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进一步优化代码，性能提高50% ~ 7KB

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# -*- coding: utf-8 -*-

'''

Created on 2012-10-5

@author: Administrator

'''

from collections import defaultdict

import itertools

import time

import gc

import operator

import profile

gc.disable()

a = [

[ 8, 0, 0, 0, 0, 0, 0, 0, 0 ],

[ 0, 0, 3, 6, 0, 0, 0, 0, 0 ],

[ 0, 7, 0, 0, 9, 0, 2, 0, 0 ],

[ 0, 5, 0, 0, 0, 7, 0, 0, 0 ],

[ 0, 0, 0, 0, 4, 5, 7, 0, 0 ],

[ 0, 0, 0, 1, 0, 6, 0, 3, 0 ],

[ 0, 0, 1, 0, 0, 0, 0, 6, 8 ],

[ 0, 0, 8, 5, 0, 0, 0, 1, 0 ],

[ 0, 9, 0, 0, 0, 0, 4, 0, 0 ]

# 0, 1, 2, 3,|4, 5, 6,|7, 8

]

#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 = [

# [0,0,2,4,5,0,7,0,0]

# ,[0,4,0,0,0,8,0,3,0]

# ,[8,0,1,0,0,3,5,0,6]

# ,[0,5,3,0,0,0,0,0,4]

# ,[7,0,0,0,0,0,0,0,2]

# ,[2,0,0,0,0,0,6,7,0]

# ,[3,0,6,5,0,0,1,0,7]

# ,[0,2,0,1,0,0,0,5,0]

# ,[0,0,7,0,2,9,4,0,0]

# ]

#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

# ]

#===============================================================================

# 得到坐标点和值，相当于稀疏矩阵:(X,Y):value

#===============================================================================

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)

#===============================================================================

# 二维数组

#===============================================================================

for k, v in exists_d.iteritems():

h_exist[k[0]][k[1]] = v

v_exist[k[1]][k[0]] = v

#===============================================================================

# 生成所有的组合

#===============================================================================

permutations = tuple(itertools.permutations(range(1, 10), 9))

h_d = {}

now = time.time()

#===============================================================================

# 取得行号与该行符合条件的组合

#===============================================================================

for hk, hv in h_exist.iteritems():

#===========================================================================

# 过滤横向，x轴与同行已知的点的值重复的组合

#===========================================================================

q = filter(lambda x:all((x[k] == v

for k, v in hv.iteritems())), permutations)

#===========================================================================

# 过滤纵向，Y轴与同列已知的点的值重复的组合

#===========================================================================

q = filter(lambda x:all((x[vk] != v

for vk , vv in v_exist.iteritems()

for k, v in vv.iteritems()

if k != hk)), q)

h_d[hk] = q

#===============================================================================

# 不全某行没有任何提示

#===============================================================================

for line_idx in range(0, 9):

if line_idx not in h_d:

h_d[line_idx] = permutations

print time.time() - now

def filter_by_column(y, resource_dict):

'''@summary: 纵向过滤重复的值

@param y: 待检测的一行排列

@param resource_dict: 列序号及组合在对应列号的值列表

'''

if not resource_dict:

return 1

return all((y[k] not in v for k, v in resource_dict.iteritems()))

sudokus = []

def check_3_lines(f, s, t):

'''@summary: 检测3行的每三列是不是符合条件

@param f: first line

'''

if len(set((f[0], f[1], f[2], s[0], s[1], s[2], t[0], t[1], t[2]))) != 9:

return 1

if len(set((f[3], f[4], f[5], s[3], s[4], s[5], t[3], t[4], t[5]))) != 9:

return 1

if len(set((f[6], f[7], f[8], s[6], s[7], s[8], t[6], t[7], t[8]))) != 9:

return 1

return 0

def check_2_lines(f, s):

'''@summary: 检测两行数据的每三列是不是有重复值，提前淘汰有重复值的组合

@param f: first line

'''

if len(set((f[0], f[1], f[2], s[0], s[1], s[2]))) != 6:

return 1

if len(set((f[3], f[4], f[5], s[3], s[4], s[5]))) != 6:

return 1

if len(set((f[6], f[7], f[8], s[6], s[7], s[8]))) != 6:

return 1

return 0

def solve_sudoku(h_d, h_idx=None, reserves=None

, solves=None, resource_dict=None):

'''

@param reserves: 已经验证过符合条件的排列

@param solves: 最终的解决方案集合

@param resource_dict: dict key in range(1,10),values 是每行同一列的数字的列表

'''

if solves is None:

solves = []

if h_idx is None :

h_idx = 0

for l0 in h_d[h_idx]:

if reserves == None:

_reserves = [l0, ]

solve_sudoku(h_d, h_idx + 1, _reserves, solves)

else:

if not filter_by_column(l0, resource_dict):

continue

if h_idx in (1 , 4, 7):

if check_2_lines(reserves[h_idx - 1], l0):

continue

elif h_idx in (2, 5, 8) :

if check_3_lines(reserves[h_idx - 2], reserves[h_idx - 1], l0):

continue

_reserves = list(reserves)

_reserves.append(l0)

if h_idx < 8:

solve_sudoku(h_d, h_idx + 1, _reserves, solves

, dict((idx, set([i[idx] for i in _reserves]))

for idx in range(0, 9)))

if h_idx == 8:

solves.append(_reserves)

print u"calc No. {num} result".format(num=len(solves))

print u"*" * 50

for i in solves[-1]:

print i

else:

if h_idx == 0:

return solves

if __name__ == '__main__':

#(8, 7, 9, 1, 2, 3, 4, 6, 5)

#(5, 4, 3, 7, 9, 6, 2, 1, 8)

#(1, 6, 2, 5, 8, 4, 7, 3, 9)

#(7, 1, 8, 3, 4, 2, 9, 5, 6)

#(9, 5, 4, 6, 1, 8, 3, 7, 2)

#(3, 2, 6, 9, 7, 5, 8, 4, 1)

#(6, 8, 1, 2, 3, 7, 5, 9, 4)

#(2, 9, 7, 4, 5, 1, 6, 8, 3)

#(4, 3, 5, 8, 6, 9, 1, 2, 7)

now = time.time()

x = solve_sudoku(h_d)

print time.time() - now

# profile.run(u"solve_sudoku(h_d)", filename=r"d:\test.txt")

# import pstats

# p = pstats.Stats(r"d:\test.txt")

# p.sort_stats("time").print_stats()