# Tag Info

## Hot answers tagged cellular-automata

87

I found a sun destroyer with just 10 cells filled! It takes 39 generations. As animation:

79

The following is a solution with 34 generations: And, yes, it was really fun to play with it :).

65

6 cells, 43 and 50 generations I found 11 unique (up to reflection) 6-cell solutions: By the first generation of evolution, they form two identical groups: The first, fourth, and last two patterns work by the same method as my previous answer, but are three generations (one pulsar period) slower. Except for some self-destructing debris they are the same ...

58

6 cells, 111 generations I found 12 unique (up to reflection) 6-cell solutions: They all work by the same method as my previous 8-cell solution, forming a house one cell below the top of the base area at generation 6. Here's an animation: There are no solutions with fewer than 6 cells (that stabilize within 200 generations). previous solution: 8 cells, ...

57

23 cells, 41 generations, humorous answer Sorry, I couldn't resist it... The pulsar dies on "magical command" after 41 generations but it takes 23 initial cells. I apologise for the crudeness of the image. It's my first time using Golly. EDIT: The next challenge is answered under my registered name Ruutsa

53

I have a solution: I shamelessly stole the upper part from BaSzAt's 8-cell sun destroyer, and, by trial and error, found a 7-cell lower part that generates a new pulsar without interfering with the upper half. Like BaSzAt's solution, it takes 44 steps to gobble up the old sun, although the new one is already fully formed after 32 generations. The center ...

47

In case anyone is wondering, here is my own solution, which takes 38 generations. (NB: When I made this puzzle, I had no idea if it's even generally solvable. It took me 4 or 5 hours, while having the freedom to change the drawing area as I like.) By the way, the bottom cells are just for decoration. The following works too: As animation:

40

I made another one, with a mere 8 cells. It takes 44 generations. As animation:

36

More humorous solutions (destroy methods) Sorry, I couldn't resist too Call 911 :) 20 cells, 38 generations: Use AI (artificial intelligence), 62 generations:

22

23 cells, 41 generations, humorous answer 2: a special word? It seems to be a special word. When said whimsically it moves stars. When said in annoyance it destroys them. Here's the animation

22

Here is my answer in text form. It's long, so I stored it on pastebin: 19 steps I don't have time to convert it to png (thanks @TheDarkTruth, see comments to the question), so maybe later. Gif animation :

21

19 cells, 41 generations, another humorous answer EDIT: Only just noticed my answer to the previous challenge was logged in under the guest name user21465. Continuing in the vein of humorous answers, here's another one. I thought it was going to be difficult and when I tried it I said ... Here's the animation

21

We have the smallest (6 cells), so to complete the picture here is (probably) the biggest "ship", 46 cells, 40 generations:

19

This is a pretty common puzzle. Warm up Answer: Advanced Answer Explanation:

17

There are only 16 different possible state combinations of the four ancestor cells, and you can find them all in the image, so there is a unique answer. The rule is as follows:

17

The pattern is self-similar, and can be formed by repeatedly scaling and rotating copies of itself: An alternate dissection that fits in a diamond:

15

I have found this solution (35 gens): Animation:

14

Not exactly answering this question, but given 9 infections still on an 8x8 board, it is actually possible to delay the inevitable until 40 days. Pretty counter-intuitive huh? Locations are A1, A5, A8, B3, B7, D1, E8, G1, H8 Also, I believe that this is the maximum for an 8x8 board with any number of initial infections. As for the edges argument I made ...

12

The nonincreasing quantity $X$ to look at this time is Lower bound: Considerations: Minimal example: Pattern code to verify in a CA simulator such as Golly: x = 9, y = 9, rule = B45678/S012345678 o7bo$2bo$7bo2$2bobo2bo2$2bobo$6bo$obobo3bo!

11

EDITED - thanks to @Meelo for pointing out my mistake. We can do it in We place the C.Coli's at a4, a8, b2, c1, e2, g1, h3, h6. Now we prove that this is optimal. Every day there are either 1 new infected cell or 2+ new infected cells. Let us have k days with single infections and l days with 2+ infections. Then k+2l<=56 and therefore k+l<=28+k/2. ...

9

21 cells, 111 generations, last humorous answer As they can be weary, this is my last humorous answer to this question. Moving stars seems to be a no brainer ... Here's the animation

9

I know it is a checkerboard but this problem may need a bit of chess:

8

8

I think the rule is:

8

I'm not sure how the rules are written but it seems to be either or Image

7

This is by no means an answer, but using @2012rcampion's interpretation, I wanted to illustrate how this fractal can be generated using top-down approach, starting from a single element: /\ /\ / \/ \ / /\ \ / / \ \ \ / \ / \/ \/ /\ /\ / \/\/\/ \ \ /\/\/\ / \/ \/ Step one: Step two: Step three: The element is ...

7

Here's a 17-cell solution that works at any distance: I bet this (or something very similar) is the solution you were expecting. It consists of: This solution takes 2×n + 32 steps to destroy the target, where n is the number of rows between the starting area and the target oscillator. In your baseline puzzle, n = 7, so the target is destroyed in 46 ...

7

Info A .rule file, for Golly users, provided by AxiomaticSystem Further features of this cellular automaton beyond the questions posed here, including new spaceships, a puffer and rakes that generate gliders, racers, and swimmers, are in this thread at the Conway Game of Life Wiki. Oscillators period 2 Phoenix 1: (added by Jeremy Dover) Beacon and Clock: (...

6

Twenty virus samples are necessary. A chessboard can be divided into two diagonal meshes with no diagonal adjacencies between them, and the virus can't spread from one to the other just like a bishop can only reach half the squares, so we can consider only the "black" squares and use the same pattern for the white. Furthermore, if we rotate the board 45 ...

6

Consider the diagonals going northwest/southeast. (I'll call these primary diagonals and the ones going the other way secondart diagonals.) Look at the top-right-most primary diagonal that contains any alive squares. None of those squares can survive, since they can have at most one cell alive in their neighborhood. This means that in each step with living ...

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