3-color immigration

For discussion of other cellular automata.
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Gliderstorm
Posts: 5
Joined: October 21st, 2022, 11:46 pm

3-color immigration

Post by Gliderstorm » November 20th, 2022, 1:25 am

Immigration can work fairly with three colors:

Code: Select all

@RULE Immigration3
@TABLE
n_states:8
neighborhood:Moore
symmetries:permute


0, 1,1,1,0,0,0,0,0, 1
0, 2,2,2,0,0,0,0,0, 2
0, 3,3,3,0,0,0,0,0, 3
0, 4,4,4,0,0,0,0,0, 4


0, 1,1,2,0,0,0,0,0, 1
0, 1,1,3,0,0,0,0,0, 1
0, 2,2,1,0,0,0,0,0, 2
0, 2,2,3,0,0,0,0,0, 2
0, 3,3,1,0,0,0,0,0, 3
0, 3,3,2,0,0,0,0,0, 3


0, 1,2,3,0,0,0,0,0, 4

0, 1,2,4,0,0,0,0,0, 4
0, 1,3,4,0,0,0,0,0, 4
0, 2,3,4,0,0,0,0,0, 4
0, 2,4,4,0,0,0,0,0, 2
0, 3,4,4,0,0,0,0,0, 3
0, 1,4,4,0,0,0,0,0, 1
0, 2,2,4,0,0,0,0,0, 2
0, 3,3,4,0,0,0,0,0, 3
0, 1,1,4,0,0,0,0,0, 1

var s1 = {1,2,3,4}
var s2 = {1,2,3,4}
var s3 = {1,2,3,4}
var s4 = {1,2,3,4}
var s5 = {1,2,3,4}
var s6 = {1,2,3,4}
var s7 = {1,2,3,4}
var s8 = {1,2,3,4}
var s0 = {1,2,3,4}

s0, 0,0,0,0,0,0,0,0, 0
s0, s1,0,0,0,0,0,0,0, 0
s0, s1,s2,s3,s4,0,0,0,0, 0
s0, s1,s2,s3,s4,s5,0,0,0, 0
s0, s1,s2,s3,s4,s5,s6,0,0, 0
s0, s1,s2,s3,s4,s5,s6,s7,0, 0
s0, s1,s2,s3,s4,s5,s6,s7,s8, 0

var a1 = {0,1,2,3,4,5,6,7}
var a2 = {0,1,2,3,4,5,6,7}
var a3 = {0,1,2,3,4,5,6,7}
var a4 = {0,1,2,3,4,5,6,7}
var a5 = {0,1,2,3,4,5,6,7}
var a6 = {0,1,2,3,4,5,6,7}
var a7 = {0,1,2,3,4,5,6,7}
var a8 = {0,1,2,3,4,5,6,7}
var a0 = {0,1,2,3,4,5,6,7}

var w1 = {5,6,7}
var w2 = {5,6,7}
var w3 = {5,6,7}

0, w1,0,0,0,0,0,0,0, 4
0, w1,w2,0,0,0,0,0,0, 4
0, w1,w2,w3,0,0,0,0,0, 4

4, w1,w2,w3,a3,a4,a5,a6,a7, 0

@COLORS
0 0 0 0
1 255 0 0
2 0 255 0
3 0 0 255
4 255 255 255
5 128 128 128
6 96 128 128
7 128 96 128
But what if all three colors meet? They make a colorless cell. In practice, colorless cells are rarely formed and very short-lived. Colorless cells are only created when no color has a majority over the other colors. This happens in only five cases (C=colorless): RGB, CCC, CRG, CRB, CGB. Three-way boundaries are very rare. A two-way boundary between, say an R and G region, is not a refuge for C cells for very long: they will quickly die out and then it is ordinary two-color life.

Here is an example three-color linepuffer. It's hard to notice the colorless cells at all despite close-mingling of all three colors.

