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.

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

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)