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

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mentioned equivalence to BB(5) champion
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     & \to & D(2207) & \to & D(3682) & \to & D(6141) & \to & \textrm{Blank} \\
     & \to & D(2207) & \to & D(3682) & \to & D(6141) & \to & \textrm{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).
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:
<math display="block">\begin{array}{lc}
  g(2(3k)+2) & = & g(6k+2) & = & \text{HALT} & = & f(3k) \\
  g(2(3k+1)+2) & = & g(6k+4) & = & 10k+14 & = & 2f(3k+1)+2 \\
  g(2(3k+2)+2) & = & g(6k+6) & = & 10k+16 & = & 2f(3k+2)+2 \\
\end{array}</math>
So the size of this machine's BLB output is tied to the size of the BB(5) champion's output.


== References ==
== References ==

Revision as of 18:25, 4 September 2024

1RB1LD_1RC1RB_1LC1LA_0RC0RD (bbch)

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.[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).

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