Enumerating all 3-tuples

[Steven D'Aprano]

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.

Clearly I'm having a "cannot brain today" moment.

Can anyone help me out here?



-- 
Steve