785 lines
24 KiB
Plaintext
785 lines
24 KiB
Plaintext
---
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title: "Counting in 2D 2: Reorienting Polynomials"
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draft: true
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format:
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html:
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html-math-method: katex
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date: "2021-03-07"
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date-modified: "2025-02-23"
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jupyter: python3
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categories:
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- algebra
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- python
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execute:
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echo: false
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---
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```{python}
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from pathlib import Path
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from IPython.display import Video, display
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from matplotlib import pyplot as plt
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import numpy as np
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from carry2d.carry import Carry
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from carry2d.expansion import add, poly_to_array
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from carry2d.anim import animate, animate_carry_count
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```
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[Previously](../cell1), I explored some very basic 2D counting systems, which turned out to include sandpiles.
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As the name suggests, these planar numbers require a discrete grid of numbers to contain them.
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What follows is a deeper investigation into this natural extension of positional systems.
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Direction of Propagation
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------------------------
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Given a carry (expressed as a polynomial in two variables or a grid) and an increasing sequence of integers,
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the expansions of the sequence will appear to propagate in a certain direction.
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Recall that from the last post, we know that $x + y = 2$ and a certain case of
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the folium of Descartes will make similar patterns:
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$$
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\begin{gather*}
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& ~ & \text{Folium} \\
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x + y - 2 & ~ &
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x^3 + y^3 - 2xy \\
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\begin{array}{|c} \hline
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\bar{2} & 1 \vphantom{2^{2^2}} \\
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1
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\end{array} & ~ &
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\begin{array}{|c} \hline
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0 & 0 & 0 & 1 \\
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0 & \bar{2} \\
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0 \\
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1
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\end{array}
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\end{gather*}
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$$
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The general direction in which we carry can be understood as the directions from the negative term of the carry
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(hereafter *pseudo-base*) to the other terms.
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These can interpreted vectors extending from the pseudo-base to every other term.
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The simplest way to get these vectors is by placing the pseudo-base at the constant position
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and reading off the exponent of the other terms.
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$$
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\begin{gather*}
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\begin{array}{|c} \hline
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\bar{2} & \textcolor{red}{1} \vphantom{2^{2^2}} \\
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\textcolor{blue}{1}
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\end{array} & ~ &
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\begin{array}{c|cc}
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0 & 0 & 0 & \textcolor{red}{1} \\ \hline
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0 & \bar{2} \vphantom{2^{2^2}} \\
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0 \\
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\textcolor{blue}{1}
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\end{array} \\
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\begin{matrix}
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\textcolor{red}{ x^1 y^0 } \implies \langle 1, 0 \rangle \\
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\textcolor{blue}{ x^0 y^1 } \implies \langle 0, 1 \rangle
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\end{matrix}
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& ~ &
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\begin{matrix}
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\textcolor{red}{ x^2 y^{-1} } \implies \langle 2, -1 \rangle \\
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\textcolor{blue}{ x^{-1} y^2 } \implies \langle -1, 2 \rangle
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\end{matrix}
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\end{gather*}
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$$
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It is clear that as we count, the expansions can extend only in a linear combination of these vectors.
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However, they cannot propagate in the reverse direction, as there is no sense of "backwards".
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### Balancing Act
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We can naively balance the carry by adding additional carry digits.
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To do this, we just need to double the pseudo-base and add reflections each vector in it.
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Equivalently, this adds the carry with the carry where we have replaced *x* and *y* with their reciprocals.
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If we balance the left carry, it yields the Laplacian.
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But what about the right?
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Unsurprisingly, it just stretches the expansions along the different "basis" vectors.
