2018

Try sub r1=0r_1 = 0 and r2=1r_2 = 1 into the intensity mapping function T:

z1=1−0z_1 = 1 - 0

z2=1−1z_2 = 1 - 1

Therefore r1=0,r2=0,z1=1,z2=0r_1 = 0, r_2 = 0, z_1 = 1, z_2 = 0.

This violate the rule of histogram equalization because the mapping must be consistant. (if r1r_1 < r2r_2 , z1z_1 should also be < z2z_2)

(The left annotation)

Note

(a) They are on the same plane, therefore same Intensity (same luminance).

(b) smallest saturation : the closest to the middle of plane.

© Hue = 0 is red.

Rules:

Look at Addivtive primaries.

When a Material Reflect {X}\{X\} Colors, and Expose to {Y}\{Y\} Colors Light,

Material would appear {X∩Y}\{X \cap Y\} Color.

(If {X}\{X\} and {Y}\{Y\} has no intersection, Black Color appear.)

Median filtering ingore out bounded windows, and use a round up approach.

Be careful the size of mask.

  • bit-depth = 4 means intensity is from 0 to 24−12^4 - 1 (i.e. 0 to 15)

Plot the table with each intensity and frequency count. Then you get the histogram.

Histogram equalization and Contrast stretching can be used to improve the contrast of an image.

Initial Threshold is the average intensity.

0×1+1×1+3×2+5×1+6×4+7×31+1+2+1+4+3=4.75\frac{0\times 1+ 1\times 1 + 3\times2 + 5\times 1 + 6\times 4 + 7\times 3}{1 + 1 + 2 + 1 + 4 + 3} = 4.75

Then separate into 2 groups using the threshold. computer the the average inensity of the 2 groups.

then new Threshold = μ1+μ22=4\frac{\mu_1 + \mu_2}{2} = 4

Since new Threshold and old Threshold are both lie on 4, we don’t need to further computer the threshold.

Point Detection can be used in Noise removing. You apply the mask, use a threshold to find out the noise. Then you can remove it and get back the original image.

  • Seed point (Starting point) is selected by the user (given bottom lef)
  • Select a threshold
  • Then start growing (follow checking order)

2019

Median filtering ingore out bounded windows, and use a round up approach.

Use a 3x5 filter:

636363666666606060\begin{array}{|l|l|l|} \hline 63 & 63 & 63 \\ \hline 66 & 66 & 66 \\ \hline 60 & 60 & 60 \\ \hline \end{array}

  • First find the pdf
  • then find s for different t values
  • Finally plot the mapping function

(a)

sub (0.25, 4), and (1,0) into y = mx + c

4 = 0.25m + c …(1)
0 = 0.75m + c …(2)

(2) - (1)

0.5m = -4

m = -8

c = 6

Therefore pdf of p(x)=−8x+6p(x) = -8x + 6

Let the mapping output be s and the mapping input be t.

s={0if t<0.25∫0.25t(−8x+6)dx=[−4x2+6x]0.25t=−4t2+6t−54 if 0.75≥t≥0.251 if t>0.75s=\left\{\begin{array}{ccc} 0 & \text {if } & t<0.25 \\ \int_{0.25}^{t}(-8 x+6) d x=\left[-4 x^{2}+6 x\right]_{0.25}^{t}=-4 t^{2}+6 t-\frac{5}{4} & \text { if } & 0.75 \geq t \geq 0.25 \\ 1 & \text { if } & t>0.75 \end{array}\right.

Then sub t with different values to get the table:

Input t 0 0.25 0.5 0.75 1
Output s 0 0 0.75 1 1

(b)
We linearly map the input range [0.25, 0.75] to output [0,1]. The mapping function is

(a)

Material Reflect Red and Blue light and the paper is exposed toYellow light which composed of red and green.

(b)

(i)

Color A because higher Intensity. (0.5 > 0.3)

(ii)

Color A’s H= pi = Cyan

Color B’s H=0.5pi = middle between Yellow and Green

So Color A’s color looks closer to blue

(a) RGB

(Brighter = Higher value of that Color)

Fig.Q4b is Red channel

Fig.Q4c is Green channel

Fig.Q4d is Blue channel

(b) CMYK

(Brighter = Higher value of that Color)

Fig Q4f is Magenta channel

Fig Q4g is Yellow channel

Fig Q4h is Cyan channel

Fig Q4e is Black channel