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arXiv:2012.01992v1 [math.CO] 3 Dec 2020

Spectral properties of the n-Queens’ Graphs

Domingos M. Cardoso1,2, Inˆes Serˆodio Costa1,2, and Rui Duarte1,2

1Centro de Investiga¸c˜ao e Desenvolvimento em Matem´atica e Aplica¸c˜oes

2Departamento de Matem´atica, Universidade de Aveiro, 3810-193, Aveiro, Portugal.

December 4, 2020

Abstract

The n-Queens’ graph, Q(n), is the graph associated to the n×nchess-

board (a generalization of the classical 8 ×8 chessboard), with n2vertices,

each one corresponding to a square of the chessboard. Two vertices of

Q(n) are adjacent if and only if they are in the same row, in the same

column or in the same diagonal of the chessboard. After a short overview

on the main combinatorial properties of Q(n), its spectral properties are

investigated. First, a lower bound on the least eigenvalue of an arbitrary

graph is obtained using clique edge partitions and a suﬃcient condition for

this lower bound be attained is deduced. For the particular case of Q(n),

we prove that for every n, its least eigenvalue is not less than −4 and it

is equal to −4 with multiplicity (n−3)2, for every n≥4. Furthermore,

n−4 is also an eigenvalue of Q(n), with multiplicity at least n−2

2when n

is even and at least n+1

2when nis odd. A conjecture about the integer

eigenvalues of Q(n) is presented. We ﬁnish this article with an algorithm

to determine an equitable partition of the n-Queens’ graph, Q(n), for

n≥3, concluding that such equitable partition has (⌈n/2⌉+1)⌈n/2⌉

2cells.

Keywords: Queens’ Graph, graph spectra, equitable partition.

MSC 2020: 05C15, 05C50, 05C69, 05C70.

1 Introduction

The problem of placing 8 queens on a chessboard such that no two queens

attack each other was ﬁrst posed in 1848 by a German chess player [4]. The

German mathematician and physicist Johann Carl Friedrich Gauss (1777–1855)

had knowledge of this problem and found 72 solutions. However, according to

[3], the ﬁrst to solve the problem by ﬁnding all 92 solutions was Nauck in [20],

in 1850. As claimed later by Gauss, this number is indeed the total number of

solutions. The proof that there is no more solutions was published by E. Pauls

in 1874 [23].

The n-Queens’ problem is a generalization of the above problem, consisting

of placing nnon attacking queens on n×nchessboard. In [23] it was also proved

that the n-Queens’ problem has a solution for every n≥4. Each solution of the

n-Queens’ problem corresponds to a permutation π= (i1i2. . . in), deﬁning

1

the entries (i1,1), (i2,2), . . . , (in, n) which are the positions of the nqueens on

the squares of the n×nchessboard. Then, we may say that the solutions of

the n-Queens’ problem are a few permutation among the n! permutations of n

elements. The corresponding permutation matrices of order nhave no two 1s

both on the main diagonal, both on the second diagonal or both on any other

diagonal parallel to one of these diagonals.

In [14] it was proved that a variant of the n-Queens’ problem (dating to

1850) called n-Queens’ completion problem is NP-Complete. In the n-Queens’

completion problem, assuming that some queens are already placed, the question

is to know how to place the rest of the queens, in case such placement be

possible. After the publication of this result, the n-Queens’ completion problem

has deserved a growing interest on its research. Probably, the motivation is

that some researchers believe in the existence of a polynomial-time algorithm to

solve this problem (see [12]). Therefore, if such an algorithm is found, then the

problem that asks whether Pis equal to NP is solved. This problem belongs

to the list of the Millenium Prize Problems stated by the Clay Mathematics

Institute which awards one million dollars to anyone who ﬁnds a solution to

any of the seven problems of the list. So far only one of these problems (the

Poincar´e conjecture) has been solved.

The graph associated to the n-Queens’ problem, called the n-Queens’ graph,

is obtained from the n×nchessboard considering its squares as vertices of the

graph with two of them adjacent if and only if they are in the same row, col-

umn or diagonal of the chessboard. It is immediate that each solution of the

n-Queens’ graph corresponds to a maximum stable set (also known as maximum

independent vertex set) of the n-Queens’ graph. Furthermore, the n-Queens’

completion problem corresponds to the determination of a maximum stable set

which includes a particular stable set when such maximum stable set exists or

the conclusion that there is no maximum stable set in such conditions.

Our focus in this article is neither the n-Queens’ problem nor the n-Queens’

completion problem. The main goal is to study the properties of the n-Queens’

graph, especially its spectral properties.

In Section 2 a short overview on the main combinatorial properties of Q(n)

is presented. Namely, the stability, clique, chromatic and domination numbers

are analyzed.

Section 3 is devoted to the investigation of the spectral properties of Q(n).

A lower bound on the least eigenvalue of an arbitrary graph is obtained using

clique edge partitions and a suﬃcient condition for this lower bound be attained

is deduced. For the particular case of Q(n), we prove that for every n, its least

eigenvalue is not less than −4 and it is equal to −4 with multiplicity (n−3)2, for

every n≥4. Furthermore, n−4 is also an eigenvalue of Q(n), with multiplicity

at least n−2

2when nis even and at least n+1

2when nis odd. We ﬁnish this

section with a conjecture about the integer eigenvalues of Q(n).

In Section 4 an algorithm to determine an equitable partition of the n-

Queens’ graph, Q(n), with n≥3, is introduced. This equitable partition has

(⌈n/2⌉+1)⌈n/2⌉

2cells.

2

From now on, the n×nchessboard is herein denoted by Tnand the associated

graph by Q(n). The n-Queens’ graph, Q(n), associated to the n×nchessboard

Tnhas n2vertices, each one corresponding to a square of the n×nchessboard.

Two vertices of Q(n) are adjacent, that is, linked by an edge if and only if

they are in the same row, in the same column or in the same diagonal of the

chessboard. The squares of Tnand the corresponding vertices in Q(n) are

labeled from the left to the right and from the top to the bottom. For instance,

the squares of T4(vertices of Q(4)) are labelled as depicted in the next Figure.

0Z0Z

Z0Z0

0Z0Z

Z0Z0

13

9

5

1

14

10

6

2

15

11

7

3

16

12

8

4

Figure 1: Labeling of T4.

The vertex set and the edge set of Q(n) are denoted by V(Q(n)) and

E(Q(n)), respectively. The order of Q(n) is the cardinality of V(Q(n)), n2,

and the size of Q(n) is the cardinality of E(Q(n)) and is denoted by e(Q(n)).

2 A short overview on the main combinatorial

properties of Q(n)

We start this section by recalling some classical concepts of graph theory.

