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## Thread for basic questions

For general discussion about Conway's Game of Life.

### Re: Thread for basic questions

Moosey wrote:
toroidalet wrote: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

Well, I just ran some crackpot numbers based on the current 16x16 methuselah results from Catagolue. If each 1K bin on average contains 5/9ths of the number of methuselahs in the previous bin, then if we ran all 2^256 possible 16x16 patterns and distributed them among 1K bins, the last bin that would be likely to have a methuselah in it would be the 329K bin.

Up to now we've found methuselahs that last as long as 47,575 ticks, but according to my very inaccurate math there should be one that lasts almost 330,000 ticks.

Slight changes in assumptions will make huge differences in the estimate of longest-lived 16x16, but it does seem pretty likely that the longest lifespan will be six digits long -- unless some lucky weird mechanism shows up that isn't like the random ash we've seen so far. Maybe there's something based on interacting natural glider guns, for example.

dvgrn
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### Re: Thread for basic questions

dvgrn wrote:-- unless some lucky weird mechanism shows up that isn't like the random ash we've seen so far. Maybe there's something based on interacting natural glider guns, for example.

Personally, my money would be on an ark that shoots backward gliders at debris, which, after a large number of ticks (possibly as many as millions, given the size of the search space), sends a glider or gliders forward into the engine, destroying it. (There are similar small symmetrical examples in Golly's pattern collection IIRC.)
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)

$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$

http://conwaylife.com/wiki/A_for_all

Aidan F. Pierce

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### Re: Thread for basic questions

A for awesome wrote:Personally, my money would be on an ark that shoots backward gliders at debris, which, after a large number of ticks (possibly as many as millions, given the size of the search space), sends a glider or gliders forward into the engine, destroying it.

Stopping both switch engines in an ark seems like a fairly tough trick shot, maybe more likely with multiple synchronized gliders (?). I'm not motivated enough to try to calculate the odds that that will happen starting from some 16x16 soup, but I suppose 2^256 is big enough that it's pretty likely to be out there somewhere.

If one switch engine gets away, then we're in a category of long-lived pattern that seems not to count, or at least apgsearch isn't designed to collect it as a methuselah.

Patterns that "go boring" with infinite growth don't seem to be getting added to the official Methuse List, even though now there's finally a non-symmetrical 16x16 soup that makes a long-lived infinite growth pattern.

So my basic question #1 is, do people think it would be okay to add a separate section to the Methuse List for neverending-population-growth long-lived patterns with small bounding boxes, say up to 32x32? Seems like there's some interesting stuff out there, that might be worth tracking as long as they don't get mixed up with regular methuselahs which are only neverending-bounding-box-growth.

Basic question #2: are there any Gottsian 32x32 starting patterns for which it's not even known if they ever go boring or not?

dvgrn
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### Re: Thread for basic questions

dvgrn wrote:
A for awesome wrote:Personally, my money would be on an ark that shoots backward gliders at debris, which, after a large number of ticks (possibly as many as millions, given the size of the search space), sends a glider or gliders forward into the engine, destroying it.

Stopping both switch engines in an ark seems like a fairly tough trick shot, maybe more likely with multiple synchronized gliders (?). I'm not motivated enough to try to calculate the odds that that will happen starting from some 16x16 soup, but I suppose 2^256 is big enough that it's pretty likely to be out there somewhere.

Couldn’t you just break one SE so that the other is destabilized?
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

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Moosey

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### Re: Thread for basic questions

Moosey wrote:Couldn’t you just break one SE so that the other is destabilized?

Theoretically, maybe. In practice that adds a whole lot of unlikeliness to the equation. It all depends on what switch engine pair we're talking about. The pair has to be firing backward gliders. Usually those arrangements protect themselves fairly well from gliders coming from behind. For example, see if you can find a forward glider that disrupts even one switch engine of the two in the 133,100-tick not-methuselah.

Run it in LifeHistory to check for targets. There are a few ticks where you could disrupt the backward-firing Herschel, where the reaction envelope just barely sticks out past the backward glider stream. On the opposite side there are no blue cells sticking out beyond the repetitive ash at all. So that's not a lot of options.

