User:MrSolis/Playground - Revision history

MrSolis at 15:36, 4 December 2025

2025-12-04T15:36:31Z

← Older revision	Revision as of 15:36, 4 December 2025
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MrSolis: Blanked the page

2025-02-17T00:57:50Z

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	~~{{unsolved\|Does Antihydra run forever?}}{{TM\|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA\|undecided}}~~
	~~[[File:Antihydra-depiction.png\|right\|thumb\|Artistic depiction of Antihydra by Jadeix]]~~
	~~'''ANTIHYDRA PAGE REVAMP (WIP)'''~~

	~~'''Antihydra''' is a [[BB(6)]] [[Cryptid]]. It is similar to [[Hydra]] in that it halts if and only if the sequence~~
	~~<math display="block">H_{n+1}=\bigg\lfloor\frac{3}{2}H_n\bigg\rfloor,H_0=8,</math>~~
	~~ever has more than twice the number of odd terms as the amount of even terms.~~
	~~== Analysis ==~~
	~~===Rules===~~
	Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E>}\;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>,
	~~<math display="block">\begin{array}{\|lll\|}\hline~~
	~~A(a,2b)& \xrightarrow{2a+3b^2+12b+11}& A(a+2,3b+2),\\~~
	~~A(0,2b+1)&\xrightarrow{3b^2+9b-1}& 0^\infty\;\textrm{<F}\;110\;1^{3b}\;0^\infty,\\~~
	~~A(a+1,2b+1)&\xrightarrow{3b^2+12b+5}& A(a,3b+3).\\\hline~~
	~~\end{array}</math>~~
	~~===Proof===~~
	Consider the partial configuration <math>P(m,n):=0\;1^m\;\textrm{E>}\;0\;1^n\;0^\infty</math>. The configuration after two steps is <math>0\;1^{m-1}\;0\;\textrm{A>}\;1^{n+1}\;0^\infty</math>. We note the following shift rule:
	~~<math display="block">\begin{array}{\|c\|}\hline\textrm{A>}\;1^s\xrightarrow{s}1^s\;\textrm{A>}\\\hline\end{array}</math>~~
	As a result, we get <math>0\;1^{m-1}\;0\;1^{n+1}\;\textrm{A>}\;0^\infty</math> after <math>n+1</math> steps. Advancing two steps produces <math>0\;1^{m-1}\;0\;1^{n+2}\;\textrm{<C}\;0^\infty</math>. A second shift rule is useful here:
	~~<math display="block">\begin{array}{\|c\|}\hline1^s\;\textrm{<C}\xrightarrow{s}\textrm{<C}\;1^s\\\hline\end{array}</math>~~
	This allows us to reach <math>0\;1^{m-1}\;0\;\textrm{<C}\;1^{n+2}\;0^\infty</math> in <math>n+2</math> steps. Moving five more steps gets us to <math>0\;1^{m-2}\;\textrm{E>}\;0\;1^{n+3}\;0^\infty</math>, which is the same configuration as <math>P(m-2,n+3)</math>. Accounting for the head movement creates the condition that <math>m\ge 4</math>. In summary:
	~~<math display="block">\begin{array}{\|c\|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}</math>~~
	With <math>A(a,b)</math> we have <math>P(b,0)</math>. As a result, we can apply this rule <math display="inline>\big\lfloor\frac{1}{2}b\big\rfloor-1</math> times (assuming <math>b\ge 4</math>), which creates two possible scenarios:
	#If <math>b\equiv0\ (\operatorname{mod}2)</math>, then in <math>\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6</math> steps we arrive at <math display="inline">P\Big(2,\frac{3}{2}b-3\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;011\;\textrm{E>}\;0\;1^{(3b)/2-3}\;0^\infty</math>. After <math>3b+4</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<C}\;00\;1^{(3b)/2}\;0^\infty</math>, which then leads to <math>0^\infty\;\textrm{<C}\;1^a\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a</math> steps. After five more steps, we reach <math>0^\infty\;1\;\textrm{E>}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty</math>, from which another shift rule must be applied:<math display="block">\begin{array}{\|c\|}\hline\textrm{E>}\;1^s\xrightarrow{s}1^s\;\textrm{E>}\\\hline\end{array}</math>Doing so allows us to get the configuration <math>0^\infty\;1^{a+3}\;\textrm{E>}\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a+2</math> steps. In six steps we have <math>0^\infty\;1^{a+2}\;011\;\textrm{E>}\;1^{(3b)/2}\;0^\infty</math>, so we use the shift rule again, ending at <math>0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E>}\;0^\infty</math>, equal to <math display="inline">A\Big(a+2,\frac{3}{2}b+2\Big)</math>, <math display="inline">\frac{3}{2}b</math> steps later. This gives a total of <math display="inline">2a+\frac{3}{4}b^2+6b+11</math> steps.
	#If <math>b\equiv1\ (\operatorname{mod}2)</math>, then in <math display="inline">\frac{3}{4}b^2-\frac{27}{4}</math> steps we arrive at <math display="inline">P\Big(3,\frac{3b-9}{2}\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;0111\;\textrm{E>}\;0\;1^{(3b-9)/2}\;0^\infty</math>. After <math>3b+2</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty</math>. If <math>a=0</math> then we have reached the undefined <code>F0</code> transition with a total of <math display="inline">\frac{3}{4}b^2+3b-\frac{19}{4}</math> steps. Otherwise, continuing for six steps gives us <math>0^\infty\;1^{a-1}\;0111\;\textrm{E>}\;1^{(3b-3)/2}\;0^\infty</math>. We conclude with the configuration <math>0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E>}\;0^\infty</math>, equal to <math display="inline">A\Big(a-1,\frac{3b+3}{2}\Big)</math>, in <math display="inline">\frac{3b-3}{2}</math> steps. This gives a total of <math display="inline">\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}</math> steps.
	~~The information above can be summarized as~~
	<math display="block">A(a,b)\rightarrow\begin{cases}A\Big(a+2,\frac{3}{2}b+2\Big)&\text{if }b\ge 2,b\equiv0\pmod{2};\\0^\infty\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty&\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a=0;\\A\Big(a-1,\frac{3b+3}{2}\Big)&\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a>0.\end{cases}</math>
	~~Substituting <math>b\leftarrow 2b</math> for the first case and <math>b\leftarrow 2b+1</math> for the other two yields the final result.~~
	~~== Trajectory ==~~
	~~11 steps are required to enter the configuration <math>A(0, 4)</math> before the [[Collatz-like]] rules are repeatedly applied. Here are the first few iterations:~~
	<math display="block">\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)\rightarrow\cdots\\\hline\end{array}</math>
	The halting problems for Antihydra and Hydra are connected by the [[Hydra function]], so the heuristic argument suggesting that machine is [[probviously]] nonhalting can be applied here. After <math>2^{31}</math> rule steps, we have <math>b=1073720884</math><ref name="bl"></ref>, so this machine, if treated as a random process, has an extremely minuscule chance of ever halting.
	~~==References==~~

