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

-- Dec 6 In-Class Exercise Thread

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-- Dec 6 In-Class Exercise Thread

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.

-- Dec 6 In-Class Exercise Thread

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-- Dec 6 In-Class Exercise Thread

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-- Dec 6 In-Class Exercise Thread

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-- Dec 6 In-Class Exercise Thread

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

-- Dec 6 In-Class Exercise Thread

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))