# Eight queens puzzle

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Eight queens puzzle
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
One solution to the eight queens puzzle

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens attack each other. Thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general n-queens problem of placing n queens on an n×n chessboard, where solutions exist for all natural numbers n with the exception of 2 and 3.[1]

## History

The puzzle was originally proposed in 1848 by the chess player Max Bezzel, and over the years, many mathematicians, including Gauss, have worked on this puzzle and its generalized n-queens problem. The first solutions were provided by Franz Nauck in 1850. Nauck also extended the puzzle to n-queens problem (on an n×n board—a chessboard of arbitrary size). In 1874, S. Günther proposed a method of finding solutions by using determinants, and J.W.L. Glaisher refined this approach.

Edsger Dijkstra used this problem in 1972 to illustrate the power of what he called structured programming. He published a highly detailed description of the development of a depth-first backtracking algorithm.2

## Constructing a solution

The problem can be quite computationally expensive as there are 4,426,165,368 (i.e., 64 choose 8) possible arrangements of eight queens on a 8×8 board, but only 92 solutions. It is possible to use shortcuts that reduce computational requirements or rules of thumb that avoids brute-force computational techniques. For example, just by applying a simple rule that constrains each queen to a single column (or row), though still considered brute force, it is possible to reduce the number of possibilities to just 16,777,216 (that is, 88) possible combinations. Generating the permutations that are solutions of the eight rooks puzzle[2] and then checking for diagonal attacks further reduces the possibilities to just 40,320 (that is, 8!). The following Python code uses this technique to calculate the 92 solutions:[3]

```from itertools import permutations

n = 8
cols = range(n)
for vec in permutations(cols):
if (n == len(set(vec[i]+i for i in cols))
== len(set(vec[i]-i for i in cols))):
print vec
```

These brute-force algorithms are computationally manageable for n = 8, but would be intractable for problems of n ≥ 20, as 20! = 2.433 * 1018. Advancements for this and other toy problems are the development and application of heuristics (rules of thumb) that yield solutions to the n queens puzzle at a small fraction of the computational requirements.

This heuristic solves N queens for any N ≥ 4. It forms the list of numbers for vertical positions (rows) of queens with horizontal position (column) simply increasing. N is 8 for eight queens puzzle.

1. If the remainder from dividing N by 6 is not 2 or 3 then the list is simply all even numbers followed by all odd numbers ≤ N
2. Otherwise, write separate lists of even and odd numbers (i.e. 2,4,6,8 - 1,3,5,7)
3. If the remainder is 2, swap 1 and 3 in odd list and move 5 to the end (i.e. 3,1,7,5)
4. If the remainder is 3, move 2 to the end of even list and 1,3 to the end of odd list (i.e. 4,6,8,2 - 5,7,9,1,3)
5. Append odd list to the even list and place queens in the rows given by these numbers, from left to right (i.e. a2, b4, c6, d8, e3, f1, g7, h5)

For N = 8 this results in the solution shown above. A few more examples follow.

• 14 queens (remainder 2): 2, 4, 6, 8, 10, 12, 14, 3, 1, 7, 9, 11, 13, 5.
• 15 queens (remainder 3): 4, 6, 8, 10, 12, 14, 2, 5, 7, 9, 11, 13, 15, 1, 3.
• 20 queens (remainder 2): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 1, 7, 9, 11, 13, 15, 17, 19, 5.
##### Solutions to the eight queens puzzle

The eight queens puzzle has 92 distinct solutions. If solutions that differ only by symmetry operations (rotations and reflections) of the board are counted as one, the puzzle has 12 unique (or fundamental) solutions.

A fundamental solution usually has 8 variants (including its original form) obtained by rotating 90, 180, or 270 degrees and then reflecting each of the four rotational variants in a mirror in a fixed position. However, should a solution be equivalent to its own 90 degree rotation (as happens to one solution with 5 queens on a 5x5 board) that fundamental solution will have only 2 variants. Should a solution be equivalent to its own 180 degree rotation it will have 4 variants. Of the 12 fundamental solutions to the problem with 8 Queens on an 8x8 board, exactly 1 is equal to its own 180 degree rotation, and none are equal to their 90 degree rotation. Thus the number of distinct solutions is 11*8 + 1*4 = 92.

The unique solutions are presented below:

 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 1
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 2
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 3
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 4
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 5
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 6
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 7
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 8
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 9
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 10
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 11
 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Unique solution 12

## Explicit solutions

Explicit solutions exist for placing n queens on an n × n board, requiring no combinatorial search whatsoever [4]. The explicit solutions exhibit stair-stepped patterns, as in the following examples for n = 8, 9 and 10:

 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Explicit solution for 8 queens
 a b c d e f g h i 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h i
Explicit solution for 9 queens
 a b c d e f g h i j 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h i j
Explicit solution for 10 queens

## Counting solutions

The following table gives the number of solutions for placing n queens on an n × n board, both unique (sequence A002562 in OEIS) and distinct (sequence A000170 in OEIS), for n=1–14, 24–26.

n: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 .. 24 25 26
unique: 1 0 0 1 2 1 6 12 46 92 341 1,787 9,233 45,752 .. 28,439,272,956,934 275,986,683,743,434 2,789,712,466,510,289
distinct: 1 0 0 2 10 4 40 92 352 724 2,680 14,200 73,712 365,596 .. 227,514,171,973,736 2,207,893,435,808,352 22,317,699,616,364,044

Note that the six queens puzzle has fewer solutions than the five queens puzzle.

