递归如何在 n_queen Python 中工作

Question

我正在尝试解决 python 中的经典 n Queen 问题。不幸的是，输出与我的预期完全不同。我了解回溯算法，但可能不清楚在编写递归时可能会出错的地方。我尝试了 1.print 一种可能的解决方案。 2. 打印所有可能的解决方案。 （这里我的解决方案被记为 n 的列表，其中 list[i] 的值是第 i 行上的选定列）

def n_queen_find1(n):
    answer = [-1] * n
    def dfs(depth,answer):   
        if depth == n:
            print ("cur valid answer is ",answer)
            return answer
        else:
            for colNum in range(n): # check every col at cur depth
                if is_safe(depth,colNum,answer):
                    answer[depth] = colNum
                    print ("now the answer is ", answer)
                    print ("now depth move to ", depth + 1)
                    dfs(depth + 1,answer)

    def is_safe(i,j,answer_cur):
        # given current queen placement , could we place the ith queen (at row i) at the col j
        for prerow in range(i):
            if answer_cur[prerow] == j or abs(prerow - i) == abs(answer_cur[prerow] - j):
                return False
        return True

    return dfs(0,answer)

我希望 n_queen_find1 在 n == 4 时输出 4 个元素的列表，例如 [1, 3, 0, 2]。在我的递归过程中，生成了正确的答案（如打印中所示）。但是，该函数什么也没返回。我添加了那些有助于调试的打印行，发现我不明白为什么在创建第一个正确答案时 dfs 没有停止并返回？

trial1 = n_queen_find1(4)
print trial1
('now the answer is ', [0, -1, -1, -1])
('now depth move to ', 1)
('now the answer is ', [0, 2, -1, -1])
('now depth move to ', 2)
('now the answer is ', [0, 3, -1, -1])
('now depth move to ', 2)
('now the answer is ', [0, 3, 1, -1])
('now depth move to ', 3)
('now the answer is ', [1, 3, 1, -1])
('now depth move to ', 1)
('now the answer is ', [1, 3, 1, -1])
('now depth move to ', 2)
('now the answer is ', [1, 3, 0, -1])
('now depth move to ', 3)
('now the answer is ', [1, 3, 0, 2])
('now depth move to ', 4)
('cur valid answer is ', [1, 3, 0, 2])
('now the answer is ', [2, 3, 0, 2])
('now depth move to ', 1)
('now the answer is ', [2, 0, 0, 2])
('now depth move to ', 2)
('now the answer is ', [2, 0, 3, 2])
('now depth move to ', 3)
('now the answer is ', [2, 0, 3, 1])
('now depth move to ', 4)
('cur valid answer is ', [2, 0, 3, 1])
('now the answer is ', [3, 0, 3, 1])
('now depth move to ', 1)
('now the answer is ', [3, 0, 3, 1])
('now depth move to ', 2)
('now the answer is ', [3, 0, 2, 1])
('now depth move to ', 3)
('now the answer is ', [3, 1, 2, 1])
('now depth move to ', 2)
None

我稍微修改了 n_queen_find1 以生成所有可能的解决方案，但我也无法解释输出。

def n_queen_findall(n):
    answer = [-1] * n
    finalAnswer = []
    def dfs(depth,answer,finalAnswer):   
        if depth == n:
            print ("cur valid answer is ",answer)            
            finalAnswer.append(answer)
            print ("after appending final answer is ", finalAnswer)
            return
        else:
            for colNum in range(n): # check every col at cur depth
                if is_safe(depth,colNum,answer):
                    answer[depth] = colNum
                    print ("now the answer is ", answer)
                    print ("now depth move to ", depth + 1)
                    dfs(depth + 1,answer,finalAnswer)

    def is_safe(i,j,answer_cur):
        # given current queen placement , could we place the ith queen (at row i) at the col j
        for prerow in range(i):
            if answer_cur[prerow] == j or abs(prerow - i) == abs(answer_cur[prerow] - j):
                return False
        return True
    dfs(0,answer,finalAnswer)
    return finalAnswer

