MIXDECOMPOSITON OF THE COMPLETE GRAPH INTO DIRECTED FACTORS OF DIAMETER 2 AND UNDIRECTED FACTORS OF DIAMETER 3. University of Split, Croatia


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1 GLASNIK MATEMATIČKI Vol. 38(58)(2003), MIXDECOMPOSITON OF THE COMPLETE GRAPH INTO DIRECTED FACTORS OF DIAMETER 2 AND UNDIRECTED FACTORS OF DIAMETER 3 Damir Vukičević University of Split, Croatia Abstract. We estimate the values, for each k, of the smallest n such that K n can be mixdecomposed into k undirected factors of diameter 3 and one directed factor of diameter 2. We find the asymptotic value of ratio of n and k, when k tends to infinity and generalize this result for mixdecompositions into p directed factors of diameter 2 and k undirected factors of diameter 3.. Introduction In this paper, we solve the following problem. There is a system of n communication devices and there are k + communication networks. There are k communication networks that allow twoway communication (the link of this network connecting two devices allows each of them to receive and send information) and the remaining communication network allows only oneway communication (the link of this network connecting two devices allows one of them only to send information and the other only to receive them). The following requirements are given: ) Every pair of different devices can be connected only by a single link. 2) For every pair of different devices and for every twoway communication network, there has to be a path of at most 3 links through which they can communicate. 3) For every pair of different devices a and b there has to be a path of at most 2 links of the oneway communication network through which a can send information to b and a path of at most 2 links of the oneway communication network through which b can send information to a Mathematics Subject Classification. 05C70. Key words and phrases. Decomposition, mixdecomposition, factor, graph. 2
2 22 D. VUKIČEVIĆ Of course, it is not always possible to meet these requirements. Take for instance the case n = 3 and k =. In this paper, we want to estimate, for each k, the smallest n such that this is possible This problem is closely connected to wellknown problems of decompositions of graphs. Decompositions of graphs into factors with given diameters have been extensively studied. The problem of decomposition of the factors of equal diameters, where diameter of each factor is at least three has been solved in [4]. The majority of the papers written about decomposition of the graphs are written about decompositions into the factors of diameter two. Denote by f (k) the smallest natural number so that complete graph with n vertices can be decomposed into k factors of diameter 2. In [5], it was proved that f (k) 7k. Then in [3] this was improved to f (k) 6k. In [7], it was further proved that this upper bound is quite close to the exact value of f (k) since f (k) 6k 7, k 664 and in [8] the correct value of f (k) was given for large values of k, to be more specific it was proved that f (k) = 6k, k 0 7. Therefore, the most interesting problem of decompositions of graphs are decompositions into factors of small diameters. The problem most closely related to ours, i.e. the problem of decompositions of complete graph into factors of diameter 2 and 3 was treated in [6]. We obtain few results similar to the results of that paper, but these constructions will be somewhat more complicated, since in this paper we have to deal with directed factors of diameter 2. Also, the organization of the system of communication networks presented here is somewhat better than one given in [6], because in this system the privileged communication network is much faster; here links of privileged network allow only oneway communication, which is much faster then twoway communication. 2. Basic definitions Let G be undirected (resp. directed graph). By V (G), we denote set of its vertices, by v (G) number of its vertices, by E (G) set of its edges (resp. directed edges) and by e (G) the number of its edges (resp. directed edges). By d G (x, y) we denote the length of the shortest path connecting x and y (resp. the length of the shortest path from x to y). We also denote diam G = max {d G (x, y) : x, y V (G)}.
