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Outrunning the Glider

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Outrunning the Glider

Postby Hdjensofjfnen » February 6th, 2019, 8:45 pm

Post any diagonal spaceships (EDIT: preferably in non-B0 rules) that can outrun the glider (c/4) below!
For starters:
2c/7d, one transition from Life: (EDIT: Shown with glider)
x = 11, y = 11, rule = B3/S234w
9bo$10bo$8b3o4$bo2bo$o3bo$4bo$3bo$3o!

c/3d, a lot of transitions from Life:
x = 3, y = 3, rule = B2e3-j4a/S2-k3-ry4q5i
obo$2bo$3o!
Last edited by Hdjensofjfnen on February 6th, 2019, 8:48 pm, edited 1 time in total.
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Outrunning the Glider

Postby muzik » February 6th, 2019, 8:47 pm

This rule also has a c/3d: viewtopic.php?f=11&t=3194&p=54127#p54076
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
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Re: Outrunning the Glider

Postby PkmnQ » February 7th, 2019, 1:00 am

Here's an alternating rule that has a c/2 that I call "Slider":
@RULE B26S2B358S3

@TABLE
n_states:3
neighborhood:Moore
symmetries:permute

#B26S2
0,1,1,0,0,0,0,0,0,2
0,1,1,1,1,1,1,0,0,2
1,0,0,0,0,0,0,0,0,0
1,1,0,0,0,0,0,0,0,0
1,1,1,0,0,0,0,0,0,2
1,1,1,1,0,0,0,0,0,0
1,1,1,1,1,0,0,0,0,0
1,1,1,1,1,1,0,0,0,0
1,1,1,1,1,1,1,0,0,0
1,1,1,1,1,1,1,1,0,0
1,1,1,1,1,1,1,1,1,0

#B358S3
0,2,2,2,0,0,0,0,0,1
0,2,2,2,2,2,0,0,0,1
0,2,2,2,2,2,2,2,2,1
2,0,0,0,0,0,0,0,0,0
2,2,0,0,0,0,0,0,0,0
2,2,2,0,0,0,0,0,0,0
2,2,2,2,0,0,0,0,0,1
2,2,2,2,2,0,0,0,0,0
2,2,2,2,2,2,0,0,0,0
2,2,2,2,2,2,2,0,0,0
2,2,2,2,2,2,2,2,0,0
2,2,2,2,2,2,2,2,2,0


And here's the ship itself:
x = 3, y = 3, rule = B26S2B358S3
A.A$A$.2A!
x = 3, y = 3, rule = B26/S2|B358/S3
2o$2bo$obo!
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Re: Outrunning the Glider

Postby wildmyron » February 7th, 2019, 5:12 am

There was some discussion about the fastest possible diagonal spaceship speeds in 2-state rules on the Moore nighbourhood, both with and without B0, some time ago but I can't remember which thread it was in. For non-totalistic rules, the limit is c:
x = 1, y = 1, rule = MAPQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
o!

For isotropic rules with B0, the limit is 3c/4. Here's a small one which moves at that speed
x = 3, y = 2, rule = B02akn3n4ak5ak6a/S01e2k4r5a6e7e
2bo$2o!

There's a nice collection of low period diagonal ships in rules with B0 here, many of which are faster than c/4.

In isotropic rules without B0, the limit is c/2:
x = 3, y = 4, rule = B2ac/S1
2bo2$b2o$o!

There are example c/2 ships with many different periods in the 5S diagonal collection. You'll also find example ships for all speeds k/P c for P/4 < k <= P/2, where P < 20 as well as many more for P > 20.

In my collections of small spaceships I have over a hundred such ships which have a minimum population of 3 cells, and there would be far more if I hadn't filtered low period ships out of most of those searches.

Now, if you restrict the rulespace to isotropic rules in which the CGoL glider works, then I suspect the speed limit is c/3, but that's just a guess on my part. Here's a 2c/7 ship racing a glider.
x = 9, y = 7, rule = B3-r4k5jy6e7c/S02-ci3ajnqr4arw5ak6e
7b2o$6b2o$6bo2$2bo$obo$b2o!

If you'd like to find more ships like this, then logic life search is your friend.
./lls -r pB2-ace3aijn4-r5-n678/S02ae3jnr4-k5678 -b 7 -p 7 -x 2 -y 2 -i -s 'D2\'

is the command I used to find the 2c/7 above.

