Counterexample: c/3dNNlk05 wrote: February 25th, 2026, 1:55 pm Conjecture: Life is omnidirectional.
There exists an elementary spaceship for every direction and every speed.
Note: I am probably not the first one to come up with this.
I meant Life has every slope. Sorry but I don't have time to edit the post.UnbihexiumFan wrote: February 26th, 2026, 1:35 pmCounterexample: c/3dNNlk05 wrote: February 25th, 2026, 1:55 pm Conjecture: Life is omnidirectional.
There exists an elementary spaceship for every direction and every speed.
Note: I am probably not the first one to come up with this.
If you remove the "every speed" part, it doesn't seem as if that's an unproven conjecture. For any (x,y) offset, we could use existing technology (e.g., slsparse) to compile a self-constructing spaceship that builds a copy of itself at some (kx, ky) distance, and then self-destructs.NNlk05 wrote: February 26th, 2026, 2:06 pm I meant Life has every slope. Sorry but I don't have time to edit the post.
Conj.2:Upper bound is 4/9(which has been demonstrated by the original self-forcing patch)Anivec wrote: May 14th, 2026, 4:25 pm Conjecture 1: a self forcing still life patch must decay fully in one generation under cell "alive<->dead" inversion, excluding cells outside of the patch and cells on the perimeter. A related statement of this is if the patch is placed on a finite grid that has a bounding box larger than the patch, and all the cells on the grid are inverted, the result decays in a single generation. The converse, "a still life patch that decays in one generation under inversion is self forcing", is not necessarily true, but it does give a hint to what one may look like.
Conjecture 2: A self forcing still life patch must have a density that is greater than or equal to some number n. This does not account for the area the patch isn't in. What are the upper and lower bounds for this number?
Conjecture 3: Is the resulting pattern of the self forcing patch after inversion a Garden of Eden?
Code: Select all
#include <iostream>
#include <string>
#include <algorithm>
#include <fstream>
#include <random>
#include <vector>
#include <ctime>
#include <cstdint>
#include <cstdlib>
#include "lifelib/pattern2.h"
int numgliders = 2;
//Modify to match your computer's number of threads, or how many instances you are running:
int cores = 12;
std::string getrle(apg::pattern pat){
std::ostringstream ss;
pat.write_rle(ss);
std::string s = ss.str();
return s;
}
void assemblecollision(apg::lifetree<uint32_t, 1>& lt, const std::vector<apg::pattern> gliders, const std::string apgcode) {
int32_t loops = 0;
apg::pattern pt(<, apgcode, "b3s23");
int ptpop = pt.totalPopulation();
apg::pattern fullpt(<, apgcode, "b3s23");
int i;
while (0 == 0) {
fullpt = apg::pattern(<, apgcode, "b3s23");
apg::pattern gliderset(<, "b!", "b3s23");
for (i = 0; i < numgliders; i++) {
int glider = rand() % 16;
int x = (rand() % 24) + 8;
int y = (rand() % 24) + 8;
switch(glider%4) {
case 0:
x = -x;
y = -y;
break;
case 1:
x = -x;
break;
case 2:
break;
case 3:
y = -y;
break;
default:
break;
}
apg::pattern newglider = gliders[glider];
newglider = newglider.shift(x, y);
gliderset += newglider;
}
fullpt += gliderset;
apg::pattern evpt = fullpt[150];
if (evpt.totalPopulation() == 0) {
if (fullpt.totalPopulation() == ptpop + numgliders * 5 and fullpt[1].totalPopulation() == ptpop + numgliders*5) {
break;
}
}
loops++;
if (loops > 1000000) {
std::cerr << "Unable to destroy " << apgcode << "!!!" << std::endl;
return;
}
}
//std::cout << getrle(fullpt) << std::endl;
}
int main(int argc, char* argv[]) {
srand(time(0));
apg::lifetree<uint32_t, 1> lt(1000);
//Assemble a vector of the 16 unique gliders:
apg::pattern glider(<, "bob$bbo$ooo!", "b3s23");
apg::pattern glider2 = glider[0];
std::vector<apg::pattern> gliders;
std::vector<std::string> transformations = {"identity", "rot90", "rot180", "rot270"};
int i, j;
for (i = 0; i < 4; i++) {
for (auto j : transformations) {
glider2 = glider[i];
glider2 = glider2.transform(j, 0, 0);
gliders.push_back(glider2);
}
}
if (argc < 3) {
std::cout << "Usage: main <file> <index>" << std::endl;
