User:Polygon/Page for testing: Difference between revisions

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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>.
{{TM|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD|halt}}
==Analysis by Shawn Ligocki==
https://www.sligocki.com/2023/05/20/bb-2-6-p3.html
<pre>
<pre>
Analysis
S is any tape configuration
Level 1
1. S D> 2^a S --> S 2^a D> S
These rules can all be verified by direct simulation:
2. S B> 1^a S --> S 1^a B> S
00 <A 212 22^n 55 → <A 212 22^n+2
3. S 1 B> 0 S --> S <A 1^2 S
4. S D> (11)^a S --> S (21)^a D> S
  S A> (11)^a S --> S (12)^a A> S
5. S (21)^a <C S --> S <C (11)^a S
  S (12)^a <A S --> S <A (11)^a S
6. S (12)^a A> 0^2 S --> S <A (11)^a+1 S


00 <A 212 22^n 2 55 → <A 212 55^n+2 2
7. S (12)^a 2 (12)^b A> 0^2 S --> S (12)^a-1 2 (12)^b+2 A> S
by:
S (12)^a 2 (12)^b A> 0^2 S
--> S (12)^a 2 (12)^b 1 B> 0 S
--> S (12)^a 2 (12)^b <A (11) S
--> S (12)^a 2 <A (11)^b+1 S
--> S (12)^a <C 1 (11)^b+1 S
--> S (12)^a-1 1 <D (11)^b+2 S
--> S (12)^a-1 2 A> (11)^b+2 S
--> S (12)^a-1 2 (12)^b+2 A> S


0^5 <A 212 22^n 52 5555 → <A 212 55 2 55^n+3 52
8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf
00 <A 212 22^n 2 52 5 → <A 212 55^n+2 52
Obtained by repeating rule 8.


Level 2
9. S (12)^a <D (11)^b 0^inf --> S (12)^a-1 <D (11)^2b+3 0^inf
Repeating the first rule above we get:
by:
0^∞ <A 212 22^n 55^k → 0^∞ 212 22^n+2k
S (12)^a <D (11)^b 0^inf
--> S (12)^a D> (11)^b 0^inf
--> S (12)^a (21)^b D> 0^inf
--> S (12)^a (21)^b 2 B> 0^inf
--> S (12)^a (21)^b 2 <B 2 0^inf
--> S (12)^a (21)^b <C 1 2 0^inf
--> S (12)^a <C (11)^b 1 2 0^inf
--> S (12)^a-1 1 <D (11)^b+1 2 0^inf
--> S (12)^a-1 2 A> (11)^b+1 2 0^inf
--> S (12)^a-1 2 (12)^b+1 A> 2 0^inf
--> S (12)^a-1 2 (12)^b+1 <C 1 0^inf
--> S (12)^a-1 2 (12)^b 1 <D 11 0^inf
--> S (12)^a-1 2 (12)^b 2 A> (11)^1 0^inf
--> S (12)^a-1 2 (12)^b 2 (12)^1 A> 0^inf
--> S (12)^a-1 2 2 (12)^2b+1 A> 0^inf
--> S (12)^a-1 2^2 <A (11)^2b+2 0^inf
--> S (12)^a-1 2 <C 1 (11)^2b+2 0^inf
--> S (12)^a-1 <D (11)^2b+3 0^inf


which let's is prove Rule 2:
10. S (12)^a <D (11)^b 0^inf --> S <D (11)^((2^(a))*b+(2^(a))*3-3) 0^inf
0^∞ <A 212 22^n 2 55 → 0^∞ <A 212 55^n+2 2
Obtained by repeating rule 9.
                    → 0^∞ <A 212 22^2n+4 2
 