Given a graph G, a stable set (resp. clique) of Gis a vertex subset where

every two vertices are not adjacent (resp. are adjacent). The stability (resp.

clique) number of G,α(G) (resp. ω(G)), is the cardinality of a stable set (resp.

clique) of maximum cardinality. A proper coloring of the vertices of Gis a func-

tion ϕ:V(G)→C, where Cis a set of colors, such that ϕ(x)6=ϕ(y) whenever

xis adjacent to y, that is, xy ∈E(G). The chromatic number of G,χ(G), is the

minimum cardinality of Cfor which there is a proper coloring ϕ:V(G)→Cof

the vertices of G. From now on, a proper coloring of the vertices of a graph G

is just called vertex coloring of G. A vertex v∈V(G), dominates itself and all

its neighbors. A vertex set S⊂V(G) is a dominating set if every vertex of Gis

dominated by at least one vertex of S. The domination number of a graph G,

γ(G), is the cardinality of a dominating set in Gwith minimum cardinality.

Regarding the diameter of Q(n), diam(Q(n)), which is the maximum dis-

tance among the distances between every pair of vertices, it is immediate that

diam(Q(n)) = 2, for n > 2 and diam(Q(n)) = 1 for n∈ {1,2}.

In this section, the size, vertex degrees and average degree, the stability,

clique, chromatic and domination numbers of Q(n) are analyzed.

3

2.1 Size, vertex degrees and average degree

Taking into account that two vertices of Q(n) are linked by an edge if and only

if they are in the same row, column or diagonal, it follows that the size of Q(n)

can be obtained by the expression

e(Q(n)) = 2(n+ 1)n

2+ 4 2

2+···+n−1

2

= 2(n+ 1)n

2+ 4n

3(1)

=n(n−1)(5n−1)

3.(2)

The expression (1) is obtained using the hockey-stick identity twice.

The degree of a vertex v,dv, is the number of its neighbors, that is, the

number of vertices adjacent to v.

By the Handshaking Lemma, the average degree of a graph Gof order nis

dG=2e(G)

n. From (2) we get

dQ(n)=2e(Q(n))

n2=2(n−1)(5n−1)

3n.(3)

A closed formula (in terms of n) for the degrees of the vertices of Q(n) can be

obtained from the structure of these graphs. Before that, we need to introduce

some additional notation for the elements of the following partition of V(Q(n)).

1.The ﬁrst peripheral vertex subset V1is the vertex subset corresponding to

the more peripheral squares of the board Tn;

2.The second peripheral vertex subset V2is the vertex subset corresponding

to the more peripheral squares of the board Tnwithout considering the

vertices in V1;

.

.

.

⌊n+1

2⌋.The ⌊n+1

2⌋-th peripheral vertex subset V⌊n+1

2⌋is the vertex subset corre-

sponding to the squares of the board Tn, without considering the vertices

in V1∪V2∪ · ·· ∪ V⌊n+1

2⌋−1.

Clearly, the vertex subsets V1, V2,...,V⌊n+1

2⌋form a partition of V(Q(n)).

For instance, for n= 5 (see the chessboard T5depicted in Figure 2, where

the labels of the squares are the labels of the corresponding vertices of Q(n)),

⌊5+1

2⌋= 3 and

1. V1={1,2,3,4,5,10,15,20,25,24,23,22,21,16,11,6},

2. V2={7,8,9,14,19,18,17,12},

3. V3={13}.

4

5Z0Z0Z

40Z0Z0

3Z0Z0Z

20Z0Z0

1Z0Z0Z

12345

12345

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

Figure 2: The chessboard T5.

Regarding the cardinality of those peripheral vertex subsets it follows that

|Vi|= (n−2(i−1))2−(n−2(i−1) −2)2

= ((n−2(i−1)) −(n−2(i−1) −2)) ((n−2(i−1)) + (n−2(i−1) −2))

= 2(2n−4(i−1) −2) = 4(n−(2i−1)), i = 1,2,...,n+ 1

2−1,(4)

V⌊n+1

2⌋=4,whether nis even,

1,otherwise. (5)

For the sake of simplicity of the proof of the next theorem, it is worth to

consider the chessboard Tnas a n×nmatrix, were the coordinates (p, q) of

a square means that it is in the p-th row (from the top to bottom) and q-th

column (from the left to right).

Theorem 2.1. Considering the partition of V(Q(n)) into the peripheral vertex

subsets V1, V2,...,V⌊n+1

2⌋, the degrees of the vertices of Q(n)are

dv= 3(n−1) + 2(j−1),for all v∈Vj, j = 1,2,...,n+ 1

2.

Proof. For v∈V1, it is immediate that dv= 3(n−1). On the other hand, for

v∈Vi, with i=⌊n+1

2⌋, it is also immediate to conclude that

dv=4(n−1) = 2(n−1) + 2(⌊n+1

2⌋ − 1),if nis odd;

3(n−1) + n−2 = 3(n−1) + 2(⌊n+1

2⌋ − 1),otherwise.

Let us consider isuch that 1 < i < ⌊n+1

2⌋and that v∈Vicorresponds to

the square of the chessboard Tndetermined by the pair of coordinates (p, q).

Then (p, q)∈Xi,i ∪Xi,(n−(i−1)) ∪Xn−(i−1),n−(i−1) ∪X(n−(i−1)),i, where

Xi,i ={(i, j)∈[n]2|i≤j≤n−i}

Xi,(n−(i−1)) ={(j, n −(i−1)) ∈[n]2|i≤j≤n−i}

Xn−(i−1),n−(i−1) ={(n−(i−1), j)∈[n]2|i+ 1 ≤j≤n−(i−1)}

Xn−(i−1),i ={(j, i)∈[n]2|i+ 1 ≤j≤n−(i−1)}.

Assuming (p, q)∈Xi,i , then p=i,i≤q≤n−iand (p, q) has 2(n−1)

neighbors corresponding to the squares of the row and column of Tnwhose

intersection is (p, q) plus the vertices corresponding to the diagonal neighbors

deﬁned by the following pairs (x, y):

5

(i+ 1, q −1),(i+ 2, q −2),...,(i+ (q−1), q −(q−1))

|{z }

q−1 lower right to left diagonal neighbors

,

(i+ 1, q + 1),(i+ 2, q + 2),...,(i+ (n−q), q + (n−q))

|{z }

n−qlower left to right diagonal neighbors

,

(i−1, q −1),(i−2, q −2),...,(i−(i−1), q −(i−1))

|{z }

i−1 upper right to left diagonal neighbors

,

(i−1, q + 1),(i−2, q + 2),...,(i−(i−1), q + (i−1))

|{z }

i−1 upper left to right diagonal neighbors

.

Therefore, the total number of neighbors of (p, q) is

d(p,q)= 2(n−1) + q−1 + n−q+ 2(i−1) = 3(n−1) + 2(i−1).

For the vertices (p, q) in any other of the subsets Xi,(n−(i−1)),X(n−(i−1)),(n−(i−1))

and X(n−(i−1)),i, by symmetry, the result is the same.

As immediate consequence, we have the following corollary of Theorem 2.1.

Corollary 2.2. For all vertices vof Q(n),

3(n−1) ≤dv≤4n−5,if nis even;

4n−4,if nis odd. (6)

Therefore, the minimum and maximum degrees of Q(n) are δ(Q(n)) =

3(n-1) and ∆(Q(n)) = 4n−5,if nis even;

4n−4,if nis odd, respectively.

