Domination of the rectangular queen's graph

The queen's graph $Q_{m \times n}$ has the squares of the $m \times n$ chessboard as its vertices; two squares are adjacent if they are in the same row, column, or diagonal of the board. A set $D$ of squares of $Q_{m \times n}$ is a dominating set for $Q_{m \times n}$ if every square of $Q_{m \times n}$ is either in $D$ or adjacent to a square in $D$. The minimum size of a dominating set of $Q_{m \times n}$ is the domination number, denoted by $\gamma(Q_{m \times n})$. Values of $\gamma(Q_{m \times n}), \, 4 \leq m \leq n \leq 18, \,$ are given here, in each case with a file of minimum dominating sets (often all of them, up to symmetry) in an online appendix at https://www.combinatorics.org/ojs/index.php/eljc/article/view/v26i4p45/HTML. In these ranges for $m$ and $n$, monotonicity fails once: $\gamma(Q_{8 \times 11}) = 6>5 = \gamma(Q_{9 \times 11}) = \gamma(Q_{10 \times 11}) = \gamma(Q_{11 \times 11})$. Lower bounds on $\gamma(Q_{m \times n})$ are given. In particular, if $m \leq n$ then $\gamma(Q_{m \times n}) \geq \min \{ m, \lceil (m+n-2)/4 \rceil \}$. A set of squares is independent if no two of its squares are adjacent. The minimum size of an independent dominating set of $Q_{m \times n}$ is the independent domination number, denoted by $i(Q_{m \times n})$. Values of $i(Q_{m \times n}), \, 4 \leq m \leq n \leq 18, \,$ are given here, in each case with some minimum dominating sets. In these ranges for $m$ and $n$, monotonicity fails twice: $i(Q_{8 \times 11}) = 6>5 = i(Q_{9 \times 11}) = i(Q_{10 \times 11}) = i(Q_{11 \times 11})$, and $i(Q_{11 \times 18}) = 9>8 = i(Q_{12 \times 18})$.


Introduction
Let m and n be positive integers. We will identify the m × n chessboard with a rectangle in the Cartesian plane, having sides parallel to the coordinate axes. We place the board so that the center of every square has integer coordinates, and refer to each square by the coordinates (x,y) of its center. Unless otherwise noted, squares have edge length one, and the board is placed so that the lower left corner has center (1, 1); sometimes it is more convenient to use squares of edge length two or to place the board with its center at the origin of the coordinate system. By symmetry it suffices to consider the case m n, which we will assume throughout: the board has at least as many columns as rows.
The square (x,y) is in column x and row y. Columns and rows will be referred to collectively as orthogonals. The difference diagonal (respectively sum diagonal ) through square (x,y) is the set of all board squares with centers on the line of slope +1 (respectively −1) through the point (x,y). The value of y − x is the same for each square (x,y) on a difference diagonal, and we will refer to the diagonal by this value. Similarly, the value of y + x is the same for each square on a sum diagonal, and we associate this value to the diagonal. Orthogonals and diagonals are collectively referred to as lines of the board.
The queens graph Q m×n has the squares of the m × n chessboard as its vertices; two squares are adjacent if they are in some line of Q m×n . A set D of squares of Q m×n is a dominating set for Q m×n if every square of Q m×n is either in D or adjacent to a square in D. The minimum size of a dominating set is the domination number, denoted by γ(Q m×n ). A set of squares is independent if no two squares in the set are adjacent.
Almost all previous work on queen domination has concerned square boards. The problem of finding values of γ(Q n×n ) has interested mathematicians for over 150 years. The first published work is that of De Jaenisch [15] in 1862, who gave minimum dominating sets and minimum independent dominating sets of Q n×n for n 8. His work was briefly summarized by Rouse Ball [21] in 1892, who considered several other questions about queen domination. In 1901, W. Ahrens [1,Chapter X] gave minimum dominating sets for Q 9×9 , and in 1902-3, K. von Szily [22,23] gave minimum dominating sets of Q n×n for 10 n 13 and n = 17. Proof that these sets were minimum had to wait for later work, described below. De Jaenisch, Ahrens, and von Szily also worked extensively to find the number of different minimum dominating sets for each n, often giving lists with one set from each symmetry class. Many of these results were collected by Ahrens in the 1910 edition [2] of his book, and can also be found in its later editions.
