In this paper, we study the minimum cycle base of the planar graphs obtained
from two 2-connected planar graphs by identifying an edge (or a cycle) of one graph with the corresponding edge (or cycle) of another, related with map geometries, i.e., Smarandache 2-dimensional manifolds

Dengju, Ma

Han, Ren

English

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International J.Math. Combin. Vol.1 (2007), No.1, 59-66Minimum Cycle Base ofGraphs Identified by Two Planar GraphsDengju Ma(School of Science, Nantong University, Jiangsu 226007, P.R.China)E-mail:jdm8691@yahoo.com.cnHan Ren(Department of Mathematics, East China Normal University, Shanghai 200062, P.R.China)E-mail:hren@math.ecnu.edu.cnAbstract: In this paper, we study the minimum cycle base of the planar graphs obtainedfrom two 2-connected planar graphs by identifying an edge (or a cycle) of one graph with thecorresponding edge (or cycle) of another, related with map geometries, i.e., Smarandache2-dimensional manifolds. Also, we give a formula for calculating the length of minimumcycle base of a planar graph N(d, λ) defined in paper [11].Key Words: graph, planar graph, cycle space, minimum cycle base.AMS(2000) : O5C10§1. IntroductionThroughout this paper we consider simple and undirected graphs. The cardinality of a set Ais |A|. Let’s begin with some terminologies and some facts about cycle bases of graphs. LetG(V,E) be a 2-connected graph with vertex set V and edge set E. The set E of all subsets of Eforms an |E|-dimensional vector space overGF (2) with vector additionX⊕Y = (X∪Y )\(X∩Y )and scalar multiplication 1 •X = X, 0 •X = ∅ for all X,Y ∈ E . A cycle is a connected graphwhose any vertex degree is 2. The set C of all cycles of G forms a subspace of (E ,⊕, •) which iscalled the cycle space of G. The dimension of the cycle space C is the Betti number of G, sayβ(G), which is equal to |E(G)| − |V (G)|+ 1. A base B of the cycle space of G is called a cyclebase of G.The length |C| of a cycle C is the number of its edges. The length l(B) of a cycle base Bis the sum of lengths of all its cycles. A minimum cycle base (or MCB in short) is a cycle basewith minimal length. A graph may has many minimum cycle bases, but every two minimumcycle bases have the same length.Let G be a 2-connected planar graph embedded in the plane. G has |E(G)| − |V (G)| + 2faces by Euler formula. There is exactly one face of G being unbounded which is called the1Received July 8, 2007. Accepted August 18, 20072Supported by NNSF of China under the granted NO.10671073 and NSF of Jiangsu’s universities under thegranted NO.07KJB11009060 Dengju Ma and Han Renexterior of G. All faces but the exterior of G are called interior faces of G. Each interior faceof G has a cycle as its boundary which is called an interior facial cycle. Also, the cycle of Gbeing incident with the exterior of G is called the exterior facial cycle.We know that if G is a 2-connected planar graph embedded in the plane, then any set of|E(G)| − |V (G)| + 1 facial cycles forms a cycle base of G. For a 2-connected planar graph, weask whether there is a minimum cycle base such that each cycle is a facial cycle. The answerisn’t confirmed. The counterexample is easy to be constructed by Lemma 1.1. Need to saythat Lemma 1.1 is a special case of Theorem A in the reference [10] which is deduced by HallTheorem.Lemma 1.1 Let B be a cycle base of a 2-connected graph G. Then B is a minimum cyclebase of G if and only if for any cycle C of G and cycle B in B, if B ∈ Int(C), then |C| ≥ |B|,where Int (C) denotes the set of cycles in B which generate C.For some special 2-connected planar graph, there exist a minimum