EDGE DOMINATION IN COMPLETE PARTITE GRAPHS

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DISCRETE

MATHEMATICS

ELSEVIER Discrete Mathematics 132 (1994) 29-35

Edge domination

in complete partite graphs*

Bor-Liang Chen, Hung-Lin Fu*

Department of Applied Mathematics, National Chiao-Tung University, Hsin-Chu, Taiwan Received 27 November 1990; revised 29 October 1992

Abstract

An edge dominating set in a graph G is a set of edges D such that every edge not in D is adjacent to an edge of D. An edge domatic partition of a graph C=(V, E) is a collection of pairwise-disjoint edge dominating sets of G whose union is E. The maximum size of an edge domatic partition of G is called the edge domatic number. In this paper, we study the edge domatic number of the complete partite graphs and give the answers for balanced complete partite graphs and complete split graphs.

1. Introduction

In this paper all graphs are finite, undirected, loopless, and without multiple edges. A balanced complete t-partite graph is a complete t-partite graph K(ml, mz, . . ..m.) where ml = m2 = ... =m,=r. Such a graph is denoted by 0:. It is also known as a regular complete partite graph. A compEete split graph is the join of a complete graph K, and an independent set 0, which we shall denote by S(n, r). (S(n, r) can be viewed as a complete (n+ 1)-partite graph K(r, 1, 1, . . . . l).)

An edge dominating set D of a graph G is a set of edges such that every edge of G not in D is adjacent to an edge in D. An edge domatic partition of a graph G=(V, E) is a collection of pairwise-disjoint edge dominating sets of G whose union is E. The edge domatic number problem is to determine the edge domatic number cd(G) of G, which is the maximum size of an edge domatic partition of G. Zelinka [7] showed that 6(G)bed(G)<6,(G)+ 1 where 6(G) is the minimum degree of G and 6,(G) is the minimum degree of the line graph of G, i.e., 6(15(G)). He also determined the values of cd(G) when G is a circuit, a complete graph, a complete bipartite graph or a tree.

* Research supported by the National Science Council of the Republic of China under grant NSC80-0208- MOO9-04.

* Corresponding author.

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30 B.-L. Chen, H.-L. Fu / Discrete Mathematics 132 (1994) 29-35

In [4], it was proved that ed(O:)<L r’(i)/r r(t- 1)/2]], and the equality holds for (i) t is odd, and (ii) t = 4 and r is even. Subsequently, Hwang [3] conjectured that the equality holds for t even. In this paper, we solve the edge domination problem of 0: by showing that for t 2 3 and r 3 2, ed(0:) = rt - 2 if t is even and r is odd, and ed(0:) = rt

otherwise. Moreover, we consider the complete split graph S(n,r) and we prove that ed(S(n, r))= n +r if II is even and n-r is a positive odd integer, and ed(S(n, r)) = n + r - 1 otherwise.

2. The edge domatic number of 0:

We start with some definitions. Let S = { 1,2, . . . , u}. A Latin square of order v based on S is a u x u array with entries from S such that in each row and each column, every element of S occurs exactly once. A Latin square L = [Eij] is said to be commutative provided that 1, = lji for every 1~ i, j < u. L is idempotent if iii = i for each ieS. It is well known that a commutative Latin square exists for all orders and an idempotent commutative Latin square of order u exists if and only if u is odd. For u=2k, let H={{I,2},{3,4},..., (2k- 1,2k}}. The elements in H are called holes. A Latin square

with holes H is a Latin square such that for each hole ~EH, the subarray formed by h x h is a subsquare based on h. Since all the holes are of size two, we also refer to such a Latin square as a Latin square with 2 x 2 holes H. It was shown by Fu [2] that a commutative Latin square of order 2k with 2 x 2 holes H (briefly CLSH (2k)) exists for each k> 3. In what follows, we will use these Latin squares to obtain the edge domatic number of 0:.

Since 0: = K,, 0,’ = K,,, and 0,’ = O,, their edge domatic numbers are either known or trivial; hence, we will consider t > 2 and r > 1.

