## Thread for basic questions

- testitemqlstudop
**Posts:**1362**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
**Contact:**

### Re: Thread for basic questions

It depends on your cpu and hardware, but less than bitcoin mining.

- Moosey
**Posts:**3277**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Thread for basic questions

Has anyone thought to look for ships like the copperhead or loafer recently--small+slow ships nobody thought to look for?

Something like a c/17 or something else that would be moderately slow, but slower than any known small-scale ships?

Something like a c/17 or something else that would be moderately slow, but slower than any known small-scale ships?

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Thread for basic questions

It's funny that you chose c/17. That's only one away from the classic hypothetical c/18 spaceship -- it's like you're channeling muzik from a few years ago.Moosey wrote:Has anyone thought to look for ships like the copperhead or loafer recently--small+slow ships nobody thought to look for?

Something like a c/17 or something else that would be moderately slow, but slower than any known small-scale ships?

My long answer then looked like this. Read a few posts down for an estimate of search time for period 14, then add ten more zeroes (or maybe a hundred, depending on your assumptions) to get to period 17.

Short answer: yes, a number of people have definitely thought to look, but fewer people have actually done much looking.

There don't seem to be all that many people willing to devote actual months or years of CPU time to probably-hopeless searches. I think Sokwe's page on the LifeWiki is a pretty good summary of what searches have actually been done. The rows in these tables give some rough indication of what people with lots of experience in spaceship-searching think is the outer edge of what might be vaguely worth trying. Notice there's nothing above period 10, let alone period 14 or 17 or 18 or 26.

It's absolutely possible that some search utility, set up in just the right way, will have a stroke of incredible luck and return a tiny c/17 ship after just a few minutes of searching. It just doesn't seem terribly likely. After someone leaves their computer running for a few weeks or months or years, on searches that are guaranteed not to complete before the Sun swallows the Earth and/or burns out ... well, enthusiasm starts to wane a little bit.

That just means that these kinds of searches will have to be done by whoever is still enthusiastic about running them!

- Moosey
**Posts:**3277**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Thread for basic questions

Would it be plausible, though, to run a quick search for small ships across many speeds and eliminate all within a slightly-larger-than-a-copperhead-sized bounding box in a reasonable amount of time?

EDIT:

EDIT:

Classical muzikdvgrn wrote:muzik from a few years ago

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Thread for basic questions

Vaguely plausible, sure. Except we might already have covered the speeds and sizes for which this kind of thing can be done.Moosey wrote:Would it be plausible, though, to run a quick search for small ships across many speeds and eliminate all within a slightly-larger-than-a-copperhead-sized bounding box in a reasonable amount of time?

You might eliminate all the bilaterally symmetric spaceships up to c/10 up to a little over Copperhead size, for example, but I think that's been done already. As you increase the period, each search is a thousand or a million times harder to complete than the last one, at the same size... so you have to shrink the bounding box fairly quickly if you want to keep running searches that will complete in your lifetime.

Otherwise you can't actually eliminate anything useful, unless you want to start keeping track of having completed "the first 0.00000000001% of a standard {pick your program} search for [bilaterally symmetric/short wide/long skinny/blobby] c/14 spaceships". Up to a certain size, maybe 7x7 with a clever distributed search, we could eventually run all possible configurations through apgsearch/Catagolue and see if anything comes out with

**any**speed. But above that size it's hard to see how to definitely eliminate things with periods above 10 or 12 or so.

The counterintuitive thing is that making a search utility that's a million times more efficient, or a computer that's a million times faster, doesn't really get you very much farther in this kind of search. If we can eliminate a Copperhead-sized c/11 spaceship with a month-long search now (let's say), then after this big hypothetical factor-of-a-million improvement you _might_ be able to eliminate a Copperhead-sized c/12 spaceship with a month-long search. But then that's it. It almost instantly feels like you're right back where you started... and c/17 is still just about as far out of reach as it was before.

### Re: Thread for basic questions

I believe Moosey was specifically asking about doing an apgsearch type search. Various such searches have been done, but I don't know up to what size bounding box. David Eppstein wrote gsearch for this purpose, and various other similar searches have been run by others.

The latest version of the 5S Project contains over 226,000 spaceships. There is also a GitHub mirror of the collection. Tabulated pages up to period 160 (out of date) are available on the LifeWiki.

- Moosey
**Posts:**3277**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Thread for basic questions

What

EDIT:

Are there any conduits that can attach to this and eventually to a block factory?
This block factory (credit: a for awesome) is the first I noticed that works:
Needless to say, these copperhead-to-Gs will have even longer repeat times than existing ones. Especially if I use a B-not-a-conduit which just cleans up the B junk:

*is*the smallest CC reflector?EDIT:

Are there any conduits that can attach to this and eventually to a block factory?

