Mini Analyses
I spend a lot of time doing “quick-and-dirty” analyses of new Busy Beaver champions. Recently, when I have time, I have been posting those analyses on this blog in detailed form. However, converting from my quick notes to a polished blog post is a lot of time and work and so most of my analyses don’t make it. However, I think there is some value in the quick notes, so I keep them in unsaved documents in my editor until I end up with too many tabs (or my computer crashes) and eventually I delete them. As a middle ground, I’ve decided to try and post unpolished notes here as a value to myself if I ever come back and want to understand these machines behavior again in the future. I may add more notes here over time.
2022
May 2022
Pavel-e197k
Shared by email 15 May 2022 by Pavel Kropitz. This TM runs >10^197,282 steps leaving >10^98,641 symbols.
1RB 1RH 1RC 1RA 1RD 0RB 1LE 0RC 0LF 0LD 0LB 1LA
Rule 1: 11 01^n D> 00 -> 01^n+2 D>
Rule 1x: 11^n D> 0^inf -> 01^2n D> 0^inf
Rule 2: 1010 01^m D> 0^inf -> 01^2m+8 D> 0^inf
Rule 2x: 1010^n 01^m D> 0^inf -> 01^((m+8) 2^n - 8) D> 0^inf
Rule 3: 000 01^n D> 000 -> 0100 10^n 11 D>
Rule 4: 100 01^n D> 0^inf -> 11 10^n+1 11^1 D> 0^inf
Rule 5: 000 01^2k+1 D> 0^inf -> 01^(20 2^k - 8) D> 0^inf
Rule 6: 000 01^2k D> 0^inf -> 011 Z> 10 10^(10 2^(5 2^k - 4) - 8) 1 0^inf (Halt(10 2^(5 2^k - 4) - 4))
Rule 7: 0^inf 01^2k+1 D> 0^inf -> Halt(10 2^(5 2^(10 2^k - 4) - 4) - 5)
At step 9, it’s in config 0^inf 01^3 D> 0^inf -> Halt(10 2^(5 2^16 - 4) - 4)
Machine Adjacent to Pavel-e197k
While analyzing 6x2 machines “adjacent” to Pavel-e197k, I found this machine which our filters do not appear to be able to categorize:
0LD 1RZ 1RC 1RA 1RD 0RB 1LE 0RC 0LF 0LD 0LB 1LA
Rule 1: 1010 01^m D> 00 -> 01^m+3 D>
Rule 1x: 1010^n 01^m D> 0^inf -> 01^m+3n D> 0^inf
Rule 2: 0^inf 100 01^m D> 0^inf -> 0^inf 10010 01^m+1 D> 0^inf
Rule 3: 0^inf 10010 01^m D> 0^inf -> 0^inf 100 10^m+3 01^2 D> 0^inf
Rule 4a: 100 01^2k D> -> 100 10^2k+4 01^2 D> -> 100 01^3k+8 D>
Rule 4b: 100 01^2k+1 D> -> 100 10^2k+5 01^2 D> -> 10010 01^3k+8 D> -> 100 10^3k+11 01^2 D>
Rule 5a: 100 10^2k 01^2 D> -> 100 01^3k+2 D>
Rule 5b: 100 10^2k+1 01^2 D> -> 10010 01^3k+2 D> -> 100 10^3k+5 01^2 D>
At step 20, it’s in config 100 01^2 D> at which point Rules 4/5 a/b will apply forever, so this is infinite (but our system is not smart enough to know that).
Machine Adjacent to e78k
While analyzing 6x2 machines “adjacent” to the short-lived 6x2 champion I announced on Friday (which runs for >10^78913 steps), I found exactly one machine (with 3 different permutations of start state) which our filters aren’t able to categorize. This is the same as my 6x2 former champion except with A1 -> 1RA.
1RB 1RA 1LC 0RF 1RA 0LD 0LC 0LE 1LD 0RA 1RE 1RZ
a0: 00001 <E 0101^a 1^3k 00 -> 1 <E 0101 1 0^4a 001^k 1
a1: ` 001 <E 0101^a 1^3k+1 00 -> 1 <E 0101^a+1 001^k 1
a2: 0001 <E 0101^a 1^3k+2 00 -> 1 <E 0101 1^4a+2 001^k 1`
Once this machine gets into a tape matching regular expression 0^inf 1 <E (0101)+ (00|1)* 0^inf it will never leave. Thus these machines are infinite.
