Question

I think I can use Cantor’s to create a unique hash n = ((x + y)*(x + y) + x – y)/2…

but can I reverse this hash? And if not, can someone provide a similar formula pair for a reversible hash?

Thanks.

Answer 1

if x and y and n are all the same data types.

n = ((x + y)*(x + y) + x - y)/2...

when x and y are near the datatype::max n will overflow and you will lose information and not be able to recover x and y.

If on the other hand if x and y are always within a range let say 0-FOO

n = Foo * x + y 

can be a recoverable hash provided again n has not overflown

   n % Foo

will give you y. Once y is known (n-y)/Foo will give you x