Code: Select all

```
x = 20, y = 51, rule = LifeHistory
9.2A$9.2A9$9.D.D$9.D2.D$8.D3.D$9.D2.D5.A$9.3D5.A.A$18.A5$15.2A$15.A.A
$17.A$8.2D2.D4.2A$9.3D$10.D13$4.2A$3.4A2$2.6A$3.4A2$2.2A2.2A$2A.A2.A.
2A$3.A2.A3$4.2A$4.2A!
```

Code: Select all

```
x = 20, y = 56, rule = LifeHistory
6.3C$6.C$5.3C5.2A.A$13.A.2A2$9.2B$9.2B9$9.D.D$9.D2.D$8.D3.D$9.D2.D5.A
$9.3D5.A.A$18.A5$15.2A$15.A.A$17.A$8.2D2.D4.2A$9.3D$10.D13$4.2A$3.4A
2$2.6A$3.4A2$2.2A2.2A$2A.A2.A.2A$3.A2.A3$4.2A$4.2A!
```

Code: Select all

```
x = 66, y = 70, rule = LifeHistory
4.2A.A$4.A.2A2$5.5A$2A2.A4.A20.2A$A2.A2.A23.A$.A.A.2A21.A.A$2A.A5.A
18.2A$3.A4.A.A$3.2A2.A2.A$8.2A$39.2A$39.A$37.A.A$37.2A13.3C$52.C$19.
2A30.3C5.2A.A$19.2A38.A.2A2$55.2B$55.2B9$29.A25.D.D$28.A.A24.D2.D$28.
A.A23.D3.D$29.A25.D2.D5.A$55.3D5.A.A$64.A5$61.2A$61.A.A$63.A$54.2D2.D
4.2A$55.3D$56.D13$50.2A$49.4A2$48.6A$49.4A2$48.2A2.2A$46.2A.A2.A.2A$
49.A2.A3$50.2A$50.2A!
```

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Thread for basic questions

Not quite sure these days. It's definitely not the one that the LifeWiki / Life Lexicon stable reflector article says is "among the smallest" -- that's 50x37. We really ought to do a thorough survey with syringes plus all known H-to-Gs. The smallest CC reflector in simeks' collection seems to be this 52x30 one:Moosey wrote:Whatisthe smallest CC reflector?

Code: Select all

```
#C syringe + HNE5T-4
x = 55, y = 33, rule = B3/S23
bo$2bo$3o$40bo$40b3o$43bo$30b2o10b2o$31bo$31bobo19b2o$21bo10b2o19bo$
19b3o29bobo$18bo32b2o$18b2o$3b2o$4bo$4bob2o43b2o$5bo2bo42bobo$6b2o45bo
$21b2o24b2o4b2o$21b2o24bobo$49bo$49b2o2$30bo3b2o$29bobo3bo$28bobo3bo$
24b2obobo3bo$24b2obo2b4obo$28bobo3bobo$24b2ob2o2bo2bobo$25bobo2b2o3bo$
13b2o10bobo$13b2o11bo!
```

Code: Select all

```
#C syringe + HSE18T9
x = 55, y = 37, rule = LifeHistory
27.A$26.A.A4.2A$26.A.A3.B2AB$25.2A.2A2.3B7.A$25.A2.B3.2B6.3A$26.AB2AB
.3B4.A$24.A.A.2A5B4.2A$24.2A4.7B.3B$31.8B$3.A26.10B$4.A25.5B2A2B$2.3A
B23.6B2A2B$3.4B22.9B.B3.2B$4.4B10.A10.8B.3B.4B$5.4B7.3A11.17B$6.4B5.A
14.17B$7.4B4.2A14.17B$2A6.9B14.16B$.A7.6B14.18B$.A.2A5.6B3.B2.2B2.21B
$2.A2.A4.19BD18B$3.2AB3.6BD13BDBD4B.7B.4B$4.9BDBD2B2A9B3D4B.7B2.4B$5.
9B2D2B2A11BD4B.7B3.4B$6.29B3.5B5.4B$6.17B.B13.7B4.4B$7.15B15.2B4.2A5.
4B$7.15B15.2AB3.A7.4B$8.13B5.A3.2A.2A.A.A5.3A5.3B$10.13B2.A.A3.A.A.2A
.A.A5.A6.2B$9.8B4.2A.A.A3.A8.2A13.B$9.6B6.2ABA4.A$9.5B8.B2.5A.A$9.B.B
5.2A.A.2A.A4.A.A$10.3B4.A.2A.A2.A.2A2.A$9.B2AB11.2A.A.2A$10.2A!
```

Code: Select all

```
#C syringe + HSE22T-8
x = 53, y = 29, rule = LifeHistory
37.A$3.A33.3A$4.A23.A11.A$2.3AB22.3A8.2A5.4B$3.4B24.A7.11B$4.4B10.A
11.2A3.B5.11B$5.4B7.3A11.8B2.11B2A$6.4B5.A16.19B2A$7.4B4.2A15.20B$2A
6.9B14.20B$.A7.6B14.21B$.A.2A5.6B3.B2.2B2.25B$2.A2.A4.19BD14B4.4B$3.
2AB3.6BD13BDBD4B.6B6.4B$4.9BDBD2B2A9B3D4B2.B.5B5.3B$5.9B2D2B2A11BD4B
7.2A6.2B$6.29B8.A8.B$6.17B.B.2B16.3A$7.15B4.B19.A$7.15B$8.13B5.A3.2A$
10.13B2.A.A3.A$9.8B4.2A.A.A3.A$9.6B6.2ABA4.A$9.5B8.B2.5A.A$9.B.B5.2A.
A.2A.A4.A.A$10.3B4.A.2A.A2.A.2A2.A$9.B2AB11.2A.A.2A$10.2A!
```

Yeah, the big problem is that the output travels back in the direction the copperhead comes from. There probably aren't any known conduits you could attach to that B output that would magically push it sideways out of the copperhead track... the conduit's catalysts are almost guaranteed to get in the way of the copperhead.Needless to say, these copperhead-to-Gs will have even longer repeat times than existing ones. Especially if I use a B-not-a-conduit which just cleans up the B junk...