3 MIXDECOMPOSITON OF THE COMPLETE GRAPH 23 We shall slightly abuse the word subgraph (resp. supergraph) by saying that A is a subgraph of B (resp. B is a supergraph of A) if B contains a subgraph isomorphic to A. For undirected graph G, we denote by d G (x) degree of vertex x, by δ (G) minimal degree of the graph and by (G) maximal degree of the graph. For a directed graph G we denote by d + (x) outdegree of vertex x and by d (x) indegree of vertex x. By + (G), we denote maximal outdegree of graph G and by (G) maximal indegree of graph G. By δ + (G) we denote minimal outdegree of graph G and by δ (G) minimal indegree of graph G. We say that vertex x is naccessiblefrom (resp. naccessibleto) y if there is a path from y to x (resp. from x to y) of length at most n, and for any set of vertices A, we say that A is naccessiblefrom (resp. naccessibleto) x if each vertex in A is naccessiblefrom (resp. naccessibleto) x. If (x, y) E (G), we say that y is outneighbor of x and that x is inneighbor of y. Let D be a directed graph. Denote by D a graph such that V ( D ) = V (D) and We also define xy E ( D ) ((x, y) E (D) (y, x) E (D)) Definition 2.. Let G be undirected graph. We say that G is mixdecomposed into undirected factors F, F 2,..., F k and directed factors D, D 2,..., D p if V (F ) = V (F 2 ) = = V (F k ) = V (D ) = V (D 2 ) = = V (D p ) = V (G) and for each edge {a, b} E (G) exactly one of the following is true: ) {a, b} E (F ) 2) {a, b} E (F 2 ). k) {a, b} E (F k ) k+) (a, b) E (D ) k+2) (a, b) E (D 2 ). k+p) (a, b) E (D p ) k+p+) (b, a) E (D ) k+p+2) (b, a) E (D 2 ). k+2p) (b, a) E (D p ) Definition 2.2. Let G be an undirected graph. Mixcomplement of G is any directed graph G such that V (G) = V (G ) and that for each a, b V (G ) exactly one of the following statements is true:
4 24 D. VUKIČEVIĆ ) {a, b} V (G) 2) (a, b) E (G ) 3) (b, a) E (G ). Definition 2.3. Let G be a directed graph. Mixcomplement of G is any undirected graph G such that V (G) = V (G ) and such that for each a, b V (G ) exactly one of the following statements is true: ) {a, b} V (G ) 2) (a, b) E (G) 3) (b, a) E (G). We define the function φ : N N by φ (k) = n if and only if n is the smallest natural number such that K n can be mixdecomposed into k undirected factors of diameter 3 and one directed factor of diameter 2. Therefore, we have to estimate the values of the function φ. 3. The value of φ () It can be easily, but tediously proved that (if the reader is interested in the details of this and other missing proofs, please send me a mail): Lemma 3.. There is no mixcomplement D of cycle with 7 vertices such that diam D 2. Now, we prove Lemma 3.2. φ () > 7. Proof. Suppose to the contrary, that it is possible to mixdecompose K n, n 7, into undirected factor F of diameter 2 and directed factor D of diameter 3. It is easy to see that δ + (D) 2 and δ (D) 2, because otherwise it would be δ (F ) = 0, which is impossible. If n 4, then this is impossible. If n = 5, then (F ) = 0. If n = 6, then (F ), hence F is disconnected. If n = 7, then (F ) 2 and diam F 3, so F is a cycle, but this is in contradiction with the previous lemma. In all the cases we have obtained a contradiction, therefore φ () > 7. Let us prove that φ () > 8. Denote by G 8,, G 8,2 and G 8,3 the graphs on Figures, 2 and 3. By a simple analysis, we can prove: Lemma 3.3. Let G be a graph such that v (G) = 8 and diam G 3, (G) 3 and δ (G) =, then G is a supergraph of one of the graphs G 8,, G 8,2 and G 8,3. Simple, but tedious analysis shows that: Lemma 3.4. There is no directed graph D such that diam D 2 and D is mixcomplement of G 8,.
5 MIXDECOMPOSITON OF THE COMPLETE GRAPH 25 v v 2 v 3 v 4 v 5 v 7 v 6 v 8 Figure v v 2 v 3 v 4 v 5 v 7 v 6 v 8 Figure 2 v v 2 v 3 v 4 v 5 v 7 v 6 v 8 Figure 3 Lemma 3.5. There is no directed graph D such that diam D 2 and D is mixcomplement of G 8,2.
