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

[ 5,3,6,#1

[ 0,2,8,#2

#

[ 0,5,#3

[ 0,4,1,#4

[ 0,9,#5

#

[ 0,#6

[ 0,3],#7

[ 0,#8

# 0,|4,|7,8

]

#a = [

# [0,#0

# [0,#1

# [0,#2

# #

# [0,#3

# [0,#4

# [0,#5

# #

# [0,#6

# [0,#7

# [0,#8

## 0,8

# ]

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

# 1,3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,l2,l3),l4):

continue

for l5 in h_d[ 5]:

if not test((l0,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,l4,l5,l6):

continue

for l7 in h_d[ 7]:

if not test((l0,l6),l7):

continue

for l8 in h_d[ 8]:

if not test((l0,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,l7,l8))

希望本文所述对大家的Python程序设计有所帮助。