Algorithms

Luc 64801111-Apr-09 6:18

Luc 648011

11-Apr-09 6:18

Hi Dave,

Don't worry, I'm using a new account to avoid all the crap gold members get to digest on the home page.
And its not 1 million posts, even CG does not reach 100K!

If all numbers are ascending and that is what you don't want, everything gets swapped.
If all numbers are descending and that is what you don't want, everything gets swapped.
I expect these two test cases to have the same speed. I understood they did.

If all numbers are ascending and that is what you want, no swaps occur.
If all numbers are descending and that is what you want, no swaps occur.
I expect these two test cases to have the same speed. I understood they did not, they were quite different, and by much more than some Windows unpredictability could explain.

Member 4194593 wrote:
all this activity as indicated by the Task Manager? Paging with 4 million bytes involved probably plays a part in it

Huh. What is your hardware? how could 4MB be cause for paging?
If this is a miniature PC, try the same code on a real one!

Smile | :)

Member 419459315-Apr-09 8:29

Member 4194593

15-Apr-09 8:29

Luc,

I found an error in the Descending Descending where I had supplied the ascending
data instead of the descending data, and I corrected that. I also modified the
timing procedure to create all of the data first, then execute the tests, and
then added a loop to run all of the tests 10 times. I added code to call
SetPriorityClass to set the priority to HIGH_PRIORITY_CLASS. I added code to
put the start of each procedure on an aligned 16 location to make sure that the
same code in the ascending and descending routines stayed on the same relative
boundaries to avoid differences in the hardware branch predictions - this would
only cause a 1 or 2 cycle difference in time, but with 1,000,000,000 *
1,000,000,000 compares it will make a difference - note that something as small
as a 26 record sort takes 109 compares and 14 exchanges.

I put the loops for the left scan and the right scan on 16 BYTE boundaries to
aid in the hardware branch prediction.

I modified the code to clean up the floating jump on condition tests (had to
re-order my register usages to free up eax, then debug and re-test).

I modified the test code to add a timing test for floats.

The following are the results from the first two iterations of the tests (almost
the same times as for DWORDS):

uilding Ascending data.
Building Descending data.
Building Random data.
                                                                         Floating tests         DWORD tests 
1024*1024 Ascending Floats sorted 1024 times Ascending, time in msec:    32359, good compare.   32068, good compare.
1024*1024 Ascending Floats sorted 1024 times Descending, time in msec:   36407, good compare.   37310, good compare.
1024*1024 Descending Floats sorted 1024 times Ascending, time in msec:   36749, good compare.   37276, good compare.
1024*1024 Descending Floats sorted 1024 times Descending, time in msec:  36674, good compare.   38611, good compare.
1024*1024 Random Floats sorted 1024 times Ascending, time in msec:      130539, good compare.  132599, good compare.
1024*1024 Random Floats sorted 1024 times Descending, time in msec:     131883, good compare.  131187, good compare.

1024*1024 Ascending Floats sorted 1024 times Ascending, time in msec:    32005, good compare.   32005, good compare.
1024*1024 Ascending Floats sorted 1024 times Descending, time in msec:   37254, good compare.   36943, good compare.
1024*1024 Descending Floats sorted 1024 times Ascending, time in msec:   37126, good compare.   35773, good compare.
1024*1024 Descending Floats sorted 1024 times Descending, time in msec:  37446, good compare.   37359, good compare.
1024*1024 Random Floats sorted 1024 times Ascending, time in msec:      130639, good compare.  133531, good compare.
1024*1024 Random Floats sorted 1024 times Descending, time in msec:     130431, good compare.  131447, good compare.

Dave.

Luc 64801115-Apr-09 8:39

Luc 648011

15-Apr-09 8:39

OK, this looks good. The timing anomaly has disappeared. Well done.

Smile | :)

Re: Rotation of Bitmap [modified]

Member 419459315-Apr-09 9:45

Member 4194593

15-Apr-09 9:45

Luc,

It was an interesting search. The only "mistake" was the use of ascending data instead of descending data for the test. The rest of the changes were to avoid the branch prediction problems - keeping all of the tight loops on a 16 byte boundary and keeping the loops down to 16 bytes or less.

Dave.

Rotation of Bitmap

CodeOfLife5-Apr-09 16:21

CodeOfLife

5-Apr-09 16:21

Hi all

I would be very happy, if anyone could provide me with a algorithm to rotate a bitmap around it's center. The problem is that it has to run in .NET micro v3.0(otherwise I would have just used that very nice Graphics that exist in full version of .NET) in which they removed the Bitmap.RotateBlt method. Speed is not a(major) concern.

Luc Pattyn5-Apr-09 16:30

Luc Pattyn

5-Apr-09 16:30

Hi,

without a method that offers free rotation, you would have to do it yourself, along these lines:

foreach row y2 in destination image
    foreach column x2 in destination image
        calculate corresponding location (x1,y1) in source image
        copy from source image at (x1,y1) to destination image (x2,y2)
    }
}

without translation, the rotation formulas would be something like

x1 = x2 * cosa - y2 * sina
y1 = x2 * sina + y2 * cosa

[EDIT: fixed a typo /]

where cosa and sina are the (constant) cosine and sine of the rotation angle.

You would need to add some constant to x1 and y1 to choose the center of rotation.

