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

From BusyBeaverWiki
Jump to navigation Jump to search
Link my old analysis
Polygon (talk | contribs)
Combined introduction with link to bbch, reworded part of the introduction
Line 1: Line 1:
{{machine|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{machine|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{TM|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{TM|1RB1LD_1RC1RB_1LC1LA_0RC0RD}} is the current [[Blanking Beaver]] BLB(4,2) champion, creating a blank tape after 32,779,477 steps. 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>
 
[[Blanking Beaver]] BLB(4,2) champion which creates a blank tape after 32,779,477 steps. 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 20:46, 10 August 2025

1RB1LD_1RC1RB_1LC1LA_0RC0RD (bbch) is the current Blanking Beaver BLB(4,2) champion, creating a blank tape after 32,779,477 steps. 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