There is currently no known formula for the exact number of solutions.

## Related problems

Using pieces other than queens
On an 8×8 board one can place 32 knights, or 14 bishops, 16 kings or eight rooks, so that no two pieces attack each other. Fairy chess pieces have also been substituted for queens. In the case of knights, an easy solution is to place one on each square of a given color, since they move only to the opposite color.
Costas array
In mathematics, a Costas array can be regarded geometrically as a set of n points lying on the squares of a nxn chessboard, such that each row or column contains only one point, and that all of the n(n − 1)/2 displacement vectors between each pair of dots are distinct. Thus, an order-n Costas array is a solution to an n-rooks puzzle.
Nonstandard boards
Pólya studied the n queens problem on a toroidal ("donut-shaped") board and showed that there is a solution on an n×n board if and only if n is not divisible by 2 or 3.[5] In 2009 Pearson and Pearson algorithmically populated three-dimensional boards (n×n×n) with n2 queens, and proposed that multiples of these can yield solutions for a four-dimensional version of the puzzle.
Domination
Given an n×n board, the domination number is the minimum number of queens (or other pieces) needed to attack or occupy every square. For n=8 the queen's domination number is 5.
Nine queens problem
Place nine queens and one pawn on an 8×8 board in such a way that queens don't attack each other. Further generalization of the problem (complete solution is currently unknown): given an n×n chess board and m > n queens, find the minimum number of pawns, so that the m queens and the pawns can be set up on the board in such a way that no two queens attack each other.
Queens and knights problem
Place m queens and m knights on an n×n board so that no piece attacks another.
Magic squares
In 1992, Demirörs, Rafraf, and Tanik published a method for converting some magic squares into n queens solutions, and vice versa.[6]
Latin squares
In an n×n matrix, place each digit 1 through n in n locations in the matrix so that no two instances of the same digit are in the same row or column.
Exact cover
Consider a matrix with one primary column for each of the n ranks of the board, one primary column for each of the n files, and one secondary column for each of the 4n-6 nontrivial diagonals of the board. The matrix has n2 rows: one for each possible queen placement, and each row has a 1 in the columns corresponding to that square's rank, file, and diagonals and a 0 in all the other columns. Then the n queens problem is equivalent to choosing a subset of the rows of this matrix such that every primary column has a 1 in precisely one of the chosen rows and every secondary column has a 1 in at most one of the chosen rows; this is an example of a generalized exact cover problem, of which sudoku is another example.

## The eight queens puzzle as an exercise in algorithm design

Finding all solutions to the eight queens puzzle is a good example of a simple but nontrivial problem. For this reason, it is often used as an example problem for various programming techniques, including nontraditional approaches such as constraint programming, logic programming or genetic algorithms. Most often, it is used as an example of a problem which can be solved with a recursive algorithm, by phrasing the n queens problem inductively in terms of adding a single queen to any solution to the problem of placing n−1 queens on an n-by-n chessboard. The induction bottoms out with the solution to the 'problem' of placing 0 queens on an 0-by-0 chessboard, which is the empty chessboard.

This technique is much more efficient than the naïve brute-force search algorithm, which considers all 648 = 248 = 281,474,976,710,656 possible blind placements of eight queens, and then filters these to remove all placements that place two queens either on the same square (leaving only 64!/56! = 178,462,987,637,760 possible placements) or in mutually attacking positions. This very poor algorithm will, among other things, produce the same results over and over again in all the different permutations of the assignments of the eight queens, as well as repeating the same computations over and over again for the different sub-sets of each solution. A better brute-force algorithm places a single queen on each row, leading to only 88 = 224 = 16,777,216 blind placements.

It is possible to do much better than this. One algorithm solves the eight rooks puzzle by generating the permutations of the numbers 1 through 8 (of which there are 8! = 40,320), and uses the elements of each permutation as indices to place a queen on each row. Then it rejects those boards with diagonal attacking positions. The backtracking depth-first search program, a slight improvement on the permutation method, constructs the search tree by considering one row of the board at a time, eliminating most nonsolution board positions at a very early stage in their construction. Because it rejects rook and diagonal attacks even on incomplete boards, it examines only 15,720 possible queen placements. A further improvement which examines only 5,508 possible queen placements is to combine the permutation based method with the early pruning method: the permutations are generated depth-first, and the search space is pruned if the partial permutation produces a diagonal attack. Constraint programming can also be very effective on this problem.

