Transcript Digital Image Processing

Digital Image Processing
Image Enhancement
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Image Enhancement
 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:
– Spatial domain: operate on the original image
g(m,n) = T[f(m,n)]
– 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)],
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Image Enhancement Techniques
Point Operations
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Image Negative
Contrast
Stretching
Compression of
dynamic range
Graylevel slicing
Image
Subtraction
Image Averaging
Histogram
operations
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Mask Operations
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Smoothing
operations
Median Filtering
Sharpening
operations
Derivative
operations
Histogram
operations
Transform Operations
Coloring Operations
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Low pass
Filtering
Hi pass Filtering
Band pass
Filtering
Homomorphic
Filtering
Histogram
operations
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False Coloring
Full color
Processing
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Point Operations
 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].
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Image Negative
T(r) = s = L-1-r, L: max grayvalue
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Negative Image
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Contrast Stretching

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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].
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Contrast Stretching
 Special cases:
– Thresholding or
binarization
r1 = r2 , s1 = 0 and s2 = 1
– Useful when we are
only interested in the
shape of the objects
and on on their actual
grayvalues.
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Contrast Stretching
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Contrast Stretching
 Special cases (cont.):
– Gamma correction:
S1 = 0, S2 = 1 and
0, r  r1

g

 r  r1 
 , r1  r  r2
T r   
 r2  r1 
1, r  r2

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Contrast Stretching
Gamma correction
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Compression of Dynamic
Range
 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)
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Compression of Dynamic
Range
Saturn Image
Mag. Spectrum
Mag. Spectrum
in log scale
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Compression of Dynamic
Range
 Graylevel Slicing: Highlight a specific
range of grayvalues.
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Compression of Dynamic
Range
 Example:
Highlighted Image (no background)
Original Image
Highlighted Image (with background)
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Compression of Dynamic
Range
 Bitplane Slicing: Display the different
bits as individual binary images.
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Compression of Dynamic
Range
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Image Subtraction
 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
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Example: Mask mode radiography
f1(m, n): Image before dye injection
f2(m, n): Image after dye injection
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g(m, n): Image after dye injection,
followed by subtraction
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Image Averaging for noise
reduction
 Noise is any random (unpredictable)
phenomenon that contaminates an image.
 Noise is inherent in most practical systems:
– Image acquisition
– Image transmission
– Image recording
 Noise is typically modeled as an additive process:
g(m,n) = f(m,n) + (m,n)
Noisy
Image
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Noise-free
Image
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Noise
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Image Averaging for noise
reduction
 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”
1
g m, n  
M
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M
 g m, n
i 1
i
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Image Averaging for noise
reduction
 In this case, we can show that:
E g m, n   f m, n 
1
Var g m, n  
Var  m, n 
M
 Therefore, as the number of observations
increases (M  ), the effect of noise
tends to zero.
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Image Averaging Example
Noise-free Image
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Noisy Image
Noise Variance = 0.05
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Image Averaging Example
M =2
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M =5
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Image Averaging Example
M =10
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M =25
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Image Averaging Example
M =50
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M =100
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Some Averaging Filters
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Histograms
 The histogram of a digital image with
grayvalues r0, r1, …, rL-1 is the discrete function
 nk
nk
prk   , where 
n
n
# pixels with value rk
total# pixelsin image
 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.
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Histograms
 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)
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Some Typical Histograms
 The shape of a histogram provides useful
information for contrast enhancement.
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Some Typical Histograms (cont.)
 The shape of a histogram provides useful
information for contrast enhancement.
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Example: Histogram Stretching
The original image displayed
The histogram of the original image
10000
5000
0
0
The stretched image
0.5
1
The stretched histogram
10000
5000
0
0
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Histogram Equalization
 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).
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Histogram Equalization
 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:
– T(r) is a monotonically increasing function for 0r1
(preserves order from black to white)
– T(r) maps [0,1] into [0,1] (preserves the range of
allowed grayvalues).
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Histogram Equalization
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Histogram Equalization
 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.
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Histogram Equalization
 If pin(r) and T(r) are known, and T-1(s) satisfies
condition 1, we can write (result from probability
theory):
dr 

pout s    pin r  
ds  r T 1 s 

 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
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Histogram Equalization
 In terms of histograms, the output image
will have all grayvalues in “equal
proportion.”
 This technique is called histogram
equalization.
 Consider the transformation
r
s  T r    pin wdw,
0  r 1
o
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Histogram Equalization
 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,
ds
 pin r 
dr
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Histogram Equalization
 Therefore, the output histogram is given by

