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	<id>https://wiki.bbchallenge.org/w/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Cosmo</id>
	<title>BusyBeaverWiki - User contributions [en]</title>
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	<updated>2026-05-12T17:15:42Z</updated>
	<subtitle>User contributions</subtitle>
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		<id>https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=4606</id>
		<title>Main Page</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=4606"/>
		<updated>2025-10-19T09:59:55Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* Notes */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;The [[Busy Beaver function]] BB (called &#039;&#039;S&#039;&#039; originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to &#039;&#039;m&#039;&#039;-symbol Turing machines:&amp;lt;ref&amp;gt;Radó, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
| BB(&#039;&#039;n&#039;&#039;, &#039;&#039;m&#039;&#039;) = Maximum number of steps taken by a halting &#039;&#039;n&#039;&#039;-state, &#039;&#039;m&#039;&#039;-symbol Turing machine starting from a blank (all 0) tape&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
The 2-symbol case BB(&#039;&#039;n&#039;&#039;, 2) is abbreviated as BB(&#039;&#039;n&#039;&#039;). The busy beaver function is not computable, but a few of its values are known:&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+ Small busy beaver values&amp;lt;ref&amp;gt;P. Michel, &amp;quot;[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]&amp;quot;.&amp;lt;/ref&amp;gt;&lt;br /&gt;
! !!2-state!!3-state !!4-state!!5-state!!6-state &lt;br /&gt;
!7-state&lt;br /&gt;
|-  &lt;br /&gt;
! 2-symbol &lt;br /&gt;
| [[BB(2)]] = 6 &lt;br /&gt;
| [[BB(3)]] = 21&lt;br /&gt;
| [[BB(4)]] = 107 &lt;br /&gt;
| [[BB(5)]] = 47,176,870 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(6)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow \uparrow \uparrow 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(7)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{11} 2 \uparrow^{11} 3&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
! 3-symbol&lt;br /&gt;
| [[BB(2,3)]] = 38 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(3,3)]] &amp;gt; &amp;lt;math&amp;gt;10^{17}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(4,3)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow\uparrow\uparrow 2^{2^{32}}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 4-symbol  &lt;br /&gt;
| [[BB(2,4)]] = 3,932,964&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,4)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{15} 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 5-symbol &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(2,5)]] &amp;gt; &amp;lt;math&amp;gt;10\uparrow\uparrow 4&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,5)]] &amp;gt; &amp;lt;math&amp;gt; f_\omega(2 \uparrow^{15} 5)&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 6-symbol &lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(2,6)]] &amp;gt; &amp;lt;math&amp;gt;10 \uparrow\uparrow\uparrow 3&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
In the above table, &amp;lt;span style=&amp;quot;background: orange&amp;quot;&amp;gt;cells are highlighted in orange&amp;lt;/span&amp;gt; when there are known [[Cryptids]] (mathematically-hard machines) in that class, and &amp;lt;span style=&amp;quot;background: #ffe4b2&amp;quot;&amp;gt;cells are highlighted in light orange&amp;lt;/span&amp;gt; when the existence of a Cryptid is given by using a known one with less states or symbols. 1-state domains are omitted in the table as [[BB(1,m)]] = 1.&lt;br /&gt;
&lt;br /&gt;
== About bbchallenge ==&lt;br /&gt;
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt.  Other efforts also include:&lt;br /&gt;
&lt;br /&gt;
* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Rocq Rocq])&lt;br /&gt;
* Maintaining [[Holdouts lists]] for small busy beaver values&lt;br /&gt;
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]&lt;br /&gt;
* Finding [[Cryptids]] (mathematically-hard machines)&lt;br /&gt;
* Searching for new [[Champions]]&lt;br /&gt;
* Building [[Accelerated Simulator]]s to simulate halting machines faster&lt;br /&gt;
* Writing papers and giving talks about busy beaver, see [[Papers &amp;amp; Talks]]&lt;br /&gt;
&lt;br /&gt;
In June 2024, bbchallenge achieved a significant milestone by proving in [https://en.wikipedia.org/wiki/Rocq Rocq] (previously known as Coq) that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.&amp;lt;ref&amp;gt;H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.&lt;br /&gt;
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== This Month in Beaver Research (TMBR) ==&lt;br /&gt;
[[:Category:This Month in Beaver Research|This Month in Beaver Research]] (TMBR, pronounced &amp;quot;timber&amp;quot;) is a monthly summary of Busy Beaver research progress. Here are the three most recent released entries:&lt;br /&gt;
&lt;br /&gt;
* [[TMBR: September 2025|TMBR September 2025]]&lt;br /&gt;
* [[TMBR: August 2025|TMBR August 2025]]&lt;br /&gt;
* [[TMBR: July 2025|TMBR July 2025]]&lt;br /&gt;
&lt;br /&gt;
[[TMBR: October 2025|TMBR October 2025]] is currently work in progress.&lt;br /&gt;
&lt;br /&gt;
== Contribute to this wiki ==&lt;br /&gt;
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;inputbox&amp;gt;&lt;br /&gt;
type=create&lt;br /&gt;
width=100&lt;br /&gt;
break=no&lt;br /&gt;
buttonlabel=Create new article&lt;br /&gt;
default=(Article title)&lt;br /&gt;
&amp;lt;/inputbox&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Notes==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=User_talk:Polygon&amp;diff=4604</id>
		<title>User talk:Polygon</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=User_talk:Polygon&amp;diff=4604"/>
		<updated>2025-10-19T09:50:31Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hi Polygon! This is Andrew, another big contributor to the wiki. Most of us are on a discord channel BBChallenge that you should join! Here&#039;s an invite link: https://discord.gg/brgtycR8 &lt;br /&gt;
&lt;br /&gt;
Your detail-oriented wiki management has not gone unnoticed and is definitely appreciated. &lt;br /&gt;
&lt;br /&gt;
I was curious if you were interested in helping with migrating results compiled on https://cosearch.bbchallenge.org to the wiki. This is a webpage built to help keep track of the results shared in the discord, but has seen less usage relative to the wiki. I was hoping you could go through its 75 &amp;quot;contributions&amp;quot; listed in Cosearch, making sure they were described in the wiki.&lt;br /&gt;
&lt;br /&gt;
What do you think? And thanks again for your wiki management!&lt;br /&gt;
&lt;br /&gt;
:Hello Andrew, I am indeed interested in helping to migrate the results to the wiki. [[User:Polygon|Polygon]] ([[User talk:Polygon|talk]]) 15:14, 27 August 2025 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hello Polygon, Tristan/cosmo here. Wanted to thank you very much for all the amazing work you&#039;re doing on the wiki, its truly amazing, thank you!!&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Papers_%26_Talks&amp;diff=3561</id>
		<title>Papers &amp; Talks</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Papers_%26_Talks&amp;diff=3561"/>
		<updated>2025-08-31T11:38:43Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Papers ==&lt;br /&gt;
Paper produced by The bbchallenge Collaboration:&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;&#039;Determination of the fifth Busy Beaver value.&#039;&#039;&#039; The bbchallenge Collaboration, J. Blanchard, Daniel Briggs, Konrad Deka, Nathan Fenner, Yannick Forster, Georgi Georgiev (Skelet), Matthew L. House, Rachel Hunter, Iijil, Maja Kądziołka, Pavel Kropitz, Shawn Ligocki, mxdys, Mateusz Naściszewski, savask, Tristan Stérin, Chris Xu, Jason Yuen, Théo Zimmermann. &#039;&#039;In preparation.&#039;&#039; [https://github.com/bbchallenge/bbchallenge-paper (github)]&lt;br /&gt;
* &#039;&#039;&#039;Turing machines deciders, part I.&#039;&#039;&#039; The bbchallenge Collaboration, J. Blanchard, K. Deka, N. Fenner, T. Guilfoyle, Iijil, M. Kądziołka, P. Kropitz, S. Ligocki, P. Michel, M. Naściszewski, and T. Stérin. Apr. 2025. https://arxiv.org/abs/2504.20563&lt;br /&gt;
Paper produced by community members:&lt;br /&gt;
* Stérin, T., Woods, D. (2024). Hardness of Busy Beaver Value BB(15). In: Kovács, L., Sokolova, A. (eds) Reachability Problems. RP 2024. Lecture Notes in Computer Science, vol 15050. Springer, Cham. https://doi.org/10.1007/978-3-031-72621-7_9 &lt;br /&gt;
** Paper: [https://arxiv.org/abs/2107.12475 arxiv version] (first released in 2021), [https://link.springer.com/chapter/10.1007/978-3-031-72621-7_9 conference version]&lt;br /&gt;
** Talk on September 25th 2024 at RP 2024 in Vienna: [https://drive.google.com/file/d/1gCoKCHvXGJeY7LvdQX9kN8u0ITdCKtJN/view?usp=sharing talk recording], [https://drive.google.com/file/d/1x3MRuKyh92XWlBHRawqDzxEdXIGSgsX_/view?usp=drive_link talk recording (compressed)], [https://docs.google.com/presentation/d/1rVEMr-AJM_kTzdI95xT2KLux4G24AoTBBCYQoPk8vS0/edit#slide=id.g2d359b31dea_0_2125 talk slides]&lt;br /&gt;
* Xu, C. (2024). Skelet #17 and the fifth Busy Beaver number. [https://arxiv.org/abs/2407.02426 Arxiv preprint]&lt;br /&gt;
&lt;br /&gt;
== Talks ==&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;To come: &#039;&#039;&#039;Le cinquième nombre Busy Beaver.&#039;&#039;&#039;&#039;&#039; Talk by Tristan Stérin at the seminar of Prof. Timothy Gowers [https://www.college-de-france.fr/fr/chaire/timothy-gowers-combinatoire-chaire-statutaire/ College de France&#039;s lectures] , October 27th 2025.&lt;br /&gt;
* &#039;&#039;&#039;Determination of the fifth Busy Beaver value.&#039;&#039;&#039; Poster by The bbchallenge Collaboration and Tristan Stérin presented at the 31st International Conference on DNA Computing and Molecular Programming [https://dna31.sciencesconf.org/ DNA31]. [https://drive.google.com/file/d/1vRq6C7BIhKdq9hSZYdQtzV5JJRuwGc6q/view?usp=sharing (poster PDF)] [https://drive.google.com/file/d/1vQhcKVgEaRBeANwmRZ-fCzoJcP6kxJ4D/view?usp=sharing (poster SVG)]&lt;br /&gt;
* &#039;&#039;&#039;Coq proof of the fifth Busy Beaver value.&#039;&#039;&#039; Talk given by Tristan Stérin at &#039;&#039;&#039;[https://msp.cis.strath.ac.uk/types2025/cfp.html Types 2025],&#039;&#039;&#039; Glasgow, Scotland, June 9th 2025. [https://docs.google.com/presentation/d/1koofTIAOdk-DsKH6bnLi_jPeHG3-9vz2EcS_N5PsU-0/edit (slides)] [https://bbchallenge.org/types2025 (links)]&lt;br /&gt;
* &#039;&#039;&#039;Formal verification of the 5th Busy Beaver value.&#039;&#039;&#039; Talk given by Maja Kądziołka and Tristan Stérin at Cambridge&#039;s seminar &amp;quot;[https://talks.cam.ac.uk/show/index/164015 Formalisation of mathematics with interactive theorem provers]&amp;quot;, March 13th 2025. [https://www.youtube.com/watch?v=5X6YVEnbLZU (recording)] [https://docs.google.com/presentation/d/1enqHCmXNYubhZyHmaj1qaYciY0bJtSJasc8gqwMLZ7c/edit?slide=id.g319e2b7dbee_0_3#slide=id.g319e2b7dbee_0_3 (slides, part 1)] [https://sakamoto.pl/~mei/bbslides/bbslides.html (interactive slides, part 2)]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Papers_%26_Talks&amp;diff=3560</id>
		<title>Papers &amp; Talks</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Papers_%26_Talks&amp;diff=3560"/>
		<updated>2025-08-31T11:38:31Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Papers ==&lt;br /&gt;
Paper produced by The bbchallenge Collaboration:&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;&#039;Determination of the fifth Busy Beaver value.&#039;&#039;&#039; The bbchallenge Collaboration, J. Blanchard, Daniel Briggs, Konrad Deka, Nathan Fenner, Yannick Forster, Georgi Georgiev (Skelet), Matthew L. House, Rachel Hunter, Iijil, Maja Kądziołka, Pavel Kropitz, Shawn Ligocki, mxdys, Mateusz Naściszewski, savask, Tristan Stérin, Chris Xu, Jason Yuen, Théo Zimmermann. &#039;&#039;In preparation.&#039;&#039; [https://github.com/bbchallenge/bbchallenge-paper (github)]&lt;br /&gt;
* &#039;&#039;&#039;Turing machines deciders, part I.&#039;&#039;&#039; The bbchallenge Collaboration, J. Blanchard, K. Deka, N. Fenner, T. Guilfoyle, Iijil, M. Kądziołka, P. Kropitz, S. Ligocki, P. Michel, M. Naściszewski, and T. Stérin. Apr. 2025. https://arxiv.org/abs/2504.20563&lt;br /&gt;
Paper produced by community members:&lt;br /&gt;
* Stérin, T., Woods, D. (2024). Hardness of Busy Beaver Value BB(15). In: Kovács, L., Sokolova, A. (eds) Reachability Problems. RP 2024. Lecture Notes in Computer Science, vol 15050. Springer, Cham. https://doi.org/10.1007/978-3-031-72621-7_9 &lt;br /&gt;
** Paper: [https://arxiv.org/abs/2107.12475 arxiv version] (first released in 2021), [https://link.springer.com/chapter/10.1007/978-3-031-72621-7_9 conference version]&lt;br /&gt;
** Talk on September 25th 2024 at RP 2024 in Vienna: [https://drive.google.com/file/d/1gCoKCHvXGJeY7LvdQX9kN8u0ITdCKtJN/view?usp=sharing talk recording], [https://drive.google.com/file/d/1x3MRuKyh92XWlBHRawqDzxEdXIGSgsX_/view?usp=drive_link talk recording (compressed)], [https://docs.google.com/presentation/d/1rVEMr-AJM_kTzdI95xT2KLux4G24AoTBBCYQoPk8vS0/edit#slide=id.g2d359b31dea_0_2125 talk slides]&lt;br /&gt;
* Xu, C. (2024). Skelet #17 and the fifth Busy Beaver number. [https://arxiv.org/abs/2407.02426 Arxiv preprint]&lt;br /&gt;
&lt;br /&gt;
== Talks ==&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;To come: &#039;&#039;&#039;Le cinquième nombre Busy Beaver.&#039;&#039;&#039;&#039;&#039; Talk by Tristan Stérin at the seminar of Prof. Timothy Gowers [https://www.college-de-france.fr/fr/chaire/timothy-gowers-combinatoire-chaire-statutaire/ College de France&#039;s lectures] , October 27th 2025.&lt;br /&gt;
* &#039;&#039;&#039;Determination of the fifth Busy Beaver value.&#039;&#039;&#039; Poster by The bbchallenge Collaborationa and Tristan Stérin presented at the 31st International Conference on DNA Computing and Molecular Programming [https://dna31.sciencesconf.org/ DNA31]. [https://drive.google.com/file/d/1vRq6C7BIhKdq9hSZYdQtzV5JJRuwGc6q/view?usp=sharing (poster PDF)] [https://drive.google.com/file/d/1vQhcKVgEaRBeANwmRZ-fCzoJcP6kxJ4D/view?usp=sharing (poster SVG)]&lt;br /&gt;
* &#039;&#039;&#039;Coq proof of the fifth Busy Beaver value.&#039;&#039;&#039; Talk given by Tristan Stérin at &#039;&#039;&#039;[https://msp.cis.strath.ac.uk/types2025/cfp.html Types 2025],&#039;&#039;&#039; Glasgow, Scotland, June 9th 2025. [https://docs.google.com/presentation/d/1koofTIAOdk-DsKH6bnLi_jPeHG3-9vz2EcS_N5PsU-0/edit (slides)] [https://bbchallenge.org/types2025 (links)]&lt;br /&gt;
* &#039;&#039;&#039;Formal verification of the 5th Busy Beaver value.&#039;&#039;&#039; Talk given by Maja Kądziołka and Tristan Stérin at Cambridge&#039;s seminar &amp;quot;[https://talks.cam.ac.uk/show/index/164015 Formalisation of mathematics with interactive theorem provers]&amp;quot;, March 13th 2025. [https://www.youtube.com/watch?v=5X6YVEnbLZU (recording)] [https://docs.google.com/presentation/d/1enqHCmXNYubhZyHmaj1qaYciY0bJtSJasc8gqwMLZ7c/edit?slide=id.g319e2b7dbee_0_3#slide=id.g319e2b7dbee_0_3 (slides, part 1)] [https://sakamoto.pl/~mei/bbslides/bbslides.html (interactive slides, part 2)]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Papers_%26_Talks&amp;diff=3559</id>
		<title>Papers &amp; Talks</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Papers_%26_Talks&amp;diff=3559"/>
		<updated>2025-08-31T11:34:32Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Papers ==&lt;br /&gt;
Paper produced by The bbchallenge Collaboration:&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;&#039;Determination of the fifth Busy Beaver value.&#039;&#039;&#039; The bbchallenge Collaboration, J. Blanchard, Daniel Briggs, Konrad Deka, Nathan Fenner, Yannick Forster, Georgi Georgiev (Skelet), Matthew L. House, Rachel Hunter, Iijil, Maja Kądziołka, Pavel Kropitz, Shawn Ligocki, mxdys, Mateusz Naściszewski, savask, Tristan Stérin, Chris Xu, Jason Yuen, Théo Zimmermann. &#039;&#039;In preparation.&#039;&#039; [https://github.com/bbchallenge/bbchallenge-paper (github)]&lt;br /&gt;
* &#039;&#039;&#039;Turing machines deciders, part I.&#039;&#039;&#039; The bbchallenge Collaboration, J. Blanchard, K. Deka, N. Fenner, T. Guilfoyle, Iijil, M. Kądziołka, P. Kropitz, S. Ligocki, P. Michel, M. Naściszewski, and T. Stérin. Apr. 2025. https://arxiv.org/abs/2504.20563&lt;br /&gt;
Paper produced by community members:&lt;br /&gt;
* Stérin, T., Woods, D. (2024). Hardness of Busy Beaver Value BB(15). In: Kovács, L., Sokolova, A. (eds) Reachability Problems. RP 2024. Lecture Notes in Computer Science, vol 15050. Springer, Cham. https://doi.org/10.1007/978-3-031-72621-7_9 &lt;br /&gt;
** Paper: [https://arxiv.org/abs/2107.12475 arxiv version] (first released in 2021), [https://link.springer.com/chapter/10.1007/978-3-031-72621-7_9 conference version]&lt;br /&gt;
** Talk on September 25th 2024 at RP 2024 in Vienna: [https://drive.google.com/file/d/1gCoKCHvXGJeY7LvdQX9kN8u0ITdCKtJN/view?usp=sharing talk recording], [https://drive.google.com/file/d/1x3MRuKyh92XWlBHRawqDzxEdXIGSgsX_/view?usp=drive_link talk recording (compressed)], [https://docs.google.com/presentation/d/1rVEMr-AJM_kTzdI95xT2KLux4G24AoTBBCYQoPk8vS0/edit#slide=id.g2d359b31dea_0_2125 talk slides]&lt;br /&gt;
* Xu, C. (2024). Skelet #17 and the fifth Busy Beaver number. [https://arxiv.org/abs/2407.02426 Arxiv preprint]&lt;br /&gt;
&lt;br /&gt;
== Talks ==&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;To come: &#039;&#039;&#039;Le cinquième nombre Busy Beaver.&#039;&#039;&#039;&#039;&#039; Talk by Tristan Stérin at the seminar of Prof. Timothy Gowers [https://www.college-de-france.fr/fr/chaire/timothy-gowers-combinatoire-chaire-statutaire/ College de France&#039;s lectures] , October 27th 2025.&lt;br /&gt;
* &#039;&#039;&#039;Coq proof of the fifth Busy Beaver value.&#039;&#039;&#039; Talk given by Tristan Stérin at &#039;&#039;&#039;[https://msp.cis.strath.ac.uk/types2025/cfp.html Types 2025],&#039;&#039;&#039; Glasgow, Scotland, June 9th 2025. [https://docs.google.com/presentation/d/1koofTIAOdk-DsKH6bnLi_jPeHG3-9vz2EcS_N5PsU-0/edit (slides)] [https://bbchallenge.org/types2025 (links)]&lt;br /&gt;
* &#039;&#039;&#039;Formal verification of the 5th Busy Beaver value.&#039;&#039;&#039; Talk given by Maja Kądziołka and Tristan Stérin at Cambridge&#039;s seminar &amp;quot;[https://talks.cam.ac.uk/show/index/164015 Formalisation of mathematics with interactive theorem provers]&amp;quot;, March 13th 2025. [https://www.youtube.com/watch?v=5X6YVEnbLZU (recording)] [https://docs.google.com/presentation/d/1enqHCmXNYubhZyHmaj1qaYciY0bJtSJasc8gqwMLZ7c/edit?slide=id.g319e2b7dbee_0_3#slide=id.g319e2b7dbee_0_3 (slides, part 1)] [https://sakamoto.pl/~mei/bbslides/bbslides.html (interactive slides, part 2)]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=TMBR:_August_2025&amp;diff=3446</id>
		<title>TMBR: August 2025</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=TMBR:_August_2025&amp;diff=3446"/>
		<updated>2025-08-27T20:15:43Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Stub}} &lt;br /&gt;
[[:Category:This Month in Beaver Research|This Month in Beaver Research]] for August 2025, featuring plenty of holdouts reduction in numerous domains, alongside an upgraded TM-visualizer and a couple new BB adjacent games.&lt;br /&gt;
&lt;br /&gt;
== Cryptids ==&lt;br /&gt;
&lt;br /&gt;
* A fast algorithm for [[Consistent Collatz]] simulation was re-discovered and popularized. Using it:&lt;br /&gt;
** apgroucher simulated [[Antihydra]] to &amp;lt;math&amp;gt;2^{38}&amp;lt;/math&amp;gt; iterations. This is actually a result from one year ago, but was rediscovered and added to the wiki. https://discord.com/channels/960643023006490684/1026577255754903572/1271528180246773883&lt;br /&gt;
** Shawn Ligocki simulated {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC}} out to one additional Collatz reset, demonstrating that (if they halt, which they probviously should) they will have sigma scores &amp;lt;math&amp;gt;&amp;gt; 10^{10^{10^7}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
== Holdouts ==&lt;br /&gt;
&lt;br /&gt;
* [[BB(6)]] holdouts:&lt;br /&gt;
** XnoobSpeakable found 9 new halting TMs in the high exponential runtime range (~10^100000) by running Enumerate.py out to extremely high parameters. https://discord.com/channels/960643023006490684/1239205785913790465/1401470301467836556&lt;br /&gt;
** Andrew Ducharme found a couple surprisingly short running halting TMs in the [[BB(6)]] holdouts list with runtime ~10^78. https://discord.com/channels/960643023006490684/1239205785913790465/1407754434523693179&lt;br /&gt;
*After the [[BB(7)#Phase 1|enumeration of BB(7)]] was completed, Andrew Ducharme ran [[BB(7)#Phase 2|several deciders]] on the holdouts list, filtering the original 86,129,304 holdouts down to 60,765,943 in 8 days.  https://drive.google.com/drive/u/0/folders/17U0BRpJHTMLtB0poBlOSZhGGp4FkCHIO&lt;br /&gt;
*[[BB(3,3)|BB(3,3):]] 9 holdouts were proven non-halting in Rocq (previously known as Coq) by mxdys. [https://wiki.bbchallenge.org/wiki/BB(3,3)#Holdouts 10 holdouts remain, 4 of them solved with moderate rigor.]https://discord.com/channels/960643023006490684/1259770474897080380/1410308974275985428&lt;br /&gt;
&lt;br /&gt;
==BB Adjacent==&lt;br /&gt;
*John Tromp introduced the &amp;lt;math&amp;gt;BB \lambda _1(n)&amp;lt;/math&amp;gt; function for [[Busy Beaver for lambda calculus#Oracle Busy Beaver|Busy Beaver for lambda calculus with an oracle]] and computed it up to &amp;lt;math&amp;gt;BB \lambda _1(22)&amp;lt;/math&amp;gt;.&lt;br /&gt;
* Instruction-Limited Greedy Busy Beaver gBBi(n) and an [[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants|Instruction-Limited variant]] of the [[Blanking Busy Beaver]] (BLBi(n)) were introduced. gBBi(n) was computed up to n = 13 and BLBi(n) was computed up to n = 7.&lt;br /&gt;
[[Category:This Month in Beaver Research]]&lt;br /&gt;
[[File:Conference poster for DNA31 by Tristan Stérin.png|thumb|396x396px|Conference poster for the [https://dna31.sciencesconf.org/ 31st International Conference on DNA Computing and Molecular Programming], [https://discord.com/channels/960643023006490684/960643023530762341/1409904231468761159 made by Tristan Stérin (cosmo)]]]&lt;br /&gt;
&lt;br /&gt;
== Misc ==&lt;br /&gt;
&lt;br /&gt;
* Iijil shared an algorithm for converting an arbitrary n-state m-symbol TM into a 2-state TM with 3(n+1)m symbols. https://gist.github.com/Iijil1/0d611dbf0a9d52984f72cb14e66a4b28&lt;br /&gt;
* Carl K updated his TM web-visualizer to support multi-symbol machines. [https://carlkcarlk.github.io/busy_beaver_blaze/v0.2.6/index.html#palette=edit&amp;amp;colors=000000%2Cff0000%2Cffff00%2Cff00ff%2C00ffff&amp;amp;run=true https://carlkcarlk.github.io/busy_beaver_blaze/v0.2.6/index.html] He also extended his series of videos showing TM simulation accompanied by classical music out to some multi-symbol TMs:&lt;br /&gt;
** [[Bigfoot]]: https://youtu.be/YvOHWbQNMoY&lt;br /&gt;
** [[Surprise in a Box|Brady&#039;s Surprise in a Box]]: https://youtu.be/vIG2CvJShRc&lt;br /&gt;
** [[1RB3LA4RB0RB2LA 1LB2LA3LA1RA1RZ|BB(2,5) champ:]] https://youtu.be/QpYBzYDdLEY&lt;br /&gt;
&lt;br /&gt;
==In the News==&lt;br /&gt;
&lt;br /&gt;
* 22 Aug 2025. Ben Brubaker. Quanta Magazine. [https://www.quantamagazine.org/busy-beaver-hunters-reach-numbers-that-overwhelm-ordinary-math-20250822/ Busy Beaver Hunters Reach Numbers That Overwhelm Ordinary Math].&lt;br /&gt;
&lt;br /&gt;
==Interesting TMs==&lt;br /&gt;
&lt;br /&gt;
[[Category:This Month in Beaver Research]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=TMBR:_August_2025&amp;diff=3417</id>
		<title>TMBR: August 2025</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=TMBR:_August_2025&amp;diff=3417"/>
		<updated>2025-08-27T16:19:53Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Stub}} &lt;br /&gt;
[[:Category:This Month in Beaver Research|This Month in Beaver Research]] for August 2025, featuring plenty of holdouts reduction in numerous domains, alongside an upgraded TM-visualizer and a couple new BB adjacent games.&lt;br /&gt;
&lt;br /&gt;
== Cryptids ==&lt;br /&gt;
&lt;br /&gt;
* A fast algorithm for [[Consistent Collatz]] simulation was re-discovered and popularized. Using it:&lt;br /&gt;
** apgroucher simulated [[Antihydra]] to &amp;lt;math&amp;gt;2^{38}&amp;lt;/math&amp;gt; iterations. This is actually a result from one year ago, but was rediscovered and added to the wiki. https://discord.com/channels/960643023006490684/1026577255754903572/1271528180246773883&lt;br /&gt;
** Shawn Ligocki simulated {{TM|1RB1RA_0RC1RC_1LD0LF_0LE1LE_1RA0LB_---0LC}} out to one additional Collatz reset, demonstrating that (if they halt, which they probviously should) they will have sigma scores &amp;lt;math&amp;gt;&amp;gt; 10^{10^{10^7}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
== Holdouts ==&lt;br /&gt;
&lt;br /&gt;
* [[BB(6)]] holdouts:&lt;br /&gt;
** XnoobSpeakable found 9 new halting TMs in the high exponential runtime range (~10^100000) by running Enumerate.py out to extremely high parameters. https://discord.com/channels/960643023006490684/1239205785913790465/1401470301467836556&lt;br /&gt;
** Andrew Ducharme found a couple surprisingly short running halting TMs in the [[BB(6)]] holdouts list with runtime ~10^78. https://discord.com/channels/960643023006490684/1239205785913790465/1407754434523693179&lt;br /&gt;
*After the [[BB(7)#Phase 1|enumeration of BB(7)]] was completed, Andrew Ducharme ran [[BB(7)#Phase 2|several deciders]] on the holdouts list, filtering the original 86,129,304 holdouts down to 60,986,231 in 7 days.  https://drive.google.com/drive/u/0/folders/17U0BRpJHTMLtB0poBlOSZhGGp4FkCHIO&lt;br /&gt;
*[[BB(3,3)]] Rocq certified holdout list down to 12. https://discord.com/channels/960643023006490684/1259770474897080380/1409402854292066335&lt;br /&gt;
&lt;br /&gt;
==BB Adjacent==&lt;br /&gt;
*John Tromp introduced the &amp;lt;math&amp;gt;BB \lambda _1(n)&amp;lt;/math&amp;gt; function for [[Busy Beaver for lambda calculus#Oracle Busy Beaver|Busy Beaver for lambda calculus with an oracle]] and computed it up to &amp;lt;math&amp;gt;BB \lambda _1(22)&amp;lt;/math&amp;gt;.&lt;br /&gt;
* Instruction-Limited Greedy Busy Beaver gBBi(n) and an [[Instruction-Limited Busy Beaver#Instruction-Limited Busy Beaver Variants|Instruction-Limited variant]] of the Blanking Busy Beaver (BLBi(n)) were introduced. gBBi(n) was computed up to n = 13 and BLBi(n) was computed up to n = 7.&lt;br /&gt;
[[Category:This Month in Beaver Research]]&lt;br /&gt;
[[File:DNA31 BB5 Poster.png|alt=Conference poster for DNA31 by Tristan Stérin (cosmo)|thumb|Conference poster for the [https://dna31.sciencesconf.org/ 31st International Conference on DNA Computing and Molecular Programming], [https://discord.com/channels/960643023006490684/960643023530762341/1409904231468761159 made by Tristan Stérin (cosmo)]]]&lt;br /&gt;
&lt;br /&gt;
== Misc ==&lt;br /&gt;
&lt;br /&gt;
* Iijil shared an algorithm for converting an arbitrary n-state m-symbol TM into a 2-state TM with 3(n+1)m symbols. https://gist.github.com/Iijil1/0d611dbf0a9d52984f72cb14e66a4b28&lt;br /&gt;
* Carl K updated his TM web-visualizer to support multi-symbol machines. [https://carlkcarlk.github.io/busy_beaver_blaze/v0.2.6/index.html#palette=edit&amp;amp;colors=000000%2Cff0000%2Cffff00%2Cff00ff%2C00ffff&amp;amp;run=true https://carlkcarlk.github.io/busy_beaver_blaze/v0.2.6/index.html] He also extended his series of videos showing TM simulation accompanied by classical music out to some multi-symbol TMs:&lt;br /&gt;
** [[Bigfoot]]: https://youtu.be/YvOHWbQNMoY&lt;br /&gt;
** [[Surprise in a Box|Brady&#039;s Surprise in a Box]]: https://youtu.be/vIG2CvJShRc&lt;br /&gt;
** [[1RB3LA4RB0RB2LA 1LB2LA3LA1RA1RZ|BB(2,5) champ:]] https://youtu.be/QpYBzYDdLEY&lt;br /&gt;
&lt;br /&gt;
==In the News==&lt;br /&gt;
&lt;br /&gt;
* 22 Aug 2025. Ben Brubaker. Quanta Magazine. [https://www.quantamagazine.org/busy-beaver-hunters-reach-numbers-that-overwhelm-ordinary-math-20250822/ Busy Beaver Hunters Reach Numbers That Overwhelm Ordinary Math].&lt;br /&gt;
&lt;br /&gt;
==Interesting TMs==&lt;br /&gt;
&lt;br /&gt;
[[Category:This Month in Beaver Research]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=File:DNA31_BB5_Poster.png&amp;diff=3416</id>
		<title>File:DNA31 BB5 Poster.png</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=File:DNA31_BB5_Poster.png&amp;diff=3416"/>
		<updated>2025-08-27T16:19:35Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;dd&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Champions&amp;diff=2862</id>
		<title>Champions</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Champions&amp;diff=2862"/>
		<updated>2025-08-09T16:18:57Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* 4-Symbol TMs */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Busy Beaver &#039;&#039;&#039;Champions&#039;&#039;&#039; are the current record holding [[Turing machine|Turing machines]] which maximize a [[Busy Beaver function]]. In this article we focus specifically on the longest running TMs. Some have been proven to be the longest running of all (and so are the ultimate champion) while others are only current champions and may be usurped in the future. For smaller domains, Pascal Michel&#039;s website is the canonical source for [https://bbchallenge.org/~pascal.michel/bbc Busy Beaver champions] and the [https://bbchallenge.org/~pascal.michel/ha History of Previous Champions].&lt;br /&gt;
&lt;br /&gt;
== 2-Symbol TMs ==&lt;br /&gt;
