On the Connectivity of Graphs Embedded in Surfaces

On the connectivity of graphs embedded in surfaces II Michael D. Plummer Department of Mathematics Vanderbilt University Nashville, TN 37240, USA [email protected] Xiaoya Zha* Department of Mathematical Sciences Middle Tennessee State University Murfreesboro, TN 37132, USA [email protected] Submitted: February 6, 2002; Accepted: September 12, 2002 MR Subject Classification: 05C10, 05C40 Abstract Let κmax (Σ) denote the maximum value for the connectivity of any graph which embeds in the topological surface Σ. The connectivity interval for Σ is the set of integers in the interval [1, κmax (Σ)]. Given an integer i in [1, κmax (Σ)] it is a trivial problem to demonstrate that there is a graph Gi with connectivity i which also embeds in Σ. We will say that one can saturate the connectivity interval in this case. Note that no restrictions have been placed on the embeddings in the above problem, however. What if we demand that the embeddings in question be 2-cell or even that they be genus embeddings? The problem of saturating the connectivity interval for 2-cell embeddings will be solved completely in the present work. In connection with the apparently much harder saturation question for genus embeddings, it will be shown that one can always saturate the subinterval [1, ⌊0.7κmax (Σ)⌋]. * work supported by NSF Grant DMS-9622780 and Middle Tennessee State University Faculty Research Grant 1999 the electronic journal of combinatorics 9 (2002), #R38 1 1. Introduction In his 1973 paper, Cook [C1] studied the relation between the connectivity of a graph and the surfaces into which it can be embedded. He proved that the following result holds for all surfaces except the plane:   √ 5 + 49 − 24χ κ(G) ≤ , 2 where κ(G) denotes the vertex connectivity of graph G and χ is the Euler characteristic of any surface in which G embeds. Moreover, Cook showed that these bounds are attained by complete graphs in all cases except the Klein bottle. In [PZ1] we explored in more detail relations between the connectivity of embedded graphs and the surfaces in which they are embedded. Let κmax (Σ) denote the maximum connectivity among all graphs which embed in Σ and let κgen (Σ) be the maximum connectivity among all graphs which genus embed in Σ. It was shown that, somewhat surprisingly, κgen is not a monotone non-decreasing function of the genus. Also for some surfaces, κgen is strictly less than κmax . In fact it was proved that such so-called Class B surfaces are not only infinite in number, but that they must occur infinitely often periodically as the genus parameter increases to infinity. It was also shown that in the case when the complete graph which attains Cook’s maximum connectivity bound actually genus embeds in the surface, that with a finite number of exceptions, it is the unique graph attaining this bound. Let us begin the present work by considering a very easy problem. If Gmax is a graph which embeds in a surface Σ and κ(Gmax ) = κmax for that surface, then by removing edges all incident with a common vertex one by one, it is trivial to construct a sequence of graphs Gr = Gmax , Gr−1 , . . . , G2 , G1 such that κ(Gi ) = i and each Gi embeds in the surface. In this case, we say that we can saturate the connectivity interval [1, κmax ]. But note that nothing was said about the nature of the embeddings along the way. In particular, it was not demanded that they be 2-cell or that they be genus embeddings. In the present paper, the focus will be on two questions: Question 1: Given a surface Σ and an integer i, does there exist a graph G with κ(G) = i such that G 2-cell embeds in Σ (i.e., each face is homeomorphic to an open disk)? Question 2: Let Σ and i be as in Question 1. Does there exist a graph G with κ(G) = i such that G genus embeds in Σ? With the aid of theorems by Duke [D1] and Stahl [S1], we will answer Question 1 completely. The solution is not difficult. Regarding the second question, it will be shown that for i ∈ [1, ⌊0.7κmax⌋], Question 2 has a positive answer. In fact the second question appears to be quite difficult. For one thing, all embeddings sought must be genus embeddings and the genus problem for graphs is known to be NP-complete. (See [T1].) Thus one needs to construct special genus embeddings with the required connectivity the genera of which are easy to determine. Also for the electronic journal of combinatorics 9 (2002), #R38 2 certain values of i in Question 2, an affirmative answer to the question turns out to imply an answer to an unsolved problem of long standing. Let O(m) denote the generalized octahedron graph which is obtained from the complete graph Km by deleting a maximum matching. When m is even and m/2 ≡ 0, 1( mod 3), the orientable genus of O(m) is known. (See [AG1] and [JR1].) Determination of the orientable genus of (O(m)) remains open for the remaining values of m. Let i in Question 2 be the connectivity of O(m) ( = m − 2). Then by Theorems 5.1 and 5.2 of [PZ1], this special case of Question 2 actually implies the answer to the problem of determining the orientable genus of O(m) for certain of the remaining values of m. This will be explained in more detail in the final section of the present paper. Most of the proofs to follow involve constructions of graphs and embeddings such that (1) the embeddings are genus embeddings and (2) the graphs have the correct connectivity. 2. Saturation of the Connectivity Interval for 2-cell Embeddings In this section, we proceed to answer Question 1 of the Introduction. As usual, γ(G) (respectively, γ(G)) will denote the orientable (respectively, non-orientable) genus of graph G. An embedding of a graph G in a surface S is said to be a 2-cell embedding if every face is homeomorphic to an open disk. The maximum orientable (respectively maximum non-orientable) genus γM (G) (respectively γ M (G)) of graph G is the largest genus of any orientable surface S (respectively, non-orientable surface N ) in which G has a 2-cell embedding. The following results due to Duke [D1] in the orientable case and to Stahl [S1] in the non-orientable case guarantee that there is a 2-cell embedding of G in all surfaces with genera values between and including those of the minimum and maximum surfaces. Theorem 2.1. If γ(G) ≤ γ ≤ γM (G) ( or if γ(G) ≤ γ ≤ γ M (G)), then there is a 2-cell embedding of G in the orientable surface of genus γ (respectively, in the non-orientable surface of genus γ). Let G be a connected graph and let T be an arbitrary spanning tree of G. The deficiency ξ(G, T ) of spanning tree T is the number of components of G − E(T ) which have an odd number of edges. The deficiency ξ(G) of graph G is the minimum of ξ(G, T ) over all spanning trees T . Finally, let β(G), called the Betti number or cyclomatic number of a connected graph G be defined by β(G) = |E(G)|−|V (G)|+1. Xuong [X1] and Edmonds [E2] have characterized the maximum orientable and non-orientable genus of a graph G in terms of its Betti number and its deficiency. Theorem 2.2. If γM (G)(γ M (G)), ξ(G) and β(G) represent the maximum orientable (non-orientable) genus of a connected graph G, the deficiency of G and the Betti number of G respectively, then γM (G) = (1/2)(β(G) − ξ(G)) and γ M (G) = β(G). the electronic journal of combinatorics 9 (2002), #R38 3 We make use of Theorem 2.2 in the next result which states that one can saturate the interval of connectivity [1, κmax ] with 2-cell embeddings for any surface orientable or non-orientable. Thus Question 1 in Section 1 has a complete affirmative answer. Theorem 2.3. If Σ is any surface, orientable or non-orientable, then for any integer i in the interval [1, κmax (Σ)], there is a graph G with connectivity i which 2-cell embeds in Σ. Proof: First consider the orientable case. Let Sg be any orientable surface, let i be any integer in the interval [1, κmax (Sg )], and let Km be the largest complete graph which embeds in Sg . Note that we do not assume that this embedding is necessarily 2-cell. If g = 0 then the surface is the sphere, κmax (sphere) = 5, and the theorem is trivial. If g = 1 then the surface is the torus, and