目录
- 简介
- 自动解数独思路
- 核心思路
- 输入
- 解析
- 打印
- 完整代码
简介
最近玩上了一款类似于数独的微信小程序游戏，名字叫数独趣味闯关，过了数独的关卡之后会给拼图，玩了几关之后摸清套路了就有点累了，但是还想集齐拼图，所以就编了个程序自动解数独。
自动解数独思路
核心思路
核心思路就是尝试所有排列组合，记录哪些位置能够确定是【□】或者【x】。例如下图所示，假设共有8个格，该行前数字为【5,1】。则有三种组合方式，以此能推断出位置【1，2，3，4】肯定为【□】，其他为待定，逐行逐列循环以得到所有位置的填充方式。
输入
目前输入只能将每行/列前数字手动输入，为了输入方便，单个数字以整形直接输入，多个数字以字符串形式输入，用空格分割。如下图所示为输入样例
row = [3, 5, 6, 9, 10, 12, '5 3 3', '10 1', '3 5 1', '2 4 1 1', '3 1 3 1', '3 1 1', '3 1', '2 1', 11]
col = [6, 8, '4 5', '5 1', '8 1', '10 1', '6 3 1', '11 1', '9 1', '6 1', '3 1 1', '3 3 1', '3 1 1', '2 1', 7]
row = [[int(i) for i in s.split()] if type(s) is str else [s] for s in row]
col = [[int(i) for i in s.split()] if type(s) is str else [s] for s in col]
解析
- 对于一个新棋盘，逐行逐列扫描，以得到每个确定的位置。对于不同个数的数字，用不同层数的循环来解析。
- 具体来说，对于某一行/列，循环得出每个可能的填充方式，用一个长度为行数的列表，列表中每个元素为一个元组来记录每个位置可能的填充，如果该位置只有一种【□】或者【x】，就将其填入到棋盘中。若该位置有两种，则视为不确定，不进行填充。
- 由于扫描一次之后棋盘上某些位置已经确定填充，若循环得到的填充方式与现有棋盘相悖，则跳过当前循环，以此方式得到最终所有位置的填充。
打印
- 为了更直观的观看以便自己在手机上通关，所以做的更整齐一些。如下图所示。
完整代码
import numpy as np
row = [3, 5, 6, 9, 10, 12, '5 3 3', '10 1', '3 5 1', '2 4 1 1', '3 1 3 1', '3 1 1', '3 1', '2 1', 11]
col = [6, 8, '4 5', '5 1', '8 1', '10 1', '6 3 1', '11 1', '9 1', '6 1', '3 1 1', '3 3 1', '3 1 1', '2 1', 7]
row = [[int(i) for i in s.split()] if type(s) is str else [s] for s in row]
col = [[int(i) for i in s.split()] if type(s) is str else [s] for s in col]
map_size = len(row)
row_maxlen = max([len(i) for i in row])
col_maxlen = max([len(i) for i in col])
map_default_str = '-'
map = np.full([map_size, map_size], map_default_str)
num = [0]
def map_print():
candi_list = [1] * map_size # col中是否有超过10的数字，需要额外多一个空格
for i in range(map_size):
for j in col[i]:
if j >= 10:
candi_list[i] = 2
f = ''
for i in range(1, col_maxlen + 1):
cur_num_str = ''
for j in range(map_size):
if len(col[j]) >= i:
cur_num_str += str(col[j][-i]) + ' ' + ' ' * candi_list[j]
if col[j][-i] >= 10:
cur_num_str = cur_num_str[:-1]
else:
cur_num_str += ' ' * 2 + ' ' * candi_list[j]
f = ' ' + ' ' * (row_maxlen - 1) + cur_num_str + '\n' + f
print(f)
f = ''
for i in range(map_size):
cur_num_str = ''
for j in range(row_maxlen):
if len(row[i]) >= row_maxlen - j:
if row[i][j + len(row[i]) - row_maxlen] >= 10:
cur_num_str = cur_num_str[:-1]
cur_num_str += str(row[i][j + len(row[i]) - row_maxlen]) + ' ' * 2
else:
cur_num_str += ' ' * 2 + ' ' * candi_list[j]
cur_num_str += ' '
for j in range(map_size):
# cur_num_str += str(int(map[i][j])) + ' '
cur_num_str += ' ' * candi_list[j] + map[i][j] + ' '
f += cur_num_str + '\n'
print(f)
def simple_scan(idx, is_row=True):
map_list_tmp = np.full(map_size, 'x') # 当前行或列的值，用'x'初始化后，循环用'o'替换
cur_list = [set() for _ in range(map_size)] # 扫描多次保存可能填充的值
if is_row:
row_col_tmp = row.copy()
map_row_col = map[idx]
else:
row_col_tmp = col.copy()
map_row_col = map[:, idx]