==========

Here are a few of the ships from the 5S collection:

c/2, period 4
x = 3, y = 3, rule = B2-ei3aq4ik5ceq/S02aei3acjn4aijny5ejkn6ein7e
o$b2o$bo!

2c/5, period 2
x = 3, y = 3, rule = B2aci3a4ny5a/S03i4ajw5a7c
2bo2$obo!

c/2, period 6
x = 3, y = 3, rule = B2ac3a/S02-ik3a
o$b2o$bo!

c/3, period 6
x = 3, y = 3, rule = B2cik3a4qt5kny6n/S01e2k3ejkn4ktw5ek6n
2bo2$obo!

3c/7, period 7
x = 3, y = 3, rule = B2-en3ak4iw5acr/S02ek3aiy5ck
2bo2$obo!

c/2, period 8
x = 3, y = 4, rule = B2acn3y4e/S12-ei3c4ey
2bo2$b2o$o!

3c/8, period 8
x = 3, y = 3, rule = B2-ek3ae4r5a/S01e3j4e5ack
2bo2$obo!
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Re: Outrunning the Glider

Postby 77topaz » February 7th, 2019, 5:33 am

wildmyron wrote:Now, if you restrict the rulespace to isotropic rules in which the CGoL glider works, then I suspect the speed limit is c/3, but that's just a guess on my part. Here's a 2c/7 ship racing a glider.
x = 9, y = 7, rule = B3-r4k5jy6e7c/S02-ci3ajnqr4arw5ak6e
7b2o$6b2o$6bo2$2bo$obo$b2o!


Surprisingly, this rule also has a third tiny diagonal spaceship - but this one has a speed of c/19:
x = 3, y = 3, rule = B3-r4k5jy6e7c/S02-ci3ajnqr4arw5ak6e
obo$2bo$3o!


EDIT: And there's a small 3c/17 diagonal too! :o
x = 5, y = 5, rule = B3-r4k5jy6e7c/S02-ci3ajnqr4arw5ak6e
o$3bo$2b2o$b2obo$3b2o!
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Re: Outrunning the Glider

Postby cvojan » February 7th, 2019, 9:04 pm

Wow... that's a lot of spaceships
'sup.
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Re: Outrunning the Glider