1 
pout s    pin r 



p
r
in

 r T 1  s 
 1r T 1 s   1, for 0  s  1
 The output probability density function is
uniform, regardless of the input.
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Histogram Equalization
 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.
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Example: Histogram Equalization
Original image Pout
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after histogram equalization
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Example: Histogram Equalization
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Histogram Equalization
 For images with discrete grayvalues, we
have
nk
pin rk   , for 0  r  1, and 0  k  L  1
n
– L: Total number of graylevels
– nk: Number of pixels with grayvalue rk
– n: Total number of pixels in the image
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Histogram Equalization
 The discrete version of the previous
transformation based on CDF is given by:
k
k
ni
sk  T rk      pin ri , for 0  k  L  1
i 0 n
i 0
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Example
 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:
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k
rk
nk
0
1
0
1/7
790
1023
2
3
4
5
6
7
2/7
3/7
4/7
5/7
6/7
1
850
656
329
245
122
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p(rk)=
nk/n
0.19
0.25
0.21
0.16
0.08
0.06
0.03
0.02
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Example(cont.)
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Example(cont.)
 Applying the previous
transformation, we
have (after rounding off
to nearest graylevel):
0
s0  T r0    pin ri   pin r0   0.19  1
i 0
7
1
s1  T r1    pin ri   pin r0   pin r1   0.44  3
i 0
7
2
s2  T r2    pin ri   pin r0   pin r1   pin r2   0.65  5
i 0
7
3
s3  T r3    pin ri   pin r0   pin r1     pin r3   0.81  6
i 0
 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).
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4
s4  T r4    pin ri   pin r0   pin r1     pin r4   0.89  6
i 0
5
s5  T r5    pin ri   pin r0   pin r1     pin r5   0.95  1
i 0
6
s6  T r6    pin ri   pin r0   pin r1     pin r6   0.98  1
i 0
7
s7  T r7    pin ri   pin r0   pin r1     pin r7   1.00  1
i 0
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Example(cont.)
 With this transformation, the output image
will have histogram
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k
sk
nk
p(sk)=
nk/n
0
1
2
1/7
3/7
5/7
790
1023
850
0.19
0.25
0.21
3
4
6/7
1
985
448
0.24
0.11
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Example(cont.)
 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.
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Example(cont.)
Original image
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Equalized image
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Example(cont.)
Histogram of original image
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Histogram of equalized image
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Example(cont.)
Original image
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Equalized image
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Example(cont.)
Histogram of original image
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Histogram of equalized image
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Example(cont.)
Original image
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Equalized image
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Example(cont.)
Histogram of original image
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Histogram of equalized image
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Example(cont.)
Original image
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Equalized image
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Example(cont.)-Histograms
Histogram of original image
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Histogram of equalized image
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Histogram Equalization
 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.”
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Histogram Specification
 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.
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Histogram Specification
 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
r
s  T r    pin wdw,
0  r 1
*
o
This gives an image with a uniform probability density.
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Histogram Specification
 If the desired output image were available,
then the following transformation would
generate an image with uniform density:
z
  Gz    pout wdw,
0  z  1 * *
o
 From the grayvalues  we can obtain the
grayvalues z by using the inverse
transformation, z = G-1().
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Histogram Specification
 If instead of using the grayvalues  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
k
ni
sk  T rk    , for 0  k  L  1 and
i 0 n
k
 k  Gzk    pout zi , for 0  k  L  1
i 0
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Histogram Specification
 If the transformation zk  G(zk) is one-toone, 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.
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Ex.: Histogram Specification
 Consider the previous 8-graylevel 64 x 64 image
histogram:
k
r
n
p(r )=n /n
k
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k
k
k
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
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Ex.: Histogram Specification
 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:
k
z
p (z )
k
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out
k
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
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Ex.: Histogram Specification
 The transformation T(r) was obtained earlier
(reproduced below):
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r i 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
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Ex.: Histogram Specification
 Next we compute
the transformation
G as before
0
 0  G z0    pout zi   pout z0   0.00  0
i 0
1
 1  G z1    pout zi   pout z0   pout z1   0.00  0
i 0
2
 2  G z 2    pout zi   pout z0   pout z1   pout z2   0.00  0
i 0
3
 3  G z3    pout zi   pout z0   pout z1     pout z3   0.15  1 7
i 0
4
 4  G z 4    pout zi   pout z0   pout z1     pout z 4   0.35  2 7
i 0
5
 5  G z5    pout zi   pout z0   pout z1     pout z5   0.65  5 7
i 0
6
 6  G z6    pout zi   pout z0   pout z1     pout z6   0.85  6 7
i 0
7
 7  G z7    pout zi   pout z0   pout z1     pout z7   1.00  1
i 0
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Ex.: Histogram Specification
 Notice that G is not invertible. But we will do the
best possible by setting
G-1(0) = ?
G-1(1/7) = 3/7
G-1(2/7) = 4/7
G-1(3/7) = 4/7
G-1(4/7) = ?
G-1(5/7) = 5/7
G-1(6/7) = 6/7
G-1(1) = 1
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(This does not matter since s0)
(This does not matter since s2/7)
(This is not defined, but we use a close match)
(This does not matter since s4/7)
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Ex.: Histogram Specification
 Combining the two transformation T and G-1, we get our
required transformation H
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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
r 7 = 1  z 7= 1
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Ex.: Histogram Specification
 Applying the transformation H to the original image yields
an image with histogram as below:
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k
zk
nk
nk/n
(actual hist.)
pout(zk)
(spec. hist.)
0
1
2
3
4
5
6
7
0
1/7
2/7
3/7
4/7
5/7
6/7
1
0
0
0
790
1023
850
985
448
0.00
0.00
0.00
0.19
0.25
0.21
0.24
0.11
0.00
0.00
0.00
0.15
0.20
0.30
0.20
0.15
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Ex.: Histogram Specification
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Ex.: Histogram Specification
 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.
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Example
Original image and its histogram
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Example
Histogram equalized image
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Actual histogram of output
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Example
Histogram specified image, Actual Histogram, and Specified Histogram
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Enhancement Using
Local Histogram
 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).
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Enhancement Using
Local Histogram
 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.
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