Re:The Mutilated Chessboard Revisited
by jstanley on Saturday September 29, 2018 @03:28PM
(#18102289)
I solved this by trial and error and found that it is a (3,3) point, but I haven't worked out a reason why.
EDIT: 6 is the largest multiple of 3 that is <= 8. So you can get 2 length-wise triminos on a side. The remaining 2 spots are filled with rotated triminos. With this scheme you can cover the entirety of the outer 2 perimeters, leaving a 4x4 grid in the centre still to cover. All 4 rows of the 4x4 grid have a horizontal trimino on the first 3 columns. The 4th column can fit 1 trimino vertically, leaving an empty space in a corner of the 4x4 grid, which is a (3,3) point of the full chessboard :)
Although I still don't have a proof that this is the only way to put 21 triminos on the board.
Note that each tromino covers one yellow square, one red square and one green square. Also, there are 21 yellow squares, 21 green squares and 22 red squares, so the square which will be left must be red.
and again there is (obviously) one more red square than yellow squares and green squares. The square which will be left empty, should be red in both of the colorings (the original coloring and its reflection). The only squares that satisfy that, are (3, 3), (6, 3), (3, 6) and (6, 6).
Let a tromino be a 3x1 piece. Since a chess-board has 64 squares, 1 necessarily remains when covering the board with (21) trominoes.
Which one is it (unique up to symmetry) and why?
EDIT: 6 is the largest multiple of 3 that is <= 8. So you can get 2 length-wise triminos on a side. The remaining 2 spots are filled with rotated triminos. With this scheme you can cover the entirety of the outer 2 perimeters, leaving a 4x4 grid in the centre still to cover. All 4 rows of the 4x4 grid have a horizontal trimino on the first 3 columns. The 4th column can fit 1 trimino vertically, leaving an empty space in a corner of the 4x4 grid, which is a (3,3) point of the full chessboard :)
Although I still don't have a proof that this is the only way to put 21 triminos on the board.
https://imgur.com/a/kuaeQJW
Note that each tromino covers one yellow square, one red square and one green square. Also, there are 21 yellow squares, 21 green squares and 22 red squares, so the square which will be left must be red.
Now flip the coloring of the board horizontally:
https://imgur.com/a/U7bnATI
and again there is (obviously) one more red square than yellow squares and green squares. The square which will be left empty, should be red in both of the colorings (the original coloring and its reflection). The only squares that satisfy that, are (3, 3), (6, 3), (3, 6) and (6, 6).