Signed Domination of Graphs and (0, 1)-Matrices Adam H. Berliner, Richard A. Brualdi, Louis Deaett, Kathleen P. Kiernan, Seth A. Meyer, and Michael W. Schroeder To Reza Koshrovshahi on his 70th birthday.
Abstract. We briefly review known results about the signed edge domination number of graphs. In the case of bipartite graphs, the signed edge domination number can be viewed in terms of its bi-adjacency matrix. This motivates the introduction of the signed domination number of a (0, 1)-matrix. We investigate the signed domination number for various classes of (0, 1)-matrices, in particular for regular and semi-regular matrices. A conjectured upper bound for the signed edge domination of a graph of order n leads to the conjecture that the signed domination number of an m by n (0, 1)-matrix is bounded above by m + n − 1, and this conjecture is the focus of much of our work. We also determine a lower bound on the signed edge domination number of regular graphs and characterize the case of equality.
1. Introduction To place our work in a larger context, we begin with the following description and review. Let G be a finite (simple) graph with vertex set V and edge set E. The open neighborhood of a vertex v is the set N (v) of vertices joined to v by an edge. The closed neighborhood of v is the set N [v] = N (v) ∪ {v}. A function g : V → {1, −1} has value given by X γs (g) = g(v). v∈V
The function g is a signed vertex domination function for G provided that X g(w) ≥ 1 for all v ∈ V . w∈N [v]
Here N [v] is the closed neighborhood of the vertex v consisting of v and all those vertices joined to v by an edge. The signed vertex domination number of G is ( ) X γsd (G) = min{γs (g)} = min g(v) g
g
v∈V
1991 Mathematics Subject Classification. Primary 05C22, 05C50, 05C69.. Key words and phrases. signed (edge) domination, (bipartite) graph, (0, 1)-matrix. 1
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 2
where the minimum is taken over all signed vertex domination functions g of G. Let L(G) be the line graph of G. Thus the set of vertices of L(G) is E, and for distinct e, f ∈ E, e and f are joined by an edge in L(G) if and only if e and f have a common vertex. A signed vertex domination function of L(G), called a signed edge domination function of G, is a function h : E → {1, −1} with value X γs0 (h) = h(e) e∈E
satisfying X
h(f ) ≥ 1 for all e ∈ E.
f ∈N [e]
Here N [e] is the closed neighborhood of the edge e in G (or the closed neighborhood of the vertex e of L(G)) consisting of e and all those edges having a common vertex with e. The signed edge domination number of G is ( ) X 0 0 γsd (G) = min γs (h) = min h(e) h
h
e∈E
where the minimum is taken over all signed edge domination functions h of G. The signed edge domination number was introduced by Xu [12] who determined 0 the smallest value of γsd (G) over all graphs G with a specified number of vertices and edges. In [14] Xu proved that for a graph G with n vertices and m edges (1.1)
0 n − m ≤ γsd (G) ≤ 2n − 4.
For every graph G, the graph H obtained from G by adding enough pendent edges to each vertex (one less than the degree of each vertex) achieves the lower bound. Xu also conjectured that (1.2)
0 γsd (G) ≤ n − 1
independent of the number m of edges. In [8] connected graphs G for which equality holds in the lower bound in (1.1) are characterized as graphs for which the degree of each vertex is at most 1 more than twice the number of pendent edges at the vertex. In [7] the conjecture (1.2) was established for Eulerian graphs (so all vertices of even degree), graphs with all vertices of odd degree, and hence for all regular graphs. In addition, the upper bound 0 (G) ≤ d3n/2e γsd was established which improves that in (1.1) and also improves an upper bound derived in [3]. In [4, 5] additional results are obtained on the signed vertex domination number of a graph. In [13] and [6] it is shown that the signed edge domination number of every tree is at least 1. Trees with signed edge domination number 1, 2, 3, or 4 have been characterized [13, 11]. In [12], it is proved that the signed edge domination of a connected graph equals its number of edges (so no −1s are possible in any dominating edge signing) if and only if the graph is either a path on at most 5 vertices or a subdivision of a star K1,n (n ≥ 3). Other results on the signed edge domination numbers of trees are contained in [15, 9]. In this paper we consider the signed edge domination number of regular graphs and certain classes of bipartite graphs, in particular, complete bipartite graphs Km,n , the complete bipartite graph Kn,n in which a perfect matching has been removed, and semi-regular bipartite graphs. The signed edge domination number
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
3
of a bipartite graph can be usefully and attractively formulated in terms of matrices by considering the bi-adjacency matrix of a bipartite graph. Let G ⊆ Km,n be a bipartite graph with a bipartition V = U ∪ W into disjoint sets U and W of sizes m and n, respectively. Let U = {u1 , u2 , . . . , um } and W = {w1 , w2 , . . . , wn }, and let A = [aij ] be the bi-adjacency matrix of G in which aij equals 1 if ui and wj are joined by an edge and equals 0 otherwise. A signing of A is an m by n matrix A0 = [a0ij ] obtained from A by replacing some of its 1s with −1s. The cross of an entry xpq of an m by n matrix X = [xij ] is the subset of positions of X defined by Cpq (X) = {(p, j) : 1 ≤ j ≤ n} ∪ {(i, q) : 1 ≤ i ≤ m}. Each cross of X contains m + n − 1 positions. (For our purposes it is convenient to think of the cross as the subset of Cpq (X) consisting of those positions containing a nonzero entry, or, when X = A0 , as the the multiset of 1s and −1s in those positions.) We say that A0 is a dominating signing of the matrix A provided the sum χ(a0pq ) of the entries in each cross of a nonzero entry a0pq of A0 is at least 1, that is, X X χ(a0pq ) = a0pq + a0pj + a0iq ≥ 1 whenever a0pq 6= 0. j6=q
i6=p
0
The value of the signing A of A is X σ(A0 ) = a0ij = σ(A) − 2σ − (A0 ), i,j 0
where σ(A ) is the sum of the entries of A0 , σ(A) is the number of 1s in A, and σ − (A0 ) is the number of −1s in A0 . 0 The signed domination number γsd (A) of the matrix A is defined as 0 γsd (A) = min {σ(A0 ) : A0 is a dominating signing of A} .
The signed domination number of A equals the signed edge domination number of G. The conjecture (1.2) for bipartite graphs G ⊆ Km,n is equivalent to the conjecture that (1.3)
0 γsd (A) ≤ m + n − 1 for every m by n (0, 1)-matrix A.
The investigation of dominating edge signings of bipartite graphs is equivalent to that of dominating signings of (0, 1)-matrices. When dealing with bipartite graphs, we generally frame our discussion in terms of dominating signings and signed domination numbers of matrices. If A is the direct sum of two matrices A1 and A2 , then 0 0 0 γsd (A) = γsd (A1 ) + γsd (A2 ). Thus there is no loss in generality in assuming that A is not a non-trivial direct sum, that is, the bipartite graph whose bi-adjacency matrix equals A is connected; in particular, we implicitly assume throughout that A does not have any rows or columns containing only 0s. Also note that if P and 0 0 Q are permutation matrices, then γsd (P AQ) = γsd (A). Let A be an m by n (0, 1)-matrix. From (1.1) we get that (1.4)
0 m + n − σ(A) ≤ γsd (A) ≤ 2(m + n) − 4.
It follows from [14] that given any bipartite graph G without isolated vertices, the bi-adjacency matrix of the graph H obtained from G by appending deg(v) − 1 pendent edges to each vertex v of G attains equality on the left side of (1.4).
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 4 0 Let A0 be a signing of A such that γsd (A) = m + n − σ(A). Then σ(A) − 0 0 2σ (A ) = γsd (A) = m + n − σ(A), and hence −
2(σ(A) − σ − (A0 )) = m + n, where σ(A) − σ − (A0 ) is the number of 1s in A0 . It follows that m and n have the same parity. Hence, if m and n have opposite parity, the lower bound in (1.4) can be improved to (1.5)
0 γsd (A) ≥ m + n + 1 − σ(A)
(m and n of opposite parity).