return 0;
}
std::string line;
std::string filename(argv[1]);
std::ifstream file(filename);
int index = std::stoi(argv[2]);
int32_t destroyed = 0;
while (std::getline(file, line)) {
if (line.length() > 3 and (destroyed % cores == index) ) {
assemblecollision(lt, gliders, line);
}
destroyed += 1;
if (destroyed%10000 == 0 and index == 0) {
std::cout << "Destroyed " << destroyed << " still lifes." << std::endl;
}
}
file.close();
}Code: Select all
#!/bin/bash
./main $1 0 &
./main $1 1 &
./main $1 2 &
./main $1 3 &
./main $1 4 &
./main $1 5 &...If you are willing to spend some money, compute comes pretty cheap these days, I just checked with one provider (vast.ai), you can get a high tier AMD CPU for about USD$0.03 an hour.PK22 wrote: June 26th, 2026, 6:41 pm *snip*
...more computing power than me to search 26+ bit SL, please let me know.
I think this is almost certainly not true. For the rule B2a3r/S that I engineered just now:MDA wrote: June 29th, 2026, 3:17 am I conjecture that, in 2-state INT rules, every puffer, rake, or replicator not based on spaceships can be corderized in at most two engines without supporting spaceships if the rule in which they exist allows spaceships.
It seems to me like the only way to solve this conjecture is to find a counterexample.
Code: Select all
x = 3, y = 2, rule = B2a3r/S
o$obo!
There no doubt exists a strict still life composed of connected fuses that collapses into a universal constructor that reconstructs itself, but this might be impractical to actually build. The problem with finding a small strict still life reflector is that almost all of our stable G-to-X reactions rely on a bait still life that can't be a part of a larger strict still life. The only baitless reaction I can recall is the following G-to-pi by Mitchell Riley (with stator modified to form a single strict still life):NNlk05 wrote: June 30th, 2026, 10:09 pm Conjecture: There exists a strict still life that when hit by a glider in a way, emits a glider 90º relative to input and regenerates itself (ie: glider reflector) .
Had anyone checked?
Code: Select all
x = 16, y = 15, rule = B3/S23
6bo$7bo$5b3o3b2ob2o$11b2obo$14bo$2o12b2o$o2bob2o8bo$b3obo3b2o3bo$5bo3b
obobo$b3ob3obob2o$o2b2o3bo$bo4b2o$2b3obo$4bobo$5bo!There is this problem that m can have infinite choices but there are only a finite number of isotropic non-totalistic rules with standard moore neighborhood. So the remaining work is to show that there are also a finite number of <=4 cell constellations to make it a spaceship in any such rules. I believe that the cells cannot be seperated too far apart, otherwise they will not interact or if they still interact it will be explosive (such as rules having a B1 or B2) and cannot form a spaceship. If this is true the conjecture is disproven since a finite number of spaceship choices cannot obtain an infinite number of orthogonal speeds.WhiteHawk wrote: April 4th, 2026, 9:28 pm conjecture: any orthogonal speed can be achieved with at most 4 cells in isotropic non-totalistic rules with standard moore neighborhood - I think this has been proven for speeds of the form 1/m (m being any integer greater than or equal to 1 - 1c/m would technically also be right, where c is speed of light)
Also, is there a similar proof for diagonals of the form 1c/m
I was thinking there might be an adjustable ship of some sort that would permit proof of this concept. I could be wrong and the low bound may be 5 cells, but I think whoever runs 5S may have some partial answerOhhhhhhhhh wrote: July 2nd, 2026, 8:40 pmThere is this problem that m can have infinite choices but there are only a finite number of isotropic non-totalistic rules with standard moore neighborhood. So the remaining work is to show that there are also a finite number of <=4 cell constellations to make it a spaceship in any such rules. I believe that the cells cannot be seperated too far apart, otherwise they will not interact or if they still interact it will be explosive (such as rules having a B1 or B2) and cannot form a spaceship. If this is true the conjecture is disproven since a finite number of spaceship choices cannot obtain an infinite number of orthogonal speeds.WhiteHawk wrote: April 4th, 2026, 9:28 pm conjecture: any orthogonal speed can be achieved with at most 4 cells in isotropic non-totalistic rules with standard moore neighborhood - I think this has been proven for speeds of the form 1/m (m being any integer greater than or equal to 1 - 1c/m would technically also be right, where c is speed of light)
