Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 253 https://internationalpubls.com A Study on Emergence of Clutch Graphs from Cycle Graphs: A Comprehensive Analysis Tamilselvi V1, Thamizhendhi G2 1 Assistant Professor of Mathematics, Vellalar College for Women(Autonomous), Erode, Tamilnadu, India, Email id: tmlselvi18@gmail.com 2 Assistant Professor of Mathematics, Sri Vasavi College, Erode, Tamilnadu, India, Email id: gkthamil@gmail.com Article History: Received: 28-10-2023 Revised: 09-12-2023 Accepted: 24-12-2023 Abstract: The clutch is one of the substantial devices for constructing vehicles in automobile engineering. In terms of network toughness against failures or disruptions, the clutch- based graph can be used to model insufficiency of primary pathways and the availability of alternative routes in communication networks. In this paper, the identification of the clutch graph Cl3n(G) from the cycle graph Cn(G) has been proposed. A clutch graph Cl3n(G) generating from a cycle graph with 3n vertices and 4n edges (n ≥ 4 and even). The notions of degree, girth, and chromatic number of the clutch graph have been discussed. Further, the existence of the bipartite and Hamiltonian graphs on the clutch graph has been examined. Keywords: Clutch graph, Cycle graph, Spanning tree, Bipartite, Hamiltonian. 1. Introduction Graph theory explores connections between vertices and edges, offering a way to under- stand and analyze various real-world systems [6]. A cycle in graph theory is a closed path that starts and ends at the same vertex. A cycle graph is a graph composed of a single cycle, forming a circular structure [7]. A clutch is a mechanical device that connects or dis- connects power transmission, enabling smooth control over energy transfer [8]. In this way, the authors motivated to define clutch based graph that is used to design networks. Clutch graph based network models effective in identifying the smooth control of data transmission between entities. The clutch graph inherits the properties of bipartite and Hamiltonian graphs. The chromatic number and girth have been explored with appropriate illustration. 2. Preliminaries Definition 2.1:[1] The cycle Cn , n ≥ 3, made up of n vertices c1, c2, . . . , cn and edges {c1, c2},{c2, c3}, . . . ,{cn−1, cn},{cn, c1}. The cycles c3 and c4 are shown in Figure 1 and Figure 2. mailto:tmlselvi18@gmail.com mailto:gkthamil@gmail.com Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 254 https://internationalpubls.com Definition 2.2:[1] A graph with a single cycle (at least 3 vertices) connected in a closed chain is referred to as a cycle graph Cn or circular graph. In Cn, the number of edges is equal to the number of vertices, and each vertex has degree 2. Definition 2.3:[2] The number of edges that are incident on a vertex is the degree of the graph. Definition 2.4:[5] The length of shortest cycle in the graph is said to be girth. Definition 2.5:[1] The least number of colors that allows to be colored differently for the adjacent vertices is said to be chromatic number. 