Oh, right. There are also other cases where you have less-than-N-way rotational symmetry without mirror symmetry. I was thinking that when you have K-way rotational symmetry, you can always pick an axis a symmetry that bisects one of the lines, but when you have multiple sets of lines, there may not be one mirror axis that works for both. For instance, for N=8, take a square of connections that each skip one node, and also connect each of these four nodes to the (previously skipped) node to its left.
What is still true is that when you have N-way symmetry, you have to take complete sets of N like connections and/or the complete set of N/2 connections through the center, and each one of those sets has mirror symmetry with respect to all the possible axes, so the combination of these sets does also.
Also when N is even, you have axes of symmetry that go through the nodes, and axes of symmetry that go between the nodes, while with odd N all the axes of symmetry are equivalent. So it is actually quite a bit more complicated for even N. Still, with some more work it should be possible to compute these directly without having to go through exponentially increasing numbers of cases.
This is the kind of problem I expect to see come up on Project Euler. (Except then I would have to compute the sum of all distinct ways to draw lines from 1 to 100 points, mod 10^8, or some other crazy calculation to show that I really understand all the intricacies of the problem. Here, I was just trying to get you started on a new way to think about the problem.)
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Date: 2011-02-05 10:45 pm (UTC)What is still true is that when you have N-way symmetry, you have to take complete sets of N like connections and/or the complete set of N/2 connections through the center, and each one of those sets has mirror symmetry with respect to all the possible axes, so the combination of these sets does also.
Also when N is even, you have axes of symmetry that go through the nodes, and axes of symmetry that go between the nodes, while with odd N all the axes of symmetry are equivalent. So it is actually quite a bit more complicated for even N. Still, with some more work it should be possible to compute these directly without having to go through exponentially increasing numbers of cases.
This is the kind of problem I expect to see come up on Project Euler. (Except then I would have to compute the sum of all distinct ways to draw lines from 1 to 100 points, mod 10^8, or some other crazy calculation to show that I really understand all the intricacies of the problem. Here, I was just trying to get you started on a new way to think about the problem.)