python实现数独算法实例
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-*-
'''
Createdon2012-10-5
@author:Administrator
'''
fromcollectionsimportdefaultdict
importitertools
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)forh_idx,yinenumerate(a)fory_idx,vinenumerate(y)ifv))
h_exist=defaultdict(dict)
v_exist=defaultdict(dict)
fork,vinexists_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={}
forhk,hvinh_exist.items():
x=filter(lambdax:all((x[k]==vfork,vinhv.items())),aa)
x=filter(lambdax:all((x[vk]!=vforvk,vvinv_exist.items()fork,vinvv.items()ifk!=hk)),x)
#printx
h_d[hk]=x
deftest(x,y):
returnall([y[i]notin[x_[i]forx_inx]foriinrange(len(y))])
deftest2(x):
returnlen(set(x))!=9
s=set(range(9))
sudokus=[]
forl0inh_d[0]:
forl1inh_d[1]:
ifnottest((l0,),l1):
continue
forl2inh_d[2]:
ifnottest((l0,l1),l2):
continue
#1,2,3行进行验证
iftest2([l0[0],l0[1],l0[2]
,l1[0],l1[1],l1[2]
,l2[0],l2[1],l2[2]
]):continue
iftest2([l0[3],l0[4],l0[5]
,l1[3],l1[4],l1[5]
,l2[3],l2[4],l2[5]
]):continue
iftest2([l0[6],l0[7],l0[8]
,l1[6],l1[7],l1[8]
,l2[6],l2[7],l2[8]
]):continue
forl3inh_d[3]:
ifnottest((l0,l1,l2),l3):
continue
forl4inh_d[4]:
ifnottest((l0,l1,l2,l3),l4):
continue
forl5inh_d[5]:
ifnottest((l0,l1,l2,l3,l4),l5):
continue
#4，5，6行进行验证
iftest2([l3[0],l3[1],l3[2]
,l4[0],l4[1],l4[2]
,l5[0],l5[1],l5[2]
]):continue
iftest2([l3[3],l3[4],l3[5]
,l4[3],l4[4],l4[5]
,l5[3],l5[4],l5[5]
]):continue
iftest2([l3[6],l3[7],l3[8]
,l4[6],l4[7],l4[8]
,l5[6],l5[7],l5[8]
]):continue
forl6inh_d[6]:
ifnottest((l0,l1,l2,l3,l4,l5,),l6):
continue
forl7inh_d[7]:
ifnottest((l0,l1,l2,l3,l4,l5,l6),l7):
continue
forl8inh_d[8]:
ifnottest((l0,l1,l2,l3,l4,l5,l6,l7),l8):
continue
#7，8，9行进行验证
iftest2([l6[0],l6[1],l6[2]
,l7[0],l7[1],l7[2]
,l8[0],l8[1],l8[2]
]):continue
iftest2([l6[3],l6[4],l6[5]
,l7[3],l7[4],l7[5]
,l8[3],l8[4],l8[5]
]):continue
iftest2([l6[6],l6[7],l6[8]
,l7[6],l7[7],l7[8]
,l8[6],l8[7],l8[8]
]):continue
printl0
printl1
printl2
printl3
printl4
printl5
printl6
printl7
printl8
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。
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