Enumerating all 3-tuples

[Steven D'Aprano]

On Mon, 12 Mar 2018 13:17:15 +0000, Robin Becker wrote:

> It's possible to generalize the cantor pairing function to triples, but
> that may not give you what you want. Effectively you can generate an
> arbitrary number of triples using an iterative method. My sample code
> looked like this
> 
> 
> import math
> 
> def cantor_pair(k1,k2):
> 	return (((k1+k2)*(k1+k2+1))>>1) + k2
> 
> def inverse_cantor_pair(z):
> 	w = int((math.sqrt(8*z+1)-1)/2.0)
> 	t = (w*(w+1))>>1
> 	j = z - t
> 	i = w - j
> 	return i, j

I definitely want to avoid any round trips through float in order to use 
sqrt.


But thanks for the code, I'll certainly steal it, er I mean study it for 
future reference :-)


-- 
Steve