Post to this thread your solutions to the Dec 6 In-Class Exercise.

Best,

Chris

Post to this thread your solutions to the Dec 6 In-Class Exercise.
Best,
Chris

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calculate Xbar.
(each pixel has a value of <1,0,0> or <0,1,0> or <0,0,1>. sum these up, and divide by 1/(3*r*c) = 9/27= 1/3

calculate std dev. the 3rd term will always evaluate to (1-1/3)^2+ (-1/3)^2 + (-1/3)^2 no matter which pixel we are looking at = 2/3.

multiplied by number of cells 9, divided by 1/(3*r*c) = (2/3)*9 * 1/(3*3*3) = 2/9

constrast = sqrt(2/9)

both images are the same, because each pixel still has the form of <1,0,0>. THe contrast is a measure between the brightest and darkest pixels. which in the two images are the same.

calculate Xbar.
(each pixel has a value of <1,0,0> or <0,1,0> or <0,0,1>. sum these up, and divide by 1/(3*r*c) = 9/27= 1/3
calculate std dev. the 3rd term will always evaluate to (1-1/3)^2+ (-1/3)^2 + (-1/3)^2 no matter which pixel we are looking at = 2/3.
multiplied by number of cells 9, divided by 1/(3*r*c) = (2/3)*9 * 1/(3*3*3) = 2/9
constrast = sqrt(2/9)
both images are the same, because each pixel still has the form of <1,0,0>. THe contrast is a measure between the brightest and darkest pixels. which in the two images are the same.

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((resource:WhatsApp Image 2017-12-06 at 5.24.43 PM.jpeg|Resource Description for WhatsApp Image 2017-12-06 at 5.24.43 PM.jpeg))

((resource:WhatsApp Image 2017-12-06 at 5.42.59 PM.jpeg|Resource Description for WhatsApp Image 2017-12-06 at 5.42.59 PM.jpeg))

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Avg. = 1/3

In both cases the contrast is 2/(3*sqrt(3))

Avg. = 1/3
In both cases the contrast is 2/(3*sqrt(3))

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))

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))