pst-bspline update (29nov08)

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-The following files have been removed in the TeX Live installation of
-the current package, typically due to duplication, lack of space, or
-missing source code.  You can find these files on CTAN.  If questions or
-concerns, email tex-live@tug.org.
-
-	bsplinePstricks.pdf
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+\documentclass[dvips,11pt]{article}
+\usepackage{amsmath}
+\usepackage{amsthm} 
+\usepackage{graphicx}
+\usepackage{pstricks}
+\usepackage{multido,pst-node,pst-bspline,pstricks-add}
+\usepackage{amssymb}
+\usepackage[parfill]{parskip}
+\usepackage{hyperref}
+\hyphenation{Post-Script}
+%\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 $B_0 \ldots B_n$ is a curve parametrized by the interval $[0,n]$, which is $C^2$-continuous (that is, has continuous curvature) and is on each interval $[k-1,k]$ given by a cubic B\'ezier curve whose control points are derived from the $(B_k)$. These curves are discussed in any reasonably modern text on Numerical Analysis. One easily accessible source is the UCLA lecture notes of Kirby Baker:
+
+\noindent\url{http://www.math.ucla.edu/~baker/149.1.02w/handouts/dd_splines.pdf}
+
+I'll focus on two special cases: (i) relaxed, uniform B-splines; (ii) periodic, uniform B-splines. Uniform refers to the condition mentioned in the first paragraph: each B\'ezier sub-curve is parametrized by an interval of  length~1. Relaxed means that the curvature at the endpoints $t=0, t=n$ is zero. Periodic means in effect that the $B_i$ repeat periodically, and the curve generated is a closed curve. 
+
+\section{Relaxed, Open B-spline} The algorithm has the following steps.
+
+\begin{itemize}
+\item The curve starts at $B_0$ and ends at $B_n$.
+\item Divide each line $B_{k-1}B_k$ 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 $0<k<n$, divide the line segment $L_kR_k$ in half, letting $S_k$ denote the midpoint. In effect, for $0< k<n$, $S_k=(B_{k-1}+4B_k+B_{k+1})/6$.
+\item Let $S_0=B_0$ and $S_n=B_n$.
+\item For $0<k\le n$, construct the cubic B\'ezier curve with control points $S_{k-1}$, $R_{k-1}$, $L_k$, $S_k$, parametrized by $k-1\le t\le k$.
+\end{itemize}
+
+The {\tt pst-Bspline} package implements this algorithm as \verb|\psBspline|, whose simplest form is, for example
+
+\noindent
+\verb|\psBspline(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)|
+
+\noindent The coordinates are the B-spline control points. Aside from the usual keywords, like {\tt linestyle}, {\tt linecolor} and {\tt arrows}, there is a Boolean keyword {\tt showframe}. The effect of {\tt showframe=true} is to show the intermediate points and lines in the algorithm described above.
+
+There is another optional argument that can be applied if you wish to be able to refer to any of the points constructed in the algorithm. By example,  
+
+\noindent
+\verb|\psBspline{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)|
+
+\noindent sets the root of the naming scheme to {\tt B}, the effect of which is that the B-spline control points will be nodes of type \verb|\pnode| with names {\tt B0}, {\tt B1} and so on, the other points being similarly named {\tt BL0}, {\tt BL1}, ... , {\tt BR0}, {\tt BR1}, ... , {\tt BS0}, {\tt BS1}, ... . For example, to draw a line between {\tt BL1} and {\tt BS4}, just use \verb|\ncline(BL1)(BS4)|. 
+
+The  algorithm is implemented entirely in PSTricks code, without any PostScript programming at all, depending for the most part on the flexibility of nodes, and above all the \verb|\multido| macro, which allows one to construct with relative ease items that look and feel like arrays. Use of \verb|\SpecialCoor| is essential.
