1RB3LA4RB0RB2LA 1LB2LA3LA1RA1RZ

1RB3LA4RB0RB2LA_1LB2LA3LA1RA1RZ (bbch)

This is the current BB(2,5) champion. It halts with sigma score (and runtime) over $10^{10^{10^{3\,314\,360}}}$. It was discovered by Daniel Yuan and shared on Discord on 24 Jun 2024 and shared on the busy-beaver-discuss email list the next day.

Analyses

dyuan01

dyuan01 24 Jun 2024 11:51am EDT:

the TM appears to halt at

[11] [01]^(a+b+c+31) [11]^(d+1) 4 <B 1

Where

a = (2^25-2^19-9)/3 = 11010045
b = [2^(a+27)-2^(a+2)-11]/3
c = [2^(a+b+29)-2^(b+2)-9]/3
d = [2^(a+b+c+31)-2^(c+2)-8]/3

dyuan01 24 Jun 2024 12:55pm EDT:

Alright, so first things first, I should establish some rules, (all variables are natural numbers unless otherwise specified)
Basic rules (all can be simulated in a fixed number of steps):
1) [01] <A -> [11] A>
2) [11] <A -> <A [33]
3) A> [33] -> [01] A>
4) [01] <A3 -> [11] 0B>
5) [11] <A3 -> <A3 [33]
6) 0B> [33] -> [01] 0B>
7) A> [21] -> <A [22]
8) A> [22] -> <A [23]
9) A> [23] -> [41] A>
10) A> 0^inf -> <A [21] 0^inf
11) 0B> [21] -> <A3 [33]
12) 0B> [22] 0^inf -> Halt at [11] 4 <B 1 0^inf
13) 0B> [23] -> [11] 0B>
14) [41] <A -> <A [23]
15) 0^inf [11] <A -> 0^inf [11] 0B>
Counter rule helper:
A) [01] [11]^x <A -> [11] [01]^x A> (Basic rules 2*, 1, 3*)
Counter rules:
1) [01] [11]^x <A [23]^y 0^inf -> [11] [01]^x <A [23]^y [21] 0^inf (Counter rule A, Basic rules 9*, 10, 14*)
2) [01] [11]^x <A [23]^y [21] -> [11] [01]^x <A [23]^y [22] (Counter rule A, Basic rules 9*, 7, 14*)
3) [01] [11]^x <A [23]^y [22] -> [11] [01]^x <A [23]^(y+1) (Counter rule A, Basic rules 9*, 8, 14*)
4) 0^inf [11]^(x+1) <A -> 0^inf [11] [01]^x 0B> (Basic rules 2*, 15, 6*)
5) [01] [11]^x <A3 -> [11] [01]^x 0B> (Split <A3 into <A 3, then use Counter rule A, then simulate a step and merge 0 B> into 0B>)
Advanced rules:
1) [01]^3 0B> 0^inf -> [11] [01]^3 <A [21] 0^inf (simulatable in finite steps)
2) [01] [11]^(x+2) 0B> 0^inf -> [11] [01]^x [11]^2 [01] <A [21] 0^inf (Uses Basic rules 2* and 3*, but otherwise can be simulated in finite and bounded steps)
Now, I will simulate some tapes when they are overflowing from certain starting positions (assume 0^inf on both ends and x,y>=2):
(Overflow rule 1)
[11]^x <A [23]^y [21]
[11] [01]^(x-1) 0B> [23]^y [21] (Counter rule 4)
[11] [01]^(x-1) [11]^y 0B> [21] (Basic rule 13*)
[11] [01]^(x-1) [11]^y <A3 [33] (Basic rule 11)
[11] [01]^(x-2) [11] [01]^y 0B> [33] (Counter rule 5)
[11] [01]^(x-2) [11] [01]^(y+1) 0B> (Basic rule 6)
[11] [01]^(x-2) [11] [01]^(y-2) [11] [01]^3 <A [21] (Advanced rule 1)
(Overflow rule 2)
[11]^x <A [23]^y
[11] [01]^(x-1) 0B> [23]^y (Counter rule 4)
[11] [01]^(x-1) [11]^y 0B> (Basic rule 13*)
[11] [01]^(x-2) [11] [01]^(y-2) [11]^2 [01] <A [21] (Advanced rule 2)
(Overflow rule 3)
[11]^x <A [23]^y [22]
[11] [01]^(x-1) 0B> [23]^y [22] (Counter rule 4)
[11] [01]^(x-1) [11]^y 0B> [22] (Basic rule 13*)
Halt at [11] [01]^(x-1) [11]^(y+1) 4 <B 1 (Basic rule 12)
The only steps missing now are the steps to get from (11|01)* <A (21) to (11)* <A (23)* (|21|22), which takes advantage of binary representations and division mod 3.
Now we start to simulate the tape
At some point, the tape reaches [11]^7 <A [23]^17 [21]. By overflow rule 1, we reach
[11] [01]^5 [11] [01]^15 [11] [01]^3 <A [21]
If we treat [11] as 0 and [01] as 1 in binary, then [11] [01]^5 [11] [01]^15 [11] [01]^3 has a binary representation of (2^25-2^19-9). This is divisible by 3, so let a = (2^25-2^19-9)/3 = 11010045 (an odd number). Then we get
[11] [01]^5 [11] [01]^15 [11] [01]^3 <A [21]
[11] [11]^5 [11] [11]^15 [11] [11]^3 <A [23]^a [21]
[11]^26 <A [23]^a [21]
We apply overflow rule 1 again to get 
[11] [01]^24 [11] [01]^(a-2) [11] [01]^3 <A [21]
This has a binary representation of 2^(a+27)-2^(a+2)-9, which is 2 mod 3. Let b = (2^(a+27)-2^(a+2)-11)/3, which is an odd number, then we get
[11] [01]^24 [11] [01]^(a-2) [11] [01]^3 <A [21]
[11]^(a+28) <A [23]^(b+1)
Apply overflow rule 2 to get 
[11] [01]^(a+26) [11] [01]^(b-1) [11]^2 [01] <A [21]
This has a binary representation of 2^(a+b+29)-2^(b+2)-7, which is also 2 mod 3. Let c = (2^(a+b+29)-2^(b+2)-9)/3, another odd number, then we get
[11]^(a+b+30) <A [23]^(c+1)
We apply overflow rule 2 again to get
[11] [01]^(a+b+28) [11] [01]^(c-1) [11]^2 [01] <A [21]
This has a binary representation of 2^(a+b+c+31)-2^(c+2)-7, which is 1 mod 3. Let d = (2^(a+b+c+31)-2^(c+2)-8)/3. We get
[11]^(a+b+c+32) <A [23]^d [22]
After applying overflow rule 3 we get that this halts at
[11] [01]^(a+b+c+31) [11]^(d+1) 4 <B 1

