On behalf of @marmot : Improving the calculation of bounding boxes fo…

…r Bezier curves

experiments/Marmot/BezierBoundingBox/BezierBoundingBox.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{tikz}
+\begin{document}
+\section*{More precise determination of the bounding boxes of
+\texttt{tikzpicture}s}
+
+\subsection*{Current status}
+
+Ti\emph{k}Z determines the bounding box of (cubic) Bezier curves by establishing the
+smallest rectangle that contains the end point and the two control points of the
+curve (cf.\ Figure~\ref{fig:BoundingBoxBezier}).
+\begin{figure}[!htb]
+\centering
+\begin{tikzpicture}[bullet/.style={circle,fill,inner sep=1pt}]
+ \draw (0,0)  .. controls (-1,1) and (1,2) .. (2,0);
+ \draw (current bounding box.south west) rectangle 
+  (current bounding box.north east);
+ \draw[red,dashed] (0,0) -- (-1,1) node[bullet,label=above:{$(x_a,y_a)$}]{}
+  (2,0) -- (1,2) node[bullet,label=above:{$(x_b,y_b)$}]{};
+ \path (0,0) node[bullet,label=below:{$(x_0,y_0)$}]{}
+ (2,0) node[bullet,label=below:{$(x_1,y_1)$}]{};
+\end{tikzpicture}\quad%
+\begin{tikzpicture}
+ \draw (0,0) .. controls (-1,1) and (1,2) .. (2,0);
+ \draw (current bounding box.south west) rectangle 
+  (current bounding box.north east);
+\end{tikzpicture}
+\caption{Ti\emph{k}Z bounding boxes and an example.}
+\label{fig:BoundingBoxBezier}
+\end{figure}
+
+This may lead to drastic overestimates of the bounding box (cf.\
+Figure~\ref{fig:BoundingBoxExample}).
+
+\begin{figure}[!htb]
+\centering
+\begin{tikzpicture}[y=0.30pt, x=0.4pt,yscale=-1]
+ \path[draw=black,fill=white]
+   (258.9527,290.5199) .. controls (173.9885,538.4766) and (568.5860,261.2969) ..
+   (306.5098,257.1141) .. controls (44.4337,252.9312) and (429.9845,542.5624) ..
+   (352.9767,292.0206) .. controls (275.9689,41.4788) and (119.6549,497.6604) ..
+   (334.1376,346.9999) .. controls (548.6203,196.3394) and (66.4622,188.6439) ..
+   (276.0276,346.0724) .. controls (485.5930,503.5010) and (343.9169,42.5633) ..
+   (258.9527,290.5199) -- cycle;
+  \draw (current bounding box.south west) rectangle 
+  (current bounding box.north east);
+ \end{tikzpicture}
+\caption{Example from \texttt{https://tex.stackexchange.com/q/43621/121799}.}
+\label{fig:BoundingBoxExample}
+\end{figure} 
+
+\clearpage
+\subsection*{Computing the bounding box}
+
+Establishing the precise bounding box has been discussed in various places, the
+following discussion uses in part the results from
+\texttt{https://pomax.github.io/bezierinfo/}. What is a cubic Bezier curve? A
+cubic Bezier curve running from $(x_0,y_0)$ to $(x_1,y_1)$ with control points
+$(x_a,y_a)$ and $(x_a,y_a)$ can be parametrized by
+\begin{equation}
+ \gamma(t)~=~
+ \begin{pmatrix} x(t)\\ y(t) \end{pmatrix}~=~
+ \begin{pmatrix}t^3 x_{1}+3 t^2 (1-t) x_{b}+(1-t)^3
+   x_{0}+3 t (1-t)^2 x_{a}\\
+   t^3 y_{1}+3
+   t^2 (1-t) y_{b}+(1-t)^3 y_{0}+3 t (1-t)^2
+   y_{a}\end{pmatrix}\;,
+\end{equation}
+where $t$ runs from 0 to 1 (and $\gamma(0)=(x_0,y_0)$ and
+$\gamma(1)=(x_1,y_1)$). Surely, the bounding box has to contain
+$(x_0,y_0)$ and $(x_1,y_1)$. If the functions $x(t)$ and $y(t)$ have extrema in
+the interval $[0,1]$, then the bounding box will in general be larger than that.
