I vaguely know that by looking at the inversions of a permutation, you can write down the reduced word expressing the permutation as a product of adjacent transpositions $s_i = (i,i+1)$. However, I am a little unclear on what order you check the inversions, or if there is a simple way to do it without multiplying your original permutation by all the transpositions you've found so far.
Is there a simple algorithmic description of how to convert a permutation's inversion set to its reduced word?
Secondly, I'd actually like to do this in the signed permutation group (whose elements are monomial matrices with nonzero entries taken from $\{1,-1\}$; a monomial matrix is like a permutation matrix except the nonzero entries don't have to be $1$).
I'm a little unclear on when you look for the sign changes, but possibly this is due in large part to not having a specific algorithm for the plain permutations.
Is there a simple algorithmic description of how to convert a signed permutation's inversion set and positions of $-1$s to its reduced word?
Example:
The permutation $\pi = (1,2,3) = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$ has inversion set $\{ \{1,3\}, \{2,3\} \}$ since $1 < 3$ but $\pi(3) < \pi(1)$ and $2 < 3$ but $\pi(3) < \pi(2)$. The $\{2,3\}$ is pretty clearly caused by $(2,3)$ and so we need to multiply by something to invert $\{(1,3)\}$. Since already know $(2,3)$ is used, that moves the $3$ to a $2$, and $\{1,2\}$ is caused by $(1,2)$. Now depending on which order you apply your functions, we get the product of $(1,2)$ and $(2,3)$ is $(1,2,3)$.
How did I know to use $\{2,3\}$ first? Should I really transform the inversion set by every adjacent transposition I discover?
What if it was $\sigma = (1,\bar 2,3) = \begin{bmatrix} . & -1 & . \\ . & . & 1 \\ 1 & . & . \end{bmatrix}$ with the same inversion set as before, but with the $2$nd column negated? When do we take of the negative?