Chessboard Puzzles Part 2 - Independence

Chessboard Puzzles: Independence

Part 2 of a 4-part Series of Papers on the Mathematics of the Chessboard

by Dan Freeman

May 1, 2014

Dan Freeman Chessboard Puzzles: IndependenceMAT 9000 Graduate Math Seminar

Table of Contents

Table of Figures...............................................................................................................................3

Introduction......................................................................................................................................4

Definition of Independence.............................................................................................................4

Rooks Independence........................................................................................................................5

Bishops Independence.....................................................................................................................7

Kings Independence.......................................................................................................................10

Knights Independence...................................................................................................................12

The 8-queens Problem...................................................................................................................14

Proof that β(Qnxn) = n.....................................................................................................................17

The n-queens Problem...................................................................................................................21

Conclusion.....................................................................................................................................23

Sources Cited.................................................................................................................................25

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Table of Figures

Image 1: Rook Movement...............................................................................................................5Image 2: Eight Independent Rooks on 8x8 Board...........................................................................6Image 3: All Six Permutations of Three Independent Rooks on 3x3 Boards..................................7Image 4: Bishop Movement.............................................................................................................7Image 5: Maximum Set of Independent Bishops on 8x8 Board......................................................8Image 6: Pairs of Independent Bishops on Either the Top and Bottom Rows or the Left and Right Columns......................................................................................................10Image 7: King Movement..............................................................................................................10Image 8: Maximum Set of Independent Kings on 8x8 Board.......................................................11Image 9: Maximum Set of Independent Kings on 9x9 Board.......................................................12Image 10: Knight Movement.........................................................................................................13Image 11: Queen Movement.........................................................................................................14Image 12: The 1 5 8 6 3 7 2 4 Solution to the 8-queens Problem.................................................15Image 13: The 4 2 8 5 7 1 3 6 Solution to the 8-queens Problem.................................................17Image 14: Ten Independent Queens on 10x10 Board...................................................................18Image 15: Eight Independent Queens on 8x8 Board Using Cloning Construction.......................19Image 16: Nine Independent Queens on 9x9 Board Using Clone Construction...........................21Image 17: Doubly Centrosymmetric Solution to the 5-queens Problem.......................................22

Table 1: Independence Number Notation........................................................................................4Table 2: Sum of Row and Column Numbers for the 1 5 8 6 3 7 2 4 Solution..............................15Table 3: Sum of Row and Column Numbers for the 1 5 8 6 3 7 2 4 Solution..............................15Table 4: Permutations of Solutions to the 8-queens Problem........................................................16Table 5: Solutions to the n-queens Problem by Symmetry Type for 4 ≤ n ≤ 15...........................23Table 6: Domination Number Formulas by Piece.........................................................................23Table 7: Independence Number Formulas by Piece......................................................................24

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Introduction

This paper analyzes the concept of chessboard independence, a similar but different idea

from domination, which was examined in my first paper on chessboard puzzles. Like

domination, independence is a problem that involves placing an optimal amount of chess pieces

of a particular type on a chessboard. However, unlike domination, which is a minimization

problem, independence is a maximization problem. Therefore, the way to attack puzzles with

independence differs somewhat from the approach used in domination. In addition, the level of

knowledge and understanding regarding formulas and patterns associated with independence will

naturally deviate from what is known about domination.

In a structure that closely parallels my first paper in this series, this paper seeks to

provide a full survey of chessboard independence among rooks, bishops, kings, knights and

queens. In doing this, I hope to be able to make the link between the two ideas of domination

and independence by comparing and contrasting what has already been established vs. what

remains as an open problem.

Definition of Independence

An independent set of chess pieces is one such that no piece in the set attacks another

piece in the set. The independence number for a certain piece and certain size mxn1chessboard is

the maximum number of independent pieces that can be placed on the board; a set of such

maximum size is called a maximum independent set [1, p. 163]. Independence numbers are

denoted by β(Pmxn) where P represents the type of chess piece, as denoted in Table 1.

Table 1 : Independence Number Notation

Piece Abbreviation

Knight NBishop BRook RQueen QKing K

1 Throughout this paper, m and n refer to arbitrary positive integers denoting the number of rows and columns of a chessboard, respectively.

