**Abbreviations**

The following abbreviations are used in this manuscript:


#### **Appendix A. Lemmas for Theorem 1**

To perform the proof of Theorem 1, we need the following lemmas.

**Lemma A1** (**Polarization lemma, Theorem 5 in [25]**)**.** *Let α and β satisfy* 0 ≤ *α* < *β*2 ≤ 1*. Then, there is a deterministic polynomial-time procedure that, on input* (*Q*0, *Q*1, 1*n*) *where Q*0 *and Q*1 *are quantum circuits, outputs descriptions of quantum circuits* (*<sup>R</sup>*0, *<sup>R</sup>*1) *(each having size polynomial in n and in the size of Q*0 *and Q*1*) such that*

$$\begin{array}{rcl} \parallel \rho\_0 - \rho\_1 \parallel\_{\rm tr} \leq \alpha & \Rightarrow & \parallel \mu\_0 - \mu\_1 \parallel\_{\rm tr} \leq 2^{-n} \\\parallel \rho\_0 - \rho\_1 \parallel\_{\rm tr} \geq \beta & \Rightarrow & \parallel \mu\_0 - \mu\_1 \parallel\_{\rm tr} \geq 1 - 2^{-n} .\end{array}$$

The proof of the following lemmas can be found in [38].

**Lemma A2** (**Joint entropy theorem** [38])**.** *Suppose pi are probabilities,* |*i are orthogonal states for a system A and ρi is any set of density operators for another system B. Then,*

$$S\left(\sum\_{i} p\_i \, |i\rangle\langle i| \otimes \rho\_i\right) = H(p\_i) + \sum\_{i} p\_i S(\rho\_i). \tag{A1}$$

**Lemma A3** (**Fannes' inequality** [38])**.** *Suppose ρ and σ are density matrices over a Hilbert space of dimension d. Suppose further that the trace distance between them satisfies t* = *ρ* − *σ* tr ≤ 1/*e. Then,*

$$|S(\rho) - S(\sigma)| \le t(\ln d - \ln t). \tag{A2}$$

**Lemma A4** (**Lemma 3.2 in [44]**)**.** *Let ρ*0 *and ρ*1 *be two density matrices, and let ρ* = 12 (*ρ*0 + *ρ*1)*. If there exists a measurement with outcome* 0 *or* 1 *such that making the measurement on ρb yields the bit b with probability at least p, then*

$$S(\rho) \ge \frac{1}{2} [S(\rho\_0) + S(\rho\_1)] + (1 - H(p)). \tag{A3}$$

In particular, by choosing the right observable we have

**Lemma A5** (**Theorem 1 in [25]**)**.** *Let ρ*0 *and ρ*1 *be two density matrices, and let ρ* = 12 (*ρ*0 + *ρ*1)*. Then,*

$$S(\rho) \ge \frac{1}{2} [S(\rho\_0) + S(\rho\_1)] + (1 - H(\frac{1}{2} + \frac{||\rho\_0 - \rho\_1||\_{\text{tr}}}{2})).\tag{A4}$$

#### **Appendix B. Proof of the Central Lemma 1**

The proof of the lemma comes straightforwardly from the following definitions and previously established theorems.

**Definition A1.** *Let* H *be a finite dimensional Hilbert space and ρi* ∈ B(H)*, i = 1,2, density operators. Their relative entropy is defined as:*

$$S(\rho\_1 \| \rho\_2) := \begin{cases} \operatorname{Tr} \rho\_1 (\log \rho\_1 - \log \rho\_2) & \text{if } \operatorname{supp}(\rho\_1) \subseteq \operatorname{supp}(\rho\_2) \\ +\infty & \text{otherwise.} \end{cases} \tag{A5}$$

Originally defined by Umegaki [45]. A relevant property of the quantum relative entropy is its monotonicity under CPTP maps, also known as Uhlmann's theorem [46].

