The eight queens puzzle is the problem of putting eight chess queens on an 8×8 chessboard such that none of them is able to capture any other using the standard chess queen's moves. The colour of the queens is meaningless in this puzzle, and any queen is assumed to be able to attack any 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 puzzle of placing n queens on an n×n chessboard.
This puzzle appeared in the popular early 1990s computer game, The 7th Guest.
For n = 8 this results in the solution shown above. A few more examples follow.
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 solutions, which are presented below:
The following table gives the number of solutions for n queens, both unique and distinct .
| n: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| unique: | 1 | 0 | 0 | 1 | 2 | 1 | 6 | 12 | 46 | 92 | 341 | 1,787 | 9,233 | 45,752 | 285,053 |
| distinct: | 1 | 0 | 0 | 2 | 10 | 4 | 40 | 92 | 352 | 724 | 2,680 | 14,200 | 73,712 | 365,596 | 2,279,184 |
Note that the 6 queens puzzle has, interestingly, fewer solutions than the 5 queens puzzle!
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, amongst 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 slightly 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. For example, the breadth-first search program below examines only 15,720 possible queen placements by constructing 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.
Constraint programming is even more effective on this problem. An 'iterative repair' algorithm 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 an 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 solves 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 column such that it conflicts with the smallest number of queens already on the board.
Note that 'iterative repair', unlike the 'breadth-first' search outlined above, does not guarantee a solution: like all hillclimbing 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 breadth-first search.
The Python functions below can generate all solutions for an n-queens problem, using a recursive breadth-first search combined with the hard-coded insights that :
# Return a list of solutions to the n-queens problem on an # n-by-width board. A solved board is expressed as a list of # column positions for queens, indexed by row. # Rows and columns are indexed from zero. def n_queens(n, width): if n == 0: return # one solution, the empty list else: return add_queen(n-1, width, n_queens(n-1, width)) # Try all ways of adding a queen to a column of row new_row, returning # a list of solutions. previous_solutions must be a list of new_row-queens # solutions. def add_queen(new_row, width, previous_solutions): solutions = * for sol in previous_solutions: # Try to place a queen on each column on row new_row. for new_col in range(width): # print 'trying', new_col, 'on row', new_row if safe_queen(new_row, new_col, sol): # No interference, so add this solution to the list. solutions.append(sol + *) return solutions # Is it safe to add a queen to sol at (new_row, new_col)? Return # true if so. sol must be a solution to the new_row-queens problem. def safe_queen(new_row, new_col, sol): # Check against each piece on each of the new_row existing rows. for row in range(new_row): if (sol* == new_col or # same column clash sol* + row == new_col + new_row or # diagonal clash sol* - row == new_col - new_row): # other diagonal return 0 return 1 for sol in n_queens(8, 8): print sol
The constraint logic programming (over finite domains) approach to this kind of problem is very efficient. The GNU Prolog program below resolved a 100 queens problem in less than a tenth of a second. It finds a permutation of the first n naturals such that the distance between any two is not the normal distance (for example, 1 is normally three away from 4).
/* Generates a list which represents a single solution with the specified length and ensures that the list has each value from 1 to N once and only once. */ nqueens(N,Ls) :- length(Ls,N), fd_domain(Ls,1,N), fd_all_different(Ls), constraint_queens(Ls), fd_labeling(Ls,*). /* Ensures that all positions are valid */ constraint_queens(*). constraint_queens(*) :- noattack(X,Xs,1), constraint_queens(Xs). /* Ensures that no queens share diagonals */ noattack(_,*,_). noattack(X,*,N) :- X#\=Y+N, X#\=Y-N, T=N+1, noattack(X,Xs,T).
The following J verb generates all solutions for an n-queens problem, using an iterative approach.
queens=: 3 : 0 z=.i.n,*n=.y. for. }.z do. b=. -. (i.n) e."1 ,. z +"1 _ ((-i.){:$z) */ _1 0 1 z=. ((+/"1 b)#z),.(,b)#(*/$b)$i.n end. )
A k-arrangement x has k queens on a k-by-n board where none of queens attack each other. To generate all k+1-arrangements leading from x, place a queen on all the places on row k which are not on the any of the columns or diagonals attacked by the queens in x. The 1-arrangements are 0, 1, 2, ..., n-1; the n-arrangements are all the solutions to the n-queens problem.
For example, 0 2, 0 3, and 0 4 are valid 2-arrangements for the 8-queens problem. The 3-arrangements leading from these are: 0 2 4 0 2 5 0 2 6 0 2 7 0 3 1 0 3 5 0 3 6 0 3 7 0 4 1 0 4 6 0 4 7
8-dronningeproblemet | Damenproblem | Problema de las n damas | Problème des huit dames | Rompicapo delle otto regine | რვა ლაზიერის ამოცანა | エイト・クイーン | Problem 8 hetmanów | Problem osmih dam | Проблем осам дама | ปัญหาควีนแปดตัว
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It uses material from the
"Eight queens puzzle".
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