DIP FINAL — Ch 8–12 (189 Q&A, pipeline-blocked + mixed review) — MASTER THIS

Audited against the supplied textbook, Courses 08–12 slides, and professor sample images. All 189 cards are retained: factual errors were corrected and repeated cards were converted into stronger calculation, transfer, and full-pipeline questions. Learn in blocks; use Practice Test for shuffled MCQ; explain why the nearest distractor is wrong.

This operation is defined as A ⊖ B = {z | (B)z ⊆ A}. What is it?

Erosion — the SE must fit ENTIRELY inside A (⊆); it shrinks objects.

This operation is defined as A ⊕ B = {z | (B̂)z ∩ A ≠ ∅}. What is it?

Dilation — the reflected SE need only OVERLAP A by one pixel (∩ ≠ ∅); it grows objects.

This operation is defined as (A ⊖ B) ⊕ B — erosion followed by dilation with the same SE. What is it?

Opening — erosion followed by dilation with the same SE. It removes foreground features that cannot survive the erosion and then regrows the remaining set. The result is anti-extensive (A∘B ⊆ A) and may still differ in size or shape from the original.

This operation is defined as (A ⊕ B) ⊖ B — dilation followed by erosion with the same SE. What is it?

Closing — dilate-then-erode; fills small holes and narrow breaks. (Opening reversed.)

This operation is defined as β(A) = A − (A ⊖ B), the original set minus its erosion. What is it?

Boundary extraction — removing the eroded interior leaves only the object's outline.

This iterative algorithm is Xₖ = (Xₖ₋₁ ⊕ B) ∩ Iᶜ, repeated until Xₖ = Xₖ₋₁. What is it?

Hole filling — the ∩ Iᶜ confines growth to background pixels of the hole; needs a seed inside it.

This iterative algorithm is Xₖ = (Xₖ₋₁ ⊕ B) ∩ I, repeated until Xₖ = Xₖ₋₁. What is it?

Connected-component extraction — ∩ I grows the seed inside its own foreground component. (Hole filling uses Iᶜ.)

This expression is Sₖ(A) = (A ⊖ kB) − (A ⊖ kB) ∘ B. What does it compute?

The k-th skeleton subset — the opening deletes every point where the SE still fits, so subtracting it from the k-times-eroded set keeps only the points too thin to hold the SE: the skeleton pixels sitting at erosion depth k.

On a grayscale image, g = (f ⊕ b) − (f ⊖ b): dilation minus erosion. What is this?

Morphological gradient — local max minus local min highlights edges; flat regions cancel to ~0.

This transform is f − (f ∘ b), the image minus its own opening. What is it?

White top-hat — f − (f∘b) returns bright structures and local intensity detail removed by the opening, typically features smaller than or poorly fitting the SE. It is not guaranteed to reproduce an object mask exactly; it can also estimate/correct slowly varying background.

An operation smooths contours, breaks thin bridges/protrusions, and removes small foreground islands. Which operation is it, and does it exactly restore every surviving object's original size?

Opening — erosion followed by dilation with the same SE. It removes foreground parts that cannot contain the SE and smooths contours. It is anti-extensive (A∘B ⊆ A): erosion and dilation are not inverses, so a surviving object's exact original size or shape is not guaranteed.

An operation smooths contours, bridges narrow foreground breaks, fills small dark gaps/holes, and is extensive. Which operation is it, and does it exactly undo dilation?

Closing — dilation followed by erosion with the same SE. It fills/bridges dark gaps smaller than the SE and is extensive (A ⊆ A•B). The erosion reduces the expansion but does not generally restore the exact original boundary; the operators are not inverses.

An image has several shapes; after an operation exactly one pixel is marked, at one specific shape. Which operation?

Hit-or-miss transform — matches foreground AND background simultaneously, marking one point per exact match.

A marker image F is dilated repeatedly, and after EVERY dilation the result is intersected with a separate mask image G (F ⊆ G), so it can never grow past G. What is this operation called?

