[Auf TeX.SX gibt es folgenden Vorschlag direkt mit TikZ statt `cryptotable` von Zarko](https://tex.stackexchange.com/a/550609):
Zarko](https://tex.stackexchange.com/a/550609), bei dem ich lediglich die Klasse entsprechend der Vorgabe ersetzt habe:
\documentclass[tikz, margin=3mm]{standalone}
\documentclass[a4paper]{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,
positioning,
quotes}
\usepackage{mathtools, amssymb}
\DeclareMathOperator{\gen}{gen}
\DeclareMathOperator{\PRF}{PRF}
\begin{document}
\begin{tikzpicture}[auto,
box/.style = {draw, minimum height=17mm, align=left, outer sep=0pt},
lbl/.style = {anchor=south east, outer sep=1mm},
pin edge = {Straight Barb-, draw=black},
every edge/.style = {draw, semithick, -Straight Barb}
]
\node (n1) [box,
label={[lbl] south east:$\mathcal{C}_{\PRF}$},
pin=$1^n$]
{$\begin{aligned}
K \leftarrow \gen(1^n) \\
b \xleftarrow{r} \{0,1\} \\
\text{Wenn } b = 0 \text{ denn } \vartheta_0 = F_K(\cdot) \\
\text{Wenn } b = 1 \text{ denn } \vartheta_1 = f(\cdot) \\[1ex]
\text{Wenn } b' = b \text{return } 1\\
\text{Wenn } b' \neq b \text{ return } 0\\[5ex]
\end{aligned}$
};
\node (n2) [box, above right=-1ex and 0pt of n1.east]
{$\vartheta_b(\cdot)$};
\node (n3) [box,
label={[lbl]south east:$\mathcal{A}$},
pin=???,
right=9mm of n1 -| n2.east]
{$\begin{aligned}
x = 0 \quad b \\
x = 1 \quad a+b \rightarrow b \\
x = 2 \quad 2a+b \\[1ex]
\text{Wenn } b' = 0 \\
\text{ sonst } b' = 1 \\
\text{ return } b' \\[5ex]
\end{aligned}$
};
\draw ([yshift=+2mm] n2.east) coordinate (aux1) edge["$x$"] (aux1 -| n3.west)
([yshift=-2mm] n2.east) coordinate (aux2) (aux2 -| n3.west) edge["$y$"] (aux2)
([yshift=-7ex] n3.west) edge["$b'$" '] ([yshift=-7ex] n1.east);
\end{tikzpicture}
\end{document}
