1RB1RC 1LC1RE 1LD0LB 1RE1LC 1LE0RF 1RZ1RA - Revision history

Polygon: Made analysis section consistent

2025-10-09T19:30:14Z

Made analysis section consistent

← Older revision		Revision as of 19:30, 9 October 2025
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	{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA\|halt}} is a long running halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].		{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA\|halt}} is a long running halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].

	Analysis by Racheline:		== Analysis by Racheline ==
	<pre>		<pre>
	a_1 = (2^179+1)/3+179		a_1 = (2^179+1)/3+179

Int-y1: add category BB(6)

2025-09-27T04:56:22Z

add category BB(6)

← Older revision		Revision as of 04:56, 27 September 2025
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	Splitting the counter into bits is simple even for very large tapes like <math>0^\infty\! <\!\!B\ 0\ (10)^{2^{a_3}+1}\ 11\ (10)^{2^{a_2}+1}\ 11\ (10)^{2^{a_1}+1}\ 11\ (10)^{2^{179}+1}\ 11\ (10)^{513}\ 11\ (10)^{17}\ 11\ (10)^5\ 11\ (10)^2\ 11\ 0^\infty</math>, because almost all bits are <math>(101010\ \|\ 111111)</math>, so we only need to compute the remainders modulo 3 of the amounts of <math>10</math>s between each pair of consecutive <math>11</math>s, and <math>2^n\ (\text{mod}\ 3)</math> only depends on the parity of <math>n</math>.		Splitting the counter into bits is simple even for very large tapes like <math>0^\infty\! <\!\!B\ 0\ (10)^{2^{a_3}+1}\ 11\ (10)^{2^{a_2}+1}\ 11\ (10)^{2^{a_1}+1}\ 11\ (10)^{2^{179}+1}\ 11\ (10)^{513}\ 11\ (10)^{17}\ 11\ (10)^5\ 11\ (10)^2\ 11\ 0^\infty</math>, because almost all bits are <math>(101010\ \|\ 111111)</math>, so we only need to compute the remainders modulo 3 of the amounts of <math>10</math>s between each pair of consecutive <math>11</math>s, and <math>2^n\ (\text{mod}\ 3)</math> only depends on the parity of <math>n</math>.

			[[Category:BB(6)]]

Polygon: Added "halt" to bbch-link

2025-08-16T18:33:02Z

Added "halt" to bbch-link

← Older revision		Revision as of 18:33, 16 August 2025
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	{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}		{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}
	{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].		{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA\|halt}} is a long running halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].

	Analysis by Racheline:		Analysis by Racheline:

Polygon: Changed "halting" to lowercase

2025-08-05T13:11:49Z

Changed "halting" to lowercase

← Older revision		Revision as of 13:11, 5 August 2025
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	{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}		{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}
	{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running ~~Halting~~ [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].		{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].

	Analysis by Racheline:		Analysis by Racheline:

Int-y1 at 02:56, 9 July 2025

2025-07-09T02:56:34Z

← Older revision		Revision as of 02:56, 9 July 2025
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	{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}		{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}

	{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].		{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].

Sligocki at 20:24, 5 February 2025

2025-02-05T20:24:02Z

← Older revision		Revision as of 20:24, 5 February 2025
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	{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}		{{machine\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}

	{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]):		{{TM\|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]). It runs for over <math>10 \uparrow\uparrow 7</math> steps. It is a halting [[shift overflow counter]].

	Analysis by Racheline:		Analysis by Racheline:

Racheline: explained TM behavior

2024-11-26T10:46:29Z

explained TM behavior

← Older revision		Revision as of 10:46, 26 November 2024
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	a_5 = (2^a_4+1)/3+a_4		a_5 = (2^a_4+1)/3+a_4

	the ~~counters~~ after overflow:		the tapes after overflow:
	(10)^17 11 (10)^5 11 (10)^3 11 (10)^2 1111		0^∞ <B 0 (10)^17 11 (10)^5 11 (10)^3 11 (10)^2 1111 0^∞
	(10)^513 11 (10)^17 11 (10)^5 11 (10)^3 11 10 1111		0^∞ <B 0 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 11 10 1111 0^∞
	(10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^4 1111		0^∞ <B 0 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^4 1111 0^∞
	(10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 1111		0^∞ <B 0 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 1111 0^∞
	(10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 11		0^∞ <B 0 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 11 0^∞
	(10)^(2^a_3+1) 11 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^2 11		0^∞ <B 0 (10)^(2^a_3+1) 11 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^2 11 0^∞
	(10)^(2^a_4+1) 11 (10)^(2^a_3+1) 11 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^4 111111 10 11		0^∞ <B 0 (10)^(2^a_4+1) 11 (10)^(2^a_3+1) 11 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^4 111111 10 11 0^∞

