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1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD (bbch) is a pentational 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 $2 ↑^{4} 5 < m a t h >, m a k i n g i t t h e B B (4, 3) c h a m p i o n . = = A n a l y s i s = = < p r e > S i s a n y t a p e c o n f i g u r a t i o n 1 . S D > 2^{a} S - - > S 2^{a} D > S [+ a s t e p s] 2 . S B > 1^{a} S - - > S 1^{a} B > S [+ a s t e p s] 3 . S A > 0^{2} S - - > S < A 1^{2} S [+ 5 s t e p s] 4 . S D > (11)^{a} S - - > S (21)^{a} D > S [+ 2 a s t e p s] S A > (11)^{a} S - - > S (12)^{a} A > S [+ 2 a s t e p s] 5 . S (21)^{a} < C S - - > S < C (11)^{a} S [+ 2 a s t e p s] S (12)^{a} < A S - - > S < A (11)^{a} S [+ 2 a s t e p s] 6 . S (12)^{a} A > 0^{2} S - - > S < A (11)^{a} + 1 S [+ 2 a + 5 s t e p s] 7 . S A > (11)^{1} 2^{b} S - - > S 2 A > (11)^{1} 2^{b} - 1 S [+ 5 s t e p s] 8 . S A > (11)^{1} 2^{b} S - - > S 2^{b} A > (11)^{1} S [+ 5 b s t e p s] 9 . S D > 0^{2} S - - > S < B 2^{2} S [+ 3 s t e p s] 10 . S 2 < D (11)^{a} 0^{2} S - - > S < D (11)^{a} + 12 S [+ 4 a + 7 s t e p s] 11 . S 2 < D (11)^{a} 2 0^{2} S - - > S < D (11)^{a} + 1 2^{2} S [+ 4 a + 7 s t e p s] 12 . S 1^{a} < A (11)^{b} 0^{2} S - - > S 1^{a} - 1 < A (11)^{b} + 12 S [+ 4 b + 7 s t e p s] 13 . S 1^{a} < A (11)^{b} 2 0^{2} S - - > S 1^{a} - 1 < A (11)^{b} + 1 2^{2} S [+ 4 b + 7 s t e p s] 14 . S (12)^{a} 1 < D (11)^{b} 0^{2} S - - > S (12)^{a} - 11 < D (11)^{b} + 2 [+ 4 b + 8 s t e p s] 15 . S (12)^{a} 1 < D (11)^{b} 0^{i} n f - - > S 1 < D (11)^{b} + 2 a 0^{i} n f [+ 4 a^{2} + 4 b a + 4 a s t e p s] 16 . S (12)^{a} 21 < D (11)^{b} 0^{i} n f - - > S (12)^{a} - 12 (12)^{b} + 21 < D (11)^{1} 0^{i} n f [+ 10 b + 28 s t e p s] 17 . S (12)^{a} 21 < D (11)^{b} 0^{i} n f - - > S (12)^{a} - 121 < D (11)^{2} b + 5 0^{i} n f 18 . S (12)^{a} 21 < D (11)^{b} 0^{i} n f - - > S 21 < D (11)^{(} 2^{a}) * b + (2^{a}) * 5 - 5 0^{i} n f 19 . S (12)^{a} 21 < D (11)^{b} 2 0^{i} n f - - > S (12)^{a} 2^{2} 1 < D (11)^{2} b - 1 0^{i} n f 20 . S (12)^{a} 1 < D (11)^{b} 2 0^{i} n f - - > S (12)^{a} 21 < D (11)^{2} b - 1 0^{i} n f 21 . S (12)^{a} 2^{2} 1 < D (11)^{b} 0^{i} n f - - > S (12)^{a} - 1 2^{2} 1 < D (11)^{2}^{(} b + 4) * 3 - 5 0^{i} n f 22 . S 1 < D (11)^{b} 2^{2} 0^{i} n f - - > S 2 (12)^{b} - 121 < D (11)^{1} 0^{i} n f 23 . S (11)^{a} 2^{2} 1 < D (11)^{b} 0^{i} n f - - > S (11)^{a} - 3 (12)^{2} b + 11 2^{2} 1 < D (11)^{1} 0^{i} n f 24 . 0^{i} n f 2^{2} 1 < D (11)^{c} 0^{i} n f - - > 0^{i} n f (11)^{c} + 1 (12)^{3} 2^{2} 1 < D (11)^{1} 0^{i} n f 25 . 0^{i} n f (11)^{2} 2^{2} 1 < D (11)^{c} 0^{i} n f - - > 0^{i} n f 1 (11)^{2} c + 8 (12)^{3} 2^{2} 1 < D (11)^{1} 0^{i} n f 26 . 0^{i} n f 1 (11)^{1} 2^{2} 1 < D (11)^{c} 0^{i} n f - - > 0^{i} n f 1 (11)^{2} c + 7 (12)^{3} 2^{2} 1 < D (11)^{1} 0^{i} n f 27 . 0^{i} n f 1 2^{2} 1 < D (11)^{c} 0^{i} n f - - > 0^{i} n f (11)^{2} c + 5 (12)^{3} 2^{2} 1 < D (11)^{1} 0^{i} n f 28 . 0^{i} n f (11)^{1} 2^{2} 1 < D (11)^{c} 0^{i} n f - - > 0^{i} n f 1 Z > 1 (11)^{2} c + 8 0^{i} n f < / p r e > L e t D (a, b, c) = 0^{i} n f (11)^{a} (12)^{b} 2^{2} 1 < D (11)^{c} 0^{i} n f L e t D_{1} (a, b, c) = 0^{i} n f 1 (11)^{a} (12)^{b} 2^{2} 1 < D (11)^{c} 0^{i} n f L e t < m a t h > f_{1} (n) = 2^{n + 4} \times 3 - 5$

