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

From BusyBeaverWiki
Jump to navigation Jump to search
Polygon (talk | contribs)
Fixed link to Blanking Busy Beaver
Simplify new commentary about TC. I think we don't need to repeat the preperiod bit.
Line 1: Line 1:
{{machine|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{machine|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{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. It is also a [[translated cycler]] with a preperiod of 32,779,478 steps, a period length of 1 step and a cycle offset of 1 symbol to the left. 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 [[User:sligocki|Shawn Ligocki]] ==
== Analysis by [[User:sligocki|Shawn Ligocki]] ==

Revision as of 20:20, 27 August 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)=01a0bD>0then: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