1RB1RD1LC 2LB1RB1LC 1RZ1LA1LD 2RB2RA2RD - Revision history

Polygon: /* Analysis by Polygon */ made the page look less weird

2026-03-30T14:08:10Z

Analysis by Polygon: made the page look less weird

← Older revision		Revision as of 14:08, 30 March 2026
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	12. S 1^a <A (11)^b 0^inf --> 1^a-1 <A (11)^b+1 2 0^inf [+4b +5 steps]		12. S 1^a <A (11)^b 0^inf --> 1^a-1 <A (11)^b+1 2 0^inf [+4b +5 steps]
	</pre>		</pre>
	Let <math>A(a, b, c)</math> = S (11)^a (12)^b <D (11)^c 0^~~inf~~		Let <math>A(a, b, c)</math> = <math>\textrm{S}\;11^{a}\;12^{b}\;\textrm{<}\textrm{D}\;11^{c}\;0^\infty</math>.
	* Rule 9: <math>A(a, b, c) \rightarrow A(a, b - 1, 2c + 3)</math>		* Rule 9: <math>A(a, b, c) \rightarrow A(a, b - 1, 2c + 3)</math>
	* Rule 10: <math>A(a, b, c) \rightarrow A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.		* Rule 10: <math>A(a, b, c) \rightarrow A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.
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	'''Trajectory:'''		'''Trajectory:'''

	The TM enters configuration <math>A(19, 2, 3)</math> with S = 2 1 after 799 steps.		The TM enters configuration <math>A(19, 2, 3)</math> with S = <math>2\;1</math> after 799 steps.

	<math>A(19, 2, 3) \rightarrow A(1, f^{9}(2), 3) \rightarrow A(1, 0, f^{10}(2)-3)</math>		<math>A(19, 2, 3) \rightarrow A(1, f^{9}(2), 3) \rightarrow A(1, 0, f^{10}(2)-3)</math>
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	Let m = <math>f^{10}(2)-3</math>		Let m = <math>f^{10}(2)-3</math>

	<math>\rightarrow</math> 0^~~inf~~ 2 1 (11)^1 <D (11)^m 0^~~inf~~		<math>\rightarrow</math> <math>0^\infty\;2\;1\;11^{1}\;\textrm{<}\textrm{D}\;11^{m}\;0^\infty</math>.
	<pre>		<pre>
	Final trajectory:		Final trajectory:

Polygon: formatting

2026-02-19T19:10:40Z

formatting

← Older revision		Revision as of 19:10, 19 February 2026
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	Let m = <math>f^{10}(2)-3</math>		Let m = <math>f^{10}(2)-3</math>

	--> 0^inf 2 1 (11)^1 <D (11)^m 0^inf		<math>\rightarrow</math> 0^inf 2 1 (11)^1 <D (11)^m 0^inf
	<pre>		<pre>
	Final trajectory:		Final trajectory:

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

2025-10-19T12:04:35Z

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

← Older revision		Revision as of 12:04, 19 October 2025
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	12. S 1^a <A (11)^b 0^inf --> 1^a-1 <A (11)^b+1 2 0^inf [+4b +5 steps]		12. S 1^a <A (11)^b 0^inf --> 1^a-1 <A (11)^b+1 2 0^inf [+4b +5 steps]
	</pre>		</pre>
	Let A(a,b,c) = S (11)^a (12)^b <D (11)^c 0^inf		Let <math>A(a, b, c)</math> = S (11)^a (12)^b <D (11)^c 0^inf
	* Rule 9: A(a, b, c) ~~-->~~ A(a, b - 1, 2c + 3)		* Rule 9: <math>A(a, b, c) \rightarrow A(a, b - 1, 2c + 3)</math>
	* Rule 10: A(a, b, c) ~~--> <math>~~A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.		* Rule 10: <math>A(a, b, c) \rightarrow A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.
	* Rule 11: A(a, 0, c) ~~-->~~ A(a - 2, c + 3, 3)		* Rule 11: <math>A(a, 0, c) \rightarrow A(a - 2, c + 3, 3)</math>

	Further: let <math>f(n) = 2^{n+1} \times 3</math>		Further: let <math>f(n) = 2^{n+1} \times 3</math>
	* If c = 3: A(a, b, 3) ~~-->~~ A(a, 0, f(b) - 3) ~~-->~~ A(a - 2, f(b), 3)		* If c = 3: <math>A(a, b, 3) \rightarrow A(a, 0, f(b) - 3) \rightarrow A(a - 2, f(b), 3)</math>

	* A(a, 0, c) ~~--> <math>~~A(a-2, c+3, 3) ~~-->~~ A(a-2, 0, f(c+3)-3)</math>		* <math>A(a, 0, c) \rightarrow A(a-2, c+3, 3) \rightarrow A(a-2, 0, f(c+3)-3)</math>

	* A(2k + d, b, 3) ~~--> <math>~~A(d, f^{k}(b), 3)</math>		* <math>A(2k + d, b, 3) \rightarrow A(d, f^{k}(b), 3)</math>

	'''Trajectory:'''		'''Trajectory:'''

	The TM enters configuration A(19, 2, 3) with S = 2 1 after 799 steps.		The TM enters configuration <math>A(19, 2, 3)</math> with S = 2 1 after 799 steps.

