修改深度优先搜索Python

时间：2017-03-17 10:17:06

标签： python recursion graph-theory depth-first-search

我希望通过图表来查找以特定方式连接的节点。

graphed = {
    1: [25, 30], 2: [11], 3: [13], 4: [17], 5: [17],
    6: [26], 7: [11, 12], 8: [10, 13], 9: [14, 26],
   10: [8, 11, 15], 11: [2, 7, 10], 12: [7, 16, 17],
   13: [3, 8, 14], 14: [9, 13, 20], 15: [10, 19],
   16: [12, 18], 17: [4, 5, 12], 18: [16, 21, 22],
   19: [15, 28, 29], 20: [14, 25], 21: [18, 23],
   22: [18, 24], 23: [21, 27], 24: [22, 27], 25: [1, 20],
   26: [6, 9], 27: [23, 24], 28: [19], 29: [19], 30: [1]
}

letters = {
    'S': [1],
    'C': [10, [11], [12], [13], [14], [15], [16], [17],
              [18], [19], [20], [21], [22], [23], [24],
              [25], [26], [27], [28], [29], [30] ],
    'O': [2, [3], [4], [5], [6]],
    'N': [7, [8], [9]]}


some_string = 'CO'

def dfs(graph, start, num, visited = None):
    # depth first search for connected nodes
    k = some_string[num]
    y = some_string[(num+1)]
    if visited is None:
        visited = []

    if start in visited:
        return

    visited.append(start)
    # if start corresponds to first letter of string, continue
    if start in letters.get(k):
        # find all nodes that it is connected too
        for each in [x for x in graph[start] if x not in visited]:
            # if any of the connected nodes correspond to
            #   the next letter of string, continue...
            if [each] in letters.get(y):
                dfs(graph, each, (num+1), visited) #recursion

    return visited

lst1 = []
for i in range(1,len(graphed)): 
     lst1.append(dfs(graphed, i, 0))
print(lst1)

目前，DFS函数返回的所有内容都是原始起始值，i在范围内。

[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10],
 [11], [12], [13], [14], [15], [16], [17], [18], [19], [20],
 [21], [22], [23], [24], [25], [26], [27], [28], [29]]

我希望如此：

[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10],
 [11, 2], [12], [13,3], [14], [15], [16], [17,4], [17,5],
 [18], [19], [20], [21], [22], [23], [24], [25], [26,6],
 [27], [28], [29]]

澄清一下：我希望程序运行dfs函数;如果起始值是'C'中的值，找到所有连接的节点并检查它们是否在'O'中;如果是的话，与他们一起做dfs;如果没有，则返回单个值。

希望这很清楚，我无法理解为什么它不起作用。感谢任何帮助。谢谢。

2 个答案:

答案 0 :(得分：1)

此代码存在一些问题。我建议你重新开始并使用增量编程：编写几行代码，调试它们，并且在它们以你想要的方式工作之前不要继续。你没有得到你的评论似乎假设的一些功能。

  1. 我不确定您访问的节点列表是否按预期运行。每次返回主程序中的 for 循环时，都会重置此项。
  2. 您的主程序没有点击最后一个节点，30。
  3. 我怀疑你的主要问题是你的节点符号。我不确定您打算使用 [7，[8]，[9]] 之类的列表，其中节点编号位于不同的嵌套级别，但您的逻辑似乎假设没有筑巢。你找不到一些东西，例如，你在列表 [10，[11]，[12]，...] 中搜索元素 11 - - 这是 False 。 [11] 是后一个列表的元素，但 11 不是。

    4. 请参阅这个可爱的debug博客以获取帮助。

    我做了一些调试，包括将一些变量名更改为更有意义的名称。这就是我目前所拥有的，严格的低技术印刷声明。

    graphed = {
    1: [25, 30], 2: [11], 3: [13], 4: [17], 5: [17],
    6: [26], 7: [11, 12], 8: [10, 13], 9: [14, 26],
   10: [8, 11, 15], 11: [2, 7, 10], 12: [7, 16, 17],
   13: [3, 8, 14], 14: [9, 13, 20], 15: [10, 19],
   16: [12, 18], 17: [4, 5, 12], 18: [16, 21, 22],
   19: [15, 28, 29], 20: [14, 25], 21: [18, 23],
   22: [18, 24], 23: [21, 27], 24: [22, 27], 25: [1, 20],
   26: [6, 9], 27: [23, 24], 28: [19], 29: [19], 30: [1]
}