if len(row_col_tmp[idx]) == 1:
for i in range(map_size - sum(row_col_tmp[idx]) + 1):
map_list_tmp[i: i + row_col_tmp[idx][0]] = 'o'
if is_require(map_list_tmp, map_row_col):
cur_list = merge_list(cur_list, map_list_tmp)
map_list_tmp = np.full(map_size, 'x')
elif len(row_col_tmp[idx]) == 2:
for i in range(map_size - sum(row_col_tmp[idx]) + 1 - len(row_col_tmp[idx]) + 1):
for j in range(i + row_col_tmp[idx][0] + 1, map_size - row_col_tmp[idx][1] + 1):
map_list_tmp[i: i + row_col_tmp[idx][0]] = 'o'
map_list_tmp[j: j + row_col_tmp[idx][1]] = 'o'
if is_require(map_list_tmp, map_row_col):
cur_list = merge_list(cur_list, map_list_tmp)
map_list_tmp = np.full(map_size, 'x')
elif len(row_col_tmp[idx]) == 3:
for i in range(map_size - sum(row_col_tmp[idx]) + 1 - len(row_col_tmp[idx]) + 1): # 8-2+1-2+1
for j in range(i + row_col_tmp[idx][0] + 1, map_size - sum(row_col_tmp[idx][1:]) + 1 - len(row_col_tmp[idx][1:]) + 1):
for k in range(j + row_col_tmp[idx][1] + 1, map_size - sum(row_col_tmp[idx][2:]) + 1 - len(row_col_tmp[idx][2:]) + 1):
map_list_tmp[i: i + row_col_tmp[idx][0]] = 'o'
map_list_tmp[j: j + row_col_tmp[idx][1]] = 'o'
map_list_tmp[k: k + row_col_tmp[idx][2]] = 'o'
if is_require(map_list_tmp, map_row_col):
cur_list = merge_list(cur_list, map_list_tmp)
map_list_tmp = np.full(map_size, 'x')
elif len(row_col_tmp[idx]) == 4:
for i in range(map_size - sum(row_col_tmp[idx]) + 1 - len(row_col_tmp[idx]) + 1): # 8-2+1-2+1
for j in range(i + row_col_tmp[idx][0] + 1, map_size - sum(row_col_tmp[idx][1:]) + 1 - len(row_col_tmp[idx][1:]) + 1):
for k in range(j + row_col_tmp[idx][1] + 1, map_size - sum(row_col_tmp[idx][2:]) + 1 - len(row_col_tmp[idx][2:]) + 1):
for l in range(j + row_col_tmp[idx][2] + 1, map_size - sum(row_col_tmp[idx][3:]) + 1 - len(row_col_tmp[idx][3:]) + 1):
map_list_tmp[i: i + row_col_tmp[idx][0]] = 'o'
map_list_tmp[j: j + row_col_tmp[idx][1]] = 'o'
map_list_tmp[k: k + row_col_tmp[idx][2]] = 'o'
map_list_tmp[l: l + row_col_tmp[idx][3]] = 'o'
if is_require(map_list_tmp, map_row_col):
cur_list = merge_list(cur_list, map_list_tmp)
map_list_tmp = np.full(map_size, 'x')
for i in range(map_size):
if is_row:
if len(cur_list[i]) == 1 and map[idx][i] == map_default_str:
map[idx][i] = list(cur_list[i])[0]
num[0] += 1
else:
if len(cur_list[i]) == 1 and map[i][idx] == map_default_str:
map[i][idx] = list(cur_list[i])[0]
num[0] += 1
def merge_list(l, n):
for i in range(len(l)):
l[i] = l[i].union(set(n[i]))
return l
def is_require(l1, l2):
for i in range(map_size):
if l2[i] != map_default_str and l1[i] != l2[i]:
return False
return True
if __name__ == '__main__':
while True:
for i in range(map_size):
simple_scan(i, is_row=True)
for j in range(map_size):
simple_scan(j, is_row=False)
map_print()
if num[0] == map_size * map_size:
break
