A straight line segment looked like a pretty good candidate, and it leads to an obvious puzzle goal: make the segments on two layers perpendicular. I still needed to choose pieces for these markings, but after a little trial and error, I landed on dominoes, with a segment centered in each square. For these, given some reasonable restriction on the allowable angles of the segments, the number of different pieces possible would land somewhere in the range of what would make for a good puzzle.

I ended up using segments that were turned either 15° or 45° off from the edges of the pieces. These admit exactly 12 different pieces, which can tile two layers of a 3×4 rectangle:

What makes this set particularly nice is that you can get two more puzzle challenges by changing the goal angle for the crossing segments. In addition to making them all perpendicular, you can make them all cross at 30° or 60°. These challenges should be easier, as there are two ways for an angle to differ from another one by 30° or 60°, but only one way to be perpendicular.

I also found a related puzzle that uses 10 dihexes. There are 13 pieces possible in this scheme, but I’ve omitted the ones with a lengthwise axis of symmetry from the puzzle:

In the end, I decided not to make either of these my exchange gift. I had a couple of prototypes made of the first puzzle, and it was clear to me that it needed to be larger than I could afford to make it and give away a few hundred copies. It also works best with a frame to hold the pieces and keep them neatly aligned, which adds considerably to the time and expense per copy. But even though I won’t be able to give this away at G4G13, I hope to be able to be able to sell a few copies at my vendor table there!

There are 13 pentapennies. A tiling with fivefold rotational symmetry may be possible, but I haven’t been able to find one. (This is problem **#27**.) However, I recently found a way to tile a figure with fourfold rotational symmetry with them:

Since I’ve had trouble with five symmetries, you’d think ten would be out of the question. But I found a repeating pattern on the plane with ten symmetries that can be tiled with the pentapennies:

Notice that there are five translation symmetries. Reflecting the pattern on a vertical axis gives five more symmetries. This pattern uses the wallpaper group *cm*. (Conway orbifold symbol: *×) We could also try to find a tilable pattern with the same amount of symmetry using the wallpaper group *p2*. (Conway orbifold symbol: 2222)

Problem **#45**: Find a tiling of the pentapennies on a repeating pattern on the plane that has at least as many symmetries as the one above, but a different wallpaper group. I don’t think going above 10 symmetries is possible, but I’d love to be surprised.

As I was pondering the possibility of designing my own edgematching puzzle, I considered that pieces with multiple colors are a bit of a pain to make with a laser cutter. However, engraving lines is relatively easy. And I found a promising set of line patterns related to the Catalan numbers. The Catalan number sequence (1, 2, 5, 14, 42, 132…) is one that comes up in a surprising variety of contexts. For example, Catalan(*n*) is the number of distinct ways that *n* pairs of matched parentheses can occur in a string. For *n* = 3, we have ()()(), (()()), ((())), ()(()), and (())().

Replacing the parentheses with diagonal lines, and stretching them until they connect, we can make the following figures:

We can place these figures around the edges of a square, and use them to do edgematching— but with a new twist as to what counts as a match. Looking at the patterns that are made when two of these figures meet at an edge, we see that one thing that distinguishes them is the number of loops they make. One, two, or three loops can be made. This gives us three different puzzle objectives: place the pieces of the puzzle so that all internal edges have the same number of loops, where that number can be one, two, or three. The fact that there are three loop numbers is promising because it means that, on average, the probability of an edge in these three puzzles being a match will be one-third, the same as in the MacMahon Squares puzzle. This table shows the possible edge patterns, colored according to the number of loops:

From the perspective of puzzle design, it’s convenient that the proportion of pattern matches of each type is different; this seems likely to result in puzzles that are correspondingly different in difficulty, which is desirable. The table also reveals symmetries in how the edge figures form the different types of patterns. If you swap both the rows and the columns of the mirror pair of figures, you get a table with the same loop numbers in the same places, as we would expect. Perhaps more surprisingly, this also works with the first and second rows and columns. Thus those two edge figures form a sort of hidden symmetric pair. If you swap all instances of one in a puzzle solution with instances of the other, you will get a second solution.

There’s still one task remaining before we have a puzzle design: choosing the pieces. The MacMahon squares puzzle used every possible piece. Since we have five edge types here, instead of three, that would give us more pieces than we’d really want for a good manual puzzle. There are probably several reasonable options here, but in the interest of prototyping, I wanted to start with a small piece set, so I decided to explore a set of nine pieces with bilateral symmetry. Here’s a solution with single loops:

And here’s a double loop solution for comparison:

I’ve already had prototypes made of this puzzle, and it feels promising. I hope to have at least one more post about variations on this design, but for now I’ll stop here.

]]>George Sicherman solved problem #44. (Tile a certain torus with the 1–5-ominoes such that the 1–4-ominoes do not touch each other. Finding a solution where none of the smaller pieces even meet at corners was optional, but well appreciated!) This solution contains four “crossroads”, or points where four polyominoes meet. These are sometimes considered an aesthetic flaw in a polyomino tiling, and whether or not you agree with that, finding solutions without them tends to be good for an extra challenge.

