Image filtering in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of 3x3 averaging filter on a grayscale image

What is the output matrix after applying the given 3x3 averaging filter to the input image matrix?

MATLAB

input_img = [10 20 30; 40 50 60; 70 80 90];
filter = ones(3,3) / 9;
output_img = conv2(input_img, filter, 'valid');
disp(output_img);

A [30 40 50; 60 70 80; 90 100 110]

B [50]

C [50 60; 80 90]

D [10 20; 40 50]

💡 Hint

Remember that 'valid' convolution returns only positions where the filter fully overlaps the image.

The 3x3 averaging filter sums all 9 pixels in the 3x3 window and divides by 9. Sum is 10+20+30+40+50+60+70+80+90=450, average is 50. Since 'valid' is used on 3x3 input and 3x3 filter, conv2 returns 1x1 output [50].

🧠 Conceptual

intermediate

Effect of zero-padding in image filtering

What is the main effect of using zero-padding when applying a filter to an image in MATLAB?

A It keeps the output image size the same as the input and preserves edge information.

B It decreases the output image size and removes noise.

C It changes the filter coefficients automatically.

D It converts the image to grayscale before filtering.

💡 Hint

Think about how padding affects the borders of the image during filtering.

Zero-padding adds zeros around the image edges so the filter can be applied at the borders. This keeps the output image size the same as the input and helps preserve edge information.

🔧 Debug

advanced

Identify the error in median filtering code

What error will this MATLAB code produce when applying a median filter to a grayscale image?

MATLAB

img = imread('cameraman.tif');
filtered_img = medfilt2(img, [3 3 3]);
imshow(filtered_img);

A Error: SIZE must be a two-element vector.

B Error: Image file not found.

C Error: Unsupported image format.

D No error, code runs correctly.

💡 Hint

Check the number of dimensions specified in the filter size argument.

The medfilt2 function expects a 2-element vector for the filter size. Passing [3 3 3] is invalid and causes an error because SIZE must be a two-element vector.

📝 Syntax

advanced

Syntax error in applying Gaussian filter

Which option contains the correct syntax to apply a Gaussian filter with standard deviation 2 to an image 'img' using MATLAB's imgaussfilt function?

A filtered_img = imgaussfilt(img, sigma=2);

B filtered_img = imgaussfilt(img, 2, 'FilterSize');

C filtered_img = imgaussfilt(img, 'sigma', 2);

D filtered_img = imgaussfilt(img, 2);

💡 Hint

Check the function signature and how parameters are passed in MATLAB.

imgaussfilt takes the image and the standard deviation as the second positional argument. Named parameters require name-value pairs as strings.

🚀 Application

expert

Number of pixels changed after edge detection filter

Given a 5x5 grayscale image matrix with all values 100 except the center pixel which is 0, how many pixels will have a non-zero value after applying the following edge detection filter? filter = [-1 -1 -1; -1 8 -1; -1 -1 -1]; output_img = conv2(img, filter, 'same');

MATLAB

img = 100*ones(5,5);
img(3,3) = 0;
filter = [-1 -1 -1; -1 8 -1; -1 -1 -1];
output_img = conv2(img, filter, 'same');
nonzero_count = nnz(output_img);

A 0

B 1

C 9

D 25

💡 Hint

Consider how the filter highlights differences around the center pixel.

The filter detects edges by emphasizing differences. The center pixel is 0 surrounded by 100s, so the 3x3 neighborhood around the center will have non-zero values after filtering, totaling 9 pixels.

Practice

1. What is the main purpose of applying an image filter using a kernel in MATLAB?

easy

A. To convert the image to grayscale

B. To modify the image by emphasizing or reducing certain features

C. To increase the image file size

D. To change the image file format

Solution

1. Step 1: Understand what a kernel does

   A kernel is a small matrix used to process an image by changing pixel values based on neighbors.

2. Step 2: Identify the effect of filtering

   Filtering modifies the image to highlight edges, blur, or sharpen features, not file format or size.

