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## Still life puzzles

For general discussion about Conway's Game of Life.

### Still life puzzles

These two have left me confused no matter what I try.
One (None exist): Do any still lifes exist from 9 to 19 cells so that every living cell has exactly two living neighbors and if not, then why(I've also apgsearched and turned up empty-handed.)?
And two (None exist): Other than some agars, is the block the only still life where every living cell has exactly three neighbors? If so, then why?
Last edited by wwei23 on August 6th, 2017, 1:15 pm, edited 3 times in total. wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
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### Re: Still life puzzles

I'm pretty sure that the block is the only finite pattern at all with very cell having exactly 3 neighbours.
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
muzik

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### Re: Still life puzzles

muzik wrote:I'm pretty sure that the block is the only finite pattern at all with very cell having exactly 3 neighbours.

I also believe that the only stable patterns with every cell having 3 neighbors consist exclusively of blocks.
Here I will attempt to construct a pattern with every cell having 3 neighbors and containing no blocks.
Set your coordinate system so that the lower left corner of the pattern's bounding box is at (0,0), with coordinates increasing going up and to the right:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$D$D$10D! Assume (0,0) is ON. For the cell to have 3 live neighbors, (1,0), (0,1), and (1,0) must be ON, but then the pattern contains a block and is invalid, so (0,0) must be OFF: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$D$2A$CA8D!

Let's say (1,0) is ON instead.
There are four possible neighbors for (1,0): (0,1), (1,1), (2,1), and (2,0), but one of these must be (0,1) because the other three neighbors constitute a block. In addition, allowing (1,1) to be ON creates B3a at (0,0) so it is OFF and the other two neighbors must therefore be ON. Similar logic applies to (0,1), resulting in a ship:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$2C$C.C$B2C7D! But (2,0) has two neighbors, and giving it a third causes (2,1) to have four neighbors and die, so this solution is also unworkable, and neither (1,0) nor (0,1) can be on in any solution. (This logic is reflection-invariant and so applies to both cases) Let's force (2,0) to be ON, then. By identical logic to the above we can immediately force these cells ON: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$D$BC.C$2B2C6D!

To prevent the contradiction in the previous case, though, (2,2) must stay OFF, forcing (1,1)'s two neighbors to be (2,1) and (2,0), creating B3a on (0,1). Therefore, (2,0) and (0,2) don't work either:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$2C$BC.C$2B2C6D! Similarly, (3,0) creates this: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$B2C$B.C.C$3B2C5D!

(4,0) doesn't work:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$B$B3C$B2.C.C$4B2C4D! This is the only way (5,0) can be done without creating birth at (2,1) or death at (3,2), but (1,1) and (1,2) are still unrescuable: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$B$B$B4C$BC2.C.C$5B2C3D!

And (6,0) has to be one of these but I don't know how to carry the logic past that:
x = 30, y = 10, rule = LifeHistoryD19.D$D19.D$D19.D$D19.D$B19.B$B19.B$B19.B$B.4C14.B5C$B2C2.C.C12.B.C2.C.C$6B2C2D10.6B2C2D! I'm pretty sure some sort of proof-by-induction is possible along these lines. If I could prove in each step that not just the end cells but each successive diagonal must be clear then the solution should just reduce to showing that each attempt to follow the instructions just creates the next level of ship and the proof would trivially follow from that and the logic used above. LifeWiki: Like Wikipedia but with more spaceships. [citation needed]  BlinkerSpawn Posts: 1858 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Still life puzzles If such a pattern with exclusively 3 neighbours did actually exist, it would have to be big: https://catagolue.appspot.com/census/bs3/C1 Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace! muzik Posts: 3300 Joined: January 28th, 2016, 2:47 pm Location: Scotland ### Re: Still life puzzles A pattern can contain a block, but it can't be JUST blocks. wwei23 Posts: 935 Joined: May 22nd, 2017, 6:14 pm Location: The (Life?) Universe ### Re: Still life puzzles I mean there's still an infinitely long shiptie or infinitely long barge, but I guess those wouldn't count. EDIT: Also I don't know if that's what you meant by 'agars'. I was picturing something like infinite chicken wire: x = 10, y = 4, rule = B3/S232o2b2o2b2o$2b2o2b2o$2o2b2o2b2o$2b2o2b2o!
This post was brought to you by the letter D, for dishes that Andrew J. Wade won't do. (Also Daniel, which happens to be me.)
Current rule interest: B2ce3-ir4a5y/S2-c3-y drc

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### Re: Still life puzzles

The still life must be finite. wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

### Re: Still life puzzles

I found a potential induction coil. Now we just need to stabilize it.
x = 5, y = 10, rule = B3/S232bo$b3o$o3bo$2ob2o$bobo$bobo$2ob2o$o3bo$b3o$2bo! Edit: Never mind: Because all the living cells of the seed have three living neighbors, no cells can be on adjacent to any of them. These cells are shown in blue: x = 7, y = 12, rule = LifeHistory2.3B$.2BA2B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$.2BA2B$2.3B!

Therefore, none of the red cells can be on, because they will cause a birth at gray. But all other surrounding cells except for the seed cells are blue, so the red cells must be off:
x = 7, y = 12, rule = LifeHistory2.3B$D2BA2BD$BF3AFB$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$BF3AFB$D2BA2BD$2.3B! So all cells forced off are shown in blue: x = 7, y = 12, rule = LifeHistory2.3B$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$2.3B!

The red cells have three on neighbors, four forced off neighbors, and one unset neighbor. To prevent a birth at red, the unset neighbor must be forced on:
x = 7, y = 12, rule = LifeHistory.A3BA$2BDAD2B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$2BDAD2B$.A3BA! So this is the pattern so far: x = 7, y = 12, rule = LifeHistory.A3BA$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$.A3BA!

Because the red cells would cause a birth at gray, and all other neighbors are either seed cells, cells forced on, or blue, the red cells must be off:
x = 7, y = 12, rule = LifeHistoryDA3BAD$BFBABFB$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$BFBABFB$DA3BAD! All blue cells must be off: x = 7, y = 12, rule = LifeHistoryBA3BAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BA3BAB!