Geodesic (conditional) dilation — D_G⁽¹⁾(F) = (F ⊕ B) ∩ G; the per-step intersection with the mask caps growth, and iterating it to stability gives reconstruction by dilation.

Rice grains under uneven lighting can't be thresholded; after an operation with a disk SE larger than any grain the background flattens and thresholding works. Which operation?

Top-hat transform (shading correction) — a disk bigger than any grain means the opening erases every grain and keeps only the smooth uneven background. Subtracting that background estimate flattens the lighting so one global threshold works — bottom-hat is the mirror trick for DARK objects on a light background.

What is the main drawback of the morphological skeleton (erosions-minus-openings formulation)?

It isn't guaranteed connected — it may also be thicker than needed and contain spurs, so heuristics/pruning are required.

What is the main drawback of the hit-or-miss transform for shape detection?

It requires an EXACT match — noise, rotation, or scale change makes it miss.

Repeating the opening (or closing) with the same SE produces no further change. What is this property called?

Idempotence — (A ∘ B) ∘ B = A ∘ B; once is enough. (Convergence belongs to iterative algorithms like reconstruction.)

A binary object has a hole strictly LARGER than the structuring element. What does closing with that SE do to the hole?

Nothing — the hole survives untouched. Closing only fills holes/gaps SMALLER than the SE: by Eq. 9-13 (the ball rolled OUTSIDE the boundary), closing keeps as background every SE translate that avoids the foreground — so an SE that fits entirely inside the hole leaves it unfilled (intuitively, the dilation can't bridge a gap wider than the SE).

A solid 5×5 square of foreground is eroded by a 3×3 SE of all 1s. What is the result?

A 3×3 square — (M−m+1)×(N−n+1). Dilation would give 7×7; empty only if object smaller than SE.

A solid 5×5 square of foreground is dilated by a 3×3 SE of all 1s. What is the result?

A 7×7 square — (M+m−1)×(N+n−1); each border grows outward by ⌊m/2⌋ = 1 pixel.

A 3×3 grayscale neighborhood has values {12, 45, 7, 200, 33, 90, 15, 60, 22}. What does grayscale erosion with a flat SE assign the center?

7 — flat-SE erosion outputs the MINIMUM of the covered values. (Dilation would give the max, 200.)

For the neighborhood {12, 45, 7, 200, 33, 90, 15, 60, 22}, what is the morphological gradient (dilation minus erosion) at the center?

193 — max − min = 200 − 7; large value signals an edge, flat areas give 0.

During skeletonization, object A becomes empty on the 3rd successive erosion. What is the last skeleton index K?

K = 2 — the largest k for which A ⊖ kB is still non-empty (the step before A vanishes).

In grayscale morphology with a flat SE, what does dilation assign the center pixel?

The MAXIMUM intensity in the neighborhood — dilation is a max filter, so dilated images look lighter. (Erosion = min.)

In morphological reconstruction, what are the roles of the marker F and the mask G?

The marker grows (is transformed); the mask stays fixed and constrains it, with F ⊆ G. Reversing the roles is the classic trap.

By erosion–dilation duality, the complement of an erosion, (A ⊖ B)ᶜ, equals what?

Aᶜ ⊕ B̂ — shrinking the foreground = growing the background (complement AND reflected SE). They are duals, not inverses.

Dilation can be written as the union of translates {w | w = a + b, a ∈ A, b ∈ B}. What is this formulation called?

Minkowski addition — dilation as a union of translates; erosion's intersection-of-translates form is Minkowski subtraction.

Mathematical morphology is fundamentally based on which mathematical framework?

Mathematical morphology is based on order/set theory: binary images are subsets of Z²; grayscale images are ordered-valued functions f:D⊆Z²→R (or a gray-level lattice) using local suprema/infima. A 3-D surface/umbra is a useful geometric interpretation, not the only formal model.

What distinguishes opening by reconstruction from a standard opening?

