BB(2, 6) > 10↑↑70

[Shawn Ligocki]

I'm excited to share a new 2-state 6-symbol TM which I believe leaves 10↑↑70 non-zero symbols on the tape upon halting!

The TM is:

1RB2LA1RA4LA5RA0LB_1LA3RA2RB1RZ3RB4LA

|     |  0  |  1  |  2  |  3  |  4  |  5  |
| :-: | :-: | :-: | :-: | :-: | :-: | :-: |
|  A  | 1RB | 2LA | 1RA | 4LA | 5RA | 0LB |
|  B  | 1LA | 3RA | 2RB | 1RZ | 3RB | 4LA |

And the exact number of non-zeros left on the tape is

(14 + 2^(-1 + 2^(2 + 2^(1 + 2^(-1 + 2^(4 + 2^2^(-1 + 2^2^(2 + 2^2^(2 + 2^2^(-1 + 2^2^(-1 + 2^2^(3 + 2^2^(3 + 2^2^(2 + 2^2^(-1 + 2^2^(-1 + 2^2^(2 + 2^2^(4 + 2^2^(-1 + 2^2^(2 + 2^2^(2 + 2^2^(-1 + 2^2^(3 + 2^2^(-1 + 2^(4 + 2^2^(-1 + 2^2^(-1 + 2^2^(-1 + 2^2^(4 + 2^2^(-1 + 2^2^(-1 + 2^2^(1 + 2^(1 + 2^(-1 + 2^(4 + 2^2^(-1 + 2^2^(2 + 2^2^(2 + 2^2^(-1 + 2^2^(2 + 2^8))))))))))))))))))))))))))))))))))))))))

For more details including a rudimentary analysis of the behavior, see my blog post. I have not yet found an elegant way to describe the behavior of this machine. Instead it appears to me so far to be reasonably chaotic, occasionally applying the function n -> 4(2^n - 1). That function is applies 69 times before the TM halts and so we end up with a score around 10↑↑69.

-Shawn

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

looks good :)

```

shawn 2x6 t70: BeaverRes {

    result: Halt(Powers(70)),
    machine: "1RB2LA1RA4LA5RA0LB_1LA3RA2RB1RZ3RB4LA",
    last_conf: "1 2 {3 1}^2 {3 2}^2 3 1 Z> 0 4 2 4^2 2^(32*(2^(64*(2^(128*(2^(32*(2^(64*(2^(128*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(64*(2^(64*(2^(64*(2^(64*(2^(64*(2^(32*(2^(128*(2^(32*(2^(128*(2^(32*(2^(64*(2^(32*(2^(64*(2^(64*(2^(128*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(128*(2^(32*(2^(128*(2^(32*(2^(64*(2^(32*(2^(64*(2^(128*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(64*(2^(64*(2^(128*(2^(32*(2^(128*(2^(32*(2^(64*(2^(32*(2^(128*(2^(32*(2^(64*(2^(128*(2^(32*(2^(128*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(64*(2^(32*(2^(128*(2^(1)) - 3)) - 6)) - 6)) - 6)) - 3)) - 6)) - 3)) - 7)) - 6)) - 7)) - 2)) - 6)) - 6)) - 4)) - 6)) - 6)) - 7)) - 6)) - 7)) - 2)) - 6)) - 6)) - 6)) - 6)) - 6)) - 6)) - 7)) - 2)) - 6)) - 6)) - 2)) - 7)) - 6)) - 7)) - 3)) - 6)) - 3)) - 6)) - 6)) - 7)) - 2)) - 6)) - 3)) - 6)) - 6)) - 7)) - 6)) - 7)) - 3)) - 6)) - 3)) - 6)) - 3)) - 6)) - 6)) - 6)) - 6)) - 6)) - 3)) - 6)) - 3)) - 6)) - 6)) - 7)) - 2)) - 6)) - 6)) - 4)) - 6)) - 3) 1",
}

shawn 2x6 t69: BeaverRes {
    result: Halt(Powers(69)),
    machine: "1RB2LB1RZ3LA2LA4RB_1LA3RB4RB1LB5LB0RA",
    last_conf: "1 3^(16*(2^(32*(2^(64*(2^(16*(2^(32*(2^(64*(2^(16*(2^(32*(2^(16*(2^(32*(2^(16*(2^(32*(2^(16*(2^(32*(2^(16*(2^(32*(2^(32*(2^(32*(2^(32*(2^(32*(2^(16*(2^(64*(2^(16*(2^(64*(2^(16*(2^(32*(2^(16*(2^(32*(2^(32*(2^(64*(2^(16*(2^(32*(2^(16*(2^(32*(2^(16*(2^(64*(2^(16*(2^(64*(2^(16*(2^(32*(2^(16*(2^(32*(2^(64*(2^(16*(2^(32*(2^(16*(2^(32*(2^(16*(2^(32*(2^(32*(2^(64*(2^(16*(2^(64*(2^(16*(2^(32*(2^(16*(2^(64*(2^(16*(2^(32*(2^(64*(2^(16*(2^(64*(2^(16*(2^(32*(2^(16*(2^(32*(2^(16*(2^(32*(2^(32*(2^(13)) - 5)) - 5)) - 5)) - 2)) - 5)) - 2)) - 6)) - 5)) - 6)) - 1)) - 5)) - 5)) - 3)) - 5)) - 5)) - 6)) - 5)) - 6)) - 1)) - 5)) - 5)) - 5)) - 5)) - 5)) - 5)) - 6)) - 1)) - 5)) - 5)) - 1)) - 6)) - 5)) - 6)) - 2)) - 5)) - 2)) - 5)) - 5)) - 6)) - 1)) - 5)) - 2)) - 5)) - 5)) - 6)) - 5)) - 6)) - 2)) - 5)) - 2)) - 5)) - 2)) - 5)) - 5)) - 5)) - 5)) - 5)) - 2)) - 5)) - 2)) - 5)) - 5)) - 6)) - 1)) - 5)) - 5)) - 3)) - 5)) - 3) 4^2 3 4 0 1 Z> 2 {3 2}^2 {1 2}^2 3 1",
}
```

