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A solution to the classic N queens problem.
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#! /usr/bin/env python
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"""N queens problem.
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The (well-known) problem is due to Niklaus Wirth.
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This solution is inspired by Dijkstra (Structured Programming). It is
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a classic recursive backtracking approach.
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"""
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N = 8 # Default; command line overrides
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class Queens:
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def __init__(self, n=N):
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self.n = n
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self.reset()
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def reset(self):
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n = self.n
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self.y = [None]*n # Where is the queen in column x
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self.row = [0]*n # Is row[y] safe?
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self.up = [0] * (2*n-1) # Is upward diagonal[x-y] safe?
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self.down = [0] * (2*n-1) # Is downward diagonal[x+y] safe?
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self.nfound = 0 # Instrumentation
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def solve(self, x=0): # Recursive solver
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for y in range(self.n):
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if self.safe(x, y):
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self.place(x, y)
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if x+1 == self.n:
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self.display()
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else:
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self.solve(x+1)
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self.remove(x, y)
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def safe(self, x, y):
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return not self.row[y] and not self.up[x-y] and not self.down[x+y]
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def place(self, x, y):
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self.y[x] = y
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self.row[y] = 1
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self.up[x-y] = 1
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self.down[x+y] = 1
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def remove(self, x, y):
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self.y[x] = None
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self.row[y] = 0
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self.up[x-y] = 0
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self.down[x+y] = 0
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silent = 0 # If set, count solutions only
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def display(self):
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self.nfound = self.nfound + 1
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if self.silent:
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return
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print '+-' + '--'*self.n + '+'
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for y in range(self.n-1, -1, -1):
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print '|',
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for x in range(self.n):
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if self.y[x] == y:
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print "Q",
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else:
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print ".",
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print '|'
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print '+-' + '--'*self.n + '+'
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def main():
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import sys
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silent = 0
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n = N
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if sys.argv[1:2] == ['-n']:
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silent = 1
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del sys.argv[1]
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if sys.argv[1:]:
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n = int(sys.argv[1])
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q = Queens(n)
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q.silent = silent
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q.solve()
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print "Found", q.nfound, "solutions."
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if __name__ == "__main__":
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main()
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