3. Main Results Definition 3.1: A Clutch Graph Cl3n(G) = (V, E) is a type of graph that can be de- rived from the cycle graph. The construction of a clutch graph Cl3n(G) involves specific steps as described below. • Start with a cycle graph Cn(G) = (Vc, Ec), with vertex set Vc = {c1, c2, . . . , cn},where (n ≥ 4, even) is the number of vertices and edge set, Ec = {(ci, ci+1)|i ∈ {1, 2, . . . , n − 1}} ∪ (cn, c1) • Add a set of vertices, Vp = {p1, p2, . . . , pn} to the existing vertex set Vc • Establish the edge set, Ecp by joining each vertex ci with its corresponding pi, Ecp = {(ci, pi)|i ∈ {1, 2, . . . , n}} • Form an edge set, Ep = {(pi, pi+1)|i ∈ {1, 3, 5, . . . , n − 1}} • Create another set of vertices, Vq = {q1, q2, . . . , qn} and add to the vertex set Vp. The resulting vertex set be Vcpq = {c1, c2, . . . , cn, p1, p2, . . . , pn, q1, q2, . . . , qn} • Built an edge set Epq = {(pi, qi)|i ∈ {1, 2, . . . , n}} connecting corresponding vertices from Vp and Vq • Finally, create an edge set, Eq = {(qi, qi+1)|i ∈ {2, 4, . . . , n}} ∪ (qn, q1)} By following these steps, the clutch graph Cl3n(G) have the vertex set V = Vc ∪ Vp ∪ Vq and edge set E = Ec ∪ Ecp ∪ Ep ∪ Epq ∪ Eq. The resulting clutch graph is characterized by with |V | = 3n vertices and |E| = 4n edges (n ≥ 4 & even). Example: The clutch graph Cl18(G) in Fig: 3 and Cl24(G) in Fig: 4 which have been constructed from the cycle graph C6(G) and C8(G) respectively. c4 c3 c3 c1 c2 c1 c2 Figure 1: C3 (G) Figure 2: C4 (G) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 255 https://internationalpubls.com Figure 3: Cl18 (G) Figure 4: Cl24 (G) Definition 3.2: The degree of the clutch graph is represented by deg Cl3n(G). The degree of the vertices ci in Vc and pi in Vp of the clutch graph are denoted by s. The degree of the vertices qi in Vq is denoted by t. 𝑑𝑒𝑔𝑐𝑖 ∈ 𝑉𝑐 𝐶𝑙3𝑛(𝑐𝑖) = s = 3 𝑑𝑒𝑔𝑝𝑖 ∈ 𝑉𝑝 𝐶𝑙3𝑛(𝑝𝑖) = s = 3, and 𝑑𝑒𝑔𝑞𝑖 ∈ 𝑉𝑞 𝐶𝑙3𝑛(𝑞𝑖) = t = 2 Definition 3.3: The girth g of a clutch graph (Cl3n(G)) is always 4. g = {(𝑐2𝑖−𝑖, 𝑝2𝑖−1, 𝑝2𝑖, 𝑐2𝑖)|𝑖 ∈ 1, 2, . . . , 𝑛 2 } Definition 3.4: The chromatic number for clutch graph χ(Cl3n(G)) is 2. 𝜒(𝐶𝑙3𝑛(𝐺)) = { 1 𝑖𝑓 {(𝑐2𝑖−1, 𝑝2𝑖, 𝑞2𝑖−1)|𝑖 ∈ {1, 2, . . . , 𝑛 2 } 2 𝑖𝑓 {(𝑐2𝑖, 𝑝2𝑖−𝑖, 𝑞2𝑖)|𝑖 ∈ {1, 2, . . . , 𝑛 2 } Here 1,2 represents the colors assign to the vertices of the clutch graph. 4. Some properties of clutch graph Theorem 4.1 For every clutch graph, the sum of the degrees of vertices is equal to twice the number of edges. Proof Let us consider a clutch graph Cl3n(G) with vertex set Vcpq where Vcpq =(Vc∪VP∪Vq). Each vertex ci in Vc and pi in VP have degrees 3 and each vertex qi in Vq have degrees 2. ∑ 𝑑𝑒𝑔(𝑐𝑖) 𝑐𝑖 ∈ 𝑉𝑐 + ∑ 𝑑𝑒𝑔(𝑝𝑖) 𝑝𝑖 ∈ 𝑉𝑝 + ∑ 𝑑𝑒𝑔(𝑞𝑖) 𝑞𝑖 ∈ 𝑉𝑞 = 3n + 3n + 2n = 8n = 2(4n) = 2e. Example: In Figure 3, the clutch graph Cl18 (G) with n = 6 vertices. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 256 https://internationalpubls.com ∑ 𝑑𝑒𝑔(𝑐𝑖) 𝑐𝑖 ∈ 𝑉𝑐 + ∑ 𝑑𝑒𝑔(𝑝𝑖) 𝑝𝑖 ∈ 𝑉𝑝 + ∑ 𝑑𝑒𝑔(𝑞𝑖) 𝑞𝑖 ∈ 𝑉𝑞 == 18 + 18 + 12 = 48 = 2(24) = 2e. Theorem 4.2 Let Cl3n (G) be a clutch graph with vertices of degrees s (or) t, then show that n(2s + t) = 2 ∈, n is the number of vertices in the cycle graph. Proof Given that Cl3n (G) is a clutch graph whose vertices have degree s or t. The vertices in Vc and VP have degrees s and Vq have degrees t respectively. By Theorem [4.1], given that, ∑ deg(ci) ci ∈ Vc + ∑ deg(pi) pi ∈ Vp + ∑ deg(qi) qi ∈ Vq = 2e 𝑛𝑠 + 𝑛𝑠 + 𝑛𝑡 = 2 ∈ 𝑛(2𝑠 + 𝑡) = 2 ∈ Hence proved Theorem 4.3 Every clutch graph Cl3n (G) is constructed with 2n vertices having of odd degree in Vc and Vp, and n vertices of even degree in Vq, results in a connected graph. Proof Given that s = 3 be the degree of each vertex ci in Vc and each vertex pi in Vp. Then total sum of odd degrees for these 2n vertices is ∑ deg(𝑐𝑖) + 2𝑛 𝑖=1 ∑ deg (𝑝𝑖) 2𝑛 𝑖=1 = 2𝑛. 𝑠 = 6𝑛 Also t = 2 be the degree of each vertex qi in Vq. The total sum of even degrees for these n vertices is ∑ deg(𝑞𝑖) = 𝑛. 𝑡 = 2𝑛.𝑛 𝑖=1 The total sum of degrees for all vertices is 6n + 2n = 8n. According to the Handshaking Lemma, in a graph ∑ deg 𝐶𝐼𝑣𝜖𝑉 3n(G)=2.|E| = 2.4n ∑ deg 𝐶𝐼𝑣∈𝑉 3n (G) = 8n The fact that ∑ deg 𝐶𝐼𝑣∈𝑉 3n (G) = 8n implies |𝐸| = 1 2 ∑ deg𝐶𝐼𝑣∈𝑉 3n (G) = 4𝑛 edges. This ensures that every vertex is incident to at least one edge, establishing connectivity in the clutch graph. Theorem 4.4 For every clutch graph, χ + g ≤ n + 2, where χ is chromatic number, g is girth and n is number of vertices. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 257 https://internationalpubls.com Example: In Fig: 3, the clutch graph Cl18 (G) with n =6 vertices, resulting in 3n=18 vertices. For every clutch graph, the chromatic number χ is 2, and the girth g is 4. Therefore, 𝜒 + 𝑔 = 2 + 4 = 6 𝑛 + 2 = 6 + 2 = 8 The example satisfies the inequality χ + g ≤ n + 2, illustrating the validity of the theorem for this clutch graph with 𝑛 = 6. Theorem 4.5 In clutch graph Cl3n (G), the number of vertices with odd degree is even. Proof The clutch graph Cl3n (G) has 3n vertices. Since, the sum of degrees of all vertices in the clutch graph is 2e, e is the number of edges. ∑ deg(𝑣𝑖) 𝑛 𝑖=1 = 2e, vi ∈ V = Vc ∪ VP ∪ Vq ∑ 𝑑𝑒𝑔(𝑐𝑖) 𝑐𝑖 ∈ 𝑉𝑐 + ∑ 𝑑𝑒𝑔(𝑝𝑖) 𝑝𝑖 ∈ 𝑉𝑝 + ∑ 𝑑𝑒𝑔(𝑞𝑖) 𝑞𝑖 ∈ 𝑉𝑞 = 2e In the clutch graph, each vertex in Vc and Vp with degree 3, and each vertex in Vq with degree 2. Let n be the number of vertices of the vertex set Vc , Vp and Vq respectively. 