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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.

enter image description here

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:

enter image description here

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?

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    $\begingroup$ 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? $\endgroup$ – Rand al'Thor May 19 '18 at 17:54
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    $\begingroup$ Wow! This looks so much fun, let me pull up some nails and a thread :D $\endgroup$ – ABcDexter May 19 '18 at 18:05
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Part 1

I count 12 ways:enter image description here 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.

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