2.2 Stability, clique, chromatic and domination numbers

2.2.1 The stability and clique numbers

For n= 2,3, it is clear that α(Q(n)) is equal to 1 and 2, respectively. Taking into

account that every solution of the n-Queens’ problem corresponds to a maximum

stable set and, as proved in [23], the n-Queens’ problem has a solution for every

n≥4, the next proposition follows.

Proposition 2.3. The stability number of the n-Queens’ graph is

α(Q(n)) =

1,if n= 2;

2,if n= 3;

n, if n≥4.

The next theorem states the clique number of the n-Queens’ graph.

Theorem 2.4. The clique number of the n-Queens’ graph is

ω(Q(n)) =

4,if n= 2;

5,if n∈ {3,4};

n, if n≥5.

6

Proof. The values of the clique number for n∈ {2,3,4}are immediate. For

instance, for n= 4, considering the 5 vertices labeled by the symbol of a Queen

in the squares of the chessboard depicted in Figure 3, it is obviuous that they

form a clique and there is no other clique with more than ﬁve vertices. Note

that there are no four (resp. three) vertices in the same row or column having

two (resp. three) common neighbors outside the row or column and there are

no two vertices in the same row or column with four common neighbors outside

the row or column.

0L0L

Z0L0

0L0L

Z0Z0

Figure 3: Five Queens attacking each other in T4.

Let us assume n≥5 and that Q(n) has a clique with n+ 1 vertices. Then,

among these vertices (squares of the chessboard) there are two, say (i, j1) and

(i, j2) (similarly (i1, j) and (i2, j)) in the same row (resp. column) of the chess-

board. The common neighbors to these vertices out of the i-th row forming (all

together) a maximal clique are the following.

1. If i= 1 (for i=nis similar) and j2=j1+k, with k≥1, those vertices

are (i+k, j1) and (i+k, j2), plus the vertex (i+k/2, j1+k/2) when kis

even.

2. If 1 < i < n and j2=j1+k, with k≥1, those vertices are (i−k, j1) and

(i−k, j2), if i−k≥1, and the vertex (i−k/2, j1+k/2) when kis even

and i−k/2≥1, or (i+k , j1) and (i+k, j2), if i+k≤nplus the vertex

(i+k/2, j1+k/2) when kis even and i+k/2≤n.

In both cases, there are at most three common neighbors to (i, j1) and (i, j2)

outside the i-th row forming (all together) a maximal clique. Choosing just

one vertex, among the common neighbors of (i, j1) and (i, j2) outside the i-th

row, the maximum number of its neighbors belonging to the i-th row is three.

Therefore, the vertices of any row (or column) form a maximum clique.

2.2.2 The chromatic number

It is well-known that the chromatic number of a graph is not less than its clique

number. Therefore, χ(Q(2)) = 4 and, since ω(Q(n)) = 5 for n∈ {3,4,5},

χ(Q(n)) = 5,(7)

as it follows from the vertex colorings presented in Figure 4, where the set of

colors is C={0,1,2,3,4}.

Remark 2.5. Each color class (subset of vertices with the same color) of the

vertex coloring of Q(5) presented in Figure 4 is a solution for the 5-Queens’

7

Z0Z

0Z0

Z0Z

3

1

4

4

2

0

0

3

10Z0Z

Z0Z0

0Z0Z

Z0Z0

4

2

0

3

0

3

1

4

1

4

2

0

2

0

3

1Z0Z0Z

0Z0Z0

Z0Z0Z

0Z0Z0

Z0Z0Z

23401

40123

12340

34012

01234

Figure 4: A vertex coloring of Q(3),Q(4) and Q(5).

problem. More generally, when χ(Q(n)) = n, each color class of a vertex color-

ing of Q(n)with ncolors is a solution for the n-Queens’ problem.

The vertex coloring of Q(5) presented in Figure 4 forms a Latin square with

the particular property that every diagonal has no repeated colors. More gen-

eral, it is immediate that χ(Q(n)) = nif and only if there is a Latin square of or-

der nformed by ndistinct color symbols, where each color appears once in each

of its diagonals. The next theorem gives a suﬃcient condition for χ(Q(n)) = n.

Its proof can be found in [7]. Even so, for the readers convenience, this proof is

herein reproduced.

Theorem 2.6. If n≡1,5 mod 6, then χ(Q(n)) = n.

Proof. Assume that nis nor even neither divisible by 3 and consider the vertex

coloring proposed in [7], where the n×nchessboard is seen as the cartesian prod-

uct {0,1,...,n−1}×{0,1,...,n−1}and the set of colors is C={0,1,...,n−1}.

Assign to each vertex (i, j ) the color j−2imod n.

If (i, j) and (p, q ) are two vertices with the same color, then j−2i≡q−2p

mod n, which is equivalent to

2(p−i)≡(q−j) mod n. (8)

1. If the two vertices are in the same row, that is, i=p, then (8) implies

j=q.

2. If the two vertices are in the same column, that is, j=q, then (8) implies

i=p, since nis odd.

3. If the two vertices are in the same diagonal, then (a) p−i=q−jor (b)

p−i=j−qand, from (8),

(a) p−i=q−j⇒2(q−j)≡(q−j) mod n⇒j=qand then

(i, j) = (p, q ).

(b) p−i=j−q⇒2(j−q)≡(q−j) mod n⇒3(j−q)≡0 mod n

and then (i, j) = (p, q ), since nis not divisible by 3.

It is known that χ(Q(n)) > n for n= 2,3,4,6,8,9,10 and thus we may

say that the suﬃcient condition, n≡1,5 mod 6, for χ(Q(n)) = n, is also

necessary when n < 12. However, in general, this condition is not necessary.

8

0Z0Z0Z0Z0Z0Z

Z0Z0Z0Z0Z0Z0

0Z0Z0Z0Z0Z0Z

Z0Z0Z0Z0Z0Z0

0Z0Z0Z0Z0Z0Z

Z0Z0Z0Z0Z0Z0

0Z0Z0Z0Z0Z0Z

Z0Z0Z0Z0Z0Z0

0Z0Z0Z0Z0Z0Z

Z0Z0Z0Z0Z0Z0

0Z0Z0Z0Z0Z0Z

Z0Z0Z0Z0Z0Z0

0 5 9 6 3 8 4 1 10 11 7 2

7 11 4 2 1 6 10 3 0 8 9 5

8 1 10 9 5 2 0 7 11 6 3 4

10 0 3 8 7 11 9 5 4 1 2 6

5 6 11 4 2 1 3 0 8 9 10 7

11 7 0 1 10 4 8 6 3 2 5 9

2 8 6 3 9 5 7 11 1 10 4 0

3 4 5 0 11 10 6 9 2 7 8 1

9 2 1 10 4 7 5 8 6 3 0 11

4 10 7 11 0 3 1 2 9 5 6 8

6 3 2 5 8 9 11 4 7 0 1 10

1 9 8 7 6 0 2 10 5 4 11 3

Figure 5: The vertex coloring of Q(12) with 12 colors presented in [28].

Counterexamples with n= 12 and n= 14 are presented in [28]. Figure 5 shows

the former counterexample.

Additional counterexamples for the suﬃcient condition n≡1,5 mod 6 be-

ing also a necessary condition for χ(Q(n)) = nare referred in [7], for n=

15,16,18,20,21,22,24,28,32.