0 It follows from (1.4) that if A is the bi-adjacency matrix of a tree, then γsd (A) ≥ 1. The signed domination number of a matrix may be negative. For example, let A be the 5 by 7 matrix 1 1 1 1 1 0 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 with 13 1s. Then 1 1 1 −1 −1 0 0 1 1 1 0 0 −1 −1 0 0 0 0 0 0 0 A = −1 0 −1 0 0 0 0 0 0 0 −1 0 0 0 0
is a dominating signing of A with σ(A0 ) = −1. It is straightforward to show that 0 γsd (A) = −1. 2. Regular Matrices and Regular Graphs The bi-adjacency matrix of the complete bipartite graph Km,n is the m by n matrix Jm,n of all 1s. In case of a bipartite graph with an equal bipartition (m = n), we write Jn in place of Jn,n . A matrix of order n is regular provided that there is an integer k such that each row and each column of A contains exactly k 1s; A is then called k-regular. An m by n matrix A is semiregular provided that there are integers k and l such that each row contains exactly k 1s and each column contains exactly l 1s; in this case, km = ln and A is called (k, l)-semiregular. The corresponding bipartite graph is also called (k, l)-semiregular. A square semiregular matrix is necessarily regular. The matrix Jn is n-regular, and the matrix Jm,n is (n, m)-semiregular. We consider the signed domination number of regular matrices in this section and improve upon the conjecture (1.3). We also establish a lower bound for the signed edge domination number of regular graphs and characterize those graphs attaining equality. In the following section, we consider semiregular matrices. Let Pn denote the full-cycle permutation matrix of order n. Thus Pn has 1s in positions (1, 2), (2, 3), . . . , (n − 1, n), (n, 1). The matrix In + Pn is the bi-adjacency matrix of a bipartite graph equal to a cycle C2n of length 2n. Theorem 2.1. Let n be an integer with n ≥ 2. Then 2n if 2n ≡ 0 mod 3, 3 0 2n+2 if 2n ≡ 1 mod 3, γsd (In + Pn ) = 3 2n+4 if 2n ≡ 2 mod 3. 3
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
5
In particular, conjecture (1.3) holds in this case. Proof. In terms of the cycle C2n , in order for a signing of In + Pn to be dominating, two −1s must be separated by at least two plus 1s. The theorem follows easily from this observation. If Q is a permutation matrix corresponding to a permutation of {1, 2. . . . , n} which decomposes into cycles of lengths n1 , n2 , . . . , nk , then 0 γsd (In + Q) =
k X
0 γsd (Ini + Pni ).
i=1
Recall that a k-regular (0, 1)-matrix A of order n can be written as A = P1 +P2 + · · · + Pk where the Pi are permutation matrices, and that such a decomposition can be obtained by recursively choosing permutation matrices in A, and replacing the 1s in A corresponding to the 1s in the permutation matrices with 0s. Since the signed domination number is invariant under arbitrary row and column permutations, we can always assume that Pk = In . The following lemma can be useful for obtaining upper bounds on the signed domination number. Lemma 2.2. Let A be a k-regular (0, 1)-matrix of order n where k is an even integer. Without loss of generality, assume that A = P1 + · · · + Pk−1 + In where P1 , . . . , Pk−1 are permutation matrices. Suppose that B ∗ A= ∗ ∗ where B is a matrix of order p such that B − Ip contains a total of q 1s and no two of these 1s belong to the same row or column. (i) If the q 1s in B −Ip belong to at most (k−2)/2 of the permutation matrices P1 , . . . , Pk−1 , then 0 γsd (A) ≤ 2(n − p).
(ii) Otherwise, let q 0 be the maximum number of 1s in B−Ip that are contained in at most (k + 2)/2 of the permutation matrices P1 , . . . , Pk−1 . Then 0 γsd (A) ≤ 2(n − q 0 ).
Proof. First consider (i). Let P1 , . . . , P(k−2)/2 be the (k − 2)/2 permutation matrices. Let A0 = −(P1 + · · · + P(k−2)/2 ) + (Pk/2 + · · · + Pk−1 ) + In∗ . Then A0 is a signing of A where In∗ is obtained from In by replacing its first p 1s with −1s. The number of −1s in each row and column of A0 is at most k/2, and, by construction, those rows and columns that contain k/2 −1s intersect in B 0 = −B in A0 . Hence the cross sums of the nonzero entries of A0 are at least 2k − 1 − 2(k − 1) = 1, and A0 is a dominating signing of A with σ(A0 ) = 2(n − p). 0 Thus γsd (A) ≤ 2(n − p). Now consider (ii). Let the permutation matrices be P1 , . . . , P(k+2)/2 . Let ∗ A0 = (P1∗ + · · · + P(k+2)/2 ) − (P(k+4)/2 + · · · + Pk−1 + In ),
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 6 ∗ where P1∗ , . . . , P(k+2)/2 are obtained from P1 , . . . , P(k+2)/2 by replacing their 1s that belong to B with −1s. Then again it is easy to check that A0 is a dominating signing 0 of A where σ(A0 ) = 2(n − q 0 ). Hence γsd (A) ≤ 2(n − q 0 ).
Theorem 2.3. Let n and k be positive integers with 2 ≤ k ≤ n. Let A be a k-regular (0, 1)-matrix of order n. Then n if k is odd, 0 2 if k = 2 and n = 2, γsd (A) ≤ 2(n − 2) if k is even and n ≥ 3. Proof. There are permutation matrices P1 , P2 , . . . , Pk such that A = P1 + P2 + · · · + Pk . First suppose that k is odd. Let A0 = −(P1 + · · · + P(k−1)/2 ) + (P(k+1)/2 + · · · + Pk ). 0 Then A0 is a dominating signing of A with σ(A0 ) = n, and hence γsd (A) ≤ n. Now suppose that k is even, where because of Theorem 2.1, we may assume that k ≥ 4. Without loss of generality we may assume that Pk = In . The matrix C = A − In is (k − 1)-regular and can be regarded as the adjacency matrix of a digraph with the indegree and outdegree of each vertex equal to k −1. This digraph has an induced directed cycle of length p ≥ 2 leading to a principal submatrix of C of order p. Hence A has a principal submatrix of order p containing exactly p 1s off of its main diagonal where these p 1s are in different rows and columns. If p ≤ (k+2)/2, 0 (A) ≤ 2(n−p) ≤ 2(n−2). we apply (ii) of Lemma 2.2 with q 0 = p and obtain that γsd Now suppose that p > (k+2)/2. We then apply (ii) of Lemma 2.2 wih q 0 = (k+2)/2 0 and obtain that γsd (A) ≤ 2n − (k + 2) ≤ 2(n − 3).
