The Modern Mathematics of Deep Learning

Local sparsity

Penguin learning project

The project uses 342 complete Palmer penguin records: 240 training-pool and 102 held-out rows. The selected training rows determine feature scaling and model parameters. Source CSV

This project trains weights with softmax, unlike the homepage lab, which uses nearest species averages. Same penguin measurements; different classification methods. Predictions can differ.

Frames and sparse signals · Encode or reconstruct signals with frames and dictionaries.

The project scales bill and flipper measurements using training-row means and scales; its feature vector has two entries.

Consider this concept alongside the project; its source assumptions may differ. These observations do not prove a theorem.

Input: 12 × 2; Held-out evaluation: 102 · #58 · 40.6 mm, 193 mm. Known answer: Adélie; Prediction: Adélie · [≈0.7383, ≈0.2444, ≈0.01737].

Used in these source sections: 7.2 A few building blocks at a time

For each circular position count nonzero coefficients in the local window across channels, then take the largest count.

These computations organize features and sparse coefficients.

Source formula example

\|\phi\|^{(n)}_{0,\infty}\coloneqq\max_{g\in G}\|((\phi_{j})_{g+k})_{j\in[C],k\in[2n-1]}\|_{0},

illustration · The invented finite inputs illustrate this expression. They do not prove a general identity, bound, or existence claim.

Symbols

$φ$: Collection of coefficient channels on a finite group.
$G$: Set of positions.
$n$: Positive integer determining a window of length 2n−1.
$\Vert ·\Vert ₀$: Number of nonzero entries.
$C$: Number of coefficient channels.

Source formula example

This source example uses its own inputs and assumptions. Editing it does not change the project.

Edit a value or the code, then run the example. This is a finite demonstration, not a proof of the paper’s theorem.

Last coefficient

This example’s data flow

You choose

input.last = 1

Calculated here

Result expression: { counts, maximum: Math.max(...counts) }

channels = [ [1, 0, input.last], [0, 1, 0], ] Uses: input.last
counts = [0, 1, 2].map((g) => channels.reduce( (sum, c) => sum + [0, 1, 2].filter((k) => c[(g + k) % 3] !== 0).length, 0, ), ) Uses: channels

Returns

Example result

counts: array (3 items): [3, 3, 3]
maximum: 3

JavaScript

Read numbers from input and return the result. JavaScript basics.

Result

Enable JavaScript to run this example.

Technical statement

The example uses n = 2, so the local window has 2n − 1 = 3 positions. The maximum is taken over every position in this finite group.

Source: Source expression