Thread for basic questions

[dvgrn]

How would I find Elbow operations to use with an overclocked silver reflector?

Now that's an irresistible question!

First figure out what the range of allowable following times is, for gliders going through a Silver reflector.

Then decide whether you want just one elbow type, or if you want to also collect elbow ops that switch between different types of elbows (blocks in different positions, honeyfarms in different positions, etc.)

Then maybe adapt some of simeks' search code, or chris_c's elbow0.py (?) to find elbow operations that are compatible with a Silver reflector. They're probably out there somewhere, but they might be kind of long.

[googoIpIex]

A (suboptimal) self destruct for a silver reflector when it is done overclocking:

Code: Select all

x = 116, y = 120, rule = B3/S23
60bo$59bobo$60b2o$75bo$74bobo$74b2o11$76bo$75bobo$75b2o2$60bo$59bobo$
60b2o4$92b2o$92bobo$93bo3$88b2o$88b2o2$43bo$42bobo45b2o21b2o$41bo2bo9b
2o34bo22bobo$42b2o10b2o32bobo23bo$46bo37b2o2b2o$20bo9bo15b3o35b2o$20b
3o5b3o18bo$23bo3bo20b2o11b2o$22b2o3b2o32b2o6$27b2o38b2o$27bobo37b2o$
27bo8b2o52b2o$36b2o34b2o16bobo$24b2o45bobo18bo5b2o$23bo2bo44bo20b2o4b
2o$18b2o4b2o44b2o4b2o$17bobo55bobo$17bo57bo$16b2o56b2o7b2o$20b2o4b2o
33b2o20b2o$19bo2bo3bo34bo$19bobo5b3o32b3o11b2o$bo18bo8bo34bo10bobo$obo
73bo$b2o58b2o$60bo2bo$12b2o46bo2bo$11bobo47b2o$12bo$75b2o$74bobo$75bo
2$73bo$73b3o$76bo$75b2o2$70b2o$49bo20b2o$48bobo$49bobo$50b2o4$95b2o$
55b2o38b2o$54bobo$55bo$41b2o$41b2o5$80b2o$80b2o5$62b2o$62bobo$63bo13$
62bo$61bobo$62bobo2b2o$63b2o2b2o!

[dvgrn]

A (suboptimal) self destruct for a silver reflector when it is done overclocking...

Looks good! Something like that should work fine. Enough runs of simeks' GoL-destroy would cut down the cost a little bit, but this is probably good enough.

You might end up wanting to over-overclock your Silver reflectors, though -- have three signals being processed at once, instead of just two. That would make startup a little trickier, but shutdown will still be able to use your same self-destruct mechanism:

Code: Select all

x = 177, y = 158, rule = B3/S23
obo$b2o$bo33$121bo$120bobo$121b2o$136bo$135bobo$135b2o6$48bo$46bobo$
47b2o3$137bo$136bobo$136b2o2$121bo$120bobo$121b2o4$153b2o$153bobo$154b
o3$149b2o$149b2o2$104bo$103bobo45b2o21b2o$102bo2bo9b2o34bo22bobo$103b
2o10b2o32bobo23bo$107bo37b2o2b2o$81bo9bo15b3o35b2o$81b3o5b3o18bo$84bo
3bo20b2o11b2o$83b2o3b2o32b2o5$84bo$85bo42b2o$83b3o42b2o$97b2o52b2o$97b
2o34b2o16bobo$85b2o45bobo18bo5b2o$84bo2bo44bo20b2o4b2o$79b2o4b2o44b2o
4b2o$78bobo55bobo$78bo57bo$77b2o56b2o7b2o$81b2o4b2o33b2o20b2o$80bo2bo
3bo34bo$80bobo5b3o32b3o11b2o$62bo18bo8bo34bo10bobo$61bobo73bo$62b2o58b
2o$121bo2bo$73b2o46bo2bo$72bobo47b2o$73bo$136b2o$135bobo$110b2o24bo$
109b2o$111bo22bo$134b3o$137bo$136b2o2$131b2o$110bo20b2o$109bobo$110bob
o$111b2o4$156b2o$116b2o38b2o$115bobo$116bo$102b2o$102b2o5$141b2o$141b
2o5$123b2o$123bobo$124bo13$123bo$122bobo$123bobo2b2o$124b2o2b2o$160b2o
$159b2o$161bo!

Then again, single-channel searches so far have almost all been limited to pairs of synchronized gliders, with longer spaces between them. Maybe that will be enough to find a universal set of recipes here, too. If you switch to triplets, you'll get less adjustability for each glider in the triplet -- but the odds are better that you'll be able to influence the active reaction before it has settled down. Not sure how that will balance out in practice.