Rows are blank if no champion has been found which surpasses a smaller size problem. Take also note that the &amp;lt;math&amp;gt; f_{x}(n) &amp;lt;/math&amp;gt; used in the lower bounds represent the [[Fast-Growing Hierarchy]]. Note that most champions above 6 states are self-reported and have not been independently verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!&lt;br /&gt;
!Runtime&lt;br /&gt;
!Champions&lt;br /&gt;
!Discovered By&lt;br /&gt;
!Verification&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; 6 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1LB_1LA1RZ|halt}} {{TM|1RB0LB_1LA1RZ|halt}} {{TM|1RB1RZ_1LB1LA|halt}} {{TM|1RB1RZ_0LB1LA|halt}} {{TM|0RB1RZ_1LA1RB|halt}}&lt;br /&gt;
|Tibor Radó&lt;br /&gt;
|Direct Simulation&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; 21 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1RZ_1LB0RC_1LC1LA|halt}}&lt;br /&gt;
|Proven by Shen Lin&lt;br /&gt;
|Direct Simulation&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; 107 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1LB_1LA0LC_1RZ1LD_1RD0RA|halt}}&lt;br /&gt;
|Allen Brady&lt;br /&gt;
|Direct Simulation&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; 47\,176\,870 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA|halt}}&lt;br /&gt;
|Heiner Marxen &amp;amp; Jürgen Buntrock in 1989&lt;br /&gt;
|Direct Simulation&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; 2\uparrow\uparrow\uparrow 5 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}}&lt;br /&gt;
|mxdys in 2025&lt;br /&gt;
|See mxdys&#039;s analysis on the TM page&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(7)]]&lt;br /&gt;
|&amp;lt;math&amp;gt;&amp;gt; 2 \uparrow^{11} 2 \uparrow^{11} 3&amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB0RA_1LC1LF_1RD0LB_1RA1LE_1RZ0LC_1RG1LD_0RG0RF|halt}}&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1369339127652159509/1370678203395604562 Pavel Kropitz in 2025]&lt;br /&gt;
|Analyzed by Shawn Ligocki (see TM page)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(8)]]&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(9)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_\omega(f_9(2)) &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1RA_0LC0LF_0RD1LC_1RA1RG_1RZ0RA_1LB1LF_1LH1RE_0LI1LH_1LB0LH|halt}}&lt;br /&gt;
|Jacobzheng in 2024&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(10)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_\omega^2(25) &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB1RA_0LC0LF_0RD1LC_1RA1RG_1RZ0RA_1LB1LF_1LH1RE_0LI1LH_0LF0LJ_1LH0LJ|halt}}&lt;br /&gt;
|Racheline in 2024&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(11)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_\omega^2(2 \uparrow\uparrow 12) &amp;gt; f_\omega^2(f_3(9)) &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1LH1LA_1LI1RG_0RD1LC_0RF1RE_1LJ0RF_1RB1RF_0LC1LH_0LC0LA_1LK1LJ_1RZ0LI_0LD1LE|halt}}&lt;br /&gt;
|Racheline in 2024&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(12)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_\omega^4(2 \uparrow\uparrow\uparrow 4-3) &amp;gt; f_\omega^4(f_4(2)) &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|0LJ0RF_1LH1RC_0LD0LG_0RE1LD_1RF1RA_1RB1RF_1LC1LG_1LL1LI_1LK0LH_1RH1LJ_1RZ1LA_1RF1LL|halt}}&lt;br /&gt;
|Racheline in 2024&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(13)&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(14)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega + 1}(65\,536) &amp;gt; g_{64} &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1LH1LA_1LI1RG_0RD1LC_0RF1RE_1LJ0RF_1RB1RF_0LC1LH_0LC0LA_1LK1LJ_1RL0LI_0LL1LE_1LM1RZ_0LN1LF_0LJ---|halt}}&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1274366178529120287 Racheline in 2024]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(15)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega + 1}(f_\omega(10^{57})) &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|0RH1LD_1RI0RC_1RB1LD_0LD1LE_1LF1RA_1RG0LE_1RB1RG_1RD1RA_0LN0RJ_1RZ0LK_0LK1LL_1RG1LM_0LL0LL_1LO1LN_0LG1LN|halt}}&lt;br /&gt;
|Jacobzheng in 2025&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(16)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega + 1}^2(10^{10^{57}}) &amp;lt;/math&amp;gt;&lt;br /&gt;
|[[User:Jacobzheng/BB(16)]]&lt;br /&gt;
|Jacobzheng in 2025&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(17)&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(18)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega + 2}(f_{\omega + 1}^3(f_{\omega}^2(60))) &amp;lt;/math&amp;gt;&lt;br /&gt;
|[[User:Jacobzheng/BB(18)]]&lt;br /&gt;
|Jacobzheng in 2025&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(19)&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(20)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega + 2}^2(21) &amp;lt;/math&amp;gt;&lt;br /&gt;
|&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1026577255754903572/1274414683331366924 Racheline in 2024]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(21)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega^2}^2(4 \uparrow\uparrow 341) &amp;lt;/math&amp;gt;&lt;br /&gt;
|&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1026577255754903572/1274471360206344213 Racheline in 2024]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(40)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega^\omega}(75\,500) &amp;lt;/math&amp;gt;&lt;br /&gt;
|[[User:Jacobzheng/BB(40)]]&lt;br /&gt;
|Jacobzheng in 2024&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(41)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\omega^\omega}^4(32) &amp;lt;/math&amp;gt;&lt;br /&gt;
|[[User:Jacobzheng/BB(41)]]&lt;br /&gt;
|Jacobzheng in 2024&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(51)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_{\varepsilon_0 + 1}(8) &amp;lt;/math&amp;gt;&lt;br /&gt;
|&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1026577255754903572/1276881449685094495 Racheline in 2024]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== 3-Symbol TMs ==&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!&lt;br /&gt;
!Runtime&lt;br /&gt;
!Champions&lt;br /&gt;
!Discovered By&lt;br /&gt;
!Verification&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,3)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; 38 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB2LB1RZ_2LA2RB1LB|halt}}&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; 10^{17} &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|0RB2LA1RA_1LA2RB1RC_1RZ1LB1LC|halt}}&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; 2 \uparrow\uparrow\uparrow 2^{2^{32}}&amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|0RB1RZ0RB_1RC1LB2LB_1LB2RD1LC_1RA2RC0LD|halt}}&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== 4-Symbol TMs ==&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!&lt;br /&gt;
!Runtime&lt;br /&gt;
!Champions&lt;br /&gt;
!Discovered By&lt;br /&gt;
!Verification&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,4)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; 3\,932\,964 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB2LA1RA1RA_1LB1LA3RB1RZ|halt}}&lt;br /&gt;
|Shawn &amp;amp; Terry Ligocki in 2005&lt;br /&gt;
|Pascal Michel, Heiner Marxen, Allen Brady&lt;br /&gt;
|-&lt;br /&gt;
|BB(3,4)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; 2 \uparrow^{15} 5 &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB3LB1RZ2RA_2LC3RB1LC2RA_3RB1LB3LC2RC|halt}}&lt;br /&gt;
|See [https://www.sligocki.com/2024/05/22/bb-3-4-a14.html blog post]&lt;br /&gt;
|See [https://www.sligocki.com/2024/05/22/bb-3-4-a14.html blog post]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== 5-Symbol TMs ==&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
!&lt;br /&gt;
!Runtime&lt;br /&gt;
!Champions&lt;br /&gt;
!Discovered By&lt;br /&gt;
!Verification&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; 10^{10^{10^{3\,314\,360}}} &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB3LA4RB0RB2LA_1LB2LA3LA1RA1RZ|halt}}&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(3,5)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; f_\omega(2 \uparrow^{15} 5) &amp;gt; f_\omega^2(15) &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB3LB4LC2RA4LB_2LC3RB1LC2RA1RZ_3RB1LB3LC2RC4LC|halt}}&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== 6-Symbol TMs ==&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
!&lt;br /&gt;
!Runtime&lt;br /&gt;
!Champions&lt;br /&gt;
!Discovered By&lt;br /&gt;
!Verification&lt;br /&gt;
|-&lt;br /&gt;
|BB(2,6)&lt;br /&gt;
|&amp;lt;math&amp;gt; &amp;gt; 10 \uparrow\uparrow 10 \uparrow\uparrow 10^{10^{115}} &amp;lt;/math&amp;gt;&lt;br /&gt;
|{{TM|1RB3RB5RA1LB5LA2LB_2LA2RA4RB1RZ3LB2LA|halt}}&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE&amp;diff=2861</id>
		<title>1RB1RA 1RC1RZ 1LD0RF 1RA0LE 0LD1RC 1RA0RE</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE&amp;diff=2861"/>
		<updated>2025-08-09T15:53:02Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE}}&lt;br /&gt;
{{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}}&lt;br /&gt;
&lt;br /&gt;
Current [[BB(6)]] Champion. Discovered by mxdys on 25 June 2025.&lt;br /&gt;
&lt;br /&gt;
It&#039;s in a family of 4 machines with the halting time and sigma score between 2↑↑2↑↑2↑↑10 and 2↑↑2↑↑2↑↑11:&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
1RB1RA_1RC---_1LD0RF_1RA0LE_0LD1RC_1RA0RE (hereafter referred to as TM1)&lt;br /&gt;
1RB---_1LC0RF_1RE0LD_0LC1RB_1RA1RE_1RE0RD (TM2)&lt;br /&gt;
1RB0LE_1RC1RB_1RD---_1LA0RF_0LA1RD_1RB0RE (TM3)&lt;br /&gt;
1RB0RF_1RC1RB_1RD---_1LE0RA_1RB0LF_0LE1RD (TM4)&lt;br /&gt;
&amp;lt;/pre&amp;gt;Coq proof: https://github.com/ccz181078/busycoq/blob/3f302b87f5fb933c46e97672ffbb6907f373fb6e/verify/SOBCv5.v#L10210-L11283&lt;br /&gt;
&lt;br /&gt;
== Analysis by mxdys ==&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
Inc2:&lt;br /&gt;
S1(len0,a0+1,2,a    ,b    ) --&amp;gt;&lt;br /&gt;
S1(len0,a0  ,1,a+b+2,2^b-1)&lt;br /&gt;
&lt;br /&gt;
Inc1:&lt;br /&gt;
S1(len0,a0+1,1,a    ,b    ) --&amp;gt;&lt;br /&gt;
S1(len0,a0  ,0,a+b+2,2^b-1)&lt;br /&gt;
&lt;br /&gt;
Inc0:&lt;br /&gt;
S1(len0,a0+1,0,a    ,b    ) --&amp;gt;&lt;br /&gt;
S1(len0,a0  ,2,a+b+1,2^b-1)&lt;br /&gt;
&lt;br /&gt;
Rst0:&lt;br /&gt;
S1(a0,0,0,a,b) --&amp;gt;&lt;br /&gt;
halt&lt;br /&gt;
&lt;br /&gt;
Rst1:&lt;br /&gt;
S1(a0,0,1,a,b) --&amp;gt;&lt;br /&gt;
S1(a0+a+2,(2^(a0+2)-1)*2^a-1,2,b,2^b-1)&lt;br /&gt;
&lt;br /&gt;
start: S1(3,7,2,6,63)&lt;br /&gt;
&lt;br /&gt;
the rules are used in the following order:&lt;br /&gt;
Inc2,Inc1,Inc0, Inc2,Inc1,Inc0, Inc2, Rst1,&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2, Rst1,&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2,Inc1, Rst0.&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
S1(len0,a0,m,a,b) = 0^inf LH LC(len0,a0) d0 10 1^m LC(a,0) &amp;lt;X 0 11100 111^(1+b) 0^inf&lt;br /&gt;
d0 = 100&lt;br /&gt;
d1 = 111&lt;br /&gt;
LC(0,0) = &amp;quot;&amp;quot;&lt;br /&gt;
LC(n+1,2x) = LC(n,x) d1&lt;br /&gt;
LC(n+1,2x+1) = LC(n,x) d0&lt;br /&gt;
for TM2, X=D, LH=111011&lt;br /&gt;
for TM3, X=E, LH=11&lt;br /&gt;
&lt;br /&gt;
TM1 is equivalent to TM2 after several steps&lt;br /&gt;
TM4 is equivalent to TM3 after several steps&lt;br /&gt;
TM1 has the highest halting time among this family&lt;br /&gt;
TM1,TM2 have the highest sigma score among this family&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
estimation of time/score:&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2&lt;br /&gt;
n mod 3 = 1:&lt;br /&gt;
S1(len0,n,2,b,2^b-1) --&amp;gt;&lt;br /&gt;
S1(len0,0,1,st2(n,b)+floor(n/3)*5+2,t2(n+1,b))&lt;br /&gt;
&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2,Inc1, Rst0&lt;br /&gt;
n mod 3 = 2:&lt;br /&gt;
S1(len0,n,2,b,2^b-1) --&amp;gt;&lt;br /&gt;
S1(len0,0,0,st2(n,b)+floor(n/3)*5+4,t2(n+1,b)) --&amp;gt;&lt;br /&gt;
halt&lt;br /&gt;
&lt;br /&gt;
Rst1:&lt;br /&gt;
S1(len0,0,1,a,b) --&amp;gt;&lt;br /&gt;
S1(len0+a+2,2^(len0+a+2)-2^a-1,2,b,2^b-1)&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
t2(0,b) = b, t2(a+1,b) = 2^t2(a,b)-1&lt;br /&gt;
st2(a,b) = t2(0,b) + t2(1,b) + ... + t2(a,b)&lt;br /&gt;
&lt;br /&gt;
S1(3,7,2,6,63) --&amp;gt;&lt;br /&gt;
S1(3,0,1,st2(7,6)+12,t2(8,6)) --&amp;gt;&lt;br /&gt;
S1(≈t2(7,6),≈t2(8,6),2,_,_) --&amp;gt;&lt;br /&gt;
S1(≈t2(7,6),0,1,≈2^^t2(8,6),_) --&amp;gt;&lt;br /&gt;
S1(≈2^^t2(8,6),≈2^^t2(8,6),2,_,_) --&amp;gt;&lt;br /&gt;
S1(≈2^^t2(8,6),0,1,≈2^^2^^t2(8,6),≈2^^2^^t2(8,6)) --&amp;gt;&lt;br /&gt;
halt with time/score ≈2^^2^^((2^)^8 6)&lt;br /&gt;
2^^^5 &amp;lt; 2^^2^^2^^10 &amp;lt; 2^^2^^((2^)^8 6) &amp;lt; 2^^2^^2^^11 &amp;lt; 2^^^6&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE&amp;diff=2860</id>
		<title>1RB1RA 1RC1RZ 1LD0RF 1RA0LE 0LD1RC 1RA0RE</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE&amp;diff=2860"/>
		<updated>2025-08-09T15:52:46Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE}}&lt;br /&gt;
{{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE|halt}}&lt;br /&gt;
&lt;br /&gt;
Current [[BB(6)]] Champion. Discovered by mxdys on 25 June 2025.&lt;br /&gt;
&lt;br /&gt;
It&#039;s in a family of 4 machines with the halting time and sigma score between 2↑↑2↑↑2↑↑10 and 2↑↑2↑↑2↑↑11:&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
1RB1RA_1RC---_1LD0RF_1RA0LE_0LD1RC_1RA0RE (hereafter referred to as TM1)&lt;br /&gt;
1RB---_1LC0RF_1RE0LD_0LC1RB_1RA1RE_1RE0RD (TM2)&lt;br /&gt;
1RB0LE_1RC1RB_1RD---_1LA0RF_0LA1RD_1RB0RE (TM3)&lt;br /&gt;
1RB0RF_1RC1RB_1RD---_1LE0RA_1RB0LF_0LE1RD (TM4)&lt;br /&gt;
&amp;lt;/pre&amp;gt;Coq proof: https://github.com/ccz181078/busycoq/blob/3f302b87f5fb933c46e97672ffbb6907f373fb6e/verify/SOBCv5.v\#L10210-L11283&lt;br /&gt;
&lt;br /&gt;
== Analysis by mxdys ==&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
Inc2:&lt;br /&gt;
S1(len0,a0+1,2,a    ,b    ) --&amp;gt;&lt;br /&gt;
S1(len0,a0  ,1,a+b+2,2^b-1)&lt;br /&gt;
&lt;br /&gt;
Inc1:&lt;br /&gt;
S1(len0,a0+1,1,a    ,b    ) --&amp;gt;&lt;br /&gt;
S1(len0,a0  ,0,a+b+2,2^b-1)&lt;br /&gt;
&lt;br /&gt;
Inc0:&lt;br /&gt;
S1(len0,a0+1,0,a    ,b    ) --&amp;gt;&lt;br /&gt;
S1(len0,a0  ,2,a+b+1,2^b-1)&lt;br /&gt;
&lt;br /&gt;
Rst0:&lt;br /&gt;
S1(a0,0,0,a,b) --&amp;gt;&lt;br /&gt;
halt&lt;br /&gt;
&lt;br /&gt;
Rst1:&lt;br /&gt;
S1(a0,0,1,a,b) --&amp;gt;&lt;br /&gt;
S1(a0+a+2,(2^(a0+2)-1)*2^a-1,2,b,2^b-1)&lt;br /&gt;
&lt;br /&gt;
start: S1(3,7,2,6,63)&lt;br /&gt;
&lt;br /&gt;
the rules are used in the following order:&lt;br /&gt;
Inc2,Inc1,Inc0, Inc2,Inc1,Inc0, Inc2, Rst1,&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2, Rst1,&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2,Inc1, Rst0.&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
S1(len0,a0,m,a,b) = 0^inf LH LC(len0,a0) d0 10 1^m LC(a,0) &amp;lt;X 0 11100 111^(1+b) 0^inf&lt;br /&gt;
d0 = 100&lt;br /&gt;
d1 = 111&lt;br /&gt;
LC(0,0) = &amp;quot;&amp;quot;&lt;br /&gt;
LC(n+1,2x) = LC(n,x) d1&lt;br /&gt;
LC(n+1,2x+1) = LC(n,x) d0&lt;br /&gt;
for TM2, X=D, LH=111011&lt;br /&gt;
for TM3, X=E, LH=11&lt;br /&gt;
&lt;br /&gt;
TM1 is equivalent to TM2 after several steps&lt;br /&gt;
TM4 is equivalent to TM3 after several steps&lt;br /&gt;
TM1 has the highest halting time among this family&lt;br /&gt;
TM1,TM2 have the highest sigma score among this family&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
estimation of time/score:&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2&lt;br /&gt;
n mod 3 = 1:&lt;br /&gt;
S1(len0,n,2,b,2^b-1) --&amp;gt;&lt;br /&gt;
S1(len0,0,1,st2(n,b)+floor(n/3)*5+2,t2(n+1,b))&lt;br /&gt;
&lt;br /&gt;
Inc2,Inc1,Inc0, ..., Inc2,Inc1,Inc0, Inc2,Inc1, Rst0&lt;br /&gt;
n mod 3 = 2:&lt;br /&gt;
S1(len0,n,2,b,2^b-1) --&amp;gt;&lt;br /&gt;
S1(len0,0,0,st2(n,b)+floor(n/3)*5+4,t2(n+1,b)) --&amp;gt;&lt;br /&gt;
halt&lt;br /&gt;
&lt;br /&gt;
Rst1:&lt;br /&gt;
S1(len0,0,1,a,b) --&amp;gt;&lt;br /&gt;
S1(len0+a+2,2^(len0+a+2)-2^a-1,2,b,2^b-1)&lt;br /&gt;
&lt;br /&gt;
where&lt;br /&gt;
t2(0,b) = b, t2(a+1,b) = 2^t2(a,b)-1&lt;br /&gt;
st2(a,b) = t2(0,b) + t2(1,b) + ... + t2(a,b)&lt;br /&gt;
&lt;br /&gt;
S1(3,7,2,6,63) --&amp;gt;&lt;br /&gt;
S1(3,0,1,st2(7,6)+12,t2(8,6)) --&amp;gt;&lt;br /&gt;
S1(≈t2(7,6),≈t2(8,6),2,_,_) --&amp;gt;&lt;br /&gt;
S1(≈t2(7,6),0,1,≈2^^t2(8,6),_) --&amp;gt;&lt;br /&gt;
S1(≈2^^t2(8,6),≈2^^t2(8,6),2,_,_) --&amp;gt;&lt;br /&gt;
S1(≈2^^t2(8,6),0,1,≈2^^2^^t2(8,6),≈2^^2^^t2(8,6)) --&amp;gt;&lt;br /&gt;
halt with time/score ≈2^^2^^((2^)^8 6)&lt;br /&gt;
2^^^5 &amp;lt; 2^^2^^2^^10 &amp;lt; 2^^2^^((2^)^8 6) &amp;lt; 2^^2^^2^^11 &amp;lt; 2^^^6&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2853</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2853"/>
		<updated>2025-08-09T06:43:20Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,728&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[:File:BB6 holdouts 2728.txt|BB6 holdouts 2728.txt]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1259770421046411285/1355593937531961365 March 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|[[:File:BB2x5 Coq holdouts 83.txt]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2852</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2852"/>
		<updated>2025-08-09T06:42:10Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,728&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[:File:BB6 holdouts 2728.txt|BB6 holdouts 2728.txt]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1259770421046411285/1355593937531961365 March 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|[[BB(2,5)|See BB(2,5)]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2851</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2851"/>
		<updated>2025-08-09T06:40:09Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* Downloadable Holdout Lists */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,728&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[:File:BB6 holdouts 2728.txt|BB6 holdouts 2728.txt]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|March 29 2025&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|[[BB(2,5)|See BB(2,5)]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2850</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2850"/>
		<updated>2025-08-09T06:39:55Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,728&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[:File:BB6 holdouts 2728.txt|BB6 holdouts 2728.txt]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1259770421046411285/1355593937531961365 March 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|[[BB(2,5)|See BB(2,5)]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2849</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2849"/>
		<updated>2025-08-09T06:39:29Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,728&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[:File:BB6 holdouts 2728.txt|BB6 holdouts 2728.txt]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1259770421046411285/1355593937531961365 March 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|[[See BB(2,5)]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2848</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2848"/>
		<updated>2025-08-09T06:38:16Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,728&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[File:BB6 holdouts 2728.txt|thumb]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1259770421046411285/1355593937531961365 March 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|See [[BB(2,5)]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2846</id>
		<title>Holdouts lists</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Holdouts_lists&amp;diff=2846"/>
		<updated>2025-08-09T06:34:59Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* Downloadable Holdout Lists */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;A &#039;&#039;&#039;holdout&#039;&#039;&#039; (or undecided machine) is a [[Turing machine]] for which it is not known whether the machine halts or not from all-0 input tape. Holdouts are the machines which [[Decider|deciders]] are unable to decide.&lt;br /&gt;
&lt;br /&gt;
Holdout lists are often shared by contributors. There is a [[#Downloadable Holdout Lists|Downloadable Holdout Lists]] table where people have added lists with no restriction or independent verification. For some of the entries there is a reference to a spreadsheet that documents what was run to achieve the result. For others, there is additional documentation on the specific BB pages.&lt;br /&gt;
&lt;br /&gt;
The table with the &amp;quot;Number of holdouts&amp;quot; is based on the holdout lists listed in the table below it. Thus, some of these numbers have not been independently verified. All the zero entries (no remaining holdouts) have been verified.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+Number of holdouts&lt;br /&gt;
!&lt;br /&gt;
!2-state&lt;br /&gt;
!3-state&lt;br /&gt;
!4-state&lt;br /&gt;
!5-state&lt;br /&gt;
!6-state&lt;br /&gt;
|-&lt;br /&gt;
!2-symbol&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|0&lt;br /&gt;
|2,891&lt;br /&gt;
|-&lt;br /&gt;
!3-symbol&lt;br /&gt;
|0&lt;br /&gt;
|6&lt;br /&gt;
|460,916,384&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!4-symbol&lt;br /&gt;
|0&lt;br /&gt;
|434,787,751&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!5-symbol&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
!6-symbol&lt;br /&gt;
|22,302,296&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
== Downloadable Holdout Lists ==&lt;br /&gt;
{| class=&amp;quot;wikitable sortable&amp;quot;&lt;br /&gt;
|+&lt;br /&gt;
!BB space&lt;br /&gt;
!Date&lt;br /&gt;
!Shared by&lt;br /&gt;
!Number of holdouts&lt;br /&gt;
!File&lt;br /&gt;
!Notes&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896 July 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|2,728&lt;br /&gt;
|[[File:BB6 holdouts 2728.txt|thumb]]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1259770421046411285/1355593937531961365, March 29 2025]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|83&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(4,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1320172124509311004  December 21, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|460,916,384&lt;br /&gt;
|[https://drive.google.com/file/d/1hKy0TuPuI62rN95j6ZLjXgE-Pue8tRsK/view?usp=drive_link 4x3_holdouts_460916384.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1HBPZ17llVE_8wCy5FvRUFQ5MJsaYXAW-?usp=drive_link Google Drive directory for 4x3 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1310648046576730124 November 25, 2024 (@icy)]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|4,319&lt;br /&gt;
|[https://docs.google.com/spreadsheets/d/1grhW_0neb2I8TfceN5-v70_3W42Z2U159r9L6FhPGf8/edit?usp=sharing Spreadsheet of holdouts]&lt;br /&gt;
|Keeping track of BB(6) progress - informal.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|November 10, 2024&lt;br /&gt;
|&lt;br /&gt;
|6&lt;br /&gt;
|[[:File:3x3 holdout 6.txt|3x3 holdout 6.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1304303803213942846 November 8, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|4,408&lt;br /&gt;
|[[:File:BB6 holdouts 4408.txt|BB6_holdouts_4408.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1303219184221683733 November 4, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|22,302,296&lt;br /&gt;
|[https://drive.google.com/file/d/1xRLIjdiqImFP2SL38gvhxVlAaX0L1cYO/view?usp=drive_link 2x6_holdouts_22302296.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1p9b5g-Id3WEMUYIwEnaKWRBGIW66ADjM?usp=drive_link Google Drive directory for 2x6 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,4)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/960643023530762341/1302767449476694188  November 3, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
|434,787,751&lt;br /&gt;
|[https://drive.google.com/file/d/1PLzN3wLw-MRgk1OFmYh4RTwNc30nflR7/view?usp=drive_link 3x4_holdouts_434787751.txt.gz]&lt;br /&gt;
|[https://drive.google.com/drive/folders/1bZxl7jg5q9IVvHQNZwItx1kPusAWznZk?usp=drive_link Google Drive directory for 3x4 TMs]&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discordapp.com/channels/960643023006490684/1239205785913790465/1280185195877634098 September 2, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5394&lt;br /&gt;
|[[:File:BB6 holdouts 5394.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1269612923127599164 August 4, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|5877&lt;br /&gt;
|[[:File:BB6 holdouts 5877.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1259131753176498216 July 6, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|7296&lt;br /&gt;
|[[:File:BB6 holdouts 7296.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1252989316175499284 June 19, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|217&lt;br /&gt;
|[[:File:2x5_holdouts_217.txt]]&lt;br /&gt;
|273 holdouts minus machines solved by CTL&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|June 15th 2024&lt;br /&gt;
|@dyuan01&lt;br /&gt;
| 273&lt;br /&gt;
|[[:File:2x5_holdouts_273.txt]]&lt;br /&gt;
|@Justin Blanchard&#039;s 499 holdouts minus machines solved by @mxdys&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1250895665719148595 June 13, 2024]&lt;br /&gt;
|@tjligocki&lt;br /&gt;
| 12,091&lt;br /&gt;
|[[:File:BB6 holdouts 12091.txt]]&lt;br /&gt;
|Work done with @Shawn Ligocki&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1249142547217907772 June 9, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|22&lt;br /&gt;
|[[:File:3x3.todo.txt]], [[:File:Mugshots small.pdf]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|BB(6)&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1239205785913790465/1248708916381220954 June 7, 2024]&lt;br /&gt;
|@mxdys&lt;br /&gt;
|12,325&lt;br /&gt;
|[[:File:BB6 holdouts 12325.txt]]&lt;br /&gt;
|Some equivalent machines are removed.&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|June 2024&lt;br /&gt;
|BBChallenge&lt;br /&gt;
|0&lt;br /&gt;
|&lt;br /&gt;
|&#039;&#039;&#039;BB(5) is SOLVED!&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(2,5)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1242679236142170203 May 22, 2024]&lt;br /&gt;
|@Justin Blanchard&lt;br /&gt;
|499&lt;br /&gt;
|[[:File:2x5.todo.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|925&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts intersect sligocki iijil 925.txt]]&lt;br /&gt;
|Intersection of @sligocki and @Iijil from below&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116351783040716830 June 8, 2023]&lt;br /&gt;
|@Iijil&lt;br /&gt;
|2,480&lt;br /&gt;
|[[:File:2023 06 08.3x3.holdouts iijil 2380.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(3,3)]]&lt;br /&gt;
|[https://discord.com/channels/960643023006490684/1084047886494470185/1116178334620070000 June 7, 2023]&lt;br /&gt;
|@sligocki&lt;br /&gt;
|2,417&lt;br /&gt;
|[[:File:2023 06 07.3x3.holdouts 2417.txt]]&lt;br /&gt;
|&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 27, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|181,851&lt;br /&gt;
|[https://drive.google.com/file/d/1YNwFCN6XJeDNKxxK5KbGHOAFOdIBvDb6/view?usp=drive_link 6x2.holdouts_181851.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(6)]]&lt;br /&gt;
|May 10, 2023&lt;br /&gt;
|@sligocki&lt;br /&gt;
|1,458,704&lt;br /&gt;
|[https://drive.google.com/file/d/14bDnBt0OwuHATFBiubc_5jub0220EyXf/view?usp=drive_link 6x2.holdouts_1458704.txt.gz]&lt;br /&gt;
|This was posted to the BBChallenge Forum (before Discord)&lt;br /&gt;
|-&lt;br /&gt;
|[[BB(5)]]&lt;br /&gt;
|[https://skelet.ludost.net/bb/nreg.html circa May 13th, 2003]&lt;br /&gt;
|Georgi Georgiev (Skelet)&lt;br /&gt;
|43&lt;br /&gt;
|[https://bbchallenge.org/skelet List of 43 holdouts]&lt;br /&gt;
|&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=File:BB6_holdouts_2728.txt&amp;diff=2845</id>
		<title>File:BB6 holdouts 2728.txt</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=File:BB6_holdouts_2728.txt&amp;diff=2845"/>
		<updated>2025-08-09T06:33:47Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;2,728 BB(6) holdouts&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD&amp;diff=2828</id>
		<title>1RB0LE 1LC1RA ---1LD 0RB1LF 1RD1LA 0LA0RD</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD&amp;diff=2828"/>
		<updated>2025-08-08T13:49:11Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD}}&lt;br /&gt;
{{TM|1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD|non}} is a non-halting [[BB(6)]] Turing machine.&lt;br /&gt;
&lt;br /&gt;
Analysis by @racheline on 29 July 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1267551868997992652 Discord link]):&lt;br /&gt;
&lt;br /&gt;
&amp;lt;pre&amp;gt;&lt;br /&gt;
1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0RD&lt;br /&gt;
A(n) = 0^inf &amp;lt;A 0 1^n&lt;br /&gt;
&lt;br /&gt;
rules:&lt;br /&gt;
A(6n) -&amp;gt; A(12n+3)&lt;br /&gt;
A(6n+1) -&amp;gt; A(12n+6)&lt;br /&gt;
A(6n+2) -&amp;gt; halt&lt;br /&gt;
A(6n+3) -&amp;gt; A(9n+9)&lt;br /&gt;
A(6n+4) -&amp;gt; halt&lt;br /&gt;
A(6n+5) -&amp;gt; A(9n+12)&lt;br /&gt;
start from A(3)&lt;br /&gt;
&lt;br /&gt;
as we can see, everything that doesn&#039;t halt goes to A(6m) or A(6m+3) for some m, so halting is unreachable&lt;br /&gt;
&lt;br /&gt;
the next two (1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_1RD0LA and 1RB0LE_1LC1RA_---1LD_0RB1LF_1RD1LA_0LA0LA) are clearly equivalent to it, so also non-halting&lt;br /&gt;
&amp;lt;/pre&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==== Incorrect probviously halting argument ====&lt;br /&gt;
Assuming that the remainders mod 6 are independent and uniformly distributed from 0 to 5 predicts that the machine must halt with probability 1. But, the machine is nonhalting.&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=TMBR:_July_2025&amp;diff=2724</id>