6 26 D. VUKIČEVIĆ Lemma 3.6. There is no directed graph D such that diam D 2 and D is mixcomplement of G 8,3. From the last four lemmas, it follows. Lemma 3.7. Let K 8 be mixdecomposed into undirected factor F of diameter at most 3 and directed factor D of diameter at most 2; then δ (F ) 2. Let us prove Lemma 3.8. Let D be directed graph with 8 vertices of diameter 2 such that δ + (D) 2, δ (D) 2 and sum of indegree and outdegree for at most 6 vertices is 5 and for the remaining vertices is 4, then exactly three vertices have indegree 3 and there is at most one edge between these 3 vertices, and also exactly three vertices have outdegree 3, and there is at most one edge between these vertices. Proof. First let us prove that each vertex of outdegree 2 has at least one outneighbor of outdegree 3. Suppose, to the contrary, that there is a vertex x of outdegree 2 such that its both outneighbors have outdegree 2. In that case, there are only 7 vertices that are 2accessiblefrom x and that is a contradiction. Now let us prove that there are at least 3 vertices with outdegree 3. Suppose to the contrary, that there are at most 2 vertices with outdegree 3. Each of these vertices can be outneighbor of at most two vertices. Since there are at least 6 vertices of outdegree 2, we have obtained a contradiction, hence there have to be at least 3 vertices of outdegree 3. We can prove analogously that there are at least 3 vertices with indegree 3, hence there are exactly three vertices with in degree 3 and 3 vertices with outdegree 3. At this point, we prove that there cannot be two edges between three vertices of outdegree three. Suppose to the contrary, that this is possible, but then only four vertices of outdegree two can have a neighbor of outdegree 3. Since, there are 5 vertices of outdegree 2, this is a contradiction. Completely analogously, it can be proved that there is at most one edge between 3 vertices that have indegree 3. Denote by G 8,4, G 8,5 and G 8,6 respectively the following graphs Figure 4
7 MIXDECOMPOSITON OF THE COMPLETE GRAPH 27 Figure 5 Figure 6 A simple analysis shows that: Lemma 3.9. Every graph G with 8 vertices such that diam G 3, δ (G) = 2, (G) 3 and at most two vertices have a degree larger than 2, is isomorphic to G 8,4, G 8,5 or G 8,6. Simple check shows that: Lemma 3.0. In complement of G 8,4 six vertices of degree 5 cannot be divided in two triples in such a way that each triple contains only one edge. Lemma 3.. In complement of G 8,5 six vertices of degree 5 cannot be divided in two triples in such a way that each triple contains only one edge. Lemma 3.2. In complement of G 8,6 six vertices of degree 5 cannot be divided in two triples in such a way that each triple contains only one edge. Combining the last five lemmas we get: Lemma 3.3. K 8 can not be mixdecomposed into undirected factor F of diameter 3 and directed factor F 2 of diameter 2, such that δ (F 2 ) 2. The last lemma and Lemma 3.7 yield the following proposition: Proposition 3.4. φ () > 8. From the last Proposition and the following decomposition of K 9 (on this sketch we draw only the directed edges of directed factor D and all the missing edges are edges of undirected factor F ): it follows that Theorem 3.5. φ () = 9.
8 28 D. VUKIČEVIĆ Figure 7 4. Estimates of φ (k) for small values of k We start with several lemmas: Lemma 4.. If there is a directed graph D of diameter 2, such that D = K n, then there is a directed graph D such that diam D = 2 and D = K n+2. Proof. Denote vertices of D by v, v 2,..., v n. We explicitly construct D. Its vertices are v,..., v n+2 and its edges are: ) (x, y) such that (x, y) E (D) 2) (x, v i ) such that n + i n + 2, (x, v n ) E (D) 3) (v i, x) such that n + i n + 2, (v n, x) E (D) 4) (v n, v n+ ), (v n+, v n+2 ), (v n+2, v n ). Lemma 4.2. There is a directed graph D (2, 3) of diameter 2 such that D (2, 3) = K 3, and there is directed graph D (2, 6) of diameter 2 such that D (2, 3) = K 6. Proof. Graphs with the required properties are given by the following sketches: Figure 8
9 MIXDECOMPOSITON OF THE COMPLETE GRAPH 29 Figure 9 The last two Lemmas yield that there are graphs D (2, n) of diameter 2 such that D (2, n) = K n for each n, 2, 4. Denote by D a mixeddecomposition of the K 0 into undirected factor F D and directed factor D D given by the following sketch (on the following sketch only the edges of the factor D D are drawn and the missing edges are edges of the factor F D ): x x 2 x 3 a a 7 a 2 a 6 a 3 a 4 a 5 Figure 0 It can be easily seen that diam F D = 3 and that diam D D = 2. Let us prove that: Lemma 4.3. φ (k) 3k + 7, k, 2, 4. Proof. We explicitly give a mixeddecomposition of K 3k+7 into undirected factors F,..., F k and directed factor D with a required properties. Denote k V (K 3k+7 ) = A i= B i A = {a, a 2,..., a 6 }, B i = {b i,, b i,2, b i,3 }.