Problem **#44a**: find a solution for problem #44 without crossroads.

Jaap Scherphuis analyzed this challenge that was included with my Agincourt puzzle: find a tiling of an 8×8 square where none of the dominoes and trominoes touch each other or the outer edge. He found that there were 32 solutions. In just one of these, none of the smaller pieces meet at a corner:

I don’t feel that I acknowledge other puzzle creators enough in this space, so I want to give a shout out to Kadon for their Mini-Iamond Ring puzzle, which contains all of the 2–5-iamonds, and includes as a challenge isolating all of the different sizes of pieces:

]]>This led from the grid of pentomino painting instructions that I posted previously. Consider a string of arrows for which the subsequences of length 4 include instructions for producing all 12 pentominoes. (This is somewhat analogous to a de Bruijn sequence.) For the case shown, the string is ←↑↖→→→→↓↓←↓←↘↗↓, although the graphic seems more illuminating than the arrow string here.

Instead of a path, we could have a tree of pentominoes:

Along with the constraints that each pentomino occurs exactly once, and no square is used more than once, I wanted to limit the number of branches per node. The root node having three branches might be considered a flaw, but this was the best I could do.

]]>Sometimes an idea languishes in one of my notebooks for a few years before I can come up with the right iteration of it. My original idea here was to use a 4×4 grid. That would give me 8 pentominoes, (perhaps 10 using diagonals) but elegance surely requires all 12 to be present.

A combination of circumstances led me back to this problem. Some friends of mine have a tradition of playing RoboRally on New Year’s Day every year. This is a board game where you use cards with arrows on them to instruct your robot to move around a grid of squares. Also, in returning to the magic 45-omino problem, I was considering grids that could be used in sparse magic squares.

It might be possible to make an interesting grid puzzle, along the lines of sudoku, using this kind of grid as a basis. Most of the grid would start empty, except for a few squares in which arrows would be given at the start. Then the solver would fill in the rest of the grid by logical deduction so that the horizontal and vertical lines contain instructions for paining all of the pentominoes as above. Since the grid would have significantly fewer squares than a sudoku, this puzzle might be quicker to solve, but that doesn’t mean that it would necessarily be less interesting.

]]>Quite a long time ago, I came up with the idea of representing the lo shu (3×3 magic square) as a set of squares in a 9×9 grid, partitioned into nine 3×3 cells. The number of squares in each cell would correspond to a number in the lo shu. The most “magic” way to arrange the cells would seem to be to have 5 squares in the set in each row, column, and main diagonal. (This can be done because the lo shu’s magic sum of 15 can be divided among three rows or columns.) Although it doesn’t affect the “magicality” of a figure, I thought it aesthetically desirable for such a figure to be connected (i.e., a single polyomino) and hole-free. There are 12 hole-free magic 45-ominoes, if my code for discovering them is correct.

A figure with the same number of squares in each row, column, and main diagonal makes an ideal canvas for a sparse magic square. But with 45 numbers to place, and 20 constraints to meet, we start to push on the edge of what’s computationally feasible. The solver I wrote (which, I admit, might not have been very good) could not find a solution. Bryce Herdt manually tweaked the output of my solver to make a *semimagic* solution, that is, one where the rows and columns add to the magic sum, but the diagonals still didn’t work.

When I discovered that the Numberjack constraint engine could easily be used to code solvers for magic figures, I tried it on this problem, but got nowhere. The solver would run for an arbitrarily long period of time without spitting out any solutions. Recently I tried it again, and this time I got solutions. Paradoxically, what made the problem easier to solve was that I added more constraints. I manually placed the numbers 1 through 9 in the 3×3 cells that they correspond to. This seems to have made the search space small enough that the solver would not be able to spend an inordinate amount of time stuck in a barren zone.

]]>It seemed to me that the problem called for a pentomino tiling of a torus, which they could use as a wallpaper-like pattern, repeated as many times as they needed. The choice of the particular torus to use is a matter of taste, but I thought it would be nice to maximize the minimum distance between two images of the same point. (I haven’t proven that I succeeded, but it’s close.) In coding the solver for this, I used a shortcut: instead of directly checking whether a given tiling had a coloring of the correct type, I checked whether each pentomino bordered exactly six others. This turns out to be a necessary condition, but not a sufficient one, so I manually checked a few such tilings until I found one that worked. This is the pattern that appears, in user-colorable form, in Visions of The Universe by Bellos and Harriss.

The hexiamonds were the other obvious set of 12 polyforms to try to tile with this coloring scheme. Here, there is one torus with maximal symmetry. Amazingly, my solver found just two tilings where every piece bordered 6 others, of which exactly one had the right coloring properties. Recall that the solution for the pentominoes on the 6×10 rectangle was also unique. It seems incredible to me that this problem type has yielded two instances that were so finely balanced as to be solvable, but only by the barest of margins.

]]>But there is in fact something highly symmetrical that these pieces can tile. And its existence follows from the fact that while 89 may not be composite, it is the sum of two squares. 89 = 25 + 64 = 5^{2} + 8^{2}.