	A(19, 2, 3) ~~--> <math>~~A(1, f^{9}(2), 3) ~~-->~~ A(1, 0, f^{10}(2)-3)</math>		<math>A(19, 2, 3) \rightarrow A(1, f^{9}(2), 3) \rightarrow A(1, 0, f^{10}(2)-3)</math>

	Let m = <math>f^{10}(2)-3</math>		Let m = <math>f^{10}(2)-3</math>
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	Score calculated in HyperCalc:		Score calculated in HyperCalc:

	(10^)^8 30,302,671.815163		<math>(10 \uparrow)^{8} 30\,302\,671.815163</math>

	Or in tetration: 10^^9.873987 (truncated)		Or in tetration: <math>10 \uparrow\uparrow 9.873987</math> (truncated)
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

Polygon: /* Analysis by Polygon */ Added link to existing user page

2025-10-09T19:32:28Z

Analysis by Polygon: Added link to existing user page

← Older revision		Revision as of 19:32, 9 October 2025
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	</pre>		</pre>

	==Analysis by Polygon==		== Analysis by [[User:Polygon\|Polygon]] ==
	<pre>		<pre>
	S is any tape configuration		S is any tape configuration

Polygon: Fixed conversion error from PT notation

2025-10-05T17:23:09Z

Fixed conversion error from PT notation

← Older revision		Revision as of 17:23, 5 October 2025
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	{{machine\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD}}		{{machine\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD}}
	{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD\|halt}} is a tetrational 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 around 10 ↑↑ 10.873987. Polygon analysed the TM by hand in September 2025, providing its score.		{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD\|halt}} is a tetrational 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 around 10 ↑↑ 9.873987. Polygon analysed the TM by hand in September 2025, providing its score.

	Pavel listed the halting tape as:		Pavel listed the halting tape as:
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	(10^)^8 30,302,671.815163		(10^)^8 30,302,671.815163

	Or in tetration: 10^^10.873987 (truncated)		Or in tetration: 10^^9.873987 (truncated)
	[[Category:BB(4,3)]]		[[Category:BB(4,3)]]

Polygon: Created page for one of the "potential BB(4,3) champions"

2025-10-05T15:54:23Z

Created page for one of the "potential BB(4,3) champions"

New page

{{machine|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD}}
{{TM|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD|halt}} is a tetrational 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 around 10 ↑↑ 10.873987. Polygon analysed the TM by hand in September 2025, providing its score.

Pavel listed the halting tape as:
<pre>
1 Z> 1^((8*<7; (6*2^((4b + 14)) - 4); (6*2^((48*2^(21) - 2)) - 4)> + 33)) 2
</pre>

==Analysis by Polygon==
<pre>
S is any tape configuration
1. S D> 2^a S --> S 2^a D> S [+a steps]
2. S B> 1^a S --> S 1^a B> S [+a steps]
3. S 1 B> 0 S --> S <A 1^2 S [+4 steps]
4. S D> (11)^a S --> S (21)^a D> S [+2a steps]
S A> (11)^a S --> S (12)^a A> S [+2a steps]
5. S (21)^a <C S --> S <C (11)^a S [+2a steps]
S (12)^a <A S --> S <A (11)^a S [+2a steps]
6. S (12)^a A> 0^2 S --> S <A (11)^a+1 S [+2a +5 steps]

7. S (12)^a 2 (12)^b A> 0^2 S --> S (12)^a-1 2 (12)^b+2 A> S [+4b +7 steps]
8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf [+4a^2 +8a +4ba steps]

9. S (12)^a <D (11)^b 0^inf --> S (12)^a-1 <D (11)^2b+3 0^inf [+4b^2 +22b +22 steps]
10. S (12)^a <D (11)^b 0^inf --> S <D (11)^((2^(a))*b+(2^(a))*3-3) 0^inf

11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf [+10b +50 steps]

12. S 1^a <A (11)^b 0^inf --> 1^a-1 <A (11)^b+1 2 0^inf [+4b +5 steps]
</pre>
Let A(a,b,c) = S (11)^a (12)^b <D (11)^c 0^inf
* Rule 9: A(a, b, c) --> A(a, b - 1, 2c + 3)
* Rule 10: A(a, b, c) --> <math>A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.
* Rule 11: A(a, 0, c) --> A(a - 2, c + 3, 3)

Further: let <math>f(n) = 2^{n+1} \times 3</math>
* If c = 3: A(a, b, 3) --> A(a, 0, f(b) - 3) --> A(a - 2, f(b), 3)

* A(a, 0, c) --> <math>A(a-2, c+3, 3) --> A(a-2, 0, f(c+3)-3)</math>

* A(2k + d, b, 3) --> <math>A(d, f^{k}(b), 3)</math>

'''Trajectory:'''

The TM enters configuration A(19, 2, 3) with S = 2 1 after 799 steps.

A(19, 2, 3) --> <math>A(1, f^{9}(2), 3) --> A(1, 0, f^{10}(2)-3)</math>

Let m = <math>f^{10}(2)-3</math>

--> 0^inf 2 1 (11)^1 <D (11)^m 0^inf
<pre>
Final trajectory:
0^inf 2 1 (11)^1 <D (11)^m 0^inf
--> 0^inf 2 1 1 2 A> (11)^m 0^inf
--> 0^inf 2 1 (12)^m+1 A> 0^inf
--> 0^inf 2 1 <A (11)^m+2 0^inf
--> 0^inf 2 1 D> (11)^m+2 0^inf
--> 0^inf (21)^m+3 D> 0^inf
--> 0^inf (21)^m+3 2 B> 0^inf
--> 0^inf (21)^m+3 2 <B 2 0^inf
--> 0^inf (21)^m+3 <C (12)^1 0^inf
--> 0^inf <C (11)^m+3 (12)^1 0^inf
--> 0^inf 1 Z> (11)^m+3 (12)^1 0^inf
Score = 2m + 9
</pre>

Score calculated in HyperCalc:

(10^)^8 30,302,671.815163

Or in tetration: 10^^10.873987 (truncated)
[[Category:BB(4,3)]]