Scaffold for the polyhedral embedding of cubic graphs

arXiv:1911.11863v1 [math.CO] 26 Nov 2019 SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. FLOR AGUILAR 1 , GABRIELA ARAUJO-PARDO 2 , AND NATALIA GARCÍA-COLÍN 3 Abstract. Let G be a cubic graph and Π be a polyhedral embedding of this graph. The extended graph, Ge , of Π is the graph whose set of vertices is V (Ge ) = V (G) and whose set of edges E(Ge ) is equal to E(G) ∪ S, where S is constructed as follows: given two vertices t0 and t3 in V (Ge ) we say [t0 t3 ] ∈ S, if there is a 3–path, (t0 t1 t2 t3 ) ∈ G that is a Π– facial subwalk of the embedding. We prove that there is a one to one correspondence between the set of possible extended graphs of G and polyhedral embeddings of G. Introduction Our motivation for studying polyhedral embeddings of cubic graphs is twofold. On one hand, it is interesting as the natural alternative point of view on combinatorial characterisations of triangulations of surfaces [1], given that the dual structure of a triangulation is precisely a polyhedral embedding of a 3-regular graph in a surface. On the other hand, graph embeddings of 3-regular graphs are interesting in their own right as a plethora of papers on the subject prove, particularly as related to Grünbaum’s conjectured generalization of the four color theorem: If a cubic graph admits a polyhedral embedding in an orientable surface, then it is 3-edge colorable [5]. More precisely, in [1] they prove that the information on the size of the pairwise intersection of triangles in a triangulation suffices in order to determine its whole combinatorial structure. However, it is known that if we only have information on the pairs of triangles that intersect edge to edge (that is, the dual graph of the triangulation) then we cannot uniquely determine the whole incidence structure of the triangulation. For example, in [5] they prove that some connected cubic graphs can be embedded into more than one surface. In like manner, in this paper we prove that some additional combinatorial information suffices to uniquely determine a polyhedral embedding of a 3-regular graph in a surface with no boundary. For a more precise statement of our main result (Theorem 2) we need to introduce some terminology. 1 Instituto de Matemáticas, UNAM Instituto de Matemáticas, UNAM 3 CONACYT Research Fellow - INFOTEC Centro de Investigación en Tecnologías de la Información y Comunicación, Mexico. Corresponding Author. E-mail addresses: 1 [email protected], 2 [email protected], 3 [email protected]. 2 1 2 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN A topological map of a graph into a surface is called a graph embedding. If we consider the graph G together with its embedding Π, we say that G is Π–embedded. Each disjoint region of the complement of the image of an embedded graph is called a face of the embedding. The closed walk in the underlying graph G that corresponds to the boundary of a face is called a Π–facial walk. The embedding Π determines a set of Π–facial walks. Each edge is either contained in two Π–facial walks or it appears twice in the same facial walk. If a Π–facial walk is a cycle, it is also called a Π–facial cycle. Two embeddings of G are equivalent if they have the same set of facial walks, up to automorphisms of G. Let P1 and P2 be distinct Π−facial walks. We say that P1 and P2 meet properly if the intersection of P1 and P2 is either empty, a single vertex, or an edge. In this paper we will consider only embeddings of cubic simple graphs whose facial walks meet properly, this latter characteristic defines a polyhedral embedding: Definition 1. Π is said to be a polyhedral embbeding of a graph G, if every Π-facial walk is a cycle and any two Π−facial cycles meet properly. The set of all the Π–facial cycles is called Π–facial cycle system. An indication that there is some non-triviality in determining polyhedral embeddings is that a significant part of Ringel and Youngs’ Map Color Theorem [6] was to determine which complete graphs have such embeddings, even though for a complete graph Kn (with n ≥ 5) a polyhedral embedding is necessarily a triangulation. Furthermore, in [2] it is proven that the decision problem about the existence of polyhedral embeddings of a graph is NP-complete. The problem remains NPcomplete even if it is restricted to the case of embeddings in orientable surfaces and it is required that the graph is 6-connected. Concerning the uniqueness of the embedding, recall that the face-width is defined as the minimum integer r such that G has r facial walks whose union contains a cycle which is noncontractible on the surface. Whitney [9] proved that every 3-connected planar graph has an essentially unique embedding in the plane. Robertson and Vitray [7] extended the previous result to an arbitrary surface of genus g by assuming that the face-width is at least 2g + 3. Seymour and Thomas [8] and Mohar [3] improved the bound on the face width to O( logloglogg g ). Moreover, Robertson and Vitray [7] proved the following result: Proposition 1. An embedding of a graph G is polyhedral if and only if G is 3connected and the embedding has face-width at least 3. It is also known that for each surface S, there is a constant ζ = ζ(S) such that every 3-connected graph admits at most ζ embeddings of face width greater than three [4], then it can be concluded that a graph may have many different polyhedral embedings in the same or different surfaces. The extended graph of an embedding. The extended graph of the polyhedral embedding Π, Ge (Π), is the graph whose set of vertices is V (Ge (Π)) = V (G) and whose set of edges E(Ge (Π)) is equal to E(G) ∪ S. We will call S the set of scaffold edges and construct it as follows: given two vertices t0 and t3 in V (G), [t0 t3 ] ∈ S, if there are vertices t1 and t2 , different from t0 and t3 , such that (t0 t1 t2 t3 ) is a Π– SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 3 facial subwalk. In such case, we say that the 3–path of G corresponding to [t0 t3 ], is (t0 t1 t2 t3 ). Notice that the scaffold edges may be double (but not triple or more). That is, if [t0 t3 ] ∈ S and its corresponding path is (t0 t1 t2 t3 ), there may be a second 3–path between t0 and t3 , internally disjoint from (t0 t1 t2 t3 ), say (t0 t′1 t′2 t3 ), that also corresponds to [t0 t3 ]. In this case we say [t0 t3 ] is a double scaffold edge, and we denote this by a double bracket [[t0 t3 ]]. It will become obvious later, in Proposition 8, that this only happens when [[t0 t3 ]] appears as a chord of a 6-cycle which is a Π–facial cycle of the embedding, or when two 4–cicles intersect in one edges. Notice that E(G) ⊂ E(Ge (Π)). As such, we will refer to the edges in E(G) as simply edges. Given two different polyhedral embeddings Π and Π′ is not obvious that their corresponding extended graphs Ge (Π) and Ge (Π′ ) are combinatorically different. Figuring this out is precisely the aim of this paper. Theorem 2. Let G be a finite cubic graph. Then there is a one to one correspondence between the set of embeddings of G, P(G) = {Π|Π is an embedding of G}, and the set of extended graphs {Ge (Π)|Π ∈ P(G)}. Preliminaries Firstly, note that as G is a cubic graph and Π is an embedding of G then every path of length two is in a Π–facial cycle. This follows as there are three faces incident to every vertex of the embedding, thus any path of length two is contained in one of the faces incident to the vertex in the center of the path. Proposition 3. Let n ≥ 5 and C = (t0 t1 ...tn−1 t0 ) be a cycle in G corresponding to a Π–facial cycle, then there is no edge (ti tj ) in E(G), with i 6= j, |i − j| ≥ 2 and this difference taken mod n. Proof. We will proceed by contradiction. Suppose that (ti tj ) ∈ E(G) where |i−j| ≥ 2 and let Cij be a facial cycle that passes through the edge (ti tj ). Since the graph has degree three, Cij contains either the edge (ti ti+1 ) or (ti−1 ti ), and one of (tj tj+1 ) or (tj−1 tj ). This contradicts the definition of polyhedrality, since the Πij –facial cycle would intersect C in two edges.  Proposition 4. Let n ≥ 5 and C = (t0 t1 ...tn−1 t0 ) be a cycle in G corresponding to a Π–facial cycle, then for all ti , tj , there is no tk ∈ V (G) such that (ti tk tj ) is a path of G, where |i − j| ≥ 2 and this difference is taken mod n. Proof. Notice that every path of length two belongs to a Π–facial cycle, and then the proof follows by using similar arguments to those in the proof of Proposition 3.  Proposition 5. Let G be a cubic graph. If C is a 3–cycle of G then C is a facial cycle in every polyhedral embedding of G. Proof. Let (t0 t1 t2 t0 ) be a cycle in G. Since every path of length two belongs to a Π–facial cycle, say (t0 t1 t2 ) is in a facial cycle C. If (t0 t2 ) 6∈ C then we would contradict Proposition 3, and the statement follows.  