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Challenges

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Re: Challenges

Postby gmc_nxtman » February 1st, 2018, 7:13 pm

x = 37, y = 6, rule = B3/S23
4b2o13b2o13b2o$3bo2bo11bo2bo11bo2bo$3bob2o11bob2o11bob2o$2obo12b2obo
12b2obo$o2bo12bo2bo12bo2bo$b2o14b2o14b2o!
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Re: Challenges

Postby vyznev » February 1st, 2018, 10:29 pm

A for awesome wrote:[nevermind]2. Find a non-explosive B1e rule.[/nevermind]
This one has been solved by @gmc_nxtman. Feel free to come up with other examples, though.

If you allow rules with B0, this is easy:

x = 20, y = 20, rule = B01/S
3obob3o2bo2b2obobo$b2o7b3o2b2o2bo$3o2bo2b3ob5obo$2bobob2obob3ob3o$ob4o
3b4o3b3o$2ob2o2bobo3b2obo$4bo2bo5bo3bo$o3b3obo2bo7bo$3obobo2b2ob4o$2bo
b2o3bobo2b2o3bo$b2ob3o2b4obo2bobo$b2obo2b3o7b3o$2bobo2bob3o2b3obo$ob2o
3b4o2bobob2o$o2bo3b3o2bo2b2obo$2o2b2obob6obob2o$3bobo2b3obob3o2bo$3o5b
5ob6o$o4bo4b2ob2obobo$5obob2o2bo2b4o!


There are even non-explosive B01e rules with natural spaceships arising from soups; a few minutes of playing with Golly turned up B01e2/S4aceikn as an example:

x = 100, y = 100, rule = B01e2/S4aceikn
3o5bobo4bobo7b5ob2ob11ob4o6b3ob2ob2ob2o5bo5bobobob8o4bobo$99o$b99o$
100o$100o$100o$b98o$b98o$99o$99o$b98o$99o$b99o$100o$100o$b99o$100o$b
99o$100o$100o$b99o$b99o$b98o$b98o$b98o$b99o$100o$99o$b98o$100o$100o$
100o$b98o$99o$99o$100o$b98o$99o$99o$99o$99o$b98o$b98o$b98o$b99o$100o$b
99o$100o$b98o$b99o$b98o$99o$b98o$b99o$b98o$b99o$99o$100o$b99o$b98o$b
99o$b99o$b98o$99o$100o$b99o$99o$100o$99o$99o$99o$b98o$b98o$100o$b99o$
100o$b99o$b99o$100o$99o$99o$99o$100o$100o$b99o$b98o$b98o$100o$b98o$99o
$b98o$b98o$b98o$100o$b98o$99o$100o$99o$99o$5ob5o4bo2b3obo3b2o2b3obo3b
2ob4o2b3o2bobo2b3o4b2ob6ob2o3bo3bo3bob2o2b6o!


dvgrn wrote:
Saka wrote:Is there a slowest possible spaceship for n cells?

Well, yes, for small N these do exist, of course. But there's not ever going to be a discoverable algorithm where you input N and get back the relevant slowest possible spaceship.

Well, I wouldn't quite say "never". Suppose somebody found a "nanoid", i.e. a reaction between a glider (or some other small spaceship) and a smallish still life or oscillator that reflected the glider 180°, with the glider and the still life / oscillator both translated the same number of cells at a 90° angle to the glider's trajectory. By having the glider bounce between two such oscillators, one could produce a family of arbitrarily slow spaceships with a very small population count; possibly even small enough that someone could enumerate the slowest possible spaceships for all lower populations.

