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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>.
List of incomplete pages:
==Analysis by Shawn Ligocki==
* [[Coq-BB5]]
https://www.sligocki.com/2023/05/20/bb-2-6-p3.html
* [[Finite Automata Reduction]]
<pre>
* [[CTL]]
Analysis
* [[Irregular Turing Machine]]
Level 1
* [[Meet-in-the-Middle Weighted Finite Automata Reduction (MITMWFAR)]]
These rules can all be verified by direct simulation:
* [[Skelet 1]]
00 <A 212 22^n 55 → <A 212 22^n+2
* [[CPS]] (CPS_LRU, CPS_LRUH)


00 <A 212 22^n 2 55 → <A 212 55^n+2 2


0^5 <A 212 22^n 52 5555 → <A 212 55 2 55^n+3 52
* [[Repeated Word List]] (RWL_mod; more detailed description for RWLAcc)
00 <A 212 22^n 2 52 5 → <A 212 55^n+2 52


Level 2
== 1RB2LA1RC3RA_1LA2RA2RB0RC_1RZ3LC1RA1RB ==
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:
{{machine|1RB2LA1RC3RA_1LA2RA2RB0RC_1RZ3LC1RA1RB}
0^∞ <A 212 22^n 2 55 → 0^∞ <A 212 55^n+2 2
{{TM|1RB2LA1RC3RA_1LA2RA2RB0RC_1RZ3LC1RA1RB} is a non-halting [[BB(4,3)]] TM discovered by Pavel Kropitz in May 2023.<ref>https://discord.com/channels/960643023006490684/1095740122139480195/1113545691994783804</ref In April 2024, Shawn Ligocki showed the TM to follow an infinite pentational rule, proving it non-halting.<ref>https://discord.com/channels/960643023006490684/1095740122139480195/1230591736829575282</ref
                    → 0^∞ <A 212 22^2n+4 2


Level 3
=== Analysis by Shawn Ligocki ===
Repeating Rule 2 we get:
https://discord.com/channels/960643023006490684/1095740122139480195/1230591736829575282
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:
<pre>
0^∞ <A 212 22^n 52 5^5 → 0^∞ <A 212 55 2 55^n+3 52 5
Let D(a, b, c, d, e) = 0^inf 1 2^a 1 3^b 1 01^c 1 2^d <A 2^2e+1 0^inf
                      → 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
Level 1: D(a, b, c, 2k+r, e)  ->  D(a, b, c, r, e+2k)
Let
Level 2: D(a, b, c, 1, e)  ->  D(a, b, 0, 1, f2(c, e))
f(n) = 6*2^(n+4)-4
  where f2(c, e) = rep(λx -> 2x+5, c)(e)  ~= 2^c
Level 3: D(a, b, 0, 1, e)  ->  D(a, 0, 0, 1, f3(b, e))
  where f3(b, e) = rep(λx -> f2(x+2, 1), b)(e)  ~= 2^^b
Level 4: D(2a+r, 0, 0, 1, e) ->  D(r, 0, 0, 1, f4(a, e))
  where f4(a, e) = rep(λx -> f3(2x+7), a)(e)  ~= 2^^^a
Level 5: D(0, 0, 0, 1, e)  ->  D(0, 0, 0, 1, f4(4e+19, f3(1, 1)))


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).
where the last rule repeats forever.
</pre>


Halting Trajectory
=== References
 
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^∞
</pre>
∞∞∞∞
→→→→

Latest revision as of 10:46, 3 April 2026

List of incomplete pages:


1RB2LA1RC3RA_1LA2RA2RB0RC_1RZ3LC1RA1RB

{{machine|1RB2LA1RC3RA_1LA2RA2RB0RC_1RZ3LC1RA1RB} {{TM|1RB2LA1RC3RA_1LA2RA2RB0RC_1RZ3LC1RA1RB} is a non-halting BB(4,3) TM discovered by Pavel Kropitz in May 2023.<ref>https://discord.com/channels/960643023006490684/1095740122139480195/1113545691994783804</ref In April 2024, Shawn Ligocki showed the TM to follow an infinite pentational rule, proving it non-halting.<ref>https://discord.com/channels/960643023006490684/1095740122139480195/1230591736829575282</ref

Analysis by Shawn Ligocki

https://discord.com/channels/960643023006490684/1095740122139480195/1230591736829575282 

Let D(a, b, c, d, e) = 0^inf 1 2^a 1 3^b 1 01^c 1 2^d <A 2^2e+1 0^inf

Level 1: D(a, b, c, 2k+r, e)  ->  D(a, b, c, r, e+2k)
Level 2: D(a, b, c, 1, e)  ->  D(a, b, 0, 1, f2(c, e))
  where f2(c, e) = rep(λx -> 2x+5, c)(e)  ~= 2^c
Level 3: D(a, b, 0, 1, e)  ->  D(a, 0, 0, 1, f3(b, e))
  where f3(b, e) = rep(λx -> f2(x+2, 1), b)(e)  ~= 2^^b
Level 4: D(2a+r, 0, 0, 1, e)  ->  D(r, 0, 0, 1, f4(a, e))
  where f4(a, e) = rep(λx -> f3(2x+7), a)(e)  ~= 2^^^a
Level 5: D(0, 0, 0, 1, e)  ->  D(0, 0, 0, 1, f4(4e+19, f3(1, 1)))


where the last rule repeats forever.

=== References

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