OCA talk:LongLife
Omniperiodic?-wwei23 10:09AM 6/27/2017 NY time
David Eppstein's writings
Are David Eppstein's writings on LongLife available anywhere?
White whale period?
This article claims that White Whale has period, 160,000,346. The glider database claims 160,000,046. I ran this in golly and neither appears to be correct. Gzaytman (talk) 18:48, 25 November 2022 (UTC)
- I think the "hollow" pattern does evolve to a period-160000346 oscillator after 12822176 ticks. The population equals 92 at T = 114375463, 274375809, ... Below are generations T = 0 (top left), 114375463 (top right), 12822175 (bottom left), 160000346 + 12822175 (bottom right). The two patterns in the bottom row converge after one tick. Confocal (talk) 07:49, 26 November 2022 (UTC)
T = 0 (top left), 114375463 (top right), 12822175 (bottom left), 160000346 + 12822175 (bottom right) (click above to open LifeViewer) |
- : Thanks. BTW, do you know who to contact to get the entry in Eppstien's database fixed? Gzaytman (talk) 18:37, 26 November 2022 (UTC)
- I made a post here: https://conwaylife.com/forums/viewtopic.php?p=154150#p154150 Confocal (talk) 19:19, 26 November 2022 (UTC)
Proof of no still lifes adapted from Gems
Couldn't the recently added and removed proof be adapted to work for LongLife, quite easily?
I'm not sure exactly what the objection was, but this part
it must have at least 5 alive neighbours; the only possibility is that the alive neighbours are the cell to the right of C and three neighbouring cells in the next row...
wasn't changed enough from the Gems version. It would need exactly 5 neighbors, and only four of them can possibly be to the right or below the leftmost "1" cell. At least one cell must be left of the leftmost "1" cell, contradicting the assumption. EDIT: patched up the wording here to remove the reference to the "x" cell, which wouldn't be needed -- the proof in LongLife is simpler. (Right?) Dvgrn (talk) 13:12, 7 December 2023 (UTC)
- The removed text states:
>"Assume (for contradiction) that a finite still life exists. Let C denote the leftmost alive cell in the first nonempty row. For the cell C to survive in the next tick, it must have at least 5 alive neighbours; the only possibility is that the alive neighbours are the cell to the right of C and three neighbouring cells in the next row: [...]"
This is inconsistent starting at the "the only possibility" part. Since C is the leftmost alive cell in the first nonempty row, there can be at most four alive neighbours, so there are no possibilities at all.
I intended to return to this page later to see what can be improved; however, the simple copy-and-paste with single-digit change ('at least 4' → 'at least 5') does not make a correct proof, and does not help to understand the puzzle to be solved.
(Comparison between relevant versions of pages OCA:Gems and OCA:LongLife: https://conwaylife.com/w/index.php?diff=142238&oldid=142199 ) Confocal (talk) 13:25, 7 December 2023 (UTC)- Okay -- looks like an easy fix, so I'll try doing it. It's even easier if the leftmost cell in the top non-empty row is used; for a while I was incorrectly looking at the top leftmost cell in the hypothetical still life's bounding box, which is also enough to prove a contradiction. Dvgrn (talk) 13:29, 7 December 2023 (UTC)