4 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Proposition 6. Let G be a cubic graph. If C is a 4–cycle of G and G 6= K4 then C is a facial cycle in every polyhedral embedding of G. Proof. If G has a polyhedral embedding and G 6= K4 , then every 4–cycle of G is induced, since G is 3-connected by Proposition 1. Let (t0 t1 t2 t3 t0 ) be a cycle in G. Since every path of length two belongs to a Π–facial cycle, say (t0 t1 t2 ) is in a facial cycle C. If the path (t2 t3 t0 ) 6∈ C then we would contradict Proposition 4, and the statement follows.  Proposition 7. Given two paths, (t0 t1 t2 t3 ), (t0 t1 t2 t3′ ), of G then either (t0 t1 t2 t3 ) or (t0 t1 t2 t3′ ), but not both, is a Π–facial subwalk. Proof. Given that every path of length two is in a Π–facial cycle then either (t0 t1 t2 t3 ) or (t0 t1 t2 t3′ ) is a Π–facial subwalk. Suppose that both (t0 t1 t2 t3 ) and (t0 t1 t2 t3′ ) are Π–facial subwalks, if each of these paths belongs to a different facial cycle, then they would intersect improperly, contradicting Definition 1. If they belong to the same facial cycle, then such facial cycle self intersects, contradicting Definition 1.  Proposition 8. Let P = (q0 = t0 , t1 , ..., tn = qm ) and Q = (q0 = t0 , q1 , ..., qm = tn ) be two internally disjoint Π–facial subwalks, such that (t0 tn ) 6∈ E(G), then P ∪ Q must be a Π−facial cycle. Proof. We will proceed by contradiction. Let CP be the Π–facial cycle associated to P and CQ be the Π–facial cycle associated to Q, where CP 6= CQ . Since G is a cubic graph, let u and v be the remaining vertices adjacent to t0 = q0 and tn = qm , respectively. Since the edge (t0 tn ) 6∈ E(G), observe that v 6= t0 and u 6= tn . Which implies that (t0 u) and (tn v) are different. Notice both CP and CQ have to contain two edges incident to t0 = q0 , but t0 = q0 has degree three; thus CP intersects CQ in at least one edge incident to t0 = q0 . The same holds for tn = qm , hence CP and CQ would have to intersect in at least two edges, contradicting Definition 1.  Proof of the main theorem In this section we will denote as Ge any extended graph in {Ge (Π)|Π ∈ P(G)}, where we emphasize that we do not claim knowledge of what embedding Ge corresponds to. The proof of the main result of the paper will be split in to two subsections: the simple case and the difficult case. The simple case. We present the simple case first as within its proof it becomes obvious why the second part requires much more detail. Theorem 9. Let Ge be an extended graph of a finite cubic graph, G, such that for all edges [t, t′ ] ∈ S there is a unique path of length three in G whose ends are t and t′ , then Ge uniquely determines the Π–facial cycle system of an embedding. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 5 Proof. From this information we will construct the Π–facial cycle system. Let [t0 t3 ] ∈ S, by hypothesis we may say that (t0 t1 t2 t3 ) is the unique path of length three between t0 and t3 , and there is a Π–facial cycle that contains the facial subwalk (t0 t1 t2 t3 ). Let t4 and t4′ be the remaining vertices adjacent to t3 , then by Proposition 7 either (t1 t2 t3 t4 ) or (t1 t2 t3 t4′ ) is a facial subwalk, but not both. Additionally, by hypothesis, only one of [t1 t4 ] or [t1 t4′ ] is in S. Hence we know with certainty if the facial cycle that contains (t0 t1 t2 t3 ) continues on to t4 or t4′ . We can continue with this procedure until we obtain the unique Π−facial cycle that contains (t0 t1 t2 t3 ). Now consider the edge (t1 t2 ) ∈ E(G), this edge must belong to another Π–facial cycle. Let t1′ and t2′ be the remaining vertices adjacent to t1 and t2 , respectively. Then, as the embedding is polyhedral, the other facial cycle containing (t1 t2 ), necessarily contains (t1′ t1 t2 t2′ ), thus [t1′ t2′ ] ∈ S. Now we may use the same procedure as before to find the rest of the edges in this facial cycle. We can find every Π–facial cycle in this manner, by selecting at every step an edge of the union of the preceding facial cycles that hasn’t appeared in two facial cycles yet. This procedure is finite, since the graph G is finite.  As a consequence of Theorem 9 we have the following: Corollary 10. Let Ge be an extended graph of a finite cubic graph, G, with no 6–cycles, then Ge uniquely determines the Π–facial cycle system of an embedding of G. The difficult case. Clearly, the difficulty arises when we have the possibility that the hypothesis of Theorem 9 does not hold for an extended graph, Ge . Namely, NOT for all scaffold edges [t, t′ ] ∈ S there is a unique path of length three in G whose ends are t and t′ . This motivates the following definitions: Definition 2. (Fork) Let Y ⊂ Ge be a subgraph with set of vertices V (Y ) = {t1 , t2 , t3 , t4 , t4′ } and set of edges E(Y ) = {(t1 t2 ), (t2 t3 ), (t3 t4 ), (t3 t4′ )}∪{[t1 t4 ], [t1 t4′ ]}. We will call such graph a fork. Figure 1. Fork 6 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Here, if we tried to reproduce the reconstruction procedure presented in the proof of Theorem 9, when arriving at a fork we would have a disjunction consisting of whether (t1 t2 t3 t4 ) is the 3–path corresponding to [t1 t4 ] or (t1 t2 t3 t4′ ) is the 3–path corresponding to [t1 t4′ ]. A great part of this section consists of proving that, in most cases, this disjuntion can be solved by looking at the rest of the structure of the graph. We will say that the disjuntion cannot be solved whenever the rest of the structure of the extended graph does not force either (t1 t2 t3 t4 ) to be the 3–path corresponding to [t1 t4 ] or (t1 t2 t3 t4′ ) to be the 3–path corresponding to [t1 t4′ ]. We will now prove some more technical lemmas which will help us discover the very particular structure of graphs for which disjunctions cannot be solved. Proposition 11. Let P1 = (ut0 t1 v), P2 = (ut2 t3 v) and P3 = (ut4 t5 v) be three internally disjoint 3–paths in G such that [uv] ∈ Ge then [[uv]] is a double scaffold edge. Furthermore, Pi ∪ Pj is a Π-facial cycle for a pair i, j ∈ {1, 2, 3}. Proof. With out loss of generality, suppose that [uv] corresponds to the path P1 , which means that (ut0 t1 v) is Π-facial subwalk. By Proposition 7, it must satisfied that either (ut0 t1 vt3 ) or (ut0 t1 vt5 ) is a Π-facial subwalk. Suppose, with out loss of generality, that (ut0 t1 vt3 ) is a Π-facial subwalk, then by Proposition 4 (ut0 t1 vt3 t2 u) is Π-facial cycle, hence [[uv]] is a double scaffold edge. See Figure 2.  Figure 2. Corollary 12. Let P1 = (ut0 t1 v), P2 = (ut2 t3 v) and P3 = (ut4 t5 v) be three internally disjoint 3–paths in G such that [uv] ∈ Ge and (t1 t0 ut4 ) is a Π-facial subwalk. Then (vt5 t4 u) is a Π-facial subwalk. Proof. Since [uv] ∈ Ge , Proposition 11 implies [[uv]] is a double scaffold edge and either (ut0 t1 vt5 t4 u), (ut0 t1 vt3 t2 u) or (ut2 t3 vt5 t4 u) is a Π-facial cycle. Since (t1 t0 ut4 ) is a Π-facial subwalk, by Proposition 7 (t1 t0 ut2 ) is not a Π-facial subwalk. This implies that (ut0 t1 vt3 t2 u) can not be a Π-facial cycle and, necessarily, either (ut0 t1 vt5 t4 u) or (ut2 t3 vt5 t4 u) is a Π-facial cycle. Notice that the path (vt5 y4 u) appears in both cases, hence, this path is a Π-facial subwalk.  Corollary 13. Let (ut0 t1 v) and (ut2 t3 v) be two internally disjoint 3–paths in G such that [uv] ∈ Ge and (ut0 t1 v) is not a Π-facial subwalk. Then (ut2 t3 v) is a Π-facial subwalk. Proof. We proceed by contradiction. If (ut2 t3 v) is not a Π-facial subwalk, then there is a third 3–path between u and v, (ut4 t5 v), internally disjoint to (ut0 t1 v) and(ut2 t3 v) which corresponds to the scaffold edge [uv]. But, by Proposition 11 the SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 7 edge [[uv]] is double. Since (ut0 t1 v) is not a Π-facial subwalk, necessarily (ut2 t3 v) and (ut4 t5 v) are Π-facial subwalks, which is a contradiction.  Definition 3. (Butterfly one and two) Let B ⊂ Ge be a subgraph of G with set of vertices V (B) = {t0 , t0′ , t1 , t2 , t3 , t4 , t4′ , t5 , t5′ }, and set of edges equal to the union of the two cycles E(Y ) = (t0 t1 t2 t3 t4 t5 t0 ) ∪ (t0′ t1 t2 t3 t4′ t5′ t0′ ), we will call this graph a butterfly. A butterfly one, B1 , is a butterfly with the additional scaffold edges [t1 t4 ], [t1 t4′ ], [t0 t3 ], [t0′ t3 ]. A butterfly two, B2 , is a butterfly with the additional scaffold edges [t1 t4 ], [t1 t4′ ], [t0 t3 ]. Figure 3. The butterfly one is shown on the left and the butterfly two is shown on the right We are ready to state and proof the main theorems of this section. The spirit is the following: if a fork Y appears as a subgraph of Ge in such a way that the disjunction cannot be solved, then Ge will contain either butterfly one or butterfly two as subgraphs and, in either case, we will be able to reconstruct the full extended graph uniquely. Lemma 14. Let Y be a fork labelled as in Definition 2 then if either (t1 t2 t3 t4 ) is the only 3–path between t1 and t4 or (t1 t2 t3 t4′ ) is the only 3–path between t1 and t4′ then the disjunction can be solved. Proof. By Definition, every scaffold edge corresponds to a 3-path. Thus, when there is only one 3-path between the end vertices of a scaffold edge necessarily such path is its corresponding 3-path.  