Remark 2.7. In design theory, a Knut Vik design Kof order ncan be deﬁned

(see [16]) as an n×narray of elements, chosen from a set of nelements such

that (i) with respect to the rows and columns the design is a Latin square of

order nand (ii) for 0≤j≤n−1, deﬁning the j-th right (resp. left) diagonal

of Kas the set of ncells {(i, j +i)|i= 0,1,...,n−1}(resp. {(i, j −i−1) |

i= 0,1,...,n−1}), where the arithmetic operations are done modulo n, each

of these diagonals contains the nelements. Such diagonals are usually called

complete diagonals. In other words, we may say that a Latin square Kis a Knut

Vik design if every right and left diagonal of Kis complete. The above sets of

cells deﬁning right and left diagonals are all broken diagonals, except for j= 0.

In [17] the author proves that a Knut Vik design of order nexists if and only if

nis not divisible by 2or 3. If Q(n)) has a coloring deﬁning a Knut Vik design

it is immediate that χ(Q(n)) = n. However, there are values of nfor which

χ(Q(n)) = nbut in the Latin squares deﬁned by the vertex colorings not all

diagonals are complete and then they are not Knut Vik designs. For instance,

in the coloring of Q(12) depicted in Figure 5 the color symbol 7appears twice

in the 5-th right diagonal (a broken diagonal) and thus at least one color symbol

is missing in the set of n cells of this diagonal.

Theorem 2.8. For n≥5,

n≤χ(Q(n)) ≤n, if n≡1,5 mod 6;

n+ 3 otherwise.

Proof. Since ω(Q(n)) = n, when n≥5, the lower bound for χ(Q(n)) follows.

9

The upper bound follows from Theorem 2.6 and taking into account that the

chromatic number of a subgraph of a graph Gis not greater than χ(G).

As a direct consequence of Theorem 2.8, we have the following corollary.

Corollary 2.9. For n≥5,χ(Q(n)) = nwhen n≡1,5 mod 6 and

χ(Q(n)) ∈

{n, n + 1},if n≡0,4 mod 6;

{n, n + 1, n + 2, n + 3},if n≡2 mod 6;

{n, n + 1, n + 2}if n≡3 mod 6.

2.2.3 The domination number

The study of domination in graphs has its historical roots connected with the

classical problems of covering chessboards with the minimum number of chess

pieces, whose mathematical approach goes back to 1862 [19]. In the 1970s, the

research on the chessboard domination problems was redirected to more gen-

eral problems of domination in graphs. Since then, this type of problem has

attracted an increasing number of researchers, turning this topic into an intense

area of research.

The domination number of the Queens’ graph, γ(Q(n)), is the most studied

problem related with this graph. For the ﬁrst values of n, the domination

number of Q(n) is well-known since a long time ago. Indeed, the values of

γ(Q(n)), for n∈ {1,...,8}, presented in Table 1, were given by Rouse Ball in

1892 [26].

n12345678

γ(Q(n))11123345

Table 1: Values of γ(Q(n)), for n∈ {1,...,8}.

The following inequalities establish two well-known (lower and upper) bounds

for the domination number

n−1

2≤γ(Q(n)) ≤2p+q, (9)

where the lower bound holds for every natural number nand the upper bound

holds when n= 3p+q, with p∈Nand q∈ {0,1,2}. The lower bound is due

to P.H. Spencer and the upper bound is due to L. Welch, both cited in [8] as

private communications. However, as it is referred in [22, Comment added in

August 25, 2003], a proof of the lower bound was published earlier in [25]. The

proof of the upper bound appears in [8].

For particular values of n, Weakley [29] improved the lower bound of (9) as

follows.

Theorem 2.10. For all k∈N∪ {0},2k+ 1 ≤γ(Q(4k+ 1)).

According to Tables 1 and 2, it is known that γ(Q(4k+ 1)) = 2k+ 1, for

k∈ {1,...,32}. However, the problem of knowing for each nonnegative integer

kwhether the lower bound in Theorem 2.10 is sharp or not remains open.

10

Conjecture 1. [29] For all k∈N∪ {0},2k+ 1 = γ(Q(4k+ 1)).

Another open problem, herein posed as a conjecture, was proposed in [8,

Problem 1].

Conjecture 2. For all n∈N,γ(Q(n)) ≤γ(Q(n+ 1)).

However, we have the following inequality.

Proposition 2.11. For n∈N,γ(Q(n+ 1)) ≤γ(Q(n)) + 1.

Proof. Let S⊂V(Q(n)) be a minimum dominating set of Q(n). Considering

that the vertices of V(Q(n)) occupy the ﬁrst nrows and ﬁrst ncolumns of

Q(n+ 1) and S′is the vertex subset of Q(n+ 1) corresponding to S, it is

immediate that S′∪ {(n+ 1, n + 1)}is a dominating set of Q(n+ 1) and thus

γ(Q(n+ 1)) ≤γ(Q(n)) + 1.

Table 2 presents the values of γ(Q(n)) computed in [22], for n≥9. Almost

all of these values satisfy the condition γ(Q(n)) = ⌈n

2⌉. In Table 2 we display

all known values or γ(Q(9i+j)), for j= 0,1,...,8 and i= 1,2,...,14.

i\j012345678

1 555678999

2 9 10 11 12 13

3 14 15 15 16 17

4 19 20 21

5 23 25 27

6 29 31

7 33 35 36

8 37 39

9 41 43 45

10 46 47 49

11 51 53

12 55 57 58

13 59 61 63

14 65 65 66

Table 2: Known values of γ(Q(9i+j)), for j= 0,1,...,8 and for i=

1,2,...,14.

Some values presented in Table 2 were published by other authors before

[22], such as in [5, 6, 15].

The domination number can be determined by solving a binary (0-1) linear

programming problem, as it is stated by the following theorem.

Theorem 2.12. Let Gbe a graph with adjacency matrix Aand consider the

binary linear programming problem

max η(x) = X

j∈V(G)

xj,

s.t. (A+In)x≤d,(10)

x∈ {0,1}n,

11

where Inis the identity matrix of order nand dis a vector whose entries are

the degrees of the vertices of G. If x∗∈ {0,1}nis an optimal solution, denoting

the components of x∗by x∗

j, for all j∈V(G),D∗={j|x∗

j= 0}is a minimum

dominating set for Gand γ(G) = n−η(x∗).

Proof. The inequalities (10) guarantee that the vertex set deﬁned by the null

components of every feasible solution xis a dominating set. Note that for each

vertex j, the variables xjin the associated inequality correspond to vertices

belonging to the closed neighborhood of jand there is at least one of these

variables xr= 0, otherwise xis not feasible. As a consequence, assuming that

D={i|xi= 0}, for every vertex jthere is at least one vertex in Dadjacent to

j. If x∗is an optimal solution of the binary linear programming problem, then

the dominating vertex subset D∗={i|x∗

i= 0}is such that |D∗|=n−η(x∗).

Let us assume that there is a vertex dominating set D′such that |D′|<|D∗|.