Of course, I won't be holding my breath waiting for someone to do that for GoL. But it might be possible for some other isotropic CA rule, if the smallest arbitrarily-slow spaceship in that rule was small enough.
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Re: Challenges

Postby A for awesome » February 1st, 2018, 10:34 pm

x = 36, y = 11, rule = B3/S23
3bo4bo4bo12b2o$2bobo2bobo2bobo12bo$bobo3bobo3bobo10bo$2bo5bo5bo11b2o$
30b2o$29bobo$29b2o$2b2o4b2o4b2o$bo2bo2bo2bo2bo2bo15bob2o$o2bo3bo2bo3bo
2bo14b2obo$b2o5b2o5b2o!
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

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Re: Challenges

Postby danny » February 21st, 2018, 3:09 am

Challenge: Find a rule as close to CGOL that
a) retains the glider
b) has an overunity reaction from only 2G's(meaning creates 3 or more gliders)
Last edited by danny on February 21st, 2018, 10:10 am, edited 1 time in total.
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Re: Challenges

Postby 77topaz » February 21st, 2018, 3:32 am

danny wrote:Challenge: Find a rule as close to CGOL that
a) retains the glider
b) has an overunity reaction from only 2G's


Just to clarify, by "overunity reaction" you mean a reaction that creates more than two gliders and nothing else?
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Re: Challenges

Postby fluffykitty » March 9th, 2018, 7:38 pm

What's close? Fewest totalistic transitions changed? Fewest isotropic transitions changed? Fewest MAP bits changed?
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Re: Challenges

Postby Majestas32 » March 9th, 2018, 7:44 pm

I assume isotropic
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Re: Challenges

Postby 77topaz » March 9th, 2018, 7:47 pm

Yeah, I think isotropic, because barely anyone knows how to use MAP strings properly, anyway. :P
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Re: Challenges

Postby Majestas32 » March 9th, 2018, 9:09 pm

mep
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Re: Challenges

Postby fluffykitty » March 9th, 2018, 11:01 pm

*MAP
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Re: Challenges

Postby dvgrn » March 9th, 2018, 11:55 pm

77topaz wrote:
danny wrote:Challenge: Find a rule as close to CGOL that
a) retains the glider
b) has an overunity reaction from only 2G's


Just to clarify, by "overunity reaction" you mean a reaction that creates more than two gliders and nothing else?

"Over-unity" by itself doesn't necessarily imply a completely clean reaction -- at least according to the current Lexicon definition, but I think I rewrote that myself recently so it might be wrong.

If the reaction doesn't have to be clean, then here's a dirty 2G->4G that I noticed by running the random-change script on the two-glider collision stamp collection for a few minutes:

x = 5, y = 12, rule = B34t/S23
3bo$2bo$2b3o7$bo$2o$obo!
#C [[ ZOOM 6 AUTOSTART PAUSE 2 ]]

Probably not good enough, since at minimum both extra gliders would probably have to be used to clean up the junk to make a repeatable reaction (assuming we had reflectors for this rule)... but technically it might match the terms of the challenge. Or it's a start anyway, and should at least have the effect of getting the terms clarified.

Modified stamp collection:

x = 844, y = 376, rule = B3/S23
294bo56bo28bo$148bo28bo29bo26bo29bo28bo29bo26bo28bo56bo$147bo28bo29bo
26bo29bo29b3o26bo27b3o26b3o25bo27bo$147b3o26b3o27b3o24b3o27b3o56b3o81b
o28b3o$406b3o4$350bo$148b3o27b3o139bo28b2o85bo$148bo29bo30bo27bo30bo
23b2o25b2o28bobo24b2o30bo26b2o$149bo29bo28b2o26b2o29b2o23bobo24bobo54b
obo28b2o26bobo$208bobo25bobo28bobo22bo83bo30bobo7$535bo$469bo32bo31bo
92bo$410bo25bo31bo32bo32b3o28bo32bo27bo$409bo25bo32b3o30b3o60bo32bo28b
3o$186bo26bo27bo26bo26bo25bo27bo26bo32b3o23b3o126b3o30b3o$156bo28bo26b
o27bo26bo26bo25bo27bo26bo$155bo29b3o24b3o25b3o24b3o24b3o23b3o25b3o24b
3o$155b3o4$404b3o25b3o89b2o$176b3o26b3o26b3o25b3o25b3o24b3o26b3o25b3o
29bo27bo21b2o33b2o30bobo30b2o60b2o$144b3o31bo28bo28bo27bo27bo26bo28bo
27bo28bo27bo23b2o33b2o31bo31b2o22b2o36b2o$146bo30bo28bo28bo27bo27bo26b
o28bo27bo80bo34bo64bo25b2o34bo$145bo435bo16$800bo$799bo38bo$745bo53b3o
35bo$744bo92b3o$744b3o3$306bo$274bo30bo$273bo15bo15b3o533b3o$273b3o12b
o459bo92bo$288b3o456b2o44b2o47bo$747bobo44b2o$793bo4$308b2o$308bobo$
266b2o21b2o17bo$265bobo21bobo$267bo21bo13$595bo$594bo$594b3o2$494bo44b
o126bo$460bo32bo44bo126bo$459bo33b3o42b3o124b3o$459b3o3$594b2o$594bobo
$594bo2$459bo35b2o35b2o$458b2o35bobo35b2o121b2o$458bobo34bo36bo124b2o$
656bo11$165bo$164bo$81bo39bo42b3o$43bo36bo39bo75bo$42bo37b3o37b3o72bo
68bo$42b3o150b3o65bo20bo$bo261b3o17bo$o282b3o$3o3$159b2o31bo$35b3o80b
2o40b2o29b2o$3b3o31bo38b2o40bobo38bo31bobo$3bo32bo40b2o39bo161bo$4bo
71bo181b2o19b2o$257bobo19bobo$259bo17$458bo$395bo61bo42bo$394bo62b3o
39bo$394b3o102b3o$503b3o$399bo103bo$398b2o104bo$398bobo54b2o$454bobo$
456bo23$106bo$105bo151bo207bo$105b3o148bo207bo$256b3o205b3o222bo$688bo
$641bo46b3o$640bo$640b3o185bo$827bo$259b3o565b3o$259bo203bo$98b2o160bo
201b2o$99b2o361bobo$98bo$641bo42b2o$640b2o41bobo$640bobo42bo2$813b2o$
814b2o$813bo12$497bo$496bo$496b3o6$499bo$498b2o$498bobo16$259bo$258bo$
258b3o264bo$162bo361bo$161bo362b3o$161b3o3$257bo$256b2o266b3o$256bobo
265bo$525bo3$153b2o$152bobo$154bo16$510bo$509bo$509b3o5$510b3o$56bo16b
o436bo$55bo16bo438bo$55b3o14b3o7$74bo$51b2o20b2o$52b2o19bobo$51bo13$
278bo239bo$277bo239bo$277b3o237b3o6$277b3o$277bo$278bo$508b2o$507bobo$
509bo24$527bo$526bo$526b3o7$524bo$523b2o$523bobo9$246bo58bo40bo$245bo
58bo40bo$245b3o56b3o38b3o3$63bo13bo$62bo13bo$62b3o11b3o168b3o$247bo43b
3o$248bo44bo$292bo56b2o$349bobo$349bo3$58b2o21b2o$58bobo20bobo$58bo22b
o!

I did notice that one single isotropic bit change turned one of the 71 collisions into a not-collision -- hadn't thought of that:

x = 19, y = 14, rule = B3-q/S23
17bo$16bo$16b3o9$2o$b2o$o!
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Re: Challenges

Postby danny » March 9th, 2018, 11:59 pm

The reaction should be clean, my mistake for not clarifying. If it has debris, it should take less than the net gain to clean up the debris.
EDIT: You should take into account 2Gs that use B2n, because that leads to more 2G reactions than are displayed in your RLE.
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Re: Challenges

Postby danny » May 15th, 2018, 3:25 pm

make a radiation hardened quine that can have a certain amount of bytes removed, but there's a gap (e.g. 1 and 3, but there is a certain configuration n=2 that makes it not a quine)

Alternatively, make it so every n=2 breaks it. Also, make sure it returns the unmodified quine, as per the definition

Any languages are fair game.
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Re: Challenges

Postby Hdjensofjfnen » July 23rd, 2018, 9:37 pm

Challenges:
1) Find a Life oscillator with a 1 cell rotor.
2) Find a c/2 in a B2a rule.
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Challenges

Postby BlinkerSpawn » July 23rd, 2018, 11:33 pm

Hdjensofjfnen wrote:Challenges:
1) Find a Life oscillator with a 1 cell rotor.
2) Find a c/2 in a B2a rule.