Smile | :)

Luc Pattyn [Forum Guidelines] [My Articles]

- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
- use the code block button (PRE tags) to preserve formatting when showing multi-line code snippets

modified on Monday, April 6, 2009 11:25 AM

Re: Rotation of Bitmap

CodeOfLife6-Apr-09 6:37

CodeOfLife

6-Apr-09 6:37

Hi thanks for the reply

I'm having trouble making the center of rotation the center of the image. If I choose the constant to add to x1 and y1 to be 32(i'm using 64x64 pixel bitmaps), part of source bitmap is not copied.

What am I doing wrong?

in advance, Thx a bunch

Re: Rotation of Bitmap

Luc Pattyn6-Apr-09 6:59

Luc Pattyn

6-Apr-09 6:59

Hi,

if you want to rotate an image around a point C that is at (CX1,CY1) in the source image and at
(CX2, CY2) in the destination image, then obviously you have to translate the coordinate system in both images like so:

(x1-CX1) = (x2-CX2) * cosa - (y2-CY2) * sina
(y1-CY1) = (x2-CX2) * sina + (y2-CY2) * cosa

Luc Pattyn [Forum Guidelines] [My Articles]

Creating RSOM from SOM

Jasmine Pomelo4-Apr-09 14:04

Jasmine Pomelo

4-Apr-09 14:04

Hi,
I'm new to recurrent SOMs. However, I've already created a SOM. In order to make it an RSOM do I simply add a leaky integrator, or are there other aspects I need to change from my original SOM?
Thanks for any help.

Re: Creating RSOM from SOM

CPallini6-Apr-09 21:39

CPallini

6-Apr-09 21:39

Member 6038196 wrote:
I'm new to recurrent SOMs.

I.e. are you new to Turkic currency [^]?
Big Grin | :-D

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Re: Creating RSOM from SOM

Jasmine Pomelo9-Apr-09 9:38

Jasmine Pomelo

9-Apr-09 9:38

lol I wish it was that simple. More like [^] Big Grin | :-D

Single Elimination - Tournament Brackets

aslamc2-Apr-09 9:32

aslamc

2-Apr-09 9:32

I posted this on C# and realized later this might be a more appropriate location for it.

Hi,
I have been searching around for a single elimination algorithm / pseudo code for tournament brackets and I haven't found a single site on it. The best I found was Wikipedia explaining single elimination. I have been reading code project articles for a while now and I know there are many experience and very knowledgeable developers here. Since this is fairly simple to code I was hoping if somebody could help me out with the code/pseudo code or an algorithm for it. I would appreciate it a lot. Looking forward to your replies.

Re: Single Elimination - Tournament Brackets

Alan Balkany3-Apr-09 4:00

3-Apr-09 4:00

You create a binary tree where the leaves represent the tournament's contestants. Each pair of contestants under an interior node play a match, and the winner then occupies that interior node.

Likewise, an interior node that has two other interior nodes under it represents the winner of the match between the winners corresponding to those two interior nodes.

The root of the binary tree represents the winner of the tournament.

Tree construction: One approach uses the Composite design pattern, an abstract Node class with two derived classes: Contestant and InteriorNode. Start with a list of Contestants. While that list has more than one element, combine two elements under a new InteriorNode, remove those elements from the list, and insert the new InteriorNode in the list. The list should be ordered by increasing depth, so that, for example, all Contestants are paired before any InteriorNodes are paired.

When the list is down to one element, that's the root of the tree.

modified on Friday, April 3, 2009 10:21 AM

Tree algo

dfreeser2-Apr-09 5:58

2-Apr-09 5:58

Is there an algorithm that does a upside-down, upside-down tree. i.e, right side up(root on bottom). Thanks.

Alan Balkany3-Apr-09 3:43

3-Apr-09 3:43

Tree algo:

1. Grasp page in both hands.
2. Turn page 180 degrees.

dfreeser3-Apr-09 5:15

3-Apr-09 5:15

Very insightful. I see why your an expert now. Would have never figured that one out myself.

Quick question. If said tree isn't on paper but on my computer screen, do I grab the computer screen and flip it 180 degrees or the computer itself.
Big Grin | :-D

Alan Balkany3-Apr-09 5:42

3-Apr-09 5:42

Seriously, it wasn't clear what your problem was. Do you just want to draw the tree upside down on the screen? If so, for a display of height H, substitute (H - y) for every y coordinate to flip the tree.

dfreeser3-Apr-09 6:02

3-Apr-09 6:02

Sorry. I want to know if is something like treeview but where the root is on the bottom and everything goes up.

Alan Balkany3-Apr-09 6:16

3-Apr-09 6:16

Oh, I thought you might be interested in an algorithm, since this is the Algorithms forum and your title was "Tree algo". Since you're looking for a suggestion on a graphics package, you might want to post your question in the Lounge.

dfreeser3-Apr-09 6:33

calculate polygon area and its centre in 3d

3-Apr-09 6:33

Sorry. I'll try there. Thanks for your help.

beko31-Mar-09 22:57

beko

31-Mar-09 22:57

Hello,

I have some 3d points (not self intersecting). the shape of the points is looking like / (forward slash) from profile, It is actually a slice of a 3d object, cut with a rotated plane.

I want to calculate the area of 3d points and its centre. For 2d case, i use the following

http://local.wasp.uwa.edu.au/~pbourke/geometry/polyarea/[^]

but, how can i extend the 2d case to 3d for area and its centre?

Thanks in advance.

Bekir.

Re: calculate polygon area and its centre in 3d

Alan Balkany1-Apr-09 3:46