4


Note that the domination number γ(Pmxn) ≤ β(Pmxn) for all pieces P and for all m and n. To

see why this is the case, suppose that a maximum independent set of pieces, call it I, fails to

dominate a chessboard. In other words, there exists at least one square on the board that is not

covered by any piece in I and hence γ(Pmxn) > β(Pmxn). Then one could simply add as many

independent pieces as are required on the uncovered squares so as to dominate the entire board.

But then this dominating set of pieces, call it D, would be an independent set larger than I,

contradicting the fact that I is a maximum independent set. Therefore, the number of pieces in a

maximum independent set I must be at least as many as the number of pieces in a minimum

dominating set D. Put more concisely, γ(Pmxn) ≤ β(Pmxn) [1, p. 164].

Rooks Independence

Recall from my first paper on chessboard domination that rooks are permitted to move

any number of squares either horizontally or vertically, as long as they do not take the place of a

friendly piece or pass through any piece (own or opponent’s) currently on the board. In Image

1, the white rook can move to any of the squares with a white circle and the black rook can move

to any of the squares with a black circle [4].

Image 1: Rook Movement

As with domination, independence among rooks is the simplest of all chess pieces and

β(Rnxn) = n. The proof for this formula is quite simple. There are n rows and n columns on an

nxn board. In order to be independent, no two rooks may lie on the same column or same row.

Therefore, an independent set of rooks may contain no more than n rooks. Furthermore, n rooks

placed along the main diagonal is an independent set since no two rooks share the same row or

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the same column. As a result, we have that β(Rnxn) ≤ n and β(Rnxn) ≥ n, which implies that β(Rnxn)

= n. This result is illustrated in Image 2.

Also in line with rooks domination is the fact that for a general rectangular mxn

chessboard, β(Rmxn) = min(m, n). The argument that is this case closely parallels that for the

square chessboard. Without loss of generality, suppose that m < n. Clearly, if there are more

than m rooks on the board, then there exists a pair of rooks that lie on the same row. Therefore,

β(Rmxn) ≤ m. Also, if m rooks are placed in a diagonal fashion on the board, then this set of rooks

is independent and β(Rmxn) ≥ m. Since we have designated m as the lesser of m and n, we have

that β(Rmxn) = m = min(m, n). A symmetric argument works if we suppose that m > n so the

proof is complete.

A formula for the number of permutations of maximum independent sets of rooks on an

nxn square board is also well-known. The number of such permutation is n!,[1, p. 179] and we

shall denote this number by βPerm(Rnxn), where the subscripted “Perm” is a shorthand indication

for the number of permutations. As with the proof for the rooks independence number, the proof

that βPerm(Rnxn) = n! is straightforward. A rook may be placed on any of the n squares of the first

column. Another rook may be placed on any of the n – 1 squares in the second column not in the

same row as the rook in the first column. Yet another rook may be placed on any of the n – 2

squares in the third column not in the same row as either of the rooks in the first two columns.

This process continues until the nth and final column is reached when just 1 square will remain

for the final rook. Therefore, the number of permutations is n*(n – 1)*(n – 2) * … *3*2*1 = n!.

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Image 2: Eight IndependentRooks on 8x8 Board


The six different permutations of sets of three independent rooks on a 3x3 board are shown in

Image 3 [1, p. 179].

Bishops Independence

Recall that bishops move diagonally any number of squares as long as they do not take

the place of a friendly piece or pass through any piece (own or opponent’s) currently on the

board. In Image 4, the white bishop can move to any of the squares with a white circle and the

black bishop can move to any of the squares with a black circle [4].

Image 4: Bishop Movement

The formula for the bishops independence number differs from the formula for the

bishops domination number. While γ(Bnxn) = n, β(Bnxn) = 2n – 2. The proof is fairly

straightforward. For any nxn chessboard, there are 2n – 1 positive diagonals (as well as 2n – 1

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Image 3: All Six Permutations of Three Independent Rooks on 3x3 Boards


negative diagonals2). Except for opposite corner squares, exactly one bishop may be placed in

each diagonal so as to have an independent set of bishops. In the case of opposing corner

squares, only one bishop may be placed on one of the two squares, in order to avoid two bishops

attacking each other from corner to corner. Therefore, the number of pieces in an independent

set of bishops must be no more than one fewer than the number of diagonals, or (2n – 1) – 1 = 2n

– 2. Additionally, one may place n – 1 bishops along the top row and n – 1 bishops along the

bottom row, as in Image 5, for a total of 2n – 2 bishops that are independent. Therefore, β(Bnxn)

= 2n – 2 [1, p. 182].

