kh_si0567 kh_si0567 - 10 months ago 100
Python Question

Lowest common ancestor in python's NetworkX

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

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:
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.

Answer Source

Rerunning Dijkstra in a loop indeed seems like an overkill.

Say we build the digraph obtained by reversing the edges:

import networkx as nx

G = nx.DiGraph() #Directed graph
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)]

Now we run BFS from each of the two nodes:

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

Finding the common vertices reachable from both nodes in the reversed graph is easy:

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

Now you can query the above easily. For example, to find the common vertex reachable from both, closest to 1, is:

>>> min(common_preds, key=lambda n: preds_1[n])