0RB1LD 1LC1RB 1LD1RE 1LA1LE 1LZ0RC: Difference between revisions

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{{machine|0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC}}
{{machine|0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC}}
{{TM|0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC|halt}} is the [[Busy Beaver Functions|num]](5) champion (the [[BB(5)]] TM which halts leaving the most consecutive ones on the tape) according to Andrés Sancho.<ref>[https://discord.com/channels/960643023006490684/1242208042460647575/1337655697348886558 Discord message] by Andrés Sancho on 8 Feb 2025</ref><ref>https://github.com/MatterAndy/BB5-contiguous-1s</ref> It halts after 15590 steps with tape


{{TM|0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC}} is the [[Busy Beaver Functions|num]](5) champion (the [[BB(5)]] TM which halts leaving the most consecutive ones on the tape) according to Andrés Sancho.<ref>[https://discord.com/channels/960643023006490684/1242208042460647575/1337655697348886558 Discord message] by Andrés Sancho on 8 Feb 2025</ref><ref>https://github.com/MatterAndy/BB5-contiguous-1s</ref> It halts after 15590 steps with tape
<math display="block">0^\infty \; \textrm{<Z} \; 1^{165} \; 0^\infty</math>


<math display="block">0^\infty <Z \; 1^{165} \; 0^\infty</math>
It is tied for the num(5) championship with {{TM|1RB1LA_1RC1LE_1RD1RE_0LA1RC_1RZ0LB|halt}} which is the [[TNF-1RB]] version of the same TM (The [[permutation]] of this TM starting at state B).
 
It is tied for the num(5) championship with {{TM|1RB1LA_1RC1LE_1RD1RE_0LA1RC_1RZ0LB}} which is the [[TNF-1RB]] version of the same TM (The [[permutation]] of this TM starting at state B).


== Analysis by Shawn Ligocki ==
== Analysis by Shawn Ligocki ==

Latest revision as of 17:51, 16 August 2025

0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC (bbch) is the num(5) champion (the BB(5) TM which halts leaving the most consecutive ones on the tape) according to Andrés Sancho.[1][2] It halts after 15590 steps with tape

It is tied for the num(5) championship with 1RB1LA_1RC1LE_1RD1RE_0LA1RC_1RZ0LB (bbch) which is the TNF-1RB version of the same TM (The permutation of this TM starting at state B).

Analysis by Shawn Ligocki

A(a, b) = $ 1^a <A 11^b $

A(a+3, b) -> A(a, b+2)
A(0, b) -> A(2b, 1)
A(1, b) -> A(0, b+1)
A(2, b) -> $ <Z 1^{2b+3} $

A(3k,   1) -> A(4k+2, 1)
A(3k+1, 1) -> A(4k+4, 1)
A(3k+2, 1) -> $ <Z 1^{4k+5} $

@13: A(3, 1)

Trajectory of "a" values starting from A(3, 1):
3 6 10 16 24 34 48 66 90 122 Halt(165)

References