Mejorando las patatas TikZ

Question

Con las bibliotecas mathy calc, es posible hacerlo un poco más sencillo. Puede invertir dos coordenadas cualesquiera para ver que también funciona para coordenadas que aún no están ordenadas (solo probé el cambio Ay Bdebería funcionar para todas).

No cambié nada entre \begin{document}y \end{document}todavía \Distanceestá allí, pero no lo uso en mi definición, por \DrawPotatolo que realmente no lo necesitas.

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc,math}

% I don't use this in my \DrawPotato
\makeatletter % from https://tex.stackexchange.com/a/412901/121799
\newcommand{\Distance}[3]{% % from https://tex.stackexchange.com/q/56353/121799
    \tikz@scan@one@point\pgfutil@firstofone($#1-#2$)\relax  
    \pgfmathsetmacro{#3}{veclen(\the\pgf@x,\the\pgf@y)/28.45274}
}
\makeatother

\tikzset{
    declare function={
        interpolator(\x,\xmin,\xmax,\rmin,\rmax)=
            1/28.4527*((\rmin+\rmax)/2+((\rmin-\rmax)/2)*cos((\x-\xmin)*(180/(\xmax-\xmin))));
    }
}

\newcommand\DrawPotato[5][]{%
    \tikzmath{
        coordinate \p,\cent;
        \p0 = {(#2)};
        \p1 = {(#3)};
        \p2 = {(#4)};
        \p3 = {(#5)};
        \cent = ({(\px0+\px1+\px2+\px3)/4},{(\py0+\py1+\py2+\py3)/4}); % Average point
        real \angl,\dist;
        int \i,\j;
        for \i in {0,1,2,3}{
            \angl{\i} = atan2(\py{\i}-\centy,\px{\i}-\centx);
            \dist{\i} = veclen(\py{\i}-\centy,\px{\i}-\centx);
        };
        for \i in {0,1,2,3}{
            for \j in {0,1,2,3}{
                if min(\angl{0},\angl{1},\angl{2},\angl{3}) == \angl{\j} then {
                    \sortang{\i} = \angl{\j};
                    \sortind{\i} = \j;
                };
            };
            \angl{\sortind{\i}} = 500; % Arbitrary large value
            \sortdis{\i} = \dist{\sortind{\i}};
        };
    }
    \draw[#1,smooth,samples=50,shift={(\cent)}] 
        plot[variable=\x,domain=\sortang{0}:\sortang{1}] %
            (\x:{interpolator(\x,{\sortang{0}},{\sortang{1}},{\sortdis{0}},{\sortdis{1}})}) --
        plot[variable=\x,domain=\sortang{1}:\sortang{2}] %
            (\x:{interpolator(\x,{\sortang{1}},{\sortang{2}},{\sortdis{1}},{\sortdis{2}})}) --
        plot[variable=\x,domain=\sortang{2}:\sortang{3}] %
            (\x:{interpolator(\x,{\sortang{2}},{\sortang{3}},{\sortdis{2}},{\sortdis{3}})}) --
        plot[variable=\x,domain=\sortang{3}:\sortang{0}+360] %
            (\x:{interpolator(\x,{\sortang{3}},{\sortang{0}+360},{\sortdis{3}},{\sortdis{0}})}) --
        cycle ;
}

\begin{document}
\begin{frame}
\frametitle{What does the current code do?}
\begin{overlayarea}{\textwidth}{\textheight}
\begin{tikzpicture}
\path[use as bounding box] (-5,-3) rectangle (7,5);
\node[text width=8cm] (text) at (0,3){%
\only<1>{input: four points, here A, B, C \& D}
\only<2>{step one: compute the barycenter of those points}
\only<3>{imagine now a circle around the barycenter with radius equal to the 
average distance}
\only<4>{the code simply adds some sine functions to the radius such that the
contour runs through the points and the distance is extremal at those points}
\only<5>{however, the code is extremely clumsy and slow, and I have the feeling
that I'm re--inventing the wheel when dealing with lists etc.}
};
\coordinate (A) at (-1,1);
\coordinate (B) at (2,0.5);
\coordinate (C) at (4,-1);
\coordinate (D) at (1,-1);
\coordinate (Center) at (barycentric cs:A=1,B=1,C=1,D=1);
\pgfmathsetmacro{\TotalDistance}{0}
\xdef\DistanceList{}
\foreach \p in {A,B,C,D}
{\draw[fill=black] (\p) circle (1pt) node[below]{\p};
}
\Distance{(Center)}{(A)}{\DistanceA}
\Distance{(Center)}{(B)}{\DistanceB}
\Distance{(Center)}{(C)}{\DistanceC}
\Distance{(Center)}{(D)}{\DistanceD}
\pgfmathsetmacro{\AverageDistance}{(\DistanceA+\DistanceB+\DistanceC+\DistanceD)/4}
\pause
\draw[fill=black] (Center) circle (1pt) node[below]{Center};
\pause
\draw[-,dashed] (Center) circle (\AverageDistance);
\pause
\only<4->{
\DrawPotato[blue]{A}{B}{C}{D}
}
\end{tikzpicture}


\end{overlayarea}
\end{frame}
\end{document}

Answer 1

Con las bibliotecas mathy calc, es posible hacerlo un poco más sencillo. Puede invertir dos coordenadas cualesquiera para ver que también funciona para coordenadas que aún no están ordenadas (solo probé el cambio Ay Bdebería funcionar para todas).

