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Mar 24, 2022, 4:39:16 AM3/24/22

to Busy Beaver Discuss

In a mail on March 16, 2022, Nick gave the following machine M with BBB(M) > 7.4 x 10^4079:

1RB 1LC 0LD 0LB 0RE 0LA 0LE 1LD 1RE 1RA (not in tree normal form).

Shawn and Nick gave an analysis of this machine, using configurations

G(n,m) = ...0 (E0) 0^n 1^m 0...

I give here some details in addition to their analysis, using these configurations.

We have, for n, k, m >= 0

(a) ...(A0)... --(4)--> G(1,1)

(b) G(n,1) --(2n + 8)--> G(1,n + 3)

(c) G(4k,m + 2) --(7k^2 + 22k + 13)--> G(7k + 5,m)

(d) G(4k + 1,m + 2) --(7k^2 + 26k + 22)--> G(7k + 8,m + 1)

(e) G(4k + 2,m + 2) --(7k^2 + 22k + 17)---> G(7k + 6,m + 1)

(f) G(4k + 3,m + 2) --(7k^2 + 40k + 51)--> G(7k + 14,m)

(g) G(n,0) = ...(E0)... is a quasihalting configuration.

Then ...(A0)... --> ...(E0)... using 8410 transitions with configurations G(n,m)

Number of configurations of type:

(a) 1

(b) 3

(c) 2093

(d) 2101

(e) 2086

(f) 2126

Pascal

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