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# Spectral properties of the $n$-Queens' Graphs

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## Abstract

The $n$-Queens' graph, $\mathcal{Q}(n)$, is the graph associated to the $n \times n$ chessboard (a generalization of the classical $8 \times 8$ chessboard), with $n^2$ vertices, each one corresponding to a square of the chessboard. Two vertices of $\mathcal{Q}(n)$ 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 $\mathcal{Q}(n)$, 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 $\mathcal{Q}(n)$, 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 $n \ge 4$. Furthermore, $n-4$ is also an eigenvalue of $\mathcal{Q}(n)$, with multiplicity at least $\frac{n-2}{2}$ when $n$ is even and at least $\frac{n+1}{2}$ when $n$ is odd. A conjecture about the integer eigenvalues of $\mathcal{Q}(n)$ is presented. We finish this article with an algorithm to determine an equitable partition of the $n$-Queens' graph, $\mathcal{Q}(n)$, for $n \ge 3$, concluding that such equitable partition has $\frac{(\lceil n/2\rceil+1)\lceil n/2\rceil}{2}$ cells.
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¸ao e Desenvolvimento em Matem´atica e Aplica¸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 (n3)2, for every n4. Furthermore,
n4 is also an eigenvalue of Q(n), with multiplicity at least n2
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
n3, 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 n4. 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.
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 (n3)2, for
every n4. Furthermore, n4 is also an eigenvalue of Q(n), with multiplicity
at least n2
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 n3, 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 vV(G), dominates itself and all
its neighbors. A vertex set SV(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+···+n1
2
= 2(n+ 1)n
2+ 4n
3(1)
=n(n1)(5n1)
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(n1)(5n1)
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 Vn+1
2is the vertex subset corre-
sponding to the squares of the board Tn, without considering the vertices
in V1V2∪ · ·· ∪ Vn+1
2⌋−1.
Clearly, the vertex subsets V1, V2,...,Vn+1
2form 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|= (n2(i1))2(n2(i1) 2)2
= ((n2(i1)) (n2(i1) 2)) ((n2(i1)) + (n2(i1) 2))
= 2(2n4(i1) 2) = 4(n(2i1)), i = 1,2,...,n+ 1
21,(4)
Vn+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,...,Vn+1
2, the degrees of the vertices of Q(n)are
dv= 3(n1) + 2(j1),for all vVj, j = 1,2,...,n+ 1
2.
Proof. For vV1, it is immediate that dv= 3(n1). On the other hand, for
vVi, with i=n+1
2, it is also immediate to conclude that
dv=4(n1) = 2(n1) + 2(n+1
2⌋ − 1),if nis odd;
3(n1) + n2 = 3(n1) + 2(n+1
2⌋ − 1),otherwise.
Let us consider isuch that 1 < i < n+1
2and that vVicorresponds to
the square of the chessboard Tndetermined by the pair of coordinates (p, q).
Then (p, q)Xi,i Xi,(n(i1)) Xn(i1),n(i1) X(n(i1)),i, where
Xi,i ={(i, j)[n]2|ijni}
Xi,(n(i1)) ={(j, n (i1)) [n]2|ijni}
Xn(i1),n(i1) ={(n(i1), j)[n]2|i+ 1 jn(i1)}
Xn(i1),i ={(j, i)[n]2|i+ 1 jn(i1)}.
Assuming (p, q)Xi,i , then p=i,iqniand (p, q) has 2(n1)
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+ (q1), q (q1))
|{z }
q1 lower right to left diagonal neighbors
,
(i+ 1, q + 1),(i+ 2, q + 2),...,(i+ (nq), q + (nq))
|{z }
nqlower left to right diagonal neighbors
,
(i1, q 1),(i2, q 2),...,(i(i1), q (i1))
|{z }
i1 upper right to left diagonal neighbors
,
(i1, q + 1),(i2, q + 2),...,(i(i1), q + (i1))
|{z }
i1 upper left to right diagonal neighbors
.
Therefore, the total number of neighbors of (p, q) is
d(p,q)= 2(n1) + q1 + nq+ 2(i1) = 3(n1) + 2(i1).
