There is a linear-time algorithm mentioned at http://en.wikipedia.org/wiki/Longest_path_problem
Here is a (very lightly tested) implementation
EDIT, this is clearly wrong, see below. +1 for future testing more than lightly before posting
import networkx as nx
def longest_path(G):
dist = {} # stores [node, distance] pair
for node in nx.topological_sort(G):
pairs = [[dist[v][0]+1,v] for v in G.pred[node]] # incoming pairs
if pairs:
dist[node] = max(pairs)
else:
dist[node] = (0, node)
node, max_dist = max(dist.items())
path = [node]
while node in dist:
node, length = dist[node]
path.append(node)
return list(reversed(path))
if __name__=='__main__':
G = nx.DiGraph()
G.add_path([1,2,3,4])
print longest_path(G)
EDIT: Corrected version (use at your own risk and please report bugs)
def longest_path(G):
dist = {} # stores [node, distance] pair
for node in nx.topological_sort(G):
# pairs of dist,node for all incoming edges
pairs = [(dist[v][0]+1,v) for v in G.pred[node]]
if pairs:
dist[node] = max(pairs)
else:
dist[node] = (0, node)
node,(length,_) = max(dist.items(), key=lambda x:x[1])
path = []
while length > 0:
path.append(node)
length,node = dist[node]
return list(reversed(path))
if __name__=='__main__':
G = nx.DiGraph()
G.add_path([1,2,3,4])
G.add_path([1,20,30,31,32,4])
# G.add_path([20,2,200,31])
print longest_path(G)
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