Dátum: utorok 25. apríla 2023, čas: 0:13:58 UTC+2, odosielateľ: slig...@gmail.com

[nichol...@gmail.com]

Wow, congratulations Shawn! I can't wait to see what Pavel finds in two or three days :)

The 6/2 champ is 10^^15, right? So this is further evidence that colors are more powerful than states, in the sense that BB(N, 2) < BB(2, N). But I don't think there's any way to prove that.

What does the final tape configuration look like?

[Shawn Ligocki]

Thanks for the confirmation, Pavel!

Nick: Pavel has the tape listed in his confirmation. Mine looks the same (except for a printing bug):

00^inf 12^1 31^2 32^2 31^1 (04) Z> 24^1 42^1 22^(-2 + 2^(-2 + 2^(2 + 2^(1 + 2^(-1 + 2^(4 + 2^2^(-1 + 2^2^(2 + 2^2^(2 + 2^2^(-1 + 2^2^(-1 + 2^2^(3 + 2^2^(3 + 2^2^(2 + 2^2^(-1 + 2^2^(-1 + 2^2^(2 + 2^2^(4 + 2^2^(-1 + 2^2^(2 + 2^2^(2 + 2^2^(-1 + 2^2^(3 + 2^2^(-1 + 2^(4 + 2^2^(-1 + 2^2^(-1 + 2^2^(-1 + 2^2^(4 + 2^2^(-1 + 2^2^(-1 + 2^2^(1 + 2^(1 + 2^(-1 + 2^(4 + 2^2^(-1 + 2^2^(2 + 2^2^(2 + 2^2^(-1 + 2^2^258))))))))))))))))))))))))))))))))))))))) 10^1 00^inf

(the printing bug is that "31 (04) Z>" really should be "31 Z> 04")

By the way, you can run this with my code by:
$ Code/Quick_Sim.py -r --exp-linear-rules --no-steps -n2 <(echo 1RB2LA1RA4LA5RA0LB_1LA3RA2RB1RZ3RB4LA)

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

On 2023-04-25 02:09, nichol...@gmail.com wrote:
> I can't wait to see what Pavel finds in two or three days :)

i don't want to make it a tradition and this is probably the last time
i can do this trick , but to satisfy my fan (who should do it instead of
me next time):

1RB3LA4LB0RB1RA3LA_2LA2RA4LA1RA5RB1RZ (^^~90)

but from what i have seen, i don't think this is the end & i wouldn't
be surprised if there's a monstrous machine hiding out there.


* originally announced on bbchallenge.org discord:
https://discord.com/channels/960643023006490684/960643023530762341/1101400867158102087

* invite link: https://discord.gg/3uqtPJA9Uv - come! we have cookies
and holdouts to solve!