So you'd have to add another sacrificial bait object to absorb the B, get an output somehow, and then clean up the leftover mess. Probably better to just start over and look for a different initial bait.

- testitemqlstudop
**Posts:**1362**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
**Contact:**

### Re: Thread for basic questions

How do I make a History rule of a non-totalistic rule, like the ones displayed in LV (B2c3aei4ajnr5acn/S2-ci3-ck4in5jkq6c7cHistory)

### Re: Thread for basic questions

This is a case where a rule-generator script has been needed for a long time, but I don't think anyone ever wrote one. At least I hope not, because if someone did then I just wasted an hour of fiddly script-writing work.testitemqlstudop wrote:How do I make a History rule of a non-totalistic rule, like the ones displayed in LV (B2c3aei4ajnr5acn/S2-ci3-ck4in5jkq6c7cHistory)

Version 1.0 of a History-rule-making script is checked in at isotropic-history-rule-gen.py.

**UPDATE:**Probably just use isotropic-markedhistory-rule-gen.py instead -- it's backward compatible with three-state History patterns, but also supports marked states 3, 4, and 5 and a state-6 impervious boundary.

I've only tested it on Just Friends and the B2c3aei4ajnr5acn/S2-ci3-ck4in5jkq6c7c rule mentioned above, so it would be good if someone could try this on a few other rules and report back if there are any discrepancies between a plain rule R and its RHistory version. An alpha version of this script subtly re-used a variable, which caused very rare incorrect behavior, so more bugs are certainly possible.

The new script and the original one probably ought to be ported to Lua as well, unless someone has done that already for isotropic-rule-gen.py.

### Re: Thread for basic questions

What are the smallest reflectors of all of these types:

0* CP

0* CC

90* CP

90* CC

180* CP

180* CC

by bounding box and regardless of period?

0* CP

0* CC

90* CP

90* CC

180* CP

180* CC

by bounding box and regardless of period?

My favourite oscillator of all time

Code: Select all

```
x = 15, y = 13, rule = B3/S23
7bo2$3b2o5b2o$b2o4bo4b2o$5b2ob2o$bobo7bobo$bo2bobobobo2bo$5obobob5o$o
4bo3bo4bo$b3obobobob3o$3bob2obo2bo$8bobo$8b2o!
```

### Re: Thread for basic questions

0* CP could be just empty space, couldn't it?Gustone wrote:What are the smallest reflectors ... by bounding box and regardless of period?

0* CC is probably some combination of bumper and bouncer -- see below.

90* CP is the 17x11 p8 bumper, I think, barely beating the p4 bumper at 19x10. Snarks are significantly bigger (23x17).

90* CC is the p8 bouncer, which is only 16x13.

180* CP is probably a pair of bumpers: two p4 bumpers fit into 29x25, and two p8 bumpers can pack into 28x25, which is a lot better than Snarks can manage even with the minimal weld (38x31), and it even edges out two p8 bouncers (30x24).

Code: Select all

```
x = 28, y = 25, rule = LifeHistory
24.2A$8.BABA12.A$9.B2AB5.B2.BA.A$10.A3B3.3B.B2A$11.4B.2BA3B$4.B7.4B2A
B2AB$4.2B7.4BA2BAB$4.3B7.4B2AB$4.4B5.A7B$5.4B3.A7B$6.4B.B3A2.3B$7.7B
3.3B$8.5B6.3B$8.6B5.4AB2.B$6.3B2A4B3.A7B2A$2.2A2.2BA2BA4B2.A3B2A2B2A$
.BABA4BABA3B4.A2.2A3B$2.B3A3.BA3B10.3B$2.2B2A5.5B$.2A2B9.2A$.3AB9.A$
4B11.3A$4B13.A$2B2AB$2.2A!
```

- testitemqlstudop
**Posts:**1362**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
**Contact:**

### Re: Thread for basic questions

Can I have one that also includes the standard LifeHistory red cells?dvgrn wrote:This is a case where a rule-generator script has been needed for a long time, but I don't think anyone ever wrote one. At least I hope not, because if someone did then I just wasted an hour of fiddly script-writing work.testitemqlstudop wrote:How do I make a History rule of a non-totalistic rule, like the ones displayed in LV (B2c3aei4ajnr5acn/S2-ci3-ck4in5jkq6c7cHistory)

Version 1.0 of a History-rule-making script is checked in at isotropic-history-rule-gen.py.

### Re: Thread for basic questions

Sure, as long as you don't mind doing the beta testing. Let's move this kind of thing to the Rule request thread, though, since this doesn't really seem like a "basic question" any more.testitemqlstudop wrote:Can I have one that also includes the standard LifeHistory red cells?

