How to transform graph states using single-qubit operations: computational complexity and algorithms

Using the two operations local complementation and vertex-deletion, we can formulate the notion of a vertex-minor of a graph.

Definition 2.5 (Vertex-minor).  A graph G' is called a vertex-minor of G if and only if there exist a sequence of local complementations and vertex-deletions that takes G to G'. Since vertex-deletions can always be performed last in such a sequence (see lemma 2.1), an equivalent definition is the following: a graph G' is called a vertex-minor of G if and only if there exist a sequence of local complementations v such that τ _v (G)[V(G')] = G'. If G' is a vertex-minor of G we write this as

$$\begin{equation}{G}^{\prime }{< }G\end{equation} \tag{ 22 }$$

and if G' is not a vertex-minor of G then

$$\begin{equation}{G}^{\prime }\nless G.\end{equation} \tag{ 23 }$$

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Vertex-minors were first studied in [7] but by the name of l-reductions. Note that if G₁ and G₂ are two LC-equivalent graphs, then G' < G₁ if and only if G' < G₂. It is interesting to consider under which conditions a graph G' is a vertex-minor of another graph G. As theorem 2.2 below states, to decide whether G' < G it is sufficient to check whether G' is LC-equivalent to at least one out of 3|V(G)|−|V(G')| graphs. To formally state the theorem we introduce the following three operations.

Definition 2.6. The graph operations X_v , Y_v and Z_v , specified with a vertex v, act on a graph G by transforming it to

$$\begin{equation}{X}_{v}\left(G\right)={\rho }_{v}\left(G\right){\backslash}v,\quad {Y}_{v}\left(G\right)={\tau }_{v}\left(G\right){\backslash}v,\quad {Z}_{v}\left(G\right)=G{\backslash}v\end{equation} \tag{ 24 }$$

When we need to specify which edge incident on v the pivot of X_v acts on, we write ${X}_{v}^{\left(u\right)}\left(G\right)={\rho }_{\left(u,v\right)}\left(G\right){\backslash}v$.◊

The three graph operations X_v , Y_v and Z_v correspond to how Pauli-X, -Y and -Z measurements act on graph states (as proven in [29]). As mentioned in section 2.2, measuring qubit v of a graph state |G⟩ in the Pauli-X, -Y or -Z basis gives a stabilizer state which is single-qubit Clifford equivalent to |X_v (G)⟩, |Y_v (G)⟩ and |Z_v (G)⟩ respectively. equations (25)–(27) show examples of how these operations can act on graphs.

The operation ${X}_{v}^{\left(u\right)}$ is the most complicated one, so we will here quickly describe what happens to a graph when ${X}_{v}^{\left(u\right)}$ is applied. One can check that after the operation ${X}_{v}^{\left(u\right)}$, the vertex u will have the neighbors that v previously had, except v itself. Furthermore, some edges between vertices in (N_v ∪ N_u )\{u, v} will be complemented, i.e. removed if present or added if not. To know which of these edges gets complemented, let us introduce the following three sets

$$\begin{equation}{V}_{vu}={N}_{v}\cap {N}_{u},\quad {V}_{v}={N}_{v}{\backslash}\left({N}_{u}\cup \left\{u\right\}\right),\quad {V}_{u}={N}_{u}{\backslash}\left({N}_{v}\cup \left\{v\right\}\right)\end{equation} \tag{ 28 }$$

which form a partition of (N_v ∪ N_u )\{u, v}. In equation (27), these sets are V₁₂ = {3}, V₁ = {4} and V₂ = {5, 6}. An edge (w₁, w₂) between vertices in (N_v ∪ N_u )\{u, v} gets complemented if and only if w₁ and w₂ belong to different sets of the partition (V_{v u} , V_v , V_u ). All other edges in the graph, i.e. edges containing a vertex not in N_v ∪ N_v , will be unchanged.

It turns out that the three operations {X_v , Y_v , Z_v } are sufficient to check whether some graph is a vertex-minor of another graph. This is formalized in the following theorem we proved in [18].

Theorem 2.2. (Theorem 3.1 in [18]). Let G and G' be two graphs and let u = (v₁, ⋯, v_l ), where l = |V(G)\V(G')| be an ordered tuple such that each element of V(G)\V(G') occurs exactly once in u. Furthermore, let ${\mathcal{P}}_{\mathbf{u}}$ denote the set of graph operations

$$\begin{equation}{\mathcal{P}}_{\mathbf{u}}=\left\{{P}_{{v}_{l}}{\circ}\cdots {\circ}{P}_{{v}_{1}}\enspace :\enspace {P}_{v}\in \left\{{X}_{v},{Y}_{v},{Z}_{v}\right\}\right\}\end{equation} \tag{ 29 }$$

Then we have that

$$\begin{equation}{G}^{\prime }{< }G{\Leftrightarrow}\exists P\in {\mathcal{P}}_{\mathbf{u}}\enspace :\enspace {G}^{\;\prime }\;\;{\sim }_{\text{LC}}\;\;P\left(G\right).\end{equation} \tag{ 30 }$$

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Note that in [18] we indexed ${\mathcal{P}}_{\mathbf{u}}$ simply with the set associated to the word u since the statement is independent of the ordering of the elements of u . A direct corollary of the above theorem is therefore:

Corollary 2.2.1. Let G and G' be two graphs. Furthermore, let u and u' be two ordered tuples such that each element of V(G)\V(G') occurs exactly once in both u and u'. Then we have that

$$\begin{equation}\exists P\in {\mathcal{P}}_{\mathbf{u}}\enspace :\enspace {G}^{\;\prime }\;\;{\sim }_{\text{LC}}\;\;P\left(G\right){\Leftrightarrow}\exists P\in {\mathcal{P}}_{{\mathbf{u}}^{\prime }}\enspace :\enspace {G}^{\;\prime }\;\;{\sim }_{\text{LC}}\;\;P\left(G\right).\end{equation} \tag{ 31 }$$

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Proof. This follows directly from theorem 2.2 since both sides in equation (31) are true if and only if G' < G.□

Note that theorem 2.2 does not give an efficient method to check if G' is a vertex-minor of G, since the set ${\mathcal{P}}_{\mathbf{u}}$ is of exponential size for all u. To study this problem further we formally define the vertex-minor problem. □

Problem 2.1 (VERTEXMINOR).  Given a graph G and a graph G' defined on a subset of V(G), decide whether G' is a vertex-minor of G. ◊

Note again that we deal with labeled graphs here. We will often consider the special case where G' is a star graph S_V' defined on a subset V' of V(G). Remember that a graph state described by a star graph is single-qubit Clifford equivalent to a GHZ-state. Thus checking if S_V' is a vertex-minor of G is equivalent to checking if |G⟩ can be transformed to GHZ-state on the qubits V' by only using LC + LPM + CC. We will give this problem a separate name.

Problem 2.2 (STARVERTEXMINOR).  Given a graph G and a vertex subset V' of V(G), decide whether S_V' is a vertex-minor of G. ◊

Note that we have not specified which star graph on V' we use. This is not ambiguous since all star graphs on V' are equivalent under local complementation. In the rest of the text we will often leave the choice of star graph open.

Let us first define double occurrence words and equivalence classes of these. This will allow us to define circle graphs.

Definition 2.7 (Double occurrence word).  A double occurrence word X is a word with letters in some set V, such that each element in V occurs exactly twice in X . Given a double occurrence word X we will write V( X ) = V for its set of letters. ◊

Definition 2.8 (Equivalence class of double occurrence words).  We say that a double occurrence word Y is equivalent to another X , i.e. Y ∼ X , if Y is equal to X , the mirror $\tilde {\boldsymbol{X}}$ or any cyclic permutation of X or $\tilde {\boldsymbol{X}}$. We denote by d _X = { Y : Y ∼ X } the equivalence class of X , i.e. the set of words equivalent to X . ◊

Next we define alternances of these equivalence classes, which will represent the edges of an alternance graph.

Definition 2.9 (Alternance).  An alternance (u, v) of the equivalence class d _X is a pair of distinct elements u, v ∈ V such that a double occurrence word of the form ⋯u⋯v⋯u⋯v⋯ is in d _X . ◊

Note that if (u, v) is an alternance of d _X then so is (v, u), since the mirror of any word in d _X is also in d _X .

Definition 2.10 (Alternance graph).  The alternance graph $\mathcal{A}\left(\boldsymbol{X}\right)$ of a double occurrence word X is a graph with vertices V( X ) and edges given exactly by the alternances of d _X , i.e.

$$\begin{equation}E\left(\mathcal{A}\left(\boldsymbol{X}\right)\right)=\left\{\left(u,v\right)\in V\left(\boldsymbol{X}\right){\times}V\left(\boldsymbol{X}\right)\enspace :\enspace \left(u,v\right)\;\text{is}\;\text{an}\;\text{alternance}\;\text{of}\;{\boldsymbol{d}}_{\boldsymbol{X}}\right\}\end{equation} \tag{ 32 }$$

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Note that since $\mathcal{A}\left(\boldsymbol{X}\right)$ only depends on the equivalence class of X , the alternance graphs $\mathcal{A}\left(\boldsymbol{X}\right)$ and $\mathcal{A}\left(\boldsymbol{Y}\right)$ are equal if X ∼ Y . Now we can formally define circle graphs (figure 3).

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Definition 2.11 (Circle graph).  A graph G which is the alternance graph of some double occurrence word X is called a circle graph.◊

There is yet another way to represent circle graphs, closely related to double occurrence words, as Eulerian tours of 4-regular multi-graphs.

Definition 2.12 (Eulerian tour).  Let F be a connected 4-regular multi-graph. An Eulerian tour U on F is a tour that visits each edge in F exactly once.◊

Any 4-regular multi-graph is Eulerian, i.e. has a Eulerian tour, since each vertex has even degree [5].

Furthermore, any Eulerian tour on a 4-regular multi-graph F traverses each vertex exactly twice, except for the vertex which is both the start and the end of the tour. Such a Eulerian tour induces therefore a double occurrence word, the letters of which are the vertices of F, and consequently a circle graph as described in the following definition.

Definition 2.13 (Induced double occurrence word).  Let F be a connected 4-regular multi-graph on k vertices V(F). Let U be a Eulerian tour on F of the form

$$\begin{equation}U={x}_{1}{e}_{1}{x}_{2}\cdots {x}_{2k-1}{e}_{2k-1}{x}_{2k}{e}_{2k}{x}_{1}.\end{equation} \tag{ 33 }$$

with x_i ∈ V. Note that every element of V occurs exactly twice in U. From a Eulerian tour U as in equation (33) we define an induced double occurrence word as

$$\begin{equation}m\left(U\right)={x}_{1}{x}_{2}\cdots {x}_{2k-1}{x}_{2k}.\end{equation} \tag{ 34 }$$

To denote the alternance graph given by the double occurrence word induced by a Eulerian tour, we will write $\mathcal{A}\left(U\right)\equiv \mathcal{A}\left(m\left(U\right)\right)$. ◊

Similarly to double occurrence words, we also introduce equivalence classes of Eulerian tours under cyclic permutation or reversal of the tour.

Definition 2.14 (Equivalence class of Eulerian tours).  Let F be a connected 4-regular multi-graph and U be an Eulerian tour on F. We say that an Eulerian tour U' on F is equivalent to U, i.e. U ∼ U', if U' is equal to U, the reversal $\tilde {U}$ or any cyclic permutation of U or $\tilde {U}$. We denote by t _U the equivalence class of U, i.e. the set of Eulerian tours on F which are equivalent to U. ◊

It is clear that if the Eulerian tours U and U' on a 4-regular multi-graph F are equivalent, then so are the double occurrence words m(U) and m(U'). Furthermore, as for double occurrence words, two equivalent Eulerian tours on a connected 4-regular multi-graph induce the same alternance graph.

We will now introduce an operation ${\tilde {\tau }}_{v}$ on double occurrence words that will be the equivalent of performing a local complementation on the corresponding alternance graph.

Definition 2.15. (${\tilde {\tau }}_{v}$). Let X be a double occurrence word and v be an element in V( X ). We can then always find sub-words A , B and C not containing v, such that X = A v B v C . Note that some of the sub-words A , B and C are possibly empty. The operation ${\tilde {\tau }}_{v}$ acting on a double occurrence word is then defined as

$$\begin{equation}\tilde {\tau }\left(\boldsymbol{A}v\boldsymbol{B}v\boldsymbol{C}\right)=\boldsymbol{A}v\tilde {\boldsymbol{B}}v\boldsymbol{C}.\end{equation} \tag{ 35 }$$

If v = (v₁, ⋯, v_l ) is a sequence of elements of V( X ) we use the notation ${\tilde {\tau }}_{\mathbf{v}}\left(\boldsymbol{X}\right)={\tilde {\tau }}_{{v}_{l}}{\circ}\cdots {\circ}{\tilde {\tau }}_{{v}_{1}}\left(\boldsymbol{X}\right)$. ◊

The operation ${\tilde {\tau }}_{v}$ in the above definition maps equivalence classes to equivalence classes, as defined in definition 2.8. That is, if X ∼ Y and v ∈ V( X ), then ${\tilde {\tau }}_{v}\left(\boldsymbol{X}\right)\sim {\tilde {\tau }}_{v}\left(\boldsymbol{Y}\right)$. For example, assume that Y is the mirror of X , i.e. $\boldsymbol{Y}=\tilde {\boldsymbol{X}}$. Then we see that

$$\begin{equation}{\tilde {\tau }}_{v}\left(\boldsymbol{X}\right)=\boldsymbol{A}v\tilde {\boldsymbol{B}}v\boldsymbol{C}\sim \tilde {\boldsymbol{A}v\tilde {\boldsymbol{B}}v\boldsymbol{C}}=\tilde {\boldsymbol{C}}v\boldsymbol{B}v\tilde {\boldsymbol{A}}={\tilde {\tau }}_{v}\left(\boldsymbol{Y}\right).\end{equation} \tag{ 36 }$$

The case when Y is obtained by a cyclic permutation of X can be checked similarly.