3n + 3n + 2n = 2e 8n = 2e Here the sum of degrees is even, and the degrees in Vq are all even, then the sum of degrees in Vc and Vp (which are all odd) must be even. It follows that, the number of vertices with odd degree is even. Theorem 4.6 Every clutch graph Cl3n (G) has 78[4n − k]k+1 spanning trees. Proof Step 1: Consider the clutch graph Cl3n (G) with the vertex set V = Vc ∪ Vp ∪ Vq and edge sets E = Ec ∪ Ecp ∪ Ep ∪ Epq ∪ Eq. Step 2: The degrees of the vertices in V are 𝑑𝑒𝑔(𝑣𝑖) = { 3, 𝑖𝑓 𝑣𝑖 ∈ 𝑉𝑐 ∪ 𝑉𝑝 2, 𝑖𝑓 𝑣𝑖 ∈ 𝑉𝑞 Then, D(Cl3n (G)) is the diagonal matrix of vertex degrees which is given by D(Cl3n (G)) = [ 𝐷𝑐𝑐 0 0 0 𝐷𝑝𝑝 0 0 0 𝐷𝑞𝑞 ] Step 3: Form an adjacency matrix based on the edges of the clutch graph 𝐴(𝐶𝑙3𝑛(𝐺)) = { 1 , if there is an edge between vertices 𝑖 and 𝑗 0, otherwise i Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 258 https://internationalpubls.com 𝐴(𝐶𝑙3𝑛(𝐺)) = [ 𝐴𝑐𝑐 𝐴𝑐𝑝 0 𝐴𝑐𝑝 𝐴𝑝𝑝 𝐴𝑝𝑞 0 𝐴𝑃𝑞 𝐴𝑞𝑞 ] Step 4: The Laplacian matrix is defined as L(Cl3n (G)) = D(Cl3n (G)) - A(Cl3n (G)). So 𝐿(𝐶𝑙3𝑛(𝐺)) = [ 𝐷𝑐𝑐 − 𝐴𝑐𝑐 − 𝐴𝑐𝑝 0 −𝐴𝑐𝑝 𝐷𝑝𝑝 − 𝐴𝑝𝑝 − 𝐴𝑝𝑞 0 − 𝐴𝑃𝑞 𝐷𝑞𝑞 − 𝐴𝑞𝑞 ] Then, by Kirchoff’s theorem, the number of distinct spanning tree of the graph is equal to any cofactors of its Laplacian matrix. Hence, the cofactor of L(Cl3n (G)) is calculated by using (−1)i+j · det(L) that remains after deleting ith row and jth column. Therefore, the clutch graph has n distinct spanning trees. Example: Let’s find the cofactors of Laplacian matrix for the clutch graph Cl12 (G) using Jupyter notebook software. Start with the construction of Cl12 (G). import matplotlib.pyplot as plt import numpy as np #Points points={ ’c1’ : (1.48, 0.41), ’c2’ : (2.44, 0.41), ’c3’ : (2.46, -0.41), ’c4’ : (1.46, -0.43), ’p1’ : (0.88, 0.79), ’q2’ : (3.46, 1.39), ’p3’ : (2.86, -0.81), ’p4’ : (0.84, -0.79), ’q1’ : (0.08, 1.37), ’q3’ : (3.46, 1.37), ’p2’ : (2.88, 0.81), ’q4’ : (0.06, -1.39) } # Plot points for point, coordinates in points.items(): plt.scatter(*coordinates, label=point) # Connect points with lines lines = [ Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 259 https://internationalpubls.com (‘c1’,’c2’,’c3’,’c4’,’c1’), (’c1’, ’p1’), (’c2’, ’p2’), (’c3’, ’p3’), (’c4’, ’p4’), (’p1’, ’p2’), (’p3’, ’p4’), (’p1’, ’q1’), (’p2’, ’q2’), (’p3’, ’q3’), (’p4’, ’q4’), (’q2’, ’q3’), (’q4’, ’q1’) ] for line in lines: plt.plot(*zip(*[points[point] for point in line])) # Label points for point, coordinates in points.items(): plt.annotate(point, coordinates,textcoords="offset points",xytext=(0, 5), ha=’center’) # Add title below the graph fig.text(0.5, 0.02, ’Graph Diagram’, ha=’center’, fontsize=12) plt.legend() plt.grid(True) plt.show() Figure 5: Cl12 (G) First create a Diagonal matrix D(Cl3n(G)). import networkx as nx import numpy as np # Create a graph G = nx.Graph() edges = [(1, 2), (2, 3), (3, 4), (4, 1), (1, 5), (2, 6), (3, 7), (4, 8), (5, 6), (7, 8), (5, 9), Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 260 https://internationalpubls.com (6, 10), (7, 11), (8, 12), (10, 11), (12, 9)] G.add_edges_from(edges) # Get the nodes from the edges nodes = set(node for edge in edges for node in edge) # Create a diagonal matrix with zeros diagonal_matrix = np.zeros((len(nodes), len(nodes))) # Assign diagonal values for i, node in enumerate(nodes): diagonal_matrix[i, i] = G.degree(node) # Assign a name to