Because each of the gray cells are on, they must have three living neighbors:
x = 7, y = 12, rule = LifeHistoryBF3BFB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BF3BFB! And they have exactly three unset neighbors each, in red: x = 7, y = 14, rule = LifeHistory3D.3D$BA3BAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BA3BAB$3D.3D!

So we can force them to be on:
x = 7, y = 14, rule = LifeHistory3A.3A$BA3BAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BA3BAB$3A.3A! The red cells are off, with three on neighbors: x = 7, y = 14, rule = LifeHistory3A.3A$BABDBAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BABDBAB$3A.3A!

And one gray unset neighbor:
x = 7, y = 14, rule = LifeHistory3AF3A$BABDBAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BABDBAB$3AF3A! To prevent a birth at red, the gray cells must be on: x = 7, y = 14, rule = LifeHistory7A$BABDBAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BABDBAB$7A!

Because some forced on cells have three living neighbors, none of their surrounding neighbors can be on, or else they die:
x = 7, y = 16, rule = LifeHistory7B$7A$BA3BAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BA3BAB$7A$7B! But that leaves two forced on cells shown in red that have only two neighbors. And because of their surrounding forced on cells that have three, they have no unset neighbors to be forced on: x = 7, y = 16, rule = LifeHistory7B$3AD3A$BA3BAB$3BA3B$2B3A2B$BA3BAB$B2AB2AB$2BABA2B$2BABA2B$B2AB2AB$BA3BAB$2B3A2B$3BA3B$BA3BAB$3AD3A$7B!

And therefore the initial pattern, the seed, cannot be stabilized.
Last edited by wwei23 on July 6th, 2017, 5:49 pm, edited 2 times in total. wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
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### Re: Still life puzzles

wwei23 wrote:Do any still lifes exist from 9 to 19 cells so that every living cell has exactly two living neighbors and if not, then why?

I've been searching B3/S2 for years to try to answer this. I'm already pretty sure the answer is "no", because the only available islands seem to be preblocks and rings of cells joined orthogonally or diagonally. I could find three at 20 cells though:
x = 22, y = 17, rule = B3/S2316bo$15bobo$4bo10bobo2bo$3bobo6b2obobobobo$3bobo6bo2bobo2bo$b2o3b2o6bo2bo$o7bo6b2o$b2o3b2o$3bobo$3bobo$4bo10b2o$14bo2bo$14bobo2bo$11b2obobob2o$11bo2bobo$13bo2bo$14b2o! gameoflifeboy

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Joined: January 15th, 2015, 2:08 am

### Re: Still life puzzles

x = 290, y = 45, rule = LifeHistory204.2A$203.A2.A$203.A.A2.A$200.2A.A.A.2A$200.A2.A.A$202.A2.A$203.2A37.2A$241.A2.A$242.2A$204.A$203.A.A34.4A$203.A.A2.A30.A4.A$200.2A.A.A.A.A30.4A$200.A2.A.A2.A$202.A2.A34.2A$203.2A34.A2.A$240.2A2$203.A36.2A$202.A.A34.A2.A$90.D3.D5.D.D7.3D7.3D7.3D17.3D7.3D7.3D7.D.D7.3D10.A36.2A$90.2D.2D5.D.D7.D10.D8.D19.D.D8.D8.D.D7.D.D7.D$90.D.D.D5.3D7.3D8.D8.3D17.2D9.D8.D.D7.D.D7.3D8.5A34.4A$90.D3.D6.D10.D8.D8.D19.D.D8.D8.D.D7.D.D9.D7.A5.A32.A4.A$90.D3.D6.D8.3D8.D8.3D17.D.D7.3D7.3D7.3D7.3D8.5A34.4A2$203.A36.2A$202.A.A34.A2.A$203.A36.2A$244.A19.2A18.2A$204.A19.2A17.A.A17.A2.A16.A2.A$203.A.A17.A2.A16.A.A18.A2.A15.A2.A$203.A.A18.A2.A13.2A3.2A13.3A4.A12.2A4.2A$201.2A3.2A13.3A4.A11.A7.A11.A8.A10.A8.A$90.4D46.3D47.3D7.A7.A11.A5.A.A11.A7.A12.A5.A.A10.A8.A$61.A4.2A3.A9.2A3.A3.D49.D.D47.D.D8.2A3.2A13.A3.A.A13.2A3.2A14.A3.A.A12.2A4.2A$41.A18.A.A4.A2.A.A7.A2.A.A.A2.D.2D46.3D47.3D10.A.A16.A2.A17.A.A17.A2.A16.A2.A$40.A.A17.A.A2.A4.A2.A6.A2.A.A2.A.D2.D46.D.D47.D12.A.A17.A.A17.A.A18.A.A16.A2.A$41.A19.A3.2A4.2A8.2A3.A.A.4D46.D.D47.D13.A19.A19.A20.A18.2A$87.A$3B7.2B8.3B7.3B7.B.B7.3B7.3B7.3B7.3B7.3B7.2B2.3B3.2B2.2B4.2B2.3B3.2B2.3B3.2B2.B.B3.2B2.3B3.2B2.3B3.2B2.3B3.2B2.3B3.2B2.3B3.3B.3B13.3B.2B14.3B.3B13.3B.3B13.3B.B.B$B.B8.B10.B9.B7.B.B7.B9.B11.B7.B.B7.B.B8.B2.B.B4.B3.B5.B4.B4.B4.B4.B2.B.B4.B2.B6.B2.B6.B4.B4.B2.B.B4.B2.B.B5.B.B.B15.B2.B16.B3.B15.B3.B15.B.B.B$B.B8.B8.3B7.3B7.3B7.3B7.3B9.B7.3B7.3B8.B2.B.B4.B3.B5.B2.3B4.B2.3B4.B2.3B4.B2.3B4.B2.3B4.B4.B4.B2.3B4.B2.3B3.3B.B.B13.3B2.B14.3B.3B13.3B.3B13.3B.3B$B.B8.B8.B11.B9.B9.B7.B.B9.B7.B.B9.B8.B2.B.B4.B3.B5.B2.B6.B4.B4.B4.B4.B4.B4.B2.B.B4.B4.B4.B2.B.B4.B4.B3.B3.B.B13.B4.B14.B3.B15.B5.B13.B5.B$3B7.3B7.3B7.3B9.B7.3B7.3B9.B7.3B7.3B7.3B.3B3.3B.3B3.3B.3B3.3B.3B3.3B3.B3.3B.3B3.3B.3B3.3B3.B3.3B.3B3.3B.3B3.3B.3B13.3B.3B13.3B.3B13.3B.3B13.3B3.B!

