改進 TikZ 土豆

Question

使用math和calc庫，可以稍微簡單一些。您可以翻轉任何兩個座標，看看它是否也適用於尚未訂購的座標（我只測試了翻轉A，B但它應該適用於所有座標）。

我沒有改變任何東西\begin{document}，\end{document}所以\Distance仍然存在，但我沒有在我的定義中使用它，\DrawPotato所以你並不真正需要它。

\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

使用math和calc庫，可以稍微簡單一些。您可以翻轉任何兩個座標，看看它是否也適用於尚未訂購的座標（我只測試了翻轉A，B但它應該適用於所有座標）。

我沒有改變任何東西\begin{document}，\end{document}所以\Distance仍然存在，但我沒有在我的定義中使用它，\DrawPotato所以你並不真正需要它。

\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}

改進 TikZ 土豆

答案1

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