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Conways' Game of Life is a well known cellular automaton "played" on an infinite grid, filled with cells that are either alive or dead. Once given an initial state, the board evolves according to rules indefinitely. Those rules are:

  • Any live cell with 2 or 3 living neighbours (the 8 cells immediately around it) lives to the next state
  • Any dead cell with exactly 3 living neighbours becomes a living cell
  • Any other cell becomes a dead cell

Let \$B_i\$ be the board state at the \$i\$th step, beginning with \$B_0\$ as the starting configuration, before any of the evolution rules are applied. We then define a "fixed board" to be a board state \$B_i\$ such that there is a board state \$B_j\$, with \$j > i\$, such that \$B_i = B_j\$ (that is, all cells in both boards are identical). In other words, if a board state is ever repeated during the evolution of the board, it is a fixed board.

Due to the nature of Game of Life, this definition means that a "fixed board" is either constant, or the only evolution is a fixed periodic set of states.

For example, the following board states are fixed, as \$B_{217} = B_{219}\$, where \$B_{217}\$ is the left board:

two game of life boards where all structures are either still lifes or oscillators

The left board state happened to be the 217th generation of this starting state, taken from this challenge.

Note that the empty board is a fixed board, and that any patterns of infinite growth, or spaceship like patterns can never result in a fixed board.

This is an challenge where each answer is a starting state, and the number of generations it takes to reach a fixed board must be strictly larger than the previous answer.

The \$n\$th answer should be a starting configuration of cells within an \$n \times n\$ bounding box such that, if the \$n-1\$th answer took \$p_{n-1}\$ generations to reach a fixed board, the \$n\$th answer takes \$p_n > p_{n-1}\$ generations to reach a fixed board. Note that the "number of generations" is defined as the lowest \$i\$ such that \$B_i\$ is a fixed board. For example, the empty board takes \$0\$ generations, as \$B_0\$ is already a fixed board.

Each answer's score is \$i\$, the number of generations to a fixed board.

I have posted the starting answer, for \$n = 2\$, with \$p = 1\$, below.

Rules

  • You must wait an hour between posting two answers
  • You may not post two answers in a row
  • You can edit your answer at any point to change its score. However, keep in mind the following:
    • If your answer is the latest, people may be working on a follow-up answer, so changing yours may invalidate their attempt
    • If your answer is in the middle of the chain, you may edit it to any score, so long as it is within the bounds of the previous and later answers.
  • The answer with the highest score (i.e. the latest) at any given point in time is the winner
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3
  • 3
    \$\begingroup\$ Suggested subtitle: The Secret Doctors Don't Want You To Know!! \$\endgroup\$
    – Arnauld
    Sep 1, 2022 at 13:44
  • 1
    \$\begingroup\$ I offer a 500 rep bounty for the first engineered pattern in this challenge. \$\endgroup\$
    – alephalpha
    Sep 7, 2022 at 10:20
  • 1
    \$\begingroup\$ One more failed attempt... bo2b3o4b2o$o9bobo3bo$o4bo6bo3b2o$5o4bo5bobo$8b2o$8b2o$9b2o$6bo3b2o6b2o$5bobobo2b2o3bobo$6bo4b3o5bo$b2o13bo$3o5bo6b2o$2o4bo2bo$4o2bo2bo$2o5bo$o$o$2obo9bob3o$2bo9b3o$13b5o! \$\endgroup\$
    – Pavgran
    Sep 8, 2022 at 19:38

20 Answers 20

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8
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\$n = 12, p = 2215\$

Evolves to beehive + blinker + r-pentomino after 46 generations

.o.ooooooo..
....oo..o..o
.o.o...oo...
.ooooo...o..
o.....ooo...
....o..ooo..
..o..oo.o...
o.oo........
...o.o..oo.o
....oo....oo
..o.oo...o.o
..ooo.o...o.

