Quiz: Image Processing and Computer Vision
Test your understanding of image representation, filtering, and computer vision concepts.
1. A grayscale image is represented as:
- A 3D tensor with RGB channels
- A 2D matrix of intensity values
- A 1D vector of pixel positions
- A list of color names
Show Answer
The correct answer is B. A grayscale image is a 2D matrix where each entry represents the intensity (brightness) of a pixel, typically ranging from 0 (black) to 255 (white).
Concept Tested: Grayscale Image
2. An RGB image is stored as:
- A single 2D matrix
- Three separate matrices (one per color channel)
- A 1D array of values
- A text file
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The correct answer is B. An RGB image consists of three 2D matrices (or a 3D tensor), one for each color channel: Red, Green, and Blue. Each pixel has three values.
Concept Tested: RGB Image
3. Image convolution:
- Increases image resolution
- Applies a kernel to extract features or transform the image
- Converts color to grayscale
- Compresses the image
Show Answer
The correct answer is B. Convolution slides a kernel (small matrix) across the image, computing weighted sums at each position. Different kernels produce different effects like blurring, sharpening, or edge detection.
Concept Tested: Image Convolution
4. A Gaussian blur filter:
- Sharpens edges in the image
- Smooths the image by averaging with Gaussian-weighted neighbors
- Detects vertical edges only
- Increases image contrast
Show Answer
The correct answer is B. Gaussian blur uses weights that follow a Gaussian distribution, giving more weight to nearby pixels. This produces smooth, natural-looking blur without blocky artifacts.
Concept Tested: Blur Filter
5. The Sobel operator is used for:
- Image compression
- Edge detection by computing image gradients
- Color space conversion
- Image resizing
Show Answer
The correct answer is B. The Sobel operator approximates image derivatives (gradients) to detect edges. It uses two kernels: one for horizontal gradients (\(G_x\)) and one for vertical gradients (\(G_y\)).
Concept Tested: Sobel Operator
6. The Fourier transform of an image reveals:
- The color distribution
- Frequency components (how quickly intensity changes)
- The number of objects
- The image dimensions
Show Answer
The correct answer is B. The Fourier transform decomposes an image into frequency components. Low frequencies represent gradual changes (smooth areas); high frequencies represent rapid changes (edges, textures).
Concept Tested: Fourier Transform
7. SVD-based image compression works by:
- Removing all color information
- Keeping only the largest singular values and discarding the rest
- Reducing the number of pixels
- Converting to a different file format
Show Answer
The correct answer is B. SVD compression keeps only the \(k\) largest singular values, which capture the most important structure. The approximation \(A_k = \sum_{i=1}^k \sigma_i u_i v_i^T\) requires less storage.
Concept Tested: Image Compression
8. In Harris corner detection, a corner is characterized by:
- Only one large eigenvalue of the structure tensor
- Two large eigenvalues of the structure tensor
- Zero eigenvalues
- Negative eigenvalues
Show Answer
The correct answer is B. At a corner, the structure tensor has two large eigenvalues, indicating significant intensity changes in both directions. An edge has one large eigenvalue; flat regions have none.
Concept Tested: Feature Detection
9. A homography transformation:
- Only handles rotation
- Maps points between two planes, handling perspective
- Is always the identity
- Only works on 1D signals
Show Answer
The correct answer is B. A homography is a 3×3 projective transformation that maps points between two planes. It can represent perspective distortion, rotation, scaling, and translation.
Concept Tested: Homography
10. The convolution theorem states that convolution in spatial domain equals:
- Addition in frequency domain
- Multiplication in frequency domain
- Division in frequency domain
- Convolution in frequency domain
Show Answer
The correct answer is B. The convolution theorem states that \(f * g = \mathcal{F}^{-1}(\mathcal{F}(f) \cdot \mathcal{F}(g))\). Convolution in spatial domain corresponds to element-wise multiplication in frequency domain.
Concept Tested: Frequency Domain