3.2 KiB
Quadtree
Recursively subdivide an AABB into 4 regions, hence Quad. They are usually denoted as North West, North East, South West, and South East (NW, NE, SW, SE). Several things can be created with this;
Linked Implementation datastructure
Array Implementation datastructure
Insertion algorithm
Query algorithm
Find Nearest Neighbor algorithm
Notes
// exclude node if point is farther away than best distance in either axis if (x < x1 - best.d || x > x2 + best.d || y < y1 - best.d || y > y2 + best.d) { return best; }
I don't know how to explain but I get it. Because of the euclidian distance and the fact that we're dealing with rectangles, the closest distance to a rectangle is a straight line in one of the x or y axis So if the point we're checking is farther away from the rectangle on either axis, then it cannot possible be the case that it is closer I still can't visualize or understand it intuitively, I more just trust that it works, maybe if I see it in action it'll click better JosephFerano — Today at 4:15 PM This is some clever math shit this guy is doing Or that he picked up https://gist.github.com/patricksurry/6478178 Gist D3JS quadtree nearest neighbor algorithm D3JS quadtree nearest neighbor algorithm. GitHub Gist: instantly share code, notes, and snippets. D3JS quadtree nearest neighbor algorithm
/ check if kid is on the right or left, and top or bottom / and then recurse on most likely kids first, so we quickly find a // nearby point and then exclude many larger rectangles later var kids = node.nodes; var rl = (2*x > x1 + x2), bt = (2*y > y1 + y2); if (kids[bt*2+rl]) best = nearest(x, y, best, kids[bt*2+rl]); if (kids[bt*2+(1-rl)]) best = nearest(x, y, best, kids[bt*2+(1-rl)]); if (kids[(1-bt)*2+rl]) best = nearest(x, y, best, kids[(1-bt)*2+rl]); if (kids[(1-bt)*2+(1-rl)]) best = nearest(x, y, best, kids[(1-bt)*2+(1-rl)]);
That's kinda neat, estimating probability with some math and then going into the index of the array where the point is more likely to be No idea why this works But I think I'm done with this Interesting, he doesn't even check if any of the rects contain the point Hmmm Oh I see It's because he's tracking whether a node is visited or not, and I'm putting in a lot of work to make sure you don't revisit the same node twice, but I don't do it with a "visited" bool, which I intentionally avoided but now seeing his solution, I realize it may have been a mistake JosephFerano — Today at 4:27 PM Actually I don't think adding "visited" is a good idea, because you have to walk through the whole thing again once you're done to uncheck all the visited bools. That's a linear walk through all the nodes, which might not be much, but it's certainly making things slower I'll have to investigate further