Next we will prove different theorems. A long of the proofs we state different claims and the procedure that we use is given a claim followed immediately by its proof. The figures that shows the following results appears in the Apendix at the end of the paper. Lemma 15. Let Y be a fork labelled as in Definition 2 such that the disjunction cannot be solved then Ge contains a subgraph isomorphic to butterfly one, B1 , or butterfly two, B2 . 8 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Proof. Since G is a cubic graph, let t0 and t0′ be the remaining vertices adjacent to t1 , different from t2 . Claim 1. The vertices t0 and t0′ are different to each t3 , t4 , t4′ . Note that the proofs of t0 and t0′ must be analogous. We only detail the proof for t0 . As G is a 3–regular graph, t0 6= t3 (otherwise t3 would have degree four). If t0 = t4 , then by Proposition 6, (t1 t2 t3 t4 t1 ) is Π-facial cycle, and (t1 t2 t3 t4 ) is forced to be the 3-path corresponding to [t1 t4 ] thus the disjunction can be solved. Hence t0 6= t4 . The case t0 6= t4′ is analogous to the case t0 6= t4 . ⊟ Applying Lemma 14 we obtaing the following: Claim 2. For both edges [t1 t4 ] and [t1 t4′ ] there exist at least two internally disjoint 3-paths between its end vertices. Without loss of generality we may say that (t1 t0 t5 t4 ) is the additional 3–path with ends [t1 t4 ], internally disjoint from (t1 t2 t3 t4 ). Note that then t5 6= t0 , t1 , t2 , t3 , t4 . Claim 3. The vertex t5 is different to the vertices t0′ , t4′ . If t5 = t4′ , Proposition 5 implies that (t3 t4 t4′ t3 ) is a Π-facial cycle; by Proposition 7 (t2 t3 t4′ t0 ) is a Π-facial subwalk; Proposition 4 implies that (t0 t1 t2 t3 t4′ t0 ) is a Πfacial cycle, so (t1 t2 t3 t4′ ) is a 3-path corresponding to [t1 t4′ ] ; hence, the disjunction could be solved, contradicting the hypothesis. Therefore t5 6= t4′ . If t5 = t0′ , using a similar arguments we would arrive to the conclusion that (t1 t2 t3 t4 ) is a 3–path corresponding to [t1 t4 ] contradicting our hypothesis. Therefore t5 6= t0′ . ⊟ Now, we will give the additional 3-path between t1 and t4′ internally disjoint to (t1 t2 t3 t4′ ). We have three cases: (Case A) when the 3−path is (t1 t0′ t5′ t4′ ), (Case B) when the 3−path is (t1 t0 x0 t4′ ) and (Case C) when the 3−path is (t1 t0 t5 t4′ ). We will deal with each case separately. Case A. Assume the 3−path is (t1 t0′ t5′ t4′ ). Since G is a cubic graph and all the previous vertices already have degree two, t5′ must be a new vertex of G, otherwise we would have a vertex with degree four. By Proposition 7, given that the 3–paths, (t0 t1 t2 t3 ) and (t0′ t1 t2 t3 ) are in G, one of these paths is a Π–facial subwalk. So we have two cases: either [t0 t3 ] and [t0′ t3 ] are in S which implies that B1 ⊂ Ge , or only of them ([t0 t3 ] or [t0′ t3 ]) is in S, which implies that B2 ⊂ Ge . Either way, the assertion of the lemma follows. This ends the proof for Case A.⊟ Case B. Assume the 3−path is (t1 t0 x0 t4′ ). First, observe that Claim 3 implies that x0 is a new vertex. Moreover, the 3−path (t0 t1 t2 t3 ) is in G and it is possible that [t0 t3 ] ∈ S or [t0 t3 ] 6∈ S. We will analyse both cases: Case B1. [t0 t3 ] ∈ S. Claim B1.1. [[t0 t3 ]] is a double scaffold edge. Since there are three internally disjoint 3−paths between t0 and t3 : (t0 t1 t2 t3 ), (t0 t5 t4 t3 ) and (t0 x0 t4′ t3 ), then [[t0 t3 ]] is a double scaffold edge, by Corollary 11. ⊟ Claim B1.2. [t1 t4 ] or [t1 t4′ ] is a double scaffold edge. Note that if either [t1 t4 ] or [t1 t4′ ] is a double scaffold edge then we necessarily SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 9 have three 3-paths internally disjoint between the pair t1 , t4 in the first instance and t1 , t4′ in the second instance, which would confirm our claim. Thus, we may now suppose that neither [t1 t4 ] nor [t1 t4′ ] are double scaffold edges, then (t0 t5 t4 t3 t4′ x0 t0 ) is a Π–facial cycle, by Corollary 11. The latter and Corollary 13 imply that (t1 t2 t3 t4 ) and (t1 t2 t3 t4′ ) are the Π–facial subwalks corresponding to [t1 t4 ] and [t1 t4′ ] respectively, which contradicts Definition 1. Hence, at least one of [t1 t4 ] or [t1 t4′ ] has to be a double scaffold edge.⊟ Claim B1.3. B1 or B2 is contained in Ge . Without loss of generality, suppose that [[t1 t4′ ]] is a double scaffold edge. Observe that (t1 t4′ ) 6∈ E(G), otherwise it would contradict the 3−regularity of G. Since there are already two internally disjoint 3–paths between t1 and t4′ : (t1 t2 t3 t4′ ) and (t1 t0 x0 t4′ ), we have to have a third one, else (t1 t2 t3 t4′ x0 t0 t1 ) would be a Π–facial cycle, by Proposition 8, which implies that (t1 t2 t3 t4′ ) is a Π–facial subwalk. That is, the disjunction can be solved, contradicting our hypothesis. Let (t1 t0′ t5′ t4′ ) be the third path. Notice that t5′ is a new vertex, otherwise it would contradict the hypothesis that G is a cubic graph. Then, if [t0′ t3 ] ∈ S then B1 ⊂ Ge or if [t0′ t3 ] 6∈ S then B2 ⊂ Ge . If we assume instead that [[t1 t4 ]] is a double scaffold edge, the proof follows in a similar way. ⊟ This completes the proof for Case B1. Case B2. [t0 t3 ] 6∈ S. This implies that (t0 t1 t2 t3 ) is not a Π–facial subwalk and, by Proposition 7, (t0′ t1 t2 t3 ) is a Π–facial subwalk, hence [t0′ t3 ] ∈ S. Claim B2.1 If (t0′ t1 t2 t3 t4 ) is a Π–facial subwalk then (t5 t0 t1 t2 ) is a Π–facial subwalk. If (t0′ t1 t2 t3 t4′ ) then (t0′ t1 t0 t5 t4 ) is a Π–facial subwalk. Suppose that (t0′ t1 t2 t3 t4 ) is a Π–facial subwalk. The later implies that (t1 t2 t3 t4′ ) is not a Π–facial subwalk. Since [t1 t4′ ] ∈ S and using Corollary 13, (t1 t0 x0 t4′ ) is a Π–facial subwalk. This Π–facial subwalk cannot contain the edge (t1 t2 ) otherwise, by Proposition 4, (t1 t0 x0 t4′ t3 t2 t1 ) would be a Π–facial cycle intersecting the Π–facial subwalk (t0′ t1 t2 t3 ) in two edges, which contradicts the definition of polyhedral embedding. Then, (t0′ t1 t0 x0 t4′ ) is a Π–facial subwalk. We know that two different Π–facial cycles pass through each edge of G; this fact together with Definition 1 and Proposition 7 imply that (t5 t0 t1 t2 ) is a Π–facial subwalk. Now, suppose that (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk. This implies that (t1 t2 t3 t4 ) is not the Π–facial subwalk corresponding to the scaffold edge [t1 t4 ]. Hence, using Corollary 13 we know that (t1 t0 t5 t4 ) is the Π–facial subwalk corresponding to the scaffold edge [t1 t4 ]. This Π–facial subwalk cannot contain the edge (t1 t2 ), otherwise (t2 t1 t0 t5 t4 t3 t2 ) would be a Π–facial cycle, by Proposition 4, and it would intersect the Π–facial subwalk (t0′ t1 t2 t3 ) in two edges, contradicting the definition of polyhedral embedding. Then, (t0′ t1 t0 t5 t4 ) is a Π–facial subwalk.