求解数独的回溯算法

如何解决求解数独的回溯算法

首先，感谢您花时间阅读本文。

我正在尝试创建一种回溯算法来解决特定的数独棋盘。但是，我遇到了一个问题，我考虑了很长时间但无法理解。下面是我的回溯算法：

# The puzzle itself
board = [
    [0,6,8,4,0],[2,1,7],[0,3,9,7,2,[7,4],5,0]
]

# A function which prints out the board in grids
def print_board():

    j = 1

    for rows in board:
        i = 1
        for elements in rows:
            if i % 3 == 0:
                print(elements,"|",end=" ")
            else:
                print(elements,end=" ")

            i += 1

        if j % 3 == 0:
            print("")
            print("-----------------------",end="")

        j += 1
        print("")

# A function which searches for the coordinate [x,y] of empty Boxes (in this case if the value = 0 it is empty)
def search_empty():

    empty_list = []

    j = 0

    for rows in board:
        i = 0
        for elements in rows:
            if elements == 0:
                empty_list.append([i,j])

            i += 1

        j += 1

    return empty_list

# A function which takes the coordinate of an empty Box and returns all the coordinates of Boxes which fall in the same grid.
def set_grid(x,y):
    if x in range(3) and y in range(3):
        return [[0,1],2],[1,2]]

    elif x in range(3,6) and y in range(3):
        return [[3,[3,[4,[5,2]]

    elif x in range(6,9) and y in range(3):
        return [[6,[6,[8,2]]

    elif x in range(3) and y in range(3,6):
        return [[0,3],5],5]]

    elif x in range(3,6) and y in range(3,6):
        return [[3,5]]

    elif x in range(6,9) and y in range(3,6):
        return [[6,5]]

    elif x in range(3) and y in range(6,9):
        return [[0,6],8],8]]

    elif x in range(3,6) and y in range(6,9):
        return [[3,8]]

    elif x in range(6,9) and y in range(6,9):
        return [[6,8]]

# A function which takes 3 inputs: lower,x and y.
# lower is the starting integer to test on,so first Box defaults to 0. If say integer = 6 is the first answer to the first empty
# Box,but 1 - 9 does not satisfy the rules of Sudoku for the second Box,we backtrack to first Box and start from integer + 1
# x and y are both coordinates of the empty Box
def conditions(lower,x,y):

        grid_list = set_grid(x,y)
        for integers in range(lower,10):
            print("Conditions running.")
            grid_test = True
            vertical_test = True
            horizontal_test = True
            
            # Grid test is to test if the current integer is present in the grid. If present,fails the grid test.
            for i in grid_list:
                [p,q] = i
                if integers == board[q][p]:
                    grid_test = False
                    break

            # Horizontal test is to test if the current integer is present in the same row. If present,fails the row test.
            for i in board[y]:
                if integers == i:
                    horizontal_test = False
                    break
            
            # Vertical test is to test if the current integer is present in the same column. If present,fails the vertical test.
            for i in range(9):
                if integers == board[i][x]:
                    vertical_test = False
                    break
            
            # If all three tests are satisfied and passed,the function returns True and the integer which satisfies all 3 rules.
            if grid_test and horizontal_test and vertical_test:
                print("Condition satisfied.")
                return True,integers

        print("Condition unsatisfied.")
        return False,0


# This is where the backtracking begins.
def trials():

    n = 0

    a = 1

    # A list which records all the "correct at that time" integers from prevIoUs empty Boxes. New "correct" integers are appended.

    history = []

    # Creates a static list of coordinates of empty Boxes.
    empty_list = search_empty()

    while True:
        ## This line is to debug
        print(history)

        # p has value of True or False,and q has value of the correct integer if True,0 if False
        [p,q] = conditions(a,empty_list[n][0],empty_list[n][1])

        if not p:
            # If p is false,we backtrack by shifting to the last empty Box.
            n -= 1

            # a is the 'lower' input for conditions() function.
            a = history[n] + 1

            [x,y] = empty_list[n]

            # Since the 'correct' answer of prevIoUs Box is not correct anymore,we are replacing it back with 0 (erasing it).
            board[y][x] = 0

            ## This line is to debug.
            print("a:",a,"old a:",history[n])

            # Removing the old 'correct' answer from the history list.
            history.remove(history[n])

        else:
            # If p is True,the 'correct' integer gets appended to the list.
            history.append(q)

            [x,y] = empty_list[n]

            # The correct answer is replacing the 0 (writing it on the empty Box).
            board[y][x] = q

            # n increments by 1 to proceed to the next empty Box.
            n += 1
        
        # When we run through the entire list of empty Boxes,we break this loop.
        if n == len(empty_list) - 1:
            print("Done!")
            break


print_board()
trials()
print_board()

