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S̄k -factorization of symmetric complete tripartite digraphs
Discrete Mathematics 211 (2000) 281–286
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Note
S k -factorization of symmetric complete tripartite digraphs Kazuhiko Ushio ∗ Department of Industrial Engineering, Faculty of Science and Technology, Kinki University, Osaka 577-8502, Japan Received 28 October 1997; revised 7 July 1998; accepted 8 February 1999
Abstract We show that a necessary and sucient condition for the existence of an Sk -factorization of the symmetric complete tripartite digraph Kn∗1 ; n2 ; n3 is (i) k is odd, k¿3 and (ii) n1 = n2 = n3 c 2000 Elsevier Science B.V. All for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. rights reserved. Keywords: Star-factorization; Symmetric complete tripartite digraph
1. Introduction Let Kn∗1 ; n2 ; n3 denote the symmetric complete tripartite digraph with partite sets V1 ; V2 ; V3 of n1 ; n2 ; n3 vertices each, and let Sk denote the evenly partite directed star K1; k−1 in which all arcs are directed away from a center-vertex to k − 1 end-vertices such that the center-vertex is in Vi and (k − 1)=2 end-vertices are in Vj1 and (k − 1)=2 end-vertices are in Vj2 with {i; j1 ; j2 } = {1; 2; 3}. A spanning subgraph F of Kn∗1 ; n2 ; n3 is called an Sk -factor if each component of F is Sk . If Kn∗1 ; n2 ; n3 is expressed as an arc-disjoint sum of S k -factors, then this sum is called an Sk -factorization of Kn∗1 ; n2 ; n3 . In this paper, it is shown that a necessary and sucient condition for the existence of such a factorization is (i) k is odd, k¿3 and (ii) n1 = n2 = n3 for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. Let Kn1 ; n2 , Kn∗1 ; n2 , Kn∗1 ; n2 ; n3 , and Kn∗1 ; n2 ;:::; nm denote the complete bipartite graph, the symmetric complete bipartite digraph, the symmetric complete tripartite digraph, and the symmetric complete multi-partite digraph, respectively, and let Cˆ k , Sˆk , Pˆ k , and Kˆ p;q denote the cycle or the directed cycle, the star or the directed star, the path or the directed path, and the complete bipartite graph or the complete bipartite digraph, ∗
Tel.: 81-6-6721-2332; fax: 81-6-6730-1320. E-mail address: [email protected] (K. Ushio) c 2000 Elsevier Science B.V. All rights reserved. 0012-365X/00/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 9 9 ) 0 0 1 5 4 - 5
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respectively, on two partite sets Vi and Vj . Then the problems of giving the necessary and sucient conditions of Cˆ k -factorization of Kn1 ; n2 , Kn∗1 ; n2 , and Kn∗1 ; n2 ; n3 have been completely solved by Enomoto et al. [2] and Ushio [13]. Sˆk -factorization of Kn1 ; n2 , Kn∗1 ; n2 , and Kn∗1 ; n2 ; n3 have been studied by Ushio and Tsuruno [17], Ushio [15], and Wang [19]. Recently, Martin [4,5] and Ushio [12] have given the necessary and sucient conditions of Sˆk -factorization of Kn1 ; n2 and Kn∗1 ; n2 . Pˆ k -factorization of Kn1 ; n2 and Kn∗1 ; n2 have been studied by Ushio and Tsuruno [16], and Ushio [10]. Kˆp;q -factorization of Kn1 ; n2 has been studied by Martin [4]. Ushio [14] gives the necessary and sucient condition of Kˆ p;q -factorization of Kn∗1 ; n2 . For graph theoretical terms, see [1,3]. The motivation or application of Sk -factorization of symmetric complete tripartite digraphs begins with star-decomposition of complete graphs and complete bipartite graphs by Yamamoto, Ikeda, Shige-eda, Ushio and Hamada [20] and is continued to star-decomposition and partite-star-decomposition and evenly partite-star-decomposition of complete multipartite graphs by Ushio et al. [18], Tazawa et al. [7], Tazawa [6], and Ushio [8]. Those decompositions are applied in design of experiments as G-designs by Ushio [9,11]. Those decompositions are also applied in database systems as combinatorial index- le organization schemes by Yamamoto et al. [21], Yamamoto et al. [22], and Yamamoto et al. [23]. Sk -factorization of symmetric complete tripartite digraphs will be generalized to S k -factorization of symmetric complete multipartite digraphs and will be applied in design of experiments as a new type of resolvable G-design and also be applied in database systems as a new type of combinatorial index- le organization scheme. 