本文實例講述了python實現數獨算法的方法。分享給大家供大家參考。具體如下：

# -*- 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 進行驗證

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 進行驗證

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 進行驗證

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))

希望本文所述對大家的Python程序設計有所幫助。

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