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21 tromino tiles (three squares glued together in a row) can together cover 63 squares on an $8 \times 8$ grid. One of the squares thus becomes non-covered. What are the possible positions where the empty square can be?

enter image description here

If this is the relevant object to consider, and it should be arranged only horizontally, being an odd number, it cannot fill any row in the 8x8 grid, without leaving 2 squares unfilled. These two squares are left unfilled for each row, and can be filled by putting a tile on ever row on the empty remaining 2 squares. This leaves one square of the tile outside the 8x8 grid. How can the question assume that 63 squares are filled out of 64, when this combination covers all 64?

Thanks

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    $\begingroup$ Why do you think it can only be arranged horizontally? Nowhere in the question does it say it cannot be arranged vertically. $\endgroup$ Jul 24, 2023 at 14:36
  • $\begingroup$ @AdamRubinson because it says "thee squares glued in a row", and since it does not impose you to put it vertically and horizontally as you say, then why not all horizontally? $\endgroup$ Jul 24, 2023 at 14:37
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    $\begingroup$ I suspect they do not intend "row" to mean "horizontal" in this context. "In a row" means, "adjacent and in a line" (rather than some other arrangement that can be made from three squares). $\endgroup$ Jul 24, 2023 at 14:38
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    $\begingroup$ @Luthier415Hz For me tt's quite obvious, from the question, that the tile can be rotated. $\endgroup$
    – leonbloy
    Jul 24, 2023 at 14:39
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    $\begingroup$ So you assume that the tromino can go outside of the $8\times 8$ grid? How interesting $\endgroup$
    – acat3
    Jul 24, 2023 at 14:47

1 Answer 1

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First, you need to understand the rules of the game. Tiles must be placed onto the grid such that they do not overlap one another, such that they align with gridlines but may be rotated vertically or horizontally, and such that they do not extend off of the playing area.

Now... consider coloring the board as follows:

chessboard with blue/yellow/green squares repeating in BYG order horizontally and BGY vertically

Recognize that however you place a tromino piece it must cover exactly one of each color. There are 22 pictured blue spaces but only 21 each of yellow and green spaces. It must be then that after you have placed 21 trominos you will have left a blue square open.

Further, if we were to have considered a different coloring of the board like the following, then the same logic applies.

chessboard with blue/yellow/green squares repeating in BYG order both horizontally vertically

As such, the only possible spaces that could have been left open must have been blue in both images.

The only such spaces that are colored blue in both images are in chess notation: C3, C6, F3, and F6.

It remains to show that there does exist such a tiling covering 63 spaces, such as the following... tiles marked as red bars.

chessboard with tromino positions marked in red

Finally, by rotating the pictured tiling we see that indeed all four of those spaces are possible as the final square remaining.

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  • $\begingroup$ Is this based on Intuition? $\endgroup$ Jul 24, 2023 at 15:26
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    $\begingroup$ @Luthier415Hz a mix of that, boardgame and puzzle enthusiasm, and experience. See for example this problem asking if there exists a sequence of moves for a new type of chess piece called a "dolphin" such that it visits each square. The same punchline occurs there. $\endgroup$
    – JMoravitz
    Jul 24, 2023 at 15:28
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    $\begingroup$ @Luthier415Hz the more general technique and takeaway is that when approached with a problem of this type, try to find something which is "always true" regardless what actual arrangement takes place. Something we call "invariants." Here, the invariant is that all ways of laying a tronimo involves covering exactly one of each color. $\endgroup$
    – JMoravitz
    Jul 24, 2023 at 15:30
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    $\begingroup$ A problem of this nature? No. This is the recommended solution and I know of no others. You could phrase it in a more dry fashion, but the underlying argument will likely be the same. $\endgroup$
    – JMoravitz
    Jul 24, 2023 at 15:32
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    $\begingroup$ @Luthier415Hz you can "reskin" a coloring argument using modulo arithmetics if you want. It's just that for most people, visualization using coloring is much easier to do (especially for lower to medium difficulty problems like this) than using purely numbers and formulas. $\endgroup$
    – justhalf
    Jul 25, 2023 at 8:28

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