Theorem 2.3 improves for regular bipartite graphs the upper bound given in [7] for regular graphs. In the proof of Theorem 2.3 for k even, let D be the digraph whose adjacency matrix is A − In . Then the indegrees and outdegrees of all vertices equal k − 1. If the girth g ≥ 3, that is, if A does not have a pair of symmetric 1s, the proof gives 0 (A) ≤ 2(n − 3). It is possible that an improved bound that depends on this girth γsd g can be obtained. If k is even, we conjecture that the bound 2(n − 2) in Theorem 2.3 can be improved. Specifically, we conjecture that if k is an even positive integer and A is a k-regular matrix of order n, then n if n is even 0 γsd (A) ≤ n − 1 if n is odd. In view of Theorems 2.7 and 2.8, neither the bound n for k odd in Theorem 2.3 nor the conjectured bounds for k even, regarded solely as functions of n, can be improved. We now obtain a lower bound for the signed domination number of a regular matrix, in fact, for a regular graph in general. Theorem 2.4. Let G = (V, E) be a k-regular graph of order n. Then (2.1)
0 γsd (G) ≥
kn . 2(2k − 1)
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
7
Thus, if A is a k-regular (0, 1)-matrix of order n (the bi-adjacency matrix of a regular graph G ⊆ Kn,n ), then kn . 2k − 1
0 γsd (A) ≥
(2.2)
Proof. Consider a signed edge domination function h of G. Since G is kregular, G has kn/2 edges, and the closed neighborhood of each edge has cardinality 2k − 1. In addition, each edge of G belongs to 2k − 1 closed neighborhoods of edges. We thus have X X X kn ·1≤ h(f ) = (2k − 1) h(e). 2 e∈E f ∈N [e]
e∈E
Hence X
h(e) ≥
e∈E
kn . 2(2k − 1)
This inequality is valid for every signed edge domination function h of G, and thus (2.1) holds. We now investigate equality in the inequalities in Theorem 2.4. Theorem 2.5. Let n and k be positive integers. Then there exists a k-regular graph G of order n such that 0 γsd (G) =
(2.3)
kn 2(2k − 1)
if and only if n is a multiple of 12 2k k = (2k − 1)Ck−1 , where Ck−1 is the (k − 1)st Catalan number. There exists a k-regular (0, 1)-matrix of order n (a k-regular bipartite graph G ⊆ Kn,n ) such that 0 γsd (A) =
(2.4) if and only if n is a multiple of
1 2k 2 k
kn 2k − 1
.
Proof. It follows from the proof of Theorem 2.4 that a k-regular graph G = (V, E) of order n satisfies (2.3) if and only if it has a signed edge domination function h such that X (2.5) h(f ) = 1 for all edges e. f ∈N [e]
We partition the vertex set V into sets V1 , V2 , . . . , Vk where Vi = {u ∈ V : u is incident with exactly i edges e with h(e) = 1} (i = 1, 2, . . . , k). Let vi = |Vi | for i = 1, 2, . . . , k. For a vertex x, we have x ∈ Vj if and only if X h(xu) = j − (k − j) = 2j − k. u∈N (x)
Let e = xu be an edge in E. We then have X X X X 1= h(f ) = h(xz) + h(uy) − h(e) = 2j − k + h(uy) − h(e). f ∈N [e]
z∈N (x)
y∈N (u)
y∈N (u)
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 8
Hence X
h(uy)
=
1 − 2j + k + h(e)
y∈N (u)
=
2(k − j + 1) − k 2(k − j) − k
if h(e) = 1 if h(e) = −1.
Thus, u is in Vk−j+1 if h(e) = 1, and u ∈ Vk−j if h(e) = −1. We conclude that if x is a vertex in Vj , then there are j vertices u in Vk−j+1 adjacent to x with h(xu) = 1, and k−j vertices u in Vk−j adjacent to x with h(xu) = −1. In particular, except for negative edges joining vertices in Vk/2 if k is even and positive edges joining vertices in V(k+1)/2 if k is odd, G is a k-partite graph with vertex partition V1 , V2 , . . . , Vk where the sets V1 , V2 , . . . , Vk can be linearly ordered so that the only edges go between consecutive sets. This is illustrated for k = 7 in Figure 1.
7, 1 V7
+
6, 1 V1
−
6, 2 V6
+
5, 2 V2
−
5, 3 V5
+
4, 3 V3
−
V4
4, 4 +
Figure 1
where, for instance,
6, 2 V6
+
V2
Figure 2
means that there are 6 edges with value +1 from each vertex of V6 to vertices of V2 and 2 edges with value +1 from each vertex of V2 to each vertex of V6 ; in particular, the subgraph of G induced on the vertex set V6 ∪ V2 is a (6, 2)-semiregular bipartite graph. The subgraph induced on V4 is a 4-regular graph. The subgraphs of G induced on two consecutive vertex sets in this linear ordering are semi-regular bipartite graphs where vk · k = v1 · 1, v1 · (k − 1) = vk−1 · 1, vk−1 · (k − 1) = v2 · 2, . . . . We now reindex the sets V1 , V2 , . . . , Vk so that they occur in this order in the linear ordering of G as a linear k-partite graph with cardinalities now denoted by
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
9
n1 , n2 , . . . , nk . We then have n1
= vk
n2
= v1 =
n3
=
n4
=
n5
=
n6
=
n7
=
.. .
=
k k vk = n1 1 1 k−1 vk−1 = v1 = 1 k−1 v2 = vk−1 = 2 k−2 vk−2 = v2 = 2 k−2 vk−1 = v3 = 3 k−3 vk−3 = v2 = 3 .. .
k−1 n2 1 k−1 n3 2 k−2 n4 2 k−2 n5 3 k−3 n6 3
Inductively, we get
(2.6) n2i
kk−1 k−1 k k−i+1k−i+1 , ··· = vk · = vk · i−1 i 1 1 i−1 i
1 ≤ i ≤ k/2,
and (2.7) n2i+1 = vk ·
k k−1 kk−1 k−i+1k−i ··· = vk · , i i 1 1 i i
1 ≤ i ≤ (k − 1)/2,
where these quantities are integers and depend only on the value of vk . In particular, ( nk =
n2·k/2 = vk ·
k k 2
n2( k−1 )+1 = vk · 2
k−1 = vk · k 2 −1 k−1 k
k−1 2
k−1 2
2·
k−1 k 2
= vk ·
k
k−1 k 2 −1
k−1 2
·2·
if k is even k−2 k−1 2
if k is odd,
and thus nk is an even integer. Now the smallest feasible value of n in order to have a k-regular graph G of order n attaining equality in (2.3) occurs when n1 = vk = 1. Moreover, since the nj are integers with nk even, such a graph G is easily constructed: (a) On two consecutive vertex sets in the linear ordering of the vertex set of G, we choose any semiregular bipartite graphs with the appropriate vertex degrees. (b) On the vertex set Vk of even order nk , we choose any regular graph of the appropriate degree. The graphs in (a) and (b) are easy to construct. It remains to show that when n1 = 1, the number of vertices is given by (2.3).
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 10
First suppose that k is even. Then, using (2.6) and (2.7), we get
n
=
k X
ni
i=1
=
k 2 X k k−1
i=1
i
i−1
k
+
2 −1 X k k−1 i=0
k 2 −1
i
i
k−1 k k−1 k−1 k + + k k 0 0 i i−1 i 2 2 −1 i=1 k 2 −1 X k k k k 1 k k = + + k k 0 0 i i 2 2 2 i=1
=
X k k − 1
+
k
=
= =
2 −1 2 X k
2 1 k + i 2 k2 i=0 k 2 1X k 2 i=0 i 1 2k . 2 k
A similar calculation works for k odd, and we conclude that n=
1 2k = (2k − 1)Ck−1 . 2 k
This completes the proof of the theorem.
Let k = 3. By Theorem 2.4, the smallest signed domination number of a 3regular (0, 1)-matrix of order n is 3n/5, and by Theorem 2.5, equality holds if and only if n is a multiple of 10. According to the proof of Theorem 2.5, matrices achieving equality are constructed as follows. Let
1 −1 −1 0 0 0 0 0 −1 −1 0 0 . E = 1 and F = 0 1 0 0 0 0 −1 −1 If X is a matrix, let ⊕p X denote the direct sum X ⊕ · · · ⊕ X (p Xs). Let H6p denote a 2-regular (0,1)-matrix of order 6p. Then for all p ≥ 1,
⊕p E A0 = O6p,p Op,p
⊕p F H6p Op,6p
O3p,3p ⊕p F T ⊕p E T
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
11
is a matrix of order n = 10p, whose underlying (0, 1)-matrix A of order n satisfies 0 γsd (A) = 3n/5. When p = 1, we have 1 −1 −1 0 0 0 0 0 0 0 1 0 0 −1 −1 0 0 0 0 0 1 0 0 0 0 −1 −1 0 0 0 0 1 1 0 0 0 0 −1 0 0 0 0 1 1 0 0 0 −1 0 0 0 . A = 0 0 1 1 0 0 0 −1 0 0 0 0 0 0 1 1 0 0 −1 0 0 0 0 0 0 1 1 0 0 −1 0 1 0 0 0 0 1 0 0 −1 0 0 0 0 0 0 0 1 1 1 In the next section, we determine the signed domination number of the m by n matrix Jm,n of all 1s for all m and n. As a prelude, we consider here the square case, in which the following property holds for all dominating signings. Lemma 2.6. Let Jn0 be an arbitrary dominating signing of Jn . Then there is a permutation matrix P such that P Jn0 has all 1s on its main diagonal. Proof. Suppose no such permutation matrix P exists. Then by the Hall-K¨onig theorem (see e.g. [10]), there exist positive integers p and q with p + q = n + 1 such that Jn0 has a p by q submatrix of all −1s. The sum of the entries in a cross of any −1 in this submatrix is at most (n − p) + (n − q) − (p + q − 1) = (n − p) + (n − q) − n = n − (p + q) = −1, a contradiction. Hence such a permutation matrix P exists.
