Arbitrary decompositions into open and closed trails

Transcription

1 MATEMATYKA DYSKRETNA Sylwia CICHACZ, Yoshimi EGAWA and Mariusz WOŹNIAK Arbitrary decompositions into open and closed trails Preprint Nr MD 030 (otrzymany dnia 30 maja 2007) Kraków 2007

2 Redaktorami serii preprintów Matematyka Dyskretna s a: Wit FORYŚ, prowadz acy seminarium S lowa, s lowa, s lowa... w Instytucie Informatyki UJ oraz Mariusz WOŹNIAK, prowadz acy seminarium Matematyka Dyskretna - Teoria Grafów na Wydziale Matematyki Stosowanej AGH.

3 Arbitrary decompositions into open and closed trails Sylwia Cichacz a,, Yoshimi Egawa b, Mariusz Woźniak a a Faculty Applied Mathematics, AGH University of Science and Technology, Al. Mickiewicza 30, Kraków, Poland b Dept. of Mathematical Information Science, Tokyo University of Science, Japan May 30, 2007 Abstract The problem of arbitrary decomposition of a graph G into closed trails i.e. a decomposition into closed trails of prescribed lengths summing up to the size of the graph G was first considered in the case of the complete graph G = K n (for odd n) in connection with vertexdistinguishing coloring of the union of cycles. Next, the same problem was investigated for other families of graphs. In this paper we consider a more general problem: arbitrary decomposition of a graph into open and closed trails. Our results generalize all known results on decomposition of a graph into closed trails as well as some results concerning decomposition of a graph into open trails. 1 Introduction A graph G is said to be even if the degrees of all its vertices are even. By Euler s theorem, a connected even graph is Eulerian (i.e. contains a closed trail passing through all its edges exactly once). Here and in subsequent sections, we identify a trail T of length r with any sequence (v 0, v 1,...,v r ) of vertices of T such that v i v i+1 are distinct edges of Corresponding author. cichacz@agh.edu.pl 1

4 T for i = 0, 1,..., r 1. Recall that we do not require the v i to be distinct. A trail T is closed if v 1 = v r+1 and T is open if v 1 v r+1. Consider a simple graph G = (V.E) whose size we denote by e(g). Let τ l = (l 1,...,l k ) and τ t = (t 1,...,t p ) be two finite sequences of positive integers. We shall write these sequences in the following short form τ = (l 1,...,l k ; t 1,...,t p ), and call it just sequence. We allow the situations where k = 0 or p = 0. In these cases we use the notation τ = ( ; t 1,...,t p ) or τ = (l 1,..., l k ; ), respectively. A sequence τ = (l 1,..., l k ; t 1,...,t p ) is called admissible for a graph G if the following conditions hold: 1. k j=1 l j + p i=1 t i = e(g), 2. there exists an open trail of length l j in G for each j and a closed trail of length t i in G for each i, 1 j k, 1 i p, 3. k, p 0, k 1 and k j=1 l j 3 for k 2. If τ = (l 1,...,l k ; t 1,...,t p, ) is an admissible sequence for G and G can be edge-disjointly decomposed into open trails L 1,...,L k of lengths l 1,...,l k respectively, and closed trails T 1,...,T p of lengths t 1,...,t p respectively, then τ is called realizable in G and the sequence (L 1,...,L k ; T 1,...,T p ) is said to be a G-realization of τ or a realization of τ in G. If for each admissible sequence τ for graph G, there is a realization of τ in G, then we say, that graph G is arbitrarily decomposable into open and closed trails. Somewhat artificial condition 3 is to avoid exceptions in the statements of theorems. In particular, it is easy to see that the possibility k = 1 would imply that any Eulerian graph G would not be arbitrarily decomposable into open and closed trails. Remark: By definition, a graph which is arbitrarily decomposable into open and closed trails is also arbitrarily decomposable into closed trails as well as arbitrarily decomposable into open trails. Problems of arbitrarily decomposable graphs into closed trails were first investigated by P.N. Balister. The motivation and application of Theorem 1 can be found in problems concerning vertex-distinguishing proper edge-coloring of graphs ([2]). Let I denote a 1-factor in G (if it exists). 2