As might be expected, as is the case with the bishops domination number, no formula is

currently known for the bishops independence number on a general rectangular board [2, p. 13].

In 1964, the Yaglom brothers proved that a maximum set of independent bishops must be

placed on the outermost edges of the board. That is, either half of the 2n – 2 bishops must be

placed on the top row and half on the bottom row, or the bishops need to be placed half and half

on the left and right edges. To begin, assume that 2n – 2 independent bishops have been placed

on the board. Label each square with the number of bishops that cover it. A square cannot be

controlled by more than two bishops as this would result in at least one pair of bishops that attack

each other. Also, a square cannot be controlled by zero bishops, since then the set of

2 Hereafter within the Bishops Independence section of this paper, the sign (positive or negative) associated with diagonals will be omitted, as this is not an important distinction.

8

Image 5: Maximum Set of Independent Bishops on 8x8 Board


independent bishops would not be maximum. Therefore, each square gets assigned a label of 1

or 2 [1, p. 183].

Now we will show that at least 2n squares on the board are labeled with a 1. The four

corner squares can only be controlled by a single bishop, whether the squares are occupied or

not. At least two of the corner squares must be unoccupied in order to have an independent set

of bishops. In addition, the 2n – 2 squares containing bishops clearly have a label of 1.

Together, the 2 unoccupied corner squares and the 2n – 2 squares containing bishops each having

a label of 1 result in there being at least (2n – 2) + 2 = 2n squares with a label of 1 [1, pp. 183-

184].

Next, let S be the sum of the labels for the entire chessboard. Since there are at least 2n

squares with a label of 1, the remaining n2 – 2n squares have a label of 1 or 2. This results in the

following inequality: S ≤ 1*(2n) + 2*(n2 – 2n) = 2n2 – 2n = n*(2n – 2). Now we seek to obtain a

lower bound for S. A bishop on the outer ring of edge squares (top, bottom, left or right edges)

covers exactly n squares. A bishop inside of the outermost ring controls at least n + 2 squares

(this number increases by 2 for each move the bishop makes closer to the center of the board).

Let a be the number of bishops in the interior of the board and b be the number of bishops on the

outermost ring. So, by construction, a + b = 2n – 2 and we have the following inequality: S ≥

b*n + a*(n+2) = n*(a + b) + 2a = n*(2n – 2) + 2a. All together, we now have S bounded below

and above, as follows: n*(2n – 2) + 2a ≤ S ≤ n*(2n – 2), which implies that n*(2n – 2) + 2a ≤

n*(2n – 2). The only way this latter inequality can possibly hold is if a = 0. In other words,

there cannot be any bishops on the interior of the board, and so all of them must be on the

outermost ring [1, p. 184].

Now that we have proved that all bishops in a maximum independent set must be placed

on the outer ring of edge squares, we are well-equipped to go about proving a formula for the

number of permutations of 2n – 2 independent bishops on an nxn square chessboard. There are

2n such arrangements. Because a bishop must be on an edge square, we need only consider the

possible arrangements of bishops on the top row. First consider the squares on the four corners

of the board. A bishop may be placed on one of two opposite corners. Since there are two pairs

of opposite corners, there are 2*2 = 4 permutations of independent bishops on the corner squares.

Next we turn our attention to the non-corner squares. For any non-corner square in the top row,

9


either that square in the top row and the square straight down from it in the bottom row contain

bishops while the two squares on the left and right edge diagonally from the top and bottom

squares are unoccupied or the top and bottom squares are left open and the squares on the left

and right edges contain bishops. These facts are illustrated in Image 6. So for each of the n – 2

columns not on the left or right edges of the board, there are two possible arrangements of

independent bishops. Combined with the fact that are two permutations for each of the columns

on the left and right edges, we conclude that βPerm(Bnxn) = 22*2n – 2 = 2n [1, pp. 184-185].