但是，当我跑的时候，我会看到以下内容：

0 0 6 | 8 4 0 | 0 0 0 | 
2 0 1 | 0 6 0 | 0 0 7 | 
0 3 9 | 0 0 0 | 0 1 0 | 
-----------------------
0 0 0 | 0 9 8 | 3 0 0 | 
0 6 0 | 0 0 0 | 0 9 0 | 
0 0 7 | 3 2 0 | 0 0 0 | 
-----------------------
0 4 0 | 0 0 0 | 1 3 0 | 
7 0 0 | 0 1 0 | 8 0 4 | 
0 0 0 | 0 3 5 | 7 0 0 | 
-----------------------
[]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition satisfied.
[5]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition satisfied.
[5,7]
Conditions running.
Condition satisfied.
[5,1]
Conditions running.
Conditions running.
Condition satisfied.
[5,2]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition unsatisifed
a: 3 old a: 2
[5,1]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition satisfied.
[5,9]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition unsatisifed
a: 10 old a: 9
[5,1]
Condition unsatisifed
a: 2 old a: 1
[5,2]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition satisfied.
[5,9]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition unsatisifed
a: 10 old a: 9
[5,2]
Condition unsatisifed
a: 3 old a: 2
[5,3]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition satisfied.
[5,3]
Condition unsatisifed
a: 4 old a: 3
[5,7]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition satisfied.
[5,9]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition unsatisifed
a: 10 old a: 9
[5,7]
Condition unsatisifed
a: 8 old a: 7
[5]
Conditions running.
Conditions running.
Condition unsatisifed
a: 6 old a: 5
[]
Conditions running.
Conditions running.
Conditions running.
Conditions running.
Condition unsatisifed

我正在努力理解 [5,7] 如何在没有执行任何条件检查的情况下跳转到 [5]（从 [5] 和 [5,7] 之间没有“条件运行”可以看出这一点。这让我相信这是算法失败的主要原因。

非常感谢任何帮助。这是我正在从事的第一个实际项目，所以如果有任何初学者错误，我全神贯注:)

解决方法

只是一些旁注，数独问题的纯回溯通常会导致大量的执行时间。

您的问题是在找到好的解决方案时您从不重置 a。一种解决方案可能是在 while 循环的 a=1 末尾添加 else。此外，您使用 history.remove(history[n]) 删除 history 中等于 history[n] 的第一项，从而导致一些错误。您应该将其替换为 del。这是更正的循环：

   while True:
        ## This line is to debug

        # p has value of True or False,and q has value of the correct integer if True,0 if False
        [p,q] = conditions(a,empty_list[n][0],empty_list[n][1])

        if not p:
            # Removing the old 'correct' answer from the history list.
            del(history[n])
            # If p is false,we backtrack by shifting to the last empty box.
            n -= 1

            # a is the 'lower' input for conditions() function.
            a = history[n] + 1
            history[n]+=1

            board[y][x] = 0
            [x,y] = empty_list[n]

            # Since the 'correct' answer of previous box is not correct anymore,we are replacing it back with 0 (erasing it).

            ## This line is to debug.


        else:
            # If p is True,the 'correct' integer gets appended to the list.
            history[n]=q
            history.append(0)
            [x,y] = empty_list[n]

            # The correct answer is replacing the 0 (writing it on the empty box).
            board[y][x] = q
            n += 1

            # n increments by 1 to proceed to the next empty box.
            a=1 
        # When we run through the entire list of empty boxes,we break this loop.
        if n == len(empty_list):
            print("Done!")
            break

这导致输出：

0 0 6 | 8 4 0 | 0 0 0 | 
2 0 1 | 0 6 0 | 0 0 7 | 
0 3 9 | 0 0 0 | 0 1 0 | 
-----------------------
0 0 0 | 0 9 8 | 3 0 0 | 
0 6 0 | 0 0 0 | 0 9 0 | 
0 0 7 | 3 2 0 | 0 0 0 | 
-----------------------
0 4 0 | 0 0 0 | 1 3 0 | 
7 0 0 | 0 1 0 | 8 0 4 | 
0 0 0 | 0 3 5 | 7 0 0 | 
-----------------------
Done!
5 7 6 | 8 4 1 | 9 2 3 | 
2 8 1 | 9 6 3 | 5 4 7 | 
4 3 9 | 2 5 7 | 6 1 8 | 
-----------------------
1 2 4 | 5 9 8 | 3 7 6 | 
3 6 8 | 1 7 4 | 2 9 5 | 
9 5 7 | 3 2 6 | 4 8 1 | 
-----------------------
6 4 5 | 7 8 9 | 1 3 2 | 
7 9 3 | 6 1 2 | 8 5 4 | 
8 1 2 | 4 3 5 | 7 6 9 | 
-----------------------

正确答案是什么。