1RB1RD1LC 2LB1RB1LC 1RZ1LA1LD 0RB2RA2RD - Revision history

Polygon: /* Analysis by Polygon */ fixed typos

2026-10-18T12:52:24Z

Analysis by Polygon: fixed typos

← Older revision		Revision as of 12:52, 18 October 2026
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	AR6-2: <math>D_1(2, 0, 3m+1) --> D(6m+12, 0, f_1^{2}(1)) = D(3(2m+4), 0, f_1^{2}(1)) --> D(0, 0, f_2(2m+4,f_1^2(1)))</math> by rules AR6 and AR8. AR6-2 leads to AR1-2.		AR6-2: <math>D_1(2, 0, 3m+1) --> D(6m+12, 0, f_1^{2}(1)) = D(3(2m+4), 0, f_1^{2}(1)) --> D(0, 0, f_2(2m+4,f_1^2(1)))</math> by rules AR6 and AR8. AR6-2 leads to AR1-2.

	The maximum ~~trajecture~~ following these rules is: AR6-2 --> AR1-2 --> AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt		The maximum trajectory following these rules is: AR6-2 --> AR1-2 --> AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt

	This TM follows the ~~trajecture~~: AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt		This TM follows the trajectory: AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt
	</div></div>		</div></div>
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

Polygon: /* Analysis by Polygon */ thats e2 not e3

2026-10-17T16:48:12Z

Analysis by Polygon: thats e2 not e3

← Older revision		Revision as of 16:48, 17 October 2026
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	<math>e_2 = f_2(\frac{2e_1+7} 3, f_1^{2}(91))</math>		<math>e_2 = f_2(\frac{2e_1+7} 3, f_1^{2}(91))</math>

	<math>D_1(1,0,~~e_3~~)</math>		<math>D_1(1,0,e_2)</math>

	<math>e_2</math> mod 3 = 1		<math>e_2</math> mod 3 = 1
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	<math> \rightarrow D_1(2e_2+7, 3, 1) \rightarrow D_1(2e_2+7, 0, f_1^{2}(91))</math>		<math> \rightarrow D_1(2e_2+7, 3, 1) \rightarrow D_1(2e_2+7, 0, f_1^{2}(91))</math>

	<math>~~2e_3~~ + 7</math>		<math>2e_2 + 7</math>

	Modulus: 2 + 7 --> 9 mod 3 = 0		Modulus: 2 + 7 --> 9 mod 3 = 0

Polygon: /* Analysis by Polygon */ added collapsible reset behavior section

2026-10-17T16:13:52Z

Analysis by Polygon: added collapsible reset behavior section

← Older revision		Revision as of 16:13, 17 October 2026
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	By changing the C0 transition from <code>C0 --> 1RZ</code> to <code>C0 --> 1LB</code>, this machine can be turned into a [[Translated cycler]] with preperiod > <math>10 \uparrow^{4} 4</math>, making {{TM\|1RB1RD1LC_2LB1RB1LC_1LB1LA1LD_0RB2RA2RD}} the current [[Non-halting Turing machine#Translated cycler preperiod\|BBS(4,3)]] champion.		By changing the C0 transition from <code>C0 --> 1RZ</code> to <code>C0 --> 1LB</code>, this machine can be turned into a [[Translated cycler]] with preperiod > <math>10 \uparrow^{4} 4</math>, making {{TM\|1RB1RD1LC_2LB1RB1LC_1LB1LA1LD_0RB2RA2RD}} the current [[Non-halting Turing machine#Translated cycler preperiod\|BBS(4,3)]] champion.

			<div class="toccolours mw-collapsible mw-collapsed">'''Reset behavior'''<div class="mw-collapsible-content">
			Let's restate the rules:

			Rule 30:
			* <math>D(a, b, c) --> D(a, 0, f_1^{b}(c))</math>
			* <math>D_1(a, b, c) --> D_1(a, 0, f_1^{b}(c))</math>

			Rule 23:
			* <math>D(a, 0, c) --> D(a-3, 2c+11, 1)</math>
			* <math>D_1(a, 0, c) --> D_1(a-3, 2c+11, 1)</math>

