algorithm which computes $\textrm{Cond}(\textbf{c})$.

Now we have a way to compute all the untwisted multiplicity trees with a given conductor $\textbf{c}$ for all the $\textbf{c} \in \mathbb{N}^r$. Suppose that we want to find also the twisted multiplicity trees with conductor $\textbf{c}$. We will call $\overline{\textrm{Cond}(\textbf{c})}$ the set of all multiplicity trees (twisted or untwisted) associated to an Arf semigroup with conductor $\textbf{c}$. Suppose that $T$ is a twisted tree in $\overline{\textrm{Cond}(\textbf{c})}$ with $\textbf{c} \in \mathbb{N}^r$. Then there exists a permutation $\sigma \in S^r$, where $S^r$ is the symmetric group, such that $\sigma(T)$ is untwisted and is clearly belongs to $ \textrm{Cond}(\sigma(\textbf{c}))$. From this it follows that: $$\overline{\textrm{Cond}(\textbf{c})}=\bigcup_{\sigma \in S^r} \left\{ \sigma^{-1}(T): T \in  \textrm{Cond}(\sigma(\textbf{c})) \right\}.$$

Let us compute $ \textrm{Cond}([3,2,4])$. In this case $r=3$, therefore we have $t=2$. First of all we compute $S_1^1([3,2,4])$. Because $t=r-1$ the definition of this set is: $$S_1^1([3,2,4])=\left\{ T_{E}=(p_1,p_{2}): E=\left\{M_1,M_2,M_3 \right\}; \right.  T_{E_{2}}=(p_1) \in \textrm{Cond}([3,2]);$$ $$\left.  M_3 \in \textrm{Cond}^*(4) ;1\leq p_{2} \leq \min(l(M_3),^*p_{2},\textrm{Comp}(M_{2},M_3)) \right\}.$$ Then to do that we need the follwing sets:

-   $\textrm{Cond}([3,2])= \left\{A_1,A_2 \right\}$ where $A_1=T_{F_1}=(1)$ and $A_2=T_{F_2}=(2)$ with $F_1=\left\{ [3],[2]\right\}$ and $F_2=\left\{ [2],[1]\right\}.$

-   $\textrm{Cond}^*(4)=\left\{[2,2],[4] \right\}$

Hence we consider:

-   $E_1=\left\{ M_1=[3],M_2=[2],M_3=[2,2] \right\}$ and we have $\min(\max(l(M_2),p_1),\textrm{Comp}(M_2,M_3),l(M_3))=\min(1,2,2)=1$. Thus we only have 