Posts Tagged ‘tetrominoes’

Some Contributed Solutions

October 2nd, 2013

I’ve had a few solutions sent in recently, so I wanted to share them with you all.

First, Abaroth noticed that my rhombic-cell pentomino tiling had just enough space to fill out into a five pointed star if the tetrominoes were also included:

tetra-penta-star

But that was just the beginning! He then proceeded to produce an entire collection of tilings with these pieces, which he calls flexominoes. One problem that can come up in tilings of this sort is that if there is a vertex with three rhombi around it, a polyomino containing all three rhombi has an ambiguous identity, since there is more than one way to “unglue” the polyomino at that point. I contributed an ambiguity-free solution to one of the patterns Abaroth found:

flexomino-8-star

Speaking of rhombuses, Abaroth has been investigating color-matching puzzles using rhombic tiles. His puzzle page has more interesting material on color matching puzzles and symmetrical polyhex tilings.

Next up, George Sicherman sent in a symmetrical tiling for the flexible tetrarhombs:

tetrarhomb-gs-sol

What’s interesting here is that although the outline of the tiling is symmetrical, the pattern of the cells isn’t. The lesson here is that being able to trade off some cell-level symmetry for more pattern-outline symmetry can give us a little variety in our choices of what we can tile.

Finally, Bryce Herdt provided a de Bruijn sequence of invertible length 5 binary words. (That is, a cyclic sequence where each word occurs once as a substring.) Since he did so in text format, I made a visualization:

debruijn-invert

A Polyformist’s Toolkit: Practical Topology

May 23rd, 2013

In polyomino puzzles, we would frequently like to tile the simplest shape possible, and a rectangle usually seems to fit the bill. But sometimes a rectangle isn’t possible. For example, we can never make a 4×5 rectangle with the five tetrominoes. One way to prove this is with a checkerboard parity argument. Four of the 5 tetrominoes must always occupy even numbers of both black and white squares if they are placed on a checkerboard. The T tetromino must occupy odd numbers of each color. Therefore a rectangle must have odd numbers of each color, but any rectangle of size 20 will have colors evenly divided, 10 and 10. A similar argument can be made to show that the 35 hexominoes cannot tile a rectangle.

The tetrominoes, and a 5×4 rectangle.

This will never work…

Rather than give up and accept that we’ll need to find a less elegant shape to tile, we have another option. If we wrap the edges of a 5×4 rectangle around to form a cylinder, (so that the cylinder is 4 squares tall and 5 squares in circumference) tiling is once again possible. To see why this might be so, imagine that you are coloring the squares as in a checkerboard. Once you got back around to where you began, you would find that in order to continue the pattern, you would need to use the opposite colors from those you already used. Note that this would not work if you wrapped the rectangle in the other direction; because the other side has even length, the checkering colors remain consistent.

The tetrominoes tiling a 5×4 cylinder a cylinder

…until we wrap the rectangle into a cylinder.

There is a video by Edo Timmermans showing how a tetromino cylinder can be made with toy magnets. He claims that there are seven distinct tilings of a cylinder with the tetrominoes, and poses an interesting puzzle involving them. A commercially produced cylindrical polyomino puzzle is Logiq Tower, designed by Marko Pavlović, which uses wooden pentomino-based pieces that form a cylinder together with some other pieces. Because these pieces are inflexible, they lack some of the allowable symmetry actions of free pentominoes.

A cylinder isn’t our only option. We could give the rectangle a half-twist before connecting the ends; this gives us a Möbius strip. We could also connect both pairs of sides instead of one; this gives a structure that is topologically equivalent to a torus or doughnut. And then we could add twists to that— well, at this point it would be nice to be systematic so we can be sure that we’ve found all of the possibilities. One thing to note is that adding more twists doesn’t actually give us more possibilities. A strip with two twists will have exactly the same tilings as a strip with no twists, and in general, a strip with an even number of half-twists will have the same tilings as the no-twist strip, and a strip with an odd number of half-twists will have the same tilings as the Möbius strip. So for each dimension, we have three options: no connection, connection without a twist, and connection with a half-twist. This gives us the following matrix of possibilities:
Topologies for polyomino tilings
Only six possibilities here, not nine, because the ones in the lower left are equivalent to the ones across the main diagonal from them. Note that the Möbius strip, Klein bottle, and projective plane are nonorientable surfaces, which means that they effectively have only one side.

