Eight queens puzzle

Chess composer Max Bezzel published the eight queens puzzle in 1848. Franz Nauck published the first solutions in 1850. Nauck also extended the puzzle to the n queens problem, with n queens on a chessboard of n×n squares. Since then, many mathematicians, including Carl Friedrich Gauss, have worked on both the eight queens puzzle and its generalized n-queens version. In 1874, S. Günther proposed a method using determinants to find solutions. == Constructing and counting solutions when n = 8 ==

The problem of finding all solutions to the 8-queens problem can be quite computationally expensive, as there are 4,426,165,368 possible arrangements of eight queens on an 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, by applying a simple rule that chooses one queen from each column, it is possible to reduce the number of possibilities to 16,777,216 (that is, 88) possible combinations. Generating permutations further reduces the possibilities to just 40,320 (that is, 8!), which can then be checked for diagonal attacks. The eight queens puzzle has 92 distinct solutions. If solutions that differ only by the symmetry operations of rotation and reflection of the board are counted as one, the puzzle has 12 solutions. These are called fundamental solutions; representatives of each are shown below. A fundamental solution usually has eight variants (including its original form) obtained by rotating 90, 180, or 270° and then reflecting each of the four rotational variants in a mirror in a fixed position. However, one of the 12 fundamental solutions (solution 12 below) is identical to its own 180° rotation, so has only four variants (itself and its reflection, its 90° rotation and the reflection of that). Thus, the total number of distinct solutions is 11×8 + 1×4 = 92. All fundamental solutions are presented below: Solution 10 has the additional property that no three queens are in a straight line. ==Existence of solutions==

Brute-force algorithms to count the number of solutions are computationally manageable for n = 8 , but would be intractable for problems of n \geq 20 , as 20! = 2.433 × 1018. If the goal is to find a single solution, one can show solutions exist for all n ≥ 4 with no search whatsoever. These solutions exhibit stair-stepped patterns, as in the following examples for n = 8, 9 and 10: The examples above can be obtained with the following formulas. Let (i, j) be the square in column i and row j on the n × n chessboard, k an integer. One approach is • 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 not greater than n. • Otherwise, write separate lists of even and odd numbers (2, 4, 6, 8 – 1, 3, 5, 7). • If the remainder is 2, swap 1 and 3 in odd list and move 5 to the end (3, 1, 7, 5). • If the remainder is 3, move 2 to the end of even list and 1,3 to the end of odd list (4, 6, 8, 2 – 5, 7, 9, 1, 3). • Append odd list to the even list and place queens in the rows given by these numbers, from left to right (a2, b4, c6, d8, e3, f1, g7, h5). For this results in fundamental solution 1 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. ==Counting solutions for other sizes n==

Exact enumeration There is no known formula for the exact number of solutions for placing n queens on an board i.e. the number of independent sets of size n in an queen's graph. The 27×27 board is the highest-order board that has been completely enumerated. The following tables give the number of solutions to the n queens problem, both fundamental and all , for all known cases. The number of placements in which furthermore no three queens lie on any straight line is known for n \leq 25 . Asymptotic enumeration In 2021, Michael Simkin proved that for large numbers n, the number of solutions of the n queens problem is approximately (0.143n)^n. More precisely, the number \mathcal{Q}(n) of solutions has asymptotic growth \mathcal{Q}(n) = ((1 \pm o(1))ne^{-\alpha})^n where \alpha is a constant that lies between 1.939 and 1.945. (Here o(1) represents little o notation.) If one instead considers a toroidal chessboard (where diagonals "wrap around" from the top edge to the bottom and from the left edge to the right), it is only possible to place n queens on an n \times n board if n \equiv 1,5 \mod 6. In this case, the asymptotic number of solutions is T(n) = ((1+o(1))ne^{-3})^n. ==Related problems==

;Higher dimensions :Find the number of non-attacking queens that can be placed in a d-dimensional chess of size n. More than n queens can be placed in some higher dimensions (the smallest example is four non-attacking queens in a 3×3×3 chess space), and it is in fact known that for any k, there are higher dimensions where nk queens do not suffice to attack all spaces. ;Using pieces other than queens :On an 8×8 board one can place 32 knights, or 14 bishops, 16 kings or 8 rooks, so that no two pieces attack each other. 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. The solution is also easy for rooks and kings. Sixteen kings can be placed on the board by dividing it into 2-by-2 squares and placing the kings at equivalent points on each square. Placements of n rooks on an n×n board are in direct correspondence with order-n permutation matrices. ;Chess variations :Related problems can be asked for chess variations such as shogi. For instance, the n+k dragon kings problem asks to place k shogi pawns and n+k mutually nonattacking dragon kings on an n×n shogi board. ;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. ;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. ;Queens and other pieces :Variants include mixing queens with other pieces; for example, placing m queens and m knights on an n×n board so that no piece attacks another or placing queens and pawns so that no two queens attack each other. ;Magic squares :In 1992, Demirörs, Rafraf, and Tanik published a method for converting some magic squares into n-queens solutions, and vice versa. ;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. ; n-queens completion :The completion problem asks whether, given an n×n chessboard on which some queens are already placed, it is possible to place a queen in every remaining row so that no two queens attack each other. This and related questions are NP-complete and #P-complete. Any placement of at most n/60 queens can be completed, while there are partial configurations of roughly n/4 queens that cannot be completed. ==Exercise in algorithm design==

to solve the problem. The ith queen is placed in the ith row of the chessboard, 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. 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 that 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×n chessboard. The induction bottoms out with the solution to the 'problem' of placing 0 queens on the chessboard, which is the empty chessboard. This process can be made more efficient by applying the observation that each column of the chessboard must contain exactly one queen, and restricting the positions at which the algorithm tries adding the ith queen to the ith row of the chessboard, in columns that are not already attacked by other queens. These choices significantly reduce the number of placements that need to be tried, allowing the algorithm to find all solutions on an 8\times 8 chessboard in a time so short as to appear instantaneous. solution to 8 queens 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 easily finds a solution to even the 1,000,000 queens problem. Unlike the backtracking search outlined above, iterative repair does not guarantee a solution: like all greedy procedures, it may get stuck on a local optimum. (In such a 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. As an alternative to backtracking, solutions can be counted by recursively enumerating valid partial solutions, one row at a time. Rather than constructing entire board positions, blocked diagonals and columns are tracked with bitwise operations. This does not allow the recovery of individual solutions. ==Sample program==

The following program is a translation of Niklaus Wirth's solution into the Python programming language, but does without the index arithmetic found in the original and instead uses lists to keep the program code as simple as possible. By using a coroutine in the form of a generator function, both versions of the original can be unified to compute either one or all of the solutions. Only 15,720 possible queen placements are examined. def queens(n: int, i: int, a: list, b: list, c: list): if i The following program is an implementation of Donald Knuth's informal description of the solution on Page 31, Section 7.2.2 Backtrack Programming from The Art of Computer Programming, Volume 4B into the Python programming language. def property(perm: list) -> bool: for k in range(0, len(perm)): for j in range(0, len(perm)): if j int: domain = list(range(0, n)) perm = for i in range(n): new_perm = list(filter(property, extend(perm, n))) perm = new_perm return len(perm) ==In popular culture==

• In the game The 7th Guest, the 8th Puzzle: "The Queen's Dilemma" in the game room of the Stauf mansion is the de facto eight queens puzzle. • In the game Professor Layton and the Curious Village, the 130th puzzle: "Too Many Queens 5" () is an eight queens puzzle. ==See also==