9

一些背景

我正在分析一个函数的控制流图，该函数基本上将传入数据映射到传出数据。大多数块都是这样的：

if (input_variable == SPECIFIC_CONSTANT) {
    output_variable = TRUE;
}
else {
    output_variable = FALSE;
}

此类代码的典型控制流图如下图所示

digraph G {
  2 -> 3 -> 5;
  2 -> 4 -> 5;
}

图形的图片

其中，3and的执行取决于but4的值是独立的。input_variable5

问题

给定一个有向图和一个起始节点，我如何找到最近的节点，以使起始节点的任何路径都通过该节点？

示例：鉴于此图，我如何找到6从2或12开始8？

我可以反转最低共同祖先算法吗？它会有效吗？喜欢

for each node in graph:
    ancestors = node.get_all_ancestors()
    lca = find_lowest_common_ancestor(ancestors)
    junction_node[lca] = node

get_junction_point(node):
    return junction_node[node]

我的编程语言是 Python，我刚刚发现了 NetworkX，但任何算法都会受到赞赏。我不习惯图论，我想我错过了基本词汇表来找到我正在寻找的东西。

谢谢你的帮助！

4

3 回答 3

2

不是最有效的解决方案，但这里有一些东西可以帮助您入门：

做一个 DFS，然后计算所有路径的交集（每个路径中存在的节点）。然后，在这些节点中，找到每条路径中最接近起点的节点：

>>> paths
[]
>>> def dfs(G, s, path):
...     if s not in G or not G[s]:
...             paths.append(path)
...     else:
...             for t in G[s]:
...                     dfs({k:[_t for _t in v if _t!=t] for k,v in G.items()}, t, path+[t])
... 
>>> dfs(G, 2, [])
>>> paths
[[3, 4, 6, 7, 12], [3, 4, 6, 8, 9, 10, 12], [3, 4, 6, 8, 9, 12], [3, 4, 6, 8, 11, 12], [3, 5, 6, 7, 12], [3, 5, 6, 8, 9, 10, 12], [3, 5, 6, 8, 9, 12], [3, 5, 6, 8, 11, 12], [4, 6, 7, 12], [4, 6, 8, 9, 10, 12], [4, 6, 8, 9, 12], [4, 6, 8, 11, 12]]
>>> for path in paths: print(path)
... 
[3, 4, 6, 7, 12]
[3, 4, 6, 8, 9, 10, 12]
[3, 4, 6, 8, 9, 12]
[3, 4, 6, 8, 11, 12]
[3, 5, 6, 7, 12]
[3, 5, 6, 8, 9, 10, 12]
[3, 5, 6, 8, 9, 12]
[3, 5, 6, 8, 11, 12]
[4, 6, 7, 12]
[4, 6, 8, 9, 10, 12]
[4, 6, 8, 9, 12]
[4, 6, 8, 11, 12]
>>> nodes = [set(L) for L in paths]
>>> commons = functools.reduce(set.intersection, nodes)
>>> commons
{12, 6}
>>> min(commons, key=lambda v: max(L.index(v) for L in paths))
6

现在，请注意如何6在某些路径中的索引 2 处和在某些其他路径中的索引 1 处显示。如果有一个节点（比如 x），它出现在路径中的索引 1 处，其中 6 出现在索引 2 处，而在索引 2 处，6 出现在索引 1 处，那么，这将是一个平局，这个算法会随意破坏。根据您的需要，您可能想要定义如何更好地处理这种情况

于 2013-11-12T17:24:03.097 回答
0

你可以这样做：

对于每个节点，找到其所有祖先和后代的列表。如果 Size(ancestors) + size(descendants) + 1 等于网络大小，则它是候选者。现在，找到这样一个具有至少一个祖先和最大后代数的节点。

祖先和后代的列表可以很容易地计算出来。如果您不确定，请告诉我，我会扩展我的答案。

于 2013-11-12T17:14:45.767 回答
0

阅读了您提出的所有解决方案，我想出了一个主意。我给我的第一个节点一个数量 1。递归地，所有孩子都收到这个数量的相等部分。反过来，他们向下发送这个数量。如果一个孩子总共收到 1（起始金额），那么它就是“连接点”。这是我的实现（开放评论！！）。

我希望 BFS 构造限制访问的节点数量。

class Node(object):
  """
    Object representing a node in a graph where we search the junction node that
    concentrates all paths starting from a start node.