An important consideration when working with these is that one-sided polyominoes don’t exist on nonorientable surfaces. With one-sided polyominoes, translation is allowed, but reflection isn’t. However, on a non-orientable surface, translating far enough leaves an object in a reflected state.

Another consideration is that coloring is harder when we leave the plane behind. On the plane, we have a theorem stating that we never need to use more than four colors to make all of the tiles differ in color from all of their neighbors. On a torus, this may require seven colors. In 2001, Roger Phillips found 18 heptominoes that could tile a 7×7 torus, and sent these tilings to MathPuzzle.com. Here’s one:

7-colored 7-omino torus

Depending on the dimensions of the torus, it may be possible for a polyomino to wrap around and touch itself. In a strict sense, this makes any coloring impossible, since we don’t let tiles of the same color touch. However, we can follow a looser standard, and allow self-touching polyominoes in our colored tilings. Patrick Hamlyn found a 3-coloring of a tiling of the 35 hexominoes in 7 3×10 tori using this scheme in 2003:

The 35 hexominoes in 7 3×10 tori, 3-colored

This problem has no solutions if the tori are replaced with rectangles or cylinders.

Problems #31-37:
Though it seems like a pretty basic problem, if anyone has counted the number of pentomino tilings of cylinders, I am not aware of it. Wrapping the short sides of the 3×20 together should not give any solutions beyond the two obtained by wrapping the solutions on the 3×20 rectangle. That leaves the 3×20 wrapped the other way, and both ways of wrapping the 4×15, the 5×12, and the 6×10 rectangles.

Problems #38-40: Find the solution counts for the 4×15, 5×12, and 6×10 tori. I don’t know if these are all computationally tractable, but I can hope. (The 3×20 will be the same as the 3×20 cylinder with long sides wrapped together.)

Even more possibilities for tiling become available when you choose parallelograms with diagonal sides to wrap around, but this post is long enough, so that will have to be a matter for another post.

Why L-topia Is Awesome

December 7th, 2010

It’s the holiday shopping season, so I figured it couldn’t hurt to write a post or two on the puzzles I am selling.

Every mathematical puzzle designer worth his or her salt has an argument for their puzzle’s awesomeness using impressive sounding mathematical justifications. This, for L-topia, is mine.

There are 12 pieces in the set. Empirically, 12 is a good number of pieces for a mathematical puzzle. There are 12 pentominoes, and 12 hexiamonds.

The shape of the pieces, an l-tetromino, has some desirable properties. It is very highly tileable. Two factors that affect the tilability of a polyomino are its size and its symmetries. Smaller and less symmetrical polyominoes are the most tilable. The l-tetromino is the smallest asymmetrical polyomino, and the only asymmetrical tetromino, so it should be the most tilable of all.

A set of 12 l-tetrominoes tiles a 6×8 rectangle in 1114 ways. That’s probably the most for any set of 12 copies of a single polyomino tiling any rectangle, but it’s not that impressive compared to other sets containing multiple shapes. For example, the 12 pentominoes can tile a 6×10 rectangle 2339 ways. 

But because the shapes are all the same, if you mark all of them in some way to distinguish them from each other, (as the holes on the L-topia pieces do) every permutation of placements of the 12 l-tetrominoes can create a distinct tiling. Now the total number of tilings is roughly 1114 · 12!. (Actually, it’s slightly less because some of the tilings of the rectangle are symmetrical: about 55 of the 1114 solutions are symmetrical by reflection or 180° rotation, so the total is about 1059 · 12! + 55 · 12! / 2, or about 520 billion.)

Well, that’s a pretty impressive number, but having an impressively large space of possibilities does not, by itself, make for a great puzzle. In this case, however, I do think it is helpful, and I’ll explain why presently.

Suppose I think of a proposition that can apply to any of the holes in the set. For example, that the hole appears in an odd numbered row. Because there are two different kinds of holes, it may be elegant to use either the opposite of that proposition, or some proposition that is complementary in some way, to apply to the second kind of hole; in the problem illustrated by the solution above, we have the round holes in odd rows, and the square holes in odd columns. Suppose the probability of the proposition being true is ½, and suppose that the probability for each hole is independent from the others. (One must take care that the placement of holes on the pieces doesn’t fatally interfere with independence; if, for example, we had asked for circles on odd rows and squares on even rows, there would have been pieces that could not have been placed legally anywhere.) Then the probability that the proposition is true for all of the holes is 1/224. Given this piece of information, we can get an expected number of tilings where the proposition is true by multiplying that probability by the total number of tilings.

