User:Polygon/Page for testing

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.

Analysis

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:

• $D(a,b,c)-->D(a,b-1,2^{b+4}\times 3-5)$
• $D_1(a,b,c)-->D_1(a,b-1,2^{b+4}\times 3-5)$

Rule 23 becomes:

• $D(a,0,c)-->D(a-3,2c+11,1)$
• $D_1(a,0,c)-->D_1(a-3,2c+11,1)$

Rule 24 becomes:

• $D(0,0,c)-->D(c+1,3,1)$

Rule 25 becomes:

• $D(2,0,c)-->D(2c+8,3,1)$

Rule 26 becomes:

• $D_1(1,0,c)-->D_1(2c+7,3,1)$

Rule 27 becomes:

• $D_1(0,0,c)-->D(2c+5,3,1)$

Rule 28 becomes:

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

Rule 29 becomes:

• $D_1(2,0,c)-->D(2c+10,2,1)$

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

• $D(a,b,c)-->D(a,0,f_1^b(c))$
• $D_1(a,b,c)-->D1(a,0,f_1^b(c))$

If a is greater than or equal to 3: $D(a,0,c)-->D(a-3,2c+11,1)-->D(a-3,0,f_1^{2c+11}(1))$ =$D(a-3,0,f_2(1,c))$

• $D(a,0,c)-->D(a-3,0,f_1^{2c+11}(1))$

This rule can also be repeated, also note that $f_1^{2c+11}(1)=f_2(1,c)$ and $f_1^{2\times f_2(a,b)+11}(1)=f_2(a+1,b)$:

• $D(3k+d,0,c)-->D(d,0,f_2(k,c))$
• $D_1(3k+d,0,c)-->D_1(d,0,f_2(k,c))$

Trajectory

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

$D(2,0,e_1)$

-->$D_1(2e_1+8,3,1)-->D_1(2e_1+8,0,f_1^2(91))$

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

$D_1(2e_1+8,0,f_1^2(91))$

--> $D_1(1,0,f_2(\frac{2e_1+7}{3},f_1^2(91)))$

$e_2=f_2(\frac{2e_1+7}{3},f_1^2(91))$

$D_1(1,0,e_3)$

$e_{2}mod3=1$

--> $D_1(2e_2+7,3,1)-->D_1(2e_2+7,0,f_1^2(91))$

2e_3 + 7

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

--> $D_1(0,0,f_2(\frac{2e_2+7}{3},f_1^2(91)))$

$e_3=f_2(\frac{2e_2+7}{3},f_1^2(91))$

$D_1(0,0,e_3)$

--> $D(2e_3+5,3,1)-->D(2e_3+5,0,f_1^2(91))$

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

--> $D(1,0,f_2(\frac{2e_3+4}{3},f_1^2(91)))$

$e_4=f_2(\frac{2e_3+4}{3},f_1^2(91))$

$D(1,0,e_4)$

--> halts with score $4e_4+18$.