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Encrypting a file means that one is altering it in order to conceal its contents.
Encoding it means one is altering it because of technical reasons related to transmitting it (e.g., the Morse code).
Regards,
Satips.
Don't walk in front of me, I may not follow;
Don't walk behind me, I may not lead;
Walk beside me, and just be my friend. - Albert Camus
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Newton Rapheson ring a bell (Goal Seek function from Excel, that's the Algo they used)? Packaged bug free code anyone?
Thanks a bunch.
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You give no hint of what language you want, maybe Numerical Recipes is a good place to start - here is the 'c' version:
Newton-Raphson Method[^]
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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right! Numerical Recipes! Left school for sometime totally forget about it!
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Excellent read.[^]
The PostScript file (year 2000) is more informative.
"This perpetual motion machine she made is a joke. It just keeps going faster and faster. Lisa, get in here! In this house, we obey the laws of thermodynamics!" - Homer Simpson
Web - Blog - RSS - Math - LinkedIn - BM
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"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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Regards,
Satips.
Don't walk in front of me, I may not follow;
Don't walk behind me, I may not lead;
Walk beside me, and just be my friend. - Albert Camus
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Great
Russell
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I have a set of X-Y-Z coordinate points defining a 3D space (actually contour data). Then supposing I want to find the elevation at point X-Y, would the following method work (or does anyone have any other, better, suggestions):
1) Flatten the XYZ data (i.e., take the X and Y coords)
2) Construct the 2D Delauney triangulation for the data set.
3) Find the triangle in which the point X-Y lies.
4) Use the XYZ coords of the vertices of the triangle to interpolate the elevation at XY.
Of course the hard part is 3, but is the principle sound?
Anyone know of any sample code I can refer to? (bearing in mind I'm not a mathematician or geometrician!)
Cheers
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Years ago when I did some work in this area we had data on a rectangular grid - made things easy
If you google on "gis interpolation" there are a lot of interesting hits,
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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I need to retrieve the first four alphanumeric characters from a string, basically,
if I have "DE WINTER", I need to skip the third character and write "DEWI".
Other than going through each substring is there some algorithm for this pattern?
Jon
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It actually depends on the language you're using: string objects usually offer a suitable Replace method.
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.
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I'm using JavaScript.
Taking this point of view, the problem is that I need to iterate through each character in the source string and replace non-alphanumeric characters with a blank string ('').
Jon
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This http://www.w3schools.com/jsref/jsref_replace.asp[^]] doesn't help you, does it?
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.
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Hiya
i am looking for algorithm for Steganography in .mp3 and .jpg....
I found few links while searching in google but all say something different...So any one here with a good algorithm for this task...
And yeh i want to implement it in C#.Net..Yeh i know there are many articles for stego in this site but they are not for .jpg or .mp3..
thanks
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Search CodeProject for it, you will find several articles.
And yes there are many ways to do it.
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I dont think there is any article for .mp3 and .jpg in code project...Thats what i mentioned in my post...
Yeah there could be many ways, just trying to know which algo is the best way if somebody has already achieved that...
anywayz...
Thanks
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I dont know for mp3, I have seen several for JPEG.
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hi !i am writting a project where i have to deal with a lot of matrices and still a biginner in programming.i want to know if there is a library for linear algebra in c and where to get it.thanks!!
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If you're using C++ I'd go for Boost's uBLAS[^] library.
If you are going to use C try the BLAS[^] library.
Steve
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Sure Sir, gonna delete this right away.....
Am a newbie here and wanted to know more on Software Architecture with feedbacks and comments on topic I have raised at my blog pchaitanya.wordpress.com.....
Any feedback would be highly appreciated....
Thanks,
Chaitanya
Chaitanya
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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.
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For a 2-dimensional plane, I've implemented a function for determining whether a circle intersects a ray. Suppose there is one ray and many, many circles. How would I reduce the number of tests by selecting the circles that are certain or likely to be hit?
ROFLOLMFAO
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Hi,
lots can be done depending on circumstances.
SCHEME 1
Assuming you are only interested in a finite and rectangular part of the 2D plane,
you could divide the area in a number of smaller rectangles. Say 16*16 rectangles.
Now a ray would intersect only some of these rectangles, so we are only
interested in those circles that partially cover those rectangles.
Assuming your circles are static, that can be precalculated and stored: for
each record, you can hold a list of crossing circles.
So now it is a matter of finding the rectangles crossed, then scanning the circles
listed for these rectangles.
WARNING: things must be organized in such a way that the same circle is not
tested again for different rectangles; a boolean flag might help.
SCHEME 2
Assuming (most of) the circles are fairly small with respect to the entire area,
if the ray does not cross the entire area, but is restricted to some X-range
(for a steep ray) or some Y-range, one could easily eliminate all circles that
lie fully outside said X or Y-range.
Hope this helps.
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