Unproven conjectures

[WhiteHawk]

Conjecture: There is an oscillator in life with period=n where the following are all true:

[*] the period is odd
[*] strict volatility > 0, volatility > 0
[*] all sparkers (if any) must have the same period as the whole oscillator.
[*] (most importantly) the oscillator's volatility does NOT equal it's strict volatility (all prime period oscillators are eliminated by this stipulation)

This conjecture may already be proven. Engineered non-interacting solutions (see Confocaloid's post below for an example, or something like a pseudo-period gun if one had no sparkers oscillating at subperiod) are also not permitted, though they technically fill the stipulations.

EDIT: An extended unproven conjecture with the same stipulations, except:

[*] the period can be even
[*] there can be a cell that oscillates at any equivalent fraction to n/2 (2n/4, 3n/6, 4n/8, 5n/10, 8n/16, 9n/18, etc.) but NOT solely oscillating at an indivisible n/2 (E.G. Twin Bees Shuttle and Gabriel's P138 do not count since the cells that make SV and V different only oscillate at n/2)

(EDIT 3: 51p384 is an example for the second conjecture)

That requirement is unclear. Can you clarify the intended meaning?

Everything clear now?

EDIT 2:

Let's use period 25 as an example.

A p25 supported by p5 sparkers would not count.

If Charity's p25 happened to contain some p5 cells completely by coincidence (which it doesn't), it would.

As another example, if a p9 billiard-table oscillator was partially supported by a p3 dependent subpattern, that would count. A p9 oscillator using Caterers instead would not.

EDIT 4:

Got one! (kind of, there are "sparkers")

Code: Select all

x = 29, y = 29, rule = B3/S23
9bo$7b5obo5b2o4b2o$bob2obo5b2o4bo2bo3bo$b2obobo2b2o3b2o2b2obo4bo$6bo4b
3o2bobo2b2o2b2o$5b2obo6bo2bobo2bo$4bobo2b6o3b2ob6o$2b3ob3o6b3o3bo5bo$
bo3bo3bo2b3o3b2obobo3bo$bobo2bobo6bo2bobobo2bob2o$2b5ob7obobo3bo2bobo
$7bo6bobobo3bobo2bo$4bo2bo2b5obobobobobobo$3bobobobo6bobobobobob2o$3b
o2bobob3o3b3obobo2bo$b2obobobobobo6bobobobo$2bobobobobobob5o2bo2bo$bo
2bobo3bobobo6bo$bobo2bo3bobob7ob5o$2obo2bobobo2bo6bobo2bobo$bo3bobob2o
3b3o2bo3bo3bo$bo5bo3b3o6b3ob3o$2b6ob2o3b6o2bobo$5bo2bobo2bo6bob2o$2b2o
2b2o2bobo2b3o4bo$2bo4bob2o2b2o3b2o2bobob2o$3bo3bo2bo4b2o5bob2obo$2b2o
4b2o5bob5o$19bo!

Now is there one which is not a rephased billiard table

[confocaloid]

[...] there are no sub-period oscillators supporting the rotor (I.E. no sparkers working to support the rotor by overpopulation or underpopulation) - this does not count still-life supports. [...]

I don't understand this part. Can you explain it? As an illustration, could you provide examples of oscillators which fail this requirement in each "interestingly different" way, despite satisfying all the other requirements listed in your post?

(As far as I can tell, the other requirements listed in your post mean that you are looking for a nontrivial oscillator with a composite odd period and with at least one cell oscillating at a subperiod.)

EDIT 2025-02-17 (in response to an edit):

[...] no cells oscillating at subperiod are part of the stator (I.E. there are no sparkers working to support the rotor by overpopulation or underpopulation) - this does not apply to still-life supports. [...]

That looks like some confusion. It's impossible for an oscillating cell to be part of the stator of an oscillator, by definition. (The stator entirely consists of cells that don't change their state.) Hence it's unclear why this is added as a requirement, and why there is a mention of the cells oscillating at subperiod (which doesn't affect the impossibility of any oscillating cells in the stator).

That requirement is unclear. Can you clarify the intended meaning?

EDIT 2: unfortunately later edits made things more confusing, without actually clarifying what is the actual idea behind that requirement.
There may be *some* interesting idea obscured here, but it is hard to tell what it is.

[...] any self-supporting part of the oscillator, including the oscillator as a whole, must either oscillate at the same period of the oscillator (I.E. any and all sparkers working to support any unstable part of the rotor by overpopulation or underpopulation must have the same period as all other sparkers (if any) and the oscillator as a whole) OR have volatility = 0 (still life supports). [...]
[..]
Everything clear now?

[hotdogPi]

Let's use period 25 as an example.

A p25 supported by p5 sparkers would not count.

If Charity's p25 happened to contain some p5 cells completely by coincidence (which it doesn't), it would.

[confocaloid]

Let's use period 25 as an example.

A p25 supported by p5 sparkers would not count.

If Charity's p25 happened to contain some p5 cells completely by coincidence (which it doesn't), it would.

Well, it might be feasible to "brute-force" a p9 oscillator containing a p3 cell in the rotor. That would probably look and feel like a single "oscillating blob".

However, if the goal is to find an engine-based oscillator with such property (where an oscillator's engine literally "just so happens" to contain a subperiod cell without this being bruteforced in some way), then the only idea I have so far is to test every discovered engine-based oscillator for this property, and hope that eventually some new discovered oscillator will answer the question.

EDIT: of course there are also constructed examples (like this p45 oscillator with an isolated p15 cell), which don't really solve the problem (obviously it isn't a coincidence that there is an isolated p15 cell in the rotor of the oscillator, and further that p15 rotor cell isn't necessary part of any oscillator engine, and further there are some additional p15 regions elsewhere):

Code: Select all

x = 49, y = 39, rule = B3/S23
37b2o$35bo4bo$34bo6bo$33bo8bo$33bo8bo$33bo8bo$34bo6bo$35bo4bo$37b2o8b
2o$10b2o34bobo$10b2o35bo2$36b2ob2o$36bo3bo$23b2o11bo3bo$10bo26bobo$9bo
b2o25bo$7b2obob2o$bo3b2o4b2o12bo11b2o$obo21b3o10b2o$2o8b2o11bo2bo$8bo
4bo9b2o$7bo6bo9bo$6bo8bo17b3o$6bo8bo9bobo4bo3bo$6bo8bo10bo4bo5bo$7bo6b
o11bo$8bo4bo5b2o9bo7bo$10b2o7b2o9bo7bo2$26bo4bo5bo$26bo5bo3bo$25bobo5b
3o2$24bo$23b2o$23bobo3bo$24bo3bobo$29b2o!

