1RB1RD1LC 2LB1RB1LC 1RZ1LA1LD 0RB2RA2RD: Difference between revisions

Revision as of 11:56, 19 October 2025

1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD (bbch) 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 $2 ↑^{4} 5$ , possibly 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:

1 Z> 1^((2*<(<(<(16*2^(92) - 3); (24*2^((24*2^(<(b + 10); (24*2^(b) - 4); 2>) - 3)) - 11); (24*2^((24*2^(<(24*2^((24*2^(<(24*2^((24*2^(92) - 3)) - 2); (24*2^(b) - 4); 92>) - 3)) - 1); (24*2^(b) - 4); 2>) - 3)) - 11)> + 8)/3; (24*2^((24*2^(<(b + 10); (24*2^(b) - 4); 2>) - 3)) - 11); (24*2^((24*2^(<1; (24*2^(b) - 4); 2>) - 3)) - 11)> + 5)/3; (24*2^((24*2^(<(b + 10); (24*2^(b) - 4); 2>) - 3)) - 11); (24*2^((24*2^(<1; (24*2^(b) - 4); 2>) - 3)) - 11)> + 19))

Analysis by Polygon

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 A> 0^2 S --> S <A 1^2 S [+5 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 A> (11)^1 2^b S --> S 2 A> (11)^1 2^b-1 S [+5 steps]
8. S A> (11)^1 2^b S --> S 2^b A> (11)^1 S [+5b steps]

9. S D> 0^2 S --> S <B 2^2 S [+3 steps]

10. S 2 <D (11)^a 0^2 S --> S <D (11)^a+1 2 S [+4a +7 steps]
11. S 2 <D (11)^a 2 0^2 S --> S <D (11)^a+1 2^2 S [+4a +7 steps]

12. S  1^a <A (11)^b 0^2 S --> S 1^a-1 <A (11)^b+1 2 S [+4b +7 steps]
13. S 1^a <A (11)^b 2 0^2 S --> S 1^a-1 <A (11)^b+1 2^2 S [+4b +7 steps]

14. S (12)^a 1 <D (11)^b 0^2 S --> S (12)^a-1 1 <D (11)^b+2 [+4b +8 steps]
15. S (12)^a 1 <D (11)^b 0^inf --> S 1 <D (11)^b+2a 0^inf [+4a^2 +4ba + 4a steps]

16. S (12)^a 2 1 <D (11)^b 0^inf --> S (12)^a-1 2 (12)^b+2 1 <D (11)^1 0^inf [+10b +28 steps]
17. S (12)^a 2 1 <D (11)^b 0^inf --> S (12)^a-1 2 1 <D (11)^2b+5 0^inf
18. S (12)^a 2 1 <D (11)^b 0^inf --> S 2 1 <D (11)^(2^a)*b+(2^a)*5-5 0^inf

19. S (12)^a 2 1 <D (11)^b 2 0^inf --> S (12)^a 2^2 1 <D (11)^2b-1 0^inf

20. S (12)^a 1 <D (11)^b 2 0^inf --> S (12)^a 2 1 <D (11)^2b-1 0^inf

21. S (12)^a 2^2 1 <D (11)^b 0^inf --> S (12)^a-1 2^2 1 <D (11)^2^(b+4)*3-5 0^inf

22. S 1 <D (11)^b 2^2 0^inf --> S 2 (12)^b-1 2 1 <D (11)^1 0^inf

23. S (11)^a 2^2 1 <D (11)^b 0^inf --> S (11)^a-3 (12)^2b+11 2^2 1 <D (11)^1 0^inf
24. 0^inf 2^2 1 <D (11)^c 0^inf --> 0^inf (11)^c+1 (12)^3 2^2 1 <D (11)^1 0^inf
25. 0^inf (11)^2 2^2 1 <D (11)^c 0^inf --> 0^inf 1 (11)^2c+8 (12)^3 2^2 1 <D (11)^1 0^inf
26. 0^inf 1 (11)^1 2^2 1 <D (11)^c 0^inf --> 0^inf 1 (11)^2c+7 (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

Let $D (a, b, c)$ = 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 $f_{1} (n) = 2^{n + 4} \times 3 - 5$

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:

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

Rule 23 becomes:

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

Rule 24 becomes:

$D (0, 0, c) \to D (c + 1, 3, 1)$

Rule 25 becomes:

$D (2, 0, c) \to D (2 c + 8, 3, 1)$

Rule 26 becomes:

$D_{1} (1, 0, c) \to D_{1} (2 c + 7, 3, 1)$

Rule 27 becomes:

$D_{1} (0, 0, c) \to D (2 c + 5, 3, 1)$

Rule 28 becomes:

$D (1, 0, c) \to halt with score 4 c + 18$

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

$D (a, b, c) \to D (a, 0, f_{1}^{b} (c))$
$D_{1} (a, b, c) \to D_{1} (a, 0, f_{1}^{b} (c))$

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

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

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

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

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

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

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

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

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

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

$e_{1}$ mod 3 = 1; 2*1 + 8 = 10 --> 10 mod 3 = 1

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

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

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

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

$e_{2}$ mod 3 = 1

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

$2 e_{3} + 7$

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

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

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

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

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

$e_{3}$ mod 3 = 1; 2*1 + 5 = 7 --> 7 mod 3 = 1

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

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

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

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

This can be bounded by:

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

@@ Line 52: / Line 52: @@
 . 0^inf (11)^1 2^2 1 <D (11)^c 0^inf --> 0^inf 1 Z> 1 (11)^2c+8 0^inf
 </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>
@@ Line 61: / Line 61: @@
 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:
-* 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:
-* D(0, 0, c) --> D(c+1, 3, 1)
+* <math>D(0, 0, c) \rightarrow D(c+1, 3, 1)</math>
 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:
-* 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:
-* 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:
-* 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:
-* <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:
-<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, 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>:
-* <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>
@@ Line 118: / Line 116: @@
 <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(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>
@@ Line 130: / Line 128: @@
 <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>
-e_3 + 7
+<math>2e_3 + 7</math>
 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>
@@ Line 145: / Line 143: @@
 <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>

1RB1RD1LC 2LB1RB1LC 1RZ1LA1LD 0RB2RA2RD: Difference between revisions

Revision as of 11:56, 19 October 2025

Analysis by Polygon

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