22 cells:
x = 46, y = 13, rule = LifeHistory3.A$2.A.A6.2A10.2A6.2A10.2A$2.A.A5.A2.A8.A2.A4.A2.A8.A2.A$3.A7.2A10.2A6.2A10.2A2$.5A5.4A6.4A6.4A6.4A$A5.A3.A4.A4.A4.A4.A4.A4.A4.A$.5A5.4A6.4A6.4A6.4A2$3.A7.2A8.2A8.2A8.2A$2.A.A6.A9.A10.A9.A$3.A9.A9.A6.A9.A$12.2A8.2A6.2A8.2A!

23 cell still life:
x = 7, y = 12, rule = LifeHistory3.A$2.A.A$3.A2$.5A$A5.A$.5A2$3.A$2.A.A$.A2.A$2.2A! 24 cells: x = 68, y = 89, rule = LifeHistory2.A$.A.A9.A18.2A8.2A9.2A8.2A$.A2.A7.A.A16.A2.A6.A2.A7.A2.A6.A2.A$2.A.A7.A.A17.A2.A6.A2.A7.A2.A6.A2.A$3.A9.A19.2A8.2A9.2A8.2A2$.5A5.5A17.4A6.4A5.4A6.4A$A5.A3.A5.A15.A4.A4.A4.A3.A4.A4.A4.A$.5A5.5A17.4A6.4A5.4A6.4A2$3.A9.A19.2A8.2A7.2A8.2A$2.A.A7.A.A19.A8.A9.A8.A$3.A8.A.A17.A12.A5.A12.A$13.A18.2A10.2A5.2A10.2A3$33.2A8.2A10.2A8.2A$32.A2.A6.A2.A8.A2.A6.A2.A$31.A2.A6.A2.A8.A2.A6.A2.A$32.2A8.2A10.2A8.2A2$32.4A6.4A6.4A6.4A$31.A4.A4.A4.A4.A4.A4.A4.A$32.4A6.4A6.4A6.4A2$32.2A8.2A8.2A8.2A$33.A8.A10.A8.A$31.A12.A6.A12.A$31.2A10.2A6.2A10.2A3$2.2A10.2A$.A2.A8.A2.A$.A2.A8.A2.A$2.2A10.2A2$2.4A6.4A$.A4.A4.A4.A$2.4A6.4A2$2.2A8.2A$.A2.A6.A2.A$2.2A8.2A19$2.2A8.2A18.2A10.2A$.A2.A6.A2.A16.A2.A8.A2.A$.A2.A6.A2.A17.A2.A6.A2.A$2.2A8.2A19.2A8.2A2$2.4A6.4A17.4A6.4A$.A4.A4.A4.A15.A4.A4.A4.A$2.4A6.4A17.4A6.4A2$2.2A8.2A19.2A8.2A$3.A8.A19.A2.A6.A2.A$.A12.A18.2A8.2A$.2A10.2A2$33.2A10.2A$2.2A8.2A18.A2.A8.A2.A$.A2.A6.A2.A18.A2.A6.A2.A$.A2.A6.A2.A19.2A8.2A$2.2A8.2A$32.4A6.4A$2.4A6.4A15.A4.A4.A4.A$.A4.A4.A4.A15.4A6.4A$2.4A6.4A$32.2A8.2A$4.2A8.2A15.A2.A6.A2.A$5.A8.A17.2A8.2A$3.A12.A$3.2A10.2A!
Last edited by wwei23 on July 2nd, 2017, 1:42 pm, edited 3 times in total. wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

### Re: Still life puzzles

Red=too few neighbors, blue means that cell must be off, green means good to go. So far:
x = 20, y = 19, rule = LifeHistory4.BD2B$4.B2AB$4.BA2B3.4BD3B$4.BA3B2.D3AB2A2B$5B2A3B.D2BABABAB$D4A2BA6BA2B2AB$BA2BABABA4B2A4BD$5BABABA2BA2B4AB$3.3BA2B4A2BA2BAB$4.3B2A4B2A3BAB$5.3BAB2ABA3B2DB$5.D2BAB2ABA3B$5.2B2A4B2A3B$5.DA2B4A2BA3B$5.BA2BA2BABABA2B$5.2B2A4BABABAD$6.2BDBD3BA2BAB$12.3B2A2B$13.2BD2B!

Two red cells with no neighbors have exactly three off neighbors, so we get this:
x = 20, y = 19, rule = LifeHistory4.BD2B$4.B2AB$4.BA2B3.4BD3B$4.BA3B2.D3AB2A2B$5B2A3B.D2BABABAB$D4A2BA6BA2B2AB$BA2BABABA4B2A4BD$5BABABA2BA2B4AB$3.3BA2B4A2BA2BAB$4.3B2A4B2A3BAB$3.BD3BAB2ABA3B2DB$3.B2A2BAB2ABA3B$3.BA2B2A4B2A3B$3.2B2A2B4A2BA3B$4.2BA2BA2BABABA2B$5.2B2A4BABABAD$6.2BABA3BA2BAB$7.2B2AD3B2A2B$8.4B.2BD2B!

Two red cells with two neighbors have one off neighbor, so we get this:
x = 20, y = 19, rule = LifeHistory4.BD2B$4.B2AB$4.BA2B3.4BD3B$4.BA3B2.D3AB2A2B$5B2A3B.D2BABABAB$D4A2BA6BA2B2AB$BA2BABABA4B2A4BD$5BABABA2BA2B4AB$3.3BA2B4A2BA2BAB$3.D3B2A4B2A3BAB$3.BA3BAB2ABA3B2DB$3.B2A2BAB2ABA3B$3.BA2B2A4B2A3B$3.2B2A2B4A2BA3B$4.2BA2BA2BABABA2B$5.2B2A4BABABAD$6.2BABA3BA2BAB$7.2B3A3B2A2B$8.4BD2BD2B!