			Rule 31 for a greater than or equal to 3:
			* <math>D(a, 0, c) --> D(a-3, 0, f_1^{2c+11}(1)) = D(a-3, 0, f_2(1,c))</math> and
			* <math>D_1(a, 0, c) --> D_1(a-3, 0, f_1^{2c+11}(1)) = D_1(a-3, 0, f_2(1,c))</math>

			Rule 32:
			* <math>D(3k+d, 0, c) --> D(d, 0, f_2(k, c))</math>
			* <math>D_1(3k+d, 0, c) --> D_1(d, 0, f_2(k, c))</math>

			Rule 24:
			* <math>D(0, 0, c) --> D(c+1, 3, 1)</math>

			Rule 28:
			* D(1, 0, c) --> halt with score 4c + 18

			Rule 25:
			* <math>D(2, 0, c) --> D_1(2c+8, 3, 1)</math>

			Rule 27:
			* <math>D_1(0, 0, c) --> D(2c+5, 3, 1)</math>

			Rule 26:
			* <math>D_1(1, 0, c) --> D_1(2c+7, 3, 1)</math>

			Rule 29:
			* <math>D_1(2, 0, c) --> D(2c+10, 2, 1)</math>


			Let's explore what happens after the remainder rules:

			AR1: <math>D(0, 0, c) --> D(c+1, 3, 1) --> D(c+1, 0, f_1^{3}(1))</math> by rules 24 and 30.

			AR2: D(1, 0, c) --> halt with score 4c + 18 by rule 28.

			AR3: <math>D(2, 0, c) --> D_1(2c+8, 3, 1) --> D(2c+8, 0, f_1^{3}(1))</math> by rules 25 and 30.

			AR4: <math>D_1(0, 0, c) --> D(2c+5, 3, 1) --> D(2c+5, 0, f_1^{3}(1))</math> by rules 27 and 30.

			AR5: <math>D_1(1, 0, c) --> D_1(2c+7, 3, 1) --> D_1(2c+7, 0, f_1^{3}(1))</math> by rules 26 and 30.

			AR6: <math>D_1(2, 0, c) --> D(2c+10, 2, 1) --> D(2c+10, 0, f_1^{2}(1))</math> by rules 29 and 30.

			Together with:

			AR7 by rule 30:
			* <math>D(a, b, c) --> D(a, 0, f_1^{b}(c))</math>
			* <math>D_1(a, b, c) --> D_1(a, 0, f_1^{b}(c))</math>

			AR8 by rule 32 for k ≠ 0:
			* <math>D(3k+d, 0, c) --> D(d, 0, f_2(k, c))</math>
			* <math>D_1(3k+d, 0, c) --> D_1(d, 0, f_2(k, c))</math>

			Let's also consider the following which was shown earlier:
			<pre>
			f_1(n) mod 3 = 1
			f_1^a(n) mod 3 = 1
			f_2(a,b) mod 3 = 1 for a ≠ 0
			</pre>

			Every one of these rules will set c to either f_1(n), f_1^a(n) or f_2(a,b) with a ≠ 0, thus c mod 3 = 1 after the first rule application.

			Thus, c can be considered to be of the form 3m+1:


			AR1-2: <math>D(0, 0, 3m+1) --> D(3m+2, 0, f_1^{3}(1)) --> D(2, 0, f_2(m,f_1^3(1)))</math> by rules AR1 and AR8. AR1-2 leads to AR3-2.

			AR2-2: D(1, 0, 3m+1) --> halt with score 12m + 22 by rule AR2. AR2-2 halts.

			AR3-2: <math>D(2, 0, 3m+1) --> D_1(6m+10, 0, f_1^{3}(1)) = D_1(3(2m+3)+1, 0, f_1^{3}(1)) --> D_1(1, 0, f_2(2m+3,f_1^3(1)))</math> by AR3 and AR8. AR3-2 leads to AR5-2.

			AR4-2: <math>D_1(0, 0, 3m+1) --> D(6m+7, 0, f_1^{3}(1)) = D(3(2m+2)+1, 0, f_1^{3}(1)) --> D(1, 0, f_2(2m+2,f_1^3(1)))</math> by rules AR4 and AR8. AR4-2 leads to AR2-2

			AR5-2: <math>D_1(1, 0, 3m+1) --> D_1(6m+9, 0, f_1^{3}(1)) = D_1(3(2m+3), 0, f_1^{3}(1)) --> D_1(0, 0, f_2(2m+3,f_1^3(1)))</math> by rules AR5 and AR8. AR5-2 leads to AR4-2.

