17 in 17: Efficient 17-bit synthesis project

[testitemqlstudop]

Yes.

[Extrementhusiast]

Yes.

Technically true, but incredibly unhelpful.

An insufficient reduction to #11, but one that could lead to further reductions:

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x = 131, y = 28, rule = B3/S23
119bobo$104bo15b2o$105bo14bo$81bo21b3o$79b2o29bobo11bo$80b2o29b2o6bo4b
obo$111bo5bobo4b2o$118b2o$75bobo$75b2o$76bo$106b3o16bobo$7bo65b3o32bo
4b2ob2o7b2o$2bo5b2o53bo9bo33bo5bob2o9bo$obo4b2o48bo4bobo9bo43bo$b2o53b
obo3bobo49b2ob2o9bo$4b2o40bo9bobo4bo50bo13bobo$5b2o40bo9bo58b2obo8b2o$
4bo40b3o67b2ob2o$127bo$44bo81b2o$44b2o80bobo$43bobo3$39b2o$40b2o$39bo!

[Kazyan]

While that still life is indeed not #70, it's fine because #70 dropped off the list anyway. Someone added a stealth improvement to the base still life for #70's synthesis, which dropped it to 16G.

An idea for #210. The two distinct reactions at the bottom (pointy end of a LoM explosion and the domino inserter) might both be replaced by inserting the red cells at generation 40. The base still life, xs17_g88b96zc952, costs 12 gliders but looks like it could be done in...8, maybe.

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x = 41, y = 47, rule = LifeHistory
30.A$28.2A$29.2A2$16.A$17.A$15.3A3$6.A.A$7.2A$7.A$33.3A$33.A$34.A2$
27.A2.2A$26.A.A2.A$26.A.A.A$23.2A.A2.A$23.A2.A$24.2A2$25.3D6.A$28.D5.
2A$25.2D6.A.A$38.3A$38.A$39.A$10.A$10.2A$9.A.A25.2A$5.2A22.A6.2A$4.A.
A21.2A8.A$6.A21.A.A$12.3A$14.A$13.A3$11.2A$10.A.A$12.A2$.2A$A.A$2.A!

[Extrementhusiast]

Another insufficient improvement, this time to #164:

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x = 33, y = 33, rule = B3/S23
22bobo$23b2o$23bo3$26bo$17bo6b2o$18bo6b2o$16b3o$24bo$25bo$23b3o4$30bo$
8bo21bobo$6bobo10b2obo7b2o$7b2o8b3ob2o$16bo$17bo$18bo$o16b2o$b2o8bo$2o
4bo2bobo15b3o$6b2o2b2o9bo5bo$5bobo13b2o5bo$20bobo3$15b3o$17bo$16bo!

EDIT: Recovered a misplaced better insufficient improvement to #11:

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x = 192, y = 29, rule = B3/S23
114bo$114bobo$114b2o2$113bo$111bobo30b2o$112b2o8bo22b2o2bo3bo$68bo54b
2o19bo3bobobobo$67bo54b2o2b2o20bobobobo$49bobo15b3o56bobo20bo3bo28bo$
4bo45b2o74bo53bobo$4bobo43bo69b3o58b2o$o3b2o108b2o4bo29b2ob2o29b2ob2o$
b2o44b2o64bo2bo4bo28bo3bo25b2o2bo3bo$2o44bobo11b3o51b2o35b2o26bobo3b2o
$46bo13bo54bobo34bobo26bo4bobo$45b2o14bo52bo2b3o31bo2b3o28bo2b3o$114b
2o4bo30b2o4bo27b2o4bo$119b2o35b2o32b2o3$47b3o$38b3o8bo$40bo7bo6b2o$39b
o15bobo$55bo$51b3o$53bo$52bo!

[Kazyan]

Idea for #266 and #267. On the left, the red cells should appear in generation 15, and a 4G inserter should be found for the double dot spark that doesn't involve the yellow glider. With everything in place, the snake inserter on the right would have room to appear.

