We will begin with the most tedious part of this plan, namely, the complexes of the form $(\widetilde{\mu}_{G^{s+1}}^{st})^{-1} \circ \mu_{j_{s+1}}^+ \circ \widetilde{\mu}_{G^s}^{st}(\Lambda^{\sigma_s})$. First let us get rid of the excessive notation. Let $G$ be any modified Brauer tree. Recall that by $\mu_G^{st}$ we denote the standard sequence of mutations for $G$. By $\mu_j^+(G)$ we will denote the modified Brauer tree of $\End_{\D^b(\Lambda)}(\mu_j^+ \circ \mu_G^{st}(\Lambda))$. Let $\tau$ be the permutation one needs to apply to the standard labeling of edges of $\mu_j^+(G)$ to obtain be the labeling emerging via $\mu_j^+ \circ \mu_G^{st}$. As before, apply $\tau$ to the indices of mutations in $(\mu_{\mu_j^+(G)}^{st})^{-1}$ and denote the resulting series of mutations by $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1}$. It is sufficient to express the complexes of the form $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}(\Lambda)$ as images of $\Lambda$ under a series of spherical twists $t_i$ modulo $\Pic(\Lambda)$ and the shift, for any modified Brauer tree $G$ and any index $j = 1, \dots, m$. Whenever we write that two tilting complexes are equal, we mean that they are isomorphic modulo $\Pic(\Lambda)$.

§.§ Double Edge type

First we are going to assume that $G$ is a Double Edge tree. Let $\varphi = \varphi_G, \psi = \psi_G$ be the maps introduced in Section <ref>. With only a few exceptions, here we shall essentially repeat the results of [33]. In each of the cases we will denote the autoequivalence $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}$ by $\mathcal{H}$ and the tilting complex $\mathcal{H}(\Lambda)$ by $\mathcal{T}$. By $\mathcal{T}_i$ we will always denote $\mathcal{H}(P_i)$. However, it will be often convenient to consider the tilting complex $\mathcal{T}$ with the labeling of summands coming from the series of mutations $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}$. In this case we will write $\mathcal{T}'_i$ instead of $\mathcal{T}_i$, meaning that $\mathcal{T}_i'$ is the summand of $\mathcal{T}$ corresponding to the edge of the modified Brauer tree of $\End(\mathcal{T})$ labeled with $i$. In addition, we will denote the tilting complex $\mu_j^+ \circ \mu_G^{st}(\Lambda)$ by $T$.

* Let $j$ be such that the mutation $\mu_j^+$ in $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}$ does not involve irregular edges. More precisely, (1) $j$ is not a double edge, i.e. $j \neq 1, 2$, and (2) $j$ is not followed by the double edge in the cyclic ordering, i.e. $j \neq 3$. This means that the series of mutations $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}$ does not “touch” irregular edges as well. Namely, for every $i$ such that there is a mutation $\mu_i^+$ in this composition, we have $\Hom_{D^b(\Lambda)}(P_i, P_1 \oplus P_2) = 0$. Thus, ${\mathcal{T}_1 = P_1, \mathcal{T}_2 = P_2}$ and $\mathcal{H}$ restricts to an autoequivalence of $\thick(\bigoplus_{k=3}^m P_k)$, which is equivalent to the bounded derived category of the Brauer star algebra with $m-2$ indecomposable projective modules. Moreover, observe that our construction of the standard series of mutations for a Double Edge tree is in an obvious way consistent with the construction of the standard series of mutations for a Brauer tree in [33]. To conclude, in this case we can directly make use of the calculations already fulfilled in [33]. Let us briefly repeat the results in our notations.

* Suppose $j$ is pendant and there is an edge $l$ following $j$ in the cyclic ordering on the same level as $j$, i.e. $\varphi(j) = \varphi(l)$. Then $\mathcal{T} = \Lambda$ and there is nothing to express.

* Suppose $j$ is pendant and there is no edge following $j$ in the cyclic ordering on the same level as $j$. In particular, $\psi(j) = j-1$. Then

$$\mathcal{T}_i = P_i, \quad i \neq j,j-1 $$

$$\mathcal{T}_{j-1} = (P_j \xrightarrow{\beta_j} P_{j-1} \xrightarrow{soc} P_{j-1})[2]$$

And we have $\mathcal{H}(\Lambda) = t_j(\Lambda)$.

* Suppose $j$ is not pendant and there is an edge $l$ following $j$ in the cyclic ordering with $\varphi(j) = \varphi(l)$. Let $h$ be the other edge following $j$ in the cyclic ordering (with $\varphi(h) = \varphi(j)+1$). Then

$$ \mathcal{T}_j = P_h$$

$$\mathcal{T}_i = (P_h \xrightarrow{soc} P_h \xrightarrow{\beta_h \dots \beta_i} P_{i-1}), \quad j+1 \leq i \leq h$$

$$ \mathcal{T}_i = P_i \text{ otherwise}$$

We have $\mathcal{H}(\Lambda) = (\mu_j^-)^2 \circ (\mu_{j+1}^-)^2 \circ \dots \circ (\mu_{h-1}^-)^2(\Lambda) = t_{j+1}^{-1} \circ \dots \circ t_h^{-1}(\Lambda)$.

* Finally, suppose $j$ is not pendant and there is no edge on the same level as $j$, following $j$ in the cyclic ordering. In particular, $\psi(j) = j-1$. Let $l$ be the other edge following $j$ in the cyclic ordering, with $\varphi(l) = \varphi(j) +1$, and let $h$ be the edge following $l$ in the cyclic ordering with $\varphi(h) = \varphi(l)+1$. Then we have

$$ \mathcal{T}_i = P_i, \quad 3 \leq i \leq j-2 \text{ and } h+1 \leq i \leq m$$

$$\mathcal{T}_{j-1} = (P_l \xrightarrow{\beta_l \dots \beta_j} P_{j-1} \xrightarrow{soc} P_{j-1})[2] $$

$$\mathcal{T}_i = (P_l \xrightarrow{\begin{pmatrix} soc \\ -\beta_l\dots\beta_j \end{pmatrix}} P_{j-1} \oplus P_l \xrightarrow{\begin{pmatrix} 0 & -soc \\ \beta_l \dots \beta_{i+1} & 0 \end{pmatrix}} P_{j-1} \oplus P_i)[2], \quad j \leq i \leq l-1$$

$$\mathcal{T}_i = (P_{i+1} \xrightarrow{\beta_{i+1} \dots \beta_{j}} P_{j-1} \xrightarrow{soc} P_{j-1})[2], \quad l \leq i \leq h-1$$

$$ \mathcal{T}_h = P_{j-1}$$

We have

$$\mathcal{H}(\Lambda) = (\mu_h^+)^2 \circ \dots \circ (\mu_{l+1}^+)^2 \circ (\mu_j^-)^2 \circ \dots \circ (\mu_{l-1}^-)^2 \circ (\mu_{l}^+)^2 \circ \dots \circ (\mu_j^+)^2(\Lambda) = t_h \circ \dots \circ t_{l+1} \circ t_j^{-1} \circ \dots \circ t_{l-1}^{-1} \circ t_l \circ \dots \circ t_j(\Lambda).$$

* Now what is left to consider are the remaining several cases that are unique to our context and cannot be derived from [33], because the mutation $\mu_j^+$ in $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}$ involves irregular edges.

* Suppose that $j=3$ and $3$ is pendant. We have

$$ T_3 = (P_{3} \xrightarrow{\begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix}} P_1 \oplus P_{2})[1] $$

The tree $\mu_j^+(G)$ is again a Double Edge tree, the levels of whose edges are the same as in $G$. Hence we have  ${(\widetilde{\mu}^{st}_{\mu_{3}^+(\Gamma)})^{-1} = (\mu_{4}^-)^{\varphi(4)} \circ \dots \circ (\mu_m^-)^{\varphi(m)}}$ and

$$\mathcal{T}_{i} = P_{i+1}, \quad 3 \leq i \leq m-1 $$

$$ \mathcal{T}_{1} = P_1,  \mathcal{T}_{2} = P_2$$

$$\mathcal{T}_{m} = (P_{3} \xrightarrow{\begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix}} P_1 \oplus P_{2})[1]. $$

By Lemma <ref>, one can immediately see that
$$\mathcal{H}(\Lambda) = \mathcal{F}_1(\Lambda) = t_m \circ \dots t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3[-2](\Lambda).$$

*  Suppose that $j=3$ and $j$ is not pendant. Let $h$ be the regular edge following $3$ in the cyclic ordering and let $l$ be the smallest integer such that $l > 3$ and $\varphi(l) = 0$ (if it exists). The complex $T_3$ is the cone of a morphism from $P_3$ to ${P_1\oplus P_2 \oplus (P_h \xrightarrow{\beta_h \dots \beta_4} P_3)[1]}$, i.e.
$${T_3 = \mathcal{T}_3' = (P_h \xrightarrow{\begin{pmatrix} \beta_h \dots \beta_4 \alpha_1 \\ \beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_1 \oplus P_2)[1]}.$$