MrSolis at 01:13, 16 February 2025

2025-02-16T01:13:04Z

← Older revision		Revision as of 01:13, 16 February 2025
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	== Analysis ==		== Analysis ==
	===Rules===		===Rules===
	Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E>}\;0^\infty</math>. Then,		Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E>}\;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>,
	<math display="block">\begin{array}{\|lll\|}\hline		<math display="block">\begin{array}{\|lll\|}\hline
	A(a,2b)& \xrightarrow{2a+3b^2+12b+11}& A(a+2,3b+2),\\		A(a,2b)& \xrightarrow{2a+3b^2+12b+11}& A(a+2,3b+2),\\
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	This allows us to reach <math>0\;1^{m-1}\;0\;\textrm{<C}\;1^{n+2}\;0^\infty</math> in <math>n+2</math> steps. Moving five more steps gets us to <math>0\;1^{m-2}\;\textrm{E>}\;0\;1^{n+3}\;0^\infty</math>, which is the same configuration as <math>P(m-2,n+3)</math>. Accounting for the head movement creates the condition that <math>m\ge 4</math>. In summary:		This allows us to reach <math>0\;1^{m-1}\;0\;\textrm{<C}\;1^{n+2}\;0^\infty</math> in <math>n+2</math> steps. Moving five more steps gets us to <math>0\;1^{m-2}\;\textrm{E>}\;0\;1^{n+3}\;0^\infty</math>, which is the same configuration as <math>P(m-2,n+3)</math>. Accounting for the head movement creates the condition that <math>m\ge 4</math>. In summary:
	<math display="block">\begin{array}{\|c\|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}</math>		<math display="block">\begin{array}{\|c\|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}</math>
	With <math>A(a,b)</math> we have <math>P(b,0)</math>. As a result, we can apply this rule <math display="inline>\big\lfloor\frac{1}{2}b\~~Big~~\rfloor-1</math> times (assuming <math>b\ge 4</math>), which creates two possible scenarios:		With <math>A(a,b)</math> we have <math>P(b,0)</math>. As a result, we can apply this rule <math display="inline>\big\lfloor\frac{1}{2}b\big\rfloor-1</math> times (assuming <math>b\ge 4</math>), which creates two possible scenarios:
	#If <math>b\equiv0\ (\operatorname{mod}2)</math>, then in <math>\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6</math> steps we arrive at <math display="inline">P\Big(2,\frac{3}{2}b-3\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;011\;\textrm{E>}\;0\;1^{(3b)/2-3}\;0^\infty</math>. After <math>3b+4</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<C}\;00\;1^{(3b)/2}\;0^\infty</math>, which then leads to <math>0^\infty\;\textrm{<C}\;1^a\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a</math> steps. After five more steps, we reach <math>0^\infty\;1\;\textrm{E>}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty</math>, from which another shift rule must be applied:<math display="block">\begin{array}{\|c\|}\hline\textrm{E>}\;1^s\xrightarrow{s}1^s\;\textrm{E>}\\\hline\end{array}</math>Doing so allows us to get the configuration <math>0^\infty\;1^{a+3}\;\textrm{E>}\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a+2</math> steps. In six steps we have <math>0^\infty\;1^{a+2}\;011\;\textrm{E>}\;1^{(3b)/2}\;0^\infty</math>, so we use the shift rule again, ending at <math>0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E>}\;0^\infty</math>, equal to <math display="inline">A\Big(a+2,\frac{3}{2}b+2\Big)</math>, <math display="inline">\frac{3}{2}b</math> steps later. This gives a total of <math display="inline">2a+\frac{3}{4}b^2+6b+11</math> steps.		#If <math>b\equiv0\ (\operatorname{mod}2)</math>, then in <math>\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6</math> steps we arrive at <math display="inline">P\Big(2,\frac{3}{2}b-3\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;011\;\textrm{E>}\;0\;1^{(3b)/2-3}\;0^\infty</math>. After <math>3b+4</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<C}\;00\;1^{(3b)/2}\;0^\infty</math>, which then leads to <math>0^\infty\;\textrm{<C}\;1^a\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a</math> steps. After five more steps, we reach <math>0^\infty\;1\;\textrm{E>}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty</math>, from which another shift rule must be applied:<math display="block">\begin{array}{\|c\|}\hline\textrm{E>}\;1^s\xrightarrow{s}1^s\;\textrm{E>}\\\hline\end{array}</math>Doing so allows us to get the configuration <math>0^\infty\;1^{a+3}\;\textrm{E>}\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a+2</math> steps. In six steps we have <math>0^\infty\;1^{a+2}\;011\;\textrm{E>}\;1^{(3b)/2}\;0^\infty</math>, so we use the shift rule again, ending at <math>0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E>}\;0^\infty</math>, equal to <math display="inline">A\Big(a+2,\frac{3}{2}b+2\Big)</math>, <math display="inline">\frac{3}{2}b</math> steps later. This gives a total of <math display="inline">2a+\frac{3}{4}b^2+6b+11</math> steps.
	#If <math>b\equiv1\ (\operatorname{mod}2)</math>, then in <math display="inline">\frac{3}{4}b^2-\frac{27}{4}</math> steps we arrive at <math display="inline">P\Big(3,\frac{3b-9}{2}\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;0111\;\textrm{E>}\;0\;1^{(3b-9)/2}\;0^\infty</math>. After <math>3b+2</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty</math>. If <math>a=0</math>, then the undefined F0 transition ~~is reached in~~ <math display="inline">\frac{3}{4}b^2+3b-\frac{19}{4}</math> steps ~~total~~. Otherwise, ~~in five~~ steps ~~the configuration is~~ <math>0^\infty\;1^{a-1}\;0111\;\textrm{<E}\;1^{(3b-3)/2}\;0^\infty</math>. ~~One final shift rule results in~~ the configuration <math>0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E>}\;0^\infty=A\~~bigg~~(a-1,\frac{3b+3}{2}\~~bigg~~)</math> ~~after~~ <math>\frac{3}{2}~~b+1~~</math> steps. This gives a total of <math display="inline">\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}</math> steps.		#If <math>b\equiv1\ (\operatorname{mod}2)</math>, then in <math display="inline">\frac{3}{4}b^2-\frac{27}{4}</math> steps we arrive at <math display="inline">P\Big(3,\frac{3b-9}{2}\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;0111\;\textrm{E>}\;0\;1^{(3b-9)/2}\;0^\infty</math>. After <math>3b+2</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty</math>. If <math>a=0</math> then we have reached the undefined <code>F0</code> transition with a total of <math display="inline">\frac{3}{4}b^2+3b-\frac{19}{4}</math> steps. Otherwise, continuing for six steps gives us <math>0^\infty\;1^{a-1}\;0111\;\textrm{E>}\;1^{(3b-3)/2}\;0^\infty</math>. We conclude with the configuration <math>0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E>}\;0^\infty</math>, equal to <math display="inline">A\Big(a-1,\frac{3b+3}{2}\Big)</math>, in <math display="inline">\frac{3b-3}{2}</math> steps. This gives a total of <math display="inline">\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}</math> steps.
	The information above can be summarized as		The information above can be summarized as
	<math display="block">A(a,b)\rightarrow\begin{cases}A\~~big~~(a+2,\frac{3}{2}b+2\~~big~~)&\text{if }b\equiv0\pmod{2}\\0^\infty\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty&\text{if }b\equiv1\pmod{2}\text{ and }a=0\\A\~~big~~(a-1,\frac{3b+3}{2}\~~big~~)&\text{~~otherwise~~}\end{cases}</math>		<math display="block">A(a,b)\rightarrow\begin{cases}A\Big(a+2,\frac{3}{2}b+2\Big)&\text{if }b\ge 2,b\equiv0\pmod{2};\\0^\infty\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty&\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a=0;\\A\Big(a-1,\frac{3b+3}{2}\Big)&\text{if }b\ge3,b\equiv1\pmod{2},\text{ and }a>0.\end{cases}</math>
	Substituting <math>b\leftarrow 2b</math> for the first case and <math>b\leftarrow 2b+1</math> for the other two yields the final result.		Substituting <math>b\leftarrow 2b</math> for the first case and <math>b\leftarrow 2b+1</math> for the other two yields the final result.
	== Trajectory ==		== Trajectory ==
	~~TODO~~		11 steps are required to enter the configuration <math>A(0, 4)</math> before the [[Collatz-like]] rules are repeatedly applied. Here are the first few iterations:
			<math display="block">\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)\rightarrow\cdots\\\hline\end{array}</math>
			The halting problems for Antihydra and Hydra are connected by the [[Hydra function]], so the heuristic argument suggesting that machine is [[probviously]] nonhalting can be applied here. After <math>2^{31}</math> rule steps, we have <math>b=1073720884</math><ref name="bl"></ref>, so this machine, if treated as a random process, has an extremely minuscule chance of ever halting.
	==References==		==References==

MrSolis at 18:59, 15 February 2025

2025-02-15T18:59:16Z

← Older revision	Revision as of 18:59, 15 February 2025
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		{{unsolved\|Does Antihydra run forever?}}{{TM\|1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA\|undecided}}
		[[File:Antihydra-depiction.png\|right\|thumb\|Artistic depiction of Antihydra by Jadeix]]
		'''ANTIHYDRA PAGE REVAMP (WIP)'''