In binary opening by reconstruction, the erosion supplies markers and reconstruction under the original mask restores each connected mask component reached by a surviving marker. Components erased by erosion do not return. In grayscale reconstruction, describe contour/peak restoration rather than claiming unconditional exact-object recovery.

Can the original set A be reconstructed from the final morphological skeleton alone? What does reconstruction actually require?

Not from an unlabeled final skeleton alone. Exact reconstruction uses every indexed subset: A = ∪ₖ[Sₖ(A) ⊕ kB]. The erosion-depth k (or an equivalent radius label at each skeleton point) must be retained; discarding that depth information loses what is needed to rebuild A.

In seeded hole filling, why use the cross-shaped (4-connected) SE instead of the 3×3 all-1s SE?

The 3×3 all-1s SE is 8-connected, so the growing marker can leak diagonally through corner-touching pixels into neighboring regions; the cross grows only through direct 4-neighbors, keeping the fill contained inside the hole.

A morphological algorithm GROWS a set using 4 hit-or-miss structuring elements rotated 90° apart, each iterated to convergence, then unions the results. What does it compute?

The convex hull — the smallest 'dent-free' (convex) region containing A: each 90°-rotated SE keeps filling in concavities until nothing changes, so the union has no indentations left. Exam contrast: thinning uses 8 SEs at 45°, convex hull uses 4 SEs at 90°.

In morphology, what exactly are the structuring element B's origin, the translation (B)z, and the reflection B̂?

B is a small pixel set with a designated ORIGIN; (B)z translates B so its origin sits at pixel z — the fit/hit test is applied there and the answer written at z; B̂ is B rotated 180° about its origin (appears in dilation and in erosion–dilation duality; B̂ = B for symmetric SEs).

A marker is built from the image BORDER: F = I on the border (0 elsewhere), then X = I − R_I^D(F). Name the operation, and how does automatic hole filling differ?

Border clearing (Eqs. 9-47/9-48) — reconstruction by dilation from the border marker regrows exactly the objects touching the border, and subtracting them from I leaves only the interior objects. Automatic hole filling (Eqs. 9-45/9-46) uses the same border-marker idea but complemented: marker F = 1 − I on the border, mask Iᶜ, and the reconstruction is COMPLEMENTED instead of subtracted — it fills holes rather than deleting border objects. It fails if the entire border is foreground (the marker becomes all 0s).

Geodesic dilation of size 1 is D_G⁽¹⁾(F) = (F ⊕ B) ∩ G. What is geodesic EROSION of size 1, and why does it use union instead of intersection?

E_G⁽¹⁾(F) = (F ⊖ B) ∪ G — erosion SHRINKS the marker, so the mask must prop it up as a floor (∪; pointwise max ∨ in grayscale), whereas dilation GROWS the marker, so the mask must cap it (∩; pointwise min ∧). Grayscale forms: (f ⊕ b) ∧ g and (f ⊖ b) ∨ g, requiring f ≤ g (binary: F ⊆ G). Iterated to stability they give reconstruction by dilation/erosion. Trap: 'geodesic erosion uses intersection with the mask' is FALSE.

Opening by reconstruction is O_R⁽ⁿ⁾(F) = R_F^D(F ⊖ nB). What is the usual value of n, and which TWO different SEs are involved (a classic trap)?

n = 1 — a single erosion builds the marker (mask = the original image). Two SEs with different jobs: the FEATURE SE (e.g. a 51×1 vertical line matched to ~51-px strokes) is used ONLY for the initial erosion that selects objects; the CONNECTIVITY SE (3×3 of 1s) drives the reconstruction/geodesic-dilation steps. Trap: 'any SE can be used for the reconstruction step' — false, that SE defines connectivity.

An image is smoothed, opened with disk SEs of increasing radius (10, 20, 25, 30), and the DIFFERENCES between successive sums of pixel values are plotted against radius, showing two peaks. Name the technique and what the peaks mean.

