Object detection (R-CNN, YOLO, SSD): Exercises

This page is your self-testing gym for Object Detection. Every problem is graded by cognitive level (L1 → L5). Read the problem, try it yourself, THEN open the solution. Every number here is machine-verified.

Before we start, one shared picture of the vocabulary so no symbol sneaks up on you.

Definition The core quantities (every symbol you'll use)

A bounding box is a rectangle described by four numbers. Two common formats:
- corner format: $(x_{1}, y_{1}, x_{2}, y_{2})$ = top-left corner and bottom-right corner.
- center format: $(x, y, w, h)$ = center point plus width and height.
Intersection over Union (IoU) measures how much two boxes overlap: the area they share divided by the area they cover together.
Confidence in YOLO = $P (object) \times IoU$ — a single number saying "there is a thing here AND my box fits it".
The hat on a symbol always means "the network's prediction", while the plain symbol means "ground truth" (the true value we are trying to match). So $\overset{w}{^}$ is the predicted width and $w$ is the true width.
An indicator mask $1^{obj}$ is just a switch that equals $1$ when a condition is true and $0$ otherwise. In YOLO, $1_{ij}^{obj} = 1$ means "box $j$ in cell $i$ is responsible for a real object" (its cell holds the object's center and it has the highest IoU); otherwise it is $0$ . Its opposite, $1_{ij}^{noobj}$ , equals $1$ exactly when the box is not responsible.
Two balancing weights rescale parts of the YOLO loss: $λ_{coord} = 5$ boosts the localization terms (position matters a lot but is rare), and $λ_{noobj} = 0.5$ shrinks the no-object confidence term (background boxes are everywhere and would otherwise drown the gradient).

The figure below turns those two box formats into one picture: the black box on the left is drawn in corner format (top-left and bottom-right dots), and the red box on the right is the same idea in center format (center dot plus the width $w$ and height $h$ arrows). Keep it in mind — every exercise below picks one of these two formats, and mixing them up is the fastest way to a wrong answer.

Figure — Object detection (R-CNN, YOLO, SSD)

Level 1 — Recognition

Exercise 1.1 — Which paradigm?

For each, name the detector family: (a) generates ~2000 region proposals with Selective Search, then classifies each; (b) divides the image into an $S \times S$ grid and predicts everything in one forward pass; (c) uses default boxes on several feature maps of different resolution.

Recall Solution

(a) R-CNN (two-stage, region-based). (b) YOLO (single-stage, grid regression). (c) SSD (single-shot, multi-scale default boxes).

Exercise 1.2 — Output tensor size

YOLO v1 uses grid $S = 7$ , $B = 2$ boxes per cell, $C = 20$ classes. What is the shape of the output tensor?

Recall Solution

Each cell outputs $B \cdot 5 + C$ numbers (5 = $x, y, w, h, confidence$ per box). $B \cdot 5 + C = 2 \cdot 5 + 20 = 30.$ Shape $= 7 \times 7 \times 30 = 1470$ numbers total.

Level 2 — Application

Exercise 2.1 — Compute an IoU

Box A (corner format) $= (10, 10, 30, 30)$ . Box B $= (20, 20, 40, 40)$ . Compute the IoU.

Recall Solution

Step 1 — intersection rectangle. Take the inner edges (see red box in figure): $x_{1} = max (10, 20) = 20, y_{1} = max (10, 20) = 20,$ $x_{2} = min (30, 40) = 30, y_{2} = min (30, 40) = 30.$ Width $= 30 - 20 = 10$ , height $= 30 - 20 = 10$ , so intersection area $= 10 \times 10 = 100$ .

Step 2 — each area. $A = 20 \times 20 = 400$ , $B = 20 \times 20 = 400$ .

Step 3 — union. Union = sum minus the double-counted overlap: $U = 400 + 400 - 100 = 700.$

Step 4 — divide. $IoU = \frac{100}{700} = \frac{1}{7} \approx 0.1429.$

Exercise 2.2 — Disjoint boxes

Box A $= (0, 0, 10, 10)$ , Box B $= (20, 20, 30, 30)$ . What is their IoU?

Recall Solution

Intersection edges: $x_{1} = max (0, 20) = 20$ , $x_{2} = min (10, 30) = 10$ . Width $= x_{2} - x_{1} = 10 - 20 = - 10 < 0$ . A negative width means the boxes do not touch, so we clamp the area to $0$ . $IoU = \frac{0}{U} = 0 .$

Exercise 2.3 — Bounding-box regression targets

Proposal $P = (P_{x}, P_{y}, P_{w}, P_{h}) = (100, 100, 50, 40)$ . Ground truth $G = (120, 90, 60, 50)$ . Compute the regression targets $t_{x}, t_{y}, t_{w}, t_{h}$ .