		'''Antihydra''' is a [[BB(6)]] [[Cryptid]]. It is similar to [[Hydra]] in that it halts if and only if the sequence
		<math display="block">H_{n+1}=\bigg\lfloor\frac{3}{2}H_n\bigg\rfloor,H_0=8,</math>
		ever has more than twice the number of odd terms as the amount of even terms.
		== Analysis ==
		===Rules===
		Let <math>A(a,b):=0^\infty\;1^a\;0\;1^b\;\textrm{E>}\;0^\infty</math>. Then,
		<math display="block">\begin{array}{\|lll\|}\hline
		A(a,2b)& \xrightarrow{2a+3b^2+12b+11}& A(a+2,3b+2),\\
		A(0,2b+1)&\xrightarrow{3b^2+9b-1}& 0^\infty\;\textrm{<F}\;110\;1^{3b}\;0^\infty,\\
		A(a+1,2b+1)&\xrightarrow{3b^2+12b+5}& A(a,3b+3).\\\hline
		\end{array}</math>
		===Proof===
		Consider the partial configuration <math>P(m,n):=0\;1^m\;\textrm{E>}\;0\;1^n\;0^\infty</math>. The configuration after two steps is <math>0\;1^{m-1}\;0\;\textrm{A>}\;1^{n+1}\;0^\infty</math>. We note the following shift rule:
		<math display="block">\begin{array}{\|c\|}\hline\textrm{A>}\;1^s\xrightarrow{s}1^s\;\textrm{A>}\\\hline\end{array}</math>
		As a result, we get <math>0\;1^{m-1}\;0\;1^{n+1}\;\textrm{A>}\;0^\infty</math> after <math>n+1</math> steps. Advancing two steps produces <math>0\;1^{m-1}\;0\;1^{n+2}\;\textrm{<C}\;0^\infty</math>. A second shift rule is useful here:
		<math display="block">\begin{array}{\|c\|}\hline1^s\;\textrm{<C}\xrightarrow{s}\textrm{<C}\;1^s\\\hline\end{array}</math>
		This allows us to reach <math>0\;1^{m-1}\;0\;\textrm{<C}\;1^{n+2}\;0^\infty</math> in <math>n+2</math> steps. Moving five more steps gets us to <math>0\;1^{m-2}\;\textrm{E>}\;0\;1^{n+3}\;0^\infty</math>, which is the same configuration as <math>P(m-2,n+3)</math>. Accounting for the head movement creates the condition that <math>m\ge 4</math>. In summary:
		<math display="block">\begin{array}{\|c\|}\hline P(m,n)\xrightarrow{2n+12}P(m-2,n+3)\text{ if }m\ge 4.\\\hline\end{array}</math>
		With <math>A(a,b)</math> we have <math>P(b,0)</math>. As a result, we can apply this rule <math display="inline>\big\lfloor\frac{1}{2}b\Big\rfloor-1</math> times (assuming <math>b\ge 4</math>), which creates two possible scenarios:
		#If <math>b\equiv0\ (\operatorname{mod}2)</math>, then in <math>\sum_{i=0}^{(b/2)-2}(2\times 3i+12)=\textstyle\frac{3}{4}b^2+\frac{3}{2}b-6</math> steps we arrive at <math display="inline">P\Big(2,\frac{3}{2}b-3\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;011\;\textrm{E>}\;0\;1^{(3b)/2-3}\;0^\infty</math>. After <math>3b+4</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<C}\;00\;1^{(3b)/2}\;0^\infty</math>, which then leads to <math>0^\infty\;\textrm{<C}\;1^a\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a</math> steps. After five more steps, we reach <math>0^\infty\;1\;\textrm{E>}\;1^{a+2}\;00\;1^{(3b)/2}\;0^\infty</math>, from which another shift rule must be applied:<math display="block">\begin{array}{\|c\|}\hline\textrm{E>}\;1^s\xrightarrow{s}1^s\;\textrm{E>}\\\hline\end{array}</math>Doing so allows us to get the configuration <math>0^\infty\;1^{a+3}\;\textrm{E>}\;00\;1^{(3b)/2}\;0^\infty</math> in <math>a+2</math> steps. In six steps we have <math>0^\infty\;1^{a+2}\;011\;\textrm{E>}\;1^{(3b)/2}\;0^\infty</math>, so we use the shift rule again, ending at <math>0^\infty\;1^{a+2}\;0\;1^{(3b)/2+2}\;\textrm{E>}\;0^\infty</math>, equal to <math display="inline">A\Big(a+2,\frac{3}{2}b+2\Big)</math>, <math display="inline">\frac{3}{2}b</math> steps later. This gives a total of <math display="inline">2a+\frac{3}{4}b^2+6b+11</math> steps.
		#If <math>b\equiv1\ (\operatorname{mod}2)</math>, then in <math display="inline">\frac{3}{4}b^2-\frac{27}{4}</math> steps we arrive at <math display="inline">P\Big(3,\frac{3b-9}{2}\Big)</math>. The matching complete configuration is <math>0^\infty\;1^a\;0111\;\textrm{E>}\;0\;1^{(3b-9)/2}\;0^\infty</math>. After <math>3b+2</math> steps this becomes <math>0^\infty\;1^a\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty</math>. If <math>a=0</math>, then the undefined F0 transition is reached in <math display="inline">\frac{3}{4}b^2+3b-\frac{19}{4}</math> steps total. Otherwise, in five steps the configuration is <math>0^\infty\;1^{a-1}\;0111\;\textrm{<E}\;1^{(3b-3)/2}\;0^\infty</math>. One final shift rule results in the configuration <math>0^\infty\;1^{a-1}\;0\;1^{(3b+3)/2}\;\textrm{E>}\;0^\infty=A\bigg(a-1,\frac{3b+3}{2}\bigg)</math> after <math>\frac{3}{2}b+1</math> steps. This gives a total of <math display="inline">\frac{3}{4}b^2+\frac{9}{2}b-\frac{1}{4}</math> steps.
		The information above can be summarized as
		<math display="block">A(a,b)\rightarrow\begin{cases}A\big(a+2,\frac{3}{2}b+2\big)&\text{if }b\equiv0\pmod{2}\\0^\infty\;\textrm{<F}\;110\;1^{(3b-3)/2}\;0^\infty&\text{if }b\equiv1\pmod{2}\text{ and }a=0\\A\big(a-1,\frac{3b+3}{2}\big)&\text{otherwise}\end{cases}</math>
		Substituting <math>b\leftarrow 2b</math> for the first case and <math>b\leftarrow 2b+1</math> for the other two yields the final result.
		== Trajectory ==
		TODO
		==References==