Granulometry — particle-size estimation WITHOUT segmenting individual particles. Opening with a size-r SE removes mainly particles of size ≈ r, so each drop in 'surface area' (sum of pixel values) is credited to particles of that size; peaks in the difference plot mark the dominant particle sizes (two here). Limits: indirect, needs smoothing first, assumes regular particles lighter than the background — overlapping irregular particles of similar size blur the peaks.

What is the main drawback of the morphological convex hull (4 hit-or-miss SEs at 90°, iterated then unioned), and the standard remedy?

It can OVERGROW the true minimal convex hull — only 4 directions are probed, so accuracy is limited to the vertical/horizontal bounds and concavities get filled past the smallest convex set containing A. Remedy: limit the growth heuristically (e.g. to A's vertical/horizontal extent) or add SEs at intermediate angles — at extra computational cost.

This thresholding rule is defined as g(x,y) = 1 if f(x,y) > T, else 0, with a single constant T for the whole image. What is it?

Basic global thresholding — one fixed T everywhere producing a binary output is the giveaway; adaptive methods recompute T per neighborhood.

This threshold is computed per pixel as T(x,y) = (mean of the pixel's neighborhood) − C, with the constant C tuned by hand. What method is it?

Adaptive (local) thresholding — the threshold follows local brightness so it survives uneven lighting; C is the hand-tuned offset.

What is the main drawback of global thresholding?

It fails under non-uniform illumination — one constant T cannot fit both bright and shadowed regions; adaptive thresholding is the fix.

A histogram has one dominant peak and a long tail; the threshold is placed at the histogram point farthest from the line joining the peak to the minimum intensity. What method is this?

Triangle thresholding — a geometric peak-to-line distance criterion, made for single-peak (skewed) histograms where Otsu's bimodal assumption fails.

A threshold k is chosen by maximizing sigma_B^2(k) = P1(k)P2(k)[m1(k) − m2(k)]^2, computed from the histogram alone. Whose criterion is this?

Otsu's method — it tries every threshold and keeps the one that pushes the two classes' means farthest apart (weighted by class sizes), i.e. maximum between-class variance, which is equivalent to making each class internally tightest (minimum within-class variance). It needs only the histogram, never the image itself.

When does Otsu's thresholding struggle the most?

Unimodal or noisy histograms, or a tiny object drowned by the background — Otsu assumes two clear histogram modes (bimodal).

In the basic iterative global thresholding algorithm, the current class means are m1 = 180 and m2 = 60. What is the next threshold T?

120 — T = (m1 + m2)/2 = (180 + 60)/2; repeat until T stops changing.

Adaptive thresholding uses a local window mean of 120 and C = 10; the pixel value is 115. Is the pixel foreground or background?

Foreground — T = mean − C = 110 and 115 > 110; always subtract C before comparing, never compare against the raw mean.

Which clustering method minimizes the total squared distance between every sample and the mean of its assigned cluster?

K-means — each round it assigns every sample to its nearest mean, then moves each mean to the average of its samples; both steps can only shrink the total squared distance, so repeating them drives that objective down until it stops changing.

K-means on 1-D data {2, 4, 10, 12} with k = 2 and initial means {2, 12}: what are the means after one assign-and-update step?

{3, 11} — {2,4} and {10,12} get assigned, then each mean is recomputed; the next pass changes nothing, so this is also convergence.

True or false: in K-means, the number of clusters can change during training.

False — k is fixed for the whole run; what varies is the result, which depends on the initial means and only reaches a local minimum.

What is the key difference between K-medoids and K-means?

K-medoids uses actual data points as cluster centers (swapped when total cost drops), making it robust to outliers — it still needs k in advance.

Which clustering algorithm repeatedly shifts each point toward the centroid of a window around it, climbing to density peaks, so the number of clusters emerges automatically?

Mean shift — hill-climbing to density modes; no k needed, but more expensive than K-means.

