之前學習xlrd和xlwt模塊的時候,突發奇想,能不能用Python寫個求解數獨的小程序呢?然後找找資料,在實驗樓上發現了現成的例子,然後花了一晚上,改動了一部分,自己寫了一遍。(這裏的數獨並不滿足對角線原則,試了下,把對角線的要求加進去,通過這種方法生成一個數獨太耗時間了,你們要是有興趣,可以改進下這裏)
直接上程序吧,下次有空過來整理整理。
import random
import itertools
from copy import deepcopy
test = [[None, None, 3, None, 6, 9, None, 1, 7],
[None, 1, None, 2, 4, 8, None, None, 9],
[4, None, 9, 7, 1, None, None, 2, 6],
[9, 7, 2, 1, 8, 4, 6, 3, 5],
[1, None, 8, None, 5, 2, 7, 9, 4],
[6, 4, 5, 9, 3, 7, None, None, None],
[None, 9, None, 4, None, None, 2, None, 8],
[None, 6, None, 8, None, 1, None, 4, 3],
[None, None, None, None, None, None, 5, 7, 1]]
'''
+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++
&&& || || 3 &&& || 6 || 9 &&& || 1 || 7 &&&
+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++
&&& || 1 || &&& 2 || 4 || 8 &&& || || 9 &&&
+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++
&&& 4 || || 9 &&& 7 || 1 || &&& || 2 || 6 &&&
+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++
&&& 9 || 7 || 2 &&& 1 || 8 || 4 &&& 6 || 3 || 5 &&&
+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++
&&& 1 || || 8 &&& || 5 || 2 &&& 7 || 9 || 4 &&&
+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++
&&& 6 || 4 || 5 &&& 9 || 3 || 7 &&& || || &&&
+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++
&&& || 9 || &&& 4 || || &&& 2 || || 8 &&&
+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++
&&& || 6 || &&& 8 || || 1 &&& || 4 || 3 &&&
+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++
&&& || || &&& || || &&& 5 || 7 || 1 &&&
+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++'''
def full_board():
board = None
while board is None:
board = attmpt_board()
return board
def attmpt_board():
nums = list(range(1,10))
board = [[None for _ in range(9)] for _ in range(9)]
for i,j in itertools.product(range(9),repeat=2):#這個函數可以將兩層甚至多層循環簡寫一下
i0,j0 = i - i % 3,j - j % 3
random.shuffle(nums)
for x in nums:
if(x not in board[i]
and all(x != row[j] for row in board)
and all(x not in row[j0:j0 + 3] for row in board[i0:i])):
board[i][j] = x
break
else:
return None
return board
def get_board(full_board,level=1): #難度等級
board = deepcopy(full_board)
omit = [0,35,60,81] #挖出的方塊數,因爲隨機挖的時候可能重複挖到某個,所以一般小於該數
for _ in range(omit[level]):
i = random.randint(0,8) #隨機取0——8中的一個數
j = random.randint(0,8)
board[i][j] = None
return board
def print_board(board):
spacer = '+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++'
for i in range(9):
if i % 3 ==0:
print(spacer.replace('-','&'))
else:
print(spacer)
print('&&& {} || {} || {} &&& {} || {} || {} &&& {} || {} || {} &&&'
.format(*(cell or ' ' for cell in board[i])))
print(spacer.replace('-','&'))
def isfull(board):
for i,j in itertools.product(range(9),repeat=2):
if board[i][j] == None:
return False
return True
def list_cell(board,x,y): #返回某個空格可以填入的數字的列表
nums = set(range(1,10))
row = set(board[x]) #所在的行已有的數字
col = set(row[y] for row in board) #列
x0,y0 = x - x % 3,y - y % 3
block = set(board[i][j] for i,j in itertools.product(range(x0,x0+3),range(y0,y0+3))) #塊
result = nums - row - col - block
return list(result)
def min_pos(board): #返回最短的可填列表所在的座標
min_,x,y = 9,-1,-1
for i,j in itertools.product(range(9),repeat=2):
if board[i][j] == None:
temp = len(list_cell(board,i,j))
if temp < min_:
min_ = temp
x,y = i,j
return x,y
def next_pos(x,y): #返回下個位置的座標
pos = 9 * x + y
x = ((pos+1) % 81)//9
y = (pos+1) % 9
return x,y
def solve_board(board):
if isfull(board):
return board
while True:
x,y = min_pos(board)
nums = list_cell(board,x,y)
if len(nums) == 1:
board[x][y] = nums[0]
else:
break
x,y = min_pos(board)
result = core_fun(board,x,y)
return result
def core_fun(board,x,y):
if isfull(board):
return board
while board[x][y]:
x,y = next_pos(x,y)
numbers = list_cell(board,x,y)
if len(numbers) == 0:
return False
for num in numbers:
board[x][y] = num
flag = core_fun(board,next_pos(x,y)[0],next_pos(x,y)[1])#一直往下遞歸
if flag == False: #遞歸到某個空任何數都不能填進去的時候
board[x][y] = None #因爲上一步操作是讓board[x][y]=n,表明賦值爲n是不對的。接着往下走,取numbers中的下一個數
else:
return board #board不斷的在改變,直到求解出答案
#這裏不好理解,這表明numbers中的所有數據用完了,還不對,怎麼可能呢?
#通過上面的numbers = list_cell(board,x,y)算出來的numbers肯定是對的啊。所以肯定到不了這一步啊
#其實吧,因爲這裏有遞歸,當前層的numbers是在上一層選了一個數字之後得出的,有可能上一層選的數字就不對
#所以這裏試將上一層的那個錯誤的數置爲空
board[x][y] = None
return False
FullBoard = full_board()
print('fullboard:\n')
print_board(FullBoard)
print()
my_board = get_board(FullBoard,3)
print('my_board:\n')
print_board(my_board)
result = solve_board(my_board)
print('\nanswer:\n')
print_board(result)
運行截圖
(可見,數獨的答案可能並不唯一,下次試試求解所有的滿足要求的答案)