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

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{{machine|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{machine|1RB1LD_1RC1RB_1LC1LA_0RC0RD}}
{{TM|1RB1LD_1RC1RB_1LC1LA_0RC0RD}} is the current [[Blanking 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 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>
 
In 2026, this machine was proven in Rocq to be the BBB(4) winner when considering beep states.<ref name=":Coq-BBB4">[https://github.com/carrino/Coq-BBB4 Coq-BBB4]</ref>


== Analysis by [[User:sligocki|Shawn Ligocki]] ==
== Analysis by [[User:sligocki|Shawn Ligocki]] ==
Let <math display="block">D(a, b) = 0^\infty \; 1^a \; 0^b \; \textrm{D>} \; 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) \\
   D(0, & b) & = & \textrm{Blank} \\
   D(0, & b) & \to & \text{Blank} \\
   D(1, & b) & \to & D(b+2, 4) \\
   D(1, & b) & \to & D(b+2, 4) \\
   D(2, & b) & \to & D(b+3, 4) \\
   D(2, & b) & \to & D(b+3, 4) \\
\end{array}</math>let <math>D(a) = D(a, 4)</math>, then we can simplify to:
\end{array}</math>let <math>D(a):= D(a, 4)</math>, then we can simplify to:


<math display="block">\begin{array}{lc}
<math display="block">\begin{array}{lc}
   D(3k  ) & \to & \textrm{Blank} \\
   D(3k  ) & \to & \text{Blank} \\
   D(3k+1) & \to & D(5k+6) \\
   D(3k+1) & \to & D(5k+6) \\
   D(3k+2) & \to & D(5k+7) \\
   D(3k+2) & \to & D(5k+7) \\
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     & \to & D(7) & \to & D(16) & \to & D(31) & \to & D(56) & \to & D(97) \\
     & \to & D(7) & \to & D(16) & \to & D(31) & \to & D(56) & \to & D(97) \\
     & \to & D(166) & \to & D(281) & \to & D(472) & \to & D(791) & \to & D(1322) \\
     & \to & D(166) & \to & D(281) & \to & D(472) & \to & D(791) & \to & D(1322) \\
     & \to & D(2207) & \to & D(3682) & \to & D(6141) & \to & \textrm{Blank} \\
     & \to & D(2207) & \to & D(3682) & \to & D(6141) & \to & \text{Blank} \\
\end{array} </math>which has the remarkable luck of applying this [[Collatz-like]] map 14 times before reaching the blanking config (expected # of applications before halting is 3).
\end{array} </math>which has the remarkable luck of applying this [[Collatz-like]] map 14 times before reaching the blanking config (expected # of applications before halting is 3).


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== Relation to other machines ==
== Relation to other machines ==


The map and trajectory are equivalent to that of the [[BB(5) champion]]. For all <math>k</math>, let <math>f(k)</math> be the number such that <math>D(k)\to D(f(k))</math>, or <math>\text{HALT}</math> if <math>D(k)\to\textrm{Blank}</math>, and let <math>g</math> be the map simulated by the BB(5) champion. Then:
The map and trajectory are equivalent to that of the [[BB(5) champion]]. For all <math>k</math>, let <math>f(k)</math> be the number such that <math>D(k)\to D(f(k))</math>, or <math>\text{HALT}</math> if <math>D(k)\to\text{Blank}</math>, and let <math>g</math> be the map simulated by the BB(5) champion. Then:


<math display="block">\begin{array}{lc}
<math display="block">\begin{array}{lc}
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== References ==
== References ==
[[Category:BB(4)]]

Latest revision as of 16:54, 3 August 2026

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]

In 2026, this machine was proven in Rocq to be the BBB(4) winner when considering beep states.[2]

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