问题
I am trying to losslessly compress an image, and in order to take advantage of regularities, I want to convert the image from RGB to Y'CbCr. (The exact details of what I mean by RGB and Y'CbCr are not important here; the RGB data consists of three bytes, and I have three bytes to store the result in.)
The conversion process itself is pretty straightforward, but there is one problem: although the transformation is mathematically invertible, in practice there will be rounding errors. Of course these errors are small and virtually unnoticeable, but it does mean that the process is not lossless any more.
My question is: does a transformation exist, that converts three eight-bit integers (representing red, green and blue components) into three other eight-bit integers (representing a colour space similar to Y'CbCr, where two components change only slightly with respect to position, or at least less than in an RGB colour space), and that can be inverted without loss of information?
回答1:
YCoCg24
Here is one color transformation I call "YCoCg24" that that converts three eight-bit integers (representing red, green and blue components) into three other eight-bit (signed) integers (representing a colour space similar to Y'CbCr), and is bijective (and therefore can be inversed without loss of information):
G R B Y Cg Co
| | | | | |
| |->-(-1)->(+) (+)<-(-/2)<-| |
| | | | | |
| (+)<-(/2)-<-| |->-(+1)->(+) |
| | | | | |
|->-(-1)->(+) | | (+)<-(-/2)<-|
| | | | | |
(+)<-(/2)-<-| | | |->-(+1)->(+)
| | | | | |
Y Cg Co G R B
forward transformation reverse transformation
or in pseudocode:
function forward_lift( x, y ):
signed int8 diff = ( y - x ) mod 0x100
average = ( x + ( diff >> 1 ) ) mod 0x100
return ( average, diff )
function reverse_lift( average, signed int8 diff ):
x = ( average - ( diff >> 1 ) ) mod 0x100
y = ( x + diff ) mod 0x100
return ( x, y )
function RGB_to_YCoCg24( red, green, blue ):
(temp, Co) = forward_lift( red, blue )
(Y, Cg) = forward_lift( green, temp )
return( Y, Cg, Co)
function YCoCg24_to_RGB( Y, Cg, Co ):
(green, temp) = reverse_lift( Y, Cg )
(red, blue) = reverse_lift( temp, Co)
return( red, green, blue )
Some example colors:
color R G B Y CoCg24
white 0xFFFFFF 0xFF0000
light grey 0xEFEFEF 0xEF0000
dark grey 0x111111 0x110000
black 0x000000 0x000000
red 0xFF0000 0xFF01FF
lime 0x00FF00 0xFF0001
blue 0x0000FF 0xFFFFFF
G, R-G, B-G color space
Another color transformation that converts three eight-bit integers into three other eight-bit integers.
function RGB_to_GCbCr( red, green, blue ):
Cb = (blue - green) mod 0x100
Cr = (red - green) mod 0x100
return( green, Cb, Cr)
function GCbCr_to_RGB( Y, Cg, Co ):
blue = (Cb + green) mod 0x100
red = (Cr + green) mod 0x100
return( red, green, blue )
Some example colors:
color R G B G CbCr
white 0xFFFFFF 0xFF0000
light grey 0xEFEFEF 0xEF0000
dark grey 0x111111 0x110000
black 0x000000 0x000000
comments
There seem to be quite a few lossless color space transforms. Several lossless color space transforms are mentioned in Henrique S. Malvar, et al. "Lifting-based reversible color transformations for image compression"; there's the lossless colorspace transformation in JPEG XR; the original reversible color transform (ORCT) used in several "lossless JPEG" proposals; G, R-G, B-G color space; etc. Malvar et al seem pretty excited about the 26-bit YCoCg-R representation of a 24-bit RGB pixel.
However, nearly all of them require more than 24 bits to store the transformed pixel color.
The "lifting" technique I use in YCoCg24 is similar to the one in Malvar et al and to the lossless colorspace transformation in JPEG XR.
Because addition is reversible (and addition modulo 0x100 is bijective), any transform from (a,b) to (x,y) that can be produced by the following Feistel network is reversible and bijective:
a b
| |
|->-F->-(+)
| |
(+)-<-G-<-|
| |
x y
where (+) indicates 8-bit addition (modulo 0x100), a b x y are all 8-bit values, and F and G indicate any arbitrary function.
details
Why do you only have 3 bytes to store the result in? That sounds like a counter-productive premature optimization. If your goal is to losslessly compress an image into as small a compressed file as possible in a reasonable amount of time, then the size of the intermediate stages is irrelevant. It may even be counter-productive -- a "larger" intermediate representation (such as Reversible Colour Transform or the 26-bit YCoCg-R) may result in smaller final compressed file size than a "smaller" intermediate representation (such as RGB or YCoCg24).
EDIT: Oopsies. Either one of "(x) mod 0x100" or "(x) & 0xff" give exactly the same results -- the results I wanted. But somehow I jumbled them together to produce something that wouldn't work.
回答2:
I did find one such solution, used by JPEG 2000. It is called a Reversible Colour Transform (RCT), and it is described at Wikipedia as well as the JPEG site (though the rounding methods are not consistent). The results are not as good as with the irreversible colour transform, however.
I also found a better method described in the paper Improved Reversible Integer-to-integer Color Transforms by Soo-Chang Pei and Jian-Jiun Ding. However, the methods described in that paper, and the method used by JPEG 2000, require extra bits to store the result. This means that the transformed values do not fit in 24 bits any more.
来源:https://stackoverflow.com/questions/10566668/lossless-rgb-to-ycbcr-transformation