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CVML Trial Exam — Deep Analysis (SS 2025)

Exam paper analysed: CVML_TrialExamSS2025.pdf — Dr.-Ing. Susana Castillo Alejandre, TU Braunschweig, 25.07.2025. Total: 110 points.

This file walks through every trial-exam question in detail: full solutions, common wrong answers, harder practice questions and final exam-style answers.

Topic Frequency Analysis

Topic Points % of Total Notes
Image Acquisition 10 9 % Colour spaces, DoF, distortion.
Digital Image Processing Basics 20 18 % Heaviest topic. Includes a numerical convolution.
Machine Learning Basics 10 9 % Confusion-matrix mathematics.
Features 10 9 % Pinhole projection + Harris/SIFT T/F.
Optical Flow 10 9 % Backward warping + 5 T/F.
Parametric Transformations & Scattered Data 10 9 % Matrix decomposition.
Epipolar Geometry & Depth 10 9 % 3-D reconstruction + disparity.
Video Matching, Morphing, View Synthesis 10 9 % Formula explanation + 3 T/F.
Camera Calibration 10 9 % Intrinsic/extrinsic + 3 T/F.
Neural Radiance Fields 10 9 % 5 T/F.

High-priority revision list (sorted by point efficiency)

  1. Chapter 02 — Filtering, Edges, Thresholding (20 points; numerical convolution + DoG concept).
  2. Chapter 08 — Epipolar/Depth (10 points but 6 sub-questions, all derivable from \(Z = b\cdot f/d\)).
  3. Chapter 04 — ML (10 points, 3 explicit numerical sub-questions on confusion matrix).
  4. Chapter 07 — Parametric Transforms (10 points, matrix decomposition is mechanical).
  5. Chapter 11 — NeRFs (10 points, all T/F — easy if memorised).

Most-common question types

  • T/F multiple statements (Chapters 4, 5, 6, 8, 9, 10, 11): always with positive/negative scoring (+1 / −1 / 0). Answer only when sure!
  • Numerical computation (Chapters 2, 4, 5, 7, 8): convolution, accuracy/specificity, perspective projection, T·R·S decomposition, 3-D reconstruction.
  • Short-answer theory (1-2 sentences each): what is X, how does X work, name examples.
  • Sketch question (Chapter 1, possibly others): barrel distortion sketch.

Strategy for the real exam

  • Read the whole exam first; mark obvious T/F questions to do quickly first.
  • Note the +1 / −1 / 0 scoring rule on T/F: if you don't know, leave blank (0 is better than −1).
  • Show all your work on numerical questions — partial credit if formula is correct but arithmetic wrong.
  • For "describe / explain" questions, write 1-2 short sentences using the precise technical term (e.g. depth of field, convolution).
  • Time budget: 90 minutes for 110 points → ≈ 50 sec/point. Move on if stuck.

Trial Exam Question 1 — Image Acquisition (10 pts)

Q1a (3 pts) — HSV transformation matching

The image has been altered by decreasing its value, decreasing its saturation, and shifting its hue. Match images 1, 2, 3 to: a. Lower saturation, b. Hue shift, c. Lower value.

  • Topic: Image acquisition / colour spaces (Chapter 01 §5.3).
  • Concepts: HSV semantics — value = brightness; saturation = colour purity; hue = colour wheel angle.

Hard wording

  • "Hue-shift" → rotation around the colour wheel; image changes colour family (red→green→blue) without becoming darker or greyer.

How to answer (general principles, since the actual images aren't extracted)

  • The image with darker overall appearance, same colours → lower value ©.
  • The image with greyish / faded colours, same brightness → lower saturation (a).
  • The image with shifted colour family (e.g. yellow becomes cyan, blue becomes orange) → hue shift (b).

Final exam-style answer

1 — c (lower value); 2 — a (lower saturation); 3 — b (hue shift) (or whatever the visual order is).

Common wrong answers

  • Confusing "value" with "saturation" → check: did the image become darker (value) or less colourful (saturation)?

Similar harder practice question

Given an image, show the result after each of the following: increase value by 0.5; decrease saturation by 0.5; shift hue by 90°. Sketch the qualitative differences.

