On Apr 23, 12:21�pm, pierrefrancescos...@gmail.com wrote:> I was wondering if there was a relatively straightforward algorithm > for compressing image using Fourier transforms. I was thinking that I > could transform an image into frequency space and then perform a> lowpass filter on the image to set all the lesser frequencies to zero,you mean this " explicitly making low amp values to zero "; then that's correct.> then perform a run-length-encoding on the data to pack the zeroes. > However, I have been running into trouble with this method,since the> decoded image largely has zero values (i.e. it is mostly black). Am I > completely off target or what?while decoding, the DC co-eff will handle this job... I think, you don't need to worry about that. At the run-time the encoded zero's in RLE will be given life with the help of main DC co-eff which resides at the starting of each such macro block. hope this is helpful to you. best wishes nimo __________ Bernard Shaw shaking hands with a lady: " I'm very glad to see you ". Lady:"Well,I'm not". Shaw: "Why don't you pretend as I do?"

# Image Compression Using Fourier Transforms

Started by ●April 23, 2009

Reply by ●April 29, 20092009-04-29

Reply by ●May 9, 20092009-05-09

I designed a lossless YIQ color transform that requires 2 or 5 multiplications per pixel. It is like this: R -= B B += R >> 1 B -= G G += B >> 1 G += round(- (0.5 - Kb - Kr) * B + (Kr - Kb) * R / 2.0) B -= round(((1 - cos(45 - theta)) / sin(45 - theta)) * R) R += round(sin(45 - theta) * B) B -= round((((1 - cos(45 - theta)) / sin(45 - theta)) - sin(45 - theta) * cos(45 - theta) / (4 - sin(45 - theta) * sin(45 - theta))) * R) Y = G I = R Q = B Patent pending. I posted a comparison between YIQ and YCbCr at, http://groups.google.com/group/media-player/browse_thread/thread/329f6ba471f802bd I also added a new transform to Libima. It can be done on a sub-region of the image so it would be good for video. A video compressor could separately transform the sub-region which is independently coded and the sub-region which is motion compensated. It is described on my website: http://www.geocities.com/repstsb/index.html

Reply by ●May 11, 20092009-05-11

<sarcasm> Hoorah, another algebraic formula protected from general use. </sarcasm> But seriously, why share something the community can't ultimately use without fear of litigation?

Reply by ●May 11, 20092009-05-11

There is a simpler algorithm: R -= B B += R >> 1 B -= G G += round((Kb + Kr) * B + (Kr - Kb) * R / 2.0) B -= round(((1 - cos(45 - theta)) / sin(45 - theta)) * R) R += round(sin(45 - theta) * B) B -= round((((1 - cos(45 - theta)) / sin(45 - theta)) - sin(45 - theta) * cos(45 - theta) / (4 - sin(45 - theta) * sin(45 - theta))) * R) Y = G I = R Q = B

Reply by ●May 11, 20092009-05-11

It is possible to do any lossless linear color transform of the kind YC1C2 using 5 multiplications. The steps are, 1. Transform the red and blue channels so that they are in the plane R + G + B = 0 2. Add multiples of the intermediate chroma channels to the green channel to obtain the luma channel. 3. Find constants a, b, c such that the following algorithm transforms the intermediate chroma channels into the desired chroma channels: B += round(a * R) R += round(b * B) B += round(c * R) The constants can be found through linear algebra representing, [1 0 0 ] A = [0 1 0 ] [0 a 1 ] And representing the first 2 steps as the matrix T, and the desired transform as the matrix M, and solving the equation, C B A T = M For example, the JPEG YCbCr transform is, [0.299, 0.587, 0.114] [0.5, -0.418688, -0.081312] [-0.168736, -0.331264, 0.5]] To make the determinant 1, we can multiply the chroma channels by the same multiple: [.299, .587, .114] [1.029, -.861, -.167] [-.347, -.681, 1.029] So the algorithm is, R -= G B -= G G += round(Kb * B + Kr * R) B += round(-0.171 * R) R += round(-0.167 * B) B += round(-0.171 * R)

Reply by ●May 11, 20092009-05-11

Mihai Cartoaje wrote:> It is possible to do any lossless linear color transform of the kind > YC1C2 using 5 multiplications.And your point is...? A somewhat more useful observation would be: You could always implement it by lifting, thus making it lossless up to a scaling you can compensate for later on, for example. This is equally trivial. (-: So long, Thomas

Reply by ●May 19, 20092009-05-19

People can apply for a license by emailing me. If I obtain a patent and a company uses it, it would allow me to write useful software so the community wins.

Reply by ●May 19, 20092009-05-19

Prior art is 7 multiplications per pixel. Are there examples of prior art using 5 multiplications per pixel?

Reply by ●May 25, 20092009-05-25

If theta is different from 45 degrees, there is a simpler algorithm: R -= B B -= G G += round((Kb + Kr) * B + Kr * R) B += round((0.5 - (1 - cos(45 - theta)) / sin(45 - theta)) * R) R += round(sin(45 - theta) * B) B -= round((((1 - cos(45 - theta)) / sin(45 - theta)) - sin(45 - theta) * cos(45 - theta) / (4 - sin(45 - theta) * sin(45 - theta))) * R) Y = G I = R Q = B

Reply by ●May 26, 20092009-05-26

Mihai Cartoaje wrote:> It is possible to do any lossless linear color transform of the kind > YC1C2 using 5 multiplications...> For example, the JPEG YCbCr transform is, > > [0.299, 0.587, 0.114] > [0.5, -0.418688, -0.081312] > [-0.168736, -0.331264, 0.5]] > > To make the determinant 1, we can multiply the chroma channels by the > same multiple: > > [.299, .587, .114] > [1.029, -.861, -.167] > [-.347, -.681, 1.029] > > So the algorithm is, > > R -= G > B -= G > G += round(Kb * B + Kr * R) > B += round(-0.171 * R) > R += round(-0.167 * B) > B += round(-0.171 * R)But in this case G is not: 0.299 * R + 0.587 *G + 0.114 * B because the G multiplier above is 1. So in the end you need more multiplications? And if you agree with me that YUV and YCbCr only differs in the scaling of the color components (UV versus CbCr), and YUV can be made like: Y = 0.299 * R + 0.587 * G + 0.114 * B U = (B - Y) * 0.493 V = (R - Y) * 0.877 than YCbCr could be made this way, only by changing the factors for U and V to become Cb and Cr, with 5 multiplications? Jens