1RB3LB1RZ2RA 2LC3RB1LC2RA 3RB1LB3LC2RC: Difference between revisions
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{{TM|1RB3LB1RZ2RA_2LC3RB1LC2RA_3RB1LB3LC2RC|halt}} is a halting [[BB(3,4)]] TM. It was [https://discord.com/channels/960643023006490684/1095740122139480195/1233140768768524358 discovered in April 2024 by Pavel Kropitz]. It achieves a score of | {{machine|1RB3LB1RZ2RA_2LC3RB1LC2RA_3RB1LB3LC2RC}} | ||
{{TM|1RB3LB1RZ2RA_2LC3RB1LC2RA_3RB1LB3LC2RC|halt}} is a halting [[BB(3,4)]] TM. It was [https://discord.com/channels/960643023006490684/1095740122139480195/1233140768768524358 discovered in April 2024 by Pavel Kropitz]. It achieves a sigma score of exactly <math>(2 \uparrow^{15} 5) + 14</math>, making it the current BB(3,4) [[champion]]. Shawn Ligocki [https://discord.com/channels/960643023006490684/1095740122139480195/1242206442530930799 confirmed the TM's score.] | |||
Another long running TM, {{TM|1RB3RB1LC2LA_2LA2RB1LB3RA_3LA1RZ1LC2RA|halt}}, is a [[permutation]] of this TM, obtained by starting it in state B and then converting to [[TNF]] and achieves a sigma score of exactly <math>(2 \uparrow^{6} 5) + 5</math>.<ref>Shawn Ligocki, 2024. https://www.sligocki.com/2024/05/22/bb-3-4-a14.html Accessed 25. October 2025.</ref> | |||
==Analysis== | |||
Let <math>B(k,n,m) = </math> <code>0^inf 3^(2*m+1) 2^k A> 1^n</code> | |||
Then <math> B(k,n,m) \to B(k,0,g_{k-1}^n(m))</math> if <math>k \geq 1</math> | |||
where <math>2g_k(m) + 4 = 2 \uparrow^{k-1} (2m+4)</math> | |||
<code> | |||
Start -> B(16,3,0)+2 -> B(16,0,(2{15}5)/2-2)+2 -> Halt | |||
</code> | |||
==References== | |||
[[Category:BB(3,4)]] | |||
Latest revision as of 13:30, 27 October 2026
1RB3LB1RZ2RA_2LC3RB1LC2RA_3RB1LB3LC2RC (bbch) is a halting BB(3,4) TM. It was discovered in April 2024 by Pavel Kropitz. It achieves a sigma score of exactly , making it the current BB(3,4) champion. Shawn Ligocki confirmed the TM's score.
Another long running TM, 1RB3RB1LC2LA_2LA2RB1LB3RA_3LA1RZ1LC2RA (bbch), is a permutation of this TM, obtained by starting it in state B and then converting to TNF and achieves a sigma score of exactly .[1]
Analysis
Let 0^inf 3^(2*m+1) 2^k A> 1^n
Then if
where
Start -> B(16,3,0)+2 -> B(16,0,(2{15}5)/2-2)+2 -> Halt
References
- ↑ Shawn Ligocki, 2024. https://www.sligocki.com/2024/05/22/bb-3-4-a14.html Accessed 25. October 2025.