⊟ Claim B2.2 The fork Y ′ given by the set of vertices {t5 , t0 , t1 , t0′ , t2 }, and the edges, {(t5 t0 ), (t0 t1 ), (t1 t2 ), (t1 t0′ ), [t5 t2 ], [t5 t0′ ]} is contained in Ge . If the statement were false, i.e., if one of the scaffold edges [t5 t2 ] or [t5 t0′ ] is not in S, then the disjunction can be solved, contradicting our hypothesis. Then, the statement follows. ⊟ 10 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN In a way similar to our treatment of the cases for fork Y , we will analyse the fork Y ′ . We need to have two internally disjoint 3–paths for each pair of vertices {t5 , t2 } and {t5 , t0′ }. Notice that for the pair {t5 , t2 } we already have two internally disjoint 3–paths: (t5 t4 t3 t2 ) and (t5 t0 t1 t2 ). Therefore, we only need to find a second 3–path between t5 and t0′ . We have three cases: Case B2.A (t5 t4 t3 t0′ ) cannot be the second 3–path between t5 and t0′ . Suppose the second 3–path is (t5 t4 t3 t0′ ). This would imply that t3 has degree four, contradicting that G is a cubic graph. ⊟. Case B2.B If (t5 t4 x4 t0′ ) is the second 3–path between t5 and t0′ then B2 is contained in Ge . Suppose the second 3–path is (t5 t4 x4 t0′ ), where x4 is the remaining vertex adjacent to t4 . Observe that x4 is a new vertex, since x4 is different to t3 , t4 , t5 , and if x4 is equal to one of t0 , t1 , t2 , t4′ , x0 , that would contradict the fact that G is a cubic graph. Finally, x4 6= t0′ , otherwise, by Proposition 4, (t0′ t1 t2 t3 t4 t0′ ) would be a Π–facial cycle, implying that (t1 t2 t3 t4 ) is a Π–facial subwalk, contradicting our hypothesis. Hence, we would have to B2 contained in Ge , given by the set of vertices: {t0 , t0′ , t1 , t2 , t3 , t4 , t4′ , x0 , x4 } and by the set of edges of the two 6–cycles: (t1 t2 t3 t4′ x0 t0 t1 ) ∪ (t1 t2 t3 t4 x4 t0′ t1 ) and the scaffold edges {[t1 t4 ], [t1 t4′ ], [t0′ t3 ]}. ⊟ Case B2.C (t5 x5 x0′ t0′ ) cannot be the second 3–path. Suppose the second 3–path is (t5 x5 x0′ t0′ ), where x5 and x0′ are the remaining vertices adjacent to t5 and t0′ respectively. Observe that x0′ and x5 are different from the vertices t0 , t0′ , t1 , t2 , t3 , t4 , t5 by a similar proof to the case A. Also x0′ and x5 are different from the vertices t4′ and x0 , otherwise contradicts the fact that G is a cubic graph. So there is a butterfly contained in G given by the union of the two 6–cycles (t0 , t1 , t2 , t3 , t4 , t5 t0 )∪(t0 t1 t0′ x0′ x5 t5 t0 ) and the set of scaffold edges {[t5 t2 ], [t5 t0′ ], [t1 t4 ], [t1 x5 ]} if is a butterfly one, or the set of scaffold edges {[t5 t2 ], [t5 t0′ ], [t1 t4 ]} if is a butterfly two. This completes the proof for Case B2 and Case B. ⊟ Case C. Assume that the 3–path is (t1 t0 t5 t4′ ). Notice that (t3 t4 t5 t4′ t3 ) is a Π–facial cycle of G. Suppose that [t0 t3 ] ∈ S. Since (t3 t4 t5 t4′ t3 ) is a Π–facial cycle, (t0 t5 t4 t3 ) cannot be a Π–facial subwalk, by Propositon 7. Using Corollary 13, (t0 t1 t2 t3 ) is the Π–facial subwalk corresponding to [t0 t3 ]. The Π–facial subwalk (t0 t1 t2 t3 ) cannot continue trough the vertex t4 , otherwise by Proposition 4 implies that (t0 t1 t2 t3 t4 t5 t0 ) is a Π–facial cycle, and it would intersect the Π–facial cycle (t3 t4 t5 t4′ t3 ) in two edges, contradicting the definition of polyhedral embedding. So (t0 t1 t2 t3 t4′ ) is a Π–facial subwalk, i.e., the disjunction can be solved, contradicting out hypothesis. Then, suppose that [t0 t3 ] 6∈ S. Since (t0 t1 t2 t3 ) is not a Π–facial subwalk, using Proposition 7 implies that (t0′ t1 t2 t3 ) is a Π–facial subwalk. Thus either (t0′ t1 t2 t3 t4 ) or (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 11 Claim C1. The remaining vertex adjacent to t2 , x2 , is a new vertex of G. We are going to proceed by contradiction: 1. (x2 6= t5 ). If x2 = t5 it would contradict the 3–regularity of G. 2. (x2 6= t4 ). If x2 = t4 then (t2 t3 t4 t2 ) is a Π–facial cycle, by Proposition 5. The latter implies that (t0′ t1 t2 t3 t4 ) cannot be a Π–facial subwalk, since it would intersect the Π–facial cycle (t2 t3 t4 t2 ) in two edges, contradicting the polyhedrality of the embedding. Then, by Proposition 7, (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk, i.e., (t1 t2 t3 t4′ ) is the Π–facial subwalk corresponding to the edge [t1 t4′ ], and the disjunction can be solved, contradicting out hypothesis. 3. (x2 6= t4′ ) This proof follows in a manner similar to the previous one. 4. (x2 6= t0′ ). If x2 = t0′ it would contradict the definition of polyhedral embedding, since (t0′ t1 t2 t3 ) is a Π–facial subwalk. 5. (x2 6= t0 ). If x2 = t0 then (t0 t1 t2 t0 ) would be a Π–facial cycle, by Proposition 5. Since there are two different Π–facial cycles passing through every edge, then Proposition 7 implies that (t3 t2 t0 t5 ) is a Π–facial subwalk. Thus, (t3 t2 t0 t5 t4 t3 ) is a Π–facial cycle, by Proposition 4. But notice that this would contradict the polyhedrality of the embedding. Hence, x2 is a new vertex. ⊟ Claim C2. If (t0′ t1 t2 t3 t4 ) is a Π–facial subwalk then (x2 t2 t3 t4′ ) is a Π–facial subwalk. Notice that (t1 t2 t3 t4 ) is the Π–facial subwalk corresponding to [t1 t4 ]. Given that two different Π–facial cycles pass through every edge and using Proposition 7 we can deduce that (x2 t2 t3 t4′ ) is a Π–facial subwalk. ⊟ Claim C3. If (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk then (t1 t2 t3 t4′ ) is a Π–facial subwalk. This statement follows immediately from the hypothesis, since (t1 t2 t3 t4′ ) is contained in the Π–facial subwalk (t0′ t1 t2 t3 t4′ ). ⊟ Claim C4. The fork Y ′ given by the set of vertices {t4′ , t3 , t2 , t1 , x2 } and the set of edges {(t3 t4′ ), (t2 t3 ), (t1 t2 ), (t2 x2 ), [t1 t4′ ], [t4′ x2 ]}. is contained in Ge . Notice that only remains to prove that [t4′ x2 ] is in S. If this scaffold edge is not in S, then (x2 t2 t3 t4′ ) is not a Π–facial subwalk. The latter and using Proposition 7 imply that (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk. Thus there is no disjunction contradicting our hypothesis. ⊟ Let x4′ be the remaining vertex adjacent to t4′ . Claim C5. (t2 t3 t4′ x4′ ) is a Π–facial subwalk. Since every path of length two is a Π–facial subwalk, then (t2 t3 t4′ ) is a Π–facial subwalk. But this Π–facial subwalk cannot continue to t5 , otherwise it would intersect the Π–facial cycle (t3 t4 t5 t4′ t3 ) in two edges, contradicting Definition 1. Hence, we have to continue towards the remaining neighbour of t4′ , x4′ , and the statement follows. ⊟ Claim C6. There are two internally disjoint 3–paths for each pair of vertices {t1 , t4′ } and {t4′ , x2 }. It follows immediately by Lemma 14. ⊟ 12 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN The two internally disjoint 3–paths between t1 and t4′ are: (t1 t2 t3 t4′ ) and (t1 t0 t5 t4′ ). The first of the two internally disjoint 3–paths between t4′ and x2 is (t4′ t3 t2 x2 ); as for the second there are three options: (Case C1) (t4′ t5 t0 x2 ), (Case C2) (t4′ t5 t4 x2 ) and (Case C3) (t4′ x4′ x2′ x2 ). We will deal with each instance separately. Case C1. Assume that the second internally disjoint 3–path between t4′ and x2 is (t4′ t5 t0 x2 ). Claim C1.1. The remaining vertex adjacent to t4′ , x4′ , is a new vertex. We are going to proceed by contradiction: 1. (x4′ 6= t0 , x4′ 6= t1 , x4′ 6= t2 , x4′ 6= t3 , x4′ 6= t5 ). If any of the inequalities was an equality instead we would have a contradiction on the fact that G is a cubic graph. 2. (x4′ 6= t0′ ). If x4′ = t0′ then by Proposition 4, (t0′ t1 t2 t3 t4′ t0′ ) is a Π–facial cycle because (t0′ t1 t2 t3 ) is a Π–facial subwalk. This implies that (t1 t2 t3 t4′ ) is the Π–facial subwalk corresponding to the edge [t1 t4′ ], i.e., the disjunction can be solved, contradicting our hypothesis. Hence x4′ 6= t0′ . 3. (x4′ 6= t4 ). If x4′ = t4 , since (t3 t4 t5 t4′ t3 ) is a Π–facial cycle this contradicts Proposition 3. Hence x4′ 6= x4 . 4. (x4′ 6= x2 ). If x4′ = x2 then the Π–facial subwalk (t0′ t1 t2 t3 ) cannot continue through t4′ , since there is a 2–path (t2 x2 t4′ ), which would contradict Proposition 4. Then (t0′ t1 t2 t3 ) continues through t4 , implying that (t1 t2 t3 t4 ) is the Π–facial subwalk corresponding to [t1 t4 ], i.e., the disjunction can be solved, contradicting our hypothesis. Hence x4′ 6= x2 . Thus, x4′ is a new vertex. ⊟ Claim C1.2. The remaining vertex adjacent to t4 , x4 , is a new vertex. We are going to proceed by contradiction: 1. (x4 6= t0 , x4 6= t1 , x4 6= t2 , x4 6= t3 , x4 6= t5 , x4 6= t4′ ). If any of the inequalities was an equality instead we would have a contradiction on the fact that G is a cubic graph. 2. (x4 6= x2 ). If x4 = x2 then by Proposition 6, (t2 t3 t4 x2 t2 ) is a Π–facial cycle. Thus the Π–facial subwalk (t0′ t1 t2 t3 ) cannot continue through t4 , otherwise it would contradict the polyhedrality of G. Then (t0′ t1 t2 t3 t4′ x4′ ) is a Π–facial subwalk. This implies that (t1 t2 t3 t4′ ) is the Π–facial subwalk corresponding to [t1 t4′ ], therefore the disjunction can be solved, contradicting our hypothesis. Hence x4 6= x2 . 3. (x4 6= t0′ ). If x4 = t0′ then using Proposition 4 it can be said that (t0′ t1 t2 t3 t4 t0′ ) is a Π–facial cycle. The latter implies that (t1 t2 t3 t4 ) is the Π–facial subwalk corresponding to [t1 t4 ], which contradicts our hypothesis. Hence x4 6= t0′ . 4. (x4 6= x4′ ). If x4 = x4′ , since (t3 t4 t5 t4′ t3 ) is a Π–facial cycle, it would contradict Proposition 4. Hence, x4 is a new vertex. ⊟ SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 13 Claim C1.3. The remaining vertex adjacent to x2 , x2′ , is a new vertex. We are going to proceed by contradiction: 1. (x2′ 6= t0 , x2′ 6= t1 , x2′ 6= t2 , x2′ 6= t3 , x2′ 6= t4 , x2′ 6= t4′ , x2′ 6= t5 ). If any of the inequalities was an equality instead we would have a contradiction on the fact that G is a cubic graph. 2. (x2′ 6= t0′ ). If x2′ = t0′ since (t0 t1 t2 x2 t0 ) is a Π–facial cycle, it would contradict Proposition 4. Hence x2′ 6= t0′ . 