EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 953-964 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Basis Number of Mycielski’s Graph for Some Cog-Graphs Barah M. Sulaiman1, Rasha S. Hasan1, Raghad A. Mustafa1,∗ 1 Department of Mathematics, College of Computer Sciences and Mathematics, University of Mosul, Mosul, Iraq Abstract. Let G = (V,E) be a simple connected graph, then the basis number of G is denoted by b(G) and is defined by the least positive integer k such that the graph G has a k− fold basis for it is cycle space. In this paper we studied the basis number of Mycielski’s graph for some cog-special graphs, and we compute the basis number of Mycielski’s graph for cog-path graph, cog-cycle graph, cog-star graph, and cog-wheel graph. 2020 Mathematics Subject Classifications: 05C10, 05C25, 05C35 Key Words and Phrases: Basis number, k − fold, Mycielski’s graph 1. Introduction Let G be a connected graph with edges sets {e1, e2, . . . , eq}. For each subset S of edges of the graph G, there is a vector (a1, a2, a3, . . . , aq) corresponding to S such that ai = 1 if ei ∈ S and ai = 0 if ei /∈ S. These vectors form a vector space of dimension q on the field Z2, called the vector space associated with the graph G and denoted by (z2) q. The vectors of (z2) q that correspond to the cycles of G generate a vector subspace called the cycles space of G and denoted by C(G). Each vector in C(G) represents either a cycle in G or the union of separate cycles with respect to the edges. A known corollary of graph theory is that a dimension of C(G) is q − p + 1 where p represents the number of vertices of graph G and q the number of edges. The method for finding the base for the cycles space of C(G) is as follows: Let T be a generating tree for the graph G; If the edge ei belongs to G−T then T + ei contains only one cycle, let it be Cei . Clearly, q − p + 1 of cycles Cei , where ei ∈ G − T for i = 1, 2, . . . , q forms the base of the cycles space C(G). The base B of cycles space C(G) is said to have a k− fold if each edge of G shows no more than k times (iterations) in the cycles that corresponding to the vectors in the base B. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4704 Email addresses: barah mahmood82@uomosul.edu.iq (B. M. Sulaiman), sallal-rasha@uomosul.edu.iq (R. S. Hasan), raghad.math@uomosul.edu.iq (R. A. Mustafa) https://www.ejpam.com 953 © 2023 EJPAM All rights reserved. B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 954 The basis number of the graph G is defined as the smallest integer k, such that C(G) have a k− fold base; It is denoted by b(G). If B is the base of the cycles space C(G) and e is an edge in G, then the fold of the edge e in B is defined the number of cycles that exist in B and containing the edge e, and is denoted by fB(e). In recent years, interest in the basic number has increased, we refer the reader to references [3–6, 9, 10, 13] for more information. In this paper, we will assume that all graphs that we encounter are finite, unguided and simple; For undefined terms, refer to the references [7][8]. There are other types of numbers that are important in graph theory such as: detour number [1] and number of domination [17], and graph theory has an important applications at the present time, see [11, 14, 15]. Mycielski’s