Answer

  • Increase value 0.5 → image becomes brighter, near-white.
  • Decrease saturation 0.5 → colours become greyer.
  • Shift hue 90° → red ⇒ green, green ⇒ blue, blue ⇒ red (180° gives complementary colours).

Q1b (2 pts) — Print/LED gap-fill

When thinking about printers, we need to use ___ color spaces, such as ___. This is in contrast to what happens with LEDs, where we need to think about ___ color spaces, such as ___.

  • Concepts: additive vs subtractive colour mixing.

Final exam-style answer

"When thinking about printers, we need to use subtractive color spaces, such as CMYK. This is in contrast to what happens with LEDs, where we need to think about additive color spaces, such as RGB."

Common wrong answers

  • Writing RGB for printers (a classic confusion).
  • Forgetting that "additive" pigments would mean light, which doesn't apply to printing.

Q1c (3 pts) — Depth of field & aperture

How does the depth of field change with different aperture sizes in a lens camera? What other effect does the aperture have on the image?

  • Concepts: aperture, depth of field.

Final exam-style answer

"A smaller aperture (larger f-number, e.g. f/16) produces a larger depth of field — more of the scene is in focus — but the image becomes darker because less light reaches the sensor. A larger aperture (smaller f-number, e.g. f/1.4) produces a shallow depth of field with a sharp focal plane and blurred foreground/background, and the image becomes brighter."

Common wrong answers

  • Saying that aperture affects only sharpness, ignoring brightness.
  • Reversing the size/depth-of-field relationship.

Similar harder practice question

Why does increasing ISO instead of opening the aperture compensate for low light, but at a different cost?

Answer

ISO amplifies the signal electronically; both desired image and sensor noise are amplified, so SNR drops → noisier image. Opening the aperture lets in more real photons → no SNR penalty (but shallower DoF).

Q1d (2 pts) — Sketch barrel distortion

Sketch a barrel distortion. Where does this occur?

  • Concepts: lens radial distortion (Chapter 01 §5.8 / Chapter 10 §4.1).

Final exam-style answer

Sketch a square grid whose horizontal and vertical lines bow outward in the centre — straight world lines become convex outward in the image. Barrel distortion occurs in wide-angle / fish-eye lenses, where light rays close to the lens edge are refracted more strongly than those near the optical axis.

Common wrong answers

  • Drawing pincushion (lines bow inward) and labelling it barrel.
  • Saying it occurs in telephoto lenses (those exhibit pincushion).

Trial Exam Question 2 — Digital Image Processing Basics (20 pts)

Q2a (2 pts) — Match images to Fourier spectra

  • Concepts: 2-D Fourier transform / power spectrum.

Without seeing the actual images, the rule is:

  • Stripes / oriented patterns → power spectrum has bright dots perpendicular to the stripe direction.
  • Smooth gradient / large blobs → power concentrated near the centre (low frequencies).
  • Noise / fine texture → power spread broadly over high frequencies.

Final exam-style answer

1 — (matching spectrum), 2 — (matching spectrum) based on these rules.

Q2b-i (2 pts) — Purpose of Gaussian filter

Final exam-style answer

"A Gaussian filter performs smoothing / low-pass filtering: it suppresses high-frequency content (typically noise) while preserving the low-frequency image structure. Because its Fourier transform is also a Gaussian, it never introduces ringing artefacts, unlike box or sinc filters."

Common wrong answers

  • Saying "Gaussian sharpens edges" — that's a high-pass filter.
  • Confusing Gaussian with median (Gaussian is linear).

Q2b-ii (2 pts) — DoG and what it approximates

Final exam-style answer

"The Difference of Gaussians (DoG) approximates the Laplacian of Gaussian (LoG) — a band-pass filter that highlights blob-like features. DoG is used because it's separable and fast (just two Gaussian convolutions and a subtraction), and it's the core of SIFT scale-space key-point detection."

Common wrong answers

  • Saying DoG = LoG (it's an approximation, not equality).
  • Forgetting to mention SIFT or band-pass behaviour.