[get_Snacked]

the tub is the only synthesisable still life (S) where all predecessors of S that are not S itself have a greater population than S.
EDIT: the snake is a counterexample. other than that, are there any other still lifes that satisfy this condition?

[confocaloid]

the tub is the only synthesisable still life (S) where all predecessors of S that are not S itself have a greater population than S.
EDIT: the snake is a counterexample. other than that, are there any other still lifes that satisfy this condition?

I think one needs to clarify that the still life must be a strict object. Otherwise two well-separated solutions would in many cases count as another solution.

Since the requirement is stated in terms of population, it might be interesting to relax the idea so that it is about constant-population strict objects (including but not limited to strict still-life objects). The blinker would count as an additional counterexample (constant population 3, every predecessor that isn't a blinker would have at least 4 alive cells) unless I'm missing something. The glider wouldn't count, it has a constant population 5, but there is a 5-bit predecessor that isn't a glider:

Code: Select all

x = 3, y = 3, rule = B3/S23
o$2bo$3o!

Tangential: viewtopic.php?p=199478#p199478

[hotdogPi]

Since the requirement is stated in terms of population, it might be interesting to relax the idea so that it is about constant-population strict objects (including but not limited to strict still-life objects). The blinker would count as an additional counterexample (constant population 3, every predecessor that isn't a blinker would have at least 4 alive cells) unless I'm missing something. The glider wouldn't count, it has a constant population 5, but there is a 5-bit predecessor that isn't a glider:

Code: Select all

x = 3, y = 3, rule = B3/S23
o$2bo$3o!

We already know that a clock predecessor must contain 8 cells or more or a clock (7 or less: itself alone or itself plus a dot anywhere in the universe).

[Tawal]

Conjecture : No glider eater can be constructed with a recovery time of three ticks or less

May be it can be considered as false :

Code: Select all

x = 10, y = 9, rule = LifeHistory
9.E$7.3E$2E4.E$E.E3.2E$.E$4.A$3.3A$2.A.A$6.A!
#C [[ STOP 1 ]]

Related link : viewtopic.php?f=2&t=315&p=204750#p204750

Edit:
It works also with dot and domino spark.
But I didn't find a p3 sparker which works.

Code: Select all

x = 20, y = 8, rule = LifeHistory
7.A11.A$5.3A9.3A$4.A7.E3.A$E3.2A6.E3.2A2$.3A9.3A$3.A11.A$2.A11.A!

[WhiteHawk]

Conjecture : No glider eater can be constructed with a recovery time of three ticks or less

It has already been proved that there are no STABLE glider eater with a recovery time of three ticks or less, proof from 2021

Conjecture: No glider eater can be constructed with a recovery time of three ticks or less.

Kayzan did the following LLS searches to prove this conjecture:ThreeTickEater.txtThreeTickEater2.txt

To convince myself I did the following searches with even less clearance. (Double posting to circumvent 3 file limit.)fast-eater.csvfast-eater2.csvWe can declare this Conjecture a Theorem.

------

Edit:
It works also with dot and domino spark.
But I didn't find a p3 sparker which works.

Code: Select all

It is also already known that Recovery Time=3 eaters exist for periodic objects (I found this within 1 minute of searching the forums)

Kayzan did the following LLS searches to prove this conjecture:

To convince myself I did the following searches with even less clearance.

Here's a periodic 3-tick eater:

Code: Select all

x = 34, y = 27, rule = B3/S23
29b2o$bo27b2o$2bo$3o$30b2o$16b2o14b2o$19b2o6bo5bo$18bob2o4b2o$14bo2bob
3o4b3obo2bo$20b2o4b2obo$14bo5bo6b2o$14b2o14b2o$16b2o3$11bo5b2o$11bo5b
2o$10b3o4$9b5o$8bob3obo$8bobobobo$7b2obobob2o$6bo2b2ob2o2bo$6b2o7b2o!

[Tawal]

[…]
It has already been proved that there are no STABLE glider eater with a recovery time of three ticks or less, proof from 2021
[…]
It is also already known that Recovery Time=3 eaters exist for periodic objects (I found this within 1 minute of searching the forums)

Kayzan did the following LLS searches to prove this conjecture:

To convince myself I did the following searches with even less clearance.

Here's a periodic 3-tick eater:

Code: Select all

x = 34, y = 27, rule = B3/S23
29b2o$bo27b2o$2bo$3o$30b2o$16b2o14b2o$19b2o6bo5bo$18bob2o4b2o$14bo2bob
3o4b3obo2bo$20b2o4b2obo$14bo5bo6b2o$14b2o14b2o$16b2o3$11bo5b2o$11bo5b
2o$10b3o4$9b5o$8bob3obo$8bobobobo$7b2obobob2o$6bo2b2ob2o2bo$6b2o7b2o!

OK for stable case.
However I'm not agree with your example.
The Glider is effectively totally deleted in 3 ticks but the sparkers don't recover theirs original state in 3 ticks.

[Timelord Missionary]

So I was thinking about the block agar, and I realized that any new row-maker would most likely be incremental, like the 3-block one. Could this partial hypothetically work for a five-block row?