letters = {
    'S': [1],
    'C': [10, [11], [12], [13], [14], [15], [16], [17],
              [18], [19], [20], [21], [22], [23], [24],
              [25], [26], [27], [28], [29], [30] ],
    'O': [2, [3], [4], [5], [6]],
    'N': [7, [8], [9]]}


node_path = 'CO'

def dfs(graph, start, num, visited = None):
    print("ENTER dfs", start, num, visited)
    # depth first search for connected nodes
    start_ltr = node_path[num]
    end_ltr = node_path[(num+1)]
    if visited is None:
        visited = []

    if start in visited:
        return

    visited.append(start)
    # if start corresponds to first letter of string, continue
    print("  search for start node", start, "in list", start_ltr, letters.get(start_ltr))
    if start in letters.get(start_ltr):
        # find all nodes that it is connected too
        for each in [x for x in graph[start] if x not in visited]:
            print("    found", each, "in", graph[start]) 
            print("    search in list", end_ltr, letters.get(end_ltr))
            # if any of the connected nodes correspond to
            #   the next letter of string, continue...
            if [each] in letters.get(end_ltr):
                dfs(graph, each, (num+1), visited) #recursion

    print("LEAVE dfs", visited)
    return visited

lst1 = []
for i in range(1,15):    # I cut back the list for less output.
     lst1.append(dfs(graphed, i, 0))
print(lst1)

    输出：

    $ python3 so.py
ENTER dfs 1 0 None
  search for start node 1 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [1]
ENTER dfs 2 0 None
  search for start node 2 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [2]
ENTER dfs 3 0 None
  search for start node 3 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [3]
ENTER dfs 4 0 None
  search for start node 4 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [4]
ENTER dfs 5 0 None
  search for start node 5 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [5]
ENTER dfs 6 0 None
  search for start node 6 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [6]
ENTER dfs 7 0 None
  search for start node 7 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [7]
ENTER dfs 8 0 None
  search for start node 8 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [8]
ENTER dfs 9 0 None
  search for start node 9 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [9]
ENTER dfs 10 0 None
  search for start node 10 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
    found 8 in [8, 11, 15]
    search in list O [2, [3], [4], [5], [6]]
    found 11 in [8, 11, 15]
    search in list O [2, [3], [4], [5], [6]]
    found 15 in [8, 11, 15]
    search in list O [2, [3], [4], [5], [6]]
LEAVE dfs [10]
ENTER dfs 11 0 None
  search for start node 11 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [11]
ENTER dfs 12 0 None
  search for start node 12 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [12]
ENTER dfs 13 0 None
  search for start node 13 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [13]
ENTER dfs 14 0 None
  search for start node 14 in list C [10, [11], [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27], [28], [29], [30]]
LEAVE dfs [14]
[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [13], [14]]

    这会让你感动吗？

答案 1 :(得分：1)

您的letters词典格式不正确。

'C': [10, [11], [12], [13], [14], [15], [16], [17],
          [18], [19], [20], [21], [22], [23], [24],
          [25], [26], [27], [28], [29], [30] ],
'O': [2, [3], [4], [5], [6]],
'N': [7, [8], [9]]}

请注意每个字母的第一个数字是如何直接在顶级列表中，但每个其他条目本身都在子列表中。因此，对于可能连接到O的每个C节点，if start in letters.get(k)将为false。

>>> 10 in letters.get("C")
True
>>> 11 in letters.get("C")
False

所以我把字典改成了这个：

letters = {
    'S': [1],
    'C': [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
    'O': [2, 3, 4, 5, 6],
    'N': [7, 8, 9]}

您必须相应地更改此行：

if [each] in letters.get(y):

到

if each in letters.get(y):

另外，为了让每次递归调用的结果实际传回，我不得不改变这一行：

dfs(graph, each, (num+1), visited) #recursion

到

return dfs(graph, each, (num+1), visited) #recursion

最后，在方法的顶部添加一个特殊情况，以便找到所有some_string时：

def dfs(graph, start, num, visited = None):
    # depth first search for connected nodes
    if (num+1 == len(some_string)):
        return visited + [start]

此时，搜索成功，因此没有更多信件可以搜索！

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