1. The Point Processing of Images

2. Basic image processing operations
In this lecture we study techniques for resolution enhancement and reduction, and adjusting color distribution (histogram and LUT operations). We operate with 8 bpp gray-scale images each constrained to the range [0, 255], unless otherwise noted.
Changing resolution
A reduced resolution version of a given image is sometimes needed for a preview purpose, for example. A preview image (or thumbnail) must be small enough to allow fast access but also with sufficient quality so that the original image is still recognizable. A smaller resolution copy is also needed if the image is embedded into another application (or printed) using smaller size as the full resolution would allow
Resolution reduction is formally defined as follows: given an image of NXM pixels, generate an image of size N/c X M/c pixels (where c is a zooming factor) so that the visual content of the image is preserved as well as possible. There are two alternative strategies

3. Sub sampling Averaging
In both cases, the input image is divided into blocks of c x c pixels. For each block, one representative pixel is generated to the output image (Figure 2.1). In sub sampling, any of the input pixels are chosen, e.g. the upper leftmost pixel in the block. In averaging, the pixel depends on the values of all input pixels in the block. It could be chosen as the average, weighted average, or the median of the pixels. Averaging results in smoother image whereas sub sampling preserves more details.
Figure 2.1

4. I = imread('liftingbody.png');
fun=inline('mean2(x)');
I2 = blkproc(I,[8 8],fun);
imtool(I),imtool(I2,[])

5. A straightforward method simply takes copies of the input pixel but this results in a jagged (blocky) image where the pixels are clearly visible. A more sophisticated method known as bilinear interpolation generates the unknown pixel values by taking the linear combination of the four nearest known pixel values (Figure 2.2). The value of a pixel xi,j at the location (i,j) can be calculated as:
where a, b, c, d are the nearest known pixel values to x; i and j define the relative distance from a to x (varying from 0 to 1). The result is a smoother image compared to straightforward copying of a single pixel value. Examples of resolution reduction and resolution increasing by interpolation is illustrated in Figures 2.3 and 2.4.
Figure 2.2: Bilinear interpolation of pixel x using the four known pixels a, b, c, d.
I = imread('pout.tif');
x=imresize(I,[20 30]);
figure,imshow(I), imshow(x)

6. Figure 2.3: Example of resolution reduction 384256  4832: original image (left), resolution reduced by averaging (middle), and by sub sampling (right).
Figure 2.4: Bilinear interpolation using two 4832 images obtained using averaging (top), sub sampling (below). Interpolated images: original (left), 9264 (middle), 192128 (right).

7. Point Processing of Images
In a digital image, point = pixel.
Point processing transforms a pixel’s value as function of its value alone;
it does not depend on the values of the pixel’s neighbors.
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8. Spatial Domain Procedures that operate directly on pixels. where (x,y)
Origin x
(x,y)
Image f(x,y) y
T is an operator on f defined over some neighborhood of (x,y)

9. Example 1: Fixed Intensity Transformation
A 4x4, 4bits/pixel image passes through
an intensity transformation
1  round(0.0667) = 0;
3  round(0.6) = 1;
6  round(2.4) = 2;
7  round(3.2667) = 3;
8  round(4.2667) = 4;
9  round(5.4) =5;
10  round(6.6667) = 7;
11  round(8.0667) = 8;
The resulting
image is:

10. Example2: A 4x4, 4bits/pixel image passes through a point-wise intensity transformation given by
where α and β are unknown parameters. Only a few pixels are available in the input and the output images, as shown below. 3 15 ? 1 5 11 8 ?
T(r) r S

11. Constant addition and negation
The simplest form of global transform are constant addition (also known as DC-shift) and negation . The former is used to enhance dark images. The latter can be used for displaying medical images and photographs on screen with monochrome positive film with the idea of using the resulting negatives as normal slides.
Constant addition: f(x) = x + c (2.3)
Negation: f(x) = c - x (2.4)
% add 50 to every pixel
I = imread('rice.png');
Iplus50 = imadd(I,50);
imview(I), imview(Iplus50)
% Subtract a constant value from the rice image:
Iq = imsubtract(I,50);
imview(I), imview(Iq)

12. Negative Transformation
An image with gray level in the range [0, L -1] where L = 2n ; n = 1, 2… ( Number of bits/ pixel )
Negative transformation :
Reversing the intensity levels of an image.
Suitable for enhancing white or gray detail
embedded in dark regions of an image, especially
when the black area dominant in size.
f(x)= (2n -1) –x

13. Point Processing of Images
Brightness and contrast adjustment
Gamma correction
Histogram equalization

14. Point Processing - gamma - brightness original + brightness + gamma
histogram mod
- contrast
original
+ contrast
histogram EQ