In [11] it was shown that the alternance graph of $\mathcal{A}\left({\tilde {\tau }}_{v}\left(\boldsymbol{X}\right)\right)$, where X is a double occurrence word and v ∈ V( X ), is the same as the graph obtained by performing a local complementation at v, i.e.

$$\begin{equation}{\tau }_{v}\left(\mathcal{A}\left(\boldsymbol{X}\right)\right)=\mathcal{A}\left({\tilde {\tau }}_{v}\left(\boldsymbol{X}\right)\right).\end{equation} \tag{ 37 }$$

Similar to the above we can also define an operation ${\overline{\tau }}_{v}$ on Eulerian tours U on 4-regular multi-graphs which also has the effect of a local complementation on the graph $\mathcal{A}\left(U\right)$.

Definition 2.16. (${\overline{\tau }}_{v}$). Let F be a connected 4-regular multi-graph. Let U be a Eulerian tour on F and v be a vertex in V. Let P_v be the first subtrail of U that starts and ends at v, i.e. U = U₁ P_v U₂, from some U₁ and U₂. We define ${\overline{\tau }}_{v}\left(U\right)$ to be the Eulerian tour obtained by traversing U₁, the reversal of P_v and then U₂, i.e. ${\overline{\tau }}_{v}\left(U\right)={U}_{1}\tilde {{P}_{v}}{U}_{2}$. When v = v₁⋯v_l is a sequence of vertices in V we write ${\overline{\tau }}_{\mathbf{v}}\left(U\right)\equiv {\overline{\tau }}_{{v}_{l}}{\circ}\cdots {\circ}{\overline{\tau }}_{{v}_{1}}\left(U\right)$. ◊

Note in particular that ${\overline{\tau }}_{v}\left(U\right)$, where U is an Eulerian tour on F, is also a Eulerian tour on F.

We have now defined τ-operations on circle graphs, $\tilde {\tau }$-operations on double occurrence words and $\overline{\tau }$-operations on Eulerian tours of 4-regular multi-graphs. They are given similar names since they are in some sense the same operation but in different representations of circle graphs. To see this note that

$$\begin{equation}m\left({\overline{\tau }}_{v}\left(U\right)\right)=m\left({U}_{1}\tilde {{P}_{v}}{U}_{2}\right)={\tilde {\tau }}_{v}\left(m\left(U\right)\right)\end{equation} \tag{ 38 }$$

where U = U₁ P_v U₂ as in definition 2.16. From equation (37) and the shorthand $\mathcal{A}\left(U\right)=\mathcal{A}\left(m\left(U\right)\right)$ we also have that

$$\begin{equation}\mathcal{A}\left({\overline{\tau }}_{v}\left(U\right)\right)=\mathcal{A}\left({\tilde {\tau }}_{v}\left(m\left(U\right)\right)\right)={\tau }_{v}\left(\mathcal{A}\left(U\right)\right).\end{equation} \tag{ 39 }$$

The operation ${\overline{\tau }}_{v}$ on Eulerian tours of 4-regular multi-graphs was introduced by Kotzig in [32], where he called it a κ-transformation.

As stated by Bouchet in [11], Kotzig [32] proved that any two Eulerian tours of a 4-regular multi-graph are related by a sequence of κ-transformations.

Theorem 2.3. (Proposition 4.1 in [11], [32]). Let U and U' be Eulerian tours on the same connected 4-regular multi-graph. Then there exists a sequence v such that τ_v (U) ∼ U'. ◊

When we are considering vertex-minors of circle graphs, it is useful to have an operation on the double occurrence word that corresponds to the deletion of a vertex in the corresponding alternance graph. Let X = A v B v C be a double occurrence word and v be an element in V( X ). We will denote by X \v the deletion of the element v, i.e.

$$\begin{equation}\boldsymbol{X}{\backslash}v\equiv \left(\boldsymbol{A}v\boldsymbol{B}v\boldsymbol{C}\right){\backslash}v=\boldsymbol{A}\boldsymbol{B}\boldsymbol{C}.\end{equation} \tag{ 40 }$$

The resulting word ABC is also a double occurrence word and furthermore we have that

$$\begin{equation}\mathcal{A}\left(\boldsymbol{X}\right){\backslash}v=\mathcal{A}\left(\boldsymbol{X}{\backslash}v\right).\end{equation} \tag{ 41 }$$

If W = {w₁, w₂, ⋯, w_l } is a subset of V, we will write X \W as the deletion of all elements in W, i.e.

$$\begin{equation}\boldsymbol{X}{\backslash}W=\left(\cdots \left(\left(\boldsymbol{X}{\backslash}{w}_{1}\right){\backslash}{w}_{2}\right)\cdots \enspace \right){\backslash}{w}_{l}.\end{equation} \tag{ 42 }$$

Connected to this we can also define an induced double occurrence sub-word X [W] = X \(V\W). The reason for calling this an induced double occurrence sub-word stems from its relation to induced subgraphs of the alternance graph as

$$\begin{equation}\mathcal{A}\left(\boldsymbol{X}\right)\left[W\right]=\mathcal{A}\left(\boldsymbol{X}\left[W\right]\right).\end{equation} \tag{ 43 }$$

Since we now have expressions for local complementation and vertex deletion on circle graphs in terms of double occurrence words, we can consider vertex-minors of circle graphs completely in terms of double occurrence words. More precisely we have the following theorem.

Theorem 2.4. Let G be a circle graph such that $G=\mathcal{A}\left(\boldsymbol{X}\right)$ for some double occurrence word X . Then G' is a vertex-minor of G if and only if there exist a sequence v of elements in V(G) = V( X ) such that

$$\begin{equation}{G}^{\prime }=\mathcal{A}\left({\tilde {\tau }}_{\mathbf{v}}\left(\boldsymbol{X}\right)\left[V\left({G}^{\prime }\right)\right]\right).\end{equation} \tag{ 44 }$$

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Proof. By using equations (43) and (39) on the right-hand side of equation (44) we have that

$$\begin{equation}\mathcal{A}\left({\tilde {\tau }}_{\mathbf{v}}\left(\boldsymbol{X}\right)\left[V\left({G}^{\prime }\right)\right]\right)=\mathcal{A}\left({\tilde {\tau }}_{\mathbf{v}}\left(\boldsymbol{X}\right)\right)\left[V\left({G}^{\prime }\right)\right]={\tau }_{\mathbf{v}}\left(\mathcal{A}\left(\boldsymbol{X}\right)\right)\left[V\left({G}^{\prime }\right)\right]\end{equation} \tag{ 45 }$$

Since G' is a vertex-minor of $G=\mathcal{A}\left(\boldsymbol{X}\right)$ if and only if there exist a sequence v of elements in V(G) such that

$$\begin{equation}{G}^{\prime }={\tau }_{\mathbf{v}}\left(G\right)\left[V\left({G}^{\prime }\right)\right]\end{equation} \tag{ 46 }$$

the theorem follows.□

We can also consider vertex minors of circle graphs in terms of their representations as Eulerian tours on connected 4-regular multi-graphs, which we will use in section 3 to prove that VERTEXMINOR is $\mathbb{N}\mathbb{P}$-complete. Theorem 2.3, together with equation (39), implies that connected 4-regular multi-graphs describe equivalence classes of circle graphs under local complementations. Bouchet pointed out this fact in [11]. We formalize this here as a theorem together with a formal proof:□

Theorem 2.5. Let U be an Eulerian tour of a connected 4-regular multi-graph F with vertices V. Then (1) any graph LC-equivalent to $\mathcal{A}\left(U\right)$ is an alternance graph of some Eulerian tour of F and (2) any alternance graph of a Eulerian tour of F is a graph LC-equivalent to $\mathcal{A}\left(U\right)$. ◊

Proof. We start by proving (1), so let us therefore assume that G is a graph LC-equivalent to $\mathcal{A}\left(U\right)$. This means, by definition, that there exist a sequence v of vertices in V such that $G={\tau }_{\mathbf{v}}\left(\mathcal{A}\left(U\right)\right)$. By using equation (39) we have that

$$\begin{equation}G=\mathcal{A}\left({\overline{\tau }}_{\mathbf{v}}\left(U\right)\right).\end{equation} \tag{ 47 }$$

which shows that G is an alternance graph induced by a Eulerian tour of F, since ${\overline{\tau }}_{\mathbf{v}}\left(U\right)$ is a Eulerian tour on F. To prove (2), assume that U' is a Eulerian tour of F. We will now prove that the alternance graph of U', $\mathcal{A}\left({U}^{\prime }\right)$, is LC-equivalent to $\mathcal{A}\left(U\right)$. By theorem 2.3, we know that there exists a sequence of ${\overline{\tau }}_{v}$-transformations that relates U and U', i.e. there exist a sequence v such that

$$\begin{equation}{\overline{\tau }}_{\mathbf{v}}\left(U\right)\sim {U}^{\prime }.\end{equation} \tag{ 48 }$$

Since these Eulerian tours are equivalent, their induced alternance graphs are equal, i.e.

$$\begin{equation}\mathcal{A}\left({\overline{\tau }}_{\mathbf{v}}\left(U\right)\right)=\mathcal{A}\left({U}^{\prime }\right).\end{equation} \tag{ 49 }$$

Finally, using equation (39) on the above equation gives

$$\begin{equation}{\tau }_{\mathbf{v}}\left(\mathcal{A}\left(U\right)\right)=\mathcal{A}\left({U}^{\prime }\right)\end{equation} \tag{ 50 }$$

which shows that $\mathcal{A}\left(U\right)$ and $\mathcal{A}\left({U}^{\prime }\right)$ are indeed LC-equivalent.□

Similarly to theorem 2.4 we can decide if a circle graph has a certain vertex-minor by considering Eulerian tours of a 4-regular graph, which is captured in the following theorem.

Theorem 2.6. Let F be a connected 4-regular multi-graph and let G be a circle graph such that $\mathcal{A}\left(U\right)$ for some Eulerian tour U on F. Then G' is a vertex-minor of G if and only if there exist a Eulerian tour U' on F such that

$$\begin{equation}{G}^{\prime }=\mathcal{A}\left(m\left({U}^{\prime }\right)\left[V\left({G}^{\prime }\right)\right]\right).\end{equation} \tag{ 51 }$$

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Proof. Lets first assume that G' is a vertex-minor of G. This means that there exists a sequence v such that G' = τ_v (G)[V(G')]. Since $G=\mathcal{A}\left(U\right)$ we have that

$$\begin{equation}{G}^{\prime }={\tau }_{\mathbf{v}}\left(\mathcal{A}\left(U\right)\right)\left[V\left({G}^{\prime }\right)\right]\end{equation} \tag{ 52 }$$

$$\begin{equation}=\mathcal{A}\left({\overline{\tau }}_{v}\left(U\right)\right)\left[V\left({G}^{\prime }\right)\right]\end{equation} \tag{ 53 }$$

$$\begin{equation}=\mathcal{A}\left(m\left({\overline{\tau }}_{v}\left(U\right)\right)\left[V\left({G}^{\prime }\right)\right]\right)\end{equation} \tag{ 54 }$$

where we used equation (39) in the first step and equation (43) in the second. We therefore see that ${U}^{\prime }={\overline{\tau }}_{v}\left(U\right)$ is a Eulerian tour on F satisfying equation (51).

To prove the converse let us assume that there exist a Eulerian tour U' on F satisfying equation (51). From theorem 2.3 we know that there exist a sequence v such that ${U}^{\prime }={\overline{\tau }}_{\mathbf{v}}\left(U\right)$. We can then replace U' by ${\overline{\tau }}_{\mathbf{v}}\left(U\right)$ in equation (51) such that

$$\begin{equation}{G}^{\prime }=\mathcal{A}\left(m\left({\overline{\tau }}_{\mathbf{v}}\left(U\right)\right)\left[V\left({G}^{\prime }\right)\right]\right)\end{equation} \tag{ 55 }$$

$$\begin{equation}=\mathcal{A}\left(m\left({\overline{\tau }}_{\mathbf{v}}\left(U\right)\right)\right)\left[V\left({G}^{\prime }\right)\right]\end{equation} \tag{ 56 }$$

$$\begin{equation}={\tau }_{\mathbf{v}}\left(\mathcal{A}\left(U\right)\right)\left[V\left({G}^{\prime }\right)\right]\end{equation} \tag{ 57 }$$

where we again made use of equations (43) and (39). From equation (57) we see that G' is indeed a vertex-minor of G, see definition 2.5, since $G=\mathcal{A}\left(U\right)$. □

As discussed in section 2.2, the question of whether a graph state |G⟩ can be transformed into a GHZ-state on the qubits V' corresponds to whether the graph G has vertex-minors on V' in the form of star or complete graphs. From the previous sections we have seen that circle graphs and their vertex-minors can be described by Eulerian tours on connected 4-regular multi-graphs. A natural question is therefore: Given a set of vertices V', what property should a connected 4-regular multi-graph F satisfy, such that S_V' is a vertex-minor of $\mathcal{A}\left(U\right)$, for some Eulerian tour U on F. As we will see in this section, a necessary and sufficient condition is that F allows for what we call a SOET with respect to V'.

The existence of an SOET on a 4-regular graph F with respect to some vertex set V' will therefore be a key technical tool when considering STARVERTEXMINOR on circle graphs, as described in section 3. We formally define an SOET as follows.

