用Python生成数独
上次做出了怎么算数独，开心坏了，鼻子翘起还没两天，脑瓜里又蹦出了个新想法：创造数独。
于是乎，又捣鼓捣鼓，终于整出来了！（吼吼吼吼吼~~~）
这个算法的基本思路是这样的：
1、定义一个空的数独出来；
2、随机将1~9填入第一行；
3、随机将第一列填满；
4、随机将第九列填满；
5、用添加了随机解法的解数独算法填写出一个随机数独谜底；
6、建立一个集合l=[0~80]，随机从谜底抠去数字，然后算能否得到解，并验证是否唯一，行就保留，不行就回归上一步；
7、直到集合l里的数字抠完，跳出循环。
啦啦啦，数独生成完毕，大概率将会是一个骨灰级难度哟~
原创文章转载要注明，代码如下，拿走不谢：
import random
import time
def pr(a):#用于输出方便阅读的结果
for i in range(9):
if i%3==0:
print('\t---------\t\t---------\t\t---------')
for j in range(9):
if j%3==0:
print('|',end='\t')
print(a[i][j],end='\t')
print('|',end='\n')
print('\t---------\t\t---------\t\t---------')
def diff(d):
print('数独难度：',end='')
if d==0:
print('极容易')
if d==1:
print('容易')
if d==2:
print('一般')
if d==3:
print('困难')
if d==4:
print('极困难')
def con(a):#用于将数据转化为符合格式的序列
if len(a)==81:
a_con=[[],[],[],[],[],[],[],[],[]]
for i in range(9):
b=a[i*9:i*9+9]
for j in range(9):
a_con[i].append(int(b[j]))
return a_con
return False
def examine_sudoku(sudoku):#检查是否符合数独规则
if len(sudoku)!=9:
return False
for i in range(9):
if len(sudoku[i])!=9 or sudoku[i].count([])>0:
return False
for j in range(1,10):
if sudoku[i].count(j)>1:
return False
return True
def complete(a):#用于检测数独是否完成
for i in range(9):
for j in range(9):
if type(a[i][j])!=int or a[i][j]==0:
return False
return True
def copy(a):#复制数独的值，用于后期假设法运算失败，找回原来的值
d=[[], [], [], [], [], [], [], [], []]
for i in range(9):
for j in range(9):
if type(a[i][j])==int:
d[i].append(a[i][j])
elif type(a[i][j])==list:
d[i].append([])
for k in a[i][j]:
d[i][j].append(k)
return d
def extract_b_c(a):#用9行表示的a，得到9列表示的b，及9宫表示的c，并对每个0所在的位置用[1,2,3,4,5,6,7,8,9]代替，作为笔记
b=[[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0]]
c=[[],[],[],[],[],[],[],[],[]]
for i in range(9):
for j in range(9):
if a[i][j]==0:
a[i][j] = [1,2,3,4,5,6,7,8,9]
b[j][i]=a[i][j]
s=ij_s(i,j)
c[s].append(a[i][j])
return b,c
def ij_s(i,j):#用i和j求得宫所在的位置s
s=i//3*3+j//3
return s
def ij_t(i,j):#用i和j求得所在宫里的位置t
t=i%3*3+j%3
return t
def st_i(s,t):#用s和t求得原i的值
i=s//3*3+t//3
return i
def st_j(s, t):#用s和t求得原j的值
j=s%3*3+t%3
return j
def play_1(a,b,c,i,j,change):#将笔记中同行、同列及同宫的笔记里删除已知数字，并将唯一可能的格子变成已知数字
s=ij_s(i,j)
for num in range(1,10):
for k in range(9):
if type(a[i][k])==list and num in a[i] and num in a[i][k]:
a[i][k].remove(num)
change=True
if len(a[i][k])==1:
n=a[i][k][0]
a[i][k]=n
b[k][i]=n
c[ij_s(i,k)][ij_t(i,k)]=n
change=True
if type(b[j][k])==list and num in b[j] and num in b[j][k]:
b[j][k].remove(num)
change=True
if len(b[j][k])==1:
n=b[j][k][0]
a[k][j]=n
b[j][k]=n
c[ij_s(k,j)][ij_t(k,j)]=n
change=True
if type(c[s][k])==list and num in c[s] and num in c[s][k]:
c[s][k].remove(num)
change=True
if len(c[s][k])==1:
n=c[s][k][0]
a[st_i(s,k)][st_j(s,k)]=n
b[st_j(s,k)][st_i(s,k)]=n
c[s][k]=n
change=True
return a,b,c,change
