       Re: Is color overlapping possible with Mathematica?

• To: mathgroup at smc.vnet.net
• Subject: [mg28947] Re: [mg28917] Is color overlapping possible with Mathematica?
• From: BobHanlon at aol.com
• Date: Sat, 19 May 2001 22:28:06 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```This is not a general solution; however, it is an approach.

Needs["Graphics`Graphics`"];
Needs["Graphics`Colors`"];
Needs["Graphics`InequalityGraphics`"];

blend[x__ /;!FreeQ[{x}, RGBColor]] :=
Module[{colorList = Cases[Flatten[{x}], RGBColor[__]]},
RGBColor[
Sequence@@(Plus@@Apply[List,colorList,{1}]/Length[colorList])]];

cond1 = 0<x<2 && 0<y<2;
cond2 = 1<x<3 && 1<y<3;
colors = {Red, Blue};

DisplayTogether[
InequalityPlot[cond1, {x}, {y}, Fills -> colors[]],
InequalityPlot[cond2, {x}, {y}, Fills -> colors[]],
InequalityPlot[cond1 && cond2, {x}, {y}, Fills -> blend[colors]]];

cond1 = 0<x<2 && 0<y<3x && x*y < 1;
cond2 = 1<(x+2y)^2+4y^2<4;
colors = {Red, LightBlue};

DisplayTogether[
InequalityPlot[cond1, {x}, {y}, Fills -> colors[]],
InequalityPlot[cond2, {x, -3, 3}, {y, -3, 3},
Fills -> colors[]],
InequalityPlot[cond1 && cond2, {x, -3, 3}, {y, -3, 3},
Fills -> blend[colors]]];

Bob Hanlon

In a message dated 2001/5/18 1:39:03 AM, liu at vtaix.cc.vt.edu writes:

>I wonder if it's possible to use Mathematica to draw to
>overlapping shapes each filled with a different color
>and the color in the overlapping region being the
>composition of the two colors but not the color of the
>shape on top.  For example, I could do
>
>In:= Show[Graphics[{RGBColor[1.0, 0.0, 0.0],
>                       Polygon[{{0, 0}, {0, 2}, {2, 2}, {2, 0}}]}]];
>In:= Show[Graphics[{RGBColor[0.0, 0.0, 1.0],
>                       Polygon[{{1, 1}, {1, 3}, {3, 3}, {3, 1}}]}]];
>In:= Show[Out,Out];
>
>But the overlapping square area would have the color of Out,
>RGBColor[0.0, 0.0, 1.0].
>
>What I am looking for is to plot a shape transparent rather than
>opaque.
>

```

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