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Histogram—Representation of Color Feature in Image Processing Yang, Li

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1 Histogram—Representation of Color Feature in Image Processing Yang, Li
Part II: Histogram—Representation of Color Feature in Image Processing Yang, Li Good afternoon, the research interest in this part focus on the histogram, the representation of color feature in image processing. Let have a look at the recent development in this field.

2 Structure: Color histogram descriptor Color cooccurrence histogram
HSV space segmentation

3 Color histogram descriptor <C, M,{Ci}, {H(Ci)}>
A weighted Euclidean distance of colors is then taken to match color histograms, if X is the query histogram and Y is the histogram of an item in the database, then the similarity between X and Y is given by:||Z||=Z'AZ Searching and locating multimedia data need a good description and representation of multimedia information. Mahood and Tanveer propose two ways of capturing color content in image called Color Histogram “and Region Color.” The descriptors are suitable for a wide variety of applications requiring image-to-image matching and object-to-image matching. *Color histogram is a record of the number of image or region pixels of a specified color. It can be described as a tuple C is chosen color space, Ci is the color quantization cell, M is the number of of color. H(Ci) = Ni/N, and Ni is the number of image pixels whose color falls in the bin Ci and N is the total number of image pixels. A is a symmetric color similarity matrix in which a(i,j) indicates the similarity of colors i and j in the histogram. This method accounts for both the perceptual distance between different pairs of colors(e.g. orange and red are less different than orange and blue), and the difference in the amounts of a given color(e.g. particular shade of red).

4 Region-Color Descriptor
The region information in both query and image can be represented using the region color descriptor with the difference that the number of regions and their respective colors are different. For a set of image regions though, the color label or indes assigned to the corresponding regions must be the same. When the goal is to find a query embedded in an image based on the color of one or more of its regions, we need a robust description of a region color that is illumination and pose-invariant to account for the different appearances of the query object in images of the database. Ri is a region in the image and Cj is its color description. The region information in both query and image can be represented using the region color descriptor with the difference that the number of regions and their respective colors are different. For a set of query color regions to be accurately detected in the set of image regions though, the color label or index assigned to the corresponding regions must be the same. In other words, any method of describing the region’s color must be illumination and pose-invariant to account for the various appearances of the query object under different illumination conditions.

5 Color Cooccurrence Histogram
Each model image is represented as a color CH. The color CH holds the number of occurrences of pairs of color pixels C1=(R1,G1,B1) and C2=(R2,G2,B2) separated by a vector in the image y plane(x, y). The color cooccurrence histogram keeps track of the number of pairs of certain colored pixels that occur at certain separation distances in image space. The color cooccurrence histogram adds geometric information to the normal color histogram. The model cooccurrence histogram are matched to subregions in test images to find the object. By adjusting the number of colors and the number of distances used in the CH, the tolerance of the algorithm to the changes in lighting, viewpoint, and the flexibility of the object can be adjusted. C2 y C1 x

6 Quantize colors into a set of representative colors C=(c1,c2,…,c )
Assumption: ignore the direction of(x, y) and keep track of only magnitude d= Quantize colors into a set of representative colors C=(c1,c2,…,c ) Quantize the distances into a set of distance ranges D={[0,1),[1,2),…,[ -1, )}. CH is represented by CH (i, j, k). Where i and j index the two colors from set C, and k indexes the distance range from set D. In searching an image for an object, the image is scanned for a rectangular window that gives a CH similar to one of the training CHs.

7 It indicates how well the image CH accounts for the model CH.
The image and model CHs are compared by computing their intersection. The intersection is: It indicates how well the image CH accounts for the model CH. If the image accounts for all the entries in the model CH, then the intersection will be equal to the sum of all the entries in in the model CH, or Imm. If the

8 The conversion form RGB to HSV is performed with the equations:
Recursive HSV-space segmentation to extract regions within the image which contain perceptually similar color The conversion form RGB to HSV is performed with the equations: Where H=H1 if B<=G; otherwise H=360-H1; Where R,G,B are the red, green, and blue component values which exist in the range [0,255]. H: hue S:saturate V:value

9 If the colors with value<25%, they are classified as black; if the color with saturation<20% and value>75%, they can be classified as white. VAL white Green 120 Yellow 60 Blue 240 Magenta 300 BRIGHT CHROMATIC CHROMATIC black hue SAT

10 White Black chromatic yes yes no Value<25 IMAGE no no bright
SAT<20 Value<25 IMAGE no no Value>75 SAT>=20 chromatic yes bright chromatic Build Hue Histogram Build Saturation Histogram Determine N peaks Determine M peaks Each image is examined to classify the pixels into one of the four categories: Black, White, Bright chromatic, and Chromatic regions. If the If the colors with value<25%, they are classified as black; if the color with saturation<20% and value>75%, they can be classified as white. All the remaining pixels fall in the chromatic region of the HSV cone, hue and saturation histogram are built and the corresponding peaks are threshhold to segment the colors. Threshhold Peak i Threshhold Peak j no no i=n? i=m ? yes yes END


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