66 lines
3.6 KiB
Org Mode
66 lines
3.6 KiB
Org Mode
#+TEST: [[file:SpatialPartitioning.org][Spacial Partitioning]]
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* Quadtree
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Recursively subdivide an AABB into 4 regions, hence Quad. They are usually
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denoted as North West, North East, South West, and South East (NW, NE, SW, SE).
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Several things can be created with this;
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** Linked Implementation :datastructure:
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** Array Implementation :datastructure:
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** Insertion :algorithm:
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** Query :algorithm:
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** Find Nearest Neighbor :algorithm:
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** Morton codes
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This is a technique that allows us to convert coordinates into an array index,
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allowing us to pack them contiguously in memory
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http://johnsietsma.com/2019/12/05/morton-order-introduction/
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** Resources
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[[http://ericandrewlewis.github.io/how-a-quadtree-works/][Visualize a Quadtree]]
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[[http://donar.umiacs.umd.edu/quadtree/][Academic Interactive Demo]]
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[[https://codepen.io/_bm/pen/ExPBMrW][Codepen (Js example)]]
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[[https://gist.github.com/patricksurry/6478178][D3.js Example]]
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** Notes
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#+begin_src js
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// exclude node if point is farther away than best distance in either axis
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if (x < x1 - best.d || x > x2 + best.d || y < y1 - best.d || y > y2 + best.d) {
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return best;
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}
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#+end_src
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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
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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
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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
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JosephFerano
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—
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Today at 4:15 PM
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This is some clever math shit this guy is doing
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Or that he picked up
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https://gist.github.com/patricksurry/6478178
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Gist
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D3JS quadtree nearest neighbor algorithm
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D3JS quadtree nearest neighbor algorithm. GitHub Gist: instantly share code, notes, and snippets.
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D3JS quadtree nearest neighbor algorithm
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// check if kid is on the right or left, and top or bottom
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// and then recurse on most likely kids first, so we quickly find a
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// nearby point and then exclude many larger rectangles later
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var kids = node.nodes;
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var rl = (2*x > x1 + x2), bt = (2*y > y1 + y2);
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if (kids[bt*2+rl]) best = nearest(x, y, best, kids[bt*2+rl]);
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if (kids[bt*2+(1-rl)]) best = nearest(x, y, best, kids[bt*2+(1-rl)]);
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if (kids[(1-bt)*2+rl]) best = nearest(x, y, best, kids[(1-bt)*2+rl]);
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if (kids[(1-bt)*2+(1-rl)]) best = nearest(x, y, best, kids[(1-bt)*2+(1-rl)]);
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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
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No idea why this works
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But I think I'm done with this
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Interesting, he doesn't even check if any of the rects contain the point
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Hmmm
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Oh I see
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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
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JosephFerano
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—
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Today at 4:27 PM
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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
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I'll have to investigate further
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