3. (x2′ 6= x4 ). If x2′ = x4 then (x4 x2 t2 t3 ) is a Π–facial subwalk, since (t0 t1 t2 x2 t0 ) is a Π–facial cycle. Using Proposition 4 and the latter imply that (x4 x2 t2 t3 t4 x4 ) is a Π–facial cycle. Therefore the Π–facial subwalk (t0′ t1 t2 t3 ) cannot continue though t4 , otherwise it would contradict the polyhedrality of the embedding. Then (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk, which implies that (t1 t2 t3 t4′ ) is the Π–facial subwalk corresponding to [t1 t4′ ] and the disjunction can be solved, contradicting our hypothesis. Hence x2′ 6= x4 . 4. (x2′ 6= x4′ ). This proof follows in a similar way that the previuos one. Thus x2′ is a new vertex of G. ⊟ Claim C1.4. Either (t1 t2 t3 t4′ ) or (t1 t0 t5 t4′ ) is a Π–facial subwalk. We know that [t1 t4 ] and [t1 t4′ ] are in S, also Proposition 7 implies that either (t1 t2 t3 t4 ) or (t1 t2 t3 t4′ ) is a Π–facial subwalk. If (t1 t2 t3 t4′ ) is a Π–facial subwalk the statement follows. If, on the other hand, (t1 t2 t3 t4 ) is not a Π–facial subwalk, Corollary 13 implies that (t1 t0 t5 t4′ ) is the Π–facial subwalk corresponding to [t1 t4′ ]. ⊟ Claim C1.5. For either choice between (t1 t2 t3 t4′ ) or (t1 t0 t5 t4′ ) being a Π–facial subwalk, there is a Π–facial subwalk P = (t1 t0′ . . . x4′ t4′ ), such that either in P ∪ (t1 t2 t3 t4′ ) is a Π–facial cycle in the first instance or P ∪ (t1 t0 t5 t4′ ) is a Π–facial cycle in the second instance. First, suppose that (t1 t2 t3 t4′ ) is a Π–facial subwalk. This walk cannot be extended as (t0 t1 t2 t3 t4′ t5 ), otherwise it would intersect the Π–facial cycles (t0 t1 t2 x2 t0 ) and (t3 t4 t5 t4′ t3 ) in two edges, contradicting the definition of polyhedral embedding. Hence the only possible way to extend this walk is (t0′ t1 t2 t3 t4′ x4′ ) and the claim follows. Suppose (t1 t0 t5 t4′ ) is a Π–facial subwalk. Similarly, this walk cannot be extended as (t2 t1 t0 t5 t4′ t3 ), otherwise it would intersect the Π–facial cycles (t0 t1 t2 x2 t0 ) and (t3 t4 t5 t4′ t3 ) in two edges, contradicting the definition of polyhedral embedding. Hence, the only possible way to extend this walk is (t0′ t1 t0 t5 t4′ x4′ ) and the claim follows. In both cases, a path P = (t1 t0′ . . . x4′ t4′ ) completes either Π–facial subwalk in to a Π–facial cycle. ⊟ Claim C1.6. The Π–facial subwalk P in the previous claim is (t1 t0′ x4′ t4′ ). We proceed by contradiction. Suppose that there is at least one vertex tp ∈ V (G) such that P = (t1 t0′ tp . . . x4′ t4′ ). We will now prove that we can deduce further structure of Ge from this assumption and arrive to a contradiction for either (t1 t2 t3 t4′ ) or (t1 t0 t5 t4′ ) being a Π–facial subwalk.⊟ Claim C1.6.1. The fork given by the set of vertices {tp , t0′ , t1 , t0 , t2 } and the set 14 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN of edges {(tp t0′ ), (t0′ t1 ), (t1 t0 ), (t1 t2 ), [tp t0 ], [tp t2 ]} has to be contained in Ge . In this instance, by Proposition 7, either (tp t0′ t1 t0 ) or (tp t0′ t1 t2 ) is a Π–facial subwalk. Therefore, if one of the scaffold edges [tp t0 ] or [tp t2 ] is not in S we can deduce whether P ∪ (t1 t2 t3 t4′ ) or P ∪ (t1 t0 t5 t4′ ) is a Π–facial cycle, by Proposition 4, and the disjunction can be solved, contradicting the hypothesis and the claim follows.⊟ Claim C1.6.2. The paths (tp x2′ x2 t0 ) and (tp x2′ x2 t2 ) are in G. As we have argued in other claims, if there is to be a disjunction, by Lemma 14, there are a pair of 3–paths: one joining the vertices {tp , t0 } and another one joining the vertices {tp , t2 }, internally disjoint to (tp t0′ t1 t0 ) and (tp t0′ t1 t2 ), respectively. Looking at the degree of the vertices involved, it is easy to see that the only possibility is that tp is adjacent to x2′ and the additional 3–paths are (tp x2′ x2 t0 ) and (tp x2′ x2 t2 ). ⊟ Claim C1.6.3. P does not continue through x2′ after tp . We will proceed by contradiction. Suppose P = (t1 t0′ tp x2′ . . . x4′ t4′ ), then P cannot continue to the vertex t2 after t1 , otherwise there is a 2–path (t2 x2 x2′ ) contradicting Proposition 4. Then, P continue to t0 after t1 and using Proposition 4 we conclude that P ∪ (t1 t0 t5 t4′ ) = (t1 t0′ tp x2′ . . . x4′ t4′ t5 t0 t1 ) is a Π– facial cycle. This implies that undoubtedly (t1 t2 t3 t4 ) is the Π–facial subwalk corresponding to the scaffold edge [t1 t4 ]. Ergo, the disjunction can be solved, contradicting our hypothesis. Hence, x2′ is not in P.⊟ Finally, we will now see what happens in each of the cases: either P ∪ (t1 t2 t3 t4′ ) or P ∪ (t1 t0 t5 t4′ ) is a Π–facial cycle. Inasmuch as both cases are symmetric, we will only prove the first one in detail. Suppose that P ∪ (t1 t2 t3 t4′ ) is a Π– facial cycle. This implies that the 3–path (t1 t2 t3 t4′ ) corresponds to the edge [t1 t4′ ] ∈ S then the 3–path that corresponds to [t1 t4 ] is (t1 t0 t5 t4 ), given that the embedding is polyhedral, we can conclude that (t0′ t1 t0 t5 t4 x4 ) is a Π–facial subwalk. Analyzing the existing facial subwalks and using extensive use of the fact that the embedding is polyhedral is easy to see that (x4 t4 t3 t2 x2 x2′ ) is a Π–facial subwalk. Notice that this walk cannot continue through the vertex tp after x2′ , otherwise it would intersect the Π–facial cycle P ∪ (t1 t2 t3 t4′ ) in two edges. Similarly, we can conclude that (x4′ t4′ t5 t0 x2 x2′ ) is a Π–facial subwalk, because it cannot continue through the vertex tp after x2′ , otherwise it would intersect the Π–facial cycle P ∪ (t1 t2 t3 t4′ ) in two edges. We have concluded that both Π–facial subwalks (x4 t4 t3 t2 x2 x2′ ) and (x4′ t4′ t5 t0 x2 x2′ ) continue after x2′ through the same vertex (not equal to tp ). This is, they will intersect in two edges, contradicting the definition of polyhedral embedding. As we said the proof of the second case is analagous, then in both cases we get to a contradiction, and we can conclude that there is no additional vertex in the path P = (t1 t0′ x4′ t4′ ), and obviously the edge (t0′ x4′ ) is in G. Thus, there is a butterfly, B2 , contained in G: it is given by the union of the two 6–cycles (t0 t1 t2 t3 t4 t5 t0 ) ∪ (t0′ t1 t2 t3 t4′ x4′ t0′ ) and the set of scaffold edges {[t1 t4 ], [t1 t4′ ], [t0′ t3 ]}. This ends the proof of Case C1 ⊟ Case C2. Assume that the second internally disjoint 3–path between t4′ and x2 is (t4′ t5 t4 x2 ). Notice that (t2 t3 t4 x2 t2 ) is a Π–facial cycle by Proposition 6. Using Claim 1 and Definition 1 we can deduce that (t0′ t1 t2 t3 t4′ ) is a Π–facial subwalk, SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 15 i.e., the disjunction can be solved, contradicting our hypothesis. Hence, this case can never occur. This ends the proof of Case C2.⊟ Case C3. Assume that the second internally disjoint 3–path between t4′ and x2 is (t4′ x4′ x2′ x2 ). Remember that Claim C5 asserts that (t2 t3 t4′ x4′ ) is a Π–facial subwalk. So, there is a butterfly two contained in Ge , B2 , whose set of vertices is {t0 , t1 , t2 , t3 , t4′ , t5 , x4′ , x2 , x2′ } and whose set of edges is given by the union of edges in the two 6–cycles, (t0 t1 t2 t3 t4′ t5 t0 ) ∪ (t2 t3 t4′ x4′ x2′ x2 t2 ), and whose set of scaffold edges is {[t4′ t1 ], [t4′ x2 ], [t2 x4′ ]}. This ends the proof for Case C3.⊟  Theorem 16. Let Ge be such that B1 ⊂ Ge (labelled as in Definition 3, Figure 3) and such that the disjunctions that arise from the two fork subgraphs of B1 induced by the vertices {t1 , t2 , t3 , t4 , t4′ } and {t0 , t0′ , t1 , t2 , t3 } cannot be solved, then G is the Petersen’s graph, P , and Ge is the union of P and the set of (single) scaffold edges S = E(K10 ) \ E(P ). Proof. As B1 ⊂ Ge (labelled as in Definition 3) then G contains the two 6-cycles (t0 t1 t2 t3 t4 t5 t0 ) and (t0′ t1 t2 t3 t4′ t5′ t0′ ). Also, [t0 t3 ], [t0′ t3 ], [t1 t4 ], [t1 t4′ ] ∈ S. Claim 1. Either (t0 t1 t2 t3 t4′ ) or (t0′ t1 t2 t3 t4 ) (but not both) is a Π-facial subwalk. Suppose, without loss of generality, that (t0 t1 t2 t3 ) is a Π-facial subwalk. If (t1 t2 t3 t4 ) is the 3–path corresponding to [t1 t4 ], by Definition 1, (t0 t1 t2 t3 t4 ) is a Π-facial subwalk and by Proposition 4, (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle. Furthermore, by Proposition 7 and Corollary 13 imply that (t1 t0′ t5′ t4′ ) is the Π-facial subwalk corresponding to [t1 t4′ ] and (t0′ t5′ t4′ t3 ) is the Π-facial subwalk corresponding to [t0′ t3 ]. It follows that, by Definition 1, (t1 t0′ t5′ t4′ t3 ) is a Π-facial subwalk; and by Proposition 4, (t0′ t1 t2 t3 t4′ t5′ t0′ ) is a Π-facial cycle. But this last statement contradicts Definition 1, as two different facial walks would intersect in two edges. Hence (t1 t2 t3 t4 ) is not the 3–path corresponding to [t1 t4 ], so (t0 t1 t2 t3 t4′ ) is a Πfacial subwalk and then by Proposition 7, the statement follows. The case when we suppose (t0′ t1 t2 t3 ) as a Π-facial subwalk is resolved similarly. ⊟ Claim 2. Neither (t5 t0 t1 t2 t3 t4′ ) nor (t0 t1 t2 t3 t4′ t5′ ) can be a Π-facial subwalk. Suppose (t5 t0 t1 t2 t3 t4′ ) is a Π-facial subwalk, by Proposition 4, (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle, and has to be the same Π-facial walk that contains (t5 t0 t1 t2 t3 t4′ ), this contradicts the definition of polyhedral embedding (Definition 1). The case for (t0 t1 t2 t3 t4′ t5′ ) follows similarly. ⊟ Claim 3. Neither (t5′ t0′ t1 t2 t3 t4 ) nor (t0′ t1 t2 t3 t4 t5 ) can be a Π-facial subwalk. The proof of this statement follows in the same way as that of Claim 2. ⊟ Claim 4. Either (t0 t1 t2 t3 t4′ ) is a Π–facial subwalk, implying that (t0′ t1 t0 t5 t4 ) and (t0′ t5′ t4′ t3 t4 ) are Π–facial subwalks, or (t0′ t1 t2 t3 t4 ) is a Π–facial subwalk, implying that (t0 t1 t0′ t5′ t4′ ) and (t0 t5 t4 t3 t4′ ) are Π–facial subwalks. By Claim 1, we know that (t0 t1 t2 t3 t4′ ) and (t0′ t1 t2 t3 t4 ) cannot be Π–facial subwalks simultaneously. 