2017-12-06
((resource:1stpage.jpeg|Resource Description for 1stpage.jpeg))
((resource:2ndpage.jpeg|Resource Description for 2ndpage.jpeg))
For second image,
For k=3, 9 (all) pixel values are zero
For k=1, only 1 pixel value is one
For k=2, 8 pixel values are one
So, mean X_ = (0 + 1 + 8)/(3*3*3) = 1/3
Now, Let’s calculate, ∑i=1 to r ∑ j=1 to c (Xi,j,k − X_) for k = 1, 2, 3
For k = 3, this value is 9 * (1/3)^2 = 1
For k = 2, this value is 8 * (1/3)^2 + (2/3)^2 = 12/9
For k = 1, this value is 8 * (2/3)^2 + (1/3)^2 = 33/9
Adding these:= 54/9 = 6
Putting this in the formula, we get
Contrast = sqrt(6/27)
Similarly, for first image,
For k=3, only 3 pixel values are one
For k=1, only 3 pixel values are one
For k=2, only 3 pixel values are one
So, mean X_ = (3 + 3 + 3)/(3*3*3) = 1/3
Now, Let’s calculate, ∑i=1 to r ∑ j=1 to c (Xi,j,k − X_) for k = 1, 2, 3
For k = 3, this value is 6 * (1/3)^2 + 3 * (2/3)^2 = 18/9
For k = 2, this value is 6 * (1/3)^2 + 3 * (2/3)^2 = 18/9
For k = 1, this value is 6 * (1/3)^2 + 3 * (2/3)^2 = 18/9
Adding these:= 54/9 = 6
Putting this in the formula, we get
Contrast = sqrt(6/27)
So, for both images contrast value is '''sqrt(6/27)'''
<nowiki>For second image,
For k=3, 9 (all) pixel values are zero
For k=1, only 1 pixel value is one
For k=2, 8 pixel values are one
So, mean X_ = (0 + 1 + 8)/(3*3*3) = 1/3
Now, Let’s calculate, ∑i=1 to r ∑ j=1 to c (Xi,j,k − X_) for k = 1, 2, 3
For k = 3, this value is 9 * (1/3)^2 = 1
For k = 2, this value is 8 * (1/3)^2 + (2/3)^2 = 12/9
For k = 1, this value is 8 * (2/3)^2 + (1/3)^2 = 33/9
Adding these:= 54/9 = 6
Putting this in the formula, we get
Contrast = sqrt(6/27)
Similarly, for first image,
For k=3, only 3 pixel values are one
For k=1, only 3 pixel values are one
For k=2, only 3 pixel values are one
So, mean X_ = (3 + 3 + 3)/(3*3*3) = 1/3
Now, Let’s calculate, ∑i=1 to r ∑ j=1 to c (Xi,j,k − X_) for k = 1, 2, 3
For k = 3, this value is 6 * (1/3)^2 + 3 * (2/3)^2 = 18/9
For k = 2, this value is 6 * (1/3)^2 + 3 * (2/3)^2 = 18/9
For k = 1, this value is 6 * (1/3)^2 + 3 * (2/3)^2 = 18/9
Adding these:= 54/9 = 6
Putting this in the formula, we get
Contrast = sqrt(6/27)
So, for both images contrast value is '''sqrt(6/27)'''
</nowiki>
2017-12-08
r=3
c=3
1/(3* r* c) = 1/27
x_mean = '(1/( 3*3*3))* 9 = 1/3
Contrast = `sqrt(1/27 * [3*(1-1/3)^2+ 3*(1-1/3)^2+ 3*(1-1/3)^2])`
(Edited: 2017-12-08) `sqrt(1/3 * 4/9)`
`sqrt(4/27)`r=3
c=3
1/(3* r* c) = 1/27
x_mean = '(1/( 3*3*3))* 9 = 1/3
Contrast = @BT@sqrt(1/27 * [3*(1-1/3)^2+ 3*(1-1/3)^2+ 3*(1-1/3)^2])@BT@
= @BT@sqrt(1/3 * 4/9)@BT@
=@BT@sqrt(4/27)@BT@
2017-12-14
r=3, c=3
1/(3rc) = 1/27
x_bar = 1/27 * 9 = 1/3
contrast of image = sqrt(1/27 * [3(1-1/3)^2+ 3(1-1/3)^2+ 3(1-1/3)^2]) = sqrt(1/3 * 4/9)
=2/(3*sqrt(3))
Hence in both the cases, constrast is as above
(Making an edit)
r=3, c=3
1/(3rc) = 1/27
x_bar = 1/27 * 9 = 1/3
contrast of image = sqrt(1/27 * [3(1-1/3)^2+ 3(1-1/3)^2+ 3(1-1/3)^2]) = sqrt(1/3 * 4/9)
=2/(3*sqrt(3))
Hence in both the cases, constrast is as above
(Making an edit)
2017-12-15
We have,
r=3, c=3
Therefore,
1/(3rc) = 1/27
x_mean = 1/27 * 9 = 1/3
Therefore,
contrast of image = `sqrt(1/27 * [3(1-1/3)^2+ 3(1-1/3)^2+ 3(1-1/3)^2]) = sqrt(1/3 * 4/9)`
`=2/(3*sqrt(3))`
(Edited: 2017-12-15) We have,
r=3, c=3
Therefore,
1/(3rc) = 1/27
x_mean = 1/27 * 9 = 1/3
Therefore,
contrast of image = @BT@sqrt(1/27 * [3(1-1/3)^2+ 3(1-1/3)^2+ 3(1-1/3)^2]) = sqrt(1/3 * 4/9)@BT@
@BT@=2/(3*sqrt(3))@BT@
((resource:20171215_180040.jpg|Resource Description for 20171215_180040.jpg))
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