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?
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?
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
andb
, we can observe thatXORing
a^b
with eithera
orb
will result in the other value (xor-inga
yieldsb
, 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
Simple math: xor
with itself.
For example, the opposite of +
is -
But when it comes to xor
, its opposite remains the same: xor
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
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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