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| {{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 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> | | {{TM|1RB1LD_1RC1RB_1LC1LA_0RC0RD}} is the [[Beeping Booping Busy Beaver]] champion for BBBB(2), pseudoquasihalting after 18 steps. It was discovered and reported by Charlie Reams in October 2026, correcting an [[wikipedia:Off-by-one error|off-by-1 error]] in previous work.<ref>[https://discord.com/channels/960643023006490684/1243312334907375676/1565879387927224474 Discord message]</ref> This is proven to be the champion for this state and symbol count.<ref>https://github.com/calr3/bbbb</ref> |
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| 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>
| | With boop transition A1 and beep transition A0, the machine boops at steps 6, 13, 18, ...; the first interval contains a single beep (therefore emitting a 1), and all later intervals contain no beeps (therefore emitting 0 infinitely thereafter). So, in the BBBB interpretation, it outputs an infinite stream of zeroes beyond this point and the first zero is emitted at transition 18, so BBBB(M) = 18. |
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| == Analysis by [[User:sligocki|Shawn Ligocki]] ==
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| 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}
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| D(a+3, & b) & \to & D(a, b+5) \\
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| D(0, & b) & \to & \text{Blank} \\
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| D(1, & b) & \to & D(b+2, 4) \\
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| D(2, & b) & \to & D(b+3, 4) \\
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| \end{array}</math>let <math>D(a):= D(a, 4)</math>, then we can simplify to:
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| <math display="block">\begin{array}{lc}
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| D(3k ) & \to & \text{Blank} \\
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| D(3k+1) & \to & D(5k+6) \\
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| D(3k+2) & \to & D(5k+7) \\
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| \end{array}</math>Starting from <math>D(2)</math> (at step 19) we get the trajectory:
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| <math display="block">\begin{array}{l}
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| D(2)
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| & \to & D(7) & \to & D(16) & \to & D(31) & \to & D(56) & \to & D(97) \\
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| & \to & D(166) & \to & D(281) & \to & D(472) & \to & D(791) & \to & D(1322) \\
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| & \to & D(2207) & \to & D(3682) & \to & D(6141) & \to & \text{Blank} \\
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| \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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| See also, previous analysis in 2021: https://www.sligocki.com/2021/07/17/bb-collatz.html
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| == Relation to other machines ==
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| 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:
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| <math display="block">\begin{array}{lc}
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| g(2(3k)+2) & = & g(6k+2) & = & \text{HALT} & = & f(3k) \\
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| g(2(3k+1)+2) & = & g(6k+4) & = & 10k+14 & = & 2f(3k+1)+2 \\
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| g(2(3k+2)+2) & = & g(6k+6) & = & 10k+16 & = & 2f(3k+2)+2 \\
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| \end{array}</math>
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| So the size of this machine's BLB output is tied to the size of the BB(5) champion's output.
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| == References == | | == References == |
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| [[Category:BB(4)]]
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1RB1LD_1RC1RB_1LC1LA_0RC0RD (bbch) is the Beeping Booping Busy Beaver champion for BBBB(2), pseudoquasihalting after 18 steps. It was discovered and reported by Charlie Reams in October 2026, correcting an off-by-1 error in previous work.[1] This is proven to be the champion for this state and symbol count.[2]
With boop transition A1 and beep transition A0, the machine boops at steps 6, 13, 18, ...; the first interval contains a single beep (therefore emitting a 1), and all later intervals contain no beeps (therefore emitting 0 infinitely thereafter). So, in the BBBB interpretation, it outputs an infinite stream of zeroes beyond this point and the first zero is emitted at transition 18, so BBBB(M) = 18.
References