Q1

The image is a $7\times 7$ matrix, meaning the total number of pixels is $N\times M=49$. The intensity values range from 0 to 7, giving us a dynamic range where $L=8$.

By counting the occurrences of each pixel intensity in the given matrix, we get the base frequency data.

1.1 Histogram

The histogram $h(r_k)$ represents the frequency of each pixel intensity $r_k$ across the image.

1.2 Cumulative Histogram

The cumulative histogram $H(r_k)$ is calculated by taking the running sum of the histogram frequencies up to each intensity level.

1.3 Histogram Equalization

To perform histogram equalization, we map the original intensity values $r_k$ to new values $s_k$ using the cumulative probability function. The mapping formula is:

Given $L-1=7$ and $N\times M=49$, our multiplier is $\frac{7}{49}=\frac{1}{7}\approx 0.1428$.

Mapping Calculations:

• $s_0=\text{round}(1\times \frac{1}{7})=\text{round}(0.14)=0$

• $s_1=\text{round}(5\times \frac{1}{7})=\text{round}(0.71)=1$

• $s_2=\text{round}(14\times \frac{1}{7})=\text{round}(2.00)=2$

• $s_3=\text{round}(23\times \frac{1}{7})=\text{round}(3.28)=3$

• $s_4=\text{round}(31\times \frac{1}{7})=\text{round}(4.43)=4$

• $s_5=\text{round}(39\times \frac{1}{7})=\text{round}(5.57)=6$

• $s_6=\text{round}(42\times \frac{1}{7})=\text{round}(6.00)=6$

• $s_7=\text{round}(49\times \frac{1}{7})=\text{round}(7.00)=7$

Equalized Histogram:

Notice that original intensities 5 and 6 both map to the new intensity 6. Therefore, we add their original frequencies together (8 + 3 = 11).

1.4 Robust Contrast Stretching (Truncation)

Truncating allows us to ignore outliers (the extreme dark and bright pixels) when setting our new minimum and maximum bounds for the contrast stretch.

Step 1: Calculate Truncation Bounds

• 10% Dark Truncation: $10\mathrm{\%}\text{ of }49\text{ pixels}=4.9\approx 5\text{ pixels}$.

  Looking at the cumulative histogram, discarding the 5 darkest pixels perfectly removes all pixels with intensity 0 and 1. The new minimum valid intensity is $r_{min}=2$.

• 20% Light Truncation: $20\mathrm{\%}\text{ of }49\text{ pixels}=9.8\approx 10\text{ pixels}$.

  Counting from the brightest end: we have 7 pixels of intensity 7, and 3 pixels of intensity 6. Discarding the top 10 pixels removes all 7s and 6s. The new maximum valid intensity is $r_{max}=5$.

Step 2: Contrast Stretching Formula

We map the valid range $[2,5]$ to the full dynamic range $[0,7]$. Any values below $r_{min}$ clip to 0, and any values above $r_{max}$ clip to 7.

Mapping Calculations:

• $r_k<2$ (Intensities 0, 1): Clipped to 0 since both will result in negatives

• $r_k=2$: $\text{round}(\frac{0}{3}\times 7)=$ 0

• $r_k=3$: $\text{round}(\frac{1}{3}\times 7)=\text{round}(2.33)=$ 2

• $r_k=4$: $\text{round}(\frac{2}{3}\times 7)=\text{round}(4.67)=$ 5

• $r_k=5$: $\text{round}(\frac{3}{3}\times 7)=\text{round}(7.00)=$ 7

• $r_k>5$ (Intensities 6, 7): Clipped to 7

Final Stretched Histogram:

We compile the frequencies of the original pixels based on their newly mapped destinations.

• Mapped to 0: Old intensities 0, 1, 2 (1 + 4 + 9 = 14)

• Mapped to 2: Old intensity 3 (9)

• Mapped to 5: Old intensity 4 (8)

• Mapped to 7: Old intensities 5, 6, 7 (8 + 3 + 7 = 18)

Q2

When a question asks to modify a histogram into a "uniformly distributed" one across a specific interval, it is asking you to perform Histogram Equalization.

• Total number of pixels: $N\times M=25$

• Target interval $[0,7]$ means our dynamic range $L=8$, so $L-1=7$.

• Our equalization multiplier will be $\frac{L-1}{N\times M}=\frac{7}{25}=0.28$.

Step 1: Calculate Frequencies and Cumulative Frequencies

By counting the occurrences of each intensity ($r_k$) in the matrix, we build our standard histogram $h(r_k)$ and cumulative histogram $H(r_k)$.

Step 2: Apply the Equalization Formula

Now we map the original intensities ($r_k$) to their new uniform intensities ($s_k$) using the formula:

• $s_0=\text{round}(0.28\times 5)=\text{round}(1.40)=\mathbf{1}$

• $s_1=\text{round}(0.28\times 9)=\text{round}(2.52)=\mathbf{3}$

• $s_2=\text{round}(0.28\times 13)=\text{round}(3.64)=\mathbf{4}$

• $s_3=\text{round}(0.28\times 17)=\text{round}(4.76)=\mathbf{5}$

• $s_4=\text{round}(0.28\times 19)=\text{round}(5.32)=\mathbf{5}$

• $s_5=\text{round}(0.28\times 21)=\text{round}(5.88)=\mathbf{6}$

• $s_6=\text{round}(0.28\times 25)=\text{round}(7.00)=\mathbf{7}$

• $s_7=\text{round}(0.28\times 25)=\text{round}(7.00)=\mathbf{7}$

Step 3: The Modified (Equalized) Histogram

To get the final uniformly distributed histogram, we assign the original pixel counts to their newly calculated intensity levels. Notice that both original intensities 3 and 4 are mapped to the new intensity 5, so their counts are added together ($4+2=6$).

