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Quantization - PowerPoint PPT Presentation

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Quantization. If too few levels of gray, (e.g. decrease halftone spot size to increase spatial resolution), then boundaries between adjacent gray levels become apparent. This can happen in color halftoning also. See demo at http://www.ctr.columbia.edu/~sfchang/course/dip/demos/Quat.html.

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  • If too few levels of gray, (e.g. decrease halftone spot size to increase spatial resolution), then boundaries between adjacent gray levels become apparent. This can happen in color halftoning also.

  • See demo at http://www.ctr.columbia.edu/~sfchang/course/dip/demos/Quat.html


  • Distance from white point

  • Adding white desaturates but does not change hue or perceptual brightness.

  • HSB model is approximate representative of this. See PhotoShop

Device independence
Device Independence

  • Calibration to standard space

    • typically CIE XYZ

  • Coordinate transforms through standard space

  • Gamut mapping

Device independence1
Device independence

  • Stone et. al. “Color Gamut Mapping and the Printing of Digital Color Images”, ACM Transactions on Graphics, 7(4) October 1998, pp. 249-292.

  • The following slides refer to their techniques.

Device to xyz
Device to XYZ

  • Sample gamut in device space on 8x8x8 mesh (7x7x7 = 343 cubes).

  • Measure (or model) device on mesh.

  • Interpolate with trilinear interpolation

    • for small mesh and reasonable function XYZ=f(device1, device2, device3) this approximates interpolating to tangent.

Xyz to device
XYZ to Device

  • Invert function XYZ=f(device1, device2, device3)

    • hard to do in general if f is ill behaved

    • At least make f monotonic by throwing out distinct points with same XYZ.

  • e.g. CMY device:

    • (continued)

Xyz to cmy

  • Invert function XYZ=f(c,m,y)

    • Given XYZ=[x,y,z] want to find CMY=[c,m,y] such that f(CMY)=XYZ

    • Consider X(c,m,y), Y(c,m,y), Z(c,m,y)

    • A continuous function on a closed region has max and min on the region boundaries, here the cube vertices. Also, if a continuous function has opposite signs on two boundary points, it is zero somewhere in between.

Xyz to cmy1

  • Given X0, find [c,m,y] such that f(c,m,y) = X0

  • if [ci,mi,yi] [cj,mj,yj] are vertices on a given cube, and U=X(c,m,y)- X0 has opposite sign on them, then it is zero in the cube. Similarly Y, Z. If find such vertices for all of X0,Y0,Z0, then the found cube contains the desired point. (and use interpolation). Doing this recursively will find the desired point if there is one.

Gamut mapping
Gamut Mapping

  • Criteria:

    • preserve gray axis of original image

    • maximum luminance contrast

    • few colors map outside destination gamut

    • hue, saturation shifts minimized

    • increase, rather than decrease saturation

    • do not violate color knowledge, e.g. sky is blue, fruit colors, skin colors

Gamut mapping1
Gamut Mapping

  • Special colors and problems

    • Highlights: this is a luminance issue so is about the gray axis

    • Colors near black: locus of these colors in image gamut must map into something reasonably similar shape else contrast and saturation is wrong

Gamut mapping2
Gamut Mapping

  • Special colors and problems

    • Highly saturated colors (far from white point): printers often incapable.

    • Colors on the image gamut boundary occupying large parts of the image. Should map inside target gamut else have to project them all on target boundary.





Gamut mapping3
Gamut Mapping

  • First try: map black points and fill destination gamut.


device gamut

image gamut


device gamut

translate Bito Bd

image gamut

bs (black shift)


device gamut

translate Bito Bd

image gamut

scale by csf


device gamut

translate Bito Bd

image gamut

scale by csf


Gamut mapping4
Gamut Mapping

Xd = Bd + csf R (Xi - Bi)

Bi = image black, Bd = destination black

R = rotation matrix

csf = contrast scaling factor

Xi = image color, Xd = destination color


Image colors near black outside of destination are especially bad: loss of detail, hue shifts due to quantization error, ...


  • Xd = Bd + csf R (Xi - Bi) + bs (Wd- Bd)

shift and scale alongdestination gray


Fig 14a, bs>0, csf small, image gamut maps entirelyinto printer gamut, but contrast is low.

Fig 14b, bs=0, csf large, more contrast, more colors inside printer gamut, butalso more outside.

Saturation control
Saturation control

  • “Umbrella transformation”

    [Rs Gs Bs] = monitor whitepoint

    [Rn Gn Bn] new RGB coordinates such that Rs + Gs + Bs =Rn + Gn + Bnand [Rn Gn Bn] maps inside destination gamut

    First map R Rs+G Gs+B Bsto R Rn+G Gn+B Bn

    Then map into printer coordinates

    Makes minor hue changes, but “relative” colors preserved. Achromatic remain achromatic.

Projective clipping
Projective Clipping

  • After all, some colors remain outside printer gamut

  • Project these onto the gamut surface:

    • Try a perpendicular projection to nearest triangular face in printer gamut surface.

    • If none, find a perpendicular projection to the nearest edge on the surface

    • If none, use closest vertex

Projective clipping1
Projective Clipping

  • This is the closest point on the surface to the given color

  • Result is continuous projection if gamut is convex, but not else.

    • Bad: want nearby image colors to be nearby in destination gamut.

Projective clipping2
Projective Clipping

  • Problems

    • Printer gamuts have worst concavities near black point, giving quantization errors.

    • Nearest point projection uses Euclidean distance in XYZ space, but that is not perceptually uniform.

      • Try CIELAB? SCIELAB?

      • Keep out of gamut distances small at cost of use of less than full printer gamut use.