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::: {layout-ncol="2"}
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$$
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\textcolor{red}{ x^{-2} y^{1} } + x^{-1} y^{2} + \textcolor{blue}{ x^{1} y^{-2} } + x^{2} y^{-1} - 4 \\
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\begin{array}{cc|ccc}
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& & & \textcolor{blue}{1} \\
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& & 0 & 0 & 1 \\ \hline
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& 0 & \bar{4} & 0 \vphantom{2^{2^2}} \\
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\textcolor{red}{1} & 0 & 0 \\
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& 1
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\end{array} \\
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\text{Added Vectors:} \\
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\begin{matrix}
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\textcolor{red}{
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\langle -2, 1 \rangle
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} \\
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\textcolor{blue}{
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\langle 1, -2 \rangle
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}
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\end{matrix}
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$$
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```{python}
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if not Path("./count_balanced_folium.mp4").exists():
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try:
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animate_carry_count(
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carry = Carry([
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[ 0, 0, 0, 1, 0],
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[ 0, 0, 0, 0, 1],
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[ 0, 0,-4, 0, 0],
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[ 1, 0, 0, 0, 0],
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[ 0, 1, 0, 0, 0],
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]),
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operation="add",
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op_val=4,
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frames=list(range(200)),
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).save("count_balanced_folium.mp4")
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except ValueError:
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pass
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display(
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Video("./count_balanced_folium.mp4"),
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)
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```
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:::
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### Half-Balanced
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The Laplacian carry is remarkably balanced, but it provokes the question of what the expansions
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look like if there are only three directions to propagate.
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The intuition from above says that it should span a half-plane,
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up from the quarter-plane that $x + y = 2$ spans.
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::: {layout-ncol="2"}
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$$
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\textcolor{red}{ x^{-1} } + x^1 + y^1 - 3 \\
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\begin{array}{c|cc} \hline
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\textcolor{red}{1} & \bar{3} & 1 \vphantom{2^{2^2}} \\
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0 & 1 & 0
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\end{array} \\
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\text{Added Vector:} \\
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\begin{matrix}
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\textcolor{red}{
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\langle -1, 0 \rangle
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}
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\end{matrix}
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$$
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```{python}
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if not Path("./count_half_laplacian.mp4").exists():
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try:
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animate_carry_count(
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carry = Carry([
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[ 1,-3, 1],
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[ 0, 1, 0],
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]),
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operation="add",
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op_val=3,
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frames=list(range(200)),
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).save("count_half_laplacian.mp4")
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except ValueError:
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pass
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display(
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Video("./count_half_laplacian.mp4"),
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)
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```
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:::
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Fortunately, this is the case.
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### 2D Carry Types
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Based on the extent of the vectors, in general there appear to be four classes:
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1. The degenerate one-dimensional case
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- The carry is equivalent to the product of a polynomial in a single variable and any power *x* and/or *y*.
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2. Quarter-planes, like $x + y = 2$
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- Arise when the inverses of the vectors are inaccessible.
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3. Half-planes, like the above "semi-"Laplacian
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- Arise when two vectors are collinear and point in opposite directions,
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but every other vector is either on one side of this line or are also collinear.
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4. Full planes, like the Laplacian
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- Arise when the inverses of both vectors are accessible.
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Quarter-plane is a bit of a misnomer.
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Technically, the fraction of the plane spanned is between 0 and 1/2 noninclusive
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(as these bounds correspond to the previous and next cases respectively).
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For example, $x + y = 2$ actually to a quarter-plane, as the first row/column
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(which contain binary expansions) are unbounded.
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However, the Folium encompasses slightly more of the plane, or...
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$$
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\frac{1}{2\pi}\arccos \left (
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\frac{\langle 2, -1 \rangle \cdot \langle -1, 2 \rangle}
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{\| \langle 2, -1 \rangle \| \| \langle -1, 2 \rangle \|}
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\right)
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= \frac{1}{2\pi} \arccos \left ( -\frac 4 5 \right )
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\approx 0.397584
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$$
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...about 39% of the plane.
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This is still smaller than the half-plane case.
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Hunting for Implicitness
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------------------------
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With these categories in mind, I would now like to shift focus.
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All systems discussed thus far have been explicit and irreducible, meaning they are
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not derived from any "simpler" implicit rules.
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Do such rules exist, and cam we find any?
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As mentioned previously, in the 1D case, we form explicit carries from implicit ones by
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multiplying the latter by cyclotomic polynomials.
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It would be convenient if the same were true in two dimensions -- we generate an explicit 2D carry from
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the product of some input polynomial and a cyclotomic polynomial in x and y.
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However, I would like to leverage the visual representation used previously,
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since working with coefficients is linear, unintuitive, and most importantly, annoying.
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<!-- TODO: ipympl version? -->
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Above is a screenshot of (my admittedly low-effort) attempt to ameliorate these shortcomings.
