ConwayLife.com

Posted: **April 20th, 2025, 7:37 pm**

LuveelVoom wrote: April 20th, 2025, 4:53 pm uh, whoops
Such that the number of lanes of interaction approaches a finite number as N goes to infinity

It sounds like you could just use the Snark always. As the area of the "black box" grows, the average density approaches 0; it should be possible to place the Snark inside the "black box" to ensure that the geometric restrictions on the input glider lane and the output glider lane are met; and the number of glider lanes such that a glider would hit the Snark (when it isn't reflecting a glider) is finite and doesn't change as the N-by-N "black box" grows.

There are only finitely many ways to hit the Snark by a glider (one of them is shown below).

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x = 26, y = 33, rule = B3/S23
15b2o$15bobo$17bo4b2o$13b4ob2o2bo2bo$13bo2bobobobob2o$16bobobobo$17b2o
bobo$21bo2$7b2o$8bo7b2o$8bobo5b2o$9b2o7$19b2o$19bo$20b3o$22bo8$3o$2bo$
bo!

Posted: **April 24th, 2025, 8:39 am**

confocaloid wrote: April 19th, 2025, 9:15 am
To complete it (to fill the gaps), one would need to show a way to construct a (sufficiently fast) glider-triggered stable "attachment" that, when triggered, would cause a normally-alive cell to be dead in precisely n generations for an arbitrary n > 8, and to show a way to construct a (sufficiently fast) glider-triggered stable "attachment" that, when triggered, would cause a normally-dead cell to be alive in precisely m generations for an arbitrary m > 28. Once such adjustable stable "attachments" are found, the rest would be an exercise in duplicating gliders using stable glider duplicators, routing glider tracks, and routing/adjusting Herschel tracks.
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Perhaps something like this could help, where a block is on for as many "f" as needed before getting deleted, which is regulated by stable delay devices. The limiting factor in this design is the 92 ticks it takes to generate the block, 4 generations to delete the block, and then a delay by whatever number of generations needed to outdo eaters - though the long block deletion would add 28 generations if used.

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x = 191, y = 126, rule = B3/S23
75bo$73b3o$49bo22bo84b2ob2ob2obo$49b3o20b2o83bo3bobob2o$52bo105b3o2bo
3b2o$51b2o39b2o66bobob2o3bo$92b2o5b2o61bobob3o$32bo66b2o60b2obobo$32b
3o41b2o87bo$35bo41bo98bo$18b2o14b2o11b2o28bobo17b2o75b3o$18b2o27b2o29b
2o17b2o74bo$103b2o55b2o11b2o4b2o$103b2o55b2o17bo$12b2o163bobo$12b2o163b
2o$16b2o$16b2o15bo140b2o$33bobo91bo45bo2bo$33b3o36b2o52bobo44bobo$35b
o23b2o11b2o52bobo7b2o36bo$11b2o47bo63b3ob2o6b2o$11b2o44b3o63bo19b2o$57b
o66b3ob2o13b2o30b2o$126bob2o45bo2bobob2o$107bo68b3ob2obo$105b3o71bo$47b
2o55bo73bo2b2o$47bo56b2o33bo39b2obo$48b3o87bobo$50bo86bo2bo$13bo4bo119b
2o9b2o$12bobo3bobo128b2o$12bobo3b2o$9b2obob3o$9bobo5bo91b2o16b2o$12b6o
91bo17b2o$14bo92bobo$16b2o23b2o64b2o$14bobo23bobo$12b3obobo21bo147b2o
$11bo5b2o19b3o52b2o91bo2bo$11b2o24bo3b2o49bobo73b2o14b5o$37b4o2bo48bo
76bo13bo$41b3o47b2o76bobo12b3o$37b4o129b2o15bo$37bo3b5o95b2o41b4o$23b
2o7b2obobob2o4bo63b2o30bo37b2o3bo3b2o$14b2o7b2o7bob2obobo2bo66bobo30b
3o34b2o4b3o2bo$15bo20bo4b2o68bo32bo42bob2o$15bobo18bob3o61b2o7b2o74bo
$16b2o19bo2bo61b2o82b2o$38b2o2$34b2o142b2o$34bo143bo$32bobo144b3o$32b
2o147bo$92b2o$93bo$18b2o73bobo16bo$17bobo74b2o14b3o$17bo91bo$16b2o91b
2o$106bo$106b2o$105bobo2$76bo$76b3o$21b2o29bo26bo$22bo27b3o25b2o$19b3o
27bo$19bo29b2o2$16bo52bo$14b3o50b3o44b2o$13bo52bo47b2o$13b2o7bo30b2o11b
2o$22bobo28b2o$22b2o$11bo97b2o$10bobo96b2o$10b2o101b2o$113b2o$22b2o$22b
2o$2o26b2o17b2o29b2o27b2o$2o26b2o17bobo28b2o11b2o14b2o$49bo41bo$49b2o
41b3o$6b2o9b2o7b2o66bo$6b2o9b2o7b2o5b2o$33b2o$2b2o$3bo49b2o$3o51bo$o50b
3o$51bo$17b2o87b2o$8bo8b2o87bo3b2o$7bobo92b2o4bo2bo$7b2o88b2o3b2o3b4o
$91bob2obobo$91b2obobo10b2o$95b2o10bobo$108b2o6$100b2o$99bo2bo$100b2o
3$111b2o$95b2o6bo7bobo$95b2o5bobo8bo$103bobob2o4b2o$101bobobobobo$101b
2obo4bo$104bob3o$104bobo$105bo!

EDIT: Here's a p17, making p19 the smallest period without a nontrivial omnifrequent oscillator.