Taking the sum of two squares may remind you of the Pythagorean Theorem, and that is exactly where I was headed. Make a right triangle where the legs have length 5 and 8, and the hypotenuse will have a length of sqrt(89). And then, naturally, if you make a square out of four sides with that length, it will have an area of 89:

So I have something that indeed has the desired area, but you might complain that having sides that slice obliquely to the square grid makes it entirely unsuitable for tiling with a set of polyominoes. But suppose we stitched the pairs of opposite sides together. That would turn the figure into a torus, which “unwraps” into a repeated, plane-filling pattern:

Which we can tile! If fact, tori are generally relatively easy to tile because they have no edges, and the edge is typically the hardest part of a pattern to tile. Having small pieces in the mix, as we do here, also tends to make tiling easier. So for a challenge, we could try something harder.

Problem **#44**:

Find a a tiling of the torus above with the 1–5-ominoes where none of the pieces of size 4 or smaller are adjacent to each other. Touching at corners is okay, but if you can find a solution without that, that’s even better. (Weird, it’s been three years since I’ve posed a numbered problem on this blog.)

This problem runs into a wall in my current setup for solving polyform tiling problems. I typically add ugly hacks to my copy of David Googer’s Polyform Puzzler. It’s reasonably handy because it’s open source and written in my language of choice, Python. But it doesn’t include a hook for pruning the search tree when you come to a configuration that doesn’t meet a desired condition. For problems with a small enough search space this doesn’t matter; you can just filter finished solutions as long as the time needed to run a complete search is reasonable. But here the high tilability is actually a curse: the solver starts in an area of the search space where the adjacency condition isn’t met, and because the pieces are so numerous and so tilable, it can stay there for an extremely long time before it decides to change out any of the tiles placed early on. (There are technical reasons why hacking in the hook I would need appears to be difficult, but I won’t get into those here.)

Coincidentally, the area of the 1–4-ominoes, 29, is also a sum of squares:

Any parallelogram can be used as the fundamental domain of a torus. Rectangle and rhombus shaped fundamental domains can have just as much symmetry as a tilted square. (Because the square is tilted, flipping it over isn’t a valid symmetry action, though rotating it still is.) But the tilted square tori still strike me as particularly pleasing and unexpected patterns for tiling.

The Lo Shu, or 3×3 magic square, was discovered in China in antiquity. It is the only way, (up to symmetry) to place the numbers 1 through 9 in a 3×3 grid such that the numbers in each row, column, and main diagonal add up to the same number (or magic sum). This fact seems to be universally known among recreational mathematicians. So when I had the chance to meet a number of them this spring at the fabulous 12th Gathering for Gardner conference, I told them that I knew a different way to do it. When they pronounced me mad, or a liar, I showed them one of these:

Mathematics is full of counterexamples that result when the simple way of understanding a conjecture is not exactly what the conjecture literally says, so this kind of cheating is totes legit.

If the fact of the uniqueness of the Lo Shu is new to you, a quick proof might be in order. First, let’s enumerate all of the sets of three numbers between 1 and 9 that sum to 15: {1, 5, 9}, {1, 6, 8}, {2, 4, 9}, {2, 5, 8}, {2, 6, 7}, {3, 4, 8}, {3, 5, 7}, {4, 5, 6}. There are eight sets, so we’ll need all of them to fill the eight lines in the magic square. The number 5 appears four times, the other odd numbers appear twice, and the even numbers appear three times. Therefore the center square, being part of four lines, must be 5, the corner squares, being part of three lines, must be the evens, and the side squares, being part of two lines, must be the other odds. Choose any corner, and put a 2 in it. That forces 8 into the opposite corner. Choose one the remaining corners, and put a 4 in it. After that, the rest of the numbers are forced. No matter what corners you choose, the result can be rotated or flipped to get the square formed by choosing any different pair of corners. Q. E. D.

Well, wait, you say, what if the magic sum isn’t 15? Quite right, 14 and 16 also both have eight sets of numbers between 1 and 9 that sum to them, so our proof is not done. I will leave it as an exercise to the reader to show that they cannot be used to form a 3×3 magic square.

And then, once the reader is satisfied, I’ll say: there *is* a way to place the numbers 1 through 9 in a 3×3 grid, *with exactly one number in each cell*, (you didn’t think I’d try the same shenanigans twice?) so that they occupy eight lines that each connect exactly three numbers that sum to 14. And having followed me this far, you are now enough of a recreational mathematician to be able to call me mad, or a liar. But you might want to have a look at this before you wager money on it:

This result was adapted from one discovered by Lee Sallows, which is described in his book, *Geometric Magic Squares*.

Well, clearly the problem here is that you’re allowing me to draw my own graphics. If you forced me to use physical number tiles as in the first image, I couldn’t get up to any fancy tricks. So if I told you that I could arrange those exact same nine number tiles in a block of three rows of three tiles each, and make it so that for every line that passes through the center of three tiles that form a connected group, the sum of those tiles is 14, I would *have* to be mad.

Mad as a loon.

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