在遞歸中混淆對象引用行為

Question

我寫了一個深度優先搜索，根據這個問題搜索迷宮中的路徑：

# for convenience
matrix = [
    ["0", "0", "0", "0", "1", "0", "0", "0"],
    ["0", "1", "1", "0", "1", "0", "1", "0"],
    ["0", "1", "0", "0", "1", "0", "1", "0"],
    ["0", "0", "0", "1", "0", "0", "1", "0"],
    ["0", "1", "0", "1", "0", "1", "1", "0"],
    ["0", "0", "1", "1", "0", "1", "0", "0"],
    ["1", "0", "0", "0", "0", "1", "1", "0"],
    ["0", "0", "1", "1", "1", "1", "0", "0"]
]
num_rows = len(matrix)
num_cols = len(matrix[0])
goal_state = (num_rows - 1, num_cols - 1)
print(goal_state)


def dfs(current_path):
    # anchor
    row, col = current_path[-1]
    if (row, col) == goal_state:
        print("Anchored!")
        return True

    # try all directions one after the other
    for direction in [(row, col + 1), (row, col - 1), (row + 1, col), (row - 1, col)]:
        new_row, new_col = direction
        if (0 <= new_row < num_rows and 0 <= new_col < num_cols and  # stay in matrix borders
                matrix[new_row][new_col] == "0" and                  # don't run in walls
                (new_row, new_col) not in current_path):             # don't run in circles
            current_path.append((new_row, new_col))  # try new direction
            print(result == current_path)
            if dfs(current_path):  # recursive call
                return True
            else:
                current_path = current_path[:-1]  # backtrack


# the result is a list of coordinates which should be stepped through in order to reach the goal
result = [(0, 0)]
if dfs(result):
    print("Success!")
    print(result)
else:
    print("Failure!")

它可以像我期望的那樣工作，除了最后兩個坐標沒有被添加到名為result的列表中。 這就是為什么我包括行print(result == current_path) ，這應該在我的期望中始終為真 - 它是同一個對象，它作為參考傳遞。 它怎么可能不平等？ 代碼應該是可執行的，但以防萬一，這是我得到的輸出：

True
True
...
True
False
False
Anchored!
Success!
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3), (2, 2), (3, 2), (3, 1), (3, 0), (4, 0), (5, 0), (5, 1), (6, 1), (6, 2), (6, 3), (6, 4), (5, 4), (4, 4), (3, 4), (3, 5), (2, 5), (1, 5), (0, 5), (0, 6), (0, 7), (1, 7), (2, 7), (3, 7), (4, 7), (5, 7), (5, 6)]

從輸出中可以清楚地看出，如果最后一個狀態是(7, 7) ，那么該函數只能錨定，這與(6, 7)一起不存在於result變量中。 我為什么感到恍惚。

編輯： result列表中存在的坐標(5, 6)不應該存在。 它是算法回溯的最后位置，即達到目標狀態之前的死胡同。 我再次不知道為什么在回溯步驟中沒有從兩個列表中正確刪除它。

Answer 1

current_path = current_path[:-1]創建一個新列表，並將名稱current_path綁定到這個新創建的列表。 在那個階段，它不再是result列表。 請查看列表方法pop() 。