15. Point Ops via Functional Mappings
, point operator J
Image:
Input
Output
I(r,c)
function, f
J(r,c)
Pixel:
The transformation of image I into image J is accomplished by replacing each input intensity, g, with a specific output intensity, k, at every location (r,c) where I(r,c) = g.
The rule that associates k with g is usually specified with a function, f, so that f (g) = k.
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16. Basic Transformation
Negative:
Log:
Inverse Log:
Power-law:

17. Point Ops via Functional Mappings
One-band Image
Either all 3 bands are mapped through the same function, f, or …
Three-band Image
… each band is mapped through a separate func-tion, fb.
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18. Point Operations using Look-up Tables
… then the LUT that implements f is a 256x1 array
whose (g +1)th value is k = f (g).
A look-up table (LUT) implements a functional mapping.
To remap a 1-band image, I, to J :

19. Point Operations using Look-up Tables
If I is 3-band, then
each band is mapped separately using the same LUT for each band or
each band is mapped using different LUTs – one for each band.

20. Point Operations = Look-up Table Ops
index
value
...
101
102
103
104
105
106 64 68 69 70 71
E.g.:
255
output value
127
127
255
input value
input
output
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21. Look-Up Tables a pixel with this value is mapped to this value
cell index
contents .
input
output
a pixel with this value
is mapped to this value
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22. How to Generate a Look-Up Table
For example:
Or in Matlab:
a = 2;
x = 0:255;
LUT = 255 ./ (1+exp(-a*(x-127)/32));
This is just one example.
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23. Point Processes: Increase Brightness
127
255
transform mapping
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24. Point Processes: Decrease Brightness
127
255
255-g
transform mapping
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25. Point Processes: Increase Contrast
127
255
transform mapping
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26. Point Processes: Decrease Contrast
127
255
transform mapping
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27. Point Processes: Contrast StretchMJ
127
255 mJ mI MI
transform mapping
I = imread('pout.tif');
J = imadjust(I,stretchlim(I),[]);
imview(I), imview(J)
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28. Example: Full-Scale Contrast Stretch
Full-scale contrast stretch of a 4x4, 4bits/pixel image
Find
4  round(0) = 0;
6  round(4.29) = 4;
7  round(6.43) = 6;
8  round(8.57) = 9;
9  round(10.71) = 11;
10  round(12.86) = 13;
11  round(15) = 15;
The resulting
image is:

29. General Idea: Make Best Use of the Dynamic Range
Piecewise- Linear Transformation Functions
Advantage:
The form of piecewise functions can be arbitrarily complex
Disadvantage:
Their specification requires considerably more user input
General Idea: Make Best Use of the Dynamic Range

30. Contrast Stretch General form:
Special case  Full-scale contrast stretch:
Typically used:

31. Point Processes: Increased Gamma
127
255
transform mapping
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32. Point Processes: Decreased Gamma
127
255 m M
transform mapping
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33. Highlighting a specific range of gray levels in an image
Gray-level slicing
Highlighting a specific range of gray levels in an image
Display a high value of all gray levels in the range of interest and a low value for all other gray levels
(a) transformation highlights range [A,B] of gray level and
reduces all others to a constant level
(b) transformation highlights range [A,B] but preserves all
other levels a b A B A B a b

34. Image Histogram 1 2 4 5
image
histogram

35. Example:
a 4x4, 4bits/pixel image 

36. Normalized Histogram
dividing each of histogram at gray level rk by the total number of pixels in the image, n
For k = 0,1,…,L-1
p(rk) gives an estimate of the probability of occurrence of gray level rk
The sum of all components of a normalized histogram is equal to 1

37. Dark image Components of histogram are concentrated on the low side of the gray scale.
Bright image
Components of histogram are concentrated on the high side of the gray scale.

38. The Histogram of a Grayscale Image
black marks pixels with intensity g
lower RHC: number of pixels with intensity g
16-level (4-bit) image

39. The Histogram of a Grayscale Image
Black marks pixels with intensity g
Plot of histogram:
number of pixels with intensity g

40. Algorithm1 : Calculation of an Image Histogram
create an array histogram with 2b elements
for all grey levels ,I, do
histogram[i]=0;
end for
for all pixel coordinates, x,y,do
increment histogram[f(x,y)] by 1
End for

41. Single -Band Histogram Calculator in Matlab
% Sample matlab code:
Function[f]= histogram(imgname)
img=imread(imgname);
figure; imshow(img);
[M,N]=size(img);
%computing histogram:
H=zeros(256,1);
for (i=1:M)
for (j=1:N)
f=double(img(i,j))+1; H(f)=H(f)+1;
end;
figure; bar(H);
Matlab: imhist( ): can compute and display histograms

42. Multi-Band Histogram Calculator in Matlab
function h=histogram(I)
[R C B]=size(I);
% allocate the histogram
h=zeros(256,1,B);
% range through the intensity values
for g=0:255
h(g+1,1,:) = sum(sum(I==g)); % accumulate
end
return;

43. Multi-Band Histogram Calculator in Matlab
function h=histogram(I)
[R C B]=size(I);
% allocate the histogram
h=zeros(256,1,B);
% range through the intensity values
for g=0:255
h(g+1,1,:) = sum(sum(I==g)); % accumulate
end
return;
Loop through all intensity levels (0-255)
Tag the elements that have value g.
The result is an RxCxB logical array that has a 1 wherever I(r,c,b) = g and 0’s everywhere else.
Compute the number of ones in each band of the image for intensity g.
Store that value in the 256x1xB histogram at h(g+1,1,b).
sum(sum(I==g)) computes one number for each band in the image.
If B==3, then h(g+1,1,:) contains 3 numbers: the number of pixels in bands 1, 2, & 3 that have intensity g.