Swapping Rules
1RB 1RF 1LC 1LB --- 0LD 1LE 1LF 1RE 1LD 0RF 0RA
Rule 1: 0100 01^a <D 01 11^b -> 0100 01^b+1 <D 01 11^a
Rule 1x: 0100 01^a <D 01 11^b -> 0100 01^a+1 <D 01 11^b -> Infinite
1RB 1LC 0RC 1LE 1RD 0LA 1LB 1RC 1RC 1LF --- 1LE
Rule 1: 01 11^a 10 C> 01^b 11 -> 01 11^b-2 10 C> 01^a+4 11
Rule 1x: 01 11^a 10 C> 01^b 11 -> 01 11^a+2 10 C> 01^b+2 11 -> Infinite
Three Stage Rules
1RB 1LE 1LC --- 1RD 1LB 1LF 0LE 0LD 1RF 0RE 0RA
C(a, b, c) = <F 10^a 0^b 11^c
Rule 1: C(a, b + 1, c) -> C(a + 1, b, c)
Rule 1x: C(a, b, c) -> C(a + b, 0, c)
Rule 2: C(a, 0, c+2) -> C(1, 2a+2, c+1) -> C(2a+3, 0, c+1)
Rule 2x: C(a, 0, c+1) -> C( (a+3) 2^c - 3, 0, 1)
Rule 3: C(a, 0, 1) -> C(1, 4, 2a+4) -> C(5, 0, 2a+4) -> C(8 2^(2a+4) - 3, 0, 1)
Rule 3x: C(a, 0, 1) -> Infinite
1RB --- 1LC 1LB 0RE 0LD 1RD 1LC 1RF 0LE 1RE 1RA
E(a, b, c) = <E 0^2a+1 1^b 01^2c+1
Rule 1: E(a, b+1, c) -> E(a+1, b, c)
Rule 1x: E(a, b, c) -> E(a+b, 0, c)
Rule 2: E(a, 0, c+1) -> E(0, 2a+6, c) -> E(2a+6, 0, c)
Rule 2x: E(a, 0, c) -> E( (a+6) 2^c - 6, 0, 0)
Rule 3: E(2k, 0, 0) -> E(0, 0, k+1) -> E(6 2^k+1 - 6, 0, 0)
Rule 3x: E(2k, 0, 0) -> Inf
1RB 1LE 1RC 1RB 1LD 0RE --- 0LA 1LF 0RD 1RB 1LA
E(a, b, c) = <E 10^a 1^b 0 01^c
Rule 1: E(a, b+2, c) -> E(a+1, b+1, c)
Rule 1x: E(a, b+1, c) -> E(a+b, 1, c)
Rule 2: E(a, 1, c+1) -> E(1, 2a+3, c) -> E(2a+3, 1, c)
Rule 2x: E(a, 1, c) -> E((a+3) 2^c - 3, 1, 0)
Rule 3: E(a, 1, 0) -> E(1, 1, a + 2) -> E(4 2^(a+2) - 3, 1, 0)
Rule 3: E(a, 1, 0) -> Infinite
1RB --- 1RC 1RA 0RD 1LE 1RE 0RC 0LC 0LF 1LA 1RC
E(a, b, c) = 11^a 0^b 01^c E>
E(a, b+1, c) -> E(a, b, c+1) E(a, b, c) -> E(a, 0, c+b)
E(a+2, 0, c) -> E(a+1, 2c + 2, 1) -> E(a+1, 0, 2c+3)
E(1, 0, c) -> E(c+3, 0, 1) -> Inf
1RB 0LC 0LA 1RD 1LA 1LB 0RE 1LA 0RB 0RF 1RE ---
Rule 1: 00 <C 01^n+1 -> 11^2 00 <C 01^n Rule 1x: 00 <C 01^n+1 -> 11^2n 00 <C 01 Rule 1+: 00 <C 01^n+1 -> <C 01^2n 001
C(a, b, c) = 11^a 00 <C 01^b 001 01^c 001
Rule 2: C(a, b+1, c) -> C(a+1, b, 2c) Rule 2x: C(a, b, c) -> C(a+b, 0, c 2^b)
Rule 3: C(a, 0, c) -> C(0, a+1, 2c) -> C(a+1, 0, c 2^b+1) Rule 3x: C(a, 0, c) -> Infinite
June 2022
Halting 6x2
1RB 0LE 0RC 1RF 1RD 0RB 1LA 1RB 1LA 0LD 1RA 1RZ
0 011^n 01 D> 0 -> 1^3n+2 01 D>
D(n, a, b, c) = 111 011^n 0^a 1^b 011^c 01 D>
Rule 1: D(a, b+3, c) -> D(a, b+1, c+2)
Rule 1x: D(a, 2k+r, c) -> D(a, r, c+2k)
Rule 2: D(a+1, 0, c) -> D(a, 3c+2, 0)
Rule 3: D(a+1, 2k) -> D(a+1, 0, 2k) -> D(a, 6k+2) = D(a, 3b+2)
Rule 3x: D(a, 2k) -> D(0, A 3^a - B)
Rule 4: D(4, 1, c+1) -> 111 011 D(2, 3c+1, 0)
Rule 5: 1 D(2, 1, c) -> 1011 D(4, 0, c) -> 1011 D(3, 3c+2, 0)
@3: D(0, 1, 1)
@36: D(0, 5, 1)
1x: D(0, 1, 5)
4: 111 011 D(2, 13)
1x: 111 011 D(2, 1, 12)
5: 111 011 011 D(3, 38) # 38 -> 116 -> 350 -> 1052
3x: 111 011 011 D(0, 1052) = 111 011 D(1, 1054)
