1RB1LD 1RC1RB 1LC1LA 0RC0RD: Difference between revisions

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{{TM|1RB1LD_1RC1RB_1LC1LA_0RC0RD}} is the current [[Blanking Busy Beaver]] BLB(4,2) and [[Beeping Busy Beaver]] BBB(4,2) champion, creating a blank tape after 32,779,477 steps and [[Quasihalt|quasihalting]] after 32,779,478 steps at which point it becomes a [[translated cycler]] with period 1. It was discovered and reported by Nick Drozd in 2021.<ref>Nick Drozd. [https://nickdrozd.github.io/2021/07/11/self-cleaning-turing-machine.html A New Record in Self-Cleaning Turing Machines]. 2021.</ref>
{{TM|1RB1LD_1RC1RB_1LC1LA_0RC0RD}} is the current [[Blanking Busy Beaver]] BLB(4,2) and [[Beeping Busy Beaver]] BBB(4,2) champion, creating a blank tape after 32,779,477 steps and [[Quasihalt|quasihalting]] after 32,779,478 steps at which point it becomes a [[translated cycler]] with period 1. It was discovered and reported by Nick Drozd in 2021.<ref>Nick Drozd. [https://nickdrozd.github.io/2021/07/11/self-cleaning-turing-machine.html A New Record in Self-Cleaning Turing Machines]. 2021.</ref>


== Analysis by Shawn Ligocki ==
== Analysis by [[User:sligocki|Shawn Ligocki]] ==
Let <math display="block">D(a, b):= 0^\infty\;1^a\;0^b\;\textrm{D}\textrm{>}\;0^\infty</math>then:<math display="block">\begin{array}{lc}
Let <math display="block">D(a, b):= 0^\infty\;1^a\;0^b\;\textrm{D}\textrm{>}\;0^\infty</math>then:<math display="block">\begin{array}{lc}
   D(a+3, & b) & \to & D(a, b+5) \\
   D(a+3, & b) & \to & D(a, b+5) \\

Revision as of 19:21, 9 October 2025

1RB1LD_1RC1RB_1LC1LA_0RC0RD (bbch) is the current Blanking Busy Beaver BLB(4,2) and Beeping Busy Beaver BBB(4,2) champion, creating a blank tape after 32,779,477 steps and quasihalting after 32,779,478 steps at which point it becomes a translated cycler with period 1. It was discovered and reported by Nick Drozd in 2021.[1]

Analysis by Shawn Ligocki

Let D(a,b):=0∞1a0bD>0∞then:D(a+3,b)→D(a,b+5)D(0,b)→BlankD(1,b)→D(b+2,4)D(2,b)→D(b+3,4)let D(a):=D(a,4), then we can simplify to:

D(3k)→BlankD(3k+1)→D(5k+6)D(3k+2)→D(5k+7)Starting from D(2) (at step 19) we get the trajectory:

D(2)→D(7)→D(16)→D(31)→D(56)→D(97)→D(166)→D(281)→D(472)→D(791)→D(1322)→D(2207)→D(3682)→D(6141)→Blankwhich has the remarkable luck of applying this Collatz-like map 14 times before reaching the blanking config (expected # of applications before halting is 3).

See also, previous analysis in 2021: https://www.sligocki.com/2021/07/17/bb-collatz.html

Relation to other machines

The map and trajectory are equivalent to that of the BB(5) champion. For all k, let f(k) be the number such that D(k)→D(f(k)), or HALT if D(k)→Blank, and let g be the map simulated by the BB(5) champion. Then:

g(2(3k)+2)=g(6k+2)=HALT=f(3k)g(2(3k+1)+2)=g(6k+4)=10k+14=2f(3k+1)+2g(2(3k+2)+2)=g(6k+6)=10k+16=2f(3k+2)+2

So the size of this machine's BLB output is tied to the size of the BB(5) champion's output.

References