Now we need to apply $(\widetilde{\mu}^{st}_{\mu^+_3(\Gamma)})^{-1}$ to $T$. First we apply the part containing mutations with indices from $h+1$ to $l-1$. For every $i$ such that $h+1 \leq i \leq l-1$ we have $\varphi_{\mu_3^+(G)}(i) = \varphi(i)-1$. Hence $\mathcal{T}_i'$ is the cone of a morphism from $\mathcal{T}_3'$ to $(P_i \xrightarrow{\beta_i \dots \beta_{h+1}} P_h)[2]$ shifted by $-1$, i.e. ${\mathcal{T}_i' = (P_i \xrightarrow{\begin{pmatrix} \beta_i \dots \beta_4 \alpha_1 \\ \beta_i \dots \beta_4 \alpha_2 \end{pmatrix}} P_1\oplus P_2)[1]}$, for every $i$ such that $h+1 \leq i \leq l-1$. Then we apply the part of $(\widetilde{\mu}^{st}_{\mu^+_3(\Gamma)})^{-1}$ containing mutations with indices from $4$ to $h-1$. For every $i$ such that $4 \leq i \leq h-1$ we have $\varphi_{\mu_3^+(G)}(i) = \varphi(i)+1$. It is not difficult to see that $\mathcal{T}_i'$ is the cone of a morphism from $\mathcal{T}_3'$ to $(P_h \xrightarrow{\beta_h \dots \beta_{i+1}} P_i)[1]$, shifted by $-1$. More precisely, for all $i$ such that $ 4 \leq i \leq h-1$ we have
$${\mathcal{T}_i = (P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h \dots \beta_4 \alpha_1 \\ -\beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_h \oplus P_1 \oplus P_{2} \xrightarrow{ \begin{pmatrix} \beta_h \dots \beta_{i+1} & \delta_1 \beta_m \dots \beta_{i+1} & \delta_2 \beta_m \dots \beta_{i+1}  \end{pmatrix}} P_i)[1]}.$$
Finally, since $\varphi_{\mu_3^+(G)}(h) = 1$, we apply $\mu_h^-$. The complex $\mathcal{T}_h'$ is the cone of a morphism from $(P_h \xrightarrow{\begin{pmatrix} \beta_h \dots \beta_4 \alpha_1 \\ \beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_1 \oplus P_2)[1]$ to $(P_h \xrightarrow{\beta_h \dots \beta_4} P_3)[1]$, shifted by $-1$.
To sum up, we have:

$$\mathcal{T}_{i} = \mathcal{T}_{i+l-3}' = P_{i+l-3}, \quad 3 \leq i \leq m-l+3 $$

$$\mathcal{T}_{m-l+4} = \mathcal{T}_3'  = (P_h \xrightarrow{\begin{pmatrix} \beta_h \dots \beta_4 \alpha_1 \\ \beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_1\oplus P_{2})[1]$$

$$\mathcal{T}_{m-l+5} = \mathcal{T}_h' = (P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h\dots\beta_4\alpha_1 \\ -\beta_h \dots \beta_4 \alpha_2 \end{pmatrix}}P_h \oplus P_1 \oplus P_2 \xrightarrow{\begin{pmatrix} \beta_h \dots \beta_4 & \delta_1 \beta_m \dots \beta_4 & \delta_2\beta_m \dots \beta_4 \end{pmatrix}} P_3)[1]$$

$$\mathcal{T}_{i} = \mathcal{T}_{i-m+l-2}' =  $$

$$(P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h\dots\beta_4\alpha_1 \\ -\beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_h \oplus P_1 \oplus P_{2} \xrightarrow{\begin{pmatrix} \beta_h \dots \beta_{i-m+l-1} & \delta_1 \beta_m \dots \beta_{i-m+l-1} & \delta_2\beta_m \dots \beta_{i-m+l-1} \end{pmatrix}} P_{i-m+l-2})[1], $$

$$ m-l+6 \leq i \leq m-l+h+1. $$

$$\mathcal{T}_{i} = \mathcal{T}_{i-m+l-1}' = (P_{i-m+l-1} \xrightarrow{\begin{pmatrix} \beta_{i-m+l-1} \dots \beta_4 \alpha_1 \\ \beta_{i-m+l-1} \dots \beta_4 \alpha_2 \end{pmatrix}} P_1 \oplus P_{2})[1], \quad m-l+h+2 \leq i \leq m.$$

We have
$\mathcal{H}(\Lambda) = t_{m-l+5}^{-1} \circ \dots \circ t_{m-l+h+1}^{-1} \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-3}[-2(l-3)](\Lambda)$ (see Section <ref> for details).

*  The cases $j=1$ and $j=2$ are identical, so we can assume without loss of generality that $j = 1$. Then $T_1 = \mathcal{T}_1' = (P_1 \xrightarrow{\delta_1\beta_m\dots\beta_{h+1}} P_h)[1]$, where $h$ is the edge which follows the double edge in the cyclic ordering. The tree $\mu_3^+(G)$ is now a Triple Tree, with central triangle formed by $1, 2$ and $h$. At least one of its three trees is empty and the other two trees are the subtrees of $G$ formed by edges from $3$ to $h-1$ and from $h+1$ to $m$ respectively. Now we have to pick an ordering of the trees in our configuration $\mu_1^+(G) = (G_1, G_2, G_3)$. We choose the empty tree to play the role of $G_1$. Let $\varphi_i$ be the level function of $G_i$, where $i=1,2,3$. Now apply $(\widetilde{\mu}^{st}_{\mu^+_1(\Gamma)})^{-1}$ to $T$. First we apply the part making the two (possibly) non-empty trees $G_2$ and $G_3$ into the stars. The levels of the edges in $\mu_1^+(\Gamma)$ are as follows

$$\varphi_2(i) = \varphi(i), \quad h+1 \leq i \leq m.$$

$$\varphi_3(i) = \varphi(i) - 1, \quad 3 \leq i \leq h-1. $$

Hence after this step we obtain a complex $\mathcal{T}''$ with $\mathcal{T}_i'' = P_i$ for $3\leq i \leq h-1$, ${\mathcal{T}_i'' = \mathcal{T}_i' = (P_i \xrightarrow{\beta_i \dots \beta_{h+1}} P_h)[1]}$ for $h+1 \leq i \leq m$ and $\mathcal{T}_h'' = \mathcal{T}_h' = P_h$. After that we apply $\mu_{3}^-\circ \dots \circ \mu_{h-1}^-$ and, finally, $\mu_{2}^-$ (see the picture below).

$$\mathcal{T}_1 = \mathcal{T}_h' = P_h$$

$$\mathcal{T}_2 = \mathcal{T}_2' = (P_1 \xrightarrow{\delta_1\beta_m \dots \beta_{h+1}} P_h \xrightarrow{\beta_h \beta_4 \alpha_2} P_2)[1].$$

$$\mathcal{T}_i = \mathcal{T}_i' = (P_h \xrightarrow{\beta_h \dots \beta_{i+1}} P_i), \quad 3 \leq i \leq h-1.$$

$$\mathcal{T}_i = \mathcal{T}_{i+1}' = (P_{i+1} \xrightarrow{\beta_{i+1} \dots \beta_{h+1}} P_h)[1], \quad h \leq i \leq m-1.$$

$$\mathcal{T}_m = \mathcal{T}_1' = (P_1 \xrightarrow{\delta_1\beta_m\dots\beta_{h+1}} P_h)[1].$$

We have $\mathcal{H}(\Lambda) = t_m\dots t_4 t_3 t_1 t_4^{-1} \dots t_h^{-1}[-1](\Lambda)$ (see Section <ref> for details).

§.§ The Triple Tree type

Now let $G$ be of the Triple Tree type and let $G_1, G_2$ and $G_3$ be the three trees forming it. We will need the following easy auxiliary lemma.

([33], Lemma 1) Let $\Phi: D^b(\Lambda) \to D^b(\Lambda)$ be a triangular equivalence. Let $\mathcal{C}$ be a subcategory of $D^b(\Lambda)$ and let $X$ be an object of $D^b(\Lambda)$. Let $f \colon X \to C$ be a minimal left approximation of $X$ with respect to $\mathcal{C}$. Then $\Phi(f) \colon \Phi(X) \to \Phi(C)$ is a minimal left approximation of $\Phi(X)$ with respect to $\Phi(\mathcal{C})$.

* Let $j$ be a regular edge of $G$ belonging to some tree $G_i$. Suppose that all edges following $j$ in the cyclic ordering also belong to the same tree $G_i$. Then, clearly, we can assume without loss of generality that the other two trees $G_d$, $d \neq i$ are empty. Then we have

$$\mu^{st}_G =  \mu^{st}_{G_i} \circ (\mu_{m-1}^+)^{3-i} \circ \dots \circ (\mu_3^+)^{3-i} \circ \mu_2^+. $$

$$ (\widetilde{\mu}^{st}_{\mu^+_j(G)})^{-1} =    \mu_2^- \circ (\mu_{m-1}^-)^{3-i} \circ \dots \circ (\mu_3^-)^{3-i} \circ (\widetilde{\mu}^{st}_{\mu^+_j(G_i)})^{-1}.$$

Applying Lemma <ref> to the  autoequivalence $\Phi$ of $D^b(\Lambda)$ induced by $\mu^{st}_{G_i} \circ \mu_j^+ \circ (\widetilde{\mu}^{st}_{\mu^+_j(G_i)})^{-1}$, one may assume that $i = 3$, hence $\mathcal{T} =    \mu_2^- \circ (\widetilde{\mu}^{st}_{\mu^+_j(G_i)})^{-1} \circ  \mu_j^+ \circ \mu^{st}_{G_i} \circ \mu_2^+(\Lambda)$. Finally, observe that $\mu_2^+$ commutes with $\Phi$, hence we can conclude that the resulting tilting complex $\mathcal{T}$ is the same as in the corresponding Double Edge case, so there is nothing to prove.

* Now let $j$ be a regular edge of $G$ belonging to some tree $G_i$, such that one of the edges following $j$ in the cyclic ordering is a central edge ($1$, $2$ or $m$).

* First suppose $j \in E(G_i)$ is pendant. If in addition $i = 3$ or $i=2$, it is clear that $\mathcal{T} = \Lambda$, since $\mu_{\mu_j^+(G)}^{st} = \mu_j^+ \circ \mu_{G}^{st}$. Thus, it remains to consider the case $i=1$, $j=3$. We have

$$(T^{st}_G)_{3} = Q_3 = \bigg(P_3 \xrightarrow{\begin{pmatrix} -\alpha_1 \\
\alpha_2 \end{pmatrix}} P_1\oplus P_{2} \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m}\bigg) [2]$$

$$(T^{st}_G)_{m} =P_{m}.$$

$$T_3 = \bigg(P_3 \xrightarrow{\begin{pmatrix} -\alpha_1 \\
\alpha_2 \end{pmatrix}} P_1\oplus P_{2} \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m}  \xrightarrow{soc} P_{m}\bigg) [3] .$$

Now we apply $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1}$. Observe that $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1}$ is just $(\mu_{G}^{st})^{-1}$ without $\mu_3^-$. Therefore,

$$\mathcal{T}_{m-1} = \mathcal{T}_3' =  \bigg(P_3 \xrightarrow{\begin{pmatrix} -\alpha_1 \\
\alpha_2 \end{pmatrix}} P_1\oplus P_{2} \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m}  \xrightarrow{soc} P_{m}\bigg) [3].  $$

$$\mathcal{T}_i = \mathcal{T}_{i+1}' = P_{i+1}, \quad 3 \leq i \leq m-2.$$

$$\mathcal{T}_m = \mathcal{T}'_m = P_m, \mathcal{T}_1 = \mathcal{T}_1' = P_1, \mathcal{T}_2 = \mathcal{T}_2' = P_2.$$

We have $\mathcal{H}(\Lambda) = t_m^2 \circ t_{m-1} \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3[-2](\Lambda)$ (see Lemma <ref>).

* Now suppose $j$ is not pendant.