3x: 111 011 D(0, 3164) = 111 D(1, 3166)
3x: 111 D(0, 9500) = D(0, 9503)
1x: D(0, 1, 9502)
4: 111 011 D(2, 28_504)
+-----+-----+
| 0 | 1 |
+---+-----+-----+
| A | 1RB | 0LE |
+---+-----+-----+
| B | 0RC | 1RF |
+---+-----+-----+
| C | 1RD | 0RB |
+---+-----+-----+
| D | 1LA | 1RB |
+---+-----+-----+
| E | 1LA | 0LD |
+---+-----+-----+
| F | 1RA | 1RZ |
+---+-----+-----+
<D 001010 100^4 101
10 <D 00 100^2 101 001^340 01
<E 0100 100^9 10 100^e10 101
<D 001010 100^23 10 100^e28 101
10 <D 00 100^82 10 100^e111 101
<E 0100 100^279 10 100^e391 101
<D 001010 100^946 10 100^e1344 101
10 <D 00 100^3195 10 100^e4561 101
<E 0100 100^10787 10 10 100^e15427 101
A(a, b) = 1011 <A 100^a 100 101 001 001^b 01
111 011 00 111^a 1101100 B> 01 001^b 01
111 011 00 111^a 1101100 00 B> 001^b 01
111 011 00 111^a 1101100 00 011^b B> 01
111 011 00 111^a 1101100 00 011^b 01101 D> 0
= 111 011 00 1^3a+2 011 D(4, 0, b+1)
111 011 00 1^3a+2 011 D(3, 3b+5, 0)
A(a, 2k+1)
111 011 00 1^3a+2 011 D(3, 6k+8, 0)
111 011 00 1^3a+2 011 D(0, (6k+9) 3^3 - 1, 0)
= 111 011 00 1^3a+2 D(1, (6k+9) 3^3 + 1, 0)
111 011 00 1^3a+2 D(0, (6k+9) 3^4 + 5, 0)
= 111 011 00 1^3a+2 D(0, (6k+9) 3^4 + 5, 0)
Rule 1: D(a, b+3, c) -> D(a, b+1, c+2)
Rule 1x: D(a, 2k+r, c) -> D(a, r, c+2k)
Rule 2: D(a+1, 0, c) -> D(a, 3c+2, 0)
Rule 3: D(a+1, 2k) -> D(a+1, 0, 2k) -> D(a, 6k+2) = D(a, 3b+2)
Rule 3x: D(a, 2k) -> D(0, (2k+1) 3^a - 1)
Rule 4: D(4, 1, c+1) -> 111 011 D(2, 3c+1, 0)
Rule 5: 1 D(2, 1, c) -> 1011 D(4, 0, c) -> 1011 D(3, 3c+2, 0)
0000 1011 <A -> 11101100 B>
000010 <D 00
0000 <E 0100
00 <D 001010
11101100 B>
B> 100 -> 111 B>
(11) B> 1010 -> 101100 B>
B> 01 -> 00 B>
B> 001 -> 011 B>
B> 01 0000 -> <A 100101
0001 D> 00
01101 D> 0
<A 100 101
+-----+-----+
| 0 | 1 |
+---+-----+-----+
| A | 1LE | 0LD |
+---+-----+-----+
| B | 0RC | 0RE |
+---+-----+-----+
| C | 0RD | 1RC |
+---+-----+-----+
| D | 1LA | 1LF |
+---+-----+-----+
| E | 1RB | 0LA |
+---+-----+-----+
| F | 1LD | 1RZ |
+---+-----+-----+
Rule 1: 00 <E 10^n 11 -> <E 10^n+2
Rule 1x: 0^inf <E 10^n 11^m -> 0^inf <E 10^n+2m
Rule 2: 0000 <E 10^n 0101 -> <E 1^2n+8 -> 0^inf <E 10^2n+8
Rule 2x: 0^inf <E 10^n 0101^m -> 0^inf <E 10^((n+8) 2^m - 8)
Rule 3: 000 <E 10^n 000 -> <E 11 0 10^n 010 -> <E 10^2 01^n 00 10
Rule 4: 00 <E 10^n 0100 -> <E 10^n+2 11 -> 0^inf <E 10^n+4
@64: <E 10^7
R 3: <E 10^2 01^7 00 10
R 2x: <E 10^(10 2^3 - 8) 0100 10 # = 72
R 4: <E 10^76 10
R 3: <E 10^2 01^77 0010
Repeat forever:
<E 10^2k+1
R 3: <E 10^2 01^2k+1 0010
R 2x: <E 10^(10 2^k - 8) 0100 10
R 4: <E 10^(10 2^k - 4) 10 = <E 10^(10 2^k - 3)
Oct 2022
5x2 Qhalt
This is the latest Quasihalting TM I’ve found in 5x2 with at least 1 undefined transition (i.e. it’s part of the BB(5, 2) search).
