Archive for February, 2010

Pentomino Layer Cake

February 27th, 2010

On the Polyforms list, Erich Friedman posed a very interesting new pentomino tiling problem:

Tile a rectangle of minimal area with pentominoes so that for each pentomino there is exactly one stratum, or cluster of one or more copies of that pentomino that reaches from one side of the rectangle to the opposite side. Pentominoes in a stratum must form a single group, connected by edges, not just corners.

Michael Reid found this 3×30 solution:

It’s not hard to prove that it is minimal. A natural extension of the problem is to find minimal solutions for 4×n and 5×n rectangles. Michael Reid found the first 5×n solution, but I improved on it with this 5×32 solution:

The 4×n problem seems to be the hardest, and initially it was not clear that it would be possible. The X pentomino has only one possible stratum, which only can only be bordered by Y, I or N, and it is also difficult to find matches for a Z stratum. Additionally, only Y, L, and P can form straight line stratum boundaries usable for the top and bottom of the rectangle. (See wikipedia’s pentomino page if you don’t know the correspondence between letters and shapes.) I did eventually find this 4×50 solution:

This solution seems rather prolifigate with its pentominoes, but finding any solution at all was a bit of a surprise.

Update: Erich Friedman’s Math Magic for April 2010 further explored this subject.

Introducing Agincourt (to the Blog)

February 25th, 2010

Agincourt is one of the lasercut acrylic puzzles which I’m selling through the store. It’s the set of all of the ways to make 2-, 3-, and 4-ominoes with arrow shaped holes in each square pointing in the same direction. The symmetry of the arrows means that you can flip over pieces without changing the arrow directions, but you can’t rotate them. Most of the puzzles I have designed for the set ask for the solver to make all pieces point the same way, but the arrows naturally suggest a scoring system to handicap the puzzle for different levels of solvers — just count the number of pieces you had to put in the wrong direction, and try to improve on your score.

Here’s a solution to the puzzle that literally comes out of the box. (The puzzle comes in the box with 4 layers of pieces in 4 × 4 squares.)

Expect more Agincourt puzzles later.

Holy Hyperbolic Heptagons!

February 19th, 2010

Recently, MathPuzzle highlighted a program called MagicTile for playing with Rubik’s Cube variants on various tessellations in spherical, Euclidean, and hyperbolic geometries. One interesting hyperbolic tessellation is the {7, 3} tessellation, composed of heptagons, with three meeting at every corner. Twenty-four of these heptagons can be wrapped up into a genus-3 (that is, topologically like a torus but with three loops instead of one) “Platonic solid” called Klein’s Quartic, which John Baez has a fascinating page about.

Polyforms based on hyperbolic tessellations appear to be a relatively unexplored area. I’ve come across a couple of references on counting the n-cell polyforms for a given tessellation, but I haven’t found evidence that anyone has actually used them for puzzles. So I set myself this one. There are ten tetrahepts on the {7, 3} tessellation. Could they tile two copies of the same shape?

An implicit constraint in my solution of this problem was that the shapes could extend no farther than the second ring around a central heptagon. I searched for a solution using plastic game counters that would not have fit on any heptagons farther out on the diagrams I used as solving boards. I also knew that I was going to make images to show on my blog, and they would be clearer if there were no relatively tiny heptagons in the solution. In fact, the decision to find a puzzle in two pieces was affected by the observation that using an extra Poincaré disk would remove the need to use the tiny outer heptagons.

This was a pretty difficult puzzle to solve by hand, so I feel like I’m being a little mean using an even harder, (and perhaps impossible) variant as the numbered puzzle for this post:

#8: Find a symmetrical shape for which two copies can be tiled by the ten {7, 3}-tetrahepts. In addition to mirror symmetry, 180° rotational symmetry around the midpoint of an edge could work. (The other modes of rotational symmetry of the tessellation won’t work for a 20 cell shape.)

I would also love to see what other puzzles people can come up with on this or any other hyperbolic tessellation.