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Thread for basic questions

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Re: Thread for basic questions

Postby dvgrn » April 4th, 2019, 8:55 pm

danny wrote:I think of something like this 5G:

x = 34, y = 22, rule = B3/S23
32bo$31bo$31b3o4$18bo4bo$19b2obo$18b2o2b3o2$8b2o$7bobo$7bo$6b2o2$5bo$
5b2o$4bobo2$bo$b2o$obo!

Huh, yes, that looks promising. It's too bad that this doesn't _quite_ work:

x = 76, y = 76, rule = LifeHistory
69.A$68.A$68.3A3$2.A$A.A$.2A59.A$60.2A$61.2A4$7.A.A$8.2A$8.A5$49.A$
48.A$48.3A3$22.A$20.A.A$21.2A19.A$40.2A$41.2A4$27.A.A$28.2A$28.A5$47.
A$46.2A$46.A.A4$33.2A$34.2A$33.A19.2A$53.A.A$53.A3$25.3A$27.A$26.A5$
67.A$66.2A$66.A.A4$13.2A$14.2A$13.A59.2A$73.A.A$73.A3$5.3A$7.A$6.A!

But your example does make it seem like there's something with fewer gliders anyway.

EDIT: I think an Official Ruling is needed on whether the object is allowed to disappear temporarily while "eating" the gliders, along the lines of the transparent debris effect. If you weren't paying attention to what happens for a few ticks in the second-round synthesis, then these four gliders would be a solution -- right?

x = 52, y = 56, rule = LifeHistory
49.A$49.A.A$49.2A3$2.A$A.A$.2A9$33.A$33.A.A$33.2A3$18.A$16.A.A$17.2A
4$22.2D$23.D$20.3D$20.D3$27.2A$27.A.A$27.A2$31.2A$30.2A$32.A10$43.2A$
43.A.A$43.A2$47.2A$46.2A$48.A!
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Re: Thread for basic questions

Postby danny » April 5th, 2019, 6:57 pm

dvgrn wrote:EDIT: I think an Official Ruling is needed on whether the object is allowed to disappear temporarily while "eating" the gliders, along the lines of the transparent debris effect. If you weren't paying attention to what happens for a few ticks in the second-round synthesis, then these four gliders would be a solution -- right?

It works, technically, but I really want to see one without
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Re: Thread for basic questions

Postby dvgrn » April 6th, 2019, 10:38 am

danny wrote:
dvgrn wrote:EDIT: I think an Official Ruling is needed on whether the object is allowed to disappear temporarily while "eating" the gliders, along the lines of the transparent debris effect. If you weren't paying attention to what happens for a few ticks in the second-round synthesis, then these four gliders would be a solution -- right?

It works, technically, but I really want to see one without

Okay -- your wish is my command:

x = 1513, y = 3230, rule = B3/S23
o$b2o$2o3$4bobo$5b2o$5bo4$10bobo$11b2o$11bo12$24bobo$25b2o$25bo20$48bo
$49bo$47b3o6$56bo$57bo$55b3o27$85bo$86bo$84b3o4$91bo$92bo$90b3o23$114b
obo$115b2o$115bo15$133bo$134bo$132b3o5$139bo$140b2o$139b2o5$146bo$147b
2o$146b2o33$182bo$183bo$181b3o13$196bo$197b2o$196b2o34$233bo$234bo$
232b3o8$243bo$244bo$242b3o12$256bo$257b2o$256b2o26$285bo$283bobo$284b
2o4$289bobo$290b2o$290bo12$305bo$306bo$304b3o22$328bo$329b2o$328b2o20$
350bo$351b2o$350b2o6$359bo$357bobo$358b2o21$382bo$383bo$381b3o10$393bo
$394b2o$393b2o25$421bo$422bo$420b3o8$431bo$429bobo$430b2o3$436bo$437bo
$435b3o8$445bo$446b2o$445b2o3$450bo$451b2o$450b2o6$457bobo$458b2o$458b
o$462bo$463bo$461b3o15$478bo$479b2o$478b2o21$502bo$500bobo$501b2o7$
511bo$512bo$510b3o488$1000bo$1001b2o$1000b2o3$1004bobo$1005b2o$1005bo
4$1010bobo$1011b2o$1011bo12$1024bobo$1025b2o$1025bo20$1048bo$1049bo$
1047b3o6$1056bo$1057bo$1055b3o27$1085bo$1086bo$1084b3o4$1091bo$1092bo$
1090b3o23$1114bobo$1115b2o$1115bo15$1133bo$1134bo$1132b3o5$1139bo$
1140b2o$1139b2o5$1146bo$1147b2o$1146b2o33$1182bo$1183bo$1181b3o13$
1196bo$1197b2o$1196b2o34$1233bo$1234bo$1232b3o8$1243bo$1244bo$1242b3o
12$1256bo$1257b2o$1256b2o26$1285bo$1283bobo$1284b2o4$1289bobo$1290b2o$
1290bo12$1305bo$1306bo$1304b3o22$1328bo$1329b2o$1328b2o20$1350bo$1351b
2o$1350b2o6$1359bo$1357bobo$1358b2o21$1382bo$1383bo$1381b3o10$1393bo$
1394b2o$1393b2o25$1421bo$1422bo$1420b3o8$1431bo$1429bobo$1430b2o3$
1436bo$1437bo$1435b3o8$1445bo$1446b2o$1445b2o3$1450bo$1451b2o$1450b2o
6$1457bobo$1458b2o$1458bo$1462bo$1463bo$1461b3o15$1478bo$1479b2o$1478b
2o21$1502bo$1500bobo$1501b2o7$1511bo$1512bo$1510b3o205$1510b3o$1512bo$
1511bo7$1501b2o$1500bobo$1502bo21$1478b2o$1479b2o$1478bo15$1461b3o$
1463bo$1462bo$1458bo$1458b2o$1457bobo6$1450b2o$1451b2o$1450bo3$1445b2o
$1446b2o$1445bo8$1435b3o$1437bo$1436bo3$1430b2o$1429bobo$1431bo8$1420b
3o$1422bo$1421bo25$1393b2o$1394b2o$1393bo10$1381b3o$1383bo$1382bo21$
1358b2o$1357bobo$1359bo6$1350b2o$1351b2o$1350bo20$1328b2o$1329b2o$
1328bo22$1304b3o$1306bo$1305bo12$1290bo$1290b2o$1289bobo4$1284b2o$
1283bobo$1285bo26$1256b2o$1257b2o$1256bo12$1242b3o$1244bo$1243bo8$
1232b3o$1234bo$1233bo34$1196b2o$1197b2o$1196bo13$1181b3o$1183bo$1182bo
33$1146b2o$1147b2o$1146bo5$1139b2o$1140b2o$1139bo5$1132b3o$1134bo$
1133bo15$1115bo$1115b2o$1114bobo23$1090b3o$1092bo$1091bo4$1084b3o$
1086bo$1085bo27$1055b3o$1057bo$1056bo6$1047b3o$1049bo$1048bo20$1025bo$
1025b2o$1024bobo12$1011bo$1011b2o$1010bobo4$1005bo$1005b2o$1004bobo3$
1000b2o$1001b2o$1000bo488$510b3o$512bo$511bo7$501b2o$500bobo$502bo21$
478b2o$479b2o$478bo15$461b3o$463bo$462bo$458bo$458b2o$457bobo6$450b2o$
451b2o$450bo3$445b2o$446b2o$445bo8$435b3o$437bo$436bo3$430b2o$429bobo$
431bo8$420b3o$422bo$421bo25$393b2o$394b2o$393bo10$381b3o$383bo$382bo
21$358b2o$357bobo$359bo6$350b2o$351b2o$350bo20$328b2o$329b2o$328bo22$
304b3o$306bo$305bo12$290bo$290b2o$289bobo4$284b2o$283bobo$285bo26$256b
2o$257b2o$256bo12$242b3o$244bo$243bo8$232b3o$234bo$233bo34$196b2o$197b
2o$196bo13$181b3o$183bo$182bo33$146b2o$147b2o$146bo5$139b2o$140b2o$
139bo5$132b3o$134bo$133bo15$115bo$115b2o$114bobo23$90b3o$92bo$91bo4$
84b3o$86bo$85bo27$55b3o$57bo$56bo6$47b3o$49bo$48bo20$25bo$25b2o$24bobo
12$11bo$11b2o$10bobo4$5bo$5b2o$4bobo3$2o$b2o$o!
#C [[ X 550 Z -3 STEP 9 ]]

... A reasonable-sized version is left as an exercise for the reader.
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Re: Thread for basic questions

Postby Moosey » April 7th, 2019, 8:17 am

Would it be possible to make a pattern which has a gun firing into a line of semi snarks and which after awhile tags another semisnark onto the end? At what rate would it grow? Olog(t)? Oslog(t)? I assume it’s the former.
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Re: Thread for basic questions

