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<h2>sTeX mode</h2>
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\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}
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