11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
by:
S (11)^a <D (11)^b 0^inf
--> S (11)^a-1 1 2 A> (11)^b 0^inf
--> S (11)^a-1 (12)^b+1 A> 0^inf
--> S (11)^a-1 <A (11)^b+2 0^inf
--> S (11)^a-1 D> (11)^b+2 0^inf
--> S (11)^a-1 (21)^b+2 D> 0^inf
--> S (11)^a-1 (21)^b+2 2 B> 0^inf
--> S (11)^a-1 (21)^b+2 2 <B 2 0^inf
--> S (11)^a-1 (21)^b+2 <C (12)^1 0^inf
--> S (11)^a-1 <C (11)^b+2 1 2 0^inf
--> S (11)^a-2 1 <A (11)^b+3 2 0^inf
--> S (11)^a-2 1 D> (11)^b+3 2 0^inf
--> S (11)^a-2 1 (21)^b+3 D> 2 0^inf
--> S (11)^a-2 1 (21)^b+3 2 D> 0^inf
--> S (11)^a-2 1 (21)^b+3 2^2 B> 0^inf
--> S (11)^a-2 1 (21)^b+3 2^2 <B 2 0^inf
--> S (11)^a-2 1 (21)^b+3 2 <C (12)^1 0^inf
--> S (11)^a-2 1 (21)^b+3 <D 1 1 2 0^inf
Note that 1 (21)^k = (12)^k 1
= S (11)^a-2 (12)^b+3 1 <D (11)^1 2 0^inf
--> S (11)^a-2 (12)^b+3 2 A> (11)^1 2 0^inf
--> S (11)^a-2 (12)^b+3 2 (12)^1 A> 2 0^inf
--> S (11)^a-2 (12)^b+3 2 (12)^1 <C 1 0^inf
--> S (11)^a-2 (12)^b+3 2 1<D (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 A> (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 A> 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 1 B> 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 <A (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 1 <C 1 (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 <A (11)^2 0^inf
--> S (11)^a-2 (12)^b+3 2 <C 1 (11)^2 0^inf
--> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
</pre>
Let A(a,b,c) = S (11)^a (12)^b <D (11)^c 0^inf
* Rule 9: A(a, b, c) --> A(a, b - 1, 2c + 3)
* Rule 10: A(a, b, c) --> <math>A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.
* Rule 11: A(a, 0, c) --> A(a - 2, c + 3, 3)
 
Further: let <math>f(n) = 2^{n+1} \times 3</math>
* If c = 3: A(a, b, 3) --> A(a, 0, f(c) - 3) --> A(a - 2, f(c), 3)
 
* A(2k + d, 0, c) --> <math>A(d, f^{k-1}(c+3), 3)</math>
<pre>
The TM enters configuration A(19, 2, 3) after 799 steps with tape:
0^inf 2 1 (11)^19 (12)^2 <D (11)^3 0^inf
</pre>
==Trajectory==
A(19, 2, 3) --> A(19, 0, 21) --> <math>A(1, f^{8}(24), 3)</math>
 
--> <math>A(1, 0, f^{9}(24) - 3)</math>
 
Lets have <math> f^{9}(24) - 3</math> = m.
<pre>
Final trajectory:
0^inf 2 1 (11)^1 <D (11)^m 0^inf
--> 0^inf 2 1 1 2 A> (11)^m 0^inf
--> 0^inf 2 1 (12)^m+1 A> 0^inf
--> 0^inf 2 1 <A (11)^m+2 0^inf
--> 0^inf 2 1 D> (11)^m+2 0^inf
--> 0^inf (21)^m+3 D> 0^inf
--> 0^inf (21)^m+3 2 B> 0^inf
--> 0^inf (21)^m+3 2 <B 2 0^inf
--> 0^inf (21)^m+3 <C (12)^1 0^inf
--> 0^inf <C (11)^m+3 (12)^1 0^inf
--> 0^inf 1 Z> (11)^m+3 (12)^1 0^inf
Score = 2m + 9
 
Score calculated in HyperCalc:
(10^)^8 30,302,671.815163
Or in tetration: 10^^10.873987 (truncated)
</pre>
</pre>

Latest revision as of 18:09, 30 September 2025

1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD (bbch) 

S is any tape configuration
1. S D> 2^a S --> S 2^a D> S
2. S B> 1^a S --> S 1^a B> S
3. S 1 B> 0 S --> S <A 1^2 S
4. S D> (11)^a S --> S (21)^a D> S
   S A> (11)^a S --> S (12)^a A> S
5. S (21)^a <C S --> S <C (11)^a S
   S (12)^a <A S --> S <A (11)^a S
6. S (12)^a A> 0^2 S --> S <A (11)^a+1 S

7. S (12)^a 2 (12)^b A> 0^2 S --> S (12)^a-1 2 (12)^b+2 A> S
by:
S (12)^a 2 (12)^b A> 0^2 S
--> S (12)^a 2 (12)^b 1 B> 0 S
--> S (12)^a 2 (12)^b <A (11) S 
--> S (12)^a 2 <A (11)^b+1 S
--> S (12)^a <C 1 (11)^b+1 S
--> S (12)^a-1 1 <D (11)^b+2 S
--> S (12)^a-1 2 A> (11)^b+2 S
--> S (12)^a-1 2 (12)^b+2 A> S

8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf
Obtained by repeating rule 8.

