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Topic: Abducted (Read 1193 times) |
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alien2
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You wake up in a room with four walls and a locked door, clueless about how you go there. The room is well illuminated thanks to a bulb in the middle of the ceiling. The remaining two ceiling objects are a camera and a cylindrical device that looks like a weapon. An intercom is on the wall. There are cubes on the floor made out of strong metal, 201 small identical ones and 2 big identical ones. The big cubes have a lid. You open the lids to find out that both cubes are empty.
A deep voice talks to you on the intercom: "I am your mysterious captor. You were brought here for my amusement. I have the following riddle for you.
Exactly 100 of the small cubes fit in each of the big cubes on the floor. You have the option of using the intercom and asserting the claim that 201 cubes are inside 2 cubes. If this assertion is true, I'll return you to your home and there will be a pouch holding gold coins on the pillow. If it is false, or 3 hrs. expire, I'll activate the pain beam weapon and you'll experience severe pain for 60 s before I let you go empty handed. I'll periodically warn you about how much time you have left. No additional objects have a role in the riddle, just the ones you can see. Good luck to you."
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BNC
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Re: Abducted
« Reply #1 on: Jan 22nd, 2014, 5:46am » |
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Is the room itself cubic?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Grimbal
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Re: Abducted
« Reply #2 on: Jan 22nd, 2014, 2:56pm » |
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I wonder what size a cube must be to be able to hold exactly 100 unit cubes and not a single more.
Is it even possible?
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towr
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Re: Abducted
« Reply #3 on: Jan 22nd, 2014, 10:37pm » |
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Volume-wise it's not a problem.
But if it's about fitting actual whole cubes in, that's hard to say. And, I'd expect, rather hard to do. There might be room for half an extra and thus one if you put the two big cubes on top of each other.
BNC probably has the right answer, though.
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Wikipedia, Google, Mathworld, Integer sequence DB
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BNC
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Re: Abducted
« Reply #4 on: Jan 22nd, 2014, 10:59pm » |
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Actually, I think Grimbal has a very good point.
The only way this would be possible, is if the large cubes are not cubic on the inside (say, different walls' thickness)
If indeed everything is cubic, then the same number of small cubes must fit in every dimension. 4X4X4 is only 64 cubes. The next size up would hold 5X5X5 – 125 small cubes.
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alien2
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Re: Abducted
« Reply #5 on: Jan 23rd, 2014, 6:22am » |
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BNC wins.
It would be easier if I had said 2001, but this isn't the essence of the question.
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BNC
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Re: Abducted
« Reply #6 on: Jan 23rd, 2014, 6:55am » |
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On second thought, I'm not so sure.
When you wake up in the cubical room, there are already 203 cubes inside 1 cube (201 small + 2 large inside the room).
Place small cubes inside a single large one, and you have 203 cubes inside 2 cubes. not 201.
And the pain beam is on!
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alien2
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Re: Abducted
« Reply #7 on: Jan 23rd, 2014, 9:40am » |
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You're right.  Grimbal, go ahead and delete this riddle too. 
Or you are told that you should assert the claim that 201 of the small cubes are inside 2 cubes..
Or your captor tells you that there are 126 small cubes onthe floor and that each of the two big cubes can hold 64 small cubes. He then tells you that there are two more small cubes on the mat and that you should assert the claim that 128 cubes are inside 2 cubes. 
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| « Last Edit: Jan 23rd, 2014, 4:40pm by alien2 » |
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Grimbal
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Re: Abducted
« Reply #8 on: Jan 25th, 2014, 8:48am » |
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Wait! It has not yet proven impossible.
A cube of size 2+1/sqrt(2) can hold at least 10 unit cubes. The maximum capacity isn't necessarily a cube (i.e. of the form n3).
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| « Last Edit: Jan 25th, 2014, 9:42am by Grimbal » |
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BNC
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Re: Abducted
« Reply #9 on: Jan 25th, 2014, 11:18pm » |
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on Jan 25th, 2014, 8:48am, Grimbal wrote:| A cube of size 2+1/sqrt(2) can hold at least 10 unit cubes. The maximum capacity isn't necessarily a cube (i.e. of the form n3). |
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Huh?
You remember we're dealing with physical cubes, right?
I don't see how the cube you mentioned can hold any more than 8 small cubes (with space left - but not enough to fit another solid cube).
