\documentclass[dvips,11pt]{amsart}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{graphicx}
\usepackage{pstricks}
\usepackage{multido,pst-node,pst-bspline,pstricks-add}
\usepackage{amssymb}
\usepackage[parfill]{parskip}
\usepackage{lmodern}
\usepackage[scaled=.82]{luximono}%requires T1+textcomp
\usepackage[T1]{fontenc}
\newcommand\textSMC[1]{{\SMC #1}}
\newcommand\acro[1]{\textSMC{#1}\@}
\DeclareRobustCommand\cs[1]{\texttt{\char`\\#1}}
\usepackage[cal=boondoxo]{mathalfa}
\usepackage{textcomp}
%\usepackage{url}
%\def\url@ttstyle{%
% \@ifundefined{selectfont}{\def\UrlFont{\tt}}{\def\UrlFont{\ttfamily\small}}}
\usepackage{hyperref}
\hyphenation{Post-Script de-fines}
%\date{} % Activate to display a given date or no date
\begin{document}
\begin{center}{\Large Cubic B-splines Using PSTricks\\[12pt]
\large Michael Sharpe\\[10pt]
msharpe@ucsd.edu}
\end{center}
A cubic, uniform B-spline curve with control points $B_0 \ldots B_n$ ($n \ge 2$) is a curve parametrized by the interval $[0,n]$, which is, except in degenerate cases, $C^2$-continuous (that is, has continuous curvature) and is on each interval $[k-1,k]$ ($0\}}] defines a clipping path 12{\tt pt} wide around the B-spline curve with control points {\tt B0}..{\tt B5}, then draws the {\tt } clipped to that path.
\item[\cs{thickBEspline\{S\}\{5\}\{30pt\}\{18pt\}\{-70\}\{\}}] defines a clipping path of width from {\tt18pt} to {\tt 30pt} and axis angle -70 degrees, then draws the {\tt } clipped to that path. The clipping path is widest where the path is parallel to the axis angle. This macro is best called from the following auxiliary macro, because the data must be prepared in a specific order---see the example provided.
\item[\cs{thickBdraw\{S\}\{30pt\}\{18pt\}\{-70\}\{\}}] prepares and runs the macro \verb|thickBEspline| using all points in the node sequence {\tt S}, apassing all parameters to that macro.
\item[\cs{thickBsplinePen\{S\}\{5\}\{.7cm\}\{.2cm\}\{-45\}\{\}}] defines a clipping path following the outline of an elliptical pen drawn along the curve interpolating the points {\tt S0..S5}. The ellipse has $a=.7$cm, $b=.2$cm and and is rotated through -70 degrees. The macro then draws the {\tt } clipped to that path. The underlying curve has relaxed endpoints.
\item[\cs{thickBsplinePenE\{S\}\{5\}\{.7cm\}\{.2cm\}\{-45\}\{\}}] defines a clipping path following the outline of an elliptical pen drawn along the curve interpolating the points {\tt S0..S5}. The ellipse has $a=.7$cm, $b=.2$cm and and is rotated through -70 degrees. The macro then draws the {\tt } clipped to that path. The underlying curve omits the intial and final segments, and is not limited to relaxed endpoints.
\end{description}
Details and examples are provided below.
\section{Relaxed, Open B-spline} The algorithm to generate such a curve from a sequence of control points $B_0$, $\cdots$, $B_n$ is as follows:
\begin{itemize}
\item The curve starts at $B_0$ and ends at $B_n$. (Important: $n \ge 2$.)
\item Divide each line segment $B_{k-1}B_k$ into equal thirds, with subdivision points labeled $R_{k-1}$, $L_k$ respectively, so that $B_k$ has $L_k$ as its immediate neighbor to the left, and $R_k$ as its immediate neighbor to the right.
