亲宝软件园·资讯

展开

python实现数独算法 python实现数独算法实例

不吃皮蛋 人气:0
想了解python实现数独算法实例的相关内容吗,不吃皮蛋在本文为您仔细讲解python实现数独算法的相关知识和一些Code实例,欢迎阅读和指正,我们先划重点:python,数独算法,下面大家一起来学习吧。

本文实例讲述了python实现数独算法的方法。分享给大家供大家参考。具体如下:

# -*- coding: utf-8 -*-
'''
Created on 2012-10-5
@author: Administrator
'''
from collections import defaultdict
import itertools
a = [
  [ 0, 7, 0, 0, 0, 0, 0, 0, 0], #0
  [ 5, 0, 3, 0, 0, 6, 0, 0, 0], #1
  [ 0, 6, 2, 0, 8, 0, 7, 0, 0], #2
  #
  [ 0, 0, 0, 3, 0, 2, 0, 5, 0], #3
  [ 0, 0, 4, 0, 1, 0, 3, 0, 0], #4
  [ 0, 2, 0, 9, 0, 5, 0, 0, 0], #5
  #
  [ 0, 0, 1, 0, 3, 0, 5, 9, 0], #6
  [ 0, 0, 0, 4, 0, 0, 6, 0, 3], #7
  [ 0, 0, 0, 0, 0, 0, 0, 2, 0], #8
#  0, 1, 2, 3,|4, 5, 6,|7, 8
  ]
#a = [
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #0
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #1
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #2
#  #
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #3
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #4
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #5
#  #
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #6
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #7
#  [0, 0, 0, 0, 0, 0, 0, 0, 0], #8
##  0, 1, 2, 3,|4, 5, 6,|7, 8
#  ]
exists_d = dict((((h_idx, y_idx), v) for h_idx, y in enumerate(a) for y_idx , v in enumerate(y) if v))
h_exist = defaultdict(dict)
v_exist = defaultdict(dict)
for k, v in exists_d.items():
 h_exist[k[ 0]][k[ 1]] = v
 v_exist[k[ 1]][k[ 0]] = v
aa = list(itertools.permutations(range(1, 10), 9))
h_d = {}
for hk, hv in h_exist.items():
 x = filter(lambda x:all((x[k] == v for k, v in hv.items())), aa)
 x = filter(lambda x:all((x[vk] != v for vk , vv in v_exist.items() for k, v in vv.items() if k != hk)), x)
# print x
 h_d[hk] = x
def test(x, y):
 return all([y[i] not in [x_[i] for x_ in x] for i in range(len(y)) ])
def test2(x):
 return len(set(x)) != 9
s = set(range(9))
sudokus = []
for l0 in h_d[0 ]:
 for l1 in h_d[ 1]:
  if not test((l0,), l1):
   continue
  for l2 in h_d[ 2]:
   if not test((l0, l1), l2):
    continue
   # 1,2,3行 进行验证
   if test2([l0[ 0], l0[ 1], l0[ 2]
      , l1[ 0], l1[ 1], l1[ 2]
      , l2[ 0], l2[ 1], l2[ 2]
      ]) : continue   
   if test2([l0[ 3], l0[ 4], l0[ 5]
      , l1[ 3], l1[ 4], l1[ 5]
      , l2[ 3], l2[ 4], l2[ 5]
      ]) : continue   
   if test2([l0[ 6], l0[ 7], l0[ 8]
      , l1[ 6], l1[ 7], l1[ 8]
      , l2[ 6], l2[ 7], l2[ 8]
      ]) : continue   
   for l3 in h_d[ 3]:
    if not test((l0, l1, l2), l3):
     continue
    for l4 in h_d[ 4]:
     if not test((l0, l1, l2, l3), l4):
      continue
     for l5 in h_d[ 5]:
      if not test((l0, l1, l2, l3, l4), l5):
       continue
      # 4,5,6行 进行验证
      if test2([l3[ 0], l3[ 1], l3[ 2]
         , l4[ 0], l4[ 1], l4[ 2]
         , l5[ 0], l5[ 1], l5[ 2]
         ]) : continue   
      if test2([l3[ 3], l3[ 4], l3[ 5]
         , l4[ 3], l4[ 4], l4[ 5]
         , l5[ 3], l5[ 4], l5[ 5]
         ]) : continue   
      if test2([l3[ 6], l3[ 7], l3[ 8]
         , l4[ 6], l4[ 7], l4[ 8]
         , l5[ 6], l5[ 7], l5[ 8]
         ]) : continue   
      for l6 in h_d[ 6]:
       if not test((l0, l1, l2, l3, l4, l5,), l6):
        continue
       for l7 in h_d[ 7]:
        if not test((l0, l1, l2, l3, l4, l5, l6), l7):
         continue
        for l8 in h_d[ 8]:
         if not test((l0, l1, l2, l3, l4, l5, l6, l7), l8):
          continue
         # 7,8,9行 进行验证
         if test2([l6[ 0], l6[ 1], l6[ 2]
            , l7[0 ], l7[1 ], l7[2 ]
            , l8[0 ], l8[1 ], l8[2 ]
            ]) : continue   
         if test2([l6[ 3], l6[ 4], l6[ 5]
            , l7[3 ], l7[4 ], l7[5 ]
            , l8[3 ], l8[4 ], l8[5 ]
            ]) : continue   
         if test2([l6[ 6], l6[ 7], l6[ 8]
            , l7[6 ], l7[7 ], l7[8 ]
            , l8[6 ], l8[7 ], l8[8 ]
            ]) : continue   
         print l0
         print l1
         print l2
         print l3
         print l4
         print l5
         print l6
         print l7
         print l8
         sudokus.append((l0, l1, l2, l3, l4, l5, l6, l7, l8))

希望本文所述对大家的Python程序设计有所帮助。

加载全部内容

相关教程
猜你喜欢
用户评论