Xu ly anh va thi giac may tinh

1.  
Prof. Duong Anh Duc

2.  To process an image so that output is “visually better” than the
input, for a specific application.
 Enhancement is therefore, very much dependent on the particular
problem/image at hand.
 Enhancement can be done in either:
o Spatial domain: operate on the original image
g (m, n) = T [f (m, n)]
o Frequency domain: operate on the DFT of the original image
G (u, v) = T [F (u, v)]
where F(u, v) = F [f(m, n)], and G(u, v) = F [g(m, n)]
2

3. 3
Image Enhancement Techniques
Point Operations Mask Operations Transform Operations Coloring Operations
• Image Negative
• Contrast
Stretching
• Compression of
dynamic range
• Graylevel slicing
• Image
Subtraction
• Image Averaging
• Histogram
operations
• Smoothing
operations
• Median Filtering
• Sharpening
operations
• Derivative
operations
• Histogram
operations
• Low pass
Filtering
• Hi pass Filtering
• Band pass
Filtering
• Homomorphic
Filtering
• Histogram
operations
• False Coloring
• Full color
Processing

4. 4
 Output pixel value g(m, n) at pixel (m, n) depends only on the input
pixel value at f(m, n) at (m, n) (and not on the neighboring pixel
values).
 We normally write s = T(r), where s is the output pixel value and r is
the input pixel value.
 T is any increasing function that maps [0,1] into [0,1].

5. 5
T(r) = s = L – 1 – r, L: max grayvalue

6. 6

7.  Increase the dynamic range of grayvalues in the input image.
 Suppose you are interested in stretching the input intensity values
in the interval [r1, r2]:
 Note that (r1 – r2) < (s1 – s2). The grayvalues in the range [r1, r2] is
stretched into the range [s1, s2]. 7

8. 8
 Special cases:
o Thresholding or binarization
 r1 = r2 , s1 = 0 and s2 = 1
o Useful when we are only
interested in the shape of the
objects and on their actual
grayvalues.

9. 9

10.  Special cases (cont.):
o Gamma correction:
s1 = 0, s2 = 1 and
10
2
21
12
1
1
,1
,
,0
rr
rrr
rr
rr
rr
rT
g

11. 11
Gamma correction

12.  When the dynamic range of the input grayvalues is large
compared to that of the display, we need to “compress” the
grayvalue range --- example: Fourier transform magnitude.
 Typically we use a log scale.
s = T(r) = c log(1+ r )
12

13. 13
Saturn Image Mag. Spectrum Mag. Spectrum
in log scale

14.  Graylevel Slicing: Highlight a specific range of grayvalues.
14

15.  Example:
15
Original Image
Highlighted Image (no background)
Highlighted Image (with background)

16.  Bitplane Slicing: Display the different bits as individual binary
images.
16

17. 17

18.  In this case, the difference between two “similar” images is
computed to highlight or enhance the differences between them:
g(m, n) = f1(m, n) – f2(m, n)
 It has applications in image segmentation and enhancement
18

19. 19
f1(m, n): Image before dye injection
f2(m, n): Image after dye injection
g(m, n): Image after dye injection,
followed by subtraction

20.  Noise is any random (unpredictable) phenomenon that
contaminates an image.
 Noise is inherent in most practical systems:
o Image acquisition
o Image transmission
o Image recording
 Noise is typically modeled as an additive process:
g (m,n) = f (m, n) + (m, n)
20
Noisy
Image
Noise-free
Image
Noise

21.  The noise h(m, n) at each pixel (m, n) is modeled as a random
variable.
 Usually, h(m, n) has zero-mean and the noise values at different
pixels are uncorrelated.
 Suppose we have M observations {gi (m, n)}, i =1, 2, …, M, we
can (partially) mitigate the effect of noise by “averaging”
21
M
i
i nmg
M
nmg
1
,
1
,

22.  In this case, we can show that:
 Therefore, as the number of observations increases (M ), the
effect of noise tends to zero.
22
nm
M
nmg
nmfnmgE
,Var
1
,Var
,,