		<title>TMBR: July 2025</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=TMBR:_July_2025&amp;diff=2724"/>
		<updated>2025-08-04T08:25:07Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* Holdout */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Lovecraft beaver.png|alt=A monstrous beaver in the style of HP Lovecraft with pink tentacles coming out of its mouth, 5 red spider eyes, horns on its head, elbows and tail, moss colored fur, sharp purple claws and webbed feet.|thumb|Lovecraftian Beaver fan art made by Lauren]]&lt;br /&gt;
The very first edition of [[:Category:This Month in Beaver Research|This Month in Beaver Research]] for July 2025.&lt;br /&gt;
&lt;br /&gt;
== Champions ==&lt;br /&gt;
* (Late June) mxdys found a pair of new [[BB(6)]] champions pushing it into the pentational values: {{TM|1RB1RA_1RC1RZ_1LD0RF_1RA0LE_0LD1RC_1RA0RE}} (scoring over 2↑↑2↑↑2↑↑10) and {{TM|1RB1LC_1LA1RE_0RD0LA_1RZ1LB_1LD0RF_0RD1RB}} (scoring over 10↑↑11010000).&lt;br /&gt;
* mxdys confirms dyuan&#039;s [[BB(2,5)]] champion {{TM|1RB3LA4RB0RB2LA_1LB2LA3LA1RA1RZ}}. [https://discord.com/channels/960643023006490684/1259770421046411285/1379877629288644722]&lt;br /&gt;
* Nick Drozd discovered a [[Instruction-Limited Busy Beaver|BBi]](8) champion running &amp;lt;math&amp;gt;&amp;gt; 6.889 \times 10^{1565}&amp;lt;/math&amp;gt; steps: {{TM|1RB1LA------_1RC3LB1RB---_2LA2LC---0LC}}. [https://discord.com/channels/960643023006490684/1084047886494470185/1398753236835635252]&lt;br /&gt;
* Shawn Ligocki discovered a [[Reversible Turing Machine|BB&amp;lt;sub&amp;gt;rev&amp;lt;/sub&amp;gt;(6)]] champion running 537,556 steps.&lt;br /&gt;
&lt;br /&gt;
== Holdouts ==&lt;br /&gt;
[[File:BB6 num holdouts over time.png|thumb|Number of [[BB(6)]] holdouts over time]]&lt;br /&gt;
* mxdys announced that [[BB(6)]] is down to 2728 holdouts on 29 July. [https://discord.com/channels/960643023006490684/1239205785913790465/1399783936019664896]&lt;br /&gt;
* A large group effort has enumerated [[BB(7)]] and filtered it down to less than 86 million holdouts. [https://discord.com/channels/960643023006490684/1369339127652159509/1399265628900294686]&lt;br /&gt;
&lt;br /&gt;
== BB Adjacent ==&lt;br /&gt;
* [[Instruction-Limited Busy Beaver]] was introduced and calculated up to BBi(7).&lt;br /&gt;
* [[Reversible Turing Machine]] Busy Beaver values were calculated up to BB_rev(5).&lt;br /&gt;
* [[Terminating Turmite]]s (Relative Movement Turing Machines) was introduced.&lt;br /&gt;
&lt;br /&gt;
== In the News ==&lt;br /&gt;
* 1 July 2025. The Quanta Podcast. [https://discord.com/channels/960643023006490684/1285212639399776256/1389643208811745310 How Amateurs Solved a Major Computer Science Puzzle].&lt;br /&gt;
* 2 July 2025. Manon Bischoff. Spektrum. [https://www.spektrum.de/news/mathematik-die-sechste-fleissige-biber-zahl-ist-gigantisch/2274249 Wie der sechste Fleißige Biber die Mathematik an ihre Grenzen bringt].&lt;br /&gt;
* 7 July 2025. Karmela Padavic-Callaghan. New Scientist. [https://www.newscientist.com/article/2487058-mathematicians-are-chasing-a-number-that-may-reveal-the-edge-of-maths/ Mathematicians are chasing a number that may reveal the edge of maths]. (Paywalled)&lt;br /&gt;
* 11 July 2025. New Scientist podcast [https://www.newscientist.com/podcasts/how-geoengineering-could-save-us-from-climate-disaster-have-we-broken-mathematics-why-exercise-reduces-cancer-risk/ episode 311]. Discusses mxdys&#039;s [[BB(6)]] pentation result &amp;quot;We’re brushing up against the edge of mathematics&amp;quot;.&lt;br /&gt;
* 11 July 2025. Darren Orf. Popular Mechanics. [https://www.popularmechanics.com/science/math/a65357535/busy-beaver-six/ Mathematicians Say There’s a Number So Big, It’s Literally the Edge of Human Knowledge].&lt;br /&gt;
* 18 July 2025 https://francis.naukas.com/2025/07/18/espeluznante-nueva-cota-inferior-para-la-funcion-castor-afanoso-bb6/&lt;br /&gt;
&lt;br /&gt;
== Interesting TMs ==&lt;br /&gt;
* {{TM|1RB1LC_1LB0RA_0LD---_1LE0LA_0LF0RG_1LG0LG_0RB1RG}}: Tetrational [[Fractal]]&lt;br /&gt;
* {{TM|1RB1RE_1LC0RA_0RF0LD_1LE1LC_1RA0RC_---0RB}}: Skeleton Finger&lt;br /&gt;
* {{TM|1RB1RG_1LC1RA_0LD0LC_0RE0RB_1RE1RF_1LD0LC_0LE---}}: [[Bouncer]] that transitions into a [[Counter]]&lt;br /&gt;
* {{TM|1RB1RE_0RC1RG_1LD0LA_0LE0LF_0RA1RC_1LC0LE_---1RE}}: Counter that transitions into a Bouncer&lt;br /&gt;
* {{TM|1RB2LA2LC---_1LA2RB2RD---_3RB1LC1RD0LA_3LA1RD1LC0RB}}: Sneaky almost-Cryptid&lt;br /&gt;
[[Category:This Month in Beaver Research|TMBR:2025-07]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2723</id>
		<title>Main Page</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2723"/>
		<updated>2025-08-04T08:24:21Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;The [[Busy Beaver function]] BB (called &#039;&#039;S&#039;&#039; originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to &#039;&#039;m&#039;&#039;-symbol Turing machines:&amp;lt;ref&amp;gt;Rado, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
| BB(&#039;&#039;n&#039;&#039;, &#039;&#039;m&#039;&#039;) = Maximum number of steps taken by a halting &#039;&#039;n&#039;&#039;-state, &#039;&#039;m&#039;&#039;-symbol Turing machine starting from a blank (all 0) tape&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
The 2-symbol case BB(&#039;&#039;n&#039;&#039;, 2) is abbreviated as BB(&#039;&#039;n&#039;&#039;). The busy beaver function is not computable, but a few of its values are known:&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+ Small busy beaver values&amp;lt;ref&amp;gt;P. Michel, &amp;quot;[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]&amp;quot;.&amp;lt;/ref&amp;gt;&lt;br /&gt;
! !!2-state!!3-state !!4-state!!5-state!!6-state &lt;br /&gt;
!7-state&lt;br /&gt;
|-  &lt;br /&gt;
! 2-symbol &lt;br /&gt;
| [[BB(2)]] = 6 &lt;br /&gt;
| [[BB(3)]] = 21&lt;br /&gt;
| [[BB(4)]] = 107 &lt;br /&gt;
| [[BB(5)]] = 47,176,870 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(6)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow \uparrow \uparrow 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(7)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{11} 2 \uparrow^{11} 3&amp;lt;/math&amp;gt; &lt;br /&gt;
|-&lt;br /&gt;
! 3-symbol&lt;br /&gt;
| [[BB(2,3)]] = 38 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(3,3)]] &amp;gt; &amp;lt;math&amp;gt;10^{17}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(4,3)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow\uparrow\uparrow 2^{2^{32}}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 4-symbol  &lt;br /&gt;
| [[BB(2,4)]] = 3,932,964&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,4)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{15} 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 5-symbol &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(2,5)]] &amp;gt; &amp;lt;math&amp;gt;10\uparrow\uparrow 4&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,5)]] &amp;gt; &amp;lt;math&amp;gt; f_\omega(2 \uparrow^{15} 5)&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 6-symbol &lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(2,6)]] &amp;gt; &amp;lt;math&amp;gt;10 \uparrow\uparrow\uparrow 3&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
In the above table, &amp;lt;span style=&amp;quot;background: orange&amp;quot;&amp;gt;cells are highlighted in orange&amp;lt;/span&amp;gt; when there are known [[Cryptids]] (mathematically-hard machines) in that class, and &amp;lt;span style=&amp;quot;background: #ffe4b2&amp;quot;&amp;gt;cells are highlighted in light orange&amp;lt;/span&amp;gt; when the existence of a Cryptid is given by using a known one with less states or symbols.&lt;br /&gt;
&lt;br /&gt;
== About bbchallenge ==&lt;br /&gt;
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt.  Other efforts also include:&lt;br /&gt;
&lt;br /&gt;
* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Coq_(software) Coq])&lt;br /&gt;
* Maintaining [[Holdouts lists]] for small busy beaver values&lt;br /&gt;
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]&lt;br /&gt;
* Finding [[Cryptids]] (mathematically-hard machines)&lt;br /&gt;
* Searching for new [[Champions]]&lt;br /&gt;
* Building [[Accelerated Simulator]]s to simulate halting machines faster&lt;br /&gt;
* Writing papers and giving talks about busy beaver, see [[Papers &amp;amp; Talks]]&lt;br /&gt;
&lt;br /&gt;
In June 2024, bbchallenge achieved a significant milestone by proving in Coq / Rocq that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.&amp;lt;ref&amp;gt;H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.&lt;br /&gt;
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== This Month in Beaver Research (TMBR) ==&lt;br /&gt;
[[:Category:This Month in Beaver Research|This Month in Beaver Research]] (TMBR, pronounced &amp;quot;timber&amp;quot;), is a monthly summary of Busy Beaver research progress. Here are the last 3 entries:&lt;br /&gt;
&lt;br /&gt;
* [[TMBR: July 2025|TMBR July 2025]]&lt;br /&gt;
&lt;br /&gt;
== Contribute to this wiki ==&lt;br /&gt;
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;inputbox&amp;gt;&lt;br /&gt;
type=create&lt;br /&gt;
width=100&lt;br /&gt;
break=no&lt;br /&gt;
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default=(Article title)&lt;br /&gt;
&amp;lt;/inputbox&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Notes==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2722</id>
		<title>Main Page</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2722"/>
		<updated>2025-08-04T08:24:06Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: Advertising TMBR on main page&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;The [[Busy Beaver function]] BB (called &#039;&#039;S&#039;&#039; originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to &#039;&#039;m&#039;&#039;-symbol Turing machines:&amp;lt;ref&amp;gt;Rado, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
| BB(&#039;&#039;n&#039;&#039;, &#039;&#039;m&#039;&#039;) = Maximum number of steps taken by a halting &#039;&#039;n&#039;&#039;-state, &#039;&#039;m&#039;&#039;-symbol Turing machine starting from a blank (all 0) tape&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
The 2-symbol case BB(&#039;&#039;n&#039;&#039;, 2) is abbreviated as BB(&#039;&#039;n&#039;&#039;). The busy beaver function is not computable, but a few of its values are known:&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+ Small busy beaver values&amp;lt;ref&amp;gt;P. Michel, &amp;quot;[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]&amp;quot;.&amp;lt;/ref&amp;gt;&lt;br /&gt;
! !!2-state!!3-state !!4-state!!5-state!!6-state &lt;br /&gt;
!7-state&lt;br /&gt;
|-  &lt;br /&gt;
! 2-symbol &lt;br /&gt;
| [[BB(2)]] = 6 &lt;br /&gt;
| [[BB(3)]] = 21&lt;br /&gt;
| [[BB(4)]] = 107 &lt;br /&gt;
| [[BB(5)]] = 47,176,870 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(6)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow \uparrow \uparrow 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(7)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{11} 2 \uparrow^{11} 3&amp;lt;/math&amp;gt; &lt;br /&gt;
|-&lt;br /&gt;
! 3-symbol&lt;br /&gt;
| [[BB(2,3)]] = 38 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(3,3)]] &amp;gt; &amp;lt;math&amp;gt;10^{17}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(4,3)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow\uparrow\uparrow 2^{2^{32}}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 4-symbol  &lt;br /&gt;
| [[BB(2,4)]] = 3,932,964&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,4)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{15} 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 5-symbol &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(2,5)]] &amp;gt; &amp;lt;math&amp;gt;10\uparrow\uparrow 4&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,5)]] &amp;gt; &amp;lt;math&amp;gt; f_\omega(2 \uparrow^{15} 5)&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 6-symbol &lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(2,6)]] &amp;gt; &amp;lt;math&amp;gt;10 \uparrow\uparrow\uparrow 3&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
In the above table, &amp;lt;span style=&amp;quot;background: orange&amp;quot;&amp;gt;cells are highlighted in orange&amp;lt;/span&amp;gt; when there are known [[Cryptids]] (mathematically-hard machines) in that class, and &amp;lt;span style=&amp;quot;background: #ffe4b2&amp;quot;&amp;gt;cells are highlighted in light orange&amp;lt;/span&amp;gt; when the existence of a Cryptid is given by using a known one with less states or symbols.&lt;br /&gt;
&lt;br /&gt;
== About bbchallenge ==&lt;br /&gt;
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt.  Other efforts also include:&lt;br /&gt;
&lt;br /&gt;
* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Coq_(software) Coq])&lt;br /&gt;
* Maintaining [[Holdouts lists]] for small busy beaver values&lt;br /&gt;
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]&lt;br /&gt;
* Finding [[Cryptids]] (mathematically-hard machines)&lt;br /&gt;
* Searching for new [[Champions]]&lt;br /&gt;
* Building [[Accelerated Simulator]]s to simulate halting machines faster&lt;br /&gt;
* Writing papers and giving talks about busy beaver, see [[Papers &amp;amp; Talks]]&lt;br /&gt;
&lt;br /&gt;
In June 2024, bbchallenge achieved a significant milestone by proving in Coq / Rocq that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.&amp;lt;ref&amp;gt;H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.&lt;br /&gt;
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Contribute to this wiki ==&lt;br /&gt;
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;inputbox&amp;gt;&lt;br /&gt;
type=create&lt;br /&gt;
width=100&lt;br /&gt;
break=no&lt;br /&gt;
buttonlabel=Create new article&lt;br /&gt;
default=(Article title)&lt;br /&gt;
&amp;lt;/inputbox&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== This Month in Beaver Research (TMBR) ==&lt;br /&gt;
[[:Category:This Month in Beaver Research|This Month in Beaver Research]] (TMBR, pronounced &amp;quot;timber&amp;quot;), is a monthly summary of Busy Beaver research progress. Here are the last 3 entries:&lt;br /&gt;
&lt;br /&gt;
* [[TMBR: July 2025|TMBR July 2025]]&lt;br /&gt;
&lt;br /&gt;
==Notes==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_Beaver_Challenge&amp;diff=2487</id>
		<title>Busy Beaver Challenge</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_Beaver_Challenge&amp;diff=2487"/>
		<updated>2025-07-15T14:57:48Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;The &#039;&#039;&#039;Busy Beaver Challenge&#039;&#039;&#039; (or &#039;&#039;&#039;bbchallenge&#039;&#039;&#039;) is an online community founded in 2022 with the goal of solving [[BB(5)]]. This goal was accomplished in 2024. Since then the community has focussed on many different goals, including searching through unsolved domains (currently [[BB(6)]], [[BB(3,3)]], [[BB(2,5)]] and [[BB(7)]]) for new [[champions]] and [[cryptids]] and reducing [[holdout]] lists. Most of the interaction happens on [https://discord.gg/wuZhtTvYU3 our public Discord].&lt;br /&gt;
&lt;br /&gt;
== See Also ==&lt;br /&gt;
&lt;br /&gt;
* Main Website: https://bbchallenge.org/&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Beaver_Math_Olympiad&amp;diff=2341</id>
		<title>Beaver Math Olympiad</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Beaver_Math_Olympiad&amp;diff=2341"/>
		<updated>2025-07-03T11:57:08Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* 5. 1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE (bbch) */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Beaver Mathematical Olympiad&#039;&#039;&#039; (BMO) is an attempt to re-formulate the halting problem for some particular Turing machines as a mathematical problem in a style suitable for a hypothetical math olympiad. &lt;br /&gt;
&lt;br /&gt;
The purpose of the BMO is twofold. First, statements where non-essential details (related to tape encoding, number of steps, etc.) are discarded are more suitable to be shared with mathematicians who perhaps are able to help. Second, it&#039;s a way to jokingly highlight how a hard question could appear deceptively simple.&lt;br /&gt;
&lt;br /&gt;
== Unsolved problems ==&lt;br /&gt;
&lt;br /&gt;
=== 1. {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}} ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;(a_n)_{n \ge 1}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;(b_n)_{n \ge 1}&amp;lt;/math&amp;gt; be two sequences such that &amp;lt;math&amp;gt;(a_1, b_1) = (1, 2)&amp;lt;/math&amp;gt; and&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;(a_{n+1}, b_{n+1}) = \begin{cases}&lt;br /&gt;
(a_n-b_n, 4b_n+2) &amp;amp; \text{if }a_n \ge b_n \\&lt;br /&gt;
(2a_n+1, b_n-a_n) &amp;amp; \text{if }a_n &amp;lt; b_n&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
for all positive integers &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;. Does there exist a positive integer &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;a_i = b_i&amp;lt;/math&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
The first 10 values of &amp;lt;math&amp;gt;(a_n, b_n)&amp;lt;/math&amp;gt; are &amp;lt;math&amp;gt;(1, 2), (3, 1), (2, 6), (5, 4), (1, 18), (3, 17), (7, 14), (15, 7), (8, 30), (17, 22)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
=== 2. [[Hydra]] and [[Antihydra]] ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;(a_n)_{n \ge 0}&amp;lt;/math&amp;gt; be a sequence such that &amp;lt;math&amp;gt;a_{n+1} = a_n+\left\lfloor\frac{a_n}{2}\right\rfloor&amp;lt;/math&amp;gt; for all non-negative integers &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
# If &amp;lt;math&amp;gt;a_0=3&amp;lt;/math&amp;gt;, does there exist a non-negative integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; such that the list of numbers &amp;lt;math&amp;gt;a_0, a_1, a_2, \dots, a_k&amp;lt;/math&amp;gt; have more than twice as many even numbers as odd numbers? ([[Hydra]])&lt;br /&gt;
# If &amp;lt;math&amp;gt;a_0=8&amp;lt;/math&amp;gt;, does there exist a non-negative integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; such that the list of numbers &amp;lt;math&amp;gt;a_0, a_1, a_2, \dots, a_k&amp;lt;/math&amp;gt; have more than twice as many odd numbers as even numbers? ([[Antihydra]])&lt;br /&gt;
&lt;br /&gt;
=== 5. {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}} ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;(a_n)_{n \ge 1}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;(b_n)_{n \ge 1}&amp;lt;/math&amp;gt; be two sequences such that &amp;lt;math&amp;gt;(a_1, b_1) = (0, 5)&amp;lt;/math&amp;gt; and&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;(a_{n+1}, b_{n+1}) = \begin{cases}&lt;br /&gt;
(a_n+1, b_n-f(a_n)) &amp;amp; \text{if } b_n \ge f(a_n) \\&lt;br /&gt;
(a_n, 3b_n+a_n+5) &amp;amp; \text{if } b_n &amp;lt; f(a_n)&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;f(x)=10\cdot 2^x-1&amp;lt;/math&amp;gt; for all non-negative integers &amp;lt;math&amp;gt;x&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Does there exist a positive integer &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;b_i = f(a_i)-1&amp;lt;/math&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
== Solved problems ==&lt;br /&gt;
&lt;br /&gt;
=== 3. {{TM|1RB0RB3LA4LA2RA_2LB3RA---3RA4RB|non-halt}} and {{TM|1RB1RB3LA4LA2RA_2LB3RA---3RA4RB|non-halt}} ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;v_2(n)&amp;lt;/math&amp;gt; be the largest integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;2^k&amp;lt;/math&amp;gt; divides &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;. Let &amp;lt;math&amp;gt;(a_n)_{n \ge 0}&amp;lt;/math&amp;gt; be a sequence such that&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;a_n = \begin{cases}&lt;br /&gt;
2 &amp;amp; \text{if } n=0 \\&lt;br /&gt;
a_{n-1}+2^{v_2(a_{n-1})+2}-1 &amp;amp; \text{if } n \ge 1&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
for all non-negative integers &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;. Is there an integer &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;a_n=4^k&amp;lt;/math&amp;gt; for some positive integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
Link to Discord discussion: https://discord.com/channels/960643023006490684/1084047886494470185/1252634913220591728&lt;br /&gt;
&lt;br /&gt;
=== 4. {{TM|1RB3RB---1LB0LA_2LA4RA3LA4RB1LB|non-halt}} ===&lt;br /&gt;
&lt;br /&gt;
Bonnie the beaver was bored, so she tried to construct a sequence of integers &amp;lt;math&amp;gt;\{a_n\}_{n \ge 0}&amp;lt;/math&amp;gt;. She first defined &amp;lt;math&amp;gt;a_0=2&amp;lt;/math&amp;gt;, then defined &amp;lt;math&amp;gt;a_{n+1}&amp;lt;/math&amp;gt; depending on &amp;lt;math&amp;gt;a_n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; using the following rules:&lt;br /&gt;
&lt;br /&gt;
* If &amp;lt;math&amp;gt;a_n \equiv 0\text{ (mod 3)}&amp;lt;/math&amp;gt;, then &amp;lt;math&amp;gt;a_{n+1}=\frac{a_n}{3}+2^n+1&amp;lt;/math&amp;gt;.&lt;br /&gt;
* If &amp;lt;math&amp;gt;a_n \equiv 2\text{ (mod 3)}&amp;lt;/math&amp;gt;, then &amp;lt;math&amp;gt;a_{n+1}=\frac{a_n-2}{3}+2^n-1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
With these two rules alone, Bonnie calculates the first few terms in the sequence: &amp;lt;math&amp;gt;2, 0, 3, 6, 11, 18, 39, 78, 155, 306, \dots&amp;lt;/math&amp;gt;. At this point, Bonnie plans to continue writing terms until a term becomes &amp;lt;math&amp;gt;1\text{ (mod 3)}&amp;lt;/math&amp;gt;. If Bonnie sticks to her plan, will she ever finish?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Solution&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
How to guess the closed-form solution: Firstly, notice that &amp;lt;math&amp;gt;a_n \approx \frac{3}{5} \times 2^n&amp;lt;/math&amp;gt;. Secondly, calculate the error term &amp;lt;math&amp;gt;a_n - \frac{3}{5} \times 2^n&amp;lt;/math&amp;gt;. The error term appears to have a period of 4. This leads to the following guess:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;a_n=\frac{3}{5}\begin{cases}&lt;br /&gt;
2^n+\frac{7}{3} &amp;amp;\text{if } n\equiv 0 \pmod{4}\\&lt;br /&gt;
2^n-2 &amp;amp;\text{if } n\equiv 1 \pmod{4}\\&lt;br /&gt;
2^n+1 &amp;amp;\text{if } n\equiv 2 \pmod{4}\\&lt;br /&gt;
2^n+2 &amp;amp;\text{if } n\equiv 3 \pmod{4}\\&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This closed-form solution can be proven correct by induction. Unfortunately, the induction may require a lot of tedious calculations.&lt;br /&gt;
&lt;br /&gt;
For all &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;, we have &amp;lt;math&amp;gt;a_{4k} \equiv 2\text{ (mod 3)}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;a_{4k+1} \equiv a_{4k+2} \equiv a_{4k+3} \equiv 0\text{ (mod 3)}&amp;lt;/math&amp;gt;. Therefore, Bonnie will never finish.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Beaver_Math_Olympiad&amp;diff=2340</id>
		<title>Beaver Math Olympiad</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Beaver_Math_Olympiad&amp;diff=2340"/>
		<updated>2025-07-03T11:55:51Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* 5. {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}} */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Beaver Mathematical Olympiad&#039;&#039;&#039; (BMO) is an attempt to re-formulate the halting problem for some particular Turing machines as a mathematical problem in a style suitable for a hypothetical math olympiad. &lt;br /&gt;
&lt;br /&gt;
The purpose of the BMO is twofold. First, statements where non-essential details (related to tape encoding, number of steps, etc.) are discarded are more suitable to be shared with mathematicians who perhaps are able to help. Second, it&#039;s a way to jokingly highlight how a hard question could appear deceptively simple.&lt;br /&gt;
&lt;br /&gt;
== Unsolved problems ==&lt;br /&gt;
&lt;br /&gt;
=== 1. {{TM|1RB1RE_1LC0RA_0RD1LB_---1RC_1LF1RE_0LB0LE|undecided}} ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;(a_n)_{n \ge 1}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;(b_n)_{n \ge 1}&amp;lt;/math&amp;gt; be two sequences such that &amp;lt;math&amp;gt;(a_1, b_1) = (1, 2)&amp;lt;/math&amp;gt; and&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;(a_{n+1}, b_{n+1}) = \begin{cases}&lt;br /&gt;
(a_n-b_n, 4b_n+2) &amp;amp; \text{if }a_n \ge b_n \\&lt;br /&gt;
(2a_n+1, b_n-a_n) &amp;amp; \text{if }a_n &amp;lt; b_n&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
for all positive integers &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;. Does there exist a positive integer &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;a_i = b_i&amp;lt;/math&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
The first 10 values of &amp;lt;math&amp;gt;(a_n, b_n)&amp;lt;/math&amp;gt; are &amp;lt;math&amp;gt;(1, 2), (3, 1), (2, 6), (5, 4), (1, 18), (3, 17), (7, 14), (15, 7), (8, 30), (17, 22)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
=== 2. [[Hydra]] and [[Antihydra]] ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;(a_n)_{n \ge 0}&amp;lt;/math&amp;gt; be a sequence such that &amp;lt;math&amp;gt;a_{n+1} = a_n+\left\lfloor\frac{a_n}{2}\right\rfloor&amp;lt;/math&amp;gt; for all non-negative integers &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
# If &amp;lt;math&amp;gt;a_0=3&amp;lt;/math&amp;gt;, does there exist a non-negative integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; such that the list of numbers &amp;lt;math&amp;gt;a_0, a_1, a_2, \dots, a_k&amp;lt;/math&amp;gt; have more than twice as many even numbers as odd numbers? ([[Hydra]])&lt;br /&gt;
# If &amp;lt;math&amp;gt;a_0=8&amp;lt;/math&amp;gt;, does there exist a non-negative integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; such that the list of numbers &amp;lt;math&amp;gt;a_0, a_1, a_2, \dots, a_k&amp;lt;/math&amp;gt; have more than twice as many odd numbers as even numbers? ([[Antihydra]])&lt;br /&gt;
&lt;br /&gt;
=== 5. {{TM|1RB0LD_1LC0RA_1RA1LB_1LA1LE_1RF0LC_---0RE|undecided}} ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;(a_n)_{n \ge 1}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;(b_n)_{n \ge 1}&amp;lt;/math&amp;gt; be two sequences such that &amp;lt;math&amp;gt;(a_1, b_1) = (0, 5)&amp;lt;/math&amp;gt; and&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;(a_{n+1}, b_{n+1}) = \begin{cases}&lt;br /&gt;
(a_n+1, b_n-f(a_n)) &amp;amp; \text{if } b_n \ge f(a_n) \\&lt;br /&gt;
(a_n, 3b_n+a_n+5) &amp;amp; \text{if } b_n &amp;lt; f(a_n)&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;f(x)=10\cdot 2^x-1&amp;lt;/math&amp;gt; for all positive integers &amp;lt;math&amp;gt;x&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Does there exist a positive integer &amp;lt;math&amp;gt;i&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;b_i = f(a_i)-1&amp;lt;/math&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
== Solved problems ==&lt;br /&gt;
&lt;br /&gt;
=== 3. {{TM|1RB0RB3LA4LA2RA_2LB3RA---3RA4RB|non-halt}} and {{TM|1RB1RB3LA4LA2RA_2LB3RA---3RA4RB|non-halt}} ===&lt;br /&gt;
&lt;br /&gt;
Let &amp;lt;math&amp;gt;v_2(n)&amp;lt;/math&amp;gt; be the largest integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;2^k&amp;lt;/math&amp;gt; divides &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;. Let &amp;lt;math&amp;gt;(a_n)_{n \ge 0}&amp;lt;/math&amp;gt; be a sequence such that&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;a_n = \begin{cases}&lt;br /&gt;
2 &amp;amp; \text{if } n=0 \\&lt;br /&gt;
a_{n-1}+2^{v_2(a_{n-1})+2}-1 &amp;amp; \text{if } n \ge 1&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
for all non-negative integers &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;. Is there an integer &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; such that &amp;lt;math&amp;gt;a_n=4^k&amp;lt;/math&amp;gt; for some positive integer &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;?&lt;br /&gt;
&lt;br /&gt;
Link to Discord discussion: https://discord.com/channels/960643023006490684/1084047886494470185/1252634913220591728&lt;br /&gt;
&lt;br /&gt;
=== 4. {{TM|1RB3RB---1LB0LA_2LA4RA3LA4RB1LB|non-halt}} ===&lt;br /&gt;
&lt;br /&gt;
Bonnie the beaver was bored, so she tried to construct a sequence of integers &amp;lt;math&amp;gt;\{a_n\}_{n \ge 0}&amp;lt;/math&amp;gt;. She first defined &amp;lt;math&amp;gt;a_0=2&amp;lt;/math&amp;gt;, then defined &amp;lt;math&amp;gt;a_{n+1}&amp;lt;/math&amp;gt; depending on &amp;lt;math&amp;gt;a_n&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; using the following rules:&lt;br /&gt;
&lt;br /&gt;
* If &amp;lt;math&amp;gt;a_n \equiv 0\text{ (mod 3)}&amp;lt;/math&amp;gt;, then &amp;lt;math&amp;gt;a_{n+1}=\frac{a_n}{3}+2^n+1&amp;lt;/math&amp;gt;.&lt;br /&gt;
* If &amp;lt;math&amp;gt;a_n \equiv 2\text{ (mod 3)}&amp;lt;/math&amp;gt;, then &amp;lt;math&amp;gt;a_{n+1}=\frac{a_n-2}{3}+2^n-1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
With these two rules alone, Bonnie calculates the first few terms in the sequence: &amp;lt;math&amp;gt;2, 0, 3, 6, 11, 18, 39, 78, 155, 306, \dots&amp;lt;/math&amp;gt;. At this point, Bonnie plans to continue writing terms until a term becomes &amp;lt;math&amp;gt;1\text{ (mod 3)}&amp;lt;/math&amp;gt;. If Bonnie sticks to her plan, will she ever finish?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Solution&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