RLE:

x = 12, y = 12, rule = B3/S23
bob7o$4b2o2bo2bo$bobo3b2o$b5o3bo$o5b3o$4bo2b3o$obo2b2obo2bo$2b2o$3bobo
2b2obo$4b2o4b2o$2bob2o3bobo$2b3obo3bo!

Try it online!

I wonder at what value of \$n\$ would engineered patterns appear. There are some cool engineered diehards at \$n=32\$ and \$n\approx96\$.

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2
6
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\$n=7, p=456\$

I can't find a parent of Bubbler's answer. So instead, I found a grandparent of its child.

. . . . # . .
. # . . # . #
. # . # # . #
. . . # . # .
# . # . # # #
. # . . . . #
. . . . . . .

Or in RLE format:

x = 7, y = 7, rule = B3/S23
4bo$bo2bobo$bob2obo$3bobo$obob3o$bo4bo!

Try it online!

Improve this answer
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6
\$\begingroup\$

\$n = 10, p = 1353\$

Search result from a simple script that runs for 1024 gens and then another 4 gens and checks that the bounding box is the same and population is different.

oo.....o..
.o.oo.o.o.
.oo.....o.
oo...o..oo
...o....oo
...ooo..o.
....o.o.o.
ooooo.o..o
..oo.ooo.o
ooo..o...o

RLE:

x = 10, y = 10, rule = B3/S23
2o5bo$bob2obobo$b2o5bo$2o3bo2b2o$3bo4b2o$3b3o2bo$4bobobo$5obo2bo$2b2ob
3obo$3o2bo3bo!

Try it online!

Also, it looks like it has no 10x10 predecessors, but it's possible I didn't set the search for that properly.

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5
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\$n = 2, p = 1\$

This has board states \$B_0, B_1, B_2\$ left-to-right below. This is the preloaded pattern.

3 live Game of Life cells, arranged as a 2 by 2 square missing the bottom right corner A 2 by 2 square of live cells A second 2 by 2 square of live cells

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2
  • 1
    \$\begingroup\$ ... why are your two images different sizes? \$\endgroup\$
    – Neil
    Sep 1, 2022 at 8:25
  • \$\begingroup\$ @Neil Because they're cropped images of different screenshots \$\endgroup\$
    – caird coinheringaahin g ♦
    Sep 1, 2022 at 13:49
5
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\$n = 6, p = 455\$

Found by spamming the random button for two hours. Interestingly, the active pattern converges to my own n=4 solution. This makes me suspect that only very few actual long-lasting patterns exist which result in a valid answer for this challenge.

...##.
.##..#
#..#..
###.#.
#.#..#
...#..

#C Generated by copy.sh/life
x = 6, y = 6, rule = B3/S23
3b2o$b2o2bo$o2bo$3obo$obo2bo$3bo!

View it live on copy.sh/life.

Improve this answer
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5
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\$n=11, p=1697\$

I did some soup search this time.

. O . O O O O O O . .
O . O . . . O O . . .
. O . O O O . . O O .
O . . . . . O . . . .
. . O O . . . O . . O
. . . O . O . O . . O
. O . . . O O . . O O
O O O O . O . O . . .
O O O O O . O . . O .
O O . O . . O . . O O
. O . O . . . . . O O

Or in RLE format:

x = 11, y = 11, rule = B3/S23
bob6o$obo3b2o$bob3o2b2o$o5bo$2b2o3bo2bo$3bobobo2bo$bo3b2o2b2o$4obobo$
5obo2bo$2obo2bo2b2o$bobo5b2o!

Try it online!