More detail and some examples from recent work on domination of Q n×n can be found in [28].
Say that two minimum dominating sets of γ(Q m×n ) are equivalent if there is an isometry of the m × n chessboard that carries one to the other.
We have computed γ(Q m×n ) for rectangular chessboards with 4 m n 18. Results are given in Table 1; for most m and n we give a file of minimum dominating sets with one from every equivalence class, unless the number of equivalence classes is large. The online appendix at Table 1: Values of γ(Q m×n ), 4 m n 18 (OEIS A274138) The computation was done with a backtracking algorithm. The backtrack condition minimizes the number of queens placed. If a solution is found with k queens, then the remaining search space is limited to at most k − 1 queens. The algorithm places a single queen in a position covering the top left cell and does a recursive call to cover all remaining cells. Some heuristics are used also to find the first solution faster: the first queen is placed in the middle of the board (actually in the closest to middle position attacking the top left cell); other possible attacking positions are only tried later. Frequently this position is part of a minimal solution. Once it is shown that there is no solution with k −1 queens, a search for other solutions with k queens is made.
Cockayne [8,Problem 1] introduced monotonicity as an open problem (see also  Table 1) are necessary to dominate it, though five queens are sufficient (and necessary) to dominate each of Q 9×11 , Q 10×11 , Q 11×11 . A possible explanation for this is given later. We note that five queens can cover all but one square of Q 8×11 . One of the 8 placements is in Figure 1.
We extend Cockayne's question to the rectangular case.
We discuss one type of internal symmetry of minimum dominating sets that frequently occurs. A foursome is a set of four squares (x + a, y , where either each of x, y, a, b is an integer or each is half an odd integer, and a and b are unequal and nonzero. The center of the foursome is the point (x, y), which need not be a square center. For examples, see Figure 1 above, the first minimum dominating sets given for Q 9×9 and Q 11×11 , as well as the first four minimum dominating sets given for Q 11×12 .
If a foursome F is flipped across any of the four lines through its center, the result is another foursome F ′ that occupies the same lines as F ; this is illustrated in Figure 2. Thus if a dominating set D of Q m×n contains F , we may replace F in D with F ′ and obtain a dominating set D ′ of the same size as D. Usually D and D ′ are not equivalent.
As an example, we analyze the minimum dominating sets of Q 11×17 , which have size 8. Up to equivalence there are 131 solutions, shown in the file. Of these, 85 have no foursomes, 41 have exactly one foursome, four (#125, 69, 70, 62) have exactly two foursomes, and one (#76) has 3 foursomes.
We may define a relation on the set S(11, 17) of minimum dominating sets of Q 11×17 by saying that two sets are related if either they are equivalent, or flipping a foursome of the first set yields a set equivalent to the second set. This relation is reflexive and symmetric, and its transitive closure gives a partition P(11, 17) of S (11,17), which may also be regarded as a partition of the set of (isometric) equivalence classes, as we will do.
For example, solution #125 has two foursomes: one centered at (12,6) with (a, b) = (4, 2), and one centered at (9, 7) with (a, b) = (3, −1). Flipping the first gives solution #124. If instead we flip the second, we get the reflection of #125 across its vertical line of symmetry. This implies that one part of the partition P (11,17) contains just the equivalence classes of 125 and 124, and we denote this part by {125, 124}. It is then straightforward to see that P (11,17)  It would be possible to reduce the size of the appendix by giving for each (m, n) one solution from each part of the partition P(m, n) rather than one solution from each isometry equivalence class. But when two solutions differ by the flip of a foursome, it is not clear which is most useful to see, so we have not done this.

Lower bounds on queen domination numbers
We begin by looking at what happens when the board is far from square.
Proof. Each queen attacks all squares in her own row, but at most three squares in any other row. Thus m − 1 queens occupy at most m − 1 rows and cover at most 3(m − 1) squares in any row that does not contain a queen. On the other hand, m queens are certainly sufficient.