    ``reference``: Attaches the real object that contains the relationships.
    ``initial_value``: Initial amount for the node. Typically 1 for the start
      node, 0 for the others.
    ``path``: Set of already traversed nodes that reached the node. Used to
      prune circular dependencies.
  """
  def __init__(self, reference, initial_value, path=set()):
    self.reference = reference
    # See dispatch() for explaination
    self.value_can_dispatch = self.value_has_received = initial_value
    self.path = path

  def receive(self, value):
    """
      Give ``value`` to the node. If the node received 1 (or more for security)
      in total, it will return True. Else it returns False.
    """
    self.value_has_received += value
    self.value_can_dispatch += value
    if self.value_has_received >= 1.:
      return True
    return False

  def dispatch(self, children):
    """
      Dispatch the value received to the children.
      Returns a filtered list of ``children`` where children involved in a
      circular dependency are removed.
      If one child signals that it has received a total of 1, the function will
      stop and return this one child.
    """
    # Filter successors that are in the path used to access this node so to cut
    # out cycles
    true_successors = [child for child in children if child not in self.path]
    # Cut the received value into equal pieces
    amount = self.value_can_dispatch/len(true_successors)
    # We transmit only the value received after the last time it was dispatched
    # because paths may lead to the same node at different iterations (path
    # through one node may be longer than through another) and thus the same
    # node can be asked to dispatch to its children more than once.
    # The total amount of received value is kept in value_has_received because
    # the node may receive the complete amount in several steps. Thus, part of
    # its value may have been dispatched before we notice that the node received
    # the total amount of 1.
    self.value_can_dispatch = Fraction(0)
    for child in true_successors:
      # If child signaled that he received 1, then raise the winner
      if child.receive(amount):
        return child
    return set(true_successors)

  def touch(self, other_node):
    """
      "Touches" a node with another, notifying that the node is reachable
      through another path than the known ones.
      It adds the elements of the new path as ancestors of the node.
    """
    self.path |= other_node.path | {other_node}

  def make_child(self, reference):
    """
      Creates a child of the node, pointing to reference. The child receives
      its path from the current node.
    """
    # This is were the algorithm can go mad. If child is accessed through two
    # paths, the algorithm will only protect recursion into the first
    # path. If the successors recurse into the second path, we will not detect
    # it. => We should update the child's path when a second path reaches it.
    return self.__class__(reference, Fraction(0), self.path | {self})

  def __repr__(self):
    return "<{} {}>".format(self.__class__.__name__, self.reference)

def find_junction_node(first_reference, get_uid, get_successors, max_iterations=100):
  """
    Find the junction node of all paths starting from ``first_reference`` in
    a directed graph. ``get_uid`` is a function accepting a reference to a node
    in your graph and returning a unique identifier for this reference.
    ``get_successors`` is a function accepting a reference to a node in your
    graph. It should return a list of references to all its the children nodes.
    It may return None if the node has no child.
    ``max_iterations`` limits the number of pass the algorithm use to find the
    junction node. If reached, the funciton raises a RuntimeError.

    Returns ``(jp, ln)`` where ``jp`` is a reference to a node in your graph
    which is the junction node and ``ln`` the list of nodes in the subgraph
    between the start node and the junction node.
  """
  # Mapping to already created nodes
  nodes = {}
  # Initialise first node with an amount of 1
  node = Node(first_reference, Fraction(1, 1))
  nodes[get_uid(first_reference)] = node
  # Initialise first iteration of DFS
  successors = set()
  successors.add(node)
  # Max iteration provides security as I'm not sure the algorithm cannot loop
  for i in range(max_iterations):
    next_successors = set()
    # Process one level of nodes
    for node in successors:
      # Find successors in data graph
      sub_references = get_successors(node.reference)
      # This happens when we reach the end of the graph, node has no children
      if sub_references is None:
        continue
      # Make a list of Node that are children of node
      children = set()
      for reference in sub_references:
        uid = get_uid(reference)
        # Does it exist?
        child = nodes.get(uid, None)
        if not child:
          child = node.make_child(reference)
          nodes[uid] = child
        else:
          child.touch(node)
        children.add(child)
      # Dispatch the value of node equally between its children
      result = node.dispatch(children)
      #print("Children of {}: {!r}".format(node, result)) # DEBUG
      # If one child received a total of 1 from its parents, it is common to
      # all paths
      if isinstance(result, Node):
        return result.reference,  [node.reference for node in result.path]
      # Else, add the filtered list of children to the set of node to process
      # in the next level
      else:
        next_successors |= result
    successors = next_successors
    # Reached end of graph by all paths without finding a junction point
    if len(successors) == 0:
      return None
  raise RuntimeError("Max iteration reached")
于 2013-11-12T18:46:14.613 回答