Today's Timewaster

:lol: No, not all morning! Although DM's thread title is quite apt seeing as how much of my time today this is wasting! :lol:
 
Larinda McRaven said:
I decided to go a different route... can you get them all facing the same way?
chain.jpg
Click to expand...
*blink* ...wow...

:notworth: :notworth: :notworth:
 
Ok, I see it is better to refresh after a couple of tries than continuing to work with the same board.

I got 1810 as my high now. :)
 
Read my mind. The first thing I did was to get them all going in one direction. Now Im calculating how to set it up so it will turn forever. Mathematically, I think it may be possible.
 
I don't know why I'm still playing this game but here are a few hints I came up with:
1) Identify peices that are blocked from interacting, and change them
2) Peices on the edges can be turned toward the center increasing the chances they will join the interaction
3) a semicircle on the edge facing out cannot possible interact
4) A complete circle cannot possibly interact, so if there is more than one circle, then it might be good to refresh to a new board.
 
chandra said:
Read my mind. The first thing I did was to get them all going in one direction. Now Im calculating how to set it up so it will turn forever. Mathematically, I think it may be possible.
Click to expand...

I'm looking forward to seeing what you come up with.

My instinct tells me just the opposite: that a finite grid must stabilize.

Consider that for a cell to keep changing, it has to see change on at least three sides. However, the corner cells in the grid will reach a stable state when the two ends of the arc point in directions where there are no more cells. For example, the upper left corner cell will stop changing when it reaches the state where its arc points to the left and up.

If the corner cells stabilize, then the cells adjacent to the corner cells will also stabilize, as they now have two sides which will see no change - the corner cell and the direction in which there are no more cells. This condition propagates along the outer edge, forming a stable square.

At this point, the outer edge might as well not exist as far as the internal cells are concerned - they're not changing, and they can't interact with their internal neighbors. Thus, the problem reduces to a smaller problem where the width and height have been reduced by 2.

The original puzzle has a 16x16 grid. This makes the base case for the argument a 2x2 grid. The 2x2 case is simply four corners, which must stabilize.

Thus, I think any arrangement of the cells must stabilize, and the search for a perpetual motion machine will prove fruitless. However, I always learn things when I'm proven wrong :-)

If the puzzle behaved differently the result may be very different. I'd guess that if the interactions wrapped around the edge of the grid, essentially creating a torus, you might be able to find an arrangement that runs forever.

- John
 
DanceMentor said:
I don't know why I'm still playing this game but here are a few hints I came up with:
1) Identify peices that are blocked from interacting, and change them
2) Peices on the edges can be turned toward the center increasing the chances they will join the interaction
3) a semicircle on the edge facing out cannot possible interact
4) A complete circle cannot possibly interact, so if there is more than one circle, then it might be good to refresh to a new board.
Click to expand...
Used a couple of these and got back up to 3,401.
 
John, beleive your right.

ANyhow, way to compicated for me if your not. After like 5 hours spent on this, (I can set it up to get around 7000 easily) I dont think it can go forever. If it can, its way to complicated for me to solve!
 
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