Improve this answer
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1
  • 3
    \$\begingroup\$ Wow, it converges to 2bo$b3o$o$o$o! after 19 generations. It's a 7-cell methuselah, and I didn't find it on LifeWiki. It's not a novel pattern in the sense that all 10-cell (and below) patterns had been explored, but it may be novel in the explicit category of gliderless methuselahs. \$\endgroup\$
    – Pavgran
    Sep 1, 2022 at 15:27
4
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n=3, p=2

. . . . .
. # . . .
. . . # .
. . # # .
. . . . .
Improve this answer
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7
  • \$\begingroup\$ It seems that this is still \$p=1\$, because it takes 1 step to reach a still life. \$\endgroup\$
    – alephalpha
    Sep 1, 2022 at 1:04
  • \$\begingroup\$ @alephalpha don't the two top left cells die in turn 1 with the centre regrowing only in step 2? \$\endgroup\$
    – loopy walt
    Sep 1, 2022 at 1:08
  • \$\begingroup\$ @loopywalt Two dots spawn at the same time, so it becomes .#. / #.# / .## \$\endgroup\$
    – Bubbler
    Sep 1, 2022 at 1:10
  • \$\begingroup\$ Indeed, this is still \$p = 1\$, not \$2\$, as \$B_0\$ is this, \$B_1\$ is a boat still life, and so are \$B_i\$ for \$i \ge 1\$ \$\endgroup\$
    – caird coinheringaahin g ♦
    Sep 1, 2022 at 1:11
  • 2
    \$\begingroup\$ For the record, a configuration with p=175: ### / .#. / #.# (caird's is 174) \$\endgroup\$
    – Bubbler
    Sep 1, 2022 at 1:30
4
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\$n=13, p=2600\$

. O O . . O . O O . . O O
O . O O . O . . . O O . .
. . O . O . . . O O . . .
O . . . O . . O . . . O .
O . O . O O . . . O . O .
O . O O . . . O . O O O .
O O . . O . O O . O O O O
. . . O . O O O O . . . .
. . . O O . . O O . O . .
. . . O . . . . O . . O O
. . . . O . . . . O . . .
. . O . O O . . O . O O O
. O . O O . O . . O O O O

Or in RLE format:

x = 13, y = 13, rule = B3/S23
b2o2bob2o2b2o$ob2obo3b2o$2bobo3b2o$o3bo2bo3bo$obob2o3bobo$ob2o3bob3o$
2o2bob2ob4o$3bob4o$3b2o2b2obo$3bo4bo2b2o$4bo4bo$2bob2o2bob3o$bob2obo2b
4o!

Try it online!

Improve this answer
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3
\$\begingroup\$

\$n = 4, p = 431\$

.XXX
....
XX.X
.X.X

View it live on copy.sh/life.

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3
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\$n=8,p=901\$

Another product of random search. The pattern this time is clearly different from the previous ones, and it almost fits in 7x7 board (the bottom left cell can be safely removed).

.#...#.#
.#.....#
.#..#.##
..#.....
.#....#.
.#..#.##
...#.#.#
#.....#.

#C Generated by copy.sh/life
x = 8, y = 8, rule = B3/S23
bo3bobo$bo5bo$bo2bob2o$2bo$bo4bo$bo2bob2o$3bobobo$o5bo!

View it live on copy.sh/life.

Improve this answer
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3
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\$n=9, p=903\$

A grandparent of Bubbler's answer.

O . O . O O . O .
O O O . O . O O .
. O . O O . . . .
O . . . . O . . .
. . O . . . . . .
. . . . O O O . .
O O O . O . . O .
O . O O O . O . O
. O . O O . . . .

Or in RLE format:

x = 9, y = 9, rule = B3/S23
obob2obo$3obob2o$bob2o$o4bo$2bo$4b3o$3obo2bo$ob3obobo$bob2o!

Try it online!

All these parents and grandparents are found using my Mathematica package, LifeFind. The code in the package isn't too complicated. You can try it on TIO.

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3
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\$n = 14, p = 3185\$

Random soup battle continues.

. O . . O . . . O . O . O O
O O . . . O O . . . . . O .
. O O O . . O . . . O . . .
. O O O O O . . . O O O O O
. O O O . . O O . . O O . O
. O O O O . O O O O . . . .
O . . . O O O O O O . . O .
. O . . . O O O . O . O O .
O . . O . . . . . . . . . O
. . . . . O . . O . . . . O
. . O . . . . O O O . . O O
. . O O O . O O . . . . O O
. . . . . . . O . O . O . .
. . . O . . . O O O . . . .