* Let $i=3$, hence $j = m_1+m_2+3$. Let $h$ be the regular edge following $j$ in the cyclic ordering and let $l+1 > m_1+m_2+3$ be the smallest index such that $\varphi_3(l+1) = 0$. If such $l$ does not exist, it means that the subtree `under' the edge $m_1 + m_2+3$ is the whole $G_3$, in which case a few steps in what follows should be skipped. Since this only simplifies the computations, we are not going to consider this case separately.

The graph $\mu_{m_1+m_2+3}^+(G) = (G_1', G_2', G_3')$ is again of the Triple Tree type.

We order the three trees forming it in the following way. Let $G_3'$ be the tree formed by the edges with indices from $l+1$ to $m-1$ and $G_2'$ be the tree containing the edge $3$. As before, we denote the level functions for $G_1'$, $G_2'$ and $G_3'$ by $\varphi_1', \varphi_2'$ and $\varphi_3'$ respectively. We have $T_{m_1+m_2+3} = (P_h \xrightarrow{\beta_h \dots \beta_4 \alpha_1} P_1)[1]$. The levels of edges in the new tree $G_2'$ are as follows:

$$\varphi_2'(m_1+m_2+3) = 0, \varphi_2'(h) = 1 $$

$$\varphi_2'(i) = \varphi_3(i)+1, \quad m_1+m_2+4 \leq i \leq h-1$$

$$\varphi_2'(i) = \varphi_3(i)- 1, \quad h+1 \leq i \leq l $$

$$\varphi_2'(i) = \varphi_2(i), m_1 + 3 \leq i \leq m_1+m_2+2 $$

Now we apply $(\widetilde{\mu}^{st}_{\mu^+_{m_1+m_2+3}(G)})^{-1}$ to $T$ in several steps. First we apply the part mutating the three trees $G_1', G_2'$ and $G_3'$ into stars, obtaining a complex $\mathcal{T}''$. We have
$$\mathcal{T}'' = \mu_h^- \circ (\mu_{h-1}^-)^{\varphi_3(h-1)+1} \circ \dots \circ (\mu_{4}^-)^{\varphi_3(4)+1} \circ (\mu_{l}^-)^{\varphi_3(t)-1} \circ \dots \circ (\mu_{h+1}^-)^{\varphi_3(h+1)-1} \circ (\mu^{st}_{G_3'})^{-1} \circ (\mu^{st}_{G_1'})^{-1} (T). $$

$$\mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_4 \alpha_1} P_1)[1], \quad h+1 \leq i \leq l,$$

$$\mathcal{T}_i'' = \bigg(P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h \dots \beta_4 \alpha_1 \end{pmatrix}} P_h \oplus P_1 \xrightarrow{\begin{pmatrix} \beta_h\dots\beta_{i+1} & \delta_1 \dots \beta_{i+1}  \end{pmatrix}} P_{i}\bigg)[1], \quad  m_1+m_2+4 \leq i \leq h-1, $$

$$\mathcal{T}_i'' = \mathcal{T}_i' = P_i, \quad l+1 \leq i \leq m-1 \text{ and } 3 \leq i \leq m_1+2.$$

The complex $\mathcal{T}_h''$ is the cone of a morphism of ${(P_h \xrightarrow{\beta_h \dots \beta_4 \alpha_1} P_1)[1]}$ to ${(P_h \xrightarrow{\beta_h \dots \beta_{m_1 +m_2+4}} P_{m_1+m_2+3})[1]}$, shifted by $-1$. That is,

$$\mathcal{T}''_h = \bigg(P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h \dots \beta_4 \alpha_1 \end{pmatrix}} P_h \oplus P_1 \xrightarrow{\begin{pmatrix} \beta_h\dots\beta_{m_1+m_2+4} & \delta_1 \dots \beta_{m_1+m_2+4}  \end{pmatrix}} P_{m_1+m_2+3}\bigg)[1]. $$

Applying the rest of $(\widetilde{\mu}^{st}_{\mu^+_{m_1+m_2+3}(G)})^{-1}$, we get

$$\mathcal{T}_{i}= \mathcal{T}_{i-1}' = \bigg(P_h \xrightarrow{soc} P_h \xrightarrow{\beta_h\dots\beta_{i}} P_{i-1}\bigg), \quad m_1+m_2+5 \leq i \leq h, $$

$$\mathcal{T}_{m_1+m_2+3} = P_h, $$

$$\mathcal{T}_{m_1+m_2+4} = \mathcal{T}_h' = (P_h \xrightarrow{soc} P_h \xrightarrow{\beta_4\dots\beta_{m_1+m_2+4}} P_{m_1+m_2+3}) $$

$$\mathcal{T}_i = \mathcal{T}_i' = P_i, \text{ otherwise.}$$

$$\mathcal{H}(\Lambda) = t_{m_1+m_2+4}^{-1} \circ \dots \circ t_{h}^{-1}(\Lambda).$$

The next two cases are very similar to the one just discussed.

* Let $i=2$, thus, $j=m_1+3$. Let $h$ be the regular edge following $j$ in the cyclic ordering and let $l+1 > m_1+3$ be the smallest index such that $\varphi_2(l+1) = 0$. Proceeding in the same way as in the previous case, we have

$$T_{m_1+3} = (P_h \xrightarrow{ \begin{pmatrix} -\beta_h \dots \beta_4 \alpha_1 \\ \beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_1\oplus P_{2} \xrightarrow{ \begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m})[2] = Q_h $$

After we apply the part of $(\widetilde{\mu}^{st}_{\mu^+_{m_1+3}(G)})^{-1}$ to $T$ mutating the new tree $G_1'$ into a star, we obtain a complex $\mathcal{T}''$:

$$\mathcal{T}''_i = Q_i, \quad h+1 \leq i \leq l.$$

$$\mathcal{T}''_i = \bigg(P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h\dots\beta_4\alpha_1 \\ -\beta_h \dots \beta_4\alpha_2 \end{pmatrix}}P_h \oplus P_1 \oplus P_{2} \xrightarrow{ \begin{pmatrix} \beta_h\dots\beta_{i+1} && \delta_1\dots \beta_{i+1} && -\delta_2\dots \beta_{i+1} \\ \beta_h \dots \alpha_1\delta_1  && \delta_1 && -\delta_2 \end{pmatrix}} P_i \oplus P_{m} \bigg)[2], $$

$$\quad m_1 +4 \leq i \leq h-1.$$

$$\mathcal{T}''_h = \bigg(P_h \xrightarrow{\begin{pmatrix} soc \\ -\beta_h\dots\beta_4\alpha_1\\ -\beta_h\dots\beta_4\alpha_2 \end{pmatrix}}P_h \oplus P_1 \oplus P_{2} \xrightarrow{ \begin{pmatrix}  \beta_h\dots\beta_{m_1+4} && \delta_1\dots \beta_{m_1+4} && -\delta_2\dots \beta_{m_1+4} \\ \beta_h \dots \alpha_1\delta_1 && \delta_1 && -\delta_2 \end{pmatrix}} P_{m_1+3} \oplus P_{m} \bigg)[2]. $$

Applying the rest of $(\widetilde{\mu}^{st}_{\mu^+_{m_1+3}(G)})^{-1}$, we get

$$\mathcal{T}_{m_1+3} = \mathcal{T}_{m_1+3}' = P_h, $$

$$\mathcal{T}_i = \mathcal{T}_{i-1}' = (P_h \xrightarrow{soc} P_h \xrightarrow{\beta_4 \dots \beta_i} P_{i-1}), \quad m_1+5 \leq i \leq h,$$

$$\mathcal{T}_{m_1+4}= \mathcal{T}_h' = (P_h \xrightarrow{soc} P_h \xrightarrow{\beta_4 \dots \beta_{m_1+4}} P_{m_1+3}),$$

$$\mathcal{T}_i = \mathcal{T}_i' = P_i \text{ otherwise.} $$

We have $\mathcal{H}(\Lambda) = t_{m_1+4}^{-1} \circ \dots \circ t_h^{-1}(\Lambda)$.

*  Let $i=1$, thus, $j=3$. The complex $T_3 = \mathcal{T}_3'$ is the cone of a morphism from $Q_3$ to $(P_h \xrightarrow{\beta_h \dots \beta_4} P_3)[3]\oplus P_{m}$, namely,

$$ T_3 = \mathcal{T}_3' = (P_h \xrightarrow{\begin{pmatrix} -\beta_h \dots \beta_4\alpha_1 \\ \beta_h \dots \beta_4\alpha_2 \end{pmatrix}} P_1\oplus P_2 \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m} \xrightarrow{soc} P_{m})[3]. $$

Then proceeding as in the previous two cases, we get:

$$\mathcal{T}_i = \mathcal{T}_{i+l-2}' = P_{i+l-2}, \quad 3 \leq i \leq m-l+1. $$

$$\mathcal{T}_{m-l+2} = \mathcal{T}_3' = (P_h \xrightarrow{\begin{pmatrix} -\beta_h \dots \beta_4\alpha_1 \\ \beta_h \dots \beta_4\alpha_2 \end{pmatrix}} P_1\oplus P_2 \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m} \xrightarrow{soc} P_{m})[3]. $$

$$\mathcal{T}_{m-l+3} = \mathcal{T}_h' = (P_h \xrightarrow{\begin{pmatrix} -soc \\ -\beta_h \dots_4\alpha_1 \\ \beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_h \oplus P_1 \oplus P_{2} \xrightarrow{\begin{pmatrix} -\beta_h \dots \beta_4 & -\delta_1 \beta_m \dots \beta_4 & -\delta_2 \beta_m \dots \beta_4 \\ 0 & \delta_1 & \delta_2 \end{pmatrix}} P_3 \oplus P_{m} \xrightarrow{soc} P_{m})[3].$$

$$\mathcal{T}_i= \mathcal{T}_{i-m+l}' =  $$
$\kern-10.5em \small{(P_h \xrightarrow{\begin{pmatrix} -soc \\ -\beta_h \dots_4\alpha_1 \\ \beta_h \dots \beta_4 \alpha_2 \end{pmatrix}} P_h\oplus P_1 \oplus P_{2} \xrightarrow{\begin{pmatrix} -\beta_h \dots \beta_{i-m+l+1} & -\delta_1 \beta_m \dots \beta_{i-m+l+1} & -\delta_2 \beta_m \dots \beta_{i-m+l+1} \\ 0 & \delta_1 & \delta_2 \end{pmatrix}} P_{i-m+l} \oplus P_{m} \xrightarrow{soc} P_{m}) [3],} $

$$m-l+4 \leq i \leq m-l+h-1. $$

$$\mathcal{T}_i = \mathcal{T}_{i-m+l+1}' = (P_{i-m+l+1} \xrightarrow{\begin{pmatrix} -\beta_{i-m+l+1} \dots \beta_4\alpha_1 \\ \beta_{i-m+l+1} \dots \beta_4\alpha_2  \end{pmatrix}} P_1\oplus P_{2} \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_{m} \xrightarrow{soc} P_{m})[3], \quad m-l+h \leq i \leq m-1 $$

$$\mathcal{T}_i = \mathcal{T}_i' = P_i, \text{ otherwise.}$$

$$\mathcal{H}(\Lambda) = t_{m-l+3}^{-1} \circ \dots \circ t_{m-l+h-1}^{-1} \circ t_m \circ  \dots  \circ t_{m-l+3} \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-2}[-2(l-2)](\Lambda).$$

See Section <ref> for details.