10 220 D. VUKIČEVIĆ Edges of F i, for each i k, are: ) b i,j a k such that x j a k E (F D ) 2) b i, b l,2, b i,2 b l,3, b i,3 b l,, l k, i l 3) b i, b i,2, b i,2 b i,3, b i,3 b i,. Since b i,, b i,2, b i,3 are all adjacent in F i and each vertex is adjacent to at least one of them, it follows that diam F i = 3. Denote by g i : {b,i, b 2,i,..., b k,i } V (D (2, k)), i =,..., 3 any bijections. Directed edges of D are: ) (a i, a j ) such that (a i, a j ) V (D D ) 2) (b i,j, a l ) such that (x j, a l ) V (D D ) 3) (a l, b i,j ) such that (a l, x j ) V (D D ) 4) (b i,j, b l,j ) such that (g j (b i,j ), g j (b l,j )) V (D D ). Let us prove that diam D 2. We need to show that d (x, y) 2 for each x, y V (D). Distinguish three cases: ) x, y A or x A, y B i, i k or x B i, y A, i k or x, y B i, i k. Note that there is a subgraph of D isomorphic to D D that contains x and y, hence the claim follows. 2) x = b i,j, y = b l,m, i l, j m, i, j, l, m k. There is a vertex a o, o 7 such that (x j, a o ),(a o, x m ) E (D D ). Therefore (b i,j, a o ), (a o, b l,m ) E (D D ). 3) x = b i,j, y = b l,j. There is a path of length at most 2 that consists of directed edges listed in 4). We have exhausted all the cases and we have proved our claim. Lemma 4.4. φ (2) 4. Proof. We explicitly give a decomposition of K 4 into undirected factors F and F 2 of diameter 3 and directed factor D of diameter 2 by the following table T :
11 MIXDECOMPOSITON OF THE COMPLETE GRAPH where T ij = k denotes ij F k, k 2; T ij = 3 denotes (i, j) E (D) and T ij = 4 denotes (j, i) E (D). Lemma 4.5. φ (4) 20. Proof. We explicitly give a decomposition of K 20 into undirected factors F, F 2, F 3 and F 4 of diameter 3 and directed factor D of diameter 2 by the following table T : where T ij = k denotes ij F k, k 4; T ij = 5 denotes (i, j) E (D) and T ij = 6 denotes (j, i) E (D). Lemma 4.6. Let D be a directed graph with n vertices of diameter 2 such that δ + (D) 2. Then D has at least 3n 7 edges. Proof. If δ + (D) 3, the claim is trivial. If not, there is a vertex x with exactly 2 outneighbours, say y and z. Since diam D 2, it follows that d + (y) + d + (z) n 3. Therefore we have v V (D) d+ (v) 3n 7. The last Lemma yields Lemma 4.7. φ (k) 2k+7+ (2k+7)
12 222 D. VUKIČEVIĆ Proof. Suppose that K n can be decomposed into k undirected factors each of diameter at most 3 and one directed factor of diameter 2. Each of undirected factors have at least n edges and directed graph by last Lemma has at least 3n 7 directed edges. Therefore, hence k n + 3n 7 n (n ) 2 n 2 (2k + 7) n + 4 0, 2k (2k + 7) 2 56 n 2. So far we have shown that Theorem , k = 2k+7+ (2k+7) , k 2 φ (k) 3k + 6 k = 3k + 8 k = 2, 4 3k + 7 k, 2, Estimates of φ (k) for large values of k The upper bounds of the last theorem are quite good for small values of k, but they are bad for large values of k. Note that from the last theorem, it follows only We shall prove that 2 lim φ (k) k lim φ (k) k 3. φ (k) lim = 2. k k This is a result analog to the result given in [6]. The techniques of proving this are similar to those of [6], but somewhat more complicated. Lemma 5.. Let t, k and q be such natural numbers that [ ( ) ( ) 2q + 2k + 4 k + 4t 2 2q + 2k + 4 k + 4t 3 ] 2k (2k ) ( ) q 3 < 4 then there is a directed graph D such that its vertices can be divided into five pairwise disjoint sets C, C 2, C 3, C 4 and C 5 in such a way that: ) C = q 2) C 2 = q 3) C 3 = k
