Searching for a way to do Bitwise XOR on images

Question

I am looking for a way to get the Bitwise XOR of two images on the command line(or in another way that can be implemented in a program or script).

This should result in the same final picture as using the XOR Blending mode in picture editors that support it (Paint.NET, Photoshop, etc)

As an example, say I have Image A:

Image A(Seattle, from the paint.net documentation

and Image B:

ImageB(Apple, from the paint.net documentation)

then the result should look like:

Image C(Xor result of above images, from the paint.net documentation)

The fun part of this is of course, that when you XOR image C with image B again, you will get an exact copy of image A.

Now, I have been looking all over the internet for a way to do this programmatically, but I have found nothing. Even ImageMagick does not support doing a bitwise XOR on images.

Does sombebody know a way to do this?

mathematical.coffee · Accepted Answer · 2011-12-14T23:20:28.220

ImageMagick can do it, although it's a bit convoluted. One way is:

convert img1 img2 -fx "(((255*u)&(255*(1-v)))|((255*(1-u))&(255*v)))/255" img_out

(img1,img2,img_out are the two input and single output file names respectively).

Explanation

It's a bit ugly (I'm sure someone with more ImageMagick-fu than me could clean it up but it works like this:

-fx "xxx" basically says "perform the operation xxx on the image". In the expression above, u and v stand for the first and second input images respectively.
Now, -fx only has bitwise AND & and bitwise OR | in the way of bitwise operators. To reconstruct bitwise XOR, we need
```
convert img1 img2 -fx "(u & NOT v) | (NOT u & v)" img_out
```
To get the NOT (there is a logical NOT but no bitwise NOT), we remember that NOT x = 255-x if x is 8-bit. So to get NOT u we can just do 255-u, assuming image u is 8-bit. Hence, the ImageMagick command would be:
```
convert img1.png img2.img -fx "((255-u)&v)|(u&(255-v))" image_xor.png
```
- The one problem here is that when ImageMagick does fx it normalises all the pixels in u and v in the range [0,1] instead of [0,255] as we expect, and doing bitwise on non-integers screws stuff up.
- Hence, we have to multiply all occurrences of u and v in the above expression by 255 (so the bitwise operations work), and divide by 255 at the very end to get back in the range [0,1] that ImageMagick expects.

This gives us the original command,

convert img1 img2 -fx "(((255*u)&(255*(1-v)))|((255*(1-u))&(255*v)))/255" img_out

Voila!

That looks amazing ;D. There was a small error in your code tough (the second u should be a v). Fixed it. It does however take AGES to run... ^^' — Qqwy, Dec 14 '11 at 14:26
ta, shame it takes so long to run though (I only tried it on those baby images you have). Writing some code would still be fastest, `-fx` in imagemagick does have a reputation for being slower. — mathematical.coffee, Dec 14 '11 at 23:18
ImageMagick-7.0.11-Q8>magick.exe 1.png 2.png -fx "u&(1-v)|(1-u)&v" img.png (tested for 1 bit and 8 bit images) — qqqq1961, May 24 '21 at 09:42

score 27 · Answer 2 · answered Oct 14 '16 at 17:53

27

I found a need for xor on an image and the G'MIC tool works for me. G'MIC is incredibly powerful, as is Image Magick, but worth a look for solving some tough image processing problems.

gmic a.png b.png -blend xor -o result.png

G'MIC also works directly on the images posted above.

gmic https://i.stack.imgur.com/Ws6e8.png https://i.stack.imgur.com/hoBIM.png -blend xor -o result.png

For help,

gmic -h -blend

answered Oct 14 '16 at 17:53

RobLabs

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It is very unfortunate that only one answer can be selected, as this gmic looks like a wonderful tool. (I selected the answer that I ended up using originally, which I should have but forgot to select two years ago). Thank you for this great answer! – Qqwy Oct 15 '16 at 22:10

score 7 · Answer 3 · edited Sep 19 '13 at 06:24

Here is how I would do in Java:

Iterate over all the pixels of the two images at once. (for loop (x) inside a for loop (y)). Of course, use a BufferedImage. You can get the color of the pixel by doing:

int color = img.getRGB(x, y);

Do the same for the other image as well and perform the xor operation on the two colors. Store the resulting value in a new BufferedImage with the same dimensions as the two input images.

Here is some sample code:

public static BufferedImage xorEffect(BufferedImage imageA, BufferedImage imageB) {
    if (imageA.getWidth() != imageB.getWidth() ||
        imageA.getHeight() != imageB.getHeight())
    {
        throw new IllegalArgumentException("Dimensions are not the same!");
    }
    BufferedImage img = new BufferedImage(imageA.getWidth(),
                                          imageA.getHeight(),
                                          BufferedImage.TYPE_INT_ARGB_PRE);

    for (int y = 0; y < imageA.getHeight(); ++y) {
        for (int x = 0; x < imageA.getWidth(); ++x) {
           int pixelA = imageA.getRGB(x, y);
           int pixelB = imageB.getRGB(x, y);
           int pixelXOR = pixelA ^ pixelB;
           img.setRGB(x, y, pixelXOR);
        }
    }
    return img;
}

To load an image from a file use:

BufferedImage imageA = ImageIO.read(new File("/home/username/image.png"));

You might want to use BufferedImage.TYPE_INT_RGB, so that alpha does not cause differences. — Alex Collins, Feb 23 '15 at 21:58
I think it should be: int pixelXOR = pixelA ^ pixelB ^ 0xFF000000; — Vladimir Zolotov, Apr 17 '22 at 22:52

score 3 · Answer 4 · answered Mar 14 '14 at 10:26

3

If you want to do yourself, do it pixel by pixel. If you want a library, I recommend OpenCV. This is very nice and open source library with huge operations supported in image processing area. It supports direct XOR using ^ operator. Good Luck.

answered Mar 14 '14 at 10:26

Pervez Alam

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score 1 · Answer 5 · answered Oct 25 '17 at 02:10

1

OpenCV has all the logical operators on images and numpy images using bitwise_# Where # can be xor, and, and, not ... See OpenCV - Intersection between two binary images

answered Oct 25 '17 at 02:10

user1270710

553
1
6
11

score 1 · Answer 6 · answered Dec 14 '11 at 13:49

1

Knowing that

A XOR B = (A AND NOT B) OR (NOT A AND B).

and that most of common image processing tools do have and, or and not operations the rest is quite easy :)

Working in Python, you could have a simple script performing the operation and even adding it as a plugin in the gimp ;)

answered Dec 14 '11 at 13:49

jlengrand

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BTW, I might have a quick and (somehow) elegant implementation using Python and opencv. I could send it here if want need it ;) – jlengrand Dec 14 '11 at 15:20

Searching for a way to do Bitwise XOR on images

6 Answers6

Explanation

Linked