# 6 nails String Art

String art is an arrangement of thread strung between nails to form geometric patterns or representational designs such as a ship's sails, sometimes with other artist material comprising the remainder of the work. Let say we have $4$ nails, and we would like to make as many different shapes as we can where nails are put in the vertices of a square, then we could have at most two different shapes as shown below: The rules while forming the shapes are simple:

• the string cannot pass from the same nail again,
• the string needs to pass from every nail.
• and the string needs to go back where it starts at the end to attach the string again.
• while counting, do not count reflected, symmetrical or rotated shapes again.

1- If we have a regular hexagon shape where nails are located on its vertices, how many different shapes could you form?

and

2- Is it possible to generalize how many shapes we can have at most for any kind of regular polygon?

• I have to ask: do you know the answer to this? Or is there a possibility it could be some big open problem in combinatorial geometry? – Rand al'Thor May 19 '18 at 17:54
• Wow! This looks so much fun, let me pull up some nails and a thread :D – ABcDexter May 19 '18 at 18:05

Part 1

I count 12 ways: There are several ways to demonstrate that these exhaustively show all of the possibilities. For instance, we may use Burnside's Lemma as follows. First, note that there are $5! = 120$ possible 6-cycles. Next, we consider the symmetry groups of the shapes; they are (numbering them 1-3 in the first row, then 4-6 in the second, and so on):
6-way rotational, reflection - Shape 8
3-way rotational, reflection - Shape 4
2-way rotational, reflection - Shape 5, 6, 10
2-way rotational, no reflection - Shape 3
no rotation, yes reflection - Shape 1, 2, 7, 11, 12
no symmetry - Shape 9
Finally, note that there are two ways to orient a cycle ("clockwise" or "counterclockwise"), 6 ways to rotate, and 2 ways to reflect. So we add: $24(1+\frac52+\frac12+\frac34+\frac16+\frac1{12})=120$, so we've accounted for all the cases.

Part 2

I then did some searching on OEIS and found that this problem matches up the following sequence: A000940. In the link there's a program that can find such numbers, but the formula appears to be recursive perhaps using the proper factors of a number - at any rate, no explicit formula is given.