alt Hacker NewsThat's pretty nice ;)
I once wrote this humdinger, that's still on my mostly dead personal website from 2010... one of my proudest bits of code besides my poker hand evaluator ;)
The question was, how do you generate a unique number for any two positive integers, where x!=y, such that f(x,y) = f(y,x) but the resulting combined id would not be generated by any other pair of integers. What I came up with was a way to generate a unique key from any set of positive integers which is valid no matter the order, but which doesn't key to any other set.
My idea was to take the radius of a circle that intersected the integer pair in cartesian space. That alone doesn't guarantee the circle won't intersect any other integer pairs... so I had to add to it the phase multiple of sine and cosine which is the same at those two points on the arc. That works out to:
(x^2+y^2)+(sin(atan(x/y))*cos(atan(x/y)))
And means that it doesn't matter which order you feed x and y in, it will generate a unique float for the pair. It reduces to:
x^2+y^2+( (x/y) / (x^2+y^2) )
To add another dimension, just add it to the process and key it to one of the first...
x^2+y^2+z^2+( (x/y) / (x^2+y^2) )+( (x/z) / (x^2+z^2) )
It looks like you have typos? (x^2+y^2)+(sin(atan(x/y))*cos(atan(x/y))) reduces to x^2+y^2+( (x/y) / (x^2/y^2 + 1) ) - not the equation given? Tho it's easier to see that this would be symmetrical if you rearrange it to: x^2+y^2+( (xy) / (x^2+y^2) )
Also, if f(x,y) = x^2+y^2+( (x/y) / (x^2+y^2) ) then f(2,1) is 5.2 and f(1,2) is 5.1? - this is how I noticed the mistake. (the other reduction gives the same answer, 5.4, for both, by symmetry, as you suggest)
There's a simpler solution which produces integer ids (though they are large): 2^x & 2^y. Another solution is to multiply the xth and yth primes.
I only looked because I was curious how you proved it unique!