44. The Histogram of a Grayscale Image
Let I be a 1-band (grayscale) image.
I(r,c) is an 8-bit integer between 0 and 255.
Histogram, hI, of I:
a 256-element array, hI
hI (g), for g = 1, 2, 3, …, 256, is an integer
hI (g) = number of pixels in I that have value g-1.

45. The Histogram of a Grayscale Image
Luminosity

46. The Histogram of a Color Image
If I is a 3-band image (truecolor, 24-bit)
then I(r,c,b) is an integer between 0 and 255.
Either I has 3 histograms:
hR(g+1) = # of pixels in I(:,:,1) with intensity value g
hG(g+1) = # of pixels in I(:,:,2) with intensity value g
hB(g+1) = # of pixels in I(:,:,3) with intensity value g
or 1 vector-valued histogram, h(g,1,b) where
h(g+1,1,1) = # of pixels in I with red intensity value g
h(g+1,1,2) = # of pixels in I with green intensity value g
h(g+1,1,3) = # of pixels in I with blue intensity value g

47. The Histogram of a Color Image
Luminosity
There is one histo-gram per color band
R, G, & B. Luminosity histogram is from 1 band = (R+G+B)/3

48. Value or Luminance Histograms
The value histogram of a 3-band (truecolor) image, I, is the histogram of the value image,
Where R, G, and B are the red, green, and blue bands of I.
The luminance histogram of I is the histogram of the luminance image,

49. Cumulative Histogram
It records the cumulative frequency distribution of grey levels in an image. The cumulative frequency of a grey level, i , is the number of times that a grey level less than or equal to i occurs in an image. Cumulative frequencies, ci are computed from the histogram counts, hi , using
The normalized value of cumulative frequency can be obtained by dividing the above equation by n

50. Example 2 3 4 5 Grey level i 1 2 3 4 5 6 7 8 9 ni cj 11 15 16 6/161 2 3 4 5 6 7 8 9 ni cj 11 15 16
6/16
5/16
4/16
1/16
11/16
15/16

51. Histogram Equalization
Sometimes the histogram of an image contains mostly dark pixels; this is the case of an insufficiently exposed photograph. The image can be enhanced by constant addition but histogram equalization is generally more efficient technique for this purpose
The idea of the method is to spread the histogram as evenly as possible over the full intensity scale. This is done by calculating cumulative sums of the pixel samples for each gray level value x in the histogram. The sum implies the number of gray levels that should be allocated to the range [0, r], and is proportional to the cumulative frequency C(r), and to the total number of gray levels L

52. Histogram Equalization
Where
B= Number of bit/ pixel ( B=1,2,………,8)
r= grey levels of the input image and
S= grey levels of the output image

53. Histogram Equalization Example
Intensity
Number of pixels
Number of pixels

54. Histogram Equalization Example

55. Histogram Equalization Example
Suppose that a 3-bit image (L=8) of size 64 × 64 pixels (MN = 4096) has the intensity distribution shown in following table.
Get the histogram equalization transformation function and give the ps(sk) for each sk.

56. input histogram
equalized histogram

57. Example A 4x4, 4bits/pixel image
First try: full-scale contrast stretch
2  round(0) = 0;
3  round(1.67) = 2;
8  round(10.00) = 10;
9  round(11.67) = 12;
10  round(13.33) = 13;
11  round(15) = 15;
The resulting
image is:

58. Example: Histogram Change
full-scale contrast stretch
original 
big gap

59. Cumulative Histogram
H(k)
Q(k)

60. Intermediate Image
intermediate image
original 

61. Full-Scale Contrast Stretch of Intermediate Image
2  round(0) = 0;
6  round(4.29) = 4;
9  round(7.50) = 8;
12  round(10.71) = 11;
14  round(12.86) = 13;
16  round(15) = 15;
final result:
histogram equalized image

62. direct full-scale contrast stretch
Histogram Comparison
original
direct full-scale contrast stretch
histogram-equalized
more equalized

63. Given an image as below, derive the intensity mapping that will as best
as possible equalize the image histogram. The image histogram is as
shown on the right. There are 8 possible grey scale levels from 0 to 7. 4 3 5