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- \usepackage{tikz}
- % Nicht schön, aber funktioniert
- \newcommand{\puzzle}[9]{
- \begin{tikzpicture}[scale=0.8]
- \draw (0,0) rectangle (3,3);
- \ifnum#1>0
- \filldraw[black!30!white,draw=black] (0+.1,0+.1) rectangle (0+1-.1,0+1-.1);
- \draw (\intcalcMod{0}{3}+.5,\intcalcDiv{0}{3}+.5) node{#7};
- \fi
- \ifnum#2>0
- \filldraw[black!30!white,draw=black] (1+.1,0+.1) rectangle (1+1-.1,0+1-.1);
- \draw (\intcalcMod{1}{3}+.5,\intcalcDiv{1}{3}+.5) node{#8};
- \fi
- \ifnum#3>0
- \filldraw[black!30!white,draw=black] (2+.1,0+.1) rectangle (2+1-.1,0+1-.1);
- \draw (\intcalcMod{2}{3}+.5,\intcalcDiv{2}{3}+.5) node{#9};
- \fi
- \ifnum#4>0
- \filldraw[black!30!white,draw=black] (0+.1,1+.1) rectangle (0+1-.1,1+1-.1);
- \draw (\intcalcMod{3}{3}+.5,\intcalcDiv{3}{3}+.5) node{#4};
- \fi
- \ifnum#5>0
- \filldraw[black!30!white,draw=black] (1+.1,1+.1) rectangle (1+1-.1,1+1-.1);
- \draw (\intcalcMod{4}{3}+.5,\intcalcDiv{4}{3}+.5) node{#5};
- \fi
- \ifnum#6>0
- \filldraw[black!30!white,draw=black] (2+.1,1+.1) rectangle (2+1-.1,1+1-.1);
- \draw (\intcalcMod{5}{3}+.5,\intcalcDiv{5}{3}+.5) node{#6};
- \fi
- \ifnum#7>0
- \filldraw[black!30!white,draw=black] (0+.1,2+.1) rectangle (0+1-.1,2+1-.1);
- \draw (\intcalcMod{6}{3}+.5,\intcalcDiv{6}{3}+.5) node{#1};
- \fi
- \ifnum#8>0
- \filldraw[black!30!white,draw=black] (1+.1,2+.1) rectangle (1+1-.1,2+1-.1);
- \draw (\intcalcMod{7}{3}+.5,\intcalcDiv{7}{3}+.5) node{#2};
- \fi
- \ifnum#9>0
- \filldraw[black!30!white,draw=black] (2+.1,2+.1) rectangle (2+1-.1,2+1-.1);
- \draw (\intcalcMod{8}{3}+.5,\intcalcDiv{8}{3}+.5) node{#3};
- \fi
- \end{tikzpicture}
- }
- % 1 2 3
- % 4 5 6
- % 7 8 9
- % In das leere Feld 0 eintragen
- \puzzle{1}{2}{3}{0}{4}{5}{7}{8}{6}
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