Code: Select all

x = 82, y = 102, rule = B3/S23
29bo9bo$29b3o5b3o$32bo3bo$31b2o3b2o4$34bob2o$31bob2obo$31bo4bo$25bo6b
obo$25b3o3b2ob2o$28bo2b2ob2o$27bo5bo$27bo3bo2$29bobo$24b2o4bo$24bo11b
o$21bo8bo4b2o$20bobo6bobo20b2o$20bobo29bo$18b2o2b2o5b3o17b2obo$19bobo
2bob2o23bo$18bo2b2obob2o4b3o12bo$19b2o3bo7bo16bo$21b2obob2o5bo14bo5b2o
$21bo2bo2bo26bo$22bobo2bobo4bo12bo4bobo$23bo4b2o3b3o10b3o3b2o$32b2ob2o
8b2ob2o$31b3ob3o6b3ob3o$22b2o8b2ob2o8b2ob2o8b2o$22bo10b3o10b3o10bo$19b
2obo11bo12bo11bob2o$18bobob2o5b2o20b2o5b2obobo$17bo11b2o20b2o11bo$16b
o4b3o10b2o10b2o10b3o4bo$15bo4bo2bo10b2o10b2o10bo2bo4bo$14bo4bobo8bo20b
o8bobo4bo$14b2o2bobo8b3o18b3o8bobo2b2o$17bobo8b2ob2o16b2ob2o8bobo$12b
4obo9b3ob3o14b3ob3o9bob4o$12bo2bob2o9b2ob2o16b2ob2o9b2obo2bo$29b3o3b2o
8b2o3b3o$30bo4bobo6bobo4bo$6bo30bo6bo30bo$6b3o13bo14b2o4b2o14bo13b3o$
o8bo11b3o34b3o11bo8bo$3o5b2o5b2o3b5o32b5o3b2o5b2o5b3o$3bo11b2o2b2o3b2o
30b2o3b2o2b2o11bo$2bo8bo8b5o32b5o8bo8bo$2bo2bo4b3o8b3o34b3o8b3o4bo2bo
$6bo2b5o8bo36bo8b5o2bo$4bo3b2o3b2o2b2o44b2o2b2o3b2o3bo$9b5o3b2o5b2o30b
2o5b2o3b5o$10b3o11bo32bo11b3o$11bo13b3o26b3o13bo$27bo26bo5$27bo26bo$11b
o13b3o26b3o13bo$10b3o11bo32bo11b3o$9b5o3b2o5b2o30b2o5b2o3b5o$4bo3b2o3b
2o2b2o44b2o2b2o3b2o3bo$6bo2b5o8bo36bo8b5o2bo$2bo2bo4b3o8b3o34b3o8b3o4b
o2bo$2bo8bo8b5o32b5o8bo8bo$3bo11b2o2b2o3b2o30b2o3b2o2b2o11bo$3o5b2o5b
2o3b5o32b5o3b2o5b2o5b3o$o8bo11b3o34b3o11bo8bo$6b3o13bo14b2o4b2o14bo13b
3o$6bo30bo6bo30bo$30bo4bobo6bobo4bo$29b3o3b2o8b2o3b3o$12bo2bob2o9b2ob
2o16b2ob2o9b2obo2bo$12b4obo9b3ob3o14b3ob3o9bob4o$17bobo8b2ob2o16b2ob2o
8bobo$14b2o2bobo8b3o18b3o8bobo2b2o$14bo4bobo8bo20bo8bobo4bo$15bo4bo2b
o10b2o10b2o10bo2bo4bo$16bo4b3o10b2o10b2o10b3o4bo$17bo11b2o20b2o11bo$18b
obob2o5b2o20b2o5b2obobo$19b2obo11bo12bo11bob2o$22bo10b3o10b3o10bo$22b
2o8b2ob2o8b2ob2o8b2o$31b3ob3o6b3ob3o$32b2ob2o8b2ob2o$28b2o3b3o10b3o3b
2o$27bobo4bo12bo4bobo$27bo26bo$26b2o5bo14bo5b2o$32bo16bo$34bo12bo$30b
o20bo$29bob2o16b2obo$29bo22bo$28b2o22b2o!

Posted: **May 10th, 2025, 10:23 pm**

Conjecture: The only reversible(that is, have a non-B0 Range 1 INT rule as its inverse) is B/S012345678. Here is a proof that S8 is forced:
Assume the contrary, and rule A is reversible and does not have S8. Let the inverse of Rule A be Rule B. Now, evolve the following pattern in rule A:

Code: Select all

x = 5, y = 5, rule = B3/S23
5o$5o$5o$5o$5o!

The centre 3x3 grid must die. Now, evolve the pattern in rule B. Since B is A’s inverse, the center cell must be born, but that would force B0. Contradiction.

Can someone disprove B1 and B2a?

Posted: **June 6th, 2025, 9:30 am**

Conjecture: IceNine is real
On one hand, stuff like universial constructors and block-clearing glider salvos exist.
On the other, nobody has figured out a way to mash them together yet.

Posted: **June 6th, 2025, 12:10 pm**

Here’s a conjecture which can probably be proved easily:
No finite pattern exists which works in more than 2^98 rules of the 2^101 rule space that is INT minus B0. The dot works in 2^98 rules: any rule with S0 and without either of B1e or B1c. Infinite patterns exist that work in more: a line on a torus works in 2^99 rules: any rule with S2i and without B3i. Does any finite pattern work in more than 2^98 rules? This conjecture can of course be extended to other neighborhoods.

Posted: **June 6th, 2025, 4:14 pm**

confocaloid wrote: April 19th, 2025, 9:15 am The following p30 oscillator has cells with "frequencies" 1 through 19 and 27 through 30; would it otherwise "count as valid solution" if additional p30 parts were added with cell "frequencies" 20 through 26?
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Nontrivial omnifrequent p30, though it would probably be better if the p3 eater was replaced with a different true-period p20 cell. Again, known periods are 1-18, 20, & 30