* Now let $j$ be one of the central edges $1,2$ or $m$. First we suppose that the tree following $j$ in the cyclic ordering is empty. Equivalently, $\mu_j^+(G)$ is a Double Edge tree.

*   Let $j = 1$, $G_2 = \varnothing$.

Let $\varphi$ be the level function for the Double Edge tree $\mu_1^+(G)$.  The levels of the edges in $\mu_{1}^+(G)$  are as follows

$$\varphi(i) = \varphi_1(i), \quad 3 \leq i \leq m_1+2, $$

$$ \varphi(i) = \varphi_3(i)+1, \quad m_1 + 3 \leq i \leq m-1,$$

$$\varphi(m) = 0$$

We have
$(\widetilde{\mu}_{\mu_1^+(G)}^{st})^{-1} =  (\mu_{m_1+3}^-)^{\varphi(m_1+3)} \circ \dots \circ (\mu_{m-1}^-)^{\varphi(m-1)} \circ (\mu_3^-)^{\varphi(3)}\circ \dots \circ (\mu_{m_1+2}^-)^{\varphi(m_1+2)}$, thus,

$$\mathcal{T}_3 = \mathcal{T}_{m}' = P_{m} $$

$$\mathcal{T}_i = \mathcal{T}_{i+m_1-1}' = (P_{m} \xrightarrow{\beta_m \dots \beta_{i+m_1}} P_{i+m_1-1}), \quad 4 \leq i \leq m-m_1. $$

$$\mathcal{T}_i = \mathcal{T}_{i-m+m_1+2}' =  Q_{i-m+m_1+2} =  $$

$$\kern-4.5em (P_{i-m+m_1+2} \xrightarrow{\begin{pmatrix} -\beta_{i-m+m_1+2} \dots \beta_4\alpha_1 \\ \beta_{i-m+m_1+2} \dots \beta_4\alpha_2 \end{pmatrix}} P_1 \oplus P_2 \xrightarrow{\begin{pmatrix}\delta_1 & \delta_2 \end{pmatrix}} P_m)[2], \quad m-m_1+1 \leq i \leq m.$$

$$\mathcal{T}_2 = \mathcal{T}_{2}' = (P_{2} \xrightarrow{\delta_2 } P_{m})[1] $$

$$\mathcal{T}_1 = \mathcal{T}_1' = (P_1 \xrightarrow{\delta_1} P_{m})[1] $$

$$\mathcal{H}(\Lambda) = t_3^{-1} \circ t_4^{-1} \circ \dots t_{m-m_1}^{-1} \circ (t_m \circ \dots \circ t_4 t_3 \circ t_1 \circ t_2 \circ t_3)^{m_1}[-2m_1+1](\Lambda).$$

The next two cases are very similar to the one just examined, so we are going to discuss them briefly.

*  Let  $j = 2, G_1 = \varnothing$.
We have $T_2 = (P_{2} \xrightarrow{\delta_2} P_{m} \xrightarrow{\beta_m \dots \beta_4 \alpha_1} P_{1})[2].$ Now let $\varphi$ be the level function of $\mu_2^+(G)$. Then
${\varphi(i) = \varphi_3(i), m_2 +3 \leq i \leq m-1}$, ${ \varphi(i) = \varphi_2(i)+1,  3 \leq i \leq m_2+2}$, ${\varphi(1) = 0}$.

$$\mathcal{T}_1 = \mathcal{T}_{m}' = P_{m}.$$

$$\mathcal{T}_2 = \mathcal{T}_{2}' = (P_{2} \xrightarrow{\delta_2} P_{m} \xrightarrow{\beta_m \dots \beta_4 \alpha_1} P_{1})[2].$$

$$\mathcal{T}_3 = \mathcal{T}_{1}' = P_{1}. $$

$$\mathcal{T}_i = \mathcal{T}_{i-1}' = P_{i-1}, \quad 4 \leq i \leq m. $$

We have $\mathcal{H}(\Lambda) = t_1^{-1}\circ t_3^{-1}\circ t_1^{-1}\circ t_4^{-1} \circ \dots \circ t_m^{-1}[2](\Lambda)$ (see Section <ref> for details).

* Let $j=m, G_3 = \varnothing$.  We have $\mathcal{T}_m = P_2[1]$. Again, let $\varphi$ be the level function of $\mu_m^+(G)$. Then $\varphi(i) = \varphi_2(i), m_1+3 \leq i \leq m-1 $, $\varphi(i) = \varphi_1(i)+1,  3\leq i \leq m_1+2$, $\varphi(2) = 0$.

$$\mathcal{T}_i = \mathcal{T}_{i-1}' = L_{i-1} = (P_{i-1} \xrightarrow{\beta_{i-1} \dots \beta_4 \alpha_1} P_1)[1],  4 \leq i \leq m $$

$$\mathcal{T}_3 = \mathcal{T}_{2}' = (P_{2} \xrightarrow{\delta_2} P_{m})[1] $$

$$ \mathcal{T}_2 = \mathcal{T}_{m}' =  P_2[1], \mathcal{T}_1 = \mathcal{T}_{1}' = P_1$$

We have $\mathcal{H}(\Lambda) = t_2 \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{m-3}[-2(m-3)](\Lambda)$ (see Lemma <ref>).

* Finally, it remains to consider the cases when $j$ is a central edge ($1, 2$ or $m$) and the tree following it in the cyclic ordering is not empty. In other words, $j$ is central and $\mu_j^+(G)$ is again of the Triple Tree type.

*  Let $j =1, G_2 \neq \varnothing$. Let $h$ be the edge of $G_2$ which follows $1$ in the cyclic ordering.

$$(T^{st}_G)_1 = P_1, (T^{st}_G)_{m} = P_{m}   $$

$$(T^{st}_G)_{h} = (P_{h} \xrightarrow{\beta_h \dots \beta_4\alpha_1} P_1)[1] $$

$$T_1 = \bigg(P_{h} \oplus P_1 \xrightarrow{\begin{pmatrix} \beta_{h}\dots \beta_4 \alpha_1 & id \\ 0 & \delta_1  \end{pmatrix}} P_1 \oplus P_{m} \bigg)[1] \sim \bigg(P_{h} \xrightarrow{\beta_h \dots \beta_4 \alpha_1 \delta_1} P_{m} \bigg)[1]$$

We denote the three trees forming $\mu_1^+(G)$ by $G_1', G_2'$ and $G_3'$, as pictured above. Namely, the tree $G_3'$ has edges with labels $3, \dots, m_1+2$ and $m_1+m_2+3, \dots, m$, the tree $G_2'$ has edges with labels $h+1, \dots, m_1+m_2+2$, and the tree $G_1'$ has edges with labels $m_1 + 3, \dots, h-1$. Edges $1, 2$ and $h$ form the central triangle of $\mu_1^+(G)$. Denote by $\varphi_1', \varphi_2'$ and $\varphi_3'$ the level functions of $G_1'$, $G_2'$ and $G_3'$ respectively. Then

$$\varphi_3'(i) = \varphi_1(i), \quad 3 \leq i \leq m_1+2 $$

$$\varphi_3'(m) = 0 $$

$$\varphi_3'(i) = \varphi_3(i) + 1, \quad m_1+m_2+3 \leq i \leq m-1  $$

$$ \varphi_1'(i) = \varphi_2(i), \quad m_1+3 \leq i \leq h-1 \quad (i \in G_1') $$

$$ \varphi_2'(i) = \varphi_2(i)-1, \quad h+1 \leq i \leq m_1 +m_2 + 2 \quad (i \in G_2')  $$

As always, first we apply to $T$ the part of $(\widetilde{\mu}_{\mu_1^+(\Gamma)}^{st})^{-1}$ mutating the three trees $G'_1$, $G'_2$, and $G'_3$ into stars. Denote the resulting complex by $\mathcal{T}'' = (\widetilde{\mu}_{G_3'}^{st})^{-1} \circ(\widetilde{\mu}_{G_2'}^{st})^{-1} \circ (\widetilde{\mu}_{G_1'}^{st})^{-1}(T) $.
We have

$$\mathcal{T}_i'' = L_i, \quad i \in G_1', \text{ i.e. } m_1 + 3 \leq i \leq h-1. $$

$$ \mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_{h+1}} P_h)[2], \quad i \in G_2', \text{ i.e. } h+1 \leq i \leq m_1 + m_2 +2. $$

$$ \mathcal{T}_i'' = \mathcal{T}_i' = Q_i, \quad 3 \leq i \leq m_1+2.$$

$$ \mathcal{T}_i'' = \mathcal{T}_i' =  (P_{m} \xrightarrow{\beta_m \dots \beta_{i+1}} P_i), \quad m_1+m_2+3 \leq i \leq m-1.$$

$$ \mathcal{T}_{m}'' = \mathcal{T}_m' = P_{m}.$$

$$\mathcal{T}_1'' = T_1, \mathcal{T}_{2}'' = T_{2}, \mathcal{T}_h'' = T_h. $$

Next we apply $\big((\mu_{m_1+3}^-)^2\circ \dots\circ (\mu_{h-1}^-)^2\big)\circ (\mu_{h+1}^- \circ \dots \circ \mu_{m_1+m_2+2}^-)$ and get

$$\mathcal{T}_i' = (P_i \xrightarrow{\beta_i \dots \beta_4 \alpha_1 \delta_1} P_{m})[1], \quad h+1 \leq i \leq m_1+m_2+2. $$

$$\mathcal{T}_i' = (P_{m} \xrightarrow{\beta_m \dots \beta_{i+1}} P_{i}), \quad m_1+3 \leq i \leq h-1.$$

Finally, we apply $\mu_h^-$.