Also, is there a similar proof for diagonals of the form 1c/m
This is really hard. There are no easy proof for a pattern to be non-glider-constructible, except Garden of Edens, but obviously these patterns do not have any constructions too.vilc wrote: June 27th, 2026, 1:41 pm Here is an attempt at giving a more formal definition of some problems on general construction vs. glider construction. Define a construction of a pattern A_0 to be a sequence of patterns (A_n) for n >= 0 such that :
- for all positive n, A_n is a predecessor of A_{n-1}
- the patterns (A_n) have uniformly bounded population
- every cell in the plane is on at most a finite number of times in the sequence (A_n)
A valid glider synthesis of A_0 implies a construction. Is the converse true? Are there patterns with a construction but no spaceship synthesis?
This definition can be generalised naturally in any rule with a "quiescent" state (a state which remains the same when surrounded by like neighbours). Can this question be answered some rules with spaceships?
Oh right, then your statement is very likely to be true. I believe there are certain rules (actually very easy to construct!) where such ships exists. For any orthogonal speed to be achieved though we might need one more cell to allow another integer distance as a piece information, since I don't think we have such brilliant rules where ships are adjustable with speed behaving according to Cantor's Matrix (that thing used to show that rationals have the same cardinal as naturals). Even with 5 cells there has to be very careful design of spaceships as the speed can be some near-half-c thing and the 3-cell ship is too slow processing rational speeds represented by the 2 single cells. One thing to clarify: does the ship itself has to be k-celled or it can be generated from a k-cell constellation? That could make a LOT of difference since explosive rules can literally create anything. I would then assume you mean the ship can be generated but the original k cells must not create anything other than the ship.WhiteHawk wrote: July 2nd, 2026, 8:43 pmI was thinking there might be an adjustable ship of some sort that would permit proof of this concept. I could be wrong and the low bound may be 5 cells, but I think whoever runs 5S may have some partial answerOhhhhhhhhh wrote: July 2nd, 2026, 8:40 pmThere is this problem that m can have infinite choices but there are only a finite number of isotropic non-totalistic rules with standard moore neighborhood. So the remaining work is to show that there are also a finite number of <=4 cell constellations to make it a spaceship in any such rules. I believe that the cells cannot be seperated too far apart, otherwise they will not interact or if they still interact it will be explosive (such as rules having a B1 or B2) and cannot form a spaceship. If this is true the conjecture is disproven since a finite number of spaceship choices cannot obtain an infinite number of orthogonal speeds.WhiteHawk wrote: April 4th, 2026, 9:28 pm conjecture: any orthogonal speed can be achieved with at most 4 cells in isotropic non-totalistic rules with standard moore neighborhood - I think this has been proven for speeds of the form 1/m (m being any integer greater than or equal to 1 - 1c/m would technically also be right, where c is speed of light)
Also, is there a similar proof for diagonals of the form 1c/m
Wrong, since arbitrarily large square patches of zebra stripes can be constructed.NNlk05 wrote: July 8th, 2026, 8:38 am Conjecture: As n tends to infinity, there exists a upper bound of the max density still life smaller then nxn constructible by gliders that is less then the still life density cap 1/2.
