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Challenges

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Challenges

Postby A for awesome » February 18th, 2017, 5:13 pm

For challenges of any type related to cellular automata.

I'll give two to start off:

1. Who can find the non-totalistic rule with the most speeds exhibited by spaceships with <=10 cells in any phase?

[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.
Last edited by A for awesome on February 18th, 2017, 6:07 pm, edited 1 time in total.
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 gameoflifeboy » February 18th, 2017, 5:31 pm

A for awesome wrote:2. Find a non-explosive B1e rule.


B1e/S12 has natural infinite growth in the form of one-cell-thick ladders, but I wouldn't call it explosive.

In fact, I think that any survival conditions can be added to the rulestring and patterns still won't explode.
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Re: Challenges

Postby A for awesome » February 18th, 2017, 6:06 pm

gameoflifeboy wrote:B1e/S12 has natural infinite growth in the form of one-cell-thick ladders, but I wouldn't call it explosive.

In fact, I think that any survival conditions can be added to the rulestring and patterns still won't explode.

That's probably true. Congratulations on solving my challenge, even though that's really not what I was imagining at all. The second statement I would call incorrect, however:
x = 113, y = 82, rule = B1e/S012
b4ob4ob2obo2bobob2ob4obo3bob3o2bobo2b4o6bo2bobob2o2b2obo4bo2bobob4obo
3bobo4bo2b3o2bo5b2o$bobob5o2bobob2ob6o2bo2bob3o5bob3ob3ob2o7b3ob2o2b2o
b3o3b7ob4ob3obo3bo3bo3b4o3bo$3b2ob2obobobobob3o3bo3b2obo6bobob3o2b8o3b
2o3bo2bo6b3o2b2obo2b2o2b3obob2ob5o3bo4bo$2b2ob2obobob2o5b2obo5bobo2b3o
2bob2ob3o3b2o2bob2ob3o2bobo2b2ob4obo3bo4b3obo3bobo2b3o3bo4bo2bo$2bobob
ob10obo4bob2o4bo2bo2bob2obob10ob3o3bo2bob3ob3ob2obob2ob2o2bo4b2o2b2o4b
3o5b2o$b4o2b5obobob2obo2bobob4ob2obob5o5bo2b3obo3b2obobob2obo2bob2o3bo
2b3obob2obo3bobo2b2o2bob2ob5o$2obob3o3bobo2b4obo2b4o2bo2bob2obo2bo3bob
2o2b2ob3o5bob2o3bobobob2o3bobobo4b3ob3obo3b2o2b2o3bo$2b4o4bo2bobob8o2b
ob2obo3b3o2bob2obob10o2bo2b3o2bo2b4o3bobobo2bob2obo4b6ob6o4bo$4bob5o4b
2o2b3obobobo2b3o3b2o2b3ob3o3b2obob6ob3obo3bo2b2o2bo2b5ob2ob3o2b2o2b4o
2bo3b3obo$3b3obo4b4o2b2ob2obobo2b4o4bo2b6o3b2obob4o3bo4bobob7o3b3obobo
3bo2bob2obo2b2obo2b5o$3o2b3o3bobo2bob2ob2o2b2ob2o3b3o2b3obo3b2o3b3ob3o
bobo2b2o3b3ob4ob3ob2o2b2obo2b2o2bo2b3o4bo4b2o$2obo2bobob2ob2o2bob3obob
4o2b2o2bo2bo5b3o2bo3bo3b2o3b4obo3bo4bo2bobobo7bo2bob5obob3ob2obo$2b3o
2b2ob2obob4o2b3o10b3obob2obob2obo3bobob2o2bobo2b6ob3o4bobobo3bobobo2bo
2bob5ob4o2bo$4bobob4o2b2o3b2o2bo2b2obo2bo5b3o2bobo3bo4bob3o5b4ob2o3bob
o3b2obo2b2o2b2ob2o3bobob2ob3obo2bo$obob6obobob2o4bob4ob3ob3ob3ob3o2bob
o2b2obo2bobo2bobo3b4obob2obob2o4bob2obob3obo6b2o5b2o$2o3bo2bo6bo4bo2bo
2bo3bo3bob3o2b4o2b2o2bob2ob3ob3ob3ob2o3bob2o4bobo2bob5o2b4o3b4obob2obo