Then, deﬁning x′with components x′

i=0,if i∈D′;

1,otherwise , it is clear that x′is

a feasible solution for the binary linear programming problem for which η(x′) =

n−|D′|. Since |D′|<|D∗|, it follows that η(x∗) = n− |D∗|< n − |D′|=η(x′),

which is a contradiction. Therefore, γ(G) = |D∗|=n−η(x∗).

As a by-product, in the case of the Queens’ graph Q(n), we have the next

corollary, where each variable xij corresponds to a vertex of Q(n) with coordi-

nates (i, j) (the square of Tnin line ifrom top to bottom and column jfrom

left to right).

Corollary 2.13. Consider the binary linear programming problem

max η(x) = X

(i,j)∈[n]2

xij ,(11)

s.t. (A+In)x≤d,(12)

x∈ {0,1}n2,(13)

where Ais the adjacency matrix of Q(n),Inis the identity matrix of order n

and dis a vector whose entries are the degrees of the vertices (p, q)of Q(n). If

x∗∈ {0,1}n2is an optimal solution, denoting the components of x∗by x∗

pq , for

all (p, q)∈[n]2,D∗={(i, j)|x∗

ij = 0}is a minimum dominating set of Q(n)

and γ(Q(n)) = n2−η(x∗).

It is computational hard to solve the binary linear programming problem

(11) – (13), even for not very large values of n. So far, by solving the binary

linear programming problem (11) – (13), we are able to conﬁrm many of the

presented values in Table 2. However, the domination number problem of the

n-Queens’ graph remains open even for several integers n≤131. For instance,

for n= 20, from the lower bound in (9) and Proposition 2.11 it follows that

γ(Q(n)) ∈ {10,11}. However, as far as we know, its precise value is unknown.

We observe that for all values displayed in Table 2, ⌊n+1

2⌋ − 1≤γ(Q(n)) ≤

⌊n+1

2⌋+ 1, if n≤16, and γ(Q(n)) = ⌊n+1

2⌋, if n≥17.

12

3 Spectral properties of Q(n)

The eigenvalues of the adjacency matrix of a graph Gare simply called the eigen-

values of Gand its spectrum is the multiset σ(G) = nµ[m1]

1, µ[m2]

2,...,µ[mp]

po,

where µ1> µ2>·· · > µpare the pdistinct eigenvalues (indexed in decreasing

order) and mi(also denoted by m(µi)) is the multiplicity of the eigenvalue µi,

for i= 1,2,...,p. When mj= 1, we just write µj. When necessary, we write

µj(G) to clarify that µjis the j-th distinct eigenvalue of the graph G. We de-

note by EG(µ) the eigenspace associated with an eigenvalue µof G.

As it is well-known, the largest eigenvalue of a graph Gis between its average

degree, dG, and its maximum degree, ∆(G), and it is equal to both if and only

if Gis regular. Therefore, from (3) and (6), we may conclude that

2(n−1)(5n−1)

3n≤µ1(Q(n)) ≤4n−5 if nis even;

4n−4 if nis odd.(14)

The characteristic polynomials and spectra of Q(2) and Q(3) are the follow-

ing.

q2(x) = (x+ 1)3(x−3) and σ(Q(2)) = n3,−1[3]o;

q3(x) = (−1 + x)(1 + x)2(−8−5x+x2)(−1 + 2x+x2)2and

σ(Q(3)) = (5 + √57

2,1,(−1 + √2)[2],−1[2] ,5−√57

2,(−1−√2)[2]).

3.1 A lower bound on the least eigenvalue of a graph

We start the study of spectral properties introducing a general result about

lower bounds on the least eigenvalues of graphs. Before that it is worth to

deﬁne a couple of parameters associated with edge clique partitions. The edge

clique partitions (ECP for short) were introduced in [21], where the content of

a graph Gwas deﬁned as the minimum number of edge disjoint cliques whose

union includes all the edges of Gand denoted by C(G). Such minimum ECP

is called in [21] content decomposition of Gand, as proved there, in general, its

determination is NP-Complete.

Deﬁnition 3.1. (Clique degree and maximum clique degree) Consider a graph

Gand an ECP, P={Ei|i∈I}and then Vi=V(G[Ei]) is a clique of Gfor

every i∈I. For any v∈V(G), the clique degree of vrelative to P, denoted

mv(P), is the number of cliques Vicontaining the vertex v, and the maximum

clique degree of Grelative to P, denoted mG(P), is the maximum of clique

degrees of the vertices of Grelative to P.

From Deﬁnition 3.1, considering an ECP, P={Ei|i∈I}, the parameters

mv(P) and mG(P) can be expressed as follows.

mv(P) = |{i∈I|v∈V(G[Ei])}| ∀v∈V(G); (15)

mG(P) = max{mv|v∈V(G)}.(16)

13

Remark 3.2. It is clear that if Pis an ECP of G, then mG(P)is not greater

than |P|. In particular, if Pis a content decomposition of G, then mG(P)≤

C(G).

Using the above deﬁned graph parameters, the next theorem states a general

lower bound on the least eigenvalue of a graph. As it will be seen later, there are

extremal graphs for which this lower bound is attained, namely the n-Queens’

graph Q(n), with n≥4, considering an ECP whose parts are the edge subsets

associated with the ncolumns, nrows, 2n−3 left to right diagonals and 2n−3

right to left diagonals of the corresponding n×nchessboard Tn.

Theorem 3.3. Let P={Ei|i∈I}be an ECP of a graph G,m=mG(P)and

mv=mv(P)for every v∈V(G). Then

1. If µis an eigenvalue of G, then µ≥ −m.

2. −mis an eigenvalue of Giﬀ there exists a vector X6=0such that

(a) P

j∈V(G[Ei])

xj= 0, for every i∈Iand

(b) ∀v∈V(G)xv= 0 whenever m6=mv.

In the positive case, Xis an eigenvector associated with the eigenvalue

−m.

Proof. Consider Aas the adjacency matrix of G.

1. Let Xbe an eigenvector of Gassociated with an eigenvalue µ. Then

(µ+m)kXk2=XT(µX +mX) = XT(A+mIn)X

=X

i∈IX

uv∈Ei

(2xuxv) + mkXk2

=X

i∈I

X

v∈V(G[Ei])

xv

2

−X

v∈V(G)

mvx2

v+mkXk2

=X

i∈I

X

v∈V(G[Ei])

xv

2

+X

v∈V(G)

(m−mv)x2

v≥0.

2. If −mis an eigenvalue of G, then, from the proof of item 1, equalities 2a

and 2b follow. Conversely, if there exists a vector X6=0for which 2a and

2b hold, then

0 = X

i∈I

X

v∈V(G[Ei])

xv

2

+X

v∈V(G)

(m−mv)x2

v

=X

i∈I

X

v∈V(G[Ei])

xv

2

−X

v∈V(G)

mvx2

v+mkXk2

=X

i∈IX

uv∈Ei

(2xuxv) + mkXk2

=XTAX+mkXk2.

14

Assuming that µis the least eigenvalue of G,−m=XTAX

kXk2≥µ. Since

µ≥ −m(by item 1), we obtain −m≥µ≥ −mand thus µ=−m. In

the positive case, it is clear that Xis an eigenvector associated to the

eigenvalue −m.