2. S k -factorization of Kn∗1 ; n2 ; n3 Notation. Given an Sk -factorization of Kn∗1 ; n2 ; n3 , let r be the number of factors, t the number of components of each factor, and b the total number of components. Among r components having vertex x in Vi , let rij be the number of components whose center-vertex is in Vj . Among t components of each factor, let ti be the number of components whose center-vertex is in Vi . Theorem 1. If Kn∗1 ; n2 ; n3 has an Sk -factorization; then (i) k is odd; k¿3 and (ii) n1 = n2 = n3 for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. Proof. Suppose that Kn∗1 ; n2 ; n3 has an Sk -factorization. Then b = 2(n1 n2 + n1 n3 + n2 n3 )= (k − 1); t = (n1 + n2 + n3 )=k; r = b=t = 2(n1 n2 + n1 n3 + n2 n3 )k=(n1 + n2 + n3 )(k − 1). By the de nition of Sk , k is odd and k¿3. Put k = 2a + 1. For a vertex x in V1 , there are n2 and n3 arcs directed away from x to V2 and V3 , respectively, and there are also n2 and n3 arcs directed away from V2 and V3 , respectively, to x. So we have r11 a = n2 , r11 a = n3 , r12 = n2 , r13 = n3 , and r11 + r12 + r13 = r. For a vertex x in V2 , there are n1 and n3 arcs directed away from x
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to V1 and V3 , respectively, and there are also n1 and n3 arcs directed away from V1 and V3 , respectively, to x. So we have r22 a = n1 , r22 a = n3 , r21 = n1 , r23 = n3 , and r21 + r22 + r23 = r. For a vertex x in V3 , there are n1 and n2 arcs directed away from x to V1 and V2 , respectively, and there are also n1 and n2 arcs directed away from V1 and V2 , respectively, to x. So we have r33 a = n1 , r33 a = n2 , r31 = n1 , r32 = n2 , and r31 + r32 + r33 = r. From these relations, we have n1 = n2 = n3 . Put n1 = n2 = n3 = n. Then r11 = r22 = r33 = n=a, r12 = r13 = r21 = r23 = r31 = r32 = n, b = 3n2 =a, t = 3n=(2a + 1), and r = n(2a + 1)=a. Moreover, in a factor, we have t1 + t2 a + t3 a = n, t1 a + t2 + t3 a = n, t1 a + t2 a + t3 = n, t1 + t2 + t3 = t. Therefore, we have ti = (ta − n)=(a − 1) = n=(2a + 1) for a¿2. So we have n1 = n2 = n3 for a = 1 and n1 = n2 = n3 ≡ 0 (mod (2a + 1)a) for a¿2. Lemma 2. Let G; H and K be digraphs. If G has a H -factorization and H has a K-factorization; then G has a K-factorization. Sr Proof. Let E(G) = i=1 E(Fi ) be an H -factorization of G. Let Hj(i) (16j6t) be the Ss components of Fi . And let E(Hj(i) ) = k=1 E(Kk(i; j) ) be a K-factorization of Hj(i) . Then S r Ss St E(G) = i=1 k=1 E( j=1 Kk(i; j) ) is a K-factorization of G. Let Kq1 ; q2 ⊕q3 denote the tripartite digraph with partite sets U1 , U2 , U3 of q1 ; q2 ; q3 vertices such that q1 q2 arcs are directed away from q1 vertices in U1 to q2 vertices in U2 and q1 q3 arcs are directed away from q1 vertices in U1 to q3 vertices in U3 . Notation. For an Sk whose center-vertex is u and end-vertices are v1 ; v2 ; : : : ; vk−1 , we denote (u; v1 ; v2 ; : : : ; vk−1 ). Lemma 3. Let k = 2a + 1. Ks; sa⊕sa has an Sk -factorization. Proof. Let U1 = {1; 2; : : : ; s}, U2 = {10 ; 20 ; : : : ; (sa)0 }, U3 = {100 ; 200 ; : : : ; (sa)00 }. Construct s S k -factors Fi (i = 1; 2; : : : ; s) as following: Fi = {(1; (A + 1; : : : ; A + a)0 ; (A + 1; : : : ; A + a)00 ) (2; (A + a + 1; : : : ; A + 2a)0 ; (A + a + 1; : : : ; A + 2a)00 ) ::: (s; (A + (s − 1)a + 1; : : : ; A + sa)0 ; (A + (s − 1)a + 1; : : : ; A + sa)00 )}; where A = (i − 1)s, and the additions are taken modulo sa with residues 1; 2; : : : ; sa. Then they comprise an Sk -factorization of Ks; sa⊕sa . ∗ Theorem 4. If Kn;∗ n; n has an Sk -factorization; then Ksn; sn; sn has an S k -factorization.
