2024/04/30

Tetrominoes are four squares connected at one or more sides. There are only five two-sided and seven one-sided tetrominoes and in these sets there is one piece, which unbalanced under checkerboard coloring. Therefore the number of nice possible constructions with these pieces is very limited. Jared McComb suggested that we can take four sets of these pieces and color them with four different colors. This way we get 20 or 28 pieces, respectively, and the number of unbalanced pieces is even. Now various constructions are possible. Furthermore we can add two conditions for a solution:

- Same colored pieces must not touch at one side
- Pieces of same shape must not touch at one side

A jagged square with a center hole made from three sets of two-sided pentominoes can also be made.

After this I explored three or four sets of tetrominoes, pentominoes or hexominoes. You can click the number of pieces in the table to see some constructions.

Basic Sets | Number of squares | Property | One Set | Three Sets | Four Sets | |||
---|---|---|---|---|---|---|---|---|

Number of Pieces | Total Area | Number of Pieces | Total Area | Number of Pieces | Total Area | |||

Tetrominoes | 4 | Two-sided | 5 | 20 | 15 | 60 | 20 | 80 |

One-sided | 7 | 28 | 21 | 84 | 28 | 112 | ||

Pentominoes | 5 | Two-sided | 12 | 60 | 36 | 180 | 48 | 240 |

One-sided | 18 | 90 | 54 | 270 | 72 | 360 | ||

Hexominoes | 6 | Two-sided | 35 | 210 | 105 | 630 | 140 | 840 |

One-sided | 60 | 360 | 180 | 1080 | 240 | 1440 |

Here is the set of 35 two-sided hexominoes. In this set there are 11 pieces, which are unbalanced under checkerboard coloring and they are blue marked in the picture.

Since the total number of squares is even and the number of unbalanced pieces is odd no rectangles can be constructed with this set. The same applies to three sets of pieces, because the number of unbalanced pieces 3*11=33 is also odd. But we have a total number of 33*6*3=630 squares. This is just enough to fill a figure like a rectangular triangle with two legs of size 35 because 35*(35+1)/2=630 and the difference between white and black squares under checkerboard coloring is even.

If the short sides of the rectangles are of length 12 or larger, I split the figures parallel to the long sides. Since the backtracking algorithm works column by column it is much faster for columns of low height, even if more border conditions are left for the second part. This way I got solutions for the rectangles 60x14, 56x15, and 42x20.

For the missing 35x24 and 30x28 rectangles the solutions are made from three parts.

We can also try to construct some square rings with square symmetry: 41^{2}-29^{2}, 31^{2}-11^{2} and 29^{2}-1^{2}

At last I found a kind of jagged square.

- One-sided pieces can't be turned upside down. So we have fewer options to place a piece.
- With only three colors you have less possibilities to prevent touching pieces of same color.
- It's harder to decide if the first part of a construction has a good chance to lead to a complete solution, even if only few pieces of good tileability are used in the first part.

If the length of the shorter sides are above 10, you can see the horizontal dividing line. This way searching was much faster. Here are 90x12, 72x15, 60x18, 54x20, 45x24 and 36x30.

Square rings are also possible: 37^{2}-17^{2}, 51^{2}-39^{2} and 33^{2}-3^{2}.

If the shorter sides are longer than 10 we have: 120x12, 96x15, 80x18, 72x20, 60x24, 48x30 and 40x36.

In addition I made two rings with square symmetry. The horizontal parts were filled column by column and the vertical parts row by row. 77^{2}-67^{2} and 38^{2}-2^{2}