OON POLYHEDRAL GRAPHS AND THEIRCOMPLEMENTS
Riccardo W. Maffucci ∗ Abstract
We find all polyhedral graphs such that their complements are still polyhedral. Theseturn out to be all self-complementary.
Keywords:
Planar graphs, -connectivity, polyhedra, complements, dual graph, classifica-tion. MSC(2010):
The problem we investigate combines two main ideas. Polyhedral graphs, or simplypolyhedra, are -connected, planar graphs. This class of graphs is closely related to -dimensional topology and geometry, and the name comes from the fact that they are -skeletons of polyhedral solids in the sense of geometry (Rademacher-Steinitz’s Theorem, seee.g. [7, Theorem 11.6]). In what follows, we assume polyhedral solids to be convex, andconsider them up to topology, i.e., up to their -skeletons being isomorphic graphs. It maybe shown using only graph theory that there are only five regular polyhedral solids, namelythe Platonic ones [8, Theorem 1.38].Polyhedral graphs have several nice properties. They are the planar graphs that can beembedded in a sphere in a unique way (an observation due to Whitney, see e.g. [7, Theorem11.5]). Specifically, the dual graph is always unique, and duals of polyhedra (in the senseof both graph theory and geometry) are also polyhedra (e.g., [7, Chapter 11]). We alsorecord that all their regions (or ‘faces’) are delimited by cycles (elementary closed walks) [5,Proposition 4.26]. ∗ [email protected] . a r X i v : . [ m a t h . C O ] F e b he other idea comes from a classical problem in graph theory, set by Harary: to findall graphs G such that a certain property is verified by both G and its complement graph G [4, Introduction]. Our problem is the following. Question 1.
Which pairs of complementary graphs
G, G are both polyhedral?
Theorem 2.
There exist exactly three polyhedral graphs such that their complements arepolyhedral. These are all self-complementary. They are depicted in Figure 1. (a) g . (b) g . (c) g . Figure 1: The only three solutions to Question 1.All solutions to Question 1 are (8 , graphs, of degree sequence , , , , , , , . This case promises to be the most interesting, due to the following.
Remark 3.
If the polyhedron G , its dual, and its complement graphs are all of same orderand size, then G is an (8 , graph. To see this, we impose the following three conditions.If G and G (cid:48) have the same order, then the number of regions and vertices of G coincide. if G, G have the same size, then q = (cid:0) p (cid:1) . The third condition is Euler’s formula for planargraphs. Solving the resulting system, we get p = 8 and q = 14 . Related problems.
Planar graphs with planar complement were investigated in [2], [4].In [9, Figure 3.1], we find the only three non-trivial, self-complementary, self-dual graphs.In this figure, graph A is g . of Figure 1, C is g . , while B is not -connected. Bythe way, g . in Figure 1 is the only self-complementary, non-self-dual polyhedron, as weshall see in section 3. In other related work, Ando-Kaneko [1] investigated the connectivityof complements of -connected graphs. We say that G is a ( p, q ) graph when G is a graph of order (number of vertices/points) p and size (number of edges/lines) q .The vertex and edge sets of G are V ( G ) and E ( G ) respectively. Vertices are denoted by v , v , . . . , v p . Their respective degrees are non-negative2ntegers d , d , . . . , d p . The degree sequence of a graph is indicated by d , d , . . . , d p , or simplyby d d . . . d p if there is no possibility of confusion.A graph of order at least is -connected if removing any set of , , or verticesproduces a connected graph. The complement of G is denoted by G , and the dual by G (cid:48) .The letters p, q, p (cid:48) , q (cid:48) indicate their orders and sizes accordingly. The number of regions of aplanar graph is r , and of its complement and dual (if these are also planar) r, r (cid:48) . The facesof a polyhedral graph are its regions. The faces are triangular, quadrilateral, pentagonal, ...if they are bounded by a cycle of length , , , ... respectively. The author was supported by Swiss National Science Foundation project 200021_184927.
It is well-known that if G is a planar graph on at least vertices, then G is non-planar[2]. It follows right away that Question 1 has a finite number of solutions.On the other hand, we will now show that if G, G are polyhedra, then p ≥ . Denote by d , d , . . . , d p the (weakly decreasing) degree sequence of G . Since the graph is -connected, in particularone has d p ≥ . (2.1)Accordingly, the degree sequence of G is p − − d p , p − − d p − , . . . , p − − d , with p − − d ≥ , i.e. d ≤ p − . (2.2)By (2.1) and (2.2), p ≥ . If p = 7 , then G is -regular, impossible due to the handshakinglemma. Therefore, p ≥ , so that ultimately p = 8 .Further, the above yields d = 4 and d = 3 . The handshaking lemma now reduces ourcases to those of Table 1, assuming w.l.o.g. that q ≤ q . It will thus suffice to inspect thecases q = 12 , , . This will be done in the next section.3eg. sequence of G size q faces r deg. sequence of G size q faces r Table 1: Possible degree sequences for solutions of Question 1.
