ConwayLife.com

mniemiec wrote: January 22nd, 2021, 12:36 pm Are you referring to the use of the "x" state? In my synthesis database, I have always "o" for "before" cells and "x" for "after" cells, and sometimes "y" for "junk" cells (for Life programs that choose to differentiate between them). Most programs (e.g. Golly using the QuickLife engine) will just treat them the same under normal Life rules, so you will see a monochromatic pattern.

bubblegum wrote: January 22nd, 2021, 1:38 pmThis is a post from 2016.

bubblegum wrote: January 22nd, 2021, 1:38 pmThis is a post from 2016.

MathAndCode wrote: January 22nd, 2021, 1:57 pmIt's on the bottom of the first page, so maybe he saw that there were new posts there but accidentally clicked to go to this thread in general instead of to go to the latest posts, scrolled down in order to go to the latest posts because he realized that he was at the top of the page, and didn't notice the page number.

XxAlphaHexagonxX wrote: January 22nd, 2021, 4:59 pm An interesting reaction?

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x = 10, y = 10, rule = LifeSuper
7.2W$6.W2.W$6.W2.W$7.3W3$.2W$W2.W$W2.W$.3W!

dvgrn wrote: January 22nd, 2021, 10:43 am True... except I still would be very careful not to call those "methuselahs", because at this pattern size, there are still probably more of them than there are electrons in the visible universe.

Hunting wrote: January 27th, 2021, 9:11 am I know I've asked this question countless times to countless people, but I can't seem to understand how to synthesis small, unnatural spaceships.
I recently tried the 19-cell c/4d in LeapLife. One of its phase contains a long boat, so I sparked the long boat to get the center of the spaceship:

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snip

And that's the farthest I've got to. That's 11 cells out of 19 cells; But I can't figure out where should the rest, especially the glider, come from, and that's not mentioning I can't insert those sparks used to activate the 11 cells. How do I do it?

g.setrule("Life")

x = 14, y = 11, rule = B2n3/S23-q
9b2o$bobo7bo$2obo7bo$2bo5b2obo$7bobobo$bo4bo2bob2o$obo2bob2o4bo$2bo2b
o7bo$2b2ob4o2b2o$3bo4b4o$3bob2o2b2o2bo!

MathAndCode wrote: January 29th, 2021, 8:44 pmIs there a way to merge two barberpoles already in a tie with gliders. (I'm wondering whether this can allow a more efficient nonapole synthesis.)

x = 95, y = 12, rule = B3/S23
5boo38boo38boo$5bobo16bob3obbo13bobo17b3o3bo13bobo$21bobbo3bo3bo35bo3b
o$7bobo10b3obob3ob4o13bobo17bobb4o13bobo$8boo11bobbobo5bo34bo4bo$10boo
12bob3obbo13bo5bo19bo17bobo$bbo7bo56bo$obo8bobo29bo7bobo37bobo$boobboo
7bo28bo10bo39bo$4bobo6boo28bo5bo3boo38boo$6bo$45b3o!

mniemiec wrote: January 30th, 2021, 2:51 amEDIT: From the "close, but no cigar" department, comes this - which does everything except one bit; unfortunately, there's no way to save that bit, so a totally different mechanism will be needed:

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x = 95, y = 12, rule = B3/S23
5boo38boo38boo$5bobo16bob3obbo13bobo17b3o3bo13bobo$21bobbo3bo3bo35bo3b
o$7bobo10b3obob3ob4o13bobo17bobb4o13bobo$8boo11bobbobo5bo34bo4bo$10boo
12bob3obbo13bo5bo19bo17bobo$bbo7bo56bo$obo8bobo29bo7bobo37bobo$boobboo
7bo28bo10bo39bo$4bobo6boo28bo5bo3boo38boo$6bo$45b3o!

x = 10, y = 10, rule = B3/S23
2o$o$bobo$4bo$3b2o$b3ob2o$4b2obo$4bo$4bo2bobo$8b2o!

Schiaparelliorbust wrote: January 30th, 2021, 12:17 pm What is the lowest population known queen bee predecessor? I need one.

pcallahan wrote: December 14th, 2019, 1:31 pm This qualifies as useless, since we already have plenty of syntheses for Gosper's p30 glider gun but I've been obsessed with the puzzle for a couple of days of finding a minimal predecessor in terms of bounding box and cell count. Maybe somebody else can improve this.