Proof. Put k = 2a + 1. Then S k can be denoted as K1;a⊕a . Since Kn;∗ n; n has a K1;a⊕a ∗ factorization, Ksn; sn; sn has a Ks; sa⊕sa -factorization. By Lemma 3, Ks; sa⊕sa has an ∗ S k -factorization. Therefore, by Lemma 2 Ksn; sn; sn has an S k -factorization.
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Theorem 5. When k = 3; Kn;∗ n; n has an Sk -factorization. Proof. Let Vi = {vi;1 ; vi;2 ; : : : ; vi; n } (i = 1; 2; 3). Construct 3n Sk -factors Fij (i = 1; 2; 3; j = 1; 2; : : : ; n) as following: F1j = {(v1;1 ; v2; j ; v3; j ); (v1; 2 ; v2; j+1 ; v3; j+1 ); : : : ; (v1; n ; v2; j+n−1 ; v3; j+n−1 )}; F2j = {(v2;1 ; v1; j ; v3; j ); (v2;2 ; v1; j+1 ; v3; j+1 ); : : : ; (v2; n ; v1; j+n−1 ; v3; j+n−1 )}; F3j = {(v3;1 ; v1; j ; v2; j ); (v3; 2 ; v1; j+1 ; v2; j+1 ); : : : ; (v3; n ; v1; j+n−1 ; v2; j+n−1 )}; where the additions are taken modulo n with residues 1; 2; : : : ; n. Then they comprise an S k -factorization of Kn;∗ n; n . Theorem 6. When k is odd; k¿5 and n ≡ 0 (mod k(k − 1)=2); Kn;∗ n; n has an Sk factorization. Proof. Put k = 2a + 1, n = s(2a + 1)a, N = (2a + 1)a. When s = 1, let V1 = {1; 2; : : : ; N }, V2 ={10 ; 20 ; : : : ; N 0 }, V3 ={100 ; 200 ; : : : ; N 00 }. For i =1; 2; : : : ; 2a+1 and j =1; 2; : : : ; 2a+1, construct (2a + 1)2 Sk -factors Fij as following: Fij = {((A + 1); (B + 1; : : : ; B + a)0 ; (C + 1; : : : ; C + a)00 ) ((A + 2); (B + a + 1; : : : ; B + 2a)0 ; (C + a + 1; : : : ; C + 2a)00 ) ::: ((A + a); (B + (a − 1)a + 1; : : : ; B + a2 )0 ; (C + (a − 1)a + 1; : : : ; C + a2 )00 ) ((B + a2 + 1)0 ; (A + a + 1; : : : ; A + 2a); (C + a2 + 1; : : : ; C + a2 + a)00 ) ((B + a2 + 2)0 ; (A + 2a + 1; : : : ; A + 3a); (C + a2 + a + 1; : : : ; C + a2 + 2a)00 ) ::: ((B + a2 + a)0 ; (A + a2 + 1; : : : ; A + a2 + a); (C + a2 + (a − 1)a + 1; : : : ; C + 2a2 )00 ) ((C + 2a2 + 1)00 ; (A + a2 + a + 1; : : : ; A + a2 + 2a); (B + a2 + a + 1; : : : ; B + a2 + 2a)0 ) ((C + 2a2 + 2)00 ; (A + a2 + 2a + 1; : : : ; A + a2 + 3a); (B + a2 + 2a + 1; : : : ; B + a2 + 3a)0 ) ::: ((C + 2a2 + a)00 ; (A + 2a2 + 1; : : : ; A + 2a2 + a); (B + 2a2 + 1; : : : ; B + 2a2 + a)0 )}; where A = (i − 1)a, B = (j − 1)a, C = (i + j − 2)a, and the additions are taken modulo N with residues 1; 2; : : : ; N .