Kings Independence

Recall that kings are allowed to move exactly one square in any direction as long as they

do not take the place of a friendly piece. In Image 7, the king can move to any of the squares

with a white circle [4].

Image 7: King Movement

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Image 6: Pairs of Independent Bishops on Either theTop and Bottom Rows or the Left and Right Columns


As with the kings domination number, a formula has been established for the kings

independence number. The formula, β(Knxn) = └½*(n + 1)┘2, is different from the kings

domination number formula (recall that that γ(Knxn) = └(n + 2) / 3┘2) but similar in that both

formulas involve the floor function and a term that gets squared. The proof is relatively

straightforward [1, p. 185].

First, note that any 2x2 square on a chessboard may contain at most one independent

king. When n is even, the board can be split into (½*n)2 squares and so β(Knxn) = (½*n)2 [1, pp.

185-186]. The case when n is even is illustrated on an 8x8 board in Image 8.

Now if n is odd, one can divide the board into (½*(n – 1)2 2x2 squares, ½*(n – 1) 1x2

rectangles, ½*(n – 1) 2x1 rectangles and one 1x1 square. Each of these areas may contain at

most one independent king, so β(Knxn) = (½*(n – 1)2 + 2*(½*(n – 1)) + 1 = (½*(n + 1)2. The

formulas for n even and n odd can be condensed into a single formula using the floor function3,

as follows: β(Knxn) = └½*(n + 1)┘2. Since n is odd, the king in the 1x1 square fixes the

arrangement of all the other kings on the board; hence such a maximum arrangement of

independent kings is unique [1, p. 186]. The case when n is odd is shown on a 9x9 board in

Image 9.

3 Throughout this paper, the symbols ‘└’ and ‘┘’ will be used to indicate the greatest integer or floor function.

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Image 8: Maximum Set of Independent Kings on 8x8 Board


Furthermore, like we were able to do with the kings domination number formula, we can

generalize kings independence to rectangular boards by expanding the squared term in the

formula into two separate factors and replacing the n variable with an m variable in one of them,

as follows: β(Kmxn) = └½*(m + 1)┘*└½*(n + 1)┘.

Knights Independence

Recall that knights move two squares in one direction (either horizontally or vertically)

and one square in the other direction as long as they don’t take the place of a friendly piece. In

Image 10, the white and black knights can move to squares with circles of the corresponding

color [4].

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Image 9: Maximum Set of Independent Kings on 9x9 Board


Image 10: Knight Movement

Unlike with knights domination, a formula is known for the knights independence

number. The formula is broken into three separate cases, as follows [1, p. 181]:

4 if n = 2

β(Nnxn) = ½*n2 if n ≥ 4, n even

½*(n2 + 1) if n odd

I will omit the proof of the above formula since it relies on a concept known as the

knight’s tour that we won’t explore until the third paper in this series. I will be sure to revisit

this proof at the appropriate point in that paper.

For the sake of completeness, it should be mentioned that a somewhat more complicated

formula (which we will not prove in this or any subsequent paper) is known for the knights

independence number on a general mxn rectangular board. The formula, which makes use of the

ceiling function4, is as follows [5]:

n if m = 1

β(Nmxn) = 2*(┌n / 4┐ + ┌(n – 1) / 4┐) if m = 2

┌½*mn┐ if m, n ≥ 3

4 Throughout this paper, the symbols ‘┌’ and ‘┐’ will be used to indicate the least integer or ceiling function.

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The 8-queens Problem

Recall that queens move horizontally, vertically and diagonally any number of squares as

long as they do not take the place of a friendly piece or pass through any piece (own or

opponent’s) currently on the board. In Image 11, the queen can move to any of the squares with

a black circle [4].

Image 11: Queen Movement

A classic chessboard puzzle is the 8-queens problem, in which one attempts to place 8

independent queens on an 8x8 chessboard. The problem was first presented by a chess player

named Max Bezzel in a German newspaper in September 1948. Less than two years later, in

June 1850, Franz Nauck posed the problem in another German newspaper. Nauck correctly

published all 92 solutions to the 8-queens problem in the same paper on September 21, 1850.

