Skelet 10 - Revision history

Sligocki: Not a stub anymore.

2026-02-21T16:13:29Z

Not a stub anymore.

← Older revision		Revision as of 16:13, 21 February 2026
Line 1:		Line 1:
	{{machine\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}~~{{Stub}}~~		{{machine\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}{{TM\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}, 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]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref>
	{{TM\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}, 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]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref>

	== Behavior ==		== Behavior ==

Polygon: Added note

2026-02-21T15:00:39Z

Added note

← Older revision		Revision as of 15:00, 21 February 2026
Line 60:		Line 60:
	Z(n) ≠ 00 w ==> L(n) <D --> L(n+1) A>		Z(n) ≠ 00 w ==> L(n) <D --> L(n+1) A>
	</pre>		</pre>

			For a proof of this TM not halting, see https://www.sligocki.com/2023/03/14/skelet-10.html

	== See also ==		== See also ==

Polygon: Added rewrites of left side rules

2026-02-21T14:53:59Z

Added rewrites of left side rules

← Older revision		Revision as of 14:53, 21 February 2026
Line 53:		Line 53:
	00 <D --> 10 B>		00 <D --> 10 B>
	00 10^k 00 <D --> 10 10 00^k A>		00 10^k 00 <D --> 10 10 00^k A>
			</pre>

			These can be rewritten as:<ref name="Sligocki"/>
			<pre>
			Z(n) = 00 w ==> L(n) <D --> L(n+1) B>
			Z(n) ≠ 00 w ==> L(n) <D --> L(n+1) A>
	</pre>		</pre>

Polygon: /* Behavior */ added missing references

2026-02-21T14:50:49Z

Behavior: added missing references

← Older revision		Revision as of 14:50, 21 February 2026
Line 22:		Line 22:
	</pre>		</pre>

	The left side counter encodes base fibonacci differently and is also reversed to big-endian (least significant digit is on the right). The two least-significant symbols are also omitted. These differences can be described using the following definitions:<ref name="Sligocki"/>		The left side counter encodes base fibonacci differently and is also reversed to big-endian (least significant digit is on the right).<ref name="Sligocki"/> The two least-significant symbols are also omitted.<ref name="Sligocki"/> These differences can be described using the following definitions:<ref name="Sligocki"/>
	<pre>		<pre>
	T(0w) = T(w)00		T(0w) = T(w)00

Polygon: Added increment effects

2026-02-21T14:49:58Z

Added increment effects

← Older revision		Revision as of 14:49, 21 February 2026
Line 2:		Line 2:
	{{TM\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}, 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]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref>		{{TM\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}, 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]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref>

	== ~~[https://www.sligocki.com/2023/03/14/skelet-10.html Analysis] by [[User:Sligocki\|Shawn Ligocki]]~~ ==		== Behavior ==

	Skelet 10 implements two [https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem base fibonacci] counters, one on the right tape side and one on the left tape side. The TM halts if these counters become desynchronised.<ref name="bbchallenge 2025"/> The right side counter can be described by the following rules:<ref name="Sligocki">https://www.sligocki.com/2023/03/14/skelet-10.html</ref>		Skelet 10 implements two [https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem base fibonacci] counters, one on the right tape side and one on the left tape side. The TM halts if these counters become desynchronised.<ref name="bbchallenge 2025"/> The right side counter can be described by the following rules:<ref name="Sligocki">https://www.sligocki.com/2023/03/14/skelet-10.html</ref>
Line 36:		Line 36:
	</pre>		</pre>

	There are now 3 possible cases for 3 possible values of the two least-significant digits of the left counter:		There are now 3 possible cases for 3 possible values of the two least-significant digits of the left counter:<ref name="Sligocki"/>
	<pre>		<pre>
	Z(n) = 00 w --> T(Z(n)) = T(w) 00 00 --> L(n) = T(w) 00		Z(n) = 00 w --> T(Z(n)) = T(w) 00 00 --> L(n) = T(w) 00
	Z(n) = 0 10^k+1 0 w --> T(Z(n)) = T(w) 00 10 1010^k+1 00 --> L(n) = T(w) 00 10^2k+1 10		Z(n) = 0 10^k+1 0 w --> T(Z(n)) = T(w) 00 10 1010^k+1 00 --> L(n) = T(w) 00 10^2k+1 10
	Z(n) = 10 10^k 0 w --> T(Z(n)) = T(w) 00 1010^k+1 --> L(n) = T(w) 00 10^2k 10		Z(n) = 10 10^k 0 w --> T(Z(n)) = T(w) 00 1010^k+1 --> L(n) = T(w) 00 10^2k 10
			</pre>
			These are affected by incrementing in the following way:<ref name="Sligocki"/>
			<pre>
			Z(n) = 00 w --> L(n) = T(w) 00 ==> Z(n+1) = 10 w --> L(n+1) = T(w) 10
			Z(n) = 0 10^k+1 0 w --> L(n) = T(w) 00 10^2k+1 10 ==> Z(n+1) = 0^2k+2 10 w --> L(n+1) = T(w) 10 10 10^2k+1
			Z(n) = 10 10^k 0 w --> L(n) = T(w) 00 10^2k 10 ==> Z(n+1) = 0^2k+1 10 w --> L(n+1) = T(w) 10 10 00^2k
	</pre>		</pre>