5 Theorem 1 ([1]). The graphs G = K n for odd n or G = K n I for even n, are arbitrarily decomposable into closed trails Let K a,b be the complete bipartite graph with bipartition (A, B) such that A = a and B = b, a b. The following theorem was proved by M. Horňák and M. Woźniak. Theorem 2 ([6]). If a, b are even, then the complete bipartite graph K a,b is arbitrarily decomposable into closed trails. Other families of graphs that are arbitrarily decomposable into closed trails are given in the following two theorems. Theorem 3 ([4]). If a is odd, then the graph K a,a I is arbitrarily decomposable into closed trails. Theorem 4 ([5]). If p is of the form p = 2 k 5, k = 0, 1,..., then the complete tripartite graph K p,p,p is arbitrarily decomposable into closed trails. We shall show that all above mentioned graphs that are arbitrarily decomposable into closed trails are actually arbitrarily decomposable into open and closed trails. These results will be given in Section 3. The main tool will be some lemmas proved in the next section. 2 Some lemmas We start with the following simple lemma. Lemma 5. Let a, b be positive integers, and G be an Eulerian graph, such that e(g) = a + b. Then, we can decompose G into two open trails A and B of lengths a and b. Proof. Write G = (x 0, x 1,...,x a+b 1, x 0 ). Let A = (x 0, x 1,..., x a ) and B = (x a,...,x a+b 1, x 0 ). Suppose that x 0 = x a. Then, x 0 x 1 x a x a+1. Hence, the trails A and B can be replaced by A and B of length a and b, respectively, defined as follows: A = (x 1,...,x a+1 ) and B = (x a+1,...,x a+b 1, x 0, x 1 ). Lemma 6. Let a, b, c be positive integers, and G be an Eulerian graph, such that e(g) = a + b + c. Then, G can be decomposed into open trails A, B and C of lengths a, b and c, respectively. 3

6 Proof. Let l = a + b + c. Write G = (x 0, x 1,...,x l 1, x 0 ). Let A = (x 0, x 1,...,x a ), B = (x a,...,x a+b ) and C = (x a+b,...,x l 1, x 0 ). If all these trails are open, we are done. Without loss of generality we may suppose that A is closed i.e. x 0 = x a. Then a 3. We shall show that: Claim 1. x a+b+1 = x a+1. Suppose that x a+b+1 x a+1. Since x 1 x a 1, we have x a+b+1 x 1 or x a+b+1 x a 1. If x a+b+1 = x 1, then let A = (x a 1, x a 2,...,x 1, x 0, x a+1 ), B = (x a+1, x a+2,...,x a+b+1 ), C = (x a+b+1, x a+b+2,...,x l 1, x 0, x a 1 ) is a decomposition with the described properties, a contradiction. We may now assume x a+b+1 x 1. Then, by putting we get a contradiction. A = (x 1, x 2,...,x a, x a+1 ), B = (x a+1, x a+2,...,x a+b+1 ), C = (x a+b+1, x a+b+2,...,x l 1, x 0, x 1 ), It follows from Claim 1 that b 3 and x a+b x a+2. Note that we can write the closed trail G as (x 0, x 1,...,x a, x l 1, x l 2,...,x a+1, x 0 ). Applying Claim 1 to the closed trail G = (x 0, x 1,...,x a, x l 1, x l 2,...,x a+1, x 0 ) with the rules of b and c replaced by each other, we obtain: Claim 2. x a+b 1 = x l 1. By Claim 2 we have c 3 and x a+b x l 2. We can now write G as (x 0, x 1,...,x a, x a+1,..., x a+b 2, x l 1, x l 2, x l 3,...,x a+b, x l 1, x 0 ). Applying Claim 1 to such a graph G we obtain: Claim 3. x l 3 = x a+1 (it is possible that l 3 = a + b + 1). Claim 3 implies x l 2 x a+2. We write G as (x 0, x 1,...,x a, x a+b+1, x a+b, x a+b 1, x a+b 2,..., x a+2, x a+b+1, x a+b+2,...,x l 1, x 0 ) 4