**Theorem A1.** *Let* H *and* K *be finite dimensional Hilbert spaces, ρi* ∈ B(H)*, i = 1,2, density operators with supp*(*ρ*1) ⊆ *supp*(*ρ*2)*. For a CPTP map* Φ : B(H) → B(K) *the following inequality holds:*

$$S(\rho\_1 \| \rho\_2) \ge S(\Phi(\rho\_1) \| \Phi(\rho\_2)). \tag{A6}$$

**Corollary A1.** *The von Neumann entropy is strong sub-additivite:*

$$\mathcal{S}\left(\rho\_{ABC} \| \rho\_{AB} \otimes \frac{id\_{\mathcal{C}}}{d\_{\mathcal{C}}}\right) \geq \mathcal{S}\left(\rho\_{BC} \| \rho\_{B} \otimes \frac{id\_{\mathcal{C}}}{d\_{\mathcal{C}}}\right). \tag{A7}$$

**Proof of Corollary A1.** Observe that setting in (A6) *ρ*1 → *ρABC*, *ρ*2 → *ρAB* ⊗ id*C*/*dC* and *φ*(·) → Tr*A*[·], we obtain which is equivalent to the non-negativity of the quantum conditional mutual information *<sup>I</sup>ρ*(*<sup>A</sup>* : *C*|*B*) ≥ 0.

The following two theorems characterize the case of the equality and they will be the core of the proof of Lemma 1.

**Theorem A2** (**Theorem 2 in [47]**)**.** *Let* Φ : B(H) → B(K) *be a CPTP map and let ρi* ∈ B(H)*, i = 1,2, and φ*(*ρi*) ∈ B(K) *be all invertible density operators. Then, the equality holds in the Uhlmann theorem iff the following equivalent conditions hold:*


*where* (*ii*) *is obtained differentiating* (*i*) *in t = 0.*

The adjoint map *φ*†(·) is understood with respect to the *Hilbert–Schmidt inner product*. **Theorem A3** (**Theorem 5.2 in [35]**)**.** *A tripartite state ρABC* ∈ B(H*ABC*) *is a QMC in the order A-B-C iff Iρ* (*A* : *C*|*B*) = 0*. Furthermore, one can always choose as recovery map the rotated Petz map:*

$$\mathcal{P}\_{B \to B\mathbb{C}}^{t}(X) := \rho\_{B\mathbb{C}}^{\frac{1+it}{2}} \left( \rho\_{B}^{-\frac{1+it}{2}} X \rho\_{B}^{-\frac{1-it}{2}} \otimes id\_{\mathbb{C}} \right) \rho\_{B\mathbb{C}}^{\frac{1-it}{2}}, \text{ for any } X \in \mathcal{B}(\mathcal{H}\_{\mathbb{B}}), \ t \in \mathbb{R}. \tag{A8}$$

#### **Proof of Theorem A3 of Lemma 1.**

**(3** ⇒ **1)** This implication comes for free from the definition of QMC. Moreover, the map P*<sup>B</sup>*→*BC*(·) is clearly CPTP. The complete positivity indeed comes for free from the Hermitianicity of *ρB* ⊗ id*C*/*dC* and *ρBC*, then of their square-roots.

**(1** ⇒ **3)** Equation (A8) for t = 0 gives exactly the Petz map in (3), so the implication comes as corollary of Theorem A3.

**(1** ⇔ **2)** This follows from the statement of Theorem A3.

**(2** ⇔ **4)** It comes as corollary of Theorem A2, using the settings in Corollary A1.

#### **Appendix C. Lemmas for Theorem 2 and 3**

We need the following Lemmas to derive the proof.

**Lemma A6.** *Let* X = {*<sup>A</sup>*, *B*, *C*, *D*} *be the labeling of parts of a finite dimensional Hilbert space* H *and* C = {*ρXY* ∈ B(H*XY*), *X*,*Y* ∈ X} *an admissible set of two-body marginals. Assume ρAB*, *ρBC* ∈ C *and* ∃ *ρ*˜*ABC* ∈ B(H*ABC*) : *ρ*˜*ABC* ∈ *Comp*(*ρAB*, *ρBC*) *such that*

$$I\_{\rho}(A:\mathbb{C}|B) = 0.\tag{A9}$$

(A10)

*(a) If the associate graph* GC *is a chain A-B-C-D (i.e.,* C = {*ρAB*, *ρBC*, *ρCD*}*) then* ∃ *ρ*˜ ∈ B(H) : *ρ*˜ ∈ *Comp*(C) s.t *<sup>I</sup>ρ*(*<sup>A</sup>* : *CD*|*B*) = 0 iff

$$\exists \, \tilde{\rho}\_{BCD} \in \mathcal{B}(\mathcal{H}\_{BCD}) \,:\, \tilde{\rho}\_{BCD} \in \text{Comp}(\rho\_{BC}, \rho\_{CD}) \text{ s.t.}\, I\_{\rho}(B:D|\mathcal{C}) = 0.$$

