How to find all the downstream nodes from a node in a graph?

조회 수: 21(최근 30일)
Hello All,
Hope you are staying safe and doing well.
I am trying to solve a problem where I would need to figure out all the nodes going downstream from a specified node in a MatLab graph.
For example:
I have this sample code which gives me the following figure
G = graph([1,2,3,4,2,6,7,1,9],[2,3,4,5,6,7,8,9,10]);
plot(G)
Now, I need to know the number/IDs/indexes of nodes going out of node 1 towads 2 and beyond. Like, it should give me that Node 2,3,4,5,6,7,and 8 are connected to a tree branch groing out of 1. Please let me know if there is a way to figure that out.
I have tried the command OUTEDGES and successors. But they only give nodes directly connected to 1.
I would appreciate your help and ideas in this problem. Thank you.

채택된 답변

Walter Roberson
Walter Roberson 2020년 6월 8일
Gc = G;
[eid, nid] = outedges(Gc,1);
Gc = rmedge(Gc, eid(nid ~= 2));
downstream_nodes = setdiff(unique(minspantree(Gc,'Root',1).Edges.EndNodes), 1);
  댓글 수: 3
Walter Roberson
Walter Roberson 2022년 3월 23일
편집: Walter Roberson 2022년 3월 23일
G = graph([1,2,3,4,2,6,7,1,9],[2,3,4,5,6,7,8,9,10]);
plot(G)
Gc = G;
[eid, nid] = outedges(Gc,2)
eid = 3×1
1 3 4
nid = 3×1
1 3 6
downstream_nodes = setdiff(nid, 1) %remove the backlink to 1
downstream_nodes = 2×1
3 6

댓글을 달려면 로그인하십시오.

추가 답변(3개)

Christine Tobler
Christine Tobler 2020년 6월 8일
You can call
nearest(G, 1, Inf)
which will find all nodes reachable from node 1 in an infinite radius. This will contain only the reachable nodes.
  댓글 수: 1
Walter Roberson
Walter Roberson 2020년 6월 8일
... Though you still need to remove the (1,9) edge first; or more generally all edges from 1 that are not connected to 2
(since the user only wants to know what is "going out of node 1 towads 2 and beyond")

댓글을 달려면 로그인하십시오.


Christine Tobler
Christine Tobler 2022년 3월 23일
Looking at this again due to the recent comment added, it might be simpler to use a directed graph instead of an undirected one: This way, the concept of "downstream" is represented by "following the arrows".
G = digraph([1,2,3,4,2,6,7,1,9],[2,3,4,5,6,7,8,9,10]);
plot(G)
nearest(G, 1, Inf)
ans = 9×1
2 9 3 6 10 4 7 5 8
nearest(G, 2, Inf)
ans = 6×1
3 6 4 7 5 8
nearest(G, 6, Inf)
ans = 2×1
7 8
  댓글 수: 3
Christine Tobler
Christine Tobler 2022년 4월 1일
Assuming we agree "downstream" means "in the direction of the edges of this digraph", the nodes are already returned by the nearest function above. As Walter mentioned, I'm also assuming that you have no cycles in this directed graph, like is the case for your example.
For edges / branches, it's not as straightforward, the easiest is probably to just get all edges connected to one of the downstream nodes:
G = digraph([1,2,3,4,2,6,7,1,9],[2,3,4,5,6,7,8,9,10]);
plot(G, 'EdgeLabel', 1:numedges(G))
node = 2;
downstreamNodes = nearest(G, node, Inf)
downstreamNodes = 6×1
3 6 4 7 5 8
downstreamEdges = outedges(G, node);
for ii=1:length(downstreamNodes)
downstreamEdges = [downstreamEdges; outedges(G, downstreamNodes(ii))];
end
downstreamEdges
downstreamEdges = 6×1
3 4 5 7 6 8

댓글을 달려면 로그인하십시오.


Steven Lord
Steven Lord 2020년 6월 8일
You want to find all nodes reachable from 1? Those nodes have a finite distance from 1. I'm removing node 1 itself from the list (which is what D > 0 is for.)
G = graph([1,2,3,4,2,6,7,1,9],[2,3,4,5,6,7,8,9,10]);
D = distances(G, 1);
reachable = find(isfinite(D) & D > 0)
In your original graph all nodes (other than 1) are reachable from 1. Let's operate on a slightly different graph, one where there is no edge between say 7 and 8.
G2 = rmedge(G, 7, 8);
D2 = distances(G2, 1);
reachable2 = find(isfinite(D2) & D2 > 0)
In this case, 8 isn't reachable from 1.
Alternately you could ask for the connected components of G or G2 and return all nodes that are in the component containing 1.
C = conncomp(G2);
find(C == C(1)) % This will include 1

범주

Find more on Graph and Network Algorithms in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by