1RB0RF 1LC1RB 0RD0LB 1RZ0LE 1RE0RA 1RD1RE: Difference between revisions
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{{machine|1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE}} | {{machine|1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE}} | ||
{{TM|1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE}} is a [[BB(6)]] TM that halts with roughly <math>2^{2^{97/3}}</math> steps. | {{TM|1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE}} is a [[BB(6)]] TM that halts with roughly <math>2^{2^{97/3}}</math> steps. | ||
Revision as of 09:58, 11 August 2025
1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE (bbch) is a BB(6) TM that halts with roughly steps.
Analysis by racheline
https://discord.com/channels/960643023006490684/1239205785913790465/1336829029948588113
It halts with roughly 2^(2^96/3)/3 1s on the tape and after roughly 2^(2^97/3) steps. it follows these rules:
A(n,m) := 111 E> 1^(3n) 0 1^m A(n,6m+1) -> A(n+2m+1,2^(n+2m)+2^((2^(2m)-1)/3)+1) A(n,3m+2) -> halt with n+m+6 1s start from A(4,13) A(4,13) -> A(9,289) -> A(106,2^105+2^((2^96-1)/3)+1) since both 105 and (2^96-1)/3 are odd, that is A(106,3m+2), which halts as soon as the machine gets to the right side of the tape here are the details of the A(n,6m+1) case: A(n,6m+1) = 111 E> 1^(3n) 0 1^(6m+1) --(translated cycles)--> 111 (001)^n 1 (001)^(2m) 0 A> 0 --(counting with 2m bits)--> 111 (001)^n 1 (111)^(2m) 0 A> 0 1^(2^(2m)-1) --(translated cycles)--> 111 (001)^(n-1) 111 (001)^(2m) 0 111 (001)^((2^(2m)-1)/3) 0 A> 0 --(counting with (2^(2m)-1)/3 bits)--> 111 (001)^(n-1) 111 (001)^(2m) 0 (111)^((2^(2m)+2)/3) 0 A> 0 1^(2^((2^(2m)-1)/3)-1) --(translated cycles)--> 111 (001)^(n-1) 111 (001)^(2m-2) 111 001 0 A> 0 1^(2^(2m)+2^((2^(2m)-1)/3)+3) --(counting with n+2m bits)--> 111^(n+2m+1) 0 A> 0 1^(2^(n+2m)+2^((2^(2m)-1)/3)) --(translated cycles)--> A(n+2m+1,2^(n+2m)+2^((2^(2m)-1)/3)+1)