Postby Macbi » April 7th, 2019, 8:49 am

Moosey wrote:Would it be possible to make a pattern which has a gun firing into a line of semi snarks and which after awhile tags another semisnark onto the end? At what rate would it grow? Olog(t)? Oslog(t)? I assume it’s the former.
It's rate of growth would be inversely proportional to two to the power of its current length. Which means it would grow like O(log t).
You would have to use slow-salvo single-channel technology, which I've done some work on before. Possibly in this very thread.
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Re: Thread for basic questions

Postby dvgrn » April 8th, 2019, 4:33 pm

Macbi wrote:
Moosey wrote:Would it be possible to make a pattern which has a gun firing into a line of semi snarks and which after awhile tags another semisnark onto the end? At what rate would it grow? Olog(t)? Oslog(t)? I assume it’s the former.
It's rate of growth would be inversely proportional to two to the power of its current length. Which means it would grow like O(log t).
You would have to use slow-salvo single-channel technology, which I've done some work on before. Possibly in this very thread.

It's a little tricky, at least assuming that you want each new semi-Snark to take (roughly) twice as as much time to be added as the previous one.

Answering The Wrong Question First
Just following the above requirement as it's written is fairly trivial: line up two semi-Snarks in their "absorb" phase with an elbow and an initial hand target, and figure out how far away the next hand target has to be to build the same thing again, lined up in a chain. Put a block in that next hand location, compile the whole thing with slsparse, and build a memory-loop gun for the recipe that comes out.

Even if you have a low-period gun firing into the upper end of the chain, as long as you start with a few dozen semi-Snarks so that the period of the whole chain is greater than the period of the loop gun, you don't have to worry about doing anything with the gliders coming out the far end, because there will never be any.

Okay, You Actually Want The Semi-Snarks To Do Something...
The problem gets a bit more painful if you want an output glider from the semi-Snark chain to trigger the construction of the next pair of semi-Snarks. You'll still have a memory-loop gun holding more or less the same recipe, but now you probably ought to add a construction for something like a one-time G-to-*WSS converter -- like knightlife's four-block one:

x = 111, y = 80, rule = LifeHistory
7.A$8.A$6.3A8$17.A$18.A$16.3A11$74.A$72.3A$71.A$63.2A6.2A$63.A.A3.4B$
64.A.5B$65.7B$37.A27.4B2A4B.B$38.A26.4B2A3B.B2A$36.3A26.11B2A$65.9B.
2B$60.2B2.10B$59.2A13B$59.2A13B$60.2B.11B$63.9B.B2A$65.7B.BA.A$47.A
16.7B5.A$48.A14.8B5.2A$46.3AB12.4B2.4B$47.4B10.4B4.4B$48.4B8.4B6.4B$
49.4B6.4B8.4B$A9.A9.A9.A9.A9.4B4.4B10.4B$51.4B2.4B12.4B$52.8B5.2A$53.
7B5.A$54.7B.BA.A$52.9B.B2A$49.2B.11B$48.2A13B$48.2A13B$49.2B2.10B$54.
9B.2B$54.11B2A$54.4B2A3B.B2A$54.4B2A4B.B$54.7B$53.A.5B$52.A.A3.4B$52.
2A6.2A47.2A$60.A48.2A$61.3A$63.A39.2A$103.2A5$109.2A$109.2A5$102.2A$
102.2A!

Then feed that MWSS through an MWSS-to-G and into a regulator driven by a gun the same period as the memory loop. Set it up to open a switch and let out one copy of the single-channel recipe, then close the switch again.

You kind of need the universal regulator mechanism, because otherwise the time when the signal returns from the end of the semi-Snark chain is too unpredictable. I mean, you could build some counting circuitry to predict it, but it's easier to just use a regulator.

Fewer Semi-Snarks To Build, But Not Really Simpler
An interesting variant of this would be to build the semi-Snark chain in a spiral, around and around the memory-loop gun. Then your recipe could be for just one semi-Snark instead of a pair of them, plus a regular Snark, plus an MWSS seed, plus the next target hand block in a place where the recipe can pick it up on the next trip around the spiral.

The MWSS could cross the spiral directly back to the center -- from one of four directions, so you'd need a little extra circuitry to collect all those possible signals and feed them into the regulator to let the next semi-Snark recipe out.