Theorem 3.3 allows to obtain a spectral lower bound for the content number

of a graph.

Corollary 3.4. Let µbe the least eigenvalue of a graph G. Then −µ≤C(G).

Proof. If Pis a content decomposition of G, then, according to Remark 3.2,

mG(P)≤C(G). By Theorem 3.3 −mG(P)≤µand so −µ≤C(G).

The following corollary is a direct consequence of Theorem 3.3.

Corollary 3.5. Let Gbe a graph of order nand let Xbe a vector of Rn\{0}.

Then X∈ EG(−m)iﬀ the conditions 2a and 2b of Theorem 3.3 hold.

3.2 Eigenvalues and eigenvectors of the n-Queens’ graph

From now on we consider the particular case of the n-Queens’ graph Q(n).

Theorem 3.6. Let n∈Nsuch that n≥4.

1. If µis an eigenvalue of Q(n), then µ≥ −4.

2. −4∈σ(Q(n)) iﬀ there exists a vector X∈Rn2\ {0}such that

(a)

n

P

j=1

x(k,j)= 0 and

n

P

i=1

x(i,k)= 0, for every k∈[n],

(b) P

i+j=k+2

x(i,j)= 0, for every k∈[2n−3],

(c) P

i−j=k+1−n

x(i,j)= 0, for every k∈[2n−3],

(d) x(1,1) =x(1,n)=x(n,1) =x(n,n)= 0.

In the positive case, Xis an eigenvector associated with the eigenvalue

−4.

Proof. The proof follows taking into account that the summations 2a-2c corre-

spond to the summations 2a in Theorem 3.3. Here, the cliques obtained from

the ECP, P, of Q(n) are the cliques with vertices associated with each of the n

columns, nrows, 2n−3 left to right diagonals and 2n−3 right to left diagonals.

Denoting the vertices of Q(n) by their coordinates (i, j) in the corresponding

chessboard Tn,m(i,j)(P) = 3,if (i, j )∈ {1, n}2;

4,otherwise and thus mQ(n)(P) = 4.

Therefore, the equalities 2d correspond to the conditions 2b in Theorem 3.3.

As a consequence of the previous Theorem we have the following result.

Corollary 3.7. Let n≥4and X∈Rn2. Then X∈ EQ(n)(−4) if and only if

the conditions 2a-2d of Theorem 3.6 hold.

15

From the computations, for several values of n≥4, it seems that n−4 and

−4 are eigenvalues of Q(n) with multiplicities

m(n−4) = (n−2)/2,if nis even;

(n+ 1)/2,if nis odd,

and m(−4) = (n−3)2. It also seems that there are other integer eigenvalues

when nis odd, and that they are all simple.

nDistinct integer eigenvalues

3 1, -1

4 0, -4

5 1, 0, -3, -4

6 2, -4

7 3, 2, 1, -2, -3, -4

8 4, -4

9 5, 4, 3, 2, -1, -2, -3, -4

10 6, -4

11 7, 6, 5, 4, 3, 0, -1, -2, -3, -4

Table 3: Distinct integer eigenvalues of Q(n), when 3 ≤n≤11.

In what follows we will see that, for n≥4, −4 is an eigenvalue of Q(n) with

multiplicity (n−3)2. From Corollary 3.7 it follows that the multiplicity of −4 as

an eigenvalue of Q(n) coincides with the corank of the coeﬃcient matrix of the

system of 6n−2 linear equations 2a-2d. Therefore, to say that m(−4) = (n−3)2

is equivalent to say that the rank of the coeﬃcient matrix of the system of 6n

linear equations 2a-2d is 6n−9 (since n2−6n+ 9 = (n−3)2).

Before we continue, we need to deﬁne a family of vectors

Fn={X(a,b)

n∈Rn2|(a, b)∈[n−3]2}

where X4is the vector presented in Table 4 and X(a,b)

nis the vector deﬁned by

01 -1 0

-1 0 0 1

10 0 -1

0-1 1 0

Table 4: The vector X4.

X(a,b)

n(i,j)=(X4(i−a+1,j−b+1) ,if (i, j)∈A×B;

0,otherwise, (17)

with A={a, a + 1, a + 2, a + 3}and B={b, b + 1, b + 2, b + 3}. For instance,

for n= 5, F5is the family of four vectors depicted in Table 5.

Theorem 3.8. −4is an eigenvalue of Q(n)with multiplicity (n−3)2and Fn

is a basis for EQ(n)(−4).

16

01 -1 0 0

-1 0 0 10

10 0 -1 0

0-1 1 0 0

00000

0 0 1 -1 0

0-1 0 0 1

010 0 -1

00-1 1 0

00000

0 0 0 0 0

01 -1 0 0

-1 0 0 10

10 0 -1 0

0-1 1 0 0

0 0 0 0 0

0 0 1 -1 0

0-1 0 0 1

010 0 -1

0 0 -1 1 0

Table 5: The vectors X(1,1)

5,X(1,2)

5,X(2,1)

5, and X(2,2)

5.

Proof. First, note that every element of Fnbelongs to EQ(n)(−4). Indeed, if

X=x(i,j)∈ Fn, then the conditions 2a-2d of Theorem 3.6 hold and hence by

Corollary 3.7 X∈ EQ(n)(−4).

Second, Fnis linearly independent and so dim EQ(n)(−4) ≥(n−3)2. For

otherwise there would be scalars α1,1,...,αn−3,n−3∈R, not all equal to zero,

such that

α1,1X(1,1)

n+···+αn−3,n−3X(n−3,n−3)

n=0.(18)

Let (n−3)(a−1) + bbe the smallest integer such that αa,b 6= 0. Since by

(17) X(a,b)

n(a,b+1) =X4(1,2) = 1, the entry (a, b + 1) of αa,bX(a,b)

nis αa,b.

Consider any other vector X(a′,b′)

nsuch that (n−3)(a′−1)+b′>(n−3)(a−1)+b

which implies (i) a′> a or (ii) a′=aand b′> b. Denoting A′={a′,...,a′+ 3}

and B′={b′,...,b′+ 3},taking in to account (17), we may conclude the

following.

(i) a′> a implies (a, b + 1) 6∈ A′×B′and thus X(a′,b′)

n(a,b+1) = 0.

(ii) For a′=aand b′> b + 1 the conclusion is the same as above. Assuming

a′=aand b′=b+ 1 it follows that X(a′,b′)

n(a,b+1) =X4(1,1) = 0.

Therefore, entry (a, b +1) of the left-hand side of (18) is αa,b 6= 0 while the same

entry on the right-hand side of (18) is 0, which is a contradiction.

Finally, we show that dim(EQ(n)(−4)) ≤(n−3)2by showing that every

element of the subspace generated by Fnis completely determined by entries

x(i,j+1) such that (i, j)∈[n−3]2.