+
+There is a closely related macro \verb|\psBsplineE| which removes the first and last B\'ezier segments, much as \verb|\psecurve| acts in relation to \verb|\pscurve|, allowing one one to draw B-splines with non-zero curvature at the endpoints.
+
+\begin{verbatim}
+\documentclass{article}
+\usepackage{pstricks}
+\usepackage{multido,pst-node,pst-bspline}
+\pagestyle{empty}
+\begin{document}
+\SpecialCoor % essential for pst-bspline package
+\psset{unit=.6in}
+\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
+\psBspline[showframe=true]{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)
+\multido{\i=0+1}{5}{\uput[20](B\i){B\i}}
+\uput[90](B5){B5}
+\uput[90](BS1){S1}
+\uput[90](BS2){S2}
+\uput[180](BS3){S3}
+\uput[270](BS4){S4}
+\uput[-45](BR1){R1}
+\uput[-45](BL2){L2}
+\end{pspicture}
+\end{document}
+\end{verbatim} 
+
+\vspace{1pc}
+\begin{center}
+\psset{unit=.6in}
+\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
+\psBspline[showframe=true]{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)
+\multido{\i=0+1}{5}{\uput[20](B\i){B\i}}
+\uput[90](B5){B5}
+\uput[90](BS1){S1}
+\uput[90](BS2){S2}
+\uput[180](BS3){S3}
+\uput[270](BS4){S4}
+\uput[-45](BR1){R1}
+\uput[-45](BL2){L2}
+\end{pspicture}
+\end{center}
+
+\begin{verbatim}
+\documentclass{article}
+\usepackage{pstricks}
+\usepackage{multido,pst-node,pst-bspline}
+\pagestyle{empty}
+\begin{document}
+\SpecialCoor % essential for pst-bspline package
+\psset{unit=.6in}
+\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
+\psBsplineE[showframe=true]{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)
+\multido{\i=0+1}{5}{\uput[20](B\i){B\i}}
+\uput[90](B5){B5}
+\uput[90](BS1){S1}
+\uput[90](BS2){S2}
+\uput[180](BS3){S3}
+\uput[270](BS4){S4}
+\uput[-45](BR1){R1}
+\uput[-45](BL2){L2}
+\end{pspicture}
+\end{document}
+\end{verbatim} 
+
+\vspace{1pc}
+\begin{center}
+\psset{unit=.6in}
+\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
+\psBsplineE[showframe=true]{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)
+\multido{\i=0+1}{5}{\uput[20](B\i){B\i}}
+\uput[90](B5){B5}
+\uput[90](BS1){S1}
+\uput[90](BS2){S2}
+\uput[180](BS3){S3}
+\uput[270](BS4){S4}
+\uput[-45](BR1){R1}
+\uput[-45](BL2){L2}
+\end{pspicture}
+\end{center}
+
+\section{Periodic B-spline} The result here is a closed curve. The algorithm is essentially the same as in the preceding case, except:
+
+\begin{itemize}
+\item Extend $B_i$ periodically with period $n+1$, so that $B_{n+1}=B_0$ and $B_{n+2}=B_1$.
+\item Construct  $R_i$, $L_i$ for $0<i<n+2$, as above.
+\item Construct $S_k$ as above (midpoint of $L_kR_k$), for $0<k<n+2$.
+\item Set $S_0=S_{n+1}$.
+\item For $0<k\le n+1$, construct the cubic B\'ezier curve with control points $S_{k-1}$, $R_{k-1}$, $L_k$, $S_k$, parametrized by $k-1\le t\le k$.