1RB---_0LC0LB_0LD1LC_1RD0RE_1LB1LA
D(a, b, c) = 0^inf 1^a D> 1^b 0 1^c 0^inf
D(a, b+2, c) -> D(a+5, b, c)
D(a, 2, 0) -> Qhalt
D(a, 2, c) -> D(a+6, c, 0)
D(a, 1, c) -> D(2, a, c+1)
0 : D(2, 1, 0) @ step 6
1 : D(2, 2, 1)
2 : D(8, 1, 0)
3 : D(2, 8, 1)
4 : D(17, 2, 1)
5 : D(23, 1, 0)
6 : D(2, 23, 1)
7 : D(57, 1, 1)
8 : D(2, 57, 2)
9 : D(142, 1, 2)
10 : D(2, 142, 3)
11 : D(352, 2, 3)
12 : D(358, 3, 0)
13 : D(363, 1, 0)
14 : D(2, 363, 1)
15 : D(907, 1, 1)
16 : D(2, 907, 2)
17 : D(2267, 1, 2)
18 : D(2, 2267, 3)
19 : D(5667, 1, 3)
20 : D(2, 5667, 4)
21 : D(14167, 1, 4)
22 : D(2, 14167, 5)
23 : D(35417, 1, 5)
24 : D(2, 35417, 6)
25 : D(88542, 1, 6)
26 : D(2, 88542, 7)
27 : D(221352, 2, 7)
28 : D(221358, 7, 0)
29 : D(221373, 1, 0)
30 : D(2, 221373, 1)
31 : D(553432, 1, 1)
32 : D(2, 553432, 2)
33 : D(1383577, 2, 2)
34 : D(1383583, 2, 0)
35 : Qhalt
2023
March 2023
6x2 Level 3 Infinite
https://bbchallenge.org/1RB1LA_1RC0LE_1RD1RB_0LA0LF_1LC1LB_—1LD
A TM Pavel shared on Discord which does Level 3 Collatz recurrence (like https://www.sligocki.com/2022/06/21/bb-6-2-t15.html) but none of the Collatz cases halt:
Let A(a, b, c) = 0^inf <A 01^a 00 10^b 1^c 0^inf
Level 1:
A(a, b+1, c) -> A(a+2, b, c)
A(a, 0, c+5) -> A(1, a+5, c)
A(a, 0, 0) -> A(1, 2, 2a)
A(a, 0, 1) -> A(1, 2, 2a+4)
A(a, 0, 2) -> A(1, 2, 2a+6)
A(a, 0, 3) -> A(1, 2, 2a+8)
A(a, 0, 4) -> A(1, 2, 2a+10)
Level 2:
A(a, 0, c+5) -> A(2a+11, 0, c)
Level 3:
A(1, 2, 5k ) -> A(1, 2, 32 2^k - 22)
A(1, 2, 5k+1) -> A(1, 2, 32 2^k - 18)
A(1, 2, 5k+2) -> A(1, 2, 32 2^k - 16)
A(1, 2, 5k+3) -> A(1, 2, 32 2^k - 14)
A(1, 2, 5k+4) -> A(1, 2, 32 2^k - 12)
Level 1 rules are actually all you need. The set \(\{A(a, b, c) : a, b, c \in \mathbb{N}\}\) is closed, so once TM enters it cannot halt.
6x2 Level 3 Avoids Halt
https://bbchallenge.org/1RB0LF_1LC1RD_1LA0RC_0RE1RB_0LC1LA_—1LE
Another TM Pavel shared on Discord which also has Level 3 Collatz recurrence and some of the Collatz cases even halt! But it skillfully avoids them:
Let C(a, b, c) = $ 1100 (01111 1100)^a 1100 0110^b 1100^c C> 1 $
@93: C(0, 0, 1)
Level 1:
C(a, b+1, c) -> C(a, b, c+2)
C(a+1, 0, c) -> C(a, c+2, 2)
C(0, 0, 9k+1) -> C(4k+1, 0, 2)
C(0, 0, 9k+4) -> C(4k+2, 1, 2)
C(0, 0, 9k+7) -> C(4k+4, 0, 1)
C(0, 0, 3k+2) -> Halt
Level 2:
C(a+1, 0, c) -> C(a, 0, 2c+6)
*: C(a, 0, c) -> C(0, 0, (c+6) 2^a - 6)
Level 3:
C(0, 0, 9k+1) -> C(0, 0, 8 2^{4k+1} - 6)
C(0, 0, 9k+4) -> C(0, 0, 10 2^{4k+2} - 6)
C(0, 0, 9k+7) -> C(0, 0, 7 2^{4k+4} - 6)
The trick is that we can rewrite the Level 3 rules as:
C(0, 0, 9k+1) -> C(0, 0, 4^{2k+2} - 6)
C(0, 0, 9k+4) -> C(0, 0, 10 4^{2k+1} - 6)
C(0, 0, 9k+7) -> C(0, 0, 7 4^{2k+2} - 6)
And since \(4^n \equiv 1 \mod 3\) (and so do 10 and 7), the right sides will always be remainder 1 (mod 3) and so always one of remainders (1, 4, 7) mod 9.