7 Applying now Claim 2 to graph G we get: Claim 4. x a+3 = x l 1 (it is possible that a + 3 = a + b 1) (see Figure 1). Figure 1: Graph G. But then we write G as (x 2, x 3,...,x a, x l 1, x l 2, a a+1, x a+2, x a+3,...,x a+b 1, x a+b, x a+b+1,...,x l 3, x 0, x 1, x 2 ). Recall that x a+2 x a+b x l 2 x a+2. Thus by symmetry, we may assume x 2 x a+b and x 2 x l 2. Now, it is easy to see that the following trails à = (x 2, x 3,...,x a, x l 1, x l 2 ), B = (x l 2, a a+1, x a+2, x a+3,...,x a+b 1, x a+b ), C = (x a+b, x a+b+1,...,x l 3, x 0, x 1, x 2 ) define a desired decomposition, a contradiction. Remark. Lemma 6 is best possible in the sense that there is an Eulerian graph which is not arbitrarily decomposable into four open trails. For instance, the graph drawn in Fig. 2 having 12 edges cannot be decomposed 5

8 x z y Figure 2: An Eulerian graph without decomposition into four open trails into four open trails of length three. For, otherwise, at least one of these trails would have each edge outside of the triangle x, y, z. Thus, it would be closed. Lemma 7. Let l 1,...,l k be positive odd integers and let G be a bipartite trail of size e(g) = l l k. Then, G can be decomposed into k open trails of lengths l 1,...,l k. Proof. Let l = l l k. Write G = (x 0, x 1,...,x l ) and define the decomposition of G as follows: L 1 = (x 0,...,x l1 ), L 2 = (x l1,...,x l1 +l 2 ),...,L k = (x l l k 1,...,x l ). Because in a bipartite graph there is no closed trail of odd size, all trails defined above are open. 3 Results We gather all results in one theorem. Theorem 8. The following families of graphs are arbitrarily decomposable into open and closed trails: 1. complete graphs K n for n odd, 2. the graphs K n I for n even, 3. complete bipartite graphs K a,b for a and b even, 4. bipartite graphs K a,a I for a odd, 5. complete tripartite graphs K p,p,p for p = 2 k 5, k = 0, 1,... 6

9 We establish the theorem by proving two lemmas. Lemma 9. Let G be an Eulerian graph, and suppose that for each integer t with 3 t e(g) 3, G has a closed trail of length t. Suppose further that G is arbitrarily decomposable into closed trails. Then G is arbitrarily decomposable into open and closed trails. Proof. The proof is by induction on k, the number of open trails in the decomposition corresponding to an admissible sequence τ = (l 1,...,l k ; t 1,...,t p ). For k = 0, the desired conclusion immediately follows from the assumption. Thus, let k 2 (recall that, by definition, k cannot be equal to one). Without loss of generality, we may assume that l 1 l 2... l k. Consider first the case where k 4. Then the sequence τ = (l 3,...,l k ; l 1 + l 2, t 1,...,t p ) is admissible except for the case where l 1 + l 2 = 2. Thus, suppose that l 1 + l 2 3. Observe that τ has less open trails lengths than τ. Hence, by the induction hypothesis, τ is realizable. Now, by using Lemma 5, we can decompose the closed trail of length l 1 +l 2 into two open trails of length l 1 and l 2. Finally, we get a realization of the sequence τ. If l 1 + l 2 = 2, we define τ by τ = (2, l 3,...,l k ; t 1,...,t p ). Again, τ has less open trails lengths than τ. As above, in order to get a τ-realization, we apply first the induction hypothesis and next we decompose the trail of length two into two edges. In the case k = 3, we replace the sequence τ by the the sequence τ = ( ; l 1 + l 2 + l 3, t 1,...,t p ) and use Lemma 6 instead of Lemma 5. Finally, if k = 2, then, by definition, l 1 + l 2 3 and the sequence τ can be replaced by the sequence τ = ( ; l 1 +l 2, t 1,...,t p ). Since τ is admissible, we can again use the induction hypothesis and Lemma 5 to get the realization of τ. Lemma 10. Let G be a bipartite Eulerian graph, and suppose that for each even integer t with 4 t e(g) 4, G has a closed trail of length t. Suppose further that G is arbitrarily decomposable into closed trails. Then G is arbitrarily decomposable into open and closed trails. 7