graph [16]: Let G be the graph, such that the set of its vertices is V = {u1, u2, u3, . . . , un}, then the Mycielski’s graph for G consists of G itself as a sub graph isomorphic with (n + 1) additional vertices, the vertex vi corresponding to ui in G, for i = 1, 2, 3, . . . , n; and another vertex w which is adjacent to each vertex vi such that these vertices form a sub graph isomorphic with star K(1,n); In addition, for each edge ui uj in G, the Mycielski’s graph includes two edges ui vj and vi uj , therefore if G is a graph of n vertices and m edges, then the Mycielski’s graph of G has 2n + 1 vertices and 3m + n edges and is denoted by µ(G). Figure (1) represents Mycielski’s graph of the cycle C3. Figure 1: µ(C3) 2. Main Results 2.1. Cog-Path Graph P c m It is a graph consists of a path Pm : u1, u2, . . . , um where m ≥ 3, with m − 1 addi- tional vertices v1, v2, . . . , vm−1 and additional edges {uivi, viui+1, i = 1, 2, . . . ,m−1}. The number of vertices of P c m is 2m− 1 and the number of its edges is 3m− 3 [2]. B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 955 2.1.1. The Basis Number for Mycielski’s Graph of the Cog-Path µ(P c m) Let the vertices of the Cog path graph P c m be u1, u2, u3, . . . , u2m−1 and the vertices opposite to them are v1, v2, v3, . . . , v2m−1 and let the other vertex be w, from the definition of the Mycielski’s graph, it becomes clear that the number of vertices of µ(P c m) is 4m − 1 and the number of its edges are 11m− 10. See Figure (2). Figure 2: µ(P c m) Theorem 1. Let Pm be a path of order m ≥ 3 then b (µ(P c m)) = 3 Proof. We can prove that for each m ≥ 3, there is a subgraph of µ(P c m) that topo- logically equivalent K3,3, according to Kurtowski’s Theorem [8], µ(P c m) is not planar, and according to McLean’s Theorem [12] we have b (µ(P c m)) ≥ 3 (1) We will prove that there is a base B for the cycles space of a graph µ(P c m) with 3-fold. Let B be a set of cycles of µ(P c m) which defined by the following formula: B = ∪5 j=1Mj ∪ {C}, where M1 = {u2i−1u2i+1v2i−1u2iv2i+1u2i−1 : i = 1, 2, 3, . . . ,m− 1}, M2 = {wv2i−1u2i+1v2iw : i = 1, 2, 3, . . . ,m− 1}, M3 = {u2i−1u2iu2i+1u2i−1 : i = 1, 2, 3, . . . ,m− 1}, M4 = {wviui+1vi+2w : i = 1, 2, 3, . . . , 2m− 3}, M5 = {uiui+1ui+2vi+1ui : i = 1, 2, 3, . . . , 2m− 3}, C = {wv1u2u1v2w}. B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 956 In order B to be the base for the cycles space of the graph µ(P c m), it must be |B| = dimC(µ(P c m)) and B must be a linearly independent set. It is known that dimC(µ(P c m)) = 7m− 8 |B| = | ∪5 j=1 Mj |+ 1 = 3(m− 1) + 2(2m− 3) + 1 = 7m− 8. It remains to show that B is linearly independent cycles. Clearly that the cycles of each M1,M2 and M3 are independent because they are separate cycles with respect to edges; and the cycles of each M4 and M5 are independent because it is represent the boundaries of the faces of a planar subgraph. Now; the cycle C is independent of M5 because contains the edges wv1 and wv2 but these edges are not available in any linear combination of cycle M5, hence M5 ∪ {C} is linearly independent. Also, any linear combination of M5∪{C} contains the edges of type ui ui+1,i = 1, 2, . . . , 2m − 2 and these edges are not available in any linear combination of cycle M4, therefore M5 ∪ {C} ∪M4 is linearly independent. Further more