Deriving graphs with a retracting-free bidirectional double tracing

A retracting-free bidirectional double tracing in a graph G is a closed walk which traverses every edge exactly once in each direction and such that no edge is succeeded by the same edge in the opposite direction. Studying the class Ω of all graphs admitting a retracting-free bidirectional double tracing was proposed by Ore (1951) and is, by now, of practical use to (bio)nanotechnology. In particular, this ﬁeld needs various molecular polyhedra that are constructed from a single chain molecule in a retracting-free bidirectional double-tracing way. A cubic graph Q ∈ Ω has 3 h edges, where h is an odd number ≥ 3 . The graph of the triangular prism is the minimum cubic graph Q ∈ Ω , having 6 vertices and 9 edges. The graph of the square pyramid is the minimum polyhedral graph G in Ω , having 5 vertices and 8 edges. We analyze some possibilities for deriving new Ω -graphs from a given graph G ∈ Ω or G (cid:54)∈ Ω using graph-theoretical operations. In particular, there was found that every noncycle Eulerian graph H admits a retracting-free bidirectional double tracing ( H ∈ Ω ) , which is a partial solution to the problem posed by Ore.


Preliminaries
In 1951, Ore [10] posed a problem, asking for a characterization of graphs that admit closed walk which traverses every edge exactly once in each direction and such that no edge is succeeded by the same edge in the opposite direction. The problem was partially solved in [15] and [1], and completely solved almost 40 years later by Thomassen [14]. The further results were obtained in [2,6,12,13]. We denote by Ω the class of all graphs about which Ore posed his question [10]. The closed walk under consideration [1,10,14,15] is called here a retracting-free bidirectional tracing [14]. Studying the class Ω [1,2,6,10,[12][13][14][15] is, in particular, of practical use to nanotechnology [3,[7][8][9]. Nanotechnologists construct various molecular polyhedra from a single chain molecule by winding it onto a polyhedron's skeleton in a retracting-free bidirectional double-tracing way [3,[7][8][9].
Let Q = (V, E) be a simple cubic graph with the vertex set V and edge set E |V | = n, |E| = m = 3n/2 . A spanning tree T of Q is a subtree covering all the vertices of Q |V (T )| = n; |E(T )| = n − 1 . Its cotree Q − E(T ) V Q − E(T ) = n; E Q − E(T ) = (n + 2)/2 is a graph Q less all edges belonging to T . Thus, the cotree Q − E(T ) of a spanning tree T in a connected graph Q is the spanning subgraph of Q containing exactly those edges of Q which are not in T [4].
Thomassen proved the following (Theorem 3.3 of [14]): Theorem 1. A connected multigraph G has a retracting-free bidirectional double tracing if and only if G has no vertex of degree 1 and has a spanning tree T such that each connected component of G − E(T ) either has an even number of edges or contains a vertex which in G has degree at least 4.
A more specific result is [11]: Corollary 1.1. A cubic graph Q admits a closed walk which traverses every edge exactly once in each direction and such that no edge is succeeded by the same edge in the opposite direction if and only if Q has a spanning tree T such that each (connected) component of the cotree Q − E(T ) has an even number of edges and is either a proper cycle or simple path of length ≥ 0.
Another corollary is [11]: Let a cubic graph Q admit a closed walk which traverses every edge exactly once in each direction and such that no edge is succeeded by the same edge in the opposite direction (Q ∈ Ω ). Then, Q has an odd number |E(Q)| = 3h of edges (where h = 3, 5, 7, . . .).
The following lemma determines the number of all simple paths in a cotree Q − E(T ) [11]: Denote by c k and p l the numbers of cycles of length k and paths of length l in Q − E(T ), respectively, where k = 4, 6, 8, . . . , 2h and l = 0, 2, 4, . . . , 2h (h ∈ N \ 2N). Some simple relations for these numbers are known, e.g. [11]: Here, we turn to the main section, where we deduce some other results.

The main part
Let n j = n j (G) (j ∈ N) be the number of vertices of valency j in a graph G. We state: Lemma 4. Let T be a spanning tree of a simple cubic graph Q on n vertices. Then, T obeys the following conditions: (i) n 1 (T ) = n 3 (T ) + 2; (ii) n 2 (T ) = n + 2 − 2n 1 (T ) n 2 (T ) is even .

