总体思路 ：
数独九行九列，一个list装一行，也就需要一个嵌套两层的list
初始会有很多数字，我可不想一个一个赋值
那就要想办法偷懒啦
然后再是穷举，如何科学的穷举
第一部分：录入
某在线数独网站的截图
要想办法，把它方便的变成嵌套的list
我的解决办法：
手打到Excel里面
然后另存为csv文件
然后就当做txt读取
l = None
with open('数独.csv','r',encoding = 'utf-8') as f:
l = f.readlines()
print(l)
'''
运行结果
---------------------------------------
['\ufeff3,,,1,,8,4,,\n', ',,1,,,2,,3,\n', '4,,,,,,1,6,\n', ',5,8,,,9,,,4\n', ',3,,,5,,,,9\n', ',9,,3,,,5,,\n', ',,3,9,1,,,4,\n', '2,,7,5,,,9,,\n', '9,,,,4,,,5,3\n']
'''
发现文本开头有个莫名其妙的\ufeff，另外它的长度是1
>>> s = '\ufeff3'
>>> len(s)
2
只好加一句 l[0] = l[0][1:]
然后去掉末尾的\n 再以逗号为界切割
l = None
with open('数独.csv','r',encoding = 'utf-8') as f:
l = f.readlines()
l[0] = l[0][1:]
l = map(lambda i : i.rstrip(),l)
l = map(lambda i : i.split(","),l)
for i in l:
print(i,'---', len(i))
'''
运行结果
---------------------------------------
['3', '', '', '1', '', '8', '4', '', ''] --- 9
['', '', '1', '', '', '2', '', '3', ''] --- 9
['4', '', '', '', '', '', '1', '6', ''] --- 9
['', '5', '8', '', '', '9', '', '', '4'] --- 9
['', '3', '', '', '5', '', '', '', '9'] --- 9
['', '9', '', '3', '', '', '5', '', ''] --- 9
['', '', '3', '9', '1', '', '', '4', ''] --- 9
['2', '', '7', '5', '', '', '9', '', ''] --- 9
['9', '', '', '', '4', '', '', '5', '3'] --- 9
'''
九行九列..完美
下一步全部处理成数字
鉴于int()无法将空字符串转化为0 所以需要新定义一个new_int
def new_int(s):
return int(s) if s else 0
l = None
with open('数独.csv','r',encoding = 'utf-8') as f:
l = f.readlines()
l[0] = l[0][1:]
l = map(lambda i : i.rstrip(),l)
l = map(lambda i : i.split(","),l)
l = [ list(map(new_int, i)) for i in l]
for i in l:
print(i)
'''
运行结果
---------------------------------------
[3, 0, 0, 1, 0, 8, 4, 0, 0]
[0, 0, 1, 0, 0, 2, 0, 3, 0]
[4, 0, 0, 0, 0, 0, 1, 6, 0]
[0, 5, 8, 0, 0, 9, 0, 0, 4]
[0, 3, 0, 0, 5, 0, 0, 0, 9]
[0, 9, 0, 3, 0, 0, 5, 0, 0]
[0, 0, 3, 9, 1, 0, 0, 4, 0]
[2, 0, 7, 5, 0, 0, 9, 0, 0]
[9, 0, 0, 0, 4, 0, 0, 5, 3]
'''
第二部分 穷举
假设81个格子有50是空的，每个格子1-9 9种可能
>>> 9**50
515377520732011331036461129765621272702107522001
显然不能傻乎乎的直接遍历
其实一个新格子并不是1-9 9种可能
它不可能是同行，同列，同区出现过的数字
这里将会用到set的加减
x,y = 0,1
whole = {1,2,3,4,5,6,7,8,9}
x_set = set(l[x])
#行
y_set = { l[i][y] for i in range(9) }
#列
block_num = big_small[(x,y)]
#查字典得到区号
block_set = { l[i][j] for i , j in small_big[block_num] }
#根据区号查该区的9个方格,然后根据位置构建set
possible = whole - x_set - y_set - block_set
下面补充下big_small和small_big两个字典
3x3的小区共9个 分别编号上0-8
0 | 1 | 2
3 | 4 | 5
6 | 7 | 8
原来在9x9的 x行y列 对应过去 就会在x//3行y//3列
对应编号就是x//3*3 + y//3
为了方便后面的使用，建立一个字典
big_small = { (x,y): (x//3)*3+(y//3) for x in range(9) for y in range(9)}
'''
---------------------------
>>> big_small[(5,5)]
4
'''
这个字典是{位置:区号}
然后反着来一下
就可以根据 区号 查包含位置的字典（这才是重点）
big_small = { (x,y): (x//3)*3+(y//3) for x in range(9) for y in range(9)}
small_big = { x:[ ] for x in range(9)}
for i , j in big_small.items():
small_big[ j ].append(i)
for i,j in small_big.items():
print(i,'-->',j)
'''
