Data augmentation for images: Exercises — AI-ML deep dive

This page is a self-test. Each problem states cleanly what to find; the full worked solution hides inside a collapsible [!recall]- callout so you can try first, then reveal. Problems climb from recognition to mastery. After each level, a [!mistake] callout steel-mans the trap most people fall into there.

Prerequisite ideas (build them if rusty): the parent note, 5.2.3-Overfitting-vs-underfitting, 6.14-Generalization-in-ML, 3.4.8-Dropout-in-neural-networks, 3.4.12-Transfer-learning-CNs.

Everything below reuses only symbols the parent defined. Where a picture helps (rotation, flip, crop), a chalkboard figure carries the geometry.

Level 1 — Recognition

Recall Solution 1.1

A transformation is label-preserving when a human still reads the same digit afterward.

(a) Horizontal flip — NO. A flipped 2 looks like a mirror 2 (not a real digit); a 3 becomes a backwards shape. Mirror-symmetry is not respected by digits.
(b) Rotation $+ 8°$ — YES. A slightly tilted 5 is still a 5. Small tilts occur naturally in handwriting.
(c) Rotation $180°$ — NO. A 6 rotated $180°$ becomes a 9. The label changes, so this breaks the constraint.
(d) Brightness $+ 20$ — YES. A brighter 7 is still a 7; intensity does not carry digit identity.

Answer: label-preserving = (b), (d); not = (a), (c).

Recall Solution 1.2

(i) = Cutout / Random Erasing — masks a rectangle to zero.
(ii) = MixUp — $\tilde{x} = λ x_{i} + (1 - λ) x_{j}$ , and crucially the label mixes too.
(iii) = Horizontal Flip — $x^{'} = W - x - 1$ .

Level 2 — Application

Recall Solution 2.1

Use the parent formula $x^{'} = W - x - 1, y^{'} = y$ . $x^{'} = 32 - 5 - 1 = 26, y^{'} = 10.$ The pixel moves to column 26, row 10 — the same distance from the right edge (5 from left → 5 from right, and $31 - 5 = 26$ ). The figure shows the original dot and its mirror partner reflected across the vertical centre line. Answer: $(26, 10)$ .

Recall Solution 2.2

$p^{'} = 1.2 \times 100 + (- 30) = 120 - 30 = 90.$ $90$ is inside $[0, 255]$ , so no clipping needed. Answer: $p^{'} = 90$ . Note the two effects fight: contrast pushed it up to 120 (it was below the mid-grey 127.5, so $α > 1$ actually darkens dark pixels only if $β$ shifts them — here the net move is down because $β$ dominated).

Recall Solution 2.3

$H^{'} = (350 + 40) mod 360 = 390 mod 360 = 30°.$ Why modulo? Hue is a circle: $0°$ and $360°$ are the same red. Adding $40°$ to $350°$ walks past the top of the wheel and lands at $30°$ . Without the wrap we'd get an illegal $390°$ . Answer: $H^{'} = 30°$ .

Level 3 — Analysis

Recall Solution 3.1

At $θ = 90°$ : $cos 90° = 0, sin 90° = 1$ . $x^{'} = 3 (0) - 0 (1) = 0, y^{'} = 3 (1) + 0 (0) = 3.$ So $(3, 0) \mapsto (0, 3)$ — the point on the +x axis swings up to the +y axis. Look at the blue quarter-turn arc in the figure. At $θ = 180°$ : $cos = - 1, sin = 0$ . $x^{'} = 3 (- 1) - 0 = - 3, y^{'} = 3 (0) + 0 (- 1) = 0.$ So $(3, 0) \mapsto (- 3, 0)$ ✓ — a half-turn flips the point to the opposite side. Answers: $(0, 3)$ then $(- 3, 0)$ .

Recall Solution 3.2

$A_{orig} = 256 \times 256 = 65536$ pixels².

Smallest area: $0.08 \times 65536 = 5242.88$ → side $= 5242.88 \approx 72.41$ px.
Largest area: $1.0 \times 65536 = 65536$ → side $= 256$ px. So crops range from about $72 \times 72$ up to $256 \times 256$ , all then resized to the network's fixed input. The smallest crop occupies only $8%$ of the image, forcing scale invariance: the model must recognise an object even when it fills a large fraction of a zoomed-in view. Answers: min side $\approx 72.4$ px, max side $= 256$ px.

Recall Solution 3.3

$\tilde{y} = λ y_{i} + (1 - λ) y_{j} = 0.7 [1, 0, 0] + 0.3 [0, 1, 0] = [0.7, 0.3, 0] .$ The target is a soft label: 70% cat, 30% dog. This is exactly the "smoother decision boundary" idea — the network is asked for a graded, calibrated answer instead of a hard 0/1. Answer: $[0.7, 0.3, 0]$ .