Postby Βεν Γ. Κυθισ » February 26th, 2019, 4:07 am

I have these two hexagonal generations spaceships:
x = 134, y = 132, rule = 0/2/256H
5.2A$4.AB2C$5.BD2E$2A4.DF2G$A2C4.FH2I$.C2E4.HJ2K$2.E2G4.JL2M$3.G2I4.L
N2O$4.I2K4.NP2Q$5.K2M4.PR2S$6.M2O4.RT2U$7.O2Q4.TV2W$8.Q2S4.VX2pA$9.S
2U4.XpB2pC$10.U2W4.pBpD2pE$11.W2pA4.pDpF2pG$12.pA2pC4.pFpH2pI$13.pC2pE
4.pHpJ2pK$14.pE2pG4.pJpL2pM$15.pG2pI4.pLpN2pO$16.pI2pK4.pNpP2pQ$17.pK
2pM4.pPpR2pS$18.pM2pO4.pRpT2pU$19.pO2pQ4.pTpV2pW$20.pQ2pS4.pVpX2qA$
21.pS2pU4.pXqB2qC$22.pU2pW4.qBqD2qE$23.pW2qA4.qDqF2qG$24.qA2qC4.qFqH
2qI$25.qC2qE4.qHqJ2qK$26.qE2qG4.qJqL2qM$27.qG2qI4.qLqN2qO$28.qI2qK4.qN
qP2qQ$29.qK2qM4.qPqR2qS$30.qM2qO4.qRqT2qU$31.qO2qQ4.qTqV2qW$32.qQ2qS
4.qVqX2rA$33.qS2qU4.qXrB2rC$34.qU2qW4.rBrD2rE$35.qW2rA4.rDrF2rG$36.rA
2rC4.rFrH2rI$37.rC2rE4.rHrJ2rK$38.rE2rG4.rJrL2rM$39.rG2rI4.rLrN2rO$
40.rI2rK4.rNrP2rQ$41.rK2rM4.rPrR2rS$42.rM2rO4.rRrT2rU$43.rO2rQ4.rTrV
2rW$44.rQ2rS4.rVrX2sA$45.rS2rU4.rXsB2sC$46.rU2rW4.sBsD2sE$47.rW2sA4.sD
sF2sG$48.sA2sC4.sFsH2sI$49.sC2sE4.sHsJ2sK$50.sE2sG4.sJsL2sM$51.sG2sI
4.sLsN2sO$52.sI2sK4.sNsP2sQ$53.sK2sM4.sPsR2sS$54.sM2sO4.sRsT2sU$55.sO
2sQ4.sTsV2sW$56.sQ2sS4.sVsX2tA$57.sS2sU4.sXtB2tC$58.sU2sW4.tBtD2tE$
59.sW2tA4.tDtF2tG$60.tA2tC4.tFtH2tI$61.tC2tE4.tHtJ2tK$62.tE2tG4.tJtL
2tM$63.tG2tI4.tLtN2tO$64.tI2tK4.tNtP2tQ$65.tK2tM4.tPtR2tS$66.tM2tO4.tR
tT2tU$67.tO2tQ4.tTtV2tW$68.tQ2tS4.tVtX2uA$69.tS2tU4.tXuB2uC$70.tU2tW
4.uBuD2uE$71.tW2uA4.uDuF2uG$72.uA2uC4.uFuH2uI$73.uC2uE4.uHuJ2uK$74.uE
2uG4.uJuL2uM$75.uG2uI4.uLuN2uO$76.uI2uK4.uNuP2uQ$77.uK2uM4.uPuR2uS$
78.uM2uO4.uRuT2uU$79.uO2uQ4.uTuV2uW$80.uQ2uS4.uVuX2vA$81.uS2uU4.uXvB
2vC$82.uU2uW4.vBvD2vE$83.uW2vA4.vDvF2vG$84.vA2vC4.vFvH2vI$85.vC2vE4.vH
vJ2vK$86.vE2vG4.vJvL2vM$87.vG2vI4.vLvN2vO$88.vI2vK4.vNvP2vQ$89.vK2vM
4.vPvR2vS$90.vM2vO4.vRvT2vU$91.vO2vQ4.vTvV2vW$92.vQ2vS4.vVvX2wA$93.vS
2vU4.vXwB2wC$94.vU2vW4.wBwD2wE$95.vW2wA4.wDwF2wG$96.wA2wC4.wFwH2wI$
97.wC2wE4.wHwJ2wK$98.wE2wG4.wJwL2wM$99.wG2wI4.wLwN2wO$100.wIwKwJ4.wNwP
2wQ$101.wK2wL4.wPwR2wS$102.wL2wN4.wRwT2wU$103.wN2wP4.wTwV2wW$104.wP2wR
4.wVwX2xA$105.wR2wT4.wXxB2xC$106.wT2wV4.xBxD2xE$107.wV2wX4.xDxF2xG$
108.wX2xB4.xFxH2xI$109.xB2xD4.xHxJ2xK$110.xD2xF4.xJxL2xM$111.xF2xH4.xL
xN2xO$112.xH2xJ4.xNxP2xQ$113.xJ2xL4.xPxR2xS$114.xL2xN4.xRxT2xU$115.xN
2xP4.xTxV2xW$116.xP2xR4.xVxX2yA$117.xR2xT4.xXyB2yC$118.xT2xV4.yByD2yE
$119.xV2xX4.yDyF2yG$120.xX2yB4.yFyH2yI$121.yB2yD4.yHyJ2yK$122.yD2yF4.
yJyL2yM$123.yF2yH4.yLyN2yO$124.yH2yJ4.yN$125.yJ2yL$126.yL2yN$127.yN!
#C [[ AUTOSTART GPS 4 GRID THEME 0 TRACK -0.125 0.125 STARS ]]
x = 5, y = 4, rule = B36/S125
2o$2ob2o$ob3o$bo!
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Re: Outrunning the Glider

Postby AlephAlpha » February 26th, 2019, 4:20 am

A glider that outruns the glider (2c/7):

x = 3, y = 3, rule = B2ce3aijkn4w5i7e/S1c2cei3aj4ainq5ry6c
bo$2bo$3o!
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Re: Outrunning the Glider

Postby Hunting » February 26th, 2019, 5:27 am

Hello, AlephAlpha!