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The left image is clickable, with the coefficients incrementing with left clicks and decrementing with right.
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The text box controls the second multiplicand in the center image by selecting a cyclotomic polynomial
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and evaluating it with a particular expression.
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The right image is the product.
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It could obviously be improved by showing the coefficients in the picture and using color to only indicate sign.
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However, there is a benefit to hiding coefficients behind solid colors.
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For instance, I don't need to recognize numerals to pinpoint what the product should look like.
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### Degenerate Expressions
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I have chosen to evaluate the right term (center image) at $x + y$ for a reason.
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Choosing *x* or *y* on their own is a poor decision because they change nothing from the one dimensional case.
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Put another way, when arranged in a grid, the coefficients are all collinear.
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At first blush, *xy* seems like a decent pick, but it has the exact same problem;
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the line simply extends along the diagonal instead.
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$x/y$ has the same problem, but it is instead collinear along the anti-diagonals.
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We allow negative exponents for the same reason we can carry at any location in an expansion
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```{python}
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import sympy
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from sympy.abc import x, y, z
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def operation_plots():
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six = z**2 - z + 1
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x6 = six.replace(z, x).expand()
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y6 = six.replace(z, y).expand()
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xy6 = six.replace(z, x*y).expand()
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x_y6 = (six.replace(z, x/y)).expand()
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image = np.zeros((lambda q: (q,q))(3), dtype=np.int32)
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for i, (poly, title) in enumerate(zip(
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[x6, y6, xy6, x_y6],
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["x", "y", "xy", "\\frac{x}{y}"]
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)):
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plt.subplot(2, 2, i + 1)
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if i != 3:
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plt.imshow(poly_to_array(poly, image))
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plt.xticks([0,1,2])
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plt.yticks([0,1,2])
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else:
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plt.imshow(poly_to_array((poly * y**2).expand(), image), extent=(-2.5, 0.5, 0.5, -2.5))
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plt.xticks([-2,-1,0])
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plt.yticks([-2,-1,0])
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plt.title(f"$\\Phi_6({title}) = {sympy.latex(poly)}$")
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plt.tight_layout()
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operation_plots()
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```
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It should go without saying that these are all degenerate, and can only span a line.
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On the other hand, expressions like the sum of *x* and *y* form triangles, 2D shapes
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from which it is possible to build our goal polynomial.
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Since $x + y$ is symmetric in *x* and *y*, it is expected that the products have a similar sense of symmetry.
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First Results
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-------------
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The first explicit product I was able to find is above.
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It can be decomposed as the folium of Descartes, but with an extra term to the upper-left of the pseudo-base.
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The implicit carry (the left multiplicand) describes the rearrangement of coefficients
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in a backwards L-shape to and from a sparser upside-down L-shape.
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Since the coefficients of this factor sum to zero, it passes through $(1, 1)$.
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$$
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\begin{gather*}
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\text{Folium} & ~ & \text{New carry} \\
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x^3 + y^3 - 2xy & ~ &
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x^3 + y^3 + 1 - 3xy \\
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\begin{array}{|c} \hline
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0 & 0 & 0 & 1 \\
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0 & \bar{2} \\
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0 \\
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1
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\end{array} & ~ &
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\begin{array}{|c} \hline
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1 & 0 & 0 & 1 \\
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0 & \bar{3} \\
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0 \\
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1
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\end{array}
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\end{gather*}
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$$
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### Cousin Systems
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My first thought upon seeing this carry was to see what makes counting in this system different
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from its "neighbors".
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By this, I mean moving the $x^3$ and $y^3$ terms further inward and outward.
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$$
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\begin{gather*}
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x + y + 1 - 3xy &
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x^2 + y^2 + 1 - 3xy \\
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\begin{array}{|c} \hline
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1 & 1 \\
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1 & \bar{3} \\
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\end{array} &
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\begin{array}{|c} \hline
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1 & 0 & 1 \\
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0 & \bar{3} \\
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1
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\end{array} \\
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\textcolor{red} {x^3 + y^3 + 1 - 3xy} &
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x^4 + y^4 + 1 - 3xy \\
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\textcolor{red} {
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\begin{array}{|c} \hline
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1 & 0 & 0 & 1 \\
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0 & \bar{3} \\
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0 \\
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1
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\end{array}
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} &
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\begin{array}{|c} \hline
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1 & 0 & 0 & 0 & 1 \\
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0 & \bar{3} \\
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0 \\
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0 \\
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1
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\end{array}
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\end{gather*}
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$$
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All polynomials besides the one under consideration (highlighted in red) are irreducible.
