| 0 | 1 |
| A | 1RB | 1LC |
| B | 1RC | 0RD |
| C | 0LB | 0RC |
| D | 0RE | 1RD |
| E | 1LE | 1LA |
g(2x+1, n, m)=
g(x, n+1, m)
g(2x, 3k + 1, m)->
g(4x+2, 5k + m, 2)
g(4x, 3k + 2, m)->
g(8x + 2, 5k + m + 1, 2)
g(4x + 2, 3k + 2, m)->
g(8x + 2, 5k + m + 1, 2)
g(0, 3k, m)-> Quasihalt
g(2x, 3k, m)->
g(x, 0, 5k + m + 1)
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Are all Busy Beavers (or Beeping Busy Beavers, etc) structured TMs?
Are all but a finite number of Busy Beavers unstructured TMs?
Your graph reduction procedure definitely captures the process I've been going through in manually structuring these things, and it's a great way to formalize the Spaghetti Code Conjecture.
I still think the SCC is false. Even if we accept the skeptical argument that the available evidence doesn't mean much, the lesson of Scott's Bigger Number Game is that theory beats tricks. What we should expect if we believe Scott's thesis is that champions will ignore tricks and exploit theory. And that indeed is what they do.
1. Has this sort of graph reduction been studied before?
2. "Repeat until nothing changes." Is there an upper bound for how long this will take? It's computable, so the procedure definitely will terminate, but when?
I ran the procedure against my own set of test cases, and most of them came back simple. One notable exception is the 5x2 program discovered by Uwe Schult in 1982. It halts after 134,467 steps, and it reigned as champion for two and a half years.
--> 1RB 0LC 1RC 1RD 1LA 0RB 0RE 1R_ 1LC 1RA
The control flow diagram is attached.
1RB 2LC 2RA 1LA 2LB 1RC 1RA 2LC 1RB
There is one more step that can be added:
4. Inline any node with just one entry point.
This catches cases like `1RB 0LC 0LC 1RA 0RA 1LD 1LA 1LC`: D is
only reached from C, so we may as well just say that the D code
belongs to the 1-branch of C.
C:if BLANK:# C0 -> 0RAERASE; RIGHT;goto Aelse:# C1 -> 1LDPRINT; LEFT;if BLANK:# D0 -> 1LAPRINT; LEFT;goto Aelse:# D1 -> 1LCPRINT; LEFT;goto C
C:if BLANK:# C0 -> 0RAERASE; RIGHT;else:# C1 -> 1LDPRINT; LEFT;if BLANK:# D0 -> 1LAPRINT; LEFT;else:# D1 -> 1LCPRINT; LEFT;goto Cgoto A
Nice! Keep the posts coming!
An important point that stuck out to me:
> All Lin Recurrent programs are PA-CTR Recurrent.
Does this apply to all Christmas trees too? It would be nice if it
did, because then we could dispense with the elaborate classifications
of Xmas, Leaning Xmas, Double Leaning Xmas, etc.
Also the Chain Recurrence section mentions BBB(2, 4), but the program
displayed is BBB(4, 2).