I will probably try to search some fast diagonal ship in RSS(Really Small Spaceship) project, or #76 at GitHub [s]Also known as "Manual LLS"[/s].
Plz correct my grammar mistakes. I'm still studying English.

Working on:

* Rule Y Orthogonoid

Favorite gun ever:
#C Favorite Gun. Found by me.
x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
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Re: Outrunning the Glider

Postby AlephAlpha » February 26th, 2019, 8:57 am

Restricting to the rules that the original glider works:

4c/15:
x = 13, y = 13, rule = B2in3-eqy4ijkqw5cikq6aik/S02-ck3-aeky4inqtwy5nqry6an7e8
11bo$12bo$10b3o8$b2o$obo$3o!


3c/11:
x = 13, y = 13, rule = B2ik3aijn4iqw5cky6c7c/S02-ci3-ciqy4ir5acq6k7c
11bo$12bo$10b3o8$2bo$b2o$3o!


2c/7:
x = 13, y = 13, rule = B3-ckqy4k/S02ae3ijknr4iw5any6n7
11bo$12bo$10b3o8$b2o$obo$3o!


3c/10:
x = 13, y = 13, rule = B3-ekqy4aqy5i6e7c8/S02ae3ajnr4cqrw5cknq6cn
11bo$12bo$10b3o8$obo$2bo$3o!


4c/13:

x = 13, y = 13, rule = B2kn3-ekqy4jt5c/S02ae3-ckqy4jqw5aik7e
11bo$12bo$10b3o8$b2o$3o$3o!


c/3:
x = 13, y = 13, rule = B3aijnr4k7c/S2ae3jnr4i5a7c
11bo$12bo$10b3o8$b2o$obo$3o!
Last edited by AlephAlpha on February 26th, 2019, 10:06 am, edited 1 time in total.
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Re: Outrunning the Glider

Postby Hunting » February 26th, 2019, 9:38 am

AlephAlpha wrote:2c/7:
x = 13, y = 13, rule = B3-ckqy4k/S02ae3ijknr4iw5any6n7
11bo$12bo$10b3o8$2bo$b2o$3o!


Hey, I think you posted the wrong pattern. The "2c/7d" is an oscillator.
Plz correct my grammar mistakes. I'm still studying English.

Working on:

* Rule Y Orthogonoid

Favorite gun ever:
#C Favorite Gun. Found by me.
x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
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Re: Outrunning the Glider

Postby wildmyron » February 26th, 2019, 9:46 am

AlephAlpha wrote:Restricting to the rules that the original glider works:

Nice work. There's a 2c/7 in a glider compatible rule in my post above ^. Here are some other speeds:

5c/16d
x = 13, y = 13, rule = B2k3aijn4qt5acjk6-ae7c/S23ajnqr4acijrt5-cenr6-a7e
12bo$10bobo$11b2o8$2bo$2bo$3o!

5c/17d
x = 13, y = 13, rule = B2i3aijnr4eknqy5aceq6-k78/S02ae3-ceq4inrtwy5-enqy6-kn7e
12bo$10bobo$11b2o8$2bo$b2o$3o!

5c/18d
x = 13, y = 13, rule = B2kn3-cekr4eijnqt5-ckn6ci7c/S2-k3-eky4nrwyz5aei6cen78
12bo$10bobo$11b2o8$2bo$b2o$3o!

5c/19d
x = 13, y = 13, rule = B2i3-kq4ckntyz5cejk6ck7c/S02aen3jnry4cinqwz5-ey6an78
12bo$10bobo$11b2o8$b2o$obo$3o!

6c/19d
x = 13, y = 13, rule = B2k3-cery4ikyz5ikqry6cek7c8/S02aek3-ci4ceinwz5iknr8
12bo$10bobo$11b2o8$b2o$3o$3o!
Last edited by wildmyron on February 26th, 2019, 10:41 am, edited 2 times in total.
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Re: Outrunning the Glider

Postby AlephAlpha » February 26th, 2019, 10:07 am

Hunting wrote:
AlephAlpha wrote:2c/7:
x = 13, y = 13, rule = B3-ckqy4k/S02ae3ijknr4iw5any6n7
11bo$12bo$10b3o8$2bo$b2o$3o!


Hey, I think you posted the wrong pattern. The "2c/7d" is an oscillator.