$2ob2obo2b3ob4o4bob2obob4obobob4o5b4ob4o7b4ob3o2b3obob2o7b3obo3bobobo
2b2ob2o2b4o$2b2ob3ob8o6b2o3b2o3bob2o2b3obobo2bo2b2ob4ob4obo2b3o4bo2bo
3b2obo4b3o2b3ob5o2bob4o$7o4bobobo2bobob2ob3o2bob3ob3o2b2obo3b2ob2ob6o
3bobob4o3bo4b2o2b2o3bobo2b2o2bob4ob2o3b3o$bob2o2b2ob3ob3ob2obobob2o3bo
2b2obo3bo6b2obo4b4ob2ob3o5bo3b3ob5obob4o2b2o3bobob2ob4o2bo$2o4b5obobob
obobobo4bo2b2o2b2o2bo2bob3o2bobob6obo7b2o3b2obob2o7b2o2bo5bobobo4bob3o
b2o$ob2ob4obobo2bo2bob3obo3b2o3b2o3bo2bob4o2bo6bo5b4o7bob4obob4o4b4o3b
o3b3ob6o$obo3bobobob3o2bobobob3obobo5bobo2bobobob2ob2o2b3o2b2o3b2o2b4o
3bobo2bo2bob2obobob2obo2bo2b3o2b3obo$b2obobo2bob3o3b3ob2ob2o2b2o4b3o2b
o2bobob2obob4o4bobo2b2obo2bo4bo2b4obob3ob2o4bo3b2obo2b2o2b2o$bo2bob2o
2b3o2b2obo3bob2o2bo3b2ob2o3b3o4b2ob3o6bo3b8o3bob6o2bo4bobo2bo2b5o2bo2b
2ob2o$2o2b4ob3o2bob3o2b3obo2b3o2bo2b3ob3obo2b3ob4o6bob3obo2bo4b2o2bobo
3b3obob2o3bo2bo2bo2bob3o$ob2o3b2o2bob2ob5o3bob3obo4b5obob3ob2ob2obobob
5obo3b2o2b5obo3bo2bob2o2b4o2bobob3ob6ob2o$bob2o3b4o9b3ob3o2bobo2bo5bo
2b3obobo3bo7bo5bobo2bo2bo5bob2o2b2obob9obobob2obo$bobo5b3o4b2obob2o5bo
bobobo3b3o3bo2b2o3b4o3bo2b2obo2b3o3b2obobob4ob3ob4o2bo4bo2bob7o$o3bobo
b4o2b3o3b5o2b9o3b4obo3b3ob2o3bobob2o2bo2bo2bo2b2o4bob3ob2obobob3ob3obo
b2obob5o$2o7bo5bo4bobo2b5o2b6obo3b2obobobo3b2ob2ob2ob2o3b3o2b3o2bo2b2o
2b3o2bo3b4o2bobobo4b2obo$bobob2obo3b3o3bobobo2bob7o3bobo3b3obo2bob2ob
3obobo2bo2bob3ob2obo2b2obob7o2bo2bobob2ob2o2b2ob2o$obo3b2o2bobo3bo3b2o
bo6bo2b3ob2ob2ob3obob2ob3o2b5o4b2ob2o4bo6b2ob3o2b4o3bob5o2bob3obo$obob
3obo2b4obob3obo2bo4bo2bo2b3ob2obobobo4b4o3b4ob2ob2o2b2o6bob3o3b3obo3bo
b2obo3b2obobo$2ob3ob2o5b2o4b2o2bo2b2o3b2ob5o4b2obo5b2ob2o2bob2obob6o3b
4o4b5obob4o2b3o2bo2bo3b2o$2bobo2bob3o2bobob2o4bob2o3b2o2b2o4bo6bobo6b
3obo2b2ob6ob3ob3ob4obo2b2ob3obo3b3o2bo$7bo2bo4bo4b2o2b6o2b4ob4obob3ob
3obob3o3b8ob3o5bob2o2bobobob4obo4bo2b4o3b2o$2o2b3ob2o3b2o3bo7bob3obobo
2b3o2bo3bob2o4bob3ob5obobo2b2ob3obobobo3b3o3bo2b8ob2ob3o2bo$obobobob2o
b3ob2o6b3o2b2obo3bo3bo2b3obo3b2obo8bobo2bob4o3b7o2b2ob3o5b5ob2o2bo3b2o
$6b2o2b3o4bo2b2o5bobo2bob4obobo6b2o2bobo2b3o4b4o2bo2bobo3b2o2b2obob2ob
6o3b2o4bob2ob2o$bo5bobo3b2o3bobob2ob2obobo2bo5bobobob2o2bob4o3bob3o2b
2o4bobo2b7o2b2ob2o4b6ob2o2b2ob4o$b3o2b2o2bob2ob5obob2o2b3o3bobob6ob3ob
ob3o2b4o3b3o6bobo2bo2b4ob2o2b2obobo2bobob2o2b2o5b2o$b2o4b4obobobo3b2o
4b4o3bo3bo5b2ob2ob5ob6obob3o2b2o2b3o3bo2bob2o2bob2ob3o2b2obo4b3obobo$
2b3o3bobo3b2o2b2obobob2ob2obobob2o6b4o2bobo4bo4b2obob3obobobo4bo5bo2bo
bo2b2o2bob4ob3o4bo$ob3obob5obob2o3bo2b3o3bob2o6bo3b3obo3b2obobobo3b2ob
2obobo2b3ob3ob2obob2o2bob5o2bo3bo3bobo$2bo4bo3bobob7ob2obobobo4b8obobo
4b3o3bob2o3bo5bobo6b2o2b3obo2b2o3b3o2bo2bob2o2b4o$2o3b2obo3b3obo2b2o3b
2obob2o3b5o3bobob2o2bobo2bob2ob3o2b2o2bo3bobo2bobobobobobobo2bo3b2ob7o
$o2bo3b2ob2obo3b2obo3bo6b4o3bo3bob6o3b4obo2bo2b3o5b2obobob4obo4bobob5o
2b5obobobo$2bobobo6bo3b3obo2b3obo2b2o7b3ob4o2b2o2b4ob2ob2obo2bobobo6bo
bo3bo2bobo2bob3o3bob2o4b2o$2b2ob2obo3bo2b2obobobob4o2b3o5b3obo3b6o3bo
