ref: 5ec91a6413dbf87a9fc0aa3e8038d81d2054bb0a
dir: /sys/src/cmd/python/Demo/scripts/queens.py/
#! /usr/bin/env python
"""N queens problem.
The (well-known) problem is due to Niklaus Wirth.
This solution is inspired by Dijkstra (Structured Programming). It is
a classic recursive backtracking approach.
"""
N = 8 # Default; command line overrides
class Queens:
def __init__(self, n=N):
self.n = n
self.reset()
def reset(self):
n = self.n
self.y = [None]*n # Where is the queen in column x
self.row = [0]*n # Is row[y] safe?
self.up = [0] * (2*n-1) # Is upward diagonal[x-y] safe?
self.down = [0] * (2*n-1) # Is downward diagonal[x+y] safe?
self.nfound = 0 # Instrumentation
def solve(self, x=0): # Recursive solver
for y in range(self.n):
if self.safe(x, y):
self.place(x, y)
if x+1 == self.n:
self.display()
else:
self.solve(x+1)
self.remove(x, y)
def safe(self, x, y):
return not self.row[y] and not self.up[x-y] and not self.down[x+y]
def place(self, x, y):
self.y[x] = y
self.row[y] = 1
self.up[x-y] = 1
self.down[x+y] = 1
def remove(self, x, y):
self.y[x] = None
self.row[y] = 0
self.up[x-y] = 0
self.down[x+y] = 0
silent = 0 # If set, count solutions only
def display(self):
self.nfound = self.nfound + 1
if self.silent:
return
print '+-' + '--'*self.n + '+'
for y in range(self.n-1, -1, -1):
print '|',
for x in range(self.n):
if self.y[x] == y:
print "Q",
else:
print ".",
print '|'
print '+-' + '--'*self.n + '+'
def main():
import sys
silent = 0
n = N
if sys.argv[1:2] == ['-n']:
silent = 1
del sys.argv[1]
if sys.argv[1:]:
n = int(sys.argv[1])
q = Queens(n)
q.silent = silent
q.solve()
print "Found", q.nfound, "solutions."
if __name__ == "__main__":
main()