What did I miss?
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BNC
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Re: Abducted
« Reply #11 on: Jan 26th, 2014, 3:52am » |
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Oh. Wow.
Thanks!
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| « Last Edit: Jan 26th, 2014, 4:11am by BNC » |
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rloginunix
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Re: Abducted
« Reply #12 on: Jan 26th, 2014, 2:03pm » |
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Don't mean to steal Grimbal's thunder in any way shape or form. Just a quick GeoGebra sketch supporting the evidence (if you find it helpful).
In this formation there are 4^2 + 3^2 = 16 + 9 = 25 unit cubes per layer. There are 4 such layers so here you have your 100 hard unit cubes in a container cube with 4 + 3*sqrt(2)/2 to a side.
Since this is a show of squares, in the next step up there will be 5^2 + 4^2 = 25 + 16 = 41 unit cubes per layer. Over 5 layers there will be 205 hard unit cubes in a container cube with 5 + 2*sqrt(2) to a side.
Also don't mean to intrude on author's rights but one can proceed further in a number of ways. It is possible to eliminate cubic captor's room and make it a challenge to pack 305 hard unit cubes into 2 container cubes.
[edit]
Moved the drawing file to this forum.
[/edit]
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| « Last Edit: Apr 5th, 2014, 8:21am by rloginunix » |
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BNC
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Re: Abducted
« Reply #13 on: Jan 26th, 2014, 8:41pm » |
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Ah, but you see - you have the container's dimension at (4 + 3*sqrt(2)/2). Calculate that: it's about 6.1.
So conventional packing will yield 216 cubes per container, not 100.
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rloginunix
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Re: Abducted
« Reply #14 on: Jan 27th, 2014, 10:18am » |
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Yes, your are right. I thought about that. With a 5-based formation the side is 7.828427125.
We need a limiting condition that doesn't give the answer away but leads to this (or similar) formation.
First cut: smallest aggregate cube where a ladybug can touch at least 4 planes of Every unit cube.
Imagine a cubic container box with "top" and "bottom" removed. Place 100 hard unit cubes inside it, remove the cubic cover, the ladybug requirement must stand. Not sure though if that's the only formation that works.
Another idea is "smallest cubic formation with largest surface area".
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rloginunix
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Re: Abducted
« Reply #15 on: Jan 27th, 2014, 10:03pm » |
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We have a problem. There's no cube.
Within one layer we do have a square. But, once we stack the layers one on top of the other, they sit tight - there's no space in between them. That's bad because we need a sqrt(2)/2 between them for the formation to be a cube.
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towr
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Re: Abducted
« Reply #16 on: Jan 27th, 2014, 10:10pm » |
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Why would that be a problem? It doesn't need to be a tight fit, there just needs to be no way to fit a single extra unit cube in.
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Wikipedia, Google, Mathworld, Integer sequence DB
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Grimbal
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Re: Abducted
« Reply #17 on: Jan 28th, 2014, 8:09am » |
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I should have given this link as example
http://www3.combinatorics.org/Surveys/ds7.html
Anyway, there is no reason to believe that stacking cubes in layers is the best way to fill a cube.
For a capacity of 100, it won't work. If you stack 5 layers then the cube must be at least 5 high and thefore could hold 125 cubes. If you stack only 4 layers then you need 25 cubes per layer, which again requires a size of 5.
So if there is a cube that can hold 100 cubes and not a single more, the optimal packing must be more complex than stacking 2D solutions.
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rloginunix
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Re: Abducted
« Reply #19 on: Jan 28th, 2014, 10:04am » |
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Trying to make the number 100 work I thought I could simplify the problem thus: "arrange 100 unit cubes into a perfect cube formation". Obviously can't do it with the above formation. 6.1 gives room for 2 more layers or 50 extra cubes for 150 total (I guess my gut was taking me towards the figure 11 in towr's link).
Inventing problems is hard!
Final try: how many unit cubes are needed to be arranged inside a cubic box with the side A to yield the surface area S? In other words find an arrangement given two parameters.
For example, A = 2 + 1/sqrt(2) and S = 50. Or A = 3 + sqrt(2) and S = 234, etc.
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riddler358
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Re: Abducted
« Reply #20 on: Feb 27th, 2014, 2:17pm » |
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changing the assertion to "more than 200 cubes are inside 2 cubes" should do the trick i guess
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