the matrix D_Cl3n = diagonal_matrix # Print the diagonal matrix with the assigned name print("D(Cl_3n):") print(D_Cl3n) which display the output as Create an Adjacency matrix A(Cl3n(G)) using import networkx as nx import numpy as np # Create a graph G = nx.Graph() G.add edges from([(1, 2), (2, 3), (3, 4), (4, 1), (1, 5), (2, 6), (3, 7), (4, 8), (5, 6), (7, 8), (5, 9), (6, 10), (7, 11), (8, 12), (10, 11), (12, 9)])\\ # Obtain the adjacency matrix adjacency matrix = nx.adjacency matrix(G).todense() # Assign a name to the matrix Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 261 https://internationalpubls.com A(Cl_{3n}(G)) = $adjacency matrix $ # Print the adjacency matrix with the assigned name print(f"A(Cl_{3n}(G))):\n{A(Cl_{3n}(G))"} results in Next, the Laplacian matrix is given by L(Cl3n (G)) = D(Cl3n (G)) − A(Cl3n (G)). import networkx as nx import numpy as np # Create a graph G = nx.Graph() edges = [(1, 2), (2, 3), (3, 4), (4, 1), (1, 5), (2, 6), (3, 7), (4, 8), (5, 6), (7, 8), (5, 9), (6, 10), (7, 11), (8, 12), (10, 11), (12, 9)] G.add_edges_from(edges) # Get the nodes from the edges nodes = set(node for edge in edges for node in edge) # Create the adjacency matrix A_Cl3n = nx.adjacency_matrix(G).todense() # Create the diagonal matrix D_Cl3n = np.diag([G.degree(node) for node in nodes]) # Calculate the Laplacian matrix L_Cl3n = D_Cl3n - A_Cl3n # Print the Laplacian matrix print("L(Cl_3n):") print(L_Cl3n) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 262 https://internationalpubls.com Now to find the cofactor of Laplacian matrix import numpy as np # Define the Laplacian Matrix L L = np.array([ [3, -1, 0, -1, -1, 0, 0, 0, 0, 0, 0, 0], [-1, 3, -1, 0, 0, -1, 0, 0, 0, 0, 0, 0], [0, -1, 3, -1, 0, 0, -1, 0, 0, 0, 0, 0], [-1, 0, -1, 3, 0, 0, 0, -1, 0, 0, 0, 0], [-1, 0, 0, 0, 3, -1, 0, 0, -1, 0, 0, 0], [0, -1, 0, 0, -1, 3, 0, 0, 0, -1, 0, 0], [0, 0, -1, 0, 0, 0, 3, -1, 0, 0, -1, 0], [0, 0, 0, -1, 0, 0, -1, 3, 0, 0, 0, -1], [0, 0, 0, 0, -1, 0, 0, 0, 2, 0, 0, -1], [0, 0, 0, 0, 0, -1, 0, 0, 0, 2, -1, 0], [0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 2, 0], [0, 0, 0, 0, 0, 0, 0, -1, -1, 0, 0, 2] ]) # Function to find the cofactor of a matrix def cofactor(matrix, row, col): minor_matrix = np.delete(np.delete(matrix, row, axis=0), col, axis=1) sign = (-1) ** (row + col) return sign * np.linalg.det(minor_matrix) # Calculate the cofactor C12 row_index = 0 col_index = 1 cofactor_C12 = cofactor(L, row_index, col_index) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 263 https://internationalpubls.com 2 print(f"Cofactor C12: {cofactor_C12}") Finally get the output C12: 1248, which is one of the cofactor of the Laplacian matrix. Hence, for n = 4 in the clutch graph Cl3n(G), the number of spanning trees is 1248. Following the same approach, we found 41262 spanning trees for n = 6 and 210240 for n = 8. Based on these results, we derived a general formula for calculating the number