

Mushroom Life - superberliner
http://a.parsons.edu/~joseph/k2/gameoflife/

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pyre
I ended up in a deadlock. No new mushrooms are growing and the existing
mushrooms are in a pattern where they won't die. Does this mean I won or lost?
;)

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RiderOfGiraffes
It sounds to me like you're not familiar with Conway's Life.

<http://en.wikipedia.org/wiki/Conways_Game_of_Life>

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andrewcooke
this made me wonder whether there's some function of the number of cells for
which, if a population stays alive for that number of cycles, it will live
forever.

for example, for a 10x10 grid, maybe any pattern that has life after 100
cycles will always have life.

i guess this might be covered in "a new kind of science" - i haven't read
that. any who has care to comment? thanks.

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camccann
The function you're looking for would be a variation of the Busy Beaver
function, I think. On an infinite grid, the question reduces to the Halting
Problem.

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andrewcooke
ah yes, you're right. thanks...

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camccann
Actually, my apologies for the overly brief comment earlier, I was on lunch
and typing on a smartphone.

I suppose you understood my point, but I'll elaborate for posterity: Conway's
Life is Turing complete, with the conceptually unbounded grid standing in for
the infinite tape of a Turing machine. As illustration, here's a page that
implements a _literal_ Turing machine using Life patterns: <http://rendell-
attic.org/gol/tm.htm>

Starting from an empty tape, the number of states in the Turing machine must
obviously be proportional to the number of live cells needed; updates to the
state of the grid are equivalent to running time for the TM; and the output
produced by the machine must be proportional to the population of live cells.

From this, we can (in a non-rigorous, hand-wavy sort of way) conclude that
asking for an upper bound on the number of cycles to decide if any grid of a
given finite size will be populated or barren is very likely equivalent to
asking for an upper bound on the running time needed to decide if a Turing
Machine of a given finite number of states will halt. This isn't actually the
Busy Beaver function, but is closely related.

Thus concludes my proof by enthusiastic hand-waving (if there isn't a fancy
Latin term for that, there should be).

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dougp
Another cool one to try the spinner 3 in a row

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ax0n
I am a complete sucker for all things Life. Awesome!

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michael_dorfman
Cute visualization!

