Bigfoot - Revision history

MrSolis: :(

2025-10-07T22:07:22Z

RobinCodes: /* Analysis */ Attempted fix for "math input error" for this page

2025-10-07T19:45:14Z

Analysis: Attempted fix for "math input error" for this page

Polygon: Added Category:BB(3,3)

2025-09-27T16:20:15Z

Added Category:BB(3,3)

← Older revision		Revision as of 16:20, 27 September 2025
Line 64:		Line 64:
	==References==		==References==
	<references/>		<references/>
	[[Category:Cryptids]]		[[Category:BB(3,3)]][[Category:Cryptids]]

Sligocki at 14:45, 24 July 2025

2025-07-24T14:45:20Z

← Older revision		Revision as of 14:45, 24 July 2025
Line 1:		Line 1:
	{{machine~~\|1RB2RA1LC_2LC1RB2RB_---2LA1LA}}{{TM~~\|1RB2RA1LC_2LC1RB2RB_---2LA1LA}}{{unsolved\|Does Bigfoot run forever?}}		{{machine\|1RB2RA1LC_2LC1RB2RB_---2LA1LA}}{{unsolved\|Does Bigfoot run forever?}}
	'''Bigfoot''' 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's]] name.<ref name="b">S. Ligocki, "[https://www.sligocki.com/2023/10/16/bb-3-3-is-hard.html BB(3, 3) is Hard (Bigfoot)] (2024). Accessed 22 July 2024.</ref>		'''Bigfoot''' ({{TM\|1RB2RA1LC_2LC1RB2RB_---2LA1LA}}) 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's]] name.<ref name="b">S. Ligocki, "[https://www.sligocki.com/2023/10/16/bb-3-3-is-hard.html BB(3, 3) is Hard (Bigfoot)] (2024). Accessed 22 July 2024.</ref>
	<div style="width: fit-content; text-align: center; margin-left: auto; margin-right: auto;">		<div style="width: fit-content; text-align: center; margin-left: auto; margin-right: auto;">
	{\|class="wikitable" style="margin-left: auto; margin-right: auto;"		{\|class="wikitable" style="margin-left: auto; margin-right: auto;"

Sligocki at 14:44, 24 July 2025

2025-07-24T14:44:17Z

← Older revision		Revision as of 14:44, 24 July 2025
Line 22:		Line 22:
	The transition table of Bigfoot.</div>		The transition table of Bigfoot.</div>

	~~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~~.		In May of 2024, Iijil [https://github.com/sligocki/sligocki.github.io/issues/8#issuecomment-2140887228 compiled] Bigfoot into a 7-state 2-symbol machine {{TM\|0RB1RB_1LC0RA_1RE1LF_1LF1RE_0RD1RD_1LG0LG_---1LB}}.

	== Analysis ==		== Analysis ==

ISquillante: /* Analysis */

2025-05-30T16:22:38Z

Analysis

← Older revision		Revision as of 16:22, 30 May 2025
Line 53:		Line 53:
	</div></div>		</div></div>
	Using the floor function, it is possible to describe the behaviour of <math>b</math> and <math>c</math> using a function that is not defined piecewise:		Using the floor function, it is possible to describe the behaviour of <math>b</math> and <math>c</math> using a function that is not defined piecewise:
	<math display="block">\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&\text{if }x\equiv i\pmod{6},\\0&\text{otherwise.}\end{cases}\end{array}</math>		<math display="block">\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{m-i}{6}\Big\rfloor-\Big\lfloor\frac{m-i-1}{6}\Big\rfloor=\begin{cases}1&\text{if }m\equiv i\pmod{6},\\0&\text{otherwise.}\end{cases}\end{array}</math>
	In effect, the halting problem for Bigfoot is about whether through enough iterations of <math>f(m,n)</math> we encounter more <math>m</math> values that are congruent to 2 modulo 6 than ones that are congruent to 1 or 4 modulo 6.		In effect, the halting problem for Bigfoot is about whether through enough iterations of <math>f(m,n)</math> we encounter more <math>m</math> values that are congruent to 2 modulo 6 than ones that are congruent to 1 or 4 modulo 6.