*(b) If the associate graph* GC *is a star centered in B (i.e.,* C = {*ρAB*, *ρBC*, *ρBD*}*)* :

GC

*then*

$$\exists \exists \tilde{\rho} \in \mathcal{B}(\mathcal{H}) : \tilde{\rho} \in \text{Comp}(\mathcal{C}) \text{ s.t } l\_{\tilde{\rho}}(A : \mathcal{C}D|B) = 0 \quad \text{iff} \tag{A12}$$


*In both the cases, ρ*˜ = *arg max ρ*<sup>∈</sup>*Comp*(C) *<sup>S</sup>*(*ρ*) *and factorizes over the elements of* C *via Petz following a constructiveorderingfor*C*.*

**Proof of Lemma A6.** We prove cases (a) and (b) together, but each direction of the equivalence at a time. We notice than one direction follows easily from the chain rule, we start with that direction (⇐) Recall the chain rule for quantum conditional mutual information:

$$\begin{split} I\_{\rho}(A:X\_1,\ldots,X\_n|B) &= I\_{\rho}(A:X\_1|B) + I\_{\rho}(A:X\_2|BX\_1) + \\ &+ \cdots + I\_{\rho}(A:X\_n|BX\_1,\ldots,X\_{n-1}) \end{split} \tag{A13}$$

and recall that the conditional mutual information is non negative. Case (a)

$$I\_{\rho}(AB:D|\mathbb{C}) = I\_{\rho}(B:D|\mathbb{C}) + I\_{\rho}(B:D|AC) = 0 \quad \Rightarrow \quad I\_{\rho}(B:D|\mathbb{C}) = 0 \tag{A14}$$

The case (b) is analogous:

$$\begin{aligned} I\_{\rho}(A\mathbb{C}:D|B) &= I\_{\rho}(A:D|B) + I\_{\rho}(A:D|B\mathbb{C}) = 0 \quad \Rightarrow & I\_{\rho}(A:D|B) = 0\\ I\_{\rho}(A\mathbb{C}:D|B) &= I\_{\rho}(\mathbb{C}:D|B) + I\_{\rho}(\mathbb{C}:D|AB) = 0 \quad \Rightarrow & I\_{\rho}(\mathbb{C}:D|B) = 0 \end{aligned} \tag{A15}$$

(⇒) (a) To prove the other direction of the statement, we show that there exists a *ρ*˜ ∈ B(H): *ρ*˜ ∈ Comp(*ρ*˜*ABC*, *ρ*˜*BCD*) and QMC on the order AB-C-D.

By hypothesis and using Lemma 1, the tripartite states can be recovered from two of its two-body marginals using the Petz recovery map:

$$\prescript{}{}{I}\_{\rho}(A:\mathbb{C}|B) = 0 \quad \text{iff} \quad \overline{\rho}\_{ABC} = \rho\_{BC}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{AB} \rho\_B^{-\frac{1}{2}} \rho\_{BC}^{\frac{1}{2}} = \rho\_{AB}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{BC} \rho\_B^{-\frac{1}{2}} \rho\_{AB'}^{\frac{1}{2}} \tag{A16}$$

$$I\_{\rho}(B:D|\mathbb{C}) = 0 \quad \text{iff} \quad \overline{\rho}\_{B\text{CD}} = \rho\_{\text{BC}}^{\frac{1}{2}} \rho\_{\text{C}}^{-\frac{1}{2}} \rho\_{\text{CD}} \rho\_{\text{C}}^{-\frac{1}{2}} \rho\_{\text{BC}}^{\frac{1}{2}} = \rho\_{\text{CD}}^{\frac{1}{2}} \rho\_{\text{C}}^{-\frac{1}{2}} \rho\_{\text{BC}} \rho\_{\text{C}}^{-\frac{1}{2}} \rho\_{\text{CD}}^{\frac{1}{2}}.\tag{A17}$$