13 MIXDECOMPOSITON OF THE COMPLETE GRAPH 223 4) C 4 = k 5) C 5 = 4 k + 4t 2 6) For any two vertices x X and y Y either (x, y) E (D ) or (y, x) E (D ), where {X, Y } {{C, C 2 }, {C, C 4 }, {C, C 5 }, {C 2, C 3 }, {C 2, C 5 }, {C }, {C 2 }} 7) For any two vertices x X and y Y neither (x, y) E (D ) nor (y, x) E (D ), where {X, Y } {{C 3 }, {C 4 }, {C 5 }, {C, C 3 }, {C 2, C 4 }, {C 3, C 4 }, {C 3, C 5 }, {C 4, C 5 }} 8) For each pair of vertices (x, y) (V (D ) V (D )) \ ((C 3 C 4 ) (C 3 C 4 )). we have d D (x, y) 2. Proof. Let D be a random graph such that V (D ) = C C 2 C 3 C 4 C 5 such that ) For any two vertices x X and y Y either (x, y) E (D ) or (y, x) E (D ), where {X, Y } {{C, C 2 }, {C, C 4 }, {C, C 5 }, {C 2, C 3 }, {C 2, C 5 }, {C }, {C 2 }} and the probability of each direction is 2. 2) For any two vertices x X and y Y neither (x, y) E (D ) nor (y, x) E (D ), where {X, Y } {{C 3 }, {C 4 }, {C 5 }, {C, C 3 }, {C 2, C 4 }, {C 3, C 4 }, {C 3, C 5 }, {C 4, C 5 }} Denote the following condition by ( ): For each pair of vertices (x, y) (V (D ) V (D )) \ ((C 3 C 4 ) (C 3 C 4 )). we have d D (x, y) 2. Let us calculate probability that D does not satisfy ( ). First we estimate the probability prob (x, y) that d D (x, y) > 2, where x y and (x, y) (V (D ) V (D )) \ ((C 3 C 4 ) (C 3 C 4 )).
14 224 D. VUKIČEVIĆ The following 2 cases, described in the following table may occur: x y prob (x, y) C C 2 ( 3 4 C C 2 2 ( 3 4 ( C C C C 4 2 ( 3 4 C C 5 2 ( 3 4 C 2 C 2 ( 3 4 C 2 C 2 2 ( 3 4 C 2 C 3 2 ( 3 ( 4 C 2 C C 2 C 5 2 ( 3 ( 4 C 3 C 3 4 C 3 C 2 2 ( 3 ( 4 C 3 C C 4 C 2 ( 3 ( 4 C 4 C 3 2 ( 4 C 4 C C 5 C 2 ( 3 4 C 5 C 2 2 ( 3 ( 4 C 5 C 3 3 ( 4 C 5 C 3 4 ( 4 C 5 C ) 2q+k+4 k +4t 4 ) 2q+4 k +4t 4 ) q ) q ) 2q ) 2q+4 k +4t 4 ) 2q+k+4 k +4t 4 ) q ) q ) 2q ) q ) q ) q ) q ) q ) q ) 2q ) 2q ) q ) q ) 2q Thus, prob (x, y) ( 3 4) q for any two different vertices x, y such that (x, y) (V (D ) V (D )) \ ((C 3 C 4 ) (C 3 C 4 )). Therefore, a probability that D does not satisfy ( ) is less then [ ( ) ( ) 2q + 2k + 4 k + 4t 2 2q + 2k + 4 k + 4t 3 ] 2k (2k ) ( ) q 3 4 Since the last expression is less than, there is a graph with the required properties. Now we can prove:
15 MIXDECOMPOSITON OF THE COMPLETE GRAPH 225 k Theorem 5.2. For any k N, we have φ (k) 2k t 2 + 2q where t is the least natural number such that ( ) 2t k. t and q is the least natural number such that [ ( ) 2q + 2k + 4 k + 4t 2 ] 2k (2k ) ( 2q + 2k + 4 ( ) q 3 <. 4 ) k + 4t 3 k Proof. We will construct a decomposition of K n, n = 2k t 2 + 2q, into undirected factors F, F 2,..., F k and directed factor F such that diam F = 2 and diam F i = 3, i k. Let V (K n ) = L D W Z U U A B B C, where } L = {l,..., l k }, D = {d,..., d k }, W = {w,..., w k, } } { } Z = {z,..., z k, U = {u,..., u k, U = u,..., u k, A = {a,..., a q }, B = {b,..., b 2t }, B = { b,..., b 2t }, C = {c,..., c q }. Let B be the set of all subsets of t elements of the set {, 2,..., 2t }. Let f be any injection f : {,..., k} B. Let us notice that for each j {,..., k} there are unique numbers q j and r j so that j = (q j ) k + r j, q j k, r j k. The edges of the factor F i, i k are ) l i d i 2) l i l j, j < i 3) d i l j, i < j k 4) l i d j, i < j k 5) d i d j, j < i 6) l i a j, j q 7) d i c j, j q 8) l i b j, l i b j, j f (i) 9) d i b j, d i b j, j {, 2,..., 2t } \ f (i) k 0) l i w j, j