Code: Select all

#CXRLE Pos=-7,0
x = 39, y = 156, rule = Immigration3
25.A$25.4A$12.A.A12.2A$12.A2.A14.A$15.2A10.4A$17.A13.A$15.4A8.A2.3A$
14.A4.A9.3A$16.A2.A10.A$16.A2.A5.A.3A$18.A6.2A2.A$12.A.4A8.3A$12.A3.A
9.A$15.A12.A$13.A.A10.A.A$29.A$14.3A9.A.A$15.2A11.A$14.3A9.A$26.3A$
13.A.A9.2A2.A$15.A9.A.3A$12.A3.A13.A$12.A.4A11.3A$18.A8.A2.3A$16.A2.A
11.A$16.A2.A7.4A$14.A4.A10.A$15.4A8.2A$17.A7.4A2.A.A$15.2A8.A5.A2.A$
12.A2.A7.A.A8.2A$12.A.A8.A12.A$21.A2.2A8.4A$9.2A.A9.3A8.A4.A$12.A7.A.
A12.A2.A$9.A.2A7.A14.A2.A$9.A9.2A16.A$7.A7.A.3A11.A.4A$8.A5.A.2A13.A
3.A$8.A5.2A18.A$8.B.A2.2A17.A.A$8.B.A.2A$8.C24.3A$8.C.A.2A20.2A$8.A.A
2.2A5.A.A4.A.A4.2A$8.A5.2A.2A.A2.A2.A2.A.5A$8.B5.2A7.2A3.A4.A$8.B.A2.
2A5.A6.A7.A$8.C.A.2A6.A.5A6.A.A$8.C8.2A17.A$8.A.A.2A6.A.5A6.A.A$8.A.A
2.2A5.A6.A7.A$8.B5.2A7.2A3.A4.A$8.B5.2A.2A.A2.A2.A2.A.5A$8.C.A2.2A5.A
.A4.A.A4.2A$8.C.A.2A20.2A$8.A24.3A$8.A.A.2A$8.B.A2.2A17.A.A$8.B5.2A
18.A$8.C.A2.A.2A14.A3.A$2D6.C.A.3A2.A13.A.4A$2D6.A8.A19.A$8.A.A.3A2.A
17.A2.A$8.B.A2.A.2A18.A2.A$8.B5.2A17.A4.A$8.C.A2.2A19.4A$8.C.A.2A22.A
$8.A25.2A$8.A.A.2A17.A2.A$8.B.A2.2A16.A.A$8.B5.2A$8.C.A2.A.2A$8.C.A.
3A2.A$8.A8.A13.A.A$8.A.A.3A2.A13.A2.A$8.B.A2.A.2A17.2A$8.B5.2A20.A$8.
C.A2.2A19.4A$8.C.A.2A19.A4.A$8.A26.A2.A$8.A.A.2A21.A2.A$8.B.A2.2A22.A
$8.B5.2A15.A.4A$8.C.A2.A.2A14.A3.A$8.C.A.3A2.A16.A$8.A8.A14.A.A$8.A.A
.3A2.A$8.B.A2.A.2A16.3A$8.B5.2A18.2A$8.C.A2.2A5.A.A4.A.A4.2A$8.C.A.2A
3.2A.A2.A2.A2.A.5A$8.A14.2A3.A4.A$8.A.A.2A6.A6.A7.A$8.B.A2.2A5.A.5A6.
A.A$8.B5.2A.2A17.A$8.C5.2A4.A.5A6.A.A$8.C.A2.2A5.A6.A7.A$8.A.A.2A9.2A
3.A4.A$8.A8.2A.A2.A2.A2.A.5A$8.B.A.2A6.A.A4.A.A4.2A$8.B.A2.2A19.2A$8.
C5.2A17.3A$8.C.A2.A.2A$8.A.A.3A2.A14.A.A$8.A8.A16.A$8.B.A.3A2.A13.A3.
A$8.B.A2.A.2A14.A.4A$8.C5.2A21.A$8.C.A2.2A20.A2.A$8.A.A.2A21.A2.A$8.A
24.A4.A$8.B.A.2A20.4A$8.B.A2.2A21.A$8.C5.2A18.2A$8.C5.A.2A13.A2.A$7.A
7.A.3A11.A.A$9.A9.2A$9.A.2A7.A$12.A7.A.A$9.2A.A9.3A$21.A2.2A$12.A.A8.
A$12.A2.A7.A.A$15.2A8.A$17.A7.4A$15.4A8.2A$14.A4.A10.A$16.A2.A7.4A$
16.A2.A11.A$18.A8.A2.3A$12.A.4A11.3A$12.A3.A13.A$15.A9.A.3A$13.A.A9.
2A2.A$26.3A$14.3A9.A$15.2A11.A$14.3A9.A.A$29.A$13.A.A10.A.A$15.A12.A$
12.A3.A9.A$12.A.4A8.3A$18.A6.2A2.A$16.A2.A5.A.3A$16.A2.A10.A$14.A4.A
9.3A$15.4A8.A2.3A$17.A13.A$15.2A10.4A$12.A2.A14.A$12.A.A12.2A$25.4A$
25.A!
Last edited by Gliderstorm on November 20th, 2022, 2:49 am, edited 5 times in total.