Using Lemma 2, we check the compatibility of the two marginals with the desired QMC showing that the operator Θ*ABCD* := *ρ*˜ 1 2 *ABCρ* − 1 2 *C ρ* 1 2 *CD* is normal:

$$
\boldsymbol{\Theta}\_{ABCD} \boldsymbol{\Theta}\_{ABCD}^{\dagger} = \tilde{\boldsymbol{\rho}}\_{ABCD}^{\frac{1}{2}} \boldsymbol{\rho}\_{\boldsymbol{C}}^{-\frac{1}{2}} \boldsymbol{\rho}\_{\boldsymbol{C}D} \boldsymbol{\rho}\_{\boldsymbol{C}}^{-\frac{1}{2}} \tilde{\boldsymbol{\rho}}\_{ABC}^{\frac{1}{2}} \tag{A18}
$$

$$\rho = \rho\_{AB}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \underbrace{\rho\_{BC}^{\frac{1}{2}} \rho\_C^{-\frac{1}{2}} \rho\_{CD} \rho\_C^{-\frac{1}{2}} \rho\_B^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{AB}^{\frac{1}{2}}}\_{\rho\_{BCD}} \rho\_B^{-\frac{1}{2}} \rho\_{AB}^{\frac{1}{2}} \tag{A19}$$

$$=\underbrace{\rho\_{AB}^{\frac{1}{2}}\rho\_{B}^{-\frac{1}{2}}\rho\_{CD}^{\frac{1}{2}}\rho\_{C}^{-\frac{1}{2}}}\_{}\rho\_{BC}\underbrace{\rho\_{C}^{-\frac{1}{2}}\rho\_{CD}^{\frac{1}{2}}\rho\_{B}^{-\frac{1}{2}}\rho\_{AB}^{\frac{1}{2}}}\_{}\tag{A20}$$

$$=\rho\_{\underline{\boldsymbol{C}}}^{\frac{1}{2}}\rho\_{\boldsymbol{C}}^{-\frac{1}{2}}\underbrace{\rho\_{AB}^{\frac{1}{2}}\rho\_{B}^{-\frac{1}{2}}\rho\_{BC}\rho\_{B}^{-\frac{1}{2}}\rho\_{AB}^{\frac{1}{2}}\rho\_{\boldsymbol{C}}^{-\frac{1}{2}}\rho\_{\boldsymbol{C}}^{\frac{1}{2}}}^{\frac{1}{2}}\rho\_{\boldsymbol{C}}^{-\frac{1}{2}}\rho\_{\boldsymbol{\tilde{C}}D}^{\frac{1}{2}}\tag{A21}$$

$$=\Theta\_{ABCD}^{\dagger}\Theta\_{ABCD}.\tag{A22}$$

Equality in Equation (A19) follows for Equation (A16) and Lemma 2. Equality in Equation (A20) follows from permuting density operators in different Hilbert spaces.

(b) Similarly to (a), we show that there exists a *ρ*˜ ∈ B(H): *ρ*˜ ∈ Comp(*ρ*˜*ABC*, *ρ*˜*CBD*, *ρ*˜*ABD*) and QMC on the order AC-B-D. Again, using Lemma 1, the tripartite states can be recovered from two of its two-body marginals using the Petz recovery map:

$$I\_{\rho}(A:\mathbb{C}|B) = 0 \quad \text{iff} \quad \tilde{\rho}\_{ABC} = \rho\_{BC}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{AB} \rho\_B^{-\frac{1}{2}} \rho\_{BC}^{\frac{1}{2}} = \rho\_{AB}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{BC} \rho\_B^{-\frac{1}{2}} \rho\_{AB}^{\frac{1}{2}} \tag{A23}$$

$$\rho I\_{\rho}(\mathbb{C}:D|B) = 0 \quad \text{iff} \quad \overline{\rho}\_{\text{CBD}} = \rho\_{\overline{B}\mathbb{C}}^{\frac{1}{2}} \rho\_{B}^{-\frac{1}{2}} \rho\_{\overline{B}D} \rho\_{B}^{-\frac{1}{2}} \rho\_{\overline{B}\mathbb{C}}^{\frac{1}{2}} = \rho\_{\overline{B}\mathbb{C}}^{\frac{1}{2}} \rho\_{B}^{-\frac{1}{2}} \rho\_{\overline{B}D} \rho\_{B}^{-\frac{1}{2}} \rho\_{\overline{B}\mathbb{C}'}^{\frac{1}{2}} \tag{A24}$$

$$I\_{\rho}(A:D|B) = 0 \quad \text{iff} \quad \overline{\rho}\_{ABD} = \rho\_{AB}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{BD} \rho\_B^{-\frac{1}{2}} \rho\_{AB}^{\frac{1}{2}} = \rho\_{AB}^{\frac{1}{2}} \rho\_B^{-\frac{1}{2}} \rho\_{BD} \rho\_B^{-\frac{1}{2}} \rho\_{AB}^{\frac{1}{2}}.\tag{A25}$$