Let S⊆[n]2be the set of indexes (p, q )∈[n]2such that the entry x(p,q)of

X∈ EQ(n)(−4) is completely determined by the entries x(i,j+1) , with (i, j )∈

[n−3]2. Clearly, [n−3] ×([n−2] \{1})⊆S. Since x(1,1) =x(n,1) = 0, it follows

17

that

x(i,1) =−

i

X

k=2

x(i+1−k,k),for every 2 ≤i≤n−2,

x(n−1,1) =−

n−2

X

k=2

x(k,1)

=x(1,2)

+x(2,2) +x(1,3)

.

.

.

+x(n−3,2) +···+x(2,n−3) +x(1,n−2)

=X

i,j≥1

i+j≤n−2

x(i,j+1)

and then [n]× {1} ⊆ S. Additionally, since x(1,n)=x(n,n)= 0 it follows that

x(i,n−1) =−

n−2

X

j=1

x(i,j)−x(i,n),for every 1 ≤i≤n−3,

x(i+1,n)=−

i

X

k=1

x(k,n−1−i+k),for every 1 ≤i≤n−3,

x(n−1,n)=−

n−2

X

i=2

x(i,n), x(n−2,n−1) =−

n−3

X

k=1

x(k,k+1) −x(n−1,n),

x(n,n−1) =−x(n−1,n), x(n−1,n−1) =−

n−2

X

i=2

x(i,n−1) −xn,n−1

and then [n]× {n−1, n} ⊆ S. Finally, since for every 2 ≤j≤n−2

x(n,j)=−

n−1

X

k=j

x(k,n+j−k),

x(n−2,j)=−

j−1

X

k=1

x(n−2−j+k,k)−x(n−1,j+1) −x(n,j+2) ,

x(n−1,j)=−

n−2

X

i=1

x(i,j)−x(n,j),

then {n−2, n −1, n} × ([n−2] \ {1})⊆S.

Deﬁnition 3.9. Given n∈N,i, j ∈[n],k∈ {2,...,2n}and ℓ∈ {−n+

1,...,n−1}, let the row vector Ri, the column vector Cj, the sum vector Sk

and the diﬀerence vector Dℓbe the vectors of Rn×ndeﬁned as follows.

Ri(x, y) = (1,if x=i

0,otherwise. Cj(x, y) = (1,if y=j

0,otherwise.

Sk(x, y) = (1,if x+y=k

0,otherwise. Dℓ(x, y) = (1,if x−y=ℓ

0,otherwise. .

(19)

18

Furthermore, (Zn={D0−Sn+1}and

Yn={Yn

i=Ci+Cn+1−i−Ri−Rn+1−i|i∈ ⌈n

2⌉}.

Some examples for n= 3 are the following.

000

000

111

010

010

010

010

100

000

100

010

001

Table 6: R3,C2,S3and D0.

Theorem 3.10. For n≥3,n−4is an eigenvalue of Q(n)with multiplic-

ity at least n−2

2,if neven;

n+1

2,otherwise. Furthermore, Ynand Znare sets of linear

independent eigenvectors of EQ(n)(n−4).

Proof. In what follows we consider only an odd integer n≥3. The neven case

is similar with the exception that Zn

06∈ EQ(n)(n−4) when nis even.

First we note that Yn

k∈ EQ(n)(n−4) for every k∈[n−1

2] and Zn

0∈ EQ(n)(n−

4). Since

Ck(p, q) = Rk(q, p) and Cn+1−k(p, q) = Rn+1−k(q, p),

we have X

p+q=i+j

(Yn

k)(p,q)= 0.

Similarly, since

Ck(p, q) = Rn+1−k(n+1−q, n+1−p) and Cn+1−k(p, q) = Rk(n+1−q, n+1−p),

we have X

p−q=i−j

(Yn

k)(p,q)= 0.

Hence

(AQ(n)Yn

k)(i,j)=X

(p,q)∈N(i,j)

(Yn

k)(p,q)

=X

p=i

(Yn

k)(p,q)+X

q=j

(Yn

k)(p,q)+

+X

p+q=i+j

(Yn

k)(p,q)+X

p−q=i−j

(Yn

k)(p,q)−4(Yn

k)(i,j)

=X

p=i

(Yn

k)(p,q)+X

q=j

(Yn

k)(p,q)−4(Yn

k)(i,j)

=

−(n−2) + (n−2) −0 = 0 if i, j ∈ {k, n + 1 −k},

−(n−2) + 2 = −(n−4) if i∈ {k, n + 1 −k}but j /∈ {k, n + 1 −k},

−2 + (n−2) = n−4 if j∈ {k, n + 1 −k}but i /∈ {k, n + 1 −k},

−2 + 2 −0 = 0 if i, j /∈ {k, n + 1 −k}

= (n−4)(Yn

k)(p,q).

19

Similarly, X

p=i

(Zn

0)(p,q)= 0 = X

q=j

(Zn

0)(p,q),

and so

(AQ(n)Zn

0)(i,j)=X

(p,q)∈N(i,j)

(Zn

0)(p,q)

=X

p=i

(Zn

0)(p,q)+X

q=j

(Zn

0)(p,q)+

+X

p+q=i+j

(Zn

0)(p,q)+X

p−q=i−j

(Zn

0)(p,q)−4(Yn

k)(i,j)

=X

p+q=i+j

(Zn

0)(p,q)+X

p−q=i−j

(Zn

0)(p,q)−4(Yn

k)(i,j)

=

(n−1) −(n−1) −0 = 0 if i=j=n+1

2,

1 + (n−1) −4 = n−4 if i=j6=n+1

2,

−(n−1) −1 + 4 = −(n−4) if i=n+ 1 −j6=n+1

2,

0 + 0 −0 = 0 if i6=j,i+j6=n+ 1, and i+jis odd,

1−1−0 = 0 if i6=j,i+j6=n+ 1, and i+jis even

= (n−4)(Zn

0)(p,q).

Note that nYn

1, Y n

2,...,Yn

n+1

2ois linearly dependent because

Yn

1+Yn

2+···+Yn

n−1

2

+1

2Yn

n+1

2

= 0.

However nYn

1, Y n

2,...,Yn

n−1

2

, Zn

0ois linearly independent. For otherwise

there would be α1,...,αn−1

2, β not simultaneously 0 such that

L:= α1Yn

1+α2Yn

2+···+αn−1

2Yn

n−1

2

+βZ n

0= 0.

Then L(1,1) =β= 0 and so

α1Yn

1+α2Yn

2+···+αn−1

2Yn

n−1

2

= 0.

Let k∈[n−1

2] be the smallest integer such that αk6= 0. Then L(n−1

2,k)=

αk= 0 contradicting our assumption.

Hence, dim EQ(n)≥n+1

2.

An interesting open problem related with the spectrum of Q(n) is the char-

acterization of its integer eigenvalues. The eigenvalues of Q(n), computed for

several values of n, suggest the following conjecture.

Conjecture 3. For n≥4, the set of integer eigenvalues of the n-Queens’ graph

Q(n)is

σ(Q(n)) ∩Z=({−4, n −4},if nis even;

−4,−3,...,n−11

2∪n−5

2,...,n−5, n −4,if nis odd.

20

Furthermore, when nis even the eigenvalue n−4has multiplicity (n−2)/2, and

when nis odd the eigenvalue n−4has multiplicity (n+1)/2and the eigenvalues

−3,−2,...,n−11

2,n−5

2,...,n−6, n −5are simple.

