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About LinkBacksUsing Burnsides Theorem
How many distinguishable regular octagons can be painted if each edge of a octagon is painted one of eleven colours and different edges can be the same colour?
the symmetric group of a regular octagon is $\displaystyle D_8,$ the dihedral group of order 16, which is generated by two permutations $\displaystyle \sigma = (1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8)$ and $\displaystyle \tau=(2 \ 8)(3 \ 7)(4 \ 6).$Originally Posted by maths900
that is $\displaystyle D_8 = \{\tau^i \sigma^j: \ \ 0 \leq i \leq 1, \ 0 \leq j \leq 7 \}.$ for each $\displaystyle \tau^i \sigma^j$ suppose $\displaystyle n(\tau \sigma^j)$ is the number of cycles in the complete decomposition of $\displaystyle \tau^i \sigma^j.$ then Burnside's lemma says
that the number of coloring for a regular octagon using 11 colors is $\displaystyle \frac{1}{|D_8|}\sum_{i,j}11^{n(\tau^i \sigma^j)}=\frac{1}{16} \sum_{i,j} 11^{n(\tau^i \sigma^j)}.$ as an example, there are 8 cycles (of length 1) in the complete
decomposition of $\displaystyle \text{id},$ the identity element of $\displaystyle D_8,$ and so $\displaystyle n(\text{id})=8.$ more examples: $\displaystyle \sigma$ and $\displaystyle \tau$ have one cycle and 5 cycles respectively and thus $\displaystyle n(\sigma)=1, \ n(\tau)=5,$ etc.
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