Q2b-iii (2 pts) — Global vs local thresholding

Final exam-style answer

"Global thresholding uses a single value τ for the whole image; it works only when foreground and background are well separated in the global histogram. Local (adaptive) thresholding computes a different τ(x, y) from a window of pixels around each (x, y), enabling correct segmentation under uneven lighting like vignetting or smooth gradients."

Common wrong answers

  • Saying global = automatic threshold — actually Otsu is automatic global.

Q2b-iv (2 pts) — Concrete-wall gradient removal

Final exam-style answer

"A linear illumination gradient is a low-frequency component, while wall texture is high frequency. Subtract a heavily-Gaussian-blurred version of the image from itself (a high-pass filter / unsharp-mask). The blurred copy approximates the gradient and cancels out, while the texture survives."

detrended = img - cv2.GaussianBlur(img, (0, 0), sigmaX=51) + 128

Q2c (5 pts) — Convolution with [-1, 0, 1]

Convolve the 8×8 image (diamond shape) with the 1×3 kernel [-1, 0, 1], zero padding outside the image.

  • Concepts: 1-D horizontal central-difference filter; convolution mirrors the kernel → effective kernel [1, 0, -1] ⇒ \(O(x, y) = I(x - 1, y) - I(x + 1, y)\).

Step-by-step result

(See Chapter 02 §4.2 for the row-by-row derivation. Final output:)

 0  0  0  0  0  0  0  0
 0  0  0  0  0  0  0  0
 0 -1 -1  0  1  1  0  0
-1 -1  0  0  0  1  1  0
-1 -1  0  0  0  1  1  0
-1 -1  0  0  1  1  0  0
 0 -1 -1  1  1  0  0  0
 0  0  0  0  0  0  0  0

Common wrong answers

  • Forgetting to mirror the kernel before applying.
  • Mixing up the sign (correlation vs convolution).
  • Wrong padding (using image-boundary repeat instead of zero).

Q2d (5 pts) — Filter category & more filters

Interpret the result. What category of filter does the convolution belong to? Name additional filters in this category.

Final exam-style answer

"This is a horizontal gradient / edge / high-pass / first-derivative filter — it produces large positive responses where intensity increases left-to-right (right boundary of the diamond), large negative where intensity decreases (left boundary), and zero in homogeneous regions. Other filters in the same category include Sobel, Prewitt, Roberts, Scharr, Laplacian, Laplacian-of-Gaussian (LoG), Difference-of-Gaussians (DoG), Gabor and Canny."

Trial Exam Question 3 — Machine Learning Basics (10 pts)

Q3a — Confusion-matrix calculations

The exam shows a 3×3 confusion matrix (extracted text doesn't preserve the numbers). Apply these formulas:

  • Accuracy = (sum of diagonal) / total samples.
  • Specificity for class 2 = TN₂ / (TN₂ + FP₂), where TN₂ = total − TP₂ − FN₂ − FP₂; FP₂ = column 2 sum minus diagonal entry; FN₂ = row 2 sum minus diagonal entry.
  • Highest sensitivity = arg max_c (TPc / (TPc + FNc)) = arg max_c (diagonal entry / row sum).

(See Chapter 04 §4.6–4.7 for a representative worked example with arithmetic.)

Q3b (2 pts) — What is an epoch?

Final exam-style answer

"An epoch is one complete pass over the entire training dataset by the network. With N training samples and batch size B, an epoch consists of ⌈N / B⌉ parameter updates."

Q3c (3 pts) — Three T/F

Statement Answer
To choose the best hyperparameters, divide data into training, validation, and test sets. T
Data should be normalised after the split into training and test data. T
To avoid overfitting, it is always better to add more layers. F

Trial Exam Question 4 — Features (10 pts)

Q4a (2 pts) — Pinhole projection of (10, 5, 3) at \(f = n = 2\)

Final exam-style answer

\[ \begin{aligned} p'_{x} = f \cdot p_{x} / p_{z} = 2 \cdot 10 / 3 = 20/3 \approx 6.667 \\ p'_{y} = f \cdot p_{y} / p_{z} = 2 \cdot 5 / 3 = 10/3 \approx 3.333 \\ \Rightarrow p' = (20/3, 10/3) \approx (6.67, 3.33) \end{aligned} \]

Q4b (3 pts) — ZMNCC failure under real lighting

Final exam-style answer

"ZMNCC is invariant to constant intensity shifts and uniform scaling, but real lighting introduces shadows, colour casts and specular highlights — all non-linear changes. Pre-process the images to reduce these effects: convert to greyscale or a perceptual colour space (Lab/Luv); apply local contrast normalisation or histogram equalisation; or compare gradients (gradient-constancy) instead of intensities. Better still, replace ZMNCC with an illumination-robust descriptor such as SIFT or the Census transform with Hamming distance."