Definition 2.17 (SOET).  Let F be a 4-regular multi-graph and let V' ⊆ V(F) be a subset of its vertices. Furthermore, let s = s₁ s₂⋯s_k be a word with letters in V' such that each element of V' occurs exactly once in s and where k = |V'|. A semi-ordered Eulerian tour U with respect to V' is a Eulerian tour such that m(U) = X ₀ s₁ X ₁ s₂⋯s_k X _k s₁ Y ₁ s₂⋯s_k Y _k for some s and where X ₀, X ₁, ⋯, X _k , Y ₁, ⋯, Y _k are words (possibly empty) with letters in V\V'. This can also be stated as m(U)[V'] = ss , for some s . ◊

Note that the multi-graph F is not assumed to be simple, so multi-edges and self-loops are allowed. An SOET is a Eulerian tour on F that traverses the elements of V' in some order once and then again in the same order. The particular order in which the SOET traverses V' will not be important here, only that it traverses V' in the same order twice. An example of a graph that allows for an SOET with respect to the set V' = {a, b, c, d} can be seen in figure 4(a). An SOET for this graph is for example m(U) = abcdaebced. The graph in figure 4(b). On the other hand does not allow for any SOET with respect to the set V' = {a, b, c, d}.

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We also formally define the SOET-decision problem, which takes a 4-regular multi-graph F and a subset V' of the vertices as input and asks to decide whether or not the graph F allows for a semi-ordered Eulerian tour with respect to the vertex set V'.

Problem 2.3 (SOET).  Let F be a 4-regular multi-graph and let V' be a subset V(F). Decide whether there exists an SOET U on F with respect to the set V'.◊

As mentioned, the reason for introducing the notion of an SOET is that a 4-regular multi-graph F allows for an SOET with respect to a subset V' ⊆ V(F) if and only if a star graph on V' is a vertex-minor of an alternance graph $\mathcal{A}\left(U\right)$ induced by a Eulerian tour U on F. This is captured in the following theorem, formulated as a corollary of theorem 2.6.

Corollary 2.6.1. Let F be a connected 4-regular multi-graph and let G be a circle graph given by the alternance graph of a Eulerian tour U on F, i.e. $G=\mathcal{A}\left(U\right)$. Then S_V' is a vertex-minor of G if and only if F allows for an SOET (see definition 2.17) with respect to V'.◊

Proof. Note first that S_V'⩽G if and only if K_V' ⩽ G, since S_V' and K_V' are LC equivalent. From theorem 2.6 we know that K_V' is a vertex-minor of G if and only if there exist an Eulerian tour U' on F such that

$$\begin{equation}{K}_{{V}^{\prime }}=\mathcal{A}\left(m\left({U}^{\prime }\right)\left[{V}^{\prime }\right]\right).\end{equation} \tag{ 58 }$$

It is easy to verify that $\mathcal{A}\left(\boldsymbol{X}\right)$ is a complete graph on V' if and only if X = s₁ s₂⋯s_k s₁ s₂⋯s_k where s = s₁ s₂⋯s_k is a word with letters in V' such that each element of V' occur exactly once in s . The result then follows, since m(U')[V'] is of this form if and only if U' is an SOET with respect to V'.□

One can see that the existence of an SOET on a 4-regular multi-graph F with respect to V', imparts an ordering on the subset of vertices V'. We will in particular be interested in vertices in V' that are 'consecutive' with respect to the SOET. Consecutiveness is defined as follows. □

Definition 2.18 (Consecutive vertices).  Let F be a 4-regular graph and U an SOET on F with respect to a subset V' ⊆ V(F). Two vertices u, v ∈ V' are called consecutive in U if there exist a sub-word u X v or v X u of m(U) such that no letter of X is in V'.◊

We also define the notion of a 'maximal sub-word' associated with two consecutive vertices.

Definition 2.19 (Maximal sub-words).  Let F be a 4-regular multi-graph and U an SOET on F with respect to a subset V' ⊆ V(F). The double occurrence word induced by U is then of the form m(U) = X ₀ s₁ X ₁ s₂⋯s_k X _k s₁ Y ₁ s₂⋯s_k Y _k , where k = |V'|, s₁, ⋯, s_k ∈ V' and X₀, ⋯, X_k , Y₁, ⋯, Y_k are words (possibly empty) with letters in V(F)\V'. For i ∈ [k − 1], we call X _i and Y _i the two maximal sub-words associated with the consecutive vertices s_i and s_i+1. Furthermore, we call X _k and Y _k X ₀ the two maximal sub-words associated with the consecutive vertices s_k and s₁. Given two consecutive vertices u and v, we will denote their two maximal sub-words as X and X ', Y and Y ' or similar.◊

In this section we introduce distance-hereditary graphs. As shown by Bouchet in [8], distance-hereditary graphs are exactly the graphs with rank-width one. These graphs have nice properties which we make use of in section 4.1.

Definition 2.23 (Distance-hereditary).  A graph G is distance-hereditary if and only if, for each connected induced subgraph G[A] and for any two vertices u, v ∈ A the distance between u and v is the same in G and in G[A], i.e.

$$\begin{equation}{d}_{G}\left(u,v\right)={d}_{G\left[A\right]}\left(u,v\right).\end{equation} \tag{ 69 }$$

◊

Trees, i.e. graphs without loops, is clearly a direct subset of distance-hereditary graphs. The simplest example of a graph which is not distance-hereditary is the five-cycle C₅. To see this pick two vertices which have distance two in C₅ and denote their unique common neighbor by v. The distance between the same vertices in the connected induced subgraph C₅[V\V] is three and thus not the same as in C₅. It turns out that distance-hereditary graphs are exactly the graphs which do not contain a vertex-minor isomorphic to C₅ [11]. We also note that distance-hereditary graphs form a strict subclass of circle graphs [11].

In [2] an equivalent property of distance-hereditary is shown: A graph is distance-hereditary if and only if it can be obtained from a single-vertex graph using the following three operations:

• Add a leaf: Let u be a vertex in a graph G. Add the vertex v and the edge (u, v) to G.
• False twin-split: Let u be a vertex in a graph G. Add the vertex v and the edges {(v, x) : x ∈ N_u }.
• True twin-split: Let u be a vertex in a graph G. Add the vertex v and the edges {(v, x) : x ∈ {u} ∪ N_u }.

Note that this implies that a distance-hereditary graph always has at least one leaf or twin, i.e. the foliage is non-empty. This fact will be a critical element of the algorithm presented in section 4.1.

In the rest of this section we prove some properties of the foliage for distance-hereditary graphs, which we make use of in section 4.1 to find an efficient algorithm for VERTEXMINOR on distance-hereditary graphs. First we show that the twin relation is in fact transitive. This is a technical lemma we will use in later theorems.

Lemma 2.2. Let G be a graph and let u be a vertex of G that is a twin and has twin-partners {t₁, t₂, ⋯, t_k }. Then all vertices in u ∪ {t₁, t₂, ⋯, t_k } are pairwise twins. ◊

Proof. Since u and t_i form a twin-pair, for i ∈ {1, 2, ⋯, k}, we have that

$$\begin{equation}{N}_{u}{\backslash}\left\{{t}_{i}\right\}={N}_{{t}_{i}}{\backslash}\left\{u\right\}\end{equation} \tag{ 70 }$$

which implies that

$$\begin{equation}{N}_{{t}_{i}}=\left({N}_{u}{\backslash}\left\{{t}_{i}\right\}\right)\cup \left\{u\right\}.\end{equation} \tag{ 71 }$$

Thus, we have that

$$\begin{equation}{N}_{{t}_{i}}{\backslash}\left\{{t}_{j}\right\}=\left(\left({N}_{u}{\backslash}\left\{{t}_{i}\right\}\right)\cup \left\{u\right\}\right){\backslash}\left\{{t}_{j}\right\}\end{equation} \tag{ 72 }$$

$$\begin{equation}=\left(\underset{{N}_{{t}_{j}}{\backslash}\left\{u\right\}}{\underbrace{\left({N}_{u}{\backslash}\left\{{t}_{j}\right\}\right)}}\cup \left\{u\right\}\right){\backslash}\left\{{t}_{i}\right\}\end{equation} \tag{ 73 }$$

$$\begin{equation}={N}_{{t}_{j}}{\backslash}\left\{{t}_{i}\right\}.\end{equation} \tag{ 74 }$$

This shows that, for i ≠ j, t_i and t_j form a twin-pair.□

Next we prove that adding leaves to a graph G or performing (true or false) twin-splits never decreases the size of the foliage T(G). □

Lemma 2.3. Assume G is a connected, distance-hereditary graph. Let G' be a graph formed by doing a twin-split on G or adding a leaf to G. Then

$$\begin{equation}\vert T\left({G}^{\prime }\right)\vert {\geqslant}\vert T\left(G\right)\vert ,\end{equation} \tag{ 75 }$$

where T(G) is the foliage of G. ◊

Proof. To prove this, let us first consider the case when |G| ⩽ 2. Since G is connected it is necessary the case that G = K₁ or G = K₂:

• If G = K₁, then G' = K₂ and |T(G')| = 2 ⩾ 0 = |T(G)|.
• If G = K₂, then G' = K₃ or G' = P₃ and |T(G')| = 3 ⩾ 2 = |T(G)|.

Let us now consider the case when |G| > 2. We consider the two cases when G' is formed by adding a leaf and performing a twin-split separately:

• Assume G' is formed by adding a leaf v to G, making u an axil of G'. Note first that if u ∉ T(G), then |T(G)| can only increase since no vertex in T(G) was affected. Let us therefore assume that u ∈ T(G). There are then three possibilities: (1) u is a leaf, (2) u is an axil but not a twin and (3) u is a twin. We consider these three cases separately:
  • (1) Assume u is a leaf in G. Then the axil of u in G, is not in T(G'), but both u and v are. Therefore |T(G)| = |T(G')|.
  • (2) Assume u is an axil but not a twin in G. Then u is also an axil in G' and we have that |T(G')| = |T(G)| + 1.
  • (3) Assume u is a twin in G.

• Assume there is only one twin-pair containing u in G. Then the twin-partner of u in G, is not in T(G'), but both u and v are. Therefore |T(G')| = |T(G)|.

• Assume there is more than one twin-pair containing u in G. Then the twin-partners of u are all pairwise twins, by lemma 2.2, and will still be in G'. Therefore |T(G')| = |T(G)| + 1.
• Assume G' is formed by twin-splitting u in G, creating v and making v and u a twin-pair. Note first that if u ∉ T(G), then |T(G)| can only increase since no vertex in T(G) was affected. Let us therefore assume that u ∈ T(G). There are then three possibilities: (1) u is a leaf, (2) u is an axil but not a twin and (3) u is a twin. We consider these three cases separately:

• (1) Assume u is a leaf in G. Then the axil of u in G is either still an axil in G' or not, depending on if u and v are true or false twins. In either case, |T(G')| ⩾ |T(G)| since v ∈ T(G').
• (2) Assume u is an axil but not a twin in G. Note that all leaves with u as an axil in G are also twins. These vertices are also twins in G' since they are all now also adjacent to v. Thus, |T(G')| = |T(G)| + 1.
• (3) Assume u is a twin in G.
  • Assume there is only one twin-pair in G containing u. Then the twin-partner of u in G, may or may not still be a twin-partner of u in G' depending on whether the considered twin-pairs are true or false. Therefore the size of T(G) either remains the same or increases by one since again v ∈ T(G').
  • Assume there are more than one twin-pair in G containing u. Then the twin-partners of u are all pairwise twins, by lemma 2.2, and will still be in G'. Therefore |T(G')| = |T(G)| + 1.

□

We now make use of the above theorem to prove that the foliage has a certain minimum size. □

Theorem 2.8. Assume G is a connected, distance-hereditary graph and 2 ⩽ k ⩽ 4, then

$$\begin{equation}\vert G\vert {\geqslant}k{\Rightarrow}\vert T\left(G\right)\vert {\geqslant}k.\end{equation} \tag{ 76 }$$

◊

Proof. First we explicitly check that the graphs on 2, 3 and 4 vertices has T(G) = 2, T(G) = 3 and T(G) = 4, respectively ⁶ . Then by lemma 2.3 and the fact that all distance-hereditary graphs can be built up by twin-splits and adding leaves [8], the result follows.□

We point out that the theorem does not hold for k > 4. Consider for example a path graph P_k on more than four vertices. It is easy to see that size of the foliage in this case is |P_k | = 4.

Finally we show that an interesting property regarding the foliage, in relation to cut-vertices ⁷ .□□

Corollary 2.8.1. Assume that G is a connected distance-hereditary graph and that v ∈ G is a cut-vertex. Denote the connected components of G\v by G₁, G₂, ⋯, G_k , where k is the number of connected components of G\v. Then for any 1 ⩽ i ⩽ k, there exist a vertex u ∈ G_i such that u ∈ T(G). ◊

Proof. Pick an arbitrary connected component G_i with vertices V_i . If G_i is just a single vertex, then this vertex is necessarily a leaf in G and is therefore in T(G). Now assume that |G_i | > 1, then by using theorem 2.8, we have that there exist at least one twin-pair not containing v or a leaf which is not v in G[{v} ∪ V_i ]. This proves the corollary. □

In this section we show that the SOET problem reduces to STARVERTEXMINOR. For this we will make use of the properties of circle graphs, discussed in section 2. In corollary 2.6.1 we showed that a 4-regular multi-graph F allows for an SOET with respect to a subset of its vertices V' ⊆ V(F) if and only if an alternance graph $\mathcal{A}\left(U\right)$ (which is a circle graph), induced by some Eulerian tour on F, has S_V' as a vertex-minor.