	An important insight is that if <math>b</math> is odd and <math>c=2</math>, then after four iterations of <math>A</math>, that will remain the case. This allows one to define a configuration that eliminates the <math>c</math> parameter and whose rules use a modulus of 81.<ref name="b"></ref>		An important insight is that if <math>b</math> is odd and <math>c=2</math>, then after four iterations of <math>A</math>, that will remain the case. This allows one to define a configuration that eliminates the <math>c</math> parameter and whose rules use a modulus of 81.<ref name="b"></ref>

	== Trajectory ==		== Trajectory ==
	After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:		After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:

MrSolis: ...

2025-05-16T19:51:16Z

...

← Older revision		Revision as of 19:51, 16 May 2025
Line 57:		Line 57:

	An important insight is that if <math>b</math> is odd and <math>c=2</math>, then after four iterations of <math>A</math>, that will remain the case. This allows one to define a configuration that eliminates the <math>c</math> parameter and whose rules use a modulus of 81.<ref name="b"></ref>		An important insight is that if <math>b</math> is odd and <math>c=2</math>, then after four iterations of <math>A</math>, that will remain the case. This allows one to define a configuration that eliminates the <math>c</math> parameter and whose rules use a modulus of 81.<ref name="b"></ref>

	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.<ref>P. Michel, "[https://bbchallenge.org/~pascal.michel/ha.html Historical survey of Busy Beavers]".</ref>
	== Trajectory ==		== Trajectory ==
	After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:		After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:

Cosmo at 19:45, 16 May 2025

2025-05-16T19:45:38Z

← Older revision		Revision as of 19:45, 16 May 2025
Line 21:		Line 21:
	\|}		\|}
	The transition table of Bigfoot.</div>		The transition table of Bigfoot.</div>

			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.

	== Analysis ==		== Analysis ==
	Let <math>A(a,b,c):=0^\infty\;12^a\;1^{2b}\;\textrm{<A}\;1^{2c}\;0^\infty</math>. Then,		Let <math>A(a,b,c):=0^\infty\;12^a\;1^{2b}\;\textrm{<A}\;1^{2c}\;0^\infty</math>. Then,

MrSolis at 11:10, 14 April 2025

2025-04-14T11:10:54Z

← Older revision		Revision as of 11:10, 14 April 2025
Line 59:		Line 59:
	After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:		After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:
	<math display="block">\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}</math>		<math display="block">\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}</math>
	There exists a heuristic argument for Bigfoot being [[probviously]] ~~nonhalting~~. By only considering the rules for which <math>a</math> changes, one may notice that the trajectory of <math>a</math> values can be approximated by a random walk in which at each step, the walker moves +1 with probability <math display="inline">\frac{2}{3}</math> or moves -1 with probability <math display="inline">\frac{1}{3}</math>, starting at position 2. If <math>P(n)</math> is the probability that the walker will reach position -1 from position <math>n</math>, then <math display="inline">P(n)=\frac{1}{3}P(n-1)+\frac{2}{3}P(n+1)</math>. Solutions to this recurrence relation come in the form <math display="inline"> P(n)=c_02^{-n}+c_1</math>, which after applying the appropriate boundary conditions reduces to <math display="inline">P(n)=2^{-(n+1)}</math>. As a result, if the walker gets to position 3999888, then the probability of it ever reaching position -1 would be <math display="inline">2^{-3999889}\approx 2.697\times 10^{-1204087}</math>.		There exists a heuristic argument for Bigfoot being [[probviously]] non-halting. By only considering the rules for which <math>a</math> changes, one may notice that the trajectory of <math>a</math> values can be approximated by a random walk in which at each step, the walker moves +1 with probability <math display="inline">\frac{2}{3}</math> or moves -1 with probability <math display="inline">\frac{1}{3}</math>, starting at position 2. If <math>P(n)</math> is the probability that the walker will reach position -1 from position <math>n</math>, then <math display="inline">P(n)=\frac{1}{3}P(n-1)+\frac{2}{3}P(n+1)</math>. Solutions to this recurrence relation come in the form <math display="inline"> P(n)=c_02^{-n}+c_1</math>, which after applying the appropriate boundary conditions reduces to <math display="inline">P(n)=2^{-(n+1)}</math>. As a result, if the walker gets to position 3999888, then the probability of it ever reaching position -1 would be <math display="inline">2^{-3999889}\approx 2.697\times 10^{-1204087}</math>.
	==References==		==References==
	<references/>		<references/>
	[[Category:Cryptids]]		[[Category:Cryptids]]