How to guess the closed-form solution: Firstly, notice that &amp;lt;math&amp;gt;a_n \approx \frac{3}{5} \times 2^n&amp;lt;/math&amp;gt;. Secondly, calculate the error term &amp;lt;math&amp;gt;a_n - \frac{3}{5} \times 2^n&amp;lt;/math&amp;gt;. The error term appears to have a period of 4. This leads to the following guess:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;a_n=\frac{3}{5}\begin{cases}&lt;br /&gt;
2^n+\frac{7}{3} &amp;amp;\text{if } n\equiv 0 \pmod{4}\\&lt;br /&gt;
2^n-2 &amp;amp;\text{if } n\equiv 1 \pmod{4}\\&lt;br /&gt;
2^n+1 &amp;amp;\text{if } n\equiv 2 \pmod{4}\\&lt;br /&gt;
2^n+2 &amp;amp;\text{if } n\equiv 3 \pmod{4}\\&lt;br /&gt;
\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This closed-form solution can be proven correct by induction. Unfortunately, the induction may require a lot of tedious calculations.&lt;br /&gt;
&lt;br /&gt;
For all &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;, we have &amp;lt;math&amp;gt;a_{4k} \equiv 2\text{ (mod 3)}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;a_{4k+1} \equiv a_{4k+2} \equiv a_{4k+3} \equiv 0\text{ (mod 3)}&amp;lt;/math&amp;gt;. Therefore, Bonnie will never finish.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Hydra&amp;diff=2339</id>
		<title>Hydra</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Hydra&amp;diff=2339"/>
		<updated>2025-07-03T10:42:00Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: Fast Hydra sim&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA}}{{unsolved|Does Hydra run forever?}}{{TM|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA|undecided}}&lt;br /&gt;
&#039;&#039;&#039;Hydra&#039;&#039;&#039; is a [[BB(2,5)]] [[Cryptid]]. Its high-level rules were first reported [https://discord.com/channels/960643023006490684/1084047886494470185/1231110668288135208 on Discord] by Daniel Yuan on 20 April 2024, who also gave Hydra said name. Later on, a 6-state, 2-symbol [[Turing machine]] was discovered and named [[Antihydra]] for having similar behaviour to Hydra, making the study of this machine important to the study of that one.&lt;br /&gt;
&lt;br /&gt;
Hydra is known to not generate Sturmian words&amp;lt;ref&amp;gt;Dubickas A. On Integer Sequences Generated by Linear Maps. &#039;&#039;Glasgow Mathematical Journal&#039;&#039;. 2009; 51(2): 243-252. doi:[https://doi.org/10.1017/S0017089508004655 10.1017/S0017089508004655]&amp;lt;/ref&amp;gt; (Corollary 4).&lt;br /&gt;
&amp;lt;div style=&amp;quot;width: fit-content; text-align: center; margin-left: auto; margin-right: auto;&amp;quot;&amp;gt;&lt;br /&gt;
{|class=&amp;quot;wikitable&amp;quot; style=&amp;quot;margin-left: auto; margin-right: auto;&amp;quot;&lt;br /&gt;
! !!0!!1!!2!!3!!4&lt;br /&gt;
|-&lt;br /&gt;
!A&lt;br /&gt;
|1RB&lt;br /&gt;
|3RB&lt;br /&gt;
| ---&lt;br /&gt;
|3LA&lt;br /&gt;
|1RA&lt;br /&gt;
|-&lt;br /&gt;
!B&lt;br /&gt;
|2LA&lt;br /&gt;
|3RA&lt;br /&gt;
|4LB&lt;br /&gt;
|0LB&lt;br /&gt;
|0LA&lt;br /&gt;
|}&lt;br /&gt;
The transition table of Hydra.&amp;lt;/div&amp;gt;&lt;br /&gt;
== Analysis ==&lt;br /&gt;
Let &amp;lt;math&amp;gt;C(a,b):=0^\infty\;\textrm{&amp;lt;A}\;2\;0^a\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;. Then,&amp;lt;ref&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/05/10/bb-2-5-is-hard.html BB(2, 5) is Hard (Hydra)] (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|lll|}\hline&lt;br /&gt;
C(2a,0)&amp;amp;\xrightarrow{6a^2+20a+4}&amp;amp;0^\infty\;3^{3a+1}\;1\;\textrm{A&amp;gt;}\;2\;0^\infty,\\&lt;br /&gt;
C(2a,b+1)&amp;amp;\xrightarrow{6a^2+23a+10}&amp;amp;C(3a+3,b),\\&lt;br /&gt;
C(2a+1,b)&amp;amp;\xrightarrow{4b+6a^2+23a+26}&amp;amp;C(3a+3,b+2).\\\hline&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Proof&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
Consider the partial configuration &amp;lt;math&amp;gt;P(m,n):=0^\infty\;3^m\;\textrm{A&amp;gt;}\;02\;0^n&amp;lt;/math&amp;gt;. After 14 steps this configuration becomes &lt;br /&gt;
&amp;lt;math&amp;gt;0^\infty\;3^{m+3}\;\textrm{&amp;lt;A}\;2\;0^{n-2}&amp;lt;/math&amp;gt;. We note the following shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline3^s\;\textrm{&amp;lt;A}\xrightarrow{s}\textrm{&amp;lt;A}\;3^s\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
Using this shift rule, we get &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;A}\;3^{m+3}\;2\;0^{n-2}&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;m+3&amp;lt;/math&amp;gt; steps. From here, we can observe that &amp;lt;math&amp;gt;\textrm{A&amp;gt;}\;0\;3^s&amp;lt;/math&amp;gt; turns into &amp;lt;math&amp;gt;3\;\textrm{A&amp;gt;}\;0\;3^{s-1}&amp;lt;/math&amp;gt; in three steps if &amp;lt;math&amp;gt;s\ge 1&amp;lt;/math&amp;gt;. By repeating this process, we acquire this transition rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{A&amp;gt;}\;0\;3^s\xrightarrow{3s}3^s\;\textrm{A&amp;gt;}\;0\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With this rule, it takes &amp;lt;math&amp;gt;3m+9&amp;lt;/math&amp;gt; steps to reach the configuration &amp;lt;math&amp;gt;0^\infty\;3^{m+3}\;\textrm{A&amp;gt;}\;02\;0^{n-2}&amp;lt;/math&amp;gt;, which is the same configuration as &amp;lt;math&amp;gt;P(m+3,n-2)&amp;lt;/math&amp;gt;. To summarize:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline P(m,n)\xrightarrow{4m+26}P(m+3,n-2)\text{ if }n\ge 2.\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With &amp;lt;math&amp;gt;C(a,b)&amp;lt;/math&amp;gt; we have &amp;lt;math&amp;gt;P(0,a)&amp;lt;/math&amp;gt;. As a result, we can apply this rule &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\big\lfloor\frac{1}{2}a\big\rfloor&amp;lt;/math&amp;gt; times, which creates two possible scenarios:&lt;br /&gt;
#If &amp;lt;math&amp;gt;a\equiv0\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math&amp;gt;\sum_{i=0}^{(a/2)-1}(4\times 3i+26)=\textstyle\frac{3}{2}a^2+10a&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(\frac{3}{2}a,0\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;3^{(3/2)a}\;\textrm{A&amp;gt;}\;02\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;, which in four steps becomes &amp;lt;math&amp;gt;0^\infty\;3^{(3/2)a+1}\;1\;\textrm{A&amp;gt;}\;3^b\;2\;0^\infty.&amp;lt;/math&amp;gt; If &amp;lt;math&amp;gt;b=0&amp;lt;/math&amp;gt; then we have reached the undefined &amp;lt;code&amp;gt;A2&amp;lt;/code&amp;gt; transition in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a^2+10a+4&amp;lt;/math&amp;gt; steps total. Otherwise, continuing for three steps gives us &amp;lt;math&amp;gt;0^\infty\;3^{(3/2)+2}\;\textrm{&amp;lt;B}\;0\;3^{b-1}\;2\;0^\infty&amp;lt;/math&amp;gt;. Another shift rule is required here:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline3^s\;\textrm{&amp;lt;B}\xrightarrow{s}\textrm{&amp;lt;B}\;0^s\\\hline\end{array}&amp;lt;/math&amp;gt;This means the configuration becomes &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;B}\;0^{(3/2)a+3}\;3^{b-1}\;2\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a+2&amp;lt;/math&amp;gt; steps, and &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;A}\;2\;0^{(3/2)a+3}\;3^{b-1}\;2\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;C\Big(\frac{3}{2}a+3,b-1\Big)&amp;lt;/math&amp;gt;, one step later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a^2+\frac{23}{2}a+10&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;a\equiv1\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a^2+7a-\frac{17}{2}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(\frac{3a-3}{2},1\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;3^{(3a-3)/2}\;\textrm{A&amp;gt;}\;020\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;, which in four steps becomes &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;\textrm{A&amp;gt;}\;0\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;, and then &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;3^b\;\textrm{A&amp;gt;}\;02\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;3b&amp;lt;/math&amp;gt; steps. After 14 steps, we see the configuration &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;3^{b+3}\;\textrm{&amp;lt;A}\;2\;0^\infty&amp;lt;/math&amp;gt;, which turns into &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;\textrm{&amp;lt;A}\;3^{b+3}\;2\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;b+3&amp;lt;/math&amp;gt; steps. In two steps we get &amp;lt;math&amp;gt;0^\infty\;3^{(3a+1)/2}\;\textrm{&amp;lt;B}\;0\;3^{b+2}\;2\;0^\infty&amp;lt;/math&amp;gt;, followed by &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;B}\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty&amp;lt;/math&amp;gt; after &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3a+1}{2}&amp;lt;/math&amp;gt; more steps. We conclude with &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;A}\;2\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;C\Big(\frac{3a+3}{2},b+2\Big)&amp;lt;/math&amp;gt;, one step later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4b+\frac{3}{2}a^2+\frac{17}{2}a+16&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
The information above can be summarized as&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;C(a,b)\rightarrow\begin{cases}0^\infty\;3^{(3/2)a+1}\;1\;\textrm{A&amp;gt;}\;2\;0^\infty&amp;amp;\text{if }a\equiv0\pmod{2}\text{ and }b=0,\\C\Big(\frac{3}{2}a+3,b-1\Big)&amp;amp;\text{if }a\equiv0\pmod{2}\text{ and }b&amp;gt;0,\\C\Big(\frac{3a+3}{2},b+2\Big)&amp;amp;\text{if }a\equiv1\pmod2.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Substituting &amp;lt;math&amp;gt;a\leftarrow 2a&amp;lt;/math&amp;gt; for the first two cases and &amp;lt;math&amp;gt;a\leftarrow 2a+1&amp;lt;/math&amp;gt; for the third yields the final result.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
In effect, the halting problem for Hydra is about whether repeatedly applying the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;f(n)=3\big\lfloor\frac{n}{2}\big\rfloor+3&amp;lt;/math&amp;gt; will at some point produce more even values of &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; than twice the number of odd values.&lt;br /&gt;
&lt;br /&gt;
An alternative version of these rules exists that makes the connection to Antihydra more apparent by using the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;H(n)=\Big\lfloor\frac{3}{2}n\Big\rfloor&amp;lt;/math&amp;gt;, or the [[Hydra function]].&amp;lt;ref&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]&amp;quot; (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Trajectory ==&lt;br /&gt;
It takes 20 steps to reach the configuration &amp;lt;math&amp;gt;C(3,0)&amp;lt;/math&amp;gt;, and from there, the [[Collatz-like]] rules are repeatedly applied. Simulating Hydra has shown that after 4000000 rule steps, we have &amp;lt;math&amp;gt;b=2005373&amp;lt;/math&amp;gt;. Here are the first few:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline C(3,0)\xrightarrow{55}C(6,2)\xrightarrow{133}C(12,1)\xrightarrow{364}C(21,0)\xrightarrow{856}C(33,2)\xrightarrow{1938}C(51,4)\xrightarrow{4367}C(78,6)\rightarrow\cdots\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
The heuristic argument that suggests Antihydra is a [[probviously]] non-halting machine can be applied here. This means that if &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; is to be thought of as moving randomly, then the probability of Hydra halting is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\Big(\frac{\sqrt{5}-1}{2}\Big)}^{2005374}\approx 4.168\times 10^{-419099}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
== Code ==&lt;br /&gt;
&lt;br /&gt;
Fast Hydra/[[Antihydra]] simulation code by Greg Kuperberg (who said it could be made faster using FLINT):&amp;lt;syntaxhighlight lang=&amp;quot;python2&amp;quot; line=&amp;quot;1&amp;quot;&amp;gt;&lt;br /&gt;
# Python script to demonstrate almost linear time hydra simulation&lt;br /&gt;
# using fast multiplication. &lt;br /&gt;
# by Greg Kuperberg&lt;br /&gt;
&lt;br /&gt;
import time&lt;br /&gt;
from gmpy2 import mpz,bit_mask&lt;br /&gt;
&lt;br /&gt;
# Straight computation of t steps of hydra&lt;br /&gt;
def simple(n,t):&lt;br /&gt;
    for s in range(t): n += n&amp;gt;&amp;gt;1&lt;br /&gt;
    return n&lt;br /&gt;
&lt;br /&gt;
# Accelerated computation of 2**e steps of hydra&lt;br /&gt;
def hydra(n,e):&lt;br /&gt;
    if e &amp;lt; 9: return simple(n,1&amp;lt;&amp;lt;e)&lt;br /&gt;
    t = 1&amp;lt;&amp;lt;(e-1)&lt;br /&gt;
    (p3t,m) = (mpz(3)**t,bit_mask(t))&lt;br /&gt;
    n = p3t*(n&amp;gt;&amp;gt;t) + hydra(n&amp;amp;m,e-1)&lt;br /&gt;
    return p3t*(n&amp;gt;&amp;gt;t) + hydra(n&amp;amp;m,e-1)&lt;br /&gt;
&lt;br /&gt;
def elapsed():&lt;br /&gt;
    (last,elapsed.mark) = (elapsed.mark,time.process_time())&lt;br /&gt;
    return elapsed.mark-last&lt;br /&gt;
elapsed.mark = 0&lt;br /&gt;
&lt;br /&gt;
(n,e) = (mpz(3),25)&lt;br /&gt;
&lt;br /&gt;
elapsed()&lt;br /&gt;
print(&#039;hydra:  steps=%d hash=%016x time=%.6fs&#039;&lt;br /&gt;
    % (1&amp;lt;&amp;lt;e,hash(hydra(n,e)),elapsed()))&lt;br /&gt;
&lt;br /&gt;
# Quadratic time algorithm for comparison&lt;br /&gt;
# print(&#039;simple: steps=%d hash=%016x time=%.6fs&#039;&lt;br /&gt;
#     % (1&amp;lt;&amp;lt;e,hash(simple(n,1&amp;lt;&amp;lt;e)),elapsed()))&lt;br /&gt;
&amp;lt;/syntaxhighlight&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[[Category:Cryptids]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Antihydra&amp;diff=2338</id>
		<title>Antihydra</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Antihydra&amp;diff=2338"/>
		<updated>2025-07-03T10:41:04Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* Code */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA}}{{unsolved|Does Antihydra run forever?}}{{TM|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA|undecided}}&lt;br /&gt;
&#039;&#039;&#039;Antihydra&#039;&#039;&#039; is a [[BB(6)]] [[Cryptid]]. Its pseudo-random behaviour was first reported [https://discord.com/channels/960643023006490684/1026577255754903572/1256223215206924318 on Discord] by mxdys on 28 June 2024, and Racheline discovered the high-level rules soon after. It was named after the 2-state, 5-symbol [[Turing machine]] called [[Hydra]] for sharing many similarities to it.&amp;lt;ref&amp;gt;[https://discord.com/channels/960643023006490684/960643023530762341/1257053002859286701 Discord conversation where the machine was named]&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Antihydra is known to not generate Sturmian words&amp;lt;ref&amp;gt;DUBICKAS A. ON INTEGER SEQUENCES GENERATED BY LINEAR MAPS. &#039;&#039;Glasgow Mathematical Journal&#039;&#039;. 2009;51(2):243-252. doi:[https://doi.org/10.1017/S0017089508004655 10.1017/S0017089508004655]&amp;lt;/ref&amp;gt; (Corollary 4).&lt;br /&gt;
&amp;lt;table style=&amp;quot;margin: auto; text-align: center;&amp;quot;&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td style=&amp;quot;width: 200px;&amp;quot;&amp;gt;[[File:Antihydra-depiction.png|200px]]&amp;lt;br&amp;gt;Artistic depiction of Antihydra by Jadeix&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&lt;br /&gt;
{|class=&amp;quot;wikitable&amp;quot; style=&amp;quot;margin-left: auto; margin-right: auto;&amp;quot;&lt;br /&gt;
! !!0!!1&lt;br /&gt;
|-&lt;br /&gt;
!A&lt;br /&gt;
|1RB&lt;br /&gt;
|1RA&lt;br /&gt;
|-&lt;br /&gt;
!B&lt;br /&gt;
|0LC&lt;br /&gt;
|1LE&lt;br /&gt;
|-&lt;br /&gt;
!C&lt;br /&gt;
|1LD&lt;br /&gt;
|1LC&lt;br /&gt;
|-&lt;br /&gt;
!D&lt;br /&gt;
|1LA&lt;br /&gt;
|0LB&lt;br /&gt;
|-&lt;br /&gt;
!E&lt;br /&gt;
|1LF&lt;br /&gt;
|1RE&lt;br /&gt;
|-&lt;br /&gt;
!F&lt;br /&gt;
| ---&lt;br /&gt;
|0RA&lt;br /&gt;
|}&lt;br /&gt;
The transition table of Antihydra.&lt;br /&gt;
&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td style=&amp;quot;width: 220px;&amp;quot;&amp;gt;[[File:Antihydra award.jpg|220px]]&amp;lt;br&amp;gt;A community trophy - to be awarded to the first person or group who solves the Antihydra problem&lt;br /&gt;
&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;/table&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
== Analysis ==&lt;br /&gt;
Let &amp;lt;math&amp;gt;A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;. Then,&amp;lt;ref name=&amp;quot;bl&amp;quot;&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]&amp;quot; (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|lll|}\hline&lt;br /&gt;
A(a,2b)&amp;amp; \xrightarrow{2a+3b^2+12b+11}&amp;amp; A(a+2,3b+2),\\&lt;br /&gt;
A(0,2b+1)&amp;amp;\xrightarrow{3b^2+9b-1}&amp;amp; 0^\infty\;\textrm{&amp;lt;F}\;110\;1^{3b}\;0^\infty,\\&lt;br /&gt;
A(a+1,2b+1)&amp;amp;\xrightarrow{3b^2+12b+5}&amp;amp; A(a,3b+3).\\\hline&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Proof&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
Consider the partial configuration &amp;lt;math&amp;gt;P(m,n):=0\;1^m\;\textrm{E&amp;gt;}\;0\;1^n\;0^\infty&amp;lt;/math&amp;gt;. The configuration after two steps is &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;\textrm{A&amp;gt;}\;1^{n+1}\;0^\infty&amp;lt;/math&amp;gt;. We note the following shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{A&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{A&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
As a result, we get &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;1^{n+1}\;\textrm{A&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt; after &amp;lt;math&amp;gt;n+1&amp;lt;/math&amp;gt; steps. Advancing two steps produces &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;1^{n+2}\;\textrm{&amp;lt;C}\;0^\infty&amp;lt;/math&amp;gt;. A second shift rule is useful here:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline1^s\;\textrm{&amp;lt;C}\xrightarrow{s}\textrm{&amp;lt;C}\;1^s\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
This allows us to reach &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;\textrm{&amp;lt;C}\;1^{n+2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;n+2&amp;lt;/math&amp;gt; steps. Moving five more steps gets us to &amp;lt;math&amp;gt;0\;1^{m-2}\;\textrm{E&amp;gt;}\;0\;1^{n+3}\;0^\infty&amp;lt;/math&amp;gt;, which is the same configuration as &amp;lt;math&amp;gt;P(m-2,n+3)&amp;lt;/math&amp;gt;. Accounting for the head movement creates the condition that &amp;lt;math&amp;gt;m\ge 4&amp;lt;/math&amp;gt;. In summary:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With &amp;lt;math&amp;gt;A(a,b)&amp;lt;/math&amp;gt; we have &amp;lt;math&amp;gt;P(b,0)&amp;lt;/math&amp;gt;. As a result, we can apply this rule &amp;lt;math display=&amp;quot;inline&amp;gt;\big\lfloor\frac{1}{2}b\big\rfloor-1&amp;lt;/math&amp;gt; times (assuming &amp;lt;math&amp;gt;b\ge 4&amp;lt;/math&amp;gt;), which creates two possible scenarios:&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv0\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math&amp;gt;\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(2,\frac{3}{2}b-3\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;1^a\;011\;\textrm{E&amp;gt;}\;0\;1^{(3b)/2-3}\;0^\infty&amp;lt;/math&amp;gt;. After &amp;lt;math&amp;gt;3b+4&amp;lt;/math&amp;gt; steps this is &amp;lt;math&amp;gt;0^\infty\;1^a\;\textrm{&amp;lt;C}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, which then leads to &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;C}\;1^a\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; steps. After five more steps, we reach &amp;lt;math&amp;gt;0^\infty\;1\;\textrm{E&amp;gt;}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, from which another shift rule must be applied:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{E&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{E&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;Doing so allows us to get the configuration &amp;lt;math&amp;gt;0^\infty\;1^{a+3}\;\textrm{E&amp;gt;}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;a+2&amp;lt;/math&amp;gt; steps. In six steps we have &amp;lt;math&amp;gt;0^\infty\;1^{a+2}\;011\;\textrm{E&amp;gt;}\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, so we use the shift rule again, ending at &amp;lt;math&amp;gt;0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A\Big(a+2,\frac{3}{2}b+2\Big)&amp;lt;/math&amp;gt;, &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}b&amp;lt;/math&amp;gt; steps later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;2a+\frac{3}{4}b^2+6b+11&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv1\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2-\frac{27}{4}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(3,\frac{3b-9}{2}\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;1^a\;0111\;\textrm{E&amp;gt;}\;0\;1^{(3b-9)/2}\;0^\infty&amp;lt;/math&amp;gt;. After &amp;lt;math&amp;gt;3b+2&amp;lt;/math&amp;gt; steps this becomes &amp;lt;math&amp;gt;0^\infty\;1^a\;\textrm{&amp;lt;F}\;110\;1^{(3b-3)/2}\;0^\infty&amp;lt;/math&amp;gt;. If &amp;lt;math&amp;gt;a=0&amp;lt;/math&amp;gt; then we have reached the undefined &amp;lt;code&amp;gt;F0&amp;lt;/code&amp;gt; transition with a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2+3b-\frac{19}{4}&amp;lt;/math&amp;gt; steps. Otherwise, continuing for six steps gives us &amp;lt;math&amp;gt;0^\infty\;1^{a-1}\;0111\;\textrm{E&amp;gt;}\;1^{(3b-3)/2}\;0^\infty&amp;lt;/math&amp;gt;. We conclude with the configuration &amp;lt;math&amp;gt;0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A\Big(a-1,\frac{3b+3}{2}\Big)&amp;lt;/math&amp;gt;, in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3b-3}{2}&amp;lt;/math&amp;gt; steps. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
The information above can be summarized as&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;A(a,b)\rightarrow\begin{cases}A\Big(a+2,\frac{3}{2}b+2\Big)&amp;amp;\text{if }b\ge 2,b\equiv0\pmod{2};\\0^\infty\;\textrm{&amp;lt;F}\;110\;1^{(3b-3)/2}\;0^\infty&amp;amp;\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a=0;\\A\Big(a-1,\frac{3b+3}{2}\Big)&amp;amp;\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a&amp;gt;0.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Substituting &amp;lt;math&amp;gt;b\leftarrow 2b&amp;lt;/math&amp;gt; for the first case and &amp;lt;math&amp;gt;b\leftarrow 2b+1&amp;lt;/math&amp;gt; for the other two yields the final result.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
In effect, the halting problem for Antihydra is about whether repeatedly applying the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;f(n)=\Big\lfloor\frac{3n}{2}\Big\rfloor+2&amp;lt;/math&amp;gt; will at some point produce more odd values of &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; than twice the number of even values.&lt;br /&gt;
&lt;br /&gt;
These rules can be modified to use the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;H(n)=\Big\lfloor\frac{3n}{2}\Big\rfloor&amp;lt;/math&amp;gt;, or the [[Hydra function]], which strengthens Antihydra&#039;s similarities to Hydra.&lt;br /&gt;
== Trajectory ==&lt;br /&gt;
11 steps are required to enter the configuration &amp;lt;math&amp;gt;A(0, 4)&amp;lt;/math&amp;gt; before the rules are repeatedly applied. So far, Antihydra has been simulated to &amp;lt;math&amp;gt;2^{31}&amp;lt;/math&amp;gt; rule steps, at which point &amp;lt;math&amp;gt;b=1073720884&amp;lt;/math&amp;gt;. Here are the first few:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline A(0,4)\xrightarrow{47}A(2,8)\xrightarrow{111}A(4,14)\xrightarrow{250}A(6,23)\xrightarrow{500}A(5,36)\xrightarrow{1209}A(7,56)\xrightarrow{2713}A(9,86)\rightarrow\cdots\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
The trajectory of &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; values resembles a random walk in which the walker can only move in step sizes +2 or -1 with equal probability, starting at position 0. If &amp;lt;math&amp;gt;P(n)&amp;lt;/math&amp;gt; is the probability that the walker will reach position -1 from position &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;, then it can be seen that &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)=\frac{1}{2}P(n-1)+\frac{1}{2}P(n+2)&amp;lt;/math&amp;gt;. Solutions to this recurrence relation come in the form &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt; P(n)=c_0{\left(\frac{\sqrt{5}-1}{2}\right)}^n+c_1+c_2{\left(-\frac{1+\sqrt{5}}{2}\right)}^n&amp;lt;/math&amp;gt;, which after applying the appropriate boundary conditions reduces to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)={\left(\frac{\sqrt{5}-1}{2}\right)}^{n+1}&amp;lt;/math&amp;gt;. This means that if the walker were to get to position 1073720884 then the probability of it ever reaching position -1 is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\left(\frac{\sqrt{5}-1}{2}\right)}^{1073720885}\approx 4.841\times 10^{-224394395}&amp;lt;/math&amp;gt;. This combined with the fact that the expected position of the walker after &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; steps is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{1}{2}k&amp;lt;/math&amp;gt; strongly suggests Antihydra [[probviously]] runs indefinitely.&lt;br /&gt;
&lt;br /&gt;
== Code ==&lt;br /&gt;
The following Python program implements the abstracted behavior of the Antihydra. Proving whether it halts or not would also solve the Antihydra problem:&amp;lt;syntaxhighlight lang=&amp;quot;python&amp;quot; line=&amp;quot;1&amp;quot;&amp;gt;&lt;br /&gt;
# Current value of the iterated Hydra function starting with initial value 8&lt;br /&gt;
h = 8&lt;br /&gt;
# (Collatz-like) condition counter that keeps track of how many odd and even numbers have been encountered&lt;br /&gt;
c = 0&lt;br /&gt;
# If c equals -1 there have been (strictly) more than twice as many odd as even numbers and the program halts&lt;br /&gt;
while c != -1:&lt;br /&gt;
    # If h is even, add 2 to c so even numbers count twice&lt;br /&gt;
    if h % 2 == 0:&lt;br /&gt;
        c += 2&lt;br /&gt;
    else:&lt;br /&gt;
        c -= 1&lt;br /&gt;
    # Add the current hydra value divided by two (integer division, rounding down) to itself (Hydra function)&lt;br /&gt;
    # Note that integer division by 2 is equivalent to one bit shift to the right (h &amp;gt;&amp;gt; 1)&lt;br /&gt;
    h += h//2&lt;br /&gt;
&amp;lt;/syntaxhighlight&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Fast [[Hydra]]/Antihydra simulation code by Greg Kuperberg (who said it could be made faster using FLINT):&amp;lt;syntaxhighlight lang=&amp;quot;python2&amp;quot; line=&amp;quot;1&amp;quot;&amp;gt;&lt;br /&gt;
# Python script to demonstrate almost linear time hydra simulation&lt;br /&gt;
# using fast multiplication. &lt;br /&gt;
# by Greg Kuperberg&lt;br /&gt;
&lt;br /&gt;
import time&lt;br /&gt;
from gmpy2 import mpz,bit_mask&lt;br /&gt;
&lt;br /&gt;
# Straight computation of t steps of hydra&lt;br /&gt;
def simple(n,t):&lt;br /&gt;
    for s in range(t): n += n&amp;gt;&amp;gt;1&lt;br /&gt;
    return n&lt;br /&gt;
&lt;br /&gt;
# Accelerated computation of 2**e steps of hydra&lt;br /&gt;
def hydra(n,e):&lt;br /&gt;
    if e &amp;lt; 9: return simple(n,1&amp;lt;&amp;lt;e)&lt;br /&gt;
    t = 1&amp;lt;&amp;lt;(e-1)&lt;br /&gt;
    (p3t,m) = (mpz(3)**t,bit_mask(t))&lt;br /&gt;
    n = p3t*(n&amp;gt;&amp;gt;t) + hydra(n&amp;amp;m,e-1)&lt;br /&gt;
    return p3t*(n&amp;gt;&amp;gt;t) + hydra(n&amp;amp;m,e-1)&lt;br /&gt;
&lt;br /&gt;
def elapsed():&lt;br /&gt;
    (last,elapsed.mark) = (elapsed.mark,time.process_time())&lt;br /&gt;
    return elapsed.mark-last&lt;br /&gt;
elapsed.mark = 0&lt;br /&gt;
&lt;br /&gt;
(n,e) = (mpz(3),25)&lt;br /&gt;
&lt;br /&gt;
elapsed()&lt;br /&gt;
print(&#039;hydra:  steps=%d hash=%016x time=%.6fs&#039;&lt;br /&gt;
    % (1&amp;lt;&amp;lt;e,hash(hydra(n,e)),elapsed()))&lt;br /&gt;
&lt;br /&gt;
# Quadratic time algorithm for comparison&lt;br /&gt;
# print(&#039;simple: steps=%d hash=%016x time=%.6fs&#039;&lt;br /&gt;
#     % (1&amp;lt;&amp;lt;e,hash(simple(n,1&amp;lt;&amp;lt;e)),elapsed()))&lt;br /&gt;
&amp;lt;/syntaxhighlight&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[[Category:Individual machines]][[Category:Cryptids]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Antihydra&amp;diff=2337</id>
		<title>Antihydra</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Antihydra&amp;diff=2337"/>
		<updated>2025-07-03T10:40:49Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* Code */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA}}{{unsolved|Does Antihydra run forever?}}{{TM|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA|undecided}}&lt;br /&gt;
&#039;&#039;&#039;Antihydra&#039;&#039;&#039; is a [[BB(6)]] [[Cryptid]]. Its pseudo-random behaviour was first reported [https://discord.com/channels/960643023006490684/1026577255754903572/1256223215206924318 on Discord] by mxdys on 28 June 2024, and Racheline discovered the high-level rules soon after. It was named after the 2-state, 5-symbol [[Turing machine]] called [[Hydra]] for sharing many similarities to it.&amp;lt;ref&amp;gt;[https://discord.com/channels/960643023006490684/960643023530762341/1257053002859286701 Discord conversation where the machine was named]&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Antihydra is known to not generate Sturmian words&amp;lt;ref&amp;gt;DUBICKAS A. ON INTEGER SEQUENCES GENERATED BY LINEAR MAPS. &#039;&#039;Glasgow Mathematical Journal&#039;&#039;. 2009;51(2):243-252. doi:[https://doi.org/10.1017/S0017089508004655 10.1017/S0017089508004655]&amp;lt;/ref&amp;gt; (Corollary 4).&lt;br /&gt;
&amp;lt;table style=&amp;quot;margin: auto; text-align: center;&amp;quot;&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td style=&amp;quot;width: 200px;&amp;quot;&amp;gt;[[File:Antihydra-depiction.png|200px]]&amp;lt;br&amp;gt;Artistic depiction of Antihydra by Jadeix&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&lt;br /&gt;
{|class=&amp;quot;wikitable&amp;quot; style=&amp;quot;margin-left: auto; margin-right: auto;&amp;quot;&lt;br /&gt;
! !!0!!1&lt;br /&gt;
|-&lt;br /&gt;
!A&lt;br /&gt;
|1RB&lt;br /&gt;
