I am looking at the number of unique cyclic paths along the edges of an octahedron. A cycle starts and ends on the same vertex and an edge can only be walked once. They are also considered equivalent under rotation, but not reflection.
For example the two smallest (length=3) paths are the clockwise and anti-clockwise paths around a single triangular face.
There are then three (length=4) paths. One is the square around the middle and then there are the clockwise and anti-clockwise paths that surround two adjacent faces.
I found a clockwise and anti-clockwise (length=5) and then six different variations of (length=6) paths.
This makes a total of 13 unique cyclic paths. Is this the correct number, or have I missed anything?
Thanks in advance.
We are going to count the cycles that go through the bottom vertex.
we first count the cycles that pass through the bottom vertex but not the top vertex.
We separate into clockwise and counter-clockwise movement along the square. We do clockwise case, there are $4$ ways to get from the bottom vertex to the square, then we must pick how soon we go back down, it can be after $1,2$ or $3$ edges of the square have been travelled, then the answer is $4\times 3=12$, however notice this is only half because we must also count clock-wise. Thus there are $24$ cycles that do not reach the top.
We now count how many paths do reach the top. To do this we separate into three type of cycles.
The first type is when you go up immediately after reaching the square, there are $4$ ways to get to the square, then you go to the top, once at the top you must go down, here things get slightly tricky, there are three consecutive spots on the square to which you can descend, however now matter where you go down, there are going to be three ways to finish the cycle from there. Thus there are $4\times3\times3=36$ cycles of this kind.
The second type is when you go up after moving $1$ edge on the square, there are $4$ ways to get to the square, from here there are $2$ directions to move. Once you move $1$ edge on that direction you go up. After going up you must go down, there are $2$ choices and no matter which one you pick there will be $2$ ways to finish the cycle from there. Thus there are $4\times 2 \times 2 \times 2=32$ cycles of this type.
the third type is when you move $2$ edges once you get to the square, notice there are $4$ ways to get on the square and then $2$ directions to move those $2$ edges on the square, after this you must go up, and from here there is only one way to finish, thus there are $4\times 2=8$ cycles of this type.
Thus there are $24+36+32+8=100$ cycles, however be careful, because we counted each cycle twice, because we counted oriented cycles, so the final answer is $50$ cycles that pass through the bottom vertex.
Using the first result there are $12$ cycles that pass through the top vertex but not the bottom.
Finally there is one cycle that does not pass through the top or bottom vertex.
So adding up there are $63$ cycles.