23. 23
Noise-free Image Noisy Image
Noise Variance = 0.05

24. 24
M =2 M =5

25. 25
M =10 M =25

26. 26
M =50 M =100

27. 27

28.  The histogram of a digital image with grayvalues r0, r1, …, rL – 1 is
the discrete function
 The function p(rk) represents the fraction of the total number of
pixels with grayvalue rk.
 Histogram provides a global description of the appearance of the
image.
28
imageinpixels#total
ewith valupixels#
where,
n
rn
n
n
rp kkk
k

29.  If we consider the grayvalues in the image as realizations of a random
variable R, with some probability density, histogram provides an
approximation to this probability density. In other words,
Pr[R = rk] p(rk)
29

30.  The shape of a histogram provides useful information for contrast
enhancement.
30

31.  The shape of a histogram provides useful information for contrast
enhancement.
31

32. The original image displayed
The stretched image
0 0.5 1
0
5000
10000
The histogram of the original image
0 0.5 1
0
5000
10000
The stretched histogram
32

33.  Idea: find a non-linear transformation
g = T (f )
to be applied to each pixel of the input image f(x, y), such that a
uniform distribution of gray levels in the entire range results for the
output image g(x, y).
33

34.  Let us assume for the moment that the input image to be
enhanced has continuous grayvalues, with r = 0 representing black
and r = 1 representing white.
 We need to design a grayvalue transformation s = T(r), based on the
histogram of the input image, which will enhance the image.
 As before, we assume that:
o T(r) is a monotonically increasing function for 0 r 1 (preserves
order from black to white)
o T(r) maps [0,1] into [0,1] (preserves the range of allowed
grayvalues).
34

35. 35

36.  Let us denote the inverse transformation by r = T – 1(s). We assume
that the inverse transformation also satisfies the above two conditions.
 We consider the grayvalues in the input image and output image
as random variables in the interval [0, 1].
 Let pin(r) and pout(s) denote the probability density of the grayvalues in
the input and output images.
36

37.  If pin(r) and T(r) are known, and T– 1(s) satisfies condition 1, we can
write (result from probability theory):
 One way to enhance the image is to design a transformation T(.)
such that the grayvalues in the output is uniformly distributed in
[0, 1], i.e.
pout(s)=1, 0 s 1
37
sTr
inout
ds
dr
rpsp
1

38.  In terms of histograms, the output image will have all grayvalues
in “equal proportion.”
 This technique is called histogram equalization.
 Consider the transformation
38
r
o
rdwwprTs 10,in

39.  Note that this is the cumulative distribution function (CDF) of pin(r) and
satisfies the previous two conditions.
 From the previous equation and using the fundamental theorem of
calculus,
39
rp
dr
ds
in

40.  Therefore, the output histogram is given by
 The output probability density function is uniform, regardless of the
input.
40
10for,11
1
1
1
s
rp
rpsp
sTr
sTrin
inout

41.  Thus, using a transformation function equal to the CDF of input
grayvalues r, we can obtain an image with uniform grayvalues.
 This usually results in an enhanced image, with an increase in the
dynamic range of pixel values.
41

42. 42
Original image pout
after histogram equalization

43. 43

44.  For images with discrete grayvalues, we have
o L: Total number of graylevels
o nk: Number of pixels with grayvalue rk
o n: Total number of pixels in the image
44
10and,10for,in Lkr
n
n
rp k
k

45.  The discrete version of the previous transformation based on CDF is
given by:
45
10for,
0
in
0
Lkrp
n
n
rTs
k
i
i
k
i
i
kk

46.  Consider an 8-level 64 x 64
image with grayvalues
(0, 1, …, 7).
 The normalized grayvalues
are (0, 1/7, 2/7, …, 1). The
normalized histogram is given in
the table:
k rk nk
p(rk)=
nk / n
0 0 790 0.19
1 1/7 1023 0.25
2 2/7 850 0.21
3 3/7 656 0.16
4 4/7 329 0.08
5 5/7 245 0.06
6 6/7 122 0.03
7 1 81 0.02
46