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x = 49, y = 26, rule = LifeHistory
5.A$6.A$4.3A$12.A$11.A$11.3A3$5.A.A40.A$6.2A34.A3.2A$6.A35.A.A2.2A$
42.2A$27.3D11.2D$27.3D11.D2.A$4.2D21.3D12.D2A$3.D.D18.9D8.2DA.A$.2A2.
D18.9D$4A4.4D9.E2.9D$2A.2A2.D3.D7.E.E5.3D14.2A$2.2A7.D8.2E5.3D14.A.A$
9.2D16.3D14.A2$19.A.A$15.2A2.2A$15.A.A2.A$15.A!

Bonus: I stared into the ancient eyes of JLS, and this is some of what we saw between the space dust:

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x = 81, y = 152, rule = LifeHistory
12.A.A$13.A.A$16.A$12.A3.A$13.A$14.2A5$12.2A9.2A$11.A2.A9.2A2.A$15.A
7.A4.2A$11.A3.A12.3A$12.A14.5A$13.2A15.A$30.A$29.A$32.2A$31.3A$3.3A9.
3A.A11.2A$5.A2.2A7.A3.A17.2A$4.A2.A2.A5.A3.3A15.A2.A$11.A9.3A18.A$7.A
3.A9.4A13.A3.A$8.A12.2A16.A$9.2A29.2A$23.2A$11.A$25.A12$23.2A7.2A7.2A
$23.2A8.A8.A$21.2A.2A4.A2.2A4.A2.2A$21.A2.2A4.2A.2A4.2A.2A2$22.2A8.A
8.A$24.A6.A.A6.A.A$22.2A9.A8.A4$4.A$.A2.A$3A$3.2A$3.A$3.A$2.A2$33.A2$
22.A7.3A41.A$20.2A.A7.3A39.2A$20.3A.A5.A2.2A38.A.A$21.A.2A6.A.A32.3A
9.3A$22.2A8.2A34.A9.A$22.2A43.A11.A2$62.2A$63.2A$62.A3$2.A$.A4.A$.A3.
A$2.3A$31.3A2$3.3A25.3A2$3.3A22.2A2$30.3A$29.A2.A$30.A.A$31.A$33.A3$
72.2A$72.A.A$63.3A6.A$65.A$64.A3$64.2A$65.2A$64.A10.2A$75.A.A$75.A3$
37.2A$36.A.2A$36.A2.A$37.2A$24.3A2$24.3A4$23.4A$22.A3.A$26.A13.3A$24.
2A13.2A.2A$27.A12.3A$41.A3$19.2A$17.2A.2A$17.4A$15.3A.A$15.2A2$32.2A$
34.A$30.A3.A$31.4A13$21.2A2.2A$21.A.A$20.A2.A$21.2A2.2A$25.A$24.A$24.
A!

[chris_c]

Another insufficient improvement, this time to #164:

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x = 33, y = 33, rule = B3/S23
22bobo$23b2o$23bo3$26bo$17bo6b2o$18bo6b2o$16b3o$24bo$25bo$23b3o4$30bo$
8bo21bobo$6bobo10b2obo7b2o$7b2o8b3ob2o$16bo$17bo$18bo$o16b2o$b2o8bo$2o
4bo2bobo15b3o$6b2o2b2o9bo5bo$5bobo13b2o5bo$20bobo3$15b3o$17bo$16bo!

I found a couple of ways of doing the sides in 3G (see gen 30) but they don't fit with the 4G in the bottom right. Are there alternatives for the bottom right in 5G maybe?

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x = 63, y = 38, rule = B3/S23
46bo$44bobo$45b2o3$2bobo$3b2o55bo$3bo43bo12bobo$45bobo12b2o$46b2o7$45b
2obo$9bo33b3ob2o$7b2o33bo$8b2o22bobo8bo$2bo30b2o9bo$2o31bo9b2o$b2o3$2b
2o7bo$b2o7b2o$3bo6bobo22b2o$34bobo$36bo6$33b2o$34b2o$33bo!