Huh. How?vilc wrote: July 8th, 2026, 9:49 amWrong, since arbitrarily large square patches of zebra stripes can be constructed.NNlk05 wrote: July 8th, 2026, 8:38 am Conjecture: As n tends to infinity, there exists a upper bound of the max density still life smaller then nxn constructible by gliders that is less then the still life density cap 1/2.
Here's a 13x13 example:NNlk05 wrote: July 8th, 2026, 7:23 pmHuh. How?vilc wrote: July 8th, 2026, 9:49 amWrong, since arbitrarily large square patches of zebra stripes can be constructed.NNlk05 wrote: July 8th, 2026, 8:38 am Conjecture: As n tends to infinity, there exists a upper bound of the max density still life smaller then nxn constructible by gliders that is less then the still life density cap 1/2.
Code: Select all
x = 15, y = 19, rule = LifeHistory
4.2A2.A2.A$4.2A2.4A$.A$.13C$.13DA$.13C$A13D$.13C$.13DA$.13C$A13D$.13C
$.13DA$.13C$A13D$.13C$13.A$3.4A2.2A$3.A2.A2.2A!
You can use the method displayed in the long^N clock syntheses (see the continuous recipe). It would look like this, with a fast component to lay blocks on the ends :
Code: Select all
x = 49, y = 199, rule = B3/S23
32b2o$31bobo$31bobobo$32bobobo$34bobo$34b2o13$16bo3bo3bo3bo3b2o$12b2ob
obobobobobobobobobo$12b2obobobobobobobobobobobo$16bo3bo3bo3bo3bobobo$
34bobo$34b2o23$16b2ob2o3b2ob2o3b2o$12b2obobobobobobobobobobo$12b2obobo
bobobobobobobobobo$16bo3bo3bo3bo3bobobo$34bobo$34b2o16$15b2o3b2ob2o3b
2ob2o$15bobobobobobobobobobo$12b2obobobobobobobobobobo$12b2obobobobobo
bobobobobobo$16bo3bo3bo3bo3bobobo$34bobo$34b2o9$25bobo$26b2o$26bo2$37b
o$28bobo5bo$28b2o6b3o$29bo$26bo7bo$24bobo6bo$25b2o6b3o5bo$40bo6bo$40b
3o3bo$46b3o4$15b2o3b2ob2o3b2ob2o$15bobobobobobobobobobo$12b2obobobobob
obobobobobo$12b2obobobobobobobobobobobo$16bo3bo3bo3bo3bobobo$34bobo6b
3o$34b2o7bo$44bo22$16b2ob2o3b2ob2o3b2ob2o$15bobobobobobobobobobob2o$
15bobobobobobobobobobo$12b2obobobobobobobobobobo$12b2obobobobobobobobo
bobobo$16bo3bo3bo3bo3bobobo$34bobo$34b2o9$21bobo$21b2o$22bo2$11bo$12bo
5bobo$10b3o6b2o$19bo$14bo7bo$15bo6bobo$7bo5b3o6b2o$bo6bo$2bo3b3o$3o4$
16b2ob2o3b2ob2o3b2ob2o$15bobobobobobobobobobob2o$15bobobobobobobobobob
o$12b2obobobobobobobobobobo$12b2obobobobobobobobobobobo$3b3o10bo3bo3bo
3bo3bobobo$5bo28bobo$4bo29b2o17$16b2ob2o3b2ob2o3b2o$12b2obobobobobobob
obobobo$12b2obobobobobobobobobobob2o$15bobobobobobobobobobob2o$15bobob
obobobobobobobo$12b2obobobobobobobobobobo$12b2obobobobobobobobobobob2o
$15bobobobobobobobobobob2o$15bobobobobobobobobobo$12b2obobobobobobobob
obobo$12b2obobobobobobobobobobobo$16bo3bo3bo3bo3bobobo$34bobo$34b2o!