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∗ Then we claim that they comprise an Sk -factorization of KN; N; N . We can see that ∗ each of them is an S k -factor, because it spans all vertices of KN; N; N . We show that they are arc-disjoint. Suppose that they are not arc-disjoint. In the following, we consider A = (i − 1)a, B=(j−1)a, C =(i+j−2)a, D=(k −1)a, E=(l−1)a, F =(k +l−2)a, 16i; j; k; l62a+1, 16x; y; z; w6a. Note that A; B; C; D; E; F; N are integral multiples of a. First, we assume that the common arc joining from V1 to V2 or from V1 to V3 appears in xth component ((A+x);(B+(x−1)a+1; : : : ; B+(x−1)a+a)0 , (C +(x−1)a+1; : : : ; C + (x−1)a+a)00 ) of Fij and yth component ((D+y); (E+(y−1)a+1; : : : ; E+(y−1)a+a)0 , (F + (y − 1)a + 1; : : : ; F + (y − 1)a + a)00 ) of Fkl . Say ((A + x); (B + (x − 1)a + z)0 ) = ((D + y); (E + (y − 1)a + w)0 ). Then A + x ≡ D + y (mod N ) and B + (x − 1)a + z ≡ E + (y − 1)a + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Say ((A + x); (C + (x − 1)a + z)00 ) = ((D + y); (F + (y − 1)a + w)00 ). Then A + x ≡ D + y (mod N ) and C + (x − 1)a + z ≡ F + (y − 1)a + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Next, we assume that the common arc joining from V2 to V1 or from V2 to V3 appears in (a + x)th component ((B + a2 + x)0 ; (A + xa + 1; : : : ; A + xa + a); (C + a2 + (x − 1)a + 1; : : : ; C + a2 + (x − 1)a + a)00 ) of Fij and (a + y)th component ((E+a2 +y)0 ; (D+ya+1; : : : ; D+ya+a); (F +a2 +(y−1)a+1; : : : ; F +a2 +(y−1)a+a)00 ) of Fkl . Say ((B + a2 + x)0 ; (A + xa + z)) = ((E + a2 + y)0 ; (D + ya + w)): Then B + a2 + x ≡ E + a2 + y (mod N ) and A + xa + z ≡ D + ya + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Say ((B + a2 + x)0 ; (C + a2 + (x − 1)a + z)00 ) = ((E + a2 + y)0 ; (F + a2 + (y − 1)a + w)00 ). Then B+a2 +x ≡ E +a2 +y (mod N ) and C +a2 +(x−1)a+z ≡ F +a2 +(y−1)a+w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Last, we assume that the common arc joining from V3 to V1 or from V3 to V2 appears in (2a + x)th component ((C + 2a2 + x)00 ; (A + a2 + xa + 1; : : : ; A + a2 + xa + a); (B + a2 + xa + 1; : : : ; B + a2 + xa + a)0 ) of Fij and (2a + y)th component ((F +2a2 +y)00 ; (D+a2 +ya+1; : : : ; D+a2 +ya+a); (E+a2 +ya+1; : : : ; E+a2 +ya+a)0 ) of Fkl . Say ((C + 2a2 + x)00 ; (A + a2 + xa + z)) = ((F + 2a2 + y)00 ; (D + a2 + ya + w)). Then C + 2a2 + x ≡ F + 2a2 + y (mod N ) and A + a2 + xa + z ≡ D + a2 + ya + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Say ((C + 2a2 + x)00 ; (B + a2 + xa + z)0 ) = ((F + 2a2 + y)00 ; (E + a2 + ya + w)0 ). Then C + 2a2 + x ≡ F + 2a2 + y (mod N ) and B + a2 + xa + z ≡ E + a2 + ya + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption.
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∗ Therefore, (2a + 1)2 Sk -factors Fij comprise an Sk -factorization of KN;N;N . Applying ∗ Theorem 4, Kn; n; n has an S k -factorization.
Main Theorem. Kn∗1 ; n2 ; n3 has an Sk -factorization if and only if (i) k is odd; k¿3 and (ii) n1 = n2 = n3 for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. References [1] G. Chartrand, L. Lesniak, Graphs & Digraphs, 2nd Edition, Wadsworth, California, 1986. [2] H. Enomoto, T. Miyamoto, K. Ushio, Ck -factorization of complete bipartite graphs, Graphs Combin. 4 (1988) 111–113. [3] F. Harary, Graph Theory, Addison-Wesley, Massachusetts, 1972. [4] N. Martin, Complete bipartite factorisations by complete bipartite graphs, Discrete Math. 167/168 (1997) 461–480. [5] N. Martin, Balanced bipartite graphs may be completely star-factored, J. Combin. Des. 5 (1997) 407–415. [6] S. Tazawa, Claw-decomposition and evenly-partite-claw-decomposition of complete multi-partite graphs, Hiroshima Math. J. 9 (1979) 503–531. [7] S. Tazawa, K. Ushio, S. Yamamoto, Partite-claw-decomposition of a complete multi-partite graph, Hiroshima Math. J. 8 (1978) 195–206. [8] K. Ushio, Bipartite decomposition of complete multipartite graphs, Hiroshima Math. J. 11 (1981) 321–345. [9] K. Ushio, On balanced claw designs of complete