Take the 2-minute tour ×
TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. It's 100% free, no registration required.

I'm trying to get rid of the middle line. Whatever changes I make it seems that I make it worse. I would appreciate it if anyone could show me how to draw this cobordism properly.

\begin{pspicture}
  \pscustom{
         \psbezier[showpoints=true](1.5,2.5)(1.5,1.1)(.4,1.6)(.5,0)
         \psline(-0.5,0)
         \psbezier[showpoints=true](-0.5,0)(-.4,1.6)(-1.5,1.1)(-1.5,2.5)
         \psline(-.5,2.5)
         \psbezier[showpoints=true](-.5,2.5)(-.6,1.5)(0.6,1.5)(.5,2.5)
         \psline(1.5,2.5)}
\end{pspicture}

cob

share|improve this question
    
You need to use liftpen. –  Please don't touch Dec 29 '13 at 10:54
    
Thanks. I've accepted your answer. Herbert's answer was also very useful. So, thank you both. –  user43334 Dec 29 '13 at 13:00
add comment

2 Answers 2

up vote 7 down vote accepted

Normal method

\documentclass[pstricks,border=12pt]{standalone}


\begin{document}
\begin{pspicture}[showgrid=true](-2,-1)(2,3)
    \pscustom
    {
        \psset{showpoints}
        \psbezier(1.5,2.5)(1.5,1.1)(.4,1.6)(.5,0)
        \psbezier(-.5,0)(-.4,1.6)(-1.5,1.1)(-1.5,2.5)
        \psbezier[liftpen=1](-.5,2.5)(-.6,1.5)(.6,1.5)(.5,2.5)
        \closepath
    }
\end{pspicture}
\end{document}

enter image description here

Abnormal method

Symmetrical properties are taken into account with \reversepath, \scale, etc. Unfortunately, \nodexn and (!N-<nodename>.x N-<nodename>.y) cannot be used together here so the code become too complicated.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\psset{saveNodeCoors}

\begin{document}

\begin{pspicture}[showgrid=true](-2,-1)(2,3)
\pstGeonode[PointName=none,PointSymbol=none]
    (.5,2.5){A}
    (.6,1.5){B}
    (-.6,1.5){C}
    (-.5,2.5){D}
    % P = (A + B)/2
    (!N-A.x N-B.x add 2 div N-A.y N-B.y add 2 div){P}
    % Q = ((A + C)/2 + B)/2
    (!N-A.x N-C.x add 2 div N-A.y N-C.y add 2 div N-B.y add 2 div exch N-B.x add 2 div exch){Q}
    % R = (A + D + 3(B + C))/8
    (!N-A.x N-D.x add N-B.x N-C.x add 3 mul add 8 div N-A.y N-D.y add N-B.y N-C.y add 3 mul add 8 div){R}
\def\path
{
    \psbezier(.5,0)(.4,1.6)(1.5,1.1)(1.5,2.5)
    \psbezier[liftpen=1](!N-A.x N-A.y)(!N-P.x N-P.y)(!N-Q.x N-Q.y)(!N-R.x N-R.y)
}%
\pscustom
{
    \psset{showpoints}
    \path   
    \reversepath
    \scale{-1 1}    
    \path
    \closepath
}
\end{pspicture}
\end{document}

enter image description here

Warning!

(<nodename>) and (!\pstGetNodeCenter{<nodename>} <nodename>.x <nodename>.y) cannot be scaled by \scale{}. That is why I use (!N-<nodename>.x N-<nodename>.y) here.

share|improve this answer
add comment

Inside \pscustom only the first \psbezier has four points, for all others the current point is the first bezier point, too:

\documentclass[border=5mm]{standalone}
\usepackage{pstricks}
\begin{document}

\begin{pspicture}(-2,0)(2,2.5)
\pscustom{
         \psbezier[showpoints](1.5,2.5)(1.5,1.1)(.4,1.6)(.5,0)% 4 points
         \psline(-0.5,0)%% is also the first bezier point for the next \psbezier
         \psbezier[showpoints](-.4,1.6)(-1.5,1.1)(-1.5,2.5)% three
         \psline(-.5,2.5)
         \psbezier[showpoints](-.6,1.5)(0.6,1.5)(.5,2.5)% three
         \closepath
}
\end{pspicture}
\end{document}

enter image description here

If you use 4 points then there is no need for the \psline but one liftpen option:

\documentclass[border=5mm]{standalone}
\usepackage{pstricks}
\begin{document}

\begin{pspicture}(-2,0)(2,2.5)
\pscustom{
         \psbezier[showpoints](1.5,2.5)(1.5,1.1)(.4,1.6)(.5,0)
         \psbezier[showpoints](-0.5,0)(-.4,1.6)(-1.5,1.1)(-1.5,2.5)
         \psbezier[showpoints,liftpen=1](-.5,2.5)(-.6,1.5)(0.6,1.5)(.5,2.5)
         \closepath
}
\end{pspicture}
\end{document}
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.