We change viewpoint slightly, focusing on the size of the dominating set rather than the dimensions of the board. For each positive integer m, let g(m) be the largest integer such that m queens can cover the (m + 1) × g(m) board. (Proposition 1 asserts g(m) 3m.) Let g * (m) be the largest integer such that m queens, no two in a row, can cover the (m + 1) × g * (m) board.
Take the board to be a rectangle in the Cartesian plane with sides parallel to the axes, board center at the origin, and squares of edge length two.
Assume that D is a dominating set of size m for the (m + 1) × 3m board. One of two cases occurs: (1) there is only one empty row, which without loss of generality has index h, 0 h m, and each other row contains exactly one square of D, or (2) there are exactly two empty rows; in this case m is even, row 0 contains two squares of D, the empty rows are indexed ±h for some h, 0 < h m, and each other row contains exactly one square of D.
In either case: if m ≡ 0 or 4 (mod 9), then h ≡ 0(mod 3); if m ≡ 1 or 3 (mod 9), then h ≡ 0 (mod 3). We can add up the squares of the column indices of the squares in row h in two ways, giving the equation This reduces to For a particular dominating set D we may regard this as a quadratic equation in h, so there are at most two empty rows. If there are two empty rows h 1 , h 2 , then (1) implies 2mh 2 As |D| = m, there is exactly one row, say l, with two queens, and all rows except h 1 , h 2 , l have just one queen. Thus But h 1 + h 2 is even since all row indices have the same parity. Thus h 1 + h 2 = 0, so l = 0 and all row indices are even, which implies m is even. So there is h, 0 < h m, such that the empty rows are ±h.
If instead there is only one empty row h, then we may assume 0 h m by flipping across the x-axis if necessary.

is satisfied for any h.)
Only one example of the second case of Theorem 2 is known: for m = 2, the set D = {(−3, 0), (3, 0)} covers the 3 × 6 board. A computer search shows that there is no other example with m 40. It seems likely that no other example exists; if for some m there is such a set D = {(x i , y i ) : 1 i m}, we have been able to show that 0 = . By computer search (see the file) we have shown that for all m such that m 40, m = 3, and m ≡ 0, 1, 2, 3, or 4 (mod 9), there exist sets of m queen squares dominating Q (m+1)×3m , with h taking all values not ruled out by Theorem 2. We believe g(m) = g * (m) = 3m holds for all m = 3 with m ≡ 0, 1, 2, 3, or 4 (mod 9).
The following is immediate from Theorem 2.
Take the board to be a rectangle in the Cartesian plane with sides parallel to the axes, board center at the origin, and squares of edge length two.
For m ≡ 6 or 7 (mod 9), assume that D is a set of m squares that dominates the Proof. Let m be a positive integer and j an integer with 0 j 4. Let D be a dominating set of m queens for the (m + 1) × (3m − j)-board, with only row h empty.
The 3m − j column indices form the set It will be useful to consider row h extended beyond the board, and to look at the congruence classes modulo 3 of the column indices. To this end, for each integer i, We write c i for the size of C i . By symmetry, Proof. The facts that c −1 = c 1 , c −1 + c 0 + c 1 = 3m − j and the c i 's differ by at most one imply the first equation, and the second is similar.
Let (x, y) be a square of D. The squares in the extension of row h covered by (x, y) are That is, a i counts every cover of any square (x, h) that is off the board (|x| > 3m − j − 1) and all but one cover of each multiply covered square (x, h) on the board. Thus each a i is nonnegative, and since each of the m squares of D covers 3 squares of the extended row h, a −1 + a 0 + a 1 = j.
Since each square in row h of the board is covered by D, we get a system of equations: As c −1 = c 1 , subtracting (6) from (8) shows For a dominating set D as hypothesized to exist, it is necessary that the total number t of squares (x, y) in D with y ≡ h (mod 3) is one less than the number r h of rows in R h . When t = r h − 1, we will say h is eligible for t.
Theorem 2 covers the case j = 0, so we pass to less wide boards, only considering m ≡ 5, 6, 7, 8 (mod 9). Let j = 1. From (10) we have a −1 = a 1 = 0 and then a 0 = 1. Here the s of Lemma 5 is 1 so , and the following analysis shows that no h is eligible for any t for m ≡ 5, 6, 7, 8 (mod 9).