MrSolis: Blanked the page

2025-02-14T16:28:36Z

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← Older revision		Revision as of 16:28, 14 February 2025
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	~~'''HYDRA PAGE REVAMP (WIP)'''~~
	~~{{unsolved\|Does Hydra run forever?}}{{TM\|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA\|undecided}}~~
	~~Hydra is a [[BB(2,5)]] [[Cryptid]]. It simulates computing the terms of the sequence~~
	~~<math display="block>H_{n+1}=\bigg\lfloor\frac{3}{2}H_n\bigg\rfloor,H_0=3,</math>~~
	~~halting if and only if there exists a point in the sequence where the number of even terms up to that point exceeds twice the number of odd terms.~~
	~~== Analysis ==~~
	~~===Rules===~~
	~~Let <math>C(a,b):=0^\infty\;\textrm{<A}\;2\;0^a\;3^b\;2\;0^\infty</math>. Then,~~
	~~<math display="block">\begin{array}{\|lll\|}\hline~~
	~~C(2a,0)&\xrightarrow{6a^2+20a+4}&0^\infty\;3^{3a+1}\;1\;\textrm{A>}\;2\;0^\infty,\\~~
	~~C(2a,b+1)&\xrightarrow{6a^2+23a+10}&C(3a+3,b),\\~~
	~~C(2a+1,b)&\xrightarrow{4b+6a^2+23a+26}&C(3a+3,b+2).\\\hline~~
	~~\end{array}</math>~~
	~~By scaling and translating these rules we acquire the [[Hydra function]] that relates it to [[Antihydra]].~~
	~~===Proof===~~
	~~Consider the partial configuration <math>P(m,n):=0^\infty\;3^m\;\textrm{A>}\;02\;0^n</math>. After 14 steps this configuration becomes~~
	~~<math>0^\infty\;3^{m+3}\;\textrm{<A}\;2\;0^{n-2}</math>. We note the following shift rule:~~
	~~<math display="block">\begin{array}{\|c\|}\hline3^s\;\textrm{<A}\xrightarrow{s}\textrm{<A}\;3^s\\\hline\end{array}</math>~~
	Using this shift rule, we get <math>0^\infty\;\textrm{<A}\;3^{m+3}\;2\;0^{n-2}</math> in <math>m+3</math> steps. From here, we can observe that <math>\textrm{A>}\;0\;3^s</math> turns into <math>3\;\textrm{A>}\;0\;3^{s-1}</math> in three steps if <math>s\ge 1</math>. By repeating this process, we acquire this rule:
	~~<math display="block">\begin{array}{\|c\|}\hline\textrm{A>}\;0\;3^s\xrightarrow{3s}3^s\;\textrm{A>}\;0\\\hline\end{array}</math>~~
	With this rule, it takes <math>3m+9</math> steps to reach the configuration <math>0^\infty\;3^{m+3}\;\textrm{A>}\;02\;0^{n-2}</math>, which is the same configuration as <math>P(m+3,n-2)</math>. To summarize:
	~~<math display="block">\begin{array}{\|c\|}\hline P(m,n)\xrightarrow{4m+26}P(m+3,n-2)\text{ if }n\ge 2.\\\hline\end{array}</math>~~
	~~With <math>C(a,b)</math> we have <math>P(0,a)</math>. As a result, we can apply this rule <math display="inline">\big\lfloor\frac{1}{2}a\big\rfloor</math> times, which creates two possible scenarios:~~
	#If <math>a\equiv0\ (\operatorname{mod}2)</math>, then in <math>\sum_{i=0}^{(a/2)-1}(4\times 3i+26)=\textstyle\frac{3}{2}a^2+10a</math> steps we arrive at <math display="inline">P\Big(\frac{3}{2}a,0\Big)</math>. The matching complete configuration is <math>0^\infty\;3^{(3/2)a}\;\textrm{A>}\;02\;3^b\;2\;0^\infty</math>, which in four steps becomes <math>0^\infty\;3^{1+(3/2)a}\;1\;\textrm{A>}\;3^b\;2\;0^\infty.</math> If <math>b=0</math> then we have reached the undefined <code>A2</code> transition in <math display="inline">\frac{3}{2}a^2+10a+4</math> steps total. Otherwise, continuing for three steps gives us <math>0^\infty\;3^{2+(3/2)a}\;\textrm{<B}\;0\;3^{b-1}\;2\;0^\infty</math>. Another shift rule is required here:<math display="block">\begin{array}{\|c\|}\hline3^s\;\textrm{<B}\xrightarrow{s}\textrm{<B}\;0^s\\\hline\end{array}</math>This means the configuration becomes <math>0^\infty\;\textrm{<B}\;0^{3+(3/2)a}\;3^{b-1}\;2\;0^\infty</math> in <math display="inline">\frac{3}{2}a+2</math> steps, and <math>0^\infty\;\textrm{<A}\;2\;0^{3+(3/2)a}\;3^{b-1}\;2\;0^\infty</math>, equal to <math display="inline">C\Big(\frac{3}{2}a+3,b-1\Big)</math>, one step later. This gives a total of <math display="inline">\frac{3}{2}a^2+\frac{23}{2}a+10</math> steps.
	#If <math>a\equiv1\ (\operatorname{mod}2)</math>, then in <math display="inline">\frac{3}{2}a^2+7a-\frac{17}{2}</math> steps we arrive at <math display="inline">P\Big(\frac{3a-3}{2},1\Big)</math>. The matching complete configuration is <math>0^\infty\;3^{(3a-3)/2}\;\textrm{A>}\;020\;3^b\;2\;0^\infty</math>, which in four steps becomes <math>0^\infty\;3^{(3a-1)/2}\;1\;\textrm{A>}\;0\;3^b\;2\;0^\infty</math>, and then <math>0^\infty\;3^{(3a-1)/2}\;1\;3^b\;\textrm{A>}\;02\;0^\infty</math> in <math>3b</math> steps. After 14 steps, we see the configuration <math>0^\infty\;3^{(3a-1)/2}\;1\;3^{b+3}\;\textrm{<A}\;2\;0^\infty</math>, which turns into <math>0^\infty\;3^{(3a-1)/2}\;1\;\textrm{<A}\;3^{b+3}\;2\;0^\infty</math> in <math>b+3</math> steps. In two steps we get <math>0^\infty\;3^{(3a+1)/2}\;\textrm{<B}\;0\;3^{b+2}\;2\;0^\infty</math>, followed by <math>0^\infty\;\textrm{<B}\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty</math> after another <math display="inline">\frac{3a+1}{2}</math> steps. We conclude with <math>0^\infty\;\textrm{<A}\;2\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty</math>, equal to <math display="inline">C\Big(\frac{3a+3}{2},b+2\Big)</math>, one step later. This gives a total of <math display="inline">4b+\frac{3}{2}a^2+\frac{17}{2}a+16</math> steps.
	~~The information above can be summarized as~~
	<math display="block">C(a,b)\rightarrow\begin{cases}0^\infty\;3^{(3/2)a+1}\;1\;\textrm{A>}\;2\;0^\infty&\text{if }a\equiv0\pmod{2}\text{ and }b=0,\\C\Big(\frac{3}{2}a+3,b-1\Big)&\text{if }a\equiv0\pmod{2}\text{ and }b>0,\\C\Big(\frac{3a+3}{2},b+2\Big)&\text{if }a\equiv1\pmod2.\end{cases}</math>
	~~Substituting <math>a\leftarrow 2a</math> for the first two cases and <math>a\leftarrow 2a+1</math> for the third yields the final result.~~
	~~== Trajectory ==~~
	~~It takes 20 steps to reach the configuration <math>C(3,0)</math>, and from there, the [[Collatz-like]] rules are repeatedly applied. Here are the first few iterations:~~
	<math display="block">\begin{array}{\|c\|}\hline0^\infty\;\textrm{A>}\;0^\infty\xrightarrow{20}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)\rightarrow\cdots\\\hline\end{array}</math>
	After 60 million rule steps, there are 29995836 even values of <math>a</math> and 30004165 odd values, giving a very high <math>b</math> value. However, this does not sufficiently prove that Hydra does not halt.
	~~===A probabilistic nonhalting argument===~~
	The trajectory of <math>b</math> values can be approximated by a random walk, where the walker can only move in step sizes +2 or -1 with equal probability, starting at position 0. The expected position of the walker after <math>k</math> steps is <math display="inline">\frac{1}{2}k</math>, and it can be shown that the probability of the walker reaching position -1 from position <math>n</math> is <math display="inline">{\Big(\frac{\sqrt{5}-1}{2}\Big)}^{n+1}</math>.
	~~<div class="toccolours mw-collapsible mw-collapsed">'''Proof'''<div class="mw-collapsible-content">~~
	Let <math>P(n)</math> denote the probability of the random walker reaching position 0 from position <math>n</math>. If the walker reaches position 0, it will do so either by first moving +2 with probability <math display="inline">\frac{1}{2}</math> or first moving -1 with probability <math display="inline">\frac{1}{2}</math>. Therefore, the recurrence relation is
	~~<math display="block">\textstyle P(n)=\frac{1}{2}P(n+2)+\frac{1}{2}P(n-1)\text{ for }n\ge1.</math>~~
	~~Bringing all terms with <math>P</math> to the left side of the equation and then substituting <math>n\leftarrow n+1</math> gives~~
	~~<math display="block">\textstyle P(n+1)-\frac{1}{2}P(n+3)-\frac{1}{2}P(n)=0.</math>~~
	This equation has the form <math>\sum_{i=0}^k a_iP(n+i)=0</math>, which can be solved using the zeroes of the characteristic polynomial <math>f(z)=\sum_{i=0}^k a_iz^i</math>. In this instance we get <math display="inline">f(z)=-\frac{1}{2}+z-\frac{1}{2}z^3</math>, whose zeroes are given by
	~~<math display="block">\textstyle z_0=\frac{\sqrt{5}-1}{2},\qquad\qquad z_1=1,\qquad\qquad z_2=-\frac{1+\sqrt{5}}{2}.</math>~~
	For each real root <math>z_i</math> with multiplicity 1, its fundamental solution is <math>c_i{\left(z_i\right)}^n</math>, and combining these fundamental solutions produces the general solution. Therefore,
	~~<math display="block">\textstyle P(n)=c_0{\left(\frac{\sqrt{5}-1}{2}\right)}^n+c_1+c_2{\left(-\frac{1+\sqrt{5}}{2}\right)}^n</math>~~
	The boundary condition <math>\lim_{n\to\infty} P(n)=0</math> means <math>c_2=0</math> (since <math display="inline">\left\vert-\frac{1+\sqrt{5}}{2}\right\vert > 1</math>) and <math>c_1=0</math>, and the boundary condition <math>P(0)=1</math> requires that <math>c_0=1</math>.

	Finally, we note that reaching position -1 from position <math>n</math>, the required condition for halting, is the same as reaching position 0 from position <math>n+1</math>, so we must increment <math>n</math>.
	~~</div></div>~~
	~~For these reasons, Hydra is considered to be a [[probviously]] nonhalting machine.~~

MrSolis at 02:08, 14 February 2025

2025-02-14T02:08:18Z

← Older revision		Revision as of 02:08, 14 February 2025
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	'''HYDRA PAGE REVAMP (WIP)'''		'''HYDRA PAGE REVAMP (WIP)'''
	~~{{machine\|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA}}~~{{unsolved\|Does Hydra run forever?}}{{TM\|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA\|undecided}}		{{unsolved\|Does Hydra run forever?}}{{TM\|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA\|undecided}}
	Hydra is a [[BB(2,5)]] [[Cryptid]]. It simulates computing the terms of the sequence		Hydra is a [[BB(2,5)]] [[Cryptid]]. It simulates computing the terms of the sequence
	<math display="block>H_{n+1}=\bigg\lfloor\frac{3}{2}H_n\bigg\rfloor,H_0=3,</math>		<math display="block>H_{n+1}=\bigg\lfloor\frac{3}{2}H_n\bigg\rfloor,H_0=3,</math>
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	Bringing all terms with <math>P</math> to the left side of the equation and then substituting <math>n\leftarrow n+1</math> gives		Bringing all terms with <math>P</math> to the left side of the equation and then substituting <math>n\leftarrow n+1</math> gives
	<math display="block">\textstyle P(n+1)-\frac{1}{2}P(n+3)-\frac{1}{2}P(n)=0.</math>		<math display="block">\textstyle P(n+1)-\frac{1}{2}P(n+3)-\frac{1}{2}P(n)=0.</math>
	This equation has the form <math>\sum_{i=0}^k a_iP(n+i)=0</math>, which can be solved using the characteristic polynomial <math>f(z)=\sum_{i=0}^k a_iz^i</math>. In this instance we get <math display="inline">f(z)=-\frac{1}{2}+z-\frac{1}{2}z^3</math>, whose ~~solutions~~ are given by		This equation has the form <math>\sum_{i=0}^k a_iP(n+i)=0</math>, which can be solved using the zeroes of the characteristic polynomial <math>f(z)=\sum_{i=0}^k a_iz^i</math>. In this instance we get <math display="inline">f(z)=-\frac{1}{2}+z-\frac{1}{2}z^3</math>, whose zeroes are given by
	<math display="block">\textstyle z_0=\frac{\sqrt{5}-1}{2},\qquad\qquad z_1=1,\qquad\qquad z_2=-\frac{1+\sqrt{5}}{2}.</math>		<math display="block">\textstyle z_0=\frac{\sqrt{5}-1}{2},\qquad\qquad z_1=1,\qquad\qquad z_2=-\frac{1+\sqrt{5}}{2}.</math>
	For each real root <math>z_i</math> with multiplicity 1, its fundamental solution is <math>c_i{\left(z_i\right)}^n</math>, and combining these fundamental solutions produces the general solution. Therefore,		For each real root <math>z_i</math> with multiplicity 1, its fundamental solution is <math>c_i{\left(z_i\right)}^n</math>, and combining these fundamental solutions produces the general solution. Therefore,

