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L. Spiro

[Tutorial] Instant-Insertion Quadtrees

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Interesting stuff! I just made a simple quadtree program last week to refresh myself on the C++ language.

I may try making this quadtree as a challenge.

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Great tutorial!

Wit this technique, are my world coordinates forced to be from 0 - 256? And objects will only be spatially sorted at a resolution of (1/256)? Edited by web383

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Great tutorial!

Wit this technique, are my world coordinates forced to be from 0 - 256? And objects will only be spatially sorted at a resolution of (1/256)?

 

That would make it very limited, wouldn't it? It's 256 in his example since it's easier to show how it works in 8 bits. You can use 16 or 32 bit types if you want, for a much larger area.

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Wit this technique, are my world coordinates forced to be from 0 - 256?

No, you just need to apply a suitable conversion factor.

And objects will only be spatially sorted at a resolution of (1/256)?

Yes. That should be plenty to significantly speed up collision detection, or any other naive O(N^2) process.

You can use 16 or 32 bit types if you want, for a much larger area.

You will run out of memory - it's not feasible to preallocate enough storage. If you really need a quad-tree with depth greater than 8, then you need a sparse representation thereof.

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You can use 16 or 32 bit types if you want, for a much larger area.

You will run out of memory - it's not feasible to preallocate enough storage. If you really need a quad-tree with depth greater than 8, then you need a sparse representation thereof.


Oh right, I mixed up the world coordinates for depth representations. I forgot for a moment that the 8-bit value represents the tree levels, not coordinates.

Converting the coordinates as you said makes more sense.

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The world size can be any value.  In the code, m_fInvRadius is used to make the conversion from world units to internal units here:
		// Now to the range from [0..255].
		a2Shifted.m_vMin.x *= m_fInvRadius;
		a2Shifted.m_vMin.y *= m_fInvRadius;
		a2Shifted.m_vMax.x *= m_fInvRadius;
		a2Shifted.m_vMax.y *= m_fInvRadius;
You can use 16-bit values and some adjustments to the internal range to get 10 levels, but that would be 349,525 nodes and 18,175,300 megabytes of memory on 32-bit machines.

8 levels is usually enough. If your world is 10,000 meters square (10 kilometers), the smallest squares are 78.125 meters in size, which is very reasonable resolution. Remember that the smaller your nodes are the more frequently moving objects have to be re-inserted, so there is a trade-off with deeper trees.


L. Spiro

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