Collatz-like

A Collatz-like function is a partial function defined piecewise depending on the remainder of an input modulo some number. The canonical example is the original Collatz function:$$\begin{array}{lll}c(2k)& =& k\\ c(2k+1)& =& 3k+2\\ \end{array}$$

A Collatz-like problem is a question about the behavior of iterating a Collatz-like function. Collatz-like problems are famously difficult.

Many Busy Beaver Champions have Collatz-like behavior, meaning that their behavior can be concisely described via the iterated values of a Collatz-like function.

Examples

BB(5,2) Champion

Consider the BB(5,2) Champion and the generalized configuration:

$$M(n)=0^{\mathrm{\infty }}\text{<mrow data-mjx-texclass="ORD"><mo><</mo>A</mrow>}{1}^n{0}^{\mathrm{\infty }}$$Pascal Michel showed that:

$$\begin{array}{lcl}{0}^{\mathrm{\infty }}\text{<mrow data-mjx-texclass="ORD"><mo><</mo>A</mrow>}{0}^{\mathrm{\infty }}& =& M(0)\\ M(3k)& \stackrel{5k^2+19k+15}{\to }& M(5k+6)\\ M(3k+1)& \stackrel{5k^2+25k+27}{\to }& M(5k+9)\\ M(3k+2)& \stackrel{6k+12}{\to }& 0^{\mathrm{\infty }}1\text{<mrow data-mjx-texclass="ORD">H<mo>></mo></mrow>}01{(001)}^{k+1}10^{\mathrm{\infty }}\\ \end{array}$$

Starting on a blank tape $C(0)$, these rules iterate 15 times before reaching the halt config.

Hydra

Consider Hydra (a Cryptid) and the generalized configuration:

$$C(a,b)=0^{\mathrm{\infty }}\text{<mrow data-mjx-texclass="ORD"><mo><</mo>B</mrow>}{0}^{3(a-2)}{3}^{b}20^{\mathrm{\infty }}$$ Daniel Yuan showed that:$$\begin{array}{l}\\ 0^{\mathrm{\infty }}\text{<mrow data-mjx-texclass="ORD">A<mo>></mo></mrow>}{0}^{\mathrm{\infty }}& & \stackrel{19}{\to }& C(3,0)\\ C(2n,& 0)& \to & \text{Halt}(9n-6)\\ C(2n,& b+1)& \to & C(3n,& b)\\ C(2n+1,& b)& \to & C(3n+1,& b+2)\\ \end{array}$$

Where $\text{<mrow data-mjx-texclass="ORD">Halt</mrow>}(n)$ is a halting configuration with $n$ non-zero symbols on the tape.

Starting from config $C(3,0)$ this simulates a pseudo-random walk along the $b$ parameter, increasing it by 2 every time $a$ is odd, decreasing by 1 every time it's even. Deciding whether or not Hydra halts requires being able to prove a detailed question about the trajectory of the Collatz-like function$$\begin{array}{lll}h(2n)& =& 3n\\ h(2n+1)& =& 3n+1\\ \end{array}$$starting from 3:

$$\begin{array}{l}\\ 3\stackrel{O}{\to }4\stackrel{E}{\to }6\stackrel{E}{\to }9\stackrel{O}{\to }13\stackrel{O}{\to }19\stackrel{O}{\to }28\stackrel{E}{\to }42\stackrel{E}{\to }63\cdots \\ \end{array}$$

Specifically, will it ever reach a point where the cumulative number of E (even transitions) applied is greater than twice the number of O (odd transitions) applied?

Tetration Machine

Consider the current BB(6,2) Champion (discovered by Pavel Kropitz in May 2022) and consider the general configuration:$$K(a,b)=0^{\mathrm{\infty }}10^{n}110^5\text{<mrow data-mjx-texclass="ORD">C<mo>></mo></mrow>}{0}^{\mathrm{\infty }}$$Shawn Ligocki showed that: $$\begin{array}{l}\\ 0^{\mathrm{\infty }}\text{<mrow data-mjx-texclass="ORD">A<mo>></mo></mrow>}{0}^{\mathrm{\infty }}& \stackrel{45}{\to }& K(5)\\ K(4k)& \to & \text{Halt}(9n-6)\\ K(4n+1)& \to & K(\frac{3^{k+3}-11}{2})\\ K(4n+2)& \to & K(\frac{3^{k+3}-11}{2})\\ K(4n+3)& \to & K(\frac{3^{k+3}+1}{2})\\ \end{array}$$

Starting from config $K(5)$, these rules iterate 15 times before reaching the halt config leaving over $10\uparrow \uparrow 15$ non-zero symbols on the tape.