Code: Select all

x = 327, y = 104, rule = B3/S23
22$187b2o2b3o6bobo3bo$189bobo5b2obo4b2o2b2o2bo2bo$187b2o2b2o3bobobobo
bobob3obo2bo$189bo3bo3b2obobo2b2o2bo2bo3bo$107b3o31b3o25b3o15b2o2b2o5b
o11bo4bo9b3o25b3o25b3o$106bo3bo29bo3bo23bo3bo24bo26bo3bo23bo3bo23bo3b
o$110bo33bo27bo55bo27bo27bo$110bo33bo27bo55bo27bo27bo$108b2o32b2o26b2o
54b2o26b2o26b2o$107bo33bo27bo34b2o19bo27bo27bo$107bo33bo27bo34b2o19bo
27bo27bo2$107bo33bo27bo55bo27bo27bo$126b2o32b2o26b2o26b2o26b2o$15b2ob
2o28b2ob2o17b2ob2o14b2ob2o25b2ob2obobo25b2ob2obobo19b2ob2obobo19b2ob2o
bobo19b2ob2obobo19b2ob2ob2obo18b2ob2ob2o$15b2ob2o28b2ob2o17b2ob2o14b2o
b2o25b2ob2obo27b2ob2obo21b2ob2obo21b2ob2obo21b2ob2obo21b2ob2obob2o18b
2ob2ob2o$26bo98b2o32b2o26bo27bo27bo27bo$15b2ob2o4b3o21b2ob2o3b2o12b2o
b2o3b2o9b2ob2o3b2o20b2ob2o3bo25b2ob2o3bo19b2ob3o22b2ob3o22b2ob3o22b2o
b2obo21b2ob2ob2o$15b2ob2o2bo2b2o8bo12b2obobobobo12b2obobobobo9b2obobo
bobo20b2obob3o3bo22b2obob3o20b2obo24b2obo24b2obo24b2ob2ob2o20b2ob2ob2o
$22bo13bo16bobo19bobo16bobo11bo15bo5bobo9bo15bo11bo27bo27bo27bo16bo10b
o$15b2ob2ob2o9b6o10b2obobob2o13b2obobob2o10b2obobob2o11bo9b2obo7b2o11b
o9b2obo14bo9b2obo14bo9b2obo14bo9b2obo14bo9b2ob2obo11bo9b2ob2ob2o$15b2o
b2ob2o13bo11b2ob2o17b2ob2o14b2ob2o11b6o8b2ob2o3b2o10b6o8b2ob2o9b6o8b2o
b3o8b6o8b2ob3o8b6o8b2ob3o8b6o8b2ob2ob2o6b6o8b2ob2ob2o$21b3o11bo73bo17b
obo13bo15b2o10bo15bo11bo15bo11bo15bo11bo15bo11bo$15b2ob2ob3o24b2ob2o17b
2ob2o14b2ob2o14bo10b2ob2o3bo14bo10b2ob2obo10bo10b2ob2obo10bo10b2ob2ob
o10bo10b2ob2obo10bo10b2ob2obo10bo10b2ob2ob2o$15b2ob2ob4o23b2ob2o5bo11b
2ob2o14b2ob2o25b2ob2o29b2ob2obo21b2ob2obob2o18b2ob2obob2o18b2ob2obob2o
18b2ob2obob2o18b2ob2ob2o$21b3o34bobo98b2o26b2obo24bobo25bobo25bobo$15b
2ob2ob3o24b2ob2o5b2o10b2ob2o14b2ob2o25b2ob2o29b2ob2o23b2ob2o23b2ob2ob
obo19b2ob2obobo19b2ob2obobo19b2ob2ob2o$15b2ob2ob3o24b2obobo16b2obo15b
2obo26b2obo30b2obo24b2obo24b2ob2ob2o20b2ob2ob2o20b2ob2ob2o20b2ob2ob2o
$21b3o28b2o21bo17bo29bo33bo27bo$22bo51b2o18bo29bo33bo27bo$23bo37b2o16b
o15bo27b2o32b2o26b2o$56b2o3bobo13b2o15b2o$55b2o4bo16b2o52bo2bo11bo$57b
o32bo41b2obo2b2o2b2o2b2o2bo$49b3o25bo12b2o6b2o32bob2ob3ob3obobobo$51b
o3b2o19b2o11bobo6bobo31bo2bo2bo3bo3b2o2bo$50bo3bobo19bobo19bo40bo3bo5b
o57b2o$56bo35b3o75bobo10bo2bo9bo2bo2bobo2b2o$58b2o32bo76bobo3bo2b2o2b
3o4bo2b2o3bobobobo$58bobo32bo44bo5bo29bobobobo2bo2bobobobobob2obo$58b
o74b2o2b2o2b2o4b2o2bo23bo2b2o3bo2bo2bo2bobob2obo$132b3obobobobobob3ob
o27bo$133bo3b2o2b2obo2bo2bo$134bo7bo5bo34b2o$141bo41b2o2$133bo2bo2bo3b
o$132b3o2bob2o2bo2b2o$133bo2b2obobobob3o$133bo2b2ob2o2bo2bo$147bo2$132b
o16bo$134b2o3bo2b2o2bob3o2b2o2bo$132bobobobo2b3obo3bo2b3obo$132bobobo
2bo2bo2bo3bo3bo2bo$138bo4bo10bo!

I realize if there technically aren't any logical inconsistencies, it would still be a lot of work

[confocaloid]

So I was thinking about the block agar, and I realized that any new row-maker would most likely be incremental, like the 3-block one. Could this partial hypothetically work for a five-block row?

Code: Select all

x = 327, y = 104, rule = B3/S23

I realize if there technically aren't any logical inconsistencies, it would still be a lot of work

This approach would obviously fail to give the entire block agar. It would remain limited to a subset of widths for which m-by-n block arrays are known to be constructible.

But I don't see any logical inconsistency or local obstruction that could prevent constructibility of 5-by-n block arrays, or probably 6-by-n block arrays. Just a lot of work needed to discover the synthesis components, probably comparable to the efforts to find syntheses for bigger spaceships or the spacefiller synthesis.

One can imagine that there could be a way to construct (by colliding gliders near an existing large block array) a "block array extender", a moving reaction on the edge of the block array ("skimmer") that would add one or more rows of blocks as it moves.
Then there might be some way to safely stop the "skimmer" at the other edge of the block array, once it adds an entire row (or several rows) of blocks, again by colliding gliders near the block array. All this without causing disintegration of the block array.
If that is possible, it could lead to a way of constructing all finite m-by-n block arrays.