Since circle graphs are a subset of all simple graphs we can then decide whether a 4-regular graph F allows for an SOET with respect to some subset V' of it is vertices by constructing the circle graph induced by an Eulerian tour on F and checking whether it has a star-vertex minor on the vertex set V'. This leads to the following theorem.

Theorem 3.2. The decision problem SOET reduces to STARVERTEXMINOR. ◊

Proof. Let (F, V') be an instance of SOET, where F is a 4-regular multi-graph and V' a subset of the vertex set of F. Also let G be a circle graph induced by any Eulerian tour U on F. From corollary 2.6.1 we see that G has S_V' as a vertex-minor if and only if F allows for an SOET with respect to the vertex set V'. Since an Eulerian tour U can be found in polynomial time [23] and since G can be efficiently constructed given U, this concludes the reduction. □

In this section we will prove that the SOET problem, as defined in problem 2.3, is $\mathbb{NP}$-complete by reducing the problem of deciding if a 3-regular graph is Hamiltonian (CubHam), a well-known $\mathbb{N}\mathbb{P}$-complete problem [25], to the SOET problem (it is in $\mathbb{N}\mathbb{P}$ by theorem 3.2 and lemma 3.1). For completeness we include the definition of a Hamiltonian graph.

Definition 3.1 (Hamiltonian).  A graph is said to be Hamiltonian if it contains a Hamiltonian cycle. A Hamiltonian cycle is a cycle that visits each vertex in the graph exactly once. ◊

We can use this to formally define the CubHam problem.

Problem 3.1 (CubHam).  Let R be a 3-regular graph. Decide whether R is Hamiltonian. ◊

The reduction of CubHam to the SOET problem is done by going though the following steps.

• (a)  
  Introduce the notion of a (4-regular) triangular-expansion Λ(R) of a 3-regular graph. This is done in definition 3.2.
• (b)  
  Argue that given a 3-regular graph R, its triangular-expansion can be constructed efficiently. This is done in lemma 3.2.
• (c)  
  Introduce the notions of skip and true skip that capture an essential behavior of SOET s on triangular-expansions of 3-regular graphs. This is done in section 3.2.4.
• (d)  
  Prove that if a 3-regular graph R is Hamiltonian then the triangular-expansion Λ(R) of R allows for an SOET with respect to the set V(R). This is done in lemma 3.4.
• (e)  
  Prove that if the triangular-expansion Λ(R) of a 3-regular graph R allows for a special kind of SOET, called an HAMSOET with respect to R (HAMSOETs are defined in definition 3.4, but can be thought of as SOETs with no true skips), then the 3-regular graph R is Hamiltonian. This is done in lemma 3.5.
• (f)  
  Prove that if the triangular-expansion Λ(R) of a 3-regular graph R allows for an SOET with respect to V(R) then it also allows for an HAMSOET with respect to R. This is done in lemma 3.6.

Performing all these steps will lead to the following theorems.

Theorem 3.3. Let R be a 3-regular graph and Λ(R) be its triangular-expansion as defined in definition 3.2. R is Hamiltonian if and only if Λ(R) allows for an SOET with respect to V(R). ◊

Proof. Let R be a 3-regular graph and let Λ(R) be its triangular-expansion as defined in definition 3.2. If R is Hamiltonian then lemma 3.4 guarantees that Λ(R) allows for an SOET with respect to the vertices V(R). In the other direction, if Λ(R) allows for an SOET with respect to the vertices V(R) then we can see from lemma 3.5 that it also allows for an HAMSOET. The existence of an HAMSOET on Λ(R) then implies, via lemma 3.5 that R has a Hamiltonian cycle and hence that it is Hamiltonian. This proves the theorem. □

Corollary 3.3.1. The SOET problem is $\mathbb{N}\mathbb{P}$-complete. ◊

Proof. The Hamiltonian cycle problem (CubHam) is $\mathbb{N}\mathbb{P}$-complete on 3-regular graphs [25]. We will reduce this problem to the SOET problem. Let R be an instance of CubHam, i.e. a 3-regular graph. From this 3-regular graph we can construct its triangular-expansion Λ(R). In lemma 3.2 it is argued that this construction can be performed in O(|V(R)|) time. We can then use theorem 3.3 to see that R is Hamiltonian if and only if Λ(R) allows for an SOET with respect to the vertex set V(R). Hence there exists an efficient reduction of CubHam to the SOET problem. This means that the SOET problem is $\mathbb{N}\mathbb{P}$-hard. Furthermore, the SOET problem is in $\mathbb{N}\mathbb{P}$ since it can be efficiently reduced to STARVERTEXMINOR by theorem 3.2, which is in $\mathbb{N}\mathbb{P}$ by lemma 3.1. Hence the SOET problem is $\mathbb{N}\mathbb{P}$-complete. □

Corollary 3.3.2. The SOET problem is $\mathbb{N}\mathbb{P}$-complete on graphs which are triangular-expansions of planar 3-regular triply-connected graphs, i.e. graphs in the set

$$\begin{equation}\left\{{\Lambda}\left(R\right):R\;\text{is}\;\text{planar,}\;3\text{-}\text{regular}\;\text{and}\;\text{triply}\text{-}\text{connected}\right\}.\end{equation} \tag{ 77 }$$

◊

Proof. The proof is the same as the proof of corollary 3.3.1 but using the fact that CubHam is $\mathbb{N}\mathbb{P}$-complete on planar triply-connected graphs. □

It now remains to prove lemmas 3.4–3.6. These lemmas will relate Hamiltonian cycles on 3-regular graphs and SOET s on 4-regular multi-graphs by using a mapping from 3-regular graphs to 4-regular multi-graphs. We call this mapping 'triangular-expansion'. We have the following definition.

Definition 3.2 (Triangular-expansion).  Let R be a 3-regular graph. A triangular-expansion Λ(R) of a 3-regular graph R is constructed from R by performing the following two steps:

• (a)  
  Replace each vertex v in R with the subgraph below

  

  where x, y and z are the neighbors of v. We will denote this triangle subgraph associated to the vertex v with T_v , i.e. T_v = G[{v, v^(x), v^(y), v^(z), ^(y), ^(z)}].
• (b)  
  Double every edge that is incident on two subgraphs T_v , T_v'.

◊

The graph Λ(R) will be called a triangular-expansion of R. A multi-graph F that is the triangular-expansion of some 3-regular graph R will also be referred to as a triangular-expanded graph. Note that the triangular-expansion is not uniquely defined, since for each vertex v ∈ R there is a choice how to orient the triangle with respect to the neighbors of v. Furthermore, the number of vertices in Λ(R) is 6⋅|V(R)| and the number of edges is 2⋅|E(R)| + 9⋅|V(R)|. In figure 6 we show an example of a 3-regular graph and its triangular-expansion.

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For a given triangle subgraph T_v in a triangular-expanded graph, we will refer to the vertices adjacent to other triangle subgraphs T_x , T_y , T_z as 'outer vertices' and label them according to the triangle subgraph they are adjacent to. Concretely we label the vertex in T_v that is adjacent to T_w as v^(w), the index signifies which triangle subgraph it connects to.

In the following lemma we argue that this construction can be made efficiently in the size of R.

Lemma 3.2. Let R be a 3-regular graph. We can construct its triangular-expansion in O(|V(R)|²) time. ◊

Proof. Let R be a 3-regular graph. Without loss of generality assume some labeling on the vertices of R, i.e. V(R) = {v₁, ⋯, v_k } where k = |V(R)|. We begin by constructing the vertex set V(Λ(R)) off the triangular-expansion Λ(R) of R.

For each i ∈ [k] construct the set ${V}_{i}=\left\{{v}_{i},{\tilde {v}}_{i}^{\left({v}_{j}\right)},{\tilde {v}}_{i}^{\left({v}_{{j}^{\prime }}\right)},{v}_{i}^{\left({v}_{j}\right)},{{v}_{i}}^{\left({v}_{{j}^{\prime }}\right)},{v}^{\left({v}_{\hat{j}}\right)}\right\}$ where v_i ∈ V(R) and v_j , v_j', v are the three unique neighbors of v_i in R. Constructing this set takes O(|V(R)|) as we must search the set of edges E(R) of R to find the neighbors of v_i and we have that |E(R)| = O(|V(R)|) since R is 3-regular. Thus constructing the set V(Λ(R)) = ∪_i∈[k] V_i takes O(|V(R)|²) time.

Now we construct the edge multi-set E(Λ(R)) of the triangular-expansion of R. For each i ∈ [k] we define the multi-set

$$\begin{equation}{E}_{i}=\left\{\right. \left({v}_{i},{\tilde {v}}_{i}^{\left({v}_{j}\right)}\right),\left({v}_{i},{\tilde {v}}_{i}^{\left({v}_{{j}^{\prime }}\right)}\right),\left({v}_{i},{v}_{i}^{\left({v}_{j}\right)}\right),\left({v}_{i},{v}_{i}^{\left({v}_{{j}^{\prime }}\right)}\right),\left({\tilde {v}}_{i}^{\left({v}_{j}\right)},{\tilde {v}}_{i}^{\left({v}_{{j}^{\prime }}\right)}\right),\left({v}_{i}^{\left({v}_{j}\right)},{\tilde {v}}_{i}^{\left({v}_{j}\right)}\right),\left({v}_{i}^{\left({v}_{{j}^{\prime }}\right)},{\tilde {v}}_{i}^{\left({v}_{{j}^{\prime }}\right)}\right),\end{equation} \tag{ 79 }$$

$$\begin{equation}\quad \left({v}_{i}^{\left({v}_{\hat{j}}\right)},{\tilde {v}}_{i}^{\left({v}_{j}\right)}\right),\left({v}_{i}^{\left({v}_{\hat{j}}\right)},{\tilde {v}}_{i}^{\left({v}_{{j}^{\prime }}\right)}\right),\left({v}_{i}^{\left({v}_{j}\right)},{v}_{j}^{\left({v}_{i}\right)}\right),\left({v}_{i}^{\left({v}_{{j}^{\prime }}\right)},{v}_{{j}^{\prime }}^{\left({v}_{i}\right)}\right),\left({v}_{i}^{\left({v}_{\hat{j}}\right)},{v}_{\hat{j}}^{\left({v}_{i}\right)}\right) \left.\right\}.\end{equation} \tag{ 80 }$$

This multi-set can be constructed in constant time. Hence the multi-set E(Λ(R)) = ∪_i∈[k] E_i can be constructed in O(|V(R)|) time. It is easy to check that the multi-graph defined by the vertex set Λ(R) and edge multi-set E(Λ(R)) is indeed the triangular-expansion of R. This completes the lemma.

□

A key insight in the behavior of the SOET problem on triangular-expanded graphs is the notion of skips. The word skip stems from the fact that since any SOET U on the triangular-expansion Λ(R) of a 3-regular graph is a Eulerian tour, it must traverse any triangle subgraph T_v of Λ(R) exactly three times. However in order for U to be a valid SOET with respect to V(R) it must traverse the vertex v exactly two of those three times. This means it must skip the vertex v exactly once while traversing T_v .

We state a more formal definition of a (true) skip in terms of maximal sub-words (see definition 2.19).

Definition 3.3 (Skip).  Let R be a 3-regular graph and let Λ(R) be its triangular-expansion. Let U be an SOET on Λ(R) with respect to V(R). Let X be a maximal sub-word of m(U) (definition 2.19) associated to vertices u, v ∈ V(R). We say the sub-trail described by X makes a skip at a vertex w ∈ V(R) \{u, v} if ${x}_{1}^{\left(w\right)}{w}^{\left({x}_{1}\right)}$ and ${x}_{2}^{\left(w\right)}{w}^{\left({x}_{2}\right)}$ are sub-words of X (up to reflection), where x₁, x₂ ∈ V(R). Furthermore, if x₁ ≠ x₂ then we say that the trail described by X makes a true skip at w or sometimes that T_w contains a true skip. ◊

Note that since X is a maximal sub-word associated to u, v and w ∉ {u, v}, w cannot be a letter of X . As stated above, there is always exactly one maximal sub-word describing a sub-trail of an SOET that makes a skip at a certain triangle subgraph, as formalized in the following lemma. One can think of this lemma as giving necessary conditions for the existence of an SOET with respect to V(R) on the triangular-expansion of a 3-regular graph R

Lemma 3.3. Let R be a 3-regular graph and let Λ(R) be its triangular-expansion. Let U be a Eulerian tour on Λ(R). Let w ∈ V(R) and let T_w be its triangle subgraph in Λ(R). If U is an SOET on Λ(R) with respect to V(R) then there exists exactly one maximal sub-trail of U that makes a skip at w. ◊

Proof. We will prove this by showing that there are exactly three maximal sub-trails of U that traverse vertices in T_w and that exactly one of these makes a skip at w. Note first that the Eulerian tour U will enter and exit the triangle subgraph T_w exactly three times, since there are six edges incident to T_w . Hence there exists exactly three distinct edge-disjoint sub-trails, t₁, t₂ and t₃ of U that exit and enter T_w , i.e. the last vertices they traverse in T_w will be ${x}_{i}^{\left(w\right)}$, where x_i for i ∈ [3] are the neighbors of w in R. Note that t₁, t₂ and t₃ each contain at least one vertex in T_w and they jointly traverse all edges in T_w (since U is a Eulerian tour).

Now consider the vertex w. The Eulerian tour U traverses this vertex exactly twice. There are now two options for the trails t₁, t₂, t₃. Either (1) one of the trails contains the vertex w exactly twice or (2) there are exactly two trails that contain the vertex w exactly once.

Now assume U is an SOET with respect to V(R). If option (1) is true the tour U traverses the vertex w twice in succession before traversing any other vertex in the set V'. This is in contradiction with the assumption that U is an SOET. Hence if U is an SOET we must have that option (2) is true, that is, exactly two of the sub-trails t₁, t₂ and t₃ must contain w.

