0RB2LA1RA 1LA2RB1RC ---1LB1LC: Difference between revisions
Jump to navigation
Jump to search
Added categories |
→Analysis: Added analysis by Pascal Michel |
||
| Line 4: | Line 4: | ||
It runs for 119,112,334,170,342,541 steps before halting and leaves 374,676,383 ones. | It runs for 119,112,334,170,342,541 steps before halting and leaves 374,676,383 ones. | ||
== Analysis == | == Analysis by Pascal Michel == | ||
<pre> | |||
Let C(n) = . . . 0 (A0) 2^n 0 . . . | |||
Then we have: | |||
. . . 0 (A0) 0 . . . --( 3 )--> C(1) | |||
C(8k + 1) --( 112k^2 + 116k + 13 )--> C(14k + 3) | |||
C(8k + 2) --( 112k^2 + 144k + 38 )--> C(14k + 7) | |||
C(8k + 3) --( 112k^2 + 172k + 54 )--> C(14k + 8) | |||
C(8k + 4) --( 112k^2 + 200k + 74 )--> C(14k + 9) | |||
C(8k + 5) --( 112k^2 + 228k + 97 )--> . . . 0 1 (H1) 2^(14k+9) 0 . . . | |||
C(8k + 6) --( 112k^2 + 256k + 139 )--> C(14k + 14) | |||
C(8k + 7) --( 112k^2 + 284k + 169 )--> C(14k + 15) | |||
C(8k + 8) --( 112k^2 + 312k + 203 )--> C(14k + 16) | |||
So we have (in 34 transitions): | |||
. . . 0 (A0) 0 . . . --( 3 )--> | |||
C( 1 ) --( 13 )--> | |||
C( 3 ) --( 54 )--> | |||
C( 8 ) --( 203 )--> | |||
C( 16 ) --( 627 )--> | |||
C( 30 ) --( 1915 )--> | |||
. . . | |||
C( 122,343,306 ) --( 26,193,799,261,043,238 )--> | |||
C( 214,100,789 ) --( 80,218,511,093,348,089 )--> | |||
. . . 0 1 (H1) 2^374676381 0 . . . | |||
</pre> | |||
[[Category:BB(3,3)]] | [[Category:BB(3,3)]] | ||
Revision as of 12:22, 7 December 2025
0RB2LA1RA_1LA2RB1RC_1RZ1LB1LC (bbch) is the current BB(3,3) champion. It was discovered by Terry and Shawn Ligocki in November 2007.
It runs for 119,112,334,170,342,541 steps before halting and leaves 374,676,383 ones.
Analysis by Pascal Michel
Let C(n) = . . . 0 (A0) 2^n 0 . . . Then we have: . . . 0 (A0) 0 . . . --( 3 )--> C(1) C(8k + 1) --( 112k^2 + 116k + 13 )--> C(14k + 3) C(8k + 2) --( 112k^2 + 144k + 38 )--> C(14k + 7) C(8k + 3) --( 112k^2 + 172k + 54 )--> C(14k + 8) C(8k + 4) --( 112k^2 + 200k + 74 )--> C(14k + 9) C(8k + 5) --( 112k^2 + 228k + 97 )--> . . . 0 1 (H1) 2^(14k+9) 0 . . . C(8k + 6) --( 112k^2 + 256k + 139 )--> C(14k + 14) C(8k + 7) --( 112k^2 + 284k + 169 )--> C(14k + 15) C(8k + 8) --( 112k^2 + 312k + 203 )--> C(14k + 16) So we have (in 34 transitions): . . . 0 (A0) 0 . . . --( 3 )--> C( 1 ) --( 13 )--> C( 3 ) --( 54 )--> C( 8 ) --( 203 )--> C( 16 ) --( 627 )--> C( 30 ) --( 1915 )--> . . . C( 122,343,306 ) --( 26,193,799,261,043,238 )--> C( 214,100,789 ) --( 80,218,511,093,348,089 )--> . . . 0 1 (H1) 2^374676381 0 . . .