For the vertices (p, q) in any other of the subsets Xi,(n(i1)),X(n(i1)),(n(i1))
and X(n(i1)),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(n1) dv4n5,if nis even;
4n4,if nis odd. (6)
Therefore, the minimum and maximum degrees of Q(n) are δ(Q(n)) =
3(n-1) and ∆(Q(n)) = 4n5,if nis even;
4n4,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
n4, 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 n4.
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 n5.
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 n5 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 k1, 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 k1, those vertices are (ik, j1) and
(ik, j2), if ik1, and the vertex (ik/2, j1+k/2) when kis even
and ik/21, or (i+k , j1) and (i+k, j2), if i+knplus the vertex
(i+k/2, j1+k/2) when kis even and i+k/2n.
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 n1,5 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,...,n1}×{0,1,...,n1}and the set of colors is C={0,1,...,n1}.
Assign to each vertex (i, j ) the color j2imod n.
If (i, j) and (p, q ) are two vertices with the same color, then j2iq2p
mod n, which is equivalent to
2(pi)(qj) 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) pi=qjor (b)
pi=jqand, from (8),
(a) pi=qj2(qj)(qj) mod nj=qand then
(i, j) = (p, q ).
(b) pi=jq2(jq)(qj) mod n3(jq)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, n1,5 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 n1,5 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 0jn1, deﬁning the j-th right (resp. left) diagonal
of Kas the set of ncells {(i, j +i)|i= 0,1,...,n1}(resp. {(i, j i1) |
i= 0,1,...,n1}), 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 n5,
nχ(Q(n)) n, if n1,5 mod 6;
n+ 3 otherwise.
Proof. Since ω(Q(n)) = n, when n5, 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 n5,χ(Q(n)) = nwhen n1,5 mod 6 and
χ(Q(n))
{n, n + 1},if n0,4 mod 6;
{n, n + 1, n + 2, n + 3},if n2 mod 6;
{n, n + 1, n + 2}if n3 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
n1
2γ(Q(n)) 2p+q, (9)
where the lower bound holds for every natural number nand the upper bound
holds when n= 3p+q, with pNand 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 kN∪ {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 kN∪ {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 nN,γ(Q(n)) γ(Q(n+ 1)).
However, we have the following inequality.
Proposition 2.11. For nN,γ(Q(n+ 1)) γ(Q(n)) + 1.
Proof. Let SV(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 Sis 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 n9. 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
jV(G)
xj,
s.t. (A+In)xd,(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 xby x
j, for all jV(G),D={j|x
j= 0}is a minimum
dominating set for Gand γ(G) = nη(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 xis an optimal solution of the binary linear programming problem, then
the dominating vertex subset D={i|x
i= 0}is such that |D|=nη(x).
Let us assume that there is a vertex dominating set Dsuch that |D|<|D|.
Then, deﬁning xwith components x
i=0,if iD;
1,otherwise , it is clear that xis
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|=nη(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)xd,(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 xby 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)) = n2η(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 n131. 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⌋ − 1γ(Q(n))
n+1
2+ 1, if n16, and γ(Q(n)) = n+1
2, if n17.
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(n1)(5n1)
3nµ1(Q(n)) 4n5 if nis even;
4n4 if nis odd.(14)
The characteristic polynomials and spectra of Q(2) and Q(3) are the follow-
ing.
q2(x) = (x+ 1)3(x3) and σ(Q(2)) = n3,1[3]o;
q3(x) = (1 + x)(1 + x)2(85x+x2)(1 + 2x+x2)2and
σ(Q(3)) = (5 + 57
2,1,(1 + 2)[2],1[2] ,557
2,(12)[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|iI}and then Vi=V(G[Ei]) is a clique of Gfor
every iI. For any vV(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|iI}, the parameters
mv(P) and mG(P) can be expressed as follows.
mv(P) = |{iI|vV(G[Ei])}| ∀vV(G); (15)
mG(P) = max{mv|vV(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 n4, considering an ECP whose parts are the edge subsets
associated with the ncolumns, nrows, 2n3 left to right diagonals and 2n3
right to left diagonals of the corresponding n×nchessboard Tn.