Thus,
which is tantamount to (i). Now, transform the R. H. S. of (ii) and obtain: where the last but one equality is due to (i), and the last side is equal to the first side, which proves (ii). This completes the proof.
The following corollary is the dual of Lemma 4.
A caterpillar, or caterpillar tree, is a tree in which all the vertices are within distance 1 of a central path [5]. Here, we deal with the simple caterpillars having all vertex degrees ≤ 3. Such a caterpillar C 2h is obtained by attaching exactly one pendent vertex to each vertex of a path P h on h vertices (of length h − 1); thus, C 2h is a tree on 2h vertices. Throughout this text, we use odd numbers h ≥ 3, which is not a restriction for a caterpillar, in general.
Imagine an h-angular prism P 2h , whose one h-angular face is consecutively labeled by the numbers 1, 2, . . . , h, while the other such face is similarly numbered by h + 1, h + 2, . . . , 2h; and two vertices g and g + h g ∈ {1, 2, . . . , h} are the ends of a vertical edge. For h > 3, consider a path of length h − 1, consecutively spanning h vertices, 2 + h, 2, 3, . . . , h − 1, 2h − 1. By attaching to this path h pendent vertices, 1 + h, 1, 3 + h, . . . , h, 2h, respectively, we mark a spanning simple caterpillar C 2h in the skeleton of a prism P 2h . In the skeleton of the missed triangular prism P 6 (h = 3), a spanning caterpillar tree C 6 takes the form of a 'double fork' (term), with a lateral edge of the prism as its middle one. Then, we may represent any prism by its plane graph (map) and make all our construction on the latter. Denote by Q 2h and T 2h a plane graph of P 2h and that of its spanning caterpillar tree C 2h ⊂ P 2h , respectively.
Here, we state a technical lemma, viz.: be the graph of an h-angular prism (h is an odd number ≥ 3) and T 2h be its spanning caterpillar tree. Then, the cotree Q 2h − T 2h is the union of a cycle of length 4 (representing a lateral square face), a simple path of length h − 3, and h − 2 isolated vertices.
Proof. Sketch the proof, which follows from the construction. It is easy to establish that the length of a simple path and the number of isolated vertices are such as stated. What remains uncovered by the path and isolated vertices comprises edges. Such quantities of vertices and edges exclude an instance of disconnected subgraph. While a simple subgraph having 4 vertices and 4 edges may be either a triangle with an attached pendent edge or a cycle of length 4. The former is excluded because vertex degrees in the cotree Q 2h − T 2h cannot exceed 2; consequently, such a connected subgraph can only be a cycle. Whence we arrive at the proof.
be a cubic graph of an h-angular prism, then Q 2h affords a retracting-free bidirectional double tracing if and only if h is an odd number ≥ 3.
Proof. The necessity follows from Corollary 1.2. By virtue of Lemma 5, all components of the cotree Q 2h − E(T 2h ) satisfy Theorem 1 (and Corollary 1.1), which is a sufficient condition. This gives the proof.
The operation of inserting a new 2-valent vertex into an edge is called a subdivision of an edge. Also, the subdivision (graph) S(G) of a graph G is the graph obtained by subdivision of every edge in G. Two graphs are homeomorphic if both can be obtained from the same graph by a sequence of subdivisions of edges [4]. The following statement is fundamental. Theorem 6. Let G 1 and G 2 be two homeomorphic simple graphs. Then, G 1 ∈ Ω iff G 2 ∈ Ω.