First, by using Lemma 2, we check the compatibility of the two marginals *ρBD* and *ρABC* with the desired QMC showing that the operator Θ*ABCD* := *ρ*˜ 1 2 *ABCρ* − 1 2 *B ρ* 1 2 *BD* is normal:

$$
\boldsymbol{\upTheta}\_{ABCD} \boldsymbol{\upTheta}\_{ABCD}^{\dagger} = \boldsymbol{\upmu}\_{ABCD}^{\frac{1}{2}} \boldsymbol{\uprho}\_{B}^{-\frac{1}{2}} \boldsymbol{\uprho}\_{BD} \boldsymbol{\uprho}\_{B}^{-\frac{1}{2}} \boldsymbol{\uprho}\_{ABC}^{\frac{1}{2}} \tag{A26}
$$

*ABC*

$$=\rho\_{AB}^{\frac{1}{2}}\rho\_B^{-\frac{1}{2}}\underbrace{\rho\_{BC}^{\frac{1}{2}}\rho\_B^{-\frac{1}{2}}\rho\_{BD}\rho\_B^{-\frac{1}{2}}\rho\_{BC}^{\frac{1}{2}}\rho\_B^{-\frac{1}{2}}\rho\_{AB}^{\frac{1}{2}}\tag{A27}$$

$$=\underbrace{\rho\_{AB}^{\frac{1}{2}}\rho\_{B}^{-\frac{1}{2}}\rho\_{BD}^{\frac{1}{2}}\rho\_{B}^{-\frac{1}{2}}\rho\_{BC}\rho\_{B}^{-\frac{1}{2}}\underbrace{\rho\_{BD}^{\frac{1}{2}}\rho\_{B}^{-\frac{1}{2}}\rho\_{AB}^{\frac{1}{2}}}\_{\rho\_{ABD}}\tag{A.28}$$

$$=\rho\_{\underline{B}D}^{\frac{1}{2}}\rho\_{\underline{B}}^{-\frac{1}{2}}\underbrace{\rho\_{AB}^{\frac{1}{2}}\rho\_{\underline{B}}^{-\frac{1}{2}}\rho\_{\underline{BC}}\rho\_{\underline{B}}^{-\frac{1}{2}}\rho\_{\underline{AB}}^{\frac{1}{2}}\rho\_{\underline{B}}^{-\frac{1}{2}}\rho\_{\underline{B}D}^{\frac{1}{2}}}^{-\frac{1}{2}}\rho\_{\underline{B}}^{\frac{1}{2}}\tag{A29}$$

$$=\Theta\_{ABCD}^{\dagger}\Theta\_{ABCD}.\tag{A30}$$

Moreover, the QMC *ρ*˜ = <sup>Θ</sup>*ABCD*Θ†*ABCD* is in Comp(*ρ*˜*ABC*, *ρ*˜*CBD*, *ρ*˜*ABD*) by using Equations (A23)–(A25).

The previous lemma can be trivially extended to the n-partite scenario, i.e., to an arbitrary chain and a star:

**Corollary A2.** *Let* X = {*<sup>X</sup>*1, ... *Xn*} *be the labeling set of the parts of a finite dimensional Hilbert space* H *and* C = {*ρXY* ∈ B(H*XY*), *X*,*Y* ∈ X} *a set of two-body marginals on it classically compatible. Assume* GC *is a star centered in some Y* ∈ X*, i.e.,* C = {*ρXiY*, *i* = 1, ... , *n* − 1} *then there exists ρ*˜ ∈ B(H) : *ρ*˜ ∈ *Comp*(C) *such that ρ*˜ *is a quantum Markov network iff*

$$I\_{\rho}(X\_{i}:X\_{j}|Y) = 0 \,\forall i \neq j \in \mathbf{1}, \ldots, n-1. \tag{A31}$$

*Moreover,*

$$\mathfrak{d} = \arg\max\_{\rho \in \text{Conv}(\mathcal{C})} \mathcal{S}(\rho) \tag{A32}$$

*and factorizes over the elements of* C *via Petz following a constructive ordering for* C*.*

