How do you check collision against 1000x1000 objects?

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Hey, i just need some information on how to check collision if theres 1000 objects on the field.
And all 1000 objects are enemies to eachother.

So, do i for example have to make two for loops like this?:

[source lang="cpp"]
for(int i = 0; i < objects.size(); ++i) //objects.size is the size of the vector(if i use that)
for(int j = 0; j < objects.size(); ++i)
collision(objects, objects[j]);
[/source]
But thats 1000x1000 = a million loops in a single statement

So, i just want to know if thats how you do it... cant see any other way of doing this so far Edited by Assassinbeast

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Yep, subdivide the problem!!!

It's a little more complicated that JUST dividing into regions (i.e. players on the boundaries of two different regions might be able to collide, or a bullet/player might cross from one region into another in one time-step). You'll need to handle these edge cases in addition to what ultramailman suggests.

Octrees/quadtrees are common ways to hierarchically subdivide your world although depending on your requirements you can optimize this in different ways.

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ahh ok, thanks for all the help and suggestions

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in addition to what was said,ie: using some form of broad phase culling of objects that can't possibly collide, your loops are naive in that they check each item with each other item... twice. Better would be to change the second loop to:

 for(int i=0; i< objects.size() ; i++) { for(int j=0; j<i; j++) { // check collisions } } 

Your implementation results in n^2 tests, the above in a little over 1/2 that.

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Stroppy Katamari already said that, and the j = i+1 method saves checking i == j as well.

You can also sort the coordinates in each axis and check 1 dimension at a time, if something is at x = 1 and moving with an x velocity of +2, it's never going to collide with anything that isn't in the interval [1, 3] in the x coordinate (need to account for the width of both objects in the x axis as well, of course). You need to have a system for reoordering the objects when they move which doesn't involve sorting the whole lot though. Space partitioning systems are just more advanced versions of that kind of thing.

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hehe
1000 * 1000 == BOOOM

The first solution ::

 class Object { bool receiveCollision; }; //player.receiveCollision = true; //block.receiveCollision = false; //items.receiveCollision = false; for(int i = 0; i < objects.size(); ++i) { Object* obj = objects; if(obj->receiveCollision) { for(int j = 0; j < objects.size(); ++i) { collisionCheck(obj, objects[j]) } } } 

The second solution ::

 enum Tags { TagPlayer, TagEnemy, TagBlock, TagWeapons}; class Object { Tags tag; std::vector<Tags> collisionTags; }; for(int i = 0; i < objects.size(); ++i) { Object* obj = objects; for(int j = 0; j < obj->collisionTags.size(); ++i) { for(int k = 0; k < objects.size(); ++k) { if(obj->collisionTags[j] == objects[k].tag ) { collisionCheck(obj, objects[k]); } } } } 

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Also, I would say that in order for a collision to take place, one of the colliding objects must have moved. No need to test 2 objects together if neither of them moved.
Unless of course you have a more advanced physics system where objects can change size.

But yeah, identifying objects that have not changed in any way and skipping those pairs, is a very useful optimisation to use at some stage in the collision detection process. It likely doesn't matter too much at what point you do it. Edited by iMalc

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In situations where you need to actually update that many objects each and every frame, well tuned spatial hashing will handily beat all other broadphase methods for large amounts (1000s) of uniformly sized objects - where the largest are within 4-8 times the radius of the smallest objects. If you can get away with only updating a subrect of your game world, that's another optimization you should consider.

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Stroppy Katamari already said that, and the j = i+1 method saves checking i == j as well.

I was ninja'd! (i was interrupted mid post at work, and didn't get to finish until after many follow up posts)

+1 on the j=i+i solution though.