κmax (torus) = 6. Negami [N1] has shown that there are 6-connected triangulations of the torus (which therefore are 2-cell embeddings). We shall refer to these graphs as “Negami graphs”. It is an easy matter to modify one of his graphs to obtain graphs with κ = 1, . . . , 5 which also 2-cell embed on the torus. So henceforth we will assume that g ≥ 2 and m ≥ 8. Suppose i is an integer in the connectivity subinterval [1, m − 1]. Let Gi be a graph obtained from Km by deleting m−i−1 edges all incident with a common vertex v. Then κ(Gi ) = m−(m−i−1)−1 = i. Clearly γ(Gi ) ≤ γ(Km ) ≤ g < γ(Km+1 ). We now show that γM (Gi ) ≥ γ(Km+1 ). By Theorem 2.2, γM (Gi ) = (1/2)(β(Gi ) − ξ(Gi )). Moreover, β(Gi ) = 1 − m + [m(m − 1)/2 − (m − i − 1)] = m(m − 1)/2 − 2m + i + 2. It is easy to construct a spanning tree T of Gi such that Gi − E(T ) consists of only one component. (Such a tree is called a splitting tree; see White [W1].) Therefore ξ(Gi ) = 0 if G−E(T ) has an even number of edges and ξ(Gi ) = 1 if G−E(T ) has an odd number of edges. Thus γM (Gi ) ≥ (1/2)(β(Gi ) − 1) = (1/2)(m(m − 1)/2 − 2m + i + 2) = m(m − 1)/4 − m + (1/2)i + 1 > m(m − 1)/4 − m + 1. So   (m − 2)(m − 3) m(m − 1) γM (Gi ) − γ(Km+1 ) > −m+1− 4 12 ≥ (m − 2)(m − 3) m(m − 1) −m+1− −1 4 12 m2 − 5m − 3 (m − 6)(m + 1) = > > 0, 6 6 since m ≥ 8. Thus γM (Gi ) ≥ γ(Km+1 ) ≥ g and hence γ(Gi ) ≤ γ(Km ) < γ(Km+1 ) ≤ γM (Gi ). Thus by Theorem 2.1, Gi has a 2-cell embedding for every genus in the interval [γ(Gi ), γM (Gi )] which includes the interval [γ(Km ), γ(Km+1)] which in turn includes the genus g. This completes the proof in the orientable case. The proof for the non-orientable case parallels that for the orientable case. the electronic journal of combinatorics 9 (2002), #R38 4 3. Constructions IA, IB and IC — the Orientable Case Starting in this Section, we will address the more difficult Question 2. We will construct some simple graphs together with genus embeddings thereof and for each of these constructions, we will derive the genus and the connectivity. Let γ(s, k) denote the orientable genus of Km where m = 12s + k, s ≥ 1 and 0 ≤ k ≤ 11. For each value of k = 0, . . . , 11 and for each s ≥ 1 we define the function lk (s) by lk (s) = 2[γ(s, k +1)−γ(s, k)] for k 6= 2, 5, and lk (s) = 2[γ(s, k + 1) − γ(s, k)] + 2 for k = 2, 5, Then by the RingelYoungs Theorem [R1], γ(s, k) = 12s2 + 2sk − 7s + ⌈(k 2 − 7k + 12)/12⌉ and it follows that  4s, if k = 0, 1, 2, 3, 6, lk (s) = 4s + 2, if k = 4, 5, 7, 8, 9, 10, 11 It is known [R1] that when k = 2, 5, the graph Km − K2 can be genus embedded into the surface Sγ(s,k)−1 . Then for each of the twelve possible values of k, i.e., k = 0, 1, . . . , 11, let Ψ denote a genus embedding of Km into surface Sγ(s,k) when k 6= 2, 5 and a genus embedding of Km − K2 into surface Sγ(s,k)−1 , when k = 2 or 5. Then let Ψ′ be a second copy of Km (respectively, Km − K2 when k = 2, 5) into a second copy of the surface Sγ(s,k) (respectively, Sγ(s,k)−1 when k = 2, 5) in identically the same way. Now choose a value for l, 1 ≤ l ≤ lk (s). We proceed to choose l faces of Ψ and l faces of Ψ′ in a certain way and then to connect the two surfaces via l “cylinders” or “tubes” joining the interiors of the faces chosen in Ψ to the interiors of the faces chosen in Ψ′ . If k = 2, 5, let v0 be any vertex of Km −K2 which is not an endvertex of the missing edge. (If k = 2 or 5, it is known that Ψ is a triangulation [R1] [RY1] [J1].) In this case let v1 , . . . , vm−1 denote the other vertices of Ψ in clockwise order about the vertex v0 . Then for each i = 1, . . . , m − 2, let fi be the face incident with v0 which contains edges v0 vi and v0 vi+1 in its boundary. Fix a value for λ, 1 ≤ λ ≤ l ≤ lk (s). We select λ of the faces fi which are consecutive about v0 and then every second face until we have chosen a total of l faces in Ψ. In particular, we define our “chosen” faces Fi by:  1 ≤ i ≤ λ, fi , Fi = f2i−λ , λ + 1 ≤ i ≤ l. Now we turn our attention to Ψ′ . In Ψ′ , denote the vertex corresponding to vi in Ψ by vi′ for i = 0, . . . , m − 1. In Ψ′ , starting from the face F1′ , select every other face incident with v0′ and call these faces F1′ , . . . , Fl′ (in any order). Name the two vertices on the boundary of Fi′ that are adjacent to v0′ by a′i and b′i , respectively. For the cases when k = 1, 6, 9, 10, we shall proceed slightly differently. We know that the embedding is not a triangulation, so select a non-triangular face F of Ψ as F1 , and the corresponding face in Ψ′ as F1′ . Since face F1 is not a triangle, we can choose four vertices v0 , v1 , v, v2 on the boundary of F1 such that edges v0 v1 and v0 v2 are adjacent boundary edges of F1 . We may name v1 and v2 such that as one traverses the boundary of F1 , one encounters the vertices v1 , v and v2 in clockwise order. In these the electronic journal of combinatorics 9 (2002), #R38 5 cases when choosing the faces Fi′ in Ψ′ we should not choose any face with edge v0′ v ′ on its boundary. It is important to have this non-triangular face so that we may add an extra edge on the first tube in our construction. This additional edge will suffice to attain the bound in Lemma 3.1 below. For the remaining cases when k = 0, 3, 4, 7, 8 and 11, choose vertices v0 , v1 , . . . , vm−1 as we did for the cases k = 2 and 5. For each i, 1 ≤ i ≤ l, insert a tube Ti joining the interiors of faces Fi and Fi′ . We now add edges to form a new graph on 2m vertices as follows. For 1 ≤ i ≤ λ, add edges v0 a′i , v0 b′i , vi v0′ , vi a′i , vi b′i , vi+1 a′i all on tube Ti . Also when k = 1, 6, 9, 10 add the “extra” edge v0 v ′ on tube T1 . When λ + 1 ≤ i ≤ l, add edges v0 a′i , v0 b′i , v2i−λ v0′ , v2i−λ a′i , v2i−λ b′i and v2i−λ+1 a′i all on tube Ti . (See Figure 3.1.) We will call the embedded graph on 2m vertices constructed above HIA (s, k, l, λ). It is important to note that all edges added across the l tubes in the above construction are distinct; i.e., graph HIA (s, k, l, λ) has no multiple edges. This is because all faces chosen in Ψ′ only have v0′ on their common boundary, and hence all a′i ’s and b′i ’s are distinct. The parameter λ which is the number of consecutive faces chosen in Ψ determines the connectivity of the resulting graph. v i 11 00 11111 00000 00 11 11111 00000 11111 00000 00000 11111 11111 00000 11111 00000 11111 00000 11111 00000 00 11 00000 11111 00 11 1 0 0 1 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00 11 00000 11111 00 11 v’ 0 a’ b’ 11 i 00 00 11 1 0 00 11 0 1 1 0 1 0 1 0 1 0 0 1 1 0 1 0 0 001 11 0 1 00 11 0 001 11 v i+1 i v’1 0 0 1 v 0 Figure 3.1(a) the electronic journal of combinatorics 9 (2002), #R38 6 v 2i- λ1 0 v 111111 000000 0 1 000000 111111 111111 000000 000000 111111 111111 000000 000000 111111 111111 000000 000000 111111 1 0 111111 000000 1 0 0 1 000000 111111 0 1 111111 000000 000000 111111 111111 000000 000000 111111 111111 000000 000000 111111 111111 000000 11 00 000000 111111 11 00 v’ 0 a’ b’ 0 1 0 1 0 1 2i- λ+1 i i 11 00 00 11 v 0 Figure 3.1(b) We proceed to determine the genus and the connectivity of the graph HIA (s, k, l, λ). We will make use of the following corollary of Euler’s formula (see [W1, pg. 62 and pg. 179]). Lemma 3.1. For all simple graphs G with p vertices and q edges, γ(G) ≥ q p − + 1, 6 2 γ(G) ≥ q − p + 2. 3 and Furthermore, equality holds in the above inequalities if and only if there is a triangular embedding of G in its surface of minimum genus. Now we compute the genus and connectivity of HIA (s, k, l, λ). Theorem 3.2. (i) The embedding of HIA (s, k, l, λ) constructed above is a genus embedding, and γ(HIA (s, k, l, λ)) =  2γ(s, k) + l − 1, 2γ(s, k) + l − 3, for k 6= 2, 5, for k = 2, 5. (ii) For all s ≥ 1, κ(HIA (s, k, l, λ)) = 2l − λ + 2. the electronic journal of combinatorics 9 (2002), #R38 7 Proof: First suppose that k 6= 2 or 5. Denote HIA (s, k, l, λ) by H. Then pH = 2m = 2(12s + k) and qH = (12s + k)(12s + k − 1) + 6l + ǫk , where ǫk =  0 if k = 6 1, 6, 9, 10, 1 if k = 1, 6, 9, 10. Then by Lemma 3.1,   q p γ(H) ≥ − +1 6 2   1 = ((12s + k)(12s + k − 1) + 6l + ǫk ) − (12s + k) + 1 6  2  k − 7k + 6 + ǫk 2 = 24s + 4sk − 14s + l + . 