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This means that these carries hide no implicit rules, and are as fundamental as the Laplacian or $x + y = 2$.
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Analyzing the carry vectors, it is clear that the pattern produced by each should be different:
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$$
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\begin{gather*}
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x^{-1} y^{0} + x^{-1} y^{-1} + x^{0} y^{-1} - 3 &
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x^{-1} y^{1} + x^{-1} y^{-1} + x^{1} y^{-1} - 3 \\
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\begin{matrix}
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\langle -1, 0 \rangle \\
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\langle -1, -1 \rangle \\
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\langle 0, -1 \rangle
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\end{matrix} &
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\begin{matrix}
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\langle -1, 1 \rangle \\
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\langle -1, -1 \rangle \\
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\langle 1, -1 \rangle
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\end{matrix} \\
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x^{-1} y^{2} + x^{-1} y^{-1} + x^{2} y^{-1} - 3 &
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x^{-1} y^{3} + x^{-1} y^{-1} + x^{3} y^{-1} - 3 \\
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\begin{matrix}
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\langle -1, 2 \rangle \\
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\langle -1, -1 \rangle \\
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\langle 2, -1 \rangle
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\end{matrix} &
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\begin{matrix}
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\langle -1, 3 \rangle \\
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\langle -1, -1 \rangle \\
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\langle 3, -1 \rangle
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\end{matrix} \\
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\end{gather*}
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$$
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The vectors in the top-left carry contain only negative numbers, so it will tend toward
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the quarter-plane case like $x + y = 2$.
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In the top-right carry, all vectors are equidistant and meet at right (and straight) angles.
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This means that it is a 45° rotation of the altered Laplacian, which spans a half-plane.
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The remaining two should both propagate into the full plane. Is this reflected in the counting videos?
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::: {}
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```{python}
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def animate_triangle_carries(**kwargs):
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dims = 80
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center = (dims // 2, dims // 2)
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carries = [
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Carry([
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[1, 1],
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[1,-3]
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]),
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Carry([
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[1, 0, 1],
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[ 0,-3, 0],
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[ 1, 0, 0]
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]),
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Carry([
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[1, 0, 0, 1],
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[ 0,-3, 0, 0],
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[0, 0, 0, 0],
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[1, 0, 0, 0]
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]),
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Carry([
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[1, 0, 0, 0, 1],
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[ 0,-3, 0, 0, 0],
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[0, 0, 0, 0, 0],
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[0, 0, 0, 0, 0],
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[1, 0, 0, 0, 0]
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]),
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]
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expansions = [np.zeros((dims, dims), dtype=np.int64) for _ in range(4)]
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subplots = [None for _ in range(4)]
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fig = plt.gcf()
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def init():
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for i in range(len(subplots)):
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plt.subplot(2, 2, i + 1)
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plt.title(f"$x^{{{i+1}}} + y^{{{i+1}}} + ...$")
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subplots[i] = plt.imshow( # type: ignore
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expansions[i],
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extent=[-center[0], dims - center[0], dims - center[1], -center[1]], # type: ignore
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)
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subplots[i].set_clim(0, 2) # type: ignore
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fig.tight_layout()
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@animate(fig, init_func=init, **kwargs)
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def ret(_):
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for i in range(len(expansions)):
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add(expansions[i], carries[i], 3, center=center)
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subplots[i].set_data(expansions[i]) # type: ignore
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return ret
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if not Path("./count_triangle_carries.mp4").exists():
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try:
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animate_triangle_carries(
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frames=list(range(200))
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).save("./count_triangle_carries.mp4")
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except ValueError:
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pass
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Video("./count_triangle_carries.mp4")
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```
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Counting in each of the $x^n + y^n + \ldots$ systems
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:::
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Indeed it is.
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However, while the discovered carry remarkably remains centered, the final one tends more toward
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propagating to the lower right.
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It has somehow exceeded a threshold that the centered one has not.
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Incidentally, the L-shapes in the implicit carry (which I noted can be interchanged) never turn up,
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and the "initial rule" is useless.