47. 47

48.  Applying the previous
transformation, we have
(after rounding off to nearest
graylevel):
 Notice that there are only five
distinct graylevels ---
(1/7, 3/7, 5/7, 6/7, 1) in the
output image. We will relabel
them as (s0, s1, …, s4).
48
100.1
198.0
195.0
7
689.0
7
681.0
7
565.0
7
344.0
7
119.0
7in1in0in
7
0
in77
6in1in0in
6
0
in66
5in1in0in
5
0
in55
4in1in0in
4
0
in44
3in1in0in
3
0
in33
2in1in0in
2
0
in22
1in0in
1
0
in11
0in
0
0
in00
rprprprprTs
rprprprprTs
rprprprprTs
rprprprprTs
rprprprprTs
rprprprprTs
rprprprTs
rprprTs
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i






49.  With this transformation, the output image will have histogram
49
k sk nk p(sk)=nk / n
0 1/7 790 0.19
1 3/7 1023 0.25
2 5/7 850 0.21
3 6/7 985 0.24
4 1 448 0.11

50. 50
 Note that the histogram of output image is only approximately, and not
exactly, uniform. This should not be surprising, since there is no result
that claims uniformity in the discrete case.

51. 51
Original image Equalized image

52. 52
Histogram of original image Histogram of equalized image

53. 53
Original image Equalized image

54. 54
Histogram of original image Histogram of equalized image

55. 55
Original image Equalized image

56. 56
Histogram of original image Histogram of equalized image

57. 57
Original image Equalized image

58. 58
Histogram of original image Histogram of equalized image

59.  Histogram equalization may not always produce desirable results,
particularly if the given histogram is very narrow. It can produce false
edges and regions. It can also increase image “graininess” and
“patchiness.”
59

60.  Histogram equalization yields an image whose pixels are (in theory)
uniformly distributed among all graylevels.
 Sometimes, this may not be desirable. Instead, we may want a
transformation that yields an output image with a prespecified
histogram. This technique is called histogram specification.
 Again, we will assume, for the moment, continuous grayvalues.
60

61.  Suppose, the input image has probability density pin(r). We want to
find a transformation z = H(r), such that the probability density of the
new image obtained by this transformation is pout(z), which is not
necessarily uniform.
 First apply the transformation
This gives an image with a uniform probability density.
61
*10,in
r
o
rdwwprTs

62.  If the desired output image were available, then the following
transformation would generate an image with uniform density:
 From the grayvalues we can obtain the grayvalues z by using the
inverse transformation, z = G– 1( ).
62
**10,out
z
o
zdwwpzG

63.  If instead of using the grayvalues n obtained from (**), we use the
grayvalues s obtained from (*) above (both are uniformly distributed!),
then the point transformation
z = H(r) = G– 1[T(r)]
will generate an image with the specified density pout(z), from an input
image with density pin(r)
 For discrete graylevels, we have
63
10for,
and10for,
0
out
0
LkzpzG
Lk
n
n
rTs
k
i
ikk
k
i
i
kk

64.  If the transformation zk G(zk) is one-to-one, the inverse
transformation sk G– 1(sk), can be easily determined, since we are
dealing with a small set of discrete grayvalues.
 In practice, this is not usually the case (i.e., zk G(zk) is not one-to-
one) and we assign grayvalues to match the given histogram, as
closely as possible.
64

65.  Consider the previous 8-graylevel 64 x 64 image histogram:
65
k rk nk p(rk)=nk/n
0 0 790 0.19
1 1/7 1023 0.25
2 2/7 850 0.21
3 3/7 656 0.16
4 4/7 329 0.08
5 5/7 245 0.06
6 6/7 122 0.03
7 1 81 0.02

66.  It is desired to transform this image into a new image, using a
transformation z=H(r)= G– 1[T(r)], with histogram as specified below:
66
k zk pout(zk)
0 0 0.00
1 1/7 0.00
2 2/7 0.00
3 3/7 0.15
4 4/7 0.20
5 5/7 0.30
6 6/7 0.20
7 1 0.15

