python - 用 Python 编写数独求解器

Question

这是我用 python 语言编写的 Sudoku Solver，当我运行这个程序时，Update 函数和 Solve 函数似乎有问题。

无论我翻了多少遍代码，我似乎都没有运气

谁能帮帮我？

---

import copy


def display (A):

    if A:
        for i in range (9):
            for j in range (9):
                if type (A[i][j]) == type ([]): print A[i][j][0],
                else: print A[i][j]
            print
        print
    else: print A

def has_conflict(A):

    for i in range(9):
        for j in range(9):
            for (x,y) in get_neighbors(i,j):
                if len(A[i][j])==1 and A[i][j]==A[x][y]: return True
    return False


def get_neighbors(x,y):

    neighbors = []
    for i in range(3):
        for j in range(3):
            a = 3*(x / 3)+i
            b = 3*(y / 3)+j
            if (x,y) != (a,b):
                neighbors += [(a,b)]

    for i in range(9):
        if (x,y) != (x,i) and (x,i) not in neighbors:
            neighbors += [(x,i)]
        if (x,y) != (i,y) and (i,y) not in neighbors:
            neighbors += [(i,y)]

    return neighbors



def update(A,x,y,value):

    B = copy.deepcopy(A)
    B[x][y] = [value]
    for (i,j) in get_neighbors(x,y):
        if B[i][j] == B[x][y]:
            if len(B[i][j]) > 1: B[i][j].remove(value)
            else: return [] 
    if has_conflict(B) == True: return []
    else: return B


def solve(A):

    for x in range (9):
        for y in range(9):
            if len(A[x][y]) == 1: return A[x][y]
            if len(A[x][y]) > 1:
                lst = update(A,x,y,A[x][y])
                if len(lst[x][y]) > 1: solve(lst)
                if lst == []: return []
                if len(lst[x][y]) == 1: return lst
            else: return A[x][y]    



A=[]

infile = open('puzzle1.txt','r')

for i in range(9):

        A += [[]]
        for j in range(9):
            num = int(infile.read(2))
            if num: A[i] += [[num]]
            else: A[i] += [[1,2,3,4,5,6,7,8,9]]

for i in range(9):

        for j in range(9):
            if len(A[i][j])==1: A = update(A, i, j, A[i][j][0])
            if A == []: break
        if A==[]: break

if A<>[]: A = solve(A)

display(A)

---

这里有一些谜题:

谜题1

0 0 0 2 6 0 7 0 1
6 8 0 0 7 0 0 9 0
1 9 0 0 0 4 5 0 0
8 2 0 1 0 0 0 4 0
0 0 4 6 0 2 9 0 0
0 5 0 0 0 3 0 2 8
0 0 9 3 0 0 0 7 4
0 4 0 0 5 0 0 3 6
7 0 3 0 1 8 0 0 0

谜题2

1 0 0 4 8 9 0 0 6
7 3 0 0 0 0 0 4 0
0 0 0 0 0 1 2 9 5
0 0 7 1 2 0 6 0 0
5 0 0 7 0 3 0 0 8
0 0 6 0 9 5 7 0 0
9 1 4 6 0 0 0 0 0
0 2 0 0 0 0 0 3 7
8 0 0 5 1 2 0 0 4

谜题3

0 2 0 6 0 8 0 0 0
5 8 0 0 0 9 7 0 0
0 0 0 0 4 0 0 0 0
3 7 0 0 0 0 5 0 0
6 0 0 0 0 0 0 0 4
0 0 8 0 0 0 0 1 3
0 0 0 0 2 0 0 0 0
0 0 9 8 0 0 0 3 6
0 0 0 3 0 6 0 9 0

---

Answer 1

我会避免诸如“四处移动代码”之类的事情。这叫做 "Programming by Coincidence" (参见 The Pragmatic Programmer)。像这样编程不会让你成为更好的程序员。

相反，您应该拿出纸笔，开始思考事情应该如何运作。然后，阅读代码并仔细写下它实际的作用。只有明白了，再回到电脑端。