|1RA&lt;br /&gt;
|-&lt;br /&gt;
!B&lt;br /&gt;
|0LC&lt;br /&gt;
|1LE&lt;br /&gt;
|-&lt;br /&gt;
!C&lt;br /&gt;
|1LD&lt;br /&gt;
|1LC&lt;br /&gt;
|-&lt;br /&gt;
!D&lt;br /&gt;
|1LA&lt;br /&gt;
|0LB&lt;br /&gt;
|-&lt;br /&gt;
!E&lt;br /&gt;
|1LF&lt;br /&gt;
|1RE&lt;br /&gt;
|-&lt;br /&gt;
!F&lt;br /&gt;
| ---&lt;br /&gt;
|0RA&lt;br /&gt;
|}&lt;br /&gt;
The transition table of Antihydra.&lt;br /&gt;
&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td style=&amp;quot;width: 220px;&amp;quot;&amp;gt;[[File:Antihydra award.jpg|220px]]&amp;lt;br&amp;gt;A community trophy - to be awarded to the first person or group who solves the Antihydra problem&lt;br /&gt;
&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;/table&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
== Analysis ==&lt;br /&gt;
Let &amp;lt;math&amp;gt;A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;. Then,&amp;lt;ref name=&amp;quot;bl&amp;quot;&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]&amp;quot; (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|lll|}\hline&lt;br /&gt;
A(a,2b)&amp;amp; \xrightarrow{2a+3b^2+12b+11}&amp;amp; A(a+2,3b+2),\\&lt;br /&gt;
A(0,2b+1)&amp;amp;\xrightarrow{3b^2+9b-1}&amp;amp; 0^\infty\;\textrm{&amp;lt;F}\;110\;1^{3b}\;0^\infty,\\&lt;br /&gt;
A(a+1,2b+1)&amp;amp;\xrightarrow{3b^2+12b+5}&amp;amp; A(a,3b+3).\\\hline&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Proof&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
Consider the partial configuration &amp;lt;math&amp;gt;P(m,n):=0\;1^m\;\textrm{E&amp;gt;}\;0\;1^n\;0^\infty&amp;lt;/math&amp;gt;. The configuration after two steps is &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;\textrm{A&amp;gt;}\;1^{n+1}\;0^\infty&amp;lt;/math&amp;gt;. We note the following shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{A&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{A&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
As a result, we get &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;1^{n+1}\;\textrm{A&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt; after &amp;lt;math&amp;gt;n+1&amp;lt;/math&amp;gt; steps. Advancing two steps produces &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;1^{n+2}\;\textrm{&amp;lt;C}\;0^\infty&amp;lt;/math&amp;gt;. A second shift rule is useful here:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline1^s\;\textrm{&amp;lt;C}\xrightarrow{s}\textrm{&amp;lt;C}\;1^s\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
This allows us to reach &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;\textrm{&amp;lt;C}\;1^{n+2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;n+2&amp;lt;/math&amp;gt; steps. Moving five more steps gets us to &amp;lt;math&amp;gt;0\;1^{m-2}\;\textrm{E&amp;gt;}\;0\;1^{n+3}\;0^\infty&amp;lt;/math&amp;gt;, which is the same configuration as &amp;lt;math&amp;gt;P(m-2,n+3)&amp;lt;/math&amp;gt;. Accounting for the head movement creates the condition that &amp;lt;math&amp;gt;m\ge 4&amp;lt;/math&amp;gt;. In summary:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With &amp;lt;math&amp;gt;A(a,b)&amp;lt;/math&amp;gt; we have &amp;lt;math&amp;gt;P(b,0)&amp;lt;/math&amp;gt;. As a result, we can apply this rule &amp;lt;math display=&amp;quot;inline&amp;gt;\big\lfloor\frac{1}{2}b\big\rfloor-1&amp;lt;/math&amp;gt; times (assuming &amp;lt;math&amp;gt;b\ge 4&amp;lt;/math&amp;gt;), which creates two possible scenarios:&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv0\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math&amp;gt;\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(2,\frac{3}{2}b-3\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;1^a\;011\;\textrm{E&amp;gt;}\;0\;1^{(3b)/2-3}\;0^\infty&amp;lt;/math&amp;gt;. After &amp;lt;math&amp;gt;3b+4&amp;lt;/math&amp;gt; steps this is &amp;lt;math&amp;gt;0^\infty\;1^a\;\textrm{&amp;lt;C}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, which then leads to &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;C}\;1^a\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; steps. After five more steps, we reach &amp;lt;math&amp;gt;0^\infty\;1\;\textrm{E&amp;gt;}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, from which another shift rule must be applied:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{E&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{E&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;Doing so allows us to get the configuration &amp;lt;math&amp;gt;0^\infty\;1^{a+3}\;\textrm{E&amp;gt;}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;a+2&amp;lt;/math&amp;gt; steps. In six steps we have &amp;lt;math&amp;gt;0^\infty\;1^{a+2}\;011\;\textrm{E&amp;gt;}\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, so we use the shift rule again, ending at &amp;lt;math&amp;gt;0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A\Big(a+2,\frac{3}{2}b+2\Big)&amp;lt;/math&amp;gt;, &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}b&amp;lt;/math&amp;gt; steps later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;2a+\frac{3}{4}b^2+6b+11&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv1\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2-\frac{27}{4}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(3,\frac{3b-9}{2}\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;1^a\;0111\;\textrm{E&amp;gt;}\;0\;1^{(3b-9)/2}\;0^\infty&amp;lt;/math&amp;gt;. After &amp;lt;math&amp;gt;3b+2&amp;lt;/math&amp;gt; steps this becomes &amp;lt;math&amp;gt;0^\infty\;1^a\;\textrm{&amp;lt;F}\;110\;1^{(3b-3)/2}\;0^\infty&amp;lt;/math&amp;gt;. If &amp;lt;math&amp;gt;a=0&amp;lt;/math&amp;gt; then we have reached the undefined &amp;lt;code&amp;gt;F0&amp;lt;/code&amp;gt; transition with a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2+3b-\frac{19}{4}&amp;lt;/math&amp;gt; steps. Otherwise, continuing for six steps gives us &amp;lt;math&amp;gt;0^\infty\;1^{a-1}\;0111\;\textrm{E&amp;gt;}\;1^{(3b-3)/2}\;0^\infty&amp;lt;/math&amp;gt;. We conclude with the configuration &amp;lt;math&amp;gt;0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A\Big(a-1,\frac{3b+3}{2}\Big)&amp;lt;/math&amp;gt;, in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3b-3}{2}&amp;lt;/math&amp;gt; steps. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
The information above can be summarized as&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;A(a,b)\rightarrow\begin{cases}A\Big(a+2,\frac{3}{2}b+2\Big)&amp;amp;\text{if }b\ge 2,b\equiv0\pmod{2};\\0^\infty\;\textrm{&amp;lt;F}\;110\;1^{(3b-3)/2}\;0^\infty&amp;amp;\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a=0;\\A\Big(a-1,\frac{3b+3}{2}\Big)&amp;amp;\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a&amp;gt;0.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Substituting &amp;lt;math&amp;gt;b\leftarrow 2b&amp;lt;/math&amp;gt; for the first case and &amp;lt;math&amp;gt;b\leftarrow 2b+1&amp;lt;/math&amp;gt; for the other two yields the final result.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
In effect, the halting problem for Antihydra is about whether repeatedly applying the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;f(n)=\Big\lfloor\frac{3n}{2}\Big\rfloor+2&amp;lt;/math&amp;gt; will at some point produce more odd values of &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; than twice the number of even values.&lt;br /&gt;
&lt;br /&gt;
These rules can be modified to use the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;H(n)=\Big\lfloor\frac{3n}{2}\Big\rfloor&amp;lt;/math&amp;gt;, or the [[Hydra function]], which strengthens Antihydra&#039;s similarities to Hydra.&lt;br /&gt;
== Trajectory ==&lt;br /&gt;
11 steps are required to enter the configuration &amp;lt;math&amp;gt;A(0, 4)&amp;lt;/math&amp;gt; before the rules are repeatedly applied. So far, Antihydra has been simulated to &amp;lt;math&amp;gt;2^{31}&amp;lt;/math&amp;gt; rule steps, at which point &amp;lt;math&amp;gt;b=1073720884&amp;lt;/math&amp;gt;. Here are the first few:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline A(0,4)\xrightarrow{47}A(2,8)\xrightarrow{111}A(4,14)\xrightarrow{250}A(6,23)\xrightarrow{500}A(5,36)\xrightarrow{1209}A(7,56)\xrightarrow{2713}A(9,86)\rightarrow\cdots\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
The trajectory of &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; values resembles a random walk in which the walker can only move in step sizes +2 or -1 with equal probability, starting at position 0. If &amp;lt;math&amp;gt;P(n)&amp;lt;/math&amp;gt; is the probability that the walker will reach position -1 from position &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;, then it can be seen that &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)=\frac{1}{2}P(n-1)+\frac{1}{2}P(n+2)&amp;lt;/math&amp;gt;. Solutions to this recurrence relation come in the form &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt; P(n)=c_0{\left(\frac{\sqrt{5}-1}{2}\right)}^n+c_1+c_2{\left(-\frac{1+\sqrt{5}}{2}\right)}^n&amp;lt;/math&amp;gt;, which after applying the appropriate boundary conditions reduces to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)={\left(\frac{\sqrt{5}-1}{2}\right)}^{n+1}&amp;lt;/math&amp;gt;. This means that if the walker were to get to position 1073720884 then the probability of it ever reaching position -1 is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\left(\frac{\sqrt{5}-1}{2}\right)}^{1073720885}\approx 4.841\times 10^{-224394395}&amp;lt;/math&amp;gt;. This combined with the fact that the expected position of the walker after &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; steps is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{1}{2}k&amp;lt;/math&amp;gt; strongly suggests Antihydra [[probviously]] runs indefinitely.&lt;br /&gt;
&lt;br /&gt;
== Code ==&lt;br /&gt;
The following Python program implements the abstracted behavior of the Antihydra. Proving whether it halts or not would also solve the Antihydra problem:&amp;lt;syntaxhighlight lang=&amp;quot;python&amp;quot; line=&amp;quot;1&amp;quot;&amp;gt;&lt;br /&gt;
# Current value of the iterated Hydra function starting with initial value 8&lt;br /&gt;
h = 8&lt;br /&gt;
# (Collatz-like) condition counter that keeps track of how many odd and even numbers have been encountered&lt;br /&gt;
c = 0&lt;br /&gt;
# If c equals -1 there have been (strictly) more than twice as many odd as even numbers and the program halts&lt;br /&gt;
while c != -1:&lt;br /&gt;
    # If h is even, add 2 to c so even numbers count twice&lt;br /&gt;
    if h % 2 == 0:&lt;br /&gt;
        c += 2&lt;br /&gt;
    else:&lt;br /&gt;
        c -= 1&lt;br /&gt;
    # Add the current hydra value divided by two (integer division, rounding down) to itself (Hydra function)&lt;br /&gt;
    # Note that integer division by 2 is equivalent to one bit shift to the right (h &amp;gt;&amp;gt; 1)&lt;br /&gt;
    h += h//2&lt;br /&gt;
&amp;lt;/syntaxhighlight&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Fast [[Hydra]]/Antihydra simulation code by Greg Kuperberg (who said it could be made faster using FLINT):&amp;lt;syntaxhighlight lang=&amp;quot;python2&amp;quot; line=&amp;quot;1&amp;quot;&amp;gt;&lt;br /&gt;
# Python script to demonstrate almost linear time hydra simulation&lt;br /&gt;
# using fast multiplication. &lt;br /&gt;
# by Greg Kuperberg&lt;br /&gt;
&lt;br /&gt;
import time&lt;br /&gt;
from gmpy2 import mpz,bit_mask&lt;br /&gt;
&lt;br /&gt;
# Straight computation of t steps of hydra&lt;br /&gt;
def simple(n,t):&lt;br /&gt;
    for s in range(t): n += n&amp;gt;&amp;gt;1&lt;br /&gt;
    return n&lt;br /&gt;
&lt;br /&gt;
# Accelerated computation of 2**e steps of hydra&lt;br /&gt;
def hydra(n,e):&lt;br /&gt;
    if e &amp;lt; 9: return simple(n,1&amp;lt;&amp;lt;e)&lt;br /&gt;
    t = 1&amp;lt;&amp;lt;(e-1)&lt;br /&gt;
    (p3t,m) = (mpz(3)**t,bit_mask(t))&lt;br /&gt;
    n = p3t*(n&amp;gt;&amp;gt;t) + hydra(n&amp;amp;m,e-1)&lt;br /&gt;
    return p3t*(n&amp;gt;&amp;gt;t) + hydra(n&amp;amp;m,e-1)&lt;br /&gt;
&lt;br /&gt;
def elapsed():&lt;br /&gt;
    (last,elapsed.mark) = (elapsed.mark,time.process_time())&lt;br /&gt;
    return elapsed.mark-last&lt;br /&gt;
elapsed.mark = 0&lt;br /&gt;
&lt;br /&gt;
(n,e) = (mpz(3),25)&lt;br /&gt;
&lt;br /&gt;
elapsed()&lt;br /&gt;
print(&#039;hydra:  steps=%d hash=%016x time=%.6fs&#039;&lt;br /&gt;
    % (1&amp;lt;&amp;lt;e,hash(hydra(n,e)),elapsed()))&lt;br /&gt;
&lt;br /&gt;
# Quadratic time algorithm for comparison&lt;br /&gt;
# print(&#039;simple: steps=%d hash=%016x time=%.6fs&#039;&lt;br /&gt;
#     % (1&amp;lt;&amp;lt;e,hash(simple(n,1&amp;lt;&amp;lt;e)),elapsed()))&lt;br /&gt;
&amp;lt;/syntaxhighlight&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[[Category:Individual machines]][[Category:Cryptids]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2292</id>
		<title>Main Page</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2292"/>
		<updated>2025-06-27T20:14:48Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: Reverted edits by Xl643 (talk) to last revision by Mxdys&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;The [[Busy Beaver function]] BB (called &#039;&#039;S&#039;&#039; originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to &#039;&#039;m&#039;&#039;-symbol Turing machines:&amp;lt;ref&amp;gt;Rado, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
| BB(&#039;&#039;n&#039;&#039;, &#039;&#039;m&#039;&#039;) = Maximum number of steps taken by a halting &#039;&#039;n&#039;&#039;-state, &#039;&#039;m&#039;&#039;-symbol Turing machine starting from a blank (all 0) tape&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
The 2-symbol case BB(&#039;&#039;n&#039;&#039;, 2) is abbreviated as BB(&#039;&#039;n&#039;&#039;). The busy beaver function is not computable, but a few of its values are known:&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+ Small busy beaver values&amp;lt;ref&amp;gt;P. Michel, &amp;quot;[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]&amp;quot;.&amp;lt;/ref&amp;gt;&lt;br /&gt;
! !!2-state!!3-state !!4-state!!5-state!!6-state &lt;br /&gt;
!7-state&lt;br /&gt;
|-  &lt;br /&gt;
! 2-symbol &lt;br /&gt;
| [[BB(2)]] = 6 &lt;br /&gt;
| [[BB(3)]] = 21&lt;br /&gt;
| [[BB(4)]] = 107 &lt;br /&gt;
| [[BB(5)]] = 47,176,870 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(6)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow \uparrow \uparrow 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(7)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{11} 2 \uparrow^{11} 3&amp;lt;/math&amp;gt; &lt;br /&gt;
|-&lt;br /&gt;
! 3-symbol&lt;br /&gt;
| [[BB(2,3)]] = 38 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(3,3)]] &amp;gt; &amp;lt;math&amp;gt;10^{17}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(4,3)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow\uparrow\uparrow 2^{2^{32}}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 4-symbol  &lt;br /&gt;
| [[BB(2,4)]] = 3,932,964&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,4)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{15} 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 5-symbol &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(2,5)]] &amp;gt; &amp;lt;math&amp;gt;10\uparrow\uparrow 4&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 6-symbol &lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(2,6)]] &amp;gt; &amp;lt;math&amp;gt;10 \uparrow\uparrow\uparrow 3&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
In the above table, &amp;lt;span style=&amp;quot;background: orange&amp;quot;&amp;gt;cells are highlighted in orange&amp;lt;/span&amp;gt; when there are known [[Cryptids]] (mathematically-hard machines) in that class, and &amp;lt;span style=&amp;quot;background: #ffe4b2&amp;quot;&amp;gt;cells are highlighted in light orange&amp;lt;/span&amp;gt; when the existence of a Cryptid is given by using a known one with less states or symbols.&lt;br /&gt;
&lt;br /&gt;
== About bbchallenge ==&lt;br /&gt;
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt.  Other efforts also include:&lt;br /&gt;
&lt;br /&gt;
* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Coq_(software) Coq])&lt;br /&gt;
* Maintaining [[Holdouts lists]] for small busy beaver values&lt;br /&gt;
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]&lt;br /&gt;
* Finding [[Cryptids]] (mathematically-hard machines)&lt;br /&gt;
* Searching for new [[Champions]]&lt;br /&gt;
* Building [[Accelerated Simulator]]s to simulate halting machines faster&lt;br /&gt;
* Writing papers and giving talks about busy beaver, see [[Papers &amp;amp; Talks]]&lt;br /&gt;
&lt;br /&gt;
In June 2024, bbchallenge achieved a significant milestone by proving in Coq / Rocq that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.&amp;lt;ref&amp;gt;H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.&lt;br /&gt;
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Contribute to this wiki ==&lt;br /&gt;
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;inputbox&amp;gt;&lt;br /&gt;
type=create&lt;br /&gt;
width=100&lt;br /&gt;
break=no&lt;br /&gt;
buttonlabel=Create new article&lt;br /&gt;
default=(Article title)&lt;br /&gt;
&amp;lt;/inputbox&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Notes==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2173</id>
		<title>Main Page</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Main_Page&amp;diff=2173"/>
		<updated>2025-06-10T22:18:01Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;The [[Busy Beaver function]] BB (called &#039;&#039;S&#039;&#039; originally) was introduced by [https://en.wikipedia.org/wiki/Tibor_Rad%C3%B3 Tibor Radó] in 1962 for 2-symbol [[Turing machines]] and later generalised to &#039;&#039;m&#039;&#039;-symbol Turing machines:&amp;lt;ref&amp;gt;Rado, T. (1962), On Non-Computable Functions. Bell System Technical Journal, 41: 877-884. https://doi.org/10.1002/j.1538-7305.1962.tb00480.x&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198537748.003.0009, accessed 8 June 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
| BB(&#039;&#039;n&#039;&#039;, &#039;&#039;m&#039;&#039;) = Maximum number of steps taken by a halting &#039;&#039;n&#039;&#039;-state, &#039;&#039;m&#039;&#039;-symbol Turing machine starting from a blank (all 0) tape&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
The 2-symbol case BB(&#039;&#039;n&#039;&#039;, 2) is abbreviated as BB(&#039;&#039;n&#039;&#039;). The busy beaver function is not computable, but a few of its values are known:&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|+ Small busy beaver values&amp;lt;ref&amp;gt;P. Michel, &amp;quot;[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]&amp;quot;.&amp;lt;/ref&amp;gt;&lt;br /&gt;
! !!2-state!!3-state !!4-state!!5-state!!6-state &lt;br /&gt;
!7-state&lt;br /&gt;
|-  &lt;br /&gt;
! 2-symbol &lt;br /&gt;
| [[BB(2)]] = 6 &lt;br /&gt;
| [[BB(3)]] = 21&lt;br /&gt;
| [[BB(4)]] = 107 &lt;br /&gt;
| [[BB(5)]] = 47,176,870 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(6)]] &amp;gt; &amp;lt;math&amp;gt;10 \uparrow \uparrow 15&amp;lt;/math&amp;gt; &lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(7)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{11} 2 \uparrow^{11} 3&amp;lt;/math&amp;gt; &lt;br /&gt;
|-&lt;br /&gt;
! 3-symbol&lt;br /&gt;
| [[BB(2,3)]] = 38 &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(3,3)]] &amp;gt; &amp;lt;math&amp;gt;10^{17}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(4,3)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow\uparrow\uparrow 2^{2^{32}}&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 4-symbol  &lt;br /&gt;
| [[BB(2,4)]] = 3,932,964&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(3,4)]] &amp;gt; &amp;lt;math&amp;gt;2 \uparrow^{15} 5&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 5-symbol &lt;br /&gt;
| style=&amp;quot;background: orange;&amp;quot; | [[BB(2,5)]] &amp;gt; &amp;lt;math&amp;gt;10\uparrow\uparrow 4&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|-&lt;br /&gt;
! 6-symbol &lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; | [[BB(2,6)]] &amp;gt; &amp;lt;math&amp;gt;10 \uparrow\uparrow\uparrow 3&amp;lt;/math&amp;gt;&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
| style=&amp;quot;background: #ffe4b2;&amp;quot; |&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
In the above table, &amp;lt;span style=&amp;quot;background: orange&amp;quot;&amp;gt;cells are highlighted in orange&amp;lt;/span&amp;gt; when there are known [[Cryptids]] (mathematically-hard machines) in that class, and &amp;lt;span style=&amp;quot;background: #ffe4b2&amp;quot;&amp;gt;cells are highlighted in light orange&amp;lt;/span&amp;gt; when the existence of a Cryptid is given by using a known one with less states or symbols.&lt;br /&gt;
&lt;br /&gt;
== About bbchallenge ==&lt;br /&gt;
[https://www.bbchallenge.org bbchallenge] is a massively collaborative research project whose general goal is to obtain more knowledge on the [[Busy Beaver function]]. In practice, it mainly consists in collaboratively building [[Deciders]], programs that automatically prove that some Turing machines do not halt.  Other efforts also include:&lt;br /&gt;
&lt;br /&gt;
* Formalising results using theorem provers (such as [https://en.wikipedia.org/wiki/Coq_(software) Coq])&lt;br /&gt;
* Maintaining [[Holdouts lists]] for small busy beaver values&lt;br /&gt;
* Proving the behavior of [[:Category:Individual Machines|Individual machines]]&lt;br /&gt;
* Finding [[Cryptids]] (mathematically-hard machines)&lt;br /&gt;
* Searching for new [[Champions]]&lt;br /&gt;
* Building [[Accelerated Simulator]]s to simulate halting machines faster&lt;br /&gt;
* Writing papers and giving talks about busy beaver, see [[Papers &amp;amp; Talks]]&lt;br /&gt;
&lt;br /&gt;
In June 2024, bbchallenge achieved a significant milestone by proving in Coq / Rocq that the 5th busy beaver value, [[BB(5)]], is equal to the lower bound found in 1989: 47,176,870.&amp;lt;ref&amp;gt;H. Marxen and J. Buntrock. Attacking the Busy Beaver 5.&lt;br /&gt;
Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Contribute to this wiki ==&lt;br /&gt;
This wiki is collaborative, feel free to contribute by editing existing pages or creating new ones:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;inputbox&amp;gt;&lt;br /&gt;
type=create&lt;br /&gt;
width=100&lt;br /&gt;
break=no&lt;br /&gt;
buttonlabel=Create new article&lt;br /&gt;
default=(Article title)&lt;br /&gt;
&amp;lt;/inputbox&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Notes==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Bigfoot&amp;diff=1915</id>
		<title>Bigfoot</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Bigfoot&amp;diff=1915"/>
		<updated>2025-05-16T19:45:38Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB2RA1LC_2LC1RB2RB_---2LA1LA}}{{TM|1RB2RA1LC_2LC1RB2RB_---2LA1LA}}{{unsolved|Does Bigfoot run forever?}}&lt;br /&gt;
&#039;&#039;&#039;Bigfoot&#039;&#039;&#039; is a [[BB(3,3)]] [[Cryptids|Cryptid]]. Its low-level behaviour was first shared [https://discord.com/channels/960643023006490684/1084047886494470185/1163168233445130270 over Discord] by savask on 14 Oct 2023, and within two days, Shawn Ligocki described the high-level rules shown below, whose attributes inspired the [[Turing machine|Turing machine&#039;s]] name.&amp;lt;ref name=&amp;quot;b&amp;quot;&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2023/10/16/bb-3-3-is-hard.html BB(3, 3) is Hard (Bigfoot)] (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;div style=&amp;quot;width: fit-content; text-align: center; margin-left: auto; margin-right: auto;&amp;quot;&amp;gt;&lt;br /&gt;
{|class=&amp;quot;wikitable&amp;quot; style=&amp;quot;margin-left: auto; margin-right: auto;&amp;quot;&lt;br /&gt;
! !!0!!1!!2&lt;br /&gt;
|-&lt;br /&gt;
!A&lt;br /&gt;
|1RB&lt;br /&gt;
|2RA&lt;br /&gt;
|1LC&lt;br /&gt;
|-&lt;br /&gt;
!B&lt;br /&gt;
|2LC&lt;br /&gt;
|1RB&lt;br /&gt;
|2RB&lt;br /&gt;
|-align=&amp;quot;center&amp;quot;&lt;br /&gt;
!C&lt;br /&gt;
| ---&lt;br /&gt;
|2LA&lt;br /&gt;
|1LA&lt;br /&gt;
|}&lt;br /&gt;
The transition table of Bigfoot.&amp;lt;/div&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Bigfoot was also [https://github.com/sligocki/sligocki.github.io/issues/8#issuecomment-2140887228 successfully compiled] to a 7-state 2-symbol machine by Iijil: {{TM|0RB1RB_1LC0RA_1RE1LF_1LF1RE_0RD1RD_1LG0LG_---1LB}} in May of 2024.&lt;br /&gt;
&lt;br /&gt;
== Analysis ==&lt;br /&gt;
Let &amp;lt;math&amp;gt;A(a,b,c):=0^\infty\;12^a\;1^{2b}\;\textrm{&amp;lt;A}\;1^{2c}\;0^\infty&amp;lt;/math&amp;gt;. Then,&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l l l|}\hline A(a,6b,c)&amp;amp;\xrightarrow{4a+(2b+1)48b+(12b+1)2c+13}&amp;amp;A(a,8b+c-1,2),\\A(a,6b+1,c)&amp;amp;\xrightarrow{4a+(6b+5)16b+(4b+1)6c+29}&amp;amp;A(a+1,8b+c-1,3),\\A(0,6b+2,c)&amp;amp;\xrightarrow{(b+1)96b+(3b+1)8c+18}&amp;amp;0^\infty\;\textrm{&amp;lt;C}\;1^{16b+2c+5}\;2\;0^\infty,\\A(a,6b+2,c)&amp;amp;\xrightarrow{4a+(2b+3)48b+(12b+7)2c+51}&amp;amp;A(a-1,8b+c+3,2)\text{ if }a\ge1,\\A(a,6b+3,c)&amp;amp;\xrightarrow{4a+(6b+7)16b+(12b+5)2c+91}&amp;amp;A(a,8b+c+1,5),\\A(a,6b+4,c)&amp;amp;\xrightarrow{4a+(2b+3)48b+(12b+7)2c+63}&amp;amp;A(a+1,8b+c+3,2),\\A(a,6b+5,c)&amp;amp;\xrightarrow{4a+(6b+11)16b+(4b+3)6c+103}&amp;amp;A(a,8b+c+5,3).\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Proof&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
For now, we will work with the slightly different configuration &amp;lt;math&amp;gt;A&#039;(a,b,c):=0^\infty\;12^a\;1^b\;\textrm{&amp;lt;A}\;1^c\;0^\infty&amp;lt;/math&amp;gt;. Consider the partial configuration &amp;lt;math&amp;gt;P(m,n):=1^m\;\textrm{&amp;lt;A}\;1^n\;0^\infty&amp;lt;/math&amp;gt;. We first require the following shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline\textrm{A&amp;gt;}\;1^s\xrightarrow{s}2^s\;\textrm{A&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
Using this shift rule, we get &amp;lt;math&amp;gt;1^{m-1}\;2^{n+1}\;\textrm{A&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt; after &amp;lt;math&amp;gt;n+1&amp;lt;/math&amp;gt; steps, followed by &amp;lt;math&amp;gt;1^{m-1}\;2^n\;\textrm{&amp;lt;C}\;122\;0^\infty&amp;lt;/math&amp;gt; four steps later. Observing that &amp;lt;math&amp;gt;22\;\textrm{&amp;lt;C}&amp;lt;/math&amp;gt; becomes &amp;lt;math&amp;gt;\textrm{&amp;lt;C}\;11&amp;lt;/math&amp;gt; in two steps leads to another shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline2^{2s}\;\textrm{&amp;lt;C}\xrightarrow{2s}\textrm{&amp;lt;C}\;1^{2s}\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
From here, there are two different scenarios depending on if &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; is even or odd, given below as histories of transitions that use the aforementioned shift rules:&lt;br /&gt;
#If &amp;lt;math&amp;gt;n\equiv0\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then what follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline 1^{m-1}\;2^n\;\textrm{&amp;lt;C}\;122\;0^\infty \xrightarrow{n} 1^{m-1}\;\textrm{&amp;lt;C}\;1^{n+1}\;22\;0^\infty\xrightarrow{4}1^{m-3}\;\textrm{&amp;lt;A}\;1^{n+3}\;22\;0^\infty\xrightarrow{n+4}\\&lt;br /&gt;
1^{m-4}\;2^{n+4}\;\textrm{A&amp;gt;}\;22\;0^\infty\xrightarrow{1}1^{m-4}\;2^{n+4}\;\textrm{&amp;lt;C}\;12\;0^\infty\xrightarrow{n+4}1^{m-4}\;\textrm{&amp;lt;C}\;1^{n+5}\;2\;0^\infty\xrightarrow{4}\\1^{m-6}\;\textrm{&amp;lt;A}\;1^{n+7}\;2\;0^\infty\xrightarrow{n+8}1^{m-7}\;2^{n+8}\;\textrm{A&amp;gt;}\;2\;0^\infty\xrightarrow{1}1^{m-7}\;2^{n+8}\;\textrm{&amp;lt;C}\;1\;0^\infty\xrightarrow{n+8}\\1^{m-7}\;\textrm{&amp;lt;C}\;1^{n+9}\;0^\infty\xrightarrow{4}1^{m-9}\;\textrm{&amp;lt;A}\;1^{n+11}\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;Therefore, we have&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline P(m,n)\xrightarrow{6n+43}P(m-9,n+11)\text{ if }m\ge9\text{ and }n\equiv0\ (\operatorname{mod}2).\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
# If &amp;lt;math&amp;gt;n\equiv1\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then what follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline 1^{m-1}\;2^n\;\textrm{&amp;lt;C}\;122\;0^\infty \xrightarrow{n-1} 1^{m-1}\;2\;\textrm{&amp;lt;C}\;1^n\;22\;0^\infty \xrightarrow{1}1^{m-1}\;\textrm{&amp;lt;A}\;1^{n+1}\;22\;0^\infty \xrightarrow{n+2}\\ 1^{m-2}\;2^{n+2}\;\textrm{A&amp;gt;}\;22\;0^\infty \xrightarrow{1} 1^{m-2}\;2^{n+2}\;\textrm{&amp;lt;C}\;12\;0^\infty \xrightarrow{n+1}1^{m-2}\;2\;\textrm{&amp;lt;C}\;1^{n+2}\;2\;0^\infty \xrightarrow{1}\\ 1^{m-2}\;\textrm{&amp;lt;A}\;1^{n+3}\;2\;0^\infty \xrightarrow{n+4} 1^{m-3}\;2^{n+4}\;\textrm{A&amp;gt;}\;2\;0^\infty\xrightarrow{1}1^{m-3}\;2^{n+4}\;\textrm{&amp;lt;C}\;1\;0^\infty\xrightarrow{n+3}\\1^{m-3}\;2\;\textrm{&amp;lt;C}\;1^{n+4}\;0^\infty\xrightarrow{1}1^{m-3}\;\textrm{&amp;lt;A}\;1^{n+5}\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;Therefore, we have&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline P(m,n)\xrightarrow{6n+19}P(m-3,n+5)\text{ if }m\ge3\text{ and }n\equiv1\ (\operatorname{mod}2).\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
From this we know that Bigfoot&#039;s behaviour depends on the value of &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; modulo 12, and with &amp;lt;math&amp;gt;A&#039;(a,b,c)&amp;lt;/math&amp;gt; we have &amp;lt;math&amp;gt;P(b,c)&amp;lt;/math&amp;gt;. The following shift rules will be useful:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|l|l|}\hline12^s\;\textrm{&amp;lt;A}\xrightarrow{2s}\textrm{&amp;lt;A}\;21^s&amp;amp;\textrm{B&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{B&amp;gt;}&amp;amp;\textrm{B&amp;gt;}\;2^s\xrightarrow{s}2^s\;\textrm{B&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