2b3o3b2obo2bobo3b2o3b2o2b6o2bobo2b2o11bo$o4b2obobo2bobob4o5bob2ob2o2bo
7b2o4b3o5bob3o2bo3b4o2b2ob3o2bob3o2b3o2bobo3bobobob3o3bo$3b6obo2b2o5bo
b4o2b2o7bo2bo4bo3b4ob4o2bo2b2o3b2ob3o2bob5ob3o2bo4b3o3b3o2bo2bo2b2o$ob
2obob3o4b3ob3obobobo3bob2o6b2obo2bo7bo2bobo2b4o5b2ob2ob2o2bob5obob3obo
b3o5bobo2b3o$2bobob3obo2bo2b2ob2o8bo3b2o3bo4b4o3bo2b3ob3ob6ob2ob2o2bo
4b2obo3bobob3o3b2o3b3ob2obobo$b2o2b2ob2o2b5obobo3bo2b3obob2obobo3bo2b
2o5b3o3bo2b2ob4ob2o2bo8bob8obo2b4ob3obo4bobo$3obo2b3o4b3obob2ob7obobob
2obo3bob4ob3obob4obob10ob2ob3o2b3o2bob4ob2ob2ob3o2bobob3obo$3obobob5ob
o4bob5obo4bo3bobo2bob2ob4obob2ob2obo2bo2b2obob2o2bob3obo2b3ob2obo2b4o
2bob3obo2b2o$bo2bobo3b2o3b3o3b2o2b2obob2o2b6obobobobo2bob2o2b4ob2o2b2o
b4obob2o3bob3o2bo2bobob3obobo4b2ob2o$2ob2o4bob3o2bobo2bob2o2b2ob5o2b3o
bobobobob3obobob2o3b2o2bobo3b2o3b2ob4o3b3ob3obob8o2b2o2bo$obo9bobo6b2o
3bobo3b4ob3obo2bo3b2obo2b4ob2o4b2o3bob2o3b2o7bo2bob4o2b3o2b3o2b2o2b2o$
2obo3bob4ob2o2b2obob2ob2obo2bob3ob2o2b7o2bo2b6o3bo6b2obo3b2o4bobobob6o
2b2o2b3o5bob2o$2ob2o2bobobo2b2ob10obobo2b3o2bobo3b3o3bo2bobo2bob3o5b2o
bo2b6ob3o5b2ob2ob2ob3o2b2o3bobobo$2bobobo3b2o2b3o2b2o2bo3bo2bo6bo5b2ob
2ob2obob2o10bob4ob3o3b3ob2ob2o4bob4o4b2obob3o$b2o2b2o5b2o2b2o2b2o4bob
2ob3o4bob2o2b4ob2obo2b2o3bo2bo2b3ob2o3bobo2bo2b2obo2bob6ob4ob3o2b2o$3o
b7o3bob3o2b4o5bo5bo2b3ob3obobob2o4b2o5b4o4bobo6bo2b2o2bobo4bobobobobob
4o2bo$o2bob2o2bo3b3ob4ob4o2bobo7bob7ob2o3b3ob5obobo4b2o3bobo3b2o3b6obo
b2obo4bobobob3o$b3obo6bobob2ob3o2b2o5bobo2b3o3b5ob4o2bob6ob2ob3o4bobo
4b4o2b2o4bobo3bo2b4ob3obobo$2o3b4o4b2o2b4obobo3b2ob2o3b6obo2b2o4b3o5bo
3bobob2o2b4o3bo5bo2bobo2bo3b2obobobobo2bo2bo$2bo3bob9o2b2obo3bobob3o4b
2obob8o5bob4obo4b5o4b3obobo2bo4bo4b2o2bo4b3ob3obo$5bo3bobo5bo8bobobo5b
2o2bob2o3bob3ob2obo3b2o2b3ob2o4b2o7b3o4b2o3bobo2b2ob2o3bo2b2o$b2o3b3o
2bobob2o5b5obob2o2bobobo3bobobobobobob4o2bo5bobo3b2o5bob2o3bob2obo3b3o
bobob3o6bo$bo5b4o2b4o3bo4b6o3b3ob2obob4o2bob5ob6o3bobobo3bo3bobob4o4b
3ob4o2b2obobob2o2bo$3ob5o2b2ob2ob3o4bo2bo2b7o2bo3bo3b5o5bob2o3b2o2b5o
2b2o3b2o2b2obob2o4bobobobo4b2ob2o$2obobob4o3bo4bob2ob2ob2obo4b4o5bo4b
2o2b4o2b3ob2ob3o2b2o3bobo3bob3o2bo6b4o2b3o3bobo$5o2bob3o3bobo2b2o5bob
3obobobobob2o3b2o3b4obo3b2o2bobobo3b2ob2o2b4ob2o2bob2o5bobo3bob2o$b4o
2bo4b3o2b4o2bo4b2o2bo2bob5o2bo2b2o2bob2o3bob2o3b4o3bo2bobo4bo3b5obo2b
3ob2obob4obo2bo$bo2bo3b3ob4o3b3ob5o2bobo3bo2bo2bobob4ob2ob3o2b3ob6o2bo
b2obobo2bobob2o2b6obo5bo3bo4bo$2b4ob2obob4obo6b2obo4b2o2b3o2bo7b2ob3o
3b2ob3o4b2o2bo3bo2b3o2bo4b2obo2b5obob4obo3bo$3o5b2o4b2o3bob7ob2obo7bob
o5bo6b2obo2bo2b2o4b2o2bo2bob3o3b4ob2ob2o2b5o3bobobo$bob3ob6obo2b4o3b2o
4bo2b2obo3bobo3bobob2o2b4ob2o2bo3bo2bob3ob2o2b4o3bo3bo2b2o4b3obo2bo3bo
$ob5o2bob2o4b2obobobo3bobob2o2b2obo2b5o3b5obo3b6o2b2o2bo2bob4o2b2obo3b
3o2bobob7ob6o$3o4b2ob2obobobo4bob2obo5b4o3bob2ob6obo3b2o3b3o2bobo2bobo
b2o4bo3bo2bo2b4o2bobobob2o4bo!
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 dvgrn » February 18th, 2017, 11:20 pm