**Proof of Corollary A2.** The proof follows by adding at each step a node to the setting of Lemma A6 (case b). Shortly, consider the constructive ordering for the graph X = {*<sup>Y</sup>*, *X*1, ... , *Xn*}. Start from graph G3, where *V*3 ≡ *Y*, *X*1, *X*2, *X*3, clearly in this case we are in the situation of Lemma A6 (b), then:

$$I\_{\rho}(X\_2:X\_1|Y) = 0 \ I\_{\rho}(X\_3:X\_1|Y) = 0 \iff I\_{\rho}(X\_3:X\_1X\_2|Y) = 0. \tag{A33}$$

Observe that the two conditions *<sup>I</sup>ρ*(*<sup>X</sup>*<sup>3</sup> : *<sup>X</sup>*1*X*1|*Y*) = 0 and *<sup>I</sup>ρ*(*<sup>X</sup>*<sup>2</sup> : *<sup>X</sup>*1|*Y*) = 0 are those required by Theorem 2 s.t. there exists a Petz-factorizable d.o. *ρ*3 over G3. Next, we add the link *X*4 − *Y* to the graph and verify that *<sup>I</sup>ρ*(*<sup>X</sup>*<sup>4</sup> : *<sup>X</sup>*1*X*2*X*3|*Y*) = 0 also holds. We need to use again Theorem 2 to construct a Petz-factorizable *ρ*4. This condition follows by applying Lemma A6 (b):

$$\begin{aligned} I\_{\rho}(X\_4 : X\_1 X\_2 X\_3 | Y) &= 0 \text{ iff }\\ I\_{\rho}(X\_3 : X\_1 X\_2 | Y) &= 0 \text{ and } I\_{\rho}(X\_4 : X\_1 X\_2 | Y) = 0. \end{aligned} \tag{A34}$$

where the first condition is the one we go<sup>t</sup> in the previous step, the second comes from Lemma A6 (b):

$$\begin{aligned} I\_{\rho}(X\_4 : X\_1 X\_2 | Y) &= 0 \text{ iff } \\ I\_{\rho}(X\_4 : X\_1 | Y) &= 0 \text{ and } I\_{\rho}(X\_4 : X\_2 | Y) = 0. \end{aligned} \tag{A35}$$

Then, we keep adding nodes and decomposing the next required condition by Theorem 2. We notice that at each step, i.e., every time we add a node, in order to have a Petz decomposable d.o. on the new graph we have to add to the previous set of 3-chains, all the new 3-chains, i.e., the ones that involve the last added node.

**Lemma A7.** *Let ρ* ∈ B(H)*, where* X = {*<sup>X</sup>*1, ... , *Xn*} *and* H = *ni*=<sup>1</sup> H*Xi , such that ρ* ∈ *Comp*(C) *with* GC *a tree (i.e., we work under Assumption 1, and take X*1 < ··· < *Xn the constructive order). If for some* - ≤ *n the following conditions hold*

$$I\_{\rho\_j}(X\_i:\overline{Y\_j}|\,Y\_j) = 0,\,\forall j = \mathfrak{3},\,\dots,\ell,\tag{A36}$$

*then, by taking any i and mi*≥ *i such that*

$$\deg X\_i|\_{\mathcal{G}\_i} = \deg X\_i|\_{\mathcal{G}\_{m\_j}} \quad \text{and} \quad \deg \mathbf{Y}\_i|\_{\mathcal{G}\_i} = \deg \mathbf{Y}\_i|\_{\mathcal{G}\_{m\_j}} \tag{A37}$$

*the following conditions also hold*

$$I\_{\rho}(X\_{i}:V\_{r\_{i}} \mid \{X\_{i}, Y\_{i}\} \mid Y\_{i}) = 0, \; \forall r\_{i}:\; i \le r\_{i} \le m\_{i} \le \ell,\tag{A38}$$