Lets assume without loss of generality that w ∈ t₂ and w ∈ t₃. Hence we have that w ∉ t₁ and thus t₁ induces a sub-word m(t₁) not containing w. The word m(t₁) can be extended to a maximal sub-word X not associated to w but describing a sub-trail traversing vertices in T_w . Therefore by definition 3.3 X describes a sub-trail making a skip at w. Furthermore, m(t₂) has an overlap with two maximal sub-words associated with y₁, w and w, y₂, respectively for some y₁, y₂ ∈ V(R). Similarly for m(t₃) and the vertices z₁, w and w, z₂. Thus, the maximal sub-words that have an overlap with m(t₂) or m(t₃) do not make a skip at w.

Finally, there is no other maximal sub-word describing a sub-trail that traverses a vertex in T_w . The reason for this is that t₁, t₂ and t₃ jointly traverse all edges in T_w , so any maximal sub-trail traversing a vertex in T_w must have an overlap with t₁, t₂ or t₃. The lemma then follows since we found exactly one maximal sub-word describing a sub-trail of U making a skip at w, i.e. the unique one that has an overlap with t₁. □

Now that we have defined the triangular-expansion Λ(R) of a 3-regular graph R and discussed skips we can finally make the central argument of the reduction given in corollary 3.3.1. We begin by proving that if a 3-regular graph is Hamiltonian then its corresponding triangular-expansion Λ(R) allows for an SOET w.r.t. the vertex set V(R). We have the following lemma.

Lemma 3.4. Let R be a 3-regular graph and Λ(R) be a triangular-expansion as defined in definition 3.2. If R is Hamiltonian, then Λ(R) allows for an SOET with respect to V(R). ◊

Proof. Let R be Hamiltonian. This means there exists a Hamiltonian cycle M on R. We will prove that there exists an SOET U with respect to V(R) on the triangular-expansion Λ(R) of R by constructing, from the cycle M on R, a tour U that visits every vertex v ∈ V(R) twice in the same order. We will then argue that this tour can always be lifted to a Eulerian tour and hence can be made into an SOET.

Note first that M induces an ordering on the vertices of R which without of loss of generality we will take to be v₁, ⋯, v_k where k = |V(R)|. Note that for all i ∈ [k − 1] the vertices v_i and v_i+1 are adjacent in R and so are v_k and v₁. Now consider, for each i ∈ [k − 1] the triangle subgraphs ${T}_{{v}_{i}}$, ${T}_{{v}_{i+1}}$ and ${T}_{{\hat{v}}_{i}}$ in the triangular-expansion Λ(R) of R where ${\hat{v}}_{i}$ is the unique vertex adjacent to v_i in R that is not v_i+1 or v_i−1. There are now three cases, depending on the orientation of the triangle subgraph ${T}_{{v}_{i}}$. Either (1) v_i is adjacent to ${v}_{i}^{\left({v}_{i-1}\right)}$ and ${v}_{i}^{\left({v}_{i+1}\right)}$ or (2) v_i is adjacent to ${v}_{i}^{\left({v}_{i-1}\right)}$ and ${v}_{i}^{\left(\hat{{v}_{i}}\right)}$ or (3) v_i is adjacent to ${v}_{i}^{\left({v}_{i+1}\right)}$ and ${v}_{i}^{\left({\hat{v}}_{i}\right)}$.

For case (1), i.e. v_i is adjacent to ${v}_{i}^{\left({v}_{i-1}\right)}$ and ${v}_{i}^{\left({v}_{i+1}\right)}$ in ${T}_{{v}_{i}}$, we can define two edge-disjoint trails on the triangular-expansion Λ(R) by their description as words X _i , X '_i on the vertices of Λ(R).

$$\begin{equation}{\boldsymbol{X}}_{i}={v}_{i}^{\left({v}_{i-1}\right)}{\tilde {v}}_{i}^{\left({v}_{i-1}\right)}{v}_{i}^{\left({\hat{v}}_{i}\right)}{\tilde {v}}_{i}^{\left({v}_{i+1}\right)}{v}_{i}{v}_{i}^{\left({v}_{i+1}\right)}{v}_{i+1}^{\left({v}_{i}\right)},\end{equation} \tag{ 81 }$$

$$\begin{equation}{\boldsymbol{X}}_{i}^{\prime }={v}_{i}^{\left({v}_{i-1}\right)}{v}_{i}{\tilde {v}}_{i}^{\left({v}_{i-1}\right)}{\tilde {v}}_{i}^{\left({v}_{i+1}\right)}{v}_{i}^{\left({v}_{i+1}\right)}{v}_{i+1}^{\left({v}_{i}\right)}.\end{equation} \tag{ 82 }$$

An illustration of the trails described by these words is given in figure 7(a). We can similarly define two edge-disjoint trails for the cases (2) and (3). We will abuse notation and refer to the words as X _i and X _i ' in all three cases. Importantly, for all three of these cases the trails described by X _i , X _i ' are edge-disjoint and both begin at ${v}_{i}^{\left({v}_{i-1}\right)}$ and end at ${v}_{i}^{\left({v}_{i+1}\right)}$. We can extend this definition to trails U₀, U₀' described by the words X ₀ , X ₀ ', also for the case of the adjacent vertices v_k and v₁. Note now that the walk U described by the word W = X ₀ X ₁ ⋯ X _k−1 X '₀ X ₁'⋯ X _k−1 is a tour and moreover that it traverses the vertices in V(R) twice in the order v₁, ⋯, v_k .

The tour U described by the word W above visits every vertex in V(R) such that W [V(R)] = v₁⋯v_k v₁⋯v_k . However it is not yet a Eulerian tour, since it has not traversed all edges in Λ(R). Note that the edges not traversed by U are precisely the triangular-expansions of the edges in R not traversed by the Hamiltonian cycle M. We can easily construct a Eulerian tour out of U by looping over all elements of the word X and whenever we encounter a vertex ${v}_{i}^{\left({\hat{v}}_{i}\right)}$ for all i ∈ [k], where ${\hat{v}}_{i}$ is defined as above, we check whether U already traverses the edges $\left({v}_{i}^{\left({\hat{v}}_{i}\right)},{\hat{v}}_{i}^{\left({v}_{i}\right)}\right)$. If so we continue the loop and if not we insert the trail $\left({v}_{i}^{\left({\hat{v}}_{i}\right)},{\hat{v}}_{i}^{\left({v}_{i}\right)}\right){\hat{v}}^{\left({v}_{i}\right)}\left({v}_{i}^{\left({\hat{v}}_{i}\right)},{\hat{v}}_{i}^{\left({v}_{i}\right)}\right){v}_{i}^{\left({\hat{v}}_{i}\right)}$ into U at this position. This procedure is illustrated in figure 7(b). Now U is Eulerian and hence an SOET. This completes the proof. □

Next we define a special type of SOET on triangular-expanded graphs, which we call HAMSOETs. These special SOETs on a triangular-expanded graph Λ(R), are closely related to Hamiltonian cycles on R. □

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Definition 3.4 (HAMSOET).  Let R be a 3-regular graph and Λ(R) its triangular-expansion. Furthermore, let U be an SOET on Λ(R) with respect to V(R). U is called an HAMSOET with respect to R if, for all vertices u, v ∈ V(R) we have that if u and v are consecutive with respect to the SOET U they are also adjacent in the graph R. ◊

Next we prove that if the triangular-expansion of a 3-regular graph R allows for an HAMSOET with respect to R then the 3-regular graph R is Hamiltonian.

Lemma 3.5. Let R be a 3-regular graph and let Λ(R) be its triangular-expansion. If Λ(R) allows for an HAMSOET with respect to R, then R is Hamiltonian. ◊

Proof. This follows by the definition of an HAMSOET. An HAMSOET U will induce a double occurrence word of the form

$$\begin{equation}m\left(U\right)={\boldsymbol{X}}_{0}{s}_{1}{\boldsymbol{X}}_{1}{s}_{2}\cdots {s}_{k}{\boldsymbol{X}}_{k}{s}_{1}{\boldsymbol{X}}_{1}^{\prime }{s}_{2}\cdots {s}_{k}{\boldsymbol{X}}_{k}^{\prime }.\end{equation} \tag{ 83 }$$

where s₁, ⋯, s_k ∈ V(R) with k = |V(R)| and where X _i and X '_i are maximal sub-words associated to s_i , s_i+1 ∈ V(R). Now consider the induced double occurrence word m(U)[V(R)]. We have

$$\begin{equation}m\left(U\right)\left[V\left(R\right)\right]={s}_{1}\cdots {s}_{k}{s}_{1}\cdots {s}_{k}.\end{equation} \tag{ 84 }$$

Consider now the sub-word s₁⋯s_k of m(U)[V(R)]. This sub-word describes a Hamiltonian cycle on R. To see this, note that each vertex in V(R) occurs exactly once in s₁ s₂⋯s_k . Furthermore, since s_i and s_i+1 are consecutive in U, they are adjacent in R for all i ∈ [k − 1], by definition of an HAMSOET. Finally, the same also holds for s_k and s₁. Hence the tour on R described by s₁⋯s_k visits each vertex in R exactly once and is hence a Hamiltonian cycle. □

We would like to prove that for any 3-regular graph R the existence of an SOET on its triangular-expansion Λ(R) implies the existence of a Hamiltonian cycle on R. So far we have proven this only when Λ(R) allows for an HAMSOET. However, a priori not all SOETs on triangular-expansions Λ(R) have to be HAMSOETs. The reason for this is that consecutive vertices in an SOET are not necessarily adjacent in R, since an SOET can contain true skips.

In the following lemma we will prove that consecutive vertices in an SOET are actually adjacent in R, except for two special cases. These special cases can be remedied since we show that for these cases we can always find a different SOET which is actually an HAMSOET.

The arguments proven below are understood the easiest when one reproduces the visual aids given in figures 9 and 10 as one follows the arguments. We have the following lemma. □

Lemma 3.6. Let R be a 3-regular graph and Λ(R) be its triangular-expansion. If Λ(R) allows for an SOET with respect to V(R), then Λ(R) allows for an HAMSOET with respect to V(R). ◊

Proof. To prove this lemma we will go through the following steps

• Step 1. Note that if two vertices u, v in V(R) (R being a 3-regular graph) are not adjacent in R but are consecutive in an SOET U with respect to V(R) on Λ(R), then the sub-trails of U described by the maximal sub-words associated to u and v must make a non-zero number of true skips in Λ(R)
• Step 2. Argue by contradiction that this non-zero number of true skips can never be greater than one. This argument leverages lemma 3.7 which states that if a sub-trail of an SOET U with respect to V(R) makes true skips at triangle subgraphs ${T}_{{w}_{1}},{T}_{{w}_{2}}$ for w₁, w₂ ∈ V(R) and w₁, w₂ are adjacent in R then w₁, w₂ must actually be consecutive in the SOET U and lemma 3.8 which, in a slightly different initial situation than lemma 3.7, also concludes that two vertices must be consecutive in an SOET U with respect to V(R).
• Step 3. Argue by contradiction that there are only two possible ways for the sub-trails described by the maximal sub-words associated to u and v to make one true skip each. This argument also leverages lemmas 3.7 and 3.8.
• Step 4. Argue that if a triangular-expansion of a 3-regular graph R allows for an SOET U as in step 3, then it also allows for an SOET U' that is an HAMSOET.

Details of step 1. Let R be such that Λ(R) allows for an SOET with respect to V(R). Let U be any such SOET. If U is also an HAMSOET then we are directly done, therefore assume that U is not an HAMSOET. Since U is not an HAMSOET there must exist at least two vertices u, v ∈ V(R) which are consecutive in U, but not adjacent in R. By definition, since u and v are consecutive and U is an SOET, there must exist exactly two different maximal sub-words X , X ' of m(U), associated to u and v. Since u and v are not adjacent in R, the trails described by X , X ' must each traverse vertices in at least one (but possibly more) triangle subgraph that is not T_u or T_v . Since X and X ' are maximal sub-words, they can not contain any vertex in V(R) as a letter and hence the sub-trails described by X , X ' must each make true skips (see definition 3.3) at at least one triangle subgraph.

Assume w.l.o.g. that the sub-trail described by X makes true skips at the triangle subgraphs ${T}_{{s}_{1}},{T}_{{s}_{2}},\cdots ,{T}_{{s}_{r}}$, in this order, and similarly for X ' and ${T}_{{s}_{1}^{\prime }},\cdots ,{T}_{{s}_{{r}^{\prime }}^{\prime }}$. The true skips that the sub-trail described by X makes, must be at different triangle subgraphs than the ones for X ', since there can only be exactly one skip per triangle subgraph, due to lemma 3.3. The triangle subgraphs ${T}_{{s}_{1}},{T}_{{s}_{2}},\cdots ,{T}_{{s}_{r}},{T}_{{s}_{1}^{\prime }},\cdots ,{T}_{{s}_{{r}^{\prime }}^{\prime }}$ are therefore pairwise different. We will call this situation a rr'-skip, which is illustrated in figure 8. Note that by assumption, r and r' are both greater than zero. We will first show that the case where either r or r' is greater than one can never occur if U is an SOET.

Details of step 2. As a visual aid for the following argument, refer to figure 9. We will make an argument by contradiction. Therefore let r be strictly greater than one. Consider the sub-trail described by the sub-word X defined above. By assumption this sub-trail makes true skips at at least two adjacent triangle subgraphs. Take these to be ${T}_{{s}_{1}}$ and ${T}_{{s}_{2}}$. As we will prove in lemma 3.7, if two adjacent triangle subgraphs ${T}_{{s}_{1}},{T}_{{s}_{2}}$ contain true skips, then s₁ and s₂ must be consecutive in the SOET U and that there is some maximal sub-word Y of m(U) associated to s₁, s₂ that contains the sub-word ${s}_{1}^{\left({s}_{2}\right)}{s}_{2}^{\left({s}_{1}\right)}$.

