# Live a longer life

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:

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
• Suggested subtitle: The Secret Doctors Don't Want You To Know!! Sep 1, 2022 at 13:44
• I offer a 500 rep bounty for the first engineered pattern in this challenge. Sep 7, 2022 at 10:20
• One more failed attempt... bo2b3o4b2o$o9bobo3bo$o4bo6bo3b2o$5o4bo5bobo$8b2o$8b2o$9b2o$6bo3b2o6b2o$5bobobo2b2o3bobo$6bo4b3o5bo$b2o13bo$3o5bo6b2o$2o4bo2bo$4o2bo2bo$2o5bo$o$o$2obo9bob3o$2bo9b3o$13b5o! Sep 8, 2022 at 19:38 ## 20 Answers # $$\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\$$.

# $$\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! # $$\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.

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

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

• ... why are your two images different sizes?
– Neil
Sep 1, 2022 at 8:25
• @Neil Because they're cropped images of different screenshots Sep 1, 2022 at 13:49

# $$\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. # $$\n=11, p=1697\$$ I did some soup search this time. . O . O O O O O O . . O . O . . . O O . . . . O . O O O . . O O . O . . . . . O . . . . . . O O . . . O . . O . . . O . O . O . . O . O . . . O O . . O O O O O O . O . O . . . O O O O O . O . . O . O O . O . . O . . O O . O . O . . . . . O O  Or in RLE format: x = 11, y = 11, rule = B3/S23 bob6o$obo3b2o$bob3o2b2o$o5bo$2b2o3bo2bo$3bobobo2bo$bo3b2o2b2o$4obobo$5obo2bo$2obo2bo2b2o$bobo5b2o!  Try it online! • Wow, it converges to 2bo$b3o$o$o$o! after 19 generations. It's a 7-cell methuselah, and I didn't find it on LifeWiki. It's not a novel pattern in the sense that all 10-cell (and below) patterns had been explored, but it may be novel in the explicit category of gliderless methuselahs. Sep 1, 2022 at 15:27 ## n=3, p=2 . . . . . . # . . . . . . # . . . # # . . . . . .  • It seems that this is still $p=1$, because it takes 1 step to reach a still life. Sep 1, 2022 at 1:04 • @alephalpha don't the two top left cells die in turn 1 with the centre regrowing only in step 2? Sep 1, 2022 at 1:08 • @loopywalt Two dots spawn at the same time, so it becomes .#. / #.# / .## Sep 1, 2022 at 1:10 • Indeed, this is still $p = 1$, not $2$, as $B_0$ is this, $B_1$ is a boat still life, and so are $B_i$ for $i \ge 1$ Sep 1, 2022 at 1:11 • For the record, a configuration with p=175: ### / .#. / #.# (caird's is 174) Sep 1, 2022 at 1:30 # $$\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! # $$\n = 4, p = 431\$$ .XXX .... XX.X .X.X  View it live on copy.sh/life. # $$\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.

# $$\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.

# $$\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! # $$\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!

# $$\n=5,p=432\$$

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


Or in RLE format:

x = 5, y = 5, rule = B3/S23
b2o$b2obo$2obo2$2b2o!  Try it online! # $$\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!

# $$\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!

# $$\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! # $$\n=15, p=3239\$$ O . O O . . . . . . O O . . . . O O . . . O . . O . . . . . . O O . O . O O . . . . . O O O O . O . . . . . O . . . O O O . . . O . O O O . O . . O . . . O O O O . O O O . . O . O O O . O . O . O O O O . O O . O O O O O O O . . . O . . . . O . . O . O . O . . . O O O . O O . . . . . O . . . . O . . . . O O . O . . . O O . . . O O O O . O O . O . . . O O O O O . O O O . O . . O O . . O O . . O . . O . . O O . O . O . . O . . . O O O O . O . . . .  x = 15, y = 15, rule = B3/S23 ob2o6b2o$b2o3bo2bo$b2obob2o5b2o$2obo5bo3b2o$o3bob3obo2bo$2b4ob3o2bobo$2obobob4ob2o$7o3bo$o2bobobo3b3o$2o5bo4bo$2b2obo3b2o3bo$3ob2obo3b4o$ob 3obo2b2o2b2o$2bo2bo2b2obobo$bo3b4obo!  Try it online! # $$\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! # $$\n=21, p=5744\$$ O . . . . . . . . . O . O . O . . . O . O . . . . O . . . . . . . O . O O . . . . . O . O . . . . . . O O . . . O O . O O . . . . O . . . O O . . O . . . . O . . . . . . . . . . . O . . . O . O . . . . . . O . . O O O . . . . . . O . . . . . . . . . . O O O . . . . . . O O . . O . . O O . . . . . . . . . O O . . O . O O . . . O . . . O O . . O . . O . . . O . . . . . . . . . . . . . . . . . O O O O . . O . . . O . . O . O O O O O O . O O O . O O O O O O . O . . O . . . O . . O O O O . . . . . . . . . . . . . . . . . O . . . O . . O . . O O . . . O . . . O O . O . . O O . . . . . . . . . O O . . O . . O O . . . . . . O O O . . . . . . . . . . O . . . . . . O O O . . O . . . . . . O . O . . . 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 = 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!


Try it online!