6 Let us denote the right side of the final equation above by A. On the other hand, it is known (cf. [R1]) that γ(K12s+k ) = 12s2 + 2sk − 7s + 2 ⌈(k − 7k + 12)/12⌉. Denote the right side of this equality by B. Then in each of the ten cases when k 6= 2, 5, it is easy to check that A = 2B + l − 1. But we know that the constructed surface is obtained by joining l tubes between two surfaces each having genus B. Therefore the genus of the constructed surface is 2B + l − 1. Now suppose k = 2 or 5. Then pH = 2(12s + k) as before, but now qH = 2((12s + k)(12s + k − 1)/2 − 1) + 6l, so   q p γ(H) ≥ − +1 6 2   2 k − 7k + 4 2 = 24s + 4sk − 14s + l + . 6 If we denote the right side of the last equality by A′ then again it is easy to show that A′ = 2B + l − 3 when k = 2 and k = 5. In these two cases the constructed surfaces are obtained by joining l tubes between two surfaces each having genus B − 1, and therefore the genus of the constructed surface is 2B + l − 3. This proves (i). Now we consider the connectivity of graph HIA (s, k, l, λ). Since 1 ≤ λ ≤ l ≤ lk (s), it is easy to check that, for all m = 12s + k and s ≥ 1.  m − 3, for k 6= 2, 5 and 2l − λ + 2 ≤ m − 4, for k = 2, 5. It now follows that given two vertices u and w both in Ψ, or both in Ψ′ , since when k 6= 2, 5, Km is (m − 1)-connected and when n = 2, 5, Km − K2 is (m − 2)-connected, there must be more than 2l − λ + 2 vertex-disjoint u − w paths in HIA (s, k, l, λ)). So we need only treat the remaining case, namely, when u is a vertex in Ψ and w is a vertex in Ψ′ . the electronic journal of combinatorics 9 (2002), #R38 8 Since M = {vi b′i |1 ≤ i ≤ λ} ∪ {v2i−λ b′i |λ + 1 ≤ i ≤ l} ∪ {v2i−λ+1 a′i |λ ≤ i ≤ l} ∪ {v0 a′1 }. 11 00 00 11 00 11 v 0 v 1 a’ 11 00 00 111 00 11 000000000000000000 111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 v 00 2 11 0000000000000000000 1111111111111111111 00 11 0000000000000000000 1111111111111111111 F 0000000000000000000 1111111111111111111 1 0000000000000000000 1111111111111111111 0 v 1 0000000000000000000 1111111111111111111 00000000000000000001010 3 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 00 11 0000000000000000000 1111111111111111111 00 11 00 v 11 0000000000000000000 1111111111111111111 0000000000000000000 4 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 00 11 v 00 11 0000000000000000000 1111111111111111111 λ 00 11 0000000000000000000 1111111111111111111 11 00 11 00 0000000000000000000 1111111111111111111 F 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 λ11 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 00 0000000000000000000 1111111111111111111 00 11 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 v 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 λ+1 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 00 11 00 11 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 v 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 λ+2 1111111111111111111 0000000000000000000 0000000000000000000 1111111111111111111 00 11 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 00 11 0000000000000000000 v 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 λ+3 0000000000000000000 1111111111111111111 00 11 0000000000000000000 1111111111111111111 00 11 0000000000000000000 1111111111111111111 v 0000000000000000000 1111111111111111111 2i- λ 00 11 0000000000000000000 1111111111111111111 00 11 F 00 11 0000000000000000000 1111111111111111111 l v 0000000000000000000 1111111111111111111 2i- λ+1 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 00 11 00 11 0000000000000000000 1111111111111111111 v 0000000000000000000 1111111111111111111 2l- λ 0000000000000000000 1111111111111111111 00 11 00 11 0000000000000000000 1111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 b’ 111111111111111111 000000000000000000 F’ 1 111111111111111111 000000000000000000 111111111111111111 1 000000000000000000 111111111111111111 00 000000000000000000 111111111111111111 00 11 a’ 11 111111111111111111 000000000000000000 000000000000000000 2 111111111111111111 111111111111111111 000000000000000000 000000000000000000 111111111111111111 111111111111111111 000000000000000000 000000000000000000 111111111111111111 00 11 000000000000000000 00 11 000000000000000000 111111111111111111 b’ 111111111111111111 000000000000000000 000000000000000000 111111111111111111 2 111111111111111111 111111111111111111 000000000000000000 000000000000000000 111111111111111111 111111111111111111 000000000000000000 000000000000000000 111111111111111111 00 11 111111111111111111 000000000000000000 00 a’11 000000000000000000 111111111111111111 000000000000000000 3 111111111111111111 000000000000000000 111111111111111111 111111111111111111 000000000000000000 000000000000000000 111111111111111111 000000000000000000 111111111111111111 00 11 000000000000000000 111111111111111111 00 11 000000000000000000 0 1 000000000000000000 111111111111111111 b’ 111111111111111111 000000000000000000 111111111111111111 0000000000000000001010 3 111111111111111111 1111111111111111111 0000000000000000000 1111111111111111111 000000000000000000 0000000000000000000 111111111111111111 1111111111111111111 000000000000000000 0000000000000000000 111111111111111111 000000000000000000 0000000000000000000 111111111111111111 a’1111111111111111111 1111111111111111111 000000000000000000 0000000000000000000 111111111111111111 λ 111111111111111111 000000000000000000 1111111111111111111 0000000000000000000 00 11 00 11 F’ 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 b’1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 λ 1111111111111111111 000000000000000000 0000000000000000000 λ 111111111111111111 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 00 11 00 11 000000000000000000 111111111111111111 1111111111111111111 0000000000000000000 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 a’ 0000000000000000000 1111111111111111111 000000000000000000 111111111111111111 00 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 i 11 00 11 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 1111111111111111111 000000000000000000 111111111111111111 0000000000000000000 00 11 00 11 000000000000000000 111111111111111111 0000000000000000000 1111111111111111111 00 b’i11 F’ 000000000000000000 111111111111111111 l 000000000000000000 111111111111111111 000000000000000000 111111111111111111 a’ 11 00 000000000000000000 111111111111111111 00 l11 000000000000000000 111111111111111111 000000000000000000 111111111111111111 00 11 000000000000000000 111111111111111111 00 11 00 11 000000000000000000 b’ 111111111111111111 1 0 0 1 0 1 00 11 v 2l- λ +1 l Figure 3.2 (See Figure 3.2.) is a matching of size 2l − λ + 2 from vertices in Ψ to vertices in Ψ′ , we have 2l − λ + 2 vertex-disjoint paths, using these matching edges to join u ∈ Ψ and v ∈ Ψ′ . Therefore κ(HIA (s, k, l, λ)) ≥ 2l − λ + 2. Since the set of all vertices of matching M lying in Ψ is a vertex cut, we have κ(HIA (s, k, l, λ)) ≤ 2l − λ + 2. Thus κ(HIA (s, k, l, λ)) = 2l − λ + 2, and the proof of the Theorem is complete. the electronic journal of combinatorics 9 (2002), #R38 9 Note that κ(HIA (s, k, l, 1)) = 2l + 1 and for all λ, 2 ≤ λ ≤ l, κ(HIA (s, k, l, λ)) ≤ 2l + 1. For saturation purposes, the value of κ(HIA (s, k, l, 1)) is not quite high enough, so we extend our Construction IA to include, for all values of k, a graph HIB (s, k, l) having κ = 2l + 2 and for k = 1, 6, 8, 9, 10 and 11, a graph HIC (s, k, l) having κ = 2l + 3. To build graphs HIB (s, k, l) and HIC (s, k, l), we modify the Construction IA in the case λ = 1 as follows. Let Ψ be constructed just as before and let the faces F1 = f1 , F2 = f3 , . . . , Fl−1 = f2l−3 which meet only at vertex v0 be as before. We proceed to modify the choice of face Fl as follows. For graph HIB (s, k, l), let Fl be the face formed by taking the two consecutive faces f2l−1 and f2l and deleting their common boundary edge v0 v2l . (It is an easy calculation to show that since s ≥ 1, there are enough faces at v0 to make the above selection in such a way that faces f2l and f1 have no common boundary edge.) We proceed to choose faces F1′ , . . . , Fl′ in exactly the same way relative to embedding ′ Ψ as F1 , . . . , Fl were chosen relative to embedding