[googoIpIex]

As for the repeat time for Overclocked Silver reflectors, the periods which work are:

248-291 work.
329-405 also work.
497- ∞ work too.

[dvgrn]

As for the repeat time for Overclocked Silver reflectors, the periods which work are:

248-291 work.
329-405 also work.
497- ∞ work too.

Periods don't matter so much in this context, though. If you're supporting a regular stream of overclocked gliders, you can't do single-channel construction with it -- and you can't leave out some gliders to make the zeroes in a binary sequence, because that will make the overclocking fail and the circuit will explode. Once the gliders start, they'll have to keep coming... but they can come at different times, and hopefully that variation will be enough to allow for a universal set of recipes. No way to know for sure without doing the search.

So for example, here's a pair of gliders coming in with a separation of 100. This is possible if the previous gliders were timed so that a reset glider comes in at just the right time between the T=0 and T=100 gliders.

Code: Select all

x = 177, y = 163, rule = B3/S23
obo$b2o$bo33$121bo$120bobo$121b2o$136bo$135bobo$135b2o11$137bo$136bobo
$136b2o2$121bo$120bobo$59bo61b2o$60bo$58b3o2$153b2o$153bobo$154bo3$
149b2o$149b2o2$104bo$103bobo45b2o21b2o$102bo2bo9b2o34bo22bobo$103b2o
10b2o32bobo23bo$107bo37b2o2b2o$81bo9bo15b3o35b2o$81b3o5b3o18bo$84bo3bo
20b2o11b2o$83b2o3b2o32b2o5$84bo$85bo42b2o$83b3o42b2o$97b2o52b2o$97b2o
34b2o16bobo$85b2o45bobo18bo5b2o$84bo2bo44bo20b2o4b2o$79b2o4b2o44b2o4b
2o$78bobo55bobo$78bo57bo$77b2o56b2o7b2o$81b2o4b2o33b2o20b2o$80bo2bo3bo
34bo$80bobo5b3o32b3o11b2o$62bo18bo8bo34bo10bobo$61bobo73bo$62b2o58b2o$
121bo2bo$73b2o46bo2bo$72bobo47b2o$73bo$136b2o$135bobo$110b2o24bo$109b
2o$111bo22bo$134b3o$137bo$136b2o2$131b2o$110bo20b2o$109bobo$110bobo$
111b2o4$156b2o$116b2o38b2o$115bobo$116bo$102b2o$102b2o5$141b2o$141b2o
5$123b2o$123bobo$124bo13$123bo$122bobo$123bobo2b2o$124b2o2b2o6$165b2o$
164b2o$166bo!

That requirement that a reset glider shows up at the right time could be pretty difficult to handle in a search, unfortunately. In extreme cases like this 100-tick separation, it might mean that some valid recipes couldn't follow some other valid recipes, because the cleanup gliders can't be timed to fit in between the incoming gliders. That's not a wrinkle that we've had to work with before.

I think it's only a problem for double- or triple-overclocked Silver reflectors, though. Maybe the thing to try first is a search with pairs of gliders going into a regular single-overclocked reflector -- find out if there's a universal elbow-op toolkit just with those. If so, it should always be possible to delay the next glider pair to a time when the reset gliders that are already in the system won't get in the way.

[googoIpIex]

Somewhat related: is there any one-time glider duplicator in which a glider exits along the same lane with the same timing as if it was unaffected?

[dvgrn]

Somewhat related: is there any one-time glider duplicator in which a glider exits along the same lane with the same timing as if it was unaffected?

Yes, the ones Guam found certainly qualify. The symmetrical one was used in calcyman's stable pseudo-Heisenburp, but the other one is only three still lifes (though it's a bit messier to clean up).

A slightly larger search along the lines of simsim314's would probably turn up a clean 3sL splitter with exact timing, or 4sL at the most -- it's just nobody has done that search yet, or organized the results into a good lookup table.

[googoIpIex]

Would a diagonal caterloopilar be possible?

[dvgrn]

Would a diagonal caterloopilar be possible?

The question has been asked a few times.

Technically the answer is certainly "yes", but it might not be as nice as the orthogonal version. If you try to build the exactly analogous design at 45 degrees, basically replacing gliders with spaceships and spaceships with gliders, you quickly discover that gliders can't fly past stable objects and spark them into action with Heisenburp effects. So you can't have re-usable "read" salvos made of gliders, like the re-usable *WSS salvos in the Caterloopillars.

