A note on properties for a complementary graph and its tree graph

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1 A note on properties for a complementary graph and its tree graph Abulimiti Yiming Department of Mathematics Institute of Mathematics & Informations Xinjiang Normal University China Masami Yasuda Department of Mathematics & Informations Faculty of Science Chiba University Chiba, Japan Abstract In this note a complementary graph in n -th (n 2) order is defined and discussed its property. Since it can not be reduced from n -th order to 2-nd, we must consider its property of n -th order graph separately. We investigate whether similar properties of the graph arising from 2-nd order case hold or not. A relation to the complete graph K 2n and also to the tree graph associated with the complementary graph are studied. Keywords :??? 0. Introduction Let G = (V, E) be a non-directed graph whose vertex set is V and edge is E. Assume its degree is p and the size is q, that is, V = p, E = q. Here we permit the graph G has multiple edges but there are Journal of Discrete Mathematical Sciences & Cryptography Vol. ( ), No., pp. 1 9 c Taru Publications

2 2 A. YIMING AND M. YASUDA no loops. The definition and notation is usual one so refer to the classical noted book [7], etc. If the connected graph G has two spanning trees T 1 and T 2, which satisfies G = T 1 T 2 with T 1 T 2 =, we call the graph as a 2-nd order complementary tree or the graph has a structure of 2-nd order complementary. It is known that Lee [3] applies this complementary graph to the theory of circuit network firstly and then Kajitani [1] discussed several dis-cussion in 2-nd order. There are few paper concerning with this topics however. In this note we will study this notion of complementary tree from 2-nd order to n-th(n 2) and investigate their properties. 1. Definition and notations Definition 1.1. If there exist n spanning trees T 1, T 2,, T n in a given connecting graph G, and each satisfies the following conditions: (i) G = T 1 T 2 T n (ii) T i T j =, i = j; i, j = 1, 2,, n, then the graph G is called as an n-th order complementary graph. Example 1. The next graph G is a sample for 2-nd order complementary graph. Example 2. The next graph G is a sample for 3-rd order complementary graph. We note that the 3-rd order graph cannot be reduced to 2-nd one because the definition require each graph must be a tree.

3 COMPLEMENTARY GRAPH 3 Now we will discuss several properties for the n-th order complimentary tree. Theorem 1.1. For an n-th order complementary graph G = (V, E) with V = p, E = q, then it holds that the equality q = n(p 1) and λ(g) n provided λ(g) means the number of connected component for G. Proof. The concernment is clear from the definition. To prove the equality, we consider each edge. Since each spanning tree of G has same p 1 edges, so there are n(p 1) edges in total. And every vertex connects to all spanning tree. The least number of cutting edge is n. Therefore the graph G has n-edges of cutting. Thus the edge connection of G is greater than n. Theorem 1.2. The rank r(g) of n-th order complementary graph G equals r(g) = p 1 and the nullity µ(g) equals µ(g) = (n 1)(p 1). Proof. Since G is connected, the incident matrix of G has rank n 1. So the rank of G is n 1. This is also seen from the definition in directly. For any graph, it is well known that the nullity µ(g) = q p + ω(g) where ω(g) denotes the number of components of G. In this note the complementary graph is assumed to be connected so ω(g) = 1. Hence µ(g) = q p + 1. Substitute for q = n(p 1), the relation µ(g) = (n 1)(p 1) is obtained. Theorem 1.3. Every edge of n-th order complementary graph G is contained a cycle of G: e i G C i, e i C i. Proof. Every edge of G is contained by a spanning tree T i for some i. There exists other edge which connecting to it. By adding the edge to the tree, it becomes loop, that is, a cycle. Definition 1.2. Let e = xy be an edge connecting between the vertex x and y in a graph G. The length of edge e becomes shrinking as a vertex. The graph obtained by this manipulation, it is called a contraction of G and denoted by G {e}. Similarly G {e i1, e i2,, e in } is called the contraction of n-th order. Alternatively by cutting edges e from the graph G, it is a removal denoted by expressed as G {e} and also G {e i1, e i2,..., e in } is the removal of n-th order.