Proof. First, let a graph G 1 belong to Ω, and let σ 1 be a retracting-free bidirectional double tracing in it. Let the tracing σ 1 traverse an edge uv first from vertex u to vertex v, then, after traversing some other edges, traverse the same edge uv in an opposite direction, from v to u. Replace the edge uv by an elementary path consecutively visiting vertices u, w 1 , w 2 , . . . , w p , v, where all intermediate vertices w j (j ∈ {1, 2, . . . , p}) are 2-valent in a derived homeomorphic graph G 2 . Evidently, G 2 also admits a retracting-free bidirectional double tracing σ 2 which, in place of traversing in opposite directions the edge uv, similarly traverses the path . While the other edges of G 2 are traversed by σ 2 in the same order as these are traversed by σ 1 in G 1 . Conversely, let be an elementary path in G 2 , consecutively visiting vertices u , w 1 , w 2 , . . . , w p , v , where all vertices w k (k ∈ {1, 2, . . . , p }) are 2-valent in G 2 . It can similarly be demonstrated that replacing an elementary path in G 2 by a single edge u v allows a retracting-free bidirectional double tracing σ 1 in an obtained homeomorphic graph G 1 . This in sum proves that G 1 ∈ Ω ⇔ G 2 ∈ Ω. Now on the contrary, let G 1 ∈ Ω, and let ξ 1 be an arbitrary bidirectional circuit where an edge uv is consecutively traversed in both directions (from u to v, then, immediately, backwards to u). Substitute for the edge uv an elementary path consecutively visiting vertices u, w 1 , w 2 , . . . , w p , v, where all inner vertices are 2-valent in G 1 . By this substitution, the circuit ξ 1 is transformed into a circuit ξ 2 of a derivative homeomorphic graph G 2 , while preserving in G 2 the same order of traversing all edges that are inherited from G 1 . Under such conditions, ξ 2 cannot avoid immediately traversing in an opposite direction one end edge (uw 1 or w p v) of . Conversely, let be an elementary path in G 2 consecutively visiting vertices u , w 1 , w 2 , . . . , w p , v , where all inner vertices are 2-valent in G 2 . If is consecutively traversed in both directions, substitution of an edge u v for also cannot avoid immediately traversing this edge in an opposite direction. Hence, we in sum conclude that G 1 ∈ Ω ⇔ G 2 ∈ Ω. Taking in account both considered cases, when G 1 , G 2 ∈ Ω and G 1 , G 2 ∈ Ω, we arrive at the overall proof.
An undirected graph H is called a minor of the graph G if H can be formed from G by deleting edges and vertices and by contracting edges. Note that two homeomorphic graphs G 1 and G 2 always have a common minor H that is their common homeomorph without 2-valent vertices. Within such a broadened context, we come here to another statement, which adds to Theorem 6. Theorem 7. Let G be a simple graph belonging to Ω, and let G(u, v) be a graph obtained by identifying two arbitrary vertices u and v of G, without forming a loop from edge uv, if any Proof. Let σ be a retracting-free bidirectional double tracing in G, which consecutively traverses edges e 1 , e 2 , . . . , e 2|E(G)| (each of |E(G)| edges is traversed in this sequence twice, in opposite directions). First, let u and v be nonadjacent vertices in G. Then, identifying vertices u and v allows us also to use the same cyclic sequence of edges for a retracting-free bidirectional double tracing σ in G(u, v), which satisfies our statement. Now let u and v be adjacent vertices in G. In this case, deleting an edge u v and identifying vertices u and v , of G, produces another connected cyclic sequence of edges, where edge u v is missing. However, missing the edge u v of σ does not change the order of traversing the other edges (borrowed from σ) in the last reduced sequence (and the direction of traversing each edge therein). Thus, such a reduced sequence also comprises a retracting-free bidirectional double tracing σ in G(u , v ), which implies that G(u , v ) ∈ Ω. The last fact in combination with that G(u, v) ∈ Ω, for nonadjacent vertices u and v, completes the proof.
The nanobiological field [3,[7][8][9] needs finding working rules that allow researchers to construct bigger graphs belonging to Ω from smaller ones. The latter graphs may or may not themselves belong to Ω; in particular, they may be Eulerian graphs. Below, we present some rigorous assertions that contribute to this topic.
Let Θ be the class of all Eulerian simple graphs. For convenience, we also introduce a combined class Φ := Θ ∪ Ω which is comprised of all Eulerian and all Ω-graphs. For these classes of graphs, we give here the following three theorems.