\item For $0}{}
\psBsplineNodesC{}{}
\psBsplineNodesE{}{}
\end{verbatim}
corresponding to the macros \verb|\psBspline|, \verb|\psBsplineC| and \verb|\psBsplineE|. The difference is that the macros with {\tt Nodes} in the name have as arguments the root node name and the last index, rather than the list of points. For example, with the above definition of {\tt P} in force, \verb|\psBsplineNodes{P}{2}| has exactly the same effect as \verb|\psBspline(2,1.5)(3,4)(5,1).|
\subsection{The \cs{bspcurvepoints} macros}
There are two macros that provide for B-spline curves essentially the same functionality as the \verb|\pscurvepoints| macro from {\tt pstricks-add}. (That macro takes as input a parametric curve and constructs as output (at the PostScript level) arrays of data associated with the curve: the positions of points along the curve, the increment from the previous point and a normal vector to the curve. The principal uses for such data are (i) the \verb|\pspolylineticks| macro from {\tt pstricks-add}, which allows placement of ticks and other marks along a curve that has been approximated by a polyline; (ii) the \cs{polyIntersections} macro from \textsf{pst-node}, which allows you to find the points of intersection of the curve (approximated by a polyline) and an arbitrary line.) Following one of the \cs{psBsplineNodes} macros, the macros
\begin{verbatim}
\bspcurvepoints{}{}{}
\bspcurvepointsE{}{}{}
\end{verbatim}
work, in the first case, for a relaxed, uniform B-spline curve, and in the second, for such a curve with its initial and final segments removed, corresponding to the output from \verb|\psBsplineE| rather than \verb|\psBspline|. In both cases, you may set the keyword {\tt plotpoints} (default value: $50$) to change the number of sample points on each B\'ezier component. This will result in the construction of PostScript arrays with indices from $0$ to $n=$ \textsf{num of segments}$\times$\textsf{(plotpoints-1)}. After running
\begin{verbatim}
\pnodes{B}(1,2)(3,-1)(4,1)(6,2)% define B0..B3
\psBsplineNodes{B}{3}% draw B-spline with control pts B0..B3
\bspcurvepoints[plotpoints=11]{B}{3}{P}% requires previous line
\end{verbatim}
the following PostScript arrays are created, each indexed from 0 to 30:
\begin{verbatim}
P.X, P.Y (position)
PNormal.X, PNormal.Y (normal vector)
PDelta.X, PDelta.Y (increment from previous position)
\end{verbatim}
and these may be used in the usual way to create nodes. For example,
\begin{verbatim}
\pnode(! P.X 8 get P.Y 8 get ){Q}
\pnode(! PNormal.X 8 get PNormal.Y 8 get ){Dir}
\psrline{*-}(Q)(1cm;{(Dir)})
\end{verbatim}
places {\tt Q} at the position on the curve with index 8, defines {\tt Dir} to be a normal vector at that point, then draws a line from {\tt Q} of length {\tt 1cm} in the direction of that normal.
\pspicture(-.5,-.5)(6.5,2.5)
\pnodes{B}(1,2)(3,-1)(4,1)(6,2)% define B0..B3
\psBsplineNodes{B}{3}% draw B-spline with control pts B0..B3
\bspcurvepoints[plotpoints=11]{B}{3}{P}
\pnode(! P.X 8 get P.Y 8 get ){Q}\psdot(Q)
\pnode(! PNormal.X 8 get PNormal.Y 8 get ){Dir}
\psrline{*-}(Q)(1cm;{(Dir)})
\endpspicture
In the next example, we use \cs{polyIntersections} from {\tt pst-plot} to locate intersections of a line and a B-spline curve.
\begin{verbatim}
\pspicture(-.5,-.5)(6.5,4.5)%
\pnodes{B}(0,4)(1,-1)(3,4)(5,0)(6,3)% B0..B5
\psBsplineNodes{B}{4}% draw B-spline with control pts B0..B4
\pnode(1,2.8){A1}\pnode(2,2){A2}%
\psdots[linecolor=red](A1)(A2)%
\bspcurvepoints[plotpoints=30]{B}{4}{P}% construct PS arrays,
\bspcurvenodes{P}{Q}% turn them into nodes
% indices 0..116 (=4*29)
\polyIntersections{N1}{N2}(A1)(A2){Q}{116}%
\psline{*-*}(N1)(N2)%
% N1, N2 are points of intersection of curve with A1A2
\Put{;75}(A1){A1}\Put{;75}(A2){A2}
\endpspicture
\end{verbatim}
\pspicture(-.5,-.5)(6.5,4.5)%
\pnodes{B}(0,4)(1,-1)(3,4)(5,0)(6,3)% B0..B5
\psBsplineNodes{B}{4}% draw B-spline with control pts B0..B4
\pnode(1,2.8){A1}\pnode(2,2){A2}%
\psdots[linecolor=red](A1)(A2)%
\bspcurvepoints[plotpoints=30]{B}{4}{P}% construct PS arrays,
\bspcurvenodes{P}{Q}% turn them into nodes
% indices 0..116 (=4*29)
\polyIntersections{N1}{N2}(A1)(A2){Q}{116}%
\psline{*-*}(N1)(N2)% N1, N2 are points of intersection of curve with A1A2
\Put{;75}(A1){A1}\Put{;75}(A2){A2}
\endpspicture
\subsection{Setting nodes on a B-spline curve}
To set a node at $t$ on a B-spline curve after\cs{bspcurvepoints}{\tt[E]}, call the macro
\begin{verbatim}
\bspNode{}{}{}{}
\end{verbatim}
For example, if I have constructed a B-spline curve using control points $B_0$,$\dots$,$B_5$, then \verb|\bspNode{B}{5}{2.1}{Q}| defines a node named $Q$ at $t=2.1$.
