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# | {{machine|1RB1RE_1LC1RB_0RA0LD_1LB1LD_---0RA}} | ||
{{TM|1RB1RE_1LC1RB_0RA0LD_1LB1LD_---0RA}}, also called '''Finned #3''', is a non-halting [[Irregular Turing Machine|irregular]] [[BB(5)]] TM. It belongs to a group of 5 sporadic machines, called finned machines, which had to be individually proven as non-halting in [[Coq-BB5]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref> It was proven to be irregular in March 2023 by lijil.<ref>https://discuss.bbchallenge.org/t/10756090-finned-3-is-irregular/137</ref> | |||
== Analysis by lijil == | |||
<pre> | |||
With the CTL-deciders proving to be very strong there is the question which machines cannot be proven by it. 10756090 is an example of such a machine. We can prove that there is no regular language that includes all configuration the TM reaches, but rejects all configurations that lead to halting: | |||
Write C(n,m) for (10)^n A0 1^m | |||
Then the TM follows these rules: | |||
C(n,m) → halt, if n+m is odd | |||
C(n,m) → C(((n+m)/2)+1,n+1), if n+m is even | |||
I’ll leave the proof as an exercise for the reader. | |||
We start in C(0,0) and C(n,n) → C(n+1,n+1) for all n, so the machine is infinite and visits C(n,n) for all n. | |||
To see how it behaves with n != m we consider the difference n-m after applying rule 2: | |||
((n+m)/2)+1-(n+1) = (n+m-2n)/2 = -(n-m)/2 | |||
So the TM divides the difference of the exponents by 2 (and flips the sign) until the difference becomes odd. Then it halts. This can only go on forever if the difference is 0 to begin with. | |||
Thus C(n,m) → halt, if n!=m | |||
It is impossible for a regular language to separate {C(n,n) for all n} and {C(n,m) for all n!=m} | |||
To finish it off with an argument by savask: | |||
“If there was a regular language containing all non-halting configs of your machine, then we could intersect it with a language (10)* A0 1* and obtain that (10)^n A0 1^n is regular” | |||
</pre> | |||
== References == | |||
[[Category:BB(5)]] | |||
Latest revision as of 17:12, 21 February 2026
1RB1RE_1LC1RB_0RA0LD_1LB1LD_---0RA (bbch), also called Finned #3, is a non-halting irregular BB(5) TM. It belongs to a group of 5 sporadic machines, called finned machines, which had to be individually proven as non-halting in Coq-BB5.[1] It was proven to be irregular in March 2023 by lijil.[2]
Analysis by lijil
With the CTL-deciders proving to be very strong there is the question which machines cannot be proven by it. 10756090 is an example of such a machine. We can prove that there is no regular language that includes all configuration the TM reaches, but rejects all configurations that lead to halting:
Write C(n,m) for (10)^n A0 1^m
Then the TM follows these rules:
C(n,m) → halt, if n+m is odd
C(n,m) → C(((n+m)/2)+1,n+1), if n+m is even
I’ll leave the proof as an exercise for the reader.
We start in C(0,0) and C(n,n) → C(n+1,n+1) for all n, so the machine is infinite and visits C(n,n) for all n.
To see how it behaves with n != m we consider the difference n-m after applying rule 2:
((n+m)/2)+1-(n+1) = (n+m-2n)/2 = -(n-m)/2
So the TM divides the difference of the exponents by 2 (and flips the sign) until the difference becomes odd. Then it halts. This can only go on forever if the difference is 0 to begin with.
Thus C(n,m) → halt, if n!=m
It is impossible for a regular language to separate {C(n,n) for all n} and {C(n,m) for all n!=m}
To finish it off with an argument by savask:
“If there was a regular language containing all non-halting configs of your machine, then we could intersect it with a language (10)* A0 1* and obtain that (10)^n A0 1^n is regular”
References
- ↑ https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value
- ↑ https://discuss.bbchallenge.org/t/10756090-finned-3-is-irregular/137