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But surely your heap is sorted by F cost? And if you find a node a second time, the G (and therefore the F) will probably be different since you would have reached it from an other node (since the one you reached it from the first time should be in your closed list)
I used a sparse "grid of grids"-of bits to test membership of open and closed lists (closed "list" was only that grid and no list), a normal grid would be easier but I had huge maps where paths would "usually" stay in roughly 1 "area", which meant that usually only up to 4 entries of the "supergrid" would be used. The whole thing was just a BitArray[,] btw. The "supertile"-size should be a power of 2 and the area divisible by 32 to be efficient this way, like 16x16, or 32x32. Testing membership of open or closed list is then only O(1), at the expense of .. not even all that much memory (just 1 bit per tile and some overhead, and usually only for a subset of tiles).
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I think I get what you are saying. Use a grid (that is the same size of the map) of bits or bools where a value of 1 or true in a particular X,Y coordinate of the bitarray[,] means that it is in the open list? I don't use a closed list by the way, and it works just as well.
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Hello! I have a problem to make a detection, I need to locate multiple objects (people) was planning to use the algorithm of Viola and Jones extended version of Lienhart (included in OpenCV), can I make a massive detection (crowd, each one people) of objects with these algorithms? (I know it is mainly used for face detection) or exist another solution whithin OpenCV?
Can anyone help me? I no need a source code, just help me to choose an algorithm to do this please!
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There's an implementation by Irshad Ali & Mathew of people detection using Viola Jones and tracking using particle filter. They have created a Haar cascade of heads to deteck humans.
They have used particle filter implementation by Rob Hess.
All the links can be found in my webpage.
https://sites.google.com/site/learningcomputervision/particle_filter[^]
I have corrected some bugs in Irshad's code and added the modified files as attachment in the webpage.
Regards,
Kaushik
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I'm a newbie and would appreciate it if someone could start me off and guide to on how how can you convert algorithm to C# (probability distribution in Wireless LAN)
Algorithm 1 x= Corr GetLocation (n, S, X, P RM)
Input:
n : Number of samples from each access point.
S : Measured signal strength vectors from k access points (S =(s1, ..., sk)).
Each si, 1 ≤ i ≤ k is a vector containing n samples from access point i.
X : Radio map locations.
RM : Radio map, whereRM[a][x](q) represents the pdf of the average of n signal
strength samples from access point a at location x ∈ X.
Output:
The location x ∈ X that maximizes P(x/S).
1: for i = 1..k do
2: Avg(i) ← average(si)
3: end for
4: Max ← 0
5: for l ∈ X do
6: P ←k
i=1
Avg(
i)+0.5
Avg(i)−0.5
RM[i][x](q) dq
7: if P > Max then
8: x ← l
9: Max ← p
10: end if
11: end for
modified on Wednesday, November 11, 2009 1:28 PM
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This is more a programming question than an algorithm question.
What languages do you know?
Silver member by constant and unflinching longevity.
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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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Alternative solution:
re-structure your "tree" and use a radial(gross) test
radial:
if((dist(a,b)^2-RadiusSqrd)<0)
{
}
(pathagarean therum with a bit of inside/outside circle, with small error...) it is fast.
Gross:
or iterativly restructure your list...
public static int blah(point P, point Q){
if(P.X>Q.x){ return 1;}
else if(P.X==Q.x){return 0}
else{return-1}
}
this way you would only need to "check" against neighbors in the tree structure.
alternativly you could use a float type(but that is basically the radial method of above)
using this method you basically only ever have a few checks,
there is one other way off hand..
I lik eto use the System.Drawing.Region (i forget the name) but it will give the union or difference of two "polygon"(regions) which is on the gpu and crazy fast.. in which case you may be able to do it brute force.. cus it is fast.. if you like this, or need more reference send me a ping and I will scrounge up the code for you.. leaving work --Hurray
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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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There is no base case, what is T(1) ?
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take a constant as C>0
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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thanks
please say the way?how we can gain to this?
valhamdolelah.
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solution =?
how do you solve it?
valhamdolelah.
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Well I cheated and put it on wolfram alpha, this wasn't an exam so
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thanks but i want to know how to solve this problems.
valhamdolelah.
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I don't know what you mean
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In the end, he does want the answer, but he is more interested in how you come up with the answer - the way to solve it. In effect, he does not want you to do his homework for him, he wants to know HOW to do his homework.
That is unusual for this kind of question, and I think he deserves kudos!
Unfortunately, I don't know how to solve it.
Silver member by constant and unflinching longevity.
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Thank you for wanting to know HOW, not just wanting the answer.
.
Are you solving for the computational order of an algorithm that takes this long to process (the Big O notation), or are you looking for a numerical order of magnitude for the result of applying this equation to an arbitrary value 'n' ( eg. T(2) = sqrt(2)+T(sqrt(2))+2)?
.
T(n)=sqrt(n)*(T(sqrt(n))+n
.
The Big O would be O(n). the two terms are added together, so each term can be taken by itself.
sqrt(n) * T(sqrt(n))becomes 'n' in the limit, so you have two O(n) terms, which is the same as O(n).
.
If you are looking for numerics, I can't really help you.
Silver member by constant and unflinching longevity.
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thanks very much for your attention
i want to know the computational order of it as you said :
sqrt(n) * T(sqrt(n))becomes 'n' in the limit, so you have two O(n) terms, which is the same as O(n).
but why in limit it becomes n?,please explain more.
also i think finding it as numerical would be interesting but i think this would be hard!!!
thanks.
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response below
Silver member by constant and unflinching longevity.
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I have to disagree. The two terms are not O(n) at all, one of them is O(n^0.5) which is asymptotically different from O(n), and the other is unknown until the recursion is solved.
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response below your entry
Silver member by constant and unflinching longevity.
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