Theorem 8. Let G 1 ∈ Φ and G 2 ∈ Ω be two simple graphs with vertices v 1 and v 2 , respectively. Also, let G be a graph obtained by connecting vertices v 1 and v 2 with a simple path of length l (l ∈ N + ). Then, G ∈ Ω.
Proof. Divide the proof into two parts. Case 1. Let G 1 ∈ Θ and G 2 ∈ Ω. We start from vertex v 1 of G 1 and traverse all edges of an arbitrary Eulerian circuit ε 'in a positive direction' to complete this circuit and return to the original vertex v 1 . Then, we move along the connecting path to vertex v 2 of G 2 and traverse its edges to exactly complete an arbitrary retracting-free bidirectional double tracing thereof and return to vertex v 2 . From v 2 , we move along the connecting path in an opposite direction into vertex v 1 from which we traverse all edges of the Eulerian circuit ε, of G 1 , now in an opposite, 'negative direction' and return to our original point v 1 . Clearly, our walk used each edge of G exactly once in each direction and without immediately passing through any edge in an opposite direction (only doing this after traversing other edges). This proves Case 1. Case 2. Let G 1 , G 2 ∈ Ω. We can traverse from vertex v 1 all edges of an arbitrary retracting-free bidirectional double tracing of a subgraph G 1 , then, move along the connecting path into point v 2 , traverse all edges of an arbitrary retracting-free bidirectional double tracing of a subgraph G 2 and through the path return in an opposite direction into our original point vertex v 1 . This proves Case 2 and completes the overall proof. Theorem 9. Let G 1 , G 2 ∈ Φ be two graphs with vertices v 1 and v 2 , respectively. Also, let G(v 1 , v 2 ) be graph obtained by identifying of vertices v 1 and v 2 of graphs G 1 and G 2 , respectively. Then, Proof. It is similar to the proof of Theorem 8, and we only sketch it. What is new is that there are now three cases: G 1 ∈ Φ, G 2 ∈ Ω; G 1 , G 2 ∈ Ω (both as in Theorem 8); and, additionally, G 1 , G 2 ∈ Θ. To prove the first two cases, it is sufficient to repeat the respective steps of the proof to Theorem 8; however, since there is no connected path between subgraphs G 1 and G 2 , the demonstration is shorter. The third case can be proven by traversing first an arbitrary Eulerian circuit of a subgraph G 1 , then, a similar circuit of G 2 ; while traversing starts and terminates at vertex v obtained by identification of vertices v 1 and v 2 of graphs G 1 and G 2 , respectively. Then, we repeat the procedure, while traversing the same Eulerian cycle in G 1 in an opposite direction and, similarly, the same Eulerian cycle in G 2 in an opposite direction, to return to vertex v. This completes a retracting-free bidirectional double tracing. Whence the proof follows.
Proof. It is due to repetitive application of Theorems 9 and 7. First, construct from all graphs G 1 , G 2 , . . . , G s an intermediate connected graph G in which any two subgraphs G j and G k (j, k ∈ {1, 2, . . . , s}) share either 0 or 1 common vertex. By virtue of Theorem 9, G ∈ Ω. Further, by continuing identification (merging) of pertinent vertices of G , eventually produce a graph G. By Theorem 7, the graph G also belongs to Ω, which is the proof.
The next theorem affords a partial solution to the problem posed by Ore.
Theorem 11. Let G be a noncycle Eulerian graph |E(G)| − |V (G)| ≥ 1 . Then, G ∈ Ω. That is, every Eulerian graph which is not a cycle admits a retracting-free bidirectional double tracing.
Proof. Recall that each noncycle Eulerian graph G is a vertex union (without identifying edges) of proper cycles. It is sufficient to note that the cycles are themselves Eulerian graphs. Hence, by virtue of Theorem 10, we conclude that G ∈ Ω, which proves the statement.
The line graph L(H) of a simple graph H is the graph whose vertex set V (L) is in one-toone correspondence with the set of edges E(H) of the graph H, with two vertices of L(H) being adjacent if and only if the corresponding edges are incident in H.
As a last statement, we consider the following theorem. Proof. Since every vertex in L(H) has an even degree, L(H) is an Eulerian graph. Hence, by Theorem 11, L(H) ∈ Ω, which is the proof.