3

Using NetworkX

I want to get lowest common ancestor from node1 and node11 in DiGraph.

The following is the code.

import networkx as nx

G = nx.DiGraph() #Directed graph
G.add_nodes_from([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15])
G.add_edges_from([(2,1),(3,1),(4,1),(5,2),(6,2),(7,3),(8,3),(9,3),(10,4),(10,12),(14,9),(15,8),(12,11),(13,11),(14,12),(15,13)])

ancestors1 = nx.ancestors(G, 1)
ancestors2 = nx.ancestors(G, 11)

src_set = set(ancestors1)
tag_set = set(ancestors2)
matched_list = list(src_set & tag_set)

dic = {}

for elem in matched_list:
    print elem
    length1 = nx.dijkstra_path_length(G, elem, 1)
    path1 = nx.dijkstra_path(G, elem, 1)
    dist1 = len(path1)
    length2 = nx.dijkstra_path_length(G, elem, 11)
    path2 = nx.dijkstra_path(G, elem, 11)
    dist2 = len(path2)
    dist_sum = dist1 + dist2
    dic[elem] = dist_sum

min_num = min(dic.values()) 
for k, v in sorted(dic.items(), key=lambda x:x[1]):
    if v != min_num:
        break
    else:
        print k, v

I have a problem with a execution speed, so I want faster execution speed.

If you have any good idea or algorithm, please tell me the idea.

Sorry for the poor English.

4

2 回答 2

6

在循环中重新运行 Dijkstra 确实看起来有点矫枉过正。

假设我们构建通过反转边缘获得的有向图：

import networkx as nx

G = nx.DiGraph() #Directed graph
G.add_nodes_from([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15])
edges = [(2,1),(3,1),(4,1),(5,2),(6,2),(7,3),(8,3),(9,3),(10,4),(10,12),(14,9),(15,8),(12,11),(13,11),(14,12),(15,13)]
G.add_edges_from([(e[1], e[0]) for e in edges])

现在我们从两个节点中的每一个运行 BFS：

preds_1 = nx.bfs_predecessors(G, 1)
preds_2 = nx.bfs_predecessors(G, 11)

在反向图中找到两个节点都可达的公共顶点很容易：

common_preds = set([n for n in preds_1]).intersection(set([n for n in preds_2]))

现在您可以轻松查询上述内容。例如，要找到两者都可达的公共顶点，最接近 1，是：

>>> min(common_preds, key=lambda n: preds_1[n])
10
于 2016-10-08T17:42:12.287 回答
1

您可以使用 networkx.algorithms.lowest_common_ancestor 中定义的函数来执行此操作。它有一个仅适用于树的更快版本，以及一个适用于任何有向无环图的较慢版本。由于您的示例是 DAG，因此我将使用后者。

# Same as in your example
import networkx as nx
G = nx.DiGraph() #Directed graph
G.add_nodes_from([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15])
G.add_edges_from([(2,1),(3,1),(4,1),(5,2),(6,2),(7,3),(8,3),(9,3),(10,4),(10,12),(14,9),(15,8),(12,11),(13,11),(14,12),(15,13)])

# Compute the lowest common ancestor of 1 and 11. This will print 15.
print nx.algorithms.lowest_common_ancestor(G, 1, 11)

请注意，这会计算由与图源的最大距离定义的最低共同祖先；根据这个定义，10 和 15 都是 (1, 11) 的最低共同祖先，而这恰好给出 15。您的原始代码（和前面的答案）还最小化了从 1 和 11 到 LCA 的路径，这给出了 10 (总和距离为 6）而不是 15（总距离为 7）。如果您也需要该属性，那么此模块可能不适合。

使用这个模块，为单对节点找到 LCA 的运行时间在图的大小上是 O(n log(n))。为所有节点对找到 LCA 的运行时间是 O(n^3)。

于 2018-07-14T06:03:09.380 回答