Antihydra - Revision history

MrSolis at 01:41, 9 May 2026

2026-05-09T01:41:12Z

← Older revision		Revision as of 01:41, 9 May 2026
Line 97:		Line 97:

	== More Analysis ==		== More Analysis ==
	The <math>A</math> recurrence is equivalent to <math display="inline">D(0)=7, D(n+1)=\lceil D(n)\frac32\rceil</math><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Antihydra.lean</ref><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Basics.lean</ref>, by <math>H(n)=D(n)+1</math>, which is of advantage because in the Odlyzko-Wilf paper<ref>https://scholar.archive.org/work/usw755p7rzgtbnfpmkliokwdfa/access/wayback/https://www.cambridge.org/core/services/aop-cambridge-core/content/view/13435C4851F0CA24F64E08357C9D64F1/S0017089500008272a.pdf/div-class-title-functional-iteration-and-the-josephus-problem-div.pdf</ref> they showed that <math display="inline">D(n)=\lfloor K\cdot(\frac32)^n\rfloor</math>, for some <math>K</math>. They showed this for starting value <math>D(0)=1</math> but the proof works perfectly for any starting value except that the constant <math>K</math> will be different<ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Constant.lean</ref>. This means that the halting problem is a question of the parities of <math>D(n)</math> (as hinted above). However, a similar question on the parity of <math display="inline">\lfloor \xi\cdot(\frac32)^n\rfloor</math> has come up with [https://en.wikipedia.org/wiki/Mahler%27s_3/2_problem Mahler's Z-number problem]. That problem is formulated as a condition on the fractional parts of <math display="inline">\xi\cdot(\frac32)^n</math> but this is just another form. The Z-number problem is a long-standing open question and it could be answered if it were proven that <math display="inline">\{ \xi\cdot(\frac32)^n\}</math>, with <math>\{x\}</math> the fractional part of <math>x</math>, is uniformly distributed (or equidistributed) over <math>(0,1(</math>. A positive answer to that could also mean that the parity vector of <math>D(n)</math> is normal and thus a random walk.		The <math>A</math> recurrence is equivalent to <math display="inline">D(0)=7, D(n+1)=\lceil D(n)\frac32\rceil</math><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Antihydra.lean</ref><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Basics.lean</ref>, by <math>H(n)=D(n)+1</math>, which is of advantage because in the Odlyzko-Wilf paper<ref>https://scholar.archive.org/work/usw755p7rzgtbnfpmkliokwdfa/access/wayback/https://www.cambridge.org/core/services/aop-cambridge-core/content/view/13435C4851F0CA24F64E08357C9D64F1/S0017089500008272a.pdf/div-class-title-functional-iteration-and-the-josephus-problem-div.pdf</ref> they showed that <math display="inline">D(n)=\lfloor K\cdot(\frac32)^n\rfloor</math>, for some <math>K</math>. They showed this for starting value <math>D(0)=1</math> but the proof works perfectly for any starting value except that the constant <math>K</math> will be different<ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Constant.lean</ref>. This means that the halting problem is a question of the parities of <math>D(n)</math> (as hinted above). However, a similar question on the parity of <math display="inline">\lfloor \xi\cdot(\frac32)^n\rfloor</math> has come up with [https://en.wikipedia.org/wiki/Mahler%27s_3/2_problem Mahler's Z-number problem]. That problem is formulated as a condition on the fractional parts of <math display="inline">\xi\cdot(\frac32)^n</math> but this is just another form. The Z-number problem is a long-standing open question and it could be answered if it were proven that <math display="inline">\{ \xi\cdot(\frac32)^n\}</math>, with <math>\{x\}</math> the fractional part of <math>x</math>, is uniformly distributed (or equidistributed) over <math>(0,1)</math>. A positive answer to that could also mean that the parity vector of <math>D(n)</math> is normal and thus a random walk.

	== Code ==		== Code ==

MrSolis: ...

2026-05-09T01:40:34Z

...

← Older revision		Revision as of 01:40, 9 May 2026
Line 114:		Line 114:
	# Add the current hydra value divided by two (integer division, rounding down) to itself (Hydra function, instead of Antihydra)		# Add the current hydra value divided by two (integer division, rounding down) to itself (Hydra function, instead of Antihydra)
	# Note that integer division by 2 is equivalent to one bit shift to the right (h >> 1)		# Note that integer division by 2 is equivalent to one bit shift to the right (h >> 1)
	h += h//2		h += h//2
	</syntaxhighlight>		</syntaxhighlight>

MrSolis: .........

2026-05-09T01:39:21Z

.........