[muzik]

Now the challenge can be stated in the form "Find a statorless p3 oscillator where every rotor cell is green, under these rules":

Probably even simpler would be "find a period-3 oscillator where every non-dead cell in its frequency map is the same colour". The only possibilities that would satisfy this condition would be a still life, a p3 phoenix, or this type of oscillator; we can rule out the first due to its "real" period not being 3, and the second has been disproven. Sadly there doesn't seem to be any program support for this type of oscillator map so far.

Since LifeViewer now supports period maps we can see this in action. Press the Frq button to observe.

Some OCA solutions from here:

Code: Select all

x = 7, y = 6, rule = B35/S135
bo$obo$2obo$3bob2o$4bobo$5bo!
[[ AUTOIDENTIFY ]]

Code: Select all

x = 23, y = 31, rule = B35/S135
11bo$11bo$bo9bo9bo$bo4bob2obob2obo4bo$o2b2ob2obo3bob2ob2o2bo$2b2o2bob
ob3obobo2b2o$5bob9obo$5b2obo2bo2bob2o$4b3o2bobobo2b3o$5b2o2bo3bo2b2o$
4b2o2bob3obo2b2o$b2obo6bo6bob2o$bobobo5bo5bobobo$2bo8bo8bo$5bo11bo$5b
2o9b2o$2bob3o9b3obo$2bo17bo$2bo17bo$3b2o13b2o$3bo2b2o7b2o2bo$6bob2o3b
2obo$5bo2bo5bo2bo$4bobo9bobo$3b3o11b3o$2bob2o11b2obo2$2b3o13b3o$3bo15b
o$3bo15bo$3bo15bo!
[[ AUTOIDENTIFY ]]

Code: Select all

x = 21, y = 24, rule = B35/S135
5b2o3bo3b2o$4bobo3bo3bobo$5bo4bo4bo$6b2ob3ob2o$8b2ob2o$6b2ob3ob2o$7bo
2bo2bo$9b3o$10bo$9bobo$10bo$2b3o2b2obob2o2b3o$2bo2bob2obob2obo2bo$4bo
5bo5bo$2bo6bobo6bo$2o8bo8b2o$bo7b3o7bo$7bo2bo2bo$6b2ob3ob2o$8b2ob2o$6b
2ob3ob2o$5bo4bo4bo$4bobo3bo3bobo$5b2o3bo3b2o!
[[ AUTOIDENTIFY ]]

Here's Life's statorless p3 - note how the frequency map shows two colours:

Code: Select all

x = 26, y = 33, rule = B3/S23
12bo$11bobo11bo$19b3o3bo$11bo2bo2b2o6bo$12bobob3o3bobo$15b2o7bo$11bo2bo7bobo$
9b2o11bo$7b3obobo8bo$o6b2o2bo2bo7bo2bo$o3b3o16b3o$o11bobo8b3o$bobo9bo$bo21b3o
$bobo19b3o$3bo18bo2bo$3bo18bo$o2bo18bo$3o19bobo$3o21bo$12bo9bobo$3o8bobo11bo$
3o16b3o3bo$o2bo7bo2bo2b2o6bo$3bo8bobob3o$3bo11b2o$bobo7bo2bo$bo7b2o$bobo3b3ob
obo$o6b2o2bo2bo$o3b3o$o11bobo$13bo!
[[ AUTOIDENTIFY ]]

[hth3]

Conjecture: There is a spaceship that is made entirely of isolated duoplets, dots, and dominoes in one of its phases.

[confocaloid]

Conjecture: There is a spaceship that is made entirely of isolated duoplets, dots, and dominoes in one of its phases.

It might be better to add some restrictions on size, to (somehow) exclude "UC solutions" involving glider predecessors with that property. Related discussion in another thread: viewtopic.php?p=155622#p155622

I feel like such a pattern exists, and could possibly be done via some weird UC encoding that somehow ends up constructing its recipe but with glider predecessors like this:

Code: Select all

x = 12, y = 5, rule = B3/S23
bobo6bo$o3bo3bo$2bo4bo2bo$bo7bo$3bo3bo3bo!

...But that feels less satisfying than a small example. Cross and star get pretty close, though.

I wonder if LLS could find an example, or a similar SAT-solver based approach. I don't think LLS allows one to specify cells that can only be the same state if they're dead.

I 'm not quite up to the task of constructing an UC-based solution, but in case these can be useful, here are ways to construct glider predecessors for both shapes. At T = 47, neither of marked (red) cells is on; at T = 48, the marked cells are the only alive cells.

Code: Select all

x = 74, y = 43, rule = LifeHistory
62.A$60.2A$61.2A8$49.2A$48.A.A$49.A$11.A32.A$10.A.A32.A27.A$4.A4.A.A
31.3A25.2A$3.A.A3.2A43.2A16.2A$4.2A48.A.A$56.A$56.2A$3.D13.2A35.D$2.D
13.2A34.2D$.AC2D13.A34.2D$A2.A$A.A43.2A3.2A3.D2A$.A44.2A3.2A4.A.A$58.
A2$2.A$.A.A48.2A$2.A.A47.2A$3.2A9$8.2A$7.2A$9.A!

[get_Snacked]

Since the requirement is stated in terms of population, it might be interesting to relax the idea so that it is about constant-population strict objects (including but not limited to strict still-life objects). The blinker would count as an additional counterexample (constant population 3, every predecessor that isn't a blinker would have at least 4 alive cells) unless I'm missing something. The glider wouldn't count, it has a constant population 5, but there is a 5-bit predecessor that isn't a glider:

Code: Select all

x = 3, y = 3, rule = B3/S23
o$2bo$3o!

well, since this version of the conjecture received multiple counterexamples immediately, i think we should go back to the original conjecture that only talks about (EDIT: strict) still lifes.

[confocaloid]

the tub is the only synthesisable still life (S) where all predecessors of S that are not S itself have a greater population than S.
EDIT: the snake is a counterexample. other than that, are there any other still lifes that satisfy this condition?

