In Vision I an encoder learned from two labelled scenes. A label is a fact about an image that a person wrote down, and people are slow. Images, on the other hand, arrive by the billion without anyone writing anything. This part asks what an encoder can learn from an image alone. The answer is the one trick behind MAE, DINO and I-JEPA: hide or transform part of the image and ask the model for what the image already contains. We keep the same 8×8 scene and the frozen encoder from Vision I, so every number in the three methods can be traced on the page.

A masked autoencoder hides patches, runs the encoder on the visible ones only, and lets a small decoder predict the hidden pixels. The loss compares the prediction with the pixels that were hidden. Our version keeps Vision I's encoder frozen and trains only the decoder: for every hidden slot a placeholder row asks the visible patches a question, reads them through the same keys and values the encoder already computes, and turns four numbers into four pixels. That is 32 trained numbers. The frames below follow the slot of patch 8, the centre of the right mug, and show the decoder after training; the next section trains it.

Training changes the decoder's 32 numbers and nothing else: the encoder stays as Vision I left it. The loss is the mean squared error over the hidden pixels, averaged over three training scenes with the same four patches hidden: A (two mugs), B (one mug) and D (book only). Press the step buttons to move along the curve and to redraw the prediction. Then look at the last frame. Scenes A and B have the same visible patches, so the decoder must give one answer for two different holes. The best single answer under a squared error is between them, and that is the grey mug.

Every value here follows the formula of section 3. Choose a scene and a training step; the table lists the four hidden patches, the true pixels, the decoder's four numbers and the sum of squared errors per patch. The footer divides the total by the number of hidden pixels, sixteen, and that quotient is the loss the optimizer saw for this scene. The visible pixels never enter: they are copied into the drawing so you can see the whole scene, but the decoder is not asked about them and is not scored on them.

A second family of targets never hides anything. Take one scene, make two views of it, a flip and a darker crop, and ask the model to give the same answer for both. No label is needed: the pair itself says "these belong together". The choice of transformations is the whole content of that request. A flip keeps "two mugs" but swaps left and right; dimming keeps the layout but changes every value. Our encoder is frozen, so the views' features do not move; what moves is a small head on top of them, and the third frame shows its two outputs being pulled together.

DINO has a student and a teacher with the same architecture. The student learns by gradient descent; the teacher is a slow moving average of the student and receives no gradient. The student sees one view and outputs a distribution over some slots; the teacher sees the other view and produces the target distribution, after subtracting a running mean of its own logits (centring) and dividing by a smaller temperature (sharpening). Here the encoder is frozen and the two "networks" are two 4 × 3 heads on the CLS row, trained on scenes A, B, D and E with two views each. The numbers in every table come from that run.

I-JEPA keeps the masked-image setup and replaces the target. A target encoder sees the whole image; its output rows at the hidden positions are the targets. A context encoder sees only the visible patches, and a predictor guesses the target rows from them. The loss is a squared error between representations, and the target encoder follows the context encoder by a moving average. Here both encoders are the frozen Vision I encoder, so the target for patch 8 is simply its output row on the content axes, and the predictor is the placeholder read of section 2 with a 4 × 2 head instead of a 4 × 4 one: 22 trained numbers.

Pretraining gives you an encoder, not a promise. The standard test is a linear probe: freeze the encoder, pool its output rows into one feature per image, and fit a line with labels for a task it was never trained on. Our task is "two mugs or one mug?". The feature is the mean of the sixteen patch rows on brightness and contrast, so the probe is a line in that plane. Twelve views of three scenes give twelve points: A (two mugs) against B and C (one mug), each as the scene itself, flipped, dimmed, and cropped. The crops are held out.

Put the three methods side by side on the one patch we followed. MAE's target is the patch's four pixels, read off the image. DINO's target is a distribution over three slots, produced by a teacher head on another view of the whole scene. I-JEPA's target is the encoder's own row for the patch, two named numbers. All three targets are made from the image without a label, and all three are compared with something the model predicts from what it was shown. The table below also records which parameters received gradients in this toy and which followed the moving average, because that distinction is what keeps the two teacher methods from cheating.

Try to explain the part without the page: one sentence, one chain of operations, three losses. The sentence is what you would say in a corridor; the chain is what you would draw on a board; the losses are what you would write in an exam. Then answer the four questions before opening them. Each answer is computed from the toy, so you can check yours against a number. The notation card at the end collects every symbol this part added to Vision I's.