Chessboard Puzzles Part 2 - Independence

Chessboard PuzzlesPart 2: Independence

Dan Freeman

March 20, 2014

Villanova UniversityMAT 9000 Graduate Math Seminar

Introduction• Last presentation, we looked at the concept

of chessboard domination• Tonight we will examine a similar but

different idea called independence• I will answer some questions from last time

and then jump right into the new material• Reminders of how pieces move will be

given as we come across them

Chess Starting Positions

King

Queen

Rook

Bishop

Knight

Pawn

Independence Defined• 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 chessboard is the maximum number of independent pieces that can be placed on the board

• Independence numbers are denoted by β(Pmxn) where P represents the type of chess piece (see legend to the right) and m and n are the number of rows and columns of the board, respectively

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

King – K

Queen – Q

Rook – R

Bishop – B

Knight – N

Independence Number Notation

Rooks Independence• Recall that rooks move horizontally and

vertically any number of squares• Independence among rooks is the simplest

of all chess pieces• As with domination, for a square nxn

chessboard, the rooks independence number is simply n. The proof is 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 diagonal is an independent set since no two rooks share the same row or the same column. As a result, β(Rnxn) = n.

Rook Permutations• The number of permutations of maximum

independent sets of rooks is n!. The proof 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. 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!.

Rook Permutations• The 3! = 6 permutations of independent

rooks are shown below for a 3x3 board

Bishops Independence

• Recall that bishops move diagonally any number of squares

• Unlike with domination, the formula is different: β(Bnxn) = 2n – 2

• The proof is straightforward:– For any nxn chessboard, there are 2n – 1

positive (negative) diagonals. Since bishops in opposite corners attack each other, β(Bnxn) is at most (2n – 1) – 1 = 2n – 2.

– One may place n – 1 bishops along the top row and n – 1 bishops along the bottom row for a total of 2n – 2 bishops that are independent. This is shown on the next slide for n = 8.

– Therefore, β(Bnxn) = 2n – 2.

Illustration that β(Bnxn) = 2n – 2

2n – 1 DiagonalsMaximum Set of

Independent Bishops

Bishop Permutations• There are 2n possible arrangements of 2n –

2 independent bishops on an nxn board• The proof goes as follows:

– For any corner square, either that square contains a bishop or the opposite corner contains a bishop for 2 possible arrangements for each column that contains a corner square, for a total of 22 = 4 arrangements.

– For any non-corner square in the top row, either that square contains a bishop, the two squares on the left and right edge diagonally from the top square do not contain a bishop and the bottom square diagonal to the side edges does contain a bishop or vice versa. Thus, there are 2 possible arrangements of bishops for each of the n – 2 non-corner columns, for a total of 2n – 2

arrangements.– In all, there are 22*2n – 2 = 2n arrangements.

Illustration that There Are 2n Bishop Permutations

Bishops on the Top and Bottom

Bishops on the Left and Right

Kings Independence• Recall that kings are allowed to move

exactly one square in any direction

• β(Knxn) = └½*(n + 1)┘2. Proof:

– 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.

– 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 + ½*(n – 1) + 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

function, as follows: β(Knxn) = └½*(n + 1)┘2.

Maximum Sets of Independent Kings on 8x8 and 9x9 Boards

Knights Independence• Recall that knights move two squares in one

direction (either horizontally or vertically) and one square in the other direction

• Unlike with knights domination, a formula is known for the knights independence number:– β(N2x2) = 4.

– For even n ≥ 4, β(Nnxn) = ½*n2.

– For odd n, β(Nnxn) = ½*(n2 + 1).

• I will omit the proof since it involves a concept we won’t explore until the third presentation known as a knight’s tour

Queens Independence

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

• In stark contrast to queens domination, queens independence has a known formula and a very simple one too:– β(Q2x2) = 1, β(Q3x3) = 2.

– For n ≠ 2 or 3, β(Qnxn) = n.

• The proof requires considering a number of different cases and so is omitted

The 8-queens Problem• The 8-queens problem asks the question:

How many different ways can one place 8 independent queens on an 8x8 board?

• In 1953, Maurice Kraitchik provided all 92 solutions to the 8-queens problem– Each solution is given as a string of 8 numbers,

where the value of each number represents the row number and the position of each number represents the column number.

The 8-queens Problem• Gauss used this permutation notation to

study the 8-queens problem• He noticed that in the permutation 1 5 8 6 3

7 2 4, the sum of the row and column numbers for each queen were all distinct and the sum of the row and the reverse column numbers were all distinct– The first fact tells us that no two queens lie on

the same negative diagonal and the second fact tells us that no two queens lie on the same positive diagonal.

The 8-queens Problem

Row 1 5 8 6 3 7 2 4Column 1 2 3 4 5 6 7 8Sum 2 7 11 10 8 13 9 12

The 1 5 8 6 3 7 2 4 SolutionSum of Row and Column Numbers

Row 1 5 8 6 3 7 2 4Column 8 7 6 5 4 3 2 1Sum 9 12 14 11 7 10 4 5

Sum of Row and Reverse Column Numbers

The n-queens Problem• The n-queens problem asks the more

general question: How many different ways can one place n independent queens on an nxn board?

• Kraitchik classified solutions to the problem based on their level of symmetry:– 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

– Doubly Centrosymmetric solutions (Q): solutions that are unchanged under any rotation but are changed by reflections and yield a total of 2 solutions

The n-queens Problem• The definitions on the previous slide tell us

that there are 8On + 4Cn + 2Qn total solutions to the n-queens problem

• Kraitchilk identified the number of solutions for 4 ≤ n ≤ 15 in the table below

n On Cn Qn Total4 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 ?

Sources Cited• J.J. Watkins. Across the Board: The

Mathematics of Chessboard Problems. Princeton, New Jersey: Princeton University Press, 2004.

• “Chess.” Wikipedia, Wikimedia Foundation. http://en.wikipedia.org/wiki/Chess