The macro \cs{bspcurvenodes\{P\}\{R\}} creates a node sequence {\tt R0}..{\tt Rn} at the locations specified by the arrays {\tt P.X}, {\tt P.Y}. (Those arrays must first have been created with one of the \cs{bspcurvepoints} macros.)
\subsection{B-spline function curves}
By this we mean an open B-spline curve which is the graph of a function $y=f(x)$ and whose orientation is toward the right. It is not analytically simple to specify a formula for $f$ in most cases, and to compute $y$ from $x$ involves (a) finding the index $k$ of the B\'ezier segment containing $x$; (b) solving the cubic $x_k(t)=x$ for $t$, $0\le t\le1$; (c) substituting in $y_k(t)$. The package provides a macro to perform these calculations after generating the data using \cs{bspcurvepoints}{\tt[E]}:
\begin{verbatim}
\bspfnNode{}{}{}{}
\end{verbatim}
\begin{verbatim}
\pspicture(-.5,-.5)(6.5,4.5)%
\pnodes{B}(0,4)(1,-1)(3,4)(5,0)(6,3)% B0..B4
\psBsplineNodes{B}{4}% draw B-spline with control pts B0..B4
% the curve is graph of a function of x
\bspcurvepoints[plotpoints=10]{B}{4}{P}% construct PS arrays
\bspFnNode{B}{4}{4.5}{QQ}% node QQ on curve at x=4.5
\psdot[linecolor=red](QQ)%
\psline[linestyle=dashed](QQ)(0,0 | QQ)
\psline[linestyle=dashed](QQ)(QQ | 0,0)
\psaxes(0,0)(-.5,-.5)(6,4)
\endpspicture
\end{verbatim}
\pspicture(-.5,-.5)(6.5,4.5)%
\pnodes{B}(0,4)(1,-1)(3,4)(5,0)(6,3)% B0..B4
\psBsplineNodes{B}{4}% draw B-spline with control pts B0..B4
% the curve is graph of a function of x
\bspcurvepoints[plotpoints=10]{B}{4}{P}% construct PS arrays
\bspFnNode{B}{4}{4.5}{QQ}% node QQ on curve at x=4.5
\psdot[linecolor=red](QQ)%
\psline[linestyle=dashed](QQ)(0,0 | QQ)
\psline[linestyle=dashed](QQ)(QQ | 0,0)
\psaxes(0,0)(-.5,-.5)(6,4)
\endpspicture
See also the penultimate example in the next section.
\section{B-spline Interpolation}
This is the inverse problem. Being given points $(S_k)_{0\le k\le n}$, the goal is to produce the B-spline control points $B_k$ leading to the points $S_k$, so that the associated B-spline curve interpolates the $S_k$.