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Phinary, the simplest implicit case in one dimension, still requires the initial rule to go
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from the expansion of 2 (10.01) to 3 (100.01).
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Isosceles Rotations
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-------------------
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One of the similar carries is a rotation of a pattern we were already familiar with,
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like the folium was to $x + y = 2$
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Does this mean we can rotate the centered pattern?
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If so, is will the carry polynomial still be factorable?
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|
Naively, the following carry might look similar...
|
|
|
|
::: {layout-ncol="2"}
|
|
$$
|
|
\begin{array}{|c} \hline
|
|
& & \textcolor{red}{1} \\
|
|
& & \bar{3} \\
|
|
\textcolor{green}{1} & & & & \textcolor{blue}{1}
|
|
\end{array} \\
|
|
\begin{matrix}
|
|
\textcolor{red}{\langle 0,-1 \rangle} \\
|
|
\textcolor{green}{\langle -2,1 \rangle} \\
|
|
\textcolor{blue}{\langle 2,1 \rangle} \\
|
|
\text{Angle to top} \approx 116°
|
|
\end{matrix}
|
|
$$
|
|
|
|
```{python}
|
|
if not Path("./count_bad_rotated_triangle.mp4").exists():
|
|
try:
|
|
animate_carry_count(
|
|
carry = Carry([
|
|
[ 0, 0, 1, 0, 0],
|
|
[ 0, 0,-3, 0, 0],
|
|
[ 1, 0, 0, 0, 1],
|
|
]),
|
|
operation="add",
|
|
op_val=3,
|
|
frames=list(range(200)),
|
|
).save("./count_bad_rotated_triangle.mp4")
|
|
except ValueError:
|
|
pass
|
|
|
|
display(
|
|
Video("./count_bad_rotated_triangle.mp4"),
|
|
)
|
|
```
|
|
:::
|
|
|
|
But there are only even numbers in the first component of each vector
|
|
Therefore, the carry below will produce the same pattern.
|
|
|
|
::: {layout-ncol="2"}
|
|
$$
|
|
\begin{array}{|c} \hline
|
|
& \textcolor{red}{1} \\
|
|
& \bar{3} \\ \textcolor{green}{1}
|
|
&& \textcolor{blue}{1}
|
|
\end{array} \\
|
|
\begin{matrix}
|
|
\textcolor{red}{\langle 0,-1 \rangle} \\
|
|
\textcolor{green}{\langle -1,1 \rangle} \\
|
|
\textcolor{blue}{\langle 1,1 \rangle} \\
|
|
\text{Angle to top} = 135°
|
|
\end{matrix}
|
|
$$
|
|
|
|
```{python}
|
|
if not Path("./count_bad_rotated_triangle_reduced.mp4").exists():
|
|
try:
|
|
animate_carry_count(
|
|
carry = Carry([
|
|
[ 0, 1, 0],
|
|
[ 0,-3, 0],
|
|
[ 1, 0, 1],
|
|
]),
|
|
operation="add",
|
|
op_val=3,
|
|
frames=list(range(200)),
|
|
).save("./count_bad_rotated_triangle_reduced.mp4")
|
|
except ValueError:
|
|
pass
|
|
|
|
display(
|
|
Video("./count_bad_rotated_triangle_reduced.mp4"),
|
|
)
|
|
```
|
|
:::
|
|
|
|
Neither of these are similar to the original.
|
|
For comparison, the largest angle in the nonrotated case is ~108°.
|
|
In fact, the shape is much closer to the similar orientation with $x^4 + y^4 + \ldots$ terms.
|
|
|
|
|
|
### Checking Magnitudes
|
|
|
|
Measuring these quantities as angles is a bit of a lie.
|
|
Since all of these triangles are isosceles, two of the vectors have equal magnitude.
|
|
Further, each can be characterized by the ratio between them and the final vector's magnitude.