There seems to be no more forcing.
EDIT:
I was wrong:
x = 20, y = 19, rule = LifeHistory4.BD2B$4.B2AB$4.BA2B3.4BD3B$4.BA3B2.D3AB2A2B$5B2A3B.D2BABABAB$D4A2BA6BA2B2AB$BA2BABABA4B2A4BD$5BABABA2BA2B4AB$3.3BA2B4A2BA2BAB$3.D3B2A4B2A3BAB$3.BA3BAB2ABA3B2DB$3.B2A2BAB2ABA3B$3.BA2B2A4B2A3B$3.2B2A2B4A2BA3B$4.2BA2BA2BABABA2B$4.D2B2A4BABABAD$6.2BABA3BA2BAB$6.D2B3A3B2A2B$8.4BD2BD3B!

EDIT:
Coolout Conjecture Counterexample, even if made internally stable:
Here red means bad cell.
x = 20, y = 19, rule = LifeHistory4.BA2B$4.B2AB$4.BA2B3.4BA3B$4.BA3B2.4AB2A2B$5B2A3B.A2BABABAB$5A2BA6BA2B2AB$BA2BABABA4B2A4BA$5BABABA2BA2B4AB$3.3BA2B4A2BA2BAB$3.A3B2A4B2A3BAB$3.BAD2BAB2ABA3B2AB$3.B2A2BAB2ABA3B$3.BA2B2A4B2A3B$3.2B2A2B4A2BA3B$4.2BA2BA2BABABA2B$4.A2B2A4BABAB2A$6.2BABAD2BA2BAB$6.A2B3A3B2A2B$8.4BA2BA3B!

Just checked LifeWiki for the puzzle of still lives where every cell has two neighbors, none for 9, 10, 11.
And Catagolue for 12, 13, none. wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
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### Re: Still life puzzles

BlinkerSpawn wrote:
muzik wrote:I'm pretty sure that the block is the only finite pattern at all with very cell having exactly 3 neighbours.

I also believe that the only stable patterns with every cell having 3 neighbors consist exclusively of blocks.
Here I will attempt to construct a pattern with every cell having 3 neighbors and containing no blocks.
Set your coordinate system so that the lower left corner of the pattern's bounding box is at (0,0), with coordinates increasing going up and to the right:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$D$D$10D! Assume (0,0) is ON. For the cell to have 3 live neighbors, (1,0), (0,1), and (1,0) must be ON, but then the pattern contains a block and is invalid, so (0,0) must be OFF: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$D$2A$CA8D!

Let's say (1,0) is ON instead.
There are four possible neighbors for (1,0): (0,1), (1,1), (2,1), and (2,0), but one of these must be (0,1) because the other three neighbors constitute a block. In addition, allowing (1,1) to be ON creates B3a at (0,0) so it is OFF and the other two neighbors must therefore be ON. Similar logic applies to (0,1), resulting in a ship:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$2C$C.C$B2C7D! But (2,0) has two neighbors, and giving it a third causes (2,1) to have four neighbors and die, so this solution is also unworkable, and neither (1,0) nor (0,1) can be on in any solution. (This logic is reflection-invariant and so applies to both cases) Let's force (2,0) to be ON, then. By identical logic to the above we can immediately force these cells ON: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$D$BC.C$2B2C6D!

To prevent the contradiction in the previous case, though, (2,2) must stay OFF, forcing (1,1)'s two neighbors to be (2,1) and (2,0), creating B3a on (0,1). Therefore, (2,0) and (0,2) don't work either:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$2C$BC.C$2B2C6D! Similarly, (3,0) creates this: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$B2C$B.C.C$3B2C5D!

(4,0) doesn't work:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$B$B3C$B2.C.C$4B2C4D! This is the only way (5,0) can be done without creating birth at (2,1) or death at (3,2), but (1,1) and (1,2) are still unrescuable: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$B$B$B4C$BC2.C.C$5B2C3D!

And (6,0) has to be one of these but I don't know how to carry the logic past that:
x = 30, y = 10, rule = LifeHistoryD19.D$D19.D$D19.D$D19.D$B19.B$B19.B$B19.B$B.4C14.B5C$B2C2.C.C12.B.C2.C.C$6B2C2D10.6B2C2D! I'm pretty sure some sort of proof-by-induction is possible along these lines. If I could prove in each step that not just the end cells but each successive diagonal must be clear then the solution should just reduce to showing that each attempt to follow the instructions just creates the next level of ship and the proof would trivially follow from that and the logic used above. Force (2,0) to be on: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DB$D2B$10D!

4 possible arrangements:
x = 70, y = 10, rule = LifeHistoryD19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$DB3A15.DB2A16.DBA.A15.DB.2A$D2BA16.D2B2A15.D2B2A15.D2B2A$10D10.10D10.10D10.10D! One on the right is a block so it doesn't count. The others have a cell in common: x = 50, y = 10, rule = LifeHistoryD19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$DB3A15.DB2A16.DBA.A$D2BA16.D2B2A15.D2B2A$10D10.10D10.10D!

Therefore:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DBA$D2BA$10D! Gray cannot be on because of B3a: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DBAF$D2BA$10D!

Yellow has two other neighbors, they must both be on:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DBAFA$D2BEA$10D! 3 arrangements: x = 50, y = 10, rule = LifeHistoryD19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D2A17.DA.A16.D.2A$DBAFA15.DBAFA15.DBAFA$D2B2A15.D2B2A15.D2B2A$10D10.10D10.10D!

First one causes B3a into blue, other two have cell in common:
x = 30, y = 10, rule = LifeHistoryD19.D$D19.D$D19.D$D19.D$D19.D$D19.D$DA.A16.D.2A$DBAFA15.DBAFA$D2B2A15.D2B2A$10D10.10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D2.A$DBAFA$D2B2A$10D!

Yellow cannot have 3 on neighbors because both would kill the one above it:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D2.A$DBAFA$D2BAE$10D! Therefore (2,0) and (0,2) must be off: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB$DB$D3B$10D!

(1,1) must be off because it forces a block:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB2E$DBAE$D3B$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB$D2B$D3B$10D!