[Extrementhusiast]

#166 in sixteen gliders:

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x = 154, y = 60, rule = B3/S23
141bo$142bo$140b3o$102bo$100bobo10bo29b2o$101b2o9bo30b2o$112b3o$101bo$
101b2o$100bobo$60bo$59bo8bo$59b3o4b2o$56bo10b2o$57b2o$56b2o$90bobo$91b
2o$59b2o30bo15bo$58bobo45bobo40bo$60bo5b2o39b2o2b2o34b3o2b2o$66bo44bo
34bo5bo$67bo44bo34bo5bo$16bo44b2o3b2o38b2o3b2o35bo3b2o$14b2o45bobobo
40bobobo38bobo$15b2o8bo38b2o43b2o39b2o$23b2o$24b2o$16bo10b3o$16b2o9bo$
15bobo10bo$94b2o$93bobo$95bo8$3o$2bo$bo14$6b3o$8bo$7bo!

[Kazyan]

#210 in 16G:

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x = 94, y = 35, rule = B3/S23
87bo$o84b2o$b2o83b2o$2o$81bo$82b2o$81b2o2$89bo$43bo45bobo$20b2o20bo46b
2o$20bobo19b3o31b2o$21bo54bobo$79bo$71bo5b2o$69bobo4bo$18b3o49b2o4bob
3o$20bo56bo2bo7bo$19bo3b2o53b2o6b2o$22b2o43b3o17b2o$24bo7bobo34bo$32b
2o34bo$33bo57b2o$91bobo$32b2o57bo$33b2o$32bo6$80bo$80b2o$79bobo!

EDIT: Insufficient improvement to #34 by improving the intermediate. The pond inserter could probably be done in 4G, which would solve this.

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x = 84, y = 28, rule = B3/S23
80bo$6bo60bobo3bo5bo$bo5bo15bobo37bo3b2o3bo6b3o$2bo2b3o15b2o36bobo4bo
3b3o$3o21bo37b2o2$18bo$18bobo$18b2o2$2b2o10bo57b2o$2bobo9bobo56bo7b2o$
2bo11b2o56bo8bobo$72b2o7bo$70b2o2bo$29b2o39bo2bobo$28b2o42b2o2bo$23b2o
5bo44b2o$22b2o$24bo$10bo17b3o$10b2o3b2o11bo$9bobo4b2o11bo$15bo2$5bo$5b
2o$4bobo!

[chris_c]

Insufficient improvement to #34 by improving the intermediate. The pond inserter could probably be done in 4G, which would solve this.

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x = 84, y = 28, rule = B3/S23
80bo$6bo60bobo3bo5bo$bo5bo15bobo37bo3b2o3bo6b3o$2bo2b3o15b2o36bobo4bo
3b3o$3o21bo37b2o2$18bo$18bobo$18b2o2$2b2o10bo57b2o$2bobo9bobo56bo7b2o$
2bo11b2o56bo8bobo$72b2o7bo$70b2o2bo$29b2o39bo2bobo$28b2o42b2o2bo$23b2o
5bo44b2o$22b2o$24bo$10bo17b3o$10b2o3b2o11bo$9bobo4b2o11bo$15bo2$5bo$5b
2o$4bobo!

Replaced the pond inserter with a 3G reaction and a single cleanup glider:

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x = 38, y = 27, rule = B3/S23
6bo$bo5bo$2bo2b3o$3o25bo$27bo$18bo8b3o$18bobo$18b2o2$2b2o10bo$2bobo9bo
bo11bo$2bo11b2o12bobo$23b3o2b2o$23bo$24bo5$10bo$10b2o$9bobo3$5bo29b2o$
5b2o28bobo$4bobo28bo!

[Kazyan]

Idea for #72; insert red cells at generation 4:

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x = 15, y = 17, rule = LifeHistory
8.A$7.A$7.3A$3.A$4.2A$3.2A2$10.A.A$5.2A4.2A$4.A.A4.A$4.A8.A$2.D2.4A3.
A$D8.A2.3A$2.2D3.2A$7.A$9.A$8.2A!

