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Zandor
Daedalian Member
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Posted: Sat Mar 09, 2013 5:09 pm Post subject: 1 |
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Chose any polygon and color a bit of it black. With an infinite amount of this tile, how many layers can you get if there are folowing rules:
You may put a tile directly on the ground.
You may put the tile on top of other tiles if
-the complete area of the tile is supported by other tiles AND
-the tile does not cover any black coloured area.
In my best try I got 5 layers by using double hexes:
Can you find better solutions?
With an uncoloured area of only 50% my solution is pretty wasteful. |
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Zag
Tired of his old title
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Posted: Sun Mar 10, 2013 2:34 am Post subject: 2 |
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This is trickier than it looks. I had, at first, thought that a triangle with the hypotenuse a stair step would allow me to make any number of layers, but as you shift the steps, either it has holes or it gets wider when you count the projections that stick out. Anyway, I wasn't able to make any good use of any more than two stairs. I think you could make one more level with this approach if you made the polygon very thin, so that you could fit a single one tilted into the one that is already tilted, but I was tired of monkeying with it by this point.
So here's my effort, also 5 layers (but, as I said, I''m pretty sure this approach would support 6 layers). I used red instead of black for the "don't cover this" color.
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Nsof
Daedalian Member
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Posted: Sun Mar 10, 2013 8:23 pm Post subject: 3 |
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I'm probably missing something
1.
use an n sided regular polygon.
color a triangle formed by its center and two adjacent vertices.
put n of this shapes on top of each other - each layer rotated by 1/n.
2.
use a square.
color a point somewhere on it.
tile the plane.
tile another layer only shifted slightly.
repeat. _________________ Will sell this place for beer |
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novice
No harm. Pun intended!
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Posted: Sun Mar 10, 2013 8:29 pm Post subject: 4 |
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| Nsof, you can't cover the black parts with either white or black. |
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Zag
Tired of his old title
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Posted: Sun Mar 10, 2013 11:20 pm Post subject: 5 |
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| All of layer N must fit completely on only the white part of layer N-1. |
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Nsof
Daedalian Member
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Posted: Mon Mar 11, 2013 8:14 pm Post subject: 6 |
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thanks and thanks I now understand I violated the second rule. _________________ Will sell this place for beer |
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Zandor
Daedalian Member
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Posted: Tue Mar 12, 2013 1:43 pm Post subject: 7 |
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Nice solution, Zag.
By continuing your idea I came up with a solution for any number of layers.
Spoiler:
1. Take an existing solution for n layers with only rectangles where the small side has been coloured
2. Now cut every rectangle in x smaller rectangles, every one of it having a coloured part
3. Make the coloured part smaller
4. If x is big enough you will be able to add another small rectangle on the top by turning it a little.
You now have a solution for n+1 layers
5. Repeat steps 2-5 |
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Zag
Tired of his old title
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Posted: Tue Mar 12, 2013 4:59 pm Post subject: 8 |
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I was trying to convince myself whether or not you always would be able to add more layers in that fashion, but was unable to be certain. I do suspect that as x (the number of slices you cut your rectangle into) goes to infinity, you can always fit one more layer, but I'm not positive of that.
For your problem, you might specify that the black part must be at least, say, 20% of the shape, which will eliminate the infinite solutions. |
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