The recipe won't go into the spiral semi-Snark chain, of course -- well, actually it _could_, perfectly well, but it also has to be fed into the parallel spiral of Snarks, so that it gets to the outside of the spiral undamaged to do its construction work, no matter how long the spiral gets.
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Re: Thread for basic questions

Postby Moosey » April 11th, 2019, 7:33 pm

What’s the smallest G predecessor consisting of isolated dots?
It is smaller than this:
x = 16, y = 16, rule = B3/S23
3.A.A$10.A$3.A3.A$5.A3.A.A3.A$3.A3.A5.A$5.A5.A3.A$7.A.A3.A$11.A$A.A.A
.A2$.A.A.A3$7.A.A2$6.A.A.A!


Are there any predecessors for this dot parent?
x = 21, y = 14, rule = B3/S23
3bobobobobobobobo$3bobobobobobobobo$2b17o$19o$2b3o2b3obob2o3b3o$5obob
2obob2ob3o$2b3o3b2o3b2o3b3o$5obob3ob3ob3o$2b3o2b4ob3o3b3o$19o$2b19o$2b
17o$3bobobobobobobobo$3bobobobobobobobo!
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Re: Thread for basic questions

Postby dvgrn » April 11th, 2019, 8:43 pm

Moosey wrote:What’s the smallest G predecessor consisting of isolated dots?
It is smaller than this:
x = 16, y = 16, rule = B3/S23...

Probably smaller than this, too, but I don't think I'm going to find out without writing an enumerator script for dot patterns:

x = 8, y = 7, rule = B3/S23
2bobo2$obobo2$bobobobo2$4bobo!

Have you tried JavaLifeSearch on that "BYE" pattern? It's big enough that I don't know if you'll get an answer in a reasonable amount of time, but it might be worth a try.
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Re: Thread for basic questions

Postby wildmyron » April 11th, 2019, 11:16 pm

Moosey wrote:Are there any predecessors for this dot parent?
x = 21, y = 14, rule = B3/S23
3bobobobobobobobo$3bobobobobobobobo$2b17o$19o$2b3o2b3obob2o3b3o$5obob
2obob2ob3o$2b3o3b2o3b2o3b3o$5obob3ob3ob3o$2b3o2b4ob3o3b3o$19o$2b19o$2b
17o$3bobobobobobobobo$3bobobobobobobobo!

JLS reports 872 predecessors in a 23x16 bounding box. I don't believe there are any in a smaller bounding box. For this search, careful selection of the search order doesn't seem to be required, but it can speed the search up somewhat (20s to find all solutions rather than 3 minutes). From my previous experience the SAT solvers are better at this kind of task. LLS can easily be set up to find a predecessor solution, and Tom Rokicki wrote a Golly Lua script to do the same using glucose. I haven't been able to get that script to read the solutions back in properly on Windows unfortunately, but it should work well in Linux and MacOS. Neither of those tools is particularly efficient at finding all solutions, but that can be done by modifying them to use iglucose instead.

And here's a randomly selected predecessor:
x = 23, y = 16, rule = B3/S23
3bo5bo3bo5bo$5bobo3bo3bobo2bo$4bo2bo2bo2bo2bo2b4o$bobo5bo3bo5b2o$4bob
2o2bobobob2o$obo2bo2bo2bo3bobo2b3o$3bo2b2o2bobobo2b3o$2bo2bobobo2b2obo
4b3o$4bo2b2o2b2o3b2o$b2o2bo2bo2bo3bobob4o$4bob2o2bo2b2ob2o2bo$o2bo5bo
2bo8b2o$2bo2bo2bo2bo3bo2b2o$4bob2o2bobob3obo$2bobobobobobobo3bo$2bobob
obobobobo2bo!
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Re: Thread for basic questions

Postby Gustone » April 12th, 2019, 11:18 am

How do I install gfind?
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Re: Thread for basic questions

Postby googoIpIex » April 12th, 2019, 11:28 am

Download gfind from here:
http://www.ics.uci.edu/~eppstein/ca/gfind.c

And then compile with gcc -O3 -o gfind gfind.c
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Re: Thread for basic questions

Postby Gustone » April 12th, 2019, 12:35 pm

googoIpIex wrote:Download gfind from here:
http://www.ics.uci.edu/~eppstein/ca/gfind.c

And then compile with gcc -O3 -o gfind gfind.c

How do i compile it? With what? What program do I use?
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Re: Thread for basic questions

Postby googoIpIex » April 12th, 2019, 12:51 pm

What OS are you using?
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Re: Thread for basic questions

Postby Ian07 » April 12th, 2019, 2:51 pm

Gustone wrote:How do i compile it? With what? What program do I use?