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x = 169, y = 95, rule = B3/S23
12b2o$12b2o4$27b2o$27b2o4$27bo14b2o$26b3o13b2o17bo$9b2o3b2o9bo3bo30bo
bo$11b3o10bob3obo29bobo$10bo3bo10b5o29b2ob3o$11bobo51bo$12bo46b2ob3o$
59b2obo$57b3o$56bobo$26bo2b2o11bo13bobo13b2o$13b3o10bobo12b3o13bo14b2o
$b2o4bobo15b2o13b5o$bobo3b2o4bobo8b2o13b2o3b2o$3bo4bo3b5o7b2obo12b5o9b
2obob2o$o2b3o5b2o3b2o7b3o12bo3bo9bo5bo26b2o$3o3bo4b2o3b2o23bobo11bo3b
o27b2o$3b3obo34bo13b3o12b3o$2bo4bo18b2o3b2o$bob2obo7bo14bo41bobo$bo2b
obobo4bobo10bo5bo11bo25b5o27b2o$2bo3b2obo3bobo2bobo6b2ob2o10b2ob2o22b
2o3b2o26b2o$3b3o3bo4bo3b2o8bobo6bo31b2o3b2o$5bo3b2o8bo9bo6bo4bo5bo$29b
o6b3o16bo45b3o$41b2obob2o6bo17b2o27b3o13b2o$54b3o13b2ob2o42b2o$70bo2b
o10b2o3b2o$73bo12b3o$44b2o24bo14bo3bo9b2o3b2o$43bo3bo23b2o13bobo11b5o
27b2o$43bo3bo9b5o25bo13b3o13bo14b2o$44b2o10bob3obo39bo14bo$47b3o7bo3b
o9b2o3b2o38bobo$47b2o9b3o5bo5b5o38b2ob2o$59bo4b2o7b3o38bo5bo26b2o$65b
2o7bo13b3o26bo29b2o$82bobo29b2o3b2o$82b2o4bobo$83bo3b5o9bo43bo$74bo11b
2o3b2o6b2o15b4o12bo13bo$86b2o3b2o7b2o13b2o2bo11b3o12bo$76bo38b2obo11b
5o$76bo52b2o3b2o$89bo26b2o12b5o9b2o3b2o$88bob2o24bo13bo3bo12bo$88bob3o
8b2o3b2o23bobo10bo5bo$89b2o3bo8b3o26bo12b2ob2o$92b2o8bo3bo9b2obob2o23b
obo$93bo9bobo4bo5bo5bo24bo12b2o$104bo4bo7bo3bo12bo12bo12bo$109b3o6b3o
11b2ob2o24bo5b2o$127bo30b3obo3bobo$126bo4bo5bo19bo4bo3bo$126b3o29b3ob
2ob2o$131b2obob2o22bobo$143bobo14bobob3o$143b2o16b2obo2bo$144bo3b3o6b
3o6b2o$147bo3bo7bo$133bo12bo5bo5bo$133bob2o9b2obob2o$134bo2bo2$136b2o
14bo$151b2o2$150bob2o$149bo2b2o$149b4o3$148b2o3b2o$151bo$148bo5bo$136b
obo10b2ob2o$135bo3bo10bobo$136bo3bo10bo$138b2o11bo$141bo2bo$139bo2bo2b
o$140bobo2bo$139b2ob5ob2o$139bo2bo3b2obo$140b2o6bo!

It's definitely not the smallest to answer my question, but it is a solution to one more period.

confocaloid wrote: April 19th, 2025, 9:15 am The following approach (with additional "attachments") should make it possible to solve all sufficiently high periods.

...

To complete it (to fill the gaps), one would need to show a way to construct a (sufficiently fast) glider-triggered stable "attachment" that, when triggered, would cause a normally-alive cell to be dead in precisely n generations for an arbitrary n > 8, and to show a way to construct a (sufficiently fast) glider-triggered stable "attachment" that, when triggered, would cause a normally-dead cell to be alive in precisely m generations for an arbitrary m > 28. Once such adjustable stable "attachments" are found, the rest would be an exercise in duplicating gliders using stable glider duplicators, routing glider tracks, and routing/adjusting Herschel tracks.
Code: Select all

I have a better idea - using 2g synthesis & 1g destruction of block. Below shows a way to use 2g-block synthesis + 1g deletion to get cells oscillating at every period between 1 and n-7 in this way. This shows that, assuming eaters that can provide cells with frequency n-6 to n-2 exist, it is almost certain that omnifrequent oscillators exist for all periods that have oscillators which can produce an infinite amount of gliders and can be configured in various ways to have gliders synthesize & destroy blocks or other SL - all Herschel Conduit supported periods (61+). The only step now for those periods is to actually build such a POC oscillator, which would probably be enormous.