MrSolis at 11:07, 14 April 2025

2025-04-14T11:07:35Z

← Older revision		Revision as of 11:07, 14 April 2025
Line 49:		Line 49:
	Substituting <math>b\leftarrow6b+k</math> where <math>k</math> is the remainder for each case yields the final result.		Substituting <math>b\leftarrow6b+k</math> where <math>k</math> is the remainder for each case yields the final result.
	</div></div>		</div></div>
	Using the floor function, it is possible to describe the behaviour of <math>b</math> and <math>c</math> ~~as one united formula~~:		Using the floor function, it is possible to describe the behaviour of <math>b</math> and <math>c</math> using a function that is not defined piecewise:
	<math display="block">\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&\text{if }x\equiv i\pmod{6},\\0&\text{otherwise.}\end{cases}\end{array}</math>		<math display="block">\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&\text{if }x\equiv i\pmod{6},\\0&\text{otherwise.}\end{cases}\end{array}</math>
	In effect, the halting problem for Bigfoot is about whether through enough iterations of <math>f(m,n)</math> we encounter more <math>m</math> values that are congruent to 2 modulo 6 than ones that are congruent to 1 or 4 modulo 6.		In effect, the halting problem for Bigfoot is about whether through enough iterations of <math>f(m,n)</math> we encounter more <math>m</math> values that are congruent to 2 modulo 6 than ones that are congruent to 1 or 4 modulo 6.
Line 59:		Line 59:
	After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:		After 69 steps, Bigfoot will reach the configuration <math>A(2,1,2)</math> before the [[Collatz-like]] rules are repeatedly applied. Simulations of Bigfoot have shown that after 24000000 rule steps, we have <math>a=3999888</math>. Here are the first few:
	<math display="block">\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}</math>		<math display="block">\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}</math>
	There exists a heuristic argument for Bigfoot being [[probviously]] nonhalting. ~~After removing~~ the ~~instances in~~ which <math>a</math> ~~does not change~~, one ~~notices~~ that the trajectory of <math>a</math> values can be approximated by a random walk in which at each step, the walker moves +1 with probability <math display="inline">\frac{2}{3}</math> or moves -1 with probability <math display="inline">\frac{1}{3}</math>, starting at position 2. If <math>P(n)</math> is the probability that the walker will reach position -1 from position <math>n</math>, then <math display="inline">P(n)=\frac{1}{3}P(n-1)+\frac{2}{3}P(n+1)</math>. Solutions to this recurrence relation come in the form <math display="inline"> P(n)=c_02^{-n}+c_1</math>, which after applying the appropriate boundary conditions reduces to <math display="inline">P(n)=2^{-(n+1)}</math>. As a result, if the walker gets to position 3999888, then the probability of it ever reaching position -1 would be <math display="inline">2^{-3999889}\approx 2.697\times 10^{-1204087}</math>.		There exists a heuristic argument for Bigfoot being [[probviously]] nonhalting. By only considering the rules for which <math>a</math> changes, one may notice that the trajectory of <math>a</math> values can be approximated by a random walk in which at each step, the walker moves +1 with probability <math display="inline">\frac{2}{3}</math> or moves -1 with probability <math display="inline">\frac{1}{3}</math>, starting at position 2. If <math>P(n)</math> is the probability that the walker will reach position -1 from position <math>n</math>, then <math display="inline">P(n)=\frac{1}{3}P(n-1)+\frac{2}{3}P(n+1)</math>. Solutions to this recurrence relation come in the form <math display="inline"> P(n)=c_02^{-n}+c_1</math>, which after applying the appropriate boundary conditions reduces to <math display="inline">P(n)=2^{-(n+1)}</math>. As a result, if the walker gets to position 3999888, then the probability of it ever reaching position -1 would be <math display="inline">2^{-3999889}\approx 2.697\times 10^{-1204087}</math>.
	==References==		==References==
	<references/>		<references/>
	[[Category:Cryptids]]		[[Category:Cryptids]]