In which algorithm is a point a 'core point' when at least minPts points lie within distance eps of it?

DBSCAN — density-based clustering with eps/minPts; it finds arbitrary-shaped clusters and explicitly labels sparse points as noise.

DBSCAN with eps = 3 and minPts = 5: a point has only 2 neighbors within eps, but one of them is a core point. How is the point classified?

Border point — too few neighbors to be core, but inside a core point's neighborhood; noise would require NO core point within eps at all.

This operation computes B(x,y) = h'(H(x,y), S(x,y)), where h' is the max-normalized 2-D Hue–Saturation histogram of an ROI. What is it?

Back projection — every pixel is scored by the ROI's H-S histogram value, producing a color-probability map.

When segmenting by color distance in HSV space, why is the V (value) channel usually dropped?

V mainly encodes illumination — matching on H and S only makes the labels robust to lighting changes.

The line parameterization x cos(theta) + y sin(theta) = rho is used by which technique?

The Hough transform — the polar (normal) form keeps parameters bounded even for vertical lines, where the slope in y = ax + b blows up to infinity.

A 3x3 kernel has all coefficients equal to 1 except a center of −8. What is it?

The Laplacian kernel — isotropic point/line detector; its coefficients sum to zero, so flat areas give exactly zero response.

Why do first-derivative edge detectors produce thicker edges than second-derivative ones?

The first derivative is nonzero along the entire intensity ramp; the second derivative fires only at the ramp's onset and end.

Moving across a single intensity edge, what does the second derivative produce?

Two opposite-sign spikes with a zero crossing between them — the zero crossing marks the edge center, and the spike signs tell dark-to-light from light-to-dark.

At an edge point, how is the direction of the edge related to the gradient vector?

Perpendicular — the gradient points ACROSS the edge (direction of maximum intensity change), so the edge itself runs orthogonal to it.

The Sobel kernels differ from the Prewitt kernels in exactly one way. What is it?

Sobel's center weight of 2 — that doubled middle coefficient adds noise-suppressing smoothing; Prewitt uses uniform 1's.

Why does the Marr–Hildreth detector fuse a Gaussian and a Laplacian into a single LoG ('Mexican hat') operator?

The Gaussian suppresses noise below the sigma scale and the isotropic Laplacian catches edges of any orientation — linearity lets both run as one convolution.

In the Canny detector, what does nonmaxima suppression do?

Thins thick gradient ridges — it keeps only local maxima along the (quantized) edge-normal direction; linking weak edges is hysteresis's job, not NMS's.

Canny hysteresis uses T_L = 0.05 with the recommended 3:1 ratio. What is T_H, and when are the in-between pixels kept?

T_H = 0.15 (3 x 0.05) — weak pixels between the thresholds survive only if 8-connected to a strong (>= T_H) pixel.

Canny gives the cleanest, thinnest edges of the classic detectors. What is its main drawback?

It is the most complex and computationally expensive — smoothing + gradient + NMS + hysteresis; industry often just smooths and thresholds the gradient instead.

What is the main drawback of the plain (unmarked) watershed transform?

Over-segmentation — every noisy local minimum of the gradient starts its own basin; markers are the standard fix.

What is the purpose of markers in marker-controlled watershed segmentation?

They restrict flooding to real objects (injected a priori knowledge), preventing over-segmentation — internal markers seed objects, external ones the background.

In a binary image of scattered fruit blobs, each blob receives its own unique integer label so the blobs can be counted. What operation is this, and how does it differ from thresholding?

Connected Component Analysis (CCA): it assigns a distinct integer ID to each connected foreground region. Thresholding only creates a binary foreground/background mask; CCA labels the separate objects in that mask. Diagonal-only contact joins components under 8-connectivity but not 4-connectivity.

What does LDA optimize, how does it differ from PCA, and what is its maximum output dimension for C classes?

Linear Discriminant Analysis uses labels and seeks directions with high between-class scatter relative to within-class scatter. PCA ignores labels and maximizes overall variance. LDA can produce at most C-1 discriminant dimensions for C classes.