The result is about 31,000. That number is tiny compared to the size of the total space of tilings, but I can say from experience that it makes for puzzles that are challenging but solvable. And it gives us wiggle room to use propositions with probabilities that are a little smaller than ½, or for which the probabilities are not entirely independent. The result is that we can come up with a wide variety of propositions to use in designing puzzles with the expectation that they will provide a good puzzle solving experience. L-topia isn’t just a puzzle, it’s a natural puzzle creation kit!

Why L-Topia isn’t awesome, and Agincourt is

Unfortunately, to be perfectly honest, being a “puzzle creation kit” interferes with L-topia’s accessability as a puzzle. Because the circular and square holes have no inherent meaning, but have to have their meanings imposed by a puzzle’s directions, you can’t simply take the pieces out of the box and start solving.

Agincourt, on the other hand, with its 64 squares with an arrow in each, practically begs to be turned into four 4×4 layers with the arrows aligned. Of course, there are other challenges to be found, but the one that literally comes out of the box is both elegant, and has a reasonable level of difficulty. (Some of the L-topia puzzles are better for hardcore puzzle solvers.)

Once again, I have both puzzles available for sale. Order soon for delivery by Christmas!

Polystick Problems from Polyomino Solutions

September 7th, 2010

Polysticks (or polylines) are connected sets of segments in a square grid. (Polysticks on other grids are possible, but haven’t seen much attention.) The tetrasticks, of which there are 16, seem to be the most natural set for puzzle making. Donald Knuth has explored tetrastick problems, and posed the problem of tiling an Aztec Diamond with the 25 one-sided tetrasticks, which was solved by Alfred Wasserman. Here’s one I’ve come up with:

Problem #15: Tile the above shape with two sets of the five tetrominos and one monomino, and tile the borders of these polyominoes with the 16 tetrasticks. Here’s an attempt I made that fell short by a few tetrasticks, but it should give you an idea of the form a solution would take:

The observation that the lines formed by the pieces in a polyomino tiling could themselves be tiled by polysticks seems obvious, but I have not seen it elsewhere. After picking the 16 tetrasticks as my puzzle pieces for the polystick stage, I had to find a set of polyominoes to use. Since one of the tetrasticks is, in fact, the outline of a 1×1 square, or monomino, the monomino had to be present. A double set of tetrominoes plus the monomino gives a good quantity of segments for our tetrasticks to cover, and gives us an area of 2 * (5*4) + 1 = 41. The perimeter of the figure to be tiled is constrained by the following formula:

2 * total segments in the polystick set – sum of perimeters of polyominoes = perimeter of entire figure

In this case, (2 * (4 * 16)) – (4 + 2 * 48) = 28

So I needed a figure to tile with area 41 and perimeter 28, and came up with the shape above.

There are 136 solutions for the tetromino tiling with the monomino in the center as shown. (See these solutions in a Java solver applet.) I’ve experimented a little with the tetrastick stage of the problem by hand, and I’m convinced that there are no tetrastick solutions for most, if not all, of these tetromino solutions. But if it doesn’t work out in the case with the monomino in the center, I suspect there are enough solutions with the monomino elsewhere for it to be very likely that one will work. Many of the tetromino solutions fail to contain a site where the “+” tetrastick can be placed that doesn’t overlap the “□”.

Another issue that surfaces in this problem is horizontal-vertical segment parity. Eleven of the tetrasticks have even parity, that is, however you place them, they will always contain even numbers of both horizontal and vertical segments. Five of them have odd parity, and will always contain three segments of one orientation and one of the other. Because there are an odd number of pieces of odd parity, the parity of the set of tetrasticks as a whole must be odd. This means, without even starting to try placing tetrasticks on a tetromino solution, we can rule out the possibility of tiling it just by counting the number of horizontal or vertical segments. (Because the total is constant, we don’t need to count both.) If that number is even, the tetrasticks can’t tile the figure. The tetromino solution that I used in my attempt above has the correct (odd) parity.

I dredged this problem up from my archive of the polyforms mailing list, where I posted it in February, 2001. It got no takers then, but I thought it an interesting enough problem to deserve a second airing. I looked for other problems of this type in preparing this post, but I didn’t find anything good. Having both the area and the perimeter of the figure to be tiled constrained by the pieces used limits the possibilities a lot.