Problem of the Month (March 2007)
The problem of finding the order of a polyomino, the fewest number of copies of the polyomino that can tile a rectangle, has been well-studied. See Mike Reid's page for example. This month we study using polyominoes to tile rectangles with identical square corners missing. What are the smallest rectangles with 1×1 corners missing that polyominoes tile? What about rectangles with 2×2 or larger corners missing?
ANSWERS
Rectangles
1-5
|
|
6
| (David Klarner)
| (T. W. Marlow)
|
|
7
| (David Klarner)
| (T. W. Marlow)
|
|
8
| (David Klarner)
|
(Mike Reid)
|
(Mike Reid)
|
|
10
|
| (William Marshall)
| (William Marshall)
|
|
11
|
|
|
|
|
| (William Marshall)
| (Mike Reid)
|
|
Rectangles Without 1×1 Corners
1-4
|
|
5
|
|
6
|
|
7
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
|
|
8
|
|
|
| (George Sicherman)
| (George Sicherman)
|
|
9
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
|
|
10
|
|
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
11 (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (Mike Reid)
|
|
Rectangles Without 2×2 Corners
1-4
| (Patrick Hamlyn)
|
|
5
|
|
6
| (Patrick Hamlyn)
| (Mike Reid)
| (Mike Reid)
| (Mike Reid)
|
|
7
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
|
|
8
|
|
|
9
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
| (Mike Reid)
|
|
10
| (Mike Reid)
| (Mike Reid)
|
|
11 (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
|
|
Rectangles Without 3×3 Corners
1-3
|
|
4
|
|
5
|
|
6
|
| (Patrick Hamlyn)
| (Mike Reid)
|
|
7
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
8
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
10
| (Joe DeVincentis)
|
|
11 (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (Mike Reid)
|
|
Rectangles Without 4×4 Corners
1-3
|
|
4
|
|
5 | (Patrick Hamlyn)
|
|
7
|
|
|
8
|
|
| (Patrick Hamlyn)
| (Mike Reid)
| (Mike Reid)
|
|
9
|
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
| (Mike Reid)
|
|
11 (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
| (George Sicherman)
|
|
Rectangles Without 5×5 Corners
1-3
|
|
4
| (Mike Reid)
|
|
5 | (Patrick Hamlyn)
|
|
6
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Joe DeVincentis)
| (Mike Reid)
|
|
7
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
| (Mike Reid)
|
|
8
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
9
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Mike Reid)
|
|
10
|
|
Rectangles Without 6×6 Corners
3 (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
4 (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
|
|
5 (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
6 (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
|
7 (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
|
|
8 (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Mike Reid)
|
|
10 (Mike Reid)
|
|
11 (Mike Reid)
|
|
Patrick Hamlyn also considered the variant where non-symmetric polyominoes could not be flipped over. Even-sized corners were not very exciting, but here are the best results for odd-sized corners:
Rectangles Without 1×1 Corners
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
Rectangles Without 3×3 Corners
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
Rectangles Without 5×5 Corners
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
(Patrick Hamlyn)
| (Patrick Hamlyn)
| (Patrick Hamlyn)
|
Ed Pegg thought these tilings were worthy of mention, even if they don't exactly meet this month's criteria:
George Sicherman found tilings of rectangles missing only one 1×1 corner...
...two opposite 1×1 corners...
...two adjacent 1×1 corners...
...and three 1×1 corners.
Brian Astle thought that the smallest square missing two opposite 1×1 corners that could be tiled by more than 2 polyominoes other than monominoes was an 18×18 square tiled with 46 copies of a heptomino. Mike Reid found that a 10×10 square could be tiled with 14 copies of a different heptomino.
Brian Astle also conjectured that no square missing opposite corners could be tiled with an odd number of polyominoes, but Mike Reid found a counterexample using 41 heptominoes to tile a 17×17 square.
Claudio Baiocchi noted that if two copies of a polyomino could tile a 4×5 rectangle, then there are trivial tilings of rectangles missing 4×4 or 5×5 corners:
Gavin Theobold noted the same trick works for other size rectangles too.
If you can extend any of these results, please
e-mail me.
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