Here are a couple of Conway's Life challenges that came to mind today, while I was trying to sort out possible new patterns for Golly's collection -- especially ones that might be amenable to having LifeViewer waypoint scripts added, to show off lifeviewer.lua.

I put together Life/Miscellaneous/Cambrian-Explosion.rle a decade ago, as a representative sample of the amazing things that had been invented up to 2006. But the pattern kind of gets boring after you run it for a while -- it just keeps expanding and gets awkward to look at.

So -- what about a pattern that goes the other way?

A. What is the largest number of distinct spaceship types that can be crashed together and mutually annihilate each other? Different phases don't count -- no two spaceships in the pattern can have the same apgcode. Bonus points if no proper subset of the spaceships can be removed and still result in an empty universe.

EDIT: Instead of bonus points, maybe that no-proper-subset rule needs to be part of the requirement. Otherwise it's easy to set up an unlimited number of spaceships with different apgcodes to collide in head-on vanish reactions, e.g.,

x = 209, y = 25, rule = B3/S23
81b2o47b2o$79bo4bo43bo4bo$b4o80bo41bo76b4o$6o73bo5bo41bo5bo69b6o$4ob2o
73b6o41b6o69b2ob4o$4b2o197b2o3$2bo80b2o43b2o76bo$79b4ob2o41b2ob4o$2b3o
b3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3o3b5o43b5o3b3ob3o
b3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3ob3o$81b3o45b3o2$81b3o45b
3o$58b3ob3ob3ob3ob3o3b5o43b5o3b3ob3ob3o$79b4ob2o41b2ob4o$83b2o43b2o2$
56b4o85b4o$55b6o83b6o$55b4ob2o18b6o41b6o10b2ob4o$59b2o18bo5bo41bo5bo
10b2o$85bo41bo$79bo4bo43bo4bo$81b2o47b2o!

That might not be good enough to prevent boring solutions, though -- e.g., there might be a way to hit something like a block with an extensible spaceship, and wind up with a block. Probably simpler to allow only one representative of each class of extensible spaceships.

B. Same question as (A), but now only glider-constructible spaceships are allowed. The initial pattern starts out with four glider salvos in the four quadrants of the pattern, but all-but-optionally-one of the gliders has to build a spaceship at some point. And as before, no two of the constructed spaceships can have the same apgcode, and only one member of each class of extensible spaceship is allowed.

For (B) I guess it's not a requirement that all the constructed spaceships have to exist simultaneously at some particular generation of the pattern, though maybe it would be nice if that were an option.
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Re: Challenges

Postby A for awesome » February 19th, 2017, 12:03 am

dvgrn wrote:I put together Life/Miscellaneous/Cambrian-Explosion.rle a decade ago, as a representative sample of the amazing things that had been invented up to 2006. But the pattern kind of gets boring after you run it for a while -- it just keeps expanding and gets awkward to look at.

So -- what about a pattern that goes the other way?