\subsection{Open curve}
We discuss first the case of an open, uniform B-spline curve with relaxed endpoints. According to the discussion above, we have to solve the equations
\begin{align*}
B_0&=S_0\\
B_0+4B_1+B_2&=6S_1\\
B_1+4B_2+B_3&=6S_2\\
\cdots&\\
B_{n-2}+4B_{n-1}+B_n&=6S_{n-1}\\
B_n&=S_n
\end{align*}
for the $B_k$. In matrix form, this becomes the tridiagonal system
\[\begin{pmatrix}4&1\\
1&4&1\\
&1&4&1\\
&&\cdots&&1\\
&&&1&4\end{pmatrix}
\begin{pmatrix}B_1\\B_2\\B_3\\ \cdots\\ B_{n-1}\end{pmatrix}=
\begin{pmatrix}6S_1-S_0\\6S_2\\6S_3\\ \cdots\\6S_{n-1}-S_{n}\end{pmatrix}
\]
The LU decomposition of the tridiagonal matrix may be seen to take the form
\[
\begin{pmatrix}1\\
m_1&1\\
&m_2&1\\
&&\cdots\\
&&&m_{n-2}&1\end{pmatrix}
\begin{pmatrix}m_1^{-1}&1\\
&m_2^{-1}&1\\
&&m_3^{-1}&1\\
&&&\cdots&1\\
&&&&m_{n-1}^{-1}\end{pmatrix}
\]
where $m_1=1/4$, $m_{k+1}=1/(4-m_k)$ for $k=1,\cdots,n-2$. The solution of the original system is therefore accomplished in two steps, introducing intermediate points $(R_k)$, by (in pseudo-code)
\begin{verbatim}
R_1=6*S_1-S_0
for i=2 to n-2
R_i=6*S_i-m_{i-1}* R_{i-1}
R_{n-1}=(6*S_{n-1}-S_n)-m_{n-2}*R_{n-2}
B_{n-1}=m_{n-1}*R_{n-1}
for i=n-2 downto 1
B_i=m_i*(R_i-B_{i+1})
\end{verbatim}
The code for the \verb|\psBsplineInterp| command uses this algorithm to solve for the $B_k$ as nodes, except that in order to save node memory, the $B$ nodes are substituted in place for the $R$ nodes, so that, for example, the first step becomes \verb|B_1=6*S_1-S_0|.
Assuming you have previously defined nodes {\tt S0} $\cdots ${\tt S4},
\begin{verbatim}
\psBsplineInterp{S}{4}
\end{verbatim}
will construct a sequence {\tt SB0} $\cdots ${\tt SB4} of nodes at the B-spline control points for the relaxed, uniform cubic B-spline interpolating the {\tt Sk}, and this curve may then be rendered with the command
\begin{verbatim}
\psBsplineNodes{SB}{4}
\end{verbatim}
If you don't care about keeping track of the internal operations and names for nodes, you may generate the curve directly with, for example,
\begin{verbatim}
\psbspline(0,0)(.5,.1)(1.5,.6)(2.5,1.4)(3.5,1.8)(4.5,1.7)%
(5.8,1.0)(7.5,.25)(10,0)
\end{verbatim}
\subsection{Closed (periodic) case}
We turn now to the periodic uniform B-spline curve interpolating $n$ points $S_0$,...,$S_{n-1}$. Extend the sequence periodically with period $n$, so that $S_n=S_0$, $S_{n+1}=S_1$, $S_{-1}=S_{n-1}$, and so on. In order to find the periodic control points $B_k$, we have to solve the $n$ equations
\begin{align*}
B_n+4B_1+B_2&=6S_1\\
B_1+4B_2+B_3&=6S_2\\
\cdots&\\
B_{n-2}+4B_{n-1}+B_n&=6S_{n-1}\\
B_{n-1}+4B_n+B_1&=6S_n
\end{align*}
for the $B_k$, $1\le k\le n$. In matrix form, this becomes the system
\[\begin{pmatrix}4&1&&&1\\
1&4&1\\
&1&4&1\\
&&\cdots&&1\\
1&&&1&4\end{pmatrix}
\begin{pmatrix}B_1\\B_2\\B_3\\ \cdots\\ B_{n}\end{pmatrix}=
\begin{pmatrix}6S_1\\6S_2\\6S_3\\ \cdots\\6S_{n}\end{pmatrix}
\]
Let $(x_k,y_k)=6S_k$. We perform Gaussian elimination on the matrix
\[\begin{pmatrix}4&1&&&1&x_1&y_1\\
1&4&1&&&x_2&y_2\\
&1&4&1&&x_3&y_3\\
&&\cdots&&1\\
1&&&1&4&x_n&y_n\end{pmatrix}
\]
As in the previous case, let $m_1=0.25$, $m_k=1/(4-m_{k-1})$ for $k\ge 2$. The factor $m_k$ will be the multiplier of row $k$ after the previous row operation, in order to normalize the row. These are the steps in the procedure.
\begin{itemize}
\item Initialize: multiply row 1 by $m_1$ so that its first entry (1,1) is 1. Replace $x_1$ by $m_1 x_1$ and $y_1$ by $m_1 y_1$. Entry $(1,n)$ is $m_1$.