Theorem 3.3. Let P={Ei|iI}be an ECP of a graph G,m=mG(P)and
mv=mv(P)for every vV(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
jV(G[Ei])
xj= 0, for every iIand
(b) vV(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
iIX
uvEi
(2xuxv) + mkXk2
=X
iI
X
vV(G[Ei])
xv
2
X
vV(G)
mvx2
v+mkXk2
=X
iI
X
vV(G[Ei])
xv
2
+X
vV(G)
(mmv)x2
v0.
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
iI
X
vV(G[Ei])
xv
2
+X
vV(G)
(mmv)x2
v
=X
iI
X
vV(G[Ei])
xv
2
X
vV(G)
mvx2
v+mkXk2
=X
iIX
uvEi
(2xuxv) + mkXk2
=XTAX+mkXk2.
14
Assuming that µis the least eigenvalue of G,m=XTAX
kXk2µ. Since
µ≥ −m(by item 1), we obtain mµ≥ −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 µC(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 µC(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 nNsuch that n4.
1. If µis an eigenvalue of Q(n), then µ≥ −4.
2. 4σ(Q(n)) iﬀ there exists a vector XRn2\ {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[2n3],
(c) P
ij=k+1n
x(i,j)= 0, for every k[2n3],
(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, 2n3 left to right diagonals and 2n3 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 n4and XRn2. 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 n4, it seems that n4 and
4 are eigenvalues of Q(n) with multiplicities
m(n4) = (n2)/2,if nis even;
(n+ 1)/2,if nis odd,
and m(4) = (n3)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 n11.
In what follows we will see that, for n4, 4 is an eigenvalue of Q(n) with
multiplicity (n3)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 6n2 linear equations 2a-2d. Therefore, to say that m(4) = (n3)2
is equivalent to say that the rank of the coeﬃcient matrix of the system of 6n
linear equations 2a-2d is 6n9 (since n26n+ 9 = (n3)2).
Before we continue, we need to deﬁne a family of vectors
Fn={X(a,b)
nRn2|(a, b)[n3]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(ia+1,jb+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 (n3)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) (n3)2. For
otherwise there would be scalars α1,1,...,αn3,n3R, not all equal to zero,
such that
α1,1X(1,1)
n+···+αn3,n3X(n3,n3)
n=0.(18)
Let (n3)(a1) + 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 (n3)(a1)+b>(n3)(a1)+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×Band 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)) (n3)2by showing that every
element of the subspace generated by Fnis completely determined by entries
x(i,j+1) such that (i, j)[n3]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 )
[n3]2. Clearly, [n3] ×([n2] \{1})S. Since x(1,1) =x(n,1) = 0, it follows
17
that
x(i,1) =
i
X
k=2
x(i+1k,k),for every 2 in2,
x(n1,1) =
n2
X
k=2
x(k,1)
=x(1,2)
+x(2,2) +x(1,3)
.
.
.
+x(n3,2) +···+x(2,n3) +x(1,n2)
=X
i,j1
i+jn2
x(i,j+1)
and then [n]× {1} ⊆ S. Additionally, since x(1,n)=x(n,n)= 0 it follows that
x(i,n1) =
n2
X
j=1
x(i,j)x(i,n),for every 1 in3,
x(i+1,n)=
i
X
k=1
x(k,n1i+k),for every 1 in3,
x(n1,n)=
n2
X
i=2
x(i,n), x(n2,n1) =
n3
X
k=1
x(k,k+1) x(n1,n),
x(n,n1) =x(n1,n), x(n1,n1) =
n2
X
i=2
x(i,n1) xn,n1
and then [n]× {n1, n} ⊆ S. Finally, since for every 2 jn2
x(n,j)=
n1
X
k=j
x(k,n+jk),
x(n2,j)=
j1
X
k=1
x(n2j+k,k)x(n1,j+1) x(n,j+2) ,
x(n1,j)=
n2
X
i=1
x(i,j)x(n,j),
then {n2, n 1, n} × ([n2] \ {1})S.
Deﬁnition 3.9. Given nN,i, j [n],k∈ {2,...,2n}and ∈ {−n+
1,...,n1}, let the row vector Ri, the column vector Cj, the sum vector Sk
and the diﬀerence vector Dbe 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. D(x, y) = (1,if xy=
0,otherwise. .