M4 ∪M5 ∪ {C}∪M3 are independent set of cycles since any linear combination of cycles M3 contains the edges of type u2i−1 u2i+1, i = 1, 2, . . . ,m − 1 and these edges are not available in M4 ∪M5 ∪{C}.Also,M3 ∪M4 ∪M5 ∪{C}∪M2 are independent set of cycles because M2 contains the edges of type v2i−1 u2i+1, i = 1, 2, . . . ,m− 1 but these edges are not available in M3 ∪M4 ∪M5 ∪ {C}. Finally; the cycles of the set B = ∪5 j=1Mj ∪ {C} are independent since any linear combination of cycles M1 contains the edges of type u2i−1v2i+1, i = 1, 2, . . . ,m − 1 while these edges are not available in any linear combination of cycles M2∪M3∪M4∪M5∪{C}. To find the fold for base B, we divide the edges of the graph µ(P c m) into: E1 = {uiui+1 : i = 1, 2, . . . , 2m− 2} E2 = {uivi+1 : i = 1, 2, . . . , 2m− 2} E3 = {v2i−1u2i : i = 1, 2, . . . ,m− 1} E4 = {v2iu2i+1 : i = 1, 2, . . . ,m− 1} E5 = {wvi : i = 1, 2, . . . , 2m− 1} E6 = {u2i−1u2i+1 : i = 1, 2, . . . ,m− 1} E7 = {u2i−1v2i+1 : i = 1, 2, . . . ,m− 1} E8 = {v2i−1u2i+1 : i = 1, 2, . . . ,m− 1} Now, we calculate the fold for a set of the edges of the graph µ(P c m), Case I: fB(µ(P c m)(e) is less than or equal to 1 when e ∈ E7. Case II: fB(µ(P c m)(e) is less than or equal to 2 for all e ∈ Ei,i = 6, 8. B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 957 Case III: fB(µ(P c m)(e) is less than or equal to 3 for all e ∈ Ei,i = 1, 2, 3, 4, 5. From the above three cases, it can be seen that the fold for each edge in the graph µ(P c m) is not more than 3 in the base B(µ(P c m)); That is b (µ(P c m)) ≤ 3 (2) From (1) and (2), we get b(µ(P c m)) = 3. 2.2. Cog-Cycle Graph Cc m It is a graph conclude from a cycle Cm : u1, u2, . . . , um where m ≥ 3, by adding m vertices and 2m edges of the form v1, v2, . . . , vm and {uivi, ui+1vi : i = 1, 2, . . . ,m}, respectively, where um+1 ≡ u1. It is clear that the number of vertices of a graph Cc m is 2m and the number of edges is 3m [2]. 2.2.1. The Basis Number for Mycielski’s Graph of the Cog-Cycle µ(Cc m) Let the vertices of the cog-cycle graph Cc m are u1, u2, . . . , u2m where m ≥ 3, and the corresponding vertices are v1, v2, . . . , v2m and let the other vertex be w. From the definition of the Mycielski’s graph we have the number of vertices of the graph µ(Cc m) is 4m+1 and the number of its edges is 11m. Theorem 2. Let Cm be a cycle of order m ≥ 3 then b(µ(Cc m)) = 3 Proof. We can prove that for each m ≥ 3, there is a subgraph of µ(Cc m) that topo- logically equivalent K3,3, according to Kurtowski’s Theorem [8] µ(Cc m) is not planar, and according to McLean’s Theorem [12] we have b (µ(Cc m)) ≥ 3 (3) We will prove that there is a base B for the cycles space of the graph µ(Cc m) with 3-fold. Let B be a set of cycles of µ(Cc m) which defined by the following formula: B = B(µ(P c m)) ∪M,where M = {M1,M2, . . . ,M8} Where B(µ(P c m)) is the base for Michelsky’s graph of the cog-path P c m, which defined in the previous theorem, also M is a set of cycles of the graph µ(Cc m) defined as the following formula: M1 = u1u2m−1u2mu1, M2 = u1u2m−1v2mu1, M3 = u1u2m−1v2m−2wv2mu1, M4 = u1u2mv2m−1u1, B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 958 M5 = u1u2mv1u3u1, M6 = u2m−2u2m−1u2mv2m−1u2m−2, M7 = v1u2m−1u2mv1, M8 = v2mu2m−1u2m−3v2m−1wv2m. In order B to be the base for the cycles space of graph µ(Cc