----------------------------------------------------
0 --> [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
1 --> [(0, 3), (0, 4), (0, 5), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)]
2 --> [(0, 6), (0, 7), (0, 8), (1, 6), (1, 7), (1, 8), (2, 6), (2, 7), (2, 8)]
3 --> [(3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (5, 0), (5, 1), (5, 2)]
4 --> [(3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5), (5, 3), (5, 4), (5, 5)]
5 --> [(3, 6), (3, 7), (3, 8), (4, 6), (4, 7), (4, 8), (5, 6), (5, 7), (5, 8)]
6 --> [(6, 0), (6, 1), (6, 2), (7, 0), (7, 1), (7, 2), (8, 0), (8, 1), (8, 2)]
7 --> [(6, 3), (6, 4), (6, 5), (7, 3), (7, 4), (7, 5), (8, 3), (8, 4), (8, 5)]
8 --> [(6, 6), (6, 7), (6, 8), (7, 6), (7, 7), (7, 8), (8, 6), (8, 7), (8, 8)]
'''
下面列下总体框架(???表示还没确定的细节)
#伪代码
l = ???
#读取文件获得未完成的数独
all_list = [l]
#这个变量用于装 待处理的数独
key_list = []
#装正确的解
while all_list:
one = all_list.pop()
#从list末尾取出一个进行处理
x, y, ed = ???(one)
#这个函数找一个未填写的格子(值为0)
#x,y将接受格子的位置,
#ed接受一个逻辑值,以处理格子全被填满的特殊情况
if ed :
key_list.append(one)
for i in one:
print(i)
#输出,保存解
continue
possible = ???(x,y,one)
#获取格子可能的数字
for i in possible:
new_one = copy.deepcopy(one)
#深度拷贝one
new_one[x][y] = i
all_list.append(new_one)
#修改副本,并加入待处理list
只有一个函数不够清晰,
def output_cell(l):
for i in range(9):
for j in range(9):
if l[ i ][ j ] :
pass
else:
return i , j ,False
else:
return None, None,True
全部的代码，
import copy
def new_int(s):
return int(s) if s else 0
def output_cell(l):
for i in range(9):
for j in range(9):
if l[ i ][ j ] :
pass
else:
return i , j ,False
else:
return None, None,True
def possible_num(x,y,l):
whole = {1,2,3,4,5,6,7,8,9}
x_set = set(l[x])
y_set = { l[i][y] for i in range(9) }
block_num = big_small[(x,y)]
block_set = { l[i][j] for i , j in small_big[block_num] }
possible = whole - x_set - y_set - block_set
return possible
big_small = { (x,y): (x//3)*3+(y//3) for x in range(9) for y in range(9)}
small_big = { x:[ ] for x in range(9)}
for i , j in big_small.items():
small_big[ j ].append(i)
l = None
with open('数独.csv','r',encoding = 'utf-8') as f:
l = f.readlines()
l[0] = l[0][1:]
l = map(lambda i : i.rstrip(),l)
l = map(lambda i : i.split(","),l)
l = [ list(map(new_int, i)) for i in l]
all_list = [l]
key_list = []
while all_list:
one = all_list.pop()
x, y, ed = output_cell(one)
if ed :
key_list.append(one)
for i in one:
print(i)
continue
possible = possible_num(x,y,one)
for i in possible:
new_one = copy.deepcopy(one)
new_one[x][y] = i
all_list.append(new_one)
else:
print('遍历结束')
print(len_num)
运行结果
'''
---------------------------
[3, 7, 5, 1, 6, 8, 4, 9, 2]
[8, 6, 1, 4, 9, 2, 7, 3, 5]
[4, 2, 9, 7, 3, 5, 1, 6, 8]
[1, 5, 8, 6, 2, 9, 3, 7, 4]
[7, 3, 4, 8, 5, 1, 6, 2, 9]
[6, 9, 2, 3, 7, 4, 5, 8, 1]
[5, 8, 3, 9, 1, 6, 2, 4, 7]
[2, 4, 7, 5, 8, 3, 9, 1, 6]
[9, 1, 6, 2, 4, 7, 8, 5, 3]
'''
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