这是我对 n_queen_findall 的测试。它也不正确。首先，当深度不等于 n 时，为什么 finalAnswers 会附加结果？二、为什么finalAnswer有重复？

trial2 = n_queen_findall(4)
print trial2
('now the answer is ', [0, -1, -1, -1])
('now depth move to ', 1)
('now the answer is ', [0, 2, -1, -1])
('now depth move to ', 2)
('now the answer is ', [0, 3, -1, -1])
('now depth move to ', 2)
('now the answer is ', [0, 3, 1, -1])
('now depth move to ', 3)
('now the answer is ', [1, 3, 1, -1])
('now depth move to ', 1)
('now the answer is ', [1, 3, 1, -1])
('now depth move to ', 2)
('now the answer is ', [1, 3, 0, -1])
('now depth move to ', 3)
('now the answer is ', [1, 3, 0, 2])
('now depth move to ', 4)
('cur valid answer is ', [1, 3, 0, 2])
('now the final answer is ', [])
('now the answer is ', [2, 3, 0, 2])
('now depth move to ', 1)
('now the answer is ', [2, 0, 0, 2])
('now depth move to ', 2)
('now the answer is ', [2, 0, 3, 2])
('now depth move to ', 3)
('now the answer is ', [2, 0, 3, 1])
('now depth move to ', 4)
('cur valid answer is ', [2, 0, 3, 1])
('now the final answer is ', [[2, 0, 3, 1]])
('now the answer is ', [3, 0, 3, 1])
('now depth move to ', 1)
('now the answer is ', [3, 0, 3, 1])
('now depth move to ', 2)
('now the answer is ', [3, 0, 2, 1])
('now depth move to ', 3)
('now the answer is ', [3, 1, 2, 1])
('now depth move to ', 2)
[[3, 1, 2, 1], [3, 1, 2, 1]]

抱歉把问题问的这么长。这个问题的原因可能很简单。我是python的初级，尤其是python后端。也许我对python通过引用传递的理解不正确？非常感谢对此的一些指导。提前非常感谢。

Answer 1

在您的dfs 中，只有当 n==4 时，您才会有 return，但当 n 是其他任何东西时，则不会。所以无论dfs为4返回什么，它们都会被父被调用者吸收，最终什么都不会返回。

因此，为了解决您的第一个问题，因为我们知道答案中的失败展示位置返回 None，（因为 for 中的 if 循环在 else 内将失败），我们在递归 dfs 之前添加一个 if检查成功。如果是这样，我们返回answer。所以只要改变这个：

dfs(depth + 1,answer)

到这里：

if dfs(depth + 1,answer): #false i.e None for failed dfs
    return answer

在我们解决第二个问题之前，让我们想想，如果我们有 5 个解决方案，我们需要 5 个数组来存储它们，对吗？你有多少？一个answer。在finalAnswer.append(answer) 中，您不会添加包含数组的较新解决方案，您只是添加旧answer 的reference，它会不断被修改，而您打算添加的任何内容都会丢失。

因此您需要在追加之前复制并创建新数组。此外，您还需要摆脱上述return。

def n_queen_find1(n):
    answer = [-1] * n
    finalAnswer = []
    def dfs(depth,answer):
        if depth == n:
            print ("cur valid answer is ",answer)
            finalAnswer.append([i for i in answer]) #answer is copied
            return answer
        else:
            for colNum in range(n): # check every col at cur depth
                if is_safe(depth,colNum,answer):
                    answer[depth] = colNum
                    #print ("now the answer is ", answer)
                    #print ("now depth move to ", depth + 1)
                    dfs(depth + 1,answer)


    def is_safe(i,j,answer_cur):
        # given current queen placement , could we place the ith queen (at row i) at the col j
        for prerow in range(i):
            if answer_cur[prerow] == j or abs(prerow - i) == abs(answer_cur[prerow] - j):
                return False
        return True

    dfs(0,answer)
    return finalAnswer

哪个打印：

cur valid answer is  [1, 3, 0, 2]
cur valid answer is  [2, 0, 3, 1]
[[1, 3, 0, 2], [2, 0, 3, 1]]