m) must be |B| = dimC(µ(Cc m)), and B must be a linearly independent set of cycles. It is known that dim C(µ(Cc m)) = 11m− (2m+ 1) + 1 = 7m, and |B| = |B(µ(P c m))|+ |M | = (7m− 8) + 8 = 7m It remains to show that B is linearly independent. It is known that B(µ(P c m)) is linearly independent because it is represent the base of the cycles space of µ(P c m). In addition, the cycles set M is linearly independent because one of them cannot be written as a linear combination of the other cycles. Finally, the set of cycles B = B(µ(P c m)) ∪ {M1,M2, . . . ,M8} is independent because any linear combination of Mi’s cycles, i = 1, 2, . . . , 8 contains at least one new edge of type u1u2m−1, u1u2m, u2m−1u2m, u1v2m−1, u1v2m, v1u2m−1, v1u2m, u2m−1v2m, v2m−1u2m, wv2m while these edges are not exist in any linear combination for cycles of B(µ(P c m)). To find the fold for the base B we divide the edges of the graph µ(Cc m) into: E1 = E(µ(P c m))− E2 E2 = {u1u3, v1u3, u2m−3u2m−1, u2m−2u2m−1, u2m−2v2m−1, v2m−2u2m−1, wv2m−2, wv2m−1} E3 = {u1u2m−1, u1u2m, u2m−1u2m, u1v2m−1, u1v2m, v1u2m−1, v1u2m, u2m−1v2m, v2m−1u2m, wv2m} Now, we calculate the fold for a set of the edges of the graph µ(Cc m), We note that fB(µ(Cc m))(e) is less than or equal to 3 for all e ∈ Ei, i = 1, 2, 3, thus the fold for each edge in the graph µ(Cc m) is not more than 3 in the base B(µ(Cc m)); That is b (µ(Cc m)) ≤ 3 (4) From (3) and (4), we get b(µ(CC m)) = 3. 2.3. Cog-Star Graph Sc m It is a graph consisted of a star graph Sm : u1, u2, . . . , um−1, um, where m ≥ 4 with m−1 of additional vertices v1, v2, . . . , vm−2, vm−1 and additional edges {uivi+1, uivi+2, i = 1, 2, . . . ,m− 1}, where vm+1 ≡ v2 [2]. It is clear that the number of vertices of a graph Sc m is 2m−1 and the number of edges is 3m− 3. B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 959 2.3.1. The Basis Number for Mycielski’s Graph of the Cog-Star µ(Sc m) Let the vertices of the cog-star graph Sc m are u1, u2, . . . , u2m−1 and the corresponding vertices are v1, v2, . . . , v2m−1 and the other vertex is w. By Mycielski’s definition, it turns out that the number of vertices of µ(Sc m) is 4m− 1 and the number of its edges is 11m− 10. Theorem 3. Let Sm be a star of order m ≥ 4 then b(µ(Sc m)) = 3. Proof. We can prove that for each m ≥ 4, there is a subgraph of µ(Sc m) that topo- logically equivalent K3,3, according to Kurtowski’s Theorem [8], µ(Sc m) is not planar and according to McLean’s Theorem [12] we have b (µ(Sc m)) ≥ 3 (5) Let B be a set of cycles of µ(Sc m) which defined by the following formula: B = B(µ(Sc m)) = ∪5 i=1Si ∪ {C1, C2, C3, C4, C5, C6} Where S1 = {uiui+1ui+2vi+1ui : i = 1, 2, 3, . . . , 2m− 3}, S2 = {wviui+1vi+2w : i = 1, 2, 3, . . . , 2m− 3}, S3 = {u2m−1viui+1vi+2u2m−1, i = 2, 4, 6, . . . , 2m− 4}, S4 = {u2m−1uiv2m−1ui+2u2m−1, i = 2, 4, 6, . . . , 2m− 6}, S5 = {u2m−1viwvi+1ui+2u2m−1, i = 2, 4, 6, . . . , 2m− 6}, C1 = u2m−2u1u2u2m−1u2m−2, C2 = u2m−2v1wv2m−1u2m−2, C3 = v2m−2u1u2m−2u2m−1v2m−2, C4 = v1u2m−2u1v2m−2wv1, C5 = v2m−2u1v2wv2m−2, C6 = u2m−1v2m−4wv2m−3u2m−4u2m−1 In order B to be the base for the cycles space of the graph µ(Sc m), it must be |B| = dim C(µ(Sc m)), and B must be a linearly independent set of cycles. It is known that dim C(µ(Sc m)) = 7m− 8, and |B| = |B(µ(Sc m))| = | ∪5 i=1 Si|+ |{C1, C2, C3, C4, C5, C6}| = (7m− 