Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  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!

    That's the correct answer.
  5. 05 Jul '06 11:30
    Originally posted by SPMars
    Excellent!

    That's the correct answer.
    Then I can trust my intuition after all.
  6. Subscriber Suzianne
    Misfit Queen
    05 Jul '06 14:46
    Originally posted by howardbradley
    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 19:46 / 2 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 21:32 / 2 edits
    Needless to say howardbradley, your argument was logically sound.

    Suzianne -- not sure where you were coming from with your post...sorry!

    The problem is very similar to "not being able to tile a chessboard with 2x1 rectangles in such a way that one pair of opposite corners is omitted". This is because opposite corners always have the same colour, whereas a 2x1 tile necessarily covers 1 white square and 1 black square.
  9. Standard member XanthosNZ
    Cancerous Bus Crash
    05 Jul '06 23:39
    A very similar set of logic comes up when discussing knight tours on various sized boards as any knight move must finish on the opposite colour square to the one it started on.