MrSolis at 00:42, 14 February 2025

2025-02-14T00:42:16Z

← Older revision	Revision as of 00:42, 14 February 2025
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		'''HYDRA PAGE REVAMP (WIP)'''
		{{machine\|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA}}{{unsolved\|Does Hydra run forever?}}{{TM\|1RB3RB---3LA1RA_2LA3RA4LB0LB0LA\|undecided}}
		Hydra is a [[BB(2,5)]] [[Cryptid]]. It simulates computing the terms of the sequence
		<math display="block>H_{n+1}=\bigg\lfloor\frac{3}{2}H_n\bigg\rfloor,H_0=3,</math>
		halting if and only if there exists a point in the sequence where the number of even terms up to that point exceeds twice the number of odd terms.
		== Analysis ==
		===Rules===
		Let <math>C(a,b):=0^\infty\;\textrm{<A}\;2\;0^a\;3^b\;2\;0^\infty</math>. Then,
		<math display="block">\begin{array}{\|lll\|}\hline
		C(2a,0)&\xrightarrow{6a^2+20a+4}&0^\infty\;3^{3a+1}\;1\;\textrm{A>}\;2\;0^\infty,\\
		C(2a,b+1)&\xrightarrow{6a^2+23a+10}&C(3a+3,b),\\
		C(2a+1,b)&\xrightarrow{4b+6a^2+23a+26}&C(3a+3,b+2).\\\hline
		\end{array}</math>
		By scaling and translating these rules we acquire the [[Hydra function]] that relates it to [[Antihydra]].
		===Proof===
		Consider the partial configuration <math>P(m,n):=0^\infty\;3^m\;\textrm{A>}\;02\;0^n</math>. After 14 steps this configuration becomes
		<math>0^\infty\;3^{m+3}\;\textrm{<A}\;2\;0^{n-2}</math>. We note the following shift rule:
		<math display="block">\begin{array}{\|c\|}\hline3^s\;\textrm{<A}\xrightarrow{s}\textrm{<A}\;3^s\\\hline\end{array}</math>
		Using this shift rule, we get <math>0^\infty\;\textrm{<A}\;3^{m+3}\;2\;0^{n-2}</math> in <math>m+3</math> steps. From here, we can observe that <math>\textrm{A>}\;0\;3^s</math> turns into <math>3\;\textrm{A>}\;0\;3^{s-1}</math> in three steps if <math>s\ge 1</math>. By repeating this process, we acquire this rule:
		<math display="block">\begin{array}{\|c\|}\hline\textrm{A>}\;0\;3^s\xrightarrow{3s}3^s\;\textrm{A>}\;0\\\hline\end{array}</math>
		With this rule, it takes <math>3m+9</math> steps to reach the configuration <math>0^\infty\;3^{m+3}\;\textrm{A>}\;02\;0^{n-2}</math>, which is the same configuration as <math>P(m+3,n-2)</math>. To summarize:
		<math display="block">\begin{array}{\|c\|}\hline P(m,n)\xrightarrow{4m+26}P(m+3,n-2)\text{ if }n\ge 2.\\\hline\end{array}</math>
		With <math>C(a,b)</math> we have <math>P(0,a)</math>. As a result, we can apply this rule <math display="inline">\big\lfloor\frac{1}{2}a\big\rfloor</math> times, which creates two possible scenarios:
		#If <math>a\equiv0\ (\operatorname{mod}2)</math>, then in <math>\sum_{i=0}^{(a/2)-1}(4\times 3i+26)=\textstyle\frac{3}{2}a^2+10a</math> steps we arrive at <math display="inline">P\Big(\frac{3}{2}a,0\Big)</math>. The matching complete configuration is <math>0^\infty\;3^{(3/2)a}\;\textrm{A>}\;02\;3^b\;2\;0^\infty</math>, which in four steps becomes <math>0^\infty\;3^{1+(3/2)a}\;1\;\textrm{A>}\;3^b\;2\;0^\infty.</math> If <math>b=0</math> then we have reached the undefined <code>A2</code> transition in <math display="inline">\frac{3}{2}a^2+10a+4</math> steps total. Otherwise, continuing for three steps gives us <math>0^\infty\;3^{2+(3/2)a}\;\textrm{<B}\;0\;3^{b-1}\;2\;0^\infty</math>. Another shift rule is required here:<math display="block">\begin{array}{\|c\|}\hline3^s\;\textrm{<B}\xrightarrow{s}\textrm{<B}\;0^s\\\hline\end{array}</math>This means the configuration becomes <math>0^\infty\;\textrm{<B}\;0^{3+(3/2)a}\;3^{b-1}\;2\;0^\infty</math> in <math display="inline">\frac{3}{2}a+2</math> steps, and <math>0^\infty\;\textrm{<A}\;2\;0^{3+(3/2)a}\;3^{b-1}\;2\;0^\infty</math>, equal to <math display="inline">C\Big(\frac{3}{2}a+3,b-1\Big)</math>, one step later. This gives a total of <math display="inline">\frac{3}{2}a^2+\frac{23}{2}a+10</math> steps.
		#If <math>a\equiv1\ (\operatorname{mod}2)</math>, then in <math display="inline">\frac{3}{2}a^2+7a-\frac{17}{2}</math> steps we arrive at <math display="inline">P\Big(\frac{3a-3}{2},1\Big)</math>. The matching complete configuration is <math>0^\infty\;3^{(3a-3)/2}\;\textrm{A>}\;020\;3^b\;2\;0^\infty</math>, which in four steps becomes <math>0^\infty\;3^{(3a-1)/2}\;1\;\textrm{A>}\;0\;3^b\;2\;0^\infty</math>, and then <math>0^\infty\;3^{(3a-1)/2}\;1\;3^b\;\textrm{A>}\;02\;0^\infty</math> in <math>3b</math> steps. After 14 steps, we see the configuration <math>0^\infty\;3^{(3a-1)/2}\;1\;3^{b+3}\;\textrm{<A}\;2\;0^\infty</math>, which turns into <math>0^\infty\;3^{(3a-1)/2}\;1\;\textrm{<A}\;3^{b+3}\;2\;0^\infty</math> in <math>b+3</math> steps. In two steps we get <math>0^\infty\;3^{(3a+1)/2}\;\textrm{<B}\;0\;3^{b+2}\;2\;0^\infty</math>, followed by <math>0^\infty\;\textrm{<B}\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty</math> after another <math display="inline">\frac{3a+1}{2}</math> steps. We conclude with <math>0^\infty\;\textrm{<A}\;2\;0^{(3a+3)/2}\;3^{b+2}\;2\;0^\infty</math>, equal to <math display="inline">C\Big(\frac{3a+3}{2},b+2\Big)</math>, one step later. This gives a total of <math display="inline">4b+\frac{3}{2}a^2+\frac{17}{2}a+16</math> steps.
		The information above can be summarized as
		<math display="block">C(a,b)\rightarrow\begin{cases}0^\infty\;3^{(3/2)a+1}\;1\;\textrm{A>}\;2\;0^\infty&\text{if }a\equiv0\pmod{2}\text{ and }b=0,\\C\Big(\frac{3}{2}a+3,b-1\Big)&\text{if }a\equiv0\pmod{2}\text{ and }b>0,\\C\Big(\frac{3a+3}{2},b+2\Big)&\text{if }a\equiv1\pmod2.\end{cases}</math>
		Substituting <math>a\leftarrow 2a</math> for the first two cases and <math>a\leftarrow 2a+1</math> for the third yields the final result.
		== Trajectory ==
		It takes 20 steps to reach the configuration <math>C(3,0)</math>, and from there, the [[Collatz-like]] rules are repeatedly applied. Here are the first few iterations:
		<math display="block">\begin{array}{\|c\|}\hline0^\infty\;\textrm{A>}\;0^\infty\xrightarrow{20}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)\rightarrow\cdots\\\hline\end{array}</math>
		After 60 million rule steps, there are 29995836 even values of <math>a</math> and 30004165 odd values, giving a very high <math>b</math> value. However, this does not sufficiently prove that Hydra does not halt.
		===A probabilistic nonhalting argument===
		The trajectory of <math>b</math> values can be approximated by a random walk, where the walker can only move in step sizes +2 or -1 with equal probability, starting at position 0. The expected position of the walker after <math>k</math> steps is <math display="inline">\frac{1}{2}k</math>, and it can be shown that the probability of the walker reaching position -1 from position <math>n</math> is <math display="inline">{\Big(\frac{\sqrt{5}-1}{2}\Big)}^{n+1}</math>.
		<div class="toccolours mw-collapsible mw-collapsed">'''Proof'''<div class="mw-collapsible-content">
		Let <math>P(n)</math> denote the probability of the random walker reaching position 0 from position <math>n</math>. If the walker reaches position 0, it will do so either by first moving +2 with probability <math display="inline">\frac{1}{2}</math> or first moving -1 with probability <math display="inline">\frac{1}{2}</math>. Therefore, the recurrence relation is
		<math display="block">\textstyle P(n)=\frac{1}{2}P(n+2)+\frac{1}{2}P(n-1)\text{ for }n\ge1.</math>
		Bringing all terms with <math>P</math> to the left side of the equation and then substituting <math>n\leftarrow n+1</math> gives
		<math display="block">\textstyle P(n+1)-\frac{1}{2}P(n+3)-\frac{1}{2}P(n)=0.</math>
		This equation has the form <math>\sum_{i=0}^k a_iP(n+i)=0</math>, which can be solved using the characteristic polynomial <math>f(z)=\sum_{i=0}^k a_iz^i</math>. In this instance we get <math display="inline">f(z)=-\frac{1}{2}+z-\frac{1}{2}z^3</math>, whose solutions are given by
		<math display="block">\textstyle z_0=\frac{\sqrt{5}-1}{2},\qquad\qquad z_1=1,\qquad\qquad z_2=-\frac{1+\sqrt{5}}{2}.</math>
		For each real root <math>z_i</math> with multiplicity 1, its fundamental solution is <math>c_i{\left(z_i\right)}^n</math>, and combining these fundamental solutions produces the general solution. Therefore,
		<math display="block">\textstyle P(n)=c_0{\left(\frac{\sqrt{5}-1}{2}\right)}^n+c_1+c_2{\left(-\frac{1+\sqrt{5}}{2}\right)}^n</math>
		The boundary condition <math>\lim_{n\to\infty} P(n)=0</math> means <math>c_2=0</math> (since <math display="inline">\left\vert-\frac{1+\sqrt{5}}{2}\right\vert > 1</math>) and <math>c_1=0</math>, and the boundary condition <math>P(0)=1</math> requires that <math>c_0=1</math>.