* final conf: ```
1^4 0 2^2 {1 0}^2 1^4 2 1^2 0 1^4 2 1^3 2 1^2 0 1 2 1 0 {2 1}^2 0 1 2^2
1 2 {1 0}^2 {1 2 1 0}^3 1 0 1 2 1 0 1^3 {2 1^2}^2 0 1 {0 2 1^2}^2 0 1^3
{0 1}^2 0 2 1^2 0 1 2 1 0 1^4 2 1^2 0 1^4 2^2 1^2 2 1^3 2 1^2 0 1 2 1 0
2 1^2 5 1 Z> 4 3^4 4^2
3^((7808*4^((122*4^((3904*4^((3904*4^((122*4^((976*4^((488*4^((3904*4^((488*4^((3904*4^((488*4^((976*4^((244*4^((3904*4^((1952*4^((1952*4^((122*4^((976*4^((1952*4^((1952*4^((122*4^((976*4^((488*4^((976*4^((488*4^((976*4^((244*4^((976*4^((244*4^((976*4^((244*4^((3904*4^((1952*4^((122*4^((3904*4^((1952*4^((122*4^((3904*4^((488*4^((976*4^((244*4^((3904*4^((122*4^((15616*4^((244*4^((976*4^((488*4^((976*4^((244*4^((976*4^((244*4^((976*4^((488*4^((3904*4^((488*4^((3904*4^((488*4^((976*4^((488*4^((3904*4^((7808*4^((488*4^((7808*4^((244*4^((976*4^((488*4^((976*4^((976*4^((3904*4^((3904*4^((122*4^((976*4^((244*4^((3904*4^((1952*4^((1952*4^((122*4^((976*4^((488*4^((976*4^((244*4^((976*4^((976*4^((488*4^((3904*4^((1952*4^((122*4^((976*4^((1952*4^((8*(31232*4^(46)
- 17)/3 + 24)/3) - 50)/9) - 49)/9) - 44)/9) - 41)/9) - 49)/9) - 29)/9)
- 52)/9) - 52)/9) - 52)/9) - 55)/9) - 50)/9) - 49)/9) - 44)/9) - 38)/9)
- 53)/9) - 52)/9) - 49)/9) - 37)/9) - 41)/9) - 55)/9) - 73)/9) - 49)/9)
- 43)/9) - 50)/9) - 52)/9) - 55)/9) - 77)/9) - 35)/9) - 83)/9) - 67)/9)
- 50)/9) - 55)/9) - 53)/9) - 67)/9) - 53)/9) - 67)/9) - 50)/9) - 52)/9)
- 52)/9) - 52)/9) - 52)/9) - 55)/9) - 47)/9) - 28)/9) - 55)/9) - 73)/9)
- 47)/9) - 52)/9) - 49)/9) - 43)/9) - 53)/9) - 61)/9) - 44)/9) - 41)/9)
- 49)/9) - 44)/9) - 41)/9) - 52)/9) - 49)/9) - 40)/9) - 52)/9) - 52)/9)
- 52)/9) - 55)/9) - 47)/9) - 31)/9) - 50)/9) - 49)/9) - 44)/9) - 35)/9)
- 56)/9) - 49)/9) - 44)/9) - 38)/9) - 53)/9) - 52)/9) - 49)/9) - 43)/9)
- 53)/9) - 67)/9) - 53)/9) - 67)/9) - 47)/9) - 25)/9) - 41)/9) - 58)/9)
- 25)/9) - 41)/9) - 59)/3) 2
```

[nichol...@gmail.com]

> but to satisfy my fan (who should do it instead of me next time)

Perhaps! For the time being, my simulator is stuck using actual numbers for block counts, rather than algebraic expressions. So the best I can run is the 6/2 machine that goes for 10^646_456_993 steps.

[Pascal Michel]

I give here more analysis on the Turing machine t70 found by Shawn Ligocki on April 24, 2023, with s(t70) > 10^^70.

I use the analysis by Shawn in his blog, where he studies the computation of t69 on ...0(B0)0...

We have, for all n>=0,

(a) ...013^n (B1) 251 --(2^(2n + 7) - 9x2^(n + 3) +18n + 49)/3)--> ...013^(2^(n + 3) - 4) (B1) 2212

(b) (B1) 2212522 --(7)--> 3443401 (Z2)

So the last two transitions of the computation of t69 on ...(B0)... are

...013^n (B1) 25152232321212310...

--((2^(2n + 7) - 9x2^(n + 3) + 18n + 49)/3) -->

...013^(2^(n + 3) - 4) (B1) 221252232321212310...

--(7)-->

...013^(2^(n + 3) - 3)443401 (Z2) 32321212310...

The leading term in s(t70) is (2^(2n + 7))/3, and we have sigma(t70) = 14 + 2^(n + 3).

So s(t70) is approximately (2/3)sigma(t70)^2.

[Shawn Ligocki]

I can confirm Pavel's TM and give a more precise tower notation estimate for it's sigma score (using "fractional tetration" as described in my Extended Up-arrow Notation post):