multi-partite graphs, Discrete Math. 38 (1982) 117–119. [10] K. Ushio, P3 -factorization of complete bipartite graphs, Discrete Math. 72 (1988) 361–366. [11] K. Ushio, G-designs and related designs, Discrete Math. 116 (1993) 299–311. [12] K. Ushio, Star-factorization of symmetric complete bipartite digraphs, Discrete Math. 167/168 (1997) 593–596. [13] K. Ushio, Cˆ k -factorization of symmetric complete bipartite and tripartite digraphs, J. Fac. Sci. Technol. Kinki Univ. 33 (1997) 221–222. [14] K. Ushio, Kˆ p;q -factorization of symmetric complete bipartite digraphs, Graph Theory, Combinatorics, Algorithms and Applications, New Issues Press, 1998, pp. 823–826. [15] K. Ushio, Sˆk -factorization of symmetric complete tripartite digraphs, Discrete Math. 197/198 (1999) 991–999. [16] K. Ushio, R. Tsuruno, P3 -factorization of complete multipartite graphs, Graphs Combin. 5 (1989) 385–387. [17] K. Ushio, R. Tsuruno, Cyclic Sk -factorization of complete bipartite graphs, Graph Theory, Combinatorics, Algorithms and Applications, SIAM, Philadelphia, PA, 1991, pp. 557–563. [18] K. Ushio, S. Tazawa, S. Yamamoto, On claw-decomposition of a complete multi-partite graph, Hiroshima Math. J. 8 (1978) 207–210. [19] H. Wang, On K1; k -factorizations of a complete bipartite graph, Discrete Math. 126 (1994) 359–364. [20] S. Yamamoto, H. Ikeda, S. Shige-eda, K. Ushio, N. Hamada, On claw-decomposition of complete graphs and complete bigraphs, Hiroshima Math. J. 5 (1975) 33–42. [21] S. Yamamoto, H. Ikeda, S. Shige-eda, K. Ushio, N. Hamada, Design of a new balanced le organization scheme with the least redundancy, Inform. Control 28 (1975) 156–175. [22] S. Yamamoto, S. Tazawa, K. Ushio, H. Ikeda, Design of a generalized balanced multiple-valued le organization scheme of order two, in: E. Lowenthal, N.B. Dale (Eds.), Proceedings of the ACM-SIGMOD International Conference on Management of Data, 1978, pp. 47–51. [23] S. Yamamoto, S. Tazawa, K. Ushio, H. Ikeda, Design of a balanced multiple-valued le organization scheme with the least redundancy, ACM Trans. Database Systems 4 (1979) 518–530.
www.elsevier.com/locate/disc
Note
S k -factorization of symmetric complete tripartite digraphs Kazuhiko Ushio ∗ Department of Industrial Engineering, Faculty of Science and Technology, Kinki University, Osaka 577-8502, Japan Received 28 October 1997; revised 7 July 1998; accepted 8 February 1999
Abstract We show that a necessary and sucient condition for the existence of an Sk -factorization of the symmetric complete tripartite digraph Kn∗1 ; n2 ; n3 is (i) k is odd, k¿3 and (ii) n1 = n2 = n3 c 2000 Elsevier Science B.V. All for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. rights reserved. Keywords: Star-factorization; Symmetric complete tripartite digraph
1. Introduction Let Kn∗1 ; n2 ; n3 denote the symmetric complete tripartite digraph with partite sets V1 ; V2 ; V3 of n1 ; n2 ; n3 vertices each, and let Sk denote the evenly partite directed star K1; k−1 in which all arcs are directed away from a center-vertex to k − 1 end-vertices such that the center-vertex is in Vi and (k − 1)=2 end-vertices are in Vj1 and (k − 1)=2 end-vertices are in Vj2 with {i; j1 ; j2 } = {1; 2; 3}. A spanning subgraph F of Kn∗1 ; n2 ; n3 is called an Sk -factor if each component of F is Sk . If Kn∗1 ; n2 ; n3 is expressed as an arc-disjoint sum of S k -factors, then this sum is called an Sk -factorization of Kn∗1 ; n2 ; n3 . In this paper, it is shown that a necessary and sucient condition for the existence of such a factorization is (i) k is odd, k¿3 and (ii) n1 = n2 = n3 for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. Let Kn1 ; n2 , Kn∗1 ; n2 , Kn∗1 ; n2 ; n3 , and Kn∗1 ; n2 ;:::; nm denote the complete bipartite graph, the symmetric complete bipartite digraph, the symmetric complete tripartite digraph, and the symmetric complete multi-partite digraph, respectively, and let Cˆ k , Sˆk , Pˆ k , and Kˆ p;q denote the cycle or the directed cycle, the star or the directed star, the path or the directed path, and the complete bipartite graph or the complete bipartite digraph, ∗
