User:Polygon/Page for testing: Difference between revisions

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List of incomplete pages:
* [[Coq-BB5]]
* [[Finite Automata Reduction]]
* [[CTL]]
* [[Irregular Turing Machine]]
* [[Meet-in-the-Middle Weighted Finite Automata Reduction (MITMWFAR)]]
* [[Skelet 1]]
* [[CPS]] (CPS_LRU, CPS_LRUH)
* [[Repeated Word List]] (RWL_mod; more detailed description for RWLAcc)
== 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
<pre>
<pre>
Let's rewrite the first rule as:
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
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):
Level 1: D(a, b, c, 2k+r, e) -> D(a, b, c, r, e+2k)
A(a, b) --> A(2^n * a + 2^(n + 2) - 4, b - 2^n * a + a - 2^(n + 2) + n + 4)
Level 2: D(a, b, c, 1, e) ->  D(a, b, 0, 1, f2(c, e))
Considering that a is always set to 4 when leaving B, this simplifies to:
  where f2(c, e) = rep(λx -> 2x+5, c)(e)  ~= 2^c
A(4, b) --> A(2^(n + 3) - 4, b - 2^(n + 3) + n + 8)
Level 3: D(a, b, 0, 1, e)  ->  D(a, 0, 0, 1, f3(b, e))
Now consider the halting configuration: A(a, a + 1):
  where f3(b, e) = rep(λx -> f2(x+2, 1), b)(e) ~= 2^^b
So, for this to be in a halting configuration, the following must be true:
Level 4: D(2a+r, 0, 0, 1, e) -> D(r, 0, 0, 1, f4(a, e))
2^(n+3)-4 + 1 = b-2^(n+3)+n+8
  where f4(a, e) = rep(λx -> f3(2x+7), a)(e) ~= 2^^^a
This simplifies to:
Level 5: D(0, 0, 0, 1, e) ->  D(0, 0, 0, 1, f4(4e+19, f3(1, 1)))
b = 2^(n+4)-n-11
 
--> For any positive integer n (and 0): A(4, 2^(n+4)-n-11) --> halt
 
Considering that all rules either keep the overall score or increase it, and that the transition from B(3a+1,b) moves the score onto b, it is guaranteed that b will always increase between A(4,b) configs.
where the last rule repeats forever.
As a result of this, once a halting value of b is passed, it cannot be reached again. However, as b increases, the distance between these halting values of b increases exponentially, making it less and less likely for the TM to halt.
</pre>
</pre>
=== References

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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