9. S (12)^a <D (11)^b 0^inf --> S (12)^a-1 <D (11)^2b+3 0^inf
by:
S (12)^a <D (11)^b 0^inf
--> S (12)^a D> (11)^b 0^inf
--> S (12)^a (21)^b D> 0^inf
--> S (12)^a (21)^b 2 B> 0^inf
--> S (12)^a (21)^b 2 <B 2 0^inf
--> S (12)^a (21)^b <C 1 2 0^inf
--> S (12)^a <C (11)^b 1 2 0^inf
--> S (12)^a-1 1 <D (11)^b+1 2 0^inf
--> S (12)^a-1 2 A> (11)^b+1 2 0^inf
--> S (12)^a-1 2 (12)^b+1 A> 2 0^inf
--> S (12)^a-1 2 (12)^b+1 <C 1 0^inf
--> S (12)^a-1 2 (12)^b 1 <D 11 0^inf
--> S (12)^a-1 2 (12)^b 2 A> (11)^1 0^inf
--> S (12)^a-1 2 (12)^b 2 (12)^1 A> 0^inf
--> S (12)^a-1 2 2 (12)^2b+1 A> 0^inf
--> S (12)^a-1 2^2 <A (11)^2b+2 0^inf
--> S (12)^a-1 2 <C 1 (11)^2b+2 0^inf
--> S (12)^a-1 <D (11)^2b+3 0^inf

10. S (12)^a <D (11)^b 0^inf --> S <D (11)^((2^(a))*b+(2^(a))*3-3) 0^inf
Obtained by repeating rule 9.

11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
by:
S (11)^a <D (11)^b 0^inf
--> S (11)^a-1 1 2 A> (11)^b 0^inf
--> S (11)^a-1 (12)^b+1 A> 0^inf
--> S (11)^a-1 <A (11)^b+2 0^inf
--> S (11)^a-1 D> (11)^b+2 0^inf
--> S (11)^a-1 (21)^b+2 D> 0^inf
--> S (11)^a-1 (21)^b+2 2 B> 0^inf
--> S (11)^a-1 (21)^b+2 2 <B 2 0^inf
--> S (11)^a-1 (21)^b+2 <C (12)^1 0^inf
--> S (11)^a-1 <C (11)^b+2 1 2 0^inf
--> S (11)^a-2 1 <A (11)^b+3 2 0^inf
--> S (11)^a-2 1 D> (11)^b+3 2 0^inf
--> S (11)^a-2 1 (21)^b+3 D> 2 0^inf
--> S (11)^a-2 1 (21)^b+3 2 D> 0^inf
--> S (11)^a-2 1 (21)^b+3 2^2 B> 0^inf
--> S (11)^a-2 1 (21)^b+3 2^2 <B 2 0^inf
--> S (11)^a-2 1 (21)^b+3 2 <C (12)^1 0^inf
--> S (11)^a-2 1 (21)^b+3 <D 1 1 2 0^inf
Note that 1 (21)^k = (12)^k 1
= S (11)^a-2 (12)^b+3 1 <D (11)^1 2 0^inf
--> S (11)^a-2 (12)^b+3 2 A> (11)^1 2 0^inf
--> S (11)^a-2 (12)^b+3 2 (12)^1 A> 2 0^inf
--> S (11)^a-2 (12)^b+3 2 (12)^1 <C 1 0^inf
--> S (11)^a-2 (12)^b+3 2 1<D (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 A> (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 A> 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 1 B> 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 <A (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 1 <C 1 (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 <A (11)^2 0^inf
--> S (11)^a-2 (12)^b+3 2 <C 1 (11)^2 0^inf
--> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf

Let A(a,b,c) = S (11)^a (12)^b <D (11)^c 0^inf

  • Rule 9: A(a, b, c) --> A(a, b - 1, 2c + 3)
  • Rule 10: A(a, b, c) --> which becomes if c = 3.
  • Rule 11: A(a, 0, c) --> A(a - 2, c + 3, 3)

Further: let

  • If c = 3: A(a, b, 3) --> A(a, 0, f(c) - 3) --> A(a - 2, f(c), 3)
  • A(2k + d, 0, c) --> 
The TM enters configuration A(19, 2, 3) after 799 steps with tape:
0^inf 2 1 (11)^19 (12)^2 <D (11)^3 0^inf

Trajectory

A(19, 2, 3) --> A(19, 0, 21) -->

-->

Lets have = m. 

Final trajectory:
0^inf 2 1 (11)^1 <D (11)^m 0^inf
--> 0^inf 2 1 1 2 A> (11)^m 0^inf
--> 0^inf 2 1 (12)^m+1 A> 0^inf
--> 0^inf 2 1 <A (11)^m+2 0^inf
--> 0^inf 2 1 D> (11)^m+2 0^inf
--> 0^inf (21)^m+3 D> 0^inf
--> 0^inf (21)^m+3 2 B> 0^inf
--> 0^inf (21)^m+3 2 <B 2 0^inf
--> 0^inf (21)^m+3 <C (12)^1 0^inf
--> 0^inf <C (11)^m+3 (12)^1 0^inf
--> 0^inf 1 Z> (11)^m+3 (12)^1 0^inf
Score = 2m + 9

Score calculated in HyperCalc:
(10^)^8 30,302,671.815163
Or in tetration: 10^^10.873987 (truncated)
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