User avatar
pzq_alex
Posts: 793
Joined: May 1st, 2021, 9:00 pm
Location: tell me if you know

Re: 3-color immigration

Post by pzq_alex » November 20th, 2022, 2:32 am

Gliderstorm wrote:
November 20th, 2022, 1:25 am
(Lifeviewer waiting on wiki approval to upload rule)
Done!
\sum_{n=1}^\infty H_n/n^2 = \zeta(3)

How much of current CA technology can I redevelop "on a desert island"?

Gliderstorm
Posts: 5
Joined: October 21st, 2022, 11:46 pm

Re: 3-color immigration

Post by Gliderstorm » November 20th, 2022, 2:38 am

Thanks!

User avatar
toroidalet
Posts: 1514
Joined: August 7th, 2016, 1:48 pm
Location: My computer
Contact:

Re: 3-color immigration

Post by toroidalet » November 20th, 2022, 5:45 pm

That's a nice solution to the 3-color Immigration problem, and better than the one I tried{a}. Here are some new spaceships using colorless cells (there don't seem to be any tricolor gliders):

Code: Select all

x = 16, y = 51, rule = Immigration3
11.B2.B$15.B$11.A3.D$12.4A7$11.A2.D$15.D$11.A3.D$12.A3D5$11.A2.D$15.D
$11.A3.D$12.2A2D3$.A$2.D$3D$11.A2.A$15.D$11.A3.D$12.2A2D7$11.A2.B$15.
D$11.A3.D$12.2A2D7$11.D2.B$15.D$11.C3.D$12.2C2D!
This idea naturally generalizes to Immigration for any rule, which gives us such wonders as Immigration Seeds:

Code: Select all

@RULE immigrationseeds
@TABLE
n_states:4
neighborhood:Moore
symmetries:permute
var a={0,1,2,3}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var h=a
var i=a

0,1,1,0,0,0,0,0,0,1
0,1,3,0,0,0,0,0,0,1
0,1,2,0,0,0,0,0,0,3
0,2,2,0,0,0,0,0,0,2
0,2,3,0,0,0,0,0,0,2
0,3,3,0,0,0,0,0,0,3

a,b,c,d,e,f,g,h,i,0

Code: Select all

x = 122, y = 95, rule = immigrationseeds
.B8.AB.CAB2CB.2C2.2C2.B2.C2.C.CB2.C.C2.2B.AC.B2.B2.B2.C3.CAB.A.BA.2A
3.BC.C2.B3.ACA.C$2.A.A5.B2.BCBC3.B.ABC3.A.CB.A2.C.B.2B.C.A2C.AC.A2.A.
2ACB.BAB.BA2.CB.AB2.CAB.2C.CA2C2A.B$4.AC3.2AB2.2B.A.B5.2C2.BC.A.A2CBC
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2B2.C2B3.A2.BC.A2.A.2C!
@RULE immigrationseeds
@TABLE
n_states:4
neighborhood:Moore
symmetries:permute
var a={0,1,2,3}
var b=a
var c=a
var d=a
var e=a
var f=a
var g=a
var h=a
var i=a

