Algorithms

Re: Whats the fastest way to search through a binary heap?

19-Nov-09 6:00

CaptainSeeSharp wrote:
I'm not quite sure what you mean. Are you saying bounds back and forth between the first and last, then second to first, then second to last, ect...?

Not sure what you mean here.. I mean, start at the root, if the value you're searching is smaller than the root then clearly the item is not in the heap - this applies recursively to every sub-heap as well

OTOH, I didn't need that feature (searching by value) in my A* implementation. Why do you need it?

C^aptainS^eeS^harp19-Nov-09 6:07

C^aptainS^eeS^harp

19-Nov-09 6:07

harold aptroot wrote:
I didn't need that feature (searching by value) in my A* implementation. Why do you need it?

I would really like to see how to make mine not need it. When mine gets the nodes around it, it checks to see if they are on the open list (that is where IndexOf is called), and if they are not then it will add whatever nodes not on the open list.

Watch the Fall of the Republic (High Quality 2:24:19)[^]

Re: Whats the fastest way to search through a binary heap?

harold aptroot19-Nov-09 6:25

Re: Whats the fastest way to search through a binary heap?

19-Nov-09 6:25

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).

C^aptainS^eeS^harp19-Nov-09 7:17

C^aptainS^eeS^harp

19-Nov-09 7:17

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.

Watch the Fall of the Republic (High Quality 2:24:19)[^]

Crowd detection (people), how to?

Gamma_ace12-Nov-09 15:53

Gamma_ace

12-Nov-09 15:53

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!

Re: Crowd detection (people), how to?

kaushik_acharya5-Nov-11 20:03

kaushik_acharya

5-Nov-11 20:03

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

Algorithm to C# [modified]

VS newbie 11-Nov-09 7:05

VS newbie

11-Nov-09 7:05

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

Re: Algorithm to C#

RichardM111-Nov-09 11:29

RichardM1

11-Nov-09 11:29

This is more a programming question than an algorithm question.
What languages do you know?

Silver member by constant and unflinching longevity.

Collision Detection with a Quad tree

Anthony Mushrow5-Nov-09 16:22

Anthony Mushrow

5-Nov-09 16:22

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[^]-

Re: Collision Detection with a Quad tree

Luc Pattyn5-Nov-09 16:40

Luc Pattyn

5-Nov-09 16:40

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.

Smile | :)

Luc Pattyn

I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

Re: Collision Detection with a Quad tree

ely_bob18-Nov-09 12:05

ely_bob

18-Nov-09 12:05

Alternative solution:
re-structure your "tree" and use a radial(gross) test

radial:

if((dist(a,b)^2-RadiusSqrd)<0)
{
//inside circle might be a collision canidate
}

(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)//IComparer
{
     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 Smile | :)

order of algorithm

khomeyni4-Nov-09 20:30

4-Nov-09 20:30

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.

harold aptroot5-Nov-09 1:32

5-Nov-09 1:32

There is no base case, what is T(1) ?

khomeyni5-Nov-09 4:45

5-Nov-09 4:45

take a constant as C>0
valhamdolelah.

harold aptroot5-Nov-09 5:10

5-Nov-09 5:10

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

khomeyni5-Nov-09 5:26

5-Nov-09 5:26

thanks
please say the way?how we can gain to this?
valhamdolelah.

harold aptroot5-Nov-09 5:30

5-Nov-09 5:30

What do you mean?

khomeyni5-Nov-09 5:49

5-Nov-09 5:49

solution =?
how do you solve it?
valhamdolelah.

harold aptroot5-Nov-09 6:45

5-Nov-09 6:45

Well I cheated and put it on wolfram alpha, this wasn't an exam so Smile | :)

khomeyni5-Nov-09 19:35

5-Nov-09 19:35

thanks but i want to know how to solve this problems.
valhamdolelah.

harold aptroot6-Nov-09 1:00

6-Nov-09 1:00

I don't know what you mean

RichardM111-Nov-09 11:33

RichardM1

11-Nov-09 11:33

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.

RichardM111-Nov-09 11:58

RichardM1

11-Nov-09 11:58

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.

khomeyni12-Nov-09 2:20