[Extrementhusiast]

#164 in sixteen gliders, due to an unlikely merger of steps:

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x = 129, y = 32, rule = B3/S23
83bo$81b2o$82b2o3$93bo22bo$91b2o21bobo$4bo87b2o21b2o$4bobo120bo$4b2o
105bobo11b2o$2bo109b2o7bobo2b2o$obo27bo29bo51bo3b2o4b2o$b2o27bo29bo55b
obo3bo$30bo9bo19bo57bo$38bobo76bo6bo$31bo7b2o20bo54bo3b2obobo$4b3o23bo
bo27bobo4b3o45bo5bob2o$4bo25b2o5b2o21b2o53b2o3bo6b2o$5bo31bobo24b2o12b
o38bobo6b2o$33b2o2bo25bo2bo10bo39b2o9bo$34b2o28b2o11b3o$33bo4$59bobo$
60b2o$60bo2$60b2o$59bobo$61bo!

[Goldtiger997]

Idea for #72; insert red cells at generation 4:

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x = 15, y = 17, rule = LifeHistory
8.A$7.A$7.3A$3.A$4.2A$3.2A2$10.A.A$5.2A4.2A$4.A.A4.A$4.A8.A$2.D2.4A3.
A$D8.A2.3A$2.2D3.2A$7.A$9.A$8.2A!

I made the base still life in 6G so if anyone can make Kazyan's blinker-predecessor in 6 gliders or less that will solve #72:

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x = 33, y = 26, rule = B3/S23
10bo$8bobo$9b2o2$22bo$21b2o$21bobo2$2o$b2o$o2$21b3o$21bo$7b3o12bo$9bo$
8bo7$30b3o$30bo$31bo!

#164 in sixteen gliders, due to an unlikely merger of steps:...

Nicely done!

[chris_c]

I made the base still life in 6G so if anyone can make Kazyan's blinker-predecessor in 6 gliders or less that will solve #72:

Made the spark with 5G in total. There might be a way to do cleanup in one less glider.

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x = 44, y = 54, rule = B3/S23
bo$2bo$3o8$2bo$obo$b2o2$22bo$21bo$15bobo3b3o$16b2o$16bo2$38bobo$24bo
13b2o$24bobo12bo$24b2o4$36bo$34b2o$35b2o3$22b2o$21bo2bo$21bob2o$18b2ob
o$16bo2bobo$16b2o2bo9$41b2o$41bobo$41bo3$31b2o$31bobo$31bo!

[Macbi]

There might be a way to do cleanup in one less glider.

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x = 27, y = 39, rule = B3/S23
5bo$o4bobo11bo$b2o2b2o12bobo$2o17b2o2$23bo$8bo12b2o$7bo14b2o$7b3o5$18b
obo$18b2o$19bo2$6b2o$5bo2bo$5bob2o$2b2obo$o2bobo$2o2bo9$24b3o$24bo$25b
o3$14b3o$14bo$15bo!

[Kazyan]

The current 17G synthesis for #135 is below, but with a cleanup glider removed in the intermediate step. This would be a 16G solution, but there's a single random blinker.

Neither of the 3G versions of the spark generated by the four marked gliders fit here, but I'm certain that a 4G version could both form the spark and delete the extraneous blinker. There's plenty of room to wave around the glider in white, but there does not appear to be a solution just by doing that, from my manual testing. A different mechanism for the spark may be needed.

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x = 163, y = 36, rule = LifeHistory
88.A$88.A.A$88.2A$77.A$77.A.A$77.2A3$2.A75.2A$A.A74.2A$.2A76.A$121.C$
84.A.A35.2C$84.2A35.2C$85.A47.3A$71.2A$71.A$72.3A70.2A$74.A70.A2.A$
81.A65.2A$19.A60.A.A65.A.2A.A$3.2A14.A.A58.A2.A55.E7.A2.A.2A$4.2A8.3A
2.2A60.2A55.2E8.2A$3.A10.A119.E3.E.E$15.A59.3A56.2E$58.2A73.E.E$59.2A
2.2A$58.A5.2A60.3E$63.A64.E$127.E3$161.2A$143.A16.2A$143.2A17.A$142.A
.A!