Looks like the most hopeful idea is to send a glider to kick back one of the sideways gliders, and cross your fingers that it will be possible to make some huge ugly chaotic explosion that can overtake both switch engines. But the kicked-back glider's collision point will be pretty far back in the ash trail; by the time the hypothetical chaos gets going, it will have an uphill battle spreading far enough to catch the switch engines. And there are only just so many possible kickbacks before the returning glider is just hitting settled ash -- no more new options.

What are the odds that the mess created by the backward glider stream will just happen to create a perfectly aligned forward glider that stops both switch engines? I can't estimate without surveying all switch-engine pairs that produce backward gliders, and trying all possible forward gliders against all of them to see if any are stoppable.

But it's going to be painfully low odds, because the kind of novelty produced by a single backward glider stream has a fairly low half-life: at any time it has good odds of building itself an eater, or drilling through the ash and going boring that way.

dvgrn
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### Re: Thread for basic questions

is there any way to reach the end result (with the unstable block) without creating the piece of junk?
x = 13, y = 14, rule = B3/S232o3b2o$o2bo2bo$b5o2$b3o$bo2bo$3bobo$4bobo$5b2o3$10b3o$10bo$11bo!
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googoIpIex

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### Re: Thread for basic questions

Does this work?
x = 16, y = 17, rule = B3/S232o3b2o6bo$o2bo2bo6bobo$b5o7b2o2$b3o$bo2bo$3b2o8$b3o$3bo$2bo!
I like making rules
fluffykitty

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### Re: Thread for basic questions

Almost, but not quite.

I'm looking for one which results in this pattern:
x = 7, y = 6, rule = B3/S232o3b2o$o2bo2bo$b5o2$b2o$b2o!
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googoIpIex

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### Re: Thread for basic questions

googoIpIex wrote:Almost, but not quite.

I'm looking for one which results in this pattern:
x = 7, y = 6, rule = B3/S232o3b2o$o2bo2bo$b5o2$b2o$b2o!

Can you use a recipe like the one for the tub-supported variant of the siamese-bookends object?

x = 21, y = 11, rule = B3/S237b2o9b2o$7b2o8bo2bo$17bo2bo$18b2o2$8b3o$16b3o$16bo$b2o14bo$obo$2bo! For example, there must be a way to build the following by adding the block at the top as a last step: x = 7, y = 7, rule = B3/S23b2o$b2o2$2bo$bobob2o$o2bob2o$b2o!

At T=2 this produces a great-grandfather of the siamese bookends, but if the single cell at the right can be continuously suppressed by ... whatever else you're adding (?) ... then it will settle into the right shape with no problem, and you have the maximum amount of space to do something with that part of the induction coil.

Of course if you're also making changes on the siamese-bookends side, this probably won't work. Seemed like it might be worth mentioning just in case.

dvgrn
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### Re: Thread for basic questions

googoIpIex wrote:Almost, but not quite.

I'm looking for one which results in this pattern:
x = 7, y = 6, rule = B3/S232o3b2o$o2bo2bo$b5o2$b2o$b2o!

Does this work?

x = 148, y = 46, rule = B3/S233bo$bobo$2b2o62b2o58b2o5bo$66b2o58b2o3b2o$132b2o3$3bobo$3b2o$4bo56b3o57b3o2$2bo$obo$b2o3$56bo88bobo$45bo11bo55bo31b2o$46bo8b3o55bo32bo$44b3o4b2o60bo$52b2o$51bo$106b2o38b2o$106b2o38b2o$82bo$80b2o$81b2o4b3o50bo$76b3o8bo19bo32bo$76bo11bo18b2o31bo$77bo28bobo3$11b2o$11bobo$11bo2$9bo60b3o57b3o$9b2o$8bobo3$120b2o$66b2o53b2o3b2o$10b2o54b2o52bo5b2o$10bobo$10bo! Goldtiger997 Posts: 531 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Thread for basic questions Goldtiger997 wrote:Anyway, I found two distinct ways of synthesizing the gun in 16 gliders... Did you try the 3G way of making that block+pi? x = 59, y = 87, rule = LifeHistory57.E$56.E$56.3E9$.E$2.2E5.E$.2E7.E34.E$8.3E32.2E$30.A.A11.2E$30.2A$31.A7$21.2D$21.2D4$22.2D17.A$21.D.D17.A.A$21.D19.2A$21.D.D$22.2D2$.A$2.2A$.2A6$33.2D.2D$.2D4.3D23.D3.D3.2D$.2D3.D3.D23.3D4.2D$6.2D.2D6$41.2A$40.2A$42.A2$20.2D$20.D.D$.2A19.D$A.A17.D.D$2.A17.2D4$21.2D$21.2D3$2.E$2.2E$.E.E2$12.A$12.2A18.3E$11.A.A18.E$33.E4$3.E$3.2E$2.E.E3$48.2E$48.E.E$48.E! A quicker 4G way of making the pi+block would be needed in the other two locations. I checked the three 2G pi collisions, x = 218, y = 66, rule = LifeHistory25.A$25.A.A$25.2A$83.A$84.A$82.3A2$157.A$155.A.A$156.2A17$2D$D.D$2.D$D.D$2D$112.2D$112.D.D$114.D74.2D$112.D.D74.D.D$112.2D77.D$189.D.D$189.2D20$29.2A$28.2A$30.A$78.3A$80.A$79.A$215.3A$215.A$216.A!