$$\mathcal{T}_1 = \mathcal{T}_h' = (P_2 \xrightarrow{ \begin{pmatrix} \delta_2\beta_m \dots \beta_{h+1} \\ -\delta_2 \end{pmatrix}} P_h \oplus P_m \xrightarrow{ \begin{pmatrix} \beta_h \dots \beta_4 \alpha_1 & \beta_m \dots \beta_4 \alpha_1 \end{pmatrix}} P_1)[1].$$

$$ \mathcal{T}_2 = \mathcal{T}_1' = (P_h \xrightarrow{\beta_h \dots \beta_4 \alpha_1 \delta_1} P_m)[2].$$

$$\mathcal{T}_i = \mathcal{T}_{i+m_1}' = (P_m \xrightarrow{\beta_m \dots \beta_{i+m_1+1}} P_{i+m_1}), \quad 3 \leq i \leq h-1-m_1.$$

$$\mathcal{T}_i = \mathcal{T}_{i+m_1+1}' = (P_{i+m_1+1} \xrightarrow{\beta_{i+m_1+1} \dots \beta_4 \alpha_1 \delta_1} P_m)[1], \quad h-m_1 \leq i \leq m_2+1.$$

$$\mathcal{T}_{m_2+2} = \mathcal{T}_m' = P_m $$

$$\mathcal{T}_i = \mathcal{T}_{i+m_1}' = (P_m \xrightarrow{\beta_m \dots \beta_{i+m_1+1}} P_{i+m_1}), \quad m_2 +3 \leq i \leq m-m_1-1.$$

$$ \mathcal{T}_i = \mathcal{T}_{i-m+m_1+3}' =  Q_{i-m+m_1+3} = $$

$$ (P_{i-m+m_1+3} \xrightarrow{\begin{pmatrix} -\beta_{i-m+m_1+3} \dots \beta_4 \alpha_1 \\ \beta_{i-m+m_1+3} \dots \beta_4 \alpha_2\end{pmatrix}} P_1 \oplus P_2 \xrightarrow{\begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_m)[2], \quad m-m_1 \leq i \leq m-1.$$

$$\mathcal{T}_m = \mathcal{T}_{2}' = (P_2 \xrightarrow{\delta_2} P_m)[1]$$

$$\mathcal{H}(\Lambda) = t_2t_{m_2+2}\dots t_4t_3t_2^{-1}t_4^{-1} \dots t_{h-m_1}^{-1} t_m \dots t_4t_3t_2t_4^{-1}\dots t_{m-m_1}^{-1}  (t_m \dots t_4 t_3t_1t_2t_3)^{m_1}[-2m_1-1](\Lambda)$$

See Section <ref> for details. The remaining two cases are going to be similar to the one we have just considered.

*  Let $j=2, G_1 \neq \varnothing$. Let $d$ be the edge of $G_1$, following $2$ in the cyclic ordering.

First we calculate $T_{2} = \mathcal{T}_{2}'$. Since  $(T^{st}_{G})_{2} = (P_{2} \xrightarrow{\delta_2} P_{m})[1]$,  $(T^{st}_{G})_d = Q_d$ and $(T^{st}_{G})_1 =  P_1$, we have $T_{2} = \mathcal{T}_{2}' = (P_d \xrightarrow{\beta_d \dots \beta_4\alpha_1} P_1 \xrightarrow{soc} P_1)[2]$.
The levels of edges in the new trees forming $\mu_{2}^+(G)$ are as follows.

$$\varphi_3'(i) = \varphi_3(i), \quad m_1+m_2+3 \leq i \leq m-1. $$

$$\varphi_3'(1) = 0 $$

$$\varphi_3'(i) = \varphi_2(i)+1, \quad m_1 + 3 \leq i \leq m_1+m_2+2. $$

$$\varphi_2'(i) = \varphi_1(i) - 1, \quad d+1\leq i \leq m_1+2.$$

$$\varphi_1'(i) = \varphi_1(i), \quad 3\leq i\leq d-1.$$

First we apply to $T$ the part of $(\widetilde{\mu}_{\mu_2^+(\Gamma)}^{st})^{-1}$ mutating the three trees $G'_1$, $G'_2$, and $G'_3$ into stars. As in the previous case, denote the corresponding complex by $\mathcal{T}''$. We immediately have

$$\mathcal{T}_i'' = \mathcal{T}_i' = P_i, \quad m_1+m_2+3 \leq i \leq m-1. $$

$$\mathcal{T}_i'' = Q_i, \quad 3 \leq i \leq d-1. $$

$$\mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_{d+1}} P_d)[3], \quad d+1 \leq i \leq m_1+2.$$

Now we compute $\mathcal{T}_i''$ for $m_1+3 \leq i \leq m_1 + m_2+2$. The complex $\mathcal{T}_{i}''$ is the cone of the morphism $(0, id)$ from $T_1 = P_1$ to $(P_i \xrightarrow{\beta_i \dots \beta_4\alpha_1} P_1)[1]$, shifted by $-1$, hence

$$\mathcal{T}_i'' = P_i, \quad m_1+3 \leq i \leq m_1+m_2+2 $$

Now we shall apply $(\mu_{d+1}^-) \circ \dots \circ (\mu_{m_1+2}^-)$. The complex
$\mathcal{T}_i'$ is the cone of the morphism $(0, id, 0, 0)$ from ${\mathcal{T}_{2}'' = (P_d \xrightarrow{\beta_d \dots \beta_4\alpha_1} P_1 \xrightarrow{soc} P_1)[2]}$ to ${\mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_{d+1}} P_d)[3]}$ for ${d+1 \leq i \leq m_1+2}$. Thus,

$$\mathcal{T}_i' = (P_i \xrightarrow{\beta_i \dots \beta_4 \alpha_1} P_1 \xrightarrow{soc} P_1)[2], \quad d+1 \leq i \leq m_1+2. $$

Next apply $(\mu_3^-)^2 \circ \dots \circ (\mu_{d-1}^-)^2$. Let $3 \leq i \leq d-1$. After one  mutation $\mu_i^{-}$ in this sequence, the summand labeled $i$ will be isomorphic to the cone of the morphism $(0,id,id)$ from $Q_d$ to $Q_i$, shifted by $-1$. This is just $(P_d \xrightarrow{\beta_d \dots \beta_{i+1}} P_i)[2]$. Applying $\mu_i^-$ again, we see that $\mathcal{T}_i'$ is the cone of the morphism $(soc,\delta_1\beta_m\dots\beta_{i+1},0)$ from $(P_d \xrightarrow{\beta_d \dots \beta_{4}\alpha_1} P_1 \xrightarrow{soc} P_1)[2]$ to $(P_d \xrightarrow{\beta_d \dots \beta_{i+1}} P_i)[2]$, shifted by $-1$.

$$\mathcal{T}_i' = (P_d \xrightarrow{\begin{pmatrix} soc \\ -\beta_d \dots \beta_{4}\alpha_1 \end{pmatrix}} P_d \oplus P_1 \xrightarrow{\begin{pmatrix} 0 & -soc  \\ \beta_d \dots \beta_{i+1} &  \delta_1\beta_m \dots \beta_{i+1}  \end{pmatrix}} P_1\oplus P_i)[2], \quad 3 \leq i \leq d-1. $$

Finally, apply $\mu_d^-$. Recall that $T_{m}=P_{m}$, $T_d = Q_d$. Hence
$${\mathcal{T}_d' = (P_d \xrightarrow{\begin{pmatrix} -\beta_d \dots \beta_4 \alpha_1 \\ \beta_d \dots \beta_4 \alpha_1 \end{pmatrix}} P_2\oplus P_1)[1]}.$$
Summarising what we obtained above:

$$\mathcal{T}_1 = \mathcal{T}_2' = (P_d \xrightarrow{\beta_d \dots \beta_4 \alpha_1} P_1 \xrightarrow{soc} P_1)[2].$$

$$\mathcal{T}_2 = \mathcal{T}_d' = (P_d \xrightarrow{\begin{pmatrix} -\beta_d \dots \beta_4 \alpha_1 \\ \beta_d \dots \beta_4 \alpha_1 \end{pmatrix}} P_2\oplus P_1)[1].$$

$$\mathcal{T}_i = \mathcal{T}_i' = (P_d \xrightarrow{\begin{pmatrix} soc \\ -\beta_d \dots \beta_{4}\alpha_1 \end{pmatrix}} P_d \oplus P_1 \xrightarrow{\begin{pmatrix} 0 & -soc  \\ \beta_d \dots \beta_{i+1} &  \delta_1\beta_m \dots \beta_{i+1}  \end{pmatrix}} P_1\oplus P_i)[2], \quad 3 \leq i \leq d-1.$$

$$\mathcal{T}_i = \mathcal{T}_{i+1}' = (P_{i+1} \xrightarrow{\beta_{i+1} \dots \beta_4 \alpha_1} P_1 \xrightarrow{soc} P_1)[2], \quad d \leq i \leq m_1+1.$$

$$\mathcal{T}_{m_1+2} = \mathcal{T}_1' = P_1.$$

$$\mathcal{T}_i = \mathcal{T}_i' = P_i, \quad m_1 +3 \leq i \leq m$$

We have $\mathcal{H}(\Lambda) = t_1\circ t_{m_1+2} \circ \dots \circ t_4 \circ t_3 \circ t_1^{-1} \circ t_4^{-1}\circ \dots \circ t_d^{-1}(\Lambda)$ (see Section <ref> for details).

*   Let $j=m, G_3 \neq \varnothing$.  Let $l$ be the edge of $G_1$, following $m$ in the cyclic ordering.

