[pstricks] multido

[Richard K. Oliver]

Thanks Herbert. Your code worked fine.

I now wish to print out the dots by diagonals.

First attempt:

\documentclass{article}
\usepackage{pst-all}
\begin{document}

\begin{pspicture}[showgrid=true](-1,-1)(10,10)
  \multido{\iA=0+1}{11}{%
    \multido{\iB=0+1}{iA+1}{%
      \psdots[linecolor=red,dotsize=10pt](\iB,\iA-\iB)}}
\end{pspicture}

\end{document}

As expected, this did not work, undoubtedly because the multido macro does 
not have this kind of flexibility.

Second attempt (using fpAdd and fpSub from Version 1.41 of the multido 
documentation 
http://www.ctan.org/tex-archive/macros/generic/multido/multido.pdf ):

\documentclass{article}
\usepackage{pst-all}
\begin{document}

\begin{pspicture}[showgrid=true](-1,-1)(10,10)
  \multido{\iA=0+1}{11}{%
    \fpAdd{\iA}{1}{\iAA}
    \multido{\iB=0+1}{iAA}{%
      \fpSub{\iA}{\iB}{\iC}
      \psdots[linecolor=red,dotsize=10pt](\iB,\iC)}}
\end{pspicture}

\end{document}

Again, this did not work. I know substitution in macros is tricky, but is 
there some way to make the number of repetitions of the inner loop depend on 
the variable of the outer loop? The error message was:

! Missing number, treated as zero.
<to be read again>
                   i
l.10 ...ots[linecolor=red,dotsize=10pt](\iB,\iC)}}

Sincerely,
R. K. Oliver









> Am 11.11.2009 06:50, schrieb Richard K. Oliver:
>> \documentclass{article}
>> \usepackage{pst-all}
>> \begin{document}
>> \begin{figure}
>> \begin{pspicture}(-1,-1)(10,10)
>> \psgrid[subgriddiv=1,griddots=0,gridlabels=7pt]
>> \multido{\i=0+1}{11}{\multido{\j=0+1}{11}{\psdots[linecolor=red,dotsize=10pt](\i,\j)}}
>
> see multi documentation for variable names
>
> \documentclass{article}
> \usepackage{pst-all}
> \begin{document}
>
> \begin{pspicture}[showgrid=true](-1,-1)(10,10)
>  \multido{\iA=0+1}{11}{ %
>    \multido{\iB=0+1}{11}{ %
>      \psdots[linecolor=red,dotsize=10pt](\iA,\iB)}}
> \end{pspicture}
>
> \end{document}
>
>
> Herbert
>
>