Morphological operations (erosion, dilation, opening, closing) in Computer Vision - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of erosion on a binary image

Given the following 5x5 binary image and a 3x3 square structuring element, what is the output after applying one iteration of erosion?

Computer Vision

import numpy as np
import cv2

image = np.array([
  [1, 1, 1, 1, 0],
  [1, 1, 1, 1, 0],
  [1, 1, 1, 1, 0],
  [1, 1, 1, 1, 1],
  [0, 0, 0, 1, 1]
], dtype=np.uint8)

kernel = np.ones((3,3), np.uint8)
eroded = cv2.erode(image, kernel, iterations=1)
print(eroded)

A

[[1 1 1 0 0]
 [1 1 1 0 0]
 [1 1 1 0 0]
 [0 0 0 1 1]
 [0 0 0 1 1]]

B

[[0 0 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]]

C

[[0 0 0 0 0]
 [0 1 1 0 0]
 [0 1 1 0 0]
 [0 0 0 0 0]
 [0 0 0 0 1]]

D

[[0 0 0 0 0]
 [0 1 1 0 0]
 [0 1 1 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]]

💡 Hint

Erosion shrinks white regions by removing pixels on object boundaries where the structuring element does not fit entirely.

Erosion with a 3x3 kernel removes pixels from the edges of white regions. Only the central 2x2 block in the top-left square remains white because the kernel fits fully there. The bottom-right white region is too small for the kernel, so it disappears.

🧠 Conceptual

intermediate

Effect of dilation on image noise

Which of the following best describes the effect of applying dilation to a binary image with small isolated noise pixels?

A Dilation enlarges white regions and can cause isolated noise pixels to grow.

B Dilation removes isolated noise pixels by shrinking white regions.

C Dilation converts white pixels to black pixels in isolated noise areas.

D Dilation has no effect on isolated noise pixels.

💡 Hint

Think about how dilation changes the size of white areas.

Dilation expands white regions by adding pixels around them. Isolated noise pixels, which are white, will grow larger after dilation, making noise more prominent.

❓ Hyperparameter

advanced

Choosing kernel size for opening operation

You want to remove small white noise dots from a binary image without affecting larger objects. Which kernel size for the opening operation is most appropriate?

A A kernel size slightly larger than the noise dots but smaller than the smallest object.

B A kernel size smaller than the noise dots but larger than the smallest object.

C A very large kernel that is bigger than the largest object in the image.

D A kernel size equal to the entire image size.

💡 Hint

Opening removes small objects smaller than the kernel size.

Opening is erosion followed by dilation. It removes small white noise smaller than the kernel. To keep larger objects intact, the kernel must be bigger than noise but smaller than objects.

❓ Metrics

advanced

Evaluating closing operation effect

After applying a closing operation on a binary image, which metric would best indicate that small black holes inside white objects have been filled?

A Increase in the number of connected white components.

B Increase in the total white pixel area inside objects.

C Decrease in the number of white pixels.

D Decrease in the total black pixel area outside objects.

💡 Hint

Closing fills small black holes inside white regions.

Closing is dilation followed by erosion. It fills small black holes inside white objects, increasing the white pixel area inside those objects.

🔧 Debug

expert

Identifying error in morphological operation code

What error will this code raise when applying dilation with an invalid kernel shape? import numpy as np import cv2 image = np.array([[0,1,0],[1,1,1],[0,1,0]], dtype=np.uint8) kernel = np.array([1,1,1]) # Incorrect kernel shape result = cv2.dilate(image, kernel)

A ValueError: invalid kernel shape

B TypeError: kernel must be a 2D array

C cv2.error: (-215:Assertion failed) !ksize.empty() in function 'dilate'

D No error, code runs successfully

💡 Hint

Check the expected shape of the kernel argument in OpenCV dilation.

OpenCV expects the kernel to be a 2D array. Passing a 1D array causes an assertion failure with a specific error message about kernel size.

Practice

1. What does the erosion operation do to the white parts of a binary image?

easy

A. It grows the white parts by adding pixels at the edges.

B. It removes noise by smoothing the edges.

C. It fills small holes inside the white parts.

D. It shrinks the white parts by removing pixels at the edges.

Solution

1. Step 1: Understand erosion effect on white pixels

   Erosion removes pixels from the boundaries of white regions, making them smaller.

2. Step 2: Compare with other operations

   Dilation grows white parts, opening removes noise, closing fills holes, so erosion must shrink white parts.

3. Final Answer:

   It shrinks the white parts by removing pixels at the edges. -> Option D

4. Quick Check:

   Erosion = Shrinks white parts [OK]

Hint: Erosion shrinks white areas by cutting edges [OK]

Common Mistakes:

• Confusing erosion with dilation
• Thinking erosion fills holes
• Mixing erosion with noise removal

2. Which of the following is the correct syntax to perform dilation using OpenCV in Python on an image img with a 3x3 kernel?

easy

A. cv2.erode(img, np.ones((3,3), np.uint8))

B. cv2.dilate(img, np.ones((3,3), np.uint8))

C. cv2.dilate(img, (3,3))

D. cv2.dilate(img, kernel=3)

Solution

1. Step 1: Recall dilation syntax in OpenCV

   Dilation requires the image and a structuring element (kernel), usually created with np.ones of desired size and type.