But it is delicate! A slight tweak, like 7 2^{4k+1} - 6 and this condition would not hold any longer!
All of the permutations of this TM fall into the same starting config C(0, 0, 1) except for starting from state B in which case we reach C(0, 0, 0) which leads to the rule:
C(0, 0, 9k+0) -> C(0, 0, 10 2^{4k+1} - 7) = C(0, 0, 5 4^{2k+1} - 7)
But that is also guaranteed to always be remainder 1 (mod 3)! and so will never halt.
A re-analysis which makes absorbs these coincidences is:
E(a, b, c) = C(2a, 3b, 3c+1) = $ 1100 {01111 1100 01111 1100}^a 1100 {0110 0110 0110}^b 1100 {1100 1100 1100}^c C> 1 $
E(a, b+1, c) -> E(a, b, c+2)
E(a+1, 0, c) -> E(a, 0, 4c+7)
E(0, 0, 3k) -> E(2k , 0, 3)
E(0, 0, 3k+1) -> E(2k+1, 0, 1)
E(0, 0, 3k+2) -> E(2k+2, 0, 0)
And Blank -> E(0, 0, 0) in 93 steps.
At which point we have a closed set, so presumably this TM is solveable by CTL with regular expression: 1100 (01111 1100 01111 1100)* 1100 (0110 0110 0110)* 1100 (1100 1100 1100)* C> 1?
6x2 Another Avoids Halt
https://bbchallenge.org/1RB1RD_1LC1LF_0LD1LB_1RE0RA_1RA1RE_—1LD
TM shared by @savask on Discord: https://discord.com/channels/960643023006490684/1026577255754903572/1091130641317372035
1RB1RD_1LC1LF_0LD1LB_1RE0RA_1RA1RE_---1LD
C(a, b, c, d) = $ 1^a 10^b 1^c 01^d <C 1 $
Level 1:
C(a, b, c+4, d) -> C(a, b, c, d+3)
C(a, b, 2, d) -> HALT!
C(a, b+1, 0, d) -> C(a, b, 2d+5, 0)
C(a, b+1, 1, d) = C(0, 0, a+1, b+d+1)
C(a, b+1, 3, d) -> C(a, b, 2d+8, 0)
C(0, 0, 0, d) -> C(2, d, 4, 0)
C(0, 0, 1, d) = C(0, 0, 0, d+1)
C(0, 0, 3, d) -> C(0, 0, 2d+7, 0)
Level 2:
C(a, b+1, 4k+0, 0) -> C(a, b, 6k+5, 0)
C(a, b+1, 4k+1, 0) -> C(0, 0, a+1, 3k+b+1)
C(a, b+1, 4k+2, 0) -> HALT!
C(a, b+1, 4k+3, 0) -> C(a, b, 6k+8, 0)
Level 3:
C(0, 0, 4k+1, 0) -> C(0, 0, 6k+131, 0) for k >= 2
C(0, 0, 4k+3, 0) -> C(0, 0, 6k+7, 0)
So, you can see the “Level 3” rules are closed, but the way that C(0, 0, 4k+1, 0) -> C(0, 0, 6k+131, 0) is proven is wild! Basically it repeats the same sequence of “Level 2” rules each time (and they happen to never hit the 4k+2 case):
Proof of C(0, 0, 4k+1, 0) -> C(0, 0, 6k+131, 0):
C(0, 0, 4k+1, 0) -> C(0, 0, 1, 3k) = C(0, 0, 0, 3k+1)
-> C(2, 3k+1, 4, 0)
-> C(2, 3k+0, 11, 0)
-> C(2, 3k-1, 20, 0)
-> C(2, 3k-2, 35, 0)
-> C(2, 3k-3, 56, 0)
-> C(2, 3k-4, 89, 0)
-> C(0, 0, 3, 3k+62)
-> C(0, 0, 6k+131, 0)
April 2023
3x3 Linear Rules
Among my 3x3 holdouts there are 8 that I can prove Linear Rules for:
Group A:
1RB2RB1RC_1LC0LA---_1RA2LC1LB
1RB0RC---_1LC2RB1RA_1LA2LA1LB
1RB2LA1LC_1RC2RC1RA_1LA0LB---
Group B:
1RB---2RA_0RC1RA0RB_2LC2LB0LA
1RB2RA---_0RC2LB2RB_1LC0LA0RB
1RB0RA---_0RC2LB2RB_1LC0LA0RB
Group C:
1RB2LB2LA_1LA2RC1LB_---2RB0LB
1RB1LA2LA_1LA2RC1LB_---2RB0LB
Group A
1RB2RB1RC_1LC0LA---_1RA2LC1LB
$ 1 11^a A> 22 _ -> $ 1 11^a+2 A> _
$ 1 11^a A> 21 _ -> $ 1 A> 22^a+1 2 _
$ 1 11^a A> 2000 _ -> $ 1 A> 22^a+2 1 _
$ 1 11^a A> 10 _ -> $ 1 A> 2 1^2a+1 _
A(a, b, c) = $ 1 11^a C> 2^b 1^c $
Level 1:
A(a, b+2, c) -> A(a+2, b, c)
A(a, 1, c+1) -> A(0, 2a+3, c)
A(a, 1, 0) -> A(0, 2a+4, 1)
A(a, 0, 1) -> A(0, 1, 2a+1)
Level 2:
A(a, 2k+r, c) -> A(a+2k, r, c)
A(0, 2k+1, c+1) -> A(0, 4k+3, c) 4k+3 = 2 (2k+1) + 1
A(0, 2k, 1) -> A(0, 1, 4k+1)
Level 3:
A(0, 1, c) -> A(0, 2^c+1 - 1, 0)
-> A(2^c+1 - 2, 1, 0)
-> A(0, 2^c+2, 1)
-> A(0, 1, 2^c+3 + 1) [Infinite Rule]
Starts in A(0, 1, 1) @4
The next two are just permutations of the first with start configs:
1RB0RC---_1LC2RB1RA_1LA2LA1LB : Starts in A(0, 1, 3) @10
1RB2LA1LC_1RC2RC1RA_1LA0LB--- : Starts in A(0, 1, 7) @38
Group B
These are not permutations of each other, but do share a lot of rules
1RB---2RA_0RC1RA0RB_2LC2LB0LA
_ 22 0^c C> $ -> _ 0^c+6 C> $
_ 01 0^c C> $ -> _ 1^c+2 0 C> $
_ 11 0^c C> $ -> _ 1 2^c+1 0^5 C> $
D(a, b, c) = $ 1^a+1 2^2b 0^2c+1 C> $
Level 1 (Closed Set):
D(a, b+1, c) -> D(a, b, c+3)
D(a+1, 0, c) -> D(a, c+1, 2)
D(0, 0, c) -> D(2c+2, 0, 0)
Level 2:
D(a, b, c) -> D(a, 0, c+3b)
D(a+1, 0, c) -> D(a, 0, 3c+5)
D(a, 0, c) -> D(0, 0, (c+5/2) 3^a - 5/2)
Level 3:
D(0, 0, c) -> D(0, 0, 5/2 (3^2c+2 - 1)) [Infinite Rule]
Starts in D(0, 0, 0) @2
Next: 1RB2RA—_0RC2LB2RB_1LC0LA0RB
1RB2RA---_0RC2LB2RB_1LC0LA0RB
_ 2 0^c C> 00 _ -> _ 0^c+3 C> _
_ 01 0^c C> 0 _ -> _ 1^c+2 0 C> _
_ 11 0^c C> 00 _ -> _ 2^c+3 0 C> _
C(a, b, c) = $ 1^a 2^b 0^c C> $
Level 1 (Closed Set):
C(a, b+1, c) -> C(a, b, c+3)
C(a+2, 0, c) -> C(a, c+3, 1)
C(1, 0, c) -> C(c+2, 0, 1)
C(0, 0, c) -> Inf Translated Cycle
Starts in C(1, 0, 1) @2
Note that we can even prove that this TM never reaches the C(0, 0, c) transition:
Level 2:
C(a, b, c) -> C(a, 0, c+3b)
C(a+2, 0, c) -> C(a, 0, 3c+10)
Level 3:
C(2k+1, 0, 1) -> C(1, 0, 6 3^k - 5)
-> C(6 3^k - 3, 0, 1) [Infinite Rule]
Next: 1RB0RA—_0RC2LB2RB_1LC0LA0RB is very similar to the previous one.
1RB0RA---_0RC2LB2RB_1LC0LA0RB
_ 2 0^c C> 00 _ -> _ 0^c+3 C> _
_ 01 0^c C> 0 _ -> _ 1^c+2 0 C> _
_ 11 0^c C> 00 _ -> _ 00 2^c+1 0 C> _
C(a, b, c) = $ 1^a 00 2^b 0^c C> $
Level 1 (Closed Set):
C(a, b+1, c) -> C(a, b, c+3)
C(a+2, 0, c) -> C(a, c+3, 1)
C(1, 0, c) -> C(c+2, 2, 1)
C(0, 0, c) -> Inf Translated Cycle
Starts in C(1, 2, 1) @27
Like the last TM we have rules:
Level 2:
C(a, b, c) -> C(a, 0, c+3b)
C(a+2, 0, c) -> C(a, 0, 3c+10)
Level 3:
C(2k+1, 2, 1) -> C(2k+1, 0, 7)
-> C(1, 0, 12 3^k - 5)
-> C(12 3^k - 3, 2, 1) [Infinite Rule]
Group C
1RB2LB2LA_1LA2RC1LB_---2RB0LB
_ 00 <A 11^n 22 _ -> _ <A 11^n+2 _
_ 0 <A 11^n 12 _ -> _ <A 22^n+1 1 _
_ 000 <A 11^n 0 _ -> _ <A 22 1 2^2n 1 _
A(a, b, c) = $ <A 11^a 22^b 1 2^c 1 $
Level 1 (Closed Set):
A(a, b+1, c) -> A(a+2, b, c)
A(a, 0, c+1) -> A(0, a+1, c)
A(a, 0, 0) -> A(0, 1, 2a+2)
Level 2:
A(a, b, c) -> A(a+2b, 0, c)
A(a, 0, c+1) -> A(2a+2, 0, c)
Level 3:
A(a, 0, c) -> A((a+2) 2^c - 2, 0, 0)
-> A(0, 1, (a+2) 2^c+1 - 2)
-> A(2, 0, (a+2) 2^c+1 - 2) [Infinite Rule]
Starts at A(0, 0, 1) @3
And the last TM is extremely similar, but with a small tweak to the A(a, 0, 0) rule.
