1. Joined
    24 Sep '06
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    3736
    04 Nov '08 04:491 edit
    this problem is too easy.

    on a chess board, 16 of these "four square objects" will fit:

    ~o
    ooo


    what is the least amount possible which may be situated with their 3-square's aligned in the same direction.

    (ignore ~)
  2. Joined
    24 Nov '05
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    1113
    11 Nov '08 01:251 edit
    I get 12
  3. Standard memberforkedknight
    Defend the Universe
    127.0.0.1
    Joined
    18 Dec '03
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    16018
    11 Nov '08 01:57
    I get 8, assuming that you mean that no others can be placed on the board with the same orientation.

    With the wide 3 squares at the bottom, the leftmost square for each piece are location in positions:
    a2, a6,
    c1, c5,
    e2, e6,
    g1, g5
  4. Joined
    31 May '07
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    696
    11 Nov '08 16:251 edit
    The answer's thirteen I think.
  5. Joined
    24 Nov '05
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    1113
    11 Nov '08 19:25
    You're right, I just got 13 as well. I'll upload a pic if someone wants it.
  6. Joined
    31 May '07
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    696
    11 Nov '08 21:251 edit
    If anyone wants to know the answer, this hopefully shouldn't spoil it too much. Here's a possibility for 13 with all the pieces represented by white pawns:



    edit: and a proof showing that there isn't going to be a higher answer than 14 (even though I currently still think the maximum is 13)
    The two edges have to have 4 blank squares. This is very easily seen from the fact that the edge can only be occupied by a piece extremity, and then the bit that juts out from it prevents the square above/below being used. So if 4 edge squares are used on one edge, another 4 must be blank. This leaves us with 56 squares remaining. 56/4=14
    Though I doubt any tile combination would slot together this perfectly.
  7. Joined
    24 Sep '06
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    3736
    12 Nov '08 11:13
    i obviously did not explain well enough.

    16*4=64

    all squares will be occupied.

    perhaps i did not do well explaining the shape.

    it is like the shape in "tetris" which has one square adjacent to the center of 3 aligned squares.
  8. Joined
    15 Feb '07
    Moves
    667
    12 Nov '08 14:29
    On a cylindrical board (where a-file is adjacent to the h-file), you could fit all 16 with the same 3-square alignment.

    I'm not sure about the normal non-looping chessboard, however.
  9. Joined
    24 Nov '05
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    1113
    12 Nov '08 15:18
    Ok, 12.
  10. Joined
    31 May '07
    Moves
    696
    14 Nov '08 07:41
    I think I understand the quesiton now.

    This figure:

    Darn attempts at explaining it 😛
    Pieces are either all black or all white, I tried to make it easy enough to see.
    It is the only way to fill a 4x8
    Which leads me to suspect that you can only have two of them for an 8x8, leaving us with 6 pieces facing the same way (up), and 12 pieces with the 3 bar in the same direction (up and down)
  11. Joined
    24 Nov '05
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    1113
    15 Nov '08 02:57
    Yeah, that's how I got it
  12. Joined
    02 Mar '06
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    17881
    16 Nov '08 07:50
    but one can fill a 4x4 section with 4 "tetris pieces" where 2 are placed vertically and 2 horizontally. for example look at the top left quadrant of a chessboard, and imagine piece 1 (a8-c8) piece 2 (d8-d6) piece 3 (d5-b5) piece 4 (a5-a7) with their "nubbins" pointing toward the middle of that 4x4 block.

    then placing pieces in a similar fashion in the other 3 quadrants, wouldn't this better answer the question? 8 pieces would have horizontal orientation and 8 would have vertical orientation, so it minimizes the "number of pieces with their long side facing in the same direction."

    unless i misunderstood the question lol which is entirely possible
  13. Joined
    31 May '07
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    696
    16 Nov '08 08:24
    Yesm you seem to be right. Don't worry. I've just misunderstood the question twice 😀 well done.
  14. Joined
    24 Nov '05
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    1113
    16 Nov '08 18:59
    Rereading the question, I see the word "least". Why, then, was I looking for the most?
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