		Finally, we note that reaching position -1 from position <math>n</math>, the required condition for halting, is the same as reaching position 0 from position <math>n+1</math>, so we must increment <math>n</math>.
		</div></div>
		For these reasons, Hydra is considered to be a [[probviously]] nonhalting machine.

MrSolis: MrSolis moved page User:MrSolis/BB5ChampRM to User:MrSolis/Playground

2025-02-08T23:18:22Z

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	~~=5-state busy beaver winner (WIP Revamp)=~~
	The 5-state busy beaver ([[BB(5)]]) winner is {{TM\|1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA\|halt}}. Discovered by Heiner Marxen and Jürgen Buntrock in 1989<ref>H. Marxen and J. Buntrock. Attacking the Busy Beaver 5. Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html</ref>, this machine proved that <math>\operatorname{BB}(5)\ge 47176870</math> and <math>\Sigma(5)\ge 4098</math> at the time.
	~~== Analysis ==~~
	~~===Rules===~~
	~~Let <math>g(x):=0^\infty\;\textrm{<A}\,1^x\;0^\infty</math>. Then<ref>Pascal Michel. Behavior of busy beavers.https://bbchallenge.org/~pascal.michel/beh#tm52a</ref>,~~
	~~<math display="block">\begin{array}{\|lll\|}\hline~~
	~~g(3x)&\xrightarrow{5x^2+19x+15}&g(5x+6),\\~~
	~~g(3x+1)&\xrightarrow{5x^2+25x+27}&g(5x+9),\\~~
	~~g(3x+2)&\xrightarrow{6x+12}&0^\infty\;1\;\textrm{Z>}\;01\;001^{x+1}\;1\;0^\infty.\\\hline~~
	~~\end{array}</math>~~
	~~===Proof===~~
	Consider the configuration <math>C(m,n):=0^\infty\;\textrm{<A}\;1^m\;001^n\;1\;0^\infty</math>. After one step this configuration becomes <math>0^\infty\;1\;\textrm{B>}\;1^m\;001^n\;1\;0^\infty</math>. We note the following shift rule:
	~~<math display="block">\begin{array}{\|c\|}\hline\textrm{B>}\;1^a\xrightarrow{a}1^a\;\textrm{B>}\\\hline\end{array}</math>~~
	Using this shift rule, we get <math>0^\infty\;1^{m+1}\;\textrm{B>}\;001^n\;1\;0^\infty</math> after <math>m</math> steps. If <math>n=0</math>, then we get <math>0^\infty\;1^{m+4}\;\textrm{<A}\;1\;0^\infty</math> four steps later. Another shift rule is needed here:
	~~<math display="block">\begin{array}{\|c\|}\hline1^{3a}\;\textrm{<A}\xrightarrow{3a}\textrm{<A}\;001^a\\\hline\end{array}</math>~~
	In this instance, <math display="inline">\Big\lfloor\frac{m+4}{3}\Big\rfloor</math> is substituted for <math>a</math>, which creates three different scenarios depending on the value of <math>m</math> modulo 3. They are as follows:
	# If <math>m+4\equiv0\ (\operatorname{mod}3)</math>, then in <math>m+4</math> steps we arrive at <math>0^\infty\;\textrm{<A}\;001^{(m+4)/3}\;1\;0^\infty</math>, which is the same configuration as <math display="inline">C\Big(0,\frac{m+4}{3}\Big)</math>.
	# If <math>m+4\equiv1\ (\operatorname{mod}3)</math>, then in <math>m+3</math> steps we arrive at <math>0^\infty\;1\;\textrm{<A}\;001^{(m+3)/3}\;1\;0^\infty</math>, which is five steps becomes <math>0^\infty\;\textrm{<A}\;111\;001^{(m+3)/3}\;1\;0^\infty</math>, equal to <math display="inline">C\Big(3,\frac{m+3}{3}\Big)</math>.
	# If <math>m+4\equiv2\ (\operatorname{mod}3)</math>, then in <math>m+2</math> steps we arrive at <math>0^\infty\;11\;\textrm{<A}\;001^{(m+2)/3}\;1\;0^\infty</math>, which in three steps halts with the configuration <math>0^\infty\;1\;\textrm{Z>}\;01\;001^{(m+2)/3}\;1\;0^\infty</math>, for a total of <math>2m+10</math> steps from <math>C(m,0)</math>.
	Returning to <math>0^\infty\;1^{m+1}\;\textrm{B>}\;001^n\;1\;0^\infty</math>, if <math>n\ge 1</math>, then in three steps it changes into <math>0^\infty\;1^{m+3}\;\textrm{<D}\;1\;001^{n-1}\;1\;0^\infty</math>. Here we can make use of one more shift rule:
	~~<math display="block">\begin{array}{\|c\|}\hline1^a\;\textrm{<D}\xrightarrow{a}\textrm{<D}\;1^a\\\hline\end{array}</math>~~
	Doing so takes us to <math>0^\infty\;\textrm{<D}\;1^{m+4}\;001^{n-1}\;1\;0^\infty</math> in <math>m+3</math> steps, which after one step becomes the configuration <math>0^\infty\;\textrm{<A}\;1^{m+5}\;001^{n-1}\;1\;0^\infty</math>, equal to <math>C(m+5,n-1)</math>. To summarize:
	~~<math display="block">\begin{array}{\|c\|}\hline C(m,n)\xrightarrow{2m+8}C(m+5,n-1)\text{ if }n\ge 1.\\\hline\end{array}</math>~~
	We have <math>g(x)=C(x-1,0)</math>. As a result, if <math>x\equiv0\ (\operatorname{mod}3)</math>, we then get <math display="inline">C\Big(0,\frac{1}{3}x+1\Big)</math> and the above rule is applied until we reach <math display="inline">C\Big(\frac{5}{3}x+5,0\Big)</math>, equal to <math display="inline">g\Big(\frac{5}{3}x+6\Big)</math>, in <math>\sum_{i=0}^{x/3}(2\times 5i+8)=\textstyle\frac{5}{9}x^2+\frac{13}{3}x+8</math> steps for a total of <math display="inline">\frac{5}{9}x^2+\frac{19}{3}x+15</math> steps from <math>g(x)</math> (with <math>g(0)</math> we see the impossible configuration <math>C(-1,0)</math>, but it reaches <math>g(6)</math> in 15 steps regardless). However, if <math>x\equiv1\ (\operatorname{mod}3)</math>, we then get <math display="inline">C\Big(3,\frac{x+2}{3}\Big)</math> which reaches <math display="inline">C\Big(3+\frac{5(x+2)}{3},0\Big)</math>, equal to <math display="inline">g\Big(\frac{5x+22}{3}\Big)</math>, in <math display="inline">\frac{5}{9}x^2+\frac{47}{9}x+\frac{74}{9}</math> steps (<math display="inline">\frac{5}{9}x^2+\frac{65}{9}x+\frac{173}{9}</math> steps total).