← Older revision		Revision as of 19:45, 7 October 2025
Line 25:		Line 25:

	== Analysis ==		== Analysis ==
	Let <math>A(a,b,c):=0^\infty\;12^a\;1^{2b}\;\~~textrm~~{<A}\;1^{2c}\;0^\infty</math>. Then,		Let <math>A(a,b,c):=0^\infty\;12^a\;1^{2b}\;<\text{A}\;1^{2c}\;0^\infty</math>. Then,
	<math display="block">\begin{array}{\|l l l\|}\hline A(a,6b,c)&\xrightarrow{4a+(2b+1)48b+(12b+1)2c+13}&A(a,8b+c-1,2),\\A(a,6b+1,c)&\xrightarrow{4a+(6b+5)16b+(4b+1)6c+29}&A(a+1,8b+c-1,3),\\A(0,6b+2,c)&\xrightarrow{(b+1)96b+(3b+1)8c+18}&0^\infty\;\~~textrm~~{<C}\;1^{16b+2c+5}\;2\;0^\infty,\\A(a,6b+2,c)&\xrightarrow{4a+(2b+3)48b+(12b+7)2c+51}&A(a-1,8b+c+3,2)\text{ if }a\ge1,\\A(a,6b+3,c)&\xrightarrow{4a+(6b+7)16b+(12b+5)2c+91}&A(a,8b+c+1,5),\\A(a,6b+4,c)&\xrightarrow{4a+(2b+3)48b+(12b+7)2c+63}&A(a+1,8b+c+3,2),\\A(a,6b+5,c)&\xrightarrow{4a+(6b+11)16b+(4b+3)6c+103}&A(a,8b+c+5,3).\\\hline\end{array}</math>		<math display="block">\begin{array}{\|l l l\|}\hline A(a,6b,c)&\xrightarrow{4a+(2b+1)48b+(12b+1)2c+13}&A(a,8b+c-1,2),\\A(a,6b+1,c)&\xrightarrow{4a+(6b+5)16b+(4b+1)6c+29}&A(a+1,8b+c-1,3),\\A(0,6b+2,c)&\xrightarrow{(b+1)96b+(3b+1)8c+18}&0^\infty\;<\text{C}\;1^{16b+2c+5}\;2\;0^\infty,\\A(a,6b+2,c)&\xrightarrow{4a+(2b+3)48b+(12b+7)2c+51}&A(a-1,8b+c+3,2)\text{ if }a\ge1,\\A(a,6b+3,c)&\xrightarrow{4a+(6b+7)16b+(12b+5)2c+91}&A(a,8b+c+1,5),\\A(a,6b+4,c)&\xrightarrow{4a+(2b+3)48b+(12b+7)2c+63}&A(a+1,8b+c+3,2),\\A(a,6b+5,c)&\xrightarrow{4a+(6b+11)16b+(4b+3)6c+103}&A(a,8b+c+5,3).\\\hline\end{array}</math>
	<div class="toccolours mw-collapsible mw-collapsed">'''Proof'''<div class="mw-collapsible-content">		<div class="toccolours mw-collapsible mw-collapsed">'''Proof'''<div class="mw-collapsible-content">
	For now, we will work with the slightly different configuration <math>A'(a,b,c):=0^\infty\;12^a\;1^b\;\textrm{<A}\;1^c\;0^\infty</math>. Consider the partial configuration <math>P(m,n):=1^m\;\textrm{<A}\;1^n\;0^\infty</math>. We first require the following shift rule:		For now, we will work with the slightly different configuration <math>A'(a,b,c):=0^\infty\;12^a\;1^b\;\textrm{<A}\;1^c\;0^\infty</math>. Consider the partial configuration <math>P(m,n):=1^m\;\textrm{<A}\;1^n\;0^\infty</math>. We first require the following shift rule:
Line 42:		Line 42:
	#If <math>b\equiv0\ (\operatorname{mod}12)</math>, then in <math display="inline">{\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</math> steps we arrive at <math display="inline">P\Big(0,16\times\frac{b}{12}+c\Big)</math>, or <math>0^\infty\;12^a\;\textrm{<A}\;1^{4b/3+c}\;0^\infty</math> when considering the complete configuration. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;\textrm{<A}\;1^{4b/3+c}\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{<A}\;21^a\;1^{4b/3+c}\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B>}\;21^a\;1^{4b/3+c}\;0^\infty\xrightarrow{2a+4b/3+c}\\0^\infty\;12^a\;1^{4b/3+c+1}\;\textrm{B>}\;0^\infty\xrightarrow{12}0^\infty\;12^a\;1^{4b/3+c-2}\;\textrm{<A}\;1^4\;0^\infty\\\hline\end{array}</math>This means that if <math display="inline">\frac{4}{3}b+c\ge2</math>, then we will reach <math display="inline">A'\Big(a,\frac{4}{3}b+c-2,4\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+4b+bc+c+13</math> steps.		