Computer search reveals (see the file) that for m 40 and m ≡ 5, 6, 7 (mod 9), all minimum dominating sets of Q m+1×3m−2 have just one empty row, and all eligible values of h actually occur. For m ≡ 8 (mod 9), Theorem 4 does not say any h are ineligible; indeed, our search has found solutions with one empty row for all the h values. The only board size in the 3 m 40 range where the minimum dominating sets found have two empty rows is Q 9×20 . Those dominating sets demonstrate numerous patterns of pairs of empty rows, as shown in this file. We next show that by "pasting together" dominating sets of a certain type, we can extend the range of values for which the bounds of Theorem 4 are known to be exact.
Say that a topless dominating set for the (m + 1) × n board is a dominating set of size m having one square in each row except the top row, which is empty. Proof. We return to our usual scheme of indexing columns and rows from the bottom left board corner. For (A): For i = 1, 2, let S i be a topless dominating set of m i queens for the (m i + 1) × n i board. On the (m 1 + m 2 + 1) × (n 1 + n 2 ) board, the squares in which the columns indexed 1 to n 1 and the rows indexed 1 to m 1 + 1 meet form a copy of the (m 1 + 1) × n 1 board. Place a copy S ′ 1 of S 1 on that. The squares in which the columns indexed n 1 + 1 to n 1 + n 2 and the rows indexed m 1 + 1 to m 1 + m 2 + 1 meet form a copy of the (m 2 + 1) × n 2 board. Place a copy S ′ 2 of S 2 , rotated by a half turn, on that. The union of the two copies is a set S 1 ⊕ S 2 of m 1 + m 2 squares that leaves only row m 2 + 1 empty on the (m 1 + m 2 + 1) × (n 1 + n 2 ) board. As S ′ 1 covers the left n 1 squares of that row and S ′ 2 covers the remainder, S 1 ⊕ S 2 dominates the (m 1 + m 2 + 1) × (n 1 + n 2 ) board. For (B): We will use topless dominating sets of the (m + 1) × (3m − j) board for (m, j) = (1,0), (5,2), (8,4), and (11,0). Assume the hypotheses; we then need to prove the existence of (minimum) dominating sets of size m for the (m + 1) × (3m − j) boards in the ranges claimed. For m < 18 the database (see the files here and here) contain the claimed dominating sets. For other board sizes of m = 9l, 9l + 1, 9l + 2, 2 l k, we are assuming there are such sets (in fact, topless). For m = 9l + 3, we use part (A) to "paste together" topless sets for m = 9l + 2 and m = 1. For m = 9l + 4, we paste together topless sets for m = 9(l −1)+2 and m = 11. Pasting together a topless set for each of m = 9l, 9l +1, 9l +2 with a topless set for m = 5 gives the result for m = 9l + 5, 9l + 6, 9l + 7. Finally, pasting together topless sets for m = 9l and m = 8 gives the result for m = 9l + 8.
As mentioned after Theorem 2, computer search gave topless dominating sets as required in Proposition 6(B) for k = 4. It follows that g * (m) equals the bound of Theorem 4 for m 44, m = 3.
We next develop a lower bound for γ(Q m×n ) for more general m, n. Raghavan and Venketesan [20] and Spencer [8,25] independently proved that It has been shown [11] that γ(Q n×n ) = (n − 1)/2 only for n = 3, 11. Both of these values are significant for our work here, as we now discuss.
A central queen on Q 3×3 shows γ(Q 3×3 ) = 1. This simple fact has a useful generalization: if C is a central sub-board of Q m×n such that every square of Q m×n has a line meeting C, then a subset of C that occupies all those lines is a dominating set of Q m×n . More than a hundred years ago, Szily [22,23] gave dominating sets of this type for Q 13×13 and Q 17×17 , which were later shown to be minimum. We found that Q 13×16 has a minimum dominating set (solution #23 in the database) of this centrally strong form and have also used this idea to produce good upper bounds of Q m×n for some m, n, as shown below.