	~~The information above can be summarized as<ref>Aaronson, S. (2020). The Busy Beaver Frontier. Page 10-11. https://www.scottaaronson.com/papers/bb.pdf</ref>~~
	<math display="block">g(x)\rightarrow\begin{cases}g\Big(\frac{5}{3}x+6\Big)&\text{if }x\equiv0\pmod{3}\\g\Big(\frac{5x+22}{3}\Big)&\text{if }x\equiv1\pmod{3}\\0^\infty\;1\;\textrm{Z>}\;01\;001^{(x+1)/3}\;1\;0^\infty&\text{if }x\equiv2\pmod{3}\end{cases}</math>
	~~Substituting <math>x\leftarrow 3x</math>, <math>x\leftarrow 3x+1</math>, and <math>x\leftarrow 3x+2</math> to each of these cases respectively gives us our final result.~~
	~~== Trajectory ==~~
	~~[[File:BB5Champ 0-365.gif\|right\|thumb\|An animation of <math>g(0)</math> becoming <math>g(34)</math> in 365 steps (''click to view'').]]~~
	~~The initial blank tape represents <math>g(0)</math>, and the [[Collatz-like]] rules are iterated 15 times before halting:~~
	~~<math display="block">\begin{array}{\|lllllllllll\|}\hline g(0)&\xrightarrow{15} &g(6)&\xrightarrow{73} &g(16)&\xrightarrow{277}\\~~
	~~g(34)&\xrightarrow{907}&g(64)&\xrightarrow{2757}&g(114)&\xrightarrow{7957}\\~~
	~~g(196)&\xrightarrow{22777}&g(334)&\xrightarrow{64407}&g(564)&\xrightarrow{180307}\\~~
	~~g(946)&\xrightarrow{504027}&g(1584)&\xrightarrow{1403967}&g(2646)&\xrightarrow{3906393}\\~~
	~~g(4416)&\xrightarrow{10861903}&g(7366)&\xrightarrow{30196527}&g(12284)&\xrightarrow{24576} &0^\infty\;1\;\textrm{Z>}\;01\;001^{4095}\;1\;0^\infty\\\hline\end{array}</math>~~
	~~== References ==~~

