Top 30 Image processing Interview Questions and Answers

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1. Question 1 of 30

   What is the main drawback of representing a digital image as a fixed grid of pixels?

   • aIt results in a fixed file size regardless of image complexity.
   • bIt causes the image to appear blurry or pixelated when scaled up significantly.
   • cIt cannot accurately capture continuous color and light variations.
   • dIt requires complex mathematical equations to define each pixel's color.
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   Answer: b · It causes the image to appear blurry or pixelated when scaled up significantly.

   The card states that the 'primary weakness is scaling' for raster images, as making them larger requires inventing new pixels, leading to blurriness or pixelation. Option D describes vector graphics, which is a common misconception about how raster images work.

   Read the full bite: Digital Images as Grids of Pixels

2. Question 2 of 30

   Which statement best describes how histogram equalization remaps grayscale intensities?

   • aIt counts pixels in each bin to detect whether the image is under- or over-exposed.
   • bIt linearly stretches the darkest and brightest pixels to fit the full 0-255 range.
   • cIt shifts all pixel values by a constant to center the histogram around the mid-tone.
   • dIt applies the cumulative distribution function to redistribute intensities toward a uniform histogram.
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   Answer: d · It applies the cumulative distribution function to redistribute intensities toward a uniform histogram.

   Histogram equalization uses the cumulative distribution function of the original histogram to remap intensities so the output approximates a uniform distribution, maximizing global contrast. The first option describes linear contrast stretching, which merely rescales the minimum and maximum values without considering the frequency of each intensity level.

   Read the full bite: Describe a grayscale histogram and its use in exposure and equalization

3. Question 3 of 30

   Which of the following best describes a fundamental limitation of an image histogram?

   • aIt only shows the overall average brightness, not the specific distribution of tones.
   • bIt cannot be used to analyze the tonal distribution of color images.
   • cIt is ineffective for identifying overexposed or underexposed areas in an image.
   • dIt fails to provide any information about the spatial arrangement of pixels in an image.
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   Answer: d · It fails to provide any information about the spatial arrangement of pixels in an image.

   An image histogram has zero spatial awareness; it summarizes pixel counts by brightness but does not indicate where those pixels are located in the image. Therefore, it cannot judge composition. The other options describe capabilities that histograms possess or misrepresent their core function.

   Read the full bite: Image Histograms: Visualizing an Image's Tonal DNA

4. Question 4 of 30

   For which application is correcting lens distortion most essential?

   • aReconstructing a 3D environment from a series of images
   • bReducing chromatic aberration in a photograph
   • cTraining a machine learning model to classify objects in images
   • dImproving the aesthetic quality of a portrait
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   Answer: a · Reconstructing a 3D environment from a series of images

   Correcting lens distortion is mandatory for applications like 3D reconstruction (photogrammetry) that rely on precise geometric measurements from images. While object classification might use images, it often doesn't require the same geometric precision, and aesthetic quality can sometimes even be enhanced by distortion.

   Read the full bite: Lens Distortion: Why Straight Lines Curve in Photos

5. Question 5 of 30

   Why is demosaicing necessary after a Bayer sensor captures an image?

   • aBecause each photosite records only one color channel, leaving missing values to estimate.
   • bBecause the raw data requires white balance and gamma correction before display.
   • cBecause the sensor merges multiple short exposures that must be aligned.
   • dBecause the sensor outputs full RGB per pixel in a compressed mosaic layout.
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   Answer: a · Because each photosite records only one color channel, leaving missing values to estimate.

   Demosaicing is required because every photosite measures only a single color channel, so the missing two channels must be interpolated from neighbors. Option D represents the common misconception that Bayer pixels already contain complete RGB data.

   Read the full bite: How does a Bayer filter capture color and what is demosaicing?

6. Question 6 of 30

   In an RGB system, what does "True color" (24-bit color depth) signify regarding bits per channel (bpc)?

   • aThe Red channel uses 24 bits, while Green and Blue share the remaining bits.
   • bThe total color palette is limited to 24 distinct colors.
   • cEach of the Red, Green, and Blue channels uses 24 bits.
   • dEach of the Red, Green, and Blue channels uses 8 bits.
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   Answer: d · Each of the Red, Green, and Blue channels uses 8 bits.

