If the signature is s(x) we know that (A) ∀ x,y ∊ Sn s(xy)= s(x)s(y) . Suppose we want to compute the signature of a permutation using the DCN method, My question is why do we need to have the permutation in DCN ? Since s(x) when x is an r-cycle is (-1)^(r-1) we can use this and (A) to compute the signature. 0
DCN means disjoint cycle notation and s(x) is defined as follow : Δ = ∏ ( xj - xi) j ⪪ i , Let each permutation g in Sn permutes the variables x1 ,......., xn in the same way with 1,........., n . then g(Δ) = Δs(g) where s(g) = +1 or -1 .
Thanks for considering my question .
After some discussion in the comments I think your question is: does decomposing the permutations as a product of DISJOINT cycles have any advantage over decomposing it as ANY product of cycles when you goal is to compute sign. The answer can be yes or no depending on your perspective:
Perspective 1: I have permutation written as a product of (non-disjoint) cycles, should I recompute it into a product of disjoint cycles in order to more easily compute the sign?
Answer: NO
The formula "sign is the product of the signs of the cycles, and a cycle of length $r$ has sign $(-1)^{r+1}$" is correct whether or not the cycles are disjoint.
(You already know this)
Perspective 2: I have a permutation in the form of a black box function, i.e. for every number I know to which number the permutation sends it. I want to compute the sign via writing the permutation as a product of cycles so I can apply the formula described above. Is there any reason for writing it as a product of DISJOINT cycles specifically?
Answer: YES.
Reason: there is a very quick and easy algorithm for writing your permutation in this form. (I can edit it in if you want, but you probably already know it).
Of course there is a difference between the two answers. The argumentation for 'NO' relies on a hard mathematical fact (the formula works in both cases) the argument for 'YES' relies on a personal preference for quick algorithms that I already know over slower or more complicated or equally easy but yet unknown to me algorithms. But other people might have other preferences. So in that sense you can argue that 'NO' is the 'better' answer. But both have some merit.