Revenge of Flexible Polyominoes

Several months ago, I felt myself at a bit of a creative ebb. I wasn’t coming up with any bold new polyform ideas, so the best I could do would be to tinker around in a space that was already well-trodden. In this state of mind, I asked myself: did Abaroth miss anything good?

When I came up with the idea of letting the cells in polyominoes be flexible rhombi, Abaroth ran with it, and made an entire gallery of tilable shapes with solutions. Some of Abaroth’s discoveries used rows of squares as seams between the “leaves” of a target shape, but he missed this nice pentomino star:

An obvious thing to want after seeing a tiling like this with five-fold dihedral symmetry is one with sixfold dihedral symmetry. So far, the attempts have run into some problems.

The tiling on the left is Abaroth’s. It contains a couple of ambiguous pentominoes in the upper right. Where the green one wraps around a degree-3 vertex, it could be “unglued” to form either an X or an F pentomino. The red one could be an L, an N, or a Y.

The tiling on the right is mine, and has a different problem. The P pentomino in the upper left is not ambiguous, but it is split. This type of flaw can only occur in a polyomino that contains a square tetromino; P is the only pentomino that does.

Problem #53: Find a flexible pentomino tiling with sixfold dihedral symmetry without ambiguous or split pentominoes.

One-sided flexible polyominoes were another area that had been missed. It turns out that there are some nice tilings here:

George Sicherman, Abaroth, and Edo Timmermans all found one-sided pentomino tilings for the above double star. This double balanced three-coloring found by Edo Timmermans is particularly nice. Remarkably, the one-sided hexominoes also admit a double star:

(Solution again by Edo Timmermans.)

It might not be clear at first that other symmetry variations on polyominoes will survive in this weird flexible world, but in fact some can. If squares can flex into rhombi, then rectangles can flex into parallelograms, and we can get tilings like the following, using the 3-, 4-, and 5-rects:

For the second post in a row, I’m going to stop with at least another post’s worth of material left to share. If I leave you in suspense, you’ll have to keep coming back, right?

Binary System, Decimal Star

If you participated in the gift exchange at the 10th Gathering for Gardner, which was held recently in Atlanta, you would have received one of these in your bag of exchange gifts:


It is common, (but by no means required) for participants to use the number of the conference as a theme in their exchange gift in some way. I considered a ten pointed star with pieces that slot together as a promising shape for a puzzle, and I recalled that there were ten distinct reversible binary sequences of length four. (In this scheme 0011 and 1100, for example, are considered equivalent because they reverse to each other.) This meant that with four slots at intersection points, if there were two possible positions for each slot, (like up and down) there would be exactly ten possible pieces, which would make an elegant puzzle set if I used one of each. Conveniently, the pieces could be flipped horizontally to physically realize the reversal of the string. Inconveniently, the pieces could also be flipped vertically, which would invert the 1’s and 0’s, and lower the number of distinct piece shapes to six. Another problem is that some configurations of pieces could not be physically assembled. If there was a triangle of pieces where each had an up slot followed by a down going around the triangle, there would be no way to fit the third piece in, because it would simultaneously need to be slotted in from above and below.

I solved both of these problems at once by changing the inner slots to all face the same direction, and to have shallow vs. deep as their two possible states instead of up and down. Now the ten pieces can be divided into two pentagonal configurations that are connected by their outer slots, and connect to each other by the inner slots. Because every triangle in the star contains the two inner slots of a piece, the triangles are all assemblable. The pentagons must also be assemblable, because there are only four pieces with up and down outer slots, so one side of the pentagon must have two slots pointing the same direction, and that side may be placed last. And because the direction of the inner slots is forced, only horizontal flipping is allowable. Here’s a photo of an assembled puzzle, along with an unassembled set of pieces:

Mathematical niftiness aside, is this a good puzzle? I think so. It has a fair number of solutions, but neither so many that you can easily stumble upon one without applying any strategy to solving the puzzle, nor so few that you have to spend a lot of time engaged in trial and error. Let me know if you have one of these and need hints for solving it.

In an upcoming post, I’ll discuss some variations on this type of puzzle.