Live data from Hacker News

1D Conway's Life glider found, 3.7B cells long

conwaylife.com

221–223 of 223 posts

Re: 1D Conway's Life glider found, 3.7B cells long

#221

Earlier quoted context omitted.

Since it's not provable with pi, then we'd have to do a more circuitous proof of every finite pattern occurring. Inspired by Champernowne's constant, I propose a Pontifier Pattern that is simple, inefficient, but provably contains every finite pattern. Starting at the origin, mark off rows of squares. the Nth row would contain NxN^2 squares of size n x n. Each square would be filled in left to right reading order wit…

minor correction: 2^(NxN) squares per row, right?

Yeah, what was I thinking?? I really need to slow down sometimes. This should contain every finite pattern, right?

Re: 1D Conway's Life glider found, 3.7B cells long

#222

Earlier quoted context omitted.

minor correction: 2^(NxN) squares per row, right?

Yeah, what was I thinking?? I really need to slow down sometimes. This should contain every finite pattern, right?

Yes, sounds like it! Though I'm thinking that the relative arrangement of patterns would also make a difference. I wonder if such a thing as "all (infinitely many) possible arrangements of all patterns" can exist

Re: 1D Conway's Life glider found, 3.7B cells long

#223
post #162
post #140

Earlier quoted context omitted.

Piet is mine - the programs are 2D images: https://esolangs.org/wiki/Piet Primarily because of the note on the "calculating pi" example program: > Richard Mitton supplies this amazing program which calculates an approximation of pi... literally by dividing a circular area by the radius twice. > Naturally, a more accurate value can be obtained by using a bigger program. https://www.dangermouse.net/esoteric/piet/sample…

One of my favorite calculations of pi is to pick random coordinates in a unit square and count how many of them are in a circle. it's so stupid and so clever at the same time. This was recreated from memory. I think it is close but I may have a bounding bug. import random def pi(count): inside = 0 for i in range(count): test_x = random.random() test_y = random.random() if test_x ** 2 + test_y ** 2

https://en.wikipedia.org/wiki/Monte_Carlo_method
Post reply on HN