+In order to determine the extrema of the curve, all
+we need to find the extrema of the functions $x(t)$ and $y(t)$ for $0\le t\le
+1$. That is, we need to find the solutions of the quadratic equations
+\begin{equation}
+ \frac{\mathrm{d}x}{\mathrm{d}t}(t)~=~0\quad\text{and}\quad
+ \frac{\mathrm{d}y}{\mathrm{d}t}(t)~=~0\;.
+\end{equation}
+Let's discuss $x$, $y$ is analogous. If the discriminant
+\begin{equation}
+ d~:=~x_{0} x_{1}-x_{0}
+   x_{b}-x_{1}
+   x_{a}+x_{a}^2-x_{a}
+   x_{b}+x_{b}^2
+\end{equation}
+is greater than 0, there are two solutions
+\begin{equation}
+ t_\pm~=~\frac{x_{0}-2
+   x_{a}+x_{b}\pm\sqrt{x_{0} x_{1}-x_{0}
+   x_{b}-x_{1}
+   x_{a}+x_{a}^2-x_{a}
+   x_{b}+x_{b}^2}}{x_{0}-x_{1}-3
+   x_{a}+3 x_{b}} \;.
+\end{equation}   
+In this case, we need to make sure that the bounding box contains, say
+$(x(t_-),y_0)$ and $(x(t_+),y_0)$. If $d\le0$, the bounding box does not need to
+be increased in the $x$ direction. As already mentioned, the analogous
+statements apply to $y(t)$. 
+
+It is rather straightforward to implement this
+prodecure in Ti\emph{k}Z. The relevant macros are \verb|\pgf@lt@curveto| (and
+\verb|\pgf@nlt@curveto|).\footnote{Some care has to be taken when squaring
+lengths since they are all measured in points, and the square can easily become
+large and trigger a \texttt{dimension too large} error. When computing the
+discriminant $d$, I thus divided these distances by some taming factor that I
+took to be 32 for no special reasons. It might very well be that there are
+better taming factors, or that one needs to dial the taming factor as a function
+of the input values.} The macro \verb|\pgf@lt@curveto| takes six arguments,
+which are $x_a$, $y_a$, $x_b$, $y_b$, $x_1$ and $y_1$ (in that order). $x_0$ and
+$y_0$ are stored in \verb|\pgf@path@lastx| and \verb|\pgf@path@lastx|,
+respectively.
+
+\subsection*{Examples}
+
+\makeatletter
+\def\pgf@lt@curveto#1#2#3#4#5#6{%
+  % extrema in x
+  \pgfmathparse{((#1/32)*(#1/32)-1*((#1/32)*(#3/32))+(#3/32)*(#3/32)-1*((#1/32)*(#5/32))+(-(#3/32)+(#5/32))*(\pgf@path@lastx/32))}%
+  \pgfutil@tempdima=\pgfmathresult pt % <- why do I need this space?
