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hello_there

sloooooowwwwwwww

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how much would it downgrade speed drawing 850 cubes to the screen and checking for collisions with all of them? i mean would it be heaps or just a little?

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Excuse me, I don''t know much about collision checking, and maybe I''m totally wrong, but:

Wouldn''t the number of checks be

850+849+848...+1 = (850+1)*(850/2) = 361675 ?

That''s a little bit less than 2^850...

/John

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FunkyTUNE!!! is correct. By doing 850*849 you're repeating *alot* of checks. Take this example, you have objects A, B and C.

What your loop will do is this:
1st interation of outer loop:
(A,B), (A,C)
2d interation of outer loop:
(B,A), (B,C)
3rd interation of outer loop:
(C,A), (C,B)

However notice that in the 2nd interation we've already checked AB, and in the third interation we've already checked AC and BC.

You want to do this:

1st
(A,B), (A,C)
2nd
(B,C)

This can be accomplished with this code (Assuming areas are Low to High):

for I=Low to High-1
for J=I+1 to High
Compare(I,J)
next J
next I


[EDIT]: Correct horrible insult to FunkyToon, ...eer tune

[edited by - michalson on May 6, 2002 8:27:08 AM]

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