This might, at first glance, appear to just be a random tiling of a bunch of dominoes and trominoes.
But what would be the point of such a thing? In fact, it’s a complete* set of dominoes and trominoes where edges may be marked, and the marked edges are exactly the ones that can occur on the border of this tiling. That thick rectangle around the tiling is part of the pieces!
*There is, in fact, one more tromino with markings that could fit in that rectangle. But if we included it, there would be a single cell in a corner that would be unfillable, since we’ve specified that the tiling has only dominoes and trominoes. Therefore, by the rule of exclusion of things that would be awkward to keep around, it has to go.
That this works at all, even with this minor fudge, feels like a pretty bit of luck. Not only do we have a perimeter and area that are compatible, we have exactly the usual number of corners for a rectangle.
With polyiamonds, I thought my luck ran out. No combination of sizes gave me compatible perimeter, area, and corners. But when I abandoned corners entirely, and focused on pieces with only one marked side, I found that a parallelogram with opposite sides marked could be made using the 2-, 3-, and 4-iamonds. Since it wouldn’t do to have any unmarked edges lying bare to the outside world, I wrapped that parallelogram into a cylinder:
And here are the pieces individually:
Sometimes when I have a novel polyform puzzle idea, I feel like I’m tapping into a rich vein of possibilities. Here, I’m not so sure. The problem is that when you move up to sets with larger sizes of polyforms, the area and border segment length are unlikely to scale in a way that gives you tilings with completely marked borders. But I would love to be surprised!
Before I started this blog, I explored polyominoes with cells individually labeled with numbers. I called these sumominoes, as I was looking at sets of all polyominoes with a given sum. Erich Friedman discovered them independently, and called them weightominoes in his July 2009 Math Magic Problem of the Month. I prefer his term for the general concept, as there is no reason they need to be grouped by sum. Both Friedman and I looked at problems where the goal was to overlap these polyominoes in a rectangle so that every cell had the same sum of labels. While I looked at a particular complete set, Friedman looked at pilings of multiple copies of the same cell numbered polyomino. (Aside: “pilings” isn’t a standard term, but it’s a concept we need a term for. We have “packings” for deficient tilings that don’t fill a space, so “pilings” for abundant tilings that fill it with overlap. Then a “tacking” is when there is both empty space and overlap, of course.)
All polyominoes with positive integer labels that sum to 4.
In this kind of problem we looked at sums of cells in the “z” direction. But we could instead look in the x and y directions. There is a common type of figure where we do this already: magic squares!
For this type of problem, excluding 0 as a potential cell label isn’t necessary. Standard numbered dominoes include 0 (blank) as a label, so we might want to do the same for physical puzzles using pips. (Pip patterns are preferable to numerals for physical puzzle pieces since they don’t have a preferred orientation.)
In fact, in the context of standard dominoes, examples have existed for some time. Here is a domino magic rectangle using a full set of double-six dominoes. The row sums are all 24, and the column sums are 21.
Solution from “The Existence of Domino Magic Squares and Rectangles”, by Michael Springfield and Wayne Goddard, graphic mine.
Of course, the fact that it can be done doesn’t make it a good puzzle, and working with a full set of dominoes might get tedious. Since I’ve been looking for simple puzzles with small piece sets, I tried to find one in this format. There are two cell numbered dominoes and four L-trominoes with a cell sum of 2. Their total area is 16, good for a 4×4 square,. and their total label sum is 12, giving a row and column sum of 3. One nice thing about a magic square type puzzle is you get an extra challenge for free. Finding a configuration with just row and column magic sums is a fairly light challenge, but getting the main diagonals to also match the magic sum is much harder. I had a small number of these made to give away at the 2022 MOVES conference:
Show Solution
Looking upward in size, there are 12 trominoes with a cell sum of 3, good for a 6×6 square with line sums of 6. I made a prototype:
But this isn’t quite as great as a puzzle, which is why I didn’t bother to conceal the solution as a spoiler. The reason is that it’s easy to make subunits where pip sums are preserved on applying a symmetry action. For example, in the solution above the three I trominoes in the upper left can be permuted in any order without changing row or column sums. Likewise, the two 2×3 rectangles formed from two L trominoes on the bottom can be flipped over one axis. This makes it much easier to turn a semimagic (row and column only) solution into one where the diagonals also work. (Subunits like these can appear in the smaller puzzles here, but there isn’t really enough room for them to dominate a solution.)
What I really want from going one step up from a 4×4 puzzle is a 5×5 one. Well, if we exclude the pieces with 3’s, we have an area of 24, which is almost right. We can make a 5×5, but it would have an unfortunate hole:
Or a fortunate one! Since my pips were lasered out holes to begin with, the big hole should clearly just count as one hole for the purpose of line sums. Now we have 25 holes, and 5 will work as the line sum. (While I normally like rounded corners for pieces since they have a softer tactile feel, if I made more of these, I’d use sharp corners and square pip holes, to visually unify the two different hole sizes.)
Show Solution
Is there anything else we can do with cell numbering? Polyhexes seem promising since there is an extra direction for sums to happen on. And perhaps we can use the numbers for something other than sums. I’m sure there are more creative discoveries waiting to be made!
Recently, while I was considering possible designs for a puzzle for my exchange gift for the next Gathering for Gardner, I thought about doing something with multiple layers of clear plastic, where interactions of markings on the layers define the puzzle. When you’re going to lasercut a large quantity of puzzles, keeping down the cost, and therefore the cut length, is paramount. So I wanted to be able to use the simplest possible markings on the pieces.
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!
