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>.
Placeholder
==Analysis by Shawn Ligocki==
{{TM|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD|halt}}
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
8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf


0^5 <A 212 22^n 52 5555 → <A 212 55 2 55^n+3 52
9. S (12)^a <D (11)^b 0^inf --> S (12)^a-1 <D (11)^2b+3 0^inf
00 <A 212 22^n 2 52 5 → <A 212 55^n+2 52
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


Level 2
10. S (12)^a <D (11)^b 0^inf --> S <D (11)^((2^(a))*b+(2^(a))*3-3) 0^inf
Repeating the first rule above we get:
11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
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
</pre>
</pre>
∞∞∞∞
→→→→

Latest revision as of 16:27, 30 September 2025

Placeholder 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
8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf

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
11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
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