Performing calculations using the xor variable

When I calculate, I follow this method:

117^196 

The result is:

177

Now I want to revert back to 117 so I need to perform a replacement

(replace)^196 = 117

What is the reverse operation of the xor operator?

Answer №1

The opposite of exclusive OR (xor) is still xor :). When you perform the xor operation twice on a pair of bits, such as (a^b)^b == a, you will get back the original value.

Proving this concept is quite simple. Let's break it down for each individual bit:

a and b, we can observe that

XORing a^b with either a or b will result in the other value (xor-ing a yields b, and vice versa)

1 2 filter output
0^0^0      = 0
0^1^0      = 1
0^1^1      = 0
1^0^0      = 1
1^0^1      = 0
1^1^1      = 1

Answer №2

Simple math: xor with itself.

For example, the opposite of + is -

But when it comes to xor, its opposite remains the same: xor

Answer №3

Take the outcome you received: 177

117 ^ 196        = 177         | () ^ 196
117 ^ 196 ^ 196  = 177 ^ 196   | self-inverse
117 ^ 0          = 177 ^ 196   | neutral element
117              = 177 ^ 196

XOR showcases three key properties. It is

  • associative
  • commutative
  • self-inverse

This demonstrates that a value acts as its own inverse:

a^a = 0

Being both commutative and associative, you can restructure and xor-expression containing an even number of the same operands like so:

a^O^b^c^O^d = O^O^a^b^c^d = 0^a^b^c^d = a^b^c^d

You could interpret it as operands appearing an even number of times "neutralizing each other".

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