Theorem 2.7. If n is a positive integer, then 0 γsd (Jn ) = n.
Proof. Let Jn0 = [xij ] be any dominating signing of Jn . It follows from Lemma 2.6 that we may assume that Jn0 has all 1s on its main diagonal. Let Jn0 have a 1s not on the main diagonal and b −1s. We calculate that n X n=n·1≤ χ(xii ) = n + 2(a − b), i=1
σ(Jn0 )
and hence a ≥ b. Thus = n + (a − b) ≥ n. If n is odd, then by Theorem 2.3, γsd (Jn ) ≤ n, and hence γsd (Jn ) = n. Now suppose that n is even. Then −Jn/2 Jn/2 00 Jm = Jn/2 2In/2 − Jn/2 0 is a dominating signing of Jn with σ(Jn00 ) = n. Hence γsd (Jn ) = n also when n is even.
Up to permutation equivalence, there is only one (n−1)-regular matrix of order n, namely Jn −In . In the next theorem we determine its signed domination number. Theorem 2.8. Let n ≥ 2 be an integer. Then n if n is even, 0 (2.8) γsd (Jn − In ) = n − 1 if n is odd.
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 12
Proof. The following signings of Jn − In demonstrate that the values of 0 0 γsd (Jn − In ) given in (2.8) are upper bounds for γsd (Jn − In ): −(Jn/2 − In/2 ) Jn/2 (n even), Jn/2 −(Jn/2 − In/2 ) −(J(n+1)/2 − I(n+1)/2 ) J(n+1)/2,(n−1)/2 (n odd). J(n−1)/2,(n+1)/2 −(J(n−1)/2 − I(n−1)/2 ) The number of nonzeros in a cross of a nonzero entry of Jn − In equals 2n − 3, and hence the number of −1s in a dominating signing can be at most n − 2. We 0 now show that the values given in (2.8) are lower bounds for γsd (Jn − In ). These conjectured values correspond to a dominating signing where the number p of −1s is given by ( p=
n(n−2) 2 (n−1)2 2
n even, n odd.
First assume that n is even. Suppose there exists a dominating signing A0 of Jn −In with p > n(n−2)/2. Then the average number of −1s per row (respectively, column) of A0 is p n(n − 2) n > = − 1. n 2n 2 Thus there exists a row (respectively, column) of A0 whose number of −1s is at least n/2. Such a row and column of A0 must intersect in a 0, for otherwise there is a cross with at least (n − 1) −1s. Because our matrix is Jn − In , there must be exactly one row and exactly one column whose number of −1s is at least n/2. Without loss of generality, let row 1 of A0 contain a ≥ n/2 −1s and let column 1 contain b ≥ n/2 −1s. We consider the crosses of the nonzero entries in row 1. Since the number of −1s in row 1 is a, and since the number of −1s in a cross of a nonzero element of A0 is at most n − 2, columns 2, 3, . . . , n of A0 can contain at most n − 2 − a −1s outside of row 1. Thus the total number of −1s in A0 is at most b + a + (n − 1)(n − 2 − a). Therefore b + a + (n − 1)(n − 2 − a) >
n(n − 2) . 2
This gives n(n − 2) − 2(n − 2)(n − 1 − a) , 2 and using a ≥ n/2, we get that b > n − 2. Thus b = n − 1, implying that the cross of each nonzero element in column 1 of A0 contains at least n − 1 −1s, a contradiction. Now assume that n is odd. Suppose that there exists a dominating signing A0 of Jn − In with p > (n − 1)2 /2. Consider the possibility that some row, say row 1, contains (n + 1)/2 (or more) −1s. Then using the crosses of the nonzero elements of row 1, we see that the total number of −1s in A0 is at most n+1 n−5 + (n − 1) + (n − 1), 2 2 where the n − 1 accounts for as many as n − 1 −1s in column 1. Then b>
n+1 n−5 (n − 1)2 + (n − 1) + (n − 1) > . 2 2 2
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
13
After simplification, this implies, since n is odd, that n = 1. Hence there are at most (n − 1)/2 −1s in each row and similarly in each column. Moreover, each row with (n − 1)/2 −1s and each column with (n − 1)/2 −1s must intersect either in a 0 or −1. It is easy to check that A0 must have at least (n + 3)/2 rows and at least (n + 3)/2 columns with (n − 1)/2 −1s, and so these rows and columns must intersect only in 0s and −1s. Since A0 is a dominating signing of Jn − In , this gives (n + 3)/2 rows (so at least one) with (n + 1)/2 −1s, a contradiction, completing the proof of this case and so of the theorem. 0 The dominating signings of Jn − In with value γsd (Jn − In ) given at the beginning of the proof of Theorem 2.8 are symmetric. Since Jn − In is the adjacency matrix of the complete graph Kn , this implies the following corollary.
Corollary 2.9. The signed edge domination number of Kn is n/2 if n is even and is (n − 1)/2 if n is odd. To conclude this section, we determine the signed domination number of a (0,1)-matrix with exactly one 0 (a bipartite graph obtained from Kn,n by removing one edge). Perhaps surprisingly, replacing a 1 by a 0 in Jn increases its signed domination number. Theorem 2.10. Let Jn# be a (0, 1)-matrix of order n ≥ 2 with exactly one 0. Then 0 γsd (Jn# ) = n + 1. Proof. Without loss of generality we assume that the 0 in Jn# occurs in position (n, n). We first demonstrate a dominating signing A0 of Jn# with σ(A0 ) = n+1. These are " # 2In/2 − Jn/2 Jn/2 (n even), and # Jn/2 −Jn/2 −In# − Pn · · · − Pn(n−3)/2 + Pn(n−1)/2 + · · · + Pnn−1 (n odd), where In# is obtained from the identity matrix In by replacing its (n, n)-entry with 0 0, and Pn is the full-cycle permutation matrix of order n. Thus γsd (Jn# ) ≤ n + 1. 0 If n = 2 or n = 3, it is easily verified that γsd (Jn# ) = n + 1. Now assume 0 # that n ≥ 4. Suppose to the contrary that γsd (Jn ) ≤ n and let A0 = [a0ij ] be a dominating signing with σ(A0 ) ≤ n. Then n2 − 1 − 2σ − (A0 ) ≤ n and hence n(n − 1) − 1 , 2 0 a non-integer. It follows that σ − (A0 ) ≥ n(n − 1)/2 and hence that γsd (Jn# ) ≤ n − 1. 0 Let the number of −1s in column n and row n of A be p and q, respectively. Let xij be the number of −1s in the cross of a0ij (1 ≤ i, j ≤ n). Then σ − (A0 ) ≥
n−1 X i=1
xin +
n−1 X
xni
=
(σ − (A0 ) − q − p) + (n − 1)p + (σ − (A0 ) − p − q) + (n − 1)q
=
2σ − (A0 ) + (n − 3)(p + q).
