Using Burnside's lemma to count number of strings

Asked Dec 02 '21 at 18:24

Active Dec 02 '21 at 18:43

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Recall that Burnside lemma states the the number of orbits of an operation of a group $G$ on a set $ X $ is given by

$$ \frac{1}{|G|}\sum_{g\in G}\lvert \text{fix}\left(g\right)\rvert,$$

where

$$ \text{fix}\left(g\right)=\left\{ x\in X:gx=x\right\} $$

and $gx=x$ is notation for $ \varphi\left(g\right)\left(x\right)=x $ where $\varphi $ is the opertaion of the group $ G $ on the set $ X $.

Now, how can I use Burnside's lemma in order to calculate different strings of length $k $ that can be built by $ n $ types of characters ?

Different strings = upto circular permutation

(If its too complicated I'd also appreciate an example for say, k=6, and I'll think by myself how to generalize).

I'd appreciate an idea. Not sure how to start.

Thanks in advance.

edited Dec 02 '21 at 18:43

asked Dec 02 '21 at 18:24

FreeZe

1

Can you clarify what you mean by "different" strings? When are two stings considered the same? – Mike Earnest Dec 02 '21 at 18:30
I guess you consider strings under circular permutations ? – TheSilverDoe Dec 02 '21 at 18:38
@TheSilverDoe Yes. – FreeZe Dec 02 '21 at 18:42
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This fact sheet on necklaces might help, supposing circular permutation means rotation. – Marko Riedel Dec 02 '21 at 18:50
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Hint : Let $A=\lbrace 1, ..., n \rbrace$ be your alphabet, and let $X=A^k$ the set of all words of $k$ letters with this alphabet. Consider the action of $\mathbb{Z}/n\mathbb{Z}$ defined by $1.(x_1, ..., x_k) = (x_2, ..., x_k, x_1)$. Then what you want to count is exactly the number of orbits for this action. Apply Burnside's formula. – TheSilverDoe Dec 02 '21 at 18:51

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