Jeff Moden (6/28/2008)
about 300-million rows
That's kinda what I thought you were going to say.
What I was suggesting is that the distance from Point A to Point B is exactly the same as the distance from Point B to Point A... wouldn't that cut your row count about in half if you did it like half a multiplication table for each "group"?
It's already one-way. The two tables have completely different data in them. No self-join at all. A to B or B to A, it's the same number of calculations in what I'm doing. And I'm already filtering for lat/long outside of the possible distances range, etc. (otherwise, it would be a full cartesian). It's a question of taking a large number of addresses in one list, and finding the four closest addresses in a different list for each one in the first list. There's no overlap in the two lists.
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