|
|
The previous two triangles have ratio:
|
|
|
|
$$
|
|
\\
|
|
\frac{m_1}{n_1} =
|
|
\frac{\| \langle -2, -1 \rangle \|}{\| \langle 0, -1 \rangle \|} =
|
|
\frac{\sqrt 5}{1} =
|
|
\sqrt 5
|
|
\\
|
|
\frac{m_2}{n_2} =
|
|
\frac{\| \langle -1, -1 \rangle \|}{\| \langle 0, -1 \rangle \|} =
|
|
\frac{\sqrt 2}{1} =
|
|
\sqrt 2
|
|
$$
|
|
|
|
The triangle we are trying to rotate has ratio:
|
|
|
|
$$
|
|
\frac{m}{n} =
|
|
\frac{\| \langle 2, -1 \rangle \|}{\| \langle -1, -1 \rangle \|} =
|
|
\frac{\sqrt 5}{\sqrt 2} =
|
|
\frac{\sqrt {10}}{2}
|
|
$$
|
|
|
|
Obviously, one cannot express $\sqrt 10$ from only one of $\sqrt 2$ or $\sqrt 5$.
|
|
Fortunately, this suggests the rotation:
|
|
|
|
::: {layout-ncol="2"}
|
|
$$
|
|
\begin{gather*}
|
|
\begin{array}{|c} \hline
|
|
& & & \textcolor{red}{\bar{1}} \\
|
|
\\
|
|
& & & \bar{3} \\
|
|
\textcolor{green}{1} & & & & & & \textcolor{blue}{1}
|
|
\end{array} ~~
|
|
\begin{matrix}
|
|
\textcolor{red}{\langle 0,-2 \rangle} \\
|
|
\textcolor{green}{\langle -3,1 \rangle} \\
|
|
\textcolor{blue}{\langle 3,1 \rangle} \\
|
|
\end{matrix} \\
|
|
\frac{\| \langle -3, 1 \rangle \|}{\| \langle 0, -2 \rangle \|}
|
|
= \frac{\sqrt {10}}{2} \\
|
|
\end{gather*}
|
|
$$
|
|
|
|
```{python}
|
|
if not Path("./count_good_rotated_triangle.mp4").exists():
|
|
try:
|
|
animate_carry_count(
|
|
carry = Carry([
|
|
[ 0, 0, 0, 1, 0, 0, 0],
|
|
[ 0, 0, 0, 0, 0, 0, 0],
|
|
[ 0, 0, 0,-3, 0, 0, 0],
|
|
[ 1, 0, 0, 0, 0, 0, 1],
|
|
]),
|
|
operation="add",
|
|
op_val=3,
|
|
frames=list(range(200)),
|
|
).save("./count_good_rotated_triangle.mp4")
|
|
except ValueError:
|
|
pass
|
|
|
|
display(
|
|
Video("./count_good_rotated_triangle.mp4"),
|
|
)
|
|
```
|
|
:::
|
|
|
|
The counting video looks very promising, but can the polynomial factored?
|
|
Fortunately, yes.
|
|
|
|
$$
|
|
\begin{gather*}
|
|
x^{6} y^{3} + x^{3} + y^{3} - 3 x^{3} y^{2} &=&
|
|
(x^{4} y^{2} - x^{3} y - x^{2} y^{2} + x^{2} - x y + y^{2}) \\
|
|
&& (x^2 y + x + y ) \\
|
|
\begin{array}{|c} \hline
|
|
& & & 1 \\
|
|
\\
|
|
& & & \bar{3} \\
|
|
1 & & & & & & \bar{1}
|
|
\end{array} &=&
|
|
\begin{array}{|c} \hline
|
|
& & 1 & \\
|
|
& \bar{1} & & \bar{1} & \\
|
|
1 & & \bar{1} & & 1 \\
|
|
\end{array} ~~\cdot~~
|
|
\begin{array}{|c} \hline
|
|
& 1 & \\
|
|
1 & & 1
|
|
\end{array} \\
|
|
& & (x y^{2} + x + y ) = \Phi_2 \left ( \frac{y}{x} + xy \right ) x
|
|
\end{gather*}
|
|
$$
|
|
|
|
As noted, the the rightmost polynomial can still be expressed from $\Phi_2$.
|
|
|
|
All powers of *x* in the product are multiples of 3.
|
|
We can divide these out to produce a similar carry, as we did before.
|
|
Doing so preserves the shape of counting, but destroys the factorization, as the result is irreducible.