EDIT: This could remove another glider from #11, which would solve it, if different 3G inserter exists for the right spark. As-is, the gliders would collide.

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x = 21, y = 15, rule = B3/S23
10bo$8b2o$9b2o3bo$15b2o$14b2o2b2o$2o16bobo$b2o15bo$o11b3o$6b2o4bo$5bo
2bo4bo$6b2o$7bobo$6bo2b3o$6b2o4bo$11b2o!

[Extrementhusiast]

EDIT: This could remove another glider from #11, which would solve it, if different 3G inserter exists for the right spark. As-is, the gliders would collide.

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RLE

Here's the longest-lasting example:

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x = 43, y = 30, rule = B3/S23
32bobo$32b2o$33bo17$18b2o$17bo2bo$18b2o11bobo$19bobo10b2o2b2o$18bo2b3o
8bo2bobo$18b2o4bo12bo3bo$23b2o15bo$40b3o$bo$b2o$obo!

[chris_c]

Improving the predecessor of #135 using some different but equivalent junk on the left:

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x = 42, y = 47, rule = B3/S23
39bobo$39b2o$40bo$28bobo$28b2o$29bo6$36bo$35bo$7bo27b3o$8b2o19b2o$7b2o
20bobo$29bo3$11bo$12bo$10b3o$29bo$28bobo$8b3o17bo2bo$10bo18b2o$9bo14bo
$24bo$24bo16$2o$b2o$o!

[Goldtiger997]

Improving the predecessor of #135 using some different but equivalent junk on the left:

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x = 42, y = 47, rule = B3/S23
39bobo$39b2o$40bo$28bobo$28b2o$29bo6$36bo$35bo$7bo27b3o$8b2o19b2o$7b2o
20bobo$29bo3$11bo$12bo$10b3o$29bo$28bobo$8b3o17bo2bo$10bo18b2o$9bo14bo
$24bo$24bo16$2o$b2o$o!

Just beat me too it. Here's was my alternate 16G solution:

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x = 91, y = 36, rule = B3/S23
15bo$13bobo17bo$14b2o11bo4bo$25b2o5b3o$26b2o2$23bo$21b2o$4bobo15b2o$5b
2o$5bo43bo$27bobo20b2o$27b2o20b2o$28bo3$73b2o$73bo2bo$24bo50b2o$23bobo
50bob2obo$23bo2bo40bo7bo2bob2o$24b2o40b2o8b2o$62bo3bobo$18b3o41b2o$61b
obo2$54b3o$6bo49bo$6b2o47bo$5bobo2$89b2o$71bo16b2o$bo69b2o17bo$b2o67bo
bo$obo!

Now we're down to only 6 17-bitters left! 4 of those are pretty similar.

[Freywa]