i=1
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 14
For i 6= n, the crosses of a0ni and a0in contain 2(n−2) nonzeros, and since χ(a0ni ) ≥ 1, we have that xin , xni ≤ n − 2 for i 6= n. Hence we have 2(n − 1)(n − 2) ≥ 2σ − (A) + (n − 3)(p + q), which, using σ − (A0 ) ≥ n(n − 1)/2, yields after computation that p+q ≤
n2 − 5n + 4 n−3
and hence p + q ≤ n − 3. Thus the cross at a0nn = 0 has at most n − 3 −1s. Let B 0 = [b0ij ] be the matrix obtained from A0 by replacing its only 0 with a −1. We claim that B 0 is a dominating signing of Jn , and for this we need only check the crosses that use position (n, n), that is the crosses of the elements in the last row and column. We have χ(b0nn ) = 2n − 1 − 2(p + q + 1) ≥ 2n − 3 − 2(n − 3) ≥ 3. Now consider χ(b0ni ) for i 6= n. Since xni ≤ n − 2, the cross of a0ni contains at most n − 2 −1s and at least n 1s. This implies that χ(b0ni ) ≥ 1 for all i 6= n and, similarly, that χ(b0in ) ≥ 1 for all i 6= n. Thus B 0 is a dominating signing of Jn , where n(n − 1) 0 2 − 0 2 + 1 = n − 2. σ(B ) = n − 2(σ (A ) + 1) ≤ n − 2 2 0 0 This contradicts Theorem 2.7. Hence γsd (Jn# ) ≥ n + 1 and so γsd (Jn# ) = n + 1.
3. Semiregular Matrices (Bipartite Graphs) In this section we consider the signed edge domination number of semiregular matrices (bipartite graphs). Recall that a bipartite graph G is a (k, l)-semiregular bipartite graph provided that in the bipartition V = U ∪ W of its vertex set, the vertices in U have degree k and the vertices in W have degree l. Let |U | = m and |V | = n so that km = ln. The bi-adjacency matrix of G is an m by n (0, 1)-matrix A = [aij ] with k 1s in each row and l 1s in each column, and is a (k, l)-semiregular matrix. The number of 1s in the cross of each 1 of A equals k + l − 1. Since 0 0 γsd (A) = γsd (AT ), there is no loss in generality in assuming that m ≤ n. This implies that k ≥ l with equality if and only if m = n. We begin with the m by n (n, m)-semiregular matrix Jm,n for which the exact signed domination number can be determined for all m and n. For brevity we 0 0 0 0 shorten γsd (Jm,n ) to γsd (m, n). Since γsd (m, n) = γsd (n, m) we assume that m ≤ n. The following theorem is formulated in a different way and in terms of bipartite graphs in [1]. It was proved independently by us using our matrix formulation. We give a differently structured and, we think, more revealing proof. Our evaluation separates into four cases according to the parities of m and n, with subcases for each. Theorem 3.1. Let m and n be positive integers with m ≤ n. 1. If m is even and n is even, then n if m ≤ n ≤ 2m − 1 0 γsd (m, n) = 2m if 2m ≤ n. (b)
(a)
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
15
2. If m is even and n is odd, then if m ≤ n < 2m (a) 2m 0 n + 1 if 2m ≤ n < 3m (b) γsd (m, n) = 3m if 3m ≤ n. (c) 3. If m is odd and n is even, then if m ≤ n < 2m (a) 2m 0 n if 2m ≤ n < 3m − 1 (b) γsd (m, n) = 3m − 1 if 3m − 1 ≤ n (c) 4. If m is odd and n is odd, then n if m ≤ n ≤ 2m − 1 0 γsd (m, n) = 2m − 1 if 2m ≤ n (b)
(a)
Proof. We break up the proof into three parts. PART I: We first establish the equalities 1(b), 2(c), 3(c), and 4(b) in the statement of the theorem. First assume that m is even. Consider the matrix −Jm/2,(n−a)/2 Jm/2,(n−a)/2 Jm/2,a 0 Jm,n = , Jm/2,(n−a)/2 −Jm/2,(n−a)/2 Jm/2,a 0 is a where a = 2 if n is even and a = 3 if n is odd. It is easy to check that Jm,n 0 0 dominating signing of Jm,n , and that σ(Jm,n ) = 2m if n is even, and σ(A ) = 3m 0 0 if n is odd. Thus γsd (m, n) ≤ 2m if n is even, and γsd (m, n) ≤ 3m if n is odd. We now show that these dominating signings cannot be improved upon if n ≥ 2m (n 0 even) and if n ≥ 3m − 1 (n odd). This will then show that γsd (m, n) = 2m if n is 0 even and n ≥ 2m, and γsd (m, n) = 3m if n is odd and n ≥ 3m − 1. Consider an arbitrary dominating signing A0 of Jm,n . Let x be the largest number of −1s in a cross of A0 and k the largest number of −1s in a row. For 0 we have k = x − (m/2) with k −1s in every row. So assuming A0 improves Jm,n 0 , we see that A0 has a row, say row 1, with k = x − (m/2) + l −1s upon Jm,n where l ≥ 1. Since A0 is a dominating signing of Jm,n , each column of the matrix obtained from A0 by deleting row 1 has at most (m/2) − l −1s. Since each of the 0 first (n − a)/2 columns of Jm,n have (m/2) − 1 −1s below row 1, each of the first (n − a)/2 columns of A0 can have at most 1 − l more −1s below row 1. Since each 0 of the second group of (n − a)/2 columns of Jm,n have m/2 −1s below row 1, each of the second group of (n − a)/2 columns of A0 must have at least l fewer −1s than m/2. The last a columns of A0 can each have as many as (m/2) − l −1s. We thus conclude that (3.1) m n−a n−a n−a a 0 σ − (A0 )−σ − (Jm,n ) ≤ l+ (1−l)+ (−l)+a − l = l+ −nl+ m. 2 2 2 2 2 0 When (3.1) is non-positive, we conclude that Jm,n has the largest number of −1s in any dominating signing of Jm,n . We compute that (3.1) is non-positive if and only if
(3.2)
n≥1+
1 − a + am 2(l − 12 )
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 16
Since l ≥ 1, (3.2) is largest when l = 1 giving 2m (3.3) n ≥ 2 − a + am = 3m − 1
if a = 2 (n even) if a = 3 (n odd).
0 Thus we cannot improve upon the dominating signing Jm,n of Jm,n if n ≥ 2m and n is even, and n ≥ 3m − 1 and n is odd.
Now assume that m is odd. Consider the dominating signing −J(m+1)/2,(n+1−a)/2 J(m+1)/2,(n−1−a)/2 J(m+1)/2,a 0 Jm,n = , J(m−1)/2,(n+1−a)/2 −J(m−1)/2,(n−1−a)/2 J(m−1)/2,a 0 of Jm,n where now a = 2 if n is odd and a = 3 if n is even. We have σ(Jm,n ) = 2m−1 0 if n is odd, and σ(Jm,n ) = 3m−1 if n is even. As before we take A0 to be an arbitrary signing of Jm,n with x the largest number of −1s in a cross and k the largest number 0 with k or k − 1 −1s of −1s in a row. For Jm,n these values satisfy k = x − m−1 2 0 in every row. We show that we cannot improve on Jm,n first if k > x − (m − 1)/2, and then if k = x − (m − 1)/2. Suppose that A0 satisfies k > x − (m − 1)/2 + l where l ≥ 1, and row 1 of A0 0 in row 1. Since A0 is a dominating has k −1s. Thus A0 has l more −1s than Jm,n signing of Jm,n , each of the first n − a columns of A0 with row 1 deleted, has l fewer 0 . Each of the last a columns of A0 can −1s than the corresponding columns of Jm,n contain no more than (m − 1)/2 − l −1s. We thus conclude that m−1 a − 0 (3.4) σ (A ) − σ(Jm,n ) ≤ l − (n − a)l + a − l = −l(n − 1) + (m − 1). 2 2
We thus compute that (3.4) is non-positive if and only if n≥1+
(3.5)
a 2 (m
− 1) l
which is largest when l = 1. This gives a m+1 (3.6) n ≥ (m − 1) + 1 = 3m−1 +1 2 2
if a = 2 (n odd) if a = 3 (n even).
