Skelet 10: Difference between revisions
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B> 10 10^k 0 --> <D 0^2k+1 10 | B> 10 10^k 0 --> <D 0^2k+1 10 | ||
</pre> | </pre> | ||
These rules are equivalent to the Zeckendorf increment rules in base fibonacci.<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"/> | The left side counter encodes base fibonacci differently and is also reversed, with the least-significant digit being on the right.<ref name="Sligocki"/> | ||
Revision as of 12:42, 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.
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]
References
- ↑ 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