During that same summer, Carl Friedrich Gauss read Nauck’s account of the problem and began

studying it on his own [1, pp. 164-165].

In all his brilliance, Gauss used permutations of the numbers 1 through 8 to analyze

solutions to the 8-queens problem. For example, the permutation 1 5 8 6 3 7 2 4 corresponds to a

solution to the 8-queens problem (shown in Image 12) in which the queen in the first column

from the left is placed in the first row from the bottom, the queen in the second column from the

left is placed in the fifth row from the bottom, and so on. Gauss devised a method for

determining whether an arrangement of 8 queens on an 8x8 board is independent. For each

queen, if the sum of its row number and column number (again, from bottom to top, from left to

right) is unique to this sum for all other queens, then it follows that no two queens lie on the

same negative diagonal. Additionally, for each queen, if the sum of its row number and reverse

column number (from bottom to top, from right to left) is distinct from that of all other queens,

14


then it follows that no two queens lie on the same positive diagonal. If these two conditions

hold, then the 8 queens are independent [1, pp. 165-166].

In the aforementioned 1 5 8 6 3 7 2 4 permutation of 8 queens, the sums of the row

numbers and column numbers and the sums of the row numbers and reverse column numbers for

each queen are shown in Tables 2 and 3, respectively.

Table 2: Sum of Row and Column Numbers for the 1 5 8 6 3 7 2 4 Solution

Row # 1 5 8 6 3 7 2 4

Column # 1 2 3 4 5 6 7 8

Sum 2 7 1110

8 13 9 12

Table 3: Sum of Row and Column Numbers for the 1 5 8 6 3 7 2 4 Solution

Row # 1 5 8 6 3 7 2 4

Reverse Column #

8 7 6 5 4 3 2 1

Sum 9 12 14 11 7 10 4 5

In his 1953 book Mathematical Recreations, Maurice Kraitchik provided all 92 solutions

to the 8-queens problem. There are 11 permutations which account for 8 different arrangements

when all rotations and reflections are considered and one additional permutation which is

15

Image 12: The 1 5 8 6 3 7 2 4 Solution to the 8-queens Problem


symmetric under rotations of 180° and hence gives rise to 4 different solutions. These 12

different permutations are shown in Table 4 [1, pp. 166-167].

Table 4: Permutations of Solutions to the 8-queens Problem

Permutation No. of Solutions

1 5 8 6 3 7 2 4 81 6 8 3 7 4 2 5 82 4 6 8 3 1 7 5 82 5 7 1 3 8 6 4 82 5 7 4 1 8 6 3 82 6 1 7 4 8 3 5 82 6 8 3 1 4 7 5 82 7 3 6 8 5 1 4 82 7 5 8 1 4 6 3 83 5 8 4 1 7 2 6 83 6 2 5 8 1 7 4 83 5 2 8 1 7 4 6 4

As an additional note on the 8-queens problem, H.E. Dudeney observed that all but one

of the 12 fundamental solutions consists of three queens lying on a straight line (though

obviously not along the same row, column or diagonal, for then the set of queens would not be

independent). For instance, in the 1 5 8 6 3 7 2 4 solution shown in Image 12, the queens on

squares (2, 5), (4, 6) and (6, 7) lie on a straight line, where the coordinates (x, y) correspond to

the xth column from the left and the yth row from the bottom5. One solution to the 8-queens

problem that does not exhibit three queens in a straight line is 4 2 8 5 7 1 3 6, shown in Image 13

[1, pp. 167-168].

5Throughout this paper, this (x, y) coordinate notation will be used.

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Proof that β ( Q nxn) = n

Before we extend the 8-queens problem to a more general n-queens problem, it would be

prudent to first examine the queens independence number for square nxn boards. In stark

contrast to queens domination, the queens independence number has a known formula and a very

simple one too: β(Q2x2) = 1, β(Q3x3) = 2 and for n ≠ 2 or 3, β(Qnxn) = n. The proof, however, is not

nearly as simple and requires considering a number of different cases.