(3,0):
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB$D2B$D3BA$10D! x = 70, y = 10, rule = LifeHistoryD19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$DB18.DB18.DB18.DB$D2B3A14.D2B2A15.D2BA.A14.D2B.2A$D3BA15.D3B2A14.D3B2A14.D3B2A$10D10.10D10.10D10.10D!

x = 50, y = 10, rule = LifeHistoryD19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$DB18.DB18.DB$D2B3A14.D2B2A15.D2BA.A$D3BA15.D3B2A14.D3B2A$10D10.10D10.10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB$D2BA$D3BA$10D!

x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB$D2BAF$D3BA$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB$D2BAFA$D3B2A$10D!

x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB2.A$D2BAFA$D3B2A$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$DB2.A$D2BAFA$D3BAE$10D!

x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$DB$DB$D2B$D4B$10D! (2,1): x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$DB$DB$D2BA$D4B$10D!

x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$DB$DBA$D2BA$D4B$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$DB$DBAF$D2BA$D4B$10D!

x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$DB2A$DBAFA$D2B2A$D4B$10D! This is where it starts to deviate. Gray would kill yellow: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D.3F$DBAE2F$DBAFEF$D2B2AF$D4B$10D!

Yellow needs one more neighbor and one is available:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D.3F$DBEA2F$DBAFAF$D2BAEF$D4B$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$DA3F$DB2A2F$DBAFAF$D2B2AF$D4BA$10D!

But this causes B3j into blue:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$DE3F$DB2A2F$DBAFAF$D2B2AF$D4BE$10D! Therefore (2,1) must be off. x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$DB$D2B$D3B$D4B$10D!

Also, the block theorem:
x = 15, y = 5, rule = LifeHistory7.D$8.D$B4.5D2.B$B7.D3.2B$3B4.D4.3B!

(4,0):
x = 130, y = 10, rule = LifeHistoryD19.D19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D19.DB$DB18.DB18.DB18.DB18.DB18.DB18.DB$D2B17.D2B17.D2B17.D2B17.D2B2.A14.D2B2.A14.D2B$D3B16.D3BA15.D3BAF14.D3BAFA13.D3BAFA13.D3BAFA13.D3B$D4BA14.D4BA14.D4BA14.D4B2A13.D4B2A13.D4BAE13.D5B$10D10.10D10.10D10.10D10.10D10.10D10.10D! (2,2): x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB$DB$D2BA$D3B$D5B$10D!

6 arrangements:
x = 110, y = 10, rule = LifeHistoryD19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D$D19.D19.D19.D19.D19.D$DB18.DB18.DB18.DB18.DB18.DB$DB3A15.DB2A16.DB2A16.DBA.A15.DBA.A15.DB.2A$D2BA16.D2B2A15.D2BA16.D2B2A15.D2BA16.D2B2A$D3B16.D3B16.D3BA15.D3B16.D3BA15.D3B$D5B14.D5B14.D5B14.D5B14.D5B14.D5B$10D10.10D10.10D10.10D10.10D10.10D! Also, the claw theorem: x = 15, y = 5, rule = LifeHistory7.D$8.D$B4.5D.B$2B6.D2.2B$4B3.D3.4B! Only 2 are valid: x = 30, y = 10, rule = LifeHistoryD19.D$D19.D$D19.D$D19.D$DB18.DB$DBA.A15.DBA.A$D2B2A15.D2BA$D3B16.D3BA$D5B14.D5B$10D10.10D!

2 cells in common:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB$DBAFA$D2BA$D3B$D5B$10D! 2 arrangements, both equally valid for now: x = 30, y = 10, rule = LifeHistoryD19.D$D19.D$D19.D$D19.D$DB18.DB$DBAFA15.DBAFA$D2B2A15.D2BAF$D3BF15.D3BA$D5B14.D5B$10D10.10D!

The first is object 3x2 43 and the second is object 3x3 337.
Object 3x2 43:
2 and 2 for yellow:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB2A$DBEFA$D2B2A$D3BF$D5B$10D! Gray kills yellow: x = 10, y = 10, rule = LifeHistoryD$D$D$D.3F$DBAE2F$DBAFEF$D2B2AF$D3BF$D5B$10D!

1 and 1 for yellow:
x = 10, y = 10, rule = LifeHistoryD$D$D$DA3F$DBEA2F$DBAFAF$D2BAEF$D3BFA$D5B$10D! B3j error: x = 10, y = 10, rule = LifeHistoryD$D$D$DA3F$DB2A2F$DBAFAF$D2B2AF$D3BFA$D5B$10D!

Object 3x2 43 invalidated.
Object 3x3 337:
2 and 2 for yellow:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB$DBEFA$D2BAF$D3BE$D5B$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB2A$DBAFA$D2BAFA$D3B2A$D5B$10D!

Applied gray:
x = 10, y = 10, rule = LifeHistoryD$D$D$D.3F$DB2A2F$DBAFA2F$D2BAFAF$D3B2AF$D5B$10D! 1 and 1 for yellow: x = 10, y = 10, rule = LifeHistoryD$D$D$DA3F$DB2A2F$DBAFA2F$D2BAFAF$D3B2AF$D5BA$10D!

B3j error:
x = 10, y = 10, rule = LifeHistoryD$D$D$DA3F$DB2A2F$DBAFA2F$D2BAFAF$D3B2AF$D5BA$10D! Object 3x3 337 invalidated. (2,2) must be off: x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB$DB$D3B$D3B$D5B$10D!

Block theorem:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$DB$D2B$D3B$D4B$D5B$10D! Claw theorem: x = 10, y = 10, rule = LifeHistoryD$D$D$DB$DB$D2B$D3B$D4B$D6B$10D!

x = 10, y = 10, rule = LifeHistoryD$D$D$DB$D2B$D2B$D3B$D5B$D6B$10D! x = 10, y = 10, rule = LifeHistoryD$D$D$DB$D2B$D3B$D4B$D5B$D6B$10D!