0 In particular, we cannot improve upon Jm,n if n ≥ 2m − 1 and n is odd, and n ≥ 3m − 1 if n is even, and k > x − (m − 1)/2 + l with l ≥ 1. Now suppose that A0 satisfies k = x − (m − 1)/2. Then A0 has at least one more 0 row with k −1s than Jm,n , that is, at least (m + 3)/2 such rows. There cannot be a column with (m + 3)/2 −1s, since if there were, there would be a cross with at least (m + 3)/2 + k − 1 = x + 1 > x −1s. We also can say that there cannot be a column with (m + 1)/2 −1s, since such a column would have to intersect one of the (m+3)/2 rows with k −1s in a 1, giving a cross with (m+1)/2+k = x+1 > x −1s. Thus every column of A0 has at most (m − 1)/2 −1s. Thus σ − (A0 ) ≤ n(m − 1)/2. Now 0 σ − (Jm,n )=
n−a+1m+1 n−a−1m−1 m(n − a) + 1 = = . 2 2 2 2 2
0 0 Thus if σ − (A0 ) ≤ σ − (Jm,n ), then we cannot improve upon the signing Jm,n . But
n(m − 1) m(n − a) + 1 ≤ 2 2
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
17
if and only if (3.7)
n ≥ am − 1 =
2m − 1 3m − 1
if a = 2 (n odd) if a = 3 (n even).
Since the restrictions on n in (3.6) are weaker than those in (3.7), this concludes the proof in the case where m is odd. PART II: We next establish the equalities 1(a), 2(b), 3(b), and 4(a) in the statement of the theorem. Let a = 2 if m and n have the same parity, and let a = 3 if m and n have opposite parity. Then in any dominating signing of Jm,n , a cross can contain at most (m + n − a)/2 −1s. We first assume that m is even. Consider the signing of Jm,n given by −Jm/2,(n−a)/2+1 Jm/2,(n+a)/2−1 A0 = Jm/2,(n−a)/2+1 X where X is to have all −1s except for one 1 in each column. We have n if a = 2 (n even), σ(A0 ) = n + a − 2 = n + 1 if a = 3 (n odd). Under these conditions, each cross of an entry within the first m/2 rows contains (m + n − a)/2 −1s. Suppose that X has at least a − 1 1s in each row. Then it is easy to check that A0 is a dominating signing of A. If n is even, then a = 2 and (n + a)/2 − 1 = n2 ≥ m 2 , and so such an X exists. If n is odd (m and n have opposite parity), then a = 3 and a − 1 = 2; so in order that X exist we only need n ≥ 2m − 1. This shows that n if n is even and n ≥ m, 0 γsd (Jm,n ) ≤ n + 1 if n is odd and n ≥ 2m − 1. If the maximum number of −1s in a row of a dominating signing of Jm,n is (n + a)/2 − 1, then maximum number of −1s in a column obtained by deleting this row is m/2 − 1. Since every such column of A0 contains m/2 − 1 −1s, it follows that σ(A0 ) is minimum over all dominating signings of Jm,n that contain a row with m/2 − 1 −1s. To complete the proof of this part, we need to show that the maximum number of −1s in a dominating signing of Jm,n occurs when there is a row with (n−a)/2+1 −1s. It follows from our analysis in Part I that this is so when n ≤ 2m − 1 if m is even, and n ≤ 3m − 1 if n is odd. PART III. To complete the proof we establish the equalities 2(a) and 3(a). 0 Thus we need to show that γsd (m, n) = 2m when m and n have opposite parity and m ≤ n < 2m. Let n = m + k where k is an integer with 0 ≤ k < m. Consider first the case where m is even and n is odd, and so k is odd. Each cross contains an even number 2m + k − 1 of entries, and so in a dominating signing of Jm,m+k there can be at most (2m + k − 3)/2 −1s in a cross. We first construct a 0 dominating signing A0 of Jm,m+k with σ(A0 ) = 2m, thus showing γsd (Jm,n ) ≤ 2m. For this we invoke the classical Gale-Ryser theorem (see e.g. [2, 10]) which asserts that there exists a (0,1)-matrix with row sum vector R = (r1 , r2 , . . . , rm ) (r1 ≥ r2 ≥ · · · ≥ rm ≥ 0) and column sum vector S = (s1 , s2 , . . . , sn ) (s1 ≥ s2 ≥ · · · ≥ sn ≥ 0) if and only if S is majorized by the conjugate R∗ of R, written S R∗ . Here R∗ = (r1∗ , r2∗ , . . . , rn∗ ) where rj∗ = |{i : ri ≥ j}| for j = 1, 2, . . . , n, and S R∗
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 18
means that
p X
si ≤
i=1
p X
ri∗
(1 ≤ p ≤ n),
i=1
with equality for p = n. We verify that the Gale-Ryser theorem guarantees the existence of a (0, 1)-matrix B of order m with m − k − 1 m−k−1 m+k−3 m + k − 3 ,..., , ,..., R= 2 2 2 2 | {z } | {z } m 2
and
S=
We have
m 2
m−2 m−2 m−2 , ,..., 2 2 2
.
m m R∗ = (m, . . . , m, , · · · . , 0, . . . , 0), | {z } |2 {z 2} m−k−1 k−1
2
from which it easily follows that S R∗ . It is easy to check that matrix −Jm/2,k Jm,m − 2B A0 = Jm/2,k is a dominating signing of Jm,m+k with 0
σ(A ) = m(m + k) − 2
m m−2 k+ m = 2m. 2 2
0 Hence γsd (Jm,m+k ) ≤ 2m. 0 We now show that γsd (Jm,m+k ) ≥ 2m. Let A00 be an arbitrary dominating signing of Jm,m+k , and suppose that σ(A00 ) ≤ 2m − 1, equivalently, σ − (A00 ) ≥ m m−2 00 2k + 2 m + 1. Each cross of a nonzero entry of A can contain at most x = (2m+k−3)/2 −1s. Let y be the largest number of −1s in a row of A00 . Without loss of generality, let row 1 contain y −1s. Since our constructed dominating signing A0 has ((m + k − 2)/2)m −1s, if σ(A00 ) is to be less than σ(A0 ), then we must have y > (m + k − 2)/2. Each column of A00 can contain at most x − y −1s outside any row with y −1s, and so we have m m−2 k+ m + 1 ≤ σ − (A00 ) ≤ y + (m + k)(x − y) 2 2 implying that m+k k+2 y≤ − , 2 m+k−1 and, since y is an integer, m+k−1 . y≤ 2 Thus the number of −1s in a column outside row 1 is at most m−2 x−y ≤ , 2 and hence each column contains at most m/2 −1s. Suppose there is a column, say column 1, containing m/2 −1s (such a column must intersect row 1 in a −1). Since σ − (A00 ) > σ − (A0 ), there must be more than m/2 rows with y −1s. Then column 1 intersects at least one of these rows in a 1, violating the domination property. Hence
SIGNED DOMINATION OF GRAPHS AND (0, 1)-MATRICES
19
each column of A00 contains at most (m − 2)/2 −1s, and σ − (A00 ) ≤ (m + k) m−2 2 , 0 0 a contradiction. Thus γsd (Jm,n ) ≥ 2m and hence γsd (Jm,n ) = 2m. Finally, we consider the case where m is odd and n is even. Thus n = m + k where k is odd and 0 ≤ k < m. Again each cross contains an even number 2m+k−1 entries, and so in a dominating signing of Jm,m+k there can be at most (2m+k−3)/2 0 −1s in a cross. First we show that γsd (Jm,m+k ) ≤ 2m by showing the existence of a dominating signing with m(n/2 − 1) −1s. It follows from the Gale-Ryser theorem that there exists a (0, 1)-matrix B of order m with row sum vector n n − 2k n n − 2k − 1, . . . , − 1 R = − 1, · · · , − 1, 2 2 2 |2 {z } | {z } m+1 2
m−1 2
with m components equal to n/2 − 1, and column sum vector m − 1 m−1 m−3 m − 3 S= ,..., ,..., , . 2 } | 2 2 } | 2 {z {z m+k 2
m−k 2
The conjugate of R is
m + 1 m+1 , . . . , m, R∗ = ,..., m, | {z } | 2 2 } {z n 2 −k−1
k
∗
and it is easy to check that S R . The matrix −J(m−1)/2,k 0 Jm,m − 2B A = Jm+1/2,k
is a dominating signing of Jm,m+k with σ(A0 ) = m(m + k) − 2
m−1 k + σ(B) = 2m. 2
0 Hence γsd (Jm,m+k ) ≤ 2m. 0 We now show that γsd (Jm,m+k ) ≥ 2m, and the proof is similar to the m even, n odd case. Suppose there is a dominating signing A00 of Jm,m+k with σ(A00 ) ≤ 2m−1 and so σ − (A00 ) ≥ (m(n − 2) + 2)/2. There must be a row, say row 1, of A00 with y > (m + k − 2)/2 −1s, and as before this yields
m+k k+2 − . 2 m+k−1 Since m + k is even, this implies that y ≤ (m + k − 2)/2, and we already have a contradiction. This completes the proof of the theorem. y≤
We now consider a general m by n (k, l)-semiregular matrix (so mk = nl, a fact used often in our calculations). We make use of a classical theorem of Vogel (see e.g. Mirsky [10]) that easily follows from the integral version of the max-flow min-cut theorem; we state a special case below. If Q = [qij ] and A = [aij ] are m by n real matrices, then we write Q ≤ A provided qij ≤ aij for all i and j.