It follows from [26] that there are exactly two minimum dominating sets for Q 11×11 . Placing the origin of our coordinate system at board center, these sets are D = {(0, 0), ±(2, 4), ±(4, −2)} (see Figure 3) and the reflection of D across the column x = 0. So up to equivalence D is the unique minimum dominating set of Q 11×11 , consisting of a foursome and a queen at its center. This amazing set has an influence on many other values of γ(Q m×n ).
It was observed by Eisenstein et al. [9] that if a dominating set D of Q n×n contains no edge squares, the facts that there are 4(n − 1) edge squares and every queen covers eight edge squares imply |D| ⌈(n − 1)/2⌉. This suggests the bound (11). A similar approach leads one to guess the bound of our next theorem, but some care is needed to handle the general case.
Thus we may take γ(Q m×n ) m − 2 and let D be a minimum dominating set of Q m×n . Since m n, there are at least two rows and at least two columns that do not contain squares of D. Let a be the index of the leftmost empty column, b the index of the rightmost empty column, c the index of the lowest empty row, d the index of the highest empty row. The board has a rectangular sub-board U with corner squares (a, c), (a, d), (b, c), and (b, d). Let E be the set of edge squares of this sub-board. We say that U is the box of D and E is the box border of D; these sets are defined for any square Removing columns a and b and rows c and d divides the board into nine regions (some possibly empty). Let C be the set of squares of D inside U; that is, C = {(x, y) ∈ D : a < x < b and c < y < d}. Let T nw be the set of squares of D in the "northwest" region of the m × n board; that is, T nw = {(x, y) ∈ D : x < a and y > d}. Similarly we label seven more subsets of D by their "geographic direction" from the central region: T n , T ne , T e , T se , T s , T sw , and T w . Let R = T nw ∪ T ne ∪ T sw ∪ T se , the set of those squares of D whose orthogonals do not meet U. Let S = T n ∪ T e ∪ T s ∪ T w , the set of those squares of D having exactly one orthogonal that meets U. Then D is the disjoint union of R, S, and C.
Since each column to the left of column a contains at least one square of D, Similarly, Adding inequalities (13)-(16) and using the definitions of R and S gives Each square in R covers at most two squares of E, as the square's orthogonals and one of its diagonals miss E. Each square in S covers at most six squares of E, as one of the square's orthogonals misses E. Each square in C covers eight squares of E. Since D is a dominating set, D covers all squares of E, so Adding two times (17) to (18) gives 6|R| + 8|S| + 8|C| 2(m + n − 2).
A diagram illustrating the proof for Q 10×17 is given in Figure 4. There are 120 pairs (m, n) satisfying 4 m n 18. Of these, the bound (12) is achieved for 40 pairs (28 with m 6), for 76 pairs the bound is exceeded by one and for the pairs (12,14), (13,17), (14,16), and (15,15) the bound is exceeded by two.
We next explore when γ(Q m×n ) = (m + n − 2)/4. From Proposition 1 it follows that for any positive integer m, if n = 3m + 2 then γ(Q m×n ) = m, and here m = (m + n − 2)/4. So we restrict to n < 3m + 2. Proof. From n < 3m + 2 we have |D| = (m + n − 2)/4 m − 2, so the box of D is defined. Since |D| = (m + n − 2)/4, in this setting we have equality in inequalities (13)- (20), so each square of the box border E is covered exactly once by D. Thus any line meeting E contains at most one square of D. Every square of D must be diagonally adjacent to four squares of E, so if any line containing a square of D does not meet E, it is an orthogonal. From (20) we see that here the set R of "corner squares" in D is empty, so every square of D has at least one orthogonal meeting E; then since (13)- (16) are equations here, each orthogonal that misses E contains exactly one square of D. Thus D is independent.
Rarely does a minimum dominating set cover each of its box border squares uniquely; see Figure 3 for an example. We also note that the minimum dominating sets #1-4 for Q 11×12 have this property. Each of these sets consists of a foursome centered at (13/2, 13/2) plus the corner squares (1, 1) and (12, 1), so is not independent.
