# How to enumerate the possible “puzzles” resulting from seven adjacent hexagonal tiles?

There are seven hexagonal tiles; one in the center and six surrounding it. Each tile may have from one to six lines. Here is an example: Each tile can be rotated (and indeed, to solve the puzzle, one must rotate each tile so that all of the lines meet), but this of course does not generate a new puzzle. Under the modest assumption that the center tile must have at least four lines, how many unique puzzles are possible?

There are five unique center tiles - three with four lines, one with five lines, and one with six lines. Also, there are four unique edge tiles; the example shows two, and aside from that there is the tile with one line and the tile with two adjacent lines. The number of lines in the center tile poses certain constraints on the admissible edge tiles - for example, the edge tile with two lines that are not adjacent cannot be present if the center tile has six lines.

In the context I encountered this problem, I only observed ten unique puzzles. I suspect there must be many more, although I do not know how to enumerate them beyond a brute force approach. Any suggestions?

• For this puzzle, I would use a brute force search. If you want to generalize it to a larger problem, the proper abstraction would depend a good deal on which parts are generalized. – Wildcard Sep 5 '17 at 6:22

Fr each of the center tiles, part of the edge tiles is determined. The remaining freedom consists in deciding about the presence of the six edges among the edge tiles, which can be done in $2^6$ ways. However, we have the constraint that an edge tile without link to the centre must be linked to at least one of its neighbours, i.e., we subtract those configurations where such a tile ends up empty. An isolated such edge tile thus reduces the count by $\frac14$, a pair of adjacent such edge tiles by $\frac 38$. Thus we have
• $64$ solutions for the six-line centre
• $\frac34\cdot 64 = 48$ for the five-line centre
• $\frac58\cdot 64 = 40$ for the four-line centre with adjacent gaps
• $\frac34\cdot\frac34\cdot 64=36$ for each of the other four-line centres
That's a total of $64+48+40+36+36=224$.
• I'm afraid I don't follow your reasoning. Where do you get $2^6$ from exactly? – Math1000 Sep 5 '17 at 6:55