   True color (24-bit color depth) means 24 bits per pixel (bpp). In an RGB system, these 24 bits are typically divided equally among the three channels (Red, Green, Blue), meaning 8 bits per channel (8 bpc). Option C is incorrect because 24 bits per channel would result in 72 bpp, not 24 bpp.

   Read the full bite: Color Depth: Bits Per Pixel vs. Bits Per Channel

7. Question 7 of 30

   A textile company needs to ensure fabric colors match a standard precisely, regardless of the display or printer used. Why would they primarily use CIELAB for this task?

   • aCIELAB is the only color model that can be directly converted to CMYK for printing.
   • bIt provides a device-independent measure of color perception, allowing for objective comparison and quality control.
   • cIts L*, a*, and b* components are easier for designers to intuitively adjust than RGB values.
   • dCIELAB is specifically designed for physical materials like fabric, unlike RGB which is for light.
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   Answer: b · It provides a device-independent measure of color perception, allowing for objective comparison and quality control.

   CIELAB's primary advantage is its device-independent nature, modeling color based on human perception rather than display output. This allows for objective measurement and comparison of colors, crucial for quality control. Option C is less accurate because while CIELAB can be used for intuitive corrections, its core strength for precise matching lies in its perceptual uniformity and device independence, not just ease of adjustment.

   Read the full bite: CIELAB Color Space: Measuring Color Beyond RGB

8. Question 8 of 30

   When implementing a box blur, why is it important to write results into a separate destination buffer rather than updating the source image in place?

   • aIt lets the algorithm handle borders without explicit boundary checks
   • bIt ensures intermediate sums can be stored in a larger data type to avoid overflow
   • cIt prevents already-blurred pixel values from being reused in later neighborhood averages
   • dIt reduces memory usage compared to allocating a second image array
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   Answer: c · It prevents already-blurred pixel values from being reused in later neighborhood averages

   Using a separate destination buffer guarantees that every neighborhood average reads only original pixel values, not values that have already been blurred and would distort subsequent averages. The overflow issue in option B is addressed by using a larger type for the accumulator during the sum, not by allocating a second image buffer.

   Read the full bite: How would you implement a simple box blur on a grayscale image?

9. Question 9 of 30

   A grayscale image looks dull because most pixels are clustered between intensity 100 and 150. After histogram equalization, what has fundamentally changed about the pixel intensities?

   • aThe darkest pixel was mapped to 0 and the brightest to 255 with linear scaling between them.
   • bThe pixels were multiplied by a constant factor to brighten the entire image uniformly.
   • cA transfer function based on the cumulative intensity distribution was used as a lookup table to spread values across the full range.
   • dThe pixels were redistributed so that every intensity level from 0 to 255 contains an equal number of pixels.
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   Answer: c · A transfer function based on the cumulative intensity distribution was used as a lookup table to spread values across the full range.

   Histogram equalization computes the cumulative distribution function from the histogram, normalizes it to the maximum intensity, and uses it as a lookup table to remap pixels across the full range. Option A describes linear contrast stretching, which only scales the min and max values without considering the actual probability distribution of intensities.

   Read the full bite: What is an image histogram and how does histogram equalization improve contrast?

10. Question 10 of 30

    For many computer vision tasks, what is the main benefit of converting a color image to grayscale?

    • aIt allows algorithms to process color information more efficiently by separating RGB channels.
    • bIt corrects for color distortions and inconsistencies present in the original image capture.
    • cIt automatically enhances the image's visual clarity for human interpretation and review.
    • dIt reduces the computational complexity by representing each pixel with a single brightness value, aiding tasks focused on shape or texture.
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    Answer: d · It reduces the computational complexity by representing each pixel with a single brightness value, aiding tasks focused on shape or texture.

    Grayscale conversion simplifies image data by reducing each pixel from three color values (RGB) to a single brightness value, making algorithms faster and easier to develop for tasks that rely on shape, texture, or contrast. Option C is incorrect because grayscale's primary purpose in computer vision is data simplification for machine processing, not enhancing visual clarity for human interpretation.