16 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Assume that (t0 t1 t2 t3 t4′ ) is a Π-facial subwalk, then by Corollary 13, (t1 t0 t5 t4 ) is the Π-facial subwalk of [t1 t4 ]. Since t1 has degree three, necessarily (t2 t1 t0 t5 t4 ) or (t0′ t1 t0 t5 t4 ) is a Π-facial subwalk. However, if (t2 t1 t0 t5 t4 ) is a Π-facial subwalk then it intersects the Π-facial subwalk (t0 t1 t2 t3 t4′ ) in two edges, contradicting Definition 1. Hence (t0′ t1 t0 t5 t4 ) is a Π-facial subwalk. Since (t0 t1 t2 t3 ) is a Π–facial subwalk, by Proposition 7, (t0′ t1 t2 t3 ) is not a Π–facial subwalk. Thus, by Corollary 13, the 3−path corresponding to [t0′ t3 ] is (t0′ t5′ t4′ t3 ). Since t3 has degree three, necessarily (t0′ t5′ t4′ t3 t2 ) or (t0′ t5′ t4′ t3 t4 ) is a Π-facial subwalk. However, if (t0′ t5′ t4′ t3 t2 ) was a Π-facial subwalk then it would intersect the Π-facial subwalk (t0 t1 t2 t3 t4′ ) in two edges, contradicting Definition 1. Hence (t0′ t5′ t4′ t3 t4 ) is a Π-facial subwalk. The other case follows similarly. ⊟ Claim 5. (t0 t1 t2 t3 t4′ t0 ) and (t0′ t1 t2 t3 t4 t0′ ) are 5–cycles of G. (Not necessarily facial cycles of the embedding.) By Claim 1, we know that either (t0 t1 t2 t3 t4′ ) or (t0′ t1 t2 t3 t4 ) is a Π–facial subwalk. Assume that (t0 t1 t2 t3 t4′ ) is a Π–facial subwalk. Then, by Claim 4, (t0′ t1 t0 t5 t4 ) and (t0′ t5′ t4′ t3 t4 ) are Π-facial subwalks and observe that they intersect in their start and end vertices. By Proposition 8, either (t0′ t5′ t4′ t3 t4 t5 t0 t1 t0′ ) is a Π-facial cycle, or the edge (t0′ t4 ) is in E(G). However, (t0′ t5′ t4′ t3 t4 t5 t0 t1 t0′ ) can not be a Π-facial cycle, since it would intersect the Π–facial subwalk (t0 t1 t2 t3 t4′ ) in two edges and this would contradict the assumption that the embedding is polyhedral. Then, if (t0 t1 t2 t3 t4′ ) is a Π-facial subwalk, the edge (t0′ t4 ) ∈ E(G), and (t0′ t1 t2 t3 t4 t0′ ) is a 5 –cycle of G. If we assume that (t0′ t1 t2 t3 t4 ) is a Π–facial subwalk, the proof follows in a similar way and we can deduce that (t0 t4′ ) ∈ E(G) and (t0 t1 t2 t3 t4′ ) is a 5–cycle of G. Notice that, as a consequence of the previous two arguments, it can not happen that both of (t0′ t4 ) and (t0 t4′ ) are not in E(G) simultaneously. Otherwise, this would contradict Claim 1. We will now argue that both (t0′ t4 ) and (t0 t4′ ) are in E(G). Suppose that (t0 t4′ ) 6∈ E(G), then (t0′ t4 ) ∈ E(G). If (t0′ t1 t2 t3 t4 ) is a Π-facial subwalk then, by Claim 4, (t0 t1 t0′ t5′ t4′ ) and (t0 t5 t4 t3 t4′ ) are Π–facial subwalks. It follows that, by Proposition 8, (t0 t1 t0′ t5′ t4′ t3 t4 t5 t0 ) is a Π–facial cycle, but it intersects the Π–facial subwalk (t0′ t1 t2 t3 t4 ) in two edges, which contradicts the definition of polyhedral embedding. Hence (t0′ t1 t2 t3 t4 ) is not a Π–facial subwalk, and, by Claim 1, (t0 t1 t2 t3 t4′ ) has to be a Π–facial subwalk, implying that the disjunction can be solved. Therefore (t0 t4′ ) ∈ E(G). The case where we begin by assuming that (t0′ t4 ) 6∈ E(G) follows analogously, thus (t0′ t4 ) ∈ E(G). These arguments prove that (t0 t4′ ) and (t0′ t4 ) are in E(G), and (t0 t1 t2 t3 t4′ t0 ) and (t0′ t1 t2 t3 t4 t0′ ) are 5–cycles of G. ⊟ Claim 6. The scaffold edges [t0 t2 ], [t1 t3 ], [t2 t0′ ], [t2 t4′ ], [t2 t4 ] are in S. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 17 By Claim 1 and 5, and Proposition 3 either (t0 t1 t2 t3 t4′ t0 ) or (t0′ t1 t2 t3 t4 t0′ ) is a Π-facial cycle. So, if any of edges [t0 t2 ], [t1 t3 ], [t2 t0′ ], [t2 t4′ ], [t2 t4 ] were not in S, we could easily discard the possibility that one of the two cycles is a Π-facial cycle, thus, we could solve the disjunction and the claim follows. ⊟ Claim 7. There is a new vertex tx adjacent to t2 , t5 and t5′ . By Claim 6, the edges [t0 t2 ], [t2 t0′ ], [t2 t4′ ], [t2 t4 ] are in S. If for any of this edges there was only one 3−path that could correspond to it, then, by Proposition 4, we could know whether (t0 t1 t2 t3 t4′ t0 ) or (t0′ t1 t2 t3 t4 t0′ ) is a Π-facial cycle, thus we could solve the disjunction. Hence there have to be additional 3–paths that can correspond to [t0 t2 ], [t2 t0′ ], [t2 t4′ ], [t2 t4 ]. In order to find a second path between t2 and t4′ of length three, which is internally disjoint to (t2 t3 t4′ ) and (t2 t1 t0 t4′ ), let tx be the remaining vertex adjacent to t2 . Observe that tx 6= t5 , otherwise by Proposition 6 (t0 t1 t2 t5 t0 ) and (t2 t3 t4 t5 t2 ) are Π– facial cycles and using Claim 1, we can deduce easily if (t0 t1 t2 t3 t4′ ) or (t0 t1 t2 t3 t4′ ) is a Π–facial subwalk, contradicting our hypothesis. Analogously tx 6= t5′ . Thus tx is a new vertex. Notice that [tx t0 ], [tx t4 ], [tx t0′ ], [tx t4′ ] are in S, otherwise, by Proposition 7, we can deduce the sub-paths in the Π-facial subwalks that correspond to [t1 t4 ] and [t1 t4′ ], which implies that we can solve the disjunction. Then, the second 3–path between t2 and t4′ contains tx and t5′ . This implies that (tx t5′ ) is an edge of G. Similarly, for [t0 t2 ] there are already two disjoint paths (t2 t1 t0 ) and (t2 t3 t4′ t0 ) of length two and three respectively. Then, the third path necessarily contains tx and t5 , which implies (as before) that (tx t5 ) is an edge of G. ⊟ Note that we now know the complete underlying cubic graph G, namely G is the Petersen’s graph. Claim 8. [t0 t0′ ], [t0 t5′ ], [t0 t4 ], [t1 t5 ], [t1 x], [t1 t5′ ], [t2 t5′ ], [t2 t5 ], [t3 x], [t3 t5′ ], [t3 t5 ], [t4′ t5 ], [t4 t4′ ], [t0 t4′ ], [t5 t0′ ], [t4 t5′ ] and [t5 t5′ ] are in S. We will use identical arguments to those in Claim 6. For example if [t0 t0′ ] is not in S, then any 3–path whith start t0 and end t0′ can’t be a facial subwalk. This implies that by process of elimination and using Proposition 7 and Proposition 4, we can deduce the Π–facial cycles, and the disjunction can be solved. ⊟ This finalizes the proof. Therefore, G is Petersen Graph and S = E(K10 \ E(P )).  Theorem 17. Let Ge be such that B2 ⊂ Ge (labelled as Definition 3, Figure 3) and such that the disjunction that arises from the fork subgraph of B2 induced by the vertices {t1 , t2 , t3 , t4 , t4′ } cannot be solved, then G is the Franklin graph, F , and Ge is the union of F and the set S given by simple scaffold edges [t0 t5 ], [t5 x5 ], [x5 x0 ], [x0 t0 ], [t1 t2 ], [t2 x2 ], [x2 t0′ ], [t0′ t1 ], [t3 t4′ ], [t4′ t5′ ], [t5′ t4 ], [t4 t3 ], and the double scaffold edges [[t0 t3 ]], [[x0 t2 ]], [[t1 t4′ ]], [[t0′ t5 ]], [[t4 x2 ]], [[x5 t5′ ]], [[t0 t5′ ]], [[t1 t4 ]], [[t2 t5 ]], [[t3 x5 ]], [[t4′ x2 ]], [[x0 t0′ ]]. Proof. Observe that in B2 , (t0 t1 t2 t3 t4 t5 t0 ) and (t0′ t1 t2 t3 t4′ t5′ t0′ ) are two 6-cycles such that [t1 t4 ], [t1 t4′ ], [t0 t3 ] ∈ Ge and [t0′ t3 ] 6∈ Ge , which implies by Proposition 18 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN 7, that (t0 t1 t2 t3 ) is Π-facial subwalk of [t0 t3 ]. Claim 1. [[t0 t3 ]], [[t1 t4 ]] and [[t2 t5 ]] are double scaffold edges. If one of [[t0 t3 ]], [[t1 t4 ]] or [[t2 t5 ]] is not a double scaffold edge, then (t1 t2 t3 t4 ) cannot be a Π-facial subwalk, otherwise by Definition 1 and Proposition 4, this would imply that (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle, but this is impossible as we started by assuming that at least one of [[t0 t3 ]], [[t1 t4 ]] or [[t2 t5 ]] is not a double scaffold edge. It follows by Proposition 7 that (t1 t2 t3 t4′ ) is the Π-facial subwalk that corresponds to the scaffold edge [t1 t4′ ], and then, by Proposition 7 and Corollary 13, the Π-facial subwalk corresponding to the edge [t1 t4 ] should be (t1 t0 t5 t4 ), which implies that the disjunction could be solved, contradicting