cycle base such thateach cycle is a facial cycle. For example, Halin graph and outerplanar graph are such graphs.A Halin graph H(T ) consists of a tree T embedded in the plane without subdivision of an edgetogether with the additional edges joining the 1-valent vertices consecutively in their order inthe planar embedding. It is clear that a Halin graphs is a 3-connected planar graph. Theexterior facial cycle is called leaf-cycle.Lemma 1.2[9,12] Let H(T ) be a Halin graph embedded in the plane such that the leaf-cycleis the exterior facial cycle. Let F denote the set of interior facial cycles of H(T ). Then F is aminimum cycle base of H(T ).A planar graph G is outerplanar if it can be embedded in the plane such that all verticeslie on the exterior facial cycle C.Lemma 1.3[6,9] Let G(V,E) be a 2-connected outerplanar graph embedded in the plane withC as its exterior facial cycle. Let F be the set of interior facial cycles. Then F is the minimumcycle base of G, and l(F) = 2|E| − |V |.Apart from the above mentioned minimum cycle bases of a Halin graph and an outerplanargraph, many peoples researched minimum cycle bases of graphs. H. Ren et al. [9] not only gavea sufficient and necessary condition for minimum cycle base of a 2-connected planar graph, butalso studied minimum cycle bases of graphs embedded in non-spherical surfaces and presentedformulae for length of minimum cycle bases of some graphs such as the generalized Petersengraphs, the circulant graphs, etc. W.Imrich et al. [4] studied the minimum cycle bases forthe cartesian and strong product of two graphs. P.Vismara [13] discussed the union of all theminimum cycle bases of a graph. What about the minimum cycle base of the graph obtainedfrom two 2-connected planar graphs by identifying some corresponding edges? This problemis related with map geometries, i.e., Smarandache 2-dimensional manifolds (see [8] for details).We will consider it in this paper.Minimum Cycle Base of Graphs Identified by Two Planar Graphs 61§2. MCB of graphs obtained by identifying an edge of planar graphsLet G1 and G2 be two graphs and Pi be a path (or a cycle) in Gi for i = 1, 2. Suppose thelength of P1 is same as that of P2. By identifying P1 with P2, we mean that the vertices of P1are identified with the corresponding vertices of P2 and the multiedges are deleted.Theorem 2.1 Let G1 and G2 be two 2-connected planar graphs embedded in the plane. Letei be an edge in E(Gi) such that ei is in the exterior facial cycle of Gi for i = 1, 2. Let G bethe graph obtained from G1 and G2 by identifying e1 and e2 such that G2 is in the exterior ofG1. If the set of interior facial cycles of Gi, say Fi, is a minimum cycle base of Gi for i = 1, 2,then F1 ∪ F2 is a minimum cycle base of G.Proof Obviously, the graph G is a 2-connected planar graph and each cycle of F1 ∪F2 isa facial cycle of G. Since |E(G)| = |E(G1)|+ |E(G2)|−1 and |V (G)| = |V (G1)|+ |V (G2)|−2, Ghas |E(G)|−|V (G)|+2 = (|E(G1)|−|V (G1)|+1)+(|E(G2)|−|V (G2)|+1)+1 = |F1|+ |F2|+1faces. So |F| = |F1|+ |F2| = |E(G)| − |V (G)|+ 1, and F is a cycle base of G.Now we prove that F is a minimum cycle base of G. Suppose F is a cycle of G andF = f1 ⊕ f2 ⊕ · · · ⊕ fq, where fj ∈ F for j = 1, 2, · · · , q. By Lemma 1.1, We need to prove|F | ≥ |fj| for j = 1, 2, · · · , q.If E(F ) ⊂ E(G1) (or E(G2)), then fj is in F1 (or F2) for j = 1, 2, · · · , q. By the fact thatFi is a minimum cycle base of Gi for i = 1, 2 and Lemma 1.1, |F | ≥ |fj| for j = 1, 2, · · · , q.Let e be the edge of G obtained by e1 identified with e2. Suppose e = {uv}. If edges ofF aren’t in G1 entirely, then F must pass through u and v. So e ∪ F can be partitioned intotwo