First, let’s get the easy cases where 1 ≤ n ≤ 3 out of the way. The n = 1 case is

completely trivial, as we can only place a single queen on the lone square of the board; hence

β(Q1x1) = 1. The n = 2 case is almost as trivial, as we note that if two or more queens are placed

on a 2x2 board, then they will necessarily be adjacent to each other; so β(Q2x2) = 1. When n = 3,

note that each queen placed along the edge of the board (that is, any square other than the center

square) will control exactly 7 of the 9 squares on the board. The two squares not under attack

will lie on the same row, column or diagonal. Thus, no more than 2 independent queens can be

placed on a 3x3 board and β(Q2x2) = 3.

For n ≥ 4, we will have to split the proof up into even and odd cases and further divide

the even case into two subcases, one which n ≡ 2 mod 6 and the other in which n ≢ 2 mod 6.

Let’s first consider the subcase where n ≢ 2 mod 6. We start by placing a queen on the (1, 2)

17

Image 13: The 4 2 8 5 7 1 3 6 Solution to the 8-queens Problem


square and make repeated knight’s moves, one square and two squares up, until you hit the top

row. At this point, we have placed ½*n queens on the board that control all of the columns on

the left half of the board and all of the even-numbered rows. Now we place a queen in the

bottom row, one column over from where we left off, and continue the same placement of

queens until we reach the right edge of the board. These ½*n queens on the right side of the

board control all of the columns on the right half of the board, as well as all of the odd-numbered

rows. We have now placed n independent queens on an nxn board where n ≢ 2 mod 6; this is

shown for n = 10 in Image 14 [1, pp. 170-171].

Now the only thing we need to worry about in the case where n ≢ 2 mod 6 is whether or

not any of the queens on the left half of the board lie on the same negative diagonal as any of the

queens on the right half of the board. Let n = 2k. The rth queen on the left side of the board is

on square (r, 2r) and the sum of these two coordinates is 3r. The sth queen on the right side of

the board is at (k + s, 2s – 1). If these two queens occupy the same negative diagonal, then (k +

s) + (2s – 1) = 3r, that is, 3s + k – 1 = 3r. If we substitute k with ½*n and isolate n, the equation

becomes n = 6*(r – s) + 2. This implies that if two queens are on the same negative diagonal,

then n ≡ 2 mod 6. Therefore, our construction above works precisely for even n where n ≢ 2

mod 6 [1, pp. 171].

18

Image 14: Ten Independent Queenson 10x10 Board


Now for n = 8, 14, 20, …, that is n ≡ 2 mod 6, we start by placing a queen at (2, 3) and

then doing repeated knight’s moves, as we did before, until we reach the second row from the

top. At this point, the leftmost column and the bottom row are uncovered. In order to account

for these uncovered squares, we revert to the second-to-last knight we placed (the one on the

fourth row from the top), and “clone” it into two copies. We move one copy straight over to the

leftmost column and the other straight down to the bottom row. Letting n = 2k, we now have

placed k queens that control all of the columns in the left half of the board as well as all of the

odd-numbered rows. In order to take care of the right half of the board, rotate the board 180°

and repeat the same knight’s move construction. This construction on an 8x8 board is illustrated

in Image 15, in which the squares on which the “cloned” queens originated are shaded in gray

with arrows pointing to the final destinations of the cloned queens [1, pp. 172-173].

In order to verify that this construction works for all n ≡ 2 mod 6, imagine that the

cloning steps described above were skipped. That is, the cloned queens in the left column and

the bottom row were removed from the board and the queen from where they came was placed

back on the board. Likewise, the cloned queens in the right column and the top row disappear

and the queen from where they originated reappears. In order to confirm that these remaining n

– 2 queens are independent, we need only check that no queen from the left half of the board

occupies the same negative diagonal as a queen from the right half of the board. The rth queen

on the left side of the board occupies square (r + 1, 2r + 1) while the sth queen on the right side

19

Image 15: Eight Independent Queens on 8x8 Board Using

Cloning Construction


of the board occupies square (k + s, 2s). If two queens lie on the same negative diagonal, then (r

+ 1) + (2r + 1) = (k + s) + (2s), that is, 3r + 2 = 3s + k. Since n = 2k, this equation resolves to n =

6*(r – s) + 4, which implies that n ≡ 4 mod 6, contradicting the fact that n ≡ 2 mod 6. Therefore,

these n – 2 queens are independent [1, pp. 172-173].