    Read the full bite: Grayscale Conversion: Seeing in Shades of Gray

11. Question 11 of 30

    Which statement best explains why the Sobel Gx kernel has its specific 3x3 weight pattern?

    • aIt approximates a second-order derivative operator combined with uniform averaging to reduce noise.
    • bIt factors into a horizontal central-difference filter and an orthogonal vertical smoothing filter.
    • cIt is the minimal 3x3 stencil needed to produce a non-zero response at a vertical step edge.
    • dIt is a non-separable 2D Gaussian derivative designed for rotationally invariant edge detection.
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    Answer: b · It factors into a horizontal central-difference filter and an orthogonal vertical smoothing filter.

    The Sobel Gx kernel is separable into a horizontal central-difference row and a vertical smoothing column, which reduces noise while estimating the partial derivative. Distractor A is wrong because, although the smoothing weights loosely approximate a Gaussian, the kernel is separable and explicitly not rotationally invariant.

    Read the full bite: How does the Sobel operator approximate image gradients for edge detection?

12. Question 12 of 30

    When applying a 3x3 convolution to the top border of a bright photo, which padding mode preserves spatial dimensions while avoiding dark vignettes and flat streaking?

    • aReplicate-padding, because it repeats the nearest border pixel outward
    • bValid padding, because it computes convolution only where the kernel fits fully
    • cReflect-padding, because it mirrors edge pixels to assume continuity across the boundary
    • dZero-padding, because zeros represent minimum intensity beyond the border
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    Answer: c · Reflect-padding, because it mirrors edge pixels to assume continuity across the boundary

    Reflect-padding mirrors edge pixels to maintain continuity across the boundary, avoiding both the dark vignettes caused by zero-padding and the flat streaking caused by replicate-padding. Replicate-padding is tempting because it avoids darkening, but it creates frozen-edge artifacts by repeating the same pixel value outward.

    Read the full bite: Zero-padding vs reflect vs replicate padding and their visual artifacts

13. Question 13 of 30

    What is a significant drawback of applying histogram equalization to an image?

    • aIt is only effective for images that are already well-exposed.
    • bIt often reduces the overall number of distinct colors present.
    • cIt can dramatically amplify existing noise, especially in dark regions.
    • dIt requires manual intervention to define the optimal brightness range.
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    Answer: c · It can dramatically amplify existing noise, especially in dark regions.

    The card explicitly states that histogram equalization's "biggest weakness is that it can dramatically amplify noise." Option D is incorrect because the process is described as an "automatic method" that calculates the cumulative distribution function.

    Read the full bite: Histogram Equalization: Spreading Out Pixel Brightness

14. Question 14 of 30

    In which scenario would applying a strong Gaussian blur be generally counterproductive or inappropriate?

    • aWhen the goal is to preserve minute details and sharp boundaries in medical scans.
    • bAs a preliminary step before detecting edges in a noisy photograph.
    • cFor creating a soft, aesthetic background effect in a user interface.
    • dWhen preparing an image for downscaling to prevent aliasing artifacts.
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    Answer: a · When the goal is to preserve minute details and sharp boundaries in medical scans.

    The card explicitly states that Gaussian blur should be avoided when preserving sharp edges and fine details, such as in medical imaging, is critical, as a strong blur destroys this high-frequency information. The other options (A, B, D) are all listed as appropriate and beneficial uses of Gaussian blur.

    Read the full bite: Gaussian Blur: Smoothing Images with Weighted Averages

15. Question 15 of 30

    When applying a median filter to an image, what is its key benefit compared to a simple averaging filter?

    • aIt ensures all noise, including Gaussian noise, is completely eliminated.
    • bIt smooths the image by replacing pixels with the average of their neighbors, leading to a clearer output.
    • cIt effectively removes impulse noise while maintaining the sharpness of image edges.
    • dIt is a linear operation, making it computationally more efficient for large images.
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    Answer: c · It effectively removes impulse noise while maintaining the sharpness of image edges.