Sorry. Edited.
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Re: Outrunning the Glider

Postby wildmyron » February 26th, 2019, 10:42 am

Some more:

c/3d, p18
x = 13, y = 13, rule = B2n3-kqr4-eijry5aciky6-ai7e8/S2-c3-eky4ejryz5aceir6kn7c8
12bo$10bobo$11b2o8$obo$b2o$3o!

3c/10, p20
x = 13, y = 13, rule = B3aijn4-aeirt5eqry6ace7e8/S2aek3-cky4eirwyz5-ckry6cn7c
12bo$10bobo$11b2o8$2bo$b2o$3o!

2c/7d, p21
x = 13, y = 13, rule = B2in3-eqy4cejwz5-jnr6-en7c8/S02-cn3-aey4cirtyz5aijky6-k8
12bo$10bobo$11b2o8$b2o$obo$3o!

3c/11d, p22
x = 13, y = 13, rule = B2ikn3-q4aejtyz5acer67e/S02-ik3-aeiy4ejnqtz5aejqy6-k8
12bo$10bobo$11b2o8$obo$2bo$3o!

7c/22d
x = 13, y = 13, rule = B2kn3-e4ceknqy5-ekny6-ac7c/S02-c3cjnr4einyz5-eikn6eik78
12bo$10bobo$11b2o8$b2o$obo$3o!

6c/23d
x = 13, y = 13, rule = B3-qr4-aqryz5cry6-cn8/S02-n3-ai4ceijntz5-cejn6-en7e8
12bo$10bobo$11b2o8$obo$2bo$3o!

7c/23d
x = 13, y = 13, rule = B2i3-ceqr4-krwy5ciqr6-i7c8/S2-c3-ekqy4iqrtw5aikq6kn7c8
12bo$10bobo$11b2o8$b2o$3o$3o!


These were all found with LLS. The shapes are mostly a function of the boundary conditions set up by the search. You can improve the search behaviour by using the make_lls_grid.py script. I used it for the 7c/22 and 7c/23 ships. As a bonus this reduces the search space and therefore improves the search speed considerably. Here's the command I used to find the 7c/23:
./make_lls_grid.py -p 23 -x 7 -y 7 -f 3 3 5 5 | ./lls -r pB2-ace3aijn4-r5-n678/S0
2ae3jnr4-k5678 -i -s 'D2\'

This took about 15 seconds to run, most of which was in preprocessing.
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Re: Outrunning the Glider

Postby Hdjensofjfnen » February 26th, 2019, 5:25 pm

Can we either prove or disprove wildmyron's conjecture that rules with CGOL gliders have a maximum diagonal speed of c/3?
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Outrunning the Glider

Postby AforAmpere » February 26th, 2019, 5:39 pm

Hdjensofjfnen wrote:Can we either prove or disprove wildmyron's conjecture that rules with CGOL gliders have a maximum diagonal speed of c/3?


It kind of already is known. In a non-B2a or B1e rule, the max diagonal speed is C/3. Therefore, there is no faster one in a rule with the glider.
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule (someone please search the rules)
- Find a C/10 in JustFriends
- Find a C/10 in Day and Night
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Re: Outrunning the Glider

Postby Hdjensofjfnen » February 26th, 2019, 7:57 pm

AforAmpere wrote:
Hdjensofjfnen wrote:Can we either prove or disprove wildmyron's conjecture that rules with CGOL gliders have a maximum diagonal speed of c/3?


It kind of already is known. In a non-B2a or B1e rule, the max diagonal speed is C/3. Therefore, there is no faster one in a rule with the glider.

Oh. In that case, time to find every single possible speed between c/3 and c/4 with the CGOL glider.
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
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Re: Outrunning the Glider

Postby Moosey » February 26th, 2019, 8:02 pm

Hdjensofjfnen wrote:
AforAmpere wrote:
Hdjensofjfnen wrote:Can we either prove or disprove wildmyron's conjecture that rules with CGOL gliders have a maximum diagonal speed of c/3?


It kind of already is known. In a non-B2a or B1e rule, the max diagonal speed is C/3. Therefore, there is no faster one in a rule with the glider.

Oh. In that case, time to find every single possible speed between c/3 and c/4 with the CGOL glider.