Ad infinitum. Proved. wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

### Re: Still life puzzles

I think there's an easier proof.
Say we've disproved a diagonal. We force the base of the next diagonal ON and build the claw using the same logic as the previous proofs:
x = 21, y = 21, rule = LifeHistoryB2$B2$B2$B$B$B$B$B$2B$B.B$B2.B$B3.B$B4.B$B5.B$B6.B$B7.BA.A$B8.B2A$15B.B.B.B! But the corner of the claw needs another neighbor, and the only way to add one without causing a birth into blue is this way: x = 21, y = 21, rule = LifeHistoryB2$B2$B2$B$B$B$B$B$2B$B.B$B2.B$B3.B$B4.B$B5.B$B6.B$B7.BA.AE$B8.B2A$15B.B.B.B!

But now the tip of the claw has the maximum three neighbors so the "wrist" of the claw must take neighbors as shown:
x = 21, y = 21, rule = LifeHistoryB2$B2$B2$B$B$B$B$B$2B$B.B$B2.B$B3.B$B4.B$B5.B$B6.B2ED$B7.BAD2A$B8.B2A$15B.B.B.B! This forces B3a into blue so the base of the diagonal is OFF. But the logic works the exact same way at the next location too: x = 21, y = 21, rule = LifeHistoryB2$B2$B2$B$B$B$B$B$2B$B.B$B2.B$B3.B$B4.B$B5.B2ED$B6.BAD2A$B7.B2A$B8.2B$15B.B.B.B!

And at the next:
x = 21, y = 21, rule = LifeHistoryB2$B2$B2$B$B$B$B$B$2B$B.B$B2.B$B3.B$B4.B2ED$B5.BAD2A$B6.B2A$B7.2B$B8.2B$15B.B.B.B! Notice that we are unable to start a ship-tie here. We can, in fact, slide the claw all the way up, forcing every cell along the new diagonal OFF. The "block theorem" finishes off the diagonal: x = 21, y = 21, rule = LifeHistoryB2$B2$B2$B$B$B$B2A$B2A$3B$B.2B$B2.2B$B3.2B$B4.2B$B5.2B$B6.2B$B7.2B$B8.2B$15B.B.B.B!

Q.E.D.
LifeWiki: Like Wikipedia but with more spaceships. [citation needed]  BlinkerSpawn

Posts: 1858
Joined: November 8th, 2014, 8:48 pm
Location: Getting a snacker from R-Bee's

### Re: Still life puzzles

This proof only works for things where there are no blocks. It may as well be an induction coil.
Here's my proof of the claw theorem:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DB$D2B$10D! Force a cell to be on: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DB$D2BA$10D!

4 arrangements:
x = 70, y = 10, rule = LifeHistoryD19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$D19.D19.D19.D$DB3A15.DB2A16.DBA.A15.DB.2A$D2BA16.D2B2A15.D2B2A15.D2B2A$10D10.10D10.10D10.10D! Only one is valid: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DBA.A$D2B2A$10D!

3 possible arrangements, first is invalid:
x = 50, y = 10, rule = LifeHistoryD19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D19.D19.D$D2A17.DA.A16.D.2A$DBA.A15.DBA.A15.DBA.A$D2B2A15.D2B2A15.D2B2A$10D10.10D10.10D! The other two have this in common: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D2.A$DBA.A$D2B2A$10D!

But the yellow cell cannot be stabilized without killing the light green cell:
x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D2.A$DBA.C$D2BAE$10D! So the original cell must be off: x = 10, y = 10, rule = LifeHistoryD$D$D$D$D$D$D$DB$D3B$10D!

Block theorem says that the only available neighbors must make a block, so it must be off. Now, what about induction coils? wwei23

Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe

### Re: Still life puzzles

WHITE CELL = ORIGIN , (X,Y)=(E-ward,N-ward)
non-block part : white cell = on
(SW edge of bounding diamond of non-block part : Y=-X)
(S-most ON cell on SW edge of bounding diamond of non-block part = origin)
blue area : no non-block part ...(a)
IF (1,0)(0,1)(1,1) are ON
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$10B.2A$11BCA$13B$14B$15B$16B$17B$18B$19B$20B$21B$22B!
white cell isn't in the non-block part
THEREFORE (-1,1) is ON
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$10BA$11BC$13B$14B$15B$16B$17B$18B$19B$20B$21B$22B!

IF (0,1) is ON
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$10B2A$11BC$13B$14B$15B$16B$17B$18B$19B$20B$21B$22B!
(-1,0)(-2,0)(-2,1)(-1,-1)(0,-1)(1,-1) are OFF
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$9BF2A$9B2FC$10B3F$14B$15B$16B$17B$18B$19B$20B$21B$22B!

blue area : ON cells = blocks (reason : (a))
anti B3a(-1,0)
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$9BF2A$7B4FC$7BF2A3F$7BF2AF3B$7B4F4B$16B$17B$18B$19B$20B$21B$22B!
anti B3q(-1,-1)
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$9BF2A$7B4FC$7BF2A4F$7BF2AF2AF$7B4F2AFB$10B4F2B$17B$18B$19B$20B$21B$22B!
anti B3n(-1,-3)
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$9BF2A$7B4FC$7BF2A4F$7BF2AF2AF$7B4F2AFB$7BF2A4F2B$7BF2AF6B$7B4F7B$19B$20B$21B$22B!
anti B3q(-1,-4)
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$9BF2A$7B4FC$7BF2A4F$7BF2AF2AF$7B4F2AFB$7BF2A4F2B$7BF2AF2AF3B$7B4F2AF4B$10B4F5B$20B$21B$22B!
... AD INFINITUM

THEREFORE (1,0)(1,1) are ON
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$10BA.A$11BCA$13B$14B$15B$16B$17B$18B$19B$20B$21B$22B!
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B$9BFAFA$9B2FCA$9B4F$14B$15B$16B$17B$18B$19B$20B$21B$22B!
(anti B3k)

IF (-1,2)(-2,2) are ON
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2A$9BFAFA$9B2FCA$9B4F$14B$15B$16B$17B$18B$19B$20B$21B$22B!
(-3,1)(-3,2) are OFF
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$8BF2A$8B2FAFA$9B2FCA$9B4F$14B$15B$16B$17B$18B$19B$20B$21B$22B!
anti B3a(-2,1) and (a)
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$8BF2A$6B4FAFA$6BF2A2FCA$6BF2A4F$6B4F4B$15B$16B$17B$18B$19B$20B$21B$22B!
B3q ERROR
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$8BF2A$6B4FAFA$6BF2ADFCA$6BF2A4F$6B4F4B$15B$16B$17B$18B$19B$20B$21B$22B!

