Enumerating all 3-tuples

Fri Mar 9 21:44:24 EST 2018

Steven D'Aprano <steve+comp.lang.python at pearwood.info> writes:
> I am trying to enumerate all the three-tuples (x, y, z) where each of x, 
> y, z can range from 1 to ∞ (infinity).
>> This is clearly unhelpful:
>> for x in itertools.count(1):
> for y in itertools.count(1):
> for z in itertools.count(1):
> print(x, y, z)
>> as it never advances beyond x=1, y=1 since the innermost loop never 
> finishes.
>> Georg Cantor to the rescue! (Well, almost...)
>> https://en.wikipedia.org/wiki/Pairing_function
>> The Russian mathematician Cantor came up with a *pairing function* that 
> encodes a pair of integers into a single one. For example, he maps the 
> coordinate pairs to integers as follows:
>> 1,1 -> 1
> 2,1 -> 2
> 1,2 -> 3
> 3,1 -> 4
> 2,2 -> 5
>> and so forth. He does this by writing out the coordinates in a grid:
>> 1,1 1,2 1,3 1,4 ...
> 2,1 2,2 2,3 2,4 ...
> 3,1 3,2 3,3 3,4 ...
> 4,1 4,2 4,3 4,4 ...
> ...
>> and then iterating over them along the diagonals, starting from the top 
> corner. That's just what I'm after, and I have this function that works 
> for 2-tuples:
>> def cantor(start=0):
> """Yield coordinate pairs using Cantor's Pairing Function.
>> Yields coordinate pairs in (Z*,Z*) over the diagonals:
>> >>> it = cantor()
> >>> [next(it) for _ in range(10)]
> [(0,0), (1,0), (0,1), (2,0), (1,1), (0,2), (3,0), (2,1), (1,2), (0,3)]
>> If ``start`` is given, it is used as the first x- and y-coordinate.
> """
> i = start
> while True:
> for j in range(start, i+1):
> yield (i-j+start, j)
> i += 1
>>> But I've stared at this for an hour and I can't see how to extend the 
> result to three coordinates. I can lay out a grid in the order I want:
>> 1,1,1 1,1,2 1,1,3 1,1,4 ...
> 2,1,1 2,1,2 2,1,3 2,1,4 ...
> 1,2,1 1,2,2 1,2,3 1,2,4 ...
> 3,1,1 3,1,2 3,1,3 3,1,4 ...
> 2,2,1 2,2,2 2,2,3 2,2,4 ...
> ...
>> and applying Cantor's diagonal order will give me what I want, but damned 
> if I can see how to do it in code.

Rather than a grid you would need a cube, and the diagonals become
planes. But I think that is an easier way (no code yet though!) unless
you are set on one particular enumeration: consider the triple as a pair
one element of which runs over the enumeration of pairs you already
have.
Start with
 1,1 <-> 1
 2,1 <-> 2
 1,2 <-> 3
 3,1 <-> 4
 2,2 <-> 5
 1,3 <-> 6
 ...
but replace the first element by the numbered pair from this same list:
 (1,1),1
 (2,1),1
 (1,1),2
 (1,2),1
 (2,1),2
 (1,1),3
 ...
If it were a sane time here I'd try to code this. Looks like fun but it
must wait...
-- 
Ben.