## sTeX mode

\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}

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