cycles, say F1 and F2. Suppose E(Fi) ⊂ E(Gi) for i = 1, 2. Then |F | > |Fi| for i = 1, 2.Suppose F1 = f1 ⊕ f2 ⊕ · · · ⊕ fp and F2 = fp+1 ⊕ fp+2 ⊕ · · · ⊕ fq. By the fact that Fi is aminimum cycle base of Gi for i = 1, 2 and Lemma 1.1, |F | > |F1| ≥ |fi| for i = 1, 2, · · · , p and|F | > |F2| ≥ |fi| for i = p+ 1, p+ 2, · · · , q.Thus we complete the proof. Applying Theorem 2.1 and the induction principle, it is easy to prove the following con-clusion.Corollary 2.1 Let G1, G2, · · · , Gk be k(k ≥ 3) 2-connected planar graphs embedded in theplane. Let ei be an edge in E(Gi) such that ei is in the exterior facial cycle of Gi for i =1, 2, · · · , k. Let G′1 be the graph obtained from G1 and G2 by identifying e1 with e2 such thatG2 is in the exterior of G1, Let G′2 be the graph obtained from G′1 and G3 by identifying e3 withsome edge in the exterior face of G′1 such that G3 is in the exterior of G′1, and so on. Let G bethe last obtained graph in the above process. If the set of interior facial cycles of Gi, say Fi, isa minimum cycle base of Gi for i = 1, 2, · · · , k, then ∪ki=1Fi is a minimum cycle base of G.62 Dengju Ma and Han Ren@@hhh lll  u uuuuuuuuuuu uu u u uuuuuuuuu uu1u2u3u4v1v2v3v4x1x2x3x4yH1 H2 HFig.2.1Remark: In Theorem 2.1, if e1 is replaced by a path with length at least two and e2 by thecorresponding path, then the conclusion of the theorem doesn’t hold. We consider the graphH shown in Fig.2.1, where H is obtained from H1 and H2 by identified P1 = u1u2u3u4 withP2 = v1v2v3v4. For the graph H , let C = x1x2x3x4x1 and D = x1yx4x1. Since |C| > |D|, theset of interior facial cycle of H isn’t its minimum cycle base by Lemma 1.1.Furthermore, if e1 is replaced by a cycle and e2 by the corresponding cycle in Theorem2.1, then the conclusion of Theorem isn’t true. The counterexample is easy to construct, whichis left to readers. But if G1 is a special planar graph, similar results to Theorem 2.1 will beshown in the next section.§3. MCB of graphs obtained by identifying a cycle of planar graphsAn r× s cylinder is the graph with r radial lines and s cycles, where r ≥ 0, s > 0. A 4× 3cylinder is shown in Fig.3.1. The innermost cycle is called the central cycle. r× s cylinder takean important role in discussion of the minor of planar graph with sufficiently large tree-widthin paper[10].&%'$u u uuu uuuuuuuFig.3.1Theorem 3.1 Let G1 be an r × s(r ≥ 4) cylinder embedded in the plane such that C is itscentral cycle. Let G2 be a planar graph embedded in the plane such that the exterior facialcycle D has the same vertices as that of C. Let G be the graph obtained from G1 and G2 byidentifying C and D such that G2 is in the interior of G1. If the set of interior facial cycles ofG2, say F2, is its a minimum cycle base, then the set of interior facial cycles of G, say F , isa minimum cycle base of G.Minimum Cycle Base of Graphs Identified by Two Planar Graphs 63Proof At first, F is a cycle base of G. We need prove F is minimal.Let F1 = F\F2. Obviously, each element of F1 has length 4. Suppose F is a cycle ofG and F = f1 ⊕ f2 ⊕ · · · ⊕ fq, where fj ∈ F for j = 1, 2, · · · , q. If we prove |F | ≥ |fj| forj = 1, 2, · · · , q, then F is a minimum cycle base of G by Lemma 1.1.Let R be the open region bounded by F , and R′ be the open region bounded by C (or D)of G1 (or G2). We consider the following four cases.Case 1 R′ ∩R = ∅.Then F is a cycle of G1 and F is generated by F1 . Since the girth of G1 is 4, |F | ≥ |fj | = 4for j = 1, 2, · · · , q.Case 2 R′ ⊂ R.Then |F | ≥ |C| ≥ 4, because the number of radial lines which F crosses can’t be less thanthe number of vertices of C. For a fixed fj , if it is in the interior