We next extend the method of proof used in [8,20,25] for the lower bound (11) to show that the dimensions of the box of D give a lower bound for |D|. x < e or x > e + m ′ − 2}. Then no square is diagonally adjacent to more than two squares of S and no square of P is orthogonally adjacent to any square of S. As the 2(m ′ − 1) squares of S are covered by D, 2(m ′ − 1) 2|P | + 4(|D| − |P |), which implies If m ′ > n ′ then we can choose e so that all columns that do not meet S are occupied, so |P | n − (m ′ − 1). If m ′ n ′ we can choose e so that S is contained in the top and bottom edges of U, and then |P | n − n ′ . In both cases, (21) implies the conclusion.
For m < n, we have little evidence that the bound (12) is good. We were not able to use the methods of the proofs of Theorem 7 and Proposition 9 to improve on this bound. Also, a computer search using a greedy algorithm for some larger m, n did not supply evidence about lower bounds for γ(Q m×n ).
The statement of Proposition 9 leads one to consider the quantity n/2. We have checked that when 4 m n 18, γ(Q m×n ) min{m − 1, ⌊n/2⌋ − 1}. This bound and the bound (12) are close only when m and n are close. So we ask the following.
Question 3. For m, n with m n, what is a good general lower bound for γ(Q m×n )? In particular, is it true that γ(Q m×n ) min{m − 1, ⌊n/2⌋ − 1}?
The first difficulty was just discussed: in general we know the value of γ(Q m×n ) only approximately.
The second difficulty is that construction of a dominating set of Q m×n generally means specifying most or all of the lines that the set is to occupy. There are some restrictions that the indices of the lines must satisfy, as we now describe.
Let D = {(x i , y i ) : 1 i l} be a set of l squares of Q m×n that occupies difference diagonals (d i ) l i=1 and sum diagonals (s i ) l i=1 . Since the square (x,y) is on the difference diagonal with index y − x and the sum diagonal with index y + x, summing over D gives The Parallelogram Law 2x 2 + 2y 2 = (y − x) 2 + (y + x) 2 gives a quadratic constraint on the line indices.
In each of the two constructions given below, we will refer to lines that must be occupied for domination as required lines and other lines as auxiliary lines.
Both constructions produce a number of minimum dominating sets, but neither can produce a dominating set of Q m×n of size less than ⌊n/2⌋. This is a little evidence for the possible bound mentioned in Question 3.

Domination by orthodox covers
This idea generalizes [27,Section 2]. Let D be a set of squares of Q m×n . If it is possible to place the origin of the coordinate system so that every even column and every even row contains a square of D, we will say D is an orthodox set. That is, an orthodox set is one that occupies at least every other column and every other row of Q m×n .
Say that square (x,y) of Q m×n is even if x + y is even, odd if x + y is odd. We divide the even squares of Q m×n into two classes: (x,y) is even-even if both x and y are even, odd-odd if both are odd. If D is an orthodox set and each odd-odd square of Q m×n is covered by some square of D, we say D is an orthodox cover. For example, solution #10 for Q 7×11 given in Table 1 is an orthodox cover; take the origin at (6,3) to see this.
It is clear from the definition that an orthodox cover dominates every even square of Q m×n , and since every odd square of Q m×n is on one even-indexed orthogonal, all odd squares are also dominated: an orthodox cover is a dominating set of Q m×n . Many orthodox covers appear in the appendix, and are labeled there as such.
Since Q m×n has at least ⌊n/2⌋ even-indexed columns, an orthodox set on Q m×n has at least ⌊n/2⌋ members. Generally, we expect that most of the squares of D will be eveneven, to help dominate the odd-odd squares diagonally. When n is considerably larger than m, there are more possibilities of placing queens on odd squares that occupy even columns. Also, it is possible sometimes to achieve a dominating set of size less than ⌊n/2⌋ by a minor modification, as is shown in solution #1 for Q 7×12 in Table 1. If the center of the square there labeled (6, 3) is taken to be the origin of the coordinate system, the dominating set shown misses being an orthodox set only by not occupying the rightmost column. Thus the three odd-odd squares in that column are not covered along their column, as they would be by an orthodox set. But the queen on a dark square covers those three squares, and the odd squares of its column, and thus completes a dominating set of size 5.
