## Mandelbrot set sketch in Scratch

Despite my personal disbelieve in and dislike of the colored blocks dragging simulator 3, I nevertheless wanted to extract functionality other than the hardcoded cat mascot path tracing from the aforementioned software; one of the most efficient visual result to build effort ratio yields a simple plot of the Mandelbrot set, formally known as

$M:=\{z\in\mathbb{C}:|\lim_{n\to\infty}\mathrm{itr}^n(z)|<\infty\}$

where the iterator is defined as

$\mathrm{itr}^n(z) := \mathrm{itr}^{n-1}(z)^2+z, \\ \mathrm{itr}^0(z) := 0.$

The render resolution is kept at a recognizable minimum as not to overburden the machine tasked with creating it. Source: mandelbrot-set.sb3

## Factoids #0

#### I) unit polynomials with non-vanishing degree

$2t+1\in\mathbb{Z}_4[t]$ is its own multiplicative inverse, showing that $R[t]^*=R^*$ does not hold in a general commutative Ring with one.

This phenomenon is uniquely characterized by the following equivalence:

$R[t]^* = R^* \iff \nexists \,0\neq a,b\in R:a\cdot b=0=a+b$

Proof. Negated replication.

Let $R\not\owns f=\sum_{i=0}^n\alpha_it^i\in R[t]^*,\alpha_n\neq0$ be a unit polynomial of non-vanishing degree $n\geq 1$. Let $g=\sum_{j=0}^m\beta_jt^j\in R[t]^*,\beta_m\neq0$ denote its multiplicative inverse, i.e. $f\cdot g=1$.

Claim. The polynomial $g$ has non-vanishing degree $m\geq 1$.
Proof. Suppose $g \in R$. Since $f\cdot g=\sum_{i=0}^n(\alpha_i\cdot g)t^i$, it follows from $\alpha_n\cdot g=0$ that $g$ is a zero divisor. However, at the same time $a_0\cdot g=1$ implies that $g$ is a unit, arriving at a contradiction.

Since both $n,m\geq 1$, one concludes $\exists 1\leq k\leq m$ as well as $\alpha_n\cdot\beta_m=0$.

Existence of the desired ring elements $a,b$ is assured by the following construction.

Let $k=1 \nearrow m$ rise discretely.

If $a:=\alpha_n\beta_{m-k}\neq 0$, implying $b:=\sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}\neq 0$, holds, since the construction arrived at this point, one finds

$a\cdot b=\alpha_n\beta_{m-k}\cdot \sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}=\sum_{i=1}^k \underbrace{\alpha_n\beta_{m-k+i}}_{=0}\cdot \alpha_{n-i}\beta_{m-k}=0.$

The above condition is met for at least one $1\leq k\leq m$, since otherwise $k=m$ would imply $\alpha_n\beta_{m-m} = 0$, which is impossible since $\alpha_n\neq 0$ and $\beta_0$ is a unit element.

By construction, $0\neq a,b$ as well as $a+b=0$ are given.

Negated implication.

Setting $f:=at+1, g:=bt+1$, one calculates

$f\cdot g=(at+1)\cdot (bt+1)=abt^2+(a+b)\cdot t+1=0t^2+0t+1=1$,

showing $R\not\owns f,g\in R[t]^*$.

q.e.d.

## Mostly Misaligned Mirrors

Recently my stochastic professor introduced me to a problem he has been pondering for over two decades: on the two-dimensional integer lattice $\mathbb{Z}^2$ one shall flip a three-sided coin for each point and uniformly place one of three mirrors, $\{\diagup,\,\cdot\,,\diagdown\}$, where $\,\cdot\,$ denotes not placing a mirror. After having populated the world, one picks their favorite integer tuple and points a beam of light in one of the four cardinal directions. With what probability does the light fall into a loop, never fully escaping?