We have $\mathcal{T}_{m}' = T_{m} = (P_{2} \xrightarrow{\delta_2 \beta_m \dots \beta_{l+1}} P_l)[1]$. The levels of edges in the new trees are as follows.

$$\varphi_2'(i) = \varphi_3(i)-1, \quad l+1 \leq i \leq m-1.$$

$$\varphi_1'(i) = \varphi_3(i), \quad m_1+m_2+3 \leq i \leq l-1. $$

$$\varphi_3'(i) = \varphi_2(i), \quad m_1 + 3 \leq i \leq m_1+m_2+2. $$

$$\varphi'_3(i) = \varphi_1(i)+1, \quad 3 \leq i \leq m_1+2.$$

$$\varphi_3'(2) = 0.$$

Applying to $T$ the part of $(\widetilde{\mu}_{\mu_m^+(\Gamma)}^{st})^{-1}$ which mutates the three trees $G'_1$, $G'_2$, and $G'_3$ into stars, we get

$$\mathcal{T}_i'' = P_i, \quad m_1+m_2+3 \leq i \leq l-1.$$

$$\mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_{l+1}} P_l)[1], \quad l+1 \leq i \leq m-1. $$

$$\mathcal{T}_i' = \mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_4 \alpha_1} P_1)[1], \quad m_1 + 3 \leq i \leq m_1+m_2+2. $$

$$\mathcal{T}_i' = \mathcal{T}_i'' = (P_i \xrightarrow{\beta_i \dots \beta_4 \alpha_1} P_1)[1], \quad 3 \leq i \leq m_1+2.$$

$$\mathcal{T}_2' = \mathcal{T}_{2}'' = (P_{2} \xrightarrow{\delta_2} P_{m})[1].$$

Then, applying the rest of $(\widetilde{\mu}_{\mu_m^+(\Gamma)}^{st})^{-1}$, i.e. ${\mu_m^-\circ(\mu_{l+1}^-)^2 \circ \dots \circ (\mu_{m-1}^-)^2 \circ \mu_{2}^{-}\circ \mu_{3}^- \circ \dots \circ \mu_{m_1+m_2+2}^-}$, we get:

$$\mathcal{T}_1 = \mathcal{T}_1' = P_1. $$

$$\mathcal{T}_2 = \mathcal{T}_m' = P_2.$$

$$\mathcal{T}_i = \mathcal{T}_{i+l-2}' = (P_1\oplus P_2 \xrightarrow{ \begin{pmatrix} \delta_1\beta_m \dots \beta_{i+l-1} \\ \delta_1\beta_m \dots \beta_{i+l-1} \end{pmatrix}} P_{i+l-2}), \quad 3 \leq i \leq m-l+1.$$

$$\mathcal{T}_{m-l+2} = \mathcal{T}_2' = (P_1\oplus P_2 \xrightarrow{ \begin{pmatrix} \delta_1 & \delta_2 \end{pmatrix}} P_m).$$

$$\mathcal{T}_i = \mathcal{T}_{i+l-2}' = P_{i-m+l}, \quad m-l+3 \leq i \leq m.$$

We have $\mathcal{H}(\Lambda) = t_2 \circ t_1 \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-2}[-2l+2](\Lambda)$ (see Section <ref> for details).

§.§ Finishing the proof

Recall that, in the notation of Section <ref>, to finish the proof of Theorem <ref>, it remains to express the complexes $(\widetilde{\mu}_{G_1}^{st})^{-1} \circ \mu_{j_1}^+(\Lambda)$, $\mu_{j_k}^+ \circ \widetilde{\mu}_{G_{k-1}}^{st}(\Lambda^{k-1})$ as images of $\Lambda$ under series of spherical twists $t_i$ (modulo $\Pic(\Lambda)$ and the shift)

* Fix any $j_1$. Let $\mathcal{H}^1$ denote the autoequivalence of $D^b(\Lambda)$  induced by $(\widetilde{\mu}_{G_1}^{st})^{-1} \circ \mu_{j_1}^+$ and let $\mathcal{T}^1$ be the tilting complex $\mathcal{H}^1(\Lambda)$. We denote $\mathcal{H}^1(P_i)$ by $\mathcal{T}_i^1$. It is easy to see that if $j_1 \neq 3$, then $\mathcal{T}^1 = \Lambda$. If $j_1 = 3$, then the sequence $(\widetilde{\mu}_{G_1}^{st})^{-1}$ is empty, so

$$ \mathcal{T}_i^1 = P_{i+1}, \quad 3 \leq i \leq m-1. $$

$$ \mathcal{T}_m^1 = (P_{3} \xrightarrow{\begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix}} P_1 \oplus P_{2})[1]. $$

By Lemma <ref> we have $\mathcal{H}^1(\Lambda) = t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3[-2](\Lambda)$.

* Next we need to deal with complexes of the form $\mu_{j_k}^+ \circ \widetilde{\mu}_{G_{k-1}}^{st}(\Lambda^{k-1})$. We can assume without loss of generality that the labeling of summands of $\Lambda^{k-1}$ is standard and write simply $\mu_{j_k}^+ \circ \mu_{G_{k-1}}^{st}(\Lambda)$. Let $\mathcal{H}^k$ autoequivalence of $D^b(\Lambda)$ induced by $\mu_{j_k}^+ \circ \mu_{G_{k-1}}^{st}$. The fact that $\End_{D^b(\Lambda)}(\mu_{j_k}^+ \circ \mu_{G_{k-1}}^{st}(\Lambda)) \cong \Lambda$ leaves us with few options. Namely, there are two possibilities.

* $\mu_{G_{k-1}}^{st} = \mu_i^+$, where $4 \leq i \leq m$. Then $j_k = i$ and $\mathcal{H}^k(\Lambda) = (\mu_i^+)^2(\Lambda) = t_i(\Lambda)$.

* $\mu_{G_{k-1}}^{st} = \mu_2^+$. Then $j_k = 2$ and we have $ = \mu_{j_k}^+ \circ \mu_{G_{k-1}}^{st} = (\mu_2^+)^2$, hence $\mathcal{H}^k$ acts on indecomposable summands in the following way.

$$\mathcal{H}^k(P_1) = P_m, \mathcal{H}^k(P_2) = (P_2 \xrightarrow{\delta_2} P_m \xrightarrow{\beta_m\dots\beta_4\alpha_1} P_1)[2] $$

$$\mathcal{H}^k(P_3) = P_1$$

$$\mathcal{H}^k(P_i) = P_{i-1}, \quad 4 \leq i \leq m$$

We have already seen this autoequivalence in case <ref>. More precisely, ${\mathcal{H}^k(\Lambda) = t_1^{-1} \circ t_3^{-1} \circ t_1^{-1} \circ t_4^{-1} \dots t_m^{-1}[2](\Lambda)}$.

§ APPENDIX

In Section <ref> we computed all tilting complexes $\mathcal{H}(\Lambda)$, $\mathcal{H}^1(\Lambda)$, $\mathcal{H}^k(\Lambda)$ of the form $(\widetilde{\mu}_{\mu_j^+(G)}^{st})^{-1} \circ \mu_j^+ \circ \mu_G^{st}(\Lambda)$, $(\widetilde{\mu}_{G}^{st})^{-1} \circ \mu_{j}^+(\Lambda)$, and $\mu_{j}^+ \circ \mu_{G_{k}}^{st}(\Lambda)$ respectively. In each case we stated how such a complex is expressed as an image of $\Lambda$ under a composition of spherical twists $\{t_i\}_{i}^m$, $\Pic(\Lambda)$ and shifts. Many expressions are straightforward to obtain, but some require explanations. In the Appendix we sketch step-by-step computations that justify these difficult cases.

The computations will be displayed as sequences of modified Brauer trees whose edges are labeled with indecomposable direct summands of tilting complexes obtained at every step. We already used this technique in the proof of Lemma <ref>, where complexes of the form $t_i^{\pm1}(\Lambda)$ were expressed as images of $\Lambda$ under compositions of mutations and shifts. However, now we will work with bigger “chunks” of mutations and all modified Brauer trees that we show will be Double Edge stars. In each of the cases we start with the modified Brauer tree of $\Lambda$ whose edges are labeled with indecomposable projective $\Lambda-$modules. Now let $\mathcal{C} = \bigoplus_{i=1}^m C_i$ be a tilting complex over with $\End_{D^b(\Lambda)}(\mathcal{C}) \cong \Lambda$ and suppose we have the Double Edge star whose edges are labeled with indecomoposable complexes $C_i$ (the double edge is labeled with two complexes). Suppose that we want to compute $t_i^{\pm1}(\mathcal{C})$. By Lemma <ref> we can express $t_i^{\pm1}$ as a composition of mutations and shifts. In turn, a sequence of mutations yields a sequence of modified Brauer trees whose edges are labeled with direct summands of tilting complexes, as explained in the proof of Lemma <ref>. Since $t_i^{\pm 1}$ is an autoequivalence of $\D^b(\Lambda)$, the last modified Brauer tree in this sequence is again the Double Edge star. In the computations below we skip all intermediate steps where the resulting modified Brauer tree is not the star and only show the trees corresponding to compositions of twists.

When there are several consequent twists in the sequence acting similarly, we often group them into one step. When there are several consequent edges in a Double Edge star which should be labeled with complexes that differ only in one index (e.g. $P_i \to P_1$ for several $i$'s), we draw a brace around these edges and write the complex only once below the brace. To know which summand of the tilting complex is assigned to a particular edge one needs to take the integer written on this edge and substitute it in place of $i$ in the complex written below the brace. For instance, in the first picture below on the left (case 2b) we have edges labeled with $P_l$, $P_{l+1}, \dots, P_m$ and edges labeled with $({P_3 \xrightarrow{} \underline{P_1 \oplus P_2}), \dots, (P_{l-1} \xrightarrow{} \underline{P_1 \oplus P_2})}$. See also the discussion after Lemma <ref>. To make the pictures more compact we omit the differentials in our complexes. Nevertheless, they can always be easily recovered from the provided data. The resulting tilting complexes with differentials are listed in Section <ref>. All computations here are modulo $\Pic(\Lambda)$ (see Theorem <ref>).

Case <ref>. $\mathcal{H}(\Lambda) = t_{m-l+5}^{-1} \circ \dots \circ t_{m-l+h+1}^{-1} \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-3}[-2(l-3)](\Lambda)$

*-7em Step 1.$(t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-3}[-2(l-3)]$, by Lemma <ref>        Step 2. $t_{m-l+5}^{-1} \circ \dots \circ t_{m-l+h+1}^{-1}$

Case <ref>.

$\mathcal{H}(\Lambda) = t_m \circ \dots t_4 \circ t_3 \circ t_1 \circ t_4^{-1}  \circ \dots \circ t_h^{-1}[-1](\Lambda)$

*-4em Step 1.$t_4^{-1} \circ \dots \circ t_h^{-1}$             Step 2. $ t_1$              Step 3. $ t_3[-1]$                 Step 4. $t_m \circ \dots \circ t_4$

Case <ref>
$\mathcal{H}(\Lambda) = t_{m-l+3}^{-1} \circ \dots \circ t_{m-l+h-1}^{-1} \circ t_m \circ  \dots  \circ t_{m-l+3} \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-2}[-2(l-2)](\Lambda)$

*-4em Step 1. $(t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-2}[-2(l-2)]$, by Lemma <ref>     Step 2. $t_m \circ  \dots  \circ t_{m-l+3}$       Step 3. $t_{m-l+3}^{-1} \circ \dots \circ t_{m-l+h-1}^{-1}$

Case <ref> $\mathcal{H}(\Lambda) = t_3^{-1} \circ t_4^{-1} \circ \dots t_{m-m_1}^{-1} \circ (t_m \circ \dots \circ t_4 t_3 \circ t_1 \circ t_2 \circ t_3)^{m_1}[-2m_1+1](\Lambda)$.

