There are 18 one-sided pentominoes. Six of them have reflection symmetry, and the other 12 are 6 sets of mirror pairs. A while back, I asked if there was a tiling with a three-coloring where the 6 with reflection symmetry share a color, and each mirror pair has one pentomino of each of the remaining two colors. Patrick Hamlyn found that there was no rectangle tiling that could be colored in this way, but there was such a tiling of the shape below:

The one-sided pentominoes have area 90, which is the area of a tilted square on a grid:

Problem **#47**: Find a tiling of a torus with this tilted square as its fundamental domain by the one-sided pentominoes with a three-coloring as described above. If possible, find a tiling with no crossroads.

Another older problem that could be adapted to a torus is the minimal 2-colored packing problem. Here’s my conjectured minimal 2-colored pentomino packing of a rectangle:

(I had forgotten that this was problem #1 on this very blog!)

Problem **#48**: Find a two-colored packing by the pentominoes of a torus with minimal area.

Obviously, you could just take the rectangular packing above and add a one unit “moat” around it to get a torus with a 14×6 rectangle as its fundamental domain, but surely we can do better.

]]>Why minimize convexity? Well, we could maximize, but the different measures converge as we approach a convexity of one, so if we want *different* puzzles, minimal convexity is a better bet. To mangle Tolstoy, all happy (i.e. convex) polyominoes are alike; every unhappy polyomino is unhappy in its own way. On to the problems:

Problem **#46**: Find the least convex connected arrangement of the tetrominoes, where convexity is defined as Area / Convex Hull Area. Here is my best attempt, with convexity = 20/55.5 ≈ 0.36036:

Problem **#47**: Find the least convex connected arrangement of the tetrominoes, where convexity is defined as Convex Hull Perimeter / Perimeter. Again, my best attempt, this time with convexity = (14 + 5√2) / 40 ≈ .5268:

You might be expecting the probabilistic definition of convexity defined in the previous post to be next, but it’s a little unsatisfactory. Checking enough segments to have a good estimate would be computationally intensive, and it’s hard to know how many segments is enough to separate any pair of polyominoes that are competing for minimal convexity. (Presumably there is some way to calculate exact values, but I don’t know it.)

But we can cheat. If we check only segments connecting the centers of squares, we have a reasonably bounded quantity of segments to check. This is at the expense of no longer having a valid convexity measure (for example, this method would find the P pentomino to be convex.) But that’s fine; what we really care about is just having a good puzzle.

Problem **#48**: Define the “convexity” of a polyomino as the proportion of segments connecting the centers of squares within the polyomino that remain entirely within the polyomino. (The border counts.) Find the least “convex” connected arrangement of the tetrominoes. My best attempt, with convexity = 51/190 ≈ .2684:

One way of looking at this figure is as a tiling of the projective hemi-rhombic triacontahedron. The projective (also known as abstract) polyhedra can be formed by identifying the opposite faces of certain polyhedra with each other. So the projective hemi-cube has three square faces, and the projective hemi-rhombic triacontahedron has 15 rhombic faces. Stitching together the opposite sides of the unshaded area in the figure is a way to form this 15 face “polyhedron”.

I came up with that one a couple of years ago, but I neglected to put up a blog post because I didn’t like the graphic enough. I suspect that it’d look really cool if the lines of the rhombic triacontahedron were properly projected onto a flat disk, but I don’t have the expertise to make that happen. I finally decided that it was worth sharing even if it doesn’t look as cool as it could.

Below is another tiling of subdivided rhombi. The significance of this figure is that four copies could be used to cover a rhombic hexecontahedron.

]]>My first impulse was to find the simplest complete set that would work as a puzzle. There are 6 different cyclic permutations of a set of four different numbers. Why not try edge-matching with cards using these?

It turns out that this is a pretty easy and not terribly interesting puzzle.

However, in exploring the space of similar puzzles, I found what I think is a particularly elegant gem. If we use cards with edge values between 0 and 8, there are 17 different sums between 0 and 16, which is the same number as the number of internal edges in a 3×4 grid of 12 cards. Coincidentally, 12 is also the number of sets of different numbers between 0 and 8 that sum to 16, so those can be the sets of numbers on our 12 cards.

The last detail is the matter of how we arrange those numbers. Making all of the cards have clockwise ascending edge values is simple enough, although it hurt my symmetry senses to have to pick a direction. And indeed, we don’t have to, because we can make the cards flippable, so that the other sides have values in counterclockwise ascending order. Luckily, the flippable cards are just what the puzzle needed to handle another issue: without them, the puzzle would be unpleasantly hard to solve.

In addition to the collect-all-the sums puzzle, simply matching the numbers makes a good puzzle with this set:

A third challenge is to make a 4×4 square with the corners removed such that every *difference* between values at the same edge between 1 and 8 occurs exactly twice. I believe I solved this at some point, but I didn’t record the solution, so I can’t show it to you right now.

As you may have noticed, the last puzzle set has higher production quality than the first two. That’s because I’ve had a prototype custom deck of cards made including several different puzzle sets. I intend to have a small run of these made to sell.

*By recently, I mean, whatever was recently last June, when I started writing this post. I don’t mean to only finish blog posts during the earlier part of the year, but it does seem to tend to work out that way.

]]>But the problem wasn’t *whether* the pentominoes are convex, but *how* convex they are. In order to answer that, we’d like a measure that gives 1 for a shape that is convex, and ranges between 0 and 1 for concave shapes, getting higher as they more closely resemble some convex shape.

There are several measures that work. (For a discussion of convexity measures, see this paper by J. Zunic and P. Rosen.) One strategy for measuring convexity is to consider a shape in relation to its *convex hull*. A shape’s convex hull is the smallest convex shape it is contained in. Naturally, a convex shape is its own convex hull. The ratio of the area of a shape to that of its convex hull makes a reasonable measure of convexity. Or instead of areas, we could instead look at perimeters. The perimeter of a shape can never be less than that of its convex hull. So the ratio of a shape’s convex hull’s perimeter to that of the shape itself can also be used as a convexity measure.

Another method of measuring convexity is the probability that the segment between two points in a shape chosen at random lies entirely within the shape. Finding exact values for this measure can involve some tricky math, which Dan Piponi worked through for the P pentomino. However, calculating an estimate by randomly picking a large number of pairs of points is also an option. And, as it happens, Rod Bogart did exactly that.

The following table shows the convexities of the pentominoes according to each of these three measures.

Pentomino, shown with convex hull | Area / Convex Hull Area | Convex Hull Perimeter / Perimeter | Probabilistic method |

.7143 | .8387 | .786 | |

1 | 1 | 1 | |

.7692 | .9302 | .822 | |

.7692 | .8875 | .778 | |

.9091 | .9414 | .946 | |

.7143 | .8726 | .788 | |

.8333 | .8333 | .708 | |

.7143 | .9024 | .748 | |

.7692 | .8536 | .772 | |

.7143 | .8047 | .840 | |

.7692 | .8875 | .854 | |

.7143 | .8726 | .720 |

*Convex hull area method data by Vincent Pantaloni. Convex hull perimeter method data mine. Probabilistic method data by Ron Bogart.*

Even though these shapes are pretty simple, the data above shows that the different convexity measures treat them differently in interesting ways. Notice that the X pentomino, for example, is tied for the least convex pentomino by the convex hull area method, but is the fourth *most* convex by the probabilistic method.

I hope I’ve shown that convexity, as applied to polyominoes, is more interesting than it might have seemed! In part two, I’ll look at how we can use these convexity measures to make some new puzzles.

]]>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.

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