玛氪宕·梦魔(Markdown Memo)，使用Markdown的云端备忘录，百度IFE的RIA启航班的不合格的作业，才……才没有什么阴谋呢！
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111 lines
4.0 KiB
111 lines
4.0 KiB
11 years ago
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<!doctype html>
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<title>CodeMirror: sTeX mode</title>
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<meta charset="utf-8"/>
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<link rel=stylesheet href="../../doc/docs.css">
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<link rel="stylesheet" href="../../lib/codemirror.css">
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<script src="../../lib/codemirror.js"></script>
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<script src="stex.js"></script>
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<style>.CodeMirror {background: #f8f8f8;}</style>
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<div id=nav>
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<a href="http://codemirror.net"><h1>CodeMirror</h1><img id=logo src="../../doc/logo.png"></a>
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<ul>
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<li><a href="../../index.html">Home</a>
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<li><a href="../../doc/manual.html">Manual</a>
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<li><a href="https://github.com/codemirror/codemirror">Code</a>
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</ul>
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<ul>
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<li><a href="../index.html">Language modes</a>
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<li><a class=active href="#">sTeX</a>
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</ul>
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</div>
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<article>
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<h2>sTeX mode</h2>
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<form><textarea id="code" name="code">
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\begin{module}[id=bbt-size]
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\importmodule[balanced-binary-trees]{balanced-binary-trees}
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\importmodule[\KWARCslides{dmath/en/cardinality}]{cardinality}
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\begin{frame}
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\frametitle{Size Lemma for Balanced Trees}
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\begin{itemize}
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\item
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\begin{assertion}[id=size-lemma,type=lemma]
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Let $G=\tup{V,E}$ be a \termref[cd=binary-trees]{balanced binary tree}
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of \termref[cd=graph-depth,name=vertex-depth]{depth}$n>i$, then the set
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$\defeq{\livar{V}i}{\setst{\inset{v}{V}}{\gdepth{v} = i}}$ of
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\termref[cd=graphs-intro,name=node]{nodes} at
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\termref[cd=graph-depth,name=vertex-depth]{depth} $i$ has
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\termref[cd=cardinality,name=cardinality]{cardinality} $\power2i$.
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\end{assertion}
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\item
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\begin{sproof}[id=size-lemma-pf,proofend=,for=size-lemma]{via induction over the depth $i$.}
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\begin{spfcases}{We have to consider two cases}
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\begin{spfcase}{$i=0$}
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\begin{spfstep}[display=flow]
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then $\livar{V}i=\set{\livar{v}r}$, where $\livar{v}r$ is the root, so
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$\eq{\card{\livar{V}0},\card{\set{\livar{v}r}},1,\power20}$.
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\end{spfstep}
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\end{spfcase}
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\begin{spfcase}{$i>0$}
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\begin{spfstep}[display=flow]
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then $\livar{V}{i-1}$ contains $\power2{i-1}$ vertexes
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\begin{justification}[method=byIH](IH)\end{justification}
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\end{spfstep}
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\begin{spfstep}
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By the \begin{justification}[method=byDef]definition of a binary
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tree\end{justification}, each $\inset{v}{\livar{V}{i-1}}$ is a leaf or has
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two children that are at depth $i$.
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\end{spfstep}
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\begin{spfstep}
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As $G$ is \termref[cd=balanced-binary-trees,name=balanced-binary-tree]{balanced} and $\gdepth{G}=n>i$, $\livar{V}{i-1}$ cannot contain
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leaves.
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\end{spfstep}
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\begin{spfstep}[type=conclusion]
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Thus $\eq{\card{\livar{V}i},{\atimes[cdot]{2,\card{\livar{V}{i-1}}}},{\atimes[cdot]{2,\power2{i-1}}},\power2i}$.
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\end{spfstep}
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\end{spfcase}
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\end{spfcases}
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\end{sproof}
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\item
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\begin{assertion}[id=fbbt,type=corollary]
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A fully balanced tree of depth $d$ has $\power2{d+1}-1$ nodes.
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\end{assertion}
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\item
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\begin{sproof}[for=fbbt,id=fbbt-pf]{}
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\begin{spfstep}
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Let $\defeq{G}{\tup{V,E}}$ be a fully balanced tree
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\end{spfstep}
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\begin{spfstep}
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Then $\card{V}=\Sumfromto{i}1d{\power2i}= \power2{d+1}-1$.
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\end{spfstep}
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\end{sproof}
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\end{itemize}
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\end{frame}
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\begin{note}
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\begin{omtext}[type=conclusion,for=binary-tree]
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This shows that balanced binary trees grow in breadth very quickly, a consequence of
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this is that they are very shallow (and this compute very fast), which is the essence of
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the next result.
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\end{omtext}
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\end{note}
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\end{module}
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%%% Local Variables:
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%%% mode: LaTeX
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%%% TeX-master: "all"
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%%% End: \end{document}
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</textarea></form>
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<script>
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var editor = CodeMirror.fromTextArea(document.getElementById("code"), {});
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</script>
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<p><strong>MIME types defined:</strong> <code>text/x-stex</code>.</p>
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<p><strong>Parsing/Highlighting Tests:</strong> <a href="../../test/index.html#stex_*">normal</a>, <a href="../../test/index.html#verbose,stex_*">verbose</a>.</p>
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</article>
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