*-6em  Step 1.$(t_m \circ \dots \circ t_4 t_3 \circ t_1 \circ t_2 \circ t_3)^{m_1}[-2m_1]$, by Lemma <ref>            Step 2. $ t_4^{-1} \circ \dots t_{m-m_1}^{-1}$            Step 3. $ t_3^{-1}[1]$

Case <ref>
$\mathcal{H}(\Lambda) = t_1^{-1}\circ t_3^{-1}\circ t_1^{-1}\circ t_4^{-1} \circ \dots \circ t_m^{-1}[2](\Lambda)$

*-5em Step 1. $ t_4^{-1} \circ \dots \circ t_m^{-1}$            Step 2. $ t_1^{-1}$            Step 3. $ t_3^{-1}[1]$               Step 4. $t_1^{-1}[1]$

Case <ref>
$\mathcal{H}(\Lambda) = t_2t_{m_2+2}\dots t_4t_3t_2^{-1}t_4^{-1} \dots t_{h-m_1}^{-1} t_m \dots t_4t_3t_2t_4^{-1}\dots t_{m-m_1}^{-1}  (t_m \dots t_4 t_3t_1t_2t_3)^{m_1}[-2m_1-1](\Lambda)$

*-4.5em Step 1. $(t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{m_1}[-2m_1]$      Step 2. $t_4^{-1}\circ \dots \circ t_{m-m_1}^{-1}$      Step 3. $t_2$        Step 4. $t_3[-1]$

*-5em Step 5. $t_m \circ \dots \circ t_4$              Step 6. $t_4^{-1} \dots t_{h-m_1}^{-1}$              Step 7. $t_2^{-1}[-1]$

*-5em Step 8. $t_3[-1]$              Step 9. $t_{m_2+2}\dots t_4$              Step 10. $t_2$

Case <ref>

$\mathcal{H}(\Lambda) = t_1 \circ t_{m_1+2} \circ \dots \circ t_4 \circ t_3 \circ t_1^{-1} \circ t_4^{-1}  \circ \dots  \circ t_d^{-1}(\Lambda)$

*-2em Step 1.$ t_4^{-1} \circ \dots \circ t_d^{-1}$                  Step 2. $ t_1^{-1}$                          Step 3. $ t_3[-1]$

*-5em Step 4.$ t_4^{-1} \circ \dots \circ t_d^{-1}$                                  Step 5. $ t_1[1]$

Case <ref>
$\mathcal{H}(\Lambda) = t_2 \circ t_1 \circ (t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-2}[-2l+2](\Lambda)$

*-5em Step 1.$(t_m \circ \dots \circ t_4 \circ t_3 \circ t_1 \circ t_2 \circ t_3)^{l-2}[-2l+4] $                    Step 2. $ t_2 \circ t_1 [-2]$

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# Document Network Embedding: Coping for Missing Content and Missing Links

Jean Dupuy  Université de Lyon, Lyon 2, ERIC EA3083 – MeetSYS
<EMAIL_ADDRESS>,  Adrien Guille  Université de Lyon, Lyon 2, ERIC
EA3083<EMAIL_ADDRESS>and  Julien Jacques  Université de Lyon,
Lyon 2, ERIC EA3083<EMAIL_ADDRESS>
(2019)

###### Abstract.

Searching through networks of documents is an important task. A promising path
to improve the performance of information retrieval systems in this context is
to leverage dense node and content representations learned with embedding
techniques. However, these techniques cannot learn representations for
documents that are either isolated or whose content is missing. To tackle this
issue, assuming that the topology of the network and the content of the
documents correlate, we propose to estimate the missing node representations
from the available content representations, and conversely. Inspired by recent
advances in machine translation, we detail in this paper how to learn a linear
transformation from a set of aligned content and node representations. The
projection matrix is efficiently calculated in terms of the singular value
decomposition. The usefulness of the proposed method is highlighted by the
improved ability to predict the neighborhood of nodes whose links are
unobserved based on the projected content representations, and to retrieve
similar documents when content is missing, based on the projected node
representations.

Document network, Document representation, Node representation, Translation

††conference: ; ; ††journalyear: 2019††ccs: Computing methodologies Machine
translation††ccs: Computing methodologies Learning latent
representations††ccs: Information systems Information retrieval

##  1\. Introduction

Searching through a network of documents, such as scientific articles
(Brochier, 2019), Web pages or social media posts (Guille et al., 2013), is a
common task and a long-standing research topic. Recent works propose to
improve the performance of information retrieval systems by leveraging dense
vector space representations of the content and the nodes (Wen et al., 2018;
Seyler et al., 2018, 2018; Zhang et al., 2019; Makarov et al., 2018). Quite
often, for various reasons, such as technical issues or legal limitations
(e.g. paywall), not all links can be observed nor can all the content be
retrieved.

These representations are learned with algorithms similar to those devised for
word embedding, e.g. DeepWalk (Perozzi et al., 2014), Node2Vec (Grover and
Leskovec, 2016) or GVNR (Brochier et al., 2019) for node representations, or
Doc2Vec (Le and Mikolov, 2014) for content. Obviously, missing content
prevents applying any document embedding techniques. Also, since network
embedding algorithms perform random walks to measure the co-occurrence
frequency between nodes, they cannot learn representations of isolated nodes.

In this paper, we address this issue by formulating a translation task, from
available content representations to node representations, and from available
node representations to content representations. Assuming that the topology of
a network of documents correlates with the content of the documents, we
propose to estimate the missing node (resp. document) representations as a
linear transformation, i.e. projection, of the content (resp. node)
representations. More specifically, we show how to construct a dictionary of
pairs of content and node representations that allows learning a projection
matrix, inspired by recent advances in machine translation with pre-trained
word embeddings (Mikolov et al., 2013b; Xing et al., 2015; L. Smith et al.,
2017). Note that a self-translation network embedding algorithm, STNE (Liu et
al., 2018), has been recently proposed, however it addresses a different task.
It learns to translate a sequence of content to a sequence of nodes to learn
richer node representations and it cannot operate on isolated nodes nor deal
with missing content.

Our approach proves relevant, as evidenced by the quality of the alignment
between the estimated representations and the true representations.
Practically speaking, our method has two main applications. On the one hand,
estimating the node representation helps in recovering the missing links. On
the other hand, estimating the content representation from the node
representation allows measuring the similarity with the content of other
documents and thus provides helpful clues about the missing content.

The rest of this paper is organized as follows. In section 2, we survey
related work on document and network embedding, as well as embedding-based
bilingual translation. We then describe our proposal to estimate the missing
representations in section 3. Next, in section 4, we evaluate the quality of
the projection and assess the improvement brought by our method in predicting
the neighborhood of nodes whose links are missing and retrieve documents
similar to those without content. Finally, we conclude and provide future
directions in section 5.

##  2\. Related Work

Word embedding, i.e., the task of learning dense vector space representations
of words, is tightly connected to the tasks of learning document and node
representations, i.e. document embedding and network embedding. In this
section, we first briefly survey works related to word embedding and then
present models adapted for document and network embedding. Lastly, we cover
recent advances in offline bilingual translation via word embeddings, on which
our work is based.

###  2.1. Word Embedding

Word embedding builds upon the distributional hypothesis. It states that
distributional similarity and meaning similarity are correlated, which allows
learning representations of words based on the contexts in which they occur,
the context of a word being co-occurring words in a short window.

Mikolov et al. (Mikolov et al., 2013a) propose two models, namely Skip-Gram
and Continuous Bag-of-words (CBOW). The first models the conditional
probability of occurrence of a word, given a word in the context. The second
models the conditional probability of occurrence of a word, given the set of
words in the context. In both cases, the probability is defined as a softmax
function parameterized by the dot products of the vector representations of
the words.

Because the softmax makes the estimation of the word representations
computationally expensive, Mikolov et al. (Mikolov et al., 2013c) propose
approximate solutions based either on the hierarchical softmax or negative
sampling, an approximation akin to noise-contrastive estimation.

###  2.2. Content Representation Learning

Le and Mikolov (Le and Mikolov, 2014) extends the Skip-Gram and CBOW models to
learn representations of documents, under the names, respectively, Distributed
Bag-of-words (DV-DBOW) and Distributed Memory (DV-DM). They rely on a simple
trick, that consists in considering each document as a distinct new vocabulary
entry, that occurs in every window in the document. Thus, these models learn
document representations that are good at predicting the words they contain.

###  2.3. Node Representation Learning

Even though the distributional hypothesis originated in linguistics, Perozzi
et al. (Perozzi et al., 2014) show that sequences of nodes generated by random
walks and sentences share some statistical properties. In particular, they
show that the frequency at which nodes appear in short random walks follows a
power-law distribution, like the frequency of words in language. Consequently,
they propose to learn node representations under the distributional
hypothesis. The proposed model, DeepWalk, consists in learning the
representations based on the Skip-Gram model with the hierarchical softmax
approximation, from a corpus of node sequences – deemed equivalent to
sentences – generated by truncated random walks.

LINE (Tang et al., 2015) refines the objective optimized by DeepWalk, trying
to preserve both the first and second order proximities in the embedding
space. The representations are learned based on the Skip-Gram model with
negative sampling rather than the hierarchical softmax approximation, using a
specific stochastic gradient descent algorithm for weighted networks, that
samples the edges according to their weights.

Node2vec (Grover and Leskovec, 2016) performs biased random walks, in order to
better balance the exploration-exploitation trade-off, arguing that the added
flexibility in exploring neighborhoods helps learning richer representations.
The representations are also learned based on the Skip-Gram model with
negative sampling.

###  2.4. Bilingual Translation

Mikolov et al. (Mikolov et al., 2013b) first propose to learn a linear
transformation from the embedding space of a source language into the
embedding space of a target language. They suggest to estimate a projection
matrix from pre-trained vectors and a bilingual dictionary. To this end they
formulate a least square problem based on the Euclidean distance between
aligned vectors, which they solve with the regular stochastic gradient descent
algorithm.