So instead of glider salvos you might end up using 2-engine Cordership salvos, or something else equally horrible. Corderships would cut the limit speed to c/12, and suppressing them at the front and back ends might be just as big a pain as constructing them. But most likely all the technical problems are perfectly solvable, and you'd end up with something that's recognizable as a diagonal Caterloopillar.

The more general solution simsim314 gave in the link above theoretically allows travel diagonally or in any rational oblique direction. But that's a radical enough redesign that I'm not sure it would really count as a "Caterloopillar" any more.

[Ian07]

Probably a dumb question but how do you synthesize these two LWSSes?

Code: Select all

x = 5, y = 14, rule = B3/S23
2bo$b3o$2obo$3o$b2o5$b3o$bo2bo$bo$bo$2bobo!

No matter what I try, there's always gliders crossing paths:

Code: Select all

x = 16, y = 22, rule = B3/S23
obo$b2o$bo2$13b2o$13bobo$13bo2$5bo$5b2o$4bobo5$7b2o$8b2o$7bo5bo$13bobo
$9b2o2b2o$9bobo$9bo!

(trying to add the Ecologist synthesis from Niemiec's database to Catagolue)

[dvgrn]

Probably a dumb question but how do you synthesize these two LWSSes?
...
No matter what I try, there's always gliders crossing paths.

I'm not sure there's a way to do those two with six gliders, but two copies of one of the standard one-sided syntheses will certainly work --

Code: Select all

x = 31, y = 39, rule = B3/S23
26bo$25bo$25b3o8$3bobo$4b2o$4bo3$15b2o$15b2o8$14b2o$14b2o2$29bo$28b2o$
28bobo7$b2o$obo$2bo!

-- and probably there's something with seven gliders. I haven't hunted really hard for six-glider options; this was just a check that there's no problem at 8G.

EDIT: Oops, well, there's no problem at 6G either:

Code: Select all

x = 48, y = 62, rule = B3/S23
o$b2o$2o31$41bo$40bo$40b3o3$45b3o$45bo$46bo2$4b2o$5b2o$4bo10$12bo$12b
2o$11bobo3$36b3o$36bo$37bo!

Start with a simultaneous construction (which doesn't work because of signal crossings _and_ interference between construction envelopes):

Code: Select all

x = 13, y = 16, rule = B3/S23
7b2o$7bobo$obo4bo$b2o$bo3bo$5b2o$4bobo3$4b2o$5b2o$4bo5bo$10bobo$6b2o2b
2o$6bobo$6bo!

Every time you move all the gliders in a recipe backward by one cell, you have to move the construction site upwards by two cells (the LWSS moves at c/2 and is being constructed 4 ticks later).

You can't move the lower gliders backward, because that just makes the interference between construction envelopes worse. So instead you move the three upper gliders backwards, until the columns containing upper crossing gliders aren't within two cells horizontally of the corresponding lower crossing gliders. (You can get a tick closer than that rule allows, but 2 cells of clearance is safe if the crossing gliders are mirror images, as they are in this case.)

[Ian07]

I haven't hunted really hard for six-glider options; this was just a check that there's no problem at 8G.

Okay, I think this is valid:

Code: Select all

x = 98, y = 95, rule = B3/S23
o$b2o$2o46$24bo$25bo$23b3o20$74bobo$29bobo42b2o$30b2o43bo$30bo15$95b3o
$95bo$96bo2$4b2o$5b2o$4bo!

[dvgrn]

Okay, I think this is valid...

Sure, why wouldn't it be? Any time the four salvos can be separated by drawing a horizontal line and a vertical line like this, rewindability is guaranteed.

If you want to tighten something like this up, leave two blank columns around one of the crossing gliders. So you can't move the glider from the upper recipe into this red zone:

Code: Select all

x = 98, y = 97, rule = LifeHistory
A$.2A$2A46$24.A$25.A$23.3A$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$
21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.
7D$21.7D46.A.A$21.7D.A.A42.2A$21.7D2.2A43.A$21.7D2.A$21.7D$21.7D$21.
7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$21.7D$
21.7D67.3A$21.7D67.A$21.7D68.A$21.7D$4.2A15.7D$5.2A14B7D$4.A16.7D$21.
7D$21.7D!
#C [[ THUMBNAIL THUMBSIZE 2 ]]

The blue line is the length you can fast-forward the crossing glider without getting into trouble. The blue line is 14 cells long, so move the three gliders in the recipe forward by 14 cells diagonally, then move them all down 28 cells so the LWSS ends up in the same place at the same time.