Level 4 — Synthesis

Recall Solution 4.1

(a) Each epoch shows all $N$ images once, each freshly augmented: $N$ views per epoch. Over $E$ epochs: $N \times E = 10, 000 \times 50 = 500, 000 augmented views .$ (b) Here $K$ (distinct views per image) $= E = 50$ , one per epoch. It is called potentially infinite because the transforms draw continuous parameters — a rotation angle $θ$ , a crop area $A$ , a brightness $β$ from continuous ranges — so there is no finite catalogue of augmentations; each epoch samples new real-valued parameters and (almost surely) never exactly repeats. Answers: (a) $500, 000$ ; (b) $K = 50$ realised, unbounded in principle.

Recall Solution 4.2

Step 1 — flip. $x_{1} = 32 - 5 - 1 = 26, y_{1} = 10.$ New point $(26, 10)$ . Step 2 — rotate $90°$ . With $cos 90° = 0, sin 90° = 1$ : $x_{2} = x_{1} cos θ - y_{1} sin θ = 26 (0) - 10 (1) = - 10,$ $y_{2} = x_{1} sin θ + y_{1} cos θ = 26 (1) + 10 (0) = 26.$ Answer: $(- 10, 26)$ . What it means: augmentations compose — the order matters. Flipping first then rotating is generally not the same as rotating first then flipping (matrix multiplication is not commutative), which is why augmentation pipelines fix a definite order.

Recall Solution 4.3

Horizontal flip — RISKY. Anatomy is left–right asymmetric (heart on the left, liver on the right); flipping can create anatomically impossible images and mislead a model that has learned true side-specific cues. Use only with caution.
Vertical flip — NO. An upside-down chest never occurs; not realistic.
$180°$ rotation — NO. Same reason; degenerate, unrealistic.
Brightness $\pm 10%$ — YES. Simulates exposure/scanner differences; label unchanged.
Small rotation $\pm 5°$ — YES. Patient positioning varies slightly; realistic and label-preserving.
MixUp — CAUTIOUS YES. Blending can improve calibration, but blended X-rays are non-physical; used, but often with small $λ$ effect. Key principle: medical imaging tightens the "could this occur naturally?" test far more than natural photos.

Level 5 — Mastery

Recall Solution 5.1

The inner expectation is the average loss over the augmentations: $E_{t \sim T} [ℓ] = \frac{1}{2} (0.8) + \frac{1}{2} (0.4) = 0.4 + 0.2 = 0.6.$ Gradient descent minimises this averaged loss, not the loss on any one view. The effect: the model can't drive one view's loss to zero by memorising its exact pixels — it must lower the loss across all transformed versions simultaneously, which pushes it toward features invariant to $T$ . That invariance is the source of the generalization gain (link to 6.14-Generalization-in-ML). Answer: $0.6$ .

Recall Solution 5.2

$α = 0.2$ : $Var = \frac{1}{4 ( 2 ( 0.2 ) + 1 )} = \frac{1}{4 ( 1.4 )} = \frac{1}{5.6} \approx 0.1786.$
$α = 4$ : $Var = \frac{1}{4 ( 2 ( 4 ) + 1 )} = \frac{1}{4 ( 9 )} = \frac{1}{36} \approx 0.0278.$ Interpretation. Both distributions are symmetric about mean $\frac{1}{2}$ . The larger variance at $α = 0.2$ means $λ$ is pushed toward the extremes $0$ or $1$ — i.e. little actual mixing (one image dominates, gentle augmentation). The small variance at $α = 4$ means $λ$ clusters near $0.5$ — heavy, near-uniform mixing (both images contribute almost equally). This is exactly why the parent recommends small $α \in [0.2, 0.4]$ : enough blending to smooth boundaries, not so much that inputs become mush. Answers: $\approx 0.1786$ and $\approx 0.0278$ .

Recall Solution 5.3

Independent per-pixel noise leaves every region partly visible, so a model can still lean on its single favourite discriminative patch. A contiguous rectangular mask removes an entire region's information at once, mimicking a real-world occluder (a tree in front of a car, a hand over a face). To keep the loss low when any region might vanish, the network must spread its evidence across multiple distributed features — precisely the robustness Cutout targets. So contiguity, not mere randomness, is the mechanism: it reproduces the structure of real occlusion. This is regularization in the same family as 3.4.8-Dropout-in-neural-networks (drop information to prevent over-reliance on any one path).

Recall Self-test checklist (reveal to grade yourself)

Can you flip a pixel with the correct $- 1$ ? ::: Yes — $x^{'} = W - 1 - x$ . Can you rotate $(3, 0)$ by $90°$ from the matrix? ::: Yes — $(0, 3)$ . Can you write a MixUp soft label at $λ = 0.7$ ? ::: $[0.7, 0.3, 0]$ for cat/dog. Do you know why the augmented objective uses an expectation, not a minimum? ::: Because we want invariance across all views, not the easiest one.