This implies that (by definition of SOET) there is some other maximal sub-word Y ' of m(U) associated to s₁ and s₂. This sub-word Y ' cannot contain the sub-word ${s}_{1}^{\left({s}_{2}\right)}{s}_{2}^{\left({s}_{1}\right)}$, as ${s}_{1}^{\left({s}_{2}\right)}{s}_{2}^{\left({s}_{1}\right)}$ already appears in both the maximal sub-word X and the maximal sub-word Y .

This implies the sub-trail described by Y ' must make a nonzero number of true skips at triangle subgraphs ${T}_{{\hat{s}}_{1}},\cdots ,{T}_{{\hat{s}}_{k}}$ in that order. Now consider the triangle subgraphs ${T}_{{s}_{1}}$ and ${T}_{{\hat{s}}_{1}}$. As we will prove below, lemma 3.8 applied to the triangle subgraphs ${T}_{{s}_{1}},{T}_{{\hat{s}}_{1}}$ implies that the vertices s₁ and ₁ must be consecutive in U. Call the maximal sub-word of m(U) associated to these vertices W ₀. Moreover, by lemma 3.8 this maximal sub-word contains the sub-word ${s}_{1}^{\left({\hat{s}}_{1}\right)}{\hat{s}}_{1}^{\left({s}_{1}\right)}$. Similarly we can see that the vertices s₂ and _k must be consecutive in U. Call the maximal sub-word of m(U) associated to these vertices W _k . By lemma 3.8 this maximal sub-word contains the sub-word ${s}_{2}^{\left({\hat{s}}_{k+1}\right)}{\hat{s}}_{k+1}^{\left({s}_{2}\right)}$.

By applying lemma 3.8 again to all triangle subgraph pairs ${T}_{{\hat{s}}_{i}},{T}_{{\hat{s}}_{i+1}}$ for i ∈ [k − 1] we come to the conclusion that _i , _i+1 for i ∈ [k − 1] must be consecutive in U. Call the maximal sub-word of m(U) associated to these vertices W _i . By lemma 3.8 these maximal sub-words contain the sub-word ${\hat{s}}_{i}^{\left({\hat{s}}_{i+1}\right)}{\hat{s}}_{i+1}^{\left({\hat{s}}_{i}\right)}$. This implies that U must traverse the vertices s₁, ₁, ⋯, _k , s₂, s₁, ₁, ⋯, _k , s₂ in order. Since by construction $\left\{{s}_{1},{s}_{2},{\hat{s}}_{1},\cdots ,\hat{{s}_{k}}\right\}\ne V\left(R\right)$ we have that U is not a valid SOET on V(R) (this can be seen by noting that e.g. ${T}_{{s}_{1}^{\prime }}$ already contains a true skip, hence s'₁ cannot be part of the set). Hence by contradiction we must have r ⩽ 1. We can make the same argument for r' yielding r' ⩽ 1.

Details of step 3. As a visual aid for the following argument, refer to figure 10. Now let r = r' = 1. This means the sub-trails described by the maximal sub-words X and X ' associated to the vertices u and v make exactly one true skip each at triangle subgraphs triangle subgraphs ${T}_{{s}_{1}},{T}_{{s}_{1}^{\prime }}$ respectively. We will now argue that there are essentially only two ways that an SOET U with the above properties can exist on Λ(R). This argument will again go in steps:

• Step 3.1. Argue that the SOET U must have a maximal sub-word associated to the vertex s₁ and some other vertex x that describes a trail traversing the edge connecting the triangle subgraphs T_u and ${T}_{{s}_{1}}$.
• Step 3.2. Argue that this sub-trail must make a true skip at the triangle subgraph T_u . Note that this is equivalent to arguing x ≠ u.
• Step 3.3. Apply the same sequence of arguments for the vertices s₁ and v and also for the vertices s'₁ and u and s'₁ and v.
• Step 3.4. Conclude from the fact that the SOET U can only make a single true skip at the triangle subgraphs T_u and T_v that s₁ and s'₁ must be consecutive in U and moreover that the maximal sub-words Z , Z ' associated to s₁ and s'₂ must make true skips at T_u and T_v respectively.

Details of step 3.1. Consider the second edge connecting the triangle subgraphs T_u and ${T}_{{s}_{1}}$. This is the edge $\left({u}^{\left({s}_{1}\right)},{s}_{1}^{\left(u\right)}\right)$. Since the SOET U is Eulerian it must traverse this edge. Note also that ${T}_{{s}_{1}}$ already contains a true skip (made by the sub-trail described by X ). This means, by lemma 3.3 that any sub-trail of U crossing the edge $\left({u}^{\left({s}_{1}\right)},{s}_{1}^{\left(u\right)}\right)$ connecting ${T}_{u},{T}_{{s}_{1}}$ must traverse the vertex s₁. Let Z be the maximal sub-word describing the sub-trail starting at s₁ and containing the sub-word ${u}^{\left({s}_{1}\right)}{s}_{1}^{\left(u\right)}$. This maximal sub-word is (by definition) associated to two vertices in V(R). One of these vertices is s₁ and will label the other one x.

We will now argue that x ≠ u and hence that the sub-trail described by Z makes a true skip at T_u . We do this by contradiction in the following argument.

Details of step 3.2. For this part of the argument assume that x = u. This means that u is consecutive to both s₁ and v. This means there must be some other maximal sub-word Z ' associated to s₁ and u. Note that this sub-word cannot contain the sub-word ${u}^{\left({s}_{1}\right)}{s}_{1}^{\left(u\right)}$ as it is already contained in the maximal sub-words Z and X '.

Now consider the unique third vertex that is adjacent to u in R (the vertex adjacent to u which is not s₁ or s'₁). Let us label this vertex w.

Since ${T}_{{s}_{1}^{\prime }}$ already contains a true skip (which implies Z ' cannot connect to u by making a true skip at ${T}_{{s}_{1}^{\prime }}$) the sub-trail described by the maximal sub-word Z ' must make a true skip at w. Now consider the maximal sub-word Y of m(U) that that describes a sub-trail traversing the unused edge between ${T}_{{s}_{1}^{\prime }}$ and T_u , i.e. Y contains the sub-word ${u}^{\left({s}_{1}^{\prime }\right)}{{s}^{\prime }}_{1}^{\left(u\right)}$. This sub-trail must be associated to s'₁ (since ${T}_{{s}_{1}^{\prime }}$ already contains a true skip) and can not not be associated to u since u is already consecutive with two vertices in V(R). This means the sub-trail described by the maximal sub-word Y must make a true skip at the triangle subgraph T_u . Since the sub-word ${u}^{\left({s}_{1}\right)}{s}_{1}^{\left(u\right)}$ is already contained in Z and X ' the maximal sub-word Y must contain the sub-word u^(w) w^(u). Since T_w already contains a true skip this implies that w must be associated to the maximal sub-word Y and hence that w and s'₁ are consecutive. This means there must be a second maximal sub-word Y ' associated to w and s'₁.

Since both edges connecting T_u and T_w have already been traversed by, the sub-trail described by the maximal sub-word Y ' must make true skips on some triangle subgraphs ${T}_{{\hat{s}}_{1}},\cdots ,{T}_{{\hat{s}}_{k}}$.

There are now two possibilities. Either (1) we have that ₁ = v (see figure 10(a)) or (2) that ₁ ≠ v (see figure 10(b)). We will now consider both of these cases:

• (a)  
  If ₁ = v we can apply lemma 3.8 to the vertices v and s₁ to conclude that ₁ = v implies that v and s₁ are consecutive. Call the maximal sub-word connecting them W . We now have that the vertices u, v and s₁ are pairwise consecutive to each other. Since {u, v, s₁} is a strict subset of V(R), U cannot be an an SOET which is a contradiction.
• (b)  
  Now assume that v ≠ ₁. By lemma 3.8 we can now conclude that w and _k must be consecutive and that s'₁ and ₁ ≠ v must be consecutive. We have to perform one last construction to prove the lemma. This construction is visualized in figure 10(b). Call the maximal sub-words associated to these vertex pairs w and _k and s'₁ and ₁ W _k and W ₀ respectively. We can again use lemma 3.7 to conclude that _i is consecutive to _i+1 for all i ∈ [k − 1]. Call the maximal sub-words associated to the vertices _i and _i+1 W _i . This implies that U must traverse the vertices s'₁, ₁, ⋯, _k , w, s'₁, ₁, ⋯, _k , w in order. Since by construction $\left\{{s}_{1}^{\prime },{s}_{w},{\hat{s}}_{1},\cdots ,\hat{{s}_{k}}\right\}\ne V\left(R\right)$ we have that U is not a valid SOET on V(R) (this can be seen by noting that e.g. T_u already contains a true skip, hence u can not be part of the set).

Hence in both cases we arrive at a contradiction. This means by contradiction that x ≠ u and thus that the sub-trail described by Z makes a true skip at T_u .

Details of step 3.3. Now similarly to step 3.1 consider the edge connecting ${T}_{{s}_{1}}$ and T_v . We can again argue that there must exist a maximal sub-word Z ' associated to the vertex s₁ and some other vertex x' that describes a sub-trail that traverses this edge. By the same argument as step 3.2 we can conclude that x' ≠ v and thus that Z ' describes a sub-trail making a true skip at T_v .

We can make the same argument for the vertex s'₁ establishing the existence of maximal sub-words , ' that describe sub-trails making true skips at T_u and T_v respectively.

Details of step 3.4. Note that the sub-trails described by Z and make true skips at the triangle subgraph T_u . Since T_u , by lemma 3.3 can only contain a single true skip and since Z , are maximal sub-words we must have that Z ∼ and thus that the vertices s₁ and s'₁ are consecutive with respect to the SOET U. However since we must also conclude that Z ' ∼ ' we have that s₁ and s'₁ are consecutive and have two maximal sub-words associated to them. Since there are no further constraints on U imposed by the the fact that the sub-words X , X' associated to the vertices v, u make true skips. Therefore SOETs with this type of behavior are in fact allowed. These SOETs can also be found explicitly, as can be sen in figure 11. If an SOET U has this type of behavior (u and v are not adjacent in R but are consecutive in the SOET U on Λ(R)) we say that the SOET U has a valid 11-skip. Next we show that if an SOET U has a valid 11-skip, and thus is not an HAMSOET, it can always be turned into an HAMSOET by applying a fixed set of local complementation.

Details of step 4. We now show that an SOET with valid 11-skips can be turned into an HAMSOET. The two possibilities for an SOET with a valid 11-skip are shown in figure 11. Note that these possibilities have the same 'local' structure, the only difference is how the rest of the SOET U is connected to the valid 11-skip. We first show the procedure that should be applied if the 11-skip is of the form in figure 11(a).

The SOET U in figure 11(a) is of the form

$$\begin{equation}m\left(U\right)={\boldsymbol{U}}_{1}{v}^{\left({s}_{1}\right)}{\boldsymbol{U}}_{2}{v}^{\left({s}_{1}\right)}{\boldsymbol{U}}_{3}{s}_{1}^{\prime \left(u\right)}{\boldsymbol{U}}_{4}{s}_{1}^{\prime \left(u\right)}{\boldsymbol{U}}_{5}\end{equation} \tag{ 85 }$$

where

$$\begin{equation}{\boldsymbol{U}}_{1}=u{u}^{\left({s}_{1}\right)}{s}_{1}^{\left(u\right)}{s}_{1}^{\left(v\right)}\end{equation} \tag{ 86 }$$

$$\begin{equation}{\boldsymbol{U}}_{2}=v{v}^{\left(X\right)}{\boldsymbol{X}}_{\boldsymbol{1}}{s}_{1}^{\left(X\right)}{s}_{1}{s}_{1}^{\left(v\right)}\end{equation} \tag{ 87 }$$

$$\begin{equation}{\boldsymbol{U}}_{3}={v}^{\left({s}_{1}^{\prime }\right)}{s}_{1}^{\prime \left(v\right)}{s}_{1}^{\prime }{s}_{1}^{\prime \left({X}^{\prime }\right)}{\boldsymbol{X}}_{\boldsymbol{1}}^{\prime }{u}^{\left({X}^{\prime }\right)}u{u}^{\left({s}_{1}^{\prime }\right)}\end{equation} \tag{ 88 }$$

$$\begin{equation}{\boldsymbol{U}}_{4}={s}_{1}^{\prime \left(v\right)}{v}^{\left({s}_{1}^{\prime }\right)}v{v}^{\left(X\right)}{\boldsymbol{X}}_{2}{s}_{1}^{\left(X\right)}{s}_{1}{s}_{1}^{\left(u\right)}{u}^{\left({s}_{1}\right)}{u}^{\left({s}_{1}^{\prime }\right)}\end{equation} \tag{ 89 }$$

$$\begin{equation}{\boldsymbol{U}}_{5}={s}_{1}^{\prime }{s}_{1}^{\prime \left({X}^{\prime }\right)}{\boldsymbol{X}}_{\boldsymbol{2}}^{\prime }{u}^{\left({X}^{\prime }\right)}\end{equation} \tag{ 90 }$$

where X ₁ is the word associated to the sub-trail connecting v^(X) and ${s}_{1}^{\left(X\right)}$ as seen in figure 11(a) and similarly for X ₁ ', X ₂ , X ₂ '. Since our goal is to make this an HAMSOET we need to have that pairs of vertices in V(R) are only consecutive w.r.t. U if they are adjacent in R. This can be done by applying $\overline{\tau }$-operations to U at ${v}^{\left({s}_{1}\right)}$ and ${s}_{1}^{\prime \left(u\right)}$. The Eulerian tour U' after these operations will be described by