Code: Select all

x = 204, y = 126, rule = B3/S23
124bob2o$124b2obo$16bob2o$16b2obo102b5o$121bo4bo12b2o$14b5o101bo2bo14b
o2bo$13bo4bo12b2o87bob2o$12bo2bo15b2o84b2obo5bo$12bob2o102bobo4bobo11b
o2bo3b2o$9b2obo5bo98bo2b2o2bo2bo10b2o2bo2bo2bo$10bobo4bobo11b2o5b2o77b
2o6b2o13bo4b2obo$9bo2b2o2bo2bo10bobo4bo2bo89b3o15b2o$9b2o6b2o13bo4b2o
bo88bo3bo11b2o3bo$31bo8b2o86bo4bo10bo2b3o$24bo4b2o6b2o3bo77bo6bo3bobo
9bob2o$22b2obo5bo4bo2b3o78b3o4bo4bo11bo2bo$12bo9b2obo3b2o4bob2o84bo5b
obobo12b2o$12b3o6bo2bo11bo2bo82bobo5b3o4bo$15bo6b2o14b2o82b2o12bo$14b
obo5b2o5bo106b2obo$14b2o13b2o91b2o12bobo2bo9bo$27b5o90bo12b2obo2bo7b3o
$14b2o14b2o11bo80bo23bo$14bo15bobo8b3o79b2o3b2o18b2o3b2o$16bo15bo7bo88b
o23bo$15b2o3b2o8b2o8b2o3b2o79b3o7bo2bob2o12bo$21bo7bo15bo80bo9bo2bobo
12b2o$18b3o8bobo15bo90bob2o$18bo11b2o14b2o93bo12b2o$30b5o105bo4b3o5bo
bo$31b2o13b2o82b2o12bobobo5bo$32bo5b2o5bobo82bo2bo11bo4bo4b3o$22b2o14b
2o6bo84b2obo9bobo3bo6bo$22bo2bo11bo2bo6b3o78b3o2bo10bo4bo$23b2obo4b2o
3bob2o9bo77bo3b2o11bo3bo$20b3o2bo4bo5bob2o88b2o15b3o$19bo3b2o6b2o4bo91b
ob2o4bo13b2o$20b2o8bo98bo2bo5b2o10bo2bo2b2o$21bob2o4bo13b2o85b2o5b2o11b
obo4bo$21bo2bo4bobo10bo2bo2b2o101bo5bob2o$22b2o5b2o11bobo4bo84b2o18b2o
bobo$43bo5bob2o82bo18bo2bo2bo$26b2o18b2obobo80b3o16bo4bo2b2o$27bo18bo
2bo2bo79bo18b5o$24b3o16bo4bo2b2o$24bo18b5o105b2obo$153bob2o$45b2obo$45b
ob2o3$153bo$43bo110bo$44b2o106b3o$43b2o3$156b2o$156b2o2$50b2o$50bobo$
50bo110b2o$44b2o114b2o$43bobo116bo$45bo103b2o$150b2o$149bo8$66bo97b2o
bo$56b2obo5b2o97bob2o$56bob2o5bobo108b2o$29bo114bob2o17b5o6bobo9bo$29b
2o5bob2o17b5o18bo63b2obo12b2o2bo4bo6bo9b3o$28bobo5b2obo12b2o2bo4bo16b
3o53b2o24bo2bo2bo18bo$52bo2bo2bo18bo45bo9bobo6b5o14bobob2o18b2o$15bo18b
5o14bobob2o18b2o44b3o9bo6bo4bo2b2o8b2obo5bo$15b3o16bo4bo2b2o8b2obo5bo
64bo18bo2bo2bo11bo4bobo18b2o$18bo18bo2bo2bo11bo4bobo18b2o42b2o18b2obo
bo12b2o2bo2bo17bo2bo$17b2o18b2obobo12b2o2bo2bo17bo2bo58bo5bob2o16b2o18b
2obo$34bo5bob2o16b2o18b2obo37b2o18bobo4bo30bo11b2o$13b2o18bobo4bo42b2o
35bo2bo17bo2bo2b2o28b3o8b2o3bo$12bo2bo17bo2bo2b2o39b2o3bo34bob2o18b2o
34bo8bo2b3o$12bob2o18b2o43bo2b3o34b2o11bo30bo22bob2o$11b2o42bo17bo4bo
b2o36bo3b2o8b3o28b3o9bobo9bo2bo$10bo3b2o39b3o14bobo4bo2bo36b3o2bo8bo32b
o14bo7b2o$11b3o2bo41bo12b2ob2o5b2o39b2obo22bo16bobo12bobo$14b2obo4bo17b
o16bobo11bo2bo46bo2bo9bobo9b3o16b2o7bo3bo2bo$13bo2bo4bobo14b3o16b2o13b
2o47b2o7bo14bo32bo$13b2o5b2ob2o12bo91bobo12bobo18b2o8bo2bo15bo$21bo2b
o11bobo18b2o27bo43bo2bo3bo7b2o18bo10bo15b3o$22b2o13b2o18bo26b3o46bo33b
o23bo$59bo23bo33bo15bo2bo8b2o19b2o3b2o18b2o3b2o$9bo27b2o19b2o3b2o18b2o
3b2o27b3o15bo10bo25bo23bo$9b3o26bo25bo23bo31bo23bo24b3o15bo10bo$12bo23b
o24b3o26bo23b2o3b2o18b2o3b2o23bo15bo2bo8b2o$6b2o3b2o18b2o3b2o23bo27b2o
24bo23bo45bo$7bo23bo81bo10bo15b3o39bo2bo3bo7b2o$5bo26b3o39b2o13b2o22b
2o8bo2bo15bo38bobo12bobo$5b2o27bo38bo2bo11bobo35bo46b2o7bo14bo$65b2o5b
2ob2o12bo23b2o7bo3bo2bo43bo2bo9bobo9b3o$5b2o13b2o43bo2bo4bobo14b3o20b
obo12bobo43b2obo22bo$5bobo11bo2bo43b2obo4bo17bo21bo14bo7b2o32b3o2bo8b
o$6bo12b2ob2o5b2o32b3o2bo42b3o9bobo9bo2bo31bo3b2o8b3o$3b3o14bobo4bo2b
o31bo3b2o43bo22bob2o33b2o11bo$3bo17bo4bob2o33b2o61bo8bo2b3o31bob2o18b
2o6b2o$27bo2b3o31bob2o18b2o6b2o29b3o8b2o3bo30bo2bo17bo2bo2b2o2bo$28b2o
3bo30bo2bo17bo2bo2b2o2bo31bo11b2o32b2o18bobo4bobo$31b2o32b2o18bobo4bo
bo13b2o6b2o18b2obo54bo5bob2o$2o6b2o18b2obo54bo5bob2o12bo2b2o2bo2bo17b
o2bo57b2obo$o2b2o2bo2bo17bo2bo57b2obo16bobo4bobo18b2o41b2o15bo2bo$bob
o4bobo18b2o41b2o15bo2bo15b2obo5bo62b2o12bo4bo$2obo5bo62b2o12bo4bo19bo
b2o79b5o$3bob2o79b5o20bo2bo15b2o$3bo2bo15b2o88bo4bo12b2o61bob2o$4bo4b
o12b2o61bob2o24b5o75b2obo$5b5o75b2obo$115b2obo$7b2obo104bob2o$7bob2o!

Posted: **June 29th, 2025, 11:18 am**

There exists a spaceship, no amount of copies of which can cleanly annihilate each other on collision.

Posted: **June 29th, 2025, 11:40 am**

Gustone wrote: June 29th, 2025, 11:18 am There exists a spaceship, no amount of copies of which can cleanly annihilate each other on collision.

Would only include B3/S23? Because it seems to me there's probably a way to make a 3-state CA to do that.

Posted: **June 29th, 2025, 3:44 pm**

An explosive rule with a fairly large spaceship can make clean mutual annihilation nearly impossible. Proving that it is definitely impossible might be more difficult.
As an example, good luck finding mutual annihilation of any number of copperheads in this rule:

Code: Select all

x = 33, y = 32, rule = B34ceq5aeijk6aen7c8/S0234ceijyz5ekn6-a7e8
29b2o$29b2o$28bo2bo$27b6o$27b2o2b2o$27bo4bo2$27bo4bo$27bob2obo$28bo2b
o2$28b4o15$3b3ob2o$2b3o4bobo$2obo4bo2bo$2obo4bo2bo$2b3o4bobo$3b3ob2o!

I used the copperhead as an example since it is rather slow and large, and the speed of light means that the backend of a copperhead cannot die out for >10 generations after a frontend collision, by which point the backend will have changed enough to begin exploding.

Posted: **June 29th, 2025, 4:56 pm**

Gustone wrote: June 29th, 2025, 11:18 am There exists a spaceship, no amount of copies of which can cleanly annihilate each other on collision.

In Conway's Game of Life, this should hinge on whether universal destruction by gliders is possible. On the extreme, one can collide groups of spaceships until they find the right glider spacing outputs to synthesize each side of an RCT, and then use the RCT to shoot down the remaining ash before self destructing.

Posted: **June 29th, 2025, 5:16 pm**

What about a hypothetical spaceship that will certainly produce lots of gliders when destroyed? As an extreme example, we could use 0E0P metacells to emulate a chaotic spaceship from another (preferably explosive) rule, and I strongly doubt you can collide such metaships together (either in a way that simulates another rule or otherwise) without releasing gliders in every direction, making cleanup almost impossible. There is also almost certainly no way that you can clean up the released gliders using other metaships without releasing gliders in every direction. There will almost certainly be gliders that will be permanently out of reach of the RCT - no mechanism could be used to catch them.

If someone were to try this (using some futuristic supercomputer, since there is no way modern desktop computers could begin to run this), make sure that the emulated spaceship cannot cleanly annihilate copies of itself.

