Re: Enumerating all 3-tuples

[Robin Becker]

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

def cantor_triple(i,j,k):
        return cantor_pair(cantor_pair(i,j),k)

def inverse_cantor_triple(z):
        j,k = inverse_cantor_pair(z)
        i,j = inverse_cantor_pair(j)
        return i,j,k

if __name__=='__main__':
        for z in xrange(100):
                i,j,k = inverse_cantor_triple(z)
                print z, repr((i,j,k))


or changing the construction

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

def cantor_triple(i,j,k):
        return cantor_pair(i,cantor_pair(j,k))

def inverse_cantor_triple(z):
        i,k = inverse_cantor_pair(z)
        j,k = inverse_cantor_pair(k)
        return i,j,k

if __name__=='__main__':
        for z in xrange(100):
                i,j,k = inverse_cantor_triple(z)
                print z, repr((i,j,k)), cantor_triple(i,j,k)==z

this give different outcomes, but both appear to be a correct mapping of 
non-negative integers to triplets.
--
Robin Becker

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