{cardinality}
\begin{frame}
\frametitle{Size Lemma for Balanced Trees}
\begin{itemize}
\item
\begin{assertion}[id=size-lemma,type=lemma]
Let $G=\tup{V,E}$ be a \termref[cd=binary-trees]{balanced binary tree}
of \termref[cd=graph-depth,name=vertex-depth]{depth}$n>i$, then the set
$\defeq{\livar{V}i}{\setst{\inset{v}{V}}{\gdepth{v} = i}}$ of
\termref[cd=graphs-intro,name=node]{nodes} at
\termref[cd=graph-depth,name=vertex-depth]{depth} $i$ has
\termref[cd=cardinality,name=cardinality]{cardinality} $\power2i$.
\end{assertion}
\item
\begin{sproof}[id=size-lemma-pf,proofend=,for=size-lemma]{via induction over the depth $i$.}
\begin{spfcases}{We have to consider two cases}
\begin{spfcase}{$i=0$}
\begin{spfstep}[display=flow]
then $\livar{V}i=\set{\livar{v}r}$, where $\livar{v}r$ is the root, so
$\eq{\card{\livar{V}0},\card{\set{\livar{v}r}},1,\power20}$.
\end{spfstep}
\end{spfcase}
\begin{spfcase}{$i>0$}
\begin{spfstep}[display=flow]
then $\livar{V}{i-1}$ contains $\power2{i-1}$ vertexes
\begin{justification}[method=byIH](IH)\end{justification}
\end{spfstep}
\begin{spfstep}
By the \begin{justification}[method=byDef]definition of a binary
tree\end{justification}, each $\inset{v}{\livar{V}{i-1}}$ is a leaf or has
two children that are at depth $i$.
\end{spfstep}
\begin{spfstep}
As $G$ is \termref[cd=balanced-binary-trees,name=balanced-binary-tree]{balanced} and $\gdepth{G}=n>i$, $\livar{V}{i-1}$ cannot contain
leaves.
\end{spfstep}
\begin{spfstep}[type=conclusion]
Thus $\eq{\card{\livar{V}i},{\atimes[cdot]{2,\card{\livar{V}{i-1}}}},{\atimes[cdot]{2,\power2{i-1}}},\power2i}$.
\end{spfstep}
\end{spfcase}
\end{spfcases}
\end{sproof}
\item
\begin{assertion}[id=fbbt,type=corollary]
A fully balanced tree of depth $d$ has $\power2{d+1}-1$ nodes.
\end{assertion}
\item
\begin{sproof}[for=fbbt,id=fbbt-pf]{}
\begin{spfstep}
Let $\defeq{G}{\tup{V,E}}$ be a fully balanced tree
\end{spfstep}
\begin{spfstep}
Then $\card{V}=\Sumfromto{i}1d{\power2i}= \power2{d+1}-1$.
\end{spfstep}
\end{sproof}
\end{itemize}
\end{frame}
\begin{note}
\begin{omtext}[type=conclusion,for=binary-tree]
This shows that balanced binary trees grow in breadth very quickly, a consequence of
this is that they are very shallow (and this compute very fast), which is the essence of
the next result.
\end{omtext}
\end{note}
\end{module}
%%% Local Variables:
%%% mode: LaTeX
%%% TeX-master: "all"
%%% End: \end{document}
</textarea></form>
<script>
var editor = CodeMirror.fromTextArea(document.getElementById("code"), {});
</script>
<p><strong>MIME types defined:</strong> <code>text/x-stex</code>.</p>
<p><strong>Parsing/Highlighting Tests:</strong> <a href="../../test/index.html#stex_*">normal</a>, <a href="../../test/index.html#verbose,stex_*">verbose</a>.</p>
</article>