If this method turns out to be insufficient somehow, we might be able to engineer a spaceship that will always release gliders in every direction.

EDIT: We could also - as an example firmly outside of plausibility - make a spaceship that is so ridiculously large that any collision involving it will end up producing IceNine. (I only came up with this after seeing my post count at 137). Since IceNine, if it exists at all, is an unstoppable quadratic growth pattern, this would make it impossible for the spaceship to be cleanly destroyed at all, let alone by copies of itself.

Posted: **July 8th, 2025, 2:34 pm**

One more period down for proving nontrivial omnifrequent oscillators exist for all periods: p37 (from 2023 stamp collection, without LOM hassler only misses F27 and F25)

Code: Select all

x = 68, y = 79, rule = B3/S23
60b2o$56b2o3bo$56bo2bo$57b4o2$55b6o$54bo5bo$54b2o2b2o3$59bobo5bo$58b2o
bo4b2o$59bo5bobo$58bo6b2o$66bo$57bobo$57b3o$57bo$58bo3bo$59bobo$60bo3$
60bo$59bobo$58bo3bo$63bo$61b3o$61bobo$54bo$54b2o6bo$53bobo5bo$53b2o4b
ob2o$37bo15bo5bobo$24bo12b3o$23bobo14bo$23bobo13b2o20b2o2b2o$22b2ob3o
32bo5bo$17b2o9bo31b6o$17b2o3b2ob3o17b2o$22b2obo21bo12b4o$8b2o30bo6bo13b
o2bo$8bo9b2o9b2o15b2o11bo3b2o$9b3o5bo2bo9bo14bo13b2o$11bo5bobo10bobo11b
o7b2o$18bo12b2o7bo11bobo$39bo14bo$15bobo19b2o15b2o$14bo22bo6bo$3b2o6b
ob2o3bob2o15bo$3bo7bo8b2o16b2o$2obo7bo5bo$o2b3o6bo3bo$b2o3bo6b3o28b2o
$3b4o37bo$3bo41b3o$bobob2o40bo$b2o2bo27bo$5bobo24b3o$6b2o7b3o13bo3bo$
14bo3bo12bob2o$13bo5bo12b2o4b2o$9b2o8bo18bobo$9b2obo3b2obo20bo$16bo23b
2o$13bobo8b2o$24bobo$12bo13bo9b2o$11bobo5bo6b2o8bobo$10bo2bo5b3o16bo$
11b2o9bo15b2o$21b2o$5bob2o$3b3ob2o3b2o$2bo9b2o$3b3ob2o$5bobo$5bobo$6b
o!

Period list: 1-18, 20, 30, 37, and hypothetical 61+*, so 39 periods left

*with unbounded-length Herschel conduits/chained RT61 glider splitters (see speed tunnel) producing as many gliders as needed to either crash into each other to form & destroy transient still-lifes or into eaters which provide remaining cell periods

EDIT: Smaller p37 where the two 58p37s on top mutually delete each other's blocks (which I think works at all phases of the oscillator). The only period provided by the top 58p37 is 27 since it turns out that the blinker factory can provide 25-gen cells

Code: Select all

x = 69, y = 62, rule = B3/S23
50bo$50b3o14b2o$53bo13bo$52b2o11bobo$65b2o2$60b2o$58bo3b2o$58bo4bo$42b
2o18b2o$43bo13b2obo$43bobo5bo5b2obo4b2o$44b2o4bob2o5bo5bobo$50bob2o13b
o$47b2o18b2o$47bo4bo$37bo9b2o3bo$24bo12b3o9b2o$23bobo14bo$23bobo13b2o
$22b2ob3o29b2o$17b2o9bo28bo$17b2o3b2ob3o17b2o11b3o$22b2obo21bo12bo$8b
2o30bo6bo$8bo9b2o9b2o15b2o$9b3o5bo2bo9bo14bo$11bo5bobo10bobo11bo7b2o$
14bo3bo12b2o7bo11bobo$11bo2bo24bo14bo$10bo4bo21b2o15b2o$9bo5b3ob2o16b
o6bo$3b2o3bobo2bob2o4bo15bo$3bo4bo3bo2bo5bo16b2o$2obo5b2obo3bo3bo$o2b
3o4b2o5bo$b2o3bo6bo30b2o$3b4o5bobo29bo$3bo41b3o$bobob2o5b3ob3o28bo$b2o
2bo27bo$5bobo8bobo13b3o$6b2o9bo13bo3bo$13bo5b2o10bob2o$10bo3bo3bob2o10b
2o4b2o$9bo5bo2bo3bo15bobo$9bo4b2obo2bobo17bo$10b2ob3o5bo18b2o$15bo4bo
3b2o$16bo2bo4bobo$12bo3bo9bo9b2o$11bobo5bo6b2o8bobo$10bo2bo5b3o16bo$11b
2o9bo15b2o$21b2o$5bob2o$3b3ob2o3b2o$2bo9b2o$3b3ob2o$5bobo$5bobo$6bo!

Posted: **July 8th, 2025, 4:08 pm**

PK22 wrote: June 29th, 2025, 5:16 pm EDIT: We could also - as an example firmly outside of plausibility - make a spaceship that is so ridiculously large that any collision involving it will end up producing IceNine. (I only came up with this after seeing my post count at 137). Since IceNine, if it exists at all, is an unstoppable quadratic growth pattern, this would make it impossible for the spaceship to be cleanly destroyed at all, let alone by copies of itself.

Even though this '''solution''' hinges on another unproven conjecture, can someone try to find a spaceship which can be extended in 2 dimensions, and that can be proven to produce infinite novelty if it breaks down after being extended enough? An extensible greyship, for example, might be suitable.

If we want to create a realistic example in Life, is there some (ideally somewhat simple) structure for a self-constructing spaceship that would make it impossible to cleanly destroy without producing escaping spaceships in all directions? I would prefer if the original conjecture ("There exists a spaceship {in Life}, no amount of copies of which can cleanly annihilate each other on collision.") had an example that could actually be run in Golly in reasonable time, and I want to try and produce something notable with slmake.

Posted: **August 1st, 2025, 2:47 pm**

LuveelVoom wrote: August 1st, 2025, 2:34 pm Caterer is the smallest P3 oscillator, so that's also proven
Mold is tied with mazing as the smallest P4 oscillator, so that's also proven

I don't think it has been outright proven that there are no other 11-cell oscillators besides pentapole yet (I could be wrong)

EDIT: Never mind, see below.

