BB(2, 6) > 10↑↑10↑↑10↑↑3

[Shawn Ligocki]

Pavel is on a rampage and has found another whopper of a new 2x6 BB champion:

1RB3RB5RA1LB5LA2LB_2LA2RA4RB1RZ3LB2LA

I have calculated it's score as > 10↑↑10↑↑10↑↑3 > 10↑↑↑3

See a full analysis at:

https://www.sligocki.com/2023/05/20/bb-2-6-p3.html

[uni...@fu-solution.com]

thank you for confirmation :)

Dátum: sobota 20. mája 2023, čas: 9:42:32 UTC+2, odosielateľ: slig...@gmail.com

[nichol...@gmail.com]

Wow! I guess Milton Green was on the right track, just not quite clever enough.

What does the final tape config look like?

The program itself is blank-free -- it never prints 0. 2s and 4s are only printed by state B, and 1s and 5s are only printed by state A (ignoring the halt instruction). Scanning colors 0, 1, 3, and 5 causes the state to switch, while scanning colors 2 and 4 causes the state to stay the same (or halt). There's nothing to say about the graph structure, because 2-state programs have no graph structure to speak of.

[Shawn Ligocki]

You can see the final tape in my blog post:

141 Z> 2^n 152

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[Pascal Michel]

I give here more analyses on the Turing machine p3 found by Pavel Kropitz on May 19, 2023, with s(p3) > 10^^10^^10^^3.

I use the analysis by Shawn Ligocki in his blog.

We have, for all n >= 0,

(a) (It's Shawn's Rule 3) ...0(A0) 212 (22)^n 52 5^5 --( 6x4^(n + 6) - 21x2^(n + 6) + 22n + 101)--> ...0(A0) 212 (22)^(6x2^(n + 4) - 4) 52

(b) 00(A0) 212 (22)^n 52 5^2 2000 --(16n + 59)--> 141(H2) 2^(2n + 7)152

So the last two transitions of the computation of p3 on ...0(A0)0... are:

...(A0) 212 (22)^n 52 5^7 2000

  --(6x4^(n + 6) - 21x2^(n + 6) + 22n + 101)-->

...(A0) 212 (22)^(6x2^'(n + 4) - 4) 52 5^2 2000

  --(16(6x2^(n + 4) - 4) + 59)-->

...0141 (H2) 2^(2(6x2^(n + 4) - 4) + 7) 152

The leading term in s(p3) is 6x4^(n + 6).

We have sigma(p3) = 6x2^(n + 5) + 6.

So s(p3) is approximately (2/3) sigma(p3)^2.