Tel.: 81-6-6721-2332; fax: 81-6-6730-1320. E-mail address: [email protected] (K. Ushio) c 2000 Elsevier Science B.V. All rights reserved. 0012-365X/00/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 9 9 ) 0 0 1 5 4 - 5
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respectively, on two partite sets Vi and Vj . Then the problems of giving the necessary and sucient conditions of Cˆ k -factorization of Kn1 ; n2 , Kn∗1 ; n2 , and Kn∗1 ; n2 ; n3 have been completely solved by Enomoto et al. [2] and Ushio [13]. Sˆk -factorization of Kn1 ; n2 , Kn∗1 ; n2 , and Kn∗1 ; n2 ; n3 have been studied by Ushio and Tsuruno [17], Ushio [15], and Wang [19]. Recently, Martin [4,5] and Ushio [12] have given the necessary and sucient conditions of Sˆk -factorization of Kn1 ; n2 and Kn∗1 ; n2 . Pˆ k -factorization of Kn1 ; n2 and Kn∗1 ; n2 have been studied by Ushio and Tsuruno [16], and Ushio [10]. Kˆp;q -factorization of Kn1 ; n2 has been studied by Martin [4]. Ushio [14] gives the necessary and sucient condition of Kˆ p;q -factorization of Kn∗1 ; n2 . For graph theoretical terms, see [1,3]. The motivation or application of Sk -factorization of symmetric complete tripartite digraphs begins with star-decomposition of complete graphs and complete bipartite graphs by Yamamoto, Ikeda, Shige-eda, Ushio and Hamada [20] and is continued to star-decomposition and partite-star-decomposition and evenly partite-star-decomposition of complete multipartite graphs by Ushio et al. [18], Tazawa et al. [7], Tazawa [6], and Ushio [8]. Those decompositions are applied in design of experiments as G-designs by Ushio [9,11]. Those decompositions are also applied in database systems as combinatorial index- le organization schemes by Yamamoto et al. [21], Yamamoto et al. [22], and Yamamoto et al. [23]. Sk -factorization of symmetric complete tripartite digraphs will be generalized to S k -factorization of symmetric complete multipartite digraphs and will be applied in design of experiments as a new type of resolvable G-design and also be applied in database systems as a new type of combinatorial index- le organization scheme. 2. S k -factorization of Kn∗1 ; n2 ; n3 Notation. Given an Sk -factorization of Kn∗1 ; n2 ; n3 , let r be the number of factors, t the number of components of each factor, and b the total number of components. Among r components having vertex x in Vi , let rij be the number of components whose center-vertex is in Vj . Among t components of each factor, let ti be the number of components whose center-vertex is in Vi . Theorem 1. If Kn∗1 ; n2 ; n3 has an Sk -factorization; then (i) k is odd; k¿3 and (ii) n1 = n2 = n3 for k = 3 and n1 = n2 = n3 ≡ 0 (mod k(k − 1)=2) for k¿5. Proof. Suppose that Kn∗1 ; n2 ; n3 has an Sk -factorization. Then b = 2(n1 n2 + n1 n3 + n2 n3 )= (k − 1); t = (n1 + n2 + n3 )=k; r = b=t = 2(n1 n2 + n1 n3 + n2 n3 )k=(n1 + n2 + n3 )(k − 1). By the de nition of Sk , k is odd and k¿3. Put k = 2a + 1. For a vertex x in V1 , there are n2 and n3 arcs directed away from x to V2 and V3 , respectively, and there are also n2 and n3 arcs directed away from V2 and V3 , respectively, to x. So we have r11 a = n2 , r11 a = n3 , r12 = n2 , r13 = n3 , and r11 + r12 + r13 = r. For a vertex x in V2 , there are n1 and n3 arcs directed away from x
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to V1 and V3 , respectively, and there are also n1 and n3 arcs directed away from V1 and V3 , respectively, to x. So we have r22 a = n1 , r22 a = n3 , r21 = n1 , r23 = n3 , and r21 + r22 + r23 = r. For a vertex x in V3 , there are n1 and n2 arcs directed away from x to V1 and V2 , respectively, and there are also n1 and n2 arcs directed away from V1 and V2 , respectively, to x. So we have r33 a = n1 , r33 a = n2 , r31 = n1 , r32 = n2 , and r31 + r32 + r33 = r. From these relations, we have n1 = n2 = n3 . Put n1 = n2 = n3 = n. Then r11 = r22 = r33 = n=a, r12 = r13 = r21 = r23 = r31 = r32 = n, b = 3n2 =a, t = 3n=(2a + 1), and r = n(2a + 1)=a. Moreover, in a factor, we have t1 + t2 a + t3 a = n, t1 a + t2 + t3 a = n, t1 a + t2 a + t3 = n, t1 + t2 + t3 = t. Therefore, we have ti = (ta − n)=(a − 1) = n=(2a + 1) for a¿2. So we have n1 = n2 = n3 for a = 1 and n1 = n2 = n3 ≡ 0 (mod (2a + 1)a) for a¿2. Lemma 2. Let G; H and K be digraphs. If G has a H -factorization and H has a K-factorization; then G has a K-factorization. Sr Proof. Let E(G) = i=1 E(Fi ) be an H -factorization of G. Let Hj(i) (16j6t) be the Ss components of Fi . And let E(Hj(i) ) = k=1 E(Kk(i; j) ) be a K-factorization of Hj(i) . Then S r Ss St E(G) = i=1 k=1 E( j=1 Kk(i; j) ) is a K-factorization of G. Let Kq1 ; q2 ⊕q3 denote the tripartite digraph with partite sets U1 , U2 , U3 of q1 ; q2 ; q3 vertices such that q1 q2 arcs are directed away from q1 vertices in U1 to q2 vertices in U2 and q1 q3 arcs are directed away from q1 vertices in U1 to q3 vertices in U3 . Notation. For an Sk whose center-vertex is u and end-vertices are v1 ; v2 ; : : : ; vk−1 , we denote (u; v1 ; v2 ; : : : ; vk−1 ). Lemma 3. Let k = 2a + 1. Ks; sa⊕sa has an Sk -factorization. Proof. Let U1 = {1; 2; : : : ; s}, U2 = {10 ; 20 ; : : : ; (sa)0 }, U3 = {100 ; 200 ; : : : ; (sa)00 }. Construct s S k -factors Fi (i = 1; 2; : : : ; s) as following: Fi = {(1; (A + 1; : : : ; A + a)0 ; (A + 1; : : : ; A + a)00 ) (2; (A + a + 1; : : : ; A + 2a)0 ; (A + a + 1; : : : ; A + 2a)00 ) ::: (s; (A + (s − 1)a + 1; : : : ; A + sa)0 ; (A + (s − 1)a + 1; : : : ; A + sa)00 )}; where A = (i − 1)s, and the additions are taken modulo sa with residues 1; 2; : : : ; sa. Then they comprise an Sk -factorization of Ks; sa⊕sa . ∗ Theorem 4. If Kn;∗ n; n has an Sk -factorization; then Ksn; sn; sn has an S k -factorization.
Proof. Put k = 2a + 1. Then S k can be denoted as K1;a⊕a . Since Kn;∗ n; n has a K1;a⊕a ∗ factorization, Ksn; sn; sn has a Ks; sa⊕sa -factorization. By Lemma 3, Ks; sa⊕sa has an ∗ S k -factorization. Therefore, by Lemma 2 Ksn; sn; sn has an S k -factorization.
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Theorem 5. When k = 3; Kn;∗ n; n has an Sk -factorization. Proof. Let Vi = {vi;1 ; vi;2 ; : : : ; vi; n } (i = 1; 2; 3). Construct 3n Sk -factors Fij (i = 1; 2; 3; j = 1; 2; : : : ; n) as following: F1j = {(v1;1 ; v2; j ; v3; j ); (v1; 2 ; v2; j+1 ; v3; j+1 ); : : : ; (v1; n ; v2; j+n−1 ; v3; j+n−1 )}; F2j = {(v2;1 ; v1; j ; v3; j ); (v2;2 ; v1; j+1 ; v3; j+1 ); : : : ; (v2; n ; v1; j+n−1 ; v3; j+n−1 )}; F3j = {(v3;1 ; v1; j ; v2; j ); (v3; 2 ; v1; j+1 ; v2; j+1 ); : : : ; (v3; n ; v1; j+n−1 ; v2; j+n−1 )}; where the additions are taken modulo n with residues 1; 2; : : : ; n. Then they comprise an S k -factorization of Kn;∗ n; n . Theorem 6. When k is odd; k¿5 and n ≡ 0 (mod k(k − 1)=2); Kn;∗ n; n has an Sk factorization. Proof. Put k = 2a + 1, n = s(2a + 1)a, N = (2a + 1)a. When s = 1, let V1 = {1; 2; : : : ; N }, V2 ={10 ; 20 ; : : : ; N 0 }, V3 ={100 ; 200 ; : : : ; N 00 }. For i =1; 2; : : : ; 2a+1 and j =1; 2; : : : ; 2a+1, construct (2a + 1)2 Sk -factors Fij as following: Fij = {((A + 1); (B + 1; : : : ; B + a)0 ; (C + 1; : : : ; C + a)00 ) ((A + 2); (B + a + 1; : : : ; B + 2a)0 ; (C + a + 1; : : : ; C + 2a)00 ) ::: ((A + a); (B + (a − 1)a + 1; : : : ; B + a2 )0 ; (C + (a − 1)a + 1; : : : ; C + a2 )00 ) ((B + a2 + 1)0 ; (A + a + 1; : : : ; A + 2a); (C + a2 + 1; : : : ; C + a2 + a)00 ) ((B + a2 + 2)0 ; (A + 2a + 1; : : : ; A + 3a); (C + a2 + a + 1; : : : ; C + a2 + 2a)00 ) ::: ((B + a2 + a)0 ; (A + a2 + 1; : : : ; A + a2 + a); (C + a2 + (a − 1)a + 1; : : : ; C + 2a2 )00 ) ((C + 2a2 + 1)00 ; (A + a2 + a + 1; : : : ; A + a2 + 2a); (B + a2 + a + 1; : : : ; B + a2 + 2a)0 ) ((C + 2a2 + 2)00 ; (A + a2 + 2a + 1; : : : ; A + a2 + 3a); (B + a2 + 2a + 1; : : : ; B + a2 + 3a)0 ) ::: ((C + 2a2 + a)00 ; (A + 2a2 + 1; : : : ; A + 2a2 + a); (B + 2a2 + 1; : : : ; B + 2a2 + a)0 )}; where A = (i − 1)a, B = (j − 1)a, C = (i + j − 2)a, and the additions are taken modulo N with residues 1; 2; : : : ; N .