    The median filter's primary advantage is its ability to remove impulse noise (like salt-and-pepper noise) without blurring important image edges, which averaging filters tend to do. It is also a non-linear operation, not a linear one.

    Read the full bite: Median Filter: Smoothing Images Without Blurring Edges

16. Question 16 of 30

    Which of the following best describes the fundamental mechanism the Sobel operator uses to detect edges?

    • aIt identifies areas where pixel values fall outside a predefined range, indicating anomalies.
    • bIt directly compares the intensity of each pixel with its eight surrounding neighbors.
    • cIt segments the image into distinct regions based on similar color or intensity profiles.
    • dIt calculates the gradient magnitude by measuring the rate of change in pixel brightness across small regions.
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    Answer: d · It calculates the gradient magnitude by measuring the rate of change in pixel brightness across small regions.

    The Sobel operator functions as a discrete differentiation operator, measuring the local rate of change in pixel brightness (the gradient) to identify edges. Option B is a tempting distractor, but Sobel uses weighted differences via convolution kernels to approximate a derivative, not just simple direct comparisons.

    Read the full bite: The Sobel Operator: Fast, Cheap Edge Detection

17. Question 17 of 30

    To effectively remove both small bright specks and small dark holes from a binary image while preserving the main object's size, which operation sequence is most appropriate?

    • aClosing followed by Opening
    • bOpening followed by Closing
    • cDilation followed by Erosion
    • dErosion followed by Dilation
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    Answer: b · Opening followed by Closing

    Opening (erosion then dilation) removes small bright specks (salt noise) without shrinking the main object. Subsequent Closing (dilation then erosion) fills small dark holes (pepper noise) without expanding the main object, making this sequence ideal for preserving overall object size.

    Read the full bite: Morphological Transformations: Shaping Images with Kernels

18. Question 18 of 30

    What is the primary advantage of the Canny edge detector over simpler methods like the Sobel filter?

    • aIt is significantly faster due to its optimized gradient calculation.
    • bIt produces clean, thin, and continuous edge maps by effectively managing noise and connecting weak edges.
    • cIt completely eliminates all noise from an image before edge detection.
    • dIt requires no parameter tuning, making it easier to apply universally.
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    Answer: b · It produces clean, thin, and continuous edge maps by effectively managing noise and connecting weak edges.

    The card highlights that Canny's design provides a more robust method that finds continuous, well-defined edges, which is achieved through its multi-stage process of noise reduction, thinning (non-maximum suppression), and connecting weak edges (hysteresis). Option A is incorrect because the card states simpler filters might be preferred if raw speed is the only priority, implying Canny is not faster.

    Read the full bite: Canny Edge Detector: Finding True Edges

19. Question 19 of 30

    How does the Laplacian of Gaussian (LoG) primarily identify the center of a blob in an image?

    • aBy finding local extrema (peaks or valleys) after applying a Gaussian blur and then the Laplacian operator.
    • bBy detecting zero-crossings in the output of the combined Gaussian and Laplacian operations.
    • cBy directly convolving the image with a Difference of Gaussians (DoG) kernel.
    • dBy identifying areas where the image intensity gradient is steepest.
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    Answer: a · By finding local extrema (peaks or valleys) after applying a Gaussian blur and then the Laplacian operator.

    The card states that after applying the Gaussian blur and Laplacian, 'the local extrema (peaks and valleys) mark their centers.' Option B is incorrect because zero-crossings indicate the edges of blobs, not their centers.

    Read the full bite: Laplacian of Gaussian (LoG) for Blob Detection

20. Question 20 of 30

    What is the main drawback of using the Harris Corner Detector in computer vision applications?

    • aIt requires a high computational cost, making it unsuitable for real-time systems.
    • bIts performance is not robust to variations in image scale or resolution.
    • cIt cannot differentiate between corners and simple edges in complex scenes.
    • dIt frequently fails to detect corners in images with low texture or uniform color.
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    Answer: b · Its performance is not robust to variations in image scale or resolution.

    The card explicitly states that a "key weakness is its sensitivity to image scale" and advises against its use when an application "must handle images at different sizes or zoom levels." Other options are either contradicted by the text (e.g., it's described as fast and effective) or not highlighted as its primary limitation.