It is not difficult to see that if D is an edge dominating set of O:, then the edges in

D must be incident with at least all the vertices in t - 1 partite sets.

Lemma 2.1 [Hwang and Chang [4]]. ed(O:)GL r’(;)/r r(t - 1)/2] J for t 2 3.

Proof. Since every edge dominating set of 0: must cover at least t - 1 parts of O:, we have that every edge dominating set of 0: has at least rr(t- 1)/2] edges. Then

Corollary 2.2. ed(O:)< rt -2 rt if t is even, and r is odd,

otherwise.

Now it is clear that if we can obtain an edge domatic partition with the size mentioned in Corollary 2.2, then we have found ed(0:).

It is easier to solve the case when t is odd. We note here that this case was solved in [4]. For completeness, we give a different proof by using special Latin squares.

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B.-L. Chen, H.-L. Fu/ Discrete Mathematics 132 (1994) 29-35 31

Proposition 2.3.

Zf t

is odd, then ed(O:)=rt.

Proof. Let V(O:) be the disjoint union of t partite sets V,, V,, . . . , l( such that l$={Uj+(i_l)rlj=l,2, ...) r}, 1 < i < t. Then {u,,, uk} is an edge of 0: if and only if uh and ok are in two different partite sets. We first find a proper edge coloring for 0: which uses rt colors. Let M = [mij] be an idempotent commutative Latin square of order t based on {1,2, . . . . t} and A = [a,] be a commutative Latin square of order I based on {1,2 ,..., I}. Define an rt x rt array L in block form [Bij]txt such that Bij= [a,+(mij- l)r]. It is easy to check that L is a commutative Latin square of order rE. (L is also referred as the direct product of A and M.) Now let L’ = [l;k]rtxrt be the array obtained from L by deleting Bii, 1 <i < t. By coloring the edge {u,,, ok} with &k, we obtain a proper edge coloring of 0: which uses rt colors. Since M is idempotent, for each color 16 i < rt, the edges colored i form an edge dominating set. We conclude that ed(O: and by Corollary 2.2, we have the proof. 0

In [4] they showed that ed(0,4)=4r if r is even. Here we obtain a more general result.

If

r is even; then ed(O:)=rt.

Proof. LetthetpartitesetsofO:beB,,B,,...,B,,B,=A,uA,,...,B,=A,,_,uA,,

suchthatIAil=r/2foreachi=l,2,..., 2t. Then the proof is similar to the idea of the proof of Proposition 2.3. The Latin square M is replaced by a CLHS (2t) and the Latin square A is replaced with a commutative Latin square of order r/2 based on (192, . . . . r/2}. Furthermore, we delete the entries obtained from the holes of M. 0

Finally, we deal with the case when t is even and r is odd. For each commutative Latin square M, we let the upper triangular part, diagonal, and lower triangular part be denoted by U(M), D(M), and L(M) respectively. The array obtained by using U(M,), D(M,) and L(M,) is denoted by (U(M,),D(M,), L(M,)) (briefly (1,2,3)) where MI, M,, M3 are three commutative Latin squares (or commutative arrays) of the same order (of the same side). Now we are ready to show that ed(0:) art -2 whenever t is even and r is odd.

If t

is even and r is odd, then ed(O:)=rt -2.

Proof. Let MI = [m(ii’)] be an idempotent Latin square of order t based on { 1,2, . . . , t} and M,=[mjf)] be a unipotent commutative Latin square of order t based on {O,1,2, . . . . t - l} such that c=O. (A Latin square L= [Iii] is unipotent is Iii =c for each i.) Let A (k) be an idempotent commutative Latin square of order r based on A,={(k-l)r+l,(k-l)r+2,..., kr}, 1 <k < t - 1, and A(‘) be an r x r array obtained by adding one more row b=(r(t-1)+1,r(t-1)+2,...,rt-2,rt-2,0) and sym- metrically one more column bT to a unipotent commutative Latin square of order

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32 B.-L. Chen, H.-L. Ful Discrete Mathematics 132 (1994) 29-35