← Older revision		Revision as of 01:39, 9 May 2026
Line 114:		Line 114:
	# Add the current hydra value divided by two (integer division, rounding down) to itself (Hydra function, instead of Antihydra)		# Add the current hydra value divided by two (integer division, rounding down) to itself (Hydra function, instead of Antihydra)
	# Note that integer division by 2 is equivalent to one bit shift to the right (h >> 1)		# Note that integer division by 2 is equivalent to one bit shift to the right (h >> 1)
	h += h//2		h += h//2
	</syntaxhighlight>		</syntaxhighlight>

DrDisentangle: /* More Analysis */ corr

2026-04-22T09:27:20Z

More Analysis: corr

← Older revision		Revision as of 09:27, 22 April 2026
Line 97:		Line 97:

	== More Analysis ==		== More Analysis ==
	The <math>A</math> recurrence is equivalent to <math display="inline">D(0)=7, D(n+1)=\lceil D(n)\frac32\rceil</math><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Antihydra.lean</ref><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Basics.lean</ref> which is of advantage because in the Odlyzko-Wilf paper<ref>https://scholar.archive.org/work/usw755p7rzgtbnfpmkliokwdfa/access/wayback/https://www.cambridge.org/core/services/aop-cambridge-core/content/view/13435C4851F0CA24F64E08357C9D64F1/S0017089500008272a.pdf/div-class-title-functional-iteration-and-the-josephus-problem-div.pdf</ref> they showed that <math display="inline">D(n)=\lfloor K\cdot(\frac32)^n\rfloor</math>, for some <math>K</math>. They showed this for starting value <math>D(0)=1</math> but the proof works perfectly for any starting value except that the constant <math>K</math> will be different<ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Constant.lean</ref>. This means that the halting problem is a question of the parities of <math>D(n)</math> (as hinted above). However, a similar question on the parity of <math display="inline">\lfloor \xi\cdot(\frac32)^n\rfloor</math> has come up with [https://en.wikipedia.org/wiki/Mahler%27s_3/2_problem Mahler's Z-number problem]. That problem is formulated as a condition on the fractional parts of <math display="inline">\xi\cdot(\frac32)^n</math> but this is just another form. The Z-number problem is a long-standing open question and it could be answered if it were proven that <math display="inline">\{ \xi\cdot(\frac32)^n\}</math>, with <math>\{x\}</math> the fractional part of <math>x</math>, is uniformly distributed (or equidistributed) over <math>(0,1(</math>. A positive answer to that could also mean that the parity vector of <math>D(n)</math> is normal and thus a random walk.		The <math>A</math> recurrence is equivalent to <math display="inline">D(0)=7, D(n+1)=\lceil D(n)\frac32\rceil</math><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Antihydra.lean</ref><ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Basics.lean</ref>, by <math>H(n)=D(n)+1</math>, which is of advantage because in the Odlyzko-Wilf paper<ref>https://scholar.archive.org/work/usw755p7rzgtbnfpmkliokwdfa/access/wayback/https://www.cambridge.org/core/services/aop-cambridge-core/content/view/13435C4851F0CA24F64E08357C9D64F1/S0017089500008272a.pdf/div-class-title-functional-iteration-and-the-josephus-problem-div.pdf</ref> they showed that <math display="inline">D(n)=\lfloor K\cdot(\frac32)^n\rfloor</math>, for some <math>K</math>. They showed this for starting value <math>D(0)=1</math> but the proof works perfectly for any starting value except that the constant <math>K</math> will be different<ref>Formalized proof at https://github.com/rwst/Antihydra-Basics/blob/main/AntihydraBasics/Constant.lean</ref>. This means that the halting problem is a question of the parities of <math>D(n)</math> (as hinted above). However, a similar question on the parity of <math display="inline">\lfloor \xi\cdot(\frac32)^n\rfloor</math> has come up with [https://en.wikipedia.org/wiki/Mahler%27s_3/2_problem Mahler's Z-number problem]. That problem is formulated as a condition on the fractional parts of <math display="inline">\xi\cdot(\frac32)^n</math> but this is just another form. The Z-number problem is a long-standing open question and it could be answered if it were proven that <math display="inline">\{ \xi\cdot(\frac32)^n\}</math>, with <math>\{x\}</math> the fractional part of <math>x</math>, is uniformly distributed (or equidistributed) over <math>(0,1(</math>. A positive answer to that could also mean that the parity vector of <math>D(n)</math> is normal and thus a random walk.

	== Code ==		== Code ==

RobinCodes: Added break between second row two figures

2026-03-30T12:29:47Z

Added break between second row two figures

← Older revision		Revision as of 12:29, 30 March 2026
Line 36:		Line 36:
	</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem		</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem
	</td><td>      </td><td></td></tr></table>		</td><td>      </td><td></td></tr></table>
	<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td><br><br>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]		<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>      </td><td><br><br>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]
	A brave busy beaver confronts the dreaded Antihydra. <br>Copyright [https://www.nicoroper.com/ Nico Roper].		A brave busy beaver confronts the dreaded Antihydra. <br>Copyright [https://www.nicoroper.com/ Nico Roper].