Q4c (5 pts) — Five T/F

Statement Answer
The diagonal entries of the Harris matrix control the shear and the rotation. F
A good image feature should have strong gradients in multiple directions. T
Template matching needs only one template for all use cases. F
SIFT descriptors are robust to illumination changes. T
When trying to preserve rotation and scale invariance, the gradient histogram is only useful if computed globally over the entire image. F

(Justifications: see Chapter 05 §5.6.)

Trial Exam Question 5 — Optical Flow (10 pts)

Q5a (5 pts) — Backward warping & disocclusion

Final exam-style answer

"Backward warping computes the warped image by, for every output pixel, looking up its (sub-pixel) source location in the input image: \(\tilde{I}(x, y) = I(x + u, y + v)\), with bilinear interpolation. With a perfect flow and interpolated values for disocclusion, disoccluded areas appear as smeared / stretched copies of the foreground edge — there are no holes (bilinear lookup always returns a value), but the lookup duplicates colours from the surrounding pixels because the flow points "into" the disoccluded region."

Q5b (5 pts) — Five T/F

Statement Answer
If \(H(x, y) = I(x + u, y + v)\), then \(H(x - u, y - v) = I(x, y)\). T
Specular surfaces are particularly suitable for computing optical flow. F
Lucas-Kanade performs well in homogeneous regions when using a big enough window. F
The Horn-Schunck algorithm imposes a global smoothness constraint. T
FlowNet performs well on occlusions and fine texture. F

(Justifications: see Chapter 06 §5.13.)

Trial Exam Question 6 — Parametric Transformations (10 pts)

Q6a (6 pts) — Decompose \(M = T \cdot R \cdot S\)

\[ \begin{aligned} M = [ 0 -1 0 \\ 1 0 20 \\ 0 0 1 ] \end{aligned} \]

with hint \(\operatorname{atan2}(1, 0) = \pi /2 = 90^{\circ}\).

Final exam-style answer

T = [ 1 0  0
      0 1 20
      0 0  1 ]            (translation by (0, 20))

R = [ cos 90°  -sin 90°  0     [ 0 -1 0
      sin 90°   cos 90°  0  =    1  0 0
        0        0       1 ]     0  0 1 ]

S = I (identity)

Verify by \(T \cdot R \cdot S = M \;\checkmark\). (See Chapter 07 §4.12 for the full derivation step by step.)

Common wrong answers

  • Reading θ as 0 instead of 90° (sign confusion of cos/sin).
  • Putting the translation before the rotation in the chain (\(R \cdot T\) ≠ \(T \cdot R\)).

Q6b (2 pts) — 3-step rotation around a point

Final exam-style answer

"To rotate a shape around a point P, perform three steps: (1) translate the entire shape by \(-P\) so that P coincides with the origin; (2) rotate the shape by the desired angle \(\theta\); (3) translate the shape back by +P. As a single matrix this is \(M = T(P) \cdot R(\theta ) \cdot T(-P)\)."

Q6c (2 pts) — Catmull-Rom vs Hermite splines

Final exam-style answer

"Both are cubic splines with the same Hermite basis. They differ in how the tangents are obtained: Hermite splines require the user to provide position and tangent vectors at each control point, which can lead to overshooting if tangents are large. Catmull-Rom splines automatically derive each tangent from the neighbouring control points (\(P'_{i} = (P_{i+1} - P_{i-1}) / 2\)), so the curve interpolates every control point and never overshoots."

