Steven D'Aprano <steve+comp.lang.pyt...@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. -- https://mail.python.org/mailman/listinfo/python-list