Ψ. Thus faces Fl and Fl′ will have at least four distinct vertices in their facial walks. Now, as before, we insert tubes Ti joining the interiors of Fi and Fi′ , for i = 1, . . . , l. For each tube Ti , i = 1, . . . , l−1 we add edges across Ti exactly as we did in Construction IA. Recall that face Fl is at least a quadrilateral and the four vertices v0 , v2l−1 , v2l and v2l+1 appear in clockwise order about the boundary of Fl . Similarly in the embedding ′ ′ ′ lie in counterclockwise order along the face boundary , v2l+1 , v2l Ψ′ , the vertices v0′ , v2l−1 of Fl′ . ′ ′ ′ , v0 v2l+1 , v2l−1 v2l+1 , Across tube Tl we insert the eight edges v0 v2l ′ ′ ′ ′ ′ v2l−1 v0 , v2l v0 , v2l v2l−1 , v2l+1 v2l−1 , v2l+1 v2l . (See Figure 3.3.) v 2l 11 00 00 11 1 0 0 1 v 2l-1 0 1 1 0 1 0 1 0 0 1 v’ 0 v’ 2l-1 v’ v’ 1 0 2l 1 0 2l+1 1 0 0 1 1 0 1 0 v 2l+1 1 0 1 0 1 0 1 0 1 0 1 0 v 0 Figure 3.3 To construct graph HIC (s, k, l), for k = 1, 6, 8, 9, 10, 11, proceed just as in the the electronic journal of combinatorics 9 (2002), #R38 10 construction of graph HIB (s, k, l) up through the selection of faces F1 through Fl−1 . But this time let Fl be the face formed by considering the three consecutive faces f2l−1 , f2l and f2l+1 and deleting the edges v0 v2l and v0 v2l+1 . (Again it is easy to show that for each value of k, k ∈ {1, 6, 8, 9, 10, 11}, faces f2l+1 and f1 have no common boundary edge.) Thus this time, Ψ and Ψ′ are both embeddings of Km with two adjacent edges removed. Thus the five vertices v0 , v2l−1 , v2l , v2l+1 , v2l+2 appear in clockwise order about the boundary of Fl . ′ ′ ′ ′ Faces F1′ , . . . , Fl′ are chosen similarly, except that v0′ , v2l−1 , v2l , v2l+1 and v2l+2 lie ′ in counter-clockwise order along the face boundary of Fl . Thus faces Fl and Fl′ will have at least five distinct vertices in their facial walks. Add tubes T1 , . . . , Tl and six edges across each of T1 , . . . , Tl−1 exactly as before. ′ ′ ′ Next, across tube Tl we insert the ten edges v0 v2l+1 , v0 v2l+2 , v2l−1 v2l+2 , v2l−1 v0′ , v2l v0′ , ′ ′ ′ ′ ′ v2l v2l−1 , v2l+1 v2l−1 , v2l+1 v2l , v2l+2 v2l , v2l+2 v2l+1 . Theorem 3.3. (i) The above embeddings of HIB (s, k, l) and HIC (s, k, l) are genus embeddings, and  2γ(s, k) + l − 1 for k 6= 2, 5, 2γ(s, k) + l − 3 for k = 2, 5, and γ(HIC (s, k, l)) = 2γ(s, k) + l − 1 for k = 1, 6, 8, 9, 10 and 11 and γ(HIB (s, k, l)) = (ii) κ(HIB (s, k, l)) = 2l + 2 and κ(HIC (s, k, l)) = 2l + 3. Proof: The proof proceeds just as in the proof of Theorem 3.2. For Part (i), the only difference is that here for HIB (s, k, l) we have deleted one edge from each of Ψ and Ψ′ , but have added two additional edges across tube T1 . Thus the total edge counts remain the same as in Theorem 3.2 in all cases. Similarly, for HIC (s, k, l), we have deleted two edges from each of Ψ and Ψ′ , but have added four additional edges across T1 and again the total edge counts remain the same as in Theorem 3.2 in all cases. For Part (ii), the only difference is that, in HIB (s, k, l) and HIC (s, k, l), the size of the matching from Ψ to Ψ′ are increased by 1 and 2, respectively. This increases the connectivity of the graphs by 1 and 2, respectively. the electronic journal of combinatorics 9 (2002), #R38 11 4. Construction II — the Orientable Case In this section we will construct graphs and orientable embeddings thereof such that the connectivities of these graphs will cover a higher range of values than that covered by the graphs and embeddings constructed in Section 3. As before, let m = 12s + k and assume s ≥ 1 and 0 ≤ k ≤ 11. Let us once again begin by genus embedding graph Km into orientable surface Sγ(s,k) when k 6= 2, 5 and genus embedding graph Km − K2 into Sγ(s,k)−1 when k = 2 or 5. Call this embedding Ψ0 . Now take another identical embedding of the same graph in the same surface and call it Ψ′0 . Again this time we will first modify the underlying graphs of the two embeddings, and then proceed to link the two embeddings of the modified graphs with a collection of l tubes. Once again we then insert new edges across these tubes so as to form one larger graph embedded in the composite surface. Let us start with k = 2 or 5. Since Ψ0 is an embedding of Km − K2 , let v0 denote any vertex not an endvertex of the missing edge. Remembering that Ψ0 is a triangulation in these cases, let us label the vertices about vertex v0 as v1 , . . . , vm−1 in clockwise order such that v1 is also not an endvertex of the missing edge. ′ Label the vertices in Ψ′0 as v0′ , . . . , vm−1 in exactly the same way as the vertices were labeled in Ψ0 . Therefore the labels appear in the clockwise order around v0 and v0′ , as indicated in Figure 4.1. This ordering is important to avoid multiple edges in the constructions. Now suppose k = 0, 3, 4 or 7. Since Ψ0 is a triangulation in these cases, choose any vertex and label it v0 . Then label the other m − 1 vertices as v1 , . . . , vm−1 clockwise about v0 . Again do exactly the same labeling in Ψ′0 . For the remaining values of k, namely, k = 1, 6, 8, 9, 10 and 11, since Ψ0 is not a triangulation, there must exist a face of embedding Ψ0 which is of size at least 4. Choose four vertices on the boundary of this face which are consecutive in a clockwise direction about this face and call them v2 , v0 , v1 and v respectively. It is important to have this non-triangular face. For each of these cases, the addition of two extra edges on one of the connecting tubes suffices to attain the genus bound in Lemma 3.1. Now in all twelve cases (for each of the twelve values of k) delete vertex v0 and denote the new face formed by F0 . Once again repeat the above labeling and subsequent vertex deletion in exactly the same way in Ψ′0 and label with F0′ that face which corresponds to F0 . Call the resulting embeddings Ψ and Ψ′ respectively. We now have a face with all other vertices in Ψ(Ψ′ ) on its boundary. This is a major difference from the constructions in Section 3. By inserting a tube between these two big faces and adding edges on this tube, the resulting graph has higher connectivity than that obtained using Construction I. When inserting edges across the various tubes, we need to take particular care to avoid multiple edges. Insert a tube (or cylinder) T0 with one end in the interior of face F0 and the other end in the interior of face F0′ . Now suppose lk (s) is still defined as in Section 3 and suppose 1 ≤ l ≤ lk (s). Here l will denote the total number of connecting tubes we will insert between Ψ and Ψ′ and the electronic journal of combinatorics 9 (2002), #R38 12 hence tube T0 will be counted by l. We proceed to locate and insert the remaining l − 1 tubes. Let us choose l − 1 faces F1 , . . . , Fl−1 clockwise about vertex v1 such that the boundary of Fi contains the adjacent edges v1 ai and v1 bi , vertices v1 , ai and bi are labeled in clockwise order about face Fi , and such that (1) no two of the Fi have a common boundary edge, (2) if k = 1, 6, 8, 9, 10, 11, then for all 1 ≤ i ≤ l − 1, {ai , bi } ∩ {v2 , v, vm−1 } = ∅, and (3) if k = 0, 2, 3, 4, 5, 7, then for all 1 ≤ i ≤ l − 1, {ai , bi } ∩ {v2 , vm−1 } = ∅. (It is easy to check that for all twelve values of k, such a selection of l − 1 faces Fi is possible, since s ≥ 1.) Note that these ai ’s and bi ’s are neighbor vertices of v1 , and they are labeled as vi ’s on the face boundary of F0 . Again, make precisely the same selection of faces in Ψ′ . Then, for each i = 1, . . . , l− 1, insert a tube Ti joining the interiors of faces Fi and Fi′ . Then for each i, 1 ≤ i ≤ l − 1, across tube Ti add the six edges v1 a′i , v1 b′i , ai a′i , ai v1′ , bi v1′ , bi b′i . We now proceed to add edges across tube T0 as follows. Suppose α is an integer vari′ ′ able such that 3 ≤ α ≤ m−2. Consider the edge set {v1 v1′ , v1 v2′ , v2 v1′ , v2 vm−1 , v2 vm−2 }∪ ′ ′ ′ ′ {vi vm−i+1 |3 ≤ i ≤ α−1}∪{vi vm−i |3 ≤ i ≤ α−1}∪{vα vj |2 ≤ j ≤ m−α+1}∪{vi v2 |α ≤ i ≤ m − 1}. In the cases when k = 1, 6, 8, 9, 10, 11 we add to this set the two additional edges v1 v ′ and vv1′ . Finally, if α < ⌊(m + 1)/2⌋ and if vα is one of the ai ’s or bi ’s, then we do not add edge vα vα′ across tube T0 because it already exists