Even though this approach yields encouraging results, Xing et al. (Xing et
al., 2015) argue that the problem is ill-posed, since the similarity between
representations is usually measured in terms of the cosine similarity, rather
than the Euclidean distance. They show how to learn unitary word embeddings
and formulate a maximization problem in terms of the dot product between some
source vectors and projected target vectors. They also require the projection
matrix to be orthogonal, so that the projected vectors remain unitary in order
to preserve the equivalence between the dot product and the cosine similarity.
Eventually, they devise a gradient descent based algorithm to learn the
projection matrix. Yet, their approach is computationally expensive, because
it requires to orthogonalize the projection matrix after each update, by
computing its singular value decomposition.

Smith et al. (L. Smith et al., 2017) show that finding the projection matrix
that solves the problem stated by Xing et al. is actually equivalent to
solving a least square problem based on the Euclidean distance, when the
vectors are unitary and the projection matrix is orthogonal. Furthermore, they
link this problem with the orthogonal Procrustes problem, which admits an
analytical solution. The projection matrix is obtained from the singular value
decomposition of the similarity matrix between the aligned source and target
vectors.

##  3\. Proposed Method

In this section, we describe our proposal, which consists in adapting
techniques devised for bilingual translation to learn a projection from the
content representations (pre-trained with, e.g. Doc2Vec) to the node
representations (pre-trained with, e.g. DeepWalk), and vice versa.

Consider a corpus of documents organized in a network, so that the content of
some documents is missing and the links of some other documents are
unobserved. We assume a set of pre-trained node representations
$A=\\{a_{i}\in\mathbb{R}^{k}\\}$ and a set of document representations
$B=\\{b_{i}\in\mathbb{R}^{k}\\}$. Our goal is to estimate the missing node
representations in $A$, from the content representations in $B$ ; conversely,
estimate the missing content representations in $B$, from the node
representations in $A$.

We posit that the topology of the network and the content of the documents
correlate. Thus, we aim at finding a projection matrix
$W\in\mathbb{R}^{k\times k}$ that maps content representations to node
representations, and node representations to content representations. More
particularly, given a document $i$, we want $W$ to maximize the cosine of the
angle $\theta_{a_{i},b_{i}W}$ between $a_{i}$ and $b_{i}W$. That is to say, we
want to find a linear transformation so that the vectors $a_{i}$ and $b_{i}W$
point in the same direction. More formally, based on a set of pairs of node
representations and document representations, $S$, we want to find a matrix
$W$ so that:

(1) $\max_{W}\sum_{i\in S}\cos(\theta_{a_{i},b_{i}W}),$

In the rest of this section, we describe how to construct $S$ and then detail
how to learn the projection matrix $W$.

###  3.1. Dictionary construction

Rather than learning the projection from all the available pairs of node
representations and content representations, we suggest to restrain to a
specific subset. The aim in doing so is (i) favor the learning of a good
projection and (ii) limit the computational cost.

Bojanowski et al. (Bojanowski et al., 2017) note that Skip-Gram, the model
behind most network embedding algorithms, struggles to learn good
representations for rare words. This also holds for network embedding
algorithms, since Perrozi et al. (Perozzi et al., 2014) show that the
frequency at which nodes occur in short random walks and the frequency of
words in language follow a similar law. Hence, we propose to avoid rare nodes
and to construct the dictionary only from the $m$ nodes that occur the most in
short random walks on the network. Formally, with $f_{i}$ the frequency of
node $i$, we construct the set $S$ of cardinality $m$ that maximizes
$\sum_{i\in S}f_{i}$.

###  3.2. Projection Learning

Following (Xing et al., 2015), we rewrite formula 1 in terms of the dot
product, and thus require the node and content representation to be unitary
vectors and constrain $W$ to be orthogonal:

(2) $\max_{W}\sum_{i\in S}a_{i}\cdot b_{i}W\text{, subject to}~{}W^{\top}W=I.$

Yet, rather than learning the representations again as Xing et al. suggest, we
simply row-normalize $A$ and $B$. Then, thanks to the proportional
relationship between the opposite of the dot product and the squared Euclidean
distance highlighted in (L. Smith et al., 2017), because
$||a_{i}-b_{i}W||^{2}=||a_{i}||^{2}+||b_{i}||^{2}-2(a_{i}\cdot b_{i}W)$, we
transform the problem into a minimization one:

(3) $\min_{W}\sum_{i\in S}||a_{i}-b_{i}W||^{2}\text{, subject
to}~{}W^{\top}W=I.$

This is a classic linear algebra problem, which solution is given in
(Schönemann, 1966). It consists in calculating the singular value
decomposition of the matrix $M=A_{S}^{\top}{B_{S}}$, with
$A_{S}\in\mathbb{R}^{m\times k}$ and $B_{S}\in\mathbb{R}^{m\times k}$ the
aligned node and content representations according to the dictionary $S$. With
$M=U\Sigma V^{\top}$, the matrix $W$ is defined as:

(4) $W=UV^{\top}.$

We estimate the node representation of an isolated node by calculating
$\tilde{a}=bW$. Similarly, we estimate the content representation of a
document whose content is missing by calculating $\tilde{b}=aW^{\top}$.

##  4\. Experiments

In this section, we evaluate our proposal on two networks of documents. First,
we focus on missing links and show that estimating the node representations
from the content representations allows to better reconstruct the neighborhood
of nodes whose links are unobserved. Second, we focus on missing content and
show that estimating the content representations from node representations
allows to retrieve a significant fraction of the similar documents.

###  4.1. Experimental setup

####  4.1.1. Data

We consider two well known networks of scientific articles, Cora (McCallum et
al., 2000) and HepTh (Leskovec et al., 2005), whose properties are summed up
in Table 1. No content nor citation links are missing. These datasets
available online111Get the data:
https://github.com/thunlp/CANE/tree/master/datasets/cora,
https://github.com/thunlp/CANE/tree/master/datasets/HepTh.

Table 1. Network properties. Name | # of documents | # of links
---|---|---
Cora | 2,211 | 5,001
HepTh | 1,038 | 1,974

####  4.1.2. Node and Content Representations

We experiment with DeepWalk (Perozzi et al., 2014) to learn the node
representations, with the following configuration: 80 random walks length 80
from each node, and a co-occurrence window of size 10. Regarding content
representation, we experiment with the Distributed Memory (DV-DM) variant of
Doc2Vec (Le and Mikolov, 2014) with a window of size 10. Here after, we report
the results for representations of size 500.

####  4.1.3. Projection Learning

We learn the projection matrix via the exact singular value decomposition,
based on a dictionary of size 1400 for Cora, and a dictionary of size 550 for
HepTh.

####  4.1.4. Similarity Metric

We measure the similarity between vector representations in terms of cosine
similarity (Mikolov et al., 2013a).

###  4.2. Missing Links: Neighborhood Reconstruction

We begin by assessing the quality of the estimated node representations. The
evaluation task consists in reconstructing the neighborhood of a document
whose links are unobserved. We adopt a leave-one-out strategy and proceed in
the following manner. First, we learn all the content representations from the
entire corpus. Then, for each document $i$, we do the following:

  1. (1)

Hide all the incoming and outgoing links of document $i$;

  2. (2)

Learn the node representations on this sub-network;

  3. (3)

Learn the projection matrix on this sub-network;

  4. (4)

Measure the similarity between the estimated node representation
$\tilde{a}_{i}=b_{i}W$ and all the other node representations $\\{a_{j}\\}$;

  5. (5)

Order the documents accordingly and measure the precision at $n$, $P@n$ w.r.t.
the true links.

We report the relative gain in average precision in Table 2, w.r.t. two
baselines:

  * •

$\cos(\theta_{b_{i},b_{j}})$: Documents ordered according to the similarity
between the content representations;

  * •

Random200: Documents ordered randomly, picked among the 200 documents with the
highest degrees.

It reveals that the estimated node representations, obtained via the linear
transformation of the content representations always lead to a better
precision in predicting the neighborhood of a document.

Table 2. Precision gain in link prediction. |  | P@5 | P@10 | P@20 | P@50
---|---|---|---|---|---
Cora | $\cos(\theta_{b_{i},b_{j}})$ | +4.4% | +23.4% | +20.1% | +20.0%
Random200 | +681.4% | +475.9% | +166.1% | +51.9%
Hepth | $\cos(\theta_{b_{i},b_{j}})$ | +69.2% | +42.9% | +5.9% | +2.6%
Random200 | +516.6% | +375.1% | +350.4% | +175.6%

###  4.3. Missing Content: Similar Document Retrieval

We now switch our attention to missing content. We assess the quality of the
estimated content representations following a similar leave-one-out
methodology. We learn all the the content representations and define the true
sets of similar document $S_{i}=\\{j|\cos(\theta_{b_{i},b_{j}})>0.2\\}$. The
evaluation task consists in reconstructing the set of similar documents for a
document whose content is unknown. We proceed as follows. First, we learn all
the node representations from the whole network. Then, for each document $i$,
we do the following:

  1. (1)

Hide document $i$ from the corpus;

  2. (2)

Retrain the content representations on this sub-corpus;

  3. (3)

Learn the projection matrix on this sub-corpus;

  4. (4)

Measure the similarity between the estimated content representation
$\tilde{b}_{i}=a_{i}W^{\top}$ and all the content representations
$\\{b_{j}\\}$;

  5. (5)

Measure the accuracy by comparing the true set of similar documents $S_{i}$
and the estimated set of similar documents
$\tilde{S}_{i}=\\{j|\cos(\theta_{\tilde{b}_{i},b_{j}})>0.2\\}$.

We report the average precision following this procedure in the first row of
Table 3. The second row shows the average precision based on the node
representations, where $\tilde{S}_{i}=\\{j|\cos(\theta_{a_{i},a_{j}})>0.2\\}$.
It reveals that estimating the missing content representation from the
corresponding node representation doubles the accuracy in document retrieval,
in comparison with the accuracy obtained by measuring the similarity based on
the node representations. In particular, we are able to recover, on average,
two-thirds of the similar documents in the Hepth corpus.

Table 3. Average accuracy in similar document retrieval. | Cora | Hepth
---|---|---
$\tilde{S}_{i}=\\{j|\cos(\theta_{\tilde{b}_{i},b_{j}})>0.2\\}$ | 0.36 | 0.65
$\tilde{S}_{i}=\\{j|\cos(\theta_{a_{i},a_{j}})>0.2\\}$ | 0.13 | 0.30

##  5\. Conclusion

In this paper, we described a way to efficiently learn a linear function to
map pre-trained node and content representations learned from a network of
documents. In presence of missing links and missing content, this method helps
in reconstructing the neighborhood of isolated documents and also helps in
recovering documents that are similar to a document whose content is unknown.
In future work, we would like to investigate how to learn a non-linear
transformation to refine the mapping between the two spaces.

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