\item Subtract new row 1 from row 2 and multiply the resulting row by $m_2$. The leading entry (2,1) becomes $1$. Entry $(2,n)$ becomes $-m_1m_2$, and $x_2, y_2$ are updated to $m_2(x_2- x_1)$, $m_2(y_2-y_1)$. The superdiagonal entry (2,3) is the only other non-zero entry, and its new value is $m_2$.
\item Subtract new row 1 from row $n$, so that its leading entry $(n,2)$ is $-m_1$.
\item Subtract new row 2 from row 3 and multiply the result by $m_3$. The leading entry (3,3) becomes $1$ and the entry $(3,n)$ becomes $m_1m_2$, with $x_3, y_3$ updating to $m_3(x_3-x_2)$, $m_3(y_3-y_2)$. The superdiagonal entry (3,4) is now $m_3$.
\item Subtract new row 2 times $-m_1$ from row $n$, whose leading entry $(n,3)$ is now $m_1m_2$.
\item Continue in this way until row $n-2$ has been subtracted as above from row $n-1$, multiplying the result by $m_{n-1}$, and a suitable multiple has been subtracted from row $n$. The leading entry of row $n-1$ (column $n-1$) is $1$ and its $n^{\text{th}}$ entry is $1-(-1)^{n}m_1\cdots m_{n-2}$. Row $n$ has leading entry in column $n-1$, equal to $1$.
\item Finally, subtract an appropriate multiple of row $n-1$ from row $n$ so that row $n$ has leading entry in column $n$. The resulting matrix is upper triangular, and we may now substitute back starting from the last row to give a complete reduction.
\end{itemize}
Here are the steps in pseudocode. We keep track of row $n$ with the array $b_k$, column $n$ with the array $c_k$. The indices for both run from 1 to $n$.
\begin{verbatim}
m(1)=0.25
for k=2 to n-1
m(k)=1/(4-m(k-1))
b(1)=1
b(n-1)=1
b(n)=4
c(n-1)=1% don't need c(n), =b(n)
%multiply first row by m1
c(1)=m(1)
x(1)=m(1)*x(1)
y(1)=m(1)*y(1)
for k=2 to n-1
%subtract normalized row k-1 from row k, renormalize row k
c(k)=m(k)*(c(k)-c(k-1))%note that initially, c(k)=0 for 1}{}
\end{verbatim}
You must previously have defined a sequence, say {\tt S0} $\cdots$ {\tt S100} of \verb|\pnode|s that you plan to interpolate with a closed curve. If you used \verb|\pnodes| to do this, it would have constructed a macro \verb|\Snodecount| to store the value $100$. Then
\begin{verbatim}
\psBsplineInterpC{S}{100}
\end{verbatim}
constructs the sequence {\tt SB0} $\cdots$ {\tt SB100} of B-spline control points (appending~{\tt B} to the root name) for a closed curve interpolating {\tt S0} $\cdots$ {\tt S100}, which may then be rendered with the command
\begin{verbatim}
\psBsplineNodesC{SB}{101}
\end{verbatim}
with any keywords options you wish.
{\bf IMPORTANT:} The macro \verb|\psBsplineInterpC| modifies the node sequence {\tt S}, setting \verb|S101=S0|, and changing \verb|\Snodecount| to take the value $101$. This is convenient when you use the construction:
\begin{verbatim}
\pnodes{S}()()()()% sets \Snodecount to 3
\psBsplineInterpC{S}{\Snodecount}% constructs B-spline control pts SB0..SB4
\psBsplineNodesC{SB}{\Snodecount}
\end{verbatim}
The following example illustrates that there is a difference between \verb|\psccurve| and B-spline interpolation, the former having a rounder appearance. Generally speaking, B-spline interpolation comes closer to minimizing the average curvature.
\begin{verbatim}
\documentclass{article}
\usepackage{pstricks}
\usepackage{pst-bspline,pstricks-add}
\begin{document}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{P}(0,1)(2,0)(5,2)(6,4)(4,5)(2,4)%
\psBsplineInterpC{P}{5}%
\psBsplineNodesC*[linecolor=gray!40]{PB}{5}%
\psccurve[linecolor=red,showpoints=true](0,1)(2,0)(5,2)(6,4)(4,5)(2,4)
\end{pspicture}
\end{document}
\end{verbatim}
\vspace{1pc}
\begin{center}
Slight difference between psccurve and B-spline interpolation\\
\vspace*{2pc}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{P}(0,1)(2,0)(5,2)(6,4)(4,5)(2,4)%
\psBsplineInterpC{P}{5}%
\psBsplineNodesC*[linecolor=gray!40]{PB}{5}%
\psccurve[linecolor=red,showpoints=true](0,1)(2,0)(5,2)(6,4)(4,5)(2,4)
\end{pspicture}
\end{center}
A B-spline curve can in many cases provide a good function interpolation mechanism, but the result is not guaranteed to be the graph of a function.