of spanning trees in the clutch graph Cl3n(G) as 78[4n − k]k+1, where k ∈ {0, 1, 2, . . . }. Therefore, the total number of spanning trees in the clutch graph Cl3n(G) can be expressed by the formula 78[4n − k]k+1. Theorem 4.7 Every clutch graph Cl3n (G) is bipartite. Proof The clutch graph has three sets of vertices: Vc (cycle vertices), Vp (vertices introduced to the cycle), and Vq (vertices introduced to Vp). Now, consider two disjoint sets V1 and V2 such that V1 ∩ V2 = ∅ and V1 ∪ V2 = V. Partitioned the vertex sets Vc, Vp, Vq in Cl3n (G) such that, each edge (vi, vj) in the clutch graph Cl3n (G) is of the form If vi ∈ V1, then vj must be in V2 If vj ∈ V2, then vi must be in V1 The partition of the vertex set is V1 = {(c2i−1, p2i, q2i−1) | i ∈ {1, 2, . . . ., 𝑛 2 }} V2 = {(c2i, p2i−1, q2i) | i ∈ {1, 2, . . . , 𝑛 2 }} The condition holds for every edge in the clutch graph. The graph is bipartite iff it has no odd cycles. Based on the results, the clutch graph has no odd cycles. So, it is proved that every clutch graph is bipartite. Theorem 4.8 Every clutch graph Cl3n (G), (n ≥ 4, even) is Hamiltonian. Proof Consider the clutch graph Cl3n (G) with vertex set V and edge set E. Let us decompose the clutch graph into two components as G1 representing the cycle graph Cn(G) and G2 representing the additional edges connecting Vp and Vq. G1 = (Vc , Ec) G2 = (Vp ∪ Vq, Ecp ∪ Epq ∪ Eq) Since G1 is a cycle graph, there exists a Hamiltonian cycle H1. Further, Connecting the edges Vp and Vq form a Hamiltonian path P2. H1 =(ci1, 𝑐𝑖2, ……., cin,ci1 ) P2 = (pj1, qj1, pj2, qj2,………,pjn, qjn ) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 1 (2024) 264 https://internationalpubls.com Combine H1 and P2 to form a Hamiltonian cycle H for Cl3n (G). H = (ci1 , ci2 , ..., cin, ci1 , pj1 , qj1 , pj2 , qj2 , ..., pjn, qjn ). As it concluded that every clutch graph with n vertices (n ≥ 4, even) has a Hamiltonian cycle. Let’s consider the case where n = 6 and construct the clutch graph Cl18 (G). The cycle graph with n = 6 has vertices c1, c2, c3, c4, c5, c6 and edges: Cn = (c1, c2), (c2, c3), (c3, c4), (c4, c5), (c5, c6), (c6, c1) • Add vertices Vc = {c1, c2, c3, c4, c5, c6}, Vp = {p1, p2, p3, p4, p5, p6}, and Vq = {q1, q2, q3, q4, q5, q6}. • Connect each ci to its corresponding pi: Ecp = {(c1, p1), (c2, p2), . . . , (c6, p6)}. • Connect pi to qi: Epq = {(p1, q1), (p2, q2), . . . , (p6, q6)}. • Connect qi to qi+1 (with q6 connecting to q1): Eq = {(q2, q3), (q4, q5), (q6, q1)}. • Connect pi to pi+1 (with p6 connecting to p1): Ep = {(p1, p2), (p3, p4), (p5, p6)}. A Hamiltonian cycle can be traversed as c1 → p1 → q1 → c2 → p2 → q2 → c3 → p3 → q3 → c4→ p4 → q4 → c5 → p5 → q5 → c6 → p6 → q6 → c1 Conclusion This paper introduces the concept of clutch graphs from cycle graphs. The clutch graph properties, such as degree, girth, and chromatic number have been discussed. 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