			AR6-2: <math>D_1(2, 0, 3m+1) --> D(6m+12, 0, f_1^{2}(1)) = D(3(2m+4), 0, f_1^{2}(1)) --> D(0, 0, f_2(2m+4,f_1^2(1)))</math> by rules AR6 and AR8. AR6-2 leads to AR1-2.

			The maximum trajecture following these rules is: AR6-2 --> AR1-2 --> AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt

			This TM follows the trajecture: AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt
			</div></div>
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

Polygon: /* Analysis by Polygon */ fixed R25 D_1 typo

2026-10-17T16:03:29Z

Analysis by Polygon: fixed R25 D_1 typo

← Older revision		Revision as of 16:03, 17 October 2026
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	Rule 25 becomes:		Rule 25 becomes:
	* <math>D(2, 0, c) --> D(2c+8, 3, 1)</math>		* <math>D(2, 0, c) --> D_1(2c+8, 3, 1)</math>

	Rule 26 becomes:		Rule 26 becomes:

Polygon: /* Analysis by Polygon */ added a ≠ 0 as a condition

2026-10-17T15:16:41Z

Analysis by Polygon: added a ≠ 0 as a condition

← Older revision		Revision as of 15:16, 17 October 2026
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	f_2(a,b) = f_1^(2*f_2(a-1,b)+11)(1)		f_2(a,b) = f_1^(2*f_2(a-1,b)+11)(1)
	Note that f_1^(2*f_2(a-1,b)+11)(1) is also of the form f_1^a(n)		Note that f_1^(2*f_2(a-1,b)+11)(1) is also of the form f_1^a(n)
	Thus f_2(a,b) mod 3 = 1		Thus f_2(a,b) mod 3 = 1 for a ≠ 0
	</pre>		</pre>

Polygon: /* Analysis by Polygon */ restructured

2026-10-17T15:03:32Z

Analysis by Polygon: restructured

← Older revision		Revision as of 15:03, 17 October 2026
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	* <math>D_1(a, b, c) --> D_1(a, b-1, 2^{c+4} \times 3 -5) = D_1(a, b-1, f_1(c))</math>		* <math>D_1(a, b, c) --> D_1(a, b-1, 2^{c+4} \times 3 -5) = D_1(a, b-1, f_1(c))</math>

	Rule 21 ~~can be repeated to construct rule 30~~:		Rule 30 is constructed by repeating rule 21:
	* <math>D(a, b, c) --> D(a, 0, f_1^{b}(c))</math>		* <math>D(a, b, c) --> D(a, 0, f_1^{b}(c))</math>
	* <math>D_1(a, b, c) --> D_1(a, 0, f_1^{b}(c))</math>		* <math>D_1(a, b, c) --> D_1(a, 0, f_1^{b}(c))</math>

	Rule 23 becomes:		Rule 23 becomes:
	* <math>D(a, 0, c) ~~\rightarrow~~ D(a-3, 2c+11, 1)</math>		* <math>D(a, 0, c) --> D(a-3, 2c+11, 1)</math>
	* <math>D_1(a, 0, c) ~~\rightarrow~~ D_1(a-3, 2c+11, 1)</math>		* <math>D_1(a, 0, c) --> D_1(a-3, 2c+11, 1)</math>


			If a is greater than or equal to 3:

			<math>D(a, 0, c) --> D(a-3, 2c+11, 1) --> D(a-3, 0, f_1^{2c+11}(1))</math> via rules 23 and 30, and

			<math>D_1(a, 0, c) --> D_1(a-3, 2c+11, 1) --> D_1(a-3, 0, f_1^{2c+11}(1))</math> via rules 23 and 30

			Thus rule 31 is:
			* <math>D(a, 0, c) --> D(a-3, 0, f_1^{2c+11}(1)) = D(a-3, 0, f_2(1,c))</math> and
			* <math>D_1(a, 0, c) --> D_1(a-3, 0, f_1^{2c+11}(1)) = D_1(a-3, 0, f_2(1,c))</math>


			This rule can also be repeated to create rule 32, note that <math>f_1^{2c+11}(1) = f_2(1,c)</math> and <math>f_1^{2 \times f_2(a,b) + 11}(1) = f_2(a+1,b)</math>:

			* <math>D(3k+d, 0, c) --> D(d, 0, f_2(k, c))</math>
			* <math>D_1(3k+d, 0, c) --> D_1(d, 0, f_2(k, c))</math>

	Rule 24 becomes:		Rule 24 becomes:
	* <math>D(0, 0, c) ~~\rightarrow~~ D(c+1, 3, 1)</math>		* <math>D(0, 0, c) --> D(c+1, 3, 1)</math>

	Rule 25 becomes:		Rule 25 becomes:
	* <math>D(2, 0, c) ~~\rightarrow~~ D(2c+8, 3, 1)</math>		* <math>D(2, 0, c) --> D(2c+8, 3, 1)</math>

	Rule 26 becomes:		Rule 26 becomes:
	* <math>D_1(1, 0, c) ~~\rightarrow~~ D_1(2c+7, 3, 1)</math>		* <math>D_1(1, 0, c) --> D_1(2c+7, 3, 1)</math>

	Rule 27 becomes:		Rule 27 becomes:
	* <math>D_1(0, 0, c) ~~\rightarrow~~ D(2c+5, 3, 1)</math>		* <math>D_1(0, 0, c) --> D(2c+5, 3, 1)</math>

	Rule 28 becomes:		Rule 28 becomes:
	* ~~<math>~~D(1, 0, c) ~~\rightarrow \text {~~halt with score } 4c + 18~~</math>~~		* D(1, 0, c) --> halt with score 4c + 18

	Rule 29 becomes:		Rule 29 becomes:
	* <math>D_1(2, 0, c) ~~\rightarrow D(2c+10, 2, 1)</math>~~		* <math>D_1(2, 0, c) --> D(2c+10, 2, 1)</math>

	~~If a is greater than or equal to 3:~~
	~~<math>D(a, 0, c) \rightarrow D(a-3, 2c+11, 1) \rightarrow D(a~~-~~3, 0, f_1^{2c+11}(1))</math>~~
	~~=<math>D(a~~-~~3, 0, f_2(1,c))</math>~~
	* <math>D(~~a, 0, c) \rightarrow D(a-3, 0, f_1^{~~2c+~~11}(1))</math>~~

	~~This rule can also be repeated, also note that <math>f_1^{2c+11}(1) = f_2(1~~,~~c)</math> and <math>f_1^{~~2 ~~\times f_2(a~~,~~b) + 11}(1) = f_2(a+~~1,b)</math>:

	* <math>D(3k+d, 0, c) \rightarrow D(d, 0, f_2(k, c))</math>
	* <math>D_1(3k+d, 0, c) \rightarrow D_1(d, 0, f_2(k, c))</math>

	The TM starts in configuration <math>D(2, 2, 1)</math>.		The TM starts in configuration <math>D(2, 2, 1)</math>.

Polygon: /* Analysis by Polygon */ fixed variable mix-up in rule 21 write-up

2026-10-17T14:44:39Z

Analysis by Polygon: fixed variable mix-up in rule 21 write-up

← Older revision		Revision as of 14:44, 17 October 2026
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	Rule 21 becomes:		Rule 21 becomes:
	* <math>D(a, b, c) ~~\rightarrow~~ D(a, b-1, 2^{b+4} \times 3 -5)</math>		* <math>D(a, b, c) --> D(a, b-1, 2^{c+4} \times 3 -5) = D(a, b-1, f_1(c))</math>
	* <math>D_1(a, b, c) ~~\rightarrow~~ D_1(a, b-1, 2^{b+4} \times 3 -5)</math>		* <math>D_1(a, b, c) --> D_1(a, b-1, 2^{c+4} \times 3 -5) = D_1(a, b-1, f_1(c))</math>

			Rule 21 can be repeated to construct rule 30:
			* <math>D(a, b, c) --> D(a, 0, f_1^{b}(c))</math>
			* <math>D_1(a, b, c) --> D_1(a, 0, f_1^{b}(c))</math>

	Rule 23 becomes:		Rule 23 becomes:
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	Rule 29 becomes:		Rule 29 becomes:
	* <math>D_1(2, 0, c) \rightarrow D(2c+10, 2, 1)</math>		* <math>D_1(2, 0, c) \rightarrow D(2c+10, 2, 1)</math>

