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)=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).

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