1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_0RB2RA2RD

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 $10{\uparrow }^{4}4$, 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

Bonus rule: 29. 0^inf 1 (11)^2 2^2 1 <D (11)^c 0^inf --> 0^inf (11)^2c+10 (12)^2 2^2 1 <D (11)^1 0^inf

Let $D(a,b,c)$ = $0^{\mathrm{\infty }}11^{a}12^b{2}^{2}1\text{<mo><</mo>}\text{D}11^c{0}^{\mathrm{\infty }}$.

Let $D_1(a,b,c)$ = $0^{\mathrm{\infty }}111^{a}12^b{2}^{2}1\text{<mo><</mo>}\text{D}11^c{0}^{\mathrm{\infty }}$.

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:

• $D(a,b,c)-->D(a,b-1,2^{c+4}\times 3-5)=D(a,b-1,f_1(c))$
• $D_1(a,b,c)-->D_1(a,b-1,2^{c+4}\times 3-5)=D_1(a,b-1,f_1(c))$

Rule 30 is constructed by repeating rule 21:

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

Rule 23 becomes:

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

If a is greater than or equal to 3:

$D(a,0,c)-->D(a-3,2c+11,1)-->D(a-3,0,f_1^{2c+11}(1))$ via rules 23 and 30, and

$D_1(a,0,c)-->D_1(a-3,2c+11,1)-->D_1(a-3,0,f_1^{2c+11}(1))$ via rules 23 and 30

Thus rule 31 is:

• $D(a,0,c)-->D(a-3,0,f_1^{2c+11}(1))=D(a-3,0,f_2(1,c))$ and
• $D_1(a,0,c)-->D_1(a-3,0,f_1^{2c+11}(1))=D_1(a-3,0,f_2(1,c))$

This rule can also be repeated to create rule 32, note that $f_1^{2c+11}(1)=f_2(1,c)$ and $f_1^{2\times f_2(a,b)+11}(1)=f_2(a+1,b)$:

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

Rule 24 becomes:

• $D(0,0,c)-->D(c+1,3,1)$

Rule 25 becomes:

• $D(2,0,c)-->D_1(2c+8,3,1)$

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:

• $D_1(2,0,c)-->D(2c+10,2,1)$

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 for a ≠ 0

$D(2,0,e_1)$

$\to D_1(2e_1+8,3,1)\to D_1(2e_1+8,0,f_1^2(91))$

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

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

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

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

$D_1(1,0,e_2)$

$e_2$ mod 3 = 1

$\to D_1(2e_2+7,3,1)\to D_1(2e_2+7,0,f_1^2(91))$

$2e_2+7$

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

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

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

$D_1(0,0,e_3)$

$\to D(2e_3+5,3,1)\to D(2e_3+5,0,f_1^2(91))$

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

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

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

$D(1,0,e_4)$

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

halting tape: 0^inf 1 Z> 1 (11)^(2e_4)+8 0^inf

This can be bounded by:

$10{\uparrow }^{4}4<2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow (7.92\times 10^{28})<e_4<\sigma <S<2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow (7.93\times 10^{28})$

This TM is also a num(4,3) contender regardless of definition, as its score consists entirely of consecutive ones.

By changing the C0 transition from C0 --> 1RZ to C0 --> 1LB, this machine can be turned into a Translated cycler with preperiod > $10{\uparrow }^{4}4$, making 1RB1RD1LC_2LB1RB1LC_1LB1LA1LD_0RB2RA2RD (bbch) the current BBS(4,3) champion.

Reset behavior

Let's restate the rules:

Rule 30:

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

Rule 23:

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

Rule 31 for a greater than or equal to 3:

• $D(a,0,c)-->D(a-3,0,f_1^{2c+11}(1))=D(a-3,0,f_2(1,c))$ and
• $D_1(a,0,c)-->D_1(a-3,0,f_1^{2c+11}(1))=D_1(a-3,0,f_2(1,c))$

Rule 32:

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

Rule 24:

• $D(0,0,c)-->D(c+1,3,1)$

Rule 28:

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

Rule 25:

• $D(2,0,c)-->D_1(2c+8,3,1)$

Rule 27:

• $D_1(0,0,c)-->D(2c+5,3,1)$

Rule 26:

• $D_1(1,0,c)-->D_1(2c+7,3,1)$

Rule 29:

• $D_1(2,0,c)-->D(2c+10,2,1)$

Let's explore what happens after the remainder rules:

AR1: $D(0,0,c)-->D(c+1,3,1)-->D(c+1,0,f_1^3(1))$ by rules 24 and 30.

AR2: D(1, 0, c) --> halt with score 4c + 18 by rule 28.

AR3: $D(2,0,c)-->D_1(2c+8,3,1)-->D(2c+8,0,f_1^3(1))$ by rules 25 and 30.

AR4: $D_1(0,0,c)-->D(2c+5,3,1)-->D(2c+5,0,f_1^3(1))$ by rules 27 and 30.

AR5: $D_1(1,0,c)-->D_1(2c+7,3,1)-->D_1(2c+7,0,f_1^3(1))$ by rules 26 and 30.

AR6: $D_1(2,0,c)-->D(2c+10,2,1)-->D(2c+10,0,f_1^2(1))$ by rules 29 and 30.

Together with:

AR7 by rule 30:

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

AR8 by rule 32 for k ≠ 0:

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

Let's also consider the following which was shown earlier:

f_1(n) mod 3 = 1
f_1^a(n) mod 3 = 1
f_2(a,b) mod 3 = 1 for a ≠ 0

Every one of these rules will set c to either f_1(n), f_1^a(n) or f_2(a,b) with a ≠ 0, thus c mod 3 = 1 after the first rule application.

Thus, c can be considered to be of the form 3m+1:

AR1-2: $D(0,0,3m+1)-->D(3m+2,0,f_1^3(1))-->D(2,0,f_2(m,f_1^3(1)))$ by rules AR1 and AR8. AR1-2 leads to AR3-2.

AR2-2: D(1, 0, 3m+1) --> halt with score 12m + 22 by rule AR2. AR2-2 halts.

AR3-2: $D(2,0,3m+1)-->D_1(6m+10,0,f_1^3(1))=D_1(3(2m+3)+1,0,f_1^3(1))-->D_1(1,0,f_2(2m+3,f_1^3(1)))$ by AR3 and AR8. AR3-2 leads to AR5-2.

AR4-2: $D_1(0,0,3m+1)-->D(6m+7,0,f_1^3(1))=D(3(2m+2)+1,0,f_1^3(1))-->D(1,0,f_2(2m+2,f_1^3(1)))$ by rules AR4 and AR8. AR4-2 leads to AR2-2

AR5-2: $D_1(1,0,3m+1)-->D_1(6m+9,0,f_1^3(1))=D_1(3(2m+3),0,f_1^3(1))-->D_1(0,0,f_2(2m+3,f_1^3(1)))$ by rules AR5 and AR8. AR5-2 leads to AR4-2.

AR6-2: $D_1(2,0,3m+1)-->D(6m+12,0,f_1^2(1))=D(3(2m+4),0,f_1^2(1))-->D(0,0,f_2(2m+4,f_1^2(1)))$ by rules AR6 and AR8. AR6-2 leads to AR1-2.

The maximum trajectory following these rules is: AR6-2 --> AR1-2 --> AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt

This TM follows the trajectory: AR3-2 --> AR5-2 --> AR4-2 --> AR2-2 --> halt