0RB2LA1RA 1LA2RB1RC ---1LB1LC: Difference between revisions
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{{machine|0RB2LA1RA_1LA2RB1RC_---1LB1LC}}{{Stub}} | |||
{{TM|0RB2LA1RA_1LA2RB1RC_1RZ1LB1LC|halt}} is the current [[BB(3,3)]] [[Champions#3-Symbol TMs|champion]]. It was discovered by [[User:Tjligocki|Terry]] and [[User:Sligocki|Shawn Ligocki]] in November 2007. | {{TM|0RB2LA1RA_1LA2RB1RC_1RZ1LB1LC|halt}} is the current [[BB(3,3)]] [[Champions#3-Symbol TMs|champion]]. It was discovered by [[User:Tjligocki|Terry]] and [[User:Sligocki|Shawn Ligocki]] in November 2007. | ||
It runs for 119,112,334,170,342,541 steps before halting and leaves 374,676,383 | It runs for 119,112,334,170,342,541 steps before halting and leaves 374,676,383 non-zero symbols. | ||
== 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)]] | |||
Latest revision as of 12:29, 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 non-zero symbols.
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 . . .