[...] i think we should go back to the original conjecture that only talks about (EDIT: strict) still lifes.

Does this 11-bit strict still life have 11-bit predecessors other than itself? I ran Logic Life Search / kissat, and it didn't find any such predecessors within three ticks back in time with the still life centered in a 14-by-14 rectangle.

Code: Select all

x = 8, y = 8, rule = B3/S23
2o$obo$3bo$4bo$5bo$6bo$7bo$6b2o!

[hth3]

Conjecture: There is a two cell thick spaceship.

[confocaloid]

Conjecture: There is a two cell thick spaceship.

Conjecture: for every positive integer n, there exists an orthogonal spaceship in CGoL such that the minimum single-phase "width"[1] equals n.

[1]: (i.e. minimum width of a single-phase bounding rectangle, measuring perpendicular to the direction of movement of the spaceship)

Related discussion: viewtopic.php?f=2&t=2040 How about a unidimensional spaceship?

[WhiteHawk]

Two conjectures and one related question, based on the recent feature in Lifeviewer which shows cell frequency:

Conjecture - there is a non-weld nontrivial oscillator for every period n in life where there is an example cell oscillating at every frequency value between n and 1 inclusive (frequency being how long cells are on related to the period, n being all eternally on cells and 1 being cells which are on for just 1 generation).

If this conjecture is false, then my backup conjecture is that there is a Life-like rule where the previous conjecture is true.

Question: Given that a period is proven to have at least 1 oscillator where conjectured property is true, what is the smallest oscillator of this period n (by minimum population) which still has this property.

To explain what I am talking about, solution to both the conjecture and the question for periods 2 through 5 are shown below (perhaps omnifrequent could be a term to describe them?) - all still-lifes automatically have this property, as do all non-phoenix p2 oscillators. To verify these examples, go to settings, select pattern, then select identify, and check frequency.

Blinker, being the smallest p2 oscillator (and the smallest p2 oscillator with cells that stay on), is sufficient to solve the conjecture for n=2.

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x = 1, y = 3, rule = B3/S23
o$o$o!

Caterer, the smallest period-3 oscillator, suffices for that period.

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x = 8, y = 6, rule = B3/S23
2bo$o3b4o$o3bo$o$3bo$b2o!

Since Mazing doesn't have any cells which stay on throughout it's evolution, Mold is the sole smallest P4 oscillator with this property.

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x = 6, y = 6, rule = B3/S23
3b2o$2bo2bo$o2bobo$4bo$ob2o$bo!

Finally, p5 is the first period not solved by it's SKOP (Pseudo-barberpole has no cell stay alive for 4 out of 5 generations), so Fumarole is the smallest oscillator which has this property.

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x = 8, y = 7, rule = B3/S23
3b2o$bo4bo$bo4bo$bo4bo$2bo2bo$obo2bobo$2o4b2o!

P6 is where it gets tricky; since neither Unix nor p6 thumb answer the question, and I don't know the 3rd smallest p6 to check. Additionally, the smallest period, according to my searches, where there is no known example to solve the conjecture is 17. Below are examples, some larger than necessary for periods 6 through 16.

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#C examples for all periods 6-16
x = 109, y = 76, rule = B3/S23
77b2o3b2o$77b2o3b2o$55b2o$2o52bobo39bo$bo52b3o39b3o$bobo60b2o10b3o3b3o
14bo$2b2o13b2ob2o33b2o7bo7b2o13b2o9b2o$18bobo14b2o17b4o4bobo7b2o2bobo
3bobo2b2o19bo$6bo10bo4bobo10b2o2b2o13bo3bo3b2o13bo5bo22b3o$5b3o10b4o2b
o15b2o17bo41b2o2bo$6bobo14b2o14bo2bo13bo2bo41bo4bo$6bobo11bo21bo2b2o12b
o17bo5bo17bob4o$7b2o10bobo4b2o12bobo2b2o7bo3bo3b2o8b2o2bobo3bobo2b2o9b
o2bo$10b2o7b2o5bo27b4o4bobo7b2o13b2o8bobo3b2o$10bobo14b3o25b2o7bo11b3o
3b3o13bo5bo$12bo16bo34b2o38bobo$12b2o40b3o48b2o$54bobo$55b2o20b2o3b2o
$77b2o3b2o3$9b2o10b2o$8bo2bo8bo2bo$8b3o2b6o2b3o31bo37b2o$11b2o6b2o34b
3o34bo$10bo10bo36bo11bo21bo2bo$10b2obo4bob2o35b2o10bobo23bo$5bo9b2o24b
2o10b2o14bobo30b2o$5bo35bobo9b2o5bo7b2ob2o4b2o11b2o2bo6bo2bo$5bobo9bo
25bo15bobo5bo6bo3bo12b5o5bob2o$5bob2o2bo4bobo10b3o11b3o7bo4bo3bo4b2ob
2ob2o3bobo8bo5b2o5bob3o$7b2o2b3obo3bo9b3o9b2o3bo5bobo3bo3bo8bobo5b3ob
o5b2o5bo8bobo$11bo3bo3bo4b3o3bo9bo2b4o4bo3bo2bo3bo4b4o2bo7b2obo5b5o12b
o$3bo2b2o7bo3bo3b5obo10bobo8bo3bo3bobo5bo3b2o8bo2bo6bo2b2o11b2o$2bobo
2bo8bobo5b4o2bo8b2ob2ob2o4bo3bo4bo7b3o11b2o$2bo2b2o10bo9bob2obo6bo6bo
5bobo15bo19bo$b2o25bo2bobo7b2ob2o7bo5b2o9bobo17bo2bo$29b2o2bo8bobo14b
2o10b2o20bo8b3o$18b3o12b2o7bobo10b2o34b2o3b2o3b3o$13b2o2bo3bo2b2o17bo
11bo39b2obo$13b2obo5bob2o30b3o35bo3bo$17bo3bo36bo31bo3bob2o$18b3o68bo
bo3b2o4b3o$90bo11b3o2$16b2o3bo16bo37bo$17bo3b2o13b3o35b3o$14b3o4b2o12b
o37bo$14bo7bo12b2o26b2o8b2o$64bo$32b2o6b2o22bobo3b3o$32bobo4bobo23b2o
2bo3bo$33b2o4bo28bo5bo$37b3o28bo5bo$36bo27bo3bo5bo$27b2o7b2o26b2o3bo3b
o$27bobo33b2obo3b3o$29bo17b2o13bo2b3o$29b2o3b3o11bo12bobobo$33bo2bo8b
3o12bobobo$33b3o9bo6b2o4b3o2bo$26b2o15bobo6bo6bob2o9b2o$26bo12b2o2b2o
5bobo7b2o8b2o2bo$27b3o8bobo9b2o9bo5b4o2bo$29bo8bobo5b2o19b6o4bo$27b2o
9b2o6bobo27bobo$27bo19b2o28bo$28bo$27b2o11bo$34b2o4b3o28b2o$30b2obobo
7bo27b2obo$30bob2obobo4b2o31bo$36b2o34bo$73bob2o$75b2o!