1RB1LA2LA_1LA2RC1LB_---2RB0LB
_ 00 <A 11^n 22 _ -> _ <A 11^n+2 _
_ 0 <A 11^n 12 _ -> _ <A 22^n+1 1 _
_ 00 <A 11^n+1 0 _ -> _ <A 22 1 2^2n+1 1 _
A(a, b, c) = $ <A 11^a 22^b 1 2^c 1 $
Level 1 (Closed Set):
A(a, b+1, c) -> A(a+2, b, c)
A(a, 0, c+1) -> A(0, a+1, c)
A(a, 0, 0) -> A(0, 1, 2a+1)
Level 2:
A(a, b, c) -> A(a+2b, 0, c)
A(a, 0, c+1) -> A(2a+2, 0, c)
Level 3:
A(a, 0, c) -> A((a+2) 2^c - 2, 0, 0)
-> A(0, 1, (a+2) 2^c+1 - 3)
-> A(2, 0, (a+2) 2^c+1 - 3) [Infinite Rule]
Starts at A(0, 0, 0) @3
6x2 Mixed Exponential and Add
1RB0RD_0RC0RA_1LD0LC_1LE1LB_0RE0LF_1LA---
$ 01^a B> 0110 -> $ 01^a+4 B>
$ 01^2a+2 B> 00 -> $ 01^3 B> 0110^a 011
$ 01^2a+1 B> 00 -> $ 01^2 B> 110 0110^a 011
B(a, b, c) = $ 0101^a B> 110 10^b+1 0110^d $
C()
Closed Set:
B(a, b, c+1) -> B(4a+5, b+1, c-1)
B(a, b, 0) -> B(7, 0, a+2b+2)
Starts in B(7, 0, 3) @405
This proof (by closed set) is complete, but if you try to evaluate the trajectory here, it’s slightly more interesting:
B(7, 0, c) -> B((7 + 5/3) 4^c - 5/3, c, 0)
-> B(7, 0, (26 4^c + 1)/3 + 2c)
So we are recursing the map c -> (26 4^c + 1)/3 + 2c which has both exponential and addition.
6x2 ExpInt close
1RB0RC_1LC1RA_1RE1RD_1LF0RB_---1LA_1LB0LD
Rules:
00^inf 10 11^f 10^g 11^h-1 01^i+2 (01) D> 00^inf
00^inf 10 11^(j - 1) 10^(k + 2) 11^1 01^(2 l + 4) (01) D> 00^inf
11 01^c D> $ -> 01^c+2 D> $
11 10^b 01^c D> $ -> 10^b+2 11^c 01 D> $
11 01^a A> 10^2 -> 01^a+3 A>
D(a, b, c, d) = $ 10 11^a 10^b 11^c 01^d D> $
Level 1:
D(a, b, c+1, d) -> D(a, b, c, d+2)
D(a, b, 0, d) -> D(a-1, b+2, d, 1)
D(0, b, 0, d) -> D(3b+5, 3, d-2b-5, 1) (Assuming d > 2b+5)
Level 2:
D(a, b, c, d) -> D(a, b, 0, d+2c)
D(a, b, 0, d) -> D(a-1, b+2, 0, 2d+1)
D(0, b, 0, d) -> D(3b+5, 3, 0, 2d-4b-9)
Level 3:
D(a, b, 0, d) -> D(0, b+2a, 0, (d+1) 2^a - 1)
D(0, b, 0, d) -> D(0, 6b+13, 0, (d-2b-4) 2^(3b+6) - 1) [Infinite rule]
Starts at D(0, 0, 3, 1) @35
The last rule is infinite (b/c \(d > b\) remains true forever), but the sequence produces progressively more and more complicated terms:
D(0, 0, 0, 7)
D(0, 13, 0, 3 2^6 - 1)
D(0, 7*13, 0, (3 2^6 - 31) 2^45 - 1)
D(0, 13(1+6+6^2), 0, ((3 2^6 - 31) 2^45 - 187) 2^279 - 1)
May 2023
Pavel’s BB(2, 6) > 10^^91
1RB3LA4LB0RB1RA3LA_2LA2RA4LA1RA5RB1RZ
This is the current 2x6 champion found by Pavel last month. Like my 10^^70 TM, it seems to have a mix of orderly and chaotic behavior. (As usual, what I am calling chaotic is really just behavior I haven’t been able to make sense of, maybe someone else will be able to make an orderly analysis at some point.)