Recall Solution

$t_{x} = \frac{G _{x} - P _{x}}{P _{w}} = \frac{120 - 100}{50} = 0.4, t_{y} = \frac{G _{y} - P _{y}}{P _{h}} = \frac{90 - 100}{40} = - 0.25.$ $t_{w} = lo g \frac{G _{w}}{P _{w}} = lo g \frac{60}{50} = lo g (1.2) \approx 0.1823,$ $t_{h} = lo g \frac{G _{h}}{P _{h}} = lo g \frac{50}{40} = lo g (1.25) \approx 0.2231.$ Why divide the center offset by $P_{w}$ and $P_{h}$ ? Picture two proposals hunting the same object: a tiny $10$ px-wide box and a huge $500$ px-wide box. Suppose both need to shift right by $20$ px to reach the target. For the tiny box, $20$ px is a huge move (twice its own width!); for the huge box, $20$ px is a tiny nudge (4% of its width). If the network had to output the raw $20$ , it would need to know the box size to judge whether that's a big or small correction. Dividing by $P_{w}$ converts the shift into "how many box-widths do I move?" — a scale-free number ( $t_{x} = 2.0$ for the tiny box, $t_{x} = 0.04$ for the huge one). Now the same predicted $t_{x}$ means the same relative adjustment no matter how big the proposal is, so one set of weights works for all box sizes. That's the same reason $t_{w}, t_{h}$ use a ratio inside the log rather than a raw pixel difference.

Why the log? Width/height are always positive; learning $lo g$ of a ratio lets the network predict any real number and exponentiate back to a guaranteed-positive size.

Level 3 — Analysis

Exercise 3.1 — Why $w, h$ in YOLO's size loss?

Recall the hat convention: $w$ is the true width and $\overset{w}{^}$ is the network's predicted width. Box P has true width $w = 10$ px, box Q has true width $w = 200$ px. Both are predicted $10$ px too large ( $\overset{w}{^} = 20$ and $\overset{w}{^} = 210$ ). Compare the raw squared error $(w - \overset{w}{^})^{2}$ against the square-root error $(w - \overset{w}{^})^{2}$ for each box.

Recall Solution

Raw error — identical for both: $(20 - 10)^{2} = 100, (210 - 200)^{2} = 100.$ Raw loss says a 10px slip on a tiny box is just as bad as on a giant box — wrong intuition.

Square-root error: $(20 - 10)^{2} = (4.472 - 3.162)^{2} \approx 1.716,$ $(210 - 200)^{2} = (14.491 - 14.142)^{2} \approx 0.1218.$ Now the small box is penalized $\approx 14 \times$ harder. That's the point: a 10px error swallows half a 10px object but is negligible on a 200px object. The $\cdot$ compresses large values so equal absolute slips produce smaller loss on big boxes.

Exercise 3.2 — Proposal counting

Faster R-CNN: image $800 \times 600$ , backbone stride $16$ , $k = 9$ anchors per position. How many raw proposals? After NMS keeps the top 300, what fraction survives?

Recall Solution

Rule for non-integer feature sizes: a strided/padded convolution reports $⌊ \frac{input}{stride} ⌋$ output cells (you cannot have half a cell), so we floor. Feature map width $= ⌊ \frac{800}{16} ⌋ = 50$ , height $= ⌊ \frac{600}{16} ⌋ = ⌊ 37.5 ⌋ = 37$ . Positions $= 50 \times 37 = 1850$ . $proposals = 1850 \times 9 = 16650.$ Surviving fraction $= \frac{300}{16650} \approx 0.01802 = 1.80%.$ Over 98% of proposals are discarded — most anchors are background or redundant overlaps, exactly what NMS + objectness filtering are for.

Level 4 — Synthesis

Exercise 4.1 — Run NMS by hand

Three "dog" boxes with confidences: $D_{1}$ conf $0.9$ , $D_{2}$ conf $0.75$ , $D_{3}$ conf $0.6$ . IoUs: $IoU (D_{1}, D_{2}) = 0.6$ , $IoU (D_{1}, D_{3}) = 0.1$ , $IoU (D_{2}, D_{3}) = 0.05$ . Run NMS with threshold $0.5$ . Which boxes survive?

Recall Solution

Step 1. Sort by confidence: $D_{1} (0.9), D_{2} (0.75), D_{3} (0.6)$ . Step 2. Pick highest, $D_{1}$ → keep it. Remove any box overlapping $D_{1}$ above $0.5$ :

$IoU (D_{1}, D_{2}) = 0.6 > 0.5$ → suppress $D_{2}$ .
$IoU (D_{1}, D_{3}) = 0.1 < 0.5$ → keep $D_{3}$ for now. Step 3. Next survivor is $D_{3} (0.6)$ → keep it. No boxes left to compare. Survivors: $D_{1} and D_{3}$ . $D_{2}$ was a near-duplicate of the stronger $D_{1}$ .