our hypothesis. Hence the claim holds. ⊟ Claim 2. For each pair of vertices {t0 , t3 }, {t1 , t4 } and {t2 , t5 }, there is a third 3-path between them whose vertices are disjoint to the cycle (t0 t1 t2 t3 t4 t5 t0 ). Suppose that for at least one of the pairs of vertices mentioned above, there are only two 3–paths, by Proposition 8, (t0 t1 t2 t3 t4 t5 t0 ) is Π-facial cycle and then (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ], i.e., we can solve the disjunction, contradicting our hypothesis. Hence the claim holds. ⊟ Since G is a cubic graph, let xi be the remaining vertex adjacent to ti for i ∈ {0, 2, 5}, then: Claim 3. The vertex x0 is different to the vertices ti for all i ∈ {0, 1, 2, 3, 4, 5, 0′ , 4′ , 5′ } and adjacent to t4′ . First, in order to prove that all vertices are different we proceed by contradiction: 1. (x0 6= t0′ .) If x0 = t0′ , then by Proposition 5, (t0 t0′ t1 t0 ) is a Π-facial cycle, this implies, by Proposition 7, that (t5 t0 t1 t2 ) is a Π-facial subwalk. Since (t0 t1 t2 t3 ) is Π-facial subwalk, by Definition 1, necessarily (t5 t0 t1 t2 t3 ) is a Π-facial subwalk, and using Proposition 4, (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle. The latter implies that (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ] and thus we can solve the disjunction, contradicting our hypothesis. Hence x0 6= t0′ . 2. (x0 6= t2 .) If x0 = t2 , by Proposition 5, (t0 t1 t2 t0 ) is a Π–facial cycle, but this contradicts, by Definition 1, the fact that (t0 t1 t2 t3 ) is a Π–facial subwalk. Hence x0 6= t2 . 3. (x0 6= t4 .) If x0 = t4 , since (t0 t1 t2 t3 ) is a Π–facial subwalk, Proposition 4 implies that (t0 t1 t2 t3 t4 t0 ) is a Π–facial cycle, and (t1 t2 t3 t4 ) is the Π–facial subwalk corresponding to [t1 t4 ]. Thus, the disjunction can be solved, contradicting our hypothesis. Hence, x0 6= t4 . 4. (x0 6= t3 .) If x0 = t3 , then t3 would have degree four and G would not be a cubic graph. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 19 5. (x0 6= t4′ .) If x0 = t4′ , since (t0 t1 t2 t3 ) is a Π-facial subwalk, Definition 1 and Proposition 4 imply that (t0 t1 t2 t3 t4′ t0 ) is a Π-facial cycle and that (t1 t2 t3 t4′ ) is the Π-facial subwalk corresponding to [t1 t4′ ]. Thus we can solve the disjunction, contradicting our hypothesis. Hence x0 6= t4′ . 6. (x0 6= t5′ .) If x0 = t5′ , then by Proposition 6, (t0 t1 t0′ t5′ t0 ) is a Π-facial cycle; the latter and Proposition 7 imply that (t2 t1 t0 t5 ) is a Π-facial subwalk, and by Definition 1, since (t0 t1 t2 t3 ) is a Π-facial subwalk, (t5 t0 t1 t2 t3 ) is a Π-facial subwalk. By Proposition 4, it follows that (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle and (t1 t2 t3 t4 ) is the Π-facial subwalk that corresponds to [t1 t4 ]. The previous statements imply that we can solve the disjunction, contradicting our hypothesis. Hence x0 6= t5′ . Finally, Claim 2 implies x0 and t4′ are adjacent.⊟ Claim 4. The vertex x2 is different to the vertices ti for all i ∈ {0, 1, 2, 3, 4, 5, 0′ , 4′ , 5′ } and x0 . We proceed by contradiction: 1. (x2 6= t0 .) If x2 = t0 then t0 would have degree four and contradict the fact that G is a cubic graph. Hence x2 6= t0 . 2. (x2 6= t0′ .) If x2 = t0′ , by Proposition 5, (t0′ t1 t2 t0′ ) is a Π–facial cycle. Thus, by the previous statement and Definition 1, (t3 t2 t0′ t5′ ) is a Π–facial subwalk. By the previous statement and Proposition 4, (t3 t2 t0′ t5′ t4′ t3 ) is a Π–facial cycle. The latter and Definition 1, imply that (t1 t2 t3 t4′ ) is not a Π–facial subwalk. This, in turn, implies that (t1 t2 t3 t4 ) is a Π–facial subwalk, and the disjunction can be solved, contradicting our hypothesis. Hence x2 6= t0′ . 3. (x2 6= t4 .) If x2 = t4 , by Proposition 5, (t2 t3 t4 t2 ) is a Π–facial cycle. The latter together with Definition 1 imply (t1 t2 t3 t4 ) is not a Π–facial subwalk, thus (t1 t2 t3 t4′ ) is a Π–facial subwalk and the disjunction can be solved, contradicting our hypothesis. Hence x2 6= t4 . 4. (x2 6= t4′ .) The proof follows identical arguments to those used in the previous case. 5. (x2 6= t5 .) If x2 = t5 , by Proposition 6, (t2 t3 t4 t5 t2 ) is a Π–facial cycle. The latter and Definition 1, imply that (t1 t2 t3 t4 ) is not a Π–facial subwalk, thus (t1 t2 t3 t4′ ) is a Π–facial subwalk and the disjunction can be solved, contradicting our hypothesis. Hence x2 6= t5 . 6. (x2 6= t5′ .) The proof is similar to the previous case. 7. (x2 6= x0 .) If x2 = x0 , then by Proposition 6, (t0 t1 t2 x0 t0 ) is a Π-facial cycle. Thus the Π–facial cycle (t0 t1 t2 x0 ) intersects to the Π–facial subwalk (t0 t1 t2 t3 ) in three vertices contradicting that we have a polyhedral embedding. ⊟ Claim 5. The vertex x5 is different to the vertices ti for all i ∈ {0, 1, 2, 3, 4, 5, 0′ , 4′ , 5′ } and x0 , x2 , and adjacent to x2 . Once more, we proceed by contradiction to prove that the vertices are different: 20 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN 1. (x5 6= t0′ .) If x5 = t0′ , then Proposition 6 implies that (t5 t0 t1 t0′ t5 ) is a Π-facial cycle. The latter and Definition 1 imply that (t1 t0 t5 t4 ) is not a Π-facial subwalk. Then, by Corollary 13, it follows that (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to the edge [t1 t4 ]. However, using Proposition 4 we can deduce that (t0 t1 t2 t3 t4 t5 t0 ) is a Π–facial cycle which intersects the Π–facial cycle (t5 t0 t1 t0′ t5 ) in two edges, contradicting the polyhedrality of the embedding. In conclusion, this case can never occur. 2. (x5 6= t2 , x5 6= t1 , x5 6= t3 , x5 6= t4′ ,.) If any of the inequalities was an equality instead we would have a contradiction on the fact that G is a cubic graph. 3. (x5 6= x0 .) If x5 = x0 , then it is only possible to find two internally disjoint 3-paths between t2 and t5 . Since [[t2 t5 ]] is a double scaffold edge and (t2 t5 ) 6∈ E(G) (otherwise contradicts that G is a cubic graph), Proposition 8 implies that (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle. In turn, the latter implies that (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ] and that (t1 t0′ t5′ t4′ ) is a Π-facial subwalk, thus we can solve the disjunction, contradicting our hypothesis. Hence x5 6= x0 . 4. (x5 6= x2 .) If x5 = x2 , Proposition 4 implies that (t0 t1 t2 t3 t4 t5 t0 ) is not a Πfacial cycle. Since (t0 t1 t2 t3 ) is a Π−facial subwalk, using Proposition 11 we can deduce that (t0 t1 t2 t3 t4′ x0 t0 ) is a Π-facial cycle; i.e., (t1 t2 t3 t4′ ) is the Π−facial subwalk that corresponds to [t1 t4′ ]. The latter and Corollary 13 imply that (t1 t0 t5 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ], and then we can solve the disjunction, contradicting our hypothesis. Hence x5 6= x2 . 5. (x5 6= t5′ .) If x5 = t5′ , it is only possible to find two 3-paths internally disjoint between them t2 and t5 . Since [[t2 t5 ]] is double and (t2 t5 ) 6∈ E(G) (otherwise contradicts that G is a cubic graph), Proposition 8 implies that (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle. Hence, (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ]. Using the latter and Corollary 13 is easy to see that (t1 t0′ t5′ t4′ ) is the Π-facial subwalk corresponding to [t1 t4′ ]. That is we can solve the disjunction, contradicting our hypothesis. Hence x5 6= x5′ . Finally, x2 and x5 are adjacent by Claim 2. ⊟ Claim 6. The vertex x4 is different to the vertices ti , for all i ∈ {0, 1, 2, 3, 4, 5, 0′ , 4′ } and x0 , x2 , x5 . We proceed by contradiction: 1. (x4 6= t0 , x4 6= t1 , x4 6= t2 , x4 6= t4′ ). If any of the inequalities was an equality instead we would have a contradiction on the fact that G is a cubic graph. 2. (x4 6= t0′ .) If x4 = t0′ then there is a 2-path (t1 t0′ t4 ) and two 3–paths, (t1 t2 t3 t4 ) and (t1 t0 t5 t4 ) between t1 and t4 . Since [[t1 t4 ]] is a double scaffold edge, and there is no edge (t1 t4 ), by Proposition 8 (t0 t1 t2 t3 t4 t5 t0 )is a Π–facial cycle. But this contradicts Proposition 4 since there is a 2–path (t1 t0′ t4 ) between t1 and t4 . Hence x4 6= x0′ . 3. (x4 6= x0 .) If x4 = x0 , Proposition 6 implies that (t0 t5 t4 x0 t0 ) is a Π-facial cycle. The latter, Proposition 7 and Corollary 13 imply that (t1 t2 t3 t4 ) is the Π-facial SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 21 subwalk corresponding to [t1 t4 ] and we can solve the disjunction, contradicting our hypothesis. Hence x4 6= x0 . 4. (x4 6= x2 .) If x4 = x2 , then Proposition 6 implies that (t2 t3 t4 x2 t2 ) is a Πfacial cycle. The latter and Proposition 7 imply that (t1 t2 t3 t4′ ) is the Π-facial subwalk corresponding to [t1 t4′ ]. Subsequently, using Corollary 13 it is possible to deduce that (t1 t0 t5 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ], that is, we can solve the disjunction, contradicting our hypothesis. Hence x4 6= x2 . 