THEREFORE (0,2) is ON
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2.A$9BFAFA$9B2FCA$9B4F$14B$15B$16B$17B$18B$19B$20B$21B$22B!
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2.A2F$9BFAFAF$9B2FCAF$9B4F$14B$15B$16B$17B$18B$19B$20B$21B$22B!
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2.A2F$9BFAFAF$9B2FCAF$9B4FA$14B$15B$16B$17B$18B$19B$20B$21B$22B!
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2.A2F$9BFAFAF$9B2FCAF$9B4FA$12B2F$15B$16B$17B$18B$19B$20B$21B$22B!
anti B3j(1,-1)
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2.A2F$9BFAFAF$9B2FCAF$9B4FA$9BF2A2F$9BF2AF2B$9B4F3B$17B$18B$19B$20B$21B$22B!
B3n ERROR
x = 22, y = 21, rule = LifeHistoryB$2B$3B$4B$5B$6B$7B$8B$9B2.A2F$9BFAFAF$9B2FCAF$9BFD2FA$9BF2A2F$9BF2AF2B$9B4F3B$17B$18B$19B$20B$21B$22B!

ONLY blocks are possible
QED
Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) :
965808 is period 336 (max = 207085118608).
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### Re: Still life puzzles

Do still lives exist in B/S4 (None exist, what about B/S46?)?
Does an eater exist that can eat a glider, a lightweight spaceship, a middleweight spaceship, and a heavyweight spaceship?
Last edited by wwei23 on August 6th, 2017, 1:16 pm, edited 1 time in total. wwei23

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### Re: Still life puzzles

wwei23 wrote:Do still lives exist in B/S4?

Interesting question. I don't immediately see how to turn a corner without resorting to an infinite agar, but there are enough possibilities that I can't instantly prove it's impossible:

Code: Select all
x = 32, y = 32, rule = B/S4:T32,3215bo$15b2o$15b2o$16bo$16b2o$16b2o$16bo$16b2o$16b2o$16bo$15b2o$15b2o$15bo$14b2o$3b2ob2o6b2o$11o4bob2o$2o7b6ob6o7b3o$12b2obo4b11o$15b2o6b2ob2o$15b2o$15bo$14b2o$14b2o$14bo$13b2o$13b2o$14bo$13b2o$13b2o$14bo$14b2o$14b2o!#C [[ THUMBNAIL THUMBSIZE 2 ]] wwei23 wrote:Does an eater exist that can eat a glider, a lightweight spaceship, a middleweight spaceship, and a heavyweight spaceship? This one I think you need to be a little more specific about. Otherwise the answer is a trivial "yes". Gliders come in at a different angle from spaceships, so maybe you want to require that the first cell that interacts has to be the same in all four cases, or something like that? Otherwise you can just weld together any glider eater, any LWSS eater, any MWSS eater, and any HWSS eater. If you want the three *WSSes at least to use the same mechanism, here's a three-bait constellation that would work. Just have to add a factory to each of the signal outputs to rebuild one of the bait objects -- and then simply add a fishhook eater to eat a glider, or change your question to disallow that somehow. Seems to me something better has been found for a universal *WSS signal converter since 2009, but I'm not finding it offhand, and it doesn't appear to be on the Big Converter List. Anyone have a link? dvgrn Moderator Posts: 5557 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI ### Re: Still life puzzles gameoflifeboy wrote:I've been searching B3/S2 for years to try to answer this. I'm already pretty sure the answer is "no", because the only available islands seem to be preblocks and rings of cells joined orthogonally or diagonally. Could Simon Ekström's still life searcher be adapted to other CAs beyond Conway Life to answer this sort of question? It seems like a good tool for the job -- unfortunately the assumption that the rule worked with is B3/S23 seems to be baked fairly deeply into it, and I can't tell off-hand where. If you speak, your speech must be better than your silence would have been. — Arabian proverb Catagolue: Apple Bottom • Life Wiki: Apple Bottom • Twitter: @_AppleBottom_ Proud member of the Pattern Raiders! Apple Bottom Posts: 1023 Joined: July 27th, 2015, 2:06 pm ### Re: Still life puzzles dvgrn wrote: wwei23 wrote:Do still lives exist in B/S4? Interesting question... there are enough possibilities that I can't instantly prove it's impossible... Code: Select all x = 18, y = 18, rule = B/S4:T18,189bo$8bobo$7b5o$6b2o3b2o$5b2o5b2o$4b2o7b2o$3b2o9b2o$2b2o11b2o$b2o13b2o$obo13bo$b2o13b2o$2b2o11b2o$3b2o9b2o$4b2o7b2o$5b2o5b2o$6b2o3b2o$7b5o$8bobo!#C [[ THUMBNAIL THUMBSIZE 2 ]]

It does look like there's an easy proof by contradiction here.

Consider the ON cell in a hypothetical still life that's farthest to the left along the top edge of the still life's bounding diamond (let's say). That cell (white) forces four ON cells below and to the right of it (green):