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 20
Theorem 3.2. Let A = [aij ] be an m by n (0, 1)-matrix, and let R = (r1 , r2 , . . . , rm ) and S = (s1 , s2 , . . . , sn ) be two sequences of nonnegative integers. Let Q = [qij ] denote an m by n (0, 1)-matrix. Then n m X X (3.8) max σ(Q) : Q ≤ A, qij ≤ ri (1 ≤ i ≤ m), qij ≤ sj (1 ≤ j ≤ n) j=1
i=1
equals min
X
i6∈I
ri +
X
sj +
j6∈J
X
aij : I ⊆ {1, 2, . . . , m}, J ⊆ {1, 2, . . . , n}
.
i∈I,j∈J
We prove several lemmas that together give the main result of this section. The first lemma is straightforward. Lemma 3.3. Let A be an m by n (0, 1)-matrix with row sum vector R = (r1 , r2 , . . . , rm ) and column sum vector S = (s1 , s2 , . . . , sn ). Then X X X aij ≥ σ(A) − ri − sj (I ⊆ {1, 2, . . . , m}, J ⊆ {1, 2, . . . , n}). i∈I,j∈J
i6∈I
j6∈J
Lemma 3.4. Let A = [aij ] be an m by n (k, l)-semiregular (0, 1)-matrix with 0 m ≤ n. If k is odd and l is even, then γsd (A) ≤ 3m. If k is even and l is odd, then 0 γsd (A) ≤ 4m. Proof. By Theorem 3.2 there exists a (0, 1)-matrix Q = [qij ] with Q ≤ A such that Q has row sums at most (k−3)/2, column sums at most l/2, and σ(Q) ≥ k−3 2 m k−3 (and so row sums equal to (k − 3)/2 and σ(Q) = 2 m) if and only if X k−3 l k−3 (m−|I|)+ (n−|J|)+ aij ≥ m (I ⊆ {1, 2, . . . , m}, J ⊆ {1, 2, . . . , n}), 2 2 2 i∈I,j∈J
that is, if and only if X l k−3 (3.9) |I| − (n − |J|) aij ≥ 2 2
(I ⊆ {1, 2, . . . , m}, J ⊆ {1, 2, . . . , n}).
i∈I,j∈J
We apply Lemma 3.3 to get (3.10) X aij ≥ nl−l(n−|J|)−k(m−|I|) = k|I|+l|J|−nl
(I ⊆ {1, 2, . . . , m}, J ⊆ {1, 2, . . . , n}).
i∈I,j∈J
So, if for some I and J the right side of (3.10) is at least as large as the right side of (3.9), then (3.9) holds for that I and J; thus (3.9) holds provided (k + 3)|I| ≥ l(n − |J|). If the right side of (3.9) is nonpositive, then (3.9) holds trivially. The right side of (3.9) is nonpositive when (k − 3)|I| ≤ l(n − |J|), and thus is surely nonpositive when (k + 3)|I| ≤ l(n − |J|). This proves that (3.9) holds, and the matrix Q exists. The matrix A0 = (A − Q) − Q = A − 2Q is a signing of A that satisfies:
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21
(i) σ(A0 ) = km − 2 k−3 2 m = 3m, and (ii) The cross sums of the nonzero entries of A0 are at least k−3 l k+l−1−2 + = 2. 2 2 0 Thus A0 is a dominating signing of A, and hence γsd (A) ≤ 3m. A similar proof works when k is even and l is odd by replacing (k − 3)/2 with (k − 4)/2 and l/2 with (l − 1)/2. This completes the proof of the lemma.
Lemma 3.5. Let A = [aij ] be an m by n (k, l)-semiregular (0, 1)-matrix with m ≤ n, and let k and l be even. Then 0 γsd (A) ≤ 2m.
Proof. Suppose that there exists a (0, 1)-matrix with Q ≤ A such that Q has row sums (k − 2)/2, column sums at most l/2, and σ(Q) = m(k − 2)/2. Then the matrix A0 = [a0ij ] = (A − Q) − Q = A − 2Q is a signing of A that satisfies: (i) σ(A0 ) = km − 2m( k2 − 1) = 2m, and (ii) The cross sums of the nonzero entries of A are at least k−2 l (k + l − 1) − 2 + = 1. 2 2 0 (A) ≤ 2m. Thus A0 is a dominating signing of A and thus γsd To complete the proof for this case, we invoke Theorem 3.2 and show that for all I ⊆ {1, 2, . . . , m} and all J ⊆ {1, 2, . . . , n} X X X k ri + sj + aij ≥ m −1 , 2 i6∈I
j6∈J
i∈I,j∈J
that is, in our case, (m − |I|)
X k l k − 1 + (n − |J|) + −1 aij ≥ m 2 2 2 i∈I,j∈J
or, after simplification, (3.11)
X
aij ≥ |I|
i∈I,j∈J
k l mk − 1 + |J| − . 2 2 2
This is certainly satisfied if I = J = ∅, and so we may assume that not both I and J are empty. From Lemma 3.4 we get that X aij ≥ mk − (m − |I|) k − (n − |J|) l i∈I,j∈J
(3.12)
= k|I| + l|J| − mk.
Thus if
k l mk − 1 + |J| − , 2 2 2
k l mk − 1 + |J| − , 2 2 2
k|I| + l|J| − mk ≥ |I| (3.11) holds. Otherwise, we have k|I| + l|J| − mk < |I|
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 22
which reduces to |I|
k l mk + 1 + |J| < , 2 2 2
and (3.11) is satisfied because the right hand side is now negative.
Lemma 3.6. Let A = [aij ] be an m by n (k, l)-semiregular (0, 1)-matrix, and let k and l be odd. Then 0 γsd (A) ≤ m + n − 1.