0,1,1,0,0,0,0,0,0,1
0,1,3,0,0,0,0,0,0,1
0,1,2,0,0,0,0,0,0,3
0,2,2,0,0,0,0,0,0,2
0,2,3,0,0,0,0,0,0,2
0,3,3,0,0,0,0,0,0,3

a,b,c,d,e,f,g,h,i,0
Here is a blueprint for synthesizing a colorless glider or a one-colored-cell glider:

Code: Select all

x = 48, y = 15, rule = Immigration3
.2A$.A$8.A$AB6.2A.AD7.2D8.2D2.A7.2D3.C$AC9.AC7.DC8.D.D.2A6.D.A.2C$31.
D3.A7.D2.C$19.B.2B5$20.2D7.2D3.A$20.DC7.D.C.2A$30.C2.A$19.C.2C!
{a}: My solution was to say that if there were 3 colors, the cell "in the middle" would transmit its color. This doesn't generalize to all rules, and is not uniquely determined.

EDIT: 8-glider synthesis of a colorless block:

Code: Select all

x = 120, y = 22, rule = Immigration3
52.B$51.B$51.3B$8.B$7.B$7.3B2$98.A$96.A.A$97.2A$39.B$38.B$3A35.3B28.
2A3.A$2.A67.2A.A.D37.2A$.A23.2A42.A3.B.D33.A2.A2.A$25.2B47.B32.A.A2.A
2.A2.AD$108.2A3.2A3.BD3$35.C$34.2C$34.C.C!
19-glider synthesis of a mostly colorless glider:

Code: Select all

x = 176, y = 45, rule = Immigration3
52.B$51.B$51.3B$8.B$7.B$7.3B91.A$102.A$100.3A21.A3.A17.B.B$67.A54.A.A
2.A.A17.2B25.C$68.A54.2A2.A.A17.B22.A2.C$39.B26.3A50.2A7.A40.D.A.3C$
38.B58.3A18.A.A26.DA2.2B16.2D$3A35.3B28.2A3.A24.A20.A26.AD.2B22.A$2.A
67.2A.A.D22.A3.BD23.BD16.2D5.B21.2A$.A23.2A42.A3.B.D26.AD3.A19.AD16.
2D26.A.A$25.2B47.B31.A$106.3A$127.B$126.B.B2.2A$35.C90.B.B2.A.A$34.2C
67.3B21.B3.A$34.C.C66.B$104.B$134.3A$134.A$135.A4$169.A$168.D$103.A
64.3D$5.2A18.A76.A.A$5.2B17.A.D75.A.A$24.B.D76.A23.DA$25.B101.AD16.A$
68.BD55.2D17.D.A$68.AD32.BD21.2D17.2D$102.AD3$102.B$101.B.B$101.B.B$
102.B!
22-glider synthesis of a completely colorless glider (added 1; it was previously invalid):

Code: Select all

x = 213, y = 44, rule = Immigration3
52.B$51.B$51.3B$8.B$7.B$7.3B2$189.A$67.A120.A$68.A33.A47.A.A35.3A$39.
B26.3A32.A49.2A$38.B62.3A47.A33.A10.A$3A35.3B28.2A3.A111.A3.A5.A.A$2.
A67.2A.A.D23.2A45.A37.3A2.A.A4.2A$.A23.2A42.A3.B.D19.BD.2A47.2A40.A2.
A$25.2B47.B20.AD3.A19.BD24.2A34.A7.2A$120.2A38.2A18.A.A$159.2A20.2A
24.C$121.A33.BD4.A46.C2.D$35.C85.2A32.2D31.BD16.3C.A.D$34.2C84.A.A59.
2A4.2D21.2D$34.C.C87.2A55.A2.A22.C$124.A.A27.A22.2A3.A.A21.2C$124.A
28.2A23.2A3.A22.C.C$153.A.A21.A8$5.2A18.A131.A$5.2B17.A.D129.A.A$24.B
.D129.A2.A$25.B131.2A$68.BD$68.AD24.BD23.BD$94.2A23.2D$155.BD$149.2A
4.2D31.D$148.A2.A35.A.D21.D$149.A.A36.2D22.D$150.A59.3D!
I wonder if you can reduce these using a reverse caber tosser.
Any sufficiently advanced software is indistinguishable from malice.

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