Q3

1. First Marked Pixel: (2) at position row 0, column 1

Because this pixel is on the top edge of the image, a $3\times 3$ window centered on it will go out of bounds. The most standard approach in image processing is to assume zero-padding for the out-of-bounds pixels.

The $3\times 3$ neighborhood (with zero-padding on top):

The values in this window are: $0,0,0,1,2,2,2,2,2$

• a) Standard Average filter: Sum the values and divide by 9.

  • $\frac{0+0+0+1+2+2+2+2+2}{9}=\frac{11}{9}\approx 1.22$

  • (Note: If your class uses integer rounding, this becomes 1. If your class ignores out-of-bounds pixels entirely instead of zero-padding, the average would be $\frac{11}{6}\approx 1.83$).

• b) Median filter: Sort the values from lowest to highest and pick the middle (5th) value.

  • Sorted: $0,0,0,1,\mathbf{2},2,2,2,2$

  • The median is 2.

• c) Min filter: Select the smallest value in the neighborhood.

  • The minimum is 0. (Or 1, if ignoring padded pixels).

2. Second Marked Pixel: (7) at position row 2, column 2

This is an inner pixel, so it has a complete set of 8 surrounding neighbors within the image boundaries. No padding is needed.

The $3\times 3$ neighborhood:

The values in this window are: $2,2,2,1,7,3,1,2,1$

• a) Standard Average filter: Sum the values and divide by 9.

  • $\frac{2+2+2+1+7+3+1+2+1}{9}=\frac{21}{9}=2.33$

  • (Note: Rounded to the nearest integer, this is 2).

• b) Median filter: Sort the 9 values from lowest to highest and pick the middle (5th) value.

  • Sorted: $1,1,1,2,\mathbf{2},2,2,3,7$

  • The median is 2.

• c) Min filter: Select the smallest value in the neighborhood.

  • The minimum is 1.

Q4

First, let's extract the $3\times 3$ neighborhood matrix centered around each pixel.

• Neighborhood for Pixel (5):

  $$\left[\begin{array}{ccc}2& 6& 6\\ 1& \mathbf{5}& 7\\ 2& 6& 6\end{array}\right]$$

• Neighborhood for Pixel (7):

  $$\left[\begin{array}{ccc}6& 6& 7\\ 5& \mathbf{7}& 5\\ 6& 6& 6\end{array}\right]$$

a) Sobel Operator

The Sobel operator calculates the gradient of the image using two kernels: one for horizontal changes ($G_y$) and one for vertical changes ($G_x$).

The total magnitude is typically calculated as $G=\sqrt{G_y^2+G_x^2}$ (or approximated by $|G_y|+|G_x|$).

1. Applying to Pixel (5):

• $G_y$: $(-1\times 2)+(1\times 6)+(-2\times 1)+(2\times 7)+(-1\times 2)+(1\times 6)$

  $G_y=-2+6-2+14-2+6=\mathbf{20}$

• $G_x$: $(-1\times 2)+(-2\times 6)+(-1\times 6)+(1\times 2)+(2\times 6)+(1\times 6)$

  $G_x=-2-12-6+2+12+6=\mathbf{0}$

• Magnitude: $\sqrt{{20}^2+0^2}=$ 20

2. Applying to Pixel (7):

• $G_y$: $(-1\times 6)+(1\times 7)+(-2\times 5)+(2\times 5)+(-1\times 6)+(1\times 6)$

  $G_y=-6+7-10+10-6+6=\mathbf{1}$

• $G_x$: $(-1\times 6)+(-2\times 6)+(-1\times 7)+(1\times 6)+(2\times 6)+(1\times 6)$

  $G_x=-6-12-7+6+12+6=\mathbf{-}\mathbf{1}$

• Magnitude: $\sqrt{1^2+(-1{)}^2}=\sqrt{2}\approx$ 1.41

b) Laplacian Operator

Got it! Thank you for the clarification. In some textbooks and courses, the "composite" Laplacian is indeed defined as adding the original image back to the standard Laplacian to sharpen it in one step. When you subtract a standard 4-neighbor Laplacian (with a $-4$ center) from the original image (which acts as a $1$ in the center), you get a new kernel with a $-5$ in the center!

We'll calculate using the composite 4-neighbour $3\times 3$ kernel:

1. Applying to Pixel (5):

For the pixel at row 1, column 1, we look at its immediate top, bottom, left, and right neighbors.

• Top neighbor: 6

• Bottom neighbor: 6

• Left neighbor: 1

• Right neighbor: 7

• Center pixel: 5

Calculation:

• Sum of the 4 neighbors: $6+6+1+7=20$

• Multiply neighbors by $-1$: $-20$

• Center multiplied by $5$: $5\times 5=25$

• True Output: $25-20=$ 5

2. Applying to Pixel (7):

For the pixel at row 1, column 2, we again look at its immediate 4-connected neighbors.

• Top neighbor: 6

• Bottom neighbor: 6

• Left neighbor: 5 (this is the pixel we just calculated in the step above)

• Right neighbor: 5

• Center pixel: 7

Calculation:

Sum of the 4 neighbors: $6+6+5+5=22$

Negate the sum: $-22$

Center multiplied by 5: $7\times (5)=35$

Output: $35-22=$ 13