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#C Smaller p16
x = 19, y = 28, rule = B3/S23
7b2o$7b2o3$7bo$5bobo$3bo2b2o$4b2o$4bo2$3b2o3bo$3b2o3bo$7b2o7b2o$16bo$
17bo$bo14b2o$2b2o11bo$2o6bo6b3o$2bo5bobo7bo$7bob2o6b2o$7b2o$2b2o10b2o
$bobo10bobo$bo14bo$2o8b2o4b2o$10bo$11b3o$13bo!

Above 16, it becomes much harder to find examples for some reason.

EDIT: P18 (EDIT 2: Nevermind, missing F16 EDIT 4: Fixed)

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x = 81, y = 63, rule = B3/S23
36bo4b2o$35bobo3bo19b2o$35bob3obo6bobo10bo3b2o$36bo3bo7b2obo10bo2bo$38b
o12bo3b2o4b2o3bo$37b3o6b4obob3obo2bo2b4o$40bo4bo4bobo4bo3bobo$35b3ob3o
4b2o2bo2b2obob4obob4o$33b2o3bo4b2obob3o2b2obo8bo2bo$31bo2b3o2bo3b2ob2o
3bo2bob2o2b2o$31b5obobo7bob2ob2obo3bobo$8bo27b2obob7o2bobo2b3o3bobo$5b
obobo16b2o3b4o2bob2o7b2o3b2o3b4ob3o$3b3obobo15bo2bobobo9b5o8bobo4bo3b
o$2bo3bobob2ob2o2b2o6bob2obo2b2o3b4o2bo2b2obo4bobo2bobo2b2o$2bo2bo2bo
2bob2o2b2o4b3obo2bo5bobo2bo5bob2o5bo3b2o$2ob2obobo2bo10bo6b2ob3obo$bo
bob2ob3o2b8o2b6o2b2o4bo$o2bobobo3bobo7b2o6bo2bo3bo5bo$2o2bo3bo2b2o2b2o
2bo3bob2obo4bo2bo3b3o9b2o2b2o$5bo2bo2bo3bo4bob2obobo2bo3bo4bo12bobo2b
o$4b3obob3obobo3bo4bo2b4o6bobo8b2obo2b2o3bo$3bobo2b2o2b2o3bo7bo7b2ob3o
bo7bo2bob2o2b6o$2bobo3bobo3b3o3bo2bob2o7bob4o8b2obo4bo6bo$3bo7b2obo2b
7obo7b2o16bo2b2ob3obo2bo$12bo2bobo7bobo3b2obo11b2o3bobo3bobob2obo$12b
ob2obo2b4o2bobobob4o2bo5b4o4b6obo3bo$11b2o2bobobo4b2o2bobo3bo3b2o3b2o
bo2bobobo4bobobo$13bo4bo3b2o2b2o2b7ob2o4b2o2b2obob2ob2o2b2o$13b2o4b2o
b3o3b2o3bo10b2o3bobobo3bobo$11b2o2bo6bo3b2o2b3o6b4o5bo2bo2b3o2bo$12bo
bo2b3o2b2o3bobo2bo5bo6bobo3bo4bobo$12bo2b2o2b3o5bob2ob2o4bobo2bobobob
ob2o4bo6bo$11b2obo7b5obo3bob3o4bobobob2obo2bo9bobo7b2o$10bo3b3o2b2o9b
2ob4ob3o2b2obo3bobobo7bobobo5bo2bo3b2o$11b3o3bo2b2o2b6obo4b3o2b2o3bo3b
o2bobo6bo3bo5bobobo2bo$13bobob2o12bobo4bobo2b3o3b2o4bo4b2ob3o5b2obo2b
obo$15b2o2b3ob4ob2ob2ob2ob2o2bo6bo6b2o2bobobo11bobob2o$11b4o3bo2b8obo
3b2o2b2obob3o2bo9bobo9b5o5bo$10bo4b2obo14bo7bobo2bob2o7b4o7b2o2bo2b2o
b2o$10b5o2bo13bo2bo3b2obobob2obo6b2o6bo3b2o2bo4bobo$8b2o6bo14bob4o6bo
bo2b2ob3o2bob2obobobo4bo5bobo$7bo2bobobob2o17b2ob2ob5o3bobo2bobo4bobo
b4obobo4bo$7b2obobobo2bo13b2o7bo4b2obo2bob2obob2obo2bo4bob2o$10bo2bob
2o8b2o5bo12b2o4bo3bobo2b2o3b3obo$10b2obo3bo6bobo5bo19b2obo2bo8bobo$12b
obob2o6bobo4bo22bob2o10bo$12bobobo7b2o10bo17bo$13bo4bo18bo15b2o$17b2o
16b3o$47b2o$47bobo$49bo$42bo5b2ob2o$43bo5bo2bo$41b3o2b3o2bo$45bo3b2o$
45b4o$48bo$45b2obobo$46bo2b2o$44bobo$44b2o!