on one of the tubes T1 , . . . , Tl−1 . Similarly, if α ≥ ⌊(m + 1)/2⌋ and if v⌊(m+1)/2⌋ is one of the ai ’s or one of the bi ’s, then ′ we do not add edge v⌊(m+1)/2⌋ v⌊(m+1)/2⌋ across tube T0 as it already exists on one of T1 , . . . , Tl−1 . (See Figure 4.1.) Call the resulting graph on 2m − 2 vertices HII (s, k, l, α). Lemma 4.1. Each graph HII (s, k, l, α) is simple. Proof: On tube T0 , the edges incident with v1 and v1′ are v1 v1′ , v1 v2′ , v2 v1′ and (possibly) vv1′ and v1 v ′ . On the other hand, on tube Ti the edges incident with v1 and v1′ are v1 a′i , v1 b′i , ai b′i ai v1′ , bi v1′ . But {ai , bi } ∩ {v2 , v, vm−1 } = ∅ and and {a′i , b′i } ∩ ′ {v2′ , v ′ , vm−1 } = ∅, so none of these edges is multiple. The remaining edges on Ti are of the form ai a′i or bi b′i . But on T0 we have constructed no such edges, because of the reversed order of vi ’s and vi′ ’s. Let us now compute the genus of HII (s, k, l, α). the electronic journal of combinatorics 9 (2002), #R38 13 00 v 11 00 111 1 0 0 1 v’ 1 0 0 1 v’ 1 v 00 11 00 11 00 11 v 00 11 00 11 00 11 0 1 0 1 0 1 00 11 00 11 00 11 0 1 0 1 0 1 11 00 00 11 1 0 0 1 00 00 11 v 11 00 α 11 1 0 0 1 00 11 00 11 00 11 0 1 0 1 0 1 11 00 00 11 1 0 0 1 00 11 0 1 0 1 0 1 00 11 00 11 00 11 0 1 0 1 0 1 v 2 3 00 11 v 00 m-2 11 v m-1 v’ m-1 v’ m-2 v’ m- α +1 v’ 3 v’ 2 Figure 4.1 Theorem 4.2. The embedding of HII (s, k, l, α) constructed above is a genus embedding and γ(HII (s, k, l, α)) =  2γ(s, k) + l − 3, if k = 2, 5, 2γ(s, k) + l − 1, if k 6= 2, 5. the electronic journal of combinatorics 9 (2002), #R38 14 Proof: For each k = 0, . . . , 11, we have embedded HII (s, k, l, α) in a surface with the required genus. Now noting that q ≥ 2[m(m−1)/2−(m−1)−1]+6(l−1)+2(m−1), when k = 2, 5; q ≥ 2[(m − 1)(m − 2)/2 + 2(m − 1) + 6(l − 1) − 1, when k = 0, 3, 4 or 7; and q ≥ 2(m − 1)(m − 2)/2 + 2(m − 1) + 2 + 6(l − 1) − 1, when k = 1, 6, 8, 9, 10 or 11; and that p = 2(m − 1) in all twelve cases, by Lemma 3.1 and an easy counting exercise we obtain  2γ(s, k) + l − 3, if k = 2, 5, γ(HII (s, k, l, α)) ≥ 2γ(s, k) + l − 1, if k 6= 2, 5, and the theorem is proved. Now consider the connectivity of graph HII (s, k, l, α). Since we have two independent parameters α and l, in general it is impossible to determine when vα is one of the ai ’s or one of the bi ’s. Moreover, the edges on the tubes Ti , i ≥ 1, may affect the connectivity of the graphs. So it is not easy to find the exact value of κ. However, as we shall see below, the exact value will not be required. To this end, it will be convenient to introduce some new terminology. Let any matching which matches vertices of Ψ with vertices of Ψ′ be called a left-right matching (or LR-matching). Let νLR (HII (s, k, l, α)) denote the size of any largest LRmatching in HII (s, k, l, α). Theorem 4.3. κ(HII (s, k, l, α)) ≥  νLR (H(s, k, l, α)) if k = 6 2, 5 min {νLR (H(s, k, l, α)), m − 2} if k = 2, 5. Proof: Denote HII (s, k, l, α) by H. Suppose u and w are two vertices of H. We want to consider the maximum number of vertex-disjoint paths between u and w. First suppose k 6= 2, 5. If u and w are both in Ψ, then there exist m − 2 disjoint paths joining them in Ψ since Ψ is an embedding of Km−1 . It is easy to construct one additional path joining u and w by using vertices in Ψ′ . Similarly, for u and w both in Ψ′ , it is easy to construct m − 1 disjoint paths joining them. So it will suffice to consider the case with u is in Ψ and w is in Ψ′ . Let Mα be a maximum LR-matching and suppose that the endvertices of Mα in Ψ are x1 , . . . , xj , j = νLR . Now j ≤ m − 1, so since Ψ is an embedding of Km−1 , there exist j ≤ m − 1 disjoint paths from u to each of the vertices x1 , . . . , xj . These paths are all single edges, unless u ∈ {x1 , . . . , xj } and in that case one of the paths is of length 0. Similarly there exist j disjoint paths from w to the endvertices of Mα in Ψ′ . Thus there are j = νLR disjoint u-v paths in H. Hence κ(H) ≥ νLR as claimed. Now suppose k = 2 or 5. Then, if νLR ≤ m − 2, proceeding exactly as in the case for k 6= 2, 5, it may be shown that there are at least νLR disjoint paths joining u and w. Finally, if νLR > m − 2, that is, if νLR = m − 1, then α = m − 2 and once again it is the electronic journal of combinatorics 9 (2002), #R38 15 easy to find m − 2 disjoint paths joining every two vertices u and w in H. The theorem follows. If we set α = m − 2 in the above theorem, we obtain the next result. Corollary 4.4. κ(HII (s, k, l, m − 2))  =m−1 ≥m−2 when k 6= 2, 5 and when k = 2, 5. ′ Proof: Equality follows in the case k 6= 2, 5 since deg (vm−1 ) = m − 1. The next result shows that if one increases parameter α by 1, the corresponding connectivity of HII (s, k, l, α) changes by at most 1. Lemma 4.5. If 3 ≤ α ≤ m − 3, then κ(HII (s, k, l, α)) − 1 ≤ κ(HII (s, k, l, α + 1)) ≤ κ(HII (s, k, l, α)) + 1. Proof: Denote HII (s, k, l, α) by HII (α). Note first that HII (α) − vα is a spanning subgraph of HII (α +1) −vα , so κ(HII (α)) −1 ≤ κ(HII (α) −vα ) ≤ κ(HII (α +1) −vα ) ≤ κ(HII (α + 1)) and the first inequality is proved. Next note that HII (α + 1) − vα+1 is a spanning subgraph of HII (α) − vα+1 and so κ(HII (α + 1)) − 1 ≤ κ(HII (α + 1) − vα+1 ) ≤ κ(HII (α) − vα+1 and so κ(HII (α + 1)) ≤ κ(HII (α) − vα+1 ) + 1 ≤ κ(HII (α)) + 1 and the second inequality is verified. Now the proof of the following theorem is immediate. Theorem 4.6. Suppose a = κ(HII (s, k, l, 3)) and  m − 1, if k 6= 2, 5 and b= m − 2, if k = 2, 5. Then for any κ0 ∈ [a, b] there is an α ∈ [3, m − 3] such that κ(HII (s, k, l, α)) = κ0 . Theorem 4.6 tells us the value of b in the interval of connectivities [a, b] which we can saturate. The next result gives us a useful upper bound on the other endpoint a. Theorem 4.7. κ(HII (s, k, l, 3)) ≤  2l + 3 when k = 1, 6, 8, 9, 10 or 11 and 2l + 2 when k = 0, 2, 3, 4, 5, 7. the electronic journal of combinatorics 9 (2002), #R38 16 Proof: The set S0 =  {v1 , v2 , v3 , v2′ , v}, when k = 1, 6, 8, 9, 10, 11, and {v1 , v2 , v3 , v2′ }, when k = 0, 2, 3, 4, 5, 7, covers all edges crossing tube T0 and hence the set S = S0 ∪ {a1 , b1 , . . . , al−1 , bl−1 } covers all edges crossing any of the total of l tubes. So set S is a cutset separating vertex vm−1 , say, from the remaining vertices of Ψ′ and has the size claimed in the statement of the theorem. Now we turn our attention to the high end of the connectivity interval [a, b] discussed above. By Theorem 4.6 we know that for each l we can saturate [2l + 3, m − 1], when k = 1, 6, 8, 9, 10, 11, over [2l + 2, m − 1] when k = 0, 3, 4, 7, and over [2l + 2, m − 2] when k = 2, 5. For saturation reasons we want to construct graphs with genus = γ(HII (s, k, l, α)) and built from HII (s, k, l, α) by modifying the arrangement of edges on tube T0 so that the resulting graphs have connectivities m − 1 when k = 2, 5 and m when k 6= 2, 5. If k = 0, 3, 4 or 7, begin just as in the case of HII (s, k, l, m − 2) as far as selecting faces and placing tubes is concerned. Suppose that k = 1, 6, 7, 8, 9, 10 or 11. In locating tubes T1 , . . . , Tl−1 , we modify our previous choices for the ai ’s and bi ’s so that {ai , bi } ∩ {v2 , vm−1 , v} = ∅. Suppose, then, that k = 2 or 5. Recall that we began the construction in this case by embedding a copy of Km − K2 and then deleted vertex v0 which was not a vertex of the missing edge. Denote this missing edge by xy. Let us now choose our l − 1 pairs {ai , bi } so that {ai , bi } ∩ {v2 , vm−1 , x} = ∅. For k = 0, 3, 4, 7 and for k = 2, 5, modify the graph HII (s, k, l, m − 2) by replacing ′ edges v1 v2′ and v2 v1′ with edges v1 vm−1 and vm−1 v1′ . When k = 1, 6, 8, 9, 10, 11, modify ′ ′ HII (s, k, l, m − 2) by replacing edges v1 v2′ , v1 v ′ , vv1′ and v2 v1′ with edges v1 vm−1 , vvm−1 , ′ ′ vm−1 v and vm−1 v1 . In each case, denote the graph obtained by HI (k, l, m). Theorem 4.8. κ(HI (k, l, m)) =  m, if k 6= 2, 5 and m − 1, if k = 2, 5. ′ ′ Proof: Among the edges across tube T0 call v1 v1′ , v2 vm−1 , v3 vm−2 , . . . , vm−1 v2′ , ′ ′ ′ ′ the horizontal matching, and call v1 vm−1 , v2 vm−2 , v3 vm−3 , . . . , vm−2 v2 , vm−1 v1′ the ′ slanted matching. (It is important to realize that the edge