Posted: **August 2nd, 2025, 7:31 pm**

WhiteHawk wrote: August 1st, 2025, 2:47 pm since all 10-bit patterns haven't been tested yet, it is possible, if extremely unlikely, that unknown p3/p4 oscillators with 10 or 11 cells exist.

Actually, p3 oscillators have been enumerated up to 20 bits (by Beluchenko), p4 oscillators have been enumerated up to 12 bits (By Koenig), and p5 oscillators have been enumerated up to 11 bits (by Koenig). A table of enumerations by bit count for certain object types is available at LifeWiki:Object counts on the wiki.

Posted: **August 3rd, 2025, 1:39 pm**

Sokwe wrote: August 2nd, 2025, 7:31 pm Actually, p3 oscillators have been enumerated up to 20 bits (by Beluchenko), p4 oscillators have been enumerated up to 12 bits (By Koenig), and p5 oscillators have been enumerated up to 11 bits (by Koenig). A table of enumerations by bit count for certain object types is available at LifeWiki:Object counts on the wiki.

Ah, thank you for letting me know! I didn't know that those periods had been proven minimal.

Still, I guess my previous unproven conjecture concerning whether (EDIT: Pentapole, not Pseudo-barberpole) is the only 11-cell oscillator still stands

EDIT:

Entity Valkyrie 2 wrote: August 14th, 2025, 8:28 pm Firstly, pseudo-barberpole is 15 cells, not 11. Also, pentapole is 11 cells.

Whoops. I did mean the other "---pole" which has something to do with the number 5.

Posted: **August 13th, 2025, 3:58 am**

Goldtiger997 wrote: February 4th, 2023, 1:18 pm
pipsqueek wrote: February 4th, 2023, 12:17 pm pretty random, but I conjecture that there exists no stable or periodic (stationary) object that cannot be hit with a glider to create debris with a larger population than the original object. for example, the block has two collisions that have a greater final population.
I think the following 8492-cell still-life (a long^4243 ship) is a counterexample:
Code: Select all
snip
The above pattern contains the collision that has the highest population debris, which was 8212.

I believe this is a smaller counterexample. The still life in question has a population of 732 and is composed of 122 ships tied together. The maximum-population collision I could find results in a final population of 731:

Code: Select all

x = 366, y = 366, rule = B3/S23
2o$obo$b2o$3b2o$3bobo$4b2o$6b2o$6bobo$7b2o$9b2o$3b2o4bobo$2bobo5b2o$4b
o7b2o$12bobo$13b2o$15b2o$15bobo$16b2o$18b2o$18bobo$19b2o$21b2o$21bobo$
22b2o$24b2o$24bobo$25b2o$27b2o$27bobo$28b2o$30b2o$30bobo$31b2o$33b2o$
33bobo$34b2o$36b2o$36bobo$37b2o$39b2o$39bobo$40b2o$42b2o$42bobo$43b2o$
45b2o$45bobo$46b2o$48b2o$48bobo$49b2o$51b2o$51bobo$52b2o$54b2o$54bobo$
55b2o$57b2o$57bobo$58b2o$60b2o$60bobo$61b2o$63b2o$63bobo$64b2o$66b2o$
66bobo$67b2o$69b2o$69bobo$70b2o$72b2o$72bobo$73b2o$75b2o$75bobo$76b2o$
78b2o$78bobo$79b2o$81b2o$81bobo$82b2o$84b2o$84bobo$85b2o$87b2o$87bobo$
88b2o$90b2o$90bobo$91b2o$93b2o$93bobo$94b2o$96b2o$96bobo$97b2o$99b2o$
99bobo$100b2o$102b2o$102bobo$103b2o$105b2o$105bobo$106b2o$108b2o$108bo
bo$109b2o$111b2o$111bobo$112b2o$114b2o$114bobo$115b2o$117b2o$117bobo$
118b2o$120b2o$120bobo$121b2o$123b2o$123bobo$124b2o$126b2o$126bobo$127b
2o$129b2o$129bobo$130b2o$132b2o$132bobo$133b2o$135b2o$135bobo$136b2o$
138b2o$138bobo$139b2o$141b2o$141bobo$142b2o$144b2o$144bobo$145b2o$147b
2o$147bobo$148b2o$150b2o$150bobo$151b2o$153b2o$153bobo$154b2o$156b2o$
156bobo$157b2o$159b2o$159bobo$160b2o$162b2o$162bobo$163b2o$165b2o$165b
obo$166b2o$168b2o$168bobo$169b2o$171b2o$171bobo$172b2o$174b2o$174bobo$
175b2o$177b2o$177bobo$178b2o$180b2o$180bobo$181b2o$183b2o$183bobo$184b
2o$186b2o$186bobo$187b2o$189b2o$189bobo$190b2o$192b2o$192bobo$193b2o$
195b2o$195bobo$196b2o$198b2o$198bobo$199b2o$201b2o$201bobo$202b2o$204b
2o$204bobo$205b2o$207b2o$207bobo$208b2o$210b2o$210bobo$211b2o$213b2o$
213bobo$214b2o$216b2o$216bobo$217b2o$219b2o$219bobo$220b2o$222b2o$222b
obo$223b2o$225b2o$225bobo$226b2o$228b2o$228bobo$229b2o$231b2o$231bobo$
232b2o$234b2o$234bobo$235b2o$237b2o$237bobo$238b2o$240b2o$240bobo$241b
2o$243b2o$243bobo$244b2o$246b2o$246bobo$247b2o$249b2o$249bobo$250b2o$
252b2o$252bobo$253b2o$255b2o$255bobo$256b2o$258b2o$258bobo$259b2o$261b
2o$261bobo$262b2o$264b2o$264bobo$265b2o$267b2o$267bobo$268b2o$270b2o$
270bobo$271b2o$273b2o$273bobo$274b2o$276b2o$276bobo$277b2o$279b2o$279b
obo$280b2o$282b2o$282bobo$283b2o$285b2o$285bobo$286b2o$288b2o$288bobo$
289b2o$291b2o$291bobo$292b2o$294b2o$294bobo$295b2o$297b2o$297bobo$298b
2o$300b2o$300bobo$301b2o$303b2o$303bobo$304b2o$306b2o$306bobo$307b2o$
309b2o$309bobo$310b2o$312b2o$312bobo$313b2o$315b2o$315bobo$316b2o$318b
2o$318bobo$319b2o$321b2o$321bobo$322b2o$324b2o$324bobo$325b2o$327b2o$
327bobo$328b2o$330b2o$330bobo$331b2o$333b2o$333bobo$334b2o$336b2o$336b
obo$337b2o$339b2o$339bobo$340b2o$342b2o$342bobo$343b2o$345b2o$345bobo$
346b2o$348b2o$348bobo$349b2o$351b2o$351bobo$352b2o$354b2o$354bobo$355b
2o$357b2o$357bobo$358b2o$360b2o$360bobo$361b2o$363b2o$363bobo$364b2o!