Trial Exam Question 7 — Epipolar Geometry & Depth (10 pts)

Q7a (2 pts) — Compute depth from stereo correspondences

Final exam-style answer

"For two rectified, parallel cameras with baseline b and focal length f, the depth Z of a 3-D point with horizontal pixel disparity \(d = x_{L} - x_{R}\) is given by triangulation: \(Z = b \cdot f / d\). Once Z is known, the X- and Y-coordinates follow from \(X = (x_{L} - o_{x}) \cdot Z / f\) and \(Y = (y_{L} - o_{y}) \cdot Z / f\) (where \((o_{x}, o_{y})\) is the principal point and units are consistent)."

Q7b — Numerical (rectified, parallel cameras)

Setup (from exam): baseline \(b = 5 cm\), focal length \(f = 5 cm\), FOV 90°, image 10 × 10 pixels, principal point \((o_{x}, o_{y}) = (4.5, 5)\). Pixel density \(m = 1 px/cm\) (because \(\tan 45^{\circ} = 1\) and the image spans \(2 \cdot f \cdot \tan 45^{\circ} = 10 cm\) over 10 pixels). Hence focal length in pixels \(f_{pix} = 5\).

p₁: left (8, 5), right (3, 5) → disparity \(d_{1} = 5 px\). p₂: left (5, 5), right (3, 5) → disparity \(d_{2} = 2 px\).

Q7b-i (2 pts) — 3-D positions

Z = b · f_pix / d        (in cm)
X = (x_L − o_x) · Z / f_pix
Y = (y_L − o_y) · Z / f_pix
  • p₁: \(Z = 5\cdot 5/5 = 5 cm\); \(X = (8 - 4.5)\cdot 5/5 = 3.5 cm\); \(Y = 0\) ⇒ (3.5, 0, 5) cm.
  • p₂: \(Z = 5\cdot 5/2 = 12.5 cm\); \(X = (5 - 4.5)\cdot 12.5/5 = 1.25 cm\); \(Y = 0\) ⇒ (1.25, 0, 12.5) cm.

Q7b-ii (2 pts) — Differences

\[ \begin{aligned} \Delta X = X_{2} - X_{1} = 1.25 - 3.5 = -2.25 cm \\ \Delta Y = 0 \\ \Delta Z = 12.5 - 5 = 7.5 cm \end{aligned} \]

Q7b-iii (2 pts) — Disparities

\[ \begin{aligned} d_{1} = 8 - 3 = 5 px \\ d_{2} = 5 - 3 = 2 px \end{aligned} \]

Common wrong answers

  • Forgetting that pixel density depends on FOV.
  • Computing \(Z = d / (b \cdot f)\) (formula upside-down).
  • Forgetting to subtract the principal point.

Q7c (2 pts) — Optical axis & epipolar line

Final exam-style answer

"Optical axis — the imaginary line passing through the centre of the lens perpendicular to the image plane; it is the camera's primary viewing direction and meets the image plane at the principal point.

Epipolar line — given a point in one image of a stereo pair, its corresponding point in the other image must lie on a line called the epipolar line. It is the projection of the viewing ray of the first camera onto the second image plane. After rectification, epipolar lines become horizontal scan-lines."

Trial Exam Question 8 — Video Matching, Morphing, View Synthesis (10 pts)

Q8a (5 pts) — Explain the morph formula

\[ \hat{x} = H_{1}^{-1} ( H_{1} x + t \cdot ( H_{2} x' - H_{1} x ) ) \]

Final exam-style answer

"The formula computes a physically valid morph in the rectified view. Step 1: rectify both correspondences — \(H_{1} x\) and \(H_{2} x'\) are the rectified-view coordinates of the matching pair x (image 1) and x' (image 2). Step 2: linearly interpolate inside the rectified view, parameterised by \(t \in [0, 1]\) — the right-hand side walks from \(H_{1} x\) (when \(t = 0\)) to \(H_{2} x'\) (when \(t = 1\)). Step 3: undo image 1's rectification by applying \(H_{1}^{-1}\), returning the result to the original image-1 frame. Linear interpolation is physically valid only in the rectified view; without the H/H⁻¹ round-trip the morph would bend straight lines incorrectly."

Q8b (2 pts) — How does rectification help in stereo?