The only piece that remains to show is that the introduction of the cloned queens does not

compromise the independence of the n queens. It suffices to check that the cloned queens do not

lie on the same negative diagonal as one of the other queens on the board. The clone on the left

edge is situated at (1, n – 3) and the rth queen on the left side of the board is located at (r + 1, 2r

+ 1). So if these two queens lie on the same negative diagonal, then (1) + (n – 3) = (r + 1) + (2r

+ 1), which simplifies to n – 2 = 3r + 2. Therefore, n = 3r + 4 ≡ 1 mod 3, contradicting the fact n

≡ 2 mod 6. If the cloned queen at (1, n – 3) lies on the same diagonal as an sth queen on the right

side of the board, then n – 2 = (k + s) + (2s) = 3s + k. Since n = 2k, we have that n – 2 = 3s +

½*n, which implies that n = 6s + 4 ≡ 4 mod 6, another contradiction. Now the cloned queen on

the bottom row is at square (k – 1, 1), and if this queen lies on the same negative diagonal as the

rth queen on the left half of the board, then (k – 1) + (1) = (r + 1) + (2r + 1). This simplifies to k

= 3r + 2, or equivalently n = 6r + 4, which, again, contradicts the fact that n ≡ 2 mod 6. Also, if

this cloned queen lies on the same negative diagonal as the sth queen on the right half of the

board, then k = 3s + k, which implies that s = 0, which is nonsense, since s is a positive integer.

This fact makes intuitive sense since the queen is at the bottom of the board on the left side and

so cannot possibly share a negative diagonal with any queen on the right side of the board.

Lastly, a symmetric argument works for the cloned queen on the right side of the board and the

one on the top row [1, pp. 173-174].

Now the proof that β(Qnxn) = n for all even n ≥ 4 is complete. This makes the proof for

odd-sized boards is easy. The construction we used for even boards, whether n ≡ 2 mod 6 or not,

leaves the main positive diagonal unoccupied. Therefore, for any even n, we simply add a row to

the top and a column to the right of the board and place a queen in the upper-right hand corner to

produce a set of n + 1 independent queens on this new (n + 1)x(n + 1) board. Image 16 shows

this construction for a 9x9 board, with the additional row and column shaded in orange. This

completes the longwinded proof [1, p. 174].

20


As a bonus, it has been shown that β(Qmxn) = min(m, n) for any mxn rectangular board

with the exception of the aforementioned 2x2 and 3x3 cases [3, p. 9].

The n -queens Problem

Now we turn our attention to the n-queens problem, that is, how many different ways can

one place n independent queens on an nxn board? While we have a value for β(Qnxn) for all

values of n, the n-queens problem itself remains far from solved. In Mathematical Recreations,

Maurice Kraitchik provides the number of solutions of arrangements of n independent queens for

small values of n and classifies these solutions based on their level of symmetry, as follows [1,

pp. 174-175]:

Ordinary solutions (O): solutions with no symmetry and thus yield a total of 8

solutions under rotation and reflection

Centrosymmetric solutions (C): solutions that are unchanged under rotations of

180° but are changed by other rotations or reflections and thus yield a total of 4

solutions

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Image 16: Nine IndependentQueens on 9x9 Board Using

Clone Construction


Doubly Centrosymmetric solutions (Q): solutions that are unchanged under any

rotation but are changed by reflections and yield a total of 2 solutions

An example of an ordinary solution to the n-queens problem is the 1 5 8 6 3 7 2 4

solution to the 8-queens problem in Image 12. A centrosymmetric solution is the 5 3 1 7 2 8 6 4

solution in which we used the cloning construction in Image 15. Finally, the 2 5 3 1 4 solution

to the 5-queens problem is doubly centrosymmetric and is shown in Image 17 [1, pp. 175].