14)+ 6 = 7m− 8, since |S1| = |S2| = 2m− 3 and |S3| = m− 2, |S4| = |S5| = m− 3. It remains to show that B is linearly independent. It is clear that each of S1, S2, S3, S4 and S5 is linearly independent because it is rep- resent the boundaries of the faces of a planar subgraph. S1 ∪ S2 is linearly independent B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 960 because any linear combination of S2 contains edges of type wvi, i = 1, 2, . . . , 2m−1, which are not found in any linear combination of S1. Also, S3∪S4 is linearly independent because any linear combination of S4 contains edges of type u2m−1ui, i = 2, 4, . . . , 2m−4, which are not found in any linear combination of S3. In addition, S3∪S4∪S5 is linearly independent since any linear combination of S5 contains edges of type wvi, i = 2, 4, . . . , 2m− 1, which are not found in any linear combination of S3∪S4. Also, (S1∪S2)∪(S3∪S4∪S5) is linearly independent because S3 ∪ S4 ∪ S5 contains edges of type u2m−1ui, i = 2, 4, . . . , 2m − 4, which are not found in any linear combination of S1 ∪ S2. Finally, (∪5 i=1Si)∪ ({C1, C2, C3, C4, C5, C6}) is linearly independent because any linear combination of {C1, C2, C3, C4, C5, C6} contains edges of type u2m−2u1, v2m−2v1, which are not found in any linear combination of ∪5 i=1Si. To find the fold for the base B we divide the edges of the graph µ(Sc m) into: E1 = {uiui+1, i = 2, 3, . . . , 2m− 3}, E2 = {uivi+1, i = 2, 3, . . . , 2m− 5} ∪ {u2m−3v2m−2} E3 = {viui+1, i = 1, 2, . . . , 2m− 3} ∪ {u2m−1vj , j = 2, 4, . . . , 2m− 6} E4 = {wvi, i = 3, 4, . . . , 2m− 5} E5 = {u2m−1ui, i = 4, 6, . . . , 2m− 6} ∪ {v2m−1uj , j = 2, 4, . . . , 2m− 4} E6 = {v2m−2u1, u2m−2u1, u2m−2v1} E7 = {u2m−2v2m−1, u1u2, wv2m−1, u2m−1u2, u1v2} E8 = {wv1, wv2m−2, u2m−2u2m−1} E9 = {wv2, wv2m−4, wv2m−3, u2m−4v2m−3, u2m−1u2m−4, u2m−1v2m−2, u2m−1v2m−4} Now, we calculate the fold for a set of the edges of the graph µ(Sc m), Case I: fB(µ(Sc m))(e) is equal to 2 for all e ∈ Ei, i = 1, 7. Case II: fB(µ(Sc m))(e) is less than or equal to 3 for all e ∈ Ei, i = 2, 3, 4, 5, 6, 8, 9. From the above two cases, it can be seen that the fold for each edge in the graph µ(Sc m) is not more than 3 in the base B(µ(Sc m)); That is b (µ(Sc m)) ≤ 3 (6) From (5) and (6), we get b(µ(Sc m)) = 3. 2.4. Cog-Wheel Graph W c m It is a graph consisted of a wheel Wm : u1, u2, . . . , um where m ≥ 4, by adding m− 1 vertices and 2m−2 edges of the form v1, v2, . . . , vm−1 and {viui, viui+1 : i = 1, 2, . . . ,m−1} respectively, where um ≡ u1. It is clear that the number of vertices of a graphW c m is 2m−1 and the number of edges is 4m− 4 [2]. B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 961 2.4.1. The Basis Number for Mycielski’s Graph of the Cog-Wheel µ(W c m) Let the vertices of the cog-wheel graph wc m are u1, u2, . . . , u2m−1 and the corresponding vertices are v1, v2, . . . , v2m−1 and the other vertex is w, since the number of vertices of cog-wheel is 2m−1 and the number of its edges is 4(m−1), then by Mycielski’s definition, it turns out that the number of vertices of µ(wc m) is 4m − 1 and the number of its edges is 14m− 13. Theorem 4. Let Wm be a wheel of order m ≥ 5 then b(µ(W c m)) = 3. Proof. We can prove that for each m ≥ 5, there is a subgraph of µ(W c m) that topo- logically equivalent K3,3, according to Kurtowski’s