The polyhedra up to faces are tabulated in [3] and [6]. In Appendix A, we collect thoseup to edges, for quick reference.There are only two (8 , polyhedra (see Figure 9). For both, the complement is non-planar – see Figure 2.Figure 2: The two (8 , polyhedral graphs, and their complements.There are eleven (8 , polyhedra, and nine of them have sequence (Figure10). In Figure 3, these are sketched together with their complement graphs. All of thecomplements are non-planar, hence we discard this case.The forty-two (8 , polyhedra may be found in Figures 12 and 13. Exactly seventeenof them have degree sequence . These are collected in Figures 4 (self-duals) and 5(non-self-duals). We find three solutions to Question 1, namely graphs g . , g . , and g . .As these lists of polyhedra are all exhaustive [3, 6], these three are the only solutionsto Question 1. The proof of Theorem 2 is now complete. We note that g . is the onlyself-complementary, non-self-dual polyhedron. As a remark of a different flavour, g . and its dual may be embedded in -dimensional space so that all edges have unit length,as in Figure 6. 4 a) g . and its complement (b) g . and its complement(c) g . and its complement (d) g . and its complement(e) g . and its complement (f) g . and its complement(g) g . and its complement (h) g . and its complement(i) g . and its complement Figure 3: The nine polyhedra of degree sequence , and their complement graphs.5 a) g . and its complement (b) g . and its complement(c) g . and its complement (d) g . and its complement(e) g . and its complement Figure 4: The self-dual polyhedra of degree sequence , and their complementgraphs. 6 a) g . and its complement (b) g . and its complement(c) g . and its complement (d) g . and its complement(e) g . and its complement (f) g . and its complement(g) g . and its complement (h) g . and its complement(i) g . and its complement (j) g . and its complement(k) g . and its complement (l) g . and its complement Figure 5: The non-self-dual polyhedra of degree sequence , and their complementgraphs. 7igure 6: Embeddings of g . and its dual in -dimensional space, with all edges of unitlength. 8 Tables of polyhedra
We choose the following ordering. Firstly, the tables are according to increasing size,rather than order. This is due to two main, related reasons. Understanding ( p, q ) polyhedraof r = 2 + q − p faces (Euler’s formula), p > r , is no harder than studying the ( r, q ) of p faces, and then passing to the duals. In this sense, the complexity grows with q rather than p . Moreover, in this way each table lists dual pairs of polyhedra together (as they have thesame size, but not necessarily the same order).Our next criteria for ordering, size being equal, is by increasing number of vertices. Wehave the inequalities q + 63 ≤ p ≤ q due to planarity ( q ≤ p − ), and -connectivity (implying δ ( G ) ≥ , hence q ≥ p/ viathe handshaking lemma).Self-duals are listed before dual pairs. All the above being equal, we sort by decreasinghighest degrees of vertices, and then by decreasing highest degrees of vertices of the dual.If all the said criteria are equal, we list arbitrarily. A.1 Size q ≤ (a) g (b) g (c) g . and itsdual (d) g . (e) g . Figure 7: The polyhedra with q ≤ .9 a) g . and its dual (b) g . and its dual Figure 8: The polyhedra with q = 11 . (a) g . and its dual (b) g . and its dual (c) g . (d) g . (e) g . (f) g . (g) g . (h) g . (i) g . and its dual Figure 9: The polyhedra with q = 12 .10 .2 Size q = 13 (a) g . and its dual (b) g . and its dual (c) g . and its dual (d) g . and its dual(e) g . and its dual (f) g . and its dual (g) g . and its dual (h) g . and its dual(i) g . and its dual (j) g . and its dual (k) g . and its dual Figure 10: The polyhedra with q = 13 .11 .3 Size q = 14 (a) g . and its dual (b) g . and its dual (c) g . and its dual (d) g . and its dual(e) g . and its dual (f) g . and its dual (g) g . and its dual (h) g . and its dual Figure 11: The polyhedra with q = 14 and p = 7 , .12 a) g . (b) g . (c) g . (d) g . (e) g . (f) g . (g) g . (h) g . (i) g . (j) g . (k) g . (l) g . (m) g . (n) g . (o) g . (p) g . Figure 12: The self-dual polyhedra with q = 14 and p = 8 .13 a) g . and itsdual (b) g . and itsdual (c) g . and itsdual (d) g . and itsdual (e) g . and itsdual(f) g . and itsdual (g) g . and itsdual (h) g . and itsdual (i) g . and itsdual (j) g . and itsdual(k) g . and itsdual (l) g . and itsdual (m) g . and itsdual Figure 13: The non-self-dual polyhedra with q = 14 and p = 8 .14 eferences [1] Kiyoshi Ando and Atsusi Kaneko. A remark on the connectivity of the complement of a3-connected graph. Discrete Mathematics , 151(1-3):39–47, 1996.[2] Joseph Battle, Frank Harary, Yukihiro Kodama, et al. Every planar graph with ninepoints has a nonplanar complement.
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