That turns out not to quite work because the lower *WSS construction is so slow, so you have to move the construction site up a bit. Move the upper recipe forward by 13 cells, then down by 26 cells, and that works fine:

Code: Select all

x = 72, y = 69, rule = LifeHistory
A$.2A$2A7$11.A$12.A$10.3A20$61.A.A$16.A.A42.2A$17.2A43.A$17.A28$69.3A
$69.A$70.A2$4.2A$5.2A$4.A!

See also the six-glider option I edited in to my last message above.

[Freywa]

Are these the smallest strict still lifes by cell count that have the corresponding symmetry types?

Code: Select all

x = 160, y = 34, rule = B3/S23
62bo4b2o5b2o$62b3o2bo4b3obo$65bobo3bo4bo$3obo35b3obobo5bo11bobo4b2o3b
2o6b2o2b3o40b2o2bobo9bo$o3bo19b2o14bo3bobo5bo10bobo6bo4bo6bobo3bo20b2o
18bobobobo9bo9b2o$o3bo18bobo14bo3b3o5bo10bo2b3o3bob3o7bobob3o5bobo11bo
bo18bobob3o5bobobo8bobo$o3bo18bo16bo5bo5bo9b2o4bo4b2o9bobobo8bo13bo19b
obo3bo6bo2bo8b2o$3obo17b2o16b3o3bob3obo31b2o2b3ob3obobo32b2o4bob3obobo
bo3$64b2o$3ob3o5bo27b3obobo5bobo9b2o18b2o2bobo9bo13bo16b2o2bobo9bobo$o
5bo5bo12b2o13bo3bobo5bobo29bobobobo6bo2bo11b5o14bobobobo9bobo8bo$o3b3o
5bo9bobobo13bo3b3o5b3o9b4ob2o13bobob3o5b3obo10bo5bo13bobob3o5bobob3o7b
obo$o3bo7bo9b2o16bo5bo7bo9bo2bob2o13bobo3bo6bo2bo11b5o14bobo3bo6bo4bo
6bobo$3ob3ob3obo27b3o3bob3o3bo6b2obo2bo16b2o4bob3o5bo13bo16b2o4bob3obo
bo3bo7bo$61b2ob4o2$66b2o$3ob3o5b3o25b2o2b3o9bo9b2o16b2o2bobo9b3o28b2o
2b3o5bo$o5bo7bo9b2o14bobo3bo6bo2bo14bo12bobobobo6bo4bo9b2o17bobobobo5b
o13bo$o3b3o5b3o7bo2bo14bobob3o5b3obo13bobo11bobob3o5b3ob3o8bo2bo16bobo
b3o5bo12bobo$o3bo7bo9b2o16bobobo8bo2bo13bobo11bobo3bo6bo2bo11b2o17bobo
bobo5bo13bo$3ob3ob3ob3o25b2o2b3ob3o5bo12b2ob2o10b2o4bob3o5b3o28b2o2b3o
b3obo3$116bo2bo$3ob3o5bobo25b2o2b3o9b3o25b2o2bobo9bobo13b4o11b2o2b3o5b
obo$o5bo5bobo7bob2o14bobo3bo6bo4bo10b2ob2o10bobobobo6bo2bobo11b2o4b2o
9bobobobo5bobo11b2o$o3b3o5b3o7b2obo14bobob3o5b3ob3o11bobobo9bobob3o5b
3ob3o10bo2bo2bo2bo8bobob3o5b3o11b2o$o3bo9bo25bobobo8bo2bo13bobobo9bobo
3bo6bo4bo10bo2bo2bo2bo8bobobobo7bo$3ob3ob3o3bo25b2o2b3ob3o5b3o10b2ob2o
10b2o4bob3o7bo11b2o4b2o9b2o2b3ob3o3bo$116b4o$116bo2bo!

[dvgrn]

Are these the smallest strict still lifes by cell count that have the corresponding symmetry types?

I can't seem to dig up a counterexample to any of those.

The closest I could come was with D2_+2. You have a couple of tied-for-first still lifes for C4_1, but only one of the two (EDIT: three?) 14-bit D2_+2s:

Code: Select all

x = 3, y = 10, rule = B3/S23
2o$o$2bo$b2o$o$o$b2o$2bo$o$2o!

[praosylen]

The closest I could come was with D2_+2. You have a couple of tied-for-first still lifes for C4_1, but only one of the two 14-bit D2_+2s:

Code: Select all

x = 3, y = 10, rule = B3/S23
2o$o$2bo$b2o$o$o$b2o$2bo$o$2o!