Only even values of &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;c&amp;lt;/math&amp;gt; are relevant, so there remain six possible scenarios:&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv0\ (\operatorname{mod}12)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\displaystyle\sum_{i=0}^{b/12-1}}(6(16i+c)+43+6(16i+11+c)+19)={\displaystyle\sum_{i=0}^{b/12-1}}4(48i+3c+32)=\frac{2}{3}b^2+\frac{8}{3}b+bc&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(0,16\times\frac{b}{12}+c\Big)&amp;lt;/math&amp;gt;, or &amp;lt;math&amp;gt;0^\infty\;12^a\;\textrm{&amp;lt;A}\;1^{4b/3+c}\;0^\infty&amp;lt;/math&amp;gt; when considering the complete configuration. What follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline0^\infty\;12^a\;\textrm{&amp;lt;A}\;1^{4b/3+c}\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{&amp;lt;A}\;21^a\;1^{4b/3+c}\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B&amp;gt;}\;21^a\;1^{4b/3+c}\;0^\infty\xrightarrow{2a+4b/3+c}\\0^\infty\;12^a\;1^{4b/3+c+1}\;\textrm{B&amp;gt;}\;0^\infty\xrightarrow{12}0^\infty\;12^a\;1^{4b/3+c-2}\;\textrm{&amp;lt;A}\;1^4\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;This means that if &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{4}{3}b+c\ge2&amp;lt;/math&amp;gt;, then we will reach &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A&#039;\Big(a,\frac{4}{3}b+c-2,4\Big)&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4a+\frac{2}{3}b^2+4b+bc+c+13&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv2\ (\operatorname{mod}12)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\displaystyle\sum_{i=0}^{(b-2)/12-1}}4(48i+3c+32)=\frac{2}{3}b^2+bc-2c-\frac{8}{3}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(2,\frac{4(b-2)}{3}+c\Big)&amp;lt;/math&amp;gt;, or &amp;lt;math&amp;gt;0^\infty\;12^a\;11\;\textrm{&amp;lt;A}\;1^{(4b-2)/3+c}\;0^\infty&amp;lt;/math&amp;gt;. What follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline0^\infty\;12^a\;11\;\textrm{&amp;lt;A}\;1^{4(b-2)/3+c}\;0^\infty\xrightarrow{4(b-2)/3+c+1}0^\infty\;12^a\;1\;2^{4(b-2)/3+c+1}\;\textrm{A&amp;gt;}\;0^\infty\xrightarrow{4}\\0^\infty\;12^a\;1\;2^{4(b-2)/3+c}\;\textrm{&amp;lt;C}\;122\;0^\infty\xrightarrow{4(b-2)/3+c}0^\infty\;12^a\;1\;\textrm{&amp;lt;C}\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;12^a\;\textrm{&amp;lt;A}\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{&amp;lt;A}\;21^a\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B&amp;gt;}\;21^a\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{4(b-2)/3+2a+c+4}0^\infty\;12^{a+1}\;1^{4(b-2)/3+c+1}\;22\;\textrm{B&amp;gt;}\;0^\infty\xrightarrow{18}\\0^\infty\;12^{a+1}\;1^{4(b-2)/3+c-2}\;\textrm{&amp;lt;A}\;1^6\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;This means that if &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{4(b-2)}{3}+c\ge 2&amp;lt;/math&amp;gt;, then we will reach &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A&#039;\Big(a+1,\frac{4b-14}{3}+c,6\Big)&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4a+\frac{2}{3}b^2+4b+bc+c+\frac{55}{3}&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv4\ (\operatorname{mod}12)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;gt;\frac{2}{3}b^2-\frac{8}{3}b+bc-4c&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(4,\frac{4(b-4)}{3}+c\Big)&amp;lt;/math&amp;gt;, or &amp;lt;math&amp;gt;0^\infty\;12^a\;1111\;\textrm{&amp;lt;A}\;1^{4(b-4)/3+c}\;0^\infty&amp;lt;/math&amp;gt;. What follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline0^\infty\;12^a\;1111\;\textrm{&amp;lt;A}\;1^{4(b-4)/3+c}\;0^\infty\xrightarrow{(8b-14)/3+2c}0^\infty\;12^a\;11\;\textrm{&amp;lt;A}\;2\;1^{4(b-4)/3+c+1}\;22\;0^\infty\xrightarrow{3}\\0^\infty\;12^a\;1\;\textrm{&amp;lt;A}\;1^{4(b-4)/3+c+3}\;22\;0^\infty\xrightarrow{4(b-4)/3+c+4}0^\infty\;12^a\;2^{4(b-4)/3+c+4}\;\textrm{A&amp;gt;}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;12^a\;2^{4(b-4)/3+c+4}\;\textrm{&amp;lt;C}\;12\;0^\infty\xrightarrow{4(b-4)/3+c+4}0^\infty\;12^a\;\textrm{&amp;lt;C}\;1^{4(b-4)/3+c+5}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;1\;\textrm{&amp;lt;A}\;1^{4(b-4)/3+c+6}\;2\;0^\infty\xrightarrow{4(b-4)/3+c+7}0^\infty\;12^{a-1}\;2^{4(b-4)/3+c+7}\;\textrm{A&amp;gt;}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;2^{4(b-4)/3+c+7}\;\textrm{&amp;lt;C}\;1\;0^\infty\xrightarrow{4(b-4)/3+c+6}0^\infty\;12^{a-1}\;2\;\textrm{&amp;lt;C}\;1^{4(b-4)/3+c+7}\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;\textrm{&amp;lt;A}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{2(a-1)}0^\infty\;\textrm{&amp;lt;A}\;21^{a-1}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B&amp;gt;}\;21^{a-1}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{2a+4(b-4)/3+c+6}0^\infty\;12^{a-1}\;1^{4(b-4)/3+c+9}\;\textrm{B&amp;gt;}\;0^\infty\xrightarrow{12}\\0^\infty\;12^{a-1}\;1^{4(b-4)/3+c+6}\;\textrm{&amp;lt;A}\;1^4\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;This means that if &amp;lt;math&amp;gt;a=0&amp;lt;/math&amp;gt;, then Bigfoot will reach the undefined &amp;lt;code&amp;gt;C0&amp;lt;/code&amp;gt; transition with the configuration &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;C}\;1^{(4b-1)/3+c}\;2\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{2}{3}b^2+\frac{8}{3}b+bc-\frac{10}{3}&amp;lt;/math&amp;gt; steps. Otherwise, it will proceed to reach &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A&#039;\Big(a-1,\frac{4b+2}{3}+c,4\Big)&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4a+\frac{2}{3}b^2+\frac{20}{3}b+bc+3c+\frac{41}{3}&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv6\ (\operatorname{mod}12)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{2}{3}b^2-\frac{16}{3}b+bc-6c+8&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(6,\frac{4(b-6)}{3}+c\Big)&amp;lt;/math&amp;gt;, or &amp;lt;math&amp;gt;0^\infty\;12^a\;111111\;\textrm{&amp;lt;A}\;1^{4(b-6)/3+c}\;0^\infty&amp;lt;/math&amp;gt;. What follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline0^\infty\;12^a\;111111\;\textrm{&amp;lt;A}\;1^{4(b-6)/3+c}\;0^\infty\xrightarrow{16b/3+4c-14}0^\infty\;12^a\;11\;\textrm{&amp;lt;C}\;1^{4(b-6)/3+c+5}\;2\;0^\infty\xrightarrow{4}\\0^\infty\;12^a\;\textrm{&amp;lt;A}\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{&amp;lt;A}\;21^a\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B&amp;gt;}\;21^a\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{2a+4(b-6)/3+c+8}0^\infty\;12^a\;1^{4(b-6)/3+c+8}\;2\;\textrm{B&amp;gt;}\;0^\infty\xrightarrow{60}\\0^\infty\;12^a\;1^{4(b-6)/3+c+2}\;\textrm{&amp;lt;A}\;1^{10}\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;This means that we will reach &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A&#039;\Big(a,\frac{4}{3}b+c-6,10\Big)&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+59&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv8\ (\operatorname{mod}12)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;gt;\frac{2}{3}b^2-8b+bc-8c+\frac{64}{3}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(8,\frac{4(b-8)}{3}+c\Big)&amp;lt;/math&amp;gt;, or &amp;lt;math&amp;gt;0^\infty\;12^a\;1^8\;\textrm{&amp;lt;A}\;1^{4(b-8)/3+c}\;0^\infty&amp;lt;/math&amp;gt;. What follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline0^\infty\;12^a\;1^8\;\textrm{&amp;lt;A}\;1^{4(b-8)/3+c}\;0^\infty\xrightarrow{(16b-62)/3+4c}0^\infty\;12^a\;11\;\textrm{&amp;lt;A}\;1^{4(b-8)/3+c+7}\;2\;0^\infty\xrightarrow{4(b-8)/3+c+8}\\0^\infty\;12^a\;1\;2^{4(b-8)/3+c+8}\;\textrm{A&amp;gt;}\;2\;0^\infty\xrightarrow{1}0^\infty\;12^a\;1\;2^{4(b-8)/3+c+8}\;\textrm{&amp;lt;C}\;1\;0^\infty\xrightarrow{4(b-8)/3+c+8}\\0^\infty\;12^a\;1\;\textrm{&amp;lt;C}\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{1}0^\infty\;12^a\;\textrm{&amp;lt;A}\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{2a}\\0^\infty\;\textrm{&amp;lt;A}\;21^a\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B&amp;gt;}\;21^a\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{2a+4(b-8)/3+c+10}\\0^\infty\;12^{a+1}\;1^{4(b-8)/3+c+9}\;\textrm{B&amp;gt;}\;0^\infty\xrightarrow{12}0^\infty\;12^{a+1}\;1^{4(b-8)/3+c+6}\;\textrm{&amp;lt;A}\;1^4\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;This means that we will reach &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A&#039;\Big(a+1,\frac{4b-14}{3}+c,4\Big)&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+\frac{29}{3}&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv10\ (\operatorname{mod}12)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{2}{3}b^2-\frac{32}{3}b+bc-10c+40&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(10,\frac{4(b-10)}{3}+c\Big)&amp;lt;/math&amp;gt;, or &amp;lt;math&amp;gt;0^\infty\;12^a\;1^{10}\;\textrm{&amp;lt;A}\;1^{4(b-10)/3+c}\;0^\infty&amp;lt;/math&amp;gt;. What follows is:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline0^\infty\;12^a\;1^{10}\;\textrm{&amp;lt;A}\;1^{4(b-  10)/3+c}\;0^\infty\xrightarrow{8b+6c-37}0^\infty\;12^a\;1\;\textrm{&amp;lt;A}\;1^{4(b-  10)/3+c+11}\;0^\infty\xrightarrow{4(b-  10)/3+c+12}\\0^\infty\;12^a\;2^{4(b-  10)/3+c+12}\;\textrm{A&amp;gt;}\;0^\infty\xrightarrow{4}0^\infty\;12^a\;2^{4(b-10)/3+c+11}\;\textrm{&amp;lt;C}\;122\;0^\infty\xrightarrow{4(b-10)/3+c+10}\\0^\infty\;12^a\;2\;\textrm{&amp;lt;C}\;1^{4(b-  10)/3+c+11}\;22\;0^\infty\xrightarrow{1}0^\infty\;12^a\;\textrm{&amp;lt;A}\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{2a}\\0^\infty\;\textrm{&amp;lt;A}\;12^a\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B&amp;gt;}\;12^a\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{2a+4(b-10)/3+c+14}\\0^\infty\;12^a\;1^{4(b-10)/3+c+13}\;22\;\textrm{B&amp;gt;}\;0^\infty\xrightarrow{18}0^\infty\;12^a\;1^{4(b-10)/3+c+10}\;\textrm{&amp;lt;A}\;1^6\;0^\infty\\\hline\end{array}&amp;lt;/math&amp;gt;This means that we will reach &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A&#039;\Big(a,\frac{4b-10}{3}+c,6\Big)&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+23&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
The information above can be summarized as&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;A&#039;(a,b,c)\rightarrow\begin{cases}A&#039;\Big(a,\frac{4}{3}b+c-2,4\Big)&amp;amp;\text{if }b\equiv0\pmod{12}\text{ and }\frac{4}{3}b+c\ge2,\\A&#039;\Big(a+1,\frac{4b-14}{3}+c,6\Big)&amp;amp;\text{if }b\equiv2\pmod{12}\text{ and }\frac{4(b-2)}{3}+c\ge2,\\0^\infty\;\textrm{&amp;lt;C}\;1^{(4b-1)/3+c}\;2\;0^\infty&amp;amp;\text{if }b\equiv4\pmod{12}\text{ and }a=0,\\A&#039;\Big(a-1,\frac{4b+2}{3}+c,4\Big)&amp;amp;\text{if }b\equiv4\pmod{12}\text{ and }a&amp;gt;0,\\A&#039;\Big(a,\frac{4}{3}b+c-6,10\Big)&amp;amp;\text{if }b\equiv6\pmod{12},\\A&#039;\Big(a+1,\frac{4b-14}{3}+c,4\Big)&amp;amp;\text{if }b\equiv8\pmod{12},\\A&#039;\Big(a,\frac{4b-10}{3}+c,6\Big)&amp;amp;\text{if }b\equiv10\pmod{12}.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Using the definitions of &amp;lt;math&amp;gt;A&#039;&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt; to transform these rules produces this:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;A(a,b,c)\rightarrow\begin{cases}A\Big(a,\frac{4}{3}b+c-1,2\Big)&amp;amp;\text{if }b\equiv0\pmod{6}\text{ and }\frac{4}{3}b+c\ge1,\\A\Big(a+1,\frac{4b-7}{3}+c,3\Big)&amp;amp;\text{if }b\equiv1\pmod{6}\text{ and }\frac{4(b-1)}{3}+c\ge1,\\0^\infty\;\textrm{&amp;lt;C}\;1^{(8b-1)/3+2c}\;2\;0^\infty&amp;amp;\text{if }b\equiv2\pmod{6}\text{ and }a=0,\\A\Big(a-1,\frac{4b+1}{3}+c,2\Big)&amp;amp;\text{if }b\equiv2\pmod{6}\text{ and }a&amp;gt;0,\\A\Big(a,\frac{4}{3}b+c-3,5\Big)&amp;amp;\text{if }b\equiv3\pmod{6},\\A\Big(a+1,\frac{4b-7}{3}+c,2\Big)&amp;amp;\text{if }b\equiv4\pmod{6},\\A\Big(a,\frac{4b-5}{3}+c,3\Big)&amp;amp;\text{if }b\equiv5\pmod{6}.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Substituting &amp;lt;math&amp;gt;b\leftarrow6b+k&amp;lt;/math&amp;gt; where &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is the remainder for each case yields the final result.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
Using the floor function, it is possible to describe the behaviour of &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;c&amp;lt;/math&amp;gt; using a function that is not defined piecewise:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\textstyle\begin{array}{c}f(m,n)=\Big(\frac{4m-3-4(\delta_1(m)-\delta_2(m)+\delta_4(m))-2(3\delta_3(m)+\delta_5(m))}{3}+n,2+\delta_1(m)+3\delta_3(m)+\delta_5(m)\Big),\\\delta_i(m)=\Big\lfloor\frac{x-i}{6}\Big\rfloor-\Big\lfloor\frac{x-i-1}{6}\Big\rfloor=\begin{cases}1&amp;amp;\text{if }x\equiv i\pmod{6},\\0&amp;amp;\text{otherwise.}\end{cases}\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
In effect, the halting problem for Bigfoot is about whether through enough iterations of &amp;lt;math&amp;gt;f(m,n)&amp;lt;/math&amp;gt; we encounter more &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; values that are congruent to 2 modulo 6 than ones that are congruent to 1 or 4 modulo 6.&lt;br /&gt;
&lt;br /&gt;
An important insight is that if &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; is odd and &amp;lt;math&amp;gt;c=2&amp;lt;/math&amp;gt;, then after four iterations of &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt;, that will remain the case. This allows one to define a configuration that eliminates the &amp;lt;math&amp;gt;c&amp;lt;/math&amp;gt; parameter and whose rules use a modulus of 81.&amp;lt;ref name=&amp;quot;b&amp;quot;&amp;gt;&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
In May 2024, Iijil shared a 7-state, 2-symbol machine, {{TM|0RB1RB_1LC0RA_1RE1LF_1LF1RE_0RD1RD_1LG0LG_---1LB}}, that has the same behaviour as Bigfoot.&amp;lt;ref&amp;gt;P. Michel, &amp;quot;[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]&amp;quot;.&amp;lt;/ref&amp;gt;&lt;br /&gt;
== Trajectory ==&lt;br /&gt;
After 69 steps, Bigfoot will reach the configuration &amp;lt;math&amp;gt;A(2,1,2)&amp;lt;/math&amp;gt; before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have &amp;lt;math&amp;gt;a=3999888&amp;lt;/math&amp;gt;. Here are the first few:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|l|}\hline A(2,1,2)\xrightarrow{49}A(3,1,3)\xrightarrow{59}A(4,2,3)\xrightarrow{109}A(3,6,2)\xrightarrow{221}A(3,9,2)\xrightarrow{379}A(3,11,5)\xrightarrow{597}A(3,18,3)\rightarrow\cdots\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
There exists a heuristic argument for Bigfoot being [[probviously]] non-halting. By only considering the rules for which &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; changes, one may notice that the trajectory of &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; values can be approximated by a random walk in which at each step, the walker moves +1 with probability &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{2}{3}&amp;lt;/math&amp;gt; or moves -1 with probability &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{1}{3}&amp;lt;/math&amp;gt;, starting at position 2. If &amp;lt;math&amp;gt;P(n)&amp;lt;/math&amp;gt; is the probability that the walker will reach position -1 from position &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;, then &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)=\frac{1}{3}P(n-1)+\frac{2}{3}P(n+1)&amp;lt;/math&amp;gt;. Solutions to this recurrence relation come in the form &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt; P(n)=c_02^{-n}+c_1&amp;lt;/math&amp;gt;, which after applying the appropriate boundary conditions reduces to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)=2^{-(n+1)}&amp;lt;/math&amp;gt;. As a result, if the walker gets to position 3999888, then the probability of it ever reaching position -1 would be &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;2^{-3999889}\approx 2.697\times 10^{-1204087}&amp;lt;/math&amp;gt;.&lt;br /&gt;
==References==&lt;br /&gt;
&amp;lt;references/&amp;gt;&lt;br /&gt;
[[Category:Cryptids]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Antihydra&amp;diff=1875</id>
		<title>Antihydra</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Antihydra&amp;diff=1875"/>
		<updated>2025-05-01T12:33:59Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA}}{{unsolved|Does Antihydra run forever?}}{{TM|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA|undecided}}&lt;br /&gt;
&#039;&#039;&#039;Antihydra&#039;&#039;&#039; is a [[BB(6)]] [[Cryptid]]. Its pseudo-random behaviour was first reported [https://discord.com/channels/960643023006490684/1026577255754903572/1256223215206924318 on Discord] by mxdys on 28 June 2024, and Racheline discovered the high-level rules soon after. It was named after the 2-state, 5-symbol [[Turing machine]] called [[Hydra]] for sharing many similarities to it.&amp;lt;ref&amp;gt;[https://discord.com/channels/960643023006490684/960643023530762341/1257053002859286701 Discord conversation where the machine was named]&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Antihydra is known to not generate Sturmian words&amp;lt;ref&amp;gt;DUBICKAS A. ON INTEGER SEQUENCES GENERATED BY LINEAR MAPS. &#039;&#039;Glasgow Mathematical Journal&#039;&#039;. 2009;51(2):243-252. doi:10.1017/S0017089508004655&amp;lt;/ref&amp;gt; (Corollary 4).&lt;br /&gt;
&amp;lt;table style=&amp;quot;margin: auto; text-align: center;&amp;quot;&amp;gt;&amp;lt;tr&amp;gt;&amp;lt;td style=&amp;quot;width: 200px;&amp;quot;&amp;gt;[[File:Antihydra-depiction.png|200px]]&amp;lt;br&amp;gt;Artistic depiction of Antihydra by Jadeix&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&lt;br /&gt;
{|class=&amp;quot;wikitable&amp;quot; style=&amp;quot;margin-left: auto; margin-right: auto;&amp;quot;&lt;br /&gt;
! !!0!!1&lt;br /&gt;
|-&lt;br /&gt;
!A&lt;br /&gt;
|1RB&lt;br /&gt;
|1RA&lt;br /&gt;
|-&lt;br /&gt;
!B&lt;br /&gt;
|0LC&lt;br /&gt;
|1LE&lt;br /&gt;
|-&lt;br /&gt;
!C&lt;br /&gt;
|1LD&lt;br /&gt;
|1LC&lt;br /&gt;
|-&lt;br /&gt;
!D&lt;br /&gt;
|1LA&lt;br /&gt;
|0LB&lt;br /&gt;
|-&lt;br /&gt;
!E&lt;br /&gt;
|1LF&lt;br /&gt;
|1RE&lt;br /&gt;
|-&lt;br /&gt;
!F&lt;br /&gt;
| ---&lt;br /&gt;
|0RA&lt;br /&gt;
|}&lt;br /&gt;
The transition table of Antihydra.&lt;br /&gt;
&amp;lt;/td&amp;gt;&amp;lt;td&amp;gt;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;amp;nbsp;&amp;lt;/td&amp;gt;&amp;lt;td style=&amp;quot;width: 220px;&amp;quot;&amp;gt;[[File:Antihydra award.jpg|220px]]&amp;lt;br&amp;gt;A community trophy - to be awarded to the first person or group who solves the Antihydra problem&lt;br /&gt;
&amp;lt;/td&amp;gt;&amp;lt;/tr&amp;gt;&amp;lt;/table&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
== Analysis ==&lt;br /&gt;
Let &amp;lt;math&amp;gt;A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;. Then,&amp;lt;ref name=&amp;quot;bl&amp;quot;&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]&amp;quot; (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|lll|}\hline&lt;br /&gt;
A(a,2b)&amp;amp; \xrightarrow{2a+3b^2+12b+11}&amp;amp; A(a+2,3b+2),\\&lt;br /&gt;
A(0,2b+1)&amp;amp;\xrightarrow{3b^2+9b-1}&amp;amp; 0^\infty\;\textrm{&amp;lt;F}\;110\;1^{3b}\;0^\infty,\\&lt;br /&gt;
A(a+1,2b+1)&amp;amp;\xrightarrow{3b^2+12b+5}&amp;amp; A(a,3b+3).\\\hline&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Proof&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
Consider the partial configuration &amp;lt;math&amp;gt;P(m,n):=0\;1^m\;\textrm{E&amp;gt;}\;0\;1^n\;0^\infty&amp;lt;/math&amp;gt;. The configuration after two steps is &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;\textrm{A&amp;gt;}\;1^{n+1}\;0^\infty&amp;lt;/math&amp;gt;. We note the following shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{A&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{A&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
As a result, we get &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;1^{n+1}\;\textrm{A&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt; after &amp;lt;math&amp;gt;n+1&amp;lt;/math&amp;gt; steps. Advancing two steps produces &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;1^{n+2}\;\textrm{&amp;lt;C}\;0^\infty&amp;lt;/math&amp;gt;. A second shift rule is useful here:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline1^s\;\textrm{&amp;lt;C}\xrightarrow{s}\textrm{&amp;lt;C}\;1^s\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
This allows us to reach &amp;lt;math&amp;gt;0\;1^{m-1}\;0\;\textrm{&amp;lt;C}\;1^{n+2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;n+2&amp;lt;/math&amp;gt; steps. Moving five more steps gets us to &amp;lt;math&amp;gt;0\;1^{m-2}\;\textrm{E&amp;gt;}\;0\;1^{n+3}\;0^\infty&amp;lt;/math&amp;gt;, which is the same configuration as &amp;lt;math&amp;gt;P(m-2,n+3)&amp;lt;/math&amp;gt;. Accounting for the head movement creates the condition that &amp;lt;math&amp;gt;m\ge 4&amp;lt;/math&amp;gt;. In summary:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With &amp;lt;math&amp;gt;A(a,b)&amp;lt;/math&amp;gt; we have &amp;lt;math&amp;gt;P(b,0)&amp;lt;/math&amp;gt;. As a result, we can apply this rule &amp;lt;math display=&amp;quot;inline&amp;gt;\big\lfloor\frac{1}{2}b\big\rfloor-1&amp;lt;/math&amp;gt; times (assuming &amp;lt;math&amp;gt;b\ge 4&amp;lt;/math&amp;gt;), which creates two possible scenarios:&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv0\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math&amp;gt;\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(2,\frac{3}{2}b-3\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;1^a\;011\;\textrm{E&amp;gt;}\;0\;1^{(3b)/2-3}\;0^\infty&amp;lt;/math&amp;gt;. After &amp;lt;math&amp;gt;3b+4&amp;lt;/math&amp;gt; steps this is &amp;lt;math&amp;gt;0^\infty\;1^a\;\textrm{&amp;lt;C}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, which then leads to &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;C}\;1^a\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; steps. After five more steps, we reach &amp;lt;math&amp;gt;0^\infty\;1\;\textrm{E&amp;gt;}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, from which another shift rule must be applied:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{E&amp;gt;}\;1^s\xrightarrow{s}1^s\;\textrm{E&amp;gt;}\\\hline\end{array}&amp;lt;/math&amp;gt;Doing so allows us to get the configuration &amp;lt;math&amp;gt;0^\infty\;1^{a+3}\;\textrm{E&amp;gt;}\;00\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;a+2&amp;lt;/math&amp;gt; steps. In six steps we have &amp;lt;math&amp;gt;0^\infty\;1^{a+2}\;011\;\textrm{E&amp;gt;}\;1^{(3b)/2}\;0^\infty&amp;lt;/math&amp;gt;, so we use the shift rule again, ending at &amp;lt;math&amp;gt;0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A\Big(a+2,\frac{3}{2}b+2\Big)&amp;lt;/math&amp;gt;, &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}b&amp;lt;/math&amp;gt; steps later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;2a+\frac{3}{4}b^2+6b+11&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;b\equiv1\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2-\frac{27}{4}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(3,\frac{3b-9}{2}\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;1^a\;0111\;\textrm{E&amp;gt;}\;0\;1^{(3b-9)/2}\;0^\infty&amp;lt;/math&amp;gt;. After &amp;lt;math&amp;gt;3b+2&amp;lt;/math&amp;gt; steps this becomes &amp;lt;math&amp;gt;0^\infty\;1^a\;\textrm{&amp;lt;F}\;110\;1^{(3b-3)/2}\;0^\infty&amp;lt;/math&amp;gt;. If &amp;lt;math&amp;gt;a=0&amp;lt;/math&amp;gt; then we have reached the undefined &amp;lt;code&amp;gt;F0&amp;lt;/code&amp;gt; transition with a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2+3b-\frac{19}{4}&amp;lt;/math&amp;gt; steps. Otherwise, continuing for six steps gives us &amp;lt;math&amp;gt;0^\infty\;1^{a-1}\;0111\;\textrm{E&amp;gt;}\;1^{(3b-3)/2}\;0^\infty&amp;lt;/math&amp;gt;. We conclude with the configuration &amp;lt;math&amp;gt;0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E&amp;gt;}\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;A\Big(a-1,\frac{3b+3}{2}\Big)&amp;lt;/math&amp;gt;, in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3b-3}{2}&amp;lt;/math&amp;gt; steps. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
The information above can be summarized as&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;A(a,b)\rightarrow\begin{cases}A\Big(a+2,\frac{3}{2}b+2\Big)&amp;amp;\text{if }b\ge 2,b\equiv0\pmod{2};\\0^\infty\;\textrm{&amp;lt;F}\;110\;1^{(3b-3)/2}\;0^\infty&amp;amp;\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a=0;\\A\Big(a-1,\frac{3b+3}{2}\Big)&amp;amp;\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a&amp;gt;0.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Substituting &amp;lt;math&amp;gt;b\leftarrow 2b&amp;lt;/math&amp;gt; for the first case and &amp;lt;math&amp;gt;b\leftarrow 2b+1&amp;lt;/math&amp;gt; for the other two yields the final result.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
In effect, the halting problem for Antihydra is about whether repeatedly applying the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;f(n)=\Big\lfloor\frac{3n}{2}\Big\rfloor+2&amp;lt;/math&amp;gt; will at some point produce more odd values of &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; than twice the number of even values.&lt;br /&gt;
&lt;br /&gt;
These rules can be modified to use the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;H(n)=\Big\lfloor\frac{3n}{2}\Big\rfloor&amp;lt;/math&amp;gt;, or the [[Hydra function]], which strengthens Antihydra&#039;s similarities to Hydra.&lt;br /&gt;
== Trajectory ==&lt;br /&gt;
11 steps are required to enter the configuration &amp;lt;math&amp;gt;A(0, 4)&amp;lt;/math&amp;gt; before the rules are repeatedly applied. So far, Antihydra has been simulated to &amp;lt;math&amp;gt;2^{31}&amp;lt;/math&amp;gt; rule steps, at which point &amp;lt;math&amp;gt;b=1073720884&amp;lt;/math&amp;gt;. Here are the first few:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline A(0,4)\xrightarrow{47}A(2,8)\xrightarrow{111}A(4,14)\xrightarrow{250}A(6,23)\xrightarrow{500}A(5,36)\xrightarrow{1209}A(7,56)\xrightarrow{2713}A(9,86)\rightarrow\cdots\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
The trajectory of &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; values resembles a random walk in which the walker can only move in step sizes +2 or -1 with equal probability, starting at position 0. If &amp;lt;math&amp;gt;P(n)&amp;lt;/math&amp;gt; is the probability that the walker will reach position -1 from position &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;, then it can be seen that &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)=\frac{1}{2}P(n-1)+\frac{1}{2}P(n+2)&amp;lt;/math&amp;gt;. Solutions to this recurrence relation come in the form &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt; P(n)=c_0{\left(\frac{\sqrt{5}-1}{2}\right)}^n+c_1+c_2{\left(-\frac{1+\sqrt{5}}{2}\right)}^n&amp;lt;/math&amp;gt;, which after applying the appropriate boundary conditions reduces to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P(n)={\left(\frac{\sqrt{5}-1}{2}\right)}^{n+1}&amp;lt;/math&amp;gt;. This means that if the walker were to get to position 1073720884 then the probability of it ever reaching position -1 is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\left(\frac{\sqrt{5}-1}{2}\right)}^{1073720885}\approx 4.841\times 10^{-224394395}&amp;lt;/math&amp;gt;. This combined with the fact that the expected position of the walker after &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; steps is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{1}{2}k&amp;lt;/math&amp;gt; strongly suggests Antihydra [[probviously]] runs indefinitely.&lt;br /&gt;
&lt;br /&gt;
== Code ==&lt;br /&gt;
The following Python program implements the abstracted behavior of the Antihydra. Proving whether it halts or not would also solve the Antihydra problem:&amp;lt;syntaxhighlight lang=&amp;quot;python&amp;quot; line=&amp;quot;1&amp;quot;&amp;gt;&lt;br /&gt;
# Keeps track of how many odd and even numbers have been encountered&lt;br /&gt;
a = 1&lt;br /&gt;
# The iterated value&lt;br /&gt;
b = 8&lt;br /&gt;
# If a equals 0 there have been (strictly) more than twice as many odd as even numbers and the program halts&lt;br /&gt;
while a != 0:&lt;br /&gt;
    # If b is even, add 2 to a so even numbers count twice&lt;br /&gt;
    if b % 2 == 0:&lt;br /&gt;
        a += 2&lt;br /&gt;
    else:&lt;br /&gt;
        a -= 1&lt;br /&gt;
    # Add the number divided by two (integer division, rounding down) to itself (Hydra function)&lt;br /&gt;
    # Note that integer division by 2 is equivalent to one bit shift to the right (a &amp;gt;&amp;gt; 1)&lt;br /&gt;
    b += b//2&lt;br /&gt;
&amp;lt;/syntaxhighlight&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[[Category:Individual machines]][[Category:Cryptids]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Hydra&amp;diff=1874</id>
		<title>Hydra</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Hydra&amp;diff=1874"/>
		<updated>2025-05-01T12:33:23Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA}}{{unsolved|Does Hydra run forever?}}{{TM|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA|undecided}}&lt;br /&gt;
&#039;&#039;&#039;Hydra&#039;&#039;&#039; is a [[BB(2,5)]] [[Cryptid]]. Its high-level rules were first reported [https://discord.com/channels/960643023006490684/1084047886494470185/1231110668288135208 on Discord] by Daniel Yuan on 20 April 2024, who also gave Hydra said name. Later on, a 6-state, 2-symbol [[Turing machine]] was discovered and named [[Antihydra]] for having similar behaviour to Hydra, making the study of this machine important to the study of that one.&lt;br /&gt;
&lt;br /&gt;
Hydra is known to not generate Sturmian words&amp;lt;ref&amp;gt;DUBICKAS A. ON INTEGER SEQUENCES GENERATED BY LINEAR MAPS. &#039;&#039;Glasgow Mathematical Journal&#039;&#039;. 2009;51(2):243-252. doi:10.1017/S0017089508004655&amp;lt;/ref&amp;gt; (Corollary 4).&lt;br /&gt;
