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:

$${\begin{array}{l}c(2k)&=&k\\c(2k+1)&=&3k+2\\\end{array}}$$

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 (1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA) and the generalize configuration:

$$M(n)=0^{\infty }\;{\textrm {<A}}\;1^{n}\;0^{\infty }$$

Pascal Michel showed that:

$${\begin{array}{lcl}0^{\infty }\;{\textrm {<A}}\;0^{\infty }&=&M(0)\\M(3k)&\xrightarrow {5k^{2}+19k+15} &M(5k+6)\\M(3k+1)&\xrightarrow {5k^{2}+25k+27} &M(5k+9)\\M(3k+2)&\xrightarrow {6k+12} &0^{\infty }\;\;1\;\;{\textrm {H>}}\;\;01\;\;{(001)}^{k+1}\;\;1\;\;0^{\infty }\\\end{array}}$$

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

Hydra

Consider Hydra (a Cryptid) 1RB3RB---3LA1RA_2LA3RA4LB0LB0LA and the generalized configuration:

$$C(a,b)=0^{\infty }\;{\textrm {<B}}\;0^{3(a-2)}\;3^{b}\;2\;0^{\infty }$$

Daniel Yuan showed that:

$${\begin{array}{l}\\0^{\infty }\;{\textrm {A>}}\;0^{\infty }&&\xrightarrow {19} &C(3,0)\\C(2n,&0)&\to &{\text{Halt}}(9n-6)\\C(2n,&b+1)&\to &C(3n,&b)\\C(2n+1,&b)&\to &C(3n+1,&b+2)\\\end{array}}$$

Where ${\textrm {Halt}}(n)$ 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

$${\begin{array}{l}h(2n)&=&3n\\h(2n+1)&=&3n+1\\\end{array}}$$

starting from 3:

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

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) 1RB0LD_1RC0RF_1LC1LA_0LE1RZ_1LF0RB_0RC0RE and consider the general configuration:

$$K(a,b)=0^{\infty }\;1\;0^{n}\;11\;0^{5}\;{\textrm {C>}}\;0^{\infty }$$

Shawn Ligocki showed that:

$${\begin{array}{l}\\0^{\infty }\;{\textrm {A>}}\;0^{\infty }&\xrightarrow {45} &K(5)\\K(4k)&\to &{\text{Halt}}(9n-6)\\K(4n+1)&\to &K({\frac {3^{k+3}-11}{2}})\\K(4n+2)&\to &K({\frac {3^{k+3}-11}{2}})\\K(4n+3)&\to &K({\frac {3^{k+3}+1}{2}})\\\end{array}}$$

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.