Also,

Code: Select all

x = 3, y = 10, rule = B3/S23
b2o$bo$2bo$b2o$o$o$b2o$2bo$bo$b2o!

[googoIpIex]

How would I install golly 2.7? I can't find any way to install it.

[fluffykitty]

https://sourceforge.net/projects/golly/ ... golly-2.7/

[Ian07]

What is the exact definition of a shuttle? I had always been under the impression that a shuttle had to use the same hassling reaction on both sides, (and once moved the p60 B-heptomino hassler article based on that) but some of the oscillators shown in the Pre-pulsar shuttle oscillators article aren't symmetrical in that regard.

[Moosey]

What is the exact definition of a shuttle? I had always been under the impression that a shuttle had to use the same hassling reaction on both sides, (and once moved the p60 B-heptomino hassler article based on that) but some of the oscillators shown in the Pre-pulsar shuttle oscillators article aren't symmetrical in that regard.

I beilieve it’s a hassler which reflects object X with reaction 1 and then does so again with reaction 2. Each displaces or reflects it the same amount, but they can take different amounts of time.
I guess it’s like
This is still a shuttle as far is the highlighted QB is concerned— it doesn’t matter that it’s being stabilized by different Junk on each side

Code: Select all

x = 41, y = 15, rule = LifeHistory
10.A17.C$9.A.A14.C.C$A7.A.2A13.C.C11.2A$3A4.2A.2A12.C2.C11.2A$3.A4.A.
2A13.C.C$2.A2.A3.A.A14.C.C$2.4A4.A17.C$21.3A$2.4A4.A17.A$2.A2.A3.A.A
14.A.A$3.A4.A.2A13.A.A$3A4.2A.2A12.A2.A11.2A$A7.A.2A13.A.A11.2A$9.A.A
14.A.A$10.A17.A!

[computer_kid]

Does anybody know where to find a list with the longest-lived methuselah possible in a particular bounding box (4x4, 16x16 or so)? thx

[dvgrn]

Does anybody know where to find a list with the longest-lived methuselah possible in a particular bounding box (4x4, 16x16 or so)?

Try the LifeWiki's List of long-lived methuselahs for starters.

This will give the longest-lived known methuselahs. For the very small bounding boxes that have been searched exhaustively, this is the same as the longest-lived possible methuselah.

For bigger bounding boxes nobody knows. For 16x16, it's 99 point lots of nines certain that the absolute longest-lived methuselah hasn't been found yet. It's hard to even guess how high the lifespans might go.

[Moosey]

Does anybody know where to find a list with the longest-lived methuselah possible in a particular bounding box (4x4, 16x16 or so)?

Try the LifeWiki's List of long-lived methuselahs for starters.

This will give the longest-lived known methuselahs. For the very small bounding boxes that have been searched exhaustively, this is the same as the longest-lived possible methuselah.

For bigger bounding boxes nobody knows. For 16x16, it's 99 point lots of nines certain that the absolute longest-lived methuselah hasn't been found yet. It's hard to even guess how high the lifespans might go.

At some point, it probably is approximately some function of the area of cells— I’d imagine for large enough areas each additional cell possible probably increases the maximum lifespan of the longest-lived possible methuselah in some way that exceeds linear but does not exceed exponential growth, but that’s my bad intuition. Technically, given current patterns that take an extremely long amount of time stabilizing, it will actually be something in the area of exponentialish to something along the lines of c^ⁿx where x is the number of additional cells, N is the number of up arrows, and c is a constant that is being up-arrow’d.

Speculation:
For a large amount of cells our methuselah lifespan function probably has a position in the fast growing hierarchy that is at most about ω, i.e. is at most f(x) = c^ˣx
(Fine, c^ˣ⁻¹x)

Though theoretically I suppose you could get a very large object to spend time attempting to find TREE(3) by trial and error or something and keep going until it does all the games, then add a relatively tiny amount of cells to get it to do TREE(4), in which case you would have a lifespan function which grows at least as fast as TREE(n), I.E. the funtion’s position in the hierarchy would be at least ϑ (Ω (superscript ω) ω)

That curlicue (ϑ) is the Theta symbol in Unicode by the way

[toroidalet]

Actually, the lifespan of bounded patterns is the busy beaver function of one of its sides (provided both sides are long enough; the other side can be constant), since there are finite Turing machines.

[Moosey]

Actually, the lifespan of bounded patterns is the busy beaver function of one of its sides (provided both sides are long enough; the other side can be constant), since there are finite Turing machines.

What would the answer be if Turing machines are not allowed? Just, hypothetically