Theorem [8] µ(W c m) is not planar and according to McLean’s Theorem [12] we have b (µ(W c m)) ≥ 3 (7) We will prove that there is a base B for the cycles space of the graph µ(W c m) of 3-fold. Let B be a set of cycles of µ(W c m) which defined by the following formula: B = B(µ(P c m)) ∪ (∪3 i=1Si) ∪ {C1, C2, C3} Where B(µ(P c m)) is the base for Michelsky’s graph of the cog-path, m ≥ 5 and S1 = {u2m−1viui+2u2m−1 : i = 1, 3, 5, . . . , 2m− 5}, S2 = {v2m−1uiui+2v2m−1 : i = 1, 3, 5, . . . , 2m− 5}, S3 = {u2m−1uivi+2u2m−1 : i = 1, 3, 5, . . . , 2m− 5}, C1 = v2m−1u1u2m−3v2m−1, C2 = u1v2m−3u2m−5u2m−1u1 C3 = u2m−3v1u2m−1u1u2m−3, In order B to be the base for the cycles space of the graph µ(W c m), it must be |B| = dim C(µ(W c m)), and B must be a linearly independent set of cycles. Clearly, dim C(µ(W c m)) = 10m− 11, and since |B| = |B(µ(P c m))|+ | ∪3 i=1 Si|+ |{C1, C2, C3}| = 7m− 8 + 3m− 6 + 3 = 10m− 11 Now, it remains to show that B is linearly independent. It is known that B(µ(P c m)) is linearly independent because it is represent the base of the cycles space of µ(P c m). Note that each of S1, S2andS3 is linearly independent be- cause it is represent the boundaries of the faces of a planar subgraph. Now, S1 ∪ S2 is linearly independent because any linear combination of S2 contains edges of type B. M. Sulaiman, R. S. Hasan, R. A. Mustafa / Eur. J. Pure Appl. Math, 16 (2) (2023), 953-964 962 uiui+2, i = 1, 3, 5, . . . , 2m− 3, which are not found in any linear combination of S1. Now, S1 ∪ S2 ∪ S3 is linearly independent because any linear combination of S3 contains edges of type uivi+2, i = 1, 3, . . . , 2m − 5, which are not found in any linear combination of S1 ∪ S2. In addition, {C1, C2, C3} is linearly independent because we cannot write any one of them as a linear combination of the others cycles. Now, (∪3 i=1Si)∪ ({C1, C2, C3}) is linearly independent because any linear combination of {C1, C2, C3} contains at least one of the edges u1u2m−3, u1v2m−3, v1u2m−3, which are not found in any linear combination of ∪3 i=1Si. Finally, the set of cycles B = B(µ(P c m))∪ (∪3 i=1Si)∪ ({C1, C2, C3}) is linearly indepen- dent because any linear combination of cycles in (∪3 i=1Si) ∪ ({C1, C2, C3}) contains edges of type u2m−1ui, i = 1, 3, . . . , 2m − 5, which are not found in any linear combination in B(µ(P c m)), therefore B(µ(W c m)) is linearly independent. To find the fold for the base B we divide the edges of the graph µ(W c m) into: E1 = {uiui+1, i = 1, 2, . . . , 2m− 2} E2 = {uivi+1, viui+1, i = 1, 2, . . . , 2m− 2} E3 = {wvi, i = 1, 2, . . . , 2m− 1} E4 = {uivi+2, i = 1, 3, . . . , 2m− 3} E5 = {viui+2, i = 1, 3, . . . , 2m− 3} E6 = {uiui+2, i = 1, 3, . . . , 2m− 3} E7 = {u2m−1ui, i = 1, 3, . . . , 2m− 5} E8 = {v2m−1ui, u2m−1vi, i = 1, 3, . . . , 2m− 5} E9 = {u1v2m−3, v1u2m−3, u1u2m−3} Now, we calculate the fold for a set of the edges of the graph µ(W c m), Case I: fB(µ(W c m))(e) is less than or equal to 2 for all e ∈ Ei, i = 8, 9. Case II: fB(µ(W c m))(e) is less than or equal to 3 for all e ∈ Ei, i = 1, 2, . . . , 7. From the above two cases, it can be seen that the fold for each edge in the graph µ(W c m) is not more than 3 in the base B(µ(W c m)); That is b(µ(W c m)) ≤ 3 (8) From (7) and (8), we get b(µ(W c m)) = 3. 3. 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