### Re: Thread for basic questions

My favourite oscillator of all time

Code: Select all

```
x = 15, y = 13, rule = B3/S23
7bo2$3b2o5b2o$b2o4bo4b2o$5b2ob2o$bobo7bobo$bo2bobobobo2bo$5obobob5o$o
4bo3bo4bo$b3obobobob3o$3bob2obo2bo$8bobo$8b2o!
```

### Re: Thread for basic questions

I don't know an official name for that particular blob. At T=12 it turns into two V sparks, same as a pentadecathlon puts out -- "V spark" is pretty official.Gustone wrote:Is there anofficialnamefor this spark?Code: Select all

`x = 3, y = 4, rule = b3/s23 bo$3o$3o$bo!`

### Re: Thread for basic questions

I may or may not have heard of it being referred to as "table spark".

Code: Select all

```
x = 4, y = 2, rule = B3/S23
4o$o2bo!
```

Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!

- A for awesome
**Posts:**2037**Joined:**September 13th, 2014, 5:36 pm**Location:**0x-1-
**Contact:**

### Re: Thread for basic questions

"Line spark" might be a good name for it, too, given gen 2:

Code: Select all

```
x = 1, y = 6, rule = B3/S23
o$o$o$o$o$o!
```

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

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

### Re: Thread for basic questions

This may be off-topic but...

I download the software Ready under gollygang umbrella, and found it is capable of generating finite Penrose tiling by subdivision. Can I modify anything so that it can be applied to custom subdivision rules and effectively leading to other aperiodic 2d tilings?

I download the software Ready under gollygang umbrella, and found it is capable of generating finite Penrose tiling by subdivision. Can I modify anything so that it can be applied to custom subdivision rules and effectively leading to other aperiodic 2d tilings?

Glimmering Garden是怎么回事呢？各向同性非总和性细胞自动机相信大家都很熟悉，但是Glimmering Garden是怎么回事呢，下面就让GUYTU6J带大家一起了解吧。

---

Someone please find a use for this:

---

Someone please find a use for this:

Code: Select all

```
x = 9, y = 7, rule = B3/S23
6bo$6bobo$5bo2bo$b2o3b2o$o2bo$bobo$2bo!
```

### Re: Thread for basic questions

Interesting question. There are two different sizes of Penrose tilings shown in the CellularAutomata/PenroseTilings folder. But the meshes for each are hard-coded into the .vtu files:GUYTU6J wrote:I download the software Ready under gollygang umbrella, and found it is capable of generating finite Penrose tiling by subdivision. Can I modify anything so that it can be applied to custom subdivision rules and effectively leading to other aperiodic 2d tilings?

Code: Select all

```
<UnstructuredGrid>
<Piece NumberOfPoints="3126" NumberOfCells="3010" >
<PointData>
</PointData>
<CellData>
<DataArray type="Float32" Name="a" format="appended" RangeMin="0" RangeMax="2" offset="0" />
</CellData>
<Points>
<DataArray type="Float32" Name="Points" NumberOfComponents="3" format="appended" RangeMin="0" RangeMax="1.0000004786" offset="84" />
</Points>
<Cells>
<DataArray type="Int64" Name="connectivity" format="appended" RangeMin="" RangeMax="" offset="14892" />
<DataArray type="Int64" Name="offsets" format="appended" RangeMin="" RangeMax="" offset="37152" />
<DataArray type="UInt8" Name="types" format="appended" RangeMin="" RangeMax="" offset="42492" />
</Cells>
</Piece>
</UnstructuredGrid>
<AppendedData encoding="base64">
{blah blah blah blah}
```