Kraitchik identified the number of solutions to the n-queens problem by symmetry type

for 4 ≤ n ≤ 15, as shown in Table 5. Per the definitions above, the total number of solutions for

each value of n is 8On + 4Cn + 2Qn. As can be inferred from the fact that the number of solutions

to the problem for n = 13, 14 and 15 are unknown, much work still needs to be done to solve this

puzzle. Also worth pointing out is that the number of solutions decreases dramatically as the

level of symmetry increases. For example, there are a whopping 1,765 distinct ordinary

solutions to the 12-queens problem, but only 18 centrosymmetric solutions and just 1 doubly

centrosymmetric solution [1, pp. 175].

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Image 17: DoublyCentrosymmetric Solutionto the 5-queens Problem


Table 5: Solutions to the n-queens Problem by Symmetry Type for 4 ≤ n ≤ 15

n On Cn Qn Total

4 0 0 1 25 1 0 1 106 0 1 0 47 4 2 0 408 11 1 0 929 42 4 0 34210 89 3 0 72411 329 12 0 2,68012 1,765 18 1 14,20013 ? 31 1 ?14 ? 103 0 ?15 ? 298 0 ?

Conclusion

Much like domination, independence as a whole remains unsolved. Although most of the

independence number formulas for the different pieces have been verified on both square and

rectangular boards, the n-queens problem continues to befuddle mathematicians. That being

said, one could argue, based on Tables 6 and 7 below, that we are closer to resolving the

independence problem than we are to answering the questions that remain regarding domination.

I would tend to agree with this sentiment, as we don’t even have a formula for the queens

domination number yet, whereas not only do we have one for independence, but it is about as

simple as it gets (simply being n for square boards in almost all cases).

Table 6: Domination Number Formulas by Piece

Piece (P) γ ( P nxn) (Square) γ ( P mxn) (Rectangular)

Rook n min(m, n)Bishop n Unknown

King └(n + 2) / 3┘2 └(m + 2) / 3┘*└(n + 2) /

3┘Knight Unknown Unknown

QueenUnknown, though upper and lower bounds exist

Unknown

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Table 7: Independence Number Formulas by Piece

Piece (P) β ( P nxn) (Square) β ( P mxn) (Rectangular)

Rook n min(n, m)Bishop 2n – 2 UnknownKing └½*(n + 1)┘

2└½*(m + 1)┘*└½*(n + 1)┘

Knight4 if n = 2;

½*n2 if n ≥ 4, n even;½*(n2 + 1) if n odd

n if m = 1, for all n;2*(┌n / 4┐ + ┌(n – 1) / 4┐)

if m = 2, for all n;┌½*mn┐ if m, n ≥ 3

Queen1 if n = 2;2 if n = 3;

n for all other n

1 if m = n = 2;2 if m = n = 3;

min(m, n) for all other m, n

I do not see someone coming up with an ultimate formula for the number of permutations

of n independent queens on an nxn board anytime in the near future. I am not sure if that will

ever transpire, but I believe that as computers become more and more powerful, we will be able

to see more trends and patterns in large chessboards and hopefully be able to develop upper and

lower bounds for βPerm(Qnxn). In addition, studying the n-queens problem on rectangular boards,

something I have not come across in my research, would certainly be worthwhile in an attempt to

better understand this elusive puzzle.

In my next paper in this series, I will turn my attention to a completely different notion in

mathematical chessboard puzzles known as the knight’s tour. This is a classic problem and

studying it will help us realize the diversity of chessboard problems and how it relates to other

seemingly unrelated concepts in mathematics.

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Sources Cited

[1] J.J. Watkins. Across the Board: The Mathematics of Chessboard Problems. Princeton, New

Jersey: Princeton University Press, 2004.

[2] J. DeMaio, W.P. Faust. Domination on the mxn Toroidal Chessboard by Rooks and Bishops.

Department of Mathematics and Statistics, Kennesaw State University.

[3] S. Wagon. Hex Graphs. Department of Mathematics, Statistics and Computer Science,

Macalester College.

[4] “Chess.” Wikipedia, Wikimedia Foundation. http://en.wikipedia.org/wiki/Chess

[5] A. Tamid. “Math Forum Discussions.” http://mathforum.org/kb/message.jspa?

messageID=363988

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