This is somewhat unrelated, but I think a subpattern like one of these might have a place in that pattern:
x = 90, y = 103, rule = B3/S23
70bobo$73bo$69bo3bo$69bo3bo$73bo$70bo2bo$71b3o6$64b6o$63bo6bo$27bo$25b
3o$22bobo5b2o$25b3o3bo$18bo4b2o2bo2bo29bo12bo$19b4o2bobobo31b12o$24b2o
bob4o$25bobo5bo$24bo3bob3o$16b7o2b4obo27b18o$15bo7b2o32bo18bo$25bob2ob
o$22bob2obob4o$25bobo5bo$22bobo3bob3o$25b4obo$12bo10b2o29bo24bo$13b10o
2bob2obo24b24o$24b2obob4o$25bobo5bo$24bo3bob3o$10b13o2b4obo21b30o$9bo
13b2o26bo30bo$25bob2obo$22bob2obob4o$25bobo5bo$22bobo3bob3o$25b4obo$6b
o16b2o23bo36bo$7b16o2bob2obo18b36o$24b2obob4o$25bobo5bo$24bo3bob3o$4b
19o2b4obo15b42o$3bo19b2o20bo42bo$25bob2obo$22bob2obob4o$25bobo5bo$22bo
bo3bob3o$25b4obo$o22b2o20bo42bo$b22o2bob2obo15b42o$24b2obob4o$25bobo5b
o$24bo3bob3o$4b19o2b4obo18b36o$3bo19b2o23bo36bo$25bob2obo$22bob2obob4o
$25bobo5bo$22bobo3bob3o$25b4obo$6bo16b2o26bo30bo$7b16o2bob2obo21b30o$
24b2obob4o$25bobo5bo$24bo3bob3o$10b13o2b4obo24b24o$9bo13b2o29bo24bo$
25bob2obo$22bob2obob4o$25bobo5bo$22bobo3bob3o$25b4obo$12bo10b2o32bo18b
o$13b10o2bob2obo27b18o$24b2obob4o$25bobo5bo$24bo3bob3o$16b7o2b4obo30b
12o$15bo7b2o35bo12bo$25bob3o$22bob2obo2bo$25bobobo$22bobo3bo$25b3o$18b
o4b2o38bo6bo$19b4o2bo38b6o$24b2o4$71b3o$70bo2bo$73bo$69bo3bo$69bo3bo$
73bo$70bobo!

The first one only retracts in one dimension, but the second one is clean. Maybe call the whole thing "Permian-Triassic Extinction" or "Great Dying"?
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 dvgrn » February 19th, 2017, 1:02 am

dvgrn wrote:
A for awesome wrote:So -- what about a pattern that goes the other way?

This is somewhat unrelated, but I think a subpattern like one of these might have a place in that pattern...
The first one only retracts in one dimension, but the second one is clean. Maybe call the whole thing "Permian-Triassic Extinction" or "Great Dying"?

Huh, maybe something along those lines, yeah. I had thought of "Gnab Gib", a reference to Douglas Adams' _Restaurant at the End of the Universe_, but didn't really like the name enough.

I don't remember that anyone ever published an inverted spacefiller -- a "space-emptier" -- though I suppose it probably wouldn't be too hard to come up with something less trivial than the following, by sticking two back ends of diamond-shaped greyships onto a big patch of agar.

x = 78, y = 40, rule = B3/S23
38bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$39bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$40bo3b
o3bo3bo3bo3bo3bo3bo3bo3bo$41bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$38bo3bo3bo3b
o3bo3bo3bo3bo3bo3bo$39bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$40bo3bo3bo3bo3bo3b
o3bo3bo3bo3bo$41bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$10bobo25bo3bo3bo3bo3bo3b
o3bo3bo3bo3bo$10bobo26bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$8bobobobo25bo3bo3b
o3bo3bo3bo3bo3bo3bo3bo$8bobobobo26bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$6bobob
obobobo21bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$6bobobobobobo22bo3bo3bo3bo3bo3b
o3bo3bo3bo3bo$4bobobobobobobobo21bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$4bobobo
bobobobobo22bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$2bobobobobobobobobobo17bo3bo
3bo3bo3bo3bo3bo3bo3bo3bo$2bobobobobobobobobobo18bo3bo3bo3bo3bo3bo3bo3b
o3bo3bo$obobobobobobobobobobobo17bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$obobobo
bobobobobobobobo18bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$obobobobobobobobobobob
o15bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$obobobobobobobobobobobo16bo3bo3bo3bo
3bo3bo3bo3bo3bo3bo$2bobobobobobobobobobo19bo3bo3bo3bo3bo3bo3bo3bo3bo3b
o$2bobobobobobobobobobo20bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$4bobobobobobobo
bo19bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$4bobobobobobobobo20bo3bo3bo3bo3bo3bo
3bo3bo3bo3bo$6bobobobobobo23bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$6bobobobobob
o24bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$8bobobobo23bo3bo3bo3bo3bo3bo3bo3bo3bo
3bo$8bobobobo24bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$10bobo27bo3bo3bo3bo3bo3bo
3bo3bo3bo3bo$10bobo28bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$38bo3bo3bo3bo3bo3bo
3bo3bo3bo3bo$39bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$40bo3bo3bo3bo3bo3bo3bo3bo
3bo3bo$41bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$38bo3bo3bo3bo3bo3bo3bo3bo3bo3bo
$39bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$40bo3bo3bo3bo3bo3bo3bo3bo3bo3bo$41bo
3bo3bo3bo3bo3bo3bo3bo3bo3bo!
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Re: Challenges

Postby toroidalet » March 10th, 2017, 1:01 am

[quote="A for awesome"]
1. Who can find the non-totalistic rule with the most speeds exhibited by spaceships with <=10 cells in any phase?/quote]
I can find 4:
x = 44, y = 11, rule = B2i3-ck/S02-i3-ck
3o13bo12b3o10bo$o14bobo10b2ob2o8bobo$bo13bobo23bobo$16bo25bo2$16bo25bo
2$16bo3$29b3o!