**New Pattern > Penrose Tiling...**in Ready:

Code: Select all

```
/// Make a planar Penrose tiling, using either rhombi (type=0) or darts and kites (type=1).
void MeshGenerators::GetPenroseTiling(int n_subdivisions,int type,vtkUnstructuredGrid* mesh,int n_chems,int data_type)
{
// Many thanks to Jeff Preshing: http://preshing.com/20110831/penrose-tiling-explained
const int RHOMBI = 0;
const int DARTS_AND_KITES = 1;
// we keep a list of the 'red' and 'blue' Robinson triangles and use 'deflation' (decomposition)
vector<Tri> red_tris[2],blue_tris[2]; // each list has two buffers
int iCurrentBuffer = 0;
vtkSmartPointer<vtkPoints> pts = vtkSmartPointer<vtkPoints>::New();
TPairIndex edge_splits; // given a pair of point indices, what is the index of the point made by splitting that edge?
// start with 10 red triangles in a wheel, to get a nice circular shape (with 5-fold rotational symmetry)
// (any correctly-tiled starting pattern will work too)
const int NT = 10;
const double angle_step = 2.0 * 3.1415926535 / NT;
const double goldenRatio = (1.0 + sqrt(5.0)) / 2.0;
const double scale = pow( goldenRatio, n_subdivisions );
pts->InsertNextPoint(0,0,0);
for(int i=0;i<NT;i++)
{
pts->InsertNextPoint(scale*cos(angle_step*i),scale*sin(angle_step*i),0);
int i1 = (i + i%2) % NT;
int i2 = (i + 1 - i%2) % NT;
double angle1 = angle_step * i1;
double angle2 = angle_step * i2;
double p1[3] = { scale*cos(angle1), scale*sin(angle1), 0 };
double p2[3] = { scale*cos(angle2), scale*sin(angle2), 0 };
switch(type) {
default:
case RHOMBI:
red_tris[iCurrentBuffer].push_back(Tri(0,0,0,p1[0],p1[1],1+i1,p2[0],p2[1],1+i2));
break;
case DARTS_AND_KITES:
red_tris[iCurrentBuffer].push_back(Tri(p1[0],p1[1],1+i1,0,0,0,p2[0],p2[1],1+i2));
break;
}
}
// subdivide
double px,py,qx,qy,rx,ry;
for(int i=0;i<n_subdivisions;i++)
{
int iTargetBuffer = 1-iCurrentBuffer;
red_tris[iTargetBuffer].clear();
blue_tris[iTargetBuffer].clear();
// subdivide the red triangles
for(vector<Tri>::const_iterator it = red_tris[iCurrentBuffer].begin();it!=red_tris[iCurrentBuffer].end();it++)
{
switch(type)
{
default:
case RHOMBI:
{
int iP = SplitEdge(*it,0,1,px,py,edge_splits,pts); // split A and B to get a new point P
red_tris[iTargetBuffer].push_back(Tri(it->p[2][0],it->p[2][1],it->index[2],px,py,iP,it->p[1][0],it->p[1][1],it->index[1]));
blue_tris[iTargetBuffer].push_back(Tri(px,py,iP,it->p[2][0],it->p[2][1],it->index[2],it->p[0][0],it->p[0][1],it->index[0]));
break;
}
case DARTS_AND_KITES:
{
int iQ = SplitEdge(*it,0,1,qx,qy,edge_splits,pts); // split A and B to get point Q
int iR = SplitEdge(*it,1,2,rx,ry,edge_splits,pts); // split B and C to get point R
blue_tris[iTargetBuffer].push_back(Tri(rx,ry,iR,qx,qy,iQ,it->p[1][0],it->p[1][1],it->index[1]));
red_tris[iTargetBuffer].push_back(Tri(qx,qy,iQ,it->p[0][0],it->p[0][1],it->index[0],rx,ry,iR));
red_tris[iTargetBuffer].push_back(Tri(it->p[2][0],it->p[2][1],it->index[2],it->p[0][0],it->p[0][1],it->index[0],rx,ry,iR));
break;
}
}
}
// subdivide the blue triangles
for(vector<Tri>::const_iterator it = blue_tris[iCurrentBuffer].begin();it!=blue_tris[iCurrentBuffer].end();it++)
{
switch(type)
{
default:
case RHOMBI:
{
int iQ = SplitEdge(*it,1,0,qx,qy,edge_splits,pts); // split B and A to get point Q
int iR = SplitEdge(*it,1,2,rx,ry,edge_splits,pts); // split B and C to get point R
red_tris[iTargetBuffer].push_back(Tri(rx,ry,iR,qx,qy,iQ,it->p[0][0],it->p[0][1],it->index[0]));
blue_tris[iTargetBuffer].push_back(Tri(rx,ry,iR,it->p[2][0],it->p[2][1],it->index[2],it->p[0][0],it->p[0][1],it->index[0]));
blue_tris[iTargetBuffer].push_back(Tri(qx,qy,iQ,rx,ry,iR,it->p[1][0],it->p[1][1],it->index[1]));
break;
}
case DARTS_AND_KITES:
{
int iP = SplitEdge(*it,2,0,px,py,edge_splits,pts); // split C and A to get point P
blue_tris[iTargetBuffer].push_back(Tri(it->p[1][0],it->p[1][1],it->index[1],px,py,iP,it->p[0][0],it->p[0][1],it->index[0]));
red_tris[iTargetBuffer].push_back(Tri(px,py,iP,it->p[2][0],it->p[2][1],it->index[2],it->p[1][0],it->p[1][1],it->index[1]));
break;
}
}
}
iCurrentBuffer = iTargetBuffer;
}
// merge triangles that have abutting open edges into quads
vtkSmartPointer<vtkCellArray> cells = vtkSmartPointer<vtkCellArray>::New();
{
vector<Tri> all_tris(red_tris[iCurrentBuffer]);
all_tris.insert(all_tris.end(),blue_tris[iCurrentBuffer].begin(),blue_tris[iCurrentBuffer].end());
TPairIndex half_quads; // for each open edge, what is the index of its triangle?
TPairIndex::const_iterator found;
for(int iTri = 0; iTri<(int)all_tris.size(); iTri++)
{
// is this the other half of a triangle we've seen previously?
pair<int,int> edge(all_tris[iTri].index[1],all_tris[iTri].index[2]);
found = half_quads.find(edge);
if(found!=half_quads.end())
{
// output a quad (no need to store the triangle)
cells->InsertNextCell(4);
cells->InsertCellPoint(all_tris[iTri].index[0]);
cells->InsertCellPoint(all_tris[iTri].index[1]);
cells->InsertCellPoint(all_tris[found->second].index[0]);
cells->InsertCellPoint(all_tris[iTri].index[2]);
}
else
{
// this triangle has not yet found its other half so store it for later
half_quads[edge] = iTri;
}
}
}
mesh->SetPoints(pts);
mesh->SetCells(VTK_POLYGON,cells);
// allocate the chemicals arrays
for(int iChem=0;iChem<n_chems;iChem++)
{
vtkSmartPointer<vtkDataArray> scalars = vtkSmartPointer<vtkDataArray>::Take( vtkDataArray::CreateDataArray( data_type ) );
scalars->SetNumberOfComponents(1);
scalars->SetNumberOfTuples(mesh->GetNumberOfCells());
scalars->SetName(GetChemicalName(iChem).c_str());
scalars->FillComponent(0,0.0f);
mesh->GetCellData()->AddArray(scalars);
}
}
```