#If <math>b\equiv0\ (\operatorname{mod}12)</math>, then in <math display="inline">{\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</math> steps we arrive at <math display="inline">P\Big(0,16\times\frac{b}{12}+c\Big)</math>, or <math>0^\infty\;12^a\;\textrm{<A}\;1^{4b/3+c}\;0^\infty</math> when considering the complete configuration. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;\textrm{<A}\;1^{4b/3+c}\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{<A}\;21^a\;1^{4b/3+c}\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B>}\;21^a\;1^{4b/3+c}\;0^\infty\xrightarrow{2a+4b/3+c}\\0^\infty\;12^a\;1^{4b/3+c+1}\;\textrm{B>}\;0^\infty\xrightarrow{12}0^\infty\;12^a\;1^{4b/3+c-2}\;\textrm{<A}\;1^4\;0^\infty\\\hline\end{array}</math>This means that if <math display="inline">\frac{4}{3}b+c\ge2</math>, then we will reach <math display="inline">A'\Big(a,\frac{4}{3}b+c-2,4\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+4b+bc+c+13</math> steps.
	#If <math>b\equiv2\ (\operatorname{mod}12)</math>, then in <math display="inline">{\displaystyle\sum_{i=0}^{(b-2)/12-1}}4(48i+3c+32)=\frac{2}{3}b^2+bc-2c-\frac{8}{3}</math> steps we arrive at <math display="inline">P\Big(2,\frac{4(b-2)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;11\;\textrm{<A}\;1^{(4b-2)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;11\;\textrm{<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>}\;0^\infty\xrightarrow{4}\\0^\infty\;12^a\;1\;2^{4(b-2)/3+c}\;\textrm{<C}\;122\;0^\infty\xrightarrow{4(b-2)/3+c}0^\infty\;12^a\;1\;\textrm{<C}\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;12^a\;\textrm{<A}\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{<A}\;21^a\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{18}\\0^\infty\;12^{a+1}\;1^{4(b-2)/3+c-2}\;\textrm{<A}\;1^6\;0^\infty\\\hline\end{array}</math>This means that if <math display="inline">\frac{4(b-2)}{3}+c\ge 2</math>, then we will reach <math display="inline">A'\Big(a+1,\frac{4b-14}{3}+c,6\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+4b+bc+c+\frac{55}{3}</math> steps.		#If <math>b\equiv2\ (\operatorname{mod}12)</math>, then in <math display="inline">{\displaystyle\sum_{i=0}^{(b-2)/12-1}}4(48i+3c+32)=\frac{2}{3}b^2+bc-2c-\frac{8}{3}</math> steps we arrive at <math display="inline">P\Big(2,\frac{4(b-2)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;11\;\textrm{<A}\;1^{(4b-2)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;11\;\textrm{<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>}\;0^\infty\xrightarrow{4}\\0^\infty\;12^a\;1\;2^{4(b-2)/3+c}\;\textrm{<C}\;122\;0^\infty\xrightarrow{4(b-2)/3+c}0^\infty\;12^a\;1\;\textrm{<C}\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;12^a\;\textrm{<A}\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{<A}\;21^a\;2\;1^{4(b-2)/3+c+1}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{18}\\0^\infty\;12^{a+1}\;1^{4(b-2)/3+c-2}\;\textrm{<A}\;1^6\;0^\infty\\\hline\end{array}</math>This means that if <math display="inline">\frac{4(b-2)}{3}+c\ge 2</math>, then we will reach <math display="inline">A'\Big(a+1,\frac{4b-14}{3}+c,6\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+4b+bc+c+\frac{55}{3}</math> steps.
