XOR Swap Explained Visually

A common riddle-like question for programmers asks them to swap the values of two integers without a temporary intermediate value. There are two common solutions that I'm aware of, addition swap and XOR swap. Here's what each looks like in C:

void addSwap (unsigned int *x, unsigned int *y) {  
    if (x != y) {  
        *x = *x + *y;  
        *y = *x - *y;  
        *x = *x - *y;  
    }  
}

void xorSwap (int *x, int *y) {  
    if (x != y) {  
        *x ^= *y;  
        *y ^= *x;  
        *x ^= *y;  
    }  
}

^{Credit: https://en.wikipedia.org/wiki/XOR_swap_algorithm}

The more interesting of the two is XOR swap. It only uses a single operation, rather than addition and subtraction in addition swap. Wikipedia tries to explain this with bit vectors but I think 1 bit black and white images work better. In the images below, black corresponds to 0 and white corresponds to 1.

A = A ^ B:

Here A becomes white anywhere A or B had a unique presence (only A or only B was black in that pixel). Everywhere else is black.
A	B	= A

B = B ^ A:

Now that A is a combination of what was unique in each of the original images, we can apply the same XOR operation to *remove* what was unique about B, leaving only what was unique about A. So at this point B has swapped and become the circle.
A	B	= B

A = A ^ B:

We can perform the same operation as in the previous step to get back the original value of B. Since B holds the circle and we want the square, we remove what was unique about the circle, leaving just the square.
A	B	= A

Here's the code used to generate the images above:

import sys  
from PIL import Image  


a = Image.open("A.png")  
b = Image.open("B.png")  

if not a.size == b.size:  
    print("Images must have the same dimensions")  
    sys.exit(1)  

if not a.mode == b.mode:  
    print("Images must be the same mode")  
    sys.exit(1)  

result = bytes([  
    a.getpixel((x, y))[0] ^ b.getpixel((x, y))[0]  
    for y in range(a.size[1])  
    for x in range(a.size[0])  
])  

Image.frombytes("P", a.size, result).save("OUT.png")