That's also impossible (at least for small ships) because there are infinite rational numbers between any two different rational numbers.
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Re: Outrunning the Glider

Postby AforAmpere » February 26th, 2019, 8:04 pm

Moosey wrote:That's also impossible (at least for small ships) because there are infinite rational numbers between any two different rational numbers.


However, there are a finite number of rules with the glider, and if there aren't fast diagonal adjustable ships with the glider, there may only be finitely many speeds.
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Re: Outrunning the Glider

Postby Gamedziner » February 26th, 2019, 11:02 pm

AforAmpere wrote:
Moosey wrote:That's also impossible (at least for small ships) because there are infinite rational numbers between any two different rational numbers.


However, there are a finite number of rules with the glider, and if there aren't fast diagonal adjustable ships with the glider, there may only be finitely many speeds.


If you limit the bounding box, the max period is less than or equal to 2^(number of cells in the bounding box).
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Re: Outrunning the Glider

Postby Entity Valkyrie 2 » February 27th, 2019, 12:25 am

This is Entity Valkyrie 2.

c/3:

x = 5, y = 5, rule = B2e3-cjnr/S2aei3
b2o$obo$3o$4bo$3bo!
Last edited by Entity Valkyrie 2 on February 27th, 2019, 12:27 am, edited 1 time in total.
Take Very Tinily.
The ENEERG-y of the EVAD is watching.

x = 47, y = 23, rule = B3/S23
obo41bo$b2o41bobo$bo42b2o8$10bobo21bo$11b2o21bobo$11bo22b2o7$22b2o$20b
o$21bo3bo$21b4o!
#C [[ THEME 6 THUMBSIZE 2 Y 10 ZOOM 32 GPS 20 AUTOSTART LOOP 40 ]]
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Re: Outrunning the Glider

Postby Βεν Γ. Κυθισ » February 27th, 2019, 12:26 am

Entity Valkyrie 2 wrote:This is Entity Valkyrie 2.

c/3:

x = 5, y = 5, rule = B2e3-cnr/S2aei3
b2o$obo$3o$4bo$3bo!

No, it becomes ash instantly,
#C [[ AUTOSTART GPS 4 GRID THEME 0 TRACK -0.125 0.125 STARS ]]
x = 5, y = 4, rule = B36/S125
2o$2ob2o$ob3o$bo!
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Re: Outrunning the Glider

Postby Entity Valkyrie 2 » February 27th, 2019, 12:29 am

Βεν Γ. Κυθισ wrote:
Entity Valkyrie 2 wrote:This is Entity Valkyrie 2.

c/3:

x = 5, y = 5, rule = B2e3-cnr/S2aei3
b2o$obo$3o$4bo$3bo!

No, it becomes ash instantly,


I forgot the "j" in B2e3-cjnr/S2aei3! Fixed.

x = 5, y = 5, rule = B2e3-cjnr/S2aei3
b2o$obo$3o$4bo$3bo!
Take Very Tinily.
The ENEERG-y of the EVAD is watching.

x = 47, y = 23, rule = B3/S23
obo41bo$b2o41bobo$bo42b2o8$10bobo21bo$11b2o21bobo$11bo22b2o7$22b2o$20b
o$21bo3bo$21b4o!
#C [[ THEME 6 THUMBSIZE 2 Y 10 ZOOM 32 GPS 20 AUTOSTART LOOP 40 ]]
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Re: Outrunning the Glider

Postby 77topaz » February 28th, 2019, 1:00 am

AlephAlpha wrote:3c/11:
x = 13, y = 13, rule = B2ik3aijn4iqw5cky6c7c/S02-ci3-ciqy4ir5acq6k7c
11bo$12bo$10b3o8$2bo$b2o$3o!


This rule also has a small 2c/8 orthogonal:
x = 2, y = 6, rule = B2ik3aijn4iqw5cky6c7c/S02-ci3-ciqy4ir5acq6k7c
2o$2o2$bo2$bo!


Unfortunately, it is explosive.
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Re: Outrunning the Glider

Postby Redstoneboi » March 12th, 2019, 6:16 am

x = 4, y = 5, rule = B3ai4a/S2e34aeiwz5iqr6a
bo$3o$2bo$2b2o$3bo!

rule that's golfed to have a small explosive seed, modified a bit giving this c/3d with population going 8-9-10
c(>^x^<c)~
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