RLE:

x = 14, y = 14, rule = B3/S23
bo2bo3bobob2o$2o3b2o5bo$b3o2bo3bo$b5o3b5o$b3o2b2o2b2obo$b4ob4o$o3b6o2b
o$bo3b3obob2o$o2bo9bo$5bo2bo4bo$2bo4b3o2b2o$2b3ob2o4b2o$7bobobo$3bo3b
3o!

Try it online!

Improve this answer
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3
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\$n = 18, p = 5231\$

Just one more random soup. 18x18 is better, but still too tight for engineered designs I tried.

. . . O . O O . O O . . O . . O O O
O . O O O . . . O O O . . . . . . O
. . . . O . O . . O . . O . . . . .
. . O . O O O . . O . O . O O . . .
O O O O O . . O . O . O O . O O O O
. . . . O . O O . . . . . . . O . .
. O . . . . . . . O . . O O O O O O
O O . . . O . . . O O O . . . O . .
O . O . O O . . . O O . . . . . . O
. . . O . . O . O . . . O . . . O O
O . O O O O O O . O O . O . . . . .
O . . . . . . . . . O O O O O . . .
. O O O . . . O O . . . . . . O . .
. . O O . O . . O O . O O . . . O .
. . O O . . O O O O . . . O . . . .
. O . . O O . . . . O . . O . O O O
. . . O . . . O . . O O O . . O O O
. . O . . . O O . . O . O . . . . .

RLE:

x = 18, y = 18, rule = B3/S23
3bob2ob2o2bo2b3o$ob3o3b3o6bo$4bobo2bo2bo$2bob3o2bobob2o$5o2bobob2ob4o$
4bob2o7bo$bo7bo2b6o$2o3bo3b3o3bo$obob2o3b2o6bo$3bo2bobo3bo3b2o$ob6ob2o
bo$o9b5o$b3o3b2o6bo$2b2obo2b2ob2o3bo$2b2o2b4o3bo$bo2b2o4bo2bob3o$3bo3b
o2b3o2b3o$2bo3b2o2bobo!

Try it online!

Improve this answer
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2
\$\begingroup\$

\$n=5,p=432\$

A parent of Bubbler's answer.

.XX..
.XX.X
XX.X.
.....
..XX.

Or in RLE format:

x = 5, y = 5, rule = B3/S23
b2o$b2obo$2obo2$2b2o!

Try it online!

Improve this answer
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2
\$\begingroup\$

\$n = 16, p = 3456\$

I've tried a few of engineered designs for that size, but they all were not as good as a random soup.

. . . O O . O . O . O O . O . .
. O . O O . . . . . . O O . O .
O O O O . . O O . . . . . O . .
. . O O . O O . . . . . . . . O
O O . . . . O . . O . O . . O .
O . . O . . O . O . O . O O . O
. O . . . O . . . . . . . . . O
O O . . . . . . O O O . . . O .
. . . . . O O O O . . O O . O O
O O O O O O . . . . . . O . O O
. . . O O . . . O . O O . . O O
O O O O O O O O . . . . . O . O
O . . . . . . . O . O . . . . .
. . . . . O . . . O . O . O . O
. . O . O . . O . O . O . . . .
O O . . O O . O . . O . . O . .

RLE:

x = 16, y = 16, rule = B3/S23
3b2obobob2obo$bob2o6b2obo$4o2b2o5bo$2b2ob2o8bo$2o4bo2bobo2bo$o2bo2bobo
bob2obo$bo3bo9bo$2o6b3o3bo$5b4o2b2ob2o$6o6bob2o$3b2o3bob2o2b2o$8o5bobo
$o7bobo$5bo3bobobobo$2bobo2bobobo$2o2b2obo2bo2bo!

Try it online!

