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Consider the following code which compiles with Xelatex:

\usepackage[usenames,dvipsnames,svgnames,table]{xcolor}
\usepackage{pstricks,psvectorian}

\usepackage{scalefnt}
\usepackage{xcolor}

    \begin{document}
    \thispagestyle{empty}
    
    \begin{pspicture}(-3,-3)(3,3)%
    \renewcommand*{\psvectorianDefaultColor}{red}%
    \psframe[fillcolor=Beige,fillstyle=solid](-3,-3)(3,3) 
    \psframe[linecolor=red](-3,-3)(3,3)
    
    \rput[tl](-3,3){\psvectorian[width=3cm]{63}}
    \rput[tr](3,3){\psvectorian[width=3cm,mirror]{63}}
    \rput[bl](-3,-3){\psvectorian[width=3cm,flip]{63}}
    \rput[br](3,-3){\psvectorian[width=3cm,flip,mirror]{63}}
    
    \rput(0,0.35){\scalefont{1.0}{\textbf{Is this Beige}}}
    \rput(0,-.35){\scalefont{1.0}{\textbf{Tinted red?}}}
    \end{pspicture}
    \begin{pspicture}(-3,-3)(3,3)%
    \renewcommand*{\psvectorianDefaultColor}{purple}%
    \psframe[fillcolor=Beige,fillstyle=solid](-3,-3)(3,3) 
    \psframe[linecolor=purple](-3,-3)(3,3)
    
    \rput[tl](-3,3){\psvectorian[width=3cm]{63}}
    \rput[tr](3,3){\psvectorian[width=3cm,mirror]{63}}
    \rput[bl](-3,-3){\psvectorian[width=3cm,flip]{63}}
    \rput[br](3,-3){\psvectorian[width=3cm,flip,mirror]{63}}
    
    \rput(0,.35){\scalefont{1.0}{\textbf{Is this Beige}}}
    \rput(0,-.35){\scalefont{1.0}{\textbf{Tinted purple?}}}
    \end{pspicture}
    \end{document}

which produces the two pspictures:

enter image description here

QUESTION: The fillcolor of both of these pictures is beige; however, it seems to me that the beige within the pspicture on the left has taken on a slight reddish tint; whereas, the beige in the picture on the right has assumed a bit of a purplish hue. Is my observation correct or is this perhaps, some sort of an optical illusion? If it is not an illusion, then what is causing this phenomenon? Is there a way to prevent this optical occurrence from happening?

Thank you.

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1 Answer 1

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You have to control the color definitions of the predefined colours. An RGB Beige is not the same as a CMYK Beige. PSTricks itself and xcolor use by default rgb, if not defined.

try

 \usepackage[usenames,dvipsnames,svgnames,table,cmyk]{xcolor}

then all colours are converted into cmyk before used.

enter image description here

However, to compare colours you have to use only the two colours and no border!

\documentclass[border=12pt]{standalone}
\usepackage[svgnames]{xcolor}
\usepackage{pstricks,psvectorian}

\begin{document}
    \begin{pspicture}(-3,-3)(3,3)%
    \psframe*[linecolor=Beige](-3,-3)(3,3) 
    \end{pspicture}%
    \begin{pspicture}(-3,-3)(3,3)%
    \psframe*[linecolor=Beige](-3,-3)(3,3) 
    \end{pspicture}
\end{document}

Then you'll see that there is no difference:

enter image description here

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  • Thank you for posting this helpful answer. May I presume then that my original code was indeed producing slightly different shades of "beige"? Thanks again.
    – DDS
    Nov 16, 2021 at 18:22
  • no, it is an optical illusion, your left red is lighter than the right one with the purple color.
    – user187802
    Nov 16, 2021 at 18:58

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