(19)
18
Furthermore, (Zn={D0Sn+1}and
Yn={Yn
i=Ci+Cn+1iRiRn+1i|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 n3,n4is an eigenvalue of Q(n)with multiplic-
ity at least n2
2,if neven;
n+1
2,otherwise. Furthermore, Ynand Znare sets of linear
independent eigenvectors of EQ(n)(n4).
Proof. In what follows we consider only an odd integer n3. The neven case
is similar with the exception that Zn
06∈ EQ(n)(n4) when nis even.
First we note that Yn
k∈ EQ(n)(n4) for every k[n1
2] and Zn
0∈ EQ(n)(n
4). Since
Ck(p, q) = Rk(q, p) and Cn+1k(p, q) = Rn+1k(q, p),
we have X
p+q=i+j
(Yn
k)(p,q)= 0.
Similarly, since
Ck(p, q) = Rn+1k(n+1q, n+1p) and Cn+1k(p, q) = Rk(n+1q, n+1p),
we have X
pq=ij
(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
pq=ij
(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)
=
(n2) + (n2) 0 = 0 if i, j ∈ {k, n + 1 k},
(n2) + 2 = (n4) if i∈ {k, n + 1 k}but j /∈ {k, n + 1 k},
2 + (n2) = n4 if j∈ {k, n + 1 k}but i /∈ {k, n + 1 k},
2 + 2 0 = 0 if i, j /∈ {k, n + 1 k}
= (n4)(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
pq=ij
(Zn
0)(p,q)4(Yn
k)(i,j)
=X
p+q=i+j
(Zn
0)(p,q)+X
pq=ij
(Zn
0)(p,q)4(Yn
k)(i,j)
=
(n1) (n1) 0 = 0 if i=j=n+1
2,
1 + (n1) 4 = n4 if i=j6=n+1
2,
(n1) 1 + 4 = (n4) if i=n+ 1 j6=n+1
2,
0 + 0 0 = 0 if i6=j,i+j6=n+ 1, and i+jis odd,
110 = 0 if i6=j,i+j6=n+ 1, and i+jis even
= (n4)(Zn
0)(p,q).
Note that nYn
1, Y n
2,...,Yn
n+1
2ois linearly dependent because
Yn
1+Yn
2+···+Yn
n1
2
+1
2Yn
n+1
2
= 0.
However nYn
1, Y n
2,...,Yn
n1
2
, Zn
0ois linearly independent. For otherwise
there would be α1,...,αn1
2, β not simultaneously 0 such that
L:= α1Yn
1+α2Yn
2+···+αn1
2Yn
n1
2
+βZ n
0= 0.
Then L(1,1) =β= 0 and so
α1Yn
1+α2Yn
2+···+αn1
2Yn
n1
2
= 0.
Let k[n1
2] be the smallest integer such that αk6= 0. Then L(n1
2,k)=
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 n4, 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,...,n11
2n5
2,...,n5, n 4,if nis odd.
20
Furthermore, when nis even the eigenvalue n4has multiplicity (n2)/2, and
when nis odd the eigenvalue n4has multiplicity (n+1)/2and the eigenvalues
3,2,...,n11
2,n5
2,...,n6, n 5are simple.
From the computations we may say that this conjecture is true for n100.
If this conjecture is true, then Q(n) has two distinct integer eigenvalues when
nis even and n1 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
(xµji) = xpc1xp1c2xp2− · ·· − cp1xcp,(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 n2 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 n3. 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
2Assign the numbers Pn
2⌉−1
i=0 i+ 1,...,(n
2+1)n
2
21 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
2lines are labeled).
3. Reﬂect the rectangle formed by the the upper n
2lines 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) = x621x5+ 73x4+ 109x3686x2+ 580x8.
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 n3, 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¸ao para a Ciˆencia
e a Tecnologia), reference UIDB/04106/2020. I.S.C. also thanks the support
of FCT - Funda¸ao para a Ciˆencia e a Tecnologia via the Ph.D. Scholarship
PD/BD/150538/2019.
23
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25
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