MrSolis at 22:41, 8 February 2025

2025-02-08T22:41:21Z

← Older revision		Revision as of 22:41, 8 February 2025
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	=5-state busy beaver winner (WIP Revamp)=		=5-state busy beaver winner (WIP Revamp)=
	The 5-state busy beaver ([[BB(5)]]) winner is {{TM\|1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA\|halt}}. Discovered by Heiner Marxen and Jürgen Buntrock in 1989<ref>H. Marxen and J. Buntrock. Attacking the Busy Beaver 5. Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html</ref>, this machine proved that <math>\operatorname{BB}(5)\ge 47176870~~\phantom{}~~</math> and <math>\Sigma(5)\ge 4098</math> at the time.		The 5-state busy beaver ([[BB(5)]]) winner is {{TM\|1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA\|halt}}. Discovered by Heiner Marxen and Jürgen Buntrock in 1989<ref>H. Marxen and J. Buntrock. Attacking the Busy Beaver 5. Bulletin of the EATCS, 40, pages 247-251, February 1990. https://turbotm.de/~heiner/BB/mabu90.html</ref>, this machine proved that <math>\operatorname{BB}(5)\ge 47176870</math> and <math>\Sigma(5)\ge 4098</math> at the time.
	== Analysis ==		== Analysis ==
	===Rules===		===Rules===
	Let <math>g(x):=0^\infty\;\textrm{<A}\,1^x\;0^\infty</math>. Then<ref>Pascal Michel. Behavior of busy beavers.https://bbchallenge.org/~pascal.michel/beh#tm52a</ref>,		Let <math>g(x):=0^\infty\;\textrm{<A}\,1^x\;0^\infty</math>. Then<ref>Pascal Michel. Behavior of busy beavers.https://bbchallenge.org/~pascal.michel/beh#tm52a</ref>,
	<math display="block">\begin{array}{\|lll\|}\hline		<math display="block">\begin{array}{\|lll\|}\hline
	g(3x)&~~\phantom{}~~\xrightarrow{5x^2+19x+15}&g(5x+6),\\		g(3x)&\xrightarrow{5x^2+19x+15}&g(5x+6),\\
	g(3x+1)&~~\phantom{}~~\xrightarrow{5x^2+25x+27}&g(5x+9),\\		g(3x+1)&\xrightarrow{5x^2+25x+27}&g(5x+9),\\
	g(3x+2)~~\phantom{}~~&\xrightarrow{6x+12}&0^\infty\;1\;\textrm{Z>}\;01\;001^{x+1}\;1\;0^\infty.\\\hline		g(3x+2)&\xrightarrow{6x+12}&0^\infty\;1\;\textrm{Z>}\;01\;001^{x+1}\;1\;0^\infty.\\\hline
	\end{array}</math>		\end{array}</math>
	===Proof===		===Proof===
	Consider the configuration <math>C(m,n):=0^\infty\;\textrm{<A}\;1^m\;001^n\;1\;0^\infty</math>. After one step this configuration becomes <math>0^\infty\;1\;\textrm{B>}\;1^m\;001^n\;1\;0^\infty~~\phantom{}~~</math>. We note the following shift rule:		Consider the configuration <math>C(m,n):=0^\infty\;\textrm{<A}\;1^m\;001^n\;1\;0^\infty</math>. After one step this configuration becomes <math>0^\infty\;1\;\textrm{B>}\;1^m\;001^n\;1\;0^\infty</math>. We note the following shift rule:
	<math display="block">\begin{array}{\|c\|}\hline\textrm{B>}\;1^a\xrightarrow{a}1^a\;\textrm{B>}\\\hline\end{array}</math>		<math display="block">\begin{array}{\|c\|}\hline\textrm{B>}\;1^a\xrightarrow{a}1^a\;\textrm{B>}\\\hline\end{array}</math>
	Using this shift rule, we get <math>0^\infty\;1^{m+1}\;\textrm{B>}\;001^n\;1\;0^\infty</math> after <math>m</math> steps. If <math>n=0</math>, then we get <math>0^\infty\;1^{m+4}\;\textrm{<A}\;1\;0^\infty</math> four steps later. Another shift rule is needed here:		Using this shift rule, we get <math>0^\infty\;1^{m+1}\;\textrm{B>}\;001^n\;1\;0^\infty</math> after <math>m</math> steps. If <math>n=0</math>, then we get <math>0^\infty\;1^{m+4}\;\textrm{<A}\;1\;0^\infty</math> four steps later. Another shift rule is needed here:
Line 16:		Line 16:
	In this instance, <math display="inline">\Big\lfloor\frac{m+4}{3}\Big\rfloor</math> is substituted for <math>a</math>, which creates three different scenarios depending on the value of <math>m</math> modulo 3. They are as follows:		In this instance, <math display="inline">\Big\lfloor\frac{m+4}{3}\Big\rfloor</math> is substituted for <math>a</math>, which creates three different scenarios depending on the value of <math>m</math> modulo 3. They are as follows:
	# If <math>m+4\equiv0\ (\operatorname{mod}3)</math>, then in <math>m+4</math> steps we arrive at <math>0^\infty\;\textrm{<A}\;001^{(m+4)/3}\;1\;0^\infty</math>, which is the same configuration as <math display="inline">C\Big(0,\frac{m+4}{3}\Big)</math>.		# If <math>m+4\equiv0\ (\operatorname{mod}3)</math>, then in <math>m+4</math> steps we arrive at <math>0^\infty\;\textrm{<A}\;001^{(m+4)/3}\;1\;0^\infty</math>, which is the same configuration as <math display="inline">C\Big(0,\frac{m+4}{3}\Big)</math>.
	# If <math>m+4\equiv1\ (\operatorname{mod}3)</math>, then in <math>m+3</math> steps we arrive at <math>0^\infty\;1\;\textrm{<A}\;001^{(m+3)/3}\;1\;0^\infty</math>, which is five steps becomes <math>~~\phantom{}~~0^\infty\;\textrm{<A}\;111\;001^{(m+3)/3}\;1\;0^\infty</math>, equal to <math display="inline">C\Big(3,\frac{m+3}{3}\Big)</math>.		# If <math>m+4\equiv1\ (\operatorname{mod}3)</math>, then in <math>m+3</math> steps we arrive at <math>0^\infty\;1\;\textrm{<A}\;001^{(m+3)/3}\;1\;0^\infty</math>, which is five steps becomes <math>0^\infty\;\textrm{<A}\;111\;001^{(m+3)/3}\;1\;0^\infty</math>, equal to <math display="inline">C\Big(3,\frac{m+3}{3}\Big)</math>.
	# If <math>m+4\equiv2\ (\operatorname{mod}3)</math>, then in <math>m+2</math> steps we arrive at <math>0^\infty\;11\;\textrm{<A}\;001^{(m+2)/3}\;1\;0^\infty</math>, which in three steps halts with the configuration <math>0^\infty\;1\;\textrm{Z>}\;01\;001^{(m+2)/3}\;1\;0^\infty</math>, for a total of <math>2m+10</math> steps from <math>C(m,0)</math>.		# If <math>m+4\equiv2\ (\operatorname{mod}3)</math>, then in <math>m+2</math> steps we arrive at <math>0^\infty\;11\;\textrm{<A}\;001^{(m+2)/3}\;1\;0^\infty</math>, which in three steps halts with the configuration <math>0^\infty\;1\;\textrm{Z>}\;01\;001^{(m+2)/3}\;1\;0^\infty</math>, for a total of <math>2m+10</math> steps from <math>C(m,0)</math>.
	Returning to <math>0^\infty\;1^{m+1}\;\textrm{B>}\;001^n\;1\;0^\infty</math>, if <math>n\ge 1</math>, then in three steps it changes into <math>0^\infty\;1^{m+3}\;\textrm{<D}\;1\;001^{n-1}\;1\;0^\infty</math>. Here we can make use of one more shift rule:		Returning to <math>0^\infty\;1^{m+1}\;\textrm{B>}\;001^n\;1\;0^\infty</math>, if <math>n\ge 1</math>, then in three steps it changes into <math>0^\infty\;1^{m+3}\;\textrm{<D}\;1\;001^{n-1}\;1\;0^\infty</math>. Here we can make use of one more shift rule:
	<math display="block">\begin{array}{\|c\|}\hline1^a\;\textrm{<D}\xrightarrow{a}\textrm{<D}\;1^a\\\hline\end{array}</math>		<math display="block">\begin{array}{\|c\|}\hline1^a\;\textrm{<D}\xrightarrow{a}\textrm{<D}\;1^a\\\hline\end{array}</math>
	Doing so takes us to <math>0^\infty\;\textrm{<D}\;1^{m+4}\;001^{n-1}\;1\;~~\phantom{}~~0^\infty</math> in <math>m+3</math> steps, which after one step becomes the configuration <math>0^\infty\;\textrm{<A}\;1^{m+5}\;001^{n-1}\;1\;0^\infty</math>, equal to <math>C(m+5,n-1)</math>. To summarize:		Doing so takes us to <math>0^\infty\;\textrm{<D}\;1^{m+4}\;001^{n-1}\;1\;0^\infty</math> in <math>m+3</math> steps, which after one step becomes the configuration <math>0^\infty\;\textrm{<A}\;1^{m+5}\;001^{n-1}\;1\;0^\infty</math>, equal to <math>C(m+5,n-1)</math>. To summarize:
	<math display="block">\begin{array}{\|c\|}\hline C(m,n)\xrightarrow{2m+8}C(m+5,n-1)\text{ if }n\ge 1.\\\hline\end{array}</math>		<math display="block">\begin{array}{\|c\|}\hline C(m,n)\xrightarrow{2m+8}C(m+5,n-1)\text{ if }n\ge 1.\\\hline\end{array}</math>
	We have <math>g(x)=C(x-1,0)</math>. As a result, if <math>x\equiv0\ (\operatorname{mod}3)</math>, we then get <math display="inline">C\Big(0,\frac{1}{3}x+1\Big)</math> and the above rule is applied until we reach <math display="inline">C\Big(\frac{5}{3}x+5,0\Big)</math>, equal to <math display="inline">g\Big(\frac{5}{3}x+6\Big)</math>, in <math>\sum_{i=0}^{x/3}(2\times 5i+8)=\textstyle\frac{5}{9}x^2+\frac{13}{3}x+8</math> steps for a total of <math display="inline">\frac{5}{9}x^2+\frac{19}{3}x+15</math> steps from <math>g(x)</math> (with <math>g(0)</math> we see the impossible configuration <math>C(-1,0)</math>, but it reaches <math>g(6)</math> in 15 steps regardless). However, if <math>x\equiv1\ (\operatorname{mod}3)</math>, we then get <math display="inline">C\Big(3,\frac{x+2}{3}\Big)</math> which reaches <math display="inline">C\Big(3+\frac{5(x+2)}{3},0\Big)</math>, equal to <math display="inline">g\Big(\frac{5x+22}{3}\Big)</math>, in <math display="inline">\frac{5}{9}x^2+\frac{47}{9}x+\frac{74}{9}</math> steps (<math display="inline">\frac{5}{9}x^2+\frac{65}{9}x+\frac{173}{9}</math> steps total).		We have <math>g(x)=C(x-1,0)</math>. As a result, if <math>x\equiv0\ (\operatorname{mod}3)</math>, we then get <math display="inline">C\Big(0,\frac{1}{3}x+1\Big)</math> and the above rule is applied until we reach <math display="inline">C\Big(\frac{5}{3}x+5,0\Big)</math>, equal to <math display="inline">g\Big(\frac{5}{3}x+6\Big)</math>, in <math>\sum_{i=0}^{x/3}(2\times 5i+8)=\textstyle\frac{5}{9}x^2+\frac{13}{3}x+8</math> steps for a total of <math display="inline">\frac{5}{9}x^2+\frac{19}{3}x+15</math> steps from <math>g(x)</math> (with <math>g(0)</math> we see the impossible configuration <math>C(-1,0)</math>, but it reaches <math>g(6)</math> in 15 steps regardless). However, if <math>x\equiv1\ (\operatorname{mod}3)</math>, we then get <math display="inline">C\Big(3,\frac{x+2}{3}\Big)</math> which reaches <math display="inline">C\Big(3+\frac{5(x+2)}{3},0\Big)</math>, equal to <math display="inline">g\Big(\frac{5x+22}{3}\Big)</math>, in <math display="inline">\frac{5}{9}x^2+\frac{47}{9}x+\frac{74}{9}</math> steps (<math display="inline">\frac{5}{9}x^2+\frac{65}{9}x+\frac{173}{9}</math> steps total).

	The information above can be summarized as<ref>Aaronson, S. (2020). The Busy Beaver Frontier. Page 10-11. https://www.scottaaronson.com/papers/bb.pdf</ref>		The information above can be summarized as<ref>Aaronson, S. (2020). The Busy Beaver Frontier. Page 10-11. https://www.scottaaronson.com/papers/bb.pdf</ref>
	<math display="block">g(x)\rightarrow\begin{cases}g\~~big~~(\frac{5}{3}x+6\~~big~~)&\text{if }x\equiv0\pmod{3}\\g\~~big~~(\frac{5x+22}{3}\~~big~~)&\text{if }x\equiv1\pmod{3}\\0^\infty\;1\;\textrm{Z>}\;01\;001^{(x+1)/3}\;1\;0^\infty&\text{if }x\equiv2\pmod{3}\end{cases}</math>		<math display="block">g(x)\rightarrow\begin{cases}g\Big(\frac{5}{3}x+6\Big)&\text{if }x\equiv0\pmod{3}\\g\Big(\frac{5x+22}{3}\Big)&\text{if }x\equiv1\pmod{3}\\0^\infty\;1\;\textrm{Z>}\;01\;001^{(x+1)/3}\;1\;0^\infty&\text{if }x\equiv2\pmod{3}\end{cases}</math>
	Substituting <math>x\leftarrow 3x</math>, <math>x\leftarrow 3x+1</math>, and <math>x\leftarrow 3x+2</math> to each of these cases respectively gives us our final result.		Substituting <math>x\leftarrow 3x</math>, <math>x\leftarrow 3x+1</math>, and <math>x\leftarrow 3x+2</math> to each of these cases respectively gives us our final result.
	== Trajectory ==		== Trajectory ==
			[[File:BB5Champ 0-365.gif\|right\|thumb\|An animation of <math>g(0)</math> becoming <math>g(34)</math> in 365 steps (''click to view'').]]
	The initial blank tape represents <math>g(0)</math>, and the [[Collatz-like]] rules are iterated 15 times before halting:		The initial blank tape represents <math>g(0)</math>, and the [[Collatz-like]] rules are iterated 15 times before halting:
	<math display="block">\begin{array}{\|lllllllllll\|}\hline g(0)&\xrightarrow{15} &g(6)&\xrightarrow{73} &g(16)&\xrightarrow{277}\\		<math display="block">\begin{array}{\|lllllllllll\|}\hline g(0)&\xrightarrow{15} &g(6)&\xrightarrow{73} &g(16)&\xrightarrow{277}\\