P20

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x = 32, y = 29, rule = B3/S23
30b2o$14b2o6bo7bo$13bo6b3o5bobo$16bo2bo8b2o$14b2o3b2o$25b2o$24bobo$16b
3o5bo$16bobo6b4o$15bo2bo10bo$15bobo7b2ob2o$15b3o8bobo$7bo18bobo$6bobo
18bo$6bobo8b3o$5b2ob2o7bobo$5bo10bo2bo$6b4o6bobo$10bo5b3o$8bobo$8b2o$
14b2o3b2o$3bob2o8bo2bo$b3ob2o5b3o6bo$o11bo6b2o$b3ob2o$3bobo$3bobo$4bo
!

EDIT 3:

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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For the conjecture, yes, though it does not answer the question of what the smallest nontrivial oscillator which has that property at that period is.

[confocaloid]

I think you meant "feature in LifeViewer" (the software tool)?

I fail to see what kind of restriction is meant to be added by saying "true-period".

For guns, the definitions are that

• a true-period gun is such that the period of the output spaceship stream is equal to the period of the gun mechanism,
• a pseudo-period gun is such that the period of the output spaceship stream is lower than (and a divisor of) the period of the gun mechanism.

Now, unlike a gun, an oscillator doesn't have any "output". So the distinction doesn't appear to be meaningful for oscillators.

Two conjectures and one related question, based on the recent feature in Life which shows cell frequency:

Conjecture - there is a true-period nontrivial oscillator [...]

[confocaloid]

That p18 doesn't count as an example, because there is no cell with "frequency" 16:

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x = 81, y = 59, rule = B3/S23
36bo4b2o$35bobo3bo19b2o$35bob3obo6bobo10bo3b2o$36bo3bo7b2obo10bo2bo$38b
o12bo3b2o4b2o3bo$37b3o6b4obob3obo2bo2b4o$40bo4bo4bobo4bo3bobo$35b3ob3o
4b2o2bo2b2obob4obob4o$33b2o3bo4b2obob3o2b2obo8bo2bo$31bo2b3o2bo3b2ob2o
3bo2bob2o2b2o$31b5obobo7bob2ob2obo3bobo$8bo27b2obob7o2bobo2b3o3bobo$5b
obobo16b2o3b4o2bob2o7b2o3b2o3b4ob3o$3b3obobo15bo2bobobo9b5o8bobo4bo3b
o$2bo3bobob2ob2o2b2o6bob2obo2b2o3b4o2bo2b2obo4bobo2bobo2b2o$2bo2bo2bo
2bob2o2b2o4b3obo2bo5bobo2bo5bob2o5bo3b2o$2ob2obobo2bo10bo6b2ob3obo$bo
bob2ob3o2b8o2b6o2b2o4bo$o2bobobo3bobo7b2o6bo2bo3bo5bo$2o2bo3bo2b2o2b2o
2bo3bob2obo4bo2bo3b3o9b2o2b2o$5bo2bo2bo3bo4bob2obobo2bo3bo4bo12bobo2b
o$4b3obob3obobo3bo4bo2b4o6bobo8b2obo2b2o3bo$3bobo2b2o2b2o3bo7bo7b2ob3o
bo7bo2bob2o2b6o$2bobo3bobo3b3o3bo2bob2o7bob4o8b2obo4bo6bo$3bo7b2obo2b
7obo7b2o16bo2b2ob3obo2bo$12bo2bobo7bobo3b2obo11b2o3bobo3bobob2obo$12b
ob2obo2b4o2bobobob4o2bo5b4o4b6obo3bo$11b2o2bobobo4b2o2bobo3bo3b2o3b2o
bo2bobobo4bobobo$13bo4bo3b2o2b2o2b7ob2o4b2o2b2obob2ob2o2b2o$13b2o4b2o
b3o3b2o3bo10b2o3bobobo3bobo$11b2o2bo6bo3b2o2b3o6b4o5bo2bo2b3o2bo$12bo
bo2b3o2b2o3bobo2bo5bo6bobo3bo4bobo$12bo2b2o2b3o5bob2ob2o4bobo2bobobob
ob2o4bo6bo$11b2obo7b5obo3bob3o4bobobob2obo2bo9bobo7b2o$10bo3b3o2b2o9b
2ob4ob3o2b2obo3bobobo7bobobo5bo2bo3b2o$11b3o3bo2b2o2b6obo4b3o2b2o3bo3b
o2bobo6bo3bo5bobobo2bo$13bobob2o12bobo4bobo2b3o3b2o4bo4b2ob3o5b2obo2b
obo$15b2o2b3ob4ob2ob2ob2ob2o2bo6bo6b2o2bobobo11bobob2o$11b4o3bo2b8obo
3b2o2b2obob3o2bo9bobo9b5o5bo$10bo4b2obo14bo7bobo2bob2o7b4o7b2o2bo2b2o
b2o$10b5o2bo13bo2bo3b2obobob2obo6b2o6bo3b2o2bo4bobo$8b2o6bo14bob4o6bo
bo2b2ob3o2bob2obobobo4bo5bobo$7bo2bobobob2o17b2ob2ob5o3bobo2bobo4bobo
b4obobo4bo$7b2obobobo2bo13b2o7bo4b2obo2bob2obob2obo2bo4bob2o$10bo2bob
2o8b2o5bo12b2o4bo3bobo2b2o3b3obo$10b2obo3bo6bobo5bo19b2obo2bo8bobo$12b
obob2o6bobo4bo22bob2o10bo$12bobobo7b2o10bo17bo$13bo4bo18bo10bo4b2o$17b
2o16b3o8b3o$45bo$42b2o2bo$41bo2b3o$40bobo4b2o$40bobob2obo$38b3obobo2b
o$37bo3bobo2bo$37b2o2bo2b2o$40b2o!

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?

