# packing

SPMars
Posers and Puzzles 04 Jul '06 10:34
1. 04 Jul '06 10:34
Is it possible to fill a 6x6x6 cube with 1x1x4 cuboids? (No overlaps or gaps, arranged vertically and horizontally.)
2. 04 Jul '06 13:11
I don't think it is possible (by intuition).

If you put a stack of 1x1x4s in the bottom, then you get 2 volumes, 2x6x6 and 2x4x6. If you fill 2x6x6 with 1x1x4 then you cant avoid a 2x2x6 hole. And this hole you can't fill with any 1x1x4.

So I don't think so, no.
3. 04 Jul '06 16:47
Originally posted by SPMars
Is it possible to fill a 6x6x6 cube with 1x1x4 cuboids? (No overlaps or gaps, arranged vertically and horizontally.)
No.

Imagine the 6x6x6 cube as composed of 27 2x2x2 cubes coloured in chess board fashion - ie no cubes of the same colour adjacent. There would be 5+4+5 (=14) of one colour and 4+5+4 (=13) of the other.

However, no matter how you place the 1x1x4 pieces they have to take up and equal amount of each colour. Clearly a contradiction.
4. 05 Jul '06 09:45
Excellent!

5. 05 Jul '06 11:30
Originally posted by SPMars
Excellent!

Then I can trust my intuition after all.
6. Suzianne
Misfit Queen
05 Jul '06 14:46
No.

Imagine the 6x6x6 cube as composed of 27 2x2x2 cubes coloured in chess board fashion - ie no cubes of the same colour adjacent. There would be 5+4+5 (=14) of one colour and 4+5+4 (=13) of the other.

However, no matter how you place the 1x1x4 pieces they have to take up and equal amount of each colour. Clearly a contradiction.
Wait... no. Why imagine the cube as composed of 27 2x2x2 pieces? Especially when the given piece is 1x1x4? A 2x2x2 piece is 8 little (1x1x1) cubes, while the given 1x1x4 piece is 4 little cubes. By halving the number of pieces, you have a 50-50 chance of creating an odd number of pieces.

Clearly, you can't fit 54 1x1x4 pieces into a cube 6x6x6.

But the logic you describe has nothing to do with it.

Using similar logic, I just as easily could have said: Imagine the 6x6x6 cube as composed of 8 3x3x3 cubes colored in chess board fashion - ie no cubes of the same color adjacent. There would be 2+2 (=4) of one color and 2+2 (=4) of the other. No matter how you place the 1x1x4 pieces they have to take up an equal amount of each color. 4=4, so therefore, it is possible.

Clearly, this isn't the case.
7. 05 Jul '06 19:462 edits
Originally posted by Suzianne
Wait... no. Why imagine the cube as composed of 27 2x2x2 pieces? Especially when the given piece is 1x1x4? A 2x2x2 piece is 8 little (1x1x1) cubes, while the given 1x1x4 piece is 4 little cubes. By halving the number of pieces, you have a 50-50 chance of creating an odd number of pieces.

Clearly, you can't fit 54 1x1x4 pieces into a cube 6x6x6.

Bu ...[text shortened]... ual amount of each color. 4=4, so therefore, it is possible.

Clearly, this isn't the case.
Hmmmmmm.

Perhaps in the land of "Know-it-all" statements like this: "Clearly, you can't fit 54 1x1x4 pieces into a cube 6x6x6." constitute proof. Not here though.

I didn't have a 50-50 chance of creating an odd number of pieces. I had a 100% chance; because I chose that subdivision precisely because it did create an odd number of pieces. And thereby more of one colour than another.

It's a technique called Reductio ad absurdum.

Those of you who don't already know everything may be interested in this article: http://en.wikipedia.org/wiki/Reductio_ad_absurdum. It has a very apposite quote from G H Hardy at the end.

PS. This statement "No matter how you place the 1x1x4 pieces they have to take up an equal amount of each color." is wrong for your 8 3x3x3 cubes. Can you see why?
That's right: you could put a 1x1x4 so that 3 of its cubes were one colour and 1 cube the other. 3 does not equal 1 (even for large values of 1.)
8. 05 Jul '06 21:322 edits