	~~By repeating rule 21, a stronger rule can be constructed:~~
	* <math>D(a, b, c) \rightarrow D(a, 0, f_1^{b}(c))</math>
	* <math>D_1(a, b, c) \rightarrow D_1(a, 0, f_1^{b}(c))</math>

	If a is greater than or equal to 3:		If a is greater than or equal to 3:

Polygon: /* Analysis by Polygon */ tape configuration formatting

2026-03-30T14:11:49Z

Analysis by Polygon: tape configuration formatting

← Older revision		Revision as of 14:11, 30 March 2026
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	Bonus rule: 29. 0^inf 1 (11)^2 2^2 1 <D (11)^c 0^inf --> 0^inf (11)^2c+10 (12)^2 2^2 1 <D (11)^1 0^inf		Bonus rule: 29. 0^inf 1 (11)^2 2^2 1 <D (11)^c 0^inf --> 0^inf (11)^2c+10 (12)^2 2^2 1 <D (11)^1 0^inf
	</pre>		</pre>
	Let <math>D(a, b, c)</math> = 0^~~inf (~~11)^a (12)^b 2^2 1 <D (11)^c 0^~~inf~~		Let <math>D(a, b, c)</math> = <math>0^\infty\;11^{a}\;12^{b}\;2^{2}\;1\;\textrm{<}\textrm{D}\;11^{c}\;0^\infty</math>.

	Let <math>D_1(a, b, c)</math> = 0^~~inf~~ 1 (11)^a (12)^b 2^2 1 <D (11)^c 0^~~inf~~		Let <math>D_1(a, b, c)</math> = <math>0^\infty\;1\;11^{a}\;12^{b}\;2^{2}\;1\;\textrm{<}\textrm{D}\;11^{c}\;0^\infty</math>.

	Let <math>f_1(n) = 2^{n+4} \times 3 -5</math>		Let <math>f_1(n) = 2^{n+4} \times 3 -5</math>

Polygon: /* Analysis by Polygon */ clarified which TM is the BBS(4,3) champion

2025-12-28T15:46:24Z

Analysis by Polygon: clarified which TM is the BBS(4,3) champion

← Older revision		Revision as of 15:46, 28 December 2025
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	This TM is also a [[Maximum Consecutive Ones Function\|num]](4,3) contender regardless of definition, as its score consists entirely of consecutive ones.		This TM is also a [[Maximum Consecutive Ones Function\|num]](4,3) contender regardless of definition, as its score consists entirely of consecutive ones.

	By changing the C0 transition from <code>C0 --> 1RZ</code> to <code>C0 --> 1LB</code>, this machine can be turned into a [[Translated cycler]] with preperiod > <math>10 \uparrow^{4} 4</math>, making it the current [[Non-halting Turing machine#Translated cycler preperiod\|BBS(4,3)]] champion.		By changing the C0 transition from <code>C0 --> 1RZ</code> to <code>C0 --> 1LB</code>, this machine can be turned into a [[Translated cycler]] with preperiod > <math>10 \uparrow^{4} 4</math>, making {{TM\|1RB1RD1LC_2LB1RB1LC_1LB1LA1LD_0RB2RA2RD}} the current [[Non-halting Turing machine#Translated cycler preperiod\|BBS(4,3)]] champion.
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

Polygon: /* Analysis by Polygon */ added relation to BBS(4,3) champion

2025-12-28T15:45:18Z

Analysis by Polygon: added relation to BBS(4,3) champion

← Older revision		Revision as of 15:45, 28 December 2025
Line 169:		Line 169:

	This TM is also a [[Maximum Consecutive Ones Function\|num]](4,3) contender regardless of definition, as its score consists entirely of consecutive ones.		This TM is also a [[Maximum Consecutive Ones Function\|num]](4,3) contender regardless of definition, as its score consists entirely of consecutive ones.

			By changing the C0 transition from <code>C0 --> 1RZ</code> to <code>C0 --> 1LB</code>, this machine can be turned into a [[Translated cycler]] with preperiod > <math>10 \uparrow^{4} 4</math>, making it the current [[Non-halting Turing machine#Translated cycler preperiod\|BBS(4,3)]] champion.
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]