4 4 A. YIMING AND M. YASUDA Example 3. The next figure is a sample of reduction. Theorem 1.4. For n-th order complementary graph G with multiple edges; e i1, e i2,, e in, the reduction G {e i1, e i2,, e in } of G is also n-th order complementary graph G. Proof. By the assumption, let G = T 1 T 2 T n with T i T j =, i = j; i, j = 1, 2,, n. Since the number of two connecting vertices is less than or equal to n, G {e i1, e i2,, e in } has no loop. Therefore, for the edges e i j T j, j = 1, 2,, n the contraction Tj = T j {e i j } has the property j Tj = ( j T j ) {e i1, e i2,, e in } = G {e i1, e i2,, e in } and T i T j =, i = j; i, j = 1, 2,, n. Thus the assertion holds from Definition 1.1. Example 4. The graph G of Example 3 is 3-rd order complementary graph. The contraction subset G s of its multiple edge {e 1, e 2, e 3 } is also 3-rd order complementary graph: 2. The case of the complete graph Now we will discuss the important class of the complete graph is included by the complementary graph. Theorem 2.1. The complete graph K 2n (n 2) is also n-th order complementary graph. Proof. We will prove it by a mathematical induction. For n = 2, the graph K 2 2 = K 4 is seen to be a 2-nd order complementary graph immediately.

5 COMPLEMENTARY GRAPH 5 Assume the case of n = k holds. That is, there exist k spanning tree T 1, T 2,, T k with the property K 2k = T 1 T 2 T k and they are satisfy T i T j =, i = j, i, j = 1, 2,, k. Let u, v be two vertices in the complete graph K 2(k+1) and others are simply denoted by 1, 2,, 2k. Since the graph is complete, each u, v connects to other 2k vertices. So the edges connecting between u and 1, 2,, 2k are denoted by e u,1, e u,2,, e u,2k, and similarly the edges connecting between v and 1, 2,, 2k are denoted by e v,1, e v,2,, e v,2k. The edge e u,v is between u and v. By the assumption the removed graph K 2k = K 2(k+1) {u, v} is k- th order complementary graph, there exists k spanning tree T 1, T 2,, T k and these satisfy that T 1 T 2 T k = K 2k = K 2(k+1) {u, v} and T i T j =, i = j; i, j = 1, 2,, k. If we introduce a new tree T i = T i {e u,2i 1, e v,2i }, i = 1, 2,, k and by letting T k+1 = {e u,2, e v,3,, e u,2k 2, e v,2k 1, e u,2k, e v,1, e u,v }, then Definition 1.1 imply that K 2(k+1) is a k + 1-th order complementary tree. Therefore, for arbitrary n N, we have proved that K 2n is n-th order complementary graph. The next graph denotes a complete graph corresponding 3-rd order complementary graph. Corollary 2.2. For the complete graph K 2n+1, there exist n spanning trees T 1, T 2,, T n with T i T j =, i = j; i, j = 1, 2,, n. Also there exists a spanning tree T which distance equals d(t T i ) = p 2 for each T i and T T i = 1 simultaneously.

6 6 A. YIMING AND M. YASUDA Proof. Let ω be an arbitrary vertex in the complete graph K 2(n+1) and other 2n vertices are denoted by 1, 2,, 2n. In this case the removal T 2n+1 {ω} = K 2n is a complete graph. By the previous Theorem 2.1, there exist n spanning trees T 1, T 2,, T n which satisfy T 1 T 2 T n = K 2n and T i T j =, i = j; i, j = 1, 2,, n. Denote each edge between a vertex ω and vertices 1, 2,, 2n by e ω,1, e ω,2,, e ω,2n respectively, each graph T i = T i {e ω,i 1 } is n spanning trees of K 2(n+1). Also T i T j =, i = j; i, j = 1, 2,, n hold and T = {e ω,1, e ω,2,, e ω,2n } is a spanning trees of K 2(n+1) too. See the following figure. Note that T T i = {e 2i 1 }, i = 1, 2,, n. Thus d(t, T i ) = 2n 1 = p 2 holds, where d(t, T i ) = 2 1 N(T T i), where N(T T i ) = T T i = (T T i ) \ (T T i ) means a symmetrical difference T and T i. 3. Relation to a tree graph The relation between a tree graph and a complementary tree is known [7]. First the definition of a tree graph corresponding to a graph is given. Definition 3.1. The tree graph corresponding to a given graph G = (V, E) is a graph denoted by T(G) = (V t, E t ) and their vertices and edges are defined as follows: Each vertex v tk of T(G) has one to one correspond to each of a spanning tree T k in G, and two vertices v ti, v t j V t has a distance d(v ti, v t j ) = 1 provided that these are adjoining and non-directed. The distance between vertices is defined by d(t, T i ) = 1 2 N(T T i), where N(T T i ) = T T i = (T T i ) \ (T T i ) means a symmetrical difference T and T i. Example 5. The next T i, i = 1, 2,, 5 illustrates all of spanning trees for G = (V, E) and its tree graph T(G) = (V t, E t ).

