Day 13: Transparent Origami

Setup

Mix.install([
  {:kino, "~> 0.4.1"},
  {:nx, "~> 0.1.0-dev", github: "elixir-nx/nx", branch: "main", sparse: "nx"}
])

* Getting nx (https://github.com/elixir-nx/nx.git - origin/main)
remote: Enumerating objects: 10883, done.        
remote: Counting objects: 100% (2872/2872), done.        
remote: Compressing objects: 100% (651/651), done.        
remote: Total 10883 (delta 2393), reused 2547 (delta 2210), pack-reused 8011        
Resolving Hex dependencies...
Dependency resolution completed:
New:
  kino 0.4.1
* Getting kino (Hex package)
==> kino
Compiling 19 files (.ex)
Generated kino app
==> nx
Compiling 23 files (.ex)
Generated nx app

:ok

input = Kino.Input.textarea("Place your input")

This seems like a problem that would be nice to solve with some matrix transformations.

There is a projection matrix M that will map to the first fold K.

$I . M = K$

I think I can derive the formula by intuition, let’s test with a 8x4 matrix

a b c d
e f g h
i j k i
l m n o
p q r s
t u v x
z w y å
∫ ç ∂ ƒ

For a horizontal fold, along the center of the y axis, the resulting matrix would be

a+∫ b+ç c+∂ d+ƒ
e+z f+w g+y h+å
i+t j+u k+v i+x
...

Considering how matrix product works, we need to transpose our original matrix then multiply by

1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0

Which we can transpose again into the desired result.

I think the issue is if we see a transform that is not symmetrical… but let’s leave that for later

$(I^T.M)^T = R$ which implies, by multiplication property, that $I.M^T = R$. So if we simply transpose our original transformation matrix we can do a simple

$R = I.N$ where $N = M^T$

For now we can clearly see that the pattern for M is

1 0 0
0 1 0
0 0 1
... ...
0 0 1
0 1 0
1 0 0

eye = Nx.eye(8)

#Nx.Tensor<
  s64[8][8]
  [
    [1, 0, 0, 0, 0, 0, 0, 0],
    [0, 1, 0, 0, 0, 0, 0, 0],
    [0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 1, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0],
    [0, 0, 0, 0, 0, 1, 0, 0],
    [0, 0, ...],
    ...
  ]
>