计算依赖图偏序的算法

发布于 2024-10-18 01:21:21 字数 1655 浏览 5 评论 0原文

我正在尝试计算依赖图的部分“拓扑排序”,准确地说,它实际上是一个 DAG(有向无环图);以便并行执行任务而不会产生依赖冲突。

我想出了这个简单的算法,因为我在谷歌上找到的东西并没有那么有用(我一直只找到本身并行运行以计算正常拓扑排序的算法)。

visit(node)
{
    maxdist = 0;
    foreach (e : in_edge(node))
    {
        maxdist = max(maxdist, 1 + visit(source(e)))
    }
    distance[node] = maxdist;
    return distance[node];
}

make_partial_ordering(node)
{
    set all node distances to +infinite;
    num_levels = visit(node, 0);
    return sort(nodes, by increasing distance) where distances < +infinite;
}

(请注意,这只是伪代码,肯定会有一些可能的小优化)

至于正确性,似乎非常明显:对于叶子(:=没有进一步依赖的节点),到叶子的最大距离始终是0(正确,因为循环由于 0 个边而被跳过)。 对于连接到节点 n1,..,nk 的任何节点,到叶子的最大距离为 1 + max{distance[n1],..,distance[nk]}。

在写下算法后,我确实找到了这篇文章:http://msdn。 microsoft.com/en-us/magazine/dd569760.aspx 但老实说,为什么他们要进行所有列表复制等等,这看起来效率真的很低……?

无论如何,我想知道我的算法是否正确,如果是,它叫什么,以便我可以阅读一些有关它的东西。

更新:我在程序中实现了该算法,它似乎对我测试的所有内容都非常有效。 从代码角度来看,它看起来像这样:

  typedef boost::graph_traits<my_graph> my_graph_traits;
  typedef my_graph_traits::vertices_size_type vertices_size_type;
  typedef my_graph_traits::vertex_descriptor vertex;
  typedef my_graph_traits::edge_descriptor edge;

  vertices_size_type find_partial_ordering(vertex v,
      std::map<vertices_size_type, std::set<vertex> >& distance_map)
  {
      vertices_size_type d = 0;

      BOOST_FOREACH(edge e, in_edges(v))
      {
          d = std::max(d, 1 + find_partial_ordering(source(e), distance_map));
      }

      distance_map[d].insert(v);

      return d;
  }

I'm trying to compute a partial "topological sort" of a dependency graph, which is actually a DAG (Directed Acyclic Graph) to be precise; so as to execute tasks without conflicting dependencies in parallel.

I came up with this simple algorithm because what I found on Google wasn't all that helpful (I keep finding only algorithms that themselves run in parallel to compute a normal topological sort).

visit(node)
{
    maxdist = 0;
    foreach (e : in_edge(node))
    {
        maxdist = max(maxdist, 1 + visit(source(e)))
    }
    distance[node] = maxdist;
    return distance[node];
}

make_partial_ordering(node)
{
    set all node distances to +infinite;
    num_levels = visit(node, 0);
    return sort(nodes, by increasing distance) where distances < +infinite;
}

(Note that this is only pseudocode and there would certainly be a few possible small optimizations)

As for the correctness, it seems pretty obvious: For the leafs (:= nodes which have no further dependency) the maximum distance-to-leaf is always 0 (correct because the loop gets skipped due to 0 edges).
For any node connected to nodes n1,..,nk the maximum distance-to-leaf is 1 + max{distance[n1],..,distance[nk]}.

I did find this article after I had written down the algorithm: http://msdn.microsoft.com/en-us/magazine/dd569760.aspx
But honestly, why are they doing all that list copying and so on, it just seems so really inefficient..?

Anyway I was wondering whether my algorithm is correct, and if so what it is called so that I can read up some stuff about it.

Update: I implemented the algorithm in my program and it seems to be working great for everything I tested.
Code-wise it looks like this:

  typedef boost::graph_traits<my_graph> my_graph_traits;
  typedef my_graph_traits::vertices_size_type vertices_size_type;
  typedef my_graph_traits::vertex_descriptor vertex;
  typedef my_graph_traits::edge_descriptor edge;

  vertices_size_type find_partial_ordering(vertex v,
      std::map<vertices_size_type, std::set<vertex> >& distance_map)
  {
      vertices_size_type d = 0;

      BOOST_FOREACH(edge e, in_edges(v))
      {
          d = std::max(d, 1 + find_partial_ordering(source(e), distance_map));
      }

      distance_map[d].insert(v);

      return d;
  }

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请远离我 2024-10-25 01:21:21

你的算法(C++)可以工作,但它的最坏情况时间复杂度非常差。它会在节点上计算到该节点的尽可能多的路径的 find_partial_ordering 。在树的情况下,路径的数量是1,但是在一般的有向无环图中,路径的数量可以是指数的。

您可以通过记忆您的find_partial_ordering结果并返回而不递归来解决此问题您已经计算了特定节点的值。然而,这仍然给你留下了一个堆栈破坏递归解决方案。

维基百科上给出了一种有效的(线性)拓扑排序算法。这不符合您的需求吗?

编辑:啊,我明白了,您想知道深度边界在哪里,以便可以在给定深度上并行化所有内容。您仍然可以使用维基百科上的算法来实现此目的(从而避免递归)。

首先,用维基百科上的算法进行拓扑排序。 现在通过按拓扑顺序访问节点来计算深度:

depths : array 1..n
for i in 1..n
    depths[i] = 0
    for j in children of i
        depths[i] = max(depths[i], depths[j] + 1)
return depths

请注意,上面没有递归,只是一个简单的 O(|V| + |E|) 算法。这与维基百科上的算法具有相同的复杂性。

Your algorithm (C++) works, but it has very bad worst-case time complexity. It computes find_partial_ordering on a node for as many paths as there are to that node. In the case of a tree, the number of paths is 1, but in a general directed acyclic graph the number of paths can be exponential.

You can fix this by memoizing your find_partial_ordering results and returning without recursing when you have already computed the value for a particular node. However, this still leaves you with a stack-busting recursive solution.

An efficient (linear) algorithm for topological sorting is given on Wikipedia. Doesn't this fit your needs?

Edit: ah, I see, you want to know where the depth boundaries are so that you can parallelize everything at a given depth. You can still use the algorithm on Wikipedia for this (and so avoid recursion).

First, Do a topological sort with the algorithm on Wikipedia. Now compute depths by visiting nodes in topological order:

depths : array 1..n
for i in 1..n
    depths[i] = 0
    for j in children of i
        depths[i] = max(depths[i], depths[j] + 1)
return depths

Notice that there's no recursion above, just a plain O(|V| + |E|) algorithm. This has the same complexity as the algorithm on Wikipedia.

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