Most study of polyforms assumes that the shapes constituting a polyform are all immutable and identical, and that they connect in a finite number of ways. I find that sets of polyforms without these characteristics can be just as interesting.
Unit coin graphs, or penny graphs, are graphs formed by the adjacencies of non-overlapping circles of unit radius. For some graphs, finding a unit coin configuration that realizes the graph can itself be an interesting challenge. In a perfect binary tree with n levels, the area usable for placing coins is proportional to n2, but the area taken up by the coins is proportional to 2n. So there is reason to hope that just before the latter quantity overwhelms the former, finding a configuration without overlaps will make a good puzzle. And here's a solution for n = 6 with rotational symmetry.
Considering coin configurations as polyforms, we can think of two unit coin configurations (or polypennies) as being equivalent if we can transform one into the other by a combination of reflection and sliding coins without changing the adjacency graph of the configuration. Two configurations with the same graph can still be considered different if it isn't possible to transform one into the other in this manner. The numbers of configuration of n pennies for n between 1 and 5 are: 1, 1, 2, 6, and 13.
How can a tiling problem be interesting when the pieces you are using to tile are so amorphous? With polyominoes we try to tile things like rectangles, but nothing so simple will work here. One possibility is to require radial symmetry in a solution. Can the pentapennies tile a shape with 5-fold radial symmetry? I haven't found such shape with a solution yet, but I'm hopeful that one may exist. It seems like a good strategy is to put the pentagonal pentapenny in the center of the configuration to be tiled. Here's a configuration that fails to be solvable, but where one can fit most of the pieces before becoming stuck.
Alternately, if one of the pennies is placed at the center of the figure to be tiled, the remaining 64 pennies could be placed around it in a figure with 4-fold radial symmetry.
Polyform cover related problems can also be fruitful with these pieces. Here are my best attempts at a minimal cover and a maximal irreducible cover for the pentapennies.
There are several polyform variations that use triangles as their base form, including polyiamonds, polypons, and polyaboloes. These are distinguished by the shape of their base triangles. What if we allow the shapes of the triangles to vary? We can then count as equivalent all configurations of triangles that are connected in the same way. There are 9 pentatriangles in this scheme.
As with the pentapennies, an interesting puzzle using the non-rigid pentatriangles is using them to tile a configuration with 5-fold radial symmetry. Here's one solution. An open problem is finding all of the configurations like this can they tile, and all of the ways they can do it.
Polyedges are a type of polyform where the individual units are line segments, usually unit segments on a square grid. What would happen if we relaxed the grid requirement, and allowed the units to be any segment at all in the plane? It seems that what we would end up with would simply be the n-edge connected graphs. There are 12 5-edge connected graphs. Can we make good tiling problems for these?
One problem is that graphs are a lot more flexible than polyominoes, so it might be difficult to keep the number of solutions down to a reasonable number. On the other hand, we can use the flexibility of graphs to our advantage, and pick a highly symmetrical graph to tile, so the number of distinct (i.e. nonisomorphic) solutions is kept down.
In the graph to the right, there are 24 symmetries via rotation and reflection, but in addition, any vertex in the inner ring can be swapped with the vertex directly to the exterior. So its total number of symmetries, (i.e. the size of its automorphism group) is 24 · 2^12 = 96K. That should help.
Other possibilities for tiling include two icosahedra, four copies of K6, and the snub cube. (The last has relatively few symmetries, but it sure is neat looking.)
Coloring is an interesting problem with these. We define as adjacent for the purpose of coloring any two pentaedges that share a vertex. It should be easy to find a solution requiring 5 colors by having 5 different pentaedges edges come together at a vertex. Requiring 6 colors should be harder but doable. 3 colorings should be harder than we're accustomed to in polyforms, but probably still doable. For the double icosahedra tiling, an interesting coloring problem would be to 3-color both or to 6-color both icosahedra.
If you have questions, comments, solutions to unsolved problems, or ideas for new, related problems, please email me.