67.  The transformation T(r) was obtained earlier (reproduced below):
67
ri sk nk p(sk)
r0 s0 = 1/7 790 0.19
r1 s1 = 3/7 1023 0.25
r2 s2 = 5/7 850 0.21
r3,r4 s3 = 6/7 985 0.24
r5, r6,r7 s4=1 448 0.11

68.  Next we compute the transformation G as before
100.1
7
685.0
7
565.0
7
235.0
7
115.0
000.0
000.0
000.0
7out1out0out
7
0
out77
6out1out0out
6
0
out66
5out1out0out
5
0
out55
4out1out0out
4
0
out44
3out1out0out
3
0
out33
2out1out0out
2
0
out22
1out0out
1
0
out11
0out
0
0
out00
zpzpzpzpzG
zpzpzpzpzG
zpzpzpzpzG
zpzpzpzpzG
zpzpzpzpzG
zpzpzpzpzG
zpzpzpzG
zpzpzG
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i





68

69.  Notice that G is not invertible. But we will do the best possible by
setting
o G– 1 (0) = ? (This does not matter since s 0)
o G– 1 (1/7) = 3/7
o G– 1 (2/7) = 4/7 (This does not matter since s 2/7)
o G– 1 (3/7) = 4/7 (This is not defined, but we use a close match)
o G– 1 (4/7) = ? (This does not matter since s 4/7)
o G– 1 (5/7) = 5/7
o G– 1 (6/7) = 6/7
o G– 1 (1) = 1
69

70.  Combining the two transformation T and G– 1, we get our required
transformation H
70
r T(r)=s s G–1(s)=z
r G–1
[T(r)]=H(r)=z
r0 = 0 1/7 0 ? r0 = 0 z3= 3/7
r1 = 1/7 3/7 1/7 3/7 r1 = 1/7 z4= 4/7
r2 = 2/7 5/7 2/7 4/7 r2 = 2/7 z5= 5/7
r3 = 3/7 6/7 3/7 4/7 r3 = 3/7 z6= 6/7
r4 = 4/7 6/7 4/7 ? r4 = 4/7 z6= 6/7
r5 = 5/7 1 5/7 5/7 r5 = 5/7 z7= 1
r6 = 6/7 1 6/7 6/7 r6 = 6/7 z7= 1
r7 = 1 1 1 1 r7 = 1 z7= 1

71.  Applying the transformation H to the original image yields an image
with histogram as below:
71
k zk nk
nk/n
(actual hist.)
pout(zk)
(spec. hist.)
0 0 0 0.00 0.00
1 1/7 0 0.00 0.00
2 2/7 0 0.00 0.00
3 3/7 790 0.19 0.15
4 4/7 1023 0.25 0.20
5 5/7 850 0.21 0.30
6 6/7 985 0.24 0.20
7 1 448 0.11 0.15

72. 72

73.  Again, the actual histogram of the output image does not exactly but
only approximately matches with the specified histogram. This is
because we are dealing with discrete histograms.
73

74. 74
Original image and its histogram

75. 75
Histogram equalized image Actual histogram of output

76. 76Histogram specified image, Actual Histogram, and Specified Histogram

77.  Used to enhance details over small portions of the image.
 Define a square or rectangular neighborhood, whose center moves
from pixel to pixel.
 Compute local histogram based on the chosen neighborhood for
each point and apply a histogram equalization or histogram
specification transformation to the center pixel.
 Non-overlapping neighborhoods can also be used to reduce
computations. But this usually results in some artifacts
(checkerboard like pattern).
77

78.  Another use of histogram information in image enhancement is the
statistical moments associated with the histogram (recall that the
histogram can be thought of as a probability density function).
 For example, we can use the local mean and variance to
determine the local brightness/contrast of a pixel. This information
can then be used to determine what, if any transformation to apply
to that pixel.
 Note that local histogram based operations are non-uniform in the
sense that a different transformation is applied to each pixel.
78