	#If <math>b\equiv4\ (\operatorname{mod}12)</math>, then in <math display="inline>\frac{2}{3}b^2-\frac{8}{3}b+bc-4c</math> steps we arrive at <math display="inline">P\Big(4,\frac{4(b-4)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;1111\;\textrm{<A}\;1^{4(b-4)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;1111\;\textrm{<A}\;1^{4(b-4)/3+c}\;0^\infty\xrightarrow{(8b-14)/3+2c}0^\infty\;12^a\;11\;\textrm{<A}\;2\;1^{4(b-4)/3+c+1}\;22\;0^\infty\xrightarrow{3}\\0^\infty\;12^a\;1\;\textrm{<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>}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;12^a\;2^{4(b-4)/3+c+4}\;\textrm{<C}\;12\;0^\infty\xrightarrow{4(b-4)/3+c+4}0^\infty\;12^a\;\textrm{<C}\;1^{4(b-4)/3+c+5}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;1\;\textrm{<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>}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;2^{4(b-4)/3+c+7}\;\textrm{<C}\;1\;0^\infty\xrightarrow{4(b-4)/3+c+6}0^\infty\;12^{a-1}\;2\;\textrm{<C}\;1^{4(b-4)/3+c+7}\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;\textrm{<A}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{2(a-1)}0^\infty\;\textrm{<A}\;21^{a-1}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{12}\\0^\infty\;12^{a-1}\;1^{4(b-4)/3+c+6}\;\textrm{<A}\;1^4\;0^\infty\\\hline\end{array}</math>This means that if <math>a=0</math>, then Bigfoot will reach the undefined <code>C0</code> transition with the configuration <math>0^\infty\;\textrm{<C}\;1^{(4b-1)/3+c}\;2\;0^\infty</math> in <math display="inline">\frac{2}{3}b^2+\frac{8}{3}b+bc-\frac{10}{3}</math> steps. Otherwise, it will proceed to reach <math display="inline">A'\Big(a-1,\frac{4b+2}{3}+c,4\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{20}{3}b+bc+3c+\frac{41}{3}</math> steps.		#If <math>b\equiv4\ (\operatorname{mod}12)</math>, then in <math display="inline">\frac{2}{3}b^2-\frac{8}{3}b+bc-4c</math> steps we arrive at <math display="inline">P\Big(4,\frac{4(b-4)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;1111\;\textrm{<A}\;1^{4(b-4)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;1111\;\textrm{<A}\;1^{4(b-4)/3+c}\;0^\infty\xrightarrow{(8b-14)/3+2c}0^\infty\;12^a\;11\;\textrm{<A}\;2\;1^{4(b-4)/3+c+1}\;22\;0^\infty\xrightarrow{3}\\0^\infty\;12^a\;1\;\textrm{<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>}\;22\;0^\infty\xrightarrow{1}\\0^\infty\;12^a\;2^{4(b-4)/3+c+4}\;\textrm{<C}\;12\;0^\infty\xrightarrow{4(b-4)/3+c+4}0^\infty\;12^a\;\textrm{<C}\;1^{4(b-4)/3+c+5}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;1\;\textrm{<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>}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;2^{4(b-4)/3+c+7}\;\textrm{<C}\;1\;0^\infty\xrightarrow{4(b-4)/3+c+6}0^\infty\;12^{a-1}\;2\;\textrm{<C}\;1^{4(b-4)/3+c+7}\;0^\infty\xrightarrow{1}\\0^\infty\;12^{a-1}\;\textrm{<A}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{2(a-1)}0^\infty\;\textrm{<A}\;21^{a-1}\;1^{4(b-4)/3+c+8}\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{12}\\0^\infty\;12^{a-1}\;1^{4(b-4)/3+c+6}\;\textrm{<A}\;1^4\;0^\infty\\\hline\end{array}</math>This means that if <math>a=0</math>, then Bigfoot will reach the undefined <code>C0</code> transition with the configuration <math>0^\infty\;\textrm{<C}\;1^{(4b-1)/3+c}\;2\;0^\infty</math> in <math display="inline">\frac{2}{3}b^2+\frac{8}{3}b+bc-\frac{10}{3}</math> steps. Otherwise, it will proceed to reach <math display="inline">A'\Big(a-1,\frac{4b+2}{3}+c,4\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{20}{3}b+bc+3c+\frac{41}{3}</math> steps.