The coding for this kind of thing isn't as painful as it might seem, though. For example, see illustrations in p29-30 of Volume 1 of the G4G13 Gift Exchange book, for some recursive subdivisions of Robert Ammann's golden-bee / chair tiling. I have the Python code for generating those meshes, if you're interested. Here's a sample -- runs in Golly, for no particularly good reason; only fifty-some lines of code, including the SVG generation:

Code: Select all

```
import golly as g
import math
import svgwrite
from svgwrite import cm, mm
def goldpt(coord1, coord2):
x1, y1 = coord1; x2, y2 = coord2
x3, y3 = x1+golden*(x2-x1), y1+golden*(y2-y1)
return (x3, y3)
level = 12
dwg = svgwrite.Drawing(filename='c:/your/path/here/goldenbee.svg', debug=True)
scale = 500.0
margin = scale/10
golden = math.sqrt(5)/2.0-0.5
top = 0
left = 0
bottom=scale
right = math.sqrt(golden)*scale
ledge = bottom - golden*(bottom-top)
vert = left+golden*(right-left)
coord1 = (left,top)
coord2 = (left,bottom)
coord3 = (right,bottom)
coord4 = (right, ledge)
coord5 = (vert, ledge)
coord6 = (vert, top)
dwg.viewbox(left-margin, top-margin, right+margin, bottom+margin)
dwg.add(dwg.g(id='demo'))
tilelist = [([coord1, coord2, coord3, coord4, coord5, coord6], 1, 0)]
while level>0:
level -= 1
newtilelist=[]
for tile in tilelist:
coord1, coord2, coord3, coord4, coord5, coord6 = tile[0]
coord7 = goldpt(coord3, coord1)
coord9 = goldpt(coord1, coord2)
coordZ = goldpt(coord3, coord4)
coord8 = goldpt(coordZ, coord9)
othercolor = tile[2]+1 -(3 if tile[2]==2 else 0)
newtilelist.append(([coord2, coord3, coord4, coord7, coord8, coord9],tile[1]+1, tile[2]))
newtilelist.append(([coord9, coord1, coord6, coord5, coord7, coord8], tile[1]+2, othercolor))
tilelist=newtilelist[:]
for tile in tilelist[::-1]:
if tile[2]==0: s="rgb(90%,90%,100%)"
elif tile[2]==1: s="rgb(70%,90%,50%)"
else: s="rgb(100%,80%,90%)"
tiles = dwg.add(dwg.g(stroke='green', fill=s, stroke_width=10.0/tile[1]))
tiles.add(dwg.polygon(tile[0]))
dwg.save()
g.show("Wrote output to goldenbee.svg.")
```

Code: Select all

`pip install svgwrite`

It shouldn't be

**too**big of a job to port something like this to C++ and compile it into Ready, but it's not trivial. At least, I don't understand the details of how to tell Ready about connectivity and so forth. The Python code just generates SVG files directly, so it doesn't have all the data available that Ready would need.

- toroidalet
**Posts:**1082**Joined:**August 7th, 2016, 1:48 pm**Location:**My computer-
**Contact:**

### Re: Thread for basic questions

Are there any septuple+ pseudo still lives (not in life)?

Here is an example of a sextuple pseudo still life, to prove it's possible:
Here's an example in a more lifelike cellular automaton:

Here is an example of a sextuple pseudo still life, to prove it's possible:

Code: Select all

```
x = 9, y = 7, rule = B2cn3c/S012345678
b7o$bo5bo$2obobob2o2$bobobobo$bo5bo$b7o!
```

Code: Select all

```
x = 15, y = 15, rule = B2cn3-kry/S0234i
10bo$5bo2b5o$2bo4bo5bo$4bobobo2bobo$3b2ob2o5b2o$bo11bo$3b2obobo2bobo$
2bobo5bobo$bobo2bobob2o$bo11bo$2o5b2ob2o$bobo2bobobo$bo5bo4bo$2b5o2bo$
4bo!
```

"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."

-Terry Pratchett

-Terry Pratchett

### Re: Thread for basic questions

It seems as if rules with B2 might have more ways to interact to cause or suppress births than rules like Life without B2, but I guess there still has to be some upper limit.toroidalet wrote:Are there any septuple+ pseudo still lives (not in life)?

Here is an example of a sextuple pseudo still life, to prove it's possible...

I don't have an answer, but while looking for prior research on the subject, I ran into some things that Gabriel Nivasch constructed just before making the minimal triple pseudo and quad pseudo.

It looks interesting -- probably not strictly relevant, but possibly worth posting here:

On July 15, 2001, Gabriel Nivasch wrote:Message #1: Blocking crossing

Here is a still life construction that Matthew Cook calls a "blocking

crossing":

..................jj.......jj......

..................jj.......jj......

...................................

..................jjjj.j.jjjj......

..............jj.j..j.jjj.j..j.jj..

..............j.jj...........jj.j..