Code: Select all

x = 157, y = 81, rule = B3/S23
16b2o$16b2o4$31b2o$31b2o4$31bo14b2o$30b3o13b2o$13b2o3b2o9bo3bo$15b3o
10bob3obo$14bo3bo10b5o$15bobo43b2o$16bo44b2o2$61bo$60bobo$30bo2b2o11bo
13bobo13b2o$17b3o10bobo12b3o13bo14b2o$11bobo15b2o13b5o$11b2o4bobo8b2o
13b2o3b2o$12bo3b5o7b2obo12b5o9b2obob2o$15b2o3b2o7b3o12bo3bo9bo5bo26b2o
$15b2o3b2o23bobo11bo3bo27b2o$46bo13b3o12b3o$30b2o3b2o$3bob2o11bo14bo
41bobo$b3ob2o10bobo10bo5bo11bo25b5o27b2o$o16bobo2bobo6b2ob2o10b2ob2o
22b2o3b2o26b2o$b3ob2o11bo3b2o8bobo6bo31b2o3b2o$3bobo17bo9bo6bo4bo5bo$
3bobo27bo6b3o16bo45b3o$4bo40b2obob2o6bo17b2o27b3o13b2o$58b3o13b2ob2o
42b2o$74bo2bo10b2o3b2o$77bo12b3o$48b2o24bo14bo3bo9b2o3b2o$47bo3bo23b2o
13bobo11b5o27b2o$47bo3bo9b5o25bo13b3o13bo14b2o$48b2o10bob3obo39bo14bo$
51b3o7bo3bo9b2o3b2o38bobo$51b2o9b3o5bo5b5o38b2ob2o$63bo4b2o7b3o38bo5bo
26b2o$69b2o7bo13b3o26bo29b2o$86bobo29b2o3b2o$86b2o4bobo$87bo3b5o9bo43b
o$78bo11b2o3b2o6b2o15b4o12bo13bo$90b2o3b2o7b2o13b2o2bo11b3o12bo$80bo
38b2obo11b5o$80bo52b2o3b2o$93bo26b2o12b5o9b2o3b2o$92bob2o24bo13bo3bo
12bo$92bob3o8b2o3b2o23bobo10bo5bo$93b2o3bo8b3o26bo12b2ob2o$96b2o8bo3bo
9b2obob2o23bobo$97bo9bobo4bo5bo5bo24bo$108bo4bo7bo3bo12bo12bo$113b3o6b
3o11b2ob2o$131bo$130bo4bo5bo$130b3o$135b2obob2o$147bobo$147b2o$148bo3b
3o$151bo3bo$137bo12bo5bo$137bob2o9b2obob2o$138bo2bo2$140b2o11bo$152bob
o$152bobo$153bo2$153b2o$153b2o!

The following approach (with additional "attachments") should make it possible to solve all sufficiently high periods. As shown, it is a p798 oscillator with frequencies 0-28, 752, 755, 757, 759, 761, 762, 766, 774, 776-778, 780, 783-787, 790-798. It uses a glider eater by Kazyan and several other known glider eaters.

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

x = 116, y = 103, rule = B3/S23
75bo$73b3o$49bo22bo$49b3o20b2o$52bo$51b2o39b2o$92b2o5b2o$32bo66b2o$32b
3o41b2o$35bo41bo$18b2o14b2o11b2o28bobo17b2o$18b2o27b2o29b2o17b2o$103b
2o$103b2o$12b2o$12b2o$16b2o$16b2o15bo$33bobo$33b3o36b2o$35bo23b2o11b2o
$11b2o47bo$11b2o44b3o$57bo2$107bo$105b3o$47b2o55bo$47bo56b2o$48b3o$50b
o$13bo4bo$12bobo3bobo$12bobo3b2o$9b2obob3o$9bobo5bo91b2o$12b6o91bo$14b
o92bobo$16b2o23b2o64b2o$14bobo23bobo$12b3obobo21bo$11bo5b2o19b3o52b2o$
11b2o24bo3b2o49bobo$37b4o2bo48bo$41b3o47b2o$37b4o$37bo3b5o$23b2o7b2obo
bob2o4bo63b2o$14b2o7b2o7bob2obobo2bo66bobo$15bo20bo4b2o68bo$15bobo18bo
b3o61b2o7b2o$16b2o19bo2bo61b2o$38b2o2$34b2o$34bo$32bobo$32b2o$92b2o$
93bo$18b2o73bobo16bo$17bobo74b2o14b3o$17bo91bo$16b2o91b2o5$76bo$76b3o$
21b2o29bo26bo$22bo27b3o25b2o$19b3o27bo$19bo29b2o2$16bo52bo$14b3o50b3o
44b2o$13bo52bo47b2o$13b2o7bo30b2o11b2o$22bobo28b2o$22b2o$11bo97b2o$10b
obo96b2o$10b2o101b2o$113b2o$22b2o$22b2o$2o26b2o17b2o29b2o27b2o$2o26b2o
17bobo28b2o11b2o14b2o$49bo41bo$49b2o41b3o$6b2o9b2o7b2o66bo$6b2o9b2o7b
2o5b2o$33b2o39b2o$2b2o70bo$3bo49b2o20b3o$3o51bo22bo$o50b3o$51bo$17b2o$
8bo8b2o$7bobo$7b2o!

[LuveelVoom]

As N goes to infinity, does there exist a glider reflector that:
-fits in an NxN box
-has an input diagonal that passes within two cells of the lower left corner of the box and an output diagonal that passes within two cells of the lower right corner
-Has a density approaching 0
-The set of diagonals where a glider can safely "pass through" the NxN box, not interacting with the reflector assuming it's not active, remains finite

[confocaloid]

As N goes to infinity it becomes larger and larger and larger, but N always remains finite.
Hence in particular the set of glider lanes, such that a glider on the lane could interact with something inside the box (0 <= x < N, 0 <= y < N) remains finite.
Hence any subset of that set of glider lanes remains finite. It sounds like you could just use the Snark always (possibly replacing it with a fixed colour-changing glider reflector, if that ends up being a problem).

As N goes to infinity, does there exist a glider reflector that:
-fits in an NxN box
-has an input diagonal that passes within two cells of the lower left corner of the box and an output diagonal that passes within two cells of the lower right corner
-Has a density approaching 0
-The set of diagonals where a glider can safely "pass through" the NxN box, not interacting with the reflector assuming it's not active, remains finite

[LuveelVoom]

uh, whoops
Such that the number of lanes of interaction approaches a finite number as N goes to infinity