16 226 D. VUKIČEVIĆ k ) d i z j, j 2) w qi u ri, z qi u ri, w qi u r i, z qi u k r i 3) d i u j, d i u j, j, j r i. In each factor F i, i k all vertices are adjacent to either l i or d i, except u ri and u r i each of which is connected by a path of length 2 to both, l i and d i. Also, l i and d i are adjacent, and there is a path of length 2 which connects u ri and u r i, hence we have diam F i = 3, i k. Let D be a digraph with a properties required in the previous lemma and let g : V (K n ) V (D ) be a bijection such that Directed edges of F are: g (A) = C g (C) = C 2 g (L) = C 3 g (D) = C 4 g (B B V Z U U ) = C 5 ) (x, y) such that (g (x), g (y)) E (D ) 2) (l i, u ri ), (u ri, d i ), ( ) ( ) d i, u r i, u ri, l i 3) (b j, d i ), ( d i, b j), j f (i) 4) (l i, b j ), ( b j, l i), j {, 2,..., t } \ f (i) 5) Edges in ( k { xy : [(x, y) is edge of D listed in )5)] E (K n ) \ E (F i ) [(y, x) is edge of D listed in )5)] i= with an arbitrary orientations. It remains to prove that diam F 2, i.e. that for arbitrary x, y, we have d (x, y) 2. Distinguish 5 cases: ) x = l i, y = d i, i k. There is a path l i u ri d i. 2) x = d i, y = l i, i k. There is a path d i u r i l i. 3) x = l i, y = d j, i j, i, j k. Since f is a bijection, there is m f (j) \ f (i) and therefore, there is a path l i b m d j. 4) x = d i, y = l j, i j, i, j k. Since f is a bijection, there is m f (i) \ f (j) and therefore, there is a path d i b ml j. } )
17 MIXDECOMPOSITON OF THE COMPLETE GRAPH 227 5) x V (K n ) \ (L D) or y V (K n ) \ (L D). There is a path of length at most 2 from x to y consisting of edges listed in ). All the cases are exhausted and the claim is proved. From the last theorem, it easily follows Corollary 5.3. lim k φ(k) k = 2. Proof. Let k N be sufficiently large. Let us find upper and lower bounds for φ (k). We have ( ) φ (k) k (φ (k) ) k φ (k) φ (k) 2k. 2 2 Let us notice that, for sufficiently large k, we have thus t therefore q ( k 2 ) k k, k. Also, for sufficiently large k, we have [ ( ) ( ) 2q + 2k + 8 k 2 2q + 2k + 8 k 3 k. It follows that ] 2k (2k ) ( ) 2k φ (k) 2k + 0 k + which proves the claim. lim k 2 lim k ( ) 3 k <, 4 2 φ (k) k k k. ( ) ( φ (k) lim ), k k k k Let us generalize the last corollary. By a simple probabilistic argument (similarly as in Lemma 5.), we show that: Lemma 5.4. Let p, r, r and k be natural numbers such that ( ( ) ) 2 r p [(r + 2r + 2k) (r + 2r + 2k ) 2k (2k )] <. 2p There is a digraph D r,r such that its vertices can be divided into five pairwise disjoint sets C (r,r ),, C (r,r ),2, C (r,r ),3, C (r,r ),4, and C (r,r ),5 in such way that: ) C(r,r ), = r
18 228 D. VUKIČEVIĆ 2) C(r,r ),2 = r 3) C (r,r ),3 = k 4) C(r,r ),2 = k 5) C(r,r ),2 = r 6) For any two vertices x X and y Y either (x, y) E (D r,r ) or (y, x) E (D r,r ), where { { } { } { } } C(r,r {X, Y } { ),, C (r,r ),2 } {, C(r,r ),, C (r,r ),4 } {, C(r,r ),, } C { (r,r ),5, C(r,r ),2, C (r,r ),3, C(r,r ),2, C (r,r ),5, C(r,r ),, C(r,r ),2} 7) For any two vertices x X and y Y neither (x, y) E (D r,r ) nor (y, x) E (D r,r ), where { } { } { } C(r,r ),, C { (r,r ),3, C(r,r ),2, } C { (r,r ),4, C(r,r ),3, } C (r,r ),4, {X, Y } C(r,r ),3, C (r,r } { ),5, C(r,r } { ),4, C (r,r }),5,, C(r,r ),4, C(r,r ),5, { C(r,r ),3 which can be decomposed into p factors P (r,r ),,..., P (r,r ),p such that 8) For each pair of vertices (x, y) (V (D r,r ) V (D r,r ))\ (( C (r,r ),3 C (r,r ),4) ( C(r,r ),3 C (r,r ),4)). and each factor P (r,r ),i, i k, we have d P(r,r ),i, (x, y) 2. Proof. Let D r,r be a random digraph such that its vertices can be decomposed into five pairwise disjoint sets C (r,r ),, C (r,r ),2, C (r,r ),3, C (r,r ),4, and C (r,r ),5 in such way that: ) For any two vertices x X and y Y either (x, y) E ( ) D r,r or (y, x) E ( ) D r,r, where { { } { } { } } C(r,r {X, Y } { ),, C (r,r ),2} {, C(r,r ),, C (r,r ),4} {, C(r,r ),, } C { (r,r ),5, C(r,r ),2, C (r,r ),3, C(r,r ),2, C (r,r ),5, C(r,r ),, C(r,r ),2} and probability of each direction is 2. 2) For any two vertices x X and y Y neither (x, y) E ( ) D r,r nor (y, x) E ( ) D r,r, where {X, Y } { } { } { } C(r,r ),, C { (r,r ),3, C(r,r ),2, } C { (r,r ),4, C(r,r ),3, } C (r,r ),4, C(r,r { ),3, C (r,r } {),5, C(r,r } {),4, C (r,r }),5, C(r,r ),3, C(r,r ),4, C(r,r ),5, Let P (r,r ),,..., P (r,r ),p be a random decomposition of random digraph (each directed edge of (x, y) has a probability p to be a directed edge of, for every i {,..., p}). Denote by ( ) the following condition: P (r,r ),i For each pair of vertices (x, y) ( V ( D r,r ) V ( D r,r )) \ (( C(r,r ),3 C (r,r ),4) ( C(r,r ),3 C (r,r ),4)). and each factorp (r,r ),i, i k, we have d P(r,r ),i, (x, y) 2.
19 MIXDECOMPOSITON OF THE COMPLETE GRAPH 229 Let us calculate the probability that D r,r does not satisfy ( ). First, we estimate the probability prob (x, y, i) that d P (x, y) > 2, where (x, y) (r,r ),i (V (D ) V (D )) \ (( ( C (r,r ),3 C (r,r ),4) C(r,r ),3 C (r,r ),4)) and i p. Analogously as in the Lemma 5., we solve this problem by observing 2 possible cases. We get ( ( ) ) 2 r prob (x, y, i). 2p for each (x, y) ( V ( D r,r ) V ( D r,r )) \ (( C(r,r ),3 C (r,r ),4) ( C(r,r ),3 C (r,r ),4)). Therefore, a probability that D r,r does not satisfy ( ) is at most ( ( ) ) 2 r p [(r + 2r + 2k) (r + 2r + 2k ) 2k (2k )]. 2p Since the last expression is less than, there is a graph with the required properties. We can now prove: Theorem 5.5. Let p and k be any natural numbers. Then K n can be mixeddecomposed into k undirected factors F, F 2,..., F k of diameter 3 and p directed factors D, D 2,..., D p of diameter 2 where n = 2k + 2r + 2p (2t ) + (2p + 2) k, where t is the smallest integer such that ( ) 2t k t and r is the smallest integer such that p [(r + 2r + 2k) (r + 2r + 2k ) 2k (2k )] ( ( ) ) 2 r <. 2p where r = 2p (2t ) + (2p + 2) k. Proof. Denote p p p p V (K n ) = L R A B C α E α X Y U α V α, α= α= α= α=
20 230 D. VUKIČEVIĆ L = {l,..., l k }, R = {r,..., r k }, A = {a,... a r }, B = {b,..., b r }, } } X = {x,..., x k, Y = {y,..., y k, and for, each α =,..., p, denote C α = { c α,..., 2t } cα, Eα = { e α,..., 2t } eα, { } { } U α = u α,..., uα k, V α = v α,..., vα k. Let us notice, as in the Theorem 5.2 that for each j {,..., k} there are unique numbers Q j and R j such that j = (Q j ) k + R j, Q j k, R j k. Also, as in the Theorem 5.2, let B be the set of all subsets of t elements of the set {, 2,..., 2t } and let f be any injection f : {,..., k} B. We explicitly give a mixeddecomposition with the required properties. The edges of F i, i =,..., k are: ) l i r i 2) l i l j, r i r j, i < j k 3) l i r j, r i l j, j < i 4) l i a j, j =,..., r 5) r i b j, j =,..., r 6) l i c α j, l ie α j, j f (i), α =,..., p 7) r i c α j, r ie α j, j {, 2,..., 2t } \ f (i), α =,..., p k 8) l i x j, r i y j, j =,..., 9) x Qi u α R i, y Qi u α R i, x Qi v α R i, y Qi v α R i, α =,..., p 0) l i u α j, r iv α j, j k, j R i, α p It can be easily checked that diam F i = 3, for each i =,..., k. Let g : V (K n ) V (D r,r ) be any bijection such that g ( X Y g (A) = C (r,r ), g (B) = C (r,r ),2 g (L) = C (r,r ),3 g (R) = C (r,r ),4 ) p (C α E α U α V α ) = C (r,r ),5 α=
