the tree $T_1=T_{E_1}=(1,1).$

-   $E_2=\left\{ M_1=[3],M_2=[2],M_3=[4] \right\}$ and we have $\min(\max(l(M_2),p_1),\textrm{Comp}(M_2,M_3),l(M_3))=\min(1,3,1)=1$. Thus we only have the tree $T_2=T_{E_2}=(1,1).$

-   $E_3=\left\{ M_1=[2],M_2=[1],M_3=[2,2] \right\}$ and we have $\min(\max(l(M_2),p_1),\textrm{Comp}(M_2,M_3),l(M_3))=\min(2,2,2)=2$. Thus we have the trees $T_3=T_{E_3}=(2,1)$ and $T_4=T_{E_3}=(2,2).$

-   $E_4=\left\{ M_1=[2],M_2=[1],M_3=[4] \right\}$ and we have $\min(\max(l(M_2),p_1),\textrm{Comp}(M_2,M_3),l(M_3))=\min(2,2,1)=1$. Thus we only have the tree $T_5=T_{E_4}=(2,1).$

Hence $S_1^1([3,2,4])=\left\{ T_1,T_2,T_3,T_4,T_5\right\}.$

Now we compute $S_1^2([3,2,4])$. $k_3=3$ is the only value such that $I(k_3) \neq \emptyset$. In fact, if we consider $A_2$ and $M_3=[3]$, we have:

-   $ E_5=\left\{ M_1=[2],M_2=[1],M_3=[3] \right\} \textrm{ and } l(M_3)+c[3]-k_3=1+4-3=2 \leq \min(\max(l(M_2),p_1),\textrm{Comp}(M_2,M_3)))=\min(2,2)=2$. Thus we have the tree $T_6=T_{E_5}=(2,2).$

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at(current bounding box.south)[$T_1$]{};

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at(current bounding box.south)[$T_2$]{};

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at(current bounding box.south)[$T_3$]{};

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at(current bounding box.south)[$T_4$]{};

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at(current bounding box.south)[$T_5$]{};

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at(current bounding box.south)[$T_6$]{};

Now we compute $S_2^1([3,2,4])$. We are in the case $t=2$ so its definition is: $$S_2^1([3,2,4])=\left\{ T_{E}=(p_1,p_2): E=\left\{M_1,M_2,M_3 \right\}; \right.  T_{E^1}=(p_2) \in \textrm{Cond}([2,4]);$$ $$\left.  M_1 \in \textrm{Cond}^*(3) ;1\leq p_{1} \leq \min(l(M_1),p^*_{1},\textrm{Comp}(M_{1},M_2)) \right\}.$$ Then to do that we need the following sets: $$\textrm{Cond}([2,4])= \left\{B_1,B_2,B_3,B_4 \right\}\textrm{ where }$$

-   $B_1=T_{G_1}=(2)$ with $G_1=\left\{ [1],[2,2]\right\};$

-   $B_2=T_{G_2}=(1)$ with $G_2=\left\{ [2],[2,2]\right\}$;

-   $B_3=T_{G_3}=(2)$ with $G_3=\left\{ [1],[3]\right\}$ ;

-   $B_4=T_{G_4}=(1)$ with $G_4=\left\{ [2],[4]\right\}$ ;

-   $\textrm{Cond}^*(3)=\left\{[3] \right\}.$

Hence we consider:

-   $E_6=\left\{ M_1=[3],M_2=[1],M_3=[2,2] \right\}$ and we have $\min(\max(p_2,l(M_2),\textrm{Comp}(M_1,M_2),l(M_1))=\min(2,2,1)=1$. Thus we 