7 COMPLEMENTARY GRAPH 7 The next theorem is a characterization of 2-nd order complementary graph by its tree graph. However we see that it is unknown for case of n-th (n 3) order. Example 6 is a counterexample of general case. Theorem 3.1. Let there are p vertices and 2(p 1) edges in the graph G = (V, E). The necessarily and sufficient condition to be G = (V, E) a 2-nd order complementary graph is that there exists pairing vertices v i, v j in the tree graph T(G) = (V t, E t ) for G = (V, E) and the shortest length between them is equal to p 1. Proof. Sufficiency : Let v i, v j are adjoin two vertices in a tree graph T(G) which is corresponding to a graph G. Since the shortest length of path is p 1, their connecting path v i v j is written as P = v i v i1,, v ip 2 v j. From Definition 3.1, there exist p spanning trees T i, T i1,, T ip 2 T j and they correspond to v i, v i1,, v ip 2, v j of the vertices in P(G). Also, by the same Definition 3.1, d(t i, T i1 ) = 1, d(t i1, T i2 ) = 1,, d(t ip n, T i j ) = 1 hold. Since the length from v i to v j is p 1, d(t i, T ik ) k, k = 1, 2,, p 2. Thus we have d(t i, T j ) = p 1. Clearly the number of edges in G equals 2(p 1), so T i T j = G and T i T j =. Therefore G = (V, E) is a 2-complementary tree. Necessity : If G = (V, E) is 2-nd order complementary graph, there exist two spanning trees T 1, T 2 such that T 1 T 2 = G and T 1 T 2 =. By converting the edge of trees, we associate with other p 2 spanning trees T i1,, T ip 2. Thus T 1, T i1,, T ip 2, T 2 and d(t 1, T i1 ) = 1, d(t i1, T i2 ) = 1, d(t 1, T i2 ) = 1, d(t ip 2, T 2 ) = 1, d(t 1, T 2 ) = 1 are hold. For a tree graph

8 8 A. YIMING AND M. YASUDA T(G) of G, there is a pairing vertex v 1, v 2. Thus there is a shortest path of length p 1 from v 1 to v 2. As a remark of this theorem, the above assertion does not hold in general because the following example shows. Example 6. The figure G has p = 3 vertices and 3(p 1) edges. This graph is not 3-complementary. But its tree graph T(G) has the shortest length p 1 = 2 from v 1 to v 8. The next theorem is a partial answer for characterization of n-th order complementary graph by using the tree graph. Theorem 3.2. There is an n-th order complementary graph for the tree graph T(C 2n ) where C 2n is a cycle of order 2n. Proof. Since C 2n has 2n spanning trees, and the distance with each others equal 1, so T(C 2n ) and K 2n is equivalent. Theorem 2.1 implies that T(C 2n ) is n-th order complementary graph. From this Theorem 3.2, we can prove the following corollary: Corollary 3.3. If G is simple, the tree graph T(G) which number of vertices is n( 3) contains K n as its subgraph. Proof. The graph G is connected however it is not tree, so it contains a cycle of the length at least three. Similarly the distance between them in the cycle equals one. Therefore at least three vertices of T(G) are adjoined with each other. The next figure shows eight vertices of a tree graph T(G) which contain K 3 and K 4.

9 COMPLEMENTARY GRAPH 9 Corollary 3.4. If G is a simple graph, there are no tree graphs which consists of only two vertices. Proof. Assume that if there exist a tree graph with two vertices. Then the graph G has two spanning trees. This is impossible. n Corollary 3.5. For n > 3, the tree graph T(C n ) contains -th order complementary graph where the notation x denotes the gamester integer 2 x. Proof. When n = 2k, a tree graph T(G) has k-complementary graph by Theorem 3.2. If n = 2k + 1, T(G) contains k-complementary graph by Corollary 3.3. These complete the proof. References [1] Y. Kakitani, Graph Theory for Networks (in Japanese), Shokodo Co. Ltd., Tokyo, Japan, [2] D. J. Kleitman, More complementary tree graphs, Discrete Math., Vol. 15 (1976), pp [3] H. B. Lee, On the differing abilities of RL structure to realize natural frequencies, IEEE Trans. Circuit Theory, Vol. CT-12 (1965), pp [4] P. M. Lin, Complementary trees in circuit theory, IEEE Trans. Circuit and Systems, Vol. CAS-27 (1980), pp [5] B. L. Liu, 2-complementary tree graph (Chinese), J. South China Normal Univ. Natur. Sci. Ed., Vol. 8 (1988), pp [6] U. G. Rothblum, On the number of complementary tree in a graph, Discrete Math., Vol. 15 (1976), pp [7] R. J. Wilson, Introduction to Graph Theory, 2nd edn., Academic Press, New York, Received October, 2004

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