I've written about polyomino covers and colorings of polyomino tilings elsewhere; I intend to use this page for miscellaneous polyomino related topics that don't merit a seperate page.
Here's a magic 45-omino:
What's magic about it?
The number of squares within the polyomino, (dark squares,) on each row, column, and main diagonal is 5, and the number within the polyomino in each 3×3 cell corresponds to a number in a 3×3 magic square, where all rows, columns, and diagonals sum to 15:
This was a fairly challenging puzzle to solve by hand, although now that I've gotten the hang of it, I've been able to find a large number of solutions. The '8' cell in the lower right is a little tricky to keep connected to the rest of the polyomino.
Sean Barrett said that while he appreciates my magic 45-omino puzzle, the only person who could actually solve it is me. That's probably a bit of an exaggeration, but it got me thinking. One way to make the puzzle easier would be to give the solver 9 3x3 square tiles with the pattern of squares in the tiles preprinted on them. But this would seem to make the puzzle too easy. However, if we have physical tiles, we can put different patterns with the same numbers of squares on the backs of the tiles. (Except the '9' which has only one possible pattern.) Ideally we would want the patterns on the tiles to not have any rotational symmetry, for increased combinatorial goodness. I've since found a pair of solutions that, (excepting the 9,) do not share any tile patterns and do not contain any rotationally symmetrical tile patterns.
I've made a prototype of this puzzle on paper, with patterns from these two solutions on either side of the tiles. Notice that the reflection of a magic square is not equivalent to the original for this version of the puzzle, since you can't reflect the patterns on the tiles. By reflecting the patterns used for the second solution relative to the first, I guarantee that whichever magic square arrangement of the tiles you choose, there will be exactly one solution using the tiles in that arrangement. I wrote the instructions on the back of the 9, which seemed like an elegant way to make use of space that would otherwise be superfluous, and to make the puzzle completely self-contained.
It might be argued that I've defeated the purpose of making a puzzle that was easier that the one I started with.
I've made a Sudoku variant using a Magic 45-omino. It's on my Grid Puzzle page.
We can label the cells of a polyomino with positive integers so that the sum of all labels is n. Call the different ways of doing this the n-sumominoes. Here are the 12 4-sumominoes:
I've used colors to denote the labels; from lightest to darkest, the shades used represent 1 through 4.
Can one place these twelve pieces in a 3 by 4 rectangle, with overlapping allowed, so that the sum of the values of all cells on a square is four? Bryce Herdt found the following solution:
The July 2009 Math Magic Problem of the Month contains an exploration of similar problems involving polyominoes with different weights associated with their squares.
These appear to be all of the unshrinkable 8-sided polyominoes. (Depending on how you want to define the set, the one with the hole could be excluded.) I call a polyomino shrinkable if there is a grid line intersecting the polyomino such that wherever the cell on one side of the line is inside the polyomino, the adjacent cell on the other side of the line is also inside. One could shrink the polyomino by replacing the pair of rows of cells on either side of the line with a single row that has the same pattern of cells within the polyomino.
Shrinking an 8-sided polyomino
The number of unshrinkable (n*2)-sided polyominoes appears to increase very rapidly with n. Consider the subset of these polyominoes which are permutations of stacks of squares of height 1 to n - 1. The number of polyominoes in just this subset is (n - 1)!/2.
You could certainly look at unshrinkable subsets of other sets of polyominoes. My motivation for looking at unshrinkable (n*2)-sided polyominoes was that it produced a finite subset of an infinite, (and thus rather unwieldy,) set. Also, since shrinking polyominoes preserves the number of sides, the unshrinkable (n*2)-sided polyominoes could be looked at as canonical members of equivalence classes, where all polyominoes that eventually shrink to the same unshrinkable polyomino are equivalent.
It's possible to arrange a set of pentominoes so that 11 of them border and surround the twelfth. First, notice that not every pentomino can serve as the central piece. The central pentomino must have at least 11 neighboring cells. The I pentomino has 12 squares bordering it, the L and V pentominoes have 11, and all of the other pentominoes have fewer. A pentomino that occupies the cell on the inside corner of the V must also occupy one of the neighbor cells next to it, so the V pentomino doesn't work.
It's a little harder to find a convex solution. When we talk about convexity in polyominoes, our definition of convexity is a little different from the standard definition. Normally, if a shape is convex, we should be able to pick any two points within it and the entire segment connecting them will lie within the shape. If we can pick two points for which this isn't true, it's concave. For polyomino "convexity", we add the restriction that the two points we choose have to be on the same vertical or horizontal line. If the line between them is diagonal, it can't affect convexity.
Here's my solution:
For another challenge, you can try to find a symmetrical, (not necessarily convex,) solution. I haven't found such a solution, and I don't know if one exists.
What if we we turned this defect into a virtue? There are 9 tetrominoes with a notch cut into a grid point on one of their sides. Unfortunately, a parity problem prevents them from fitting into a 6×6 square, but since 7 of them are asymmetrical, the one-sided notch-tetrominoes could fit in an 8×8 square.
Is it possible to fit the 16 one-sided notch-tetrominoes in an 8×8 square so that each of the notches is at the point of a T interection, and each T intersection is at a notch point? Gunther Stertenbrink found that it is not. He found two solutions where all of the notches are at a T intersection, however, in both solutions there were extra T intersections not at notch points within the tiling. Are there other notchomino problems with elegant solutions? I was working on the 4×11 rectangle with notch 2-, 3-, and 4-ominoes, and found near solutions with one too many T intersections.
The square tetromino can be used to make two overlapping tilings of the plane such that all segments in the grid are covered by exactly one of the two tilings:
Finding the two hexominoes that have the same property was a nice puzzle:
It's pretty easy to prove that these are the only hexominoes that will work.
Some directions for further exploration:
If a pentomino chain isn't cyclic, it can tile a 7x7 square. Is it possible to make such a tiling where each square of overlap is a knight's move away from the last? Two pentominoes (the I and the X) do not contain a pair of squares a knight's move apart, therefore they must go on the ends of the chain.
If you have questions, comments, solutions to unsolved problems, or ideas for new, related problems, please email me.