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Topic: Bricks in a box (Read 887 times) |
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James Fingas
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Re: Bricks in a box
« Reply #1 on: Oct 22nd, 2003, 6:24am » |
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I'm guessing the answer is "no", just because that's always the answer for this kind of question...
But you probably want a proof, right?
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aero_guy
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Re: Bricks in a box
« Reply #2 on: Oct 22nd, 2003, 6:35am » |
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Yeah, break one in half.
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towr
wu::riddles Moderator
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Re: Bricks in a box
« Reply #3 on: Oct 22nd, 2003, 6:40am » |
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on Oct 22nd, 2003, 6:35am, aero_guy wrote:Is it still a brick if it's broken in half..
more to the point, is it stil a 1x1x4 brick ? 
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James Fingas
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Re: Bricks in a box
« Reply #4 on: Oct 22nd, 2003, 7:16am » |
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on Oct 22nd, 2003, 6:35am, aero_guy wrote:
I know you need to break at least one brick into at least two pieces, but can you actually put them in after breaking only one brick in half? Breaking two of them in half would certainly be sufficient...
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Doc, I'm addicted to advice! What should I do?
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aero_guy
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Re: Bricks in a box
« Reply #5 on: Oct 22nd, 2003, 9:42am » |
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Dang, you are right, you need to break two.
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Dudidu
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Re: Bricks in a box
« Reply #6 on: Oct 22nd, 2003, 10:59am » |
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Everyone hi,
I think that breaking bricks is forbidden (since otherwise it would have been mentioned in the question that reformation of the bricks is legal).
p.s - Is my assumption seems unreasonable to anyone 
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aero_guy
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Re: Bricks in a box
« Reply #7 on: Oct 22nd, 2003, 4:36pm » |
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Very often we will abuse puzzles by stating answers that, while possible, were obviously not intended. This is an example. We all know you are not supposed to break the brick, but we have seen several of these types of puzzles, are pretty sure it can't be done, but are a little too lazy to completely prove why right now(at least I am). Thus the cheap answer.
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aero_guy
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Re: Bricks in a box
« Reply #8 on: Oct 22nd, 2003, 5:34pm » |
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OK, I think I have a proof, though I may have messed up some math. Here it is:
Notice each 10x10x1 layer consists of either the entirety of a 4x1 block or one block of it. Name the number of each of these single blocks in each layer bi (i goes from 1 to 10). Thus, if the first layer consisted of 24 full blocks and 4 blocks stood on end, b1=4
We notice there are seven ways these vertical blocks can be arranged, 1-4, 2-4...
The number of blocks in the first position is b1, the number of blocks in the second is b2-b1... and on until we see the number of blocks in the seventh position is b7-b6-b5-b4.
We can also start from the number in the seventh position equalling b10 and working back to the sixth being b9-b10... until the first is b4-b5-b6-b7.
Now equate the number of blocks in each of the seven positions from each of the two methods to give seven equations:
b1=b4-b5-b6-b7
...
b7-b6-b5-b4=b10
now solve them simultaneously until a bunch of terms drop out and I was left with:
2b3+b5+b7=0
If the b values can only take on positive integers the only possible solutions is that all 3 equal 0. If this is so then there are no blocks in the vertical position and each level must be made independent of the others which is obviously impossible. Thus there is no solution. I am sure there is an easier way to do it that is more elegant, but I cannot think of it.
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| « Last Edit: Oct 22nd, 2003, 5:37pm by aero_guy » |
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Rezyk
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Re: Bricks in a box
« Reply #9 on: Oct 22nd, 2003, 5:39pm » |
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Here's another way:
Fill the box with 63 2x2x2 cubes of cheese and 62 2x2x2 cubes of jello in an alternating pattern. Then attempt to figure out where a brick might go such that it is oriented orthogonally and does not replace equal parts cheese and jello.
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| « Last Edit: Oct 22nd, 2003, 5:44pm by Rezyk » |
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towr
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Re: Bricks in a box
« Reply #10 on: Oct 22nd, 2003, 11:31pm » |
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that's a very nice solution.. And in line with those of similar puzzles..
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Sir Col
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Re: Bricks in a box
« Reply #11 on: Oct 23rd, 2003, 12:29am » |
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Indeed, a splendid solution, Rezyk! As towr mentioned, I too was searching for a solution in line with the 2-d 'cover a chessboard' problems, but I couldn't find an equivalent form.
*makes mental note to self for similar future problems*
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Barukh
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Re: Bricks in a box
« Reply #12 on: Oct 23rd, 2003, 2:35am » |
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Joining the praise for Rezyk's solution, I also want to note that this puzzle is just a special case of the following nice theorem (proved in 1969 by the Dutch mathematician G. de Bruijn):
A box can be packed with the brick a x ab x abc iff the box has dimensions ap x abq x abcr for some natural numbers p, q, r.
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