5 cell c/4 diagonal, 8 cell 3c/14, 10 cell 6c/29, 7 cell c/6.
Challenge: find the fastest diagonal spaceship possible.
Also the fastest ship in a rule with no survival conditions or b2a
x = 4, y = 2, rule = B3/S23
ob2o$2obo!

(Check Gen 2)
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Re: Challenges

Postby A for awesome » March 10th, 2017, 10:15 am

toroidalet wrote:Challenge: find the fastest diagonal spaceship possible.

3c/4 diagonal spaceship here. I think this is, in fact, the fastest possible.
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 dvgrn » March 12th, 2017, 9:43 pm

A glider-to-c/3-spaceship converter whose RLE fits in a forum posting and so can be run in LifeViewer.

(No fair just writing to Nathaniel and asking him to bump up the maximum number of bytes in a post by 10K.)
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Re: Challenges

Postby BlinkerSpawn » March 23rd, 2017, 3:49 pm

A for awesome wrote:3c/4 diagonal spaceship here. I think this is, in fact, the fastest possible.

I think so as well, and here's why:
First, we look at rules without B0.
A fast spaceship in a rule without B1c will have its leading edge travel orthogonally, forcing a maximum diagonal speed of c/2.
B1c, however, forces lightspeed diagonal expansion in all directions, preventing the existence of spaceships, or interesting non-replicator OFF-background patterns of any kind for that matter.
Therefore, any diagonal ship faster than c/2 must have B0.
By similar reasoning to the non-B0 case, at least one of the even-step rule and the odd-step rule must have B1c but both cannot or all patterns will explode.
Therefore, the fastest possible speed is the fastest speed at which the pattern does not experience B1c in two consecutive cycles.
The leading edge must then alternate between moving orthogonally and diagonally every generation.
This gives two maximum-speed paths for the leading edge, in which the leading edge travels one cell along the path each generation:
x = 25, y = 12, rule = B3/S23
o18bo$o18bo$b2o17bo$3bo16bo$3bo17bo$4b2o15bo$6bo15bo$6bo15bo$7b2o14bo$
9bo13bo$9bo14bo$10b2o12bo!

As you can see, the left path corresponds to 3c/4 diagonal, and the right corresponds to (2,1)c/2 as mentioned by A for Awesome below the post he linked.
Therefore, 3c/4 diagonal is the fastest possible diagonal speed, and if (2,1)c/2 is proven infeasible, the fastest possible speed in general.
(EDIT 3-26) Realized that theoretically, you could have any ordering of the two travel directions, giving a speed of (2h+k,h+2k)/(2h+2k) for any integer h,k > 0. This equates to a Euclidean speed of 1 + (h^2+k^2)/(2h+2k)^2, so 3c/4 diagonal is actually the lowest-speed path out of all paths of this type and (2,1)c/2 is the highest. However, as "compound" fast paths are harder to search for and have higher period (>=6, although that may not be considered "high" anymore), I still expect 3c/4 to remain the fastest speed.
LifeWiki: Like Wikipedia but with more spaceships. [citation needed]

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Re: Challenges

Postby Saka » March 29th, 2017, 8:24 am

Easy challenge: Find which ones are hand-made and which ones are ctrl+5 (Hint: there are 2 of each)
x = 72, y = 48, rule = B/S012345678
11bob2ob6o2bobo16b3ob2ob2obo4bo$14b3obob2ob2obo17bob2obobob3o2bo$2b2o
7bobo2bo2bobo2bobo16bo2b4o5b2obo$bobo10bo2bob3obo2bo16bobo4bo2b3ob2o7b
3o$o2bo7b4obobob2ob4o18b5o2b3ob2o7bo3bo$3bo7bobo2bo2bobo2bobo16bo3b2o
5bob2o6bo5bo$3bo7bob6o2bob4o19b2ob4ob2o2bo5bo5bo$3bo7b2obobobob3ob2o
18b2ob6ob3obo11bo$3bo9b2ob2o3bobob2o16bobo4bo4b3o12bo$3bo7b3ob2obo2bob
o2bo16b2o4b3o2b3obo10bo$3bo7bobo2b3o2bobob2o16b5obobo2bo3bo9bo$3bo7b2o
bo2bob3obobo19bob2obobob5o8bo$3bo7bobob2obo2bo2b2o17bo2bob5o3bobo7bo$b
5o5bobob2obobobob3o16b3o4b4o3b2o6bo$11bob4ob3ob3obo16b2ob3o3b2obo8b7o$
11bobobob3obob2obo18b2o4bobo2b3o17$12bobo3b3o4b2o16b2obobo2b4obobo$14b
2obob3o2b3o17b2o2b2o2b2obob2o$5o6bo2bo2b4obob2o20b2o2bo2bo3bo5bo5bo$5b
o6b2o2bobo3bo3bo17bob9o8bo5bo$5bo5b2o3bobo3bo2b2o16bo3b4o6bo5bo5bo$5bo
6b4o2b3obo2b2o18b2o6bo2bo6bo5bo$5bo6bo2b3o2b7o17b2o2b5ob4o5bo5bo$5o8b
2obob3ob2obo20b2ob5o2b2o5b9o$5bo6bo2bobob2ob5o17bobo3b2o2bo14bo$5bo5bo
2bobo2bob3o22b2o5b2obo12bo$5bo5bobobo2bob3obo20bo3b3obob2o12bo$5bo5bob
3ob5o2bo18bob5o2bob2obo11bo$5bo5b2o2bo5bo2bo18bo4b2ob2ob4o11bo$5o6b2ob
2o2b2o4b3o16bob2o7bob2o11bo$12bo3b2o2b7o16bo2b2ob4o3bo$18bo3bobobo17bo
4b2o2b3o2bo!
If you're the person that uploaded to Sakagolue illegally, please PM me.
x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!