v⌊(m+1)/2⌋ v⌊(m+1)/2⌋ will be present, whether it is drawn on one of the tubes T1 , . . . , Tl−1 or on the tube T0 .) First, suppose k = 0, 3, 4 or 7. Let u and w be two vertices of HI (k, l, m). If both vertices lie in Ψ, then it is easy to find m vertex-disjoint paths joining them, where two of these paths use vertices of Ψ′ . By symmetry, if both lie in Ψ′ , there are m vertex-disjoint paths joining them as well. the electronic journal of combinatorics 9 (2002), #R38 17 Now suppose u lies in Ψ and w lies in Ψ′ ; say, without loss of generality that u = vi and w = vj′ . (i) First suppose j ≤ m − i. There are three subcases. (i-1) Suppose j = m − i, that is, suppose edge vi vj′ is an edge of the slanted matching. Then there are m − 3 u − w paths of length 3, using the m − 3 horizontal matching edges incident with v1 , . . . , vi−1 , vi+2 , . . . , vm−1 . To these we add the three additional ′ u − w paths vi vj+1 vj′ , vi vj′ , vi vi+1 vj′ to obtain a total of m vertex-disjoint u − w paths. (i-2) Suppose j = m − i − 1. There are m − 4 u − w paths of length 3, using the m − 4 horizontal matching edges incident with v1 , . . . , vi−1 , vi+3 , . . . , vm−1 . To these we ′ ′ add the four additional u − w paths vi vj+2 vj′ , vi vj+1 vj′ , vi vi+1 vj′ and vi vi+2 vj′ , to obtain our m vertex-disjoint u − w paths. (i-3) Suppose j ≤ m − i − 2; say, j = m − t, for some t ≥ i + 2. In this case, there are m − 4 u − w paths of length 3, using the horizontal matching edges incident with vertices v1 , . . . , vi−1 , vt+2 , vt+3 , . . . , vm−1 together with the slanted matching edges incident with vertices vi+1 , . . . , vt−1 . To these paths we add the additional four paths ′ ′ vj′ , vi vm−i vj′ , vi vt′ vj′ and vi vt+1 vj′ . of length 2: vi vm−i+1 (ii) Next, suppose j = m − i + 1. In this case we use the m − 3 slanted matching edges incident with the m − 3 vertices v1 , v2 , . . . , vi−2 , vi+1 , . . . , vm−1 , together with the three paths vi vi−1 vj′ , vi vj′ and ′ vi vj+1 vj′ to get m vertex-disjoint u − w paths. (iii) Finally, suppose j > m − i + 1. This is the same as Case (i), if one performs the same cyclic relabeling on both sides. On the other hand, however, deg HI (k,l,m) (vm−1 ) = m, so κ(HI (k, l, m)) = m as claimed. It then follows that κ(HI (k, l, m)) = m in the cases k = 1, 6, 8, 9, 10 and 11, for observe that the “extra vertex” v is just one of the vertices v1 , . . . , vm−1 appearing at a different place on the facial walk of the face produced by the deletion of vertex v0 . Next, suppose k = 2 or 5. We need to show that κ(HI (k, l, m)) = m − 1. Notice that, when k = 2 or 5, as opposed to the cases when k 6= 2 or 5, there is only one edge missing in each of Ψ and Ψ′ , and these two missing edges are on different sides of the LR−matching on tube T0 . Since κ(HI (k, l, m)) = m for k 6= 2, 5, by a similar argument, it is not hard to check that κ(HI (k, l, m)) = m − 1 for k = 2, 5. Theorem 4.9. The embedding of HI (k, l, m) constructed above is a genus embedding and  2γ(s, k) + l − 3, if k = 2, 5, γ(HI (k, l, m)) = 2γ(s, k) + l − 1, if k 6= 2, 5. Proof: The proof is very similar to that of Theorem 4.2. the electronic journal of combinatorics 9 (2002), #R38 18 5. The Interval of Saturation–the Orientable Case Before presenting the main theorem of this section, it is necessary to have the following results for surfaces of small orientable genus. Lemma 5.1. Let g be an integer such that 0 ≤ g ≤ 9. Then for every integer i ∈ [1, κmax (Sg )], there is a graph G with g =√ γ(G) and κ(G) = i. For g = 10, 11, and every integer i ∈ [1, ( 2/2)κmax (Sg )], there is a graph G with g = γ(G) and κ(G) = i. Proof: When g = 0, it is well known that there are planar graphs having vertex connectivity 5,4,3,2 and 1. Secondly, we claim that it follows from the results of Negami [N1] that for every surface with orientable genus ≥ 1 (or non-orientable genus ≥ 2), there are graphs of that genus which have connectivity 6, have a vertex of degree 6, and which triangulate the surface. To see this in the orientable case, one need only start Negami’s construction with a 6-regular 6-connected graph which triangulates the torus and has sufficiently many vertices so that not all vertices of the beginning graph are subsequently affected by Negami’s construction. Similarly, if one begins with a 6-connected 6-regular triangulation of the Klein bottle which has sufficiently many vertices, at least one vertex of degree 6 will remain unmodified by the Negami construction. Moreover, since the Negami graphs are all triangulations, one may remove up to five edges incident with a vertex of degree 6 without lowering the genus. Therefore we have the following proposition. Proposition: Starting with a 6-connected triangulation of genus γ ≥ 1 (or genus γ ≥ 2) which has a vertex of degree 6, there are graphs with connectivity 5,4,3,2 and 1 all of which have genus γ (or γ). First consider the (orientable) genera 2 and 3. Since γ(K8 ) = 2, κ(K8 ) = 7 = κ(S2 ), γ(K9) = γ(K9 − K2 ) = 3, κ(K9 ) = 8 = κmax (S3 ), κ(K9 − K2 ) = 7, the lemma is true for γ = 2 and 3. Next consider the case when γ = 4. Recall that γ(K10 ) = 4. Again by edge counting it is easily seen that the genus of each of K10 − K2 and K10 − 2K2 (where 2K2 denotes two edges having a common vertex) is also 4. So there are genus 4 graphs having connectivities 1, . . . , 9, where κgen (S4 ) = κmax (S4 ) = 9. Similarly, there are graphs with κ = 1, . . . , 10 having γ = 5 and κgen (S5 ) = κmax (S5 ) = 10 as well as graphs with κ = 1, . . . , 11 having γ = 6 where κgen (S6 ) = κmax (S6 ) = 11. For S7 our approach is somewhat different. Embed K9 and K10 , choose a vertex v0 from K9 and a vertex v0′ from the K10 , delete each, and insert cylinder T0 , just as in Construction II (see Section 4). (We do not add any additional tubes this time.) Insert ′ the following edges: vi vi′ and vi vi+1 for i = 1, . . . , 8; v1 v ′ and v8 v1′ . It is then easy to check that the resulting graph is 9-connected and a simple edge count, together with Lemma 3.1, shows that the genus of this graph is 7 as desired. Modify this embedding the electronic journal of combinatorics 9 (2002), #R38 19 by inserting different edges on tube T0 can produce graphs with κ = 7 and 8 with genus 7. Consider next S8 . The octahedron K7(2) , obtained from the complete graph K14 by deleting a perfect matching has connectivity 12 and by [JR1,AG1], genus 8. If an arbitrary vertex is selected and up to five edges incident with this vertex are removed, an edge count shows that each of the resulting graphs still has genus 8. In this way graphs having connectivity 7, . . . , 11 are created. Thus the full connectivty interva for the surface S8 can be saturated. The graph K14 − K2 triangulates the surface S9 by [RY1] and [J1] and has connectivity 12. Again by a simple edge count it is easy to show that up to five edges can be deleted from some fixed vertex of this graph without lowering the genus of the resulting graph. So we can get genus 9 graphs having connectivity 1, . . . , 12 and thus can saturate the full connectivity interval for the surface S9 . For surface S10 we return to a modification of Construction II. Embed two copies of K11 , delete a vertex from each and construct the tube T0 just as in Construction ′ ′ II. Across T0 construct edges vi vi′ and vi vi+1 for i = 1, . . . , 9 as well as edges v10 v10 and v10 v1′ . The resulting graph has κ = 11 and an edge count verifies that the genus is indeed 10. A modification of the edge construction across T0 similar to that done above for the genus 7 case yields genus 10 graphs having connectivity 1, . . . , 11 and since √ ( 2/2)κmax (S10 ) = 9, we are done. Surface S11 is handled in much the same way as S10 . Embed K11 in S5 , K12 in S6 , delete a vertex from each and add tube T0 as well as edges across T0 as before. As before, graphs √ having connectivity 1, . . . , 11 can be constructed all of which have genus 11. But ( 2/2)κmax (S11 ) = 9 and again we are done. We now need to construct genus embeddings whose connectivity cover the lower range of values in connectivity intervals. In doing so, we will make use of a graph