21 MIXDECOMPOSITON OF THE COMPLETE GRAPH 23 Directed edges of D α, α =,..., p are: ) (x, y) such that ( ) g (x), g (x) P (r,r ),α 2) ( ) ( ) ( ) ( ) l i, u α R i, u α Ri, r i, ri, vr α i, v α Ri, l i 3) ( ) ( ) l i, c α j, e α j, l i, j {, 2,..., 2t } \ f (i), i =,..., k 4) ( c α j, r ( ) i), ri, e α j, j f (i), i =,..., k. Edges in k E (F i ) p E ( D α ) E (K n ) \ { i= α=2 } xy : [(x, y) is edge of D listed in )4)] [(y, x) is edge of D listed in )4)] are directed edges of D with an arbitrary orientations. Let us prove that diam D α 2, α =,..., p, i.e. that d Dα (x, y) 2, for each x, y V (K n ). Distinguish 5 cases: ) x = l i, y = r i. There is a path l i u α R i r i. 2) x = r i, y = l i. There is a path r i v α R i l i. 3) x = l i, y = r j, i j. Since f is a bijection, there is m f (j) \ f (i) and therefore, there is a path l i c α m r j. 4) x = r i, y = l j, i j. Since f is a bijection, there is m f (i) \ f (j) and therefore, there is a path r i e α m l j. 5) x V (K n ) \ (L R) or y V (K n ) \ (L D). There is a path of length at most 2 from x to y consisting of edges listed in ). All the cases are exhausted and the claim is proved. Corollary 5.6. Let p be a fixed natural number and let Φ (p, k) be the smallest natural number such that K Φ(p,k) can be decomposed into p directed factors of diameter 2 and k undirected factors of diameter 3. Then Φ (p, k) lim = 2. k k Proof. From the last theorem, it follows that Φ (p, k) 2k + 2r + 2p (2t ) + (2p + 2) where t is the smallest integer such that ( ) 2t k t k,
22 232 D. VUKIČEVIĆ and r is the smallest integer such that p [(r + 2r + 2k) (r + 2r + 2k ) 2k (2k )] ( where r = 2p (2t ) + (2p + 2) k. k Note that, for sufficiently large k, we have t and r φ (k) Φ (p, k) 2k + (6p + 4) k φ (k) lim k k which proves the claim. ( ) ) 2 r <. 2p k, hence k Φ (p, k) (6p + 4) lim lim 2 +, k k k k Acknowledgements. The author would like to thank D. Veljan, D. Svrtan, G. Igally, and G. Duplančić for useful advises and help. References [] B. Bollobás: Extremal Graph Theory, Academic Press, London, 978. [2] J. Bosák: Decompositions of Graphs, Kluwer Academic Publishers, Dordecht, 989. [3] J. Bosák: Disjoint Factors of Diameter Two in Complete Graphs, Journal Of Combinatorial Theory (B) 6 (974), [4] D. Palumbíny: On Decompositions of Complete Graphs into Factors with Equal Diameters, Bollettino U.M.I. (4) 7 (973), [5] N. Sauer: On Factorization of Complete Graphs into Factors of Diameter Two, Journal of Combinatorial Theory 9 (970), [6] D. Vukičević: Decomposition of Complete Graph into Factors of Diameter Two and Three, Discussiones Mathematicae Graph Theory 23 (2003), [7] Š. Znám: Decomposition of Complete Graphs into Factors of Diameter Two, Mathematica Slovaca 30 (980), [8] Š. Znám: On a Conjecture of a Bollobás and Bosák, Journal of Graph Theory 6 (962), D. Vukičević Department of Mathematics University of Split Teslina 2, 2000 Split Croatia Received: Revised:
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