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>.
==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).
</pre>
</pre>
∞∞∞∞
→→→→

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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