**Proof of Lemma A7.** Take that Equation (A36) with *j* = *ρri* . By Theorem 2, we know that *ρri* factorizes via Petz over its two-body marginals according to C*ri* . Then, set

$$
\Delta\_k := \rho\_{X\_k Y\_k}^{\frac{1}{2}} \rho\_{Y\_k}^{-\frac{1}{2}}, \tag{A39}
$$

it follows that the factorization via Petz can be written as follows:

$$
\rho\_{r\_i} = \Delta\_{r\_i} \Delta\_{r\_i - 1} \dots \Delta\_{i} \dots \dots \rho\_{X\_1 X\_2} \dots \Delta\_{i} \dots \Delta\_{r\_i - 1} \Delta\_{r\_i \epsilon} \tag{A40}
$$

where, in general, [<sup>Δ</sup>*i*, Δ*j*] = 0. Note that, from the definition of *mi* it must be the case that [<sup>Δ</sup>*ri* , <sup>Δ</sup>*s*] = 0 ∀*s* : *i* ≤ *s* ≤ *ri*. This follows since Equation (A37) imposes that no more nodes are connected to *Xi* and *Yi* when adding nodes from step *i* to *mi*; and therefore, the additional Δ*k*'s operate on different Hilbert spaces. Then, Equation (A40) is to be written as:

$$
\rho\_k = \Delta\_i \dots \Delta\_{r\_i} \dots \rho\_{X\_1 X\_2} \dots \Delta\_{r\_i} \dots \Delta\_i. \tag{A41}
$$

Now consider a new, but equivalent, constructive ordering <

$$X\_1 <' \cdots <' X\_{i-1} <' X\_{r\_i} <' X\_{i+1} <' \cdots <' X\_{r\_i - 1} <' X\_{i \prime} \tag{A42}$$

obtained from the order < by exchanging *ri* with *i*. By using Theorem 2 (ii) with the order <sup>&</sup>lt;, we ge<sup>t</sup> in C*ri*the condition

$$I\_{\mathbb{P}\_{r'\_i}}\left(\mathcal{X}\_{r'\_i} \colon \overline{Y\_{r'\_i}} | \mathcal{Y}\_{r'\_i}\right) = 0. \tag{A43}$$

Which for the usual order < can be stated as:

$$I\_{\rho}(X\_i:V\_{r\_i}\backslash\{X\_i,\mathcal{Y}\_i\}|Y\_i) = 0.\tag{A44}$$

The latter equality is valid for all *ri* : *i* ≤ *ri* ≤ *mi* ≤ -, since the only property used was the fact that deg*Xi*|G*i* = deg*Xi*|G*ri*.

#### **Appendix D. Number of 3-Chains**

**Proof of Lemma 3.** We make the proof by counting, for each node *Xi*, how many 3-chains *Xj* − *Xi* − *Xk* can be formed, and summing all of them afterwards.

For a spanning tree, the lower bound is the number of 3-chains in a *n*-chain (all nodes have degree 2, with exception of the root and the leaf). In this case, every node is the central node of only one 3-chain, aside for the root and the leaf; thus, #*c* = *n* − 2. The upper bound is derived by counting the number of 3-chains in a *n*-star (there is a root and all the remaining nodes are leaves). The root, say *Y*, has deg*Y* = *n* − 1, and the remaining nodes (enumerate them as *X*1, ... *Xn*−1), have degree one. In this case, consider the first edge *X*1*Y*, it can be linked through Y to more n-2 nodes, which also gives the number of 3-chains it can be part of. The next edge *X*2*Y*, it can be connected through *Y* to *n* − 3 nodes to form *n* − 3 different chains (the chain *X*2 − *Y* − *X*1 is the same as *X*1 − *Y* − *X*2, which has been already counted for). It is now clear that the number of 3-chains in an *n*-star is

$$\#c\_{i} = \sum\_{k=2}^{n} (n-k) = \sum\_{k=1}^{n-2} k = \frac{1}{2}(n-1)(n-2). \tag{A45}$$

The number of chains in a *n*-star is also the number of 3-chains that a node contributes in a complete graph. Then, to obtain the number of 3-chains in a complete graph it is enough to multiply Equation (A45) by the number of nodes, and so #*c* = *n*#*ci* = 12*n*(*n* − <sup>1</sup>)(*n* − <sup>2</sup>).

Another way of obtaining this value consists in using well-known formulas from combinatorial calculus, and observing that the number of 3-chains in a complete graph of *n* vertices is the number of *simple dispositions*, i.e., the number of ordered sequences of length 3 without repetitions in a set of *n* elements, divided by two. The factor 2 comes from the symmetry of the 3-chains; that is, *A* − *B* − *C* is the same 3-chain as *C* − *B* − *A*. Then, once again,

$$\#\varepsilon = \frac{1}{2} \frac{n!}{(n-3)!} = \frac{1}{2} n(n-1)(n-2) \tag{A46}$$
