The Modern Mathematics of Deep Learning

Sparse hierarchy

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

At each level a dictionary transforms a sparse coefficient vector into the vector at the previous level.

These computations organize features and sparse coefficients.

Source formula example

\phi^{(\ell-1)}=D^{(\ell)}\phi^{(\ell)}\quad\text{and}\quad\|\phi^{(\ell)}\|^{(n_{\ell})}_{0,\infty}\leq k_{\ell},\quad\ell\in[L].

illustration · This finite hierarchy satisfies the displayed relations for the chosen values, without proving coefficient recovery.

Symbols

$D⁽ˡ⁾$: Dictionary matrix at layer ℓ.
$φ⁽ˡ⁾$: Coefficient vector at layer ℓ.
$kₗ$: Upper bound on local nonzero count.
$L$: Positive number of layers.

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.

Deep coefficient

This example’s data flow

You choose

input.value = 2

Initial values in code

D1 = [ [1, 0], [0, 1], ]
D2 = [ [1, 0], [0, 1], ]

Calculated here

Result expression: { D1, D2, deep, middle, data, deepLocalSparsity: localCount(deep), middleLocalSparsity: localCount(middle), bound: 1, }

deep = [input.value, 0] Uses: input.value
middle = multiply(D2, deep) Uses: multiply, D2, deep
data = multiply(D1, middle) Uses: multiply, D1, middle

Returns

Example result

deep: array (2 items): [2, 0]
middle: array (2 items): [2, 0]
data: array (2 items): [2, 0]
deepLocalSparsity: 1
middleLocalSparsity: 1
bound: 1
D1: array (2 items): [array (2 items): [1, 0], array (2 items): [0, 1]]
D2: array (2 items): [array (2 items): [1, 0], array (2 items): [0, 1]]

JavaScript

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

Result

Enable JavaScript to run this example.

Technical statement

The two toy dictionaries are 2 by 2 matrices, there is one channel, two circular positions, and n₁ = n₂ = 1, so each local window has one entry. Both layers satisfy the chosen sparsity bound k₁ = k₂ = 1. Both dictionaries are identity matrices, representing delta-filter circular convolutions with window length 1 < 2 positions.

Source: Source expression