1RB1RD1LC 2LB1RB1LC 1RZ1LA1LD 0RB2RA2RD - Revision history

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
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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.
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

Polygon: /* Analysis by Polygon */ mentioned it being a num(4,3) contender

2025-12-26T13:27:01Z

Analysis by Polygon: mentioned it being a num(4,3) contender

← Older revision		Revision as of 13:27, 26 December 2025
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	<math>10 \uparrow^{4} 4 < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.92 \times 10^{28}) < e_4 < \sigma < S < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.93 \times 10^{28})</math>		<math>10 \uparrow^{4} 4 < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.92 \times 10^{28}) < e_4 < \sigma < S < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.93 \times 10^{28})</math>

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

Polygon: /* Analysis by Polygon */ Added rule 29

2025-10-26T10:26:42Z

Analysis by Polygon: Added rule 29

← Older revision		Revision as of 10:26, 26 October 2025
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	27. 0^inf 1 2^2 1 <D (11)^c 0^inf --> 0^inf (11)^2c+5 (12)^3 2^2 1 <D (11)^1 0^inf		27. 0^inf 1 2^2 1 <D (11)^c 0^inf --> 0^inf (11)^2c+5 (12)^3 2^2 1 <D (11)^1 0^inf
	28. 0^inf (11)^1 2^2 1 <D (11)^c 0^inf --> 0^inf 1 Z> 1 (11)^2c+8 0^inf		28. 0^inf (11)^1 2^2 1 <D (11)^c 0^inf --> 0^inf 1 Z> 1 (11)^2c+8 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> = 0^inf (11)^a (12)^b 2^2 1 <D (11)^c 0^inf
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	Rule 28 becomes:		Rule 28 becomes:
	* <math>D(1, 0, c) \rightarrow \text {halt with score } 4c + 18</math>		* <math>D(1, 0, c) \rightarrow \text {halt with score } 4c + 18</math>

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

	By repeating rule 21, a stronger rule can be constructed:		By repeating rule 21, a stronger rule can be constructed:

Polygon: Used a closer lower bound

2025-10-25T08:48:16Z

Used a closer lower bound

← Older revision		Revision as of 08:48, 25 October 2025
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	{{machine\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD}}		{{machine\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD}}
	{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD\|halt}} is a hexational halting [[BB(4,3)]] TM. It was discovered in May 2024 by Pavel Kropitz as one of seven long running TMs and achieves a score of over <math>2 \uparrow^{4} 5</math>, making it the current BB(4,3) [[champion]]. Polygon analysed the TM by hand in October 2025, providing its score.		{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD\|halt}} is a hexational halting [[BB(4,3)]] TM. It was discovered in May 2024 by Pavel Kropitz as one of seven long running TMs and achieves a score of over <math>10 \uparrow^{4} 4</math>, making it the current BB(4,3) [[champion]]. Polygon analysed the TM by hand in October 2025, providing its score.

	Pavel listed the halting tape as:		Pavel listed the halting tape as:
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	This can be bounded by:		This can be bounded by:

	<math>2 \uparrow^{4} 5 < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.92 \times 10^{28}) < e_4 < \sigma < S < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.93 \times 10^{28})</math>		<math>10 \uparrow^{4} 4 < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.92 \times 10^{28}) < e_4 < \sigma < S < 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow\uparrow (7.93 \times 10^{28})</math>
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

RobinCodes: Removed "possibly"

2025-10-23T17:04:02Z

Removed "possibly"

← Older revision		Revision as of 17:04, 23 October 2025
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	{{machine\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD}}		{{machine\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD}}
	{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD\|halt}} is a hexational halting [[BB(4,3)]] TM. It was discovered in May 2024 by Pavel Kropitz as one of seven long running TMs and achieves a score of over <math>2 \uparrow^{4} 5</math>, ~~possibly~~ making it the current BB(4,3) [[champion]]. Polygon analysed the TM by hand in October 2025, providing its score.		{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD\|halt}} is a hexational halting [[BB(4,3)]] TM. It was discovered in May 2024 by Pavel Kropitz as one of seven long running TMs and achieves a score of over <math>2 \uparrow^{4} 5</math>, making it the current BB(4,3) [[champion]]. Polygon analysed the TM by hand in October 2025, providing its score.

	Pavel listed the halting tape as:		Pavel listed the halting tape as:

Polygon: /* Analysis by Polygon */ Added halting tape

2025-10-20T18:31:55Z

Analysis by Polygon: Added halting tape

← Older revision		Revision as of 18:31, 20 October 2025
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	--> halts with score <math>4e_4 + 18</math>.		--> halts with score <math>4e_4 + 18</math>.
			<pre>
			halting tape: 0^inf 1 Z> 1 (11)^(2e_4)+8 0^inf
			</pre>

	This can be bounded by:		This can be bounded by:

Polygon: /* Analysis by Polygon */ formatting of an arrow I missed earlier

2025-10-19T14:31:12Z

Analysis by Polygon: formatting of an arrow I missed earlier

← Older revision		Revision as of 14:31, 19 October 2025
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	<math>e_2</math> mod 3 = 1		<math>e_2</math> mod 3 = 1

	~~-->~~ <math>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_3 + 7</math>

Polygon: /* Analysis by Polygon */ Some formatting changes to make the page look less chaotic

2025-10-19T11:56:39Z

Analysis by Polygon: Some formatting changes to make the page look less chaotic

← Older revision		Revision as of 11:56, 19 October 2025
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	28. 0^inf (11)^1 2^2 1 <D (11)^c 0^inf --> 0^inf 1 Z> 1 (11)^2c+8 0^inf		28. 0^inf (11)^1 2^2 1 <D (11)^c 0^inf --> 0^inf 1 Z> 1 (11)^2c+8 0^inf
	</pre>		</pre>
	Let D(a, b, c) = 0^inf (11)^a (12)^b 2^2 1 <D (11)^c 0^inf		Let <math>D(a, b, c)</math> = 0^inf (11)^a (12)^b 2^2 1 <D (11)^c 0^inf

