Skelet 10: Difference between revisions
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The left side counter encodes base fibonacci differently and is also reversed, with the least-significant digit being on the right.<ref name="Sligocki"/> | The left side counter encodes base fibonacci differently and is also reversed, with the least-significant digit being on the right.<ref name="Sligocki"/> | ||
== See also == | |||
* [[Skelet 1]] | |||
* [[Skelet 17]] | |||
* [[Finned 3]] | |||
== References == | == References == | ||
[[Category:BB(5)]] | [[Category:BB(5)]] | ||
Revision as of 12:49, 21 February 2026
1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE (bbch), called Skelet #10, was one of Skelet's 43 holdouts and one of the last holdouts in BB(5). It is a double fibonacci counter. It is one of the few TMs to have required an individual proof of non-halting in Coq-BB5.[1]
Behavior
Skelet 10 implements two base fibonacci counters, one on the right tape side and one on the left tape side. The TM halts if these counters become desynchronised.[1] The right side counter can be described by the following rules:[2]
A> 0 10^k 0 --> <D 0^2k 10 B> 10 10^k 0 --> <D 0^2k+1 10
These rules are equivalent to the Zeckendorf increment rules in base fibonacci.[2]
The left side counter encodes base fibonacci differently and is also reversed, with the least-significant digit being on the right.[2]
See also
References
- ↑ 1.0 1.1 https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value
- ↑ 2.0 2.1 2.2 https://www.sligocki.com/2023/03/14/skelet-10.html