$$\begin{equation}m\left({U}^{\prime }\right)=m\left({\overline{\tau }}_{\left({v}^{\left({s}_{1}\right)},{s}_{1}^{\prime \left(u\right)}\right)}\left(U\right)\right)={\boldsymbol{U}}_{1}{v}^{\left({s}_{1}\right)}\tilde {{\boldsymbol{U}}_{2}}{v}^{\left({s}_{1}\right)}{\boldsymbol{U}}_{3}{s}_{1}^{\prime \left(u\right)}\tilde {{\boldsymbol{U}}_{4}}{s}_{1}^{\prime \left(u\right)}{\boldsymbol{U}}_{5}\end{equation} \tag{ 91 }$$

where the over-set tilde indicates the mirror-inverting of a sub-word. Note that neither (u, v) or (s₁, s'₁) are consecutive anymore, but instead (u, s₁) and (v, s'₁) are now consecutive w.r.t. U'. To make sure that this procedure works we need to check two things: (1) U' is an SOET and (2) there are no additional consecutive pairs of vertices in U' that are not adjacent in R. To do this, lets look at the order U and U' traverse the vertices in V(R), i.e. we will look at m(U)[V(R)] and m(U')[V(R)]. Since U is an SOET we must have that X ₁[V(R)] = X ₂[V(R)] and similarly X'₁[V(R)] = X '₂[V(R)]. Lets denote these words by X _V = X ₁[V(R)] and X '_V = X'₁[V(R)]. We then have that the double occurrence word of m(U) induced by V(R) is

$$\begin{equation}m\left(U\right)\left[V\left(R\right)\right]=uv{\boldsymbol{X}}_{V}{s}_{1}{s}_{1}^{\prime }{\boldsymbol{X}}_{V}^{\prime }uv{\boldsymbol{X}}_{V}{s}_{1}{s}_{1}^{\prime }{\boldsymbol{X}}_{V}^{\prime }\end{equation} \tag{ 92 }$$

and similarly for U' we have

$$\begin{equation}m\left({U}^{\prime }\right)\left[V\left(R\right)\right]=u{s}_{1}\tilde {{\boldsymbol{X}}_{V}}v{s}_{1}^{\prime }{\boldsymbol{X}}_{V}^{\prime }u{s}_{1}\tilde {{\boldsymbol{X}}_{V}}v{s}_{1}^{\prime }{\boldsymbol{X}}_{V}^{\prime }\end{equation} \tag{ 93 }$$

It is therefore clear that the Eulerian tour U' is an SOET. Furthermore the only consecutive pairs of vertices in U' which where not consecutive in U are (u, s₁) and (v, s'₁). Since (u, s₁) and (v, s'₁) are edges of R we see that we can iteratively apply this procedure to any valid 11-skip as in figure 11(a) and turn the SOET into an HAMSOET. Similarly the SOET in figure 11(b) can be turned into an HAMSOET by applying τ-operations to the vertices ${s}_{1}^{\left(u\right)}$ and ${v}^{\left({s}_{1}^{\prime }\right)}$. One can explicitly check this by applying the operations to U in figure 11(b) which is given by

$$\begin{equation}m\left(U\right)={\boldsymbol{U}}_{\boldsymbol{1}}{s}_{1}^{\left(u\right)}{\boldsymbol{U}}_{\boldsymbol{2}}{s}_{1}^{\left(u\right)}{\boldsymbol{U}}_{\boldsymbol{3}}{v}^{\left({s}_{1}^{\prime }\right)}{\boldsymbol{U}}_{\boldsymbol{4}}{v}^{\left({s}_{1}^{\prime }\right)}{\boldsymbol{U}}_{\boldsymbol{5}}\end{equation} \tag{ 94 }$$

where

$$\begin{equation}{\boldsymbol{U}}_{1}=u{u}^{\left({s}_{1}\right)}\end{equation} \tag{ 95 }$$

$$\begin{equation}{\boldsymbol{U}}_{2}={s}_{1}^{\left(v\right)}{v}^{\left({s}_{1}\right)}v{v}^{\left(X\right)}{\boldsymbol{X}}_{1}{s}_{1}^{\left(X\right)}{s}_{1}\end{equation} \tag{ 96 }$$

$$\begin{equation}{\boldsymbol{U}}_{3}={u}^{\left({s}_{1}\right)}{u}^{\left({s}_{1}^{\prime }\right)}{s}_{1}^{\prime \left(u\right)}{s}_{1}^{\prime }{s}_{1}^{\prime \left({X}^{\prime }\right)}{\boldsymbol{X}}_{1}^{\prime }{u}^{\left({X}^{\prime }\right)}u{u}^{\left({s}_{1}^{\prime }\right)}{s}_{1}^{\prime \left(u\right)}{s}_{1}^{\prime \left(v\right)}\end{equation} \tag{ 97 }$$

$$\begin{equation}{\boldsymbol{U}}_{4}=v{v}^{\left(X\right)}{\boldsymbol{X}}_{2}{s}_{1}^{\left(X\right)}{s}_{1}{s}_{1}^{\left(v\right)}{v}^{\left({s}_{1}\right)}\end{equation} \tag{ 98 }$$

$$\begin{equation}{\boldsymbol{U}}_{5}={s}_{1}^{\prime \left(v\right)}{s}_{1}^{\prime }{s}_{1}^{\prime \left({X}^{\prime }\right)}{\boldsymbol{X}}_{2}^{\prime }\end{equation} \tag{ 99 }$$

with everything defined similarly to the case of figure 11(b). Going through a similar argument as above we can show that we can also turn the SOET U into an HAMSOET. This completes the lemma. □

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Lemma 3.7. Let R be a 3-regular graph and Λ(R) be its triangular-expansion. Also let u, v be adjacent vertices on R. Let U be an SOET on Λ(R) with respect to V(R). Let X be a maximal sub-word of m(U) not associated to u and/or v containing u^(v) v^(u) and describing a sub-trail that makes true skips at T_u and T_v . Then u and v are consecutive in U and moreover m(U) contains a sub-word of the form

$$\begin{equation}u{\boldsymbol{Z}}_{1}{u}^{\left(v\right)}{v}^{\left(u\right)}{\boldsymbol{Z}}_{2}v,\quad {\boldsymbol{Z}}_{1}\subset V\left({T}_{u}\right),\enspace {\boldsymbol{Z}}_{2}\subset V\left({T}_{v}\right),\end{equation} \tag{ 100 }$$

◊

Proof. The situation described in the lemma is described graphically in figure 12. Because U is an SOET the sub-word v^(u) u^(v) must be contained exactly twice in m(U). Note that at most one of these instances can be contained in the maximal sub-word X . The other instance must be contained in a different maximal sub-word Z. This maximal sub-word will be associated to two vertices w₁, w₂. Note that since v^(u) u^(v) ∈ Z and v^(u) u^(v) ∈ X either we must have that w₁ = u or that the sub-trail described by the maximal sub-word Z makes a true skip at T_u . Since T_u already contains a true skip (made by the sub-trail described by X ) we must have that w₁ = u. We can make the same argument for the vertex v. This means the maximal sub-word Z must be associated to u and v and hence that u, v must be consecutive in U and moreover we have that

$$\begin{equation}\boldsymbol{Z}={\boldsymbol{Z}}_{1}{u}^{\left(v\right)}{v}^{\left(u\right)}{\boldsymbol{Z}}_{2},\quad {\boldsymbol{Z}}_{1}\subset V\left({T}_{u}\right){\backslash}\left\{u\right\},\enspace {\boldsymbol{Z}}_{2}\subset V\left({T}_{v}\right){\backslash}\left\{v\right\}.\end{equation} \tag{ 101 }$$

□

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Lemma 3.8. Let R be a 3-regular graph and Λ(R) be its triangular-expansion. Also let u, v be adjacent vertices on R. Also take x₁, x₂ to be the vertices adjacent to u in R such that x₁ ≠ v, x₂ ≠ v. Let U be an SOET on Λ(R) with respect to V(R). Let Y be a maximal sub-word of m(U) not associated to u and/or v containing ${u}^{\left({x}_{1}\right)}{x}_{1}^{\left(u\right)}$ and ${u}^{\left({x}_{2}\right)}{x}_{2}^{\left(u\right)}$ and describing a sub-trail making a true skip at T_u . Also let X be a maximal sub-word associated to u and a vertex x₃ ≠ v that describes a sub-trail making a true skip at v. Then u, v are consecutive and moreover m(U) contains a sub-word of the form

$$\begin{equation}u{\boldsymbol{Z}}_{1}{u}^{\left(v\right)}{v}^{\left(u\right)}{\boldsymbol{Z}}_{2}v,\quad {\boldsymbol{Z}}_{1}\subset V\left({T}_{u}\right),\enspace {\boldsymbol{Z}}_{2}\subset V\left({T}_{v}\right),\end{equation} \tag{ 102 }$$

◊

Proof. The situation described in the lemma is described graphically in figure 13. Because U is an SOET the sub-word v^(u) u^(v) must be contained exactly twice in m(U). Note that at most one of these instances can be contained in the maximal sub-word Y and none can be contained in the maximal sub-word X . This means there must be a maximal sub-word Z of m(U) (different from X and X ) containing v^(u) u^(v). This maximal sub-word must again be associated with two vertices $x,\hat{x}$. If these vertices are not u, v then the sub-trail described by Z must make true skips at T_u , T_v or both. Since both of these triangle subgraphs already contain true skips this is not possible and hence Z must be associated to u and v which means they are consecutive. Moreover, by construction of Z we have

$$\begin{equation}\boldsymbol{Z}={\boldsymbol{Z}}_{1}{u}^{\left(v\right)}{v}^{\left(u\right)}{\boldsymbol{Z}}_{2},\quad {\boldsymbol{Z}}_{1}\subset V\left({T}_{u}\right){\backslash}\left\{u\right\},\enspace {\boldsymbol{Z}}_{2}\subset V\left({T}_{v}\right){\backslash}\left\{v\right\}.\end{equation} \tag{ 103 }$$

□

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We first give a rough sketch of the idea behind the algorithm. Remember that the task of the algorithm is to find a sequence of local complementations τ _v such that the induced subgraph of τ _v (G) on the vertices V' is a star graph.

The algorithm starts by choosing a one vertex c in V' which will become the center of the star graph on V'. It then proceeds by picking different vertices v ∈ V' and making them adjacent to c by performing local complementations. After every vertex that is made adjacent the algorithm will check if the induced subgraph on the neighborhood of c is a star graph. If it is not it will attempt to turn it into a star graph by local complementations. If it fails at doing so it will raise an error and if it succeeds it will pick another vertex in V' and repeat the procedure until all vertices in V' are in the neighborhood of c. We will often call this process of making a vertex v adjacent to c 'adding' the vertex v to the star graph. To understand when the algorithm might fail we now zoom in on the situation where all but one vertex of V' has been added to the neighborhood of c. Let us call this vertex f. At this point in the algorithm the induced subgraph G[V'{f}] is already a star graph (be previous successful iterations of this procedure).

The task is now to turn G[V'] into a star graph by making f adjacent to the center c of G[V'\{f}] but to no other vertex of V', and at the same time not change any edges in G[V'\{f}]. This will be done in two steps, which are explained further below:

• (a)  
  Make f and c adjacent, without changing any edges in G[V'\{f}]. The star graph S_V' is then a subgraph of the graph, but not necessarily an induced subgraph, since f could be also be adjacent to other vertices in V' than c. We will call these edges between f and vertices in V'\{f} bad edges. This first step is the task of algorithm 2 below. Interestingly, this step always succeeds if the graph is connected, even if the graph is not distance-hereditary.
• (b)  
  Remove the bad edges, without changing any other edges between vertices in V'. The removal of the bad edges is the task of algorithm 1 below. Algorithm 1 tries to remove the bad edges by checking a few cases. Thus, one of the main results of this section is to prove that these cases provide a necessary condition for S_V' being a vertex-minor of G.

We will now describe the two main steps above of the algorithm in more detail. Let us denote the vertices as above and furthermore the current leaves in G[V'\{f}] as V'\{c, f} = {l₁, ⋯, l_k }.

Details of step 1: The vertices c and f are made adjacent by performing local complementations along the shortest path P from c and f, see figure 14(a). The operations along the path P will in fact be either pivots, i.e. ρ_(u,v) = τ_v ◦τ_u ◦τ_v , or single local complementations depending on the situation. The reason for this is to not remove edges between c and the l_i 's. The details of the operations along the path P are given in algorithm 2 together with the proof in section section 4.1.3.

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As mentioned above, by making f and c adjacent we might have also added bad edges between f and some of the vertices ${\left\{{l}_{i}\right\}}_{i}$, see figure 14(b). Let us denote the set of vertices which are incident to a bad edge by B and the set of vertices not incident to a bad edge, apart from c, by L = (V'\{c})\B. We call such a graph as the induced subgraph on V' a star–star graph, see definition 4.1.

Next, we must remove the bad edges in order to turn G[V'] into a star graph. Let G now be the graph with f and c adjacent but with possibly some bad edges.