1) Impossible. The rotor cell would have 3 neighbors in order to be born but would then have S3.
2) Minimal example:
x = 3, y = 2, rule = B2a3e/S1c2c3e
obo$bo!
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Re: Challenges

Postby Hdjensofjfnen » July 24th, 2018, 8:39 pm

Challenge: Find a 3-glider synthesis of a single ship.
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Challenges

Postby Redstoneboi » July 25th, 2018, 9:04 am

x = 19, y = 8, rule = B3/S23
18bo$16b2o$17b2o2$bo$2bo9bo$3o7b2o$11b2o!
c(>^w^<c)~*
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Re: Challenges

Postby Hdjensofjfnen » July 28th, 2018, 10:28 pm

Challenges:
Find an oscillator in Life with period p, with n cells in its rotor, where p > n.
Find an oscillator in Life, with period greater than 2, in which one of the phases is the stator (plus no cells).
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Challenges

Postby danny » July 29th, 2018, 6:00 am

Hdjensofjfnen wrote:Challenges:
Find an oscillator in Life with period p, with n cells in its rotor, where p > n.
Find an oscillator in Life, with period greater than 2, in which one of the phases is the stator (plus no cells).

1. Phoenices
2. Cuphook
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Re: Challenges

Postby Hdjensofjfnen » July 29th, 2018, 2:37 pm

danny wrote:
Hdjensofjfnen wrote:Challenges:
Find an oscillator in Life with period p, with n cells in its rotor, where p > n.
Find an oscillator in Life, with period greater than 2, in which one of the phases is the stator (plus no cells).

1. Phoenices
2. Cuphook


The smallest phoenix is this:
x = 8, y = 8, rule = B3/S23
3bo4b$3bobo2b$bo6b$6b2o$2o6b$6bob$2bobo3b$4bo!


...and that has a 12 cell rotor. 2 is not greater than 12.

Good job with #2, though.
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Challenges

Postby cordership3 » July 30th, 2018, 12:21 pm

For #1, the p(2^44497 - 1) oscillator should work.
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Re: Challenges

Postby gameoflifemaniac » July 31st, 2018, 12:40 pm

cordership3 wrote:For #1, the p(2^44497 - 1) oscillator should work.

Yes, it does, but you answered the question as if it would sound like this:
Find an oscillator in Life with period p with n cells, where p > n.
There is a smaller one for sure.
EDIT: Gabriel's p138 and Karel's p177 work.
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Re: Challenges

Postby Hdjensofjfnen » July 31st, 2018, 8:18 pm

This puffer is very picky. It needs every single non-totalistic birth and survival condition except one. What is it?
x = 4, y = 5, rule = B356/S23
bo$2o$2b2o$bobo$2bo!
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Challenges

Postby Hdjensofjfnen » August 3rd, 2018, 11:47 pm

Here's a superunity reaction with 2 fivebitters:
x = 4, y = 11, rule = B2en3-eknq/S234q
2o$2bo$2o6$2bo$bobo$bobo!


EDIT: I should also mention this rule has a lot of them:
x = 16, y = 16, rule = B2en3-eknq/S234q
obobob4ob2ob2o$ob7o3b4o$4b5o$2bo2b2ob3o4bo$bo6bo3b2o$ob2ob2o4bo2bo$2b
2o11bo$3o5b2ob2o2bo$o2b4obobo2b2o$bo3b5obobo$bobo2bo2bob4o$2bo3b4o4b2o
$bob2o2b2o5bo$o2b2obo3b2obo$2b2obo2b4o2bo$o2b2o!
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Challenges

Postby Naszvadi » August 4th, 2018, 4:01 am

Hdjensofjfnen wrote:This puffer is very picky. It needs every single non-totalistic birth and survival condition except one. What is it?
x = 4, y = 5, rule = B356/S23
bo$2o$2b2o$bobo$2bo!


non-totalistic


Works from B356/S23 till B34c56/S236i8. So there are three possible answers: B4c? S6i? S8?

This is picky too. Pick one :P
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