Python基于生成器迭代实现的八皇后问题示例

Python  /  管理员 发布于 7年前   149

本文实例讲述了Python基于生成器迭代实现的八皇后问题。分享给大家供大家参考，具体如下：

问题：有一个棋盘和8个要放到上面的皇后，唯一的要求是皇后之间不能形成威胁。也就是说，必须把他们防止成每个皇后都不能吃掉其他皇后的状态。

# -*- coding: utf-8 -*-#python 2.7.13__metaclass__ = typedef confict(state, nextX):  nextY = len(state)  for i in range(nextY):    if abs(state[i] - nextX) in (0, nextY - i):      return True  return Falsedef queens(num=8, state=()):  for pos in range(num):    if not confict(state, pos):      if len(state) == num -1:        yield (pos,)      else:        for result in queens(num, state + (pos,)):          yield (pos,) + resultprint list(queens()) #打印输出

运行结果：

[(0, 4, 7, 5, 2, 6, 1, 3), (0, 5, 7, 2, 6, 3, 1, 4), (0, 6, 3, 5, 7, 1, 4, 2), (0, 6, 4, 7, 1, 3, 5, 2), (1, 3, 5, 7, 2, 0, 6, 4), (1, 4, 6, 0, 2, 7, 5, 3), (1, 4, 6, 3, 0, 7, 5, 2), (1, 5, 0, 6, 3, 7, 2, 4), (1, 5, 7, 2, 0, 3, 6, 4), (1, 6, 2, 5, 7, 4, 0, 3), (1, 6, 4, 7, 0, 3, 5, 2), (1, 7, 5, 0, 2, 4, 6, 3), (2, 0, 6, 4, 7, 1, 3, 5), (2, 4, 1, 7, 0, 6, 3, 5), (2, 4, 1, 7, 5, 3, 6, 0), (2, 4, 6, 0, 3, 1, 7, 5), (2, 4, 7, 3, 0, 6, 1, 5), (2, 5, 1, 4, 7, 0, 6, 3), (2, 5, 1, 6, 0, 3, 7, 4), (2, 5, 1, 6, 4, 0, 7, 3), (2, 5, 3, 0, 7, 4, 6, 1), (2, 5, 3, 1, 7, 4, 6, 0), (2, 5, 7, 0, 3, 6, 4, 1), (2, 5, 7, 0, 4, 6, 1, 3), (2, 5, 7, 1, 3, 0, 6, 4), (2, 6, 1, 7, 4, 0, 3, 5), (2, 6, 1, 7, 5, 3, 0, 4), (2, 7, 3, 6, 0, 5, 1, 4), (3, 0, 4, 7, 1, 6, 2, 5), (3, 0, 4, 7, 5, 2, 6, 1), (3, 1, 4, 7, 5, 0, 2, 6), (3, 1, 6, 2, 5, 7, 0, 4), (3, 1, 6, 2, 5, 7, 4, 0), (3, 1, 6, 4, 0, 7, 5, 2), (3, 1, 7, 4, 6, 0, 2, 5), (3, 1, 7, 5, 0, 2, 4, 6), (3, 5, 0, 4, 1, 7, 2, 6), (3, 5, 7, 1, 6, 0, 2, 4), (3, 5, 7, 2, 0, 6, 4, 1), (3, 6, 0, 7, 4, 1, 5, 2), (3, 6, 2, 7, 1, 4, 0, 5), (3, 6, 4, 1, 5, 0, 2, 7), (3, 6, 4, 2, 0, 5, 7, 1), (3, 7, 0, 2, 5, 1, 6, 4), (3, 7, 0, 4, 6, 1, 5, 2), (3, 7, 4, 2, 0, 6, 1, 5), (4, 0, 3, 5, 7, 1, 6, 2), (4, 0, 7, 3, 1, 6, 2, 5), (4, 0, 7, 5, 2, 6, 1, 3), (4, 1, 3, 5, 7, 2, 0, 6), (4, 1, 3, 6, 2, 7, 5, 0), (4, 1, 5, 0, 6, 3, 7, 2), (4, 1, 7, 0, 3, 6, 2, 5), (4, 2, 0, 5, 7, 1, 3, 6), (4, 2, 0, 6, 1, 7, 5, 3), (4, 2, 7, 3, 6, 0, 5, 1), (4, 6, 0, 2, 7, 5, 3, 1), (4, 6, 0, 3, 1, 7, 5, 2), (4, 6, 1, 3, 7, 0, 2, 5), (4, 6, 1, 5, 2, 0, 3, 7), (4, 6, 1, 5, 2, 0, 7, 3), (4, 6, 3, 0, 2, 7, 5, 1), (4, 7, 3, 0, 2, 5, 1, 6), (4, 7, 3, 0, 6, 1, 5, 2), (5, 0, 4, 1, 7, 2, 6, 3), (5, 1, 6, 0, 2, 4, 7, 3), (5, 1, 6, 0, 3, 7, 4, 2), (5, 2, 0, 6, 4, 7, 1, 3), (5, 2, 0, 7, 3, 1, 6, 4), (5, 2, 0, 7, 4, 1, 3, 6), (5, 2, 4, 6, 0, 3, 1, 7), (5, 2, 4, 7, 0, 3, 1, 6), (5, 2, 6, 1, 3, 7, 0, 4), (5, 2, 6, 1, 7, 4, 0, 3), (5, 2, 6, 3, 0, 7, 1, 4), (5, 3, 0, 4, 7, 1, 6, 2), (5, 3, 1, 7, 4, 6, 0, 2), (5, 3, 6, 0, 2, 4, 1, 7), (5, 3, 6, 0, 7, 1, 4, 2), (5, 7, 1, 3, 0, 6, 4, 2), (6, 0, 2, 7, 5, 3, 1, 4), (6, 1, 3, 0, 7, 4, 2, 5), (6, 1, 5, 2, 0, 3, 7, 4), (6, 2, 0, 5, 7, 4, 1, 3), (6, 2, 7, 1, 4, 0, 5, 3), (6, 3, 1, 4, 7, 0, 2, 5), (6, 3, 1, 7, 5, 0, 2, 4), (6, 4, 2, 0, 5, 7, 1, 3), (7, 1, 3, 0, 6, 4, 2, 5), (7, 1, 4, 2, 0, 6, 3, 5), (7, 2, 0, 5, 1, 4, 6, 3), (7, 3, 0, 2, 5, 1, 6, 4)]

输出列表长度：

print len(list(queens()))# 输出：92

更多关于Python相关内容感兴趣的读者可查看本站专题：《Python数学运算技巧总结》、《Python数据结构与算法教程》、《Python函数使用技巧总结》、《Python字符串操作技巧汇总》、《Python入门与进阶经典教程》及《Python文件与目录操作技巧汇总》

希望本文所述对大家Python程序设计有所帮助。