	#If <math>b\equiv6\ (\operatorname{mod}12)</math>, then in <math display="inline">\frac{2}{3}b^2-\frac{16}{3}b+bc-6c+8</math> steps we arrive at <math display="inline">P\Big(6,\frac{4(b-6)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;111111\;\textrm{<A}\;1^{4(b-6)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;111111\;\textrm{<A}\;1^{4(b-6)/3+c}\;0^\infty\xrightarrow{16b/3+4c-14}0^\infty\;12^a\;11\;\textrm{<C}\;1^{4(b-6)/3+c+5}\;2\;0^\infty\xrightarrow{4}\\0^\infty\;12^a\;\textrm{<A}\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{<A}\;21^a\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{60}\\0^\infty\;12^a\;1^{4(b-6)/3+c+2}\;\textrm{<A}\;1^{10}\;0^\infty\\\hline\end{array}</math>This means that we will reach <math display="inline">A'\Big(a,\frac{4}{3}b+c-6,10\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+59</math> steps.		#If <math>b\equiv6\ (\operatorname{mod}12)</math>, then in <math display="inline">\frac{2}{3}b^2-\frac{16}{3}b+bc-6c+8</math> steps we arrive at <math display="inline">P\Big(6,\frac{4(b-6)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;111111\;\textrm{<A}\;1^{4(b-6)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;111111\;\textrm{<A}\;1^{4(b-6)/3+c}\;0^\infty\xrightarrow{16b/3+4c-14}0^\infty\;12^a\;11\;\textrm{<C}\;1^{4(b-6)/3+c+5}\;2\;0^\infty\xrightarrow{4}\\0^\infty\;12^a\;\textrm{<A}\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{2a}0^\infty\;\textrm{<A}\;21^a\;1^{4(b-6)/3+c+7}\;2\;0^\infty\xrightarrow{1}\\0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{60}\\0^\infty\;12^a\;1^{4(b-6)/3+c+2}\;\textrm{<A}\;1^{10}\;0^\infty\\\hline\end{array}</math>This means that we will reach <math display="inline">A'\Big(a,\frac{4}{3}b+c-6,10\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+59</math> steps.
	#If <math>b\equiv8\ (\operatorname{mod}12)</math>, then in <math display="inline>\frac{2}{3}b^2-8b+bc-8c+\frac{64}{3}</math> steps we arrive at <math display="inline">P\Big(8,\frac{4(b-8)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;1^8\;\textrm{<A}\;1^{4(b-8)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;1^8\;\textrm{<A}\;1^{4(b-8)/3+c}\;0^\infty\xrightarrow{(16b-62)/3+4c}0^\infty\;12^a\;11\;\textrm{<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>}\;2\;0^\infty\xrightarrow{1}0^\infty\;12^a\;1\;2^{4(b-8)/3+c+8}\;\textrm{<C}\;1\;0^\infty\xrightarrow{4(b-8)/3+c+8}\\0^\infty\;12^a\;1\;\textrm{<C}\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{1}0^\infty\;12^a\;\textrm{<A}\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{2a}\\0^\infty\;\textrm{<A}\;21^a\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{12}0^\infty\;12^{a+1}\;1^{4(b-8)/3+c+6}\;\textrm{<A}\;1^4\;0^\infty\\\hline\end{array}</math>This means that we will reach <math display="inline">A'\Big(a+1,\frac{4b-14}{3}+c,4\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+\frac{29}{3}</math> steps.		