Enumerating all 3-tuples
Ben Bacarisse
ben.usenet at bsb.me.uk
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.
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