No cambié nada entre \begin{document}y \end{document}todavía \Distanceestá allí, pero no lo uso en mi definición, por \DrawPotatolo que realmente no lo necesitas.

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc,math}

% I don't use this in my \DrawPotato
\makeatletter % from https://tex.stackexchange.com/a/412901/121799
\newcommand{\Distance}[3]{% % from https://tex.stackexchange.com/q/56353/121799
    \tikz@scan@one@point\pgfutil@firstofone($#1-#2$)\relax  
    \pgfmathsetmacro{#3}{veclen(\the\pgf@x,\the\pgf@y)/28.45274}
}
\makeatother

\tikzset{
    declare function={
        interpolator(\x,\xmin,\xmax,\rmin,\rmax)=
            1/28.4527*((\rmin+\rmax)/2+((\rmin-\rmax)/2)*cos((\x-\xmin)*(180/(\xmax-\xmin))));
    }
}

\newcommand\DrawPotato[5][]{%
    \tikzmath{
        coordinate \p,\cent;
        \p0 = {(#2)};
        \p1 = {(#3)};
        \p2 = {(#4)};
        \p3 = {(#5)};
        \cent = ({(\px0+\px1+\px2+\px3)/4},{(\py0+\py1+\py2+\py3)/4}); % Average point
        real \angl,\dist;
        int \i,\j;
        for \i in {0,1,2,3}{
            \angl{\i} = atan2(\py{\i}-\centy,\px{\i}-\centx);
            \dist{\i} = veclen(\py{\i}-\centy,\px{\i}-\centx);
        };
        for \i in {0,1,2,3}{
            for \j in {0,1,2,3}{
                if min(\angl{0},\angl{1},\angl{2},\angl{3}) == \angl{\j} then {
                    \sortang{\i} = \angl{\j};
                    \sortind{\i} = \j;
                };
            };
            \angl{\sortind{\i}} = 500; % Arbitrary large value
            \sortdis{\i} = \dist{\sortind{\i}};
        };
    }
    \draw[#1,smooth,samples=50,shift={(\cent)}] 
        plot[variable=\x,domain=\sortang{0}:\sortang{1}] %
            (\x:{interpolator(\x,{\sortang{0}},{\sortang{1}},{\sortdis{0}},{\sortdis{1}})}) --
        plot[variable=\x,domain=\sortang{1}:\sortang{2}] %
            (\x:{interpolator(\x,{\sortang{1}},{\sortang{2}},{\sortdis{1}},{\sortdis{2}})}) --
        plot[variable=\x,domain=\sortang{2}:\sortang{3}] %
            (\x:{interpolator(\x,{\sortang{2}},{\sortang{3}},{\sortdis{2}},{\sortdis{3}})}) --
        plot[variable=\x,domain=\sortang{3}:\sortang{0}+360] %
            (\x:{interpolator(\x,{\sortang{3}},{\sortang{0}+360},{\sortdis{3}},{\sortdis{0}})}) --
        cycle ;
}

\begin{document}
\begin{frame}
\frametitle{What does the current code do?}
\begin{overlayarea}{\textwidth}{\textheight}
\begin{tikzpicture}
\path[use as bounding box] (-5,-3) rectangle (7,5);
\node[text width=8cm] (text) at (0,3){%
\only<1>{input: four points, here A, B, C \& D}
\only<2>{step one: compute the barycenter of those points}
\only<3>{imagine now a circle around the barycenter with radius equal to the 
average distance}
\only<4>{the code simply adds some sine functions to the radius such that the
contour runs through the points and the distance is extremal at those points}
\only<5>{however, the code is extremely clumsy and slow, and I have the feeling
that I'm re--inventing the wheel when dealing with lists etc.}
};
\coordinate (A) at (-1,1);
\coordinate (B) at (2,0.5);
\coordinate (C) at (4,-1);
\coordinate (D) at (1,-1);
\coordinate (Center) at (barycentric cs:A=1,B=1,C=1,D=1);
\pgfmathsetmacro{\TotalDistance}{0}
\xdef\DistanceList{}
\foreach \p in {A,B,C,D}
{\draw[fill=black] (\p) circle (1pt) node[below]{\p};
}
\Distance{(Center)}{(A)}{\DistanceA}
\Distance{(Center)}{(B)}{\DistanceB}
\Distance{(Center)}{(C)}{\DistanceC}
\Distance{(Center)}{(D)}{\DistanceD}
\pgfmathsetmacro{\AverageDistance}{(\DistanceA+\DistanceB+\DistanceC+\DistanceD)/4}
\pause
\draw[fill=black] (Center) circle (1pt) node[below]{Center};
\pause
\draw[-,dashed] (Center) circle (\AverageDistance);
\pause
\only<4->{
\DrawPotato[blue]{A}{B}{C}{D}
}
\end{tikzpicture}


\end{overlayarea}
\end{frame}
\end{document}

Mejorando las patatas TikZ

Respuesta1

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