5. (x4 6= x5 .) If x4 = x5 , using Proposition 5 we deduce that (t4 t5 x5 t4 ) is a Π– facial cycle. In addition to this, using Proposition 7 we can prove that (t0 t5 t4 t3 ) is a Π-facial subwalk, ussing that (t0 t3 ) 6∈ E(G) and applying Proposition 8, we have (t0 t1 t2 t3 t4 t5 t0 ) is a Π-facial cycle. The latter implies that (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ]. Using that (t1 t2 t3 t4 ) is the Π-facial subwalk corresponding to [t1 t4 ] and Corollary 13 we arrive to the conclusion that (t1 t0′ t5′ t4′ ) is the Π-facial subwalk corresponding to [t1 t4′ ]. Thus, we can solve the disjuntion, contradicting our hypothesis. Hence x4 6= x5 . ⊟ Claim 7. [[t1 t4′ ]] is a double scaffold edge. Since [t1 t4′ ] ∈ S and between t1 and t4′ there are three internally disjoint 3-paths, (t1 t2 t3 t4′ ), (t1 t0′ t5′ t4′ ) and (t1 t0 x0 t4′ ), by Proposition 11, necessarily [[t1 t4′ ]] is double. ⊟ Claim 8. (t1 t0 x0 t4′ ) is a Π-facial subwalk. Since (t0 t1 t2 t3 ) is a Π-facial subwalk and [[t1 t4′ ]] is a double scaffold edge (by Claim 7), then Corollary 12, implies that (t1 t0 x0 t4′ ) is a Π-facial subwalk.⊟ Claim 9. [[t0 t5′ ]], [[x0 t0′ ]], [[t0 t3 ]] and [[x0 t2 ]] are double scaffold edges and there exist three internally disjoint paths between each pair of end vertices. If some edge, [t0 t5′ ], [x0 t0′ ], [t0 t3 ] or [x0 t2 ] does not belong to S, or it is not double or they are all double but for one of them there exist only two 3-paths between its end vertices, then by Proposition 8 we know if (t1 t2 t3 t4′ ) is Π-facial subwalk or not. This would implies that the disjunction can be solved, contradicting the hypothesis. ⊟ Let x5′ be the remaining vertex adjacent to t5′ . Claim 10. (t0 t1 t0′ t5′ ) is a Π-facial subwalk. Notice that for the double scaffold edge [[t0 t5′ ]] there exist three internally disjoint 3−paths, (t0 x0 t4′ t5′ ), (t0 t5 x5′ t5′ ) and (t0 t1 t0′ t5′ ). Since (t1 t0 x0 t4′ ) is a Π-facial subwalk, then by Corollary 12, necessarily (t0 x0 t4′ t5′ t0′ t1 t0 ) or (t0 t5 x5 t5′ t0′ t1 t0 ) is a Π-facial cycle, i.e., (t0 t1 t0′ t5′ ) is a Π-facial subwalk. ⊟ Claim 11. (t1 t0 t5 t4 ) is a Π-facial subwalk. Since [[t1 t4 ]] is a double scaffold edge, there exist three internally disjoint 3−paths : (t1 t0 t5 t4 ), (t1 t0′ x4 t4 ) and (t1 t2 t3 t4 ). Since (t0 t1 t2 t3 ) is a Π-facial subwalk, then by Corollary 12, (t1 t0 t5 t4 ) is a Π-facial subwalk. ⊟ 22 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Claim 12. x4 = t5′ . If x4 6= t5′ , then (t1 t0 t5 t4 x4 t0′ t1 ) is not a Π-facial cycle because (t0 t1 t0′ t5′ ) is a Π-facial subwalk, and this would contradict Definition 1. In addition, Claim 11 implies that (t1 t0 t5 t4 t3 t2 t1 ) is a Π-facial cycle and (t1 t0′ t5′ t4′ ) is the Π-facial subwalk of [t1 t4′ ]. Thus the disjunction could be solved, contradicting our hypothesis. ⊟ Claim 13. (t2 t3 t4 t5 ) is a Π-facial subwalk. Notice that for the double scaffold edge [[t2 t5 ]] there are three internally disjoint 3-paths (t2 t1 t0 t5 ), (t2 x2 x5 t5 ) and (t2 t3 t4 t5 ). Since (t0 t1 t2 t3 ) is a Π-facial subwalk, by Corollary 12, (t2 t3 t4 t5 ) is a Π-facial subwalk.⊟ Claim 14. [[t3 x5 ]], [[t4 x2 ]], [[t0 t3 ]] and [[t1 t4 ]] are double scaffold edges in S and there exist three internally disjoint paths between its end vertices. If one of [t3 x5 ], [t4 x2 ], [t0 t3 ] or [t1 t4 ] does not belong in S, or it is not a double scaffold edge, or they are all double but for one of them there exist only two 3-paths between its end vertices, then by Proposition 8, we know if (t1 t2 t3 t4 ) is a Π–facial subwalk or not. Consequently, using Proposition 11, we could conclude if the Π– facial subwalk corresponding to the edge [t1 t4 ] is either (t1 t2 t3 t4 ) or (t1 t0′ t5′ t4 ) thus we could solve the disjunction, contradicting our hypothesis. ⊟ Observe that for the edges [[t0 t3 ]] and [[t1 t4 ]] there already exist three internally disjoint paths between their end vertices. For [[t3 x5 ]] and [[t4 x2 ]] we need to find additional edges of G to complete their corresponding 3-paths. Claim 15. (x0 x5 ) ∈ E(G) Note that for [[t3 x5 ]] there are already two 3-paths, (t3 t2 x2 x5 ) and (t3 t4 t5 x5 ). Since the only adjacent vertex to t3 that is not in either of the paths already mentioned is t4′ , then t4′ must belong to the third 3-path. The third 3-path can continue through either t5′ or x0 . However, the vertex t5′ is adjacent to t4 , t4′ and t0′ , then the third 3-path necessarily has to continue to x0 . Since x0 has degree two thus far, the only option left now is that it is adjacent to x5 . ⊟ Claim 16. (x2 x0′ ) ∈ E(G) The proof of this statement follows in a manner similar to the proof of Claim 15. ⊟ Claim 17. [[x2 t4′ ]], [[t5′ x5 ]] and [[t5 t0′ ]] are double scaffold edges. Notice that if [x2 t4′ ] 6∈ S, by Proposition 7, (t1 t2 t3 t4′ ) is the Π-facial subwalk corresponding to [t1 t4′ ], and then the disjunction can be solved, contradicting the hypothesis. Hence [x2 t4′ ] ∈ S. Since there are three internally disjoint 3-paths between its end vertices, by Proposition 11, [[x2 t4′ ]]has to be a double scaffold edge. The proof follows similarly for [[t5′ x5 ]] and [[t5 t0′ ]]. ⊟ Claim 18. The edges [t1 t0′ ], [t0′ x2 ], [x2 t2 ], [t1 t2 ], [t3 t4′ ], [t4′ t5′ ], [x4 t4 ], [t3 t4 ], [x0 t0 ], [t0 t5 ], [t5 x5 ] and [x5 x0 ] are in S. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 23 Since (t1 t2 x2 t0′ t1 ), (t0 t5 x5 x0 t0 ) and (t3 t4 t5′ t4′ t3 ) are Π-facial cycles (by Proposition 6), the edges [t1 t0′ ], [t0′ x2 ], [x2 t2 ], [t1 t2 ], [t3 t4′ ], [t4′ t5′ ], [x4 t4 ], [t3 t4 ], [x0 t0 ], [t0 t5 ], [t5 x5 ] and [x5 x0 ] are in S. Claim 18, completes the proof of the Theorem.  Conclusion The main aim of the project, that this paper initiates, is to develop an algorithmic procedure for constructing all the extended graphs of a given cubic graph. Thus, the next step is characterizing when a set of scaffold edges build on top of a cubic graph is indeed an extended graph. 24 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Theorem 15. CASE A Claim 1. Claim 2. Claim 3. CASE A. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. Theorem 15. CASE B Claim 1. Claim 2. Claim 3. CASE B. Case B1. Claim B1.1. Claim B1.2. Claim B1.3. Case B2. Claim B2.1 Claim B2.1 Claim B2.2. Case B2.A. Case B2.B. Case B2.C. 25 26 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Theorem 15. CASE C Claim 1. Claim 2. Claim 3. CASE C. Claim C1. Claim C2. Claim C3. Claim C4. Claim C5. Claim C6. Case C1. Claim C1.1. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. Claim C1.2 Claim C1.3 Claim C1.4 Claim C1.4 Claim C1.5 Claim C1.5 Claim C1.6 Claim C1.6.1 Claim C1.6.2 Claim C1.6.3 Case C2 Case C3. 27 28 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Theorem 16 Claim 1. Claim 2. Claim 3. Claim 4. Claim 5. Claim 7. Claim 6. Claim 8. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 29 Theorem 17 Claim 1. Claim 2. Claim 3. Claim 4. Claim 5. Claim 6. Claim 7. Claim 8. Claim 9. 30 F. AGUILAR, G.ARAUJO-PARDO, AND N. GARCÍA-COLÍN Claim 10. Claim 11. Claim 12. Claim 13. Claim 14. Claim 15. Claim 16. Claim 17. Claim 18. SCAFFOLD FOR THE POLYHEDRAL EMBEDDING OF CUBIC GRAPHS. 31 Acknowledgements. Research supported by PAPIIT-México (Projects IN104915, IN106318), CONACyT-México (Project 166306) and CIC-UNAM (Academic Exchange Project: “Gráficas y Triangulaciones”. 2019). References [1] J. Arocha, J., N. García-Colín and I. Hubard. Reconstructing surface triangulations by their intersection matrices. Discussiones Mathematicae Graph Theory, 35: 483-491, 2015. doi:10.7151/dmgt.1816. [2] Bojan Mohar. Existence of Polyhedral Embeddings of Graphs. Combinatorica, 21: 395–401, 2001. doi:10.1007/s004930100003. [3] Bojan Mohar. Uniqueness and minimality of large face-width embeddings of graphs. Combinatorica, 15: 541–556, 1995. [4] Bojan Mohar and Neil Robertson. Flexibility of Polyhedral Embeddings of Graphs in Surfaces. Journal of Combinatorial Theory, 83: 38-57, 2001. doi:10.1006/jctb.2001.2036 [5] Bojan Mohar and Andrej Vodopivec. On polyhedral embeddings of cubic graphs. Combinatorics, Probability and Computing, 15: 877-893, 2006. doi:10.1017/S0963548306007607. [6] Gerhard Ringel. Map Color Theorem. Springer-Verlag, 1974. [7] Neil Robertson and Richar Vitray. Representativity of surface embeddings. Paths, Flows, and VLSI-Layout. Springer-Verlag, 1990 . pp. 293–328. [8] Paul Seymour and Robin Thomas. Uniqueness of highly representative surface embeddings. J. Graph Theory, 23: 337–349, 1966. [9] Hassler Whitney. 2-isomorphic graphs. Amer. Math. J., 55: 245–254, 1933. .