\begin{verbatim}
\begin{pspicture}(-.5,-.5)(6,4)
\psdots(0,3.5)(1,.5)(3,2.5)(4,0)(5,2)(6,.5)%
\pnodes{S}(0,3.5)(1,.5)(3,2.5)(4,0)(5,2)(6,.5)% S0..S5
\psBsplineInterp{S}{5}% construct SB0..SB5
\psBsplineNodes{SB}{5}% draw B-spline with control pts SB0..SB5
\bspcurvepoints[plotpoints=10]{SB}{5}{P}
% construct the PS arrays
\bspFnNode{SB}{5}{4.5}{QQ}% node QQ on curve at x=4.5
\psdot[linecolor=red](QQ)%
\psaxes(0,0)(-.5,-.5)(6,4)
\end{pspicture}
\end{verbatim}
\vspace{1pc}
\begin{center}
\begin{pspicture}(-.5,-.5)(6,4)
\psdots(0,3.5)(1,.5)(3,2.5)(4,0)(5,2)(6,.5)%
\pnodes{S}(0,3.5)(1,.5)(3,2.5)(4,0)(5,2)(6,.5)% S0..S5
\psBsplineInterp{S}{5}%SB0..SB5
\psBsplineNodes{SB}{5}% draw B-spline with control pts SB0..SB5
\bspcurvepoints[plotpoints=10]{SB}{5}{P}
% construct the PS arrays
\bspFnNode{SB}{5}{4.5}{QQ}% node QQ on curve at x=4.5
\psdot[linecolor=red](QQ)%
\psaxes(0,0)(-.5,-.5)(6,4)
\end{pspicture}
\end{center}
\vspace{12pt}
\begin{verbatim}
\documentclass{article}
\usepackage{pstricks}
\usepackage{pst-bspline,pstricks-add}
\begin{document}
\psset{unit=.25in}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{P}(0,1)(2,0)(5,2)(6,4)(4,5)(2,4)
\pnode(3,3){C}
\multido{\ra=0+.05,\rb=1+.05,\i=30+1}{40}{%
\psBsplineC*[linecolor=blue!\i!brown]{B}%
([nodesep=\ra]{C}P0)([nodesep=\ra]{C}P1)%
([nodesep=\ra]{C}P2)([nodesep=\ra]{C}P3)%
([nodesep=\ra]{C}P4)([nodesep=\ra]{C}P5)}
\end{pspicture}
\end{document}
\end{verbatim}
\vspace{1pc}
\begin{center}
\psset{unit=.25in}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{P}(0,1)(2,0)(5,2)(6,4)(4,5)(2,4)
\pnode(3,3){C}
\multido{\ra=0+.05,\rb=1+.05,\i=30+1}{40}{%
\psBsplineC*[linecolor=blue!\i!brown]{B}%
([nodesep=\ra]{C}P0)([nodesep=\ra]{C}P1)%
([nodesep=\ra]{C}P2)([nodesep=\ra]{C}P3)%
([nodesep=\ra]{C}P4)([nodesep=\ra]{C}P5)}
\end{pspicture}
\end{center}
\section{Thick B-spline curves}
Inspired by the package {\tt pst-thick}, we provide a similar option for curves generated as B-spline interpolations. The new macro that accomplishes this is
\begin{verbatim}
\thickBspline#1#2#3#4
%#1=root | #2=nsegments | #3=thickness | #4=items to clip
\end{verbatim}
which expects the following data.
\begin{itemize}
\item
A node sequence. This can be constructed with a command like
\begin{verbatim}
\pnodes{S}(0,0)(5,1)(4,4)(1,3)%
\end{verbatim}
which declares nodes S0..S3, and sets the macro \verb|\Snodecount| to 3.
\item An interpolation command, such as
\begin{verbatim}
\psBsplineInterp{S}{\Snodecount}%
\end{verbatim}
which creates a framework of B-spline control points {\tt SB0..SB3}.