How does split-and-merge segmentation work, and what condition decides the final regions?

Start with the whole image. If its homogeneity predicate Q is false, split it into quadrants recursively; then merge adjacent regions only when their union satisfies Q. The result should be a connected partition whose individual regions satisfy Q and whose adjacent unions do not. Blocky boundaries are a common limitation.

In the formal definition of image segmentation, what must the predicate Q return for the union of two adjacent regions?

FALSE — if the merged pair still satisfied Q they should have been one region; regions must also be complete, connected, and disjoint.

A segmentation rule assigns g(x,y) = a if f(x,y) > T2, b if T1 < f(x,y) <= T2, c if f(x,y) <= T1 — a piecewise rule with more than one threshold. What is it?

Multi-level (multiple) thresholding — two or more thresholds T1 < T2 split the image into three or more classes; basic global thresholding is the single-T, two-class special case.

What is the search complexity of exhaustive multi-level Otsu with L gray levels and k thresholds, and of single-threshold (bimodal) Otsu?

O(L^k) — nested loops over every T1 < … < Tk combination (256³ for L=256, k=3) — versus O(L) for the single-threshold case; the exponential blow-up in k is why exhaustive multi-level Otsu doesn't scale.

A gradient image is treated as a topographic surface; water rises from the local minima, and a dam is built wherever two rising pools would merge. What segmentation method is this?

The watershed transform — each catchment basin becomes one region and the dams (watershed lines) become object boundaries; it runs on the GRADIENT image, whose ridges are the intensity edges the water is dammed at.

In OpenCV marker-controlled watershed, a flooding pixel's labeled neighbors carry basin labels {2, 3}. What label does it get, and what do 0 and −1 mean in the output?

−1 — floods from two different basins met, so it becomes a watershed boundary; one basin label among neighbors → join that basin, only-unknown neighbors → postpone. Conventions: unknown region = 0, basins = positive labels, final boundaries = −1, flooding lowest-gray-first via priority queue.

In the Hough transform, how does an edge pixel actually 'detect' a line?

Each edge point casts votes for every discretized (ρ, θ) line through it — a sinusoid in the accumulator array — and collinear points' sinusoids cross at one cell, so accumulator PEAKS are the detected lines; the bounded polar parameters are what keep the array finite.

A texture operator compares each of a pixel's 8 neighbors with the center, writes 1 where the neighbor is at least as bright, and reads the 8 bits (weighted by powers of 2) as a code from 0 to 255. What is it?

Local Binary Pattern (LBP) — one 8-bit code per pixel neighborhood; GLCM instead counts pixel pairs over the whole image.

A binary letter 'A' comes out noticeably thicker after a morphological operation. Which operation was applied?

Dilation — it grows the foreground by the structuring element; erosion is the opposite and would thin the letter.

One global threshold T works under uniform light. A left-to-right illumination gradient is added. What should change?

Use adaptive/local T(x,y) or correct/flatten the background first. The illumination gradient breaks the single-global-threshold assumption; do not merely retune one constant T.

For finite real logits, what exact conditions must a softmax output vector satisfy?

Every exact component satisfies 0 < pᵢ < 1 and Σpᵢ = 1. A negative entry or wrong sum is invalid. Exact 0/1 entries occur only as a limit or through finite-precision underflow/rounding; this is why the professor's all-positive sum-one option is the safe choice.

For a normalized GLCM with entries p_ij, which texture feature is defined as Σ_i Σ_j (i − j)²·p_ij?

Contrast — the (i−j)² weight rewards off-diagonal pairs (large gray-level jumps); homogeneity is the mirror-image with weight 1/(1+|i−j|).

Which GLCM texture feature is defined as Σ_i Σ_j p_ij / (1 + |i − j|)?

Homogeneity — the weight peaks at i = j, so mass near the diagonal (smooth texture) scores high; contrast rewards the off-diagonal mass instead.