Details of step 2: In this step we will remove the bad edges, if we can. A situation where bad edges can be removed, as we will show, is when there exists a vertex u ∉ V', which is adjacent to all vertices in B but not to any vertex in L. The existence of such a vertex u is thus a sufficient condition for the removal of bad edges. When G is a distance hereditary graph, it turns out that this condition is also necessary, that is if no such vertex u exists, then the star graph on V' is not a vertex-minor of G, and we can stop the algorithm. This is shown in detail in section 4.1.3. For this statement to hold L cannot be empty, but this can always be achieved by performing a local complementation at c first if needed, which is done in line 14 in algorithm 1. Assume that there indeed exist such a vertex u, i.e.

$$\begin{equation}\left(L\ne \varnothing \right)\enspace \wedge \enspace \left(u\notin {V}^{\prime }\right)\enspace \wedge \enspace \left(B\subseteq {N}_{u}\right)\enspace \wedge \enspace \left(L\cap {N}_{u}=\varnothing \right)\end{equation} \tag{ 104 }$$

see figure 15. Now u can be adjacent to c or not. Let us consider these cases separately:

• Case 1 u and c are not adjacent:Remember that f is the center of the induced star graph G[B]. If a local complementation is performed at u, the bad edges are removed but new ones will be created between the vertices in B\{f}. These new bad edges will then form a complete graph on B\{f} and we call such a graph on the vertices V'\{f} a complete-star graph, see definition 4.2.

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Performing the same step again, i.e. doing a local complementation at another vertex adjacent to all vertices in B\{f}, will remove all bad edges. We have then produced the star graph on V' in two steps.

• Case 2: u and c are adjacent:In this case, if a local complementation is performed at u, some edges between c and vertices in L will be removed, which is not desired. We can solve this by finding another vertex h adjacent to both u and c but not to any other vertex in V', by which we can remove the edge (u, c), see figure 16. In the following section we show that if there is no vertex h of this form, the star graph is not a vertex-minor of G and we can stop the algorithm.

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To prove that the algorithm is correct we need to show that cases checked by algorithm 2 to remove the bad edges actually provides a necessary condition for S_V' being a vertex-minor of G. To be precise, we will show that a necessary ⁸ condition for the star graph on V' being a vertex-minor of G is

$$\begin{equation}\mathcal{P}\left(B,L,c\right)=\exists u\in V{\backslash}{V}^{\prime }:\left(B\subseteq {N}_{u}\enspace \wedge \enspace L\cap {N}_{u}=\varnothing \enspace \wedge \enspace \left(\left(u,c\right)\notin E\enspace \vee \enspace \exists h:\left(h\in {N}_{u}\cap {N}_{c}{\backslash}\bigcup _{x\in {V}^{\prime }{\backslash}\left\{c\right\}} {N}_{x}\right)\right)\right),\end{equation} \tag{ 105 }$$

where V' = B ∪ L ∪ {c} and L is assumed to be nonempty. It is important to note here that this condition is only valid if the graph is in the correct form, i.e. the induced subgraph on V' forms a star–star graph or a complete-star graph.

We formally state that equation (105) is a necessary condition for the star graph on V' being a vertex-minor of G in theorem 4.2, which we prove in section 4.1.3. The theorem uses the notion of star–star and complete-star graphs which we formally define as:

Definition 4.1 (Star–star graph).  A graph G = (V, E) is called a star–star graph if there exist two subsets B and L and a vertex c, such that {B, L, {c}} form a partition of V and |B| > 1. Furthermore N_l = {c} ∀ l ∈ L and c ∈ N_b ∀ b ∈ B. Finally G[B] = S_B . Such a graph is denoted SS_(B,L,c). ◊

Definition 4.2 (Complete-star graph).  A graph G = (V, E) is called a complete-star graph if there exist two subsets B and L and a vertex c, such that {B, L, {c}} form a partition of V and |B| > 1. Furthermore N_l = {c}, ∀ l ∈ L and c ∈ N_b , ∀ b ∈ B. Finally G[B] = K_B . Such a graph is denoted KS_(B,L,c). Note that if |B| = 2, G is also a star–star graph.◊

Theorem 4.2. Let G be a distance-hereditary graph on the vertices V and let V' be a subset of V. Furthermore, let V' = B ∪ L ∪ {c} be a partition of V' and let S_V' be a star graph on the vertices V'. Then the following statements hold

• If G[V'] = SS_(B,L,c) is a star–star graph and |B| = 2, then
  $$\begin{equation}\mathcal{P}\left(B,L,c\right){\Leftrightarrow}{S}_{{V}^{\prime }}{< }G.\end{equation} \tag{ 106 }$$
• If G[V'] = SS_(B,L,c) is a star–star graph then
  $$\begin{equation}{\neg}\mathcal{P}\left(B,L,c\right){\Rightarrow}{S}_{{V}^{\prime }}\nless G.\end{equation} \tag{ 107 }$$
• If G[V'] = SS_(B,L,c) is a star–star graph and f is the center of the star graph G[B], then
  $$\begin{equation}\mathcal{P}\left(B,L,c\right){\Leftrightarrow}{\text{KS}}_{\left(B{\backslash}\left\{f\right\},L\cup \left\{f\right\},c\right)}{< }G.\end{equation} \tag{ 108 }$$
• If G[V'] = KS_(B,L,c) is a complete-star graph then
  $$\begin{equation}\mathcal{P}\left(B,L,c\right){\Leftrightarrow}{S}_{{V}^{\prime }}{< }G.\end{equation} \tag{ 109 }$$

◊

Theorem 4.2 implicitly gives a necessary and sufficient condition for when S_V' is a vertex-minor of G, if G[V'] is a star–star graph. More precisely, if the induced subgraph on V' is a star–star graph and $\mathcal{P}\left(B,L,c\right)$ is true then we know that local complementations can be performed to turn the induced subgraph on V'\{f} into a complete-star graph, see equation (108). Then, if $\mathcal{P}\left(B{\backslash}\left\{f\right\},L\cup \left\{f\right\},c\right)$ is again true, then a star graph can be created on V' by performing further local complementations, see equation (109). If in any of these two steps, $\mathcal{P}\left(B,L,c\right)$ or $\mathcal{P}\left(B{\backslash}\left\{f\right\},L\cup \left\{f\right\},c\right)$ is false, then S_V' is not a vertex-minor of G, see equations (107) and (109). In section 4.1.3 we prove these statements.

The algorithm described in the previous section checks if a star graph with vertex set V' is a vertex-minor of a distance-hereditary graph G. Here we show that the runtime of this algorithm is $\mathcal{O}\left(\vert {V}^{\prime }\vert \vert V\left(G\right){\vert }^{3}\right)$. We will represent subsets of a base-set as unsorted binary lists ⁹ , where 1 indicates that an element in the base-set is in the represented set and 0 that an element in the base-set is not in the represented set. This will be the case both for sets of vertices and sets of edges. The base-set for sets of vertices will be the set of vertices V(G) of the input graph G and the base-set for edges-sets will be V(G) × V(G). Thus, we assume that the input graph G is given as an unsorted binary list, of length |V(G)|², indicating which edges are in E(G). This allows us to check if an edge (u, v) is in the graph or not in constant time. Furthermore, we assume that the input-set V' is also represented as an unsorted binary list, of length |V(G)|, indicating which of the vertices of G are in V'. We also assume that the size of |V'| is given together with its representation, which allows us to faster create representations of subsets of V'.

Sets used internally by the algorithm (B, L and U) will also be represented as unsorted binary lists together with the size of the sets. The sizes of the sets will be updated accordingly whenever an element is added. Note that B and L are subsets of V' and will therefore be represented as unsorted binary lists, of length |V'|, indicating which elements of V' are in these sets. However, U is not a subset of V' and will therefore be represented by an unsorted binary list of length |V|. Thus, given a vertex v, checking if v is in a set of vertices V can be done in constant time and adding a vertex to a set can be done in constant time (flipping the bit at the corresponding position). Furthermore, iterating over elements in a set can be done in linear time with respect to the base-set, i.e. $\mathcal{O}\left(\vert V\left(G\right)\vert \right)$ for V' and U and $\mathcal{O}\left(\vert {V}^{\prime }\vert \right)$ for B and L.

As described, the full algorithm starts by calling algorithm 1, which in turn calls algorithm 2, which again calls algorithm 1 and so on. We will see that the computation that dominates the runtime is updating the graph τ _v (G) whenever v is concatenated, as in line 13 of algorithm 1 and line 6 of algorithm 2. We will assume that both algorithms 1 and 2 have access to a common graph which they can update to τ _v (G), whenever v is concatenated, to prevent this from being done for the whole sequence v every time. Note that τ _v (G) takes up the same amount of space regardless of v . Each local complementation in the sequence can be performed in time $\mathcal{O}\left(\vert V\left(G\right){\vert }^{2}\right)$ [10]. Since algorithms 1 and 2 increase the length of v by $\mathcal{O}\left(1\right)$ and $\mathcal{O}\left(\vert V\left(G\right)\vert \right)$ respectively each call, the runtime to update the graph τ _v (G) is $\mathcal{O}\left(\vert V\left(G\right){\vert }^{3}\right)$. We will now show that all other parts of both algorithms 1 and 2 takes time less than $\mathcal{O}\left(\vert V\left(G\right){\vert }^{3}\right)$, which will imply that the total runtime is $\mathcal{O}\left(\vert {V}^{\prime }\vert \vert V\left(G\right){\vert }^{3}\right)$ since algorithm 2 is called $\mathcal{O}\left(\vert {V}^{\prime }\vert \right)$ times ¹⁰ . Let us start by going through the runtime of algorithm 1 line by line:

• Line 6 (and 15, 24): checking if there is only one element (or none) in V' (in B, in U) can be done in constant time, since we keep track of the sizes of these sets.
• Line 11: finding a vertex c ∈ V' adjacent to all vertices in V' (except itself) can be done in time $\mathcal{O}\left(\vert {V}^{\prime }{\vert }^{2}\right)$ by checking for each vertex v in V' if τ _v (G) contains all edges in the set {(v, w) : w ∈ V'\{v}}. Let c be the first such vertex v.
• Line 13 and 14: constructing the sets B and L can be done in time $\mathcal{O}\left(\vert {V}^{\prime }{\vert }^{2}\right)$ by checking, for each vertex v in V'\{c}, if τ _v (G) contains at least one edge from the set $\left\{\right. \left(v,w\right):w\in {V}^{\prime }{\backslash}\left\{c\right\}$. If this is the case, v will be added to the array representing B, otherwise v will be added to L.
• Line 19: checking if B = V'\{c} can be done in time $\mathcal{O}\left(\vert {V}^{\prime }\vert \right)$ by checking if all entries of the list representing B are 1, except at position c.
• Line 23: constructing the set U can be done in time $\mathcal{O}\left(\vert {V}^{\prime }\vert \vert V\left(G\right)\vert \right)$ by checking, for each u in V(G)\V', that τ _v (G) contains all edges in the set {(u, w) : w ∈ B} and no edges in the set {(u, w) : w ∈ L}. If this is the case, u will be added to the array representing U.
• Line 28–38: the body of this for-loop will be executed $\mathcal{O}\left(\vert V\left(G\right)\vert \right)$ since there are at most $\mathcal{O}\left(\vert V\left(G\right)\vert \right)$ elements in U.
  • Line 29: checking if (u, c) is an edge in τ _v (G) can be done in constant time.
  • Line 33: finding a vertex h which is adjacent to both u and c but to no other vertex in V' can be done in time $\mathcal{O}\left(\vert {V}^{\prime }\vert \vert V\left(G\right)\vert \right)$ (or determining that there is none), by first finding the neighbors of u, i.e. all the vertices h such that (u, h) is an edge in τ _v (G) and then, for each neighbor h of u, checking if h is also adjacent to c but to no other vertex in V'. This is done by checking if (h, c) is an edge in τ _v (G) and that no element of the set {(h, w) : w ∈ V'\{c}} is.

Thus, the total runtime of algorithm 1, except for the recursive call to algorithm 2 in line 10, is $\mathcal{O}\left(\vert {V}^{\prime }\vert \vert V\left(G\right){\vert }^{2}\right)$ (from line 33 in the for-loop.).

The runtime of each command in algorithm 2 is:

• Line 5: picking the vertex f can be done in constant time (pick the first entry).
• Line 7: finding a shortest path between f and c can be done in time $\mathcal{O}\left(\vert V\left(G\right){\vert }^{2}\right)$ by using Dijkstra's algorithm [19].
• Line 8–15: the body of this for-loop will be executed $\mathcal{O}\left(\vert V\left(G\right)\vert \right)$ since the shortest path P is necessarily shorter than the number of vertices in τ _v (G).
  • Line 9: checking if f is adjacent to any vertex in V'\{c} can be done in time $\mathcal{O}\left(\vert {V}^{\prime }\vert \right)$ by checking if any of the edges $\left\{\right. \left(\enspace f,w\right):w\in {V}^{\prime }{\backslash}\left\{c\right\}$ are in τ _v (G).
  • Line 10: the column with entry 1 in line 9 can be used for v here and thus only adds a constant time to the runtime.

Thus, the total runtime of algorithm 2, except for the recursive call to algorithm 1 in line 6, is $\mathcal{O}\left(\vert V\left(G\right){\vert }^{2}\right)$ (from line 7 in the for-loop).

To further substantiate the efficiency of the algorithm we give actual run-times for an implementation of the algorithm in figure 1.

In this section we prove that the algorithm presented in the previous section works, i.e. it gives a sequence of local complementations v such that τ _v (G)[V'] = S_V', given a distance-hereditary graph G, if such a sequence exists. It is relatively easy to show that the algorithm gives the desired results when it does not return an error, which we show in section 4.1.3. The hard part is to prove that, when the algorithm gives an error-flag it is in fact not possible to produce the star graph, i.e. the star graph is not a vertex-minor of G, which is done in section 4.1.3. The notation will be the same as in the previous section, c is a vertex in V' and is adjacent to the rest of the vertices in V'. The vertices in G[V'\{c}] with degree greater than 0, i.e. the vertices incident on some bad edge, are denoted as the set B.

4.1.3.1.