Polygon: Added cases for least-significant digits

2026-02-21T14:40:38Z

Added cases for least-significant digits

← Older revision		Revision as of 14:40, 21 February 2026
Line 2:		Line 2:
	{{TM\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}, 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]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref>		{{TM\|1LC0LA_---0LC_0RD1LA_1LB1RE_1RD0RE}}, 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]].<ref name="bbchallenge 2025">https://arxiv.org/pdf/2509.12337 Determination of the fifth Busy Beaver value</ref>

	== ~~Behavior~~ ==		== [https://www.sligocki.com/2023/03/14/skelet-10.html Analysis] by [[User:Sligocki\|Shawn Ligocki]] ==

	Skelet 10 implements two [https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem base fibonacci] counters, one on the right tape side and one on the left tape side. The TM halts if these counters become desynchronised.<ref name="bbchallenge 2025"/> The right side counter can be described by the following rules:<ref name="Sligocki">https://www.sligocki.com/2023/03/14/skelet-10.html</ref>		Skelet 10 implements two [https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem base fibonacci] counters, one on the right tape side and one on the left tape side. The TM halts if these counters become desynchronised.<ref name="bbchallenge 2025"/> The right side counter can be described by the following rules:<ref name="Sligocki">https://www.sligocki.com/2023/03/14/skelet-10.html</ref>
Line 34:		Line 34:
	where v is any valid Zeckendorf expression.		where v is any valid Zeckendorf expression.
	--> This removes the two least-significant tape symbols.		--> This removes the two least-significant tape symbols.
			</pre>

			There are now 3 possible cases for 3 possible values of the two least-significant digits of the left counter:
			<pre>
			Z(n) = 00 w --> T(Z(n)) = T(w) 00 00 --> L(n) = T(w) 00
			Z(n) = 0 10^k+1 0 w --> T(Z(n)) = T(w) 00 10 1010^k+1 00 --> L(n) = T(w) 00 10^2k+1 10
			Z(n) = 10 10^k 0 w --> T(Z(n)) = T(w) 00 1010^k+1 --> L(n) = T(w) 00 10^2k 10
	</pre>		</pre>

Polygon: big-endian

2026-02-21T14:19:17Z

big-endian

← Older revision		Revision as of 14:19, 21 February 2026
Line 22:		Line 22:
	</pre>		</pre>

	The left side counter encodes base fibonacci differently and is also reversed~~, with the~~ least-significant digit ~~being~~ on the right. The two least-significant symbols are also omitted. These differences can be described using the following definitions:<ref name="Sligocki"/>		The left side counter encodes base fibonacci differently and is also reversed to big-endian (least significant digit is on the right). The two least-significant symbols are also omitted. These differences can be described using the following definitions:<ref name="Sligocki"/>
	<pre>		<pre>
	T(0w) = T(w)00		T(0w) = T(w)00

Polygon: Added T and L notations

2026-02-21T14:18:12Z

Added T and L notations

@@ Line 22: / Line 22: @@
 </pre>
-The left side counter encodes base fibonacci differently and is also reversed, with the least-significant digit being on the right. It can be described by the following rules:<ref name="Sligocki"/>
+The left side counter encodes base fibonacci differently and is also reversed, with the least-significant digit being on the right. The two least-significant symbols are also omitted. These differences can be described using the following definitions:<ref name="Sligocki"/>
 <pre>
 <D -->         10 B>

Polygon: Added left side rules

2026-02-21T14:05:12Z

Added left side rules

← Older revision		Revision as of 14:05, 21 February 2026
Line 22:		Line 22:
	</pre>		</pre>

	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. It can be described by the following rules:<ref name="Sligocki"/>
			<pre>
			00 <D --> 10 B>
			00 10^k 00 <D --> 10 10 00^k A>
			</pre>

	== See also ==		== See also ==

Polygon: Moving little-endian to increment rules

2026-02-21T13:57:40Z

Moving little-endian to increment rules

← Older revision		Revision as of 13:57, 21 February 2026
Line 13:		Line 13:
	Z(n) = 0 10^k 0 w ==> Z(n+1) = 0^2k 10 w		Z(n) = 0 10^k 0 w ==> Z(n+1) = 0^2k 10 w
	Z(n) = 10 10^k 0 w =0> Z(n+1) = 0^2k+1 10 w		Z(n) = 10 10^k 0 w =0> Z(n+1) = 0^2k+1 10 w
			Where Z(n) represents n in the little-endian (least significant digit is on the left) representation of Zeckendorf notation.
	</pre>		</pre>
	and can be restated as:<ref name="Sligocki"/>		and can be restated as:<ref name="Sligocki"/>
Line 18:		Line 19:
	Z(n) = 0 w ==> A> Z(n) --> <D Z(n+1)		Z(n) = 0 w ==> A> Z(n) --> <D Z(n+1)
	Z(n) = 1 w ==> B> Z(n) --> <D Z(n+1)		Z(n) = 1 w ==> B> Z(n) --> <D Z(n+1)
	Where Z(n) = 0 w means that the least-significant bit in Z(n) is 0 ~~and Z(n) represents n in the little-endian (least significant digit is on the left) representation of Zeckendorf notation~~.		Where Z(n) = 0 w means that the least-significant bit in Z(n) is 0.
	</pre>		</pre>