User:Polygon/Page for testing

1RB2LB0LB_2LC2LA0LA_2RD1LC1RZ_1RA2LD1RD (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 $3\uparrow \uparrow \uparrow 88574$. Polygon analysed the TM by hand in October 2025, providing its score.

Pavel listed the halting tape as:

1 Z> 1^(162*3^((3*<(243*3^(6) - 5)/2; (<(54*3^((3b + 11)/2) - 2); (54*3^((3b + 14)/2) - 6); (54*3^(7) - 6)> + 1); (<(54*3^((3*<(54*3^(7) - 3); (54*3^((3b + 14)/2) - 6); (54*3^((81*3^(7) - 2)) - 6)> + 14)/2) - 2); (54*3^((3b + 14)/2) - 6); (54*3^(7) - 6)> + 1)> + 11)/2)) 2

Analysis by Polygon

S is any tape configuration

1. S 1^a <C S --> S <C 1^a S [+a steps]
2. S 1^a <D S --> S <D 2^a S [+a steps]
3. S D> 2^a S --> S 1^a D> S [+a steps]

4. S (11)^a <A S --> S <A (22)^a S [+2a steps]
   S (11)^a <B S --> S <B (22)^a S [+2a steps]

5. 0^inf 2 (11)^a A> (22)^b S --> 0^inf 2 (11)^a+3 A> (22)^b-1 S [+8a +24 steps]
6. 0^inf 2 (11)^a A> (22)^b S --> 0^inf 2 (11)^a+3b A> S

7. 0^inf 2 (11)^a A> 0 (22)^b S --> 0^inf 2 1 (11)^1 A> (22)^a+2 0 (22)^b-1 S [+6a +28 steps]

8. 0^inf 2 (11)^a A> 2 0 2 S --> 0^inf 2 1 (11)^a+3 A> S [+8a +27 steps]

9. 0^inf 2 1 (11)^a A> (22)^b S --> 0^inf 2 1 (11)^3a+4 A> (22)^b-1 S
10. 0^inf 2 1 (11)^a A> (22)^b S --> 0^inf 2 1 (11)^g_1^b(a) A> S

11-1. 0^inf 2 1 (11)^a A> 0 (22)^b S --> 0^inf 2 (11)^3a+4 A> 0 (22)^b-1 2 S
11-2. 0^inf 2 1 (11)^a A> 0 (22)^b S --> 0^inf 2 1 (11)^1 A> (22)^3a+6 0 (22)^b-2 2 S

12. 0^inf 2 (11)^a A> 0 11 S --> 0^inf 2 1 (11)^1 A> 0 (22)^a+2 2 S [+6a +31 steps]

13. 0^inf 2 1 (11)^a A> 0 2^b S --> 0^inf 2 1 (11)^g_2(a) A> 0 2^b-3 S
14. 0^inf 2 1 (11)^a A> 0 2^3k+v S --> 0^inf 2 1 (11)^(g_2)^k(a) A> 0 2^v S

15. 0^inf 2 1 <A S --> 0^inf 1 D> 2^3 S [+8 steps]

16. 0^inf 2 1 (11)^a A> 0 2 1 2 0^inf --> 0^inf 2 1 (11)^1 A> 0 (22)^1 (11)^3a+7 2 0^inf (may be irrelevant)

17. 0^inf 2 1 (11)^a A> 0 (22)^1 1 S --> 0^inf 2 1 (11)^1 A> (22)^3a+6 2 S
18. 0^inf 2 1 (11)^a A> 0 (22)^1 1 S --> 0^inf 2 1 (11)^g_2(a) A> 2 S

19. 0^inf 2 1 (11)^a A> 2 1^3 S --> 0^inf 2 1 (11)^1 A> 0 (22)^3a+5 2 S
19*. 0^inf 2 1 (11)^a A> 2 1^2 S --> 0^inf 1 (11)^3a+5 D> S
19**. 0^inf 2 1 (11)^a A> 2 1 S --> 0^inf <B (22)^3a+4 S
19*** 0^inf 2 1 (11)^a A> 2 S --> 0^inf (11)^3a+3 1 B> S

20. 0^inf 2 1 (11)^a A> 0 (22)^1 1^b S --> 0^inf 2 1 (11)^g_3(a) A> 0 (22)^1 1^b-4 S
21. 0^inf 2 1 (11)^a A> 0 (22)^1 1^4k+v S --> 0^inf 2 1 (11)^g_3^k(a) A> 0 (22)^1 1^v S

22. 0^inf 2 1 (11)^a A> 0 (22)^1 1^3 2 0^inf --> 0^inf 2 1 (11)^1 A> 0 (22)^1 1^6*g_2(a)+12 2 0^inf
23. 0^inf 2 1 (11)^a A> 0 (22)^1 1^2 2 0^inf --> 0^inf 1 Z> (11)^3*g_2(a)+6 2 0^inf
Bonus rules which are not relevant for this TMs behavior:
24. 0^inf (11)^a A> 0 (22)^1 1 2 0^inf --> 0^inf 1 Z> (11)^3*g_2(a)+5 2 0^inf
25. 0^inf 2 1 (11)^a A> 0 (22)^1 2 0^inf --> 0^inf 1 Z> (11)^g_2(a)+1 2 0^inf

The following functions were used in these rules:

g_1(n) = 3n + 4

Note that $(3^k-2)\times 3+4=3^{k+1}-2$

And 1 = 3^1 - 2

It follows that $g_1^n(1)=3^{n+1}-2$

$g_2(n)=3^{3n+7}-2$

$g_3(n)=g_2^{2\times (g_2(n)+3)}(1)$

Further:

Let L(a,b) = 0^inf 2 1 (11)^a A> 0 (22)^1 1^b 2 0^inf

• L(a,4k+v) --> L(g_3^k(a),v) by rule 21
• L(a,0) --> 0^inf 1 Z> (11)^g_2(a)+1 2 0^inf by rule 25
• L(a,1) --> 0^inf 1 Z> (11)^3*g_2(a)+5 2 0^inf by rule 24
• L(a,2) --> 0^inf 1 Z> (11)^3*g_2(a)+6 2 0^inf by rule 23
• L(a,3) --> L(1,6*g_2(a)+12) by rule 22

The TM reaches configuration L(1,3) after running for 34 steps.

L(1,3) --> L(1,6*g_2(1)+12) by rule 22, this can be simplified to L(1,354294), then:

--> L(g_3^88573(1),2) by rule 21

--> 0^inf 1 Z> $(11{)}^{3\times (g_2(g_3^{88573}(1))+6}$ 2 0^inf by rule 22

So $\sigma =6\times g_2(g_3^{88573}(1))+14$.

This can be bunded by:

$3\uparrow \uparrow \uparrow 88574<\sigma <S<3\uparrow \uparrow \uparrow 88575$