    Read the full bite: Harris Corner Detector: Finding Sharp Changes in Images

21. Question 21 of 30

    What unique characteristic makes a point suitable for detection by a corner detection algorithm?

    • aIts pixel intensity remains constant regardless of small shifts in any direction.
    • bIt exhibits a significant change in pixel intensity when shifted along a single axis.
    • cIt represents a large, homogeneous area with consistent color and texture.
    • dShifting a small window over it in any direction causes a noticeable change in pixel intensity.
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    Answer: d · Shifting a small window over it in any direction causes a noticeable change in pixel intensity.

    The card explains that a corner is identified because moving a small window over it in any direction causes a significant change in what is seen. Option B describes an edge, where intensity changes sharply in only one direction, not multiple.

    Read the full bite: Corner Detection: Finding Stable Points in Images

22. Question 22 of 30

    What is the fundamental mechanism the Hough Transform uses to detect shapes in an image?

    • aIt directly compares predefined shape templates against image segments.
    • bIt analyzes the statistical distribution of pixel intensities across the entire image.
    • cIt segments the image into regions and then classifies each region by shape.
    • dIt identifies local maxima in a parameter space where edge pixels have cast votes.
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    Answer: d · It identifies local maxima in a parameter space where edge pixels have cast votes.

    The card explains that the Hough Transform works by having each edge pixel 'vote' in a 'parameter space,' and the 'points where many of these curves intersect are local maxima—peaks of votes.' Option A describes template matching, which the card states the Hough Transform was created to overcome due to noise and imperfections.

    Read the full bite: Hough Transform: Finding Shapes by Voting

23. Question 23 of 30

    For which task would Scale-Space representation be most advantageous?

    • aMinimizing processing time for real-time applications on low-power devices.
    • bEnhancing sharp edges in an image where feature sizes are consistent.
    • cIdentifying objects whose exact size and appearance are already known and fixed.
    • dDetecting features in an image without prior knowledge of their specific scale or size.
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    Answer: d · Detecting features in an image without prior knowledge of their specific scale or size.

    Scale-space representation's primary purpose is to analyze image structures at all possible sizes, making it ideal for detecting features when their scale is unknown. Options B and D describe scenarios where the card explicitly states scale-space is unnecessary or computationally expensive, respectively. Option B describes a different image processing goal not directly addressed by scale-space's core benefit.

    Read the full bite: Scale-Space: Analyzing Images at Multiple Scales

24. Question 24 of 30

    The primary purpose of subtracting a heavily blurred image from a lightly blurred image in the Difference of Gaussians (DoG) technique is to:

    • aIsolate features of a particular size by effectively removing both fine-grained noise and large, uniform regions.
    • bReduce the overall pixel count and data size for more efficient storage and transmission.
    • cSharpen the image by reversing the effects of the initial blurring operations.
    • dGenerate a precise mathematical representation of the image's frequency spectrum.
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    Answer: a · Isolate features of a particular size by effectively removing both fine-grained noise and large, uniform regions.

    The core idea of DoG is to isolate mid-sized details by canceling out both fine noise and large, uniform areas, as stated in the card's mental model. While DoG acts as a band-pass filter, it is an approximation and not intended for a precise mathematical representation of the frequency spectrum, making option D incorrect.

    Read the full bite: Difference of Gaussians: Finding Edges by Subtracting Blurs

25. Question 25 of 30

    What is the core principle behind HOG's ability to describe object shape robustly against lighting variations?

    • aIt calculates the average pixel intensity within regions and compares them across the image.
    • bIt summarizes local gradient orientations and normalizes these summaries within overlapping blocks.
    • cIt identifies distinct texture patterns and uses their frequency as shape descriptors.
    • dIt analyzes the distribution of color channels and adjusts for global illumination shifts.
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    Answer: b · It summarizes local gradient orientations and normalizes these summaries within overlapping blocks.

    The card explains that HOG computes gradient directions to capture an object's shape and normalizes these summaries within overlapping blocks to achieve robustness against lighting changes. The other options describe mechanisms not central to HOG's operation, such as color analysis, average intensity, or texture patterns.