From the computations we may say that this conjecture is true for n≤100.

If this conjecture is true, then Q(n) has two distinct integer eigenvalues when

nis even and n−1 distinct integer eigenvalues when nis odd.

4 Equitable partitions

The eigenvalues µof a graph Gwhich have an associated eigenspace EG(µ) not

orthogonal to the all-one n×1 vector jare said to be main. The remaining

eigenvalues are referred as non-main. The concept of main (non-main) eigen-

value was introduced in [9] and further investigated in many papers since then.

A recent overview was published in [24]. As consequence, from the proof of

Theorem 3.8 and the deﬁnition of the family of eigenvectors Fnwe may con-

clude that −4 is a non-main eigenvalue. The main eigenvalues of a graph Gare

strictly related with the concept of equitable partition of the vertex set of Gand

with the corresponding divisor (or quociente) matrix B. The main eigenvalues

of Gare also eigenvalues of the divisor matrix of any equitable partition of G.

Deﬁnition 4.1 (Equitable partition).Given a graph G, the partition V(G) =

V1˙

∪V2˙

∪... ˙

∪Vkis an equitable partition if every vertex in Vihas the same num-

ber of neighbors in Vj, for all i, j ∈ {1,2,...,k}. An equitable partition of V(G)

is also called equitable partition of Gand the vertex subsets V1, V2,...,Vkare

called the cells of the equitable partition.

Every graph has a trivial equitable partition in which each cell is a singleton.

Deﬁnition 4.2 (Divisor (or quotient) matrix).Consider an equitable partition

πof G,V(G) = V1˙

∪V2˙

∪... ˙

∪Vk, where each vertex in Vihas bij neighbors in

Vj(for all i, j ∈ {1,2,...,k}). The matrix Bπ= (bij )is called the divisor (or

quociente) matrix of π.

The following results for graphs with equitable partitions are fundamental.

Lemma 4.3. [10, 24] If Gis a graph with pdistinct main eigenvalues µj1,...,µjp,

then the main characteristic polynomial of G,

MG(x) =

p

Y

i=1

(x−µji) = xp−c1xp−1−c2xp−2− · ·· − cp−1x−cp,(20)

has integer coeﬃcients.

Theorem 4.4. [11] Let Gbe a graph with adjacency matrix Aand let πbe a

partition of V(G)with characteristic matrix C.

1. If πis equitable, with divisor matrix Bπ, then AC =CBπ.

2. The partition πis equitable if and only if the column space of Cis A-

invariant.

3. The characteristic polynomial of any divisor matrix of Gdivides the char-

acteristic polynomial of Gand it is divisible by MG(x).

21

Since the largest eigenvalue of any connected graph Gwith n≥2 vertices

is a main eigenvalue (see [9, Th. 03 and 04]), an immediate consequence of

Theorem 4.4-3 is that the largest eigenvalue of Gis an eigenvalue of any of its

divisor matrices.

A very interesting property of the n-Queens’ graphs consists that they admit

equitable partitions with divisor matrices of order (⌈n

2⌉+1)⌈n

2⌉

2, for n≥3. These

equitable partitions can be obtained applying the following algorithm, where the

squares of the chessboard Tn, corresponding to the vertices of Q(n), are labeled

with the numbers of the cells they belong. Therefore, the squares belonging to

the same cell have the same number.

Algorithm 1[for the determination of an equitable partition of Q(n)]

Require: The n×nchess board, Tn.

Ensure: The labels of the squares of Tnwhich are the numbers of the cells of an

equitable partition.

Part I: Labeling procedure.

1 Assign the the number 0 to the ﬁrst square (the top left square).

2 Assign the numbers 1 and 2 to the ﬁrst and second squares of the

second column (from the top to bottom), respectively.

.

.

.

⌈n

2⌉Assign the numbers P⌈n

2⌉−1

i=0 i+ 1,...,(⌈n

2⌉+1)⌈n

2⌉

2−1 to the ﬁrst ⌈n

2⌉

squares of the ⌈n

2⌉-th column (from top to bottom).

Part II: Geometric reﬂection procedure.

1. Reﬂect the triangle formed by the squares labeled by the above steps,

using the vertical cathetus of the triangle as mirror line (after this

reﬂection, we have two right triangles sharing the same vertical line).

2. Reﬂect both triangles, each one using its hypotenuse as mirror line

(after this reﬂections all the squares in the top ⌈n

2⌉lines are labeled).

3. Reﬂect the rectangle formed by the the upper ⌊n

2⌋lines taking the

horizontal middle line of the chessboard as mirror line (after this

reﬂection all the squares have labels which are the numbers of their

cells).

End Algorithm.

Example 4.5. Let us apply the Algorithm 1 to the determination of an equitable

partition of the vertices of Q(6), corresponding to the squares in T6, the divisor

matrix Band its characteristic polynomial.

22

0 1 3

2 4

5

0 1 3 3 1 0

2 4 4 2

55

0133 1 0

12 4 4 2 1

3 4 55 4 3

0133 1 0

1244 2 1

3 4 55 4 3

3 4 5 5 4 3

1 2 4 4 2 1

0 1 3 3 1 0

Table 7: Labeling of the squares of Tnby application of Algorithm 1.

The divisor matrix of the equitable partition with cells numbered from 0to 5

presented in Table 7 is the matrix

B=

012345

0 342402

1 242241

2 243242

3 221442

4 042443

5 222463

.

The characteristic polynomial of Bis the polynomial

pB(x) = x6−21x5+ 73x4+ 109x3−686x2+ 580x−8.

According to Theorem 4.4-3, the eigenvalues of the divisor matrix Bobtained

in the above example, which are the ones belonging to

σ(B) = {16.24,3.63,2.85,1.17,0.01,−2.91},

are also eigenvalues of Q(6) and the largest eigenvalue of Bis also the largest

eigenvalue of Q(6). Moreover, taking into account Lemma 4.3, since pB(x) is

irreducible in Z[x], we may conclude pB(x) = MQ(6)(x).

Theorem 4.6. Every n-Queens’ graph, Q(n), with n≥3, has an equitable

partition with (⌈n

2⌉+1)⌈n

2⌉

2cells, determined by Algorithm 1.

Proof. The application of Algorithm 1 produces a partition πof the vertices

(squares) of Q(n) (Tn) into (⌈n

2⌉+1)⌈n

2⌉

2cells. Furthermore, from Algorithm 1,

we may conclude that each cell of πhas 8, 4 or 1 vertices (squares) and, by

symmetry, each cell induces a regular graph and each vertex in a particular cell

has the same number of neighbors in any other cell.

Acknowledgments. This work is supported by the Center for Research

and Development in Mathematics and Applications (CIDMA) through the Por-

tuguese Foundation for Science and Technology (FCT - Funda¸c˜ao para a Ciˆencia

e a Tecnologia), reference UIDB/04106/2020. I.S.C. also thanks the support

of FCT - Funda¸c˜ao para a Ciˆencia e a Tecnologia via the Ph.D. Scholarship

PD/BD/150538/2019.

23

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