Two foreground pixels touch only at one corner. How many connected components are there under 4-connectivity and under 8-connectivity?

Two components under 4-connectivity; one component under 8-connectivity. Diagonal contact is excluded by N4 but included by N8.

A binary image contains 3 connected foreground components and 4 holes in total. What is its Euler number?

E = C − H = 3 − 4 = −1. Count all connected components and all holes before subtracting.

Feature extraction consists of which two steps?

Feature detection followed by feature description — first find the features (e.g., corners), then assign quantitative attributes to them.

Representing a shape's boundary as a sequence of direction numbers 0–7 along the contour is called what?

Freeman chain code — each boundary step becomes a direction digit; the polar signature is a distance-vs-angle function, not a digit string.

Define specificity. How is it different from recall?

Specificity = TN/(TN+FP): among the actual negatives, what fraction was correctly rejected. Recall (sensitivity) = TP/(TP+FN): among the actual positives, what fraction was found. Specificity answers the negative-class question; recall answers the positive-class question.

Taking the first difference of a chain code (counting direction changes) achieves which invariance?

Rotation invariance — direction changes ignore absolute orientation; starting-point invariance is the sibling trick (rotate the circular code to minimum magnitude).

φ1 = η20 + η02 is the first of a famous set of seven shape descriptors. What are they called?

Hu moment invariants — seven functions of normalized central moments, invariant to translation, scale, and rotation.

What is the main drawback of color/intensity histogram features?

They discard all spatial information — a striped and a checkered patch with the same color mix give identical histograms.

For a linear hard-margin SVM, what are the decision function and geometric margin width, and which training points determine the boundary?

Classify by sign(w^T x + b). The two supporting hyperplanes are w^T x+b=+1 and =-1, so geometric margin width is 2/||w||. Support vectors lie on those margin boundaries (or inside for soft margin) and determine w,b; far-away points usually do not.

M(x,y) = √(Gx² + Gy²), where Gx and Gy come from derivative kernels, defines what quantity?

Gradient magnitude — strength of the local intensity change; gradient direction is the sibling, tan⁻¹(Gy/Gx).

What is the main drawback of HOG as an object-detection descriptor?

Not invariant to rotation or scale — the fixed cell grid means other scales require an image-pyramid search.

A candidate pixel is tested against a circle of 16 surrounding pixels, with a quick pre-test on the 4 compass pixels 1, 5, 9, 13. Which detector is this?

FAST corner detection — a corner needs n contiguous circle pixels all brighter or all darker than center ± t; LBP uses the 8 immediate neighbors, not a 16-pixel circle.

A pyramid of progressively more blurred and down-sampled copies of an image is the core structure of which method?

SIFT scale space (octaves) — Gaussian blur plus downsampling by 2 per octave is exactly what gives SIFT its scale invariance.

D(x, y, σ) = L(x, y, kσ) − L(x, y, σ), where L is the Gaussian-smoothed image, defines what function?

Difference of Gaussians (DoG) — subtracting a more-blurred copy from a less-blurred one cancels everything except detail at that blur scale, cheaply approximating the Laplacian-of-Gaussian blob detector. SIFT takes keypoints where this response peaks (its extrema).

A SIFT descriptor uses a 16×16 neighborhood split into 4×4 subregions, each with an 8-bin orientation histogram. What is its dimensionality?

128 — 4 × 4 × 8; the 36-bin histogram is the separate keypoint-orientation step, not the descriptor.

The transform y = A(x − m_x), where the rows of A are eigenvectors of the covariance matrix ordered by descending eigenvalue, is called what?

PCA (Hotelling transform) — subtract the mean, then rotate onto axes derived from the data's own covariance; the DCT uses a fixed basis instead.

When reducing dimensionality with PCA, which components do you keep?

Those with the largest eigenvalues — they carry the most variance; the reconstruction error equals the sum of the discarded eigenvalues.