operation which will be called splitting. Suppose graph G is n-connected, where n ≥ 2, and suppose v ∈ V (G) with deg v = n. Partition the neighborhood NG (v) into A ∪ B where A ∩ B = ∅ and |A| ≥ 1 and |B| ≥ 1. Form a new graph G′ from G by deleting vertex v, inserting two new vertices v1 and v2 such that v1 is adjacent to each vertex in A, vertex v2 is adjacent to each vertex in B and v1 is adjacent to v2 . Suppose |A| = a and |B| = b. We then say that graph G′ is obtained from graph G by an (a, b)-split at vertex v. Lemma 5.2. Let G be a graph with κ(G) ≥ 2 and let v be a vertex in G. Let G′ be a graph obtained from graph G by an (a, b)-split at vertex v and suppose r = min {1 + a, 1 + b} ≤ κ(G). Then κ(G′ ) = r. Proof. Note that |NG′ (vi )| ≥ r, for i = 1, 2. Now let S ′ be a minimum vertex cutset in G′ and suppose |S ′ | < r. If {v1 , v2 } ∩ S ′ = ∅, then clearly S ′ is a cutset in G since v1 and v2 lie in the same component of G′ − S ′ . So κ(G) ≤ |S ′ | < r ≤ κ(G), a contradiction. If {v1 , v2 } ⊆ S ′ , then S = S ′ − v1 − v2 + v is a vertex cut in G and hence κ(G) = |S| = |S ′ | − 1 < r − 1 ≤ κ(G) − 1, again a contradiction. the electronic journal of combinatorics 9 (2002), #R38 20 So we may assume that |{v1 , v2 } ∩ S ′ | = 1. Without loss of generality, assume / S ′ . Let C be the component of G′ − S ′ containing vertex v2 . If v1 ∈ S ′ and v2 ∈ |C| ≥ 2, then S = S ′ − v1 + v is a cutset in G and κ(G) ≤ |S| = |S ′ | < r ≤ κ(G) and once more we have a contradiction. So |C| = 1; that is, V (C) = {v2 }. But then NG′ (v2 ) ⊆ S ′ and hence r ≤ |NG′ (v2 )| ≤ |S ′ | < r, and yet again we have a contradiction. So |S ′ | ≥ r and hence κ(G′ ) = |S ′ | ≥ r. But G′ has a vertex of degree r, namely either vertex v1 or v2 , and so κ(G′ ) = r as claimed. We now are prepared to assemble Lemma 5.2 and the results of the previous two sections in order to prove the following theorem. Theorem 5.3. Suppose g is any non-negative integer. For each orientable surface Sg define the quantity mg by m0 = 5, m1 = 6 and for g ≥ 2 by p   5 + 1 + 48⌊ g2 ⌋ mg = + 1. 2 Write mg = 12s + k, where 0 ≤ k ≤ 11. Then for all integers i in the interval [1, mg ] when k 6= 2, 5 (respectively, in the interval [1, mg − 1] when k = 2 or 5), there exists a graph G with γ(G) = g and κ(G) = i. Proof: If g ≤ 11, the result is immediate by Lemma 5.1. So assume g ≥ 12 and hence s ≥ 1. Cook [C1] has shown that if g/2 > 0, κmax (S⌊ g2 ⌋ ) =  5+ p  1 + 48⌊ g2 ⌋ = mg − 1, 2 and this bound is realized by the complete graph Km = Kmg . Let mg = m = 12s + k ≥ 12 and define l by  g − 2γ(s, k) + 1, if k 6= 2, 5, l= g − 2γ(s, k) + 3, if k = 2, 5. 1. Suppose k = 0, 3, 4 or 7. By performing splitting (see Lemma 5.2) on graph HI (k, l, m), we can saturate the interval [1, 6s + ⌈(k + 1)/2⌉, by Theorems 3.2 and 3.3 we can saturate [l + 2, 2l + 2], by Theorems 4.7 and 4.6 we can saturate [2l + 3, m − 1], and by Theorem 4.8, we can extend the last interval to [2l + 3, m]. 2. Suppose k = 1, 6, 8, 9, 10 or 11. By performing splitting on graph HI (k, l, m), again we can saturate the interval [1, 6s + ⌈(k + 1)/2⌉], by Theorem 3.2 we can saturate interval [l + 2, 2l + 1] and by Theorem 3.3 we can extend this to [l + 2, 2l + 3]. Again by Theorems 4.7 and 4.6 we can saturate [2l + 3, m − 1] and by Theorem 4.8 we can extend this to [2l + 3, m]. the electronic journal of combinatorics 9 (2002), #R38 21 3. Finally suppose k = 2 or 5. In this case, performing splitting on graph HI (k, l, m) yields saturation of [1, 6s + ⌊(k + 1)/2⌋], Theorem 3.2 yields saturation of [l + 2, 2l + 1] and Theorem 3.3 allows us to extend this to [l +2, 2l +2]. Then Theorems 4.7 and 4.6 allow saturation of [2l +2, m−2] and Theorem 4.8 allows us to extend this to [2l + 2, m − 1]. It is routine to check that for k 6= 2, 5, we have 6s + ⌈(k + 1)/2⌉ ≥ l + 1 and when k = 2, 5, we have 6s + ⌊(k + 1)/2⌋ ≥ l + 1. But then we have shown that when k 6= 2, 5, we can saturate [1, m] and when k = 2 or 5, we can saturate [1, m − 1]. Finally note that by Theorems 3.2, 3.3, 4.2 and 4.9, the genus of each graph constructed for this saturation is g. This completes the proof of the theorem. Theorem 5.4. For every integer g ≥ 0, let mg be defined as in Theorem 5.3. Then when 0 ≤ g√≤ 9, for every integer i in [1, κmax (Sg )], and when g ≥ 10, for every integer i in [1, ⌊( 2/2)κmax (Sg )⌋], when mg 6≡ 2, 5( mod 12) and for every integer i in [1, ⌊0.7κmax (Sg )⌋] when mg ≡ 2, 5( mod 12), there exists a graph G with connectivity i which genus embeds in surface Sg . Proof: The result is immediate by Lemma 5.1 when g ≤ 11, so suppose g ≥ 12. First suppose that mg 6≡ 2, 5( mod 12). Let us consider the ratio   √ 5+ 1+48⌊ g2 ⌋ +1 2 mg  √  = . κmax 5+ 1+48g 2 √ We want to show that mg /κmax ≥ 2/2. Case 1. Assume g is odd. Let g = 2t + 1. Then    √  √ 5+ 1+48t 7+ 1+48t +1 2 2 mg  =  √ . =  √ κmax 5+ 49+96t 5+ 49+96t 2 2 Let us write 48t+1 = a2 +b, where a, b are non-negative integers such that 0 ≤ b < 2a+1. But then √ 2 a+6 mg √ ≥ , ≥ κmax 2 5 + 2a2 + 4a + 47 whenever a ≥ 6. Moreover when 1 ≤ a ≤ 5, it follows that g ≤ 1; i.e., Sg is the torus. But we already know that we can saturate the entire connectivity interval for the torus. So Case 1 is proved. the electronic journal of combinatorics 9 (2002), #R38 22 Case 2. Suppose g is even. Say g = 2t. Then mg =  √ 5+ 1+48t 2   √ 5+ 1+96t 2 +1  . But 1 + 96t < 49 + 96t and so we are done by the argument for Case 1. Now suppose mg ≡ 2, 5( mod 12). Case 3. First suppose g is odd. Then proceeding as in Case 1, it may be shown that   √ 5+ 1+48t 2 mg − 1  ≥ 0.7, = √ κmax 5+ 49+96t 2 for a ≥ 66. When a < 66, it follows that g < 2t + 1 ≤ 187. However, the number of values of g that must be checked can be reduced from 187 to just 9 by recalling that we are in the case in which mg ≡ 2, 5( mod 12). It follows easily then that the ratio (mg − 1)/κmax ≥ 0.7 in eight of these cases. It remains only to consider the single case when g = 35. It will suffice to construct a graph G which has κ = 17 and genus 35. But this construction may be carried out in a manner similar to that above. Case 4. Finally, suppose g is even. Letting g = 2t and 1 + 48t = a2 + b and proceeding as in Case 3, it may be shown that mg /κmax ≥ 0.7 for all a ≥ 48; i.e., for all g ≥ 95. But there are only six values of g ≤ 94 for which mg ≡ 2, 5( mod 12) and one easily checks each of these values directly to show that (mg − 1)/κmax ≥ 0.7. This completes the proof of Case 4 and the corollary follows. The next result follows immediately from Theorem 5.4. Corollary 5.5. For any orientable surface Sg , (i) κgen (Sg ) = κmax √ (Sg ), when 0 ≤ g ≤ 11, (ii) κgen (Sg ) ≥ ⌈( 2/2)κmax (Sg )⌉, when g ≥ 12 and mg 6≡ 2, 5( mod 12), and (iii) κgen (Sg ) ≥ ⌈0.7κmax (Sg )⌉, when g ≥ 12 and mg ≡ 2, 5( mod 12). the electronic journal of combinatorics 9 (2002), #R38 23 6. The Non-orientable Case The constructions in the non-orientable case parallel in spirit those of the orientable case. Therefore, here we will primarily stress the differences between the non-orientable and the orientable cases. (1) We shall conform to the tradition of considering the six congruence classes modulo 6 for the non-orientable cases and to that end let m = 6s + k where s and k are non-negative integers with 0 ≤ k ≤ 5. Let γ(s, k) = γ(K6s+k ). (2) For each pair of s and k, the base embedding Ψ is a triangulation of Nγ (s,k) when k = 0, 1, 3, 4, and K6s+k − K2 into Nγ (s,k)−1 when k = 2, 5 [J1, K1, R1, R2, RY1]. (3) Let ¯lk (s) = 2[γ(s, k + 1) − γ(s, k)] + 1, for k 6= 2, 5, and ¯lk (s) = 2[γ(s, k + 1) − γ(s, k)] + 3, for k = 2, 5. This number is an upper bound of the number of crosscaps we need to construct the desired non-orientable surfaces. (4) For k = 2, 5, the base embeddings are not necessarily non-orientable [R1]. In each of these cases, we need at least two tubes to construct the desired surfaces and we always twist the last tube to obtain a non-orientable surface. (5) If a given non-orientable surface has odd Euler characteristic, when constructing embeddings in this surface we need to add a crosscap on a connecting tube to guarantee that the resulting surface is non-orientable. Also we need to put extra edges on this crosscap that are not parallel to other edges to obtain a genus embedding. The crosscap is added on the last tube for the construction analogous to Construction I and is added on the first tube for the construction analogous to Construction II. Using arguments similar to those in the orientable cases, we obtain the following results. The proofs are omitted here, but may be checked in an Appendix available on the world wide web [PZ2]. Theorem 6.1. Suppose h is any positive integer. For each non-orientable surface Nh define the quantity mh by m1 = 6 and for h ≥ 2 by q  5 + 1 + 24⌊ h ⌋  2 mh = + 1. 