Let $f_{2} (a, b) = f_{1}^{2 \times f_{2} (a - 1, b) + 11} (1)$ , where $f_{2} (0, b) = b$

Rule 21 becomes:

$D (a, b, c) - - > D (a, b - 1, 2^{b + 4} \times 3 - 5)$
$D_{1} (a, b, c) - - > D_{1} (a, b - 1, 2^{b + 4} \times 3 - 5)$

Rule 23 becomes:

$D (a, 0, c) - - > D (a - 3, 2 c + 11, 1)$
$D_{1} (a, 0, c) - - > D_{1} (a - 3, 2 c + 11, 1)$

Rule 24 becomes:

$D (0, 0, c) - - > D (c + 1, 3, 1)$

Rule 25 becomes:

$D (2, 0, c) - - > D (2 c + 8, 3, 1)$

Rule 26 becomes:

$D_{1} (1, 0, c) - - > D_{1} (2 c + 7, 3, 1)$

Rule 27 becomes:

$D_{1} (0, 0, c) - - > D (2 c + 5, 3, 1)$

Rule 28 becomes:

D(1, 0, c) --> halt with score 4c + 18

Rule 29 becomes:

$D_{1} (2, 0, c) - - > D (2 c + 10, 2, 1)$

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

$D (a, b, c) - - > D (a, 0, f_{1}^{b} (c))$
$D_{1} (a, b, c) - - > D 1 (a, 0, f_{1}^{b} (c))$

If a is greater than or equal to 3: $D (a, 0, c) - - > D (a - 3, 2 c + 11, 1) - - > D (a - 3, 0, f_{1}^{2 c + 11} (1))$ = $D (a - 3, 0, f_{2} (1, c))$

$D (a, 0, c) - - > D (a - 3, 0, f_{1}^{2 c + 11} (1))$

This rule can also be repeated, also note that $f_{1}^{2 c + 11} (1) = f_{2} (1, c)$ and $f_{1}^{2 \times f_{2} (a, b) + 11} (1) = f_{2} (a + 1, b)$ :

$D (3 k + d, 0, c) - - > D (d, 0, f_{2} (k, c))$
$D_{1} (3 k + d, 0, c) - - > D_{1} (d, 0, f_{2} (k, c))$

The TM starts in configuration D(2, 2, 1).

D(2, 2, 1) -->

$D (2, 0, f_{1}^{2} (1)) = D (2, 0, f_{1} (91))$

$e_{1} = f_{1} (91) = 2^{95} \times 3 - 5$

f_1(n) = 2^(n+4)*3 - 5
Note that the times three means that this expression of of the form 3k - 5 which can be rewritten as 3(k-1)-2 which can again be rewritten as 3(k-2)+1.
Next, 3k+1 mod 3 = 1
So f_1(n) mod 3 = 1
Thus f_1^a(n) mod 3 = 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)
Thus f_2(a,b) mod 3 = 1

$D (2, 0, e_{1})$

--> $D_{1} (2 e_{1} + 8, 3, 1) - - > D_{1} (2 e_{1} + 8, 0, f_{1}^{2} (91))$

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

$D_{1} (2 e_{1} + 8, 0, f_{1}^{2} (91))$

--> $D_{1} (1, 0, f_{2} (\frac{2 e_{1} + 7}{3}, f_{1}^{2} (91)))$

$e_{2} = f_{2} (\frac{2 e_{1} + 7}{3}, f_{1}^{2} (91))$

$D_{1} (1, 0, e_{3})$

$e_{2} m o d 3 = 1$

--> $D_{1} (2 e_{2} + 7, 3, 1) - - > D_{1} (2 e_{2} + 7, 0, f_{1}^{2} (91))$

2e_3 + 7

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

--> $D_{1} (0, 0, f_{2} (\frac{2 e_{2} + 7}{3}, f_{1}^{2} (91)))$

$e_{3} = f_{2} (\frac{2 e_{2} + 7}{3}, f_{1}^{2} (91))$

$D_{1} (0, 0, e_{3})$

--> $D (2 e_{3} + 5, 3, 1) - - > D (2 e_{3} + 5, 0, f_{1}^{2} (91))$

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

--> $D (1, 0, f_{2} (\frac{2 e_{3} + 4}{3}, f_{1}^{2} (91)))$

$e_{4} = f_{2} (\frac{2 e_{3} + 4}{3}, f_{1}^{2} (91))$

$D (1, 0, e_{4})$

--> halts with score $4 e_{4} + 18$ .

This can be bounded by:

$2 ↑^{4} 5 < 2 ↑ ↑ ↑ 2 ↑ ↑ ↑ 2 ↑ ↑ ↑ (7.92 \times 1 0^{28}) < e_{4} < σ < S < 2 ↑ ↑ ↑ 2 ↑ ↑ ↑ 2 ↑ ↑ ↑ (7.93 \times 1 0^{28})$

Revision as of 10:46, 5 October 2025 view source Polygon (talk \| contribs) 763 edits →Analysis: Added approximate score ← Older edit		Revision as of 10:50, 5 October 2025 view source Polygon (talk \| contribs) 763 edits Added score to the introduction Newer edit →
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	{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD\|halt}} is a pentational halting [[BB(4,3)]] TM. It was discovered in May 2024 by Pavel Kropitz as one of seven long running TMs.		{{TM\|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD\|halt}} is a pentational 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 BB(4,3) champion.

	==Analysis==		==Analysis==

User:Polygon/Page for testing: Difference between revisions

Revision as of 10:50, 5 October 2025

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