of C then |fj | ≤ |C| ≤ |F |by Lemma 1.1, because F2 is a minimum cycle base of G2. If fj is in the exterior of C, then|fj | = 4. So |fi| ≤ |F | for j = 1, 2, · · · , q.Case 3 R ⊂ R′.Then F is a cycle of G2. By Lemma 1.1, |F | ≥ |fj| for j = 1, 2, · · · , q.Case 4 R′ ∩R 6= ∅ and R′ is not in the interior of R.Then F must has at least one edge in E(G2)\E(C) and at least three edges in E(G1). So|F | ≥ 4. No loss of generality, suppose f1, f2, · · · , fp are cycles of {f1, f2, · · · , fq} that are inthe exterior of C. Since |fj| = 4, |F | ≥ |fj | for j = 1, 2, · · · , p.Next we prove |F | ≥ |fj | for j = p+ 1, p+ 2, · · · , q, where fj is in the interior of C.uu u u u u uuuuuR1 R2Riviui Qi1 Qi2PiyxFig. 3.2Let R” = R\ (R′∩R). R” may be the union of several regions. Let R” = R1∪R2∪· · ·∪Rlsatisfying the condition that Ri ∩ Rj is empty or a point for i 6= j, 1 ≤ i, j ≤ l. Let Bi be theboundary of Ri for i = 1, 2, · · · , l. Then Bi is a cycle in the exterior of C. For a fixed Bi, theremay be many vertices of Bi in V (F ) ∩ V (C), which can be found in Fig.3.2. We select twovertices ui and vi of Bi satisfying the following conditions:64 Dengju Ma and Han Ren(1) ui and vi are in C;(2) there is a path of Bi, say Pi, such that its endvertices are ui and vi and Pi is in theexterior of C;(3) if Mi is the path of Bi deleted E(Pi), and if M′i is the path of C such that itsendvertices are ui and vi and M′i is internally disjoint from Bi, then Mi is in the interior of thecycle which is the union of M ′i and Pi.Note that Mi may contains many disjoint paths of C, suppose they are Qi1, Qi2, · · · , Qit.Let x, y be two vertices in Pi, which are adjacent to ui, vi respectively.Obviously, x, y are in G1. Let P′i be the subpath of Pi between x and y. Considering thenumber of radial lines (including radial line x, y lie on) which P ′i crosses is not less than thenumber of vertices of ∪tj=1Qij , |Pi| > |P ′i | ≥∑tj=1 |Qij |.Since R′ ∩R may be the union of some regions, we suppose R′ ∩R = D1 ∪D2 ∪ · · · ∪Ds.Let A1, A2, · · · , As be boundaries of D1, D2, · · · , Ds respectively. For a fixed Ai, its edges maybe partitioned into two groups, one containing edges of F , denoted as AFi , another containingedges of C, denoted as ACi . Then∑si=1|Ai| =∑si=1|AFi |+∑si=1|ACi |=∑si=1|AFi |+∑li=1∑tj=1|Qij |<∑si=1|AFi |+∑si=1|Pi|< |F |Hence |F | > |Ai| for i = 1, 2, · · · , s. Since any Ai is a cycle of G2 and F1 is a minimumcycle base of G2, |Ai| ≥ |fj | for j = i1, i2, · · · , in, by lemma 2.1, where {i1, i2, · · · , in} ⊂{p+ 1, p+ 2, · · · , q}. Hence, |F | > |fp+j | for i = 1, 2, · · · , q − p.By the previous discussion and Lemma 1.1, F is a minimum cycle base of G. Since the minimum cycle base of a cycle is itself, a minimum cycle base of an r× s(r ≥ 4)cylinder embedded in the plane is the set of its interior facial cycles by Theorem 3.1, and thelength of its MCB is r + 4r(s− 1) = r(4s− 3).By Lemmas 1.2, 1.3 and Theorem 3.1. we get two corollaries following.Corollary 3.1 Assume an r × s(r ≥ 4) cylinder, a Halin graph H(T ) are embedded in theplane with C the central cycle and C′ the leaf-cycle of H(T ) containing the same vertices as C,respectively. Let G be the graph obtained from the r × s cylinder and H(T ) by identifying Cand C′ such that H(T ) is in the interior of the r × s cylinder. Then a minimum cycle base ofG is the set of interior facial cycles of G.Corollary 3.2 Assume an r×s(r ≥ 4) cylinder, a 2-connected outplanar graph H be embeddedin the plane with C the central cycle and C′ the exterior facial cycle containing same verticesas C of H containing the same vertices as C, respectively. Let G be the graph obtained