The orderly behavior is built upon these two rules:
01 11 0211^n B> 00 -> 11 0211^n+1 B>
Repeated: 01^k 11 0211^n B> $ -> 11 0211^n+k B> $
01^3 11 01 21 0211^n B> $ -> 11 01 21 0211^4n+10 B> $
Repeated: 01^3k+r 11 01 21 0211^n B> $ -> 01^r 11 01 21 0211^{((3n+10) 4^k - 10)/3} B> $
Most notably, a single simplification of the last rule basically accounts for all of the 10^^91 growth:
01^3k+r 11 01 21 0211^17 B> $ -> 01^r 11 01 21 0211^{(61 4^k - 10)/3} B> $
Which is applied ~90 times thus leading to ~4^^90 score.
Other rules I discovered while analyzing:
Hand Rules:
01 11 0211^n B> 00 -> 11 0211^n+1 B>
01 0211^n B> 00 -> 12 11 0211^n B>
12 11 0211^n B> 00 -> 21 0211^n+1 B>
12 11 11 0211^n B> 00 -> 21 01 0211^n+1 B>
01 11 02 11 11 0211^n B> 00 -> 11 02 11 01 0211^n+1 B>
01 11^a 02 21 11 0211^n B> 0000 -> 11 01^a+1 0211^n+2 B>
01 1551 11 02 11 11 0211^n B> 00 -> 11 01^2 02 11 01 0211^n+1 B>
01 11^a+1 0211^n B> 00 -> 11 01^a 0211^n+1 B>
0 11 11 0211^n B> 000 -> 11 11 01^2n 0211 B>
1 21 11 0211^n B> 000 -> 21 11 01^2n 0211 B>
0 11^a|1 01 21 0211^n B> 000 -> 11 01^a-1 1551 01^2n 0211 B>
01 01 21 0211^n B> 0000 -> 12 11 01^2n 11 0211 B>
01 11 01 21 0211^n B> 0000 -> 11 02 11 01^2n 11 0211 B>
01 21 21 0211^n B> 0000 -> 11 11 01^2n 11 0211 B>
11 21 0211^n B> 0000 -> 21 01^2n 11 0211 B>
Simplified:
01 11 02 11 11 0211^n B> $ -> 11 02 21 11 0211^2n+5 B> $
01 11^a 02 21 11 0211^n B> $ -> 11 01^a 21 0211^n+3 B> $
01 1551 11 02 11 11 0211^n B> 00 -> 11 01 21 11 0211^4n+17 B> $
01 11^a+2 0211^n B> $ -> 11 01^a 21 0211^n+2 B> $
0 11 11 0211^n|3 B> $ -> 11 11 01^2n-5 11 01 21 0211^17 B> $
1 21 11 0211^n|3 B> $ -> 21 11 01^2n-5 11 01 21 0211^17 B> $
0 11^a|1 01 21 0211^n|3 B> $ -> 11 01^a-1 1551 01^2n-5 11 01 21 0211^17 B> $
01 01 21 0211^n B> $ -> 21 21 0211^2n+3 B> $
01 11 01 21 0211^n B> $ -> 11 02 11 11 0211^2n+1 B> $
01 21 21 0211^n B> $ -> 11 11 11 0211^2n+1 B> $
11 21 0211^n B> $ -> 21 11 0211^2n+1 B> $
01^3 21 0211^n B> $ -> 11 11 11 0211^4n+7 B> $
01^4 21 0211^n B> $ -> 11 01 21 0211^4n+9 B> $
01^3 11 01 21 0211^n B> $ -> 11 01 21 0211^4n+10 B> $
01^3k 11 01 21 0211^n B> $ -> 11 01 21 0211^{(n+10/3) 4^k - 10/3} B> $
Start:
21 0211 B>
11 01^3 11 0211 B>
11 11 0211^4 B>
11 11 01^3 11 01 21 0211^17 B>
11 11 11 01 21 0211^78 B> [@q279]
11 01^2 1551 01^151 11 01 21 0211^17 B> [@q390+s5]
151 = 3 (50) + 1
a = (61 4^50 - 10)/3
11 01^2 1551 01 11 01 21 0211^a B>
11 01^2 1551 11 02 11 11 0211^2a+1 B>
11 01 11 01 21 11 0211^8a+21 B>
11 01 11 0 21 11 01^16a+37 11 01 21 0211^17 B>
4^3 % 9 = 1 52 % 3 = 1 4^52 % 9 = 4^1
61 4^52 - 49 % 9 = 7 4 - 4 % 9 = 24 % 9 = 6
16a+37 = (61 4^52 - 49)/3 = 3 k_a + 2
b = (61 4^k_a - 10)/3
11 01 11 0 21 11 01^2 11 01 21 0211^b B>
But this process doesn’t show any sign of simplifying over time. There may be endlessly many rules like this …