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 [[Maximum Consecutive Ones Function|num(5)]] champion (the [[BB(5)|five-state, two-symbol]] 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 165 consecutive ones on the tape.
{{TM|0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC|halt}} is the [[Maximum Consecutive Ones Function|num(5)]] champion (the [[BB(5)|five-state, two-symbol]] TM which halts leaving the most consecutive ones on the tape) according to Andrés Sancho.<ref>{{DiscordArchive|1242208042460647575|1337655697348886558}} by Andrés Sancho on 8 Feb 2025</ref><ref>https://github.com/MatterAndy/BB5-contiguous-1s</ref> It halts after 15590 steps with 165 consecutive ones on the tape.


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|halt}} which is the [[TNF-1RB]] version of the same TM (The [[permutation]] of this TM starting at state B).

Latest revision as of 12:31, 17 October 2026

0RB1LD_1LC1RB_1LD1RE_1LA1LE_1LZ0RC (bbch) is the num(5) champion (the five-state, two-symbol TM which halts leaving the most consecutive ones on the tape) according to Andrés Sancho.[1][2] It halts after 15590 steps with 165 consecutive ones on the 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

Let A(a):=0∞1a<A110∞. Then, A(3a)→A(4a+2)A(3a+1)→A(4a+4)A(3a+2)→0∞<Z14a+50∞

Trajectory

This Turing machine starts with A(3) after 13 steps and halts after 10 rule applications: A(3)→A(6)→A(10)→A(16)→A(24)→A(34)→A(48)→A(66)→A(90)→A(122)→0∞<Z11650∞

References