Would definitely like independent confirmation of this -- I went through the head-on and side-on collisions by hand, but it's possible I missed something.

Assuming this is a valid counterexample, a smaller counterexample could potentially be constructed by putting a "cap" on both sides (such as adding an induction coil, or replacing the last ship with something else) that, by chance, inhibits large reactions from blooming at one of the ends.

EDIT: Capping the ends with long^2 ships results in what I believe is a smaller counterexample at 542 cells. The maximum-population collision I could find results in a final population of exactly 542.

Code: Select all

x = 271, y = 271, rule = B3/S23
2o$obo$bobo$2bobo$3b2o$5b2o$5bobo$6b2o$8b2o$8bobo$9b2o$11b2o$11bobo$
12b2o$14b2o$7b2o5bobo$6bobo6b2o$8bo8b2o$17bobo$18b2o$20b2o$20bobo$21b
2o$23b2o$23bobo$24b2o$26b2o$26bobo$27b2o$29b2o$29bobo$30b2o$32b2o$32bo
bo$33b2o$35b2o$35bobo$36b2o$38b2o$38bobo$39b2o$41b2o$41bobo$42b2o$44b
2o$44bobo$45b2o$47b2o$47bobo$48b2o$50b2o$50bobo$51b2o$53b2o$53bobo$54b
2o$56b2o$56bobo$57b2o$59b2o$59bobo$60b2o$62b2o$62bobo$63b2o$65b2o$65bo
bo$66b2o$68b2o$68bobo$69b2o$71b2o$71bobo$72b2o$74b2o$74bobo$75b2o$77b
2o$77bobo$78b2o$80b2o$80bobo$81b2o$83b2o$83bobo$84b2o$86b2o$86bobo$87b
2o$89b2o$89bobo$90b2o$92b2o$92bobo$93b2o$95b2o$95bobo$96b2o$98b2o$98bo
bo$99b2o$101b2o$101bobo$102b2o$104b2o$104bobo$105b2o$107b2o$107bobo$
108b2o$110b2o$110bobo$111b2o$113b2o$113bobo$114b2o$116b2o$116bobo$117b
2o$119b2o$119bobo$120b2o$122b2o$122bobo$123b2o$125b2o$125bobo$126b2o$
128b2o$128bobo$129b2o$131b2o$131bobo$132b2o$134b2o$134bobo$135b2o$137b
2o$137bobo$138b2o$140b2o$140bobo$141b2o$143b2o$143bobo$144b2o$146b2o$
146bobo$147b2o$149b2o$149bobo$150b2o$152b2o$152bobo$153b2o$155b2o$155b
obo$156b2o$158b2o$158bobo$159b2o$161b2o$161bobo$162b2o$164b2o$164bobo$
165b2o$167b2o$167bobo$168b2o$170b2o$170bobo$171b2o$173b2o$173bobo$174b
2o$176b2o$176bobo$177b2o$179b2o$179bobo$180b2o$182b2o$182bobo$183b2o$
185b2o$185bobo$186b2o$188b2o$188bobo$189b2o$191b2o$191bobo$192b2o$194b
2o$194bobo$195b2o$197b2o$197bobo$198b2o$200b2o$200bobo$201b2o$203b2o$
203bobo$204b2o$206b2o$206bobo$207b2o$209b2o$209bobo$210b2o$212b2o$212b
obo$213b2o$215b2o$215bobo$216b2o$218b2o$218bobo$219b2o$221b2o$221bobo$
222b2o$224b2o$224bobo$225b2o$227b2o$227bobo$228b2o$230b2o$230bobo$231b
2o$233b2o$233bobo$234b2o$236b2o$236bobo$237b2o$239b2o$239bobo$240b2o$
242b2o$242bobo$243b2o$245b2o$245bobo$246b2o$248b2o$248bobo$249b2o$251b
2o$251bobo$252b2o$254b2o$254bobo$255b2o$257b2o$257bobo$258b2o$260b2o$
260bobo$261b2o$263b2o$263bobo$264b2o$266b2o$266bobo$267bobo$268bobo$
269b2o!

Better terminations, or even an elementary counterexample, can potentially be found using a dedicated script.

Posted: **August 14th, 2025, 8:28 pm**

WhiteHawk wrote: August 3rd, 2025, 1:39 pm Still, I guess my previous unproven conjecture concerning whether pseudo-barberpole is the only 11-cell oscillator still stands

Firstly, pseudo-barberpole is 15 cells, not 11. Also, pentapole is 11 cells.

Posted: **August 21st, 2025, 6:24 pm**

Probviously true, but intractable: in the long run (and at large range), an arbitrarily sparse but non-zero-density Life universe will have no moving objects as all spaceships and puffers will collide into something eventually.

In Life, there is a known 4-to-6-glider reaction: for any number N, there exists K>N such that a 4-to-K glider reaction is possible and the K gliders at the end fly off to infinity without colliding.

Across all range-1 MAP rules, there exists a rule in which the evolution of the single cell will 1) peak at a finite maximal population and 2) that population will exceed 2^100.

Posted: **August 21st, 2025, 7:24 pm**

Define a region to be a finite set of cells in the universe and a patch to be an assignment of values to the cells in a region.
Define a forcing patch P to be a patch on a region R such that in any predecessor of a universe containing P, the intersection with R is patch on P independent of the predecessor.

Conjecture: There exists a region R and a set S of patches on R satisfying the following properties:

For any patch P in S and any predecessor of a universe containing P, the intersection with R is a patch that is also in S.
S contains no forcing patch.
For every patch P in S, there exist universes containing P that can be rewound arbitrarily far.
S does not contain the empty patch.

Without condition 2, the self-forcing patches used in the unsynthesizable still lives and unsynthesizable oscillator would form one-element sets satisfying the definition. Without condition 3, a set of patches obtained by disturbing köynnös in various ways an evolving them would saitsfy the definition.

These properties are interesting because imply that any pattern containing a patch of S is unsynthesizable despite potentially not containing a self-forcing patch and potentially being arbitrarily rewindable. A few more related conjectures:

Conjecture: There exists an unsynthesizable still life that does not contain a self-forcing patch.
Conjecture: There exists a patch P such that there exist universes containing P that can be rewound arbitrarily far, but no single universe containing P can be rewound arbitrarily far.