+\end{itemize}
+
+The macro in this case is \verb|\psBsplineC|, where the {\tt C} stands for Closed. The code, being implemented as a \verb|\pscustom| object, does not accept the {\tt doubleline} keyword, but does accept, for example,
+ 
+\noindent\verb|fillstyle=solid,fillcolor=gray|
+
+\begin{verbatim}
+\documentclass{article}
+\usepackage{pstricks}
+\usepackage{multido,pst-node,pst-bspline}
+\pagestyle{empty}
+\begin{document}
+\SpecialCoor % essential for pst-bspline package
+\psset{unit=.6in}
+\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
+\psBsplineC[showframe=true]{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)
+\multido{\i=0+1}{5}{\uput[20](B\i){B\i}}
+\uput[90](B5){B5}\uput[90](BS1){S1}
+\uput[90](BS2){S2}\uput[180](BS3){S3}
+\uput[270](BS4){S4}\uput[-45](BR1){R1}
+\uput[-45](BL2){L2}
+\end{pspicture}
+\end{document}
+\end{verbatim} 
+
+\vspace{1pc}
+\begin{center}
+\psset{unit=.6in}
+\begin{pspicture}[showgrid=true](-.5,-.5)(6,5)
+\psBsplineC[showframe=true]{B}(.5,.5)(2,0)(5,2)(6,4)(4,5)(2,4)
+\multido{\i=0+1}{5}{\uput[20](B\i){B\i}}
+\uput[90](B5){B5}
+\uput[90](BS1){S1}
+\uput[90](BS2){S2}
+\uput[180](BS3){S3}
+\uput[270](BS4){S4}
+\uput[-45](BR1){R1}
+\uput[-45](BL2){L2}
+\end{pspicture}
+\end{center}
+\section{Related constructions}
+There are in addition three additional macros that draw similar curves, but organized in a slightly different way. They are particularly useful when there is a sequence of points already defined as \verb|\pnode|s. Here is a simple way to define such a sequence.
+\subsection{The {\tt pnodes} macro}
+The line
+\begin{verbatim}
+\pnodes{P}(2,1.5)(3,4)(5,1)
+\end{verbatim}
+defines a sequence of \verb|\pnode|s with the node root {\tt P}: {\tt P0}=(2,1.5), {\tt P1}=(3,4) and {\tt P0}=(5,1). The sequence may be any (reasonable) length. The macro leaves an entry in the console saying that it has defined nodes {\tt P0 .. P2}. The three new macros are:
+\begin{verbatim}
+\psBsplineNodes{<node root>}{<top index>}
+\psBsplineNodesC{<node root>}{<top index>}
+\psBsplineNodesE{<node root>}{<top index>}
+\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).|
+
+\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}
+\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<k<n-1
+  x(k)=m(k)*(x(k)-x(k-1))
+  y(k)=m(k)*(y(k)-y(k-1))
+  %subtract normalized row k-1 times b(k-1) from row n
+  b(k)=b(k)-b(k-1)*m(k-1)
+  b(n)=b(n)-c(k-1)*b(k-1)
+  x(n)=x(n)-x(k-1)*b(k-1)
+  y(n)=y(n)-y(k-1)*b(k-1)
+% subtract row n-1 times b(n-1) from row n, renormalize by 1/b(n)
+b(n)=b(n)-b(n-1)*c(n-1)
+x(n)=(x(n)-x(n-1)*b(n-1))/b(n)
+y(n)=(y(n)-y(n-1)*b(n-1))/b(n)
+%work back
+x(n-1)=x(n-1)-c(n-1)*x(n)
+y(n-1)=y(n-1)-c(n-1)*y(n)
+for k=n-2 downto 1
+   x(k)=x(k)-m(k)* x(k+1)-c(k)*x(n)
+   y(k)=y(k)-m(k)* y(k+1)-c(k)*y(n)
+\end{verbatim}
+This algorithm is implemented in \TeX/PostScript code in {\tt pst-Bspline.tex} and may be invoked using the macro 
+\begin{verbatim}
+\psBsplineInterpC{<node root>}{<index>}
+\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. 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}{100}
+\end{verbatim}
+with any keywords options you wish.
+
+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}
+
+\newpage
+\begin{verbatim}
+\documentclass{article}
+\usepackage{graphicx}
+\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)%
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