Proof. This proof proceeds in a similar way to the proof of Lemma 3.5. We now apply Theorem 3.2 with ri = k−1 for all i and sj = l−1 2 2 for all j. If the maximum in (3.8) is at least ((k − 1)m − n + 1)/2, then there is a matrix Q with row l−1 sums at most k−1 2 and column sums at most 2 with σ(Q) ≥ ((k − 1)m − n + 1)/2. 0 Then matrix A = A − 2Q is a signing of A with cross sums at least k−1 l−1 + =1 (k + l − 1) − 2 2 2 and σ(A0 ) ≤ km − 2
(k − 1)m − n + 1 = m + n − 1. 2
0 Hence γsd (A) ≤ m + n − 1. Now we need to show that X k−1 l − 1 mk − 1 (3.13) aij ≥ |I| + |J| − . 2 2 2 i∈I,j∈J
This holds if I = J = ∅, and so we now assume that not both I and J are empty. To verify (3.13), we again invoke (3.12). If k|I| + l|J| − mk ≥ |I|
l − 1 mk − 1 k−1 + |J| − , 2 2 2
then (3.13) holds. Otherwise, we have k|I| + l|J| − mk < |I|
k−1 l − 1 mk − 1 + |J| − 2 2 2
which reduces to |I|
k+1 l+1 mk + 1 + |J| < , 2 2 2
or, equivalently, |I|
k−1 l − 1 mk − 1 + |J| − < 1 − (|I| + |J|). 2 2 2
Thus the right hand side of (3.13) is negative, and hence (3.13) holds.
Lemma 3.7. Let A = [aij ] be an m by n (k, l)-semiregular (0, 1)-matrix, and let k be even and l be odd. Then 2m if m ≤ n ≤ 2m 0 γsd (A) ≤ n if 2m ≤ n ≤ 3m.
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23
Proof. We first assume that m < n ≤ 2m. In this case we show there exists l−1 a matrix Q with row sums at most k−2 2 and column sums at most 2 , and with k−2 0 σ(Q) = m 2 if m < n ≤ 2m. Then the matrix A = A − 2Q is a dominating signing of A with σ(A0 ) = 2m. In order for Q to exist, we must have X k−2 l − 1 mk − n (3.14) aij ≥ |I| + |J| − . 2 2 2 i∈I,j∈J
If k|I| + l|J| − mk ≥ |I|
l − 1 mk − n k−2 + |J| − , 2 2 2
we are done. Otherwise, k|I| + l|J| − mk < |I|
l − 1 mk − n k−2 + |J| − , 2 2 2
and this simplifies to |I|
k+2 l+1 mk + n + |J| < . 2 2 2
This can be rewritten as k−2 l − 1 mk − n |I| + |J| − < n − 2|I| − |J|. 2 2 2 Thus (3.14) holds trivially if 2|I| + |J| ≥ n. Suppose that 2|I| + |J| < n, that is, |J| < n − 2|I|. Then we get k−2 k l − 1 mk − n |I| + |J| − < |I| −l . 2 2 2 2 Since n ≤ 2m, we have l ≥ k2 , and the right hand side of (3.14) is nonpositive, and thus (3.14) holds. Now we assume that 2m < n ≤ 3m. In this case we show we can find a Q l−1 l−1 with row sums at most k−2 2 and column sums at most 2 , and with σ(Q) ≥ n 2 . 0 0 Then the matrix A = A − 2Q is a dominating signing of A with σ(A ) = n. In order for Q to exist, we must have X k−2 l − 1 m(k − 2) (3.15) aij ≥ |I| + |J| − . 2 2 2 i∈I,j∈J
As before we are done unless k|I| + l|J| − mk < |I|
k−2 l − 1 m(k − 2) + |J| − , 2 2 2
and this can be rewritten as k−2 l − 1 m(k − 2) + |J| − < −2|I| − |J| + 2m. |I| 2 2 2 So if 2|I| + |J| ≥ 2m, we are done. Now suppose that 2|I| + |J| < 2m, that is, |J| < 2(m − |I|). Then using this inequality, we get k−2 l − 1 m(k − 2) |I| − m + |J| − < (k − 2l). 2 2 2 2 Since n > 2m, mk = nl implies that k ≥ 2l. Since I ⊆ {1, 2, . . . , m}, |I| − m ≤ 0. Thus the expression on the right side of (3.16) is nonpositive, and we are done. (3.16)
|I|
ADAM H. BERLINER, RICHARD A. BRUALDI, LOUIS DEAETT, KATHLEEN P. KIERNAN, SETH A. MEYER, AND MICHAEL W. SCH 24
Using the previous lemmas, we can now show that Xu’s conjecture holds for semiregular (0, 1)-matrices most of the time. Theorem 3.8. Let A = [aij ] be an m by n (k, l)-semiregular (0, 1)-matrix. Then, except possibly in the case that m < n ≤ 2m, k is odd, and l is even, 0 γsd (A) ≤ m + n − 1.
Proof. The theorem follows by combining Theorem 2.7 and Lemmas 3.4, 3.5, 3.6, and 3.7. 0 As indicated in Theorem 3.8, we have been unable to show that γsd (A) ≤ m + n − 1 in the case that m < n ≤ 2m where k is odd and l is even. It seems a new technique may be needed to resolve this case. As pointed out by the referee, Corollaries 6 and 8 in [7] show that if A is an m by n (0,1)-matrix all of whose row 0 and columns sums are even (respectively, odd), then γsd (A) ≤ m + n − 1. We do not know to what extent this result can be extended to the case where all the row sums have one parity and the column sums have the other parity. As a final remark, we note that we know only two instances of families of m 0 by n (0, 1)-matrices A for which γsd (A) = m + n − 1 with equality. These are the matrices A = Jn,n+1 (see Theorem 3.1) and the bi-adjacency matrices of the bipartite graphs obtained from a star Sn by subdividing each edge (as already remarked, a dominating edge signing of such a graph contains no −1s). Note that both of these matrices are n by n + 1.
Acknowledgement: We are very much indebted to an anonymous referee who read our paper most carefully and who pointed out many typos and other small corrections. References [1] S. Akbari, S. Bolouki, P. Hatami, M. Siami, On the signed edge domination number of graphs, Discrete Math., 309 (2009), 587–594. [2] R.A. Brualdi, Combinatorial Matrix Classes, Cambrige University Press, Cambridge, 2006. [3] X. Fu, Y. Yang, B, Jiang, A note on the signed edges domination number in graphs, Discrete Applied Math., 156 (2008), 2790–2792. [4] R. Haas, T.B. Wexler, Signed domination numbers of a graph and its complement, Discrete Math., 283 (2004), 87–92. [5] R. Haas, T.B. Wexler, Bounds on the signed domination number of a graph, The Ninth Quadrennial International Conference on Graph Theory, Combinatorics, Algorithms, and Applications, Electron. Notes Discrete Math., 11 (9 pp.), Elsevier, Amsterdam, 2002. [6] H. Karami, A. Khodkar, S.M. Sheikholeslami, Signed edge domination numbers in trees, Ars Combinatoria, 93 (2009), 451–457. [7] H. Karami, A. Khodkar, S.M. Sheikholeslami, An improved upper bound for signed edge domination in graphs, Utilitas Math., 78 (2009), 121–128. [8] H. Karami, S.M. Sheikholeslami, A. Khodkar, Some notes on signed edge domination in graphs, Graphs and Combinatorics, 24 (2008), 29–35. [9] H. Karami, S.M. Sheikholeslami, A. Khodkar, Lower bounds on signed edge total domination numbers in graphs, Czechoslovak Math. J., 58(133) (2008), 595–603. [10] L. Mirsky, Transversal Theory, Academic Press, New York 1971, 209–211. [11] X. Pi and H. Liu, On the characterization of trees with signed edge domination numbers 1, 2, 3, or 4, Discrete Math., 309 (2009), 1179–1782. [12] B. Xu, On signed edge domination numbers of graphs, Discrete Math., 239(2001), 179–189.
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[13] B. Xu, On the lower bounds of signed edge domination numbers in graphs, J. E. China Jiatong University 1 (2004), 110-114 (in Chinese). [14] B. Xu, On edge domination numbers of graphs, Discrete Math. 294 (2005), 311-316. [15] B. Zelinka, On signed edge domination number in trees, Math. Bohem., 127 (2002), 49–55. St. Olaf College, Northfield, MN, USA E-mail address:
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