&amp;lt;div style=&amp;quot;width: fit-content; text-align: center; margin-left: auto; margin-right: auto;&amp;quot;&amp;gt;&lt;br /&gt;
{|class=&amp;quot;wikitable&amp;quot; style=&amp;quot;margin-left: auto; margin-right: auto;&amp;quot;&lt;br /&gt;
! !!0!!1!!2!!3!!4&lt;br /&gt;
|-&lt;br /&gt;
!A&lt;br /&gt;
|1RB&lt;br /&gt;
|3RB&lt;br /&gt;
| ---&lt;br /&gt;
|3LA&lt;br /&gt;
|1RA&lt;br /&gt;
|-&lt;br /&gt;
!B&lt;br /&gt;
|2LA&lt;br /&gt;
|3RA&lt;br /&gt;
|4LB&lt;br /&gt;
|0LB&lt;br /&gt;
|0LA&lt;br /&gt;
|}&lt;br /&gt;
The transition table of Hydra.&amp;lt;/div&amp;gt;&lt;br /&gt;
== Analysis ==&lt;br /&gt;
Let &amp;lt;math&amp;gt;C(a,b):=0^\infty\;\textrm{&amp;lt;A}\;2\;0^a\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;. Then,&amp;lt;ref&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/05/10/bb-2-5-is-hard.html BB(2, 5) is Hard (Hydra)] (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|lll|}\hline&lt;br /&gt;
C(2a,0)&amp;amp;\xrightarrow{6a^2+20a+4}&amp;amp;0^\infty\;3^{3a+1}\;1\;\textrm{A&amp;gt;}\;2\;0^\infty,\\&lt;br /&gt;
C(2a,b+1)&amp;amp;\xrightarrow{6a^2+23a+10}&amp;amp;C(3a+3,b),\\&lt;br /&gt;
C(2a+1,b)&amp;amp;\xrightarrow{4b+6a^2+23a+26}&amp;amp;C(3a+3,b+2).\\\hline&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;div class=&amp;quot;toccolours mw-collapsible mw-collapsed&amp;quot;&amp;gt;&#039;&#039;&#039;Proof&#039;&#039;&#039;&amp;lt;div class=&amp;quot;mw-collapsible-content&amp;quot;&amp;gt;&lt;br /&gt;
Consider the partial configuration &amp;lt;math&amp;gt;P(m,n):=0^\infty\;3^m\;\textrm{A&amp;gt;}\;02\;0^n&amp;lt;/math&amp;gt;. After 14 steps this configuration becomes &lt;br /&gt;
&amp;lt;math&amp;gt;0^\infty\;3^{m+3}\;\textrm{&amp;lt;A}\;2\;0^{n-2}&amp;lt;/math&amp;gt;. We note the following shift rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline3^s\;\textrm{&amp;lt;A}\xrightarrow{s}\textrm{&amp;lt;A}\;3^s\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
Using this shift rule, we get &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;A}\;3^{m+3}\;2\;0^{n-2}&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;m+3&amp;lt;/math&amp;gt; steps. From here, we can observe that &amp;lt;math&amp;gt;\textrm{A&amp;gt;}\;0\;3^s&amp;lt;/math&amp;gt; turns into &amp;lt;math&amp;gt;3\;\textrm{A&amp;gt;}\;0\;3^{s-1}&amp;lt;/math&amp;gt; in three steps if &amp;lt;math&amp;gt;s\ge 1&amp;lt;/math&amp;gt;. By repeating this process, we acquire this transition rule:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline\textrm{A&amp;gt;}\;0\;3^s\xrightarrow{3s}3^s\;\textrm{A&amp;gt;}\;0\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With this rule, it takes &amp;lt;math&amp;gt;3m+9&amp;lt;/math&amp;gt; steps to reach the configuration &amp;lt;math&amp;gt;0^\infty\;3^{m+3}\;\textrm{A&amp;gt;}\;02\;0^{n-2}&amp;lt;/math&amp;gt;, which is the same configuration as &amp;lt;math&amp;gt;P(m+3,n-2)&amp;lt;/math&amp;gt;. To summarize:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline P(m,n)\xrightarrow{4m+26}P(m+3,n-2)\text{ if }n\ge 2.\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
With &amp;lt;math&amp;gt;C(a,b)&amp;lt;/math&amp;gt; we have &amp;lt;math&amp;gt;P(0,a)&amp;lt;/math&amp;gt;. As a result, we can apply this rule &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\big\lfloor\frac{1}{2}a\big\rfloor&amp;lt;/math&amp;gt; times, which creates two possible scenarios:&lt;br /&gt;
#If &amp;lt;math&amp;gt;a\equiv0\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math&amp;gt;\sum_{i=0}^{(a/2)-1}(4\times 3i+26)=\textstyle\frac{3}{2}a^2+10a&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(\frac{3}{2}a,0\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;3^{(3/2)a}\;\textrm{A&amp;gt;}\;02\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;, which in four steps becomes &amp;lt;math&amp;gt;0^\infty\;3^{(3/2)a+1}\;1\;\textrm{A&amp;gt;}\;3^b\;2\;0^\infty.&amp;lt;/math&amp;gt; If &amp;lt;math&amp;gt;b=0&amp;lt;/math&amp;gt; then we have reached the undefined &amp;lt;code&amp;gt;A2&amp;lt;/code&amp;gt; transition in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a^2+10a+4&amp;lt;/math&amp;gt; steps total. Otherwise, continuing for three steps gives us &amp;lt;math&amp;gt;0^\infty\;3^{(3/2)+2}\;\textrm{&amp;lt;B}\;0\;3^{b-1}\;2\;0^\infty&amp;lt;/math&amp;gt;. Another shift rule is required here:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline3^s\;\textrm{&amp;lt;B}\xrightarrow{s}\textrm{&amp;lt;B}\;0^s\\\hline\end{array}&amp;lt;/math&amp;gt;This means the configuration becomes &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;B}\;0^{(3/2)a+3}\;3^{b-1}\;2\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a+2&amp;lt;/math&amp;gt; steps, and &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;A}\;2\;0^{(3/2)a+3}\;3^{b-1}\;2\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;C\Big(\frac{3}{2}a+3,b-1\Big)&amp;lt;/math&amp;gt;, one step later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a^2+\frac{23}{2}a+10&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
#If &amp;lt;math&amp;gt;a\equiv1\ (\operatorname{mod}2)&amp;lt;/math&amp;gt;, then in &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3}{2}a^2+7a-\frac{17}{2}&amp;lt;/math&amp;gt; steps we arrive at &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;P\Big(\frac{3a-3}{2},1\Big)&amp;lt;/math&amp;gt;. The matching complete configuration is &amp;lt;math&amp;gt;0^\infty\;3^{(3a-3)/2}\;\textrm{A&amp;gt;}\;020\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;, which in four steps becomes &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;\textrm{A&amp;gt;}\;0\;3^b\;2\;0^\infty&amp;lt;/math&amp;gt;, and then &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;3^b\;\textrm{A&amp;gt;}\;02\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;3b&amp;lt;/math&amp;gt; steps. After 14 steps, we see the configuration &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;3^{b+3}\;\textrm{&amp;lt;A}\;2\;0^\infty&amp;lt;/math&amp;gt;, which turns into &amp;lt;math&amp;gt;0^\infty\;3^{(3a-1)/2}\;1\;\textrm{&amp;lt;A}\;3^{b+3}\;2\;0^\infty&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;b+3&amp;lt;/math&amp;gt; steps. In two steps we get &amp;lt;math&amp;gt;0^\infty\;3^{(3a+1)/2}\;\textrm{&amp;lt;B}\;0\;3^{b+2}\;2\;0^\infty&amp;lt;/math&amp;gt;, followed by &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;B}\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty&amp;lt;/math&amp;gt; after &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;\frac{3a+1}{2}&amp;lt;/math&amp;gt; more steps. We conclude with &amp;lt;math&amp;gt;0^\infty\;\textrm{&amp;lt;A}\;2\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty&amp;lt;/math&amp;gt;, equal to &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;C\Big(\frac{3a+3}{2},b+2\Big)&amp;lt;/math&amp;gt;, one step later. This gives a total of &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;4b+\frac{3}{2}a^2+\frac{17}{2}a+16&amp;lt;/math&amp;gt; steps.&lt;br /&gt;
The information above can be summarized as&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;C(a,b)\rightarrow\begin{cases}0^\infty\;3^{(3/2)a+1}\;1\;\textrm{A&amp;gt;}\;2\;0^\infty&amp;amp;\text{if }a\equiv0\pmod{2}\text{ and }b=0,\\C\Big(\frac{3}{2}a+3,b-1\Big)&amp;amp;\text{if }a\equiv0\pmod{2}\text{ and }b&amp;gt;0,\\C\Big(\frac{3a+3}{2},b+2\Big)&amp;amp;\text{if }a\equiv1\pmod2.\end{cases}&amp;lt;/math&amp;gt;&lt;br /&gt;
Substituting &amp;lt;math&amp;gt;a\leftarrow 2a&amp;lt;/math&amp;gt; for the first two cases and &amp;lt;math&amp;gt;a\leftarrow 2a+1&amp;lt;/math&amp;gt; for the third yields the final result.&lt;br /&gt;
&amp;lt;/div&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
In effect, the halting problem for Hydra is about whether repeatedly applying the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;f(n)=3\big\lfloor\frac{n}{2}\big\rfloor+3&amp;lt;/math&amp;gt; will at some point produce more even values of &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; than twice the number of odd values.&lt;br /&gt;
&lt;br /&gt;
An alternative version of these rules exists that makes the connection to Antihydra more apparent by using the function &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;H(n)=\Big\lfloor\frac{3}{2}n\Big\rfloor&amp;lt;/math&amp;gt;, or the [[Hydra function]].&amp;lt;ref&amp;gt;S. Ligocki, &amp;quot;[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]&amp;quot; (2024). Accessed 22 July 2024.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Trajectory ==&lt;br /&gt;
It takes 20 steps to reach the configuration &amp;lt;math&amp;gt;C(3,0)&amp;lt;/math&amp;gt;, and from there, the [[Collatz-like]] rules are repeatedly applied. Simulating Hydra has shown that after 4000000 rule steps, we have &amp;lt;math&amp;gt;b=2005373&amp;lt;/math&amp;gt;. Here are the first few:&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{|c|}\hline C(3,0)\xrightarrow{55}C(6,2)\xrightarrow{133}C(12,1)\xrightarrow{364}C(21,0)\xrightarrow{856}C(33,2)\xrightarrow{1938}C(51,4)\xrightarrow{4367}C(78,6)\rightarrow\cdots\\\hline\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
The heuristic argument that suggests Antihydra is a [[probviously]] non-halting machine can be applied here. This means that if &amp;lt;math&amp;gt;b&amp;lt;/math&amp;gt; is to be thought of as moving randomly, then the probability of Hydra halting is &amp;lt;math display=&amp;quot;inline&amp;quot;&amp;gt;{\Big(\frac{\sqrt{5}-1}{2}\Big)}^{2005374}\approx 4.168\times 10^{-419099}&amp;lt;/math&amp;gt;.&lt;br /&gt;
==References==&lt;br /&gt;
[[Category:Cryptids]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1405</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1405"/>
		<updated>2025-01-10T11:46:25Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Reading through old Busy Beaver papers, there is a common recurring theme in which someone working on solving BB(n-1) believes BB(n) to be impossible. We document here the quotes demonstrating this phenomenon.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) and BB(4) ==&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate Σ(3) and S(3), there is &#039;&#039;&#039;no evidence that any presently known approach will yield the answer&#039;&#039;&#039;, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), &#039;&#039;&#039;the situation seems to be entirely hopeless at present&#039;&#039;&#039;.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Radó, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&amp;lt;ref&amp;gt;Radó, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963. &amp;lt;/ref&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Quoted in: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&amp;lt;ref&amp;gt;Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) and BB(6) ==&lt;br /&gt;
&lt;br /&gt;
* &amp;quot;Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have&amp;quot;&lt;br /&gt;
* &amp;quot;Prediction 5. It will never be proved that 4(5) = 1,915 and S(5) = 2, 358, 064. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Brady, Allen H, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024.&amp;lt;ref name=&amp;quot;:0&amp;quot;&amp;gt;Brady, Allen H, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024.  &amp;lt;/ref&amp;gt;   &lt;br /&gt;
&lt;br /&gt;
Arnold Oberschelp who reviewed the above paper by Brady, went even further: &lt;br /&gt;
&lt;br /&gt;
&amp;quot;The hard case is for n = 5. When computers became cheaper and faster many high-scoring and long-running (5 x 2)-machines were found, the best reported here (by Uhing) showing Σ(5) ≥ 1,915 and S(5) &amp;gt; 2.3 x 10&amp;lt;sup&amp;gt;6&amp;lt;/sup&amp;gt;. The author points out that when looking at such tricky machines found by computer one can verify that they stop but one has virtually no idea why they stop. And since whatever bounds on space and time are fixed for a computer search, there are left in the search space immensely many machines that must be treated individually, it will probably never be possible to prove mathematically that they will not stop. Even if one does get the exact value for n = 5, one might never be able to prove it rigorously. The case n = 6 is quite intractable.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Oberschelp, A, review published in The Journal of Symbolic Logic / Volume 56 / Issue 03 / September 1991, pp 1091 - 1091, [https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/allen-h-brady-the-busy-beaver-game-and-the-meaning-of-life-the-universal-turing-machine-a-halfcentury-survey-edited-by-rolf-herken-kammerer-unverzagt-hamburg-and-berlin-and-oxford-university-press-oxford-and-new-york-1988-pp-259277/CB4DC8CF087389ADDD6B42CB0625A9A4 URL]&amp;lt;ref&amp;gt;Oberschelp, A, review published in The Journal of Symbolic Logic / Volume 56 / Issue 03 / September 1991, pp 1091 - 1091, [https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/allen-h-brady-the-busy-beaver-game-and-the-meaning-of-life-the-universal-turing-machine-a-halfcentury-survey-edited-by-rolf-herken-kammerer-unverzagt-hamburg-and-berlin-and-oxford-university-press-oxford-and-new-york-1988-pp-259277/CB4DC8CF087389ADDD6B42CB0625A9A4 URL]&amp;lt;/ref&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&amp;lt;ref&amp;gt;Machlin, Rona, Stout, Quentin F. (1990/06).&amp;quot;The complex behavior of simple machines.&amp;quot; Physica D: Nonlinear Phenomena 42(1-3): 85-98.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/262/mode/2up?view=theater Pages 263-264]&amp;lt;ref name=&amp;quot;:0&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/274/mode/2up?view=theater Page 274]&lt;br /&gt;
&lt;br /&gt;
Repeated: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252 &amp;lt;ref name=&amp;quot;:0&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/274/mode/2up?view=theater Page 274-275] &amp;lt;ref name=&amp;quot;:0&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation) &amp;lt;ref&amp;gt;Lin, Shen. &amp;quot;Computer studies of Turing machine problems /.&amp;quot; Doctoral dissertation, Ohio State University, 1963. &amp;lt;nowiki&amp;gt;http://rave.ohiolink.edu/etdc/view?acc_num=osu1486554418657614&amp;lt;/nowiki&amp;gt;&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1404</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1404"/>
		<updated>2025-01-10T11:45:43Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Reading through old Busy Beaver papers, there is a common recurring theme in which someone working on solving BB(n-1) believes BB(n) to be impossible. We document here the quotes demonstrating this phenomenon.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) and BB(4) ==&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate Σ(3) and S(3), there is &#039;&#039;&#039;no evidence that any presently known approach will yield the answer&#039;&#039;&#039;, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), &#039;&#039;&#039;the situation seems to be entirely hopeless at present&#039;&#039;&#039;.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Radó, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&amp;lt;ref&amp;gt;Radó, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963. &amp;lt;/ref&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Quoted in: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&amp;lt;ref&amp;gt;Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) and BB(6) ==&lt;br /&gt;
&lt;br /&gt;
* &amp;quot;Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have&amp;quot;&lt;br /&gt;
* &amp;quot;Prediction 5. It will never be proved that 4(5) = 1,915 and S(5) = 2, 358, 064. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Brady, Allen H, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024.&amp;lt;ref name=&amp;quot;:0&amp;quot;&amp;gt;Brady, Allen H, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024.  &amp;lt;/ref&amp;gt;   &lt;br /&gt;
&lt;br /&gt;
Arnold Oberschelp who reviewed the above paper by Brady, went even further: &lt;br /&gt;
&lt;br /&gt;
&amp;quot;The hard case is for n = 5. When computers became cheaper and faster many high-scoring and long-running (5 x 2)-machines were found, the best reported here (by Uhing) showing Σ(5) ≥ 1,915 and S(5) &amp;gt; 2.3 x 10&amp;lt;sup&amp;gt;6&amp;lt;/sup&amp;gt;. The author points out that when looking at such tricky machines found by computer one can verify that they stop but one has virtually no idea why they stop. And since whatever bounds on space and time are fixed for a computer search, there are left in the search space immensely many machines that must be treated individually, it will probably never be possible to prove mathematically that they will not stop. Even if one does get the exact value for n = 5, one might never be able to prove it rigorously. The case n = 6 is quite intractable.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Oberschelp, A, review published in The Journal of Symbolic Logic / Volume 56 / Issue 03 / September 1991, pp 1091 - 1091, [https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/allen-h-brady-the-busy-beaver-game-and-the-meaning-of-life-the-universal-turing-machine-a-halfcentury-survey-edited-by-rolf-herken-kammerer-unverzagt-hamburg-and-berlin-and-oxford-university-press-oxford-and-new-york-1988-pp-259277/CB4DC8CF087389ADDD6B42CB0625A9A4 URL]&amp;lt;ref&amp;gt;Oberschelp, A, review published in The Journal of Symbolic Logic / Volume 56 / Issue 03 / September 1991, pp 1091 - 1091, [https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/allen-h-brady-the-busy-beaver-game-and-the-meaning-of-life-the-universal-turing-machine-a-halfcentury-survey-edited-by-rolf-herken-kammerer-unverzagt-hamburg-and-berlin-and-oxford-university-press-oxford-and-new-york-1988-pp-259277/CB4DC8CF087389ADDD6B42CB0625A9A4 URL]&amp;lt;/ref&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&amp;lt;ref&amp;gt;Machlin, Rona, Stout, Quentin F. (1990/06).&amp;quot;The complex behavior of simple machines.&amp;quot; Physica D: Nonlinear Phenomena 42(1-3): 85-98.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/262/mode/2up?view=theater Pages 263-264]&amp;lt;ref name=&amp;quot;:0&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/274/mode/2up?view=theater Page 274]&lt;br /&gt;
&lt;br /&gt;
Repeated: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252 &amp;lt;ref name=&amp;quot;:0&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/274/mode/2up?view=theater Page 274-275] &amp;lt;ref name=&amp;quot;:0&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation) &amp;lt;ref&amp;gt;Lin, Shen. &amp;quot;Computer studies of Turing machine problems /.&amp;quot; Doctoral dissertation, Ohio State University, 1963. &amp;lt;nowiki&amp;gt;http://rave.ohiolink.edu/etdc/view?acc_num=osu1486554418657614&amp;lt;/nowiki&amp;gt;&amp;lt;/ref&amp;gt;&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1403</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1403"/>
		<updated>2025-01-10T11:39:42Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Reading through old Busy Beaver papers, there is a common recurring theme in which someone working on solving BB(n-1) believes BB(n) to be impossible. We document here the quotes demonstrating this phenomenon.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) and BB(4) ==&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate Σ(3) and S(3), there is &#039;&#039;&#039;no evidence that any presently known approach will yield the answer&#039;&#039;&#039;, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), &#039;&#039;&#039;the situation seems to be entirely hopeless at present&#039;&#039;&#039;.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Radó, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963. &lt;br /&gt;
&lt;br /&gt;
Quoted in: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) and BB(6) ==&lt;br /&gt;
&lt;br /&gt;
* &amp;quot;Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have&amp;quot;&lt;br /&gt;
* &amp;quot;Prediction 5. It will never be proved that 4(5) = 1,915 and S(5) = 2, 358, 064. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&amp;quot;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Brady, Allen H, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024  &lt;br /&gt;
&lt;br /&gt;
Arnold Oberschelp who reviewed the above paper by Brady, went even further: &lt;br /&gt;
&lt;br /&gt;
&amp;quot;The hard case is for n = 5. When computers became cheaper and faster many high-scoring and long-running (5 x 2)-machines were found, the best reported here (by Uhing) showing Σ(5) ≥ 1,915 and S(5) &amp;gt; 2.3 x 10&amp;lt;sup&amp;gt;6&amp;lt;/sup&amp;gt;. The author points out that when looking at such tricky machines found by computer one can verify that they stop but one has virtually no idea why they stop. And since whatever bounds on space and time are fixed for a computer search, there are left in the search space immensely many machines that must be treated individually, it will probably never be possible to prove mathematically that they will not stop. Even if one does get the exact value for n = 5, one might never be able to prove it rigorously. The case n = 6 is quite intractable.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source:&#039;&#039;&#039; Oberschelp, A, review published in The Journal of Symbolic Logic / Volume 56 / Issue 03 / September 1991, pp 1091 - 1091, [https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/allen-h-brady-the-busy-beaver-game-and-the-meaning-of-life-the-universal-turing-machine-a-halfcentury-survey-edited-by-rolf-herken-kammerer-unverzagt-hamburg-and-berlin-and-oxford-university-press-oxford-and-new-york-1988-pp-259277/CB4DC8CF087389ADDD6B42CB0625A9A4 URL]  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/262/mode/2up?view=theater Pages 263-264]&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/274/mode/2up?view=theater Page 274]&lt;br /&gt;
&lt;br /&gt;
Repeated: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot; [https://archive.org/details/universalturingm0000unse/page/274/mode/2up?view=theater Page 274-275]&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Source&#039;&#039;&#039;: Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation)&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1336</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=1336"/>
		<updated>2024-11-27T10:14:43Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* BB(3) to BB(4) */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;People working on BB(n-1) believed BB(n) was impossible:&lt;br /&gt;
&lt;br /&gt;
== BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Source: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(2) to BB(3) and BB(4) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate S(3) and Σ(3), there is no evidence that any known approach will yield the answer, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), the situation seems to be entirely hopeless at present.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
Source: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Original source:&#039;&#039;&#039; Radò, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
Source: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Original source:&#039;&#039;&#039; Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024&lt;br /&gt;
&lt;br /&gt;
[[File:BradyBB(5)isHopeless.png|alt=Allen H. Brady (who proved BB(4) = 107) is saying that ever proving BB(5) is unlikely|thumb|Original source where Allen H. Brady says that solving BB(5) rigorously is unlikely to happen (1990).]]&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation)&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Skelet_17&amp;diff=1335</id>
		<title>Skelet 17</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Skelet_17&amp;diff=1335"/>
		<updated>2024-11-27T01:09:45Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{machine|1LB---_0RC1LE_0RD1RC_1LA1RB_0LB0LA}}{{TM|1RB---_0LC1RE_0LD1LC_1RA1LB_0RB0RA|non-halt}}&lt;br /&gt;
&lt;br /&gt;
Skelet #17 was one of [[Skelet&#039;s 43 holdouts]] and one of the last holdouts in BB(5). &lt;br /&gt;
&lt;br /&gt;
The first step towards its resolution was made by savask, who showed the connection to [[wikipedia:Gray_code|Gray Code]]: https://docs.bbchallenge.org/other/skelet17_savasks_analysis.pdf &lt;br /&gt;
&lt;br /&gt;
Building upon this work, Chris Xu produced a full proof of its nonhalting that can be found here: https://chrisxudoesmath.com/papers/skelet17.pdf &lt;br /&gt;
&lt;br /&gt;
Adapting the above, a formal proof of its nonhalting by mxdys can be found here: https://github.com/ccz181078/Coq-BB5/blob/main/CoqBB5/Skelet17.md&lt;br /&gt;
&lt;br /&gt;
== TM Behavior ==&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable floatright&amp;quot; style=&amp;quot;text-align:center;&amp;quot;&lt;br /&gt;
! rowspan=&amp;quot;2&amp;quot;|&lt;br /&gt;
|-&lt;br /&gt;
! 5 !! 4 !! 3 !! 2 !! 1 !! 0 !! value !! length !! &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
    | 0 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 0  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 1 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 1  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 1* || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#F22&amp;quot;|{{mono|-1}} || 2  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 2 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 1  || 2  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 3 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 0  || 2  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 3* || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 0  || 2  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 4 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || 0  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 5 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || 1  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 6 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 2  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 7 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 3  || 3  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 7* || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 0  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 7** || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#F22&amp;quot;|{{mono|-1}} || 7  || 6  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 8 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || 3  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 9 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 4  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 10 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 5  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 10* || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#F22&amp;quot;|{{mono|-1}} || 6  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 11 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || 3  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 12 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || 2  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 13 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 1  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 14 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 0  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 14* || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 7  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 14** || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || style=&amp;quot;background:#F22&amp;quot;|{{mono|-1}} || 7  || 6  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 15 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || 3  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 16 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || 4  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 17 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || 5  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 18 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 6  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 19 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 7  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 19* || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || style=&amp;quot;background:#F22&amp;quot;|{{mono|-1}} || 8  || 5  || +1&lt;br /&gt;
|-&lt;br /&gt;
    | 20 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || 4  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 21 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || 3  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 22 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || 2  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 23 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|3}} || style=&amp;quot;background:#DDD&amp;quot;|{{mono|1}} || 1  || 4  || -1&lt;br /&gt;
|-&lt;br /&gt;
    | 24 || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#111&amp;quot;|{{mono|-}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|2}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|4}} || style=&amp;quot;background:#0FF&amp;quot;|{{mono|0}} || 0  || 4  || -1&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
