python实现自动解数独小程序

#编辑数独题目，将题目输入列表中
question = [6,0,7,0,0,0,9,0,3,
  0,0,8,0,0,7,0,0,0,
  3,0,0,0,8,2,0,7,5,
  0,1,2,3,0,5,0,0,0,
  0,0,6,0,0,0,5,0,0,
  0,0,0,4,0,6,7,1,0,
  2,6,0,7,4,0,0,0,8,
  0,0,0,8,0,0,6,0,0,
  7,0,5,0,0,0,1,0,9]
 
# print(question[0])
 
#返回当前数独为0的空格中所有可能取值
def count_candidate_number(j):
 exist_all_number = [] #当前横竖大方格内所有出现的数字集
 candidate_number = [] #该方格内所有的数字候选集
 SD_Row = int(j) // 9 #行
 SD_Column = int(j) % 9 #列
 
 #用迭代器写
 exist_all_number_part1 = [question[i+SD_Row*9] for i in range(9)] #横-出现的所有数字集
 exist_all_number_part2 = [question[i*9+SD_Column] for i in range(9)] #竖-出现的所有数字集
 exist_all_number_part3 = [question[((j//9)//3)*27+((j % 9)//3)*3+i] for i in range(3)]+[question[((j//9)//3)*27+((j % 9)//3)*3+9+i] for i in range(3)]+[question[((j//9)//3)*27+((j % 9)//3)*3+18+i] for i in range(3)] #大方块-出现的所有数字集
 exist_all_number = list(set(exist_all_number_part1+exist_all_number_part2+exist_all_number_part3))  #对出现所有的数字集组合及去重
 # print(exist_all_number)
 
 #用循环写
 # for i in range(9):
 # if question[i+SD_Row*9] not in exist_all_number:
 #  exist_all_number.append(question[i+SD_Row*9])
 # if question[i*9 + SD_Cloumn] not in exist_all_number:
 #  exist_all_number.append(question[i*9 + SD_Cloumn])
 # # print(exist_all_number)
 
 #迭代器写
 candidate_number = [i for i in range(1, 10) if i not in exist_all_number] #对可能取值进行迭代输出
 
 #用循环写
 # for i in range(1,10):
 # if i not in exist_all_number:
 #  candidate_number.append(i)
 # print(candidate_number)
 
 return candidate_number

#对数组中每个为0的空格列出所有可能的取值数集，并放置于字典中
def all_possible_candidate_number():
 all_possible_candidate_number = {i:count_candidate_number(i) for i in range(81) if question[i] == 0}
 return all_possible_candidate_number
 # print(all_possible_candidate_number)

def main_count():
 answer_sudoku = question
 candidate_number_dic = {}
 while True:
 candidate_number_dic = all_possible_candidate_number() #在每次循环之前刷当前每个为0的空格，所有的取值集合
 if candidate_number_dic == {}:    #如果为空，则证明没有为0的空格，则为求解
  answer_sudoku = question    #对answer_sudoku赋值，并打印
  print("已求解",answer_sudoku)
  break
 else:
  for eachkey,eachValue in candidate_number_dic.items(): #对字典中位数为1的取值集合，既确定该数字变为当前应取值
  if len(eachValue) == 1:
   answer_sudoku[eachkey] = eachValue[0]
   print(eachkey,eachValue[0])   #打印对应键值及对应数值
  pass
 
if __name__ == '__main__':
 main_count()

D:\pythonwokr\venv\Scripts\python.exe D:/pythonwokr/数独.py
已求解 [6, 2, 7, 5, 1, 4, 9, 8, 3, 5, 4, 8, 9, 3, 7, 2, 6, 1, 3, 9, 1, 6, 8, 2, 4, 7, 5, 4, 1, 2, 3, 7, 5, 8, 9, 6, 9, 7, 6, 1, 2, 8, 5, 3, 4, 8, 5, 3, 4, 9, 6, 7, 1, 2, 2, 6, 9, 7, 4, 1, 3, 5, 8, 1, 3, 4, 8, 5, 9, 6, 2, 7, 7, 8, 5, 2, 6, 3, 1, 4, 9]
 
Process finished with exit code 0