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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.
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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.
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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.
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
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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
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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]
- 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
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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.
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Member 6038196 wrote: I'm new to recurrent SOMs.
I.e. are you new to Turkic currency [^]?
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]
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lol I wish it was that simple. More like [^]
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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.
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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
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Is there an algorithm that does a upside-down, upside-down tree. i.e, right side up(root on bottom). Thanks.
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Tree algo:
1. Grasp page in both hands.
2. Turn page 180 degrees.
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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.
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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.
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Sorry. I want to know if is something like treeview but where the root is on the bottom and everything goes up.
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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.
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Sorry. I'll try there. Thanks for your help.
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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.
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If you could rotate the points so they all have the same Z coordinate, you could then ignore the Z coordinate and apply the 2D formulas for the area and centroid.
Rotation will not change the area. Apply the reverse rotation to the calculated centroid to get the centroid of the original 3D points.
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Thank you,
i will try that and i think it will solve the problem but it will create a bit of overhead.
Bekir.
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What you want is a 2D coordinate system in the plane of intersection, then you can use your equation for area.
Assume that your original points are represented as vectors {Vk} from some origin O (this is your original 3D coordinate system).
As you have the equation of the plane, choose one point on it as the new origin Op, and choose two orthogonal unit vectors (U1, U2) that lie in the plane as the new axes. (i.e. choose two orthogonal vectors orthogonal to the normal of the plane)
Then in your new coordinate system, the 2D coordinates of the point Vk are:
( (Vk-Op).U1, (Vk-Op).U2 )
i.e. you simply project the 3D vector from your new origin onto your new axes.
Note - this includes Alan's solution, the vectors U1 and U2 would be rows of his rotation matrix. It's just a different way of looking at things.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Does anyone have a stable Quicksort? I have looked at many, and they all seem to have a problem in that they swap elements that are not equal to the pivot element before knowing whether the swapped elements match any other element. see the following example:
014447222563 where '3' is the pivot
You will swap the '4's and '2's leaving:
012227444563
and the final swap:
012223444567
however, the swaps of the '2's and '4's (done one at a time from the outside) put them out of order before you even get to see that they are duplicates.
The only way I know of is to supply an additional key element in the record (sorted along with the key) which contains the input record number of the key. This is time expensive, rebuilding the records just to sort them, only to unbuild them when the sort completes. It is easily done, however, with indirect sorting, where the indirect array is an array of structures containing both the pointer to the actual record and also the input record number. Swapping the indirect pointers as records keeps the original record number with its record definition where it can be checked when an equal comparison is encountered. Of course, indirect sorting is inefficient when huge arrays (> 2GB)are involved because the underlying data remains scattered in memory, and paging in huge blocks of memory to compare a small record is a terrible waste.
Dave Augustine
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If you are sorting numbers then consider using a bucket sort which is stable. It's also faster than QuickSort in some circumstances.
Greetings - Gajatko
Portable.NET is part of DotGNU, a project to build a complete Free Software replacement for .NET - a system that truly belongs to the developers.
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gajatko,
Thank you for the reply, but I was specifically interested in a Stable Quicksort.
Off topic, I see that you answered this over 3 hours ago. I checked the forums no more than 5 minutes ago and this message was not there, I got your reply on my Email and answered it from there. Guess I'll have to tell Chris.
Dave.
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Have a look at STL's stable_sort[^]. Perhaps this is what you're after.
Steve
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