I- 1 based on {O,r(t- l)+ l,r(t- 1)+2, . . . . rt -2). Similar to the idea of the proof of Proposition 2.3, construct an array L in block form [Bij]1x1 such that B,= (U(/lck)), D(ACh’), L(Atd’)) where rnij)= k, mf’=h and m$‘=d, 1 <i< j< t. Now we claim that the array L’ obtained from L by deleting the diagonal blocks corresponds to an edge domatic partition of 0: with size rt - 2. Since, we use rt - 2 entries in L’, it suffices to show that for each i, 1 < i<rt -2, the edges colored i form

an edge dominating set. (It will be helpful to look at the example 0: in Fig. 1.) By the construction of L’, for each ieA,, 1 <k< t - 1, i occurs in each row and each column except possibly the jth row (column) where jeA,. This implies that for each edge {u, u} in E(O:) not colored i, there exists an edge colored i which dominates {u, v}. Hence, the set of edges colored i form an edge dominating set. As to the entry in {r(t-l)+l,r(t-1)+2,..., rt-2}, the proof is similar. Thus, we have the proof. 0

By combining Propositions 2.3-2.5 and the known results we have the following theorem.

A(‘) A(z) A’3’ ,4(4)

1423 0231 M 3241 2013 132 465 7 9 8 0 10 10 “4132 IV,: 3102 321 654 987 10 0 10 2314 1320 213 546 879 10 10 0 ___- --- <l,O,l> <4,2,3> :I --- <3,2,4> <2,0,2> _--- --- <4,3,2> <1,1,4> __-- --- <2,1,3> <3,3,1> __- --- 4 10 10 7 65 198 9 5 10 10 8 4 627 8 7 6 10 10 9 543 4 98 1 10 10 7 3 2 10 57 3 210 981 10 10 6 2 1 3 879 7 10 10 1 3 2 465 ed(0;) >_ 10 6 8 10 10 2 1 354 5 4 9 10 10 3 2 1 6 165798 4 3 2 924387 6 5 1 873219 5 4 6 Fig. 1

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B.-L. Chen. H.-L. Fu/ Discrete Marhemaiics 132 (1994) 29-35 33 Theorem 2.6.

0

if

t=l,

r if t=2,

ed(O:)= t-l if I = 1 and t is even, rt-2 if t>3, t is even,

r>2 and r is odd, rt otherwise.

3. The edge domatic number of S(n,r)

In what follows, we will use V(D) to denote the set of vertices which are incident with the edges of an edge dominating set D. The following result is easy to see.

Proposition 3.1. D is an edge dominating set ofS(n, r) ifand only ifeither V(K,) c V(D)

or V(S(n,r)\v) E V(D)for some vE V(K,).

By a direct counting, we have the following proposition.

Proposition 3.2. n + r k ed(S(n, r)) > n + r - 1.

Proof. Let {D1, Dz, . . . . DI} be an edge domatic partition of S(n, r). By Proposition 3.1, the degree sum of all vertices in K, on the edge-induced subgraph (Di) is at least

n- 1, i.e.,

However, in at most n of the 1 edge dominating sets the degree sum equals to n- 1. This implies that

n(n-1)+(1-n)n< i C degCD,,vdn(n+r-l). i = 1 UE V(K,)

Hence 16 n + r. To prove the other inequality, assume V(0,) = {ul, u2, . . . , u,}. If n is even, let {D;,D;, . . . . DA_,} be a l-factorization of K, and let Df+(“_ I)= {(v,ui):

txV(K,)} for each i, 1 bi<r, so {D;,D;, . . . . DA+,_ 1} is an edge domatic partition of

S(n,r). If n is odd, let (D;,D;, . . . , DA} be a l-factorization of K,+ {v,} and let

Df+,,= {(v,Ui): txV(K,)} for each i, l<i<r-1.

Then {D;,D;, . . . , D:+,_ 1} is an edge domatic partition of S(n, r). Hence,

ed(S(n,r))>n+r-1. 0

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34 B.-L. Chen. H.-L. Fu/ Discrete Mathematics 132 (1994) 29-35

Proposition 3.3. If ed(S(n, r))= n+r, then there exists an edge domatic partition

{D1,Dz, . . . . Dn+,) such that

1 degCDi,u=n-1 if l<i<n and 1 deg(,,,v=n ifn+l<j<n+r.