Minimum dominating set #147 for Q 12×16 is an orthodox cover with a single queen on a dark square, at (6,10). The squares covered only by this queen are the dark squares in its column and (1,10). Replacing (6,10) with the white square (6, 5) covers those squares and, adding a row 0 to the board, also the square (1, 0). In fact, the full set now covers all of row 0 and is thus a minimum dominating set of Q 13×16 ; it is solution #15 for Q 13×16 , rotated by a half-turn.
There are many ways to create orthodox covers, and we will only give one example. An approach is to regard Q m×n as the union of overlapping copies of Q m×m ; for odd m, this allows us to use [27,Theorem 1], which gives sufficient conditions for an orthodox set on Q m×m to be an orthodox cover. Example 1. An orthodox cover implying γ(Q 13×19 ) 10. We take the origin of the coordinate system to be the center of Q 13×19 , and regard Q 13×19 as the union of two copies of Q 13×13 , centered at (±3, 0). From [27,Theorem 1], if we regard the center of Q 13×13 as the origin, an orthodox set on Q 13×13 dominates if the set occupies the sum and difference diagonals with indices in {−6, −2, 0, 2, 6}. Asking this on both copies of Q 13×13 , we wish to have our orthodox set occupy the sum and difference diagonals which (on Q 13×19 ) have indices in {−6, −2, 0, 2, 6}±3, which is {±1, ±3, ±5, ±9}, so there will be two auxiliary difference diagonal indices d 1 , d 2 and two auxiliary sum diagonal indices s 1 , s 2 . The required column indices are ±1, ±3, ±5, ±7, ±9, so there will be no auxiliary column indices. The required row indices are 0, ±2, ±4, ±6, so there will be three auxiliary row indices r 1 , r 2 , r 3 .

Domination by centrally strong sets
We begin by considering a board C which is to be a central sub-board of a larger board B. Say that C has m 1 rows and n 1 columns, with m 1 n 1 and m 1 , n 1 not both even. It is convenient here to have the board squares of side length two, and place C with its center at the origin. Thus if, for example, m 1 is odd and n 1 is even, then each square has center (x,y) with x an odd integer and y an even one.
We then wish to choose a nonnegative integer k and a set D of squares of B (actually, all or almost all in C) such that D contains at least one square from the extension of each orthogonal of C to B, and D contains exactly one square from the extension of each difference diagonal of C, except none from the highest k and the lowest k extended difference diagonals; similarly for sum diagonals. Let Then m n, and it is straightforward to verify that if C is taken to be the central m 1 ×n 1 sub-board of the m × n board B, then D is a dominating set of Q m×n and |D| = g. Such a D will be called a centrally strong set, as it generalizes the idea discussed for square boards in [27, page 234]. We note that our definition requires each square of D to have both diagonals among the required ones, and thus both have indices of absolute value at most m 1 + n 1 − 2k − 2, but this does not imply D ⊆ C. If in fact D ⊆ C, we say that D is a strict centrally strong set. A number of strict centrally strong sets occur in the appendix, and are labeled there as such. We note that these sets can only occur when m n < 2m; this follows from (24) and the fact that since there will be n 1 − 2k − 1 auxiliary row indices, this quantity is nonnegative.
One merit of this construction is that a single centrally strong D gives an upper bound for γ(Q m×n ) for several pairs (m, n) since D is confined to a small central region of the m × n board, especially if D is strict. For example, there is a strict centrally strong set D = {±(−5, 0), ±(−3, 4), ±(−1, 6), ±(1, 2), ±(3, 6)} with m 1 = 9, n 1 = 6, and k = 2, and |D| = 10, which shows that γ(Q m×n ) 10 when 9 m 17 and 6 n 20. For some of these pairs (m, n), this bound is poor, but for the six pairs with m + n 35, combining with the bound (12) gives 9 γ(Q m×n ) 10, and 10 is a useful upper bound for some of the smaller boards also.
The simplest centrally strong sets occur with m 1 1, n 1 = 1 and k = 0, where we get m 1 queens occupying all squares of the m 1 × 1 board C, and the following bound (which we have stated in terms of m = m 1 + 2). For 3 m 10 at least, this bound gives the exact value of γ(Q m×(2m−3) ).