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∗ Then we claim that they comprise an Sk -factorization of KN; N; N . We can see that ∗ each of them is an S k -factor, because it spans all vertices of KN; N; N . We show that they are arc-disjoint. Suppose that they are not arc-disjoint. In the following, we consider A = (i − 1)a, B=(j−1)a, C =(i+j−2)a, D=(k −1)a, E=(l−1)a, F =(k +l−2)a, 16i; j; k; l62a+1, 16x; y; z; w6a. Note that A; B; C; D; E; F; N are integral multiples of a. First, we assume that the common arc joining from V1 to V2 or from V1 to V3 appears in xth component ((A+x);(B+(x−1)a+1; : : : ; B+(x−1)a+a)0 , (C +(x−1)a+1; : : : ; C + (x−1)a+a)00 ) of Fij and yth component ((D+y); (E+(y−1)a+1; : : : ; E+(y−1)a+a)0 , (F + (y − 1)a + 1; : : : ; F + (y − 1)a + a)00 ) of Fkl . Say ((A + x); (B + (x − 1)a + z)0 ) = ((D + y); (E + (y − 1)a + w)0 ). Then A + x ≡ D + y (mod N ) and B + (x − 1)a + z ≡ E + (y − 1)a + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Say ((A + x); (C + (x − 1)a + z)00 ) = ((D + y); (F + (y − 1)a + w)00 ). Then A + x ≡ D + y (mod N ) and C + (x − 1)a + z ≡ F + (y − 1)a + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Next, we assume that the common arc joining from V2 to V1 or from V2 to V3 appears in (a + x)th component ((B + a2 + x)0 ; (A + xa + 1; : : : ; A + xa + a); (C + a2 + (x − 1)a + 1; : : : ; C + a2 + (x − 1)a + a)00 ) of Fij and (a + y)th component ((E+a2 +y)0 ; (D+ya+1; : : : ; D+ya+a); (F +a2 +(y−1)a+1; : : : ; F +a2 +(y−1)a+a)00 ) of Fkl . Say ((B + a2 + x)0 ; (A + xa + z)) = ((E + a2 + y)0 ; (D + ya + w)): Then B + a2 + x ≡ E + a2 + y (mod N ) and A + xa + z ≡ D + ya + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Say ((B + a2 + x)0 ; (C + a2 + (x − 1)a + z)00 ) = ((E + a2 + y)0 ; (F + a2 + (y − 1)a + w)00 ). Then B+a2 +x ≡ E +a2 +y (mod N ) and C +a2 +(x−1)a+z ≡ F +a2 +(y−1)a+w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Last, we assume that the common arc joining from V3 to V1 or from V3 to V2 appears in (2a + x)th component ((C + 2a2 + x)00 ; (A + a2 + xa + 1; : : : ; A + a2 + xa + a); (B + a2 + xa + 1; : : : ; B + a2 + xa + a)0 ) of Fij and (2a + y)th component ((F +2a2 +y)00 ; (D+a2 +ya+1; : : : ; D+a2 +ya+a); (E+a2 +ya+1; : : : ; E+a2 +ya+a)0 ) of Fkl . Say ((C + 2a2 + x)00 ; (A + a2 + xa + z)) = ((F + 2a2 + y)00 ; (D + a2 + ya + w)). Then C + 2a2 + x ≡ F + 2a2 + y (mod N ) and A + a2 + xa + z ≡ D + a2 + ya + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption. Say ((C + 2a2 + x)00 ; (B + a2 + xa + z)0 ) = ((F + 2a2 + y)00 ; (E + a2 + ya + w)0 ). Then C + 2a2 + x ≡ F + 2a2 + y (mod N ) and B + a2 + xa + z ≡ E + a2 + ya + w (mod N ). From these congruences, we have i = k, j = l, x = y, z = w. This contradicts the assumption.
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∗ Therefore, (2a + 1)2 Sk -factors Fij comprise an Sk -factorization of KN;N;N . Applying ∗ Theorem 4, Kn; n; n has an S k -factorization.
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