# Finding all shortest paths between given (specific) pair of vertices

I am working with graphs in sage and need a method of finding all shortest paths between some pair (or all pairs) of vertices.

Note that it is important to have all shortest paths registred, not just one, as seen in many Bellman-Ford/Dijkstra implementations (for instance Graph.shortest_path_all_pairs or networkx.algorithms.shortest_paths.all_pairs_shortest_path), and not just a number of those paths.

I am also satisfied with only a list of "optimal" predecessors... as long as the list is complete.

edit retag close merge delete

Sort by » oldest newest most voted

You can easily find the list of all predecessors : just call the "distance_all_pairs" method, and you can then get all predecessors of x in a shortest u-v path as the list of all neighbors p of x such that d(u,p)+1+d(x,v)=d(u,v).

Nathann

more

That is a good approach. Little more work on my side, but it gets the job done.

Would you show a minimum working example so that the answer implementation is clear(er)?

I do not know how to paste the code here, therefore I wrote an answer below.

There seem not to be an existing method in Sage for doing that out of the box, so you should write your own algorithm. If you are lazy, you can still use the following slow brute force method, depending on the size of your graph:

sage: G = graphs.GrotzschGraph()
sage: allp = G.all_paths(1,4)
sage: m = min(len(p) for p in allp) ; m
3
sage: [p for p in allp if len(p) == m]
[[1, 0, 4], [1, 10, 4]]

more

Thank you, but this is too much brute force for me. Anyway, it is good to know all_paths function.

The method, suggested by Nathann is the following:

 for v in G.vertices():
for u in G.vertices():
predec_list[u][v]=[i for i in G.neighbors(v) if all_dist[u][i]==all_dist[u][v]-1]


Given such list (p_l), it is easy to recursively construct a list of shortest paths from vertex a to vertex b, as follows:

def s_p_d(p_l,a,b):
r=[]
if p_l[a,b]==[a]:
r.append([a,b])
else:
for i in p_l[a,b]:
for j in s_p_d(p_l,a,i):
j.append(b)
r.append(j)
return r


rickhg12hs: here is my code snippet. This works for me :) Make sure to have networkx imported.

def list_of_predecessors(G):
all_dist=G.distance_all_pairs()
predec_list= {}
for v in G.vertices():
for u in G.vertices():
predec_list[u,v]=[i for i in G.neighbors(v) if all_dist[u][i]==all_dist[u][v]-1]
return predec_list

def shortest_path_d(G):
a_p= { "init":0}
p_l = list_of_predecessors(G)
for u in G.vertices():
for v in G.vertices():
a_p[(u,v)]=s_p_d(p_l,u,v)
del a_p["init"]
return a_p

def s_p_d(p_l,a,b):
r=[]
if p_l[a,b]==[a]:
r.append([a,b])
else:
for i in p_l[a,b]:
for j in s_p_d(p_l,a,i):
j.append(b)
r.append(j)
return r

more

Sorry, I'm a graph theory noob. Where does all_dist come from? [The above snippet crashes.]

Hello! The all_dist dictionary comes from "distance_all_pairs" method, mentioned by Nathann. If you add something like "all_dist = G.distance_all_pairs()" in the beginning, it should work. If you wish, I can publish full code here, but I would rather not, since I am not really sage master myself :)

Defining all_dist as above isn't quite enough to make the code work (still crashes). Wish @Nathann would post a complete minimum working example (MWE). Perhaps @tmonteil would be kind enough to provide a MWE for the method of @Nathann.

1

Took me a while to see that this produces the shortest path for all pairs. Nice!