An alternative to exhaustive search is an 'iterative repair' algorithm, which typically starts with all queens on the board, for example with one queen per column. It then counts the number of conflicts (attacks), and uses a heuristic to determine how to improve the placement of the queens. The 'minimum-conflicts' heuristic — moving the piece with the largest number of conflicts to the square in the same column where the number of conflicts is smallest — is particularly effective: it finds a solution to the 1,000,000 queen problem in less than 50 steps on average. This assumes that the initial configuration is 'reasonably good' — if a million queens all start in the same row, it will obviously take at least 999,999 steps to fix it. A 'reasonably good' starting point can for instance be found by putting each queen in its own row and column so that it conflicts with the smallest number of queens already on the board.

Note that 'iterative repair', unlike the 'backtracking' search outlined above, does not guarantee a solution: like all hillclimbing (i.e., greedy) procedures, it may get stuck on a local optimum (in which case the algorithm may be restarted with a different initial configuration). On the other hand, it can solve problem sizes that are several orders of magnitude beyond the scope of a depth-first search.

## An animated version of the recursive solution

This animation uses backtracking to solve the problem. A queen is placed in a column that is known not to cause conflict. If a column is not found the program returns to the last good state and then tries a different column.

## Sample program

The following is a Pascal program by Niklaus Wirth.[7] It finds one solution to the eight queens problem.

```program eightqueen1(output);
var i : integer; q : boolean;
a : array[ 1 .. 8] of boolean;
b : array[ 2 .. 16] of boolean;
c : array[ -7 .. 7] of boolean;
x : array[ 1 .. 8] of integer;
procedure try( i : integer; var q : boolean);
var j : integer;
begin j := 0;
repeat j := j + 1; q := false;
if a[ j] and b[ i + j] and c[ i - j] then
begin x[ i] := j;
a[ j] := false; b[ i + j] := false; c[ i - j] := false;
if i < 8 then
begin try( i + 1, q);
if not q then
begin a[ j] := true; b[ i + j] := true; c[ i - j] := true;
end
end else q := true
end
until q or (j = 8);
end;
begin
for i := 1 to 8 do a[  i] := true;
for i := 2 to 16 do b[  i] := true;
for i := -7 to 7 do c[  i] := true;
try( 1, q);
if q then
for i := 1 to 8 do write( x[ i]:4);
writeln
end.
```

## References

1. ^ Hoffman, et al. "Construction for the Solutions of the m Queens Problem". Mathematics Magazine, Vol. XX (1969), p. 66-72. [1]
2. ^ Rooks Problem from Wolfram MathWorld
3. ^ Hettinger, Raymond. "Eight queens. Six lines.". Retrieved 25 June 2011.
4. ^ Explicit Solutions to the N-Queens Problem for all N, Bo Bernhardsson (1991), Department of Automatic Control, Land Institute of Technology, Sweden.
5. ^ G. Polya, Uber die “doppelt-periodischen” Losungen des n-Damen-Problems, George Polya: Collected papers Vol. IV, G-C. Rota, ed., MIT Press, Cambridge, London, 1984, pp. 237–247
6. ^ O. Demirörs, N. Rafraf, and M.M. Tanik. Obtaining n-queens solutions from magic squares and constructing magic squares from n-queens solutions. Journal of Recreational Mathematics, 24:272–280, 1992
7. ^ Wirth, 1976, p. 145
• Bell, Jordan and Stevens, Brett, A survey of known results and research areas for n-queens, Discrete Mathematics, Vol. 309, Issue 1, 6 January 2009, 1-31.
• Watkins, John J. (2004). Across the Board: The Mathematics of Chess Problems. Princeton: Princeton University Press. ISBN 0-691-11503-6.
• O.-J. Dahl, E. W. Dijkstra, C. A. R. Hoare Structured Programming, Academic Press, London, 1972 ISBN 0-12-200550-3 see pp 72–82 for Dijkstra's solution of the 8 Queens problem.
• Three Dimensional NxN-Queens Problems, L.Allison, C.N.Yee, & M.McGaughey (1988), Department of Computer Science, Monash University, Australia.
• S. Nudelman, The Modular N-Queens Problem in Higher Dimensions, Discrete Mathematics, vol 146, iss. 1-3, 15 November 1995, pp. 159–167.
• On The Modular N-Queen Problem in Higher Dimensions, Ricardo Gomez, Juan Jose Montellano and Ricardo Strausz (2004), Instituto de Matematicas, Area de la Investigacion Cientifica, Circuito Exterior, Ciudad Universitaria, Mexico.
• J. Barr and S. Rao, The n-Queens Problem in Higher Dimensions, Elemente der Mathematik, vol 61 (2006), pp. 133–137.
• Wirth, Niklaus (1976), Algorithms + Data Structures = Programs, Prentice-Hall, ISBN 0-13-022418-9

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