# Analysis of the (5,2)-machine M with BBB(M) > 2.1 10^18

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### Pascal Michel

Feb 23, 2022, 4:14:19 AM2/23/22
in a mail on February 4, 2022, Shawn gave an analysis of the (5,2)-machine  M found by Nick with BBB(M) > 2.1 10^18 (Nick's mail on January 30, 2022):
1RB 0LC  1RC 0LA  1LD 0RB  0RE 0RD  1LE 0LA
He used configurations f(m,n) = 1^m <E 1^n 0^Inf.
I give here a more detailed analysis using these configurations.
Let D(x,m,n) be the configuration defined by
D(x,0,n) = ...0 bin(x) (E0) 1^n 0...
and, if m > 0, D(x,m,n) = ...0 bin(x) 1^(m - 1) (E1) 1^n 0...
where bin(x) is the number x written in binary notation, and x ix even if m > 0, because
D(2x + 1,m,n) = D(x,m + 1,n).

Then we have
(a) ...0(A0)0... --(10)--> D(0,1,4)
(b0) if k > 0, D(x,3k,n) --(5k^2 + (2n + 8)k - 1)--> D(x,0,5k + n - 1)
(b1) if k > 0, D'x,3k + 1,n) --(5k^2 + (2n + 8)k)--> D(x,1,5k + n)
(b2) if k > 0, D(x,3k + 2,n) --(5k^2 + (2n + 8)k)--> D(x,2,5k + n)
(c0) if x > 0, D(2x,0,n) --(1)--> D(x,0,n + 1)
(c1) D(2x + 1,0,n) --(1)--> D(x,1,n + 1)
(d) D(2x,1,2k) --(4k + 11)--> D(x,2k + 1,4)
(e) D(2x,1,2k + 1) --(4k + 14)--> D(x,2k + 2,5)
(f) D(4x,2,2k + 2) --(4k + 25)--> D(8x + 4,2k + 2,5)
(g) D(4x,2,2k + 1) --(4k + 22)--> D(8x + 4,2k + 1,4)
(h) if  k > 0, D(0,3k,5) --(5k^2 + 13k - 6)--> ...0(D0)0... last state D

Then ...0(A0)0... --> last state D using 111 transitions with configurations D(x,m,n).
Among these transitions, 39 are of type (b0), (b1) or (b2).

Pascal