UE W.) UE OK”)

Proof. This is a direct result of inequality (1) which turns to be an equality. 0 Proposition 3.4. Zf r 2 n or n-r is a positive even integer, then ed(S(n, r)) = n + r - 1. Proof. First, we consider the case r > n. Let D be an edge dominating set of S(n, r) such that CVEVCK.) degCD, v is minimum. Since 1”. VCK_) deg<,> v B n, there exists

ed(S(n,r))G C degsCn,,)v

UE W,) :’ UE c UK,) deg<D> v<n+r-1.

Now if n > r and n-r is even, for each dominating set D either IDE YCK.) deg(,, v > n, or c veV(K.)deg<D1v=n-l and &,vC4,degCD,v~r+l. Let {D1,Dz ,..., Dt} be an edge domatic partition of S(n, r). By counting the number of edges in E(S(n, r))- E(K,), we conclude that there are at most n- 1 edge dominating sets Di such c v.VCK.JdegCDi) v =n- 1 assume that there are kdn- 1 such edge dominating Thus, that sets. < ii 1 deg,,,,,v-(n-1)2 n +(n-l)=n+r-1, 0~ WGJ }I’ 1

By Proposition 3.2, we conclude the proof. 0

Proposition 3.5. Zf n is odd, r is even and r c n, then ed(S(n, r)) = n + r - 1.

Proof. If ed(S(n, r))= n +r, then by Proposition 3.3, there exists an edge domatic partition {D1,D2 ,..., Dn+,} such that CveVCK.)deg<,,,u=n-l if l<i<n, and c UE Y(K.) deg<,,, v = n, n + 1 d i < n + 1. By Proposition 3.1, Diy 1 < i < n, is incident to every vertex in 0,. Moreover, r < n; thus all the edges joining the vertices in V(K,) and 0, are in Uy= IDi. Hence, for each n < j< n + r, Dj is a set of edges in K, which is an edge dominating set of S(n,r). By the fact that n is odd, we must have &d’(KJ de&D,> v > n. This contradicts Proposition 3.3. Therefore,

ed(S(n,r))=n+r-1. 0

Proposition 3.6. Zf n is even and n-r is a positive odd integer, then ed(S(n, r)) = n + r. Proof. It is well known that an idempotent commutative Latin square of order v can be embedded in an idempotent commutative Latin square of odd order t 2 20 + 1 [l].

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B.-L. Chen. H.-L. Fuj Discrete Mathematics 132 (1994) 29-35 35

Hence, we can embed an idempotent commutative Latin square A of order r into an idempotent commutative Latin square L = [l,] of order n + r > 2r + 1. Deleting the entries of A and Iii, r + 1 d i < n + r, we obtain an array which corresponds to an edge domatic partition of S(n, r). By the fact that for each color 1 < i<n + r, the edges colored i form an edge dominating set, this implies that ed(S(n, r)) an +r. By Proposition 3.2, we conclude the proof. 0

Combining Propositions 3.4-3.6, and the known results, we have the following theorem.

Theorem 3.7.

ifr=O and n is odd,

if n is even and n-r is an odd positive integer, otherwise.

Acknowledgment

The authors would like to express their thanks to the referees for their helpful comments.

References

Cl] A.B. Cruse, On embedding incomplete symmetric latin squares, J. Combin. Theory Ser. A 16 (1974) 18-20.

[2] C-M. Fu, The intersection problem of pentagon systems, Ph.D. Thesis, Auburn University, USA, 1987. [3] S.F. Hwang, private communication.

[4] S.F. Hwang and G.J. Chang, The edge domination problem and the edge domatic number problem, submitted.

[S] S. Mitchell and ST. Hedetniemi, Edge domination in trees, in: Proc. 8th S-E Conf. Combin.; Graph theory and computing, Congr. Numer. 19 (1977) 489-509.

[6] M. Yannakakis and F. Gavril, Edge dominating sets in graphs, SIAM J. Appl. Math. 38 (1980) 364-372.