Final exam-style answer

"Rectification re-projects the two stereo images onto a common image plane perpendicular to the baseline, so that corresponding points share the same y-coordinate and epipolar lines become horizontal scan-lines. The dense-correspondence search reduces from 2-D to 1-D along a single scan-line, dramatically simplifying block matching, dynamic programming and graph-cut algorithms."

Q8c (3 pts) — Three T/F

Statement Answer
Feature-based matching does not work for video sequences. F
Cross-dissolve simply blends images without geometric alignment. T
Physically valid morphs require view rectification. T

Trial Exam Question 9 — Camera Calibration (10 pts)

Q9a (2 pts) — Extract intrinsics from P

Final exam-style answer

"Take the left 3×3 sub-matrix of the projection matrix P. Because that block equals \(K \cdot R\) where K is upper triangular and R is orthonormal, decompose it via RQ factorisation: the upper-triangular factor is the intrinsic matrix K; the orthonormal factor is the rotation R. The translation is then recovered from the last column: \(T = K^{-1} \cdot P[:, 4]\)."

Q9b (5 pts) — Two parameter types & how often they change

Final exam-style answer

"Cameras have intrinsic (internal) parameters and extrinsic (external) parameters.

  • Intrinsic parameters describe the camera itself; examples are the focal length f and the principal point \((o_{x}, o_{y})\) (others: pixel densities, skew, radial-distortion coefficients). They change only when the lens or sensor changes (e.g. zooming).
  • Extrinsic parameters describe the camera's pose in the world; examples are the rotation matrix R and the translation vector T (or equivalently the camera position and orientation). They change every time the camera moves."

Q9c (3 pts) — Three T/F

Statement Answer
The extrinsic matrix converts world coordinates to pixel coordinates directly. F
Radial distortion is corrected using affine transformations. F
Depth can be computed from disparity using triangulation. T

(Justifications: see Chapter 10 §5.12.)

Trial Exam Question 10 — Neural Radiance Fields (10 pts)

Q10 — Five T/F

Statement Answer
NeRFs learn a mapping from 3-D coordinates and view direction to color and density. T
NeRFs rely on voxel grids for storing scene geometry. F
NeRF training requires multiple posed images of a static scene. T
NeRFs explicitly reconstruct a mesh before rendering. F
NeRFs use ray marching to accumulate color along sampled points. T

Common Wrong-Answer Traps Across the Exam

  • Forgetting kernel mirroring during convolution → wrong sign in the gradient image.
  • Confusing additive (RGB) and subtractive (CMYK) colour spaces.
  • Reversing aperture / depth-of-field relationship.
  • Mixing pixel and metric units in \(Z = b\cdot f/d\).
  • Reading Cos/Sin without considering the angle's sign convention (especially for negative rotations).
  • Mismatched ordering in \(M = T \cdot R \cdot S\) (matrix multiplication is non-commutative).
  • Treating extrinsic matrix as full projection matrix.
  • Saying linear morphing in image space is physically valid — it isn't, only after rectification.
  • NeRFs use voxels / meshes — they do not.

Strategy for the Real Exam

  1. First pass — collect easy points. Do all T/F sections first. Skip if uncertain (0 better than -1).
  2. Second pass — numerical questions (Q2c, Q3a, Q4a, Q6a, Q7b). Show every intermediate step.
  3. Third pass — short-answer theory (Q1c, Q1d, Q2b, Q4b, Q6b, Q6c, Q7a, Q7c, Q8a, Q8b, Q9a, Q9b). Use 1–3 sentences with the precise technical term.
  4. Last pass — review marked questions; make sure formulas are dimensionally consistent.

Time budget (for a real 90-minute exam, total = 100 pts)

Topic Weight Approx time Notes
Image Acquisition 10 7 min Q1c, Q1d need short paragraphs
Filtering / Edges 20 18 min Numerical convolution is slow
ML 10 8 min Confusion-matrix maths
Features 10 7 min Numerical projection + T/F
Optical Flow 10 7 min T/F-heavy
Parametric 10 9 min Decomposition
Epipolar/Depth 10 10 min Multiple sub-questions
Morphing 10 7 min Formula explanation
Calibration 10 7 min Theory + T/F
NeRFs 10 4 min All T/F, fastest

Total: 84 min + 6 min buffer.