在包含超过 1 个循环的字典中查找所有循环

Question

给定以下循环字典：

{'E': ['F'], 'F': ['C', 'G'], 'C': ['E'], 'G': ['H'], 'H': ['I'], 'I': ['J', 'D'], 'J': ['D'], 'D': ['E']}

其中键代表父节点，值是它指向的所有子节点（这表示有向图）。

可以看出有3个循环：

1. E -> F -> C
2. E -> F -> G -> H -> I -> J -> D
3. E -> F -> G -> H -> I -> D

我正在尝试找出一种能够提取所有循环的方法或函数，以便我可以输出或返回三个循环：

EFC

EFGHIJD

EFGHID

我正在寻找 Python 中的解决方案！谢谢:)

Answer 1

获取一个起始节点的所有周期的一种方法是：

def cycles(graph, start):
    results = []
    paths = [(start,)]
    while paths:
        path = paths.pop()
        childs = set(graph.get(path[-1], []))
        if start in childs:
            results.append(path)
        paths.extend(
                  (path + (child,))
                  for child in childs.difference(path)
              )
    return results

使用您的图表和起始节点E，您将获得（print(cycles(g, 'E')) 和g 您的字典）：

[('E', 'F', 'G', 'H', 'I', 'D'),
 ('E', 'F', 'G', 'H', 'I', 'J', 'D'),
 ('E', 'F', 'C')]

如果您想要一个图表中的所有循环，您将不得不做更多的工作：一个单一的起始节点通常不会导致一个图表的所有循环。实现这一目标的一种方法是遍历所有节点，应用cycles，然后收集所有结果：

result = [cycle for node in g for cycle in cycles(g, node)]

[('E', 'F', 'G', 'H', 'I', 'J', 'D'),
 ('E', 'F', 'G', 'H', 'I', 'D'),
 ('E', 'F', 'C'),
 ('F', 'G', 'H', 'I', 'J', 'D', 'E'),
 ('F', 'G', 'H', 'I', 'D', 'E'),
 ('F', 'C', 'E'),
 ('C', 'E', 'F'),
 ('G', 'H', 'I', 'J', 'D', 'E', 'F'),
 ('G', 'H', 'I', 'D', 'E', 'F'),
 ('H', 'I', 'J', 'D', 'E', 'F', 'G'),
 ('H', 'I', 'D', 'E', 'F', 'G'),
 ('I', 'J', 'D', 'E', 'F', 'G', 'H'),
 ('I', 'D', 'E', 'F', 'G', 'H'),
 ('J', 'D', 'E', 'F', 'G', 'H', 'I'),
 ('D', 'E', 'F', 'G', 'H', 'I'),
 ('D', 'E', 'F', 'G', 'H', 'I', 'J')]

但这种方式会导致一些冗余：('E', 'F', 'C')、('C', 'E', 'F') 和 ( 'F', 'C', 'E') 本质上是相同的（等等）。如果你想摆脱它，你可以这样做：

results = []
d = {node: set(g.get(node, [])) for node in g}
nodes = list(d)
while nodes:
    node = nodes.pop()
    results.extend(cycles(d, node))
    del d[node]
    for other_node in d:
        d[other_node].discard(node)

print(results):

[('D', 'E', 'F', 'G', 'H', 'I'),
 ('D', 'E', 'F', 'G', 'H', 'I', 'J'),
 ('C', 'E', 'F')]

Answer 2

您可以使用递归生成器函数：

d = {'E': ['F'], 'F': ['C', 'G'], 'C': ['E'], 'G': ['H'], 'H': ['I'], 'I': ['J', 'D'], 'J': ['D'], 'D': ['E']}
def get_cycles(n, o = None, c = []):
   if n == o: #check if current node being examined is the same as the start
      yield c #if the condition is met, then a cycle is found and the path is yielded back
   else:
      #get the children of the current node from d and iterate over only those that have not been encountered before (except if the node is the same as the start)
      for i in filter(lambda x:x not in c or x == o, d.get(n, [])):
         #recursively find the children of this new node `i`, saving the running path (c + [n])
         yield from get_cycles(i, o = n if o is None else o, c = c + [n])

print(list(get_cycles('E')))
print(list(get_cycles('D')))

输出：

[['E', 'F', 'C'], ['E', 'F', 'G', 'H', 'I', 'J', 'D'], ['E', 'F', 'G', 'H', 'I', 'D']]
[['D', 'E', 'F', 'G', 'H', 'I', 'J'], ['D', 'E', 'F', 'G', 'H', 'I']]