为什么我的 Lifo Queue 中的项目会发生变化而不参考它们？

Question

我正在研究图形着色问题，并且我有一个字典，其中包含每个节点可以采用的可能颜色值 (0,1,2,3)。我在这里尝试了一些蛮力方法，只是为了了解算法。我遇到了一个我无法纠正的简单问题。基本上，每当我选择颜色时，我都需要跟踪每个字典，所以如果它最终不起作用，我可以回溯到那个点。下面是输出出错的地方。我已插入所有这些打印语句来尝试诊断问题。我将节点 9 设置为颜色 2，并将替代选择排队，以防我不得不回溯到该选择。

设置 9 等于 2

排队 9 的替代选择（以下是添加到 LifoQueue 的替代选择，这是关键）

 {0: [0], 1: [0], 2: [1], 3: [2], 4: [1], 5: [0], 6: [2], 7: [1], 8: [0], 9: [3], 10: [2, 3], 11: 
  [3], 12: [0, 2, 3], 13: [1, 3], 14: [0, 3], 15: [0, 3], 16: [0, 3], 17: [0, 1, 3], 18: [2, 3], 19: 
  [1, 3]}

然后代码使用该选项根据预定义的边从其他节点中消除可能的颜色，节点 9 设置为 2

边 9 到 10

 {0: [0], 1: [0], 2: [1], 3: [2], 4: [1], 5: [0], 6: [2], 7: [1], 8: [0], 9: [2], 10: [3], 11: [3], 
  12: [0, 2, 3], 13: [1, 3], 14: [0, 3], 15: [0, 3], 16: [0, 3], 17: [0, 1, 3], 18: [2, 3], 19: [1, 
  3]}

边 9 到 12

 {0: [0], 1: [0], 2: [1], 3: [2], 4: [1], 5: [0], 6: [2], 7: [1], 8: [0], 9: [2], 10: [3], 11: [3], 
  12: [0, 3], 13: [1, 3], 14: [0, 3], 15: [0, 3], 16: [0, 3], 17: [0, 1, 3], 18: [2, 3], 19: [1, 3]}

将 10 设置为 3（此处无法替代队列，因为节点 10 只有一个选项）

边缘 10 到 11

 {0: [0], 1: [0], 2: [1], 3: [2], 4: [1], 5: [0], 6: [2], 7: [1], 8: [0], 9: [2], 10: [3], 11: [], 
  12: [0, 3], 13: [1, 3], 14: [0, 3], 15: [0, 3], 16: [0, 3], 17: [0, 1, 3], 18: [2, 3], 19: [1, 3]}

在这里你可以看到节点 11 的选项现在是空白的，这告诉我——充其量——我所做的最后一个选择是错误的，所以这会触发回溯到放入队列中的最后一个项目，其中，如前所述，是节点 9 的值为 3 的字典。问题是，当我在这里执行 q.get() 时，这是我得到的输出（与我添加到上面的队列中的内容进行比较：

 {0: [0], 1: [0], 2: [1], 3: [2], 4: [1], 5: [0], 6: [2], 7: [1], 8: [0], 9: [3], 10: [3], 11: [], 
  12: [0, 3], 13: [1, 3], 14: [0, 3], 15: [0, 3], 16: [0, 3], 17: [0, 1, 3], 18: [2, 3], 19: [1, 3]}

如果我不接触队列，队列条目会如何变化？下面是相关代码。

'''

# edges are the constraint store (?)
edges = [some pre-defined list of tuples defining the edges]

# building a solution template
solution = [0]*node_count

# global constraint of four colors 
colors = [0,1,2,3]

# dictionary covering possible colors for each node
sols = {node:colors.copy() for node in range(0,node_count)}

#initiating a queue to store checkpoints for backtracking
q = queue.LifoQueue()


#placing the first entry in the queue
q.put((0,sols))

place = 0
while place < node_count:
    backtracked = False
    cursor = q.get()
    print('cursor is ', cursor)
    place = cursor[0]
    dic = cursor[1]
    val = dic[place][0]

    #any time a value is chosen from multiple, it needs to be queued
    if len(dic[place]) > 1:
        dic_copy = dic.copy()
        dic_copy[place].remove(val)
        q.put((place,dic_copy))
        
    #choosing a value
    solution[place] = val
    dic[place] = [solution[place]]
    
    print('Setting ',place,'equal to ',val)
    for edge in edges:
        if backtracked == False:
            if edge[0] == place:                
                if solution[place] in dic[edge[1]]:
                    #in case of failure, store copy
                    vals = dic[edge[1]].copy()
                    dic[edge[1]].remove(solution[place])
                    if len(dic[edge[1]]) == 0:
                        backtracked = True
                            
    if backtracked == False:
        q.put((place+1, dic))
    else:
        print('backtracking')

'''

Answer 1

使用 deepcopy 而不是我使用的常规副本解决了这个问题。