Improve this answer
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2
\$\begingroup\$

\$n=17, p=3696\$

O . O O . O . . O . . . O O O . .
. . . O . . . . O O O . O . O O O
. . . . O . O . O . . O . O . . O
. . . . O . . O O . . . O . O O .
O O . O O . . . . O . . O . . O .
O O O . O . . O O . O O O . O O .
. O O . . O . . . O O O O . O . .
O O O O . . O . O O . O O . O . .
. . . . O O O . O O . . O O O O O
O O O . . . . . O O O O . . . . .
O O O O O O O . O . O O . O O . O
O . O . . O . . O O . . . . O . O
O . O O O O O . . O . . . O O O .
. O O . . O . O O . . . O . O . .
O O . . O O . . . . . O O O . . .
O . . . O . . . . . O . O O . . .
. O . . O . . . . O O . O O O . O

x = 17, y = 17, rule = B3/S23
ob2obo2bo3b3o$3bo4b3obob3o$4bobobo2bobo2bo$4bo2b2o3bob2o$2ob2o4bo2bo2b
o$3obo2b2ob3ob2o$b2o2bo3b4obo$4o2bob2ob2obo$4b3ob2o2b5o$3o5b4o$7obob2o
b2obo$obo2bo2b2o4bobo$ob5o2bo3b3o$b2o2bob2o3bobo$2o2b2o5b3o$o3bo5bob2o
$bo2bo4b2ob3obo!

Try it online!

Improve this answer
\$\endgroup\$
2
\$\begingroup\$

\$n = 20, p = 5627\$

I've tried random symmetric soups this time. Still can't fit engineered solutions into such small a space. This pattern is a 2-generation predecessor of a (variant of a) symmetric pattern found by search.

. . O . . . O . . . . . . O . . O . O .
. . . . O O . . O O . O O O . . . . O .
. O O . O . O O O O . . . O . O . O . .
O . . . . . O . O . . O . O . O . . . O
O . O . O . . . . . O . . . . . O O . O
. . O . . . . O O . . O O . O . O . . .
O O . O . O . . . O . O O . O O . O . O
. O O . . O . . O O O O . . O . . O O O
. O O . . . . O . O . . . O . O . . . .
. O . . . O . . O O O . . O . . . O . .
. . . O O . O . O . . O . . O O O O . O
O . . O O . . . . O . . . . O . . . . .
. . . . . O O . . O . O . O O O . . . .
O . O . O O . . O O O . . . . . . O O .
. . . O . O . O O . O . O . . O . . . O
O . O O . . O . . O . . . . . O O . . O
O . . . O . . . O . . O . . O O . . . .
. . O . O . O . . . O O O O . O . O O O
. O . . . . O O O . O O . . O . . . . .
. O . O . . O . . . . . . O . O . O . O

RLE:

x = 20, y = 20, rule = B3/S23
2bo3bo6bo2bobo$4b2o2b2ob3o4bo$b2obob4o3bobobo$o5bobo2bobobo3bo$obobo5b
o5b2obo$2bo4b2o2b2obobo$2obobo3bob2ob2obobo$b2o2bo2b4o2bo2b3o$b2o4bobo
3bobo$bo3bo2b3o2bo3bo$3b2obobo2bo2b4obo$o2b2o4bo4bo$5b2o2bobob3o$obob
2o2b3o6b2o$3bobob2obobo2bo3bo$ob2o2bo2bo5b2o2bo$o3bo3bo2bo2b2o$2bobobo
3b4obob3o$bo4b3ob2o2bo$bobo2bo6bobobobo!

Try it online!