I see this as a partial order on (max dimension, min dimension, cell count), so all of these are incomparable. Ideally, though I'd like something fitting in a 36x3 bounding box or smaller with 25 or fewer cells.

First attempt and the only one to shrink the width below 36. I imagine the cell count of 31 can be reduced, but the bounding box is too big to be interesting.

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x = 32, y = 7, rule = B3/S23
2o$o2b2o$bobob2o20b2obo$bo25bo2b2o$2obobo21b2ob2o$3bo23bobo$28b4o!

Next 21 cells with a 36x5 bounding box.

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x = 36, y = 5, rule = B3/S23
26bo$8bobo16b2ob2o2bo$9bo17bobo4b2o$2o4b2o$bo6b3o!

Next I can shave off the top row on the right but I need to add a cell. So 22 cells with a 36x4 bounding box.

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x = 36, y = 4, rule = B3/S23
8bobo15b3o2bo2b2o$2o7bo15bobob2o4bo$bo4b2o$8b3o!

Finally, I can shave it further, but I need to add more cells. So 26 cells with a 36x3 bounding box. (This is also a little more interesting in that the left block stabilizer appears spontaneously.)

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x = 36, y = 3, rule = B3/S23
o2bo2bobo$b4ob7o12bobobo4b2o$26b3ob2o3bo!

One observation is that starting patterns for the queen bee that lie entirely on one side of the center row (including the row) make it possible to avoid extending the bounding box beyond that determined by block placement.

MathAndCode wrote: January 30th, 2021, 12:13 pmIf the traffic light predecessor can be modified so that generation eleven looks like this, then the bit can be added from the other side. ...

mniemiec wrote: January 31st, 2021, 6:22 amInteresting idea; unfortunately, I think you'll find that it difficult to find a predecessor for this. Every way I could find to do this also adds additional cells to the attached blinkers; if you suppress those, you get one extra cell next to both blinkers. But it's possible that a more inventive indirect approach might work. Lifenthusiasts have often been known to take a problem that has previously been declared impossible and saying "hold my beer".

x = 12, y = 12, rule = B3/S23
2o$obo2$2bobo$5bo2bo$4b2o$2b3ob2o$5b2obo$5bo$5bo2bobo$11bo$10b2o!

x = 4, y = 13, rule = B3/S23
o$obo$2o8$b3o$bo$2bo!

MathAndCode wrote: January 31st, 2021, 4:21 pmI found a solution based on my idea that only requires the addition of what I think is a two-glider spark. ... Edit: Here's the relevant two-glider collision: ...

x = 12, y = 12, rule = B3/S23
2o$obo2$2bobo$5bo$4b2o$2b3ob2o$4b3obo$5bo$5bo2bobo$11bo$10b2o!

mniemiec wrote: January 31st, 2021, 6:31 pm

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x = 12, y = 12, rule = B3/S23
2o$obo2$2bobo$5bo$4b2o$2b3ob2o$4b3obo$5bo$5bo2bobo$11bo$10b2o!

x = 12, y = 12, rule = B3/S23
2o$o$bobo2$3bobo$4b2o$2b2o2b2o$6bo$5bobobo$5bo$9bobo$10b2o!

ColorfulGalaxy wrote: January 31st, 2021, 1:33 am Any fenceposts for this?

x = 26, y = 40, rule = B3/S23
22b2o$3b3o12b2obobo$18bo$2b2o2bo12bo3bo2$2bo2bob2o11b2obobo$4bo19b2o$
4bo3b2o9b2o2$9b2o7b2o2$10b2o5b2o2$11b2o3b2o2$12b2ob2o2$12bobo2$11bobo
2$10bobo2$9bobo2$8bobo$2obo$ob2o3bobo$4b2o$b2o3b2o2bo$bo3b6o$2bobo6b2o
$obobo4bo3bo$2o3b5o3bo$7b2o2b2o$5b3ob2o$4bo5bo$4bobo2bo$5b2o2bobo$10b
2o!

hkoenig wrote: January 31st, 2021, 11:48 am

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x = 19, y = 28, rule = B3/S23
2$10bo$10bobo$10bobobo2$12bobobo$17bo$11bobo2b2o2$10bobo2$9bobo2$8bobo
2$7bobo2$2b3obobo2$b2o2b2o2$bo2bo$3bo$3bo!

As for stretchers, consider that the slope is not exactly diagonal, and what sort of spaceship would be needed to do the stretching.