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

Answer 1

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:

1. -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.

2. 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
   

3. 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
   

4. • 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!