Calculating all paths from a given node in a digraph

51 views (last 30 days)
Niels de Vries
Niels de Vries on 4 Sep 2018
Answered: Pierre Harouimi on 29 Dec 2021 at 6:48
Hey all,
I am using the digraph function and trying to find all paths from a given source node, i was wondering if there already exist a object function to do this such as the shortest path object functions.
As an example:
The output i would like:
[1 2 4]
[1 2 5]
[1 2 6]
[1 3 7 9]
[1 3 7 8]

Accepted Answer

Guillaume on 4 Sep 2018
I've not tested it thoroughly but I think this should work:
function paths = getpaths(g)
%return all paths from a DAG.
%the function will error in toposort if the graph is not a DAG
paths = {}; %path computed so far
endnodes = []; %current end node of each path for easier tracking
for nid = toposort(g) %iterate over all nodes
if indegree(g, nid) == 0 %node is a root, simply add it for now
paths = [paths; nid]; %#ok<AGROW>
endnodes = [endnodes; nid]; %#ok<AGROW>
%find successors of current node and replace all paths that end with the current node with cartesian product of paths and successors
toreplace = endnodes == nid; %all paths that need to be edited
s = successors(g, nid);
if ~isempty(s)
[p, tails] = ndgrid(paths(toreplace), s); %cartesian product
paths = [cellfun(@(p, t) [p, t], p(:), num2cell(tails(:)), 'UniformOutput', false); %append paths and successors
endnodes = [tails(:); endnodes(~toreplace)];

Sign in to comment.

More Answers (2)

Walter Roberson
Walter Roberson on 4 Sep 2018
Mathworks does not provide any function for that purpose. Perhaps the graph theory toolbox in the File Exchange?
Your text asks for "all paths", and your example is a digraph that happens to have "in degree" 1 for all nodes. In the special case of a digraph with "in degree" 1 for all nodes, then "all paths" becomes the same as all shortest path tree .
This routine will not work for cases where the in degree is more than 1, such as if node 3 also pointed to node 9: in that case the "shortest" path choices in the routine would prune out some of the paths.
Niels de Vries
Niels de Vries on 4 Sep 2018
It indeed is, backtracking is needed, but i am not sure how to implement something like that in Matlab. An example digraph is added, and unfortunately couldn't find something in the file exchange

Sign in to comment.




Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!