Btw, d > 2^c > 2^2^b > 2^2^2^a > 2^2^2^2^23 > 2^2^2^2^2^4 = 2^2^2^2^2^2^2 = 2^^7, so that's probably the lower bound that should be used in steps and sigma for this TM.

LegionMammal978 24 Jun 2024 9:11pm EDT:

Assuming your values are correct, HyperCalc and my own manual calculations agree that σ ≈ 2^2^(2.8138364628466968⋅10^3314361) ≈ 10^10^(8.470491782098934⋅10^3314360). Equivalently, this is (2 ↑)⁷ 1.127778888301767 or (10 ↑)⁴ 6.5203998005419574.

Shawn Ligocki

Shawn Ligocki 25 Jun 2025 1:35am EDT:

At a high level, this TM satisfies the following rules:

Counter Rules:

• 01 11^x <A 23^y 00 -> 11 01^x <A 23^y 21
• 01 11^x <A 23^y 21 -> 11 01^x <A 23^y 22
• 01 11^x <A 23^y 22 -> 11 01^x <A 23^y 23

In other words, each iteration, the left counts up 1 in a binary counter (using "01" for 0 and "11" for 1) while the right keeps track of the number of iterations modulo 3 (using the final value of "21", "22" or "23") and grows every 3 iterations.

This continues until the counter "overflows" and depending upon the mod on the right side it has one of 3 possible behaviors.

Overflow Rules:

• 0^inf 11^x+2 <A 23^y+2 0^inf    -> 0^inf 11 01^x 11 01^y 11^2 01 <A 21 0^inf
• 0^inf 11^x+2 <A 23^y+2 21 0^inf -> 0^inf 11 01^x 11 01^y 11 01^3 <A 21 0^inf
• 0^inf 11^x+1 <A 23^y+1 22 0^inf -> 0^inf 11 01^x 11^y+2 1 Z> 1 0^inf

I have confirmed all of the above rules in my inductive validator.

At step 2430 it is in config "0^inf 11^7 <A 23^17 21 0^inf" and Daniel reports that it proceeds to overflow 5 times with the last one hitting the halt rule (I have not confirmed this behavior yet myself). The penultimate config reported by Daniel is:

0^inf [11] [01]^(a+b+c+31) [11]^(d+1) 4 <B 1 0^inf

Where

a = (2^25-2^19-9)/3 = 11010045
b = [2^(a+27)-2^(a+2)-11]/3
c = [2^(a+b+29)-2^(b+2)-9]/3
d = [2^(a+b+c+31)-2^(c+2)-8]/3

and Mathew House on Discord reported that this makes the sigma value for this TM > (10 ↑)^4 6.5

racheline

racheline 16 Jan 2025 12:18 PM EST:

(58,6) -> (3a+1,25) -> (3b+3,a+25) -> (3c+3,a+b+25) -> (3d+2,a+b+c+25) -> halt
where
a = 11010047
b = ((2^25-1)*2^a-5)/3
c = ((2^(a+25)-1)*2^b-3)/3
d = ((2^(a+b+25)-1)*2^c-2)/3
it halts with exactly a+b+c+25+2d+2 = a+b+c+2d+27 ones on the tape (and no other nonzero symbols)