We next consider the effect of (22) and (23) on the search for centrally strong sets. Symmetry and the requirement that each difference diagonal contains exactly one square of D imply that the sum of the difference diagonal indices of D is zero. Similarly the sum of the sum diagonal indices of D is zero, and then (22) implies that (x,y)∈D x = 0 and (x,y)∈D y = 0. As we require a centrally strong set to occupy all (extended) columns of the sub-board, we regard the n 1 indices of these columns as required column indices; by symmetry their sum is zero. As C has n 1 columns and g occupied squares, there will be g − n 1 = m 1 − 2k − 1 auxiliary column indices, each having parity opposite to that of n 1 . Since (x,y)∈D x = 0 and all required column indices sum to zero, so do the auxiliary column indices. Similarly there will be m 1 required row indices and g − m 1 = n 1 − 2k − 1 auxiliary row indices, with sum zero, each having parity opposite to that of m 1 . If D is strict, then all indices of occupied columns have absolute value at most n 1 − 1 and all indices of occupied rows have absolute value at most m 1 − 1. (We have required that m 1 , n 1 not both be even because if they were, there would be an odd number of auxiliary row indices, each odd, so their sum could not be even, thus not zero.) Using , we see that the sum of the squares of the indices of all occupied diagonals of C is 4 g+1 3 , the sum of the squares of the required column indices is 2 n 1 +1 3 and the sum of the squares of the required row indices is 2 m 1 +1 3 . Letting orth denote the sum of the squares of the auxiliary column indices and auxiliary row indices, the quadratic constraint (23) gives Combined with Proposition 10, part (a) of the following proposition shows how small a centrally strong set can be. In parts (b) and (c), we limit the values of m 1 , n 1 , k that need be considered when constructing centrally strong sets.
We say that a value of k for which there exists a centrally strong set on Q m 1 ×n 1 is feasible for (m 1 , n 1 ).
Proposition 11. (a) For any centrally strong set D with n 1 > 1, |D| n/2. (b) For any (m 1 , n 1 ), it is only necessary to use the largest feasible k to determine all upper bounds for γ(Q m×n ) implied by centrally strong sets from (m 1 , n 1 ). (c) If k is feasible for (m 1 , n 1 ) and k + 1 is feasible for (m 1 , n 1 + 2), the latter gives the more useful result.
(b): Suppose for some integer h > 0 that both k and k − h are feasible for (m 1 , n 1 ). Then the triple m 1 , n 1 , k gives a dominating set D of size g on Q m×n , where m, n, g are determined by (24), and similarly the triple m 1 , n 1 , k − h gives a dominating set D ′ of size g + 2h on Q (m+2h)×(n+2h) . However, by repeating 2h times the process of adding an edge row and edge column to the board and the new corner square to the dominating set, we can construct from D a dominating set of Q (m+2h)×(n+2h) of the same size as D ′ .
(c): Using (24), if m 1 , n 1 , k gives a dominating set of size g for Q m×n , then m 1 , n 1 + 2, k + 1 gives a dominating set of size g for Q (m+2)×n .
We give two infinite families of strict centrally strong sets, each including a minimum dominating set found by von Szily [22,23].
In our approach described above, all orthogonal indices would be even here; we have divided by two, thus returning to a board with squares of edge length one.
Above we have described two approaches to the construction of dominating sets. In both, once a set of lines to be occupied by the dominating set is specified, it is necessary to see whether one can find such a dominating set. A fast backtrack search idea of Hitotumatu and Noshita [14], explained and amplified by Knuth [17], was used byÖstergård and Weakley [19] to find values and bounds of γ(Q n×n ) up to n = 120. This approach and also the algorithm of Neuhaus [18] can be applied to rectangular boards as well. But as mentioned, neither of our constructions can produce a dominating set of size less than ⌊n/2⌋ for Q m×n (with m n). Thus a resolution of Question 3 would be needed to determine whether extensive search based on these constructions is useful.
The complexity of computing minimum dominating set of queens is another open question [10, Section 5]. Backtracking algorithms, dynamic programming, and treewidth technique are analyzed extensively by Fernau [10,.