frame-em.tex

% SPDX-License-Identifier: CC-BY-4.0
% Copyright 2018 Toni Dietze
\documentclass[beamer]{standalone}
\input{preamble.tex}
\begin{document}
\begin{standaloneframe}{\jobname}
\centering
\begin{tikzpicture}[anchor=base, level distance=2.5em]
	\node (t) {\(σ\)}
		child { node {\(α\)}
			edge from parent node[left] {\only<2->{\(\nt{?}\)}}
		}
		child { node {\(σ\)}
			child { node {\(β\)}
				edge from parent node[left] {\only<2->{\(\nt{?}\)}}
			}
			child { node {\(β\)}
				edge from parent node[right] {\only<2->{\(\nt{?}\)}}
			}
			edge from parent node[right] {\only<2->{\(\nt{?}\)}}
		};
	\node[above=0 of t.north] {\only<2->{\(\nt{?}\)}};

	% coordinates of tree triangle
	\path coordinate (top)   at ([yshift=2em]t.north)
			coordinate (right) at ([xshift=2em]t-2-2.south east)
			let \p1 = (top), \p2 = (right), \p3 = (2 * \x1 - \x2, \y2) in
			coordinate (left) at (\p3);

	% draw tree triangle
	\draw (top) -- (right) -- (left) -- cycle;

	% draw subsequent tree triangles
	\coordinate (s) at (0, 0);
	\foreach \s in {1, 2, ..., 4}
		\draw
			coordinate (sOld) at (s)
			coordinate (s) at (\s*0.5em, \s*0.125em)
			(intersection cs: first line = {([shift={(sOld)}]top) -- ([shift={(sOld)}]right)}
							, second line = {([shift={(s)}]top)    -- ([shift={(s)}]left)})
			-- ([shift={(s)}]top)
			-- ([shift={(s)}]right)
			-- (intersection cs: first line = {([shift={(sOld)}]top) -- ([shift={(sOld)}]right)}
								, second line = {([shift={(s)}]right)  -- ([shift={(s)}]left)});
\end{tikzpicture}

\begin{overprint}
\onslide<1-3>
	\begin{description}
	\item[input]
		corpus $c\colon \trees{Σ} → ℝ_{≥ 0}$, ta $\mathscr{A}$ over $Σ$\strut
	\item[output]
		$\displaystyle \argmax_{\substack{\text{prob.\ wta} \\ ℳ = (\mathscr{A}, ι, δ)}} \lklhd{c}{⟦ℳ⟧}$ \only<3->{\qquad \alert{$→$ difficult!}}
	\end{description}
\onslide<4->
	\begin{description}
	\item[input]
		corpus $c\colon \trees{Σ} → ℝ_{≥ 0}$, prob.\ wta $ℳ_0 = (\mathscr{A}, ι_0, δ_0)$ over $Σ$\strut
	\item[output]
		sequence $ℳ_0, ℳ_1, …$, such~that~$\lklhd{c}{⟦ℳ_0⟧} ≤ \lklhd{c}{⟦ℳ_1⟧} ≤ …$
	\item[algorithm]
% 		\begin{spacing}{1.6}
			\begin{algorithmic}
				\For{$i \gets 0, 1, …$}
					\hfill \mycite{1977DempsterLairdRubin}
					\State
						{\usebeamercolor[fg]{enumerate item} E-Step:} $\displaystyle \forall d \in \operatorname{D}_G\colon c'_i(t, r) \gets c(t) \cdot \frac{⟦ℳ_i⟧(t, r)}{⟦ℳ_i⟧(t)}$
					\State
						{\usebeamercolor[fg]{enumerate item} M-Step:} $\displaystyle ℳ_{i + 1} \gets \argmax_{\substack{\text{prob.\ wta} \\ ℳ = (\mathscr{A}, ι, δ)}} \lklhd{c_i'}{⟦ℳ⟧}$
						\vphantom{\({A^A}^A\)}
				\EndFor
			\end{algorithmic}
% 		\end{spacing}
	\end{description}
\end{overprint}
\end{standaloneframe}
\end{document}