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

\$n=15, p=3239\$

O . O O . . . . . . O O . . .
. O O . . . O . . O . . . . .
. O O . O . O O . . . . . O O
O O . O . . . . . O . . . O O
O . . . O . O O O . O . . O .
. . O O O O . O O O . . O . O
O O . O . O . O O O O . O O .
O O O O O O O . . . O . . . .
O . . O . O . O . . . O O O .
O O . . . . . O . . . . O . .
. . O O . O . . . O O . . . O
O O O . O O . O . . . O O O O
O . O O O . O . . O O . . O O
. . O . . O . . O O . O . O .
. O . . . O O O O . O . . . .

x = 15, y = 15, rule = B3/S23
ob2o6b2o$b2o3bo2bo$b2obob2o5b2o$2obo5bo3b2o$o3bob3obo2bo$2b4ob3o2bobo$
2obobob4ob2o$7o3bo$o2bobobo3b3o$2o5bo4bo$2b2obo3b2o3bo$3ob2obo3b4o$ob
3obo2b2o2b2o$2bo2bo2b2obobo$bo3b4obo!

Try it online!

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

\$n=19, p=5232\$

A grandparent of the child of Pavgran's answer.

. . . . . O . . . . . O O . O . . . .
O . O O . . . . . O . . . . . O O . O
. . . . O O . O O . . O O . O O . O .
O O . . . . . . . O . . . . O O . . .
. . . . . O O . . O O O . . O . . . O
O O O . O O . . . . . O . . . O . . .
. O . . O . O . O . . O . . . O . O .
. O . . O . . . O . . . . . O . . O .
O O . O O O . O O . O . O . O . . O O
. . O . O . . . O . O O O . . . O O .
O . . . . . . . O . O O O O . . . . O
O . O . . . O O . O . . O . . O . O .
. O . O O O . O . . . . . . O O O . .
. . O O . O O O . . O O . O . . O . O
O . . . . O . . O . . . . . . O O . .
. . . O . . O O O . . . O . . . O . .
O . O O O O . . . . O . . O O O O . O
. . . . . . . O O O O O . O . O . . .
O . . . O O . O O . . . . . . . . . O

x = 19, y = 19, rule = B3/S23
5bo5b2obo$ob2o5bo5b2obo$4b2ob2o2b2ob2obo$2o7bo4b2o$5b2o2b3o2bo3bo$3ob
2o5bo3bo$bo2bobobo2bo3bobo$bo2bo3bo5bo2bo$2ob3ob2obobobo2b2o$2bobo3bob
3o3b2o$o7bob4o4bo$obo3b2obo2bo2bobo$bob3obo6b3o$2b2ob3o2b2obo2bobo$o4b
o2bo6b2o$3bo2b3o3bo3bo$ob4o4bo2b4obo$7b5obobo$o3b2ob2o9bo!

Try it online!

Improve this answer
\$\endgroup\$
0
\$\begingroup\$

\$n=21, p=5744\$

O . . . . . . . . . O . O . O . . . O . O
. . . . O . . . . . . . O . O O . . . . .
O . O . . . . . . O O . . . O O . O O . .
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. O O O . . . . . . O . . . . . . . . . .
O O O . . . . . . O O . . O . . O O . . .
. . . . . . O O . . O . O O . . . O . . .
O O . . O . . O . . . O . . . . . . . . .
. . . . . . . . O O O O . . O . . . O . .
O . O O O O O O . O O O . O O O O O O . O
. . O . . . O . . O O O O . . . . . . . .
. . . . . . . . . O . . . O . . O . . O O
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. . . . . O O . O . . . . . . . O . . . .
O . O . . . O . O . O . . . . . . . . . O

x = 21, y = 21, rule = B3/S23
o9bobobo3bobo$4bo7bob2o$obo6b2o3b2ob2o$2bo3b2o2bo4bo$6bo3bobo6bo$b3o6b
o$3o6b2o2bo2b2o$6b2o2bob2o3bo$2o2bo2bo3bo$8b4o2bo3bo$ob6ob3ob6obo$2bo
3bo2b4o$9bo3bo2bo2b2o$3bo3b2obo2b2o$3b2o2bo2b2o6b3o$10bo6b3o$bo6bobo3b
o$5bo4bo2b2o3bo$2b2ob2o3b2o6bobo$5b2obo7bo$obo3bobobo9bo!

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