#If <math>b\equiv8\ (\operatorname{mod}12)</math>, then in <math display="inline">\frac{2}{3}b^2-8b+bc-8c+\frac{64}{3}</math> steps we arrive at <math display="inline">P\Big(8,\frac{4(b-8)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;1^8\;\textrm{<A}\;1^{4(b-8)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;1^8\;\textrm{<A}\;1^{4(b-8)/3+c}\;0^\infty\xrightarrow{(16b-62)/3+4c}0^\infty\;12^a\;11\;\textrm{<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>}\;2\;0^\infty\xrightarrow{1}0^\infty\;12^a\;1\;2^{4(b-8)/3+c+8}\;\textrm{<C}\;1\;0^\infty\xrightarrow{4(b-8)/3+c+8}\\0^\infty\;12^a\;1\;\textrm{<C}\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{1}0^\infty\;12^a\;\textrm{<A}\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{2a}\\0^\infty\;\textrm{<A}\;21^a\;2\;1^{4(b-8)/3+c+9}\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{12}0^\infty\;12^{a+1}\;1^{4(b-8)/3+c+6}\;\textrm{<A}\;1^4\;0^\infty\\\hline\end{array}</math>This means that we will reach <math display="inline">A'\Big(a+1,\frac{4b-14}{3}+c,4\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+\frac{29}{3}</math> steps.
	#If <math>b\equiv10\ (\operatorname{mod}12)</math>, then in <math display="inline">\frac{2}{3}b^2-\frac{32}{3}b+bc-10c+40</math> steps we arrive at <math display="inline">P\Big(10,\frac{4(b-10)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;1^{10}\;\textrm{<A}\;1^{4(b-10)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;1^{10}\;\textrm{<A}\;1^{4(b- 10)/3+c}\;0^\infty\xrightarrow{8b+6c-37}0^\infty\;12^a\;1\;\textrm{<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>}\;0^\infty\xrightarrow{4}0^\infty\;12^a\;2^{4(b-10)/3+c+11}\;\textrm{<C}\;122\;0^\infty\xrightarrow{4(b-10)/3+c+10}\\0^\infty\;12^a\;2\;\textrm{<C}\;1^{4(b- 10)/3+c+11}\;22\;0^\infty\xrightarrow{1}0^\infty\;12^a\;\textrm{<A}\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{2a}\\0^\infty\;\textrm{<A}\;12^a\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{18}0^\infty\;12^a\;1^{4(b-10)/3+c+10}\;\textrm{<A}\;1^6\;0^\infty\\\hline\end{array}</math>This means that we will reach <math display="inline">A'\Big(a,\frac{4b-10}{3}+c,6\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+23</math> steps.		#If <math>b\equiv10\ (\operatorname{mod}12)</math>, then in <math display="inline">\frac{2}{3}b^2-\frac{32}{3}b+bc-10c+40</math> steps we arrive at <math display="inline">P\Big(10,\frac{4(b-10)}{3}+c\Big)</math>, or <math>0^\infty\;12^a\;1^{10}\;\textrm{<A}\;1^{4(b-10)/3+c}\;0^\infty</math>. What follows is:<math display="block">\begin{array}{\|l\|}\hline0^\infty\;12^a\;1^{10}\;\textrm{<A}\;1^{4(b- 10)/3+c}\;0^\infty\xrightarrow{8b+6c-37}0^\infty\;12^a\;1\;\textrm{<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>}\;0^\infty\xrightarrow{4}0^\infty\;12^a\;2^{4(b-10)/3+c+11}\;\textrm{<C}\;122\;0^\infty\xrightarrow{4(b-10)/3+c+10}\\0^\infty\;12^a\;2\;\textrm{<C}\;1^{4(b- 10)/3+c+11}\;22\;0^\infty\xrightarrow{1}0^\infty\;12^a\;\textrm{<A}\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{2a}\\0^\infty\;\textrm{<A}\;12^a\;1^{4(b-10)/3+c+12}\;22\;0^\infty\xrightarrow{1}0^\infty\;1\;\textrm{B>}\;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>}\;0^\infty\xrightarrow{18}0^\infty\;12^a\;1^{4(b-10)/3+c+10}\;\textrm{<A}\;1^6\;0^\infty\\\hline\end{array}</math>This means that we will reach <math display="inline">A'\Big(a,\frac{4b-10}{3}+c,6\Big)</math> in <math display="inline">4a+\frac{2}{3}b^2+\frac{4}{3}b+bc-c+23</math> steps.
	The information above can be summarized as		The information above can be summarized as