3. Final Answer:

   To modify the image by emphasizing or reducing certain features -> Option B

4. Quick Check:

   Image filtering changes features = D [OK]

Hint: Filters change image features, not file type or size [OK]

Common Mistakes:

• Confusing filtering with file format conversion
• Thinking filtering changes image size
• Assuming filtering only converts to grayscale

2. Which of the following is the correct syntax to apply a filter kernel h to an image img using MATLAB's imfilter function?

easy

A. filtered_img = filter(h, img);

B. filtered_img = filter(img, h);

C. filtered_img = imfilter(h, img);

D. filtered_img = imfilter(img, h);

Solution

1. Step 1: Recall MATLAB's imfilter syntax

   The correct syntax is imfilter(image, kernel), where the image is first and kernel second.

2. Step 2: Check each option

   filtered_img = imfilter(img, h); matches the correct syntax. Options B and D use 'filter' which is incorrect. filtered_img = imfilter(h, img); reverses arguments.

3. Final Answer:

   filtered_img = imfilter(img, h); -> Option D

4. Quick Check:

   imfilter(image, kernel) = A [OK]

Hint: imfilter(image, kernel) is the correct order [OK]

Common Mistakes:

• Using 'filter' instead of 'imfilter'
• Swapping image and kernel arguments
• Missing semicolon or parentheses

3. Given the following MATLAB code:

img = [1 2 3; 4 5 6; 7 8 9];
h = [1 0 -1; 1 0 -1; 1 0 -1];
result = imfilter(img, h, 'replicate');

What is the value of result(2,2)?

medium

A. 0

B. 3

C. 6

D. 9

Solution

1. Step 1: Understand the filter and padding

   The kernel h detects vertical edges. 'replicate' pads edges by repeating border values.

2. Step 2: Calculate convolution at (2,2)

   Neighborhood of (2,2):
   1 2 3
   4 5 6
   7 8 9
   Apply kernel:
   (1*1)+(0*2)+(-1*3)+(1*4)+(0*5)+(-1*6)+(1*7)+(0*8)+(-1*9) = 1+0-3+4+0-6+7+0-9 = (1-3)+(4-6)+(7-9) = -2 -2 -2 = -6
   But imfilter flips kernel, so actual sum is 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   imfilter result center = 6 [OK]

Hint: Remember imfilter flips kernel before applying [OK]

Common Mistakes:

• Not flipping the kernel before convolution
• Ignoring padding method 'replicate'
• Calculating sum incorrectly

4. Consider this MATLAB code snippet:

img = imread('peppers.png');
h = fspecial('average', 3);
filtered_img = imfilter(img, h, 'same');

What is the error in this code?

medium

A. No error; the code runs correctly

B. imfilter requires grayscale images only

C. The kernel size in fspecial must be odd

D. The 'same' option is invalid for imfilter

Solution

1. Step 1: Check fspecial usage

   fspecial('average', 3) creates a 3x3 averaging filter, which is valid.

2. Step 2: Check imfilter usage

   imfilter supports RGB images and 'same' option returns output same size as input.

3. Final Answer:

   No error; the code runs correctly -> Option A

4. Quick Check:

   fspecial + imfilter with 'same' = A [OK]

Hint: imfilter works with color images and 'same' option [OK]

Common Mistakes:

• Assuming imfilter only works on grayscale
• Thinking 'same' is invalid option
• Believing kernel size must be odd but 3 is odd and valid

5. You want to sharpen an image img in MATLAB by combining the original image with a filtered version using the Laplacian kernel h = fspecial('laplacian', 0.2);. Which code snippet correctly applies this sharpening?

hard

A. sharpened = img - imfilter(img, h, 'replicate');

B. sharpened = img + imfilter(img, h, 'replicate');

C. sharpened = imfilter(img, h, 'replicate') - img;

D. sharpened = imfilter(img, h, 'replicate') + img * 0.2;

Solution

1. Step 1: Understand Laplacian sharpening

   Sharpening = original image minus the Laplacian filtered image to enhance edges.

2. Step 2: Match code to formula

   sharpened = img - imfilter(img, h, 'replicate'); subtracts the filtered image from original, correctly applying sharpening.

3. Final Answer:

   sharpened = img - imfilter(img, h, 'replicate'); -> Option A

4. Quick Check:

   Sharpen = img - Laplacian filtered img = B [OK]

Hint: Sharpen by subtracting Laplacian filtered image [OK]

Common Mistakes:

• Adding instead of subtracting Laplacian result
• Multiplying original image incorrectly
• Confusing Laplacian with averaging filter