Given two quaternions, $a+bi+cj+dk$ and $e+fi+gj+hk$, their product (w.r.t. their given order) would normally be given by $Q_1+Q_2i+Q_3j+Q_4k=(ae-bf-cg-dh)+(af+be+ch-dg)i+(ag-bh+ce+df)j+(ah+bg-cf+de)k$. This takes a total of $16$ real multiplications and $12$ real additions to accomplish.
I'm looking to reduce the number of multiplications. Here's what I have so far.
$P_1=(c+d)(g+h)$, $P_2=(a+b)(e+f)$, $P_3=(c+d)(e+f)$, $P_4=(a+b)(g+h)$, $Q_1=(ae+dg)+(ch-bf)-P_1$, $Q_2=P_2+(ch-bf)-(ae+dg)$, $Q_3=P_3+(ag-de)-(bh+cf)$, $Q_4=P_4-(ag-de)-(bh+cf)$.
As far as I can tell, this requires 12 real multiplications and 16 real additions, assuming certain quantities are reused.
Are there any well-known methods for doing this using less real multiplications, or any obvious simplifications that I could make to my formulas?
EDIT: You can calculate $(R_1,R_2)=(ae+dg,ag-de)$ and $(R_3,R_4)=(cf+bh,ch-bf)$ by this method: $S_1=(a+d)(e+g)$, $S_2=(c+b)(f+h)$, $(R_1,R_2)=(S_1-ag-de,ag-de)$, $(R_3,R_4)=(S_2-ch-bf, ch-bf)$. This brings the total number of real multiplications down to 10, and the real additions up to 22.
There's an algorithm here that uses eight multiplications, as long as you don't count multiplications by $2$ and by $-1$. This paper also mentions a proof that it can't be done with less than seven. Here is a review of an article which shows it can't be done with less than ten multiplications if you count multiplying by fixed constants, and gives such an algorithm, but I can't see the paper (due to paywalls I don't know if either of these links will work.)
The algorithm in the first paper is as follows: $(a_0+a_1i+a_2j+a_3k)(b_0+b_1i+b_2j+b_3k)=c_0+c_1 i+c_2j+c_3k$, where $$\begin{pmatrix} -c_0&c_1&c_2&c_3\end{pmatrix}=1/4\begin{pmatrix} n_0p_0&n_1p_1&n_2p_2&n_3p_3\end{pmatrix} A-2\begin{pmatrix} a_0b_0&a_3b_2&a_2b_1&a_1b_3\end{pmatrix} $$ $$A=\begin{pmatrix} 1&1&1&1\\ 1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{pmatrix}$$ $$\begin{pmatrix} n_0&n_1&n_2&n_3\end{pmatrix}=\begin{pmatrix} a_0&a_1&a_2&a_3\end{pmatrix} A$$$$\begin{pmatrix} p_0&p_1&p_2&p_3\end{pmatrix}=\begin{pmatrix} b_0&b_1&b_2&b_3\end{pmatrix}A$$ So the only "real" multiplications being down are between $n$s and $p$s and between $a$s and $b$s in the final formula.