from ther×s cylinder and H by identifying C and C′ such that H is in the interior of the r×s cylinder.Then a minimum cycle base of G is the set of interior facial cycles of G. Furthermore, thelength of a MCB of G is r(4s− 5) + 2|E(H)| .Minimum Cycle Base of Graphs Identified by Two Planar Graphs 65Proof Let F be the set of interior facial cycles of G. By Theorem 3.1, F is a minimumcycle base of G. F can be partitioned into two groups F1 and F2, where F1 is the set ofinterior facial cycles of H and F2 the set of 4-cycles. Then the length of a MCB of G isl(F) = l(F1) + l(F2) = 4r(s− 1) + 2|E(H)| − |V (H)| = (4s− 5)r + 2|E(H)|. As application of Corollary 3.1, we find a formula for the length of minimum cycle base ofa planar graph N(d, λ), which can be found in paper[10].When λ ≥ 1 is an integer, the graph Yλ is tree as shown in Fig.3.3. Thus Yλ has 3× 2λ−11-valent vertices and Yλ has 3× 2λ− 2 vertices. If 1-valent vertices of Yλ are connected in theirorder in the planar embedding, we obtain a special Halin graph, denoted by H(λ).Suppose a (3 × 2λ−1) × d cylinder is embedded in the plane such that its central cycle Chas 3×2λ−1 vertices. The graph obtained from (3×2λ−1)×d cylinder and H(λ) with leaf-cycleC′ containing 3 × 2λ−1 vertices by identifying C and C′ such that H(λ) is in the interior of(3 × 2λ−1) × d cylinder is denoted as N(d, λ). N. Roberterson and P.D. Seymour[10] provedthat for all d ≥ 1, λ ≥ 1 the graph N(d, λ) has tree-width ≤ 3d+ 1.bb""uu uu ZZ aaLL##"" QQu uuuuu u uuuY1 Y2Fig.3.3Theorem 3.2 The length of minimum cycle base of N(d, λ)(λ ≥ 2) is 3(d− 1)× 2λ+1 + 9×2λ − 3× 2λ−1 − 6.Proof Let F be the set of interior facial cycles of N(d, λ). Then F is a minimum cyclebase of N(d, λ) by Corollary 3.1.Let F1 be a subset of F which is the set of interior facial cycles of N(1, λ) (a Halin graph).Then F1 consists of 3 (2λ+1)-cycles and 3×2j (2λ−2j−1)-cycles for j = 0, 1, 2, · · · , λ−2.Let F2 = F \ F1. Then each cycle of F2 has length 4. Since the leaf-cycle of N(1, λ) has3× 2λ−1 vertices, there are 3(d− 1)× 2λ−1 4-cycles in F2 all together. The length of F isl(F) =∑λ−2j=03× 2j−1(2λ− 2j − 1) + 3(2λ+ 1) + 4× 3(d− 1)× 2λ−1= 3[∑λ−2j=0λ2j+1 − 2∑λ−2j=0j2j −∑λ−2j=02j ] + (6λ+ 3) + 3(d− 1)× 2λ+1= 3[(λ2λ − 2λ)− 2(λ− 3)2λ−1 − 4− 2λ−1 + 1]+ (6λ+ 3) + 3(d− 1)× 2λ+1= 3(d− 1)× 2λ+1 + 9× 2λ − 3× 2λ−1 − 6Hence, the length of minimum cycle base ofN(d, λ) is 3(d−1)×2λ+1+9×2λ−3×2λ−1−6.66 Dengju Ma and Han RenReference[1] Bondy J.A. and Murty U.S.R., Graph Theory with Application, Macmillan, London, Else-vier, New York, 1976.[2] Downs G.M., Gillet V.J., Holliday J.D. and Lynch M.F., Review of ring perception algo-rithms for chemical graph, J. 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A polynomial-time algorithm to find the shortest cycle base of a graph, SIAM.J.Comput.,

Automorphisms groups of Maps, Surfaces and Smarandache geometries,

Generating cycle spaces for graphs on surfaces with small genera,

Graph Theory with Application,

Graphs on Surfaces, Johns Hopkins

Minimum cycle bases of Halin graphs,

Minimum cycle bases of outerplanar graphs,

Minimum Cycle Bases of Product Graphs,

On the connectivities of tree graphs,

Planar tree-width,

Review of ring perception algorithms for chemical graph,

Short cycle structures for graphs on surfaces and an open problem of Mohar and Thomassen, Science in China: Series A math.,

Union of all the minimum cycle bases of a graph,

Minimum Cycle Base of Graphs Identified by Two Planar Graphs

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Minimum Cycle Base of Graphs Identified by Two Planar Graphs

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