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Apologies for the shouting but this is important.
When answering a question please:
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Insults, slap-downs and sarcasm aren't welcome. Let's work to help developers, not make them feel stupid.
cheers,
Chris Maunder
The Code Project Co-founder
Microsoft C++ MVP
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For those new to message boards please try to follow a few simple rules when posting your question.- Choose the correct forum for your message. Posting a VB.NET question in the C++ forum will end in tears.
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cheers,
Chris Maunder
The Code Project Co-founder
Microsoft C++ MVP
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Just a brief question. In a little demo I've got going I can have up to 300 objects floating around the screen and checking every object against every other object is painfully slow.
Using a quad tree to split the screen up and group the objects greatly speeds up checking for collisions but unfortunately rebuilding the quad tree every frame eats up quite a lot of the performance that you gain from using it.
Does anybody have any idea's on what I could do, bearing in mind that all 300 objects will constantly be moving and are all fairly close together, so trying to only update objects that have moved would cause the entire tree to be jiggled about a little bit, perhaps costing more time that a full rebuild.
My current favourite word is: Sammidge!-SK Genius Game Programming articles start - here[ ^]-
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why use a tree that is bound to become invalid all the time?
I would see the screen as a 2D matrix (say N*N) with N around typical screen size divided by typical object size (and not larger than say 10), so each object typically spans two rows and two columns. Each screen cell could be represented by a List containing references to the objects that (partially) lie inside it, and a timestamp.
One time step would:
- move some objects, checking them against the 2D matrix cells; resulting in possibly removing them from some lists, and possibly adding them to some lists; store current time in cells objects got added to.
- then foreach cell marked with current time, check the listed objects for overlap.
Possible optimization: give each object a list of cells it currently is in; this would speed up the removal step.
Luc Pattyn
I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages
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in the name of god
hello
how can we solve this problem to find the order of it:
T(n)=sqrt(n)*(T(sqrt(n))+n
order =?
valhamdolelah.
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Ok then I think:
T(n) = (n * C)/(e^2) + (n * ln(ln(n))) / ln(2)
edit: that would be O(n log(log n)) I think
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It's ugly, ugly, ugly....
You measure democracy by the freedom it gives its dissidents, not the freedom it gives its assimilated conformists.
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in the name of god
thanks you are right
i found myself my answer after testing it but if you can please send me a beautiful of it.
thanks.
valhamdolelah.
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Hi,
Is it possible to use Bug-1 algorithm without knowing the distance to the target ?
In every place (a matrix) I just know the way to this (N, NE, E, SE..), but not the distance.
modified on Tuesday, November 3, 2009 5:00 PM
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If you now the direction from two points and the distance between them, then it's pretty simple trig to work out the distance to the third; Triangulation[^]
Panic, Chaos, Destruction.
My work here is done.
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I need to make a divide&conquer algorithm for finding the dominant element in an array (dominant element = element which exists at least n/2 times in an array within the size n) . All you can do between those elements is compare them (no < relation or > relation) .
And also I need an algorithm which does it in O(n) (doesn't have to be d&c)
Thanks ahead for any help...
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Well, this may not win any awards for style, but have you considered just using a hash table to aggregate the count?
It would be O(n) to hash everything, since I believe hash tables are theoretically in constant time... Then a subsequent O(n) to find the largest value in the hash table.
There's probably a more clever way to do it, but that would get the job done.
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in the name of god
hello
if i know your question true you can do:
1-for i=0 n
B[A[i]]++;
for counting the number of repeating of them.
and then searching for max with O(n);
valhamdolelah.
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That only works if you know the number range... What if the numbers are completely unknown, and can be as high as 2^16... Are you going to have an array that large?
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khomeyni wrote: in the name of god
hello
Are you really in the US as your profile says? And if so, do you talk like this on the job?
You measure democracy by the freedom it gives its dissidents, not the freedom it gives its assimilated conformists.
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no i cant yet change it i am from Iran i will change it soon and also yes we talk like this on job but not as this intensity 
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