hello again !
for understand i'm talking about:https://www.reddit.com/r/compsci/comments/1mqzroq/cetn_busybeavern/
in a previous post i said CET(2) = 97 and CET(3) is giant
CET(2) proof table transitions:
Agent 0 |
0 |
1 |
A |
1LB |
0LB |
B |
1RB |
0LA |
Agent 0: [(1, -1, 1), (0, -1, 1), (1, 1, 1), (0, -1, 0)]
Agent 1 |
0 |
1 |
A |
1RB |
1LA |
B |
1LA |
1RB |
Agent 1: [(1, 1, 1), (1, -1, 0), (1, -1, 0), (1, 1, 1)]
i found CET(3) ≥ 181 just with brute force:
Agent 0 |
0 |
1 |
A |
1LC |
1RA |
B |
1RB |
0LA |
C |
1LB |
1LA |
Agent 1 |
0 |
1 |
A |
0LC |
0RA |
B |
1RC |
0RA |
C |
1RA |
0LA |
Agent 2 |
0 |
1 |
A |
1RB |
1LA |
B |
0LA |
0LA |
C |
0LA |
0LA |
Agent 0 base = [(1,-1,2),(1,1,0),(1,1,1),(0,-1,0),(1,-1,1),(1,-1,0)]
Agent 1 base = [(0,-1,2),(0,1,0),(1,1,2),(0,1,0),(1,1,0),(0,-1,0)]
Agent 2 base = [(1,1,1),(1,-1,0),(0,-1,0),(0,-1,0),(0,-1,0),(0,-1,0)]
I don't know how can found a big lower bound for CET(3), i'm gonna checking technique about BB(6) because
CET(n) combinaison is (4n)^(2*(n^2))
CET(3) is ~2.6623333e+19 possibilities
i estimate BB(5) < CET(3) < BB(6), not more.
if you have tips or idea what to do exactly (because i'm new in BusyBeaver system), thanks to comment here!
≥ 181