10 Proof. Let τ = (l 1,...,l k ; t 1,..., t p ) be an admissible sequence for G. We proceed by induction on k. If k = 0, then the realization of τ follows from the assumption. Suppose k 2. Let us observe that all the numbers t j are even and, since the size of G is even too, the number of odd terms in the first part of the sequence τ is even. Therefore, we may assume without loss of generality that l 1 + l 2 is even. We may further assume that if k = 4 and l 1 + l 2 4, then we also have l 3 + l 4 4. Consider now the sequence τ defined as follows: for k 4, τ = (l 3,...,l k ; l 1 + l 2, t 1,...,t p ) or (l 1 + l 2, l 3,...,l k ; t 1,...,t p ) according as l 1 +l 2 4 or l 1 +l 2 = 2. τ = ( ; l 1 +l 2 +l 3, t 1,...,t p ) for k = 3 and τ = ( ; l 1 + l 2, t 1,...,t p ) for k = 2. In all cases, the sequence τ is admissible and has less open lengths terms than τ. Therefore, the realization of τ in G follows from induction and the realization of τ can be easily obtained from the realization of τ by applying Lemma 5 or Lemma 6. Proof of Theorem 8. For 1., 2. and 5., the theorem follows from Theorems 1 and 4 and Lemma 9; for 4., it follows from Theorem 3 and Lemma 10. Thus is suffices to consider the case where G = K a,b with a, b even. If a, b 4, the desired conclusion follows from Theorem 2 and Lemma 10. We may assume a = 2. As above, let τ = (l 1,...,l k ; t 1,...,t p ) be an admissible sequence for G = K 2,b. We again proceed by induction on k. If k = 0, then the realization of τ follows from Theorem 2. Suppose k 2. Let us observe that G contains only closed trails of length divisible by four. Hence, all the numbers t j satisfy t j 0 (mod 4). As above, since the size of G is divisible by four, the number of odd terms in the first part of the sequence τ is even. Denote by l the sum of all odd terms l i. If all these terms are odd, then l 0 (mod 4) and the sequence τ = ( ; l, t 1,...,t p ) is admissible. Hence, by induction and Lemma 7 we can obtain a realization of τ. If there are some even terms l i, we define τ by τ = ( l, l 1,...,l k ; t 1,...,t p ) with l i even, and we proceed as above. Consequently, we may assume that all l i terms are even. Thus, for each l i either l i 0 (mod 4) or l i 2 (mod 4). Let us observe that the number of terms with the second possibility is even. Therefore, it is possible to choose the first two terms in τ in such a way that l 1 + l 2 0 (mod 4). Then, the sequence τ defined as follows: 8

11 x y Figure 3: An graph arbitrarily decomposable into closed trails which is not arbitrarily decomposable into open and closed trails τ = (l 3,...,l k ; l 1 + l 2, t 1,...,t p ) for k 4 and τ = ( ; l 1 + l 2, t 1,...,t p ) for k = 2 is admissible. If k = 3 then of course we have l 1 + l 2 + l 3 0 (mod 4) and the sequence τ = ( ; l 1 + l 2 + l 3, t 1,...,t p ) is admissible, too. The remaider part of the proof is analogous to the proof of Lemma 10. As a corollary of the above theorem we get the following theorem concerning decomposition into open trails. Some of these results were obtained earlier by using more direct methods (see for instance [3]). Theorem 11. The following families of graphs are arbitrarily decomposable into open trails: 1 complete graphs K n for n odd, 2 the graphs K n I for n even, 3 complete bipartite graphs K a,b for a and b even, 4 bipartite graphs K a,a I for a odd, 5 complete tripartite graphs K p,p,p for p = 2 k 5, k = 0, 1,... We finish this section by giving (in Figure 3) an example of a graph G which is arbitrarily decomposable into closed trails (this fact was first observed in [5]) but is not arbitrarily decomposable into open and closed trails. Indeed, let us consider the sequence (2, 3; 3, 3). It is evidently admissible for G but is not realizable. It is sufficient to observe that each triangle of G has to contain the edge xy. 9

12 References [1] P.N. Balister, Packing Circuits into K N, Combin. Probab. Comput. 10 (2001) [2] P.N. Balister, B. Bollobàs and R.H. Schelp, Vertex distinguishing colorings of graphs with (G) = 2, Discrete Mathematics 252 (2002) [3] S. Cichacz and A. Görlich, Decomposition of complete bipartite graphs into open trails, Preprint MD 016 (2005), [4] S. Cichacz and M. Horňák, Decomposition of bipartite graphs into closed trails, P.J.Šafárik University, IM Preprint, series A, 2/2007 (2007), [5] M. Horňák and Z. Kocková, On complete tripartite graphs arbitrarily decomposable into closed trails, Tatra Mt. Math. Publ., (to appear). [6] M. Horňák and M. Woźniak, Decomposition of complete bipartite even graphs into closed trails, Czechoslovak Mathematical Journal, 128 (2003)