	Let D_1(a, b, c) = 0^inf 1 (11)^a (12)^b 2^2 1 <D (11)^c 0^inf		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>f_1(n) = 2^{n+4} \times 3 -5</math>		Let <math>f_1(n) = 2^{n+4} \times 3 -5</math>
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	Rule 21 becomes:		Rule 21 becomes:
	* <math>D(a, b, c)~~</math> --> <math>~~D(a, b-1, 2^{b+4} \times 3 -5)</math>		* <math>D(a, b, c) \rightarrow D(a, b-1, 2^{b+4} \times 3 -5)</math>
	* <math>D_1(a, b, c)~~</math> --> <math>~~D_1(a, b-1, 2^{b+4} \times 3 -5)</math>		* <math>D_1(a, b, c) \rightarrow D_1(a, b-1, 2^{b+4} \times 3 -5)</math>

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

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

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

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

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

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

	By repeating rule 21, a stronger rule can be constructed:		By repeating rule 21, a stronger rule can be constructed:
	* <math>D(a, b, c)~~</math> --> <math>~~D(a, 0, f_1^{b}(c))</math>		* <math>D(a, b, c) \rightarrow D(a, 0, f_1^{b}(c))</math>
	* <math>D_1(a, b, c)~~</math> --> <math>~~D_1(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:
	<math>D(a, 0, c)~~</math> --> <math>~~D(a-3, 2c+11, 1)~~</math> --> <math>~~D(a-3, 0, f_1^{2c+11}(1))</math>		<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-3, 0, f_2(1,c))</math>
	* <math>D(a, 0, c)~~</math> --> <math>~~D(a-3, 0, f_1^{2c+11}(1))</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>:		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)~~</math> --> <math>~~D(d, 0, f_2(k, c))</math>		* <math>D(3k+d, 0, c) \rightarrow D(d, 0, f_2(k, c))</math>
	* <math>D_1(3k+d, 0, c)~~</math> --> <math>~~D_1(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 D(2, 2, 1).		The TM starts in configuration <math>D(2, 2, 1)</math>.

	D(2, 2, 1) ~~-->~~		<math>D(2, 2, 1) \rightarrow D(2, 0, f_1^{2}(1)) = D(2, 0, f_1(91))</math>

	~~<math>~~D(2, 0, f_1^{2}(1)) = D(2, 0, f_1(91))</math>

	<math>e_1 = f_1(91) = 2^{95} \times 3 -5</math>		<math>e_1 = f_1(91) = 2^{95} \times 3 -5</math>
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	<math>D(2, 0, e_1)</math>		<math>D(2, 0, e_1)</math>

	~~-->~~<math>D_1(2e_1+8, 3, 1)~~</math> --> <math>~~D_1(2e_1+8, 0, f_1^{2}(91))</math>		<math> \rightarrow D_1(2e_1+8, 3, 1) \rightarrow D_1(2e_1+8, 0, f_1^{2}(91))</math>

	e_1 mod 3 = 1; 2*1 + 8 = 10 --> 10 mod 3 = 1		<math>e_1</math> mod 3 = 1; 2*1 + 8 = 10 --> 10 mod 3 = 1

	<math>D_1(2e_1+8, 0, f_1^{2}(91))</math>		<math>D_1(2e_1+8, 0, f_1^{2}(91))</math>

	~~-->~~ <math>D_1(1,0,f_2(\frac{2e_1+7} 3, f_1^{2}(91)))</math>		<math> \rightarrow D_1(1,0,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>e_2 = f_2(\frac{2e_1+7} 3, f_1^{2}(91))</math>
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	<math>D_1(1,0,e_3)</math>		<math>D_1(1,0,e_3)</math>

	<math>e_2 mod 3 = 1~~</math>~~		<math>e_2</math> mod 3 = 1

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

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

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

	~~-->~~ <math>D_1(0, 0, f_2(\frac{2e_2+7} 3, f_1^{2}(91)))</math>		<math> \rightarrow D_1(0, 0, f_2(\frac{2e_2+7} 3, f_1^{2}(91)))</math>

	<math>e_3 = f_2(\frac{2e_2+7} 3, f_1^{2}(91))</math>		<math>e_3 = f_2(\frac{2e_2+7} 3, f_1^{2}(91))</math>
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	<math>D_1(0, 0, e_3)</math>		<math>D_1(0, 0, e_3)</math>

	~~-->~~ <math>D(2e_3+5, 3, 1)~~</math> --> <math>~~D(2e_3+5, 0, f_1^{2}(91))</math>		<math> \rightarrow D(2e_3+5, 3, 1) \rightarrow D(2e_3+5, 0, f_1^{2}(91))</math>

	e_3 mod 3 = 1; 2*1+5 = 7 --> 7 mod 3 = 1		<math>e_3</math> mod 3 = 1; 2*1 + 5 = 7 --> 7 mod 3 = 1

	~~-->~~ <math>D(1, 0, f_2(\frac{2e_3+4} 3, f_1^{2}(91)))</math>		<math> \rightarrow D(1, 0, f_2(\frac{2e_3+4} 3, f_1^{2}(91)))</math>

	<math>e_4 = f_2(\frac{2e_3+4} 3, f_1^{2}(91))</math>		<math>e_4 = f_2(\frac{2e_3+4} 3, f_1^{2}(91))</math>