Exercise 4.2 — Full YOLO class-specific confidence

Cell (3,4) box 1 predicts confidence $0.85$ . The cell predicts $P (dog ∣ object) = 0.92$ and $P (cat ∣ object) = 0.05$ . Compute the final class-specific score for dog and for cat. Which class label does this box emit if the decision threshold is $0.5$ ?

Recall Solution

Final score $= P (C_{i} ∣ object) \times confidence$ : $dog : 0.92 \times 0.85 = 0.782, cat : 0.05 \times 0.85 = 0.0425.$ Dog $= 0.782 > 0.5$ → emit "dog". Cat $= 0.0425 < 0.5$ → suppressed. The confidence $0.85$ ("is something here + does my box fit") multiplies the per-class belief $0.92$ ("given something, it's a dog"), giving one honest number.

Level 5 — Mastery

Exercise 5.1 — Design an SSD anchor budget

An SSD head attaches to feature maps of sizes $38 \times 38$ , $19 \times 19$ , $10 \times 10$ , $5 \times 5$ , $3 \times 3$ , $1 \times 1$ . The number of default boxes per cell for these layers is $4, 6, 6, 6, 4, 4$ respectively. Compute the total number of default boxes. Then explain why the $38 \times 38$ map uses fewer boxes per cell yet contributes the most boxes overall — and which object sizes it targets.

Recall Solution

Multiply cells × boxes-per-cell for each layer: $3 8^{2} \cdot 4 = 5776, 1 9^{2} \cdot 6 = 2166, 1 0^{2} \cdot 6 = 600,$ $5^{2} \cdot 6 = 150, 3^{2} \cdot 4 = 36, 1^{2} \cdot 4 = 4.$ Total: $5776 + 2166 + 600 + 150 + 36 + 4 = 8732 .$ Why the interpretation: the $38 \times 38$ map is high resolution / early → each cell sees a small receptive patch → it detects small objects, and there are $1444$ cells, so even at only 4 boxes/cell it dominates the count. The $1 \times 1$ map is low resolution / deep → one cell sees the whole image → detects large objects, needing just 4 boxes. This resolution ladder is exactly why SSD (unlike single-scale YOLO v1) handles a wide range of object sizes. See Anchor Boxes.

The figure below draws that resolution ladder: read it left-to-right as "deeper in the network". The leftmost red box is the $38 \times 38$ map — biggest grid, smallest receptive patch per cell, so it hunts small objects; the rightmost $1 \times 1$ map is a single cell seeing the whole image, so it hunts large objects. This is the picture behind the $8732$ count you just computed.

Exercise 5.2 — Weighted no-object confidence loss

A YOLO cell has $B = 2$ predicted boxes but no object. Their confidence predictions are $\hat{C}_{1} = 0.3$ , $\hat{C}_{2} = 0.1$ (targets $= 0$ ). With $λ_{noobj} = 0.5$ , compute this cell's no-object confidence loss. Then argue why $λ_{noobj} < 1$ is necessary.

Recall Solution

No-object confidence loss $= λ_{noobj} \sum_{j} (C_{j} - \hat{C}_{j})^{2} = 0.5 [(0 - 0.3)^{2} + (0 - 0.1)^{2}]$ $= 0.5 (0.09 + 0.01) = 0.5 \times 0.10 = 0.05.$ Why down-weight: in a $7 \times 7$ grid almost every cell is background. If each contributed full-weight confidence loss, the gradient would be swamped by "push confidence to 0" signals, and the handful of object cells would be drowned out. $λ_{noobj} = 0.5$ rebalances so the rare positive cells still steer learning — the mirror image of $λ_{coord} = 5$ which boosts the rare localization signal.

Recall Self-check (cloze)

IoU intersection uses $max$ on the ::: top-left / start edges and $min$ on the bottom-right / end edges. Width or height of the intersection must be ::: clamped to zero (negative → no overlap). YOLO uses $w, h$ so that equal absolute errors hurt ::: small boxes more than large boxes. NMS is run ::: per class, keeping the highest-confidence box and suppressing overlaps above the IoU threshold. SSD detects small objects on ::: high-resolution (early) feature maps and large objects on deep low-resolution maps.

See also: Non-Maximum Suppression · Intersection over Union (IoU) · Anchor Boxes · Image Classification · Semantic Segmentation · Instance Segmentation