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#CLL state-numbering golly
x = 603, y = 213, rule = B3/S23
218bo198b3o$217b2o197bo3bo$218bo201bo$218bo200bo$218bo199bo$218bo
198bo$217b3o196b5o4$17b3o17bo19b3o17b3o19bo16b5o16b3o16b5o16b3o17b
3o17b3o17bo19b3o17b3o19bo16b5o16b3o16b5o16b3o17b3o17b3o17bo19b3o
17b3o19bo16b5o16b3o16b5o16b3o17b3o$16bo3bo15b2o18bo3bo15bo3bo17b2o
16bo19bo3bo19bo15bo3bo15bo3bo15bo3bo15b2o18bo3bo15bo3bo17b2o16bo
19bo3bo19bo15bo3bo15bo3bo15bo3bo15b2o18bo3bo15bo3bo17b2o16bo19bo3b
o19bo15bo3bo15bo3bo$16bo3bo16bo22bo19bo16bobo16bo19bo22bo16bo3bo
15bo3bo15bo3bo16bo22bo19bo16bobo16bo19bo22bo16bo3bo15bo3bo15bo3bo
16bo22bo19bo16bobo16bo19bo22bo16bo3bo15bo3bo$16bobobo16bo21bo18b2o
16bo2bo17b3o16b4o19bo17b3o17b4o15bobobo16bo21bo18b2o16bo2bo17b3o
16b4o19bo17b3o17b4o15bobobo16bo21bo18b2o16bo2bo17b3o16b4o19bo17b3o
17b4o$16bo3bo16bo20bo21bo15b5o19bo15bo3bo17bo17bo3bo19bo15bo3bo16b
o20bo21bo15b5o19bo15bo3bo17bo17bo3bo19bo15bo3bo16bo20bo21bo15b5o
19bo15bo3bo17bo17bo3bo19bo$16bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3b
o17bo17bo3bo15bo3bo15bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3bo17bo17b
o3bo15bo3bo15bo3bo16bo19bo18bo3bo18bo16bo3bo15bo3bo17bo17bo3bo15bo
3bo$17b3o16b3o17b5o16b3o19bo17b3o17b3o18bo18b3o17b3o17b3o16b3o17b
5o16b3o19bo17b3o17b3o18bo18b3o17b3o17b3o16b3o17b5o16b3o19bo17b3o
17b3o18bo18b3o17b3o10$b3o$o3bo$o3bo$obobo$o3bo$o3bo$b3o14$bo$2o$bo
$bo$bo$bo$3o13$235b2o3b2o$b3o231bo2bo2bo$o3bo232b2obo$4bo233bob2o$
3bo233bo$2bo233bo$bo234b2o$5o13$235b2o363bo$b3o231bo2bob2o357bobo$
o3bo232b2obo358bo2bo$4bo233bo2bo355b2obobo$2b2o233bo2b2o355bo2b2o$
4bo231bo361bo$o3bo231b2o357b3o$b3o591bo13$136b2o$3bo132bo$2b2o134b
o$bobo133b2o$o2bo132bo$5o130bo2b3o$3bo132b2obo$3bo137bo$140b2o13$
5o$o$o$b3o$4bo$o3bo$b3o13$535b2o3b2o$b3o531bo2bo2bo$o3bo532b2obo$o
537bob2o$4o531b3o$o3bo530bo$o3bo$b3o13$535b2o$5o530bo2bob2o$4bo
532b2obo$3bo534bo2bo$3bo531b3o2b2o$2bo532bo$2bo$2bo14$b3o$o3bo$o3b
o$b3o$o3bo$o3bo$b3o14$b3o$o3bo$o3bo$b4o$4bo$o3bo$b3o!

Right, let's push.

[Goldtiger997]

#266 in 16 gliders:

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x = 251, y = 57, rule = B3/S23
162bo$163bo$161b3o79bobo$o64bobo98bo77b2o$b2o63b2o97bo78bo4bo$2o64bo
98b3o80bo$61bo186b3o$62bo182bo$60b3o98b2o6b3o72bobo$161bo2bo4bo72bo2bo
$162b3o5bo71b3o$247b3o$8bo74b2o77b3o77b3o2bo$6bobo73bo2bo75bobo77bobo
4bo$7b2o74b2o76bo3bo75bo3bo$160b2o2b2o74b2o2b2o$10bo$9bo69b3o$9b3o67bo
$72b3o5bo10bo$74bo16bo$73bo17bo7$93bo$92b2o$92bobo11$64b3o$66bo$65bo
11$122bo$121b2o$121bobo!

I suspect #112 can be solved in a similar way.

[Hdjensofjfnen]

The final 5.