2. Step 2: Check options for correct usage

   cv2.dilate(img, np.ones((3,3), np.uint8)) uses cv2.dilate with a 3x3 kernel created by np.ones and correct dtype, which is valid syntax.

3. Final Answer:

   cv2.dilate(img, np.ones((3,3), np.uint8)) -> Option B

4. Quick Check:

   Dilation syntax = cv2.dilate(image, kernel) [OK]

Hint: Use np.ones((3,3), np.uint8) as kernel for dilation [OK]

Common Mistakes:

• Using erode instead of dilate
• Passing kernel size tuple directly
• Using wrong kernel datatype

3. Given the following Python code using OpenCV:

import cv2
import numpy as np
img = np.array([[0,0,0,0,0],
                [0,255,255,255,0],
                [0,255,0,255,0],
                [0,255,255,255,0],
                [0,0,0,0,0]], dtype=np.uint8)
kernel = np.ones((3,3), np.uint8)
eroded = cv2.erode(img, kernel)
print(eroded)

What will be the printed output?

medium

A. [[ 0 0 0 0 0]
[ 0 255 0 255 0]
[ 0 0 0 0 0]
[ 0 255 0 255 0]
[ 0 0 0 0 0]]

B. [[ 0 0 0 0 0]
[ 0 255 255 255 0]
[ 0 255 255 255 0]
[ 0 255 255 255 0]
[ 0 0 0 0 0]]

C. [[ 0 0 0 0 0]
[ 0 0 0 0 0]
[ 0 0 0 0 0]
[ 0 0 0 0 0]
[ 0 0 0 0 0]]

D. [[255 255 255 255 255]
[255 255 255 255 255]
[255 255 255 255 255]
[255 255 255 255 255]
[255 255 255 255 255]]

Solution

1. Step 1: Understand erosion on the given image

   Erosion removes pixels at edges of white regions. The center pixel (0) surrounded by 255s will cause erosion to shrink the white area.

2. Step 2: Apply 3x3 kernel erosion

   Since the kernel covers neighbors, any pixel with a zero neighbor becomes zero. The white cross shape will erode to a smaller cross with zeros at the center and edges.

3. Final Answer:

   White cross with zeros at center and edges as shown in option A -> Option A

4. Quick Check:

   Erosion shrinks white, so edge pixels vanish but some inner pixels remain [OK]

Hint: Erosion removes edge pixels, shrinking white areas [OK]

Common Mistakes:

• Assuming erosion keeps center pixels
• Confusing erosion with dilation output
• Ignoring zero pixels in kernel neighborhood

4. You wrote this code to perform opening on an image img but it does not remove noise as expected:

kernel = np.ones((3,3), np.uint8)
opened = cv2.dilate(cv2.erode(img, kernel), kernel)

What is the error and how to fix it?

medium

A. Kernel size is too small; increase kernel size to remove noise.

B. The order is reversed; opening is dilation followed by erosion, so swap the calls.

C. Use cv2.morphologyEx with cv2.MORPH_OPEN instead for correct opening.

D. The order is reversed; opening is erosion followed by dilation, so code is correct.

Solution

1. Step 1: Check the definition of opening

   Opening is erosion followed by dilation. The code applies erosion then dilation, which is correct in order.

2. Step 2: Identify practical issue

   Manual calls may work but can be error-prone; using cv2.morphologyEx with MORPH_OPEN is recommended for correct and optimized opening.

3. Final Answer:

   Use cv2.morphologyEx with cv2.MORPH_OPEN instead for correct opening. -> Option C

4. Quick Check:

   Use built-in morphologyEx for opening [OK]

Hint: Use cv2.morphologyEx with MORPH_OPEN for opening [OK]

Common Mistakes:

• Swapping erosion and dilation order
• Not using built-in morphology functions
• Assuming kernel size fixes logic errors

5. You have a noisy binary image with small black holes inside white objects. Which morphological operation should you apply to fill these holes without changing the object shapes much?

hard

A. Closing

B. Dilation

C. Opening

D. Erosion

Solution

1. Step 1: Understand the problem of black holes inside white objects

   Black holes are small dark spots inside white regions that need to be filled.

2. Step 2: Choose operation that fills holes without shrinking objects

   Closing is dilation followed by erosion; it fills small holes and gaps while preserving object shape.

3. Final Answer:

   Closing -> Option A

4. Quick Check:

   Closing fills holes inside white objects [OK]

Hint: Closing fills holes inside white objects [OK]

Common Mistakes:

• Using erosion which shrinks objects
• Using opening which removes noise but not holes
• Confusing dilation alone with closing