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{{TM|1RB3RB5RA1LB5LA2LB_2LA2RA4RB1RZ3LB2LA|halt}} is the current [[BB(2,6)]] [[champion]]. It was discovered on the 19th of May 2023 by Pavel Kropitz. It halts with a score > <math>10 \uparrow\uparrow 10 \uparrow\uparrow 10^{10^{115}}</math>.
==Analysis by Shawn Ligocki==
https://www.sligocki.com/2023/05/20/bb-2-6-p3.html
<pre>
<pre>
Analysis
Let's rewrite the first rule as:
Level 1
A(a, b) --> A(2a + 4, b - a - 3) if b >= a + 3
These rules can all be verified by direct simulation:
Then, after n applications of rule 1 (assuming b is large enough):
00 <A 212 22^n 55 → <A 212 22^n+2
A(a, b) --> A(2^n * a + 2^(n + 2) - 4, b - 2^n * a + a - 2^(n + 2) + n + 4)
 
Considering that a is always set to 4 when leaving B, this simplifies to:
00 <A 212 22^n 2 55 → <A 212 55^n+2 2
A(4, b) --> A(2^(n + 3) - 4, b - 2^(n + 3) + n + 8)
 
Now consider the halting configuration: A(a, a + 1):
0^5 <A 212 22^n 52 5555 → <A 212 55 2 55^n+3 52
So, for this to be in a halting configuration, the following must be true:
00 <A 212 22^n 2 52 5 → <A 212 55^n+2 52
2^(n+3)-4 + 1 = b-2^(n+3)+n+8
 
This simplifies to:
Level 2
b = 2^(n+4)-n-11
Repeating the first rule above we get:
0^∞ <A 212 22^n 55^k → 0^∞ 212 22^n+2k
 
which let's us prove Rule 2:
0^∞ <A 212 22^n 2 55 → 0^∞ <A 212 55^n+2 2
                    → 0^∞ <A 212 22^2n+4 2
 
Level 3
Repeating Rule 2 we get:
0^∞ <A 212 22^n 2 55^k → 0^∞ <A 212 22^(n+4)*((2^k)-4) 2
 
which let's us prove Rule 3:
0^∞ <A 212 22^n 52 5^5 → 0^∞ <A 212 55 2 55^n+3 52 5
                      → 0^∞ <A 212 22^2 2 55^n+3 52 5
                      → 0^∞ <A 212 22^(6*2^(n+3)-2) 52 5
                      → 0^∞ <A 212 55^(6*2^(n+3)-2) 52
                      → 0^∞ <A 212 22^(6*2^(n+4)-4) 52
 
Level 4
Let
f(n) = 6*2^(n+4)-4
 
Repeating Rule 3 we get the Tetration Rule:
0^∞ <A 212 22^n 52 5^5k → 0^∞ <A 212 22^f^k(n) 52
 
This rule will be the main contributor to the score since f^k(n) > 2^^k. In fact, this rule will apply 3 times, which is how we end up with 3 tetrations in the final score (>10^^10^^10^^3).
 
Halting Trajectory
 
With these high-level rules, we are now ready to describe the halting trajectory for this TM starting from a blank tape:
          191
0^∞ <A 0^∞ → 0^∞ <A 212 22^2 52 5^13 2 0^∞
 
This is our first application of the Tetration Rule. Here calculating the remainder is trivial:
A1 = 13 = 5k1 + r1
r1 = 3
k1 = (A1 - r1)/5 = 2
 
continuing the trajectory:
...→ 0^∞ <A 212 22^f^2(2) 52 5^3 2 0^∞
  → 0^∞ <A 212 55 2 55^f^2(2)+4 2 0^∞
  → 0^∞ <A 212 22^2 2 55^f^2(2)+4 2 0^∞
  → 0^∞ <A 212 22^(6*2^(f^2(2)+4)-4) 2 2 0^∞
  = 0^∞ <A 212 22^f^3(2)+1 0^∞
  → 0^∞ <A 212 55 52 5^(2*f^3(2)+5) 22 0^∞
  → 0^∞ <A 212 22^2 52 5^(2*f^3(2)+5) 22 0^∞
 
This is our second application of the Tetration Rule. Here calculating the remainder requires using Euler’s totient theorem (as described in BB(6, 2) > 10↑↑15):
A2 = 2*f^3(2)+5 = 5k2 + r2
r2 = 4
k2 = (A2 - r2)/5 = (2f^3(2)+1)/5
 
continuing the trajectory:
...→ 0^∞ <A 212 22^f^k2(2) 52 5^4 22 0^∞
  → 0^∞ <A 212 55 2 55^(f^k2(2))+3 52 22 0^∞
  → 0^∞ <A 212 22^2 2 55^(f^k2(2))+3 52 22 0^∞
  → 0^∞ <A 212 22^(6*(2^(f^k2(2))+3)-4) 2 52 22 0^∞
  → 0^∞ <A 212 55^(6*(2^(f^k2(2))+3)-1) 52 0^∞
  → 0^∞ <A 212 22^(6*(2^(f^k2(2))+4)-2) 52 0^∞
  = 0^∞ <A 212 22^f^k2+1(2)+2 52 0^∞
  → 0^∞ <A 212 55 52 5^(2*(f^k2+1(2))+13) 2 0^∞
  → 0^∞ <A 212 22^2 52 5^(2*(f^k2+1(2))+13) 2 0^∞
 
This is our third and final application of the Tetration Rule. Here calculating the remainder requires a minor arithmetic miracle (see next section):
A3 = 2*f^k2+1(2)+13 = 5k3 + r3
r3 = 2
k3 = (A3 - r3)/5 = (2*f^k2+1(2)+11)/5
 
finishing the trajectory:
0^∞ <A 212 22^f^k3(2) 52 5^2 2 0^∞ → 0^∞ 141 Z> 2^(2*f^k3(2)+8) 152 0^∞
 
And we see that it halts with a score of
Sigma(p3) = 2*f^k3(2)+14
 
Remainder Miracle
 
Calculating the remainder
2*f^k2+1(2)+13 (mod 5)
is no simple task given that this is a power tower with height k2 > 10^10^100 (a googolplex) and that the Euler’s totient theorem method for computing remainders is worse than linear on power tower heights!
 
But, as it turns out, there is a miraculous shortcut to this computation in this specific case!
 
The miracle can be summarized succinctly by the following two facts:
4|f(n) and 2^4 ≡ 1 (mod 5)
 
Specifically:
f^2(n) = 6*2^(f(n)+4)-4
      = 6*2^(4x)-4
      ≡ 6 - 4 ≡ 2 (mod 5)
 
and since this is true for all n, we have:
f^k2+1(2) ≡ 2 (mod 5)
 
and this remainder becomes trivial to compute!
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Revision as of 20:33, 25 October 2025

Let's rewrite the first rule as:
A(a, b) --> A(2a + 4, b - a - 3) if b >= a + 3
Then, after n applications of rule 1 (assuming b is large enough):
A(a, b) --> A(2^n * a + 2^(n + 2) - 4, b - 2^n * a + a - 2^(n + 2) + n + 4)
Considering that a is always set to 4 when leaving B, this simplifies to:
A(4, b) --> A(2^(n + 3) - 4, b - 2^(n + 3) + n + 8)
Now consider the halting configuration: A(a, a + 1):
So, for this to be in a halting configuration, the following must be true:
2^(n+3)-4 + 1 = b-2^(n+3)+n+8
This simplifies to:
b = 2^(n+4)-n-11
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