Code: Select all

x = 89, y = 25, rule = B3/S23
b2o17b2o3b2o13b2o18b2o23bo$bo18bo2bo2bo13bo2bob2o13bo2bob2o17bobo$3bo
18b2obo16b2obo16b2obo18bo2bo$2b2o19bob2o16bo2bo16bo2bo15b2obobo$bo20bo
19bo2b2o13b3o2b2o15bo2b2o$o2b3o15bo19bo18bo22bo$b2obo16b2o18b2o37b3o$
6bo73bo$5b2o12$3obobo13bobob3o13bobob3o11b3ob3ob3o9b3ob3ob3o$o3bobo13b
obo3bo13bobo3bo13bobo5bo11bobobo3bo$3ob3o13bobob3o13bobob3o11b3ob3o3bo
9b3ob3ob3o$obo3bo13bobobo15bobo3bo11bo3bobo3bo9bo5bo3bo$3o3bo13bobob3o
13bobob3o11b3ob3o3bo9b3ob3ob3o!

I'll guess that #293 will be the last one standing.

EDIT: Comforting that Kazyan and I use the same exact number font.

[Freywa]

#266 in 16 gliders:

I was expecting a method that added the requisite ten bits onto the snake, so that two still lifes could be solved in one go, for both pairs in the last six that have this feature.

The final 5.

I had pushed an RLE of the last six stills only a few posts above. Why would you want to be relentlessly up-to-date?

[praosylen]

A possible longshot approach for #267:

Code: Select all

x = 26, y = 11, rule = B3/S23
20bo$20b3o$19b3o$17b3obo2b2o$4bo16b2o$3bobo15bobo$3bo2bo14bo2bo$b2o2b
2o3bobobo2bob2o2b2o$o17bo$bobo15bobo$2b2o!

Since the precursor SL takes 7G, all for separate transformations would have to be done in 9G (clearly a tall order -- 11G seems like a more likely lower bound, 1/3/4/3 clockwise from bottom, with 13G probably more realistic). I wouldn't be surprised if the precursor SL could be reduced to 6G or even 5G with luck, though -- which would open an outside chance of coming in under 17G for the SL.

[Kazyan]

I was also expecting a solution that could solve both Tweedledee and Tweedledum at the same time by flipping a snake insertion, but I was never able to get anywhere close by trying methods like that. So Goldtiger pulls through again. Sure enough, #113 is probably doable, since a base can be made in 11G and there's a 4G budget for the spark in a relatively simple activation step:

Code: Select all

x = 72, y = 28, rule = B3/S23
23bo$22bo$22b3o4$29bo$28bo40b2o$28b3o39bo$12b2ob2o7bo44bo$12b2ob2o7bob
o38b2obo$6bo17b2o40bob2o$bo5bo56bobobo2bo$2bo2b3o56b2o4b2o$3o9bo$13bo
47b2o$11b3o46bobo3b2o2bo$3b3o56bo3bobobo$5bo60bo4bo$4bo3$31b3o$31bo$
32bo$14b2o$13bobo$15bo!

[chris_c]

Sure enough, #113 is probably doable, since a base can be made in 11G and there's a 4G budget for the spark in a relatively simple activation step:

Code: Select all

x = 72, y = 28, rule = B3/S23
23bo$22bo$22b3o4$29bo$28bo40b2o$28b3o39bo$12b2ob2o7bo44bo$12b2ob2o7bob
o38b2obo$6bo17b2o40bob2o$bo5bo56bobobo2bo$2bo2b3o56b2o4b2o$3o9bo$13bo
47b2o$11b3o46bobo3b2o2bo$3b3o56bo3bobobo$5bo60bo4bo$4bo3$31b3o$31bo$
32bo$14b2o$13bobo$15bo!

Yeah that's doable. 16G in total:

Code: Select all

x = 87, y = 33, rule = B3/S23
23bo$22bo$22b3o4$29bo$28bo48b2o$28b3o47bo$12b2ob2o7bo52bo$12b2ob2o7bob
o46b2obo$6bo17b2o48bob2o$bo5bo64bobobo2bo$2bo2b3o64b2o4b2o$3o9bo$13bo$
11b3o$3b3o$5bo$4bo68b3o$75bo$63b3o8bo5b2o$31b3o31bo13b2o$31bo32bo16bo
3b2o$32bo51b2o$14b2o70bo$13bobo$15bo3$66bo$66b2o$65bobo!