and the two methods you used in your 16-glider synthesis (yellow gliders, first pattern), and there were always conflicts with incoming gliders one way or another. So maybe this 3G option (green gliders, first pattern) isn't as hopeful as it looks.

dvgrn
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### Re: Thread for basic questions

Does this help?
x = 29, y = 37, rule = LifeHistory.C$2BC$3CB$.4B$2.4B$3.4B$4.4B$5.4B$6.4B4.2B$7.4B2.4B$8.4B.4B$9.10B$10.9B2.BC$9.11B2CB$8.13B2C$8.9BC5B$9.9BC4B$10.6B3C5B$10.15B$9.16B$9.16B$9.16B$10.15B$10.14B$15.11B$16.11B$16.12B$17.11B$16.13B$17.8BDBDB$17.8BDBD$17.2B2D4B3D$18.B2D6B$18.8B$19.6B$20.4B$21.B!

Macbi

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### Re: Thread for basic questions

Moosey wrote:
testitemqlstudop wrote:With the stdin symmetry (compile with --symmetry stdin)

And then what? How would I upload a spider or something as a soup?

(I didn’t ask the original question, but I’m curious now.)

When running apgluxe, paste in the RLE for spider.

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### Re: Thread for basic questions

Yeah it does help:
x = 64, y = 80, rule = B3/S2361bobo$61b2o$62bo14$34bo11bobo$34bobo9b2o$34b2o11bo3$49b3o$49bo$50bo7$46bo$44b2o$45b2o4$18bo$19bo$17b3o3$44b3o$44bo$45bo4$17b2o$18b2o$17bo7$13bo$14bo$12b3o3$16bo11b2o$16b2o9bobo$15bobo11bo14$bo$b2o\$obo!
Is there a way to submit to the catalogue in such a way that the two cleanup gliders are separate?

Macbi

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### Re: Thread for basic questions

Macbi wrote:Is there a way to submit to the catalogue in such a way that the two cleanup gliders are separate?

I don't think the synthesis submission on Catagolue supports linear growth, but I may be mistaken
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wildmyron

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### Re: Thread for basic questions

After this discovery by nick gotts, I’ve started to wonder:
Why don’t we have bunnies of higher order? Why isn’t there a bunnies 12? Why does it only go to bunnies 11?

What is the longest lived 12-cell pattern known?

Obviously it lives at least as long as bunnies 11, since you can add a dot to it so that it evolves exactly the same— so that throws out a lot of possibilities.
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My CA rules can be found here

Also, the tree game
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Moosey

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### Re: Thread for basic questions

Does 3D.lua support custom rules, and if so, under what limitations and how would one go about making them?
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### Re: Thread for basic questions

BlinkerSpawn wrote:Does 3D.lua support custom rules ... ?

No, but you could always write a variant of 3D.lua that supported whatever rules you wanted. Best to start with the 3D.lua script included in Golly 3.2 rather than the highly optimized version in 3.3.

Andrew
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### Re: Thread for basic questions

This is probably a really stupid question, but I'm pretty new to 1D automata. can someone explain to me why W90 doesn't work?
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