2 Write mh = 6s + k, where 0 ≤ k ≤ 5. Then for all integers i in the interval [1, mh ] when k 6= 2, 5 (respectively, in the interval [1, mh − 1] when k = 2 or 5), there exists a graph G with γ(G) = h and κ(G) = i. Theorem 6.2. For every integer h > 0, let mh be defined as in Theorem 6.1. Then when 1 ≤ h√≤ 7, for every integer i in [1, κmax (Nh )], and when h ≥ 8, for every integer i in [1, ⌊( 2/2)κmax (Nh )⌋], when m 6≡ 2, 5( mod 6) and for every integer i in [1, ⌊0.7κmax (Nh )⌋] when m ≡ 2, 5( mod 6), there exists a graph G with connectivity i which genus embeds in surface Nh . the electronic journal of combinatorics 9 (2002), #R38 24 Theorem 6.3. For any non-orientable surface Nh , (i) κgen (Nh ) = κmax √ (Nh ), when 1 ≤ h ≤ 7, (ii) κgen (Nh ) ≥ ⌈( 2/2)κmax (Nh )⌉, when h ≥ 8 and mh 6≡ 2, 5( mod 6), and (iii) κgen (Nh ) ≥ ⌈0.7κmax (Nh )⌉, when h ≥ 8 and mh ≡ 2, 5( mod 6). 7. Concluding Remarks It has been shown above that for genus embeddings, one can always saturate a subinterval of the connectivity interval of a surface Σ, namely √ the subinterval [1, ⌊0.7κmax (Σ)⌋] (or sometimes on the subinterval [1, ⌊( 2/2)κmax (Σ)⌋]). But what about saturation of the remaining subinterval [⌊0.7κmax (Σ)⌋ + 1, κgen (Σ)]? For some surfaces (such as S36 ), saturation is not possible because κgen < κmax for these surfaces. In [PZ1] these are called Class B surfaces and it is shown there that such surfaces must occur periodically infinitely often as the value of the genus goes to infinity. In the Introduction it was claimed that a complete answer to Question 2 would imply the solution to the problem of determining the genus of the octahedron O(m) for some values of m for which this problem is still currently not solved. We conclude with further explanation of this claim. (We will restrict our discussion to the orientable case.) We recall some terminology and a few results from [PZ1]. Call surfaces for which κgen = κmax , Class A. Thus since it is clear that κgen ≤ κmax for all surfaces, the set of all surfaces partitions into those which are Class A and those which are Class B. The Class A and Class B genera are nicely arranged in the following sense. Every finite interval of genera of the form [γ(Km ) + 1, γ(Km+1)] can be written as the disjoint union of two subintervals [κ(Km ) + 1, gA − 1] ∪ [gA , γ(Km+1)] where the first of the two subintervals consists entirely of Class B genera and the second entirely of Class A genera. Moreover, it was shown that if m ≥ 30, then the Class B subinterval is not empty. The value of gA is called the breaking point of the interval [γ(Km ) + 1, γ(Km+1)]. We now show that if we knew the complete answer to Question 2, then we could compute the genus of the octahedron O(m) for at least some of the values of m for which it is presently unknown. Write m = 12s + k where 0 ≤ k ≤ 11. In [PZ1] it was shown that for k = 1, . . . , 7 and 11, gA for the interval [γ(Km) + 1, γ(Km+1 )] must equal to γ(O(m + 1)). (We stress here that this does not mean that we know the numerical value of γ(O(m + 1))!) Set i = κ(O(m + 1)) in Question 2, where m = 12s + k and assume that k has one of the eight values 1, . . . , 7 or 11. (Clearly i = m − 1.) Inspect the set of genera 1, 2, 3, . . . , in increasing order. Among the genera in the interval [1, γ(Km) − 1], none has an (m − 1)-connected graph which genus embeds there, since κgen ≤ κmax ≤ m − 2 for all these genera. The genus γ(Km ) does have an (m − 1)-connected graph which genus embeds (namely, Km !). Then in the next subinterval [γ(Km ) + 1, gA − 1], none of the genera admits the genus embedding of any the electronic journal of combinatorics 9 (2002), #R38 25 (m − 1)-connected graph. That is because the genera in this subinterval all correspond to Class B surfaces and hence κgen < κmax = m − 1 for these genera. The next genus encountered is gA and since it is the genus of a Class A surface, it does admit the genus embedding of an (m − 1)-connected graph. (We do not necessarily know the identity of such a graph, but that is irrelevant for our purposes here.) So starting at genus 1 and allowing the genus to increase by one at each step, since we know the answer to Question 2 at each step, we simply record the second genus encountered where the answer to Question 2 is “yes”; that is, the second genus encountered which admits the genus embedding of some graph having connectivity m − 1 (the first genus encountered for κ = m − 1 is γ(Km )). We know that the numerical value of this genus must be gA . Since we have assumed that m ≡ 1, . . . , 7( mod 12) or m ≡ 11( mod 12), by Theorem 5.1 of [PZ1] we know that this numerical value for gA must be the same as the value of γ(O(m + 1)). In this way, we have determined the numerical value of γ(O(m + 1)). We conclude with the following conjecture: Conjecture: For any integer i ∈ [1, κgen(Σ)], there is a graph which genus embeds in Σ and has connectivity i. References [AG1] S. Alpert and J.L. Gross, Branched coverings of graph embeddings, Bull. Amer. Math. Soc., 79, 1973, 942-945. [C1] R.J. Cook, Heawood’s theorem and connectivity, Mathematika, 20, 1973, 201207. [D1] R.A. Duke, The genus, regional number, and Betti number of a graph, Canad. J. Math., 18, 1966, 817-822. [E1] J. Edmonds, On the surface duality of linear graphs, J. Res. Nat. Bur. Standards Sect. B, 121-135. [E2] ,A combinatorial representation for polyhedral surfaces, Notices Amer. Math. Soc., 7, 1960, 646. [GT1] J.L. Gross and T.W. Tucker, Topological Graph Theory, John Wiley & Sons, New York, 1987. [J1] M. Jungerman, Orientable cascades and related embedding techniques, Ph.D. Thesis, Dept. of Mathematics, University of California at Santa Cruz, Santa Cruz, California, 1975. the electronic journal of combinatorics 9 (2002), #R38 26 [J2] M. Jungerman, The non-orientable genus of O(n), Abstract No. 7, Graph Theory Newslett., 5, 1975, 86. [J3] M. Jungerman, The non-orientable genus of the symmetric quadripartite graph, J. Combin. Theory Ser. B, 26, 1979, 154-158. [JR1] M. Jungerman and G. Ringel, The genus of the n-octahedron: regular cases, J. Graph Theory, 2, 1978, 69-75. [K1] V.P. Korzhik, A non-orientable triangular embedding of Kn −K2 , n ≡ 8( mod 2), Discrete Math., 141, 1995, 195-211. [N1] S. Negami, Construction of graphs which are not uniquely and not faithfully embeddable in surfaces, Yokohama Math. J., 33, 1985, 67-91. [PZ1] M.D. Plummer and X. Zha, On the connectivity of graphs embedded in surfaces, J. Combin. Theory Ser. B, 72, 1998, 208-228. [PZ2] M.D. Plummer and X. Zha, On the connectivity of graphs embedded in surfaces II: Appendix, available from: http://www.math.vanderbilt.edu/ plummemd/ [R1] G. Ringel, Map Color Theorem, Springer-Verlag, New York, 1974. [R2] G. Ringel, Wie man die geschlossen nichtorientierbaren Flächen in möglichst wenig Dreiecke zerlegen kann, Math. Ann., 130, 1955, 317-326. [RY1] G. Ringel and J.W.T. Youngs, Lösung des Problems der Nachbargebiete auf orientierbaren Flächen, Archiv der Mathematik (Basel), 20, 1969, 190-201. [S1] S. Stahl, Generalized embedding schemes, J. Graph Theory, 2, 1978, 41-52. [T1] C. Thomassen, The graph genus problem is NP-complete, J. Algorithms, 10, 1989, 568-576. [W1] A.T. White, Graphs, Groups and Surfaces, North-Holland, Amsterdam, 1984. [X1] N.H. Xuong, How to determine the maximum genus of a graph, J. Combin. Theory Ser. B, 26, 1979, 217-225. [Y1] J.W.T. Youngs, Minimal embeddings and the genus of a graph, J. Math. Mech., 12, 1963, 303-315. the electronic journal of combinatorics 9 (2002), #R38 27