(Check gen 2)
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Re: Challenges

Postby _zM » March 29th, 2017, 1:11 pm

I guess 1 and 3.
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Re: Challenges

Postby Saka » March 30th, 2017, 3:58 am

_zM wrote:I guess 1 and 3.

50% Right!
If you're the person that uploaded to Sakagolue illegally, please PM me.
x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!

(Check gen 2)
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Re: Challenges

Postby Gamedziner » March 30th, 2017, 6:41 am

Saka wrote:Easy challenge: Find which ones are hand-made and which ones are ctrl+5 (Hint: there are 2 of each)
x = 72, y = 48, rule = B/S012345678
11bob2ob6o2bobo16b3ob2ob2obo4bo$14b3obob2ob2obo17bob2obobob3o2bo$2b2o
7bobo2bo2bobo2bobo16bo2b4o5b2obo$bobo10bo2bob3obo2bo16bobo4bo2b3ob2o7b
3o$o2bo7b4obobob2ob4o18b5o2b3ob2o7bo3bo$3bo7bobo2bo2bobo2bobo16bo3b2o
5bob2o6bo5bo$3bo7bob6o2bob4o19b2ob4ob2o2bo5bo5bo$3bo7b2obobobob3ob2o
18b2ob6ob3obo11bo$3bo9b2ob2o3bobob2o16bobo4bo4b3o12bo$3bo7b3ob2obo2bob
o2bo16b2o4b3o2b3obo10bo$3bo7bobo2b3o2bobob2o16b5obobo2bo3bo9bo$3bo7b2o
bo2bob3obobo19bob2obobob5o8bo$3bo7bobob2obo2bo2b2o17bo2bob5o3bobo7bo$b
5o5bobob2obobobob3o16b3o4b4o3b2o6bo$11bob4ob3ob3obo16b2ob3o3b2obo8b7o$
11bobobob3obob2obo18b2o4bobo2b3o17$12bobo3b3o4b2o16b2obobo2b4obobo$14b
2obob3o2b3o17b2o2b2o2b2obob2o$5o6bo2bo2b4obob2o20b2o2bo2bo3bo5bo5bo$5b
o6b2o2bobo3bo3bo17bob9o8bo5bo$5bo5b2o3bobo3bo2b2o16bo3b4o6bo5bo5bo$5bo
6b4o2b3obo2b2o18b2o6bo2bo6bo5bo$5bo6bo2b3o2b7o17b2o2b5ob4o5bo5bo$5o8b
2obob3ob2obo20b2ob5o2b2o5b9o$5bo6bo2bobob2ob5o17bobo3b2o2bo14bo$5bo5bo
2bobo2bob3o22b2o5b2obo12bo$5bo5bobobo2bob3obo20bo3b3obob2o12bo$5bo5bob
3ob5o2bo18bob5o2bob2obo11bo$5bo5b2o2bo5bo2bo18bo4b2ob2ob4o11bo$5o6b2ob
2o2b2o4b3o16bob2o7bob2o11bo$12bo3b2o2b7o16bo2b2ob4o3bo$18bo3bobobo17bo
4b2o2b3o2bo!

I guess 1 and 2 are hand-made.
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Re: Challenges

Postby Saka » March 30th, 2017, 6:50 am

Gamedziner wrote:
Saka wrote:Easy challenge: Find which ones are hand-made and which ones are ctrl+5 (Hint: there are 2 of each)
x = 72, y = 48, rule = B/S012345678
11bob2ob6o2bobo16b3ob2ob2obo4bo$14b3obob2ob2obo17bob2obobob3o2bo$2b2o
7bobo2bo2bobo2bobo16bo2b4o5b2obo$bobo10bo2bob3obo2bo16bobo4bo2b3ob2o7b
3o$o2bo7b4obobob2ob4o18b5o2b3ob2o7bo3bo$3bo7bobo2bo2bobo2bobo16bo3b2o
5bob2o6bo5bo$3bo7bob6o2bob4o19b2ob4ob2o2bo5bo5bo$3bo7b2obobobob3ob2o
18b2ob6ob3obo11bo$3bo9b2ob2o3bobob2o16bobo4bo4b3o12bo$3bo7b3ob2obo2bob
o2bo16b2o4b3o2b3obo10bo$3bo7bobo2b3o2bobob2o16b5obobo2bo3bo9bo$3bo7b2o
bo2bob3obobo19bob2obobob5o8bo$3bo7bobob2obo2bo2b2o17bo2bob5o3bobo7bo$b
5o5bobob2obobobob3o16b3o4b4o3b2o6bo$11bob4ob3ob3obo16b2ob3o3b2obo8b7o$
11bobobob3obob2obo18b2o4bobo2b3o17$12bobo3b3o4b2o16b2obobo2b4obobo$14b
2obob3o2b3o17b2o2b2o2b2obob2o$5o6bo2bo2b4obob2o20b2o2bo2bo3bo5bo5bo$5b
o6b2o2bobo3bo3bo17bob9o8bo5bo$5bo5b2o3bobo3bo2b2o16bo3b4o6bo5bo5bo$5bo
6b4o2b3obo2b2o18b2o6bo2bo6bo5bo$5bo6bo2b3o2b7o17b2o2b5ob4o5bo5bo$5o8b
2obob3ob2obo20b2ob5o2b2o5b9o$5bo6bo2bobob2ob5o17bobo3b2o2bo14bo$5bo5bo
2bobo2bob3o22b2o5b2obo12bo$5bo5bobobo2bob3obo20bo3b3obob2o12bo$5bo5bob
3ob5o2bo18bob5o2bob2obo11bo$5bo5b2o2bo5bo2bo18bo4b2ob2ob4o11bo$5o6b2ob
2o2b2o4b3o16bob2o7bob2o11bo$12bo3b2o2b7o16bo2b2ob4o3bo$18bo3bobobo17bo
4b2o2b3o2bo!