	halt at roughly 2^a_5 ≈ 2^2^2^2^2^(2^179/3) ≈ 10^10^10^10^10^10^52.8		halt at roughly 2^a_5 ≈ 2^2^2^2^2^(2^179/3) ≈ 10^10^10^10^10^10^52.8
	</pre>		</pre>
			Between overflows, the right side of the TM is a binary counter, whose bits are either <math>(101010\ \|\ 111111)</math>, <math>(10101110\ \|\ 11111011)</math>, <math>(1011101010\ \|\ 1110111111)</math> or <math>(11101010\ \|\ 10111111)</math>, except for the least significant bit which is <math>(1010\ \|\ 1111)</math>, where <math>(x\ \|\ y)</math> means the bit looks like <math>x</math> when it represents a 0 on the counter, and it looks like <math>y</math> when it represents a 1. The counter is split into bits in a greedy way from left to right. For example, when the tape is <math>0^\infty\! <\!\!B\ 0\ (10)^{17}\ 11\ (10)^5\ 11\ (10)^3\ 11\ (10)^2\ 1111\ 0^\infty</math>, the counter is split into bits as <math>(1010)\ (101010)^5\ (11101010)\ (10101110)^2</math>, and we have <math>101111\ 0^\infty</math> left over at the right end.

			The left side is simply a bouncer, which expands leftwards by 2 cells each time the counter increases by 1. To accelerate the TM, we can skip from one overflow to the next by splitting the counter into bits, setting all the bits to represent 1 instead of 0, running the TM (and accelerating its passes through the bits) until it reaches the left end again (which it will do in state B), and then changing the <math>0^\infty\! <\!\!B\ 0\ (10)^2\ 11</math> on the left end into <math>0^\infty\! <\!\!B\ 0\ (10)^{2^n+1}\ 11</math> where <math>n</math> is the number of bits, because the bouncer's size would have increased by <math>2\cdot(2^n-1)</math> cells during the counting we skipped.

			Splitting the counter into bits is simple even for very large tapes like <math>0^\infty\! <\!\!B\ 0\ (10)^{2^{a_3}+1}\ 11\ (10)^{2^{a_2}+1}\ 11\ (10)^{2^{a_1}+1}\ 11\ (10)^{2^{179}+1}\ 11\ (10)^{513}\ 11\ (10)^{17}\ 11\ (10)^5\ 11\ (10)^2\ 11\ 0^\infty</math>, because almost all bits are <math>(101010\ \|\ 111111)</math>, so we only need to compute the remainders modulo 3 of the amounts of <math>10</math>s between each pair of consecutive <math>11</math>s, and <math>2^n\ (\text{mod}\ 3)</math> only depends on the parity of <math>n</math>.

Sligocki: Created page with "{{machine|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} {{TM|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting BB(6) TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]): Analysis by Racheline:

 a_1 = (2^179+1)/3+179 a_2 = (2^a_1+1)/3+a_1 a_3 = (2^a_2+1)/3+a_2 a_4 = (2^a_3+1)/3+a_3-1 a_5 = (2^a_4+1)/3+a_4  the counters after overflow: (10)^17 11 (10)^5..."

2024-11-25T20:21:40Z

Created page with "{{machine|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} {{TM|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting BB(6) TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]): Analysis by Racheline: <pre> a_1 = (2^179+1)/3+179 a_2 = (2^a_1+1)/3+a_1 a_3 = (2^a_2+1)/3+a_2 a_4 = (2^a_3+1)/3+a_3-1 a_5 = (2^a_4+1)/3+a_4 the counters after overflow: (10)^17 11 (10)^5..."

New page

{{machine|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}}

{{TM|1RB1RC_1LC1RE_1LD0LB_1RE1LC_1LE0RF_1RZ1RA}} is a long running Halting [[BB(6)]] TM analyzed by Racheline on 25 Nov 2024 ([https://discord.com/channels/960643023006490684/1239205785913790465/1310651468881334394 Discord Link]):

Analysis by Racheline:
<pre>
a_1 = (2^179+1)/3+179
a_2 = (2^a_1+1)/3+a_1
a_3 = (2^a_2+1)/3+a_2
a_4 = (2^a_3+1)/3+a_3-1
a_5 = (2^a_4+1)/3+a_4

the counters after overflow:
(10)^17 11 (10)^5 11 (10)^3 11 (10)^2 1111
(10)^513 11 (10)^17 11 (10)^5 11 (10)^3 11 10 1111
(10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^4 1111
(10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 1111
(10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^3 11
(10)^(2^a_3+1) 11 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^5 11 (10)^2 11
(10)^(2^a_4+1) 11 (10)^(2^a_3+1) 11 (10)^(2^a_2+1) 11 (10)^(2^a_1+1) 11 (10)^(2^179+1) 11 (10)^513 11 (10)^17 11 (10)^4 111111 10 11

halt at roughly 2^a_5 ≈ 2^2^2^2^2^(2^179/3) ≈ 10^10^10^10^10^10^52.8
</pre>