On 03/03/2007 05:32 PM, Martin Rubey wrote:
> Ralf Hemmecke <[EMAIL PROTECTED]> writes:
>
>>>> Summation over cycles is still not gone.
>>> But over permutations. That's a lot less!
>> Arrrggghhh. Of course my sentence should have read:
>>
>> "Summation over permutations is still not gone."
>
> But it is!?
OK, before I go further, just propose the code for what you have
described. An I guess it will be wrong in any computer language.
Let me try to point you to your error (or at least, what I think is an
error).
The CIS of F is defined in BLL Def 1.2.6 as
Z_F = \sum_{n\geq0} \frac{1}{n!} f_n
where f_n = \sum_{\sigma\in S_n} \fix F[\sigma] x^\sigma.
Remark 1.2.10 then says that one can rewrite f_n to
f_n = \sum_{k_1+2k_2+...+nk_n = n} \fix F[k] x^k/z_k.
where z_k = \prod_{i=1}^n i^{k_i} k_i!.
Now, if H = F \factorialcompose G, a similar formula as above, of
course, holds for H.
>>> Similarly, if $G$ is a species, the longest cylce of $G[\sigma]$ can only be
>>> shorter than the the longest cycle of $\sigma$. (Check this...!)
>>> is enough... (apart from the fact, that the final sentence is wrong)
>> Oh, I have not guessed that
>
> guessed what?
>
>> and the function name "cycleTypeSpecies" tells me nothing at all. Not even
>> from the documentation you give I understand the input/output specification.
>
> OK, I try again, maybe this is better.
>
> Martin
>
> Definition~4 in Section~2.2 of \cite{BLL} goes as follows:
> \begin{dfn}
> Z_F \square Z_G = \sum_{n\ge0}\frac{1}{n!}
> \sum_{\sigma\in S_n}
> \fix F[(G[\sigma])_1,(G[\sigma])_2,\dots]
> x_1^{\sigma_1}x_2^{\sigma_2}\dots
> \end{dfn}
>
> We will see that $\fix F[(G[\sigma])_1,(G[\sigma])_2,\dots]$ in fact depends
> only on the cycle type of $\sigma$. Thus we can rewrite the above to avoid
> summation over all permutations:
> \begin{equation*}
> Z_F \square Z_G = \sum_{n_1+2n_2+\dots}
> \fix(F\square G)[n_1, n_2, \dots]
> \frac{x_1^{n_1}x_2^{n_2}\dots}
> {1^{n_1}n_1! 2^{n_2}n_2!\dots},
> \end{equation*}
That is basically your last equation. I don't see that I gained more
insight than I knew before.
I like to refer to a mail I sent to you yesterday.
Given:
Z_F = \sum_{n=0}^\infty \sum_{k_1+2k_2+...+nk_n=n} a(k) x^k
Z_G = \sum_{n=0}^\infty \sum_{k_1+2k_2+...+nk_n=n} b(k) x^k
Z_H = \sum_{n=0}^\infty \sum_{k_1+2k_2+...+nk_n=n} c(k) x^k
My problem is: Give a "nice" formula for c(k) which only depends on the
a(i) and b(i) (and k).
All what I have seen up to now where some little pieces of a puzzle that
I feel unable to solve.
That is why I suggest that you try to write some sketchy code for c(k)
that does not involve any permutation. Only cycle types are allowed.
Even a mathematical formula for c(k) would be OK, but no permutations.
Best regards
Ralf
PS: I think before I implement cartesian product, it is certainly better
to have Permutations as a basic species. So I'll try this one next.
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