I guess 1 and 2 are hand-made.
50%
If you're the person that uploaded to Sakagolue illegally, please PM me.
x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!

(Check gen 2)
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Re: Challenges

Postby _zM » March 30th, 2017, 10:37 am

1 and 4?
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Re: Challenges

Postby A for awesome » March 30th, 2017, 2:39 pm

_zM wrote:1 and 4?

If not, then necessarily 2 and 3.
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 drc » March 30th, 2017, 3:43 pm

Obviously 1 and 3.
This post was brought to you by the letter D, for dishes that Andrew J. Wade won't do. (Also Daniel, which happens to be me.)
Current rule interest: B2ce3-ir4a5y/S2-c3-y
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Re: Challenges

Postby A for awesome » March 30th, 2017, 3:49 pm

drc wrote:Obviously 1 and 3.

Already been guessed.
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 Gamedziner » March 30th, 2017, 4:34 pm

This somehow feels similar to Mastermind (a board game)
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Re: Challenges

Postby Saka » March 30th, 2017, 8:04 pm

_zM wrote:1 and 4?

You won! I used 2 different methods to make them:
1. Line by line (For 1)
2. Random clicking (For 4)
If you're the person that uploaded to Sakagolue illegally, please PM me.
x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!

(Check gen 2)
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Re: Challenges

Postby muzik » April 1st, 2017, 6:30 am

Seems there indeed is a (2,1)c/2 (compared with c/2 orthogonal and diagonal from the same rule):

x = 8, y = 25, rule = B012356/S2347
4b3o$4bo2bo$4bo2bo2$4bo2bo$4bo2bo$4b3o4$5b3o$5bobo$4obo$ob2obobo$4o3bo
4$4b3o$4bobo$7bo$b2o2bobo$bo2b2obo$b2o$3b3o!
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
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Re: Challenges

Postby A for awesome » April 1st, 2017, 8:09 am

muzik wrote:Seems there indeed is a (2,1)c/2 (compared with c/2 orthogonal and diagonal from the same rule):

x = 8, y = 25, rule = B012356/S2347
4b3o$4bo2bo$4bo2bo2$4bo2bo$4bo2bo$4b3o4$5b3o$5bobo$4obo$ob2obobo$4o3bo
4$4b3o$4bobo$7bo$b2o2bobo$bo2b2obo$b2o$3b3o!

That's actually (2,1)c/4.
Last edited by A for awesome on November 5th, 2017, 4:04 pm, edited 1 time in total.
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 83bismuth38 » April 14th, 2017, 12:23 pm

Find a Polythlonia rule, which is a rule that allows all (or most) polythlons, one of which is the pentadecathlon.
Find a Beconia rule, same as above except with beacons. except the beacon blocks get bigger.
Find a Toadia rule, same as above but they get longer instead of just bigger in general. I have examples for this one:
a baby toad:
x = 2, y = 3, rule = B3/S2e3
bo$2o$o!

A regular toad:
x = 2, y = 4, rule = B3/S23
bo$2o$2o$o!

And a big toad:
x = 2, y = 5, rule = B34-air/S34-ai
bo$2o$2o$2o$o!
Last edited by 83bismuth38 on April 14th, 2017, 2:27 pm, edited 1 time in total.
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Re: Challenges

Postby 83bismuth38 » April 14th, 2017, 2:25 pm

Insane challenge: make a rule where a blinker and a big blinker co-exist:
x = 1, y = 5, rule = B3-a/S1258
o$o$o$o$o!

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

or at least, any variation of the big blinker. (central single-celled startor, rotor on outside, period rotates 90 degrees)
also, side quest thing, see if it can be formed into a Blinkeria rule.
x = 8, y = 10, rule = B3/S23
3b2o$3b2o$2b3o$4bobo$2obobobo$3bo2bo$2bobo2bo$2bo4bo$2bo4bo$2bo!

No football of any dui mauris said that.
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