The wiki has a tutorial on compiling C code.
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Re: Thread for basic questions

Postby Gustone » April 12th, 2019, 4:26 pm

googoIpIex wrote:What OS are you using?

Windows 7? Ultimate? 64 bit?
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Re: Thread for basic questions

Postby Hunting » April 14th, 2019, 8:56 am

Really stupid question but I just can't use Rhombic's script because I don't have python on my dad's computer.

Uh, what will the rule behaves if I construct a MAP rule, which is same as CGOL, except when a cell has the left-upper neighbour on and other neighbours off, the center cell state is toggled.
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x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
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Re: Thread for basic questions

Postby Moosey » April 14th, 2019, 9:00 am

Hunting wrote:Really stupid question but I just can't use Rhombic's script because I don't have python on my dad's computer.

Uh, what will the rule behaves if I construct a MAP rule, which is same as CGOL, except when a cell has the left-upper neighbour on and other neighbours off, the center cell state is toggled.

Explodes in the lower right direction very quickly (grows at C diagonal) and the other directions slowly.
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Re: Thread for basic questions

Postby Hunting » April 15th, 2019, 12:18 am

Moosey wrote:
Hunting wrote:Really stupid question but I just can't use Rhombic's script because I don't have python on my dad's computer.

Uh, what will the rule behaves if I construct a MAP rule, which is same as CGOL, except when a cell has the left-upper neighbour on and other neighbours off, the center cell state is toggled.

Explodes in the lower right direction very quickly (grows at C diagonal) and the other directions slowly.

Oh, could you please show me the rule?
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Re: Thread for basic questions

Postby Moosey » April 15th, 2019, 8:01 am

Hunting wrote:
Moosey wrote:
Hunting wrote:Really stupid question but I just can't use Rhombic's script because I don't have python on my dad's computer.

Uh, what will the rule behaves if I construct a MAP rule, which is same as CGOL, except when a cell has the left-upper neighbour on and other neighbours off, the center cell state is toggled.

Explodes in the lower right direction very quickly (grows at C diagonal) and the other directions slowly.

Oh, could you please show me the rule?

No. I don’t have it. Right now.
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Re: Thread for basic questions

Postby Naszvadi » April 22nd, 2019, 7:10 am

Moore-neighbourhood, E^2 lattice:
There is a nontrivial set of cellular automata rules between isotropic nontotalistic and outer-totalistic set, where only the number of diagonal and orthogonal neigbours count. Using Hensel-notation, only the following rulestring parts are valid in a birth or survival section:
  • 0 diagonal, 0 orthogonal: 0
  • 1 d., 0 o.: 1c
  • 0 d., 1 o.: 1e
  • 2 d., 0 o.: 2cn
  • 1 d., 1 o.: 2ak
  • 0 d., 2 o.: 2ei
  • 3 d., 0 o.: 3c
  • 2 d., 1 o.: 3inqy
  • 1 d., 2 o.: 3akjr
  • 0 d., 3 o.: 3e
  • ...
and so on.
Is it worth to list already investigated nontotalistic rules if they fit into the above category?
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Re: Thread for basic questions

Postby googoIpIex » April 23rd, 2019, 9:32 am

Gustone wrote:
googoIpIex wrote:What OS are you using?

Windows 7? Ultimate? 64 bit?


If you have windows 7, then I would use Cygwin (google it). When the installer asks you what modules to install, select gcc.
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Re: Thread for basic questions

Postby googoIpIex » April 23rd, 2019, 4:01 pm

is there any converter which changes a boat bit to face the other way?
x = 38, y = 9, rule = B3/S23
16bo2$bo16bo$obobo26b2obo$b2ob3o5bobobobobo9bobob3o$7bo23bo5bo$6b2o10b
o17b2o2$16bo!
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Re: Thread for basic questions

Postby dvgrn » April 23rd, 2019, 4:19 pm

googoIpIex wrote:is there any converter which changes a boat bit to face the other way?

Is it okay to cheat?

x = 27, y = 32, rule = B3/S23
21bo$21bobo$21b2o7$11bo$9bobo$10b2o3$15bo$14bobobo$15b2ob3o$21bo$20b2o
6$b2o$obo$2bo3$25b2o$24b2o$26bo!
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Re: Thread for basic questions

Postby googoIpIex » April 23rd, 2019, 9:49 pm

What is the fastest repeat time hNN to X?
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Re: Thread for basic questions

Postby googoIpIex » April 26th, 2019, 3:33 pm

How would I find Elbow operations to use with an overclocked silver reflector?
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