It can be shown that when the machine head is at the right end of the tape, the complete tape configuration is of the following form (using [[Directed head notation]]):&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;0^\infty \; 10^{n_1} \; 1 \; 10^{n_2} \; \dots 1 \; 10^{n_k} \textrm{ B&amp;gt; } \; 0^\infty&amp;lt;/math&amp;gt;,&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;n_i&amp;lt;/math&amp;gt; represents a non-negative integer. Essentially, Skelet 17 builds &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;-length lists of non-negative numbers delimited with individual cells imprinted with a 1.&lt;br /&gt;
&lt;br /&gt;
The most common transformation (frequency tends to 1 as the number of steps increases) occuring between such configurations is described by Increment rule, similar to the Lucal form of Gray Code: the value at the last index is decremented and the value to the left of the rightmost index with odd value is incremented.&lt;br /&gt;
&lt;br /&gt;
It is useful to consider the following &#039;&#039;state variables&#039;&#039;: &lt;br /&gt;
&lt;br /&gt;
* Gray Code value (denoted by &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt;): the integer that is obtained by calculating the parity of each number (except the last) in the list, considering the result as a Gray Code bitstring corresponding to some integer and &amp;quot;recovering&amp;quot; this integer.&lt;br /&gt;
* Increment value (denoted by &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;): the parity of the entire sequence, mapped to &amp;lt;math&amp;gt;\{-1, 1\}&amp;lt;/math&amp;gt;. Dirung Increment steps, Skelet 17 counts forwards when &amp;lt;math&amp;gt;\sigma = +1&amp;lt;/math&amp;gt; or backwards if &amp;lt;math&amp;gt;\sigma = -1&amp;lt;/math&amp;gt;.&lt;br /&gt;
* Length of the sequence (including leading zeros).&lt;br /&gt;
&lt;br /&gt;
When started on a blank tape, the machine visits configurations of the form &amp;lt;math&amp;gt;(0, 2, 4, ..., 2^{2k+2}, 0)&amp;lt;/math&amp;gt; infinitely often (the step numbers where it occurs approximately equal power of 16: 14, 251, 4088, 65537, 1048646, 16777607, 268437188, etc).&lt;br /&gt;
&lt;br /&gt;
=== Analysis by bisimulation ===&lt;br /&gt;
&lt;br /&gt;
Chris Xu found it useful to divide some tape transformations into two phases, one of them is &amp;quot;virtual&amp;quot; or &amp;quot;invisible&amp;quot;, as it does not actually occur on the tape. Indeed, the intermediate step involves negative numbers (the last number being decremented to -1 from 0), which is of course cannot correspond to the &amp;lt;math&amp;gt;10^{a_i}&amp;lt;/math&amp;gt; written on the tape.&lt;br /&gt;
&lt;br /&gt;
Such skipped intermediate steps are denoted by asterisk in the table to preserve the &amp;quot;real&amp;quot; step counts (e.g. steps 7* and 7**).&lt;br /&gt;
&lt;br /&gt;
[[Category:Stub]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Halting_Segment&amp;diff=1234</id>
		<title>Halting Segment</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Halting_Segment&amp;diff=1234"/>
		<updated>2024-11-11T20:27:41Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: Created page with &amp;quot;See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 5] of bbchallenge deciders&amp;#039; write-up.&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 5] of bbchallenge deciders&#039; write-up.&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Bouncer&amp;diff=1233</id>
		<title>Bouncer</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Bouncer&amp;diff=1233"/>
		<updated>2024-11-11T20:27:06Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:1RB0LB 1LA0RA.png|alt=1RB0LB_1LA0RA|thumb|A close-up of the bouncer {{TM|1RB0LB_1LA0RA}} with 2 states, the smallest number of states for which a bouncer can appear.]]&lt;br /&gt;
A &#039;&#039;&#039;bouncer&#039;&#039;&#039; is a Turing machine whose tape head, roughly speaking, alternates back and forth between the two edges of the tape in a linear fasion, growing the tape along one or both edges with each iteration. A bouncer is a possible classification of [[Non-halting Turing machine|non-halting Turing machines]].&lt;br /&gt;
&lt;br /&gt;
See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 7] of bbchallenge&#039;s deciders write-up.&lt;br /&gt;
&lt;br /&gt;
[[Category:Zoology]]&lt;br /&gt;
[[Category:Stub]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Finite_Automata_Reduction&amp;diff=1232</id>
		<title>Finite Automata Reduction</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Finite_Automata_Reduction&amp;diff=1232"/>
		<updated>2024-11-11T20:26:05Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 6] of bbchallenge&#039;s deciders write-up.&lt;br /&gt;
[[Category:Deciders]]&lt;br /&gt;
[[Category:Stub]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Backward_Reasoning&amp;diff=1231</id>
		<title>Backward Reasoning</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Backward_Reasoning&amp;diff=1231"/>
		<updated>2024-11-11T20:25:03Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: Created page with &amp;quot;See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 4] of bbchallenge&amp;#039;s deciders write-up.&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 4] of bbchallenge&#039;s deciders write-up.&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Decider&amp;diff=1230</id>
		<title>Decider</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Decider&amp;diff=1230"/>
		<updated>2024-11-11T20:24:46Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* List of Deciders */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[Category:Deciders]]&lt;br /&gt;
A &#039;&#039;&#039;Decider&#039;&#039;&#039; (or a &#039;&#039;&#039;Filter&#039;&#039;&#039;) is a program which attempts to decide whether or not a given [[Turing machine]] (TM) will halt. Since the Halting Problem is uncomputable, no decider can decide all TMs, instead deciders categorize each TM into one of three categories: Halting, Proven Infinite, or [[Holdout]].&lt;br /&gt;
&lt;br /&gt;
There are a wide variety of methods used to construct deciders. Some broad categories are: [[Accelerated Simulator]]s, [[Backwards Reasoning]] and [[Closed Set]] methods.&lt;br /&gt;
&lt;br /&gt;
== List of Deciders ==&lt;br /&gt;
&lt;br /&gt;
* [[Translated Cycler]]&lt;br /&gt;
* [[Backward Reasoning]]&lt;br /&gt;
* [[Closed Position Set (CPS)]]&lt;br /&gt;
* [[Closed Tape Language (CTL)]]&lt;br /&gt;
* [[Finite Automata Reduction (FAR)]]&lt;br /&gt;
* [[Inductive Proof System]]&lt;br /&gt;
* [[Bouncer]]&lt;br /&gt;
* [[Halting Segment]]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
https://bbchallenge.org/method#deciders&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Decider&amp;diff=1229</id>
		<title>Decider</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Decider&amp;diff=1229"/>
		<updated>2024-11-11T20:24:31Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* List of Deciders */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[Category:Deciders]]&lt;br /&gt;
A &#039;&#039;&#039;Decider&#039;&#039;&#039; (or a &#039;&#039;&#039;Filter&#039;&#039;&#039;) is a program which attempts to decide whether or not a given [[Turing machine]] (TM) will halt. Since the Halting Problem is uncomputable, no decider can decide all TMs, instead deciders categorize each TM into one of three categories: Halting, Proven Infinite, or [[Holdout]].&lt;br /&gt;
&lt;br /&gt;
There are a wide variety of methods used to construct deciders. Some broad categories are: [[Accelerated Simulator]]s, [[Backwards Reasoning]] and [[Closed Set]] methods.&lt;br /&gt;
&lt;br /&gt;
== List of Deciders ==&lt;br /&gt;
&lt;br /&gt;
* [[Translated Cycler]]&lt;br /&gt;
* [[Backward Reasoning|Backwards Reasoning]]&lt;br /&gt;
* [[Closed Position Set (CPS)]]&lt;br /&gt;
* [[Closed Tape Language (CTL)]]&lt;br /&gt;
* [[Finite Automata Reduction (FAR)]]&lt;br /&gt;
* [[Inductive Proof System]]&lt;br /&gt;
* [[Bouncer]]&lt;br /&gt;
* [[Halting Segment]]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
https://bbchallenge.org/method#deciders&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Translated_cycler&amp;diff=1227</id>
		<title>Translated cycler</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Translated_cycler&amp;diff=1227"/>
		<updated>2024-11-11T20:22:29Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Translated_cycler_44394115_annotated.svg|right|thumb|Example &amp;quot;Translated cycler&amp;quot;: 45-step space-time diagram of bbchallenge&#039;s machine {{TM|1RB0RE_0LC1RC_0RD1LA_1LE---_1LB1RC}}. The same bounded pattern is being translated to the right forever. The text annotations illustrate the main idea for recognising &amp;quot;Translated Cyclers&amp;quot;: find two configurations that break a record (i.e. visit a memory cell that was never visited before) in the same state (here state D) such that the content of the memory tape at distance L from the record positions is the same in both record configurations. Distance L is defined as being the maximum distance to record position 1 that was visited between the configuration of record 1 and record 2.]]&lt;br /&gt;
&lt;br /&gt;
A &#039;&#039;&#039;translated cycler&#039;&#039;&#039; (also known as a &#039;&#039;&#039;partial recurrent&#039;&#039;&#039; or &#039;&#039;&#039;Lin recurrent&#039;&#039;&#039; TM) is a non-halting [[Turing machine]] which eventually exhibits a traveling cyclic behavior. Specifically, a TM has ented a translated cycle once it begins repeating a fixed sequence of transition rules in such a way that it will continue repeating them forever. It is, by far, the most common type of non-halting behavior. For example, 95% of all infinite [[BB(6)]] TMs are translated cyclers (which have cycled at least once within the first 1000 steps) and this number is relatively consistent across other BB &amp;quot;domains&amp;quot; (Say BB(5), BB(3,3), etc.).&lt;br /&gt;
&lt;br /&gt;
See [https://github.com/bbchallenge/bbchallenge-proofs/blob/build-latex-pdf/deciders/correctness-deciders.pdf Section 3] of bbchallenge&#039;s deciders write-up for a formal presentation of translated cyclers.&lt;br /&gt;
&lt;br /&gt;
== History ==&lt;br /&gt;
Translated cycling behavior was first described by Shen Lin in his proof of [[BB(3)]] where he called it &amp;quot;partial recurrence&amp;quot;.&amp;lt;ref&amp;gt;Lin, Shen; Radó, Tibor (April 1965). &amp;quot;Computer Studies of Turing Machine Problems&amp;quot;. &#039;&#039;Journal of the ACM&#039;&#039;. &#039;&#039;&#039;12&#039;&#039;&#039; (2): 196–212. &amp;lt;nowiki&amp;gt;https://doi.org/10.1145/321264.321270&amp;lt;/nowiki&amp;gt;&amp;lt;/ref&amp;gt; He describes an algorithm for detecting it which appears to be the first documented example of a [[decider]]. This behavior has been given many names over the years. For example, Nick Drozd calls this Lin recurrence in honor of Shen Lin.&amp;lt;ref&amp;gt;Nick Drozd. 2021. [https://nickdrozd.github.io/2021/02/24/lin-recurrence-and-lins-algorithm.html Lin Recurrence and Lin&#039;s Algorithm]&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Record breaking ==&lt;br /&gt;
One way to detect translated cycling is by analyzing [[record breaking]] configurations. This is the algorithm used by [[bbchallenge]]. A Turing machine is a translated cycler if it has two configurations that break a record (i.e. visit a memory cell that was never visited before) in the same state such that the content of the memory tape at distance L from the record positions is the same in both record configurations. Distance L is defined as being the maximum distance to record position 1 that was visited between the configuration of record 1 and record 2.&lt;br /&gt;
&lt;br /&gt;
Here are the properties of the translated cycler shown in the figure:&lt;br /&gt;
&lt;br /&gt;
* L = 2. After the translated cycler reaches record 1, the translated cycler moves at most L = 2 symbols to the left.&lt;br /&gt;
* The &#039;&#039;&#039;cycle period&#039;&#039;&#039; is 10 steps. This is the number of steps from record 1 to record 2.&lt;br /&gt;
* The &#039;&#039;&#039;cycle offset&#039;&#039;&#039; is 2 symbols to the right. In other words, after each cycle, the TM moves 2 places to the right.&lt;br /&gt;
* The &#039;&#039;&#039;cycle start time&#039;&#039;&#039; is 6 steps. This is the position of record 1.&lt;br /&gt;
&lt;br /&gt;
Translated cyclers are close to [[Cycler|Cyclers]] in the sense that they are only repeating a pattern but there is added complexity as they are able to translate the pattern in space at the same time, hence the decider for Cyclers cannot directly apply here.&lt;br /&gt;
&lt;br /&gt;
== Infinite shift rule ==&lt;br /&gt;
A translated cycle can be seen as an infinite [[shift rule]]. For example, consider the TM {{TM|1RB0RE_0LC1RC_0RD1LA_1LE---_1LB1RC|non-halt}} in the image at the right. It performs the following [[transition rule]]:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{l}&lt;br /&gt;
\\&lt;br /&gt;
10 \; \textrm{D&amp;gt;} \; 00 \xrightarrow{10} 00 \; 10 \; \textrm{D&amp;gt;} \\&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
which can be repeated to form the shift rule:&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;\begin{array}{l}&lt;br /&gt;
\\&lt;br /&gt;
10 \; \textrm{D&amp;gt;} \; {00}^n \xrightarrow{10n} {00}^n \; 10 \; \textrm{D&amp;gt;} \\&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Furthermore, on step 6, this TM is in config&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display=&amp;quot;block&amp;quot;&amp;gt;0^\infty \; 1 \; 10 \; \textrm{D&amp;gt;} \; 0^\infty&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, we can see that for all &amp;lt;math&amp;gt;n \ge 0&amp;lt;/math&amp;gt; we can apply the shift rule and so this TM can never halt.&lt;br /&gt;
&lt;br /&gt;
== Notable translated cyclers ==&lt;br /&gt;
&lt;br /&gt;
[[Skelet 1]] is a translated cycler that has a period of 8,468,569,863 steps, an offset of 107,917 symbols to the right, and a start time of at least &amp;lt;math&amp;gt;10^{24}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
== Sources ==&lt;br /&gt;
&amp;lt;references /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[Category:Zoology]]&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=993</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=993"/>
		<updated>2024-10-03T09:54:21Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* BB(3) to BB(4) */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;People working on BB(n-1) believed BB(n) was impossible:&lt;br /&gt;
&lt;br /&gt;
== BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Source: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(3) to BB(4) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate S(3) and Σ(3), there is no evidence that any known approach will yield the answer, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), the situation seems to be entirely hopeless at present.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
Source: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Original source:&#039;&#039;&#039; Radò, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
Source: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Original source:&#039;&#039;&#039; Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024&lt;br /&gt;
&lt;br /&gt;
[[File:BradyBB(5)isHopeless.png|alt=Allen H. Brady (who proved BB(4) = 107) is saying that ever proving BB(5) is unlikely|thumb|Original source where Allen H. Brady says that solving BB(5) rigorously is unlikely to happen (1990).]]&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation)&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=992</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=992"/>
		<updated>2024-10-03T09:50:00Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;People working on BB(n-1) believed BB(n) was impossible:&lt;br /&gt;
&lt;br /&gt;
== BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Source: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(3) to BB(4) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate S(3) and Σ(3), there is no evidence that any known approach will yield the answer, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), the situation seems to be entirely hopeless at present.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
Source: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First-hand souce:&#039;&#039;&#039; Radò, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
Source: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Original source:&#039;&#039;&#039; Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024&lt;br /&gt;
&lt;br /&gt;
[[File:BradyBB(5)isHopeless.png|alt=Allen H. Brady (who proved BB(4) = 107) is saying that ever proving BB(5) is unlikely|thumb|Original source where Allen H. Brady says that solving BB(5) rigorously is unlikely to happen (1990).]]&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation)&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=991</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=991"/>
		<updated>2024-10-03T09:49:45Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* BB(4) to BB(5) */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;People working on BB(n-1) believed BB(n) was impossible:&lt;br /&gt;
&lt;br /&gt;
== BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Source: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
&lt;br /&gt;
== BB(3) to BB(4) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate S(3) and Σ(3), there is no evidence that any known approach will yield the answer, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), the situation seems to be entirely hopeless at present.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
&lt;br /&gt;
Source: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First-hand souce:&#039;&#039;&#039; Radò, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&lt;br /&gt;
&lt;br /&gt;
== BB(4) to BB(5) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
Source: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Original source:&#039;&#039;&#039; Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024&lt;br /&gt;
&lt;br /&gt;
[[File:BradyBB(5)isHopeless.png|alt=Allen H. Brady (who proved BB(4) = 107) is saying that ever proving BB(5) is unlikely|thumb|Original source where Allen H. Brady says that solving BB(5) rigorously is unlikely to happen.]]&lt;br /&gt;
&lt;br /&gt;
== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;&lt;br /&gt;
&lt;br /&gt;
=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
&lt;br /&gt;
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
&lt;br /&gt;
Source: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252&lt;br /&gt;
&lt;br /&gt;
== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
&lt;br /&gt;
Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
&lt;br /&gt;
Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
&lt;br /&gt;
But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
&lt;br /&gt;
For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
&lt;br /&gt;
== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
&lt;br /&gt;
Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation)&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
	<entry>
		<id>https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=990</id>
		<title>Busy beaver lack of hope recurrence</title>
		<link rel="alternate" type="text/html" href="https://wiki.bbchallenge.org/w/index.php?title=Busy_beaver_lack_of_hope_recurrence&amp;diff=990"/>
		<updated>2024-10-03T09:47:35Z</updated>

		<summary type="html">&lt;p&gt;Cosmo: /* BB(3) to BB(4) */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;People working on BB(n-1) believed BB(n) was impossible:&lt;br /&gt;
&lt;br /&gt;
== BB(3) ==&lt;br /&gt;
&#039;&#039;&#039;1.5th-hand report:&#039;&#039;&#039; &amp;quot;Suppose that, just to gain experience, we simplify the situation by merely asking whether a given 3-card machine will ever stop if started (with its card 1) on an all-zero tape. This particular question has been studied extensively by the authors in connection with the subject of sequential circuits. Many computer programs were written to answer this question; these programs grew larger and larger as more and more criteria for stoppers were covered. These programs were the results of co-operative efforts of experienced mathematicians and skilled programmers, and were run on some of the finest existing computers. Yet this extremely primitive-looking problem was still unsolved when this paper was presented, and &#039;&#039;&#039;probably most of the participants in the studies felt that perhaps it would always remain so&#039;&#039;&#039;. But since then, this problem has been solved by T. Rado and one of his graduate students, S. Lin.&amp;quot;&lt;br /&gt;
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Source: R. W. House &amp;amp; T. Rado, An Approach to Artificial Intelligence, IEEE Special Publication S-142, January 1963.&lt;br /&gt;
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== BB(3) to BB(4) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;In any case, even though skilled mathematicians and experienced programmers attempted to evaluate S(3) and Σ(3), there is no evidence that any known approach will yield the answer, even if we avail ourselves of high-speed computers and elaborate programs. As regards Σ(4), S(4), the situation seems to be entirely hopeless at present.&amp;quot;—Tibor Rado, 1963. &lt;br /&gt;
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Source: Brady, Allen H.. [https://docs.bbchallenge.org/papers/Brady1983.pdf “The determination of the value of Rado’s noncomputable function Σ(4) for four-state Turing machines.”] Mathematics of Computation 40 (1983): 647-665.&lt;br /&gt;
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&#039;&#039;&#039;First-hand souce:&#039;&#039;&#039; Radò, T. [https://docs.bbchallenge.org/papers/Rado1963.pdf &amp;quot;On a simple source for non-computable functions&amp;quot;], Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press of the polytechnique Institue of Brooklyn 1963.&lt;br /&gt;
[[File:Capture d’écran 2024-10-03 à 11.45.55.png|thumb|Original source where Tibor Radò says that solving BB(4) is hopeless at present (1962).]]&lt;br /&gt;
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== BB(4) to BB(5) ==&lt;br /&gt;
&#039;&#039;&#039;Second-hand report:&#039;&#039;&#039; &amp;quot;Brady predicted that there will never be a proof of the values of Sigma(5) and S(5). We are just slightly more optimistic, and are lead to recast a parable due to Erdos (who spoke in the context of determining Ramsey numbers): suppose a vastly superior alien force lands and announces that they will destroy the planet unless we provide a value of the S function, along with a proof of its correctness. If they ask for S(5) we should put all of our mathematicians, computer scientists, and computers to the task, but if they ask for S(6) we should immediately attack because the task is hopeless.&amp;quot; &lt;br /&gt;
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Source: From Machlin &amp;amp; Stout (1990) https://web.eecs.umich.edu/~qstout/abs/busyb.html&lt;br /&gt;
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&#039;&#039;&#039;Original source:&#039;&#039;&#039; Brady, Allen H, and the Meaning of Life, &#039;The Busy Beaver Game and the Meaning of Life&#039;, in Rolf Herken (ed.), &#039;&#039;The Universal Turing Machine: A Half-Century Survey&#039;&#039; (Oxford, 1990; online edn, Oxford Academic, 31 Oct. 2023), &amp;lt;nowiki&amp;gt;https://doi.org/10.1093/oso/9780198537748.003.0009&amp;lt;/nowiki&amp;gt;, accessed 26 Sept. 2024&lt;br /&gt;
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[[File:BradyBB(5)isHopeless.png|alt=Allen H. Brady (who proved BB(4) = 107) is saying that ever proving BB(5) is unlikely|thumb|Original source where Allen H. Brady says that solving BB(5) rigorously is unlikely to happen.]]&lt;br /&gt;
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== BB(5) ==&lt;br /&gt;
=== Brady on level of intelligence required ===&lt;br /&gt;
Even though it might appear now that the five-state problem is within grasp, there is a distinct possibility that the limit of practical solvability has in fact been reached. While we can follow Uhing&#039;s current champion machines [resp. σ = 1,915, s = 2,133,492, and σ = 1,471, s = 2,358,064] until they halt, it is not clear at all how the machines work. Any cleverness in their construction is not the result of human creation, so there is a conspicuous absence of documentation! In light of Green&#039;s results it was easy to accept that the turning point for the Busy Beaver Game might occur at k = 6, but such magnitudes as have now been produced for k = 5 had never been anticipated. Any hope for solving the problem at this level will require computer programs endowed with a level of intelligence that we have not seen in anything done previously by a machine. Can it be decided by a computer program or will it be necessary to assign one mathematician per unresolved five-state Turing Machine?&lt;br /&gt;
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Source: Allen Brady, 1988, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;&lt;br /&gt;
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=== Brady&#039;s Prediction 5 ===&lt;br /&gt;
&#039;&#039;Prediction 5.&#039;&#039; It will never be proved that S(5) = 2,358,064 and Σ(5) = 1,915. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)&lt;br /&gt;
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Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the &#039;&#039;Goldbach Conjecture&#039;&#039;. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.&lt;br /&gt;
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Source: Allen Brady, 1990, &amp;quot;The Busy Beaver Game and the Meaning of Life&amp;quot;, pages 251-252&lt;br /&gt;
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== BB(6) ==&lt;br /&gt;
=== Brady&#039;s Prediction 6 ===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Prediction 6.&#039;&#039; From known results for k = 5 a six-state machine will be constructed for which it can be &amp;quot;proved&amp;quot; that its shift number (and thus a lower bound for S(6)) is an incomprehensibly large value which is in itself difficult to describe. &lt;br /&gt;
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Reason: It is now clear that determining Σ(6) and S(6) is intractable. At this level one can speculate with impunity, and we shall.&lt;br /&gt;
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Some students of the author were readily convinced after extensive examination and computer testing that Uhing&#039;s champion machine for S(5) would never halt. Seeking assurance one student ran her simulator to a point just short of two million moves! From an amusing experience such as this, one is led to consider the possibility that someday a machine of six states (or a just a few more) will be presented by a group of mathematicians along with a &amp;quot;proof&amp;quot; that it will never halt. Suppose then an efficient simulator for the machine were built on the leading but slightly jagged edge of technology and run for an extensive period of time. And then suppose it were to halt! The mathematicians, with solid reasoning to back them up, could make a valid claim that the machine malfunctioned.&lt;br /&gt;
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But now suppose that instead of building a machine, another group of equivalently qualified thinkers, supported by a great body of mathematical knowledge, countered with a different &amp;quot;proof&amp;quot; that after some unimaginable number of moves the proposed machine would in fact halt. Their number would be so large that building a simulator to check the result would be inconceivable. What then? (It is only speculation, of course!)&lt;br /&gt;
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For k = 6 the problem transcends mechanism. One reaches a point where it becomes impossible to distinguish between the finite and the infinite. Is there a point at which it will transcend logic? Rado&#039;s question remains open.&lt;br /&gt;
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== BB(1963) ==&lt;br /&gt;
&lt;br /&gt;
Should one attempt to apply the method described above to the problem BB-1963, for example, then difficulties of prohibitive character are bound to arise. In the first place, the number of cases becomes astronomical, and the storage and execution for the computer programs involved will defeat any efforts to use existing computers. Even if we assume that somehow we managed to squeeze through the computer the portion of our approach involving partial recurrence patterns, the number of &amp;quot;holdouts&amp;quot; may be expected to be enormous. Over and beyond such &amp;quot;physical&amp;quot; difficulties, there is the basic fact of the non-computability of Σ(n), which implies that no single finite computer program exists that will furnish the value of Σ(n) for every n. Accordingly, there seems to be at present no justification for the assumption that Σ(n) is effectively calculable for each individual n. Evidently, these comments suggests a number of questions relating to the BB-n problem which seem to be beyond the reach of presently known methods, Thus it appears that clarification of the idea of a &amp;quot;given&amp;quot; non-negative integer may be a fruitful and certainly difficult enterprise.&lt;br /&gt;
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Lin, 1963, &amp;quot;Computer Studies of Turing Machine Problems&amp;quot; (a PhD dissertation)&lt;/div&gt;</summary>
		<author><name>Cosmo</name></author>
	</entry>
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