	</td></tr></table>		</td></tr></table>
	== Analysis ==		== Analysis ==

RobinCodes: Finalizing figures

2026-03-30T12:29:24Z

Finalizing figures

← Older revision		Revision as of 12:29, 30 March 2026
Line 36:		Line 36:
	</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem		</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem
	</td><td>      </td><td></td></tr></table>		</td><td>      </td><td></td></tr></table>
	<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]		<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td><br><br>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]
	A brave busy beaver confronts the dreaded Antihydra.		A brave busy beaver confronts the dreaded Antihydra. <br>Copyright [https://www.nicoroper.com/ Nico Roper].
	Copyright [https://www.nicoroper.com/ Nico Roper].
	</td></tr></table>		</td></tr></table>
	== Analysis ==		== Analysis ==

RobinCodes: Attempted adding soft line break no. 2

2026-03-30T12:26:37Z

Attempted adding soft line break no. 2

← Older revision		Revision as of 12:26, 30 March 2026
Line 37:		Line 37:
	</td><td>      </td><td></td></tr></table>		</td><td>      </td><td></td></tr></table>
	<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]		<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]
	A brave busy beaver confronts the dreaded Antihydra.  ~~Copyright~~ [https://www.nicoroper.com/ Nico Roper].		A brave busy beaver confronts the dreaded Antihydra.
			Copyright [https://www.nicoroper.com/ Nico Roper].
	</td></tr></table>		</td></tr></table>
	== Analysis ==		== Analysis ==

RobinCodes: Attempt to add soft line break

2026-03-30T12:25:38Z

Attempt to add soft line break

← Older revision		Revision as of 12:25, 30 March 2026
Line 37:		Line 37:
	</td><td>      </td><td></td></tr></table>		</td><td>      </td><td></td></tr></table>
	<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]		<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]
	A brave busy beaver confronts the dreaded Antihydra.		A brave busy beaver confronts the dreaded Antihydra.  Copyright [https://www.nicoroper.com/ Nico Roper].

	~~Copyright~~ [https://www.nicoroper.com/ Nico Roper].
	</td></tr></table>		</td></tr></table>
	== Analysis ==		== Analysis ==

RobinCodes: Smaller image, better line break (?)

2026-03-30T12:24:09Z

Smaller image, better line break (?)

← Older revision		Revision as of 12:24, 30 March 2026
Line 36:		Line 36:
	</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem		</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem
	</td><td>      </td><td></td></tr></table>		</td><td>      </td><td></td></tr></table>
	<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|~~356x356px~~]]		<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|302x302px]]
	A brave busy beaver confronts the dreaded Antihydra.		A brave busy beaver confronts the dreaded Antihydra.
	Copyright [https://www.nicoroper.com/ Nico Roper].</td></tr></table>
			Copyright [https://www.nicoroper.com/ Nico Roper].
			</td></tr></table>
	== Analysis ==		== Analysis ==
	Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E}\textrm{>}\;0^\infty</math>. Then,<ref name="bl">S. Ligocki, "[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]" (2024). Accessed 22 July 2024.</ref>		Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E}\textrm{>}\;0^\infty</math>. Then,<ref name="bl">S. Ligocki, "[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]" (2024). Accessed 22 July 2024.</ref>

RobinCodes: Consistency

2026-03-30T12:23:29Z

Consistency

← Older revision		Revision as of 12:23, 30 March 2026
Line 36:		Line 36:
	</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem		</td><td>      </td><td style="width: 220px;">[[File:Antihydra award.jpg\|220px]]<br>A community trophy - to be awarded to the first person or group who solves the Antihydra problem
	</td><td>      </td><td></td></tr></table>		</td><td>      </td><td></td></tr></table>
	<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|~~thumb\|~~A brave busy beaver confronts the dreaded Antihydra. Copyright [https://www.nicoroper.com/ Nico Roper].]]</td></tr></table>		<table style="margin: auto; text-align: center;"><tr><td>[[File:Antihydra-depiction.png\|200px]]<br>Artistic depiction of Antihydra by Jadeix</td><td>[[File:Nico-BB-vs-Antihydra.jpg\|356x356px]]
			A brave busy beaver confronts the dreaded Antihydra.
			Copyright [https://www.nicoroper.com/ Nico Roper].</td></tr></table>
	== Analysis ==		== Analysis ==
	Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E}\textrm{>}\;0^\infty</math>. Then,<ref name="bl">S. Ligocki, "[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]" (2024). Accessed 22 July 2024.</ref>		Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E}\textrm{>}\;0^\infty</math>. Then,<ref name="bl">S. Ligocki, "[https://www.sligocki.com/2024/07/06/bb-6-2-is-hard.html BB(6) is Hard (Antihydra)]" (2024). Accessed 22 July 2024.</ref>