Code: Select all
x = 3, y = 2, rule = B/S4History.CA$3A!#C [[ THUMBNAIL ]] But then the cell to the southwest (white, below) can't have four ON cells around it, without the cell to its right (yellow) getting overcrowded -- unless ON cells are added to the W, NW, or N, which contradicts the initial assumption: Code: Select all x = 3, y = 2, rule = B/S4History..2A$.CEA$2A!#C [[ THUMBNAIL ]] Therefore there is no top edge of the still life's bounding diamond -- the hypothetical still life must be infinitely large. Q.E.D., right? dvgrn Moderator Posts: 5557 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI ### Re: Still life puzzles dvgrn wrote:Q.E.D., right? I believe so. It's also interesting to note that this disproves the existence of P1 photons in any 3D outer-totalistic rule in the logical extension of the Moore neighborhood containing B5 and none of B0234678 (B1 is disallowed for obvious reasons). The existence of higher-period photons would depend upon the existence of oscillators in 2D B5/S4. I suspect, but cannot prove at the moment, that they are impossible. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1783 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Still life puzzles dvgrn wrote: wwei23 wrote:Does an eater exist that can eat a glider, a lightweight spaceship, a middleweight spaceship, and a heavyweight spaceship? This one I think you need to be a little more specific about. Otherwise the answer is a trivial "yes". Gliders come in at a different angle from spaceships, so maybe you want to require that the first cell that interacts has to be the same in all four cases, or something like that? The XWSSes have to be on the same path. The glider must hit the same spot. wwei23 Posts: 935 Joined: May 22nd, 2017, 6:14 pm Location: The (Life?) Universe ### Re: Still life puzzles wwei23 wrote:The XWSSes have to be on the same path. The glider must hit the same spot. An eater with that constraint can almost certainly be built somehow -- or a multi-input converter, with the same signal output for any of the four inputs. However, it would probably take several thousand ticks for the Giant Multi-Eater to recover after any meal. Mostly for that reason, nobody may actually want to complete a construction along these lines. It seems like the kind of thing that might remain forever in the "We Could If We Wanted To But It Would Be Big And Ugly" category. If you want a reasonable-sized eater that recovers reasonably quickly, the answer might be "no" at the moment. But it's vaguely possible that some existing weird still lifes with very slow eater2-like action might be sufficiently omnivorous. Anyway, "yes" could possibly be only a Bellman search away. One more question: does this glider x = 97, y = 24, rule = B3/S23bo$2bo$3o7$11b2o$10bo2bo$10bo2bo$11b2o! strike in the "same spot" as these *WSSes? x = 100, y = 100, rule = B3/S237bo$8bo$4bo3bo$5b4o$17b2o$16bo2bo$16bo2bo$17b2o$20$7bo$8bo$3bo4bo$4b5o$17b2o$16bo2bo$16bo2bo$17b2o$20$7bo$8bo$2bo5bo$3b6o$17b2o$16bo2bo$16bo2bo$17b2o! dvgrn
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### Re: Still life puzzles

dvgrn wrote:
wwei23 wrote:The XWSSes have to be on the same path. The glider must hit the same spot.

An eater with that constraint can almost certainly be built somehow -- or a multi-input converter, with the same signal output for any of the four inputs.

However, it would probably take several thousand ticks for the Giant Multi-Eater to recover after any meal. Mostly for that reason, nobody may actually want to complete a construction along these lines. It seems like the kind of thing that might remain forever in the "We Could If We Wanted To But It Would Be Big And Ugly" category.

If you want a reasonable-sized eater that recovers reasonably quickly, the answer might be "no" at the moment. But it's vaguely possible that some existing weird still lifes with very slow eater2-like action might be sufficiently omnivorous. Anyway, "yes" could possibly be only a Bellman search away.

One more question: does this glider

x = 97, y = 24, rule = B3/S23bo$2bo$3o7$11b2o$10bo2bo$10bo2bo$11b2o!

strike in the "same spot" as these *WSSes?

x = 100, y = 100, rule = B3/S237bo$8bo$4bo3bo$5b4o$17b2o$16bo2bo$16bo2bo$17b2o$20$7bo$8bo$3bo4bo$4b5o$17b2o$16bo2bo$16bo2bo$17b2o$20$7bo$8bo$2bo5bo$3b6o$17b2o$16bo2bo$16bo2bo\$17b2o!

Basically first interaction must be the same. wwei23

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### Re: Still life puzzles

wwei23 wrote:
dvgrn wrote:One more question: does this glider... strike in the "same spot" as these *WSSes?

Basically first interaction must be the same.

Hmm. Okay, that makes it a little tougher. Now I'm fairly sure that no such eater currently exists, and also that the easiest way to find one would be to build something very large that recovers very slowly.

If I had to construct one of these, I'd probably look for a small constellation that exploded when struck by any *WSS, as the pond does -- but that also exploded in a different way when struck by a glider (with the same initial interaction).

Then it's "just" a matter of adding catalysts to channel the two explosions, getting a different signal output in each case, and using each signal appropriately to send glider salvos back to clean up any leftover junk and rebuild the initial constellation.

It would certainly be much nicer to find a bait object that explodes the same way no matter what hits it, or even better a true four-way omnivorous eater -- generated by Bellman, let's say. But it seems as if that's likely to be a lot harder to find. Gliders just plain have a lot less "weight" behind them than the weightships do. dvgrn
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### Re: Still life puzzles

wwei23 wrote:Do any still lifes exist from 9 to 19 cells so that every living cell has exactly two living neighbors and if not, then why(I've also apgsearched and turned up empty-handed.)?

No. I just did some searches in rule B3/S23 (where all still-lifes must necessarily be of that form), and the counts for various populations are:
4=1; 6=2; 7=1; 8=2; 20=4; 21=1; 22=34; 23=34; 24=57; 25=25; 26=97; 27=88; 28=165; 29=91; 30=129; 31=121; 32=265.

The reason there is a huge gap between 8 and 20 is that this constraint requires that all still-lifes be formed of one or more loops of living cells. This can be stable loops like beehives and ponds, unstable loops that require external stabilization, or L-trominos which also require stabilization. The smallest island that requires such stabilization is 8 bits (a long beehive) but that would require at least another copy of itself on both sides. A 10-bit long long beehive requires at least two beehives, for 22. A 12-bit long beehive can be supported by 2 tubs, for 20, but at that point, we already have the small lake.
wwei23 wrote:Other than some agars, is the block the only still life where every living cell has exactly three neighbors? If so, then why?

A brute-force search reveals none (other than the block) up to 36 bits. If you try to construct one manually, there are many that work as wicks or agars, but none appear to have finite stabilizations. It may be possible to prove this, although it appears that such a proof would be fairly complicated.

Some years ago, I went through a similar process for the rule B34/S34, in which P2 oscillators are plentiful, but the block appears to be the only still-life. I was able to show that other than the block, all still-lifes must necessarily have an exterior that consists of crenelations (straight lines with 1-bit protrusions every 2 or 3 cells), with Life ships or long ships at each corner. No other forms are possible. The smallest such still-life is 36 bits, and an exhaustive search did indeed find that one, plus one at 44; the next ones are two at 50 and one at 51.
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