def play_2(a,b,c,i,j,change):#唯一候选数法
def num_abc():
num_a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
num_b = [1, 2, 3, 4, 5, 6, 7, 8, 9]
num_c = [1, 2, 3, 4, 5, 6, 7, 8, 9]
for k in range(1,10):
if k in a[i]:
num_a.remove(k)
if k in b[j]:
num_b.remove(k)
if k in c[s]:
num_c.remove(k)
return num_a,num_b,num_c
s=ij_s(i,j)
num_a,num_b,num_c=num_abc()
for n in num_a:
t=0
for k in range(9):
if type(a[i][k])==list and n in a[i][k]:
t+=1
if t==1:
for k in range(9):
if type(a[i][k]) == list and n in a[i][k]:
a[i][k] = n
b[k][i] = n
c[ij_s(i, k)][ij_t(i, k)] = n
num_a, num_b, num_c = num_abc()
change=True
for n in num_b:
t=0
for k in range(9):
if type(b[j][k])==list and n in b[j][k]:
t+=1
if t==1:
for k in range(9):
if type(b[j][k])==list and n in b[j][k]:
a[k][j] = n
b[j][k] = n
c[ij_s(k, j)][ij_t(k, j)] = n
num_a, num_b, num_c = num_abc()
change=True
for n in num_c:
t=0
for k in range(9):
if type(c[s][k])==list and n in c[s][k]:
t+=1
if t==1:
for k in range(9):
if type(c[s][k])==list and n in c[s][k]:
a[st_i(s,k)][st_j(s,k)]=n
b[st_j(s,k)][st_i(s,k)]=n
c[s][k]=n
change=True
return a,b,c,change
def play_3(a,b,c,i,j,change):#链数删减法，专业用语是这样的，有2链数、3链数的，此方法扩展到n链数（帮助排除不可能项的）
list_a=[]
list_b=[]
list_c=[]
s=ij_s(i,j)
for k in range(9):
if type(a[i][k]) == list:
list_a.append(a[i][k])
for k in range(9):
if type(b[j][k]) == list:
list_b.append(b[j][k])
for k in range(9):
if type(c[s][k]) == list:
list_c.append(c[s][k])
for k in range(3,len(list_a)+1):
l=[]
n=0
for m in list_a:
if len(m)1:
k=random.randint(0,len(l)-1)
else:
k=0
x=False
p=l[k]//9
q=l[k]%9
del l[k]
if co[p][q]==0:
continue
co[p][q]=0
cop=copy(co)
cop_b, cop_c = extract_b_c(cop)
while change == True:
change = False
cop, cop_b, cop_c, change, examine, a_co= play_all_1(cop, cop_b, cop_c, change, examine, a_co)
if complete(cop)==True:
a = copy(co)
else:
if ti==0:
print('一般难度已生成，正在精细推敲骨灰难度......')
ti=1
for i in range(9):
for j in range(9):
if type(cop[i][j])==list and len(cop[i][j])==2:
a_copy=copy(cop)
n=cop[i][j][0]
a_copy[i][j]=a_copy[i][j][1]
a_copy_b, a_copy_c = extract_b_c(a_copy)
cop[i][j]=n
cop_b[j][i]=n
cop_c[ij_s(i,j)][ij_t(i,j)]=n
change=True
while change == True:
change = False
cop, cop_b, cop_c, change, examine, a_co,d=play_all(cop, cop_b, cop_c, change, examine, a_co,d)
while change == True:
change = False
a_copy, a_copy_b, a_copy_c, change, examine, a_copy2,d=play_all(a_copy, a_copy_b, a_copy_c, change, examine, a_copy2,d)
if (complete(cop)==True and complete(a_copy)==False) or (complete(cop)==False and complete(a_copy)==True):
if cop==com or cop==com:
a = copy(co)
print('数独生成完毕，生成数独所用时间为：',time.time()-t)
pr(a)
t=time.time()
a,d=complate_sudoku(a)
print('解数独所用时间为：',time.time()-t)
diff(d)
pr(a)