.....................hh.ii.........

...............ff.h...h.i...i.ee...

..............f.f.hhhh...iiii.e.e..

..............f..................e.

............fff.....hh...ii.......e

...........f.......h.h...i.i.....ee

............fff....h.......i....e..

..............f...hh.......ii...e..

...............ff.............ee...

................f.gg...g...gg.e....

...............f..g.g.g.g.g.g..e...

...............ff...g.g.g.g...ee...

............ff.f...g.g...gg....e.ee

............f..f...g...........e.ee

......ff.ff.f..ff.gg.........ee....

......f.ff.f...........e.ee.e......

....ff.........dd.ee..e.ee.ee......

ff.f...........d...e..e............

ff.f....dd...d.d...e.ee............

...ff...d.d.d.d...ee...............

...f..d.d.d.d.d.d..e...............

....f.dd...d...dd.e................

...ff.............ee...............

..f...cc.......bb...e..............

..f....c.......b....eee............

ff.....c.c...b.b.......e...........

f.......cc...bb.....eee............

.f..................e..............

..f.f.cccc...bbbb.e.e..............

...ff.c...c.b...b.ee...............

.........cc.bb.....................

..a.aa...........aa.a..............

..aa.a..a.aaa.a..a.aa..............

......aaaa.a.aaaa..................

...................................

......aa.......aa..................

......aa.......aa..................

If we want to decompose it into sets that are independently stable, we find

that:

1.- a requires b or c.

2.- If a and b are in the same set, they require e and d.

3.- If a and c are in the same set, they require f and d.

4.- Therefore: a requires d and {e or f}.

5.- Similarly, j requires g and {e or f}.

6.- Of the 4 pieces d, e, f, and g, a single set cannot contain exactly 3 of

them.

It follows that, if the pattern is to be decomposed at all, there are only two

ways of doing it: Either a goes together with e, and j with f; or a goes

together with f, and j with e. (In the first case, c and i have to belong to a

third set. In the second case, b and h have to belong to a third set.)

If we think of the imaginary lines that serve as "frontiers" between the

different sets, then this pattern allows two possibilities: either a line going

diagonally down from NW to SE, or a line going diagonally up from SW to NE.

That's why this pattern is a "blocking crossing".

In his paper "Still Life Theory"

(http://www.paradise.caltech.edu/~cook/W ... index.html), Cook gives a

diagram of the structure of a blocking crossing (p. 17), although he doesn't

show an explicit pattern. This pattern has a simpler structure than Cook's

diagram.

In his paper, Cook uses blocking crossings in a method of constructing a still

life whose decomposition is equivalent to solving a given CNF expression. It

follows that the problem decomposing a still life is NP-complete.

The above pattern has 290 cells. It can probably be improved, since I didn't

spend an enormous amount of time tinkering with it.

On July 16, 2001, Gabriel Nivasch wrote:Message #2: Blocking crossing (2)

I wrote:

> The above pattern has 290 cells.

It can be brought down to 284 cells by replacing the two instances of

o..... o....

oo.... oo...

..o... ..o..

..ooo. ..o..

.....o ...oo

..ooo. ....o

..o... ...o.

o.o... o.o..

oo.... by oo... .

In Matthew's construction, there are many times when a border-line intersects

another one twice, using two blocking crossings, in order to allow using only

one of them, like in this diagram:

-- --

\ /\ /

x x

/ \/ \

-- --

Instead of two blocking crossings, we can use a simpler arrangement that

produces the same effect:

..........aa.......

.........a.a.......

....bb..a...aa.....

....b..a......a.aa.

.....b.aa.....aa.a.

....bb.............

bb.b...xx....x.dd..

bb.b....x..xxx.d.d.

...bb...x.x......d.

...b..x.x.xx....d..

....b.xx........dd.

...bb.............d

....b.yy........dd.

...b..y.y.yy....d..

...bb...y.y......d.

bb.b....y..yyy.d.d.

bb.b...yy....y.dd..

....bb.............

.....b.cc.....cc.c.

....b..c......c.cc.

....bb..c...cc.....

.........c.c.......

..........cc.......

Piece b requires x or y. If b and x go together, they need a and d. If b and y

go together, they need c and d. Therefore, there are two possibilities: Either

b, a, and d go together and are separated from a; or b, c, and d go together

and are separated from c.

So instead of using two blocking crossings, we use half of one.

This arrangement could be called a "blocking parallel pair".

### Re: Thread for basic questions

I now have something for this. I suppose it should be integrated into my sss.py module, but for the time being, here's a standalone Python script (with some relevant parts extracted from sss.py). Apologies for the naming convention mess and for the slightly convoluted way it does things - I cobbled it together from the script I wrote for the recent automated process I ran on 5S.wildmyron wrote:I don't have anything for this.testitemqlstudop wrote:2. How can I merge two minrule/maxrule pairs? (Also script pls)

The example patterns are the T-ship from T-Life and a p2 from DRH-Oscillators (CGoL) which doesn't work in T-Life.