|
|
|
|
::: {layout-ncol="2"}
|
|
$$
|
|
\begin{array}{|c} \hline
|
|
& \textcolor{red}{1} & \\
|
|
\\
|
|
& \bar{3} & \\
|
|
\textcolor{green}{1} & & \textcolor{blue}{1}
|
|
\end{array} ~~
|
|
\begin{matrix}
|
|
\textcolor{red}{\langle 0,2 \rangle} \\
|
|
\textcolor{green}{\langle -1,-1 \rangle} \\
|
|
\textcolor{blue}{\langle 1,-1 \rangle} \\
|
|
\end{matrix} \\
|
|
\frac{| \langle -1, -1 \rangle |}{| \langle 0,2 \rangle |}
|
|
= \frac{\sqrt 2}{2}
|
|
$$
|
|
|
|
```{python}
|
|
if not Path("./count_good_rotated_triangle_reduced.mp4").exists():
|
|
try:
|
|
animate_carry_count(
|
|
carry = Carry([
|
|
[ 0, 1, 0],
|
|
[ 0, 0, 0],
|
|
[ 0,-3, 0],
|
|
[ 1, 0, 1],
|
|
]),
|
|
operation="add",
|
|
op_val=3,
|
|
frames=list(range(200)),
|
|
).save("./count_good_rotated_triangle_reduced.mp4")
|
|
except ValueError:
|
|
pass
|
|
|
|
display(
|
|
Video("./count_good_rotated_triangle_reduced.mp4"),
|
|
)
|
|
```
|
|
:::
|
|
|
|
All of these similar systems is share something in common: the pseudo-base is located at
|
|
the centroid of the triangle formed by the remaining digits.
|
|
This is most obvious from the fact that the sum of all three vectors equals $\vec 0$ in all three cases.
|
|
The "threshold we exceeded" when comparing the unrotated system's neighbors was the centroid transitioning
|
|
between being closer to the upper point or the lower two points.
|
|
|
|
The position of the centroid is unchanged so long as the lower two points have the same midpoint as above.
|
|
Therefore, all triangular barycentric carries should share this similar pattern.
|
|
|
|
|
|
Closing
|
|
-------
|
|
|
|
Carries intrinsically possess a few invariants.
|
|
For one, they must apply at any location in an expansion, a form of translational symmetry.
|
|
No matter if we multiply or divide by *x* or *y*, the polynomial retains its general appearance,
|
|
as well as whether or not it is irreducible.
|
|
The above method shows that rotational symmetry also exists for polynomials beyond interchanging *x* and *y*.
|
|
|
|
Carries seem to possess a symmetry which polynomials do not: the ability to contract and expand bases,
|
|
meaning that in some sense, $P(x) \sim P(x^k)$ for any *k*.
|
|
This preserves the counting system despite expansions and contractions which pay no regard for reducibility.
|
|
This is also the reason that systems like base-$\sqrt 2$ are so uninteresting:
|
|
for integers, it's just spaced out binary.
|
|
|
|
Three questions that immediately come to mind are:
|
|
|
|
- Not all integers can be written as the sum of two squares.
|
|
Are there any (reducible) polynomials which have no nontrivial rotation?
|
|
- The [Brahmagupta-Fibonacci](https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity) identity
|
|
implies that such numbers are closed under multiplication.
|
|
This means that it is always possible to reorient the triangle in the method shown above.
|
|
- After compacting the counting system, can reducibility be recovered for other scaling factors
|
|
than the one started with?
|
|
What about from an irrational scaling (as by a rotation)?
|
|
- I am not equipped to answer this.
|
|
- Some geometry is impossible in the 2D square lattice -- for example, it contains no equilateral triangles.
|
|
Equilateral triangles do however exist in the 3D square lattice.
|
|
Are there 2D systems which *must* be embedded in 3D dimensions, or can they always be rotated and reduced?
|
|
- Again, I am not equipped to answer this.
|
|
|
|
Regardless, it is pleasing that the exponents in polynomials can easily be interpreted as
|
|
the components of vectors.
|
|
The primary geometrical approach to such objects is to plot the curve or surface containing points which
|
|
(at least approximately) satisfies the equation.
|
|
Here however, basic coordinate geometry exists without needing to define evaluation.
|
|
Moreover, geometry, linear algebra, and polynomial number theory all at once have a role to play in
|
|
the story of these bizarre fractal patterns.
|