Nonzeros: ~10↑↑91.18  =  ((295 + 61 * 2^((-19 + 61 * 2^((-32 + 61 * 2^((-53 + 61 * 2^((-19 + 61 * 2^((-32 + 61 * 2^((-49 + 61 * 2^((-98 + 61 * 2^((-43 + 61 * 2^((-98 + 61 * 2^((-43 + 61 * 2^((-50 + 61 * 2^((-53 + 61 * 2^((-77 + 61 * 2^((-43 + 61 * 2^((-22 + 61 * 2^((-34 + 61 * 2^((-80 + 61 * 2^((-67 + 61 * 2^((-16 + 61 * 2^((-34 + 61 * 2^((-80 + 61 * 2^((-55 + 61 * 2^((-26 + 61 * 2^((-49 + 61 * 2^((-74 + 61 * 2^((-59 + 61 * 2^((-77 + 61 * 2^((-59 + 61 * 2^((-53 + 61 * 2^((-53 + 61 * 2^((-77 + 61 * 2^((-19 + 61 * 2^((-34 + 61 * 2^((-80 + 61 * 2^((-19 + 61 * 2^((-34 + 61 * 2^((-104 + 61 * 2^((-43 + 61 * 2^((-50 + 61 * 2^((-53 + 61 * 2^((-77 + 61 * 2^((-31 + 61 * 2^((-128 + 61 * 2^((-29 + 61 * 2^((-29 + 61 * 2^((-49 + 61 * 2^((-74 + 61 * 2^((-59 + 61 * 2^((-77 + 61 * 2^((-59 + 61 * 2^((-77 + 61 * 2^((-55 + 61 * 2^((-98 + 61 * 2^((-43 + 61 * 2^((-98 + 61 * 2^((-43 + 61 * 2^((-74 + 61 * 2^((-55 + 61 * 2^((-98 + 61 * 2^((-103 + 61 * 2^((2 + 61 * 2^((-118 + 61 * 2^((-38 + 61 * 2^((-77 + 61 * 2^((-55 + 61 * 2^((-50 + 61 * 2^((-53 + 61 * 2^((-101 + 61 * 2^((-47 + 61 * 2^((-19 + 61 * 2^((-56 + 61 * 2^((-53 + 61 * 2^((-77 + 61 * 2^((-43 + 61 * 2^((-22 + 61 * 2^((-34 + 61 * 2^((-80 + 61 * 2^((-55 + 61 * 2^((-74 + 61 * 2^((-59 + 61 * 2^((-77 + 61 * 2^((-59 + 61 * 2^((-13 + 61 * 2^((-62 + 61 * 2^((-19 + 61 * 2^((-34 + 61 * 2^((-80 + 61 * 2^((-55 + 61 * 2^((-74 + 61 * 2^105)/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/9))/3)

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[Shawn Ligocki]

After poking around at Pavel's 10^^91 TM a bit (1RB3LA4LB0RB1RA3LA_2LA2RA4LA1RA5RB1RZ) I have only been able to detect one clear pattern. This TM applies the following rule 86 times in my code and this is the one rule that accounts for the massive score for this machine:

... 01^3 11 01 21 0211^n B> 0^inf -> ... 11 01 21 0211^4n+10 B> 0^inf

With repeated application we get:

... 01^3k+r 11 01 21 0211^n B> 0^inf -> ... 01^r 11 01 21 0211^{((3n+10) 4^k - 10)/3} B> 0^inf

From which you can see how applying this rule ~90 times will end up with ~4^^90

However, I cannot detect any notable structure in the reset behavior between applications of this rule. Similar to my analysis for my 2^^70 TM, this TM seems to mix orderly collatz behavior with chaotic behavior.

[nichol...@gmail.com]

I am delighted to report that my simulator is now able to verify Shawn's 10^^70 TM. The other big tetrational records remain out of reach, but the 10^^70 result is nice because the floor divisions can be distributed cleanly.

Here is the final number of marks I get:

(14 + (2 ** (-1 + (2 ** (2 + (2 ** (1 + (2 ** (-1 + (2 ** (4 + (2 ** (2 ** (-1 + (2 ** (2 ** (2 + (2 ** (2 ** (2 + (2 ** (2 ** (-1 + (2 ** (2 ** (-1 + (2 ** (2 ** (3 + (2 ** (2 ** (3 + (2 ** (2 ** (2 + (2 ** (2 ** (-1 + (2 ** (2 ** (-1 + (2 ** (2 ** (2 + (2 ** (2 ** (4 + (2 ** (2 ** (-1 + (2 ** (2 ** (2 + (2 ** (2 ** (2 + (2 ** (2 ** (-1 + (2 ** (2 ** (3 + (2 ** (2 ** (-1 + (2 ** (4 + (2 ** (2 ** (-1 + (2 ** (2 ** (-1 + (2 ** (2 ** (-1 + (2 ** (2 ** (4 + (2 ** (2 ** (-1 + (2 ** (2 ** (-1 + (2 ** (2 ** (1 + (2 ** (1 + (2 ** (-1 + (2 ** (4 + (2 ** (2 ** (-1 + (2 ** (2 ** (2 + (2 ** (2 ** (2 + (2 ** (2 ** (-1 + (2 ** (~10^78))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

AFAICT this is equivalent to others posted, with slightly different normalization.