\item Create the interpolating curve and the B\'ezier control points for its components, with names like {\tt SBR0..SBR2,SBL1..SBL3} etc, using
\begin{verbatim}
\psBsplineNodes[linestyle=none]{SB}{\Snodecount}%
\end{verbatim}
(The {\tt [linestyle=none]} may be omitted if you want the curve to show.)
\item Create a clipping path of specified thickness around the interpolating curve and place graphics to be clipped:
\begin{verbatim}
\thickBspline[plotpoints=50,linestyle=none]{S}{3}{20pt}%
{\psline[linecolor=red,linestyle=solid](0,0)(6,6)}%
\end{verbatim}
(The {\tt [linestyle=none]} controls whether the clipping path is rendered, and {\tt plotpoints} controls the number of subdivisions of each B\'ezier component. Its default value is 50.)
\end{itemize}
The clipping path is drawn by default positively oriented so that objects are clipped to its interior. By specifying the keyword {\tt reverseclip}, the clipping path will be reversed so that objects are clipped to the exterior.
\begin{verbatim}
\documentclass{article}
\usepackage{pstricks}
\usepackage{pst-bspline,pstricks-add}
\begin{document}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{S}(1,0)(5,1)(4,4)(1,3)%
\psdots(1,0)(5,1)(4,4)(1,3)%
\psBsplineInterp{S}{\Snodecount}%
\psBsplineNodes[linestyle=none]{SB}{\Snodecount}%
\thickBspline[plotpoints=50,linestyle=none]{S}{3}{20pt}%
{\psframe[fillstyle=crosshatch](-1,-1)(6,6)}%
\end{pspicture}
\end{document}
\end{verbatim}
\vspace{12pt}
\begin{center}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{S}(1,0)(5,1)(4,4)(1,3)%
\psdots(1,0)(5,1)(4,4)(1,3)%
\psBsplineInterp{S}{\Snodecount}%
\psBsplineNodes[linestyle=none]{SB}{\Snodecount}%
\thickBspline[plotpoints=50,linestyle=none]{S}{\Snodecount}{20pt}{\psframe[fillstyle=crosshatch](-1,-1)(6,6)}%
\end{pspicture}
\end{center}
\vspace{12pt}
The \verb|\thickBspline| macro works as expected in the closed (periodic) case, taking advantage of automatic incrementing of the nodecount. Note that \verb|\thickBspline| interprets thickness as visual, unaffected by possible differences between {\tt xunit} and {\tt yunit}.
\begin{verbatim}
\documentclass{article}
\usepackage{pstricks}
\usepackage{pst-bspline,pstricks-add}
\begin{document}
\psset{yunit=1.5cm}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{S}(1.5,0)(5,1)(4,4)(1,3)%
\psBsplineInterpC{S}{3}%
% defines nodes SB0, SB1, SB2, SB3, SB4 --- the Bspline control points
% increments \Snodecount by 1 for future macros
% Don't use C form of \psBsplineNodes with this new \Snodecount
\psBsplineNodes[linestyle=none,showpoints=false]{SB}{\Snodecount}%
% Constructs the Bezier control points SBR0, SBL1, SBR1, etc
\thickBspline[linestyle=none]{S}{\Snodecount}{22pt}%
{\psframe[fillstyle=vlines](-1,-1)(6,6)}%
\end{pspicture}
\end{document}
\end{verbatim}
\vspace{12pt}
\begin{center}
\psset{yunit=1.5cm}
\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
\pnodes{S}(1.5,0)(5,1)(4,4)(1,3)%
\psBsplineInterpC{S}{3}%
% defines nodes SB0, SB1, SB2, SB3 --- the Bspline control points
% increments \Snodecount by 1 for future macros
\psBsplineNodes[linestyle=none,showpoints=false]{SB}{\Snodecount}%
% Constructs the Bezier control points SBR0, SBL1, SBR1, etc
\thickBspline[linestyle=none]{S}{\Snodecount}{22pt}%
{\psframe[fillstyle=vlines](-1,-1)(6,6)}%
%{\psframe[fillstyle=solid, fillcolor=lightgray](-1,-1)(6,6)}%
\end{pspicture}
\end{center}
\vspace{12pt}
There are three variants of the \verb|\thickBspline| macro. The first is
\begin{verbatim}
\thickBEspline#1#2#3#4#5#6
% #1=root | #2=nsegments | #3=maxthickness |
% #4=minthickness | #5=axis angle | #6=items to clip
\end{verbatim}
works just like \verb|\thickBspline|, but instead of providing a thickness, you provide three parameters specifying maxthickness, minthickness and an axis angle.
