Strange resultat using imagesc() on a three-value matrix

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I want to have a three-color-only visual on a matrix that contains many values which are one of three values. Here is the simplest way I could repeat the issue:
r=rand(1,2000);
r(r>0.6)=70;r(r<=0.3)=0;r((r>0) & (r<70))=35;
figure;imagesc(r);
Zooming on some part of the image, I see more than three colors: uploaded_files/82867/sh_244.jpg
I would expect the visual to be in accordance with the values in the matrix. I mean, a visual with only 3 colors. Strangely, that does not happen if I replace the value 2000 by 1000 or by 2100. Is it my video card creating this bug? Or something else?
"Ver" returns:
  • MATLAB Version: 8.5.0.197613 (R2015a)
  • Operating System: Microsoft Windows 7 Édition Familiale Premium Version 6.1 (Build 7601: Service Pack 1)
  • Java Version: Java 1.7.0_60-b19 with Oracle Corporation Java HotSpot™ Client VM mixed mode
  1 Comment
JFLabrie
JFLabrie on 13 Jul 2017
An even simpler code that leads to the same issue:
r=round(rand(1,2000));figure;imagesc(r);

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Accepted Answer

JFLabrie
JFLabrie on 14 Jul 2017
The following command solved the issue:
set(gcf,'Renderer', 'painters')

More Answers (2)

Image Analyst
Image Analyst on 13 Jul 2017
I do not see that. I see only 3 colors in your first example, and two colors in your second example, as expected. My guess is it's your video adapter and how it's trying to zoom in, like by pixel replication (which would not introduce any new colors), or bicubic interpolation (which would introduce more colors).

Walter Roberson
Walter Roberson on 13 Jul 2017
  1 Comment
JFLabrie
JFLabrie on 14 Jul 2017
I tried the following command without success:
set(gcf,'GraphicsSmoothing','off')
But your suggestion lead me to have a look at the Figure Properties. Then I tried the following command that solved the issue immediately:
set(gcf,'Renderer', 'painters')

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