Code: Select all

```
# isorule-union.py
# Merge the min-max rule ranges for two patterns in isotropic 2-state CA rules
#
# Usage: modify the rule range variables at the top of this script and run in Python
from string import replace
# T-ship
patt1RangeStr = 'B3aeijry/S2ac3aeiy - B2in34ceijknqrtwyz5acijknqry678/S2acekn34ceijknqrwyz5acejknqry6aeikn78'
# A p2 oscillator from CGoL which does not work in T-Life
patt2RangeStr = 'B3jnq/S2ak3ay - B2cin34aceinqrty5aceiknry678/S0234ceijnrtwyz5aceijknry678'
Hensel = [
['0'],
['1c', '1e'],
['2a', '2c', '2e', '2i', '2k', '2n'],
['3a', '3c', '3e', '3i', '3j', '3k', '3n', '3q', '3r', '3y'],
['4a', '4c', '4e', '4i', '4j', '4k', '4n', '4q', '4r', '4t', '4w', '4y', '4z'],
['5a', '5c', '5e', '5i', '5j', '5k', '5n', '5q', '5r', '5y'],
['6a', '6c', '6e', '6i', '6k', '6n'],
['7c', '7e'],
['8']
]
transList = [t for tlist in Hensel for t in tlist]
def parseTransitions(ruleTrans):
ruleElem = []
if not ruleTrans:
return ruleElem
context = ruleTrans[0]
bNonTot = False
bNegate = False
for ch in ruleTrans[1:] + '9':
if ch in '0123456789':
if not bNonTot:
ruleElem += Hensel[int(context)]
context = ch
bNonTot = False
bNegate = False
elif ch == '-':
bNegate = True
ruleElem += Hensel[int(context)]
else:
bNonTot = True
if bNegate:
ruleElem.remove(context + ch)
else:
ruleElem.append(context + ch)
return ruleElem
def rulestringopt(a):
result = ''
context = ''
lastnum = ''
lastcontext = ''
for i in a:
if i in 'BS':
context = i
result += i
elif i in '012345678':
if (i == lastnum) and (lastcontext == context):
pass
else:
lastcontext = context
lastnum = i
result += i
else:
result += i
result = replace(result, '4aceijknqrtwyz', '4')
result = replace(result, '3aceijknqry', '3')
result = replace(result, '5aceijknqry', '5')
result = replace(result, '2aceikn', '2')
result = replace(result, '6aceikn', '6')
result = replace(result, '1ce', '1')
result = replace(result, '7ce', '7')
return result
# convert iso-rule string to iso-rule integer
# format: ruleInt = sum(t * 2^i) for i in [ 0, len(BtransList + StransList) )
# where t = 1 if the transition (BtransList + StransList)[i] is in the rule, 0 otherwise
def ruleStr2RuleInt(ruleStr):
if not (ruleStr[0] == 'B' and '/S' in ruleStr):
return
# Parse rule string to list of transitions for Birth and Survival
Bstr, Sstr = ruleStr.split('/')
Btrans = parseTransitions(Bstr.lstrip('B'))
Strans = parseTransitions(Sstr.lstrip('S'))
# Calculate the rule integer
ruleInt = 0
Sidx = len(transList)
for idx, trans in enumerate(transList):
if trans in Btrans:
ruleInt += 2**idx
if trans in Strans:
ruleInt += 2**(idx + Sidx)
return ruleInt
# convert iso-rule integer to iso-rule string
def ruleInt2RuleStr(ruleInt):
Sidx = len(transList)
mask = 2**(Sidx)-1
BInt = ruleInt & mask
SInt = (ruleInt >> Sidx) & mask
# print BInt, SInt
B_elem = []
S_elem = []
for idx, trans in enumerate(transList):
mask = 2**idx
if BInt & mask:
B_elem.append(trans)
if SInt & mask:
S_elem.append(trans)
ruleStr = rulestringopt('B' + ''.join(B_elem) + '/S' + ''.join(S_elem))
return ruleStr
def ruleRangeUnion(ruleInt1, ruleInt2):
if (ruleInt1[0] & ~ruleInt2[1]) or (ruleInt2[0] & ~ruleInt1[1]):
return (0, 0)
return (ruleInt1[0] | ruleInt2[0], ruleInt1[1] & ruleInt2[1])
patt1Range = [ruleStr2RuleInt(r.strip()) for r in patt1RangeStr.split('-')]
patt2Range = [ruleStr2RuleInt(r.strip()) for r in patt2RangeStr.split('-')]
ruleRange = ruleRangeUnion(patt1Range, patt2Range)
print(' - '.join([ruleInt2RuleStr(r) for r in ruleRange]))
```

The latest version of the 5S Project contains over 226,000 spaceships. There is also a GitHub mirror of the collection. Tabulated pages up to period 160 (out of date) are available on the LifeWiki.

- gameoflifemaniac
**Posts:**1026**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Thread for basic questions

This is a phi spark.Gustone wrote:Is there anofficialnamefor this spark?Code: Select all

`x = 3, y = 4, rule = b3/s23 bo$3o$3o$bo!`

I was so socially awkward in the past and it will haunt me for my entire life.

### Re: Thread for basic questions

Not quite. Notice the definitive phi-shaped phase never shows up.gameoflifemaniac wrote:This is a phi spark.Gustone wrote:Is there anofficialnamefor this spark?Code: Select all

`x = 3, y = 4, rule = b3/s23 bo$3o$3o$bo!`

In a phi spark, the terminal three-bit V sparks end up with four blank rows between them, instead of six blank rows as in this case.