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Update bbox library #856
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tallmarmot authored and hmenke committed May 24, 2020
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183 changes: 116 additions & 67 deletions doc/generic/pgf/text-en/pgfmanual-en-library-bbox.tex
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% Copyright 2019 by an anonymous contributor
% Copyright 2020 by an anonymous contributor
%
% This file may be distributed and/or modified
%
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%
% See the file doc/generic/pgf/licenses/LICENSE for more details.


\section{Bounding Boxes for B\'ezier Curves}


\begin{pgflibrary}{bbox}
This library provides methods to determine tight bounding boxes for
B\'ezier curves.
B\'ezier curves. This library loads and uses the
|fpu| library.
\end{pgflibrary}

\subsection{Bounding box without the library}

\subsection{Current Status}

\tikzname\ determines the bounding box of (cubic) B\'ezier curves by
establishing the smallest rectangle that contains the end point and the two
control points of the curve. This may lead to drastic overestimates of the
bounding box.
\tikzname\ 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.

\begin{codeexample}[]
\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);
\begin{codeexample}[width=5cm]
\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}
\end{codeexample}

\subsection{Computing the Bounding Box}
As one can see from this illustration, this may lead to drastic overestimates of
the bounding box.

\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
\url{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
$(x_a,y_a)$ and $(x_b,y_b)$ can be parametrized by
\begin{equation}
\gamma(t) =
\begin{pmatrix} x(t)\\ y(t) \end{pmatrix} =
\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
Expand All @@ -55,70 +70,104 @@ \subsection{Computing the Bounding Box}
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\;.
\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
% (*parametrization of x:*)
% myx = x0 (1 - t)^3 + 3 xa (1 - t)^2 t + 3 xb (1 - t) t^2 + x1 t^3
% (*d1\ne0 condition for t1 and t2 to exist*)
% === (*case d1\ne0*) ==
% d1 = x0 - x1 - 3 xa + 3 xb
% (*square root, d2=0 \[Rule] only one solution,d2<0 \[Rule] no solution*)
% d2 = x0*x1 - x1*xa + xa*xa - x0*xb - xa xb + xb*xb
% (*first t*)
% t1 = (x0 - 2*xa + xb - sqrt(d2))/(x0 - x1 - 3*xa + 3*xb)
% = (x0 - 2*xa + xb - sqrt(d2))/d1
% (*second t*)
% t2 = (x0 - 2*xa + xb + sqrt(d2))/(x0 - x1 - 3*xa + 3*xb)
% = (x0 - 2*xa + xb + sqrt(d2))/d1
% === (*case d1=0*) ==
% (*2nd condition for extra condition: d3\ne0*)
% d3 = x1 + xa - 2 xb
% (*third t*)
% t3 = (x1 + 2*xa - 3*xb)/(2*d3)% d3 = x1 + 3 xa - 3 xb - x0
Let's discuss $x$ first. If the discriminant
\begin{equation}
d := (x_a-x_b)^2+(x_1-x_b)(x_0-x_a)
d~:=~x_0\,x_1 - x_1\,x_a + x_a\,x_a - x_0\,x_b - x_a x_b + x_b\,x_b
\end{equation}
is greater than 0, there are two solutions
\begin{equation}
t_\pm = \frac{x_{0}-2
x_{a}+x_{b}\pm\sqrt{d}}{x_{0}-x_{1}-3(x_{a}- x_{b})} \;.
\end{equation}
t_\pm~=~\frac{x_{0}-2x_{a}+x_{b}\pm\sqrt{d}}{%
x_{0}-x_{1}-3(x_{a}- x_{b})} \;.
\end{equation}
If the denominator $x_{0}-x_{1}-3(x_{a}- x_{b})$ vanishes, one may use the
l'Hospital rule to determine the solutions.
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. One can plug $t_\pm$ back into
\eqref{eq:gammaBezier}, this yields
\begin{subequations}
\begin{align}
x_- &=
\!\begin{aligned}[t]
\frac{1}{(x_0 - x_1 - 3x_a + 3x_b)^2}
\Bigl[
& x_0^2x_1 + x_0x_1^2 - 3x_0x_1x_a + 6x_1x_a^2
+ 2x_a^3 - 3(x_0 + x_a)(x_1 + x_a)x_b \\
& + 3(2x_0 - x_a)x_b^2 + 2x_b^3
- 2\sqrt{d}(x_0x_1 - x_1x_a + x_a^2 - (x_0 + x_a)x_b + x_b^2)
\Bigr],
\end{aligned} \\
x_+ &=
\!\begin{aligned}[t]
\frac{1}{(x_0 - x_1 - 3x_a + 3x_b)^2}
\Bigl[
& x_0^2x_1 + x_0x_1^2 - 3x_0x_1x_a + 6x_1x_a^2
+ 2x_a^3 - 3(x_0 + x_a)(x_1 + x_a)x_b \\
& + 3(2x_0 - x_a)x_b^2 + 2x_b^3
+ 2\sqrt{d}(x_0x_1 - x_1x_a + x_a^2 - (x_0 + x_a)x_b + x_b^2)
\Bigr].
\end{aligned}
\end{align}
\end{subequations}
As already mentioned, the analogous
statements apply to $y(t)$.

This procedure is implemented in the |bbox| library. It installs a single key
by which the tight bounding box algorithm can be turned on and off.
be increased in the $x$ direction. On the other hand, if there are solutions,
one needs include the points $\bigl(x(t_\pm),y_0\bigr)$ with $x(t)$ from
\eqref{eq:gammaBezier} in the bounding box.

The analogous statements apply to $y(t)$.

\subsection{Using the library}

\begin{key}{/pgf/bezier bounding box=\meta{boolean} (default true)}
Turn the tight bounding box algorithm on and off.
Turn the tight bounding box algorithm on and off. The initial value is
|false|.

\emph{Caveat:} As can be seen from the derivations, the necessary
computations involve the squaring of lengths, which can easily lead to
|dimension too large| errors. The library tries to account for large
numbers by appropriate normalization, such that it works in most cases, but
errors may still occur.
\emph{Caveat:} As can be seen from the derivations, the necessary
computations involve the squaring of lengths and taking ratios of lengths,
which can easily lead to |dimension too large| errors. The library uses
|fpu| to account for that, but errors may still occur.
\end{key}

\begin{codeexample}[]
\begin{tikzpicture}[bezier bounding box=true]
\draw (0,0) .. controls (-1,1) and (1,2) .. (2,0);
\draw (current bounding box.south west) rectangle
(current bounding box.north east);

\begin{codeexample}[width=5cm]
\begin{tikzpicture}[bezier bounding box,%
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}
\end{codeexample}

A few comments are in order.
\begin{enumerate}
\item For paths with arrow heads one may need to load the
\texttt{bending} library. This is because otherwise the quick arrow head
distorts the path, and this happens after the bounding box has been computed.
Even worse, arrow heads could get deformed.
\item If you shorten a path by some negative length, the bounding box will not
be accurate either. However, this has nothing to do with curves, it also
applies to straight lines. So this is not specific to the |bbox| library but
something that one may want to keep in mind.
\item Let us also note that the computations can lead to |dimension too large| errors.
These errors do not come directly from the computations done by the library,
which uses |fpu| for its computations, but from the aftermath. Many of these
problems can be avoided by using the |fpu| library also for computing
reciprocals.
\end{enumerate}

\begin{key}{/pgf/use fpu reciprocal\meta{boolean} (default true)}
This changes the computation of reciprocals from the standard pgf routine to
an |fpu| variant. The initial value is |false|.
\end{key}

\endinput


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