Posted: **August 22nd, 2025, 3:22 am**

Welcome to the forums!

andrewthelifer wrote: August 21st, 2025, 6:24 pm Probviously true, but intractable: in the long run (and at large range), an arbitrarily sparse but non-zero-density Life universe will have no moving objects as all spaceships and puffers will collide into something eventually.

(...)

Across all range-1 MAP rules, there exists a rule in which the evolution of the single cell will 1) peak at a finite maximal population and 2) that population will exceed 2^100.

1). It seems probvious at first, given an infinite Sparse Life universe, there will be something in the path of every spaceship or puffer eventually, but if something like IceNine exists (which was first theorised to be an unstoppable spacefiller, but given what we know it is likely an ash-clearing quadratic replicator if it exists at all), then it can survive all/most collisions, but it remains a moving object. If IceNine does not exist, then there is no reason why a Snark loop (or whatever glider reflector turns out to be most common) protected by enough blocks/eaters in an extremely lucky way might survive, and the gliders inside are technically 'moving objects'.

2). I am not certain - even though MAP rules have more freedom than INT rules, the only way the dot is exceeding its initial population of 1 is if it has a birth condition allowing it to grow at lightspeed, in which case it will grow infinitely and unstoppably, and therefore will not 'peak' at a finite population.

Speaking of Sparse Life, is it currently possible to emulate a finite Sparse Life universe in such a way that the behaviour listed on the wiki emerges?

Posted: **August 22nd, 2025, 5:03 am**

PK22 wrote: August 22nd, 2025, 3:22 am Welcome to the forums!
[...]
1). It seems probvious at first, given an infinite Sparse Life universe, there will be something in the path of every spaceship or puffer eventually, but if something like IceNine exists (which was first theorised to be an unstoppable spacefiller, but given what we know it is likely an ash-clearing quadratic replicator if it exists at all), then it can survive all/most collisions, but it remains a moving object. If IceNine does not exist, then there is no reason why a Snark loop (or whatever glider reflector turns out to be most common) protected by enough blocks/eaters in an extremely lucky way might survive, and the gliders inside are technically 'moving objects'.

2). I am not certain - even though MAP rules have more freedom than INT rules, the only way the dot is exceeding its initial population of 1 is if it has a birth condition allowing it to grow at lightspeed, in which case it will grow infinitely and unstoppably, and therefore will not 'peak' at a finite population.

0) Thanks for the warm welcome.

1) Yes, I do count gliders in reflective circuits as "moving", and any other spaceships, in which case a sparse world tending to reflectors infinitely often would be enough to beat my conjecture.

2) Even in INT rules, growing dots can close off their bounding box, which is enough to fit the parameter that there is a finite maximal population. toroidalet explicitly demonstrated an INT rule with maxpop = 3156 and later in the thread a 3-state MAP rule with maxpop > 2^7,000,000. Given the size of the rulespace it seems likely some rule could chain up several mechanisms, potentially interacting with each other, to cultivate the dot to 2^100 (or more) before its bounding box would close up.

Posted: **August 22nd, 2025, 6:57 am**

Background: a 0-SMOS is a spaceship, an n+1-SMOS is in at least one phase solely more than one n-SMOS about to collide.

Conjecture: For any k, INT contains a k-SMOS which isn't a higher degree of SMOS.

(The extra restriction is due to SMOIs aka failed replicators which proceed: small phase -> two or more small-phase copies -> a crash before they can fully replicate which produces the original small phase at some displacement. These are w-SMOSes and not counted.)

Posted: **August 22nd, 2025, 7:07 am**

andrewthelifer wrote: August 22nd, 2025, 6:57 am (The extra restriction is due to SMOIs aka failed replicators which proceed: small phase -> two or more small-phase copies -> a crash before they can fully replicate which produces the original small phase at some displacement. These are w-SMOSes and not counted.)

They are engineered spaceships, e.g. the c/18 in HighLife:

Code: Select all

x = 165, y = 275, rule = B36/S23
60b2o$60b2o3$63bo3bobo$62bobo3b2o$62bobo3bo$63bo$45bo$45b2o$45bobo$46b
3o2$40b3o$41bobo$42b2o$37bo5bo$37b2o17bo$37bobo16b2o$38b3o15bobo$57b3o
$32b3o$33bobo15b3o$34b2o16bobo$35bo17b2o$54bo2$23b3o13$30b3o4$71bo$70b
3o$70bob2o$71b3o$71b3o21bobo$71b2o23b2o$96bo42$123bobo$124b2o$124bo12$
68b2o$67b2o$68b2obo7bo$69b3o6b3o$70bo7bob2o$79b3o$79b3o$79b2o9$52bo$51b
3o$51bob2o$54b2o$53b2o10$151bobo$152b2o$152bo19$42bo$43bo$41b3o$17b3o
$17bo2bo$17bo3bo$18bo2bo$19b3o8$87bo$86b3o$86bob2o$87b3o$b3o83b3o$bo2b
o82b2o$bo3bo$2bo2bo$3b3o3$158b3o$161bo$o155bobo3bo$o162bo$o153bobo7bo
$157bo6bo$152bobo3bo3bobo$159bo$150bobo7bobo$153bo$148bobo3bo3bobo$155b
o$146bobo7bobo$149bo$144bobo3bo3bobo$144bo6bo$144bo7bobo$145bo$70bo75b
o3bobo$71bo75bo$69b3o76b3o13$127b3o$127bo2bo$127bo3bo$128bo2bo$123b3o
3b3o$123bo2bo$123bo3bo$124bo2bo$125b3o12$95bo$94b3o$94bob2o$95b3o$95b
3o$95b2o16$73bo$72b3o$71b2ob2o$72b2ob2o$69bo3b2ob2o$68b3o3b3o$67b2ob2o
3bo$68b2ob2o$69b2ob2o$70b3o$71bo5$56bo$56bo$56bo!

Posted: **August 22nd, 2025, 7:17 am**

pifricted wrote: August 22nd, 2025, 7:07 am They are engineered spaceships, e.g. the c/18 in HighLife:[...]

This isn't a SMOI, because there's no small phase that becomes two of itself at any point. (actual SMOI by macbi for comparison)

I suspect universal construction allows arbitrary k-SMOSes, because self-making constructors could be coordinated to "meet up" and signal appropriate collisions to reproduce a distant ancestor, restarting the cycle.

ConwayLife.com

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