    Read the full bite: Histogram of Oriented Gradients (HOG)

26. Question 26 of 30

    According to the card, what is the primary reason for combining Local Binary Patterns (LBP) with other descriptors like HOG?

    • aTo allow LBP to process images that contain only global structural information, rather than local textures.
    • bTo achieve significantly improved performance in computer vision tasks, such as object detection.
    • cTo overcome LBP's inherent limitation in extracting any meaningful texture features from an image.
    • dTo decrease the total number of features generated, thereby speeding up the machine learning model's training process.
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    Answer: b · To achieve significantly improved performance in computer vision tasks, such as object detection.

    The card explicitly states that combining LBP with other descriptors like HOG is done "For even better results, especially in object detection." This directly supports option B. Option C is incorrect because LBP is designed to provide powerful texture descriptions; combining it enhances performance, it doesn't compensate for an inability to extract features.

    Read the full bite: Local Binary Patterns (LBP) for Texture Classification

27. Question 27 of 30

    Which scenario would most likely lead to inaccurate results if only a single homography is used?

    • aStitching together multiple images to create a panorama of a distant mountain range.
    • bCorrecting the skewed perspective of a photograph of a flat poster.
    • cAligning two images of a chessboard taken from slightly different angles.
    • dOverlaying a virtual object onto a flat floor in an augmented reality application.
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    Answer: a · Stitching together multiple images to create a panorama of a distant mountain range.

    A homography is designed to map points between images of a flat surface. A distant mountain range, despite appearing far, still presents significant depth and parallax, which a single homography cannot accurately model, as explicitly stated in the 'When Not To Use It' section. The other options describe scenarios involving flat surfaces where a homography is appropriate.

    Read the full bite: Homography: Mapping Flat Surfaces Between Images

28. Question 28 of 30

    A security system uses frame differencing to detect intruders. Which scenario would most likely cause it to trigger a false alarm?

    • aMultiple people entering the scene at the same time from different directions.
    • bThe system being unable to identify the specific type of object causing the motion.
    • cAn intruder moving very slowly, almost imperceptibly.
    • dThe camera experiencing slight, continuous vibrations from nearby traffic.
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    Answer: d · The camera experiencing slight, continuous vibrations from nearby traffic.

    Frame differencing is highly susceptible to camera jitter or shake, which it misinterprets as motion across the entire scene, leading to false positives. While very slow-moving objects are a limitation, they typically result in missed detections (false negatives) rather than false alarms.

    Read the full bite: Frame Differencing: The Simplest Way to See Motion

29. Question 29 of 30

    Which situation would most likely cause background subtraction to perform poorly?

    • aA camera detecting customer movement in a retail store.
    • bA fixed security camera monitoring a building entrance.
    • cA camera mounted on a moving vehicle detecting pedestrians.
    • dA stationary camera counting vehicles on a road.
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    Answer: c · A camera mounted on a moving vehicle detecting pedestrians.

    The card explicitly states that background subtraction is the "wrong tool when the camera is moving, as there is no stable background to model." A camera mounted on a moving vehicle fits this description. The other options describe scenarios with fixed cameras, which are ideal for this technique.

    Read the full bite: Background Subtraction: Finding What's Moving in Video

30. Question 30 of 30

    How does the Horn-Schunck method primarily address the ambiguity of local motion estimation (the aperture problem)?

    • aBy assuming that the intensity of a pixel does not change as it moves.
    • bBy tracking distinct, high-contrast features across multiple frames.
    • cBy enforcing a global constraint that neighboring pixels should have similar motion vectors.
    • dBy analyzing motion in different frequency domains to resolve ambiguities.
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    Answer: c · By enforcing a global constraint that neighboring pixels should have similar motion vectors.

    The card explicitly states that the "global 'smoothness' constraint solves the aperture problem" by assuming neighboring pixels move similarly. While brightness constancy (option A) is part of the method, it's the global smoothness that integrates local information to resolve ambiguity.

    Read the full bite: Horn-Schunck: Assuming Smooth Motion to See Clearly