Cha cha in C2_2, found on July 11 by TechnobladeNeverDies:
Code: Select all
x = 31, y = 32, rule = B3/S23
bbbbbbbbbbbbbbboobobobbooooooob$
bbbbbbbbbbbbbbbooobbbooobboooob$
bbbbbbbbbbbbbbbooobooobobobbboo$
bbbbbbbbbbbbbbbbooboobooboooobo$
bbbbbbbbbbbbbbbbobbbooobooboboo$
bbbbbbbbbbbbbbboobobobboobooboo$
bbbbbbbbbbbbbbbbboobobbobobbooo$
bbbbbbbbbbbbbbboobbobbboobboooo$
bbbbbbbbbbbbbbbbbobobooboobbobo$
bbbbbbbbbbbbbbboobbobooooobbbbb$
bbbbbbbbbbbbbbbbooobooooboobobb$
bbbbbbbbbbbbbbboooobobbobbboobo$
bbbbbbbbbbbbbbboobbooooobobboob$
bbbbbbbbbbbbbbboobbooooobooboob$
bbbbbbbbbbbbbbbbbobobooooobbbbo$
bbbbbbbbbbbbbbbbboboboobobooboo$
oobooboboobobobbbbbbbbbbbbbbbbb$
obbbbooooobobobbbbbbbbbbbbbbbbb$
booboobooooobboobbbbbbbbbbbbbbb$
boobbobooooobboobbbbbbbbbbbbbbb$
oboobbbobboboooobbbbbbbbbbbbbbb$
bbobooboooobooobbbbbbbbbbbbbbbb$
bbbbbooooobobboobbbbbbbbbbbbbbb$
obobbooboobobobbbbbbbbbbbbbbbbb$
oooobboobbbobboobbbbbbbbbbbbbbb$
ooobbobobboboobbbbbbbbbbbbbbbbb$
oobooboobboboboobbbbbbbbbbbbbbb$
ooboboobooobbbobbbbbbbbbbbbbbbb$
obooooboobooboobbbbbbbbbbbbbbbb$
oobbbobobooobooobbbbbbbbbbbbbbb$
boooobbooobbbooobbbbbbbbbbbbbbb$
booooooobboboboobbbbbbbbbbbbbbb!
Note that the two halves of the soup does not interact before this oscillator forms, and soup
c_B7IJkdjoXSK4168814188 (which is the seed of the C2_2 soup shown), in asymmetric form, still allows the oscillator to form and survive.
The definition of "natural" is that there exists a known seed that, through SHA-256, gets mapped to an asymmetric soup that forms a specific object. Should this oscillator be marked as "natural", or shall we wait for a C1/G1 soup for this?
Edit on August 27: Given the comments on this case, I think the former scenario is true. This oscillator should be marked natural.