An axis angle of -45 means that lines normal to angle -45 will be minthickness and lines normal to 45 will be maxthickness. You may get unexpected results if your initial curve is too curvy. This macro is best called from \verb|\thickBdraw|, as shown in the following example.
\begin{verbatim}
\psset{yunit=.7cm}
\begin{pspicture}[showgrid=false](-1,-1)(6,7)
\psgrid%
\pnodes{S}(0,1)(0,5)(3,5)(4,2)(5,0)%
\thickBdraw{S}{30pt}{16pt}{-70}{\psframe[fillstyle=solid,fillcolor=yellow](-1,-1)(6,7)}%
\end{pspicture}
\end{verbatim}
\begin{center}
\psset{yunit=.7cm}
\begin{pspicture}[showgrid=false](-1,-1)(6,7)
\psgrid%
\pnodes{S}(0,1)(0,5)(3,5)(4,2)(5,0)%
\thickBdraw{S}{30pt}{16pt}{-70}{\psframe[fillstyle=solid,fillcolor=yellow](-1,-1)(6,7)}%
\end{pspicture}
\psset{yunit=1cm}
\end{center}
A second variant \verb|\thickBsplinePen| simulates the effect of an elliptical pen run along the specified curve. Note that you will receive an error message if xunit differs from yunit---the calculations in the general case are quite nasty. The macro is called with something like
\begin{verbatim}
\thickBsplinePen{S}{\Snodecount}{2cm}{.5cm}{10}{%
\psframe[fillstyle=crosshatch](-1,-1)(7,6)}%
\end{verbatim}
where {\tt S0..} is an established node sequence with highest index \verb|\Snodecount| and the pen has semi-major axis {\tt 2cm}, semi-minor axis {\tt.5cm}, and the major axis is rotated 10 degrees from the positive x axis. This gives results that look like those generated in \textsf{metafont/post} except that the curve ends are not drawn by the pen---if you want them, draw them separately using \verb|\psellipse|. Note that the underlying curve has relaxed ends.
\begin{verbatim}
\begin{pspicture}[showgrid=false](-1,-1)(7,7)
\psgrid%
\pnodes{S}(1,0)(4,1)(5,2)(3,3)(1,5)(5,6)%
\thickBsplinePen[linecolor=green]{S}{\Snodecount}{.7cm}{.1cm}{-45}{\psframe[fillstyle=solid,fillcolor=green](-1,-1)(7,7)}%
\psBsplineNodes[linestyle=solid]{SB}{\Snodecount}%
\end{pspicture}
\end{verbatim}
\vspace{12pt}
\begin{center}
\begin{pspicture}[showgrid=false](-1,-1)(7,7)
\psgrid%
\pnodes{S}(1,0)(4,1)(5,2)(3,3)(1,5)(5,6)%
\thickBsplinePen[linecolor=green]{S}{\Snodecount}{.7cm}{.1cm}{-45}{\psframe[fillstyle=solid,fillcolor=green](-1,-1)(7,7)}%
\psBsplineNodes[linestyle=solid]{SB}{\Snodecount}%
\end{pspicture}
\end{center}
A third variant \verb|\thickBsplinePenE| is quite similar but omits the first and last segments of the underlying curve, permitting the endpoints to not be relaxed.
\begin{verbatim}
\begin{pspicture}[showgrid=false](-1,-1)(7,7)
\psgrid%
\pnodes{S}(0,4)(1,0)(4,.5)(5,2)(3,3)(1,5)(5,6)(5,3)%
\thickBsplinePenE[linecolor=green]{S}{\Snodecount}{.7cm}{.1cm}{30}{\psframe[fillstyle=solid,fillcolor=green](-1,-1)(7,7)}%
\psBsplineNodesE[linestyle=solid]{SB}{\Snodecount}%
\end{pspicture}
\end{verbatim}
\begin{center}
\begin{pspicture}[showgrid=false](-1,-1)(7,7)
\psgrid%
\pnodes{S}(0,4)(1,0)(4,.5)(5,2)(3,3)(1,5)(5,6)(5,3)%
\thickBsplinePenE[linecolor=green]{S}{\Snodecount}{.7cm}{.1cm}{30}{\psframe[fillstyle=solid,fillcolor=green](-1,-1)(7,7)}%
\psBsplineNodesE[linestyle=solid]{SB}{\Snodecount}%
\end{pspicture}
\end{center}
\end{document}