+  \ifdim\pgfutil@tempdima>0pt%
+   \pgfmathsetmacro{\tone}{min(1,max(0,(\pgf@path@lastx-2*#1+#3-32*sqrt(\pgfutil@tempdima))/(\pgf@path@lastx-#5-3*#1+3*#3)))}%
+   \pgfmathparse{\pgf@path@lastx*pow(1-\tone,3)+3*#1*pow(1-\tone,2)*\tone+3*#3*(1-\tone)*\tone*\tone+#5*pow(\tone,3)}%
+   \pgf@protocolsizes{\pgfmathresult pt}{#6}%
+   \pgfmathsetmacro{\tone}{min(1,max(0,(\pgf@path@lastx-2*#1+#3+32*sqrt(\pgfutil@tempdima))/(\pgf@path@lastx-#5-3*#1+3*#3)))}%
+   \pgfmathparse{\pgf@path@lastx*pow(1-\tone,3)+3*#1*pow(1-\tone,2)*\tone+3*#3*(1-\tone)*\tone*\tone+#5*pow(\tone,3)}%
+   \pgf@protocolsizes{\pgfmathresult pt}{#6}%
+  \fi% 
+  % extrema in y
+  \pgfmathparse{((#2/32)*(#2/32)-1*((#2/32)*(#4/32))+(#4/32)*(#4/32)-1*((#2/32)*(#6/32))+(-(#4/32)+(#6/32))*(\pgf@path@lasty/32))}%
+  \pgfutil@tempdima=\pgfmathresult pt % <- why do I need this space?
+  \ifdim\pgfutil@tempdima>0pt%
+   \pgfmathsetmacro{\tone}{min(1,max(0,(\pgf@path@lasty-2*#2+#4-32*sqrt(\pgfutil@tempdima))/(\pgf@path@lasty-#6-3*#2+3*#4)))}%
+   \pgfmathparse{\pgf@path@lasty*pow(1-\tone,3)+3*#2*pow(1-\tone,2)*\tone+3*#4*(1-\tone)*\tone*\tone+#6*pow(\tone,3)}%
+   \pgf@protocolsizes{#5}{\pgfmathresult pt}%
+   \pgfmathsetmacro{\tone}{min(1,max(0,(\pgf@path@lasty-2*#2+#4+32*sqrt(\pgfutil@tempdima))/(\pgf@path@lasty-#6-3*#2+3*#4)))}%
+   \pgfmathparse{\pgf@path@lasty*pow(1-\tone,3)+3*#2*pow(1-\tone,2)*\tone+3*#4*(1-\tone)*\tone*\tone+#6*pow(\tone,3)}%
+   \pgf@protocolsizes{#5}{\pgfmathresult pt}%
+  \fi% 
+  \pgf@protocolsizes{\the\pgf@path@lastx}{\the\pgf@path@lasty}%
+  \pgf@protocolsizes{#5}{#6}%
+  \pgfsyssoftpath@curveto{\the#1}{\the#2}{\the#3}{\the#4}{\the#5}{\the#6}%
+}
+\let\pgf@nlt@curveto\pgf@lt@curveto
+\makeatother 
+
+\begin{figure}[!htb]
+\centering
+\begin{tikzpicture}
+ \draw (0,0) .. controls (-1,1) and (1,2) .. (2,0);
+ \draw (current bounding box.south west) rectangle 
+  (current bounding box.north east);
+\end{tikzpicture}
+\caption{Tight bounding box for figure~\ref{fig:BoundingBoxBezier}.}
+\end{figure}
+
+\begin{figure}[!htb]
+\centering
+\begin{tikzpicture}[y=0.40pt, x=0.4pt,yscale=-1]
+ \path[draw=black,fill=white]
+   (258.9527,290.5199) .. controls (173.9885,538.4766) and (568.5860,261.2969) ..
+   (306.5098,257.1141) .. controls (44.4337,252.9312) and (429.9845,542.5624) ..
+   (352.9767,292.0206) .. controls (275.9689,41.4788) and (119.6549,497.6604) ..
+   (334.1376,346.9999) .. controls (548.6203,196.3394) and (66.4622,188.6439) ..
+   (276.0276,346.0724) .. controls (485.5930,503.5010) and (343.9169,42.5633) ..
+   (258.9527,290.5199) -- cycle;
+  \draw (current bounding box.south west) rectangle 
+  (current bounding box.north east);
+\end{tikzpicture}
+\caption{Tight bounding box for figure~\ref{fig:BoundingBoxExample}.}
+\end{figure}
+\end{document}