Tag: polyrects

Revenge of Flexible Polyominoes

~ munizao ~ Leave a comment

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?

Sparse and Magic Squares

~ munizao ~ Leave a comment

A while back, I had the insight that the 3×3 magic square could be transformed into a sparse square, or a set of cells within a square in a grid where every row and column contains the same number of cells. The converse transformation of turning a sparse square into a magic figure is not original to me, but applying it to the sparse square from the first transformation gave a pleasing result. And then I stopped there.

Or I almost did:

Right, so if I had set the 1 where the 34 is, and the 6 where the 26 is before running the solver, (and likewise the 4 and 9) that would have been more elegant.

This is a figure I came up with before my blogging hiatus that never made it into a post. The rows, columns, main diagonals, and 3×3 blocks all sum to 115. This could be seen as a version of my doubly transformed magic square on a slightly degenerate 3×3 magic square whose entries are all 5’s.

But it was hard to see where to go from there. The next size up, the 4×4 square, was unsuitable. Its magic sum is 34, and since we’d be using a 4×4 block, we wouldn’t be able to evenly split the squares from each row and column into four lines. And 5×5 seemed big enough to be a mess to work with. So I was done.

But it turns out we can do something nice with the 4×4 after all. Suppose we number the squares starting with 0 instead of 1. We still can’t use 4×4 blocks, but since the highest value is now 15, 3×5 blocks look like a possibility. And indeed, our magic sum is now 30, of which 3 and 5 are both factors, so it works:

There was an interesting bug in the code I used to produce this. I was looking for figures that are single, hole free polyrects. One way to help filter these out from other solutions is to check for 2×2 blocks with checkered corners. If one is present, you must have either a hole or more than one polyrect. But by accident the constraint I added was more broad; it applied to any 2×2 block with two dark and two light rects. I was not able to get a single polyrect solution with this constraint, but I did find something at least as interesting. Notice that every 3×5 block is the same, after swapping dark and light, as the complimentary block with the number that you get by subtracting from 15. This was not an explicit constraint in my code, so it’s interesting that this property managed to emerge. I still haven’t found a single hole-free polyrect solution without the buggy constraint, but I’m confident they are out there. My solver code is just too inefficient to find them in a reasonable amount of time.

Once I was back on the trail of magic sparse squares, I came back to the magic partition square I discussed in a previous post. (There are four ways to partition the numbers from 1 to 8 into two subsets of four numbers that sum to 18. Each of the 8 subsets is present as a row or column in the square.) Since this figure uses smaller blocks, my solver had no trouble with it.

My work on magic sparse squares is ongoing, and I hope to have more to share with you soon. I’ve put my solver code up on GitHub. I know I’ve had a long hiatus without coming up with new material, but I expect to have a few more blog posts in the coming months, and I hope you’ll enjoy the shiny objects that I find and share.