# How to create a triangulation from a list of edges and list of nodes?

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Dominik Mattioli am 29 Jul. 2018
Beantwortet: Johannes Korsawe am 14 Okt. 2022
I'm beginning with a triangulation so that I know this graph is actually composed of non-intersecting edges that define a set of triangles.
dt = delaunayTriangulation(rand(101,2)); % Keep it small for now.
edges = unique(sort(... % Unique mx2 list of
[dt.ConnectivityList(:,1:2);... % all edges in dt.
dt.ConnectivityList(:,2:3);...
dt.ConnectivityList(:,[3 1])],...
2),'rows');
figure; triplot(dt);
% Convert dt to graph, plot.
g = graph(edges(:,1),edges(:,2));
figure; g.plot;
% We know that g.edges describe dt, how can we reconstruct it?
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### Antworten (2)

Christine Tobler am 9 Aug. 2018
The problem is that a graph is a more general data structure than a Delaunay triangulation. Many graphs do not represent a triangulation at all, so there is not direct way of doing this.
If you also save the locations of each point in the original triangulation, you could write an algorithm that goes through each point in the graph and finds all the triangles it's a part of.
Could you tell me why you want to retrieve the triangulation from the graph?
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Dominik Mattioli am 14 Aug. 2018
Bearbeitet: Dominik Mattioli am 14 Aug. 2018
Your suggestion is the most intuitive, however, I imagine that being expensive for very large graphs unless it is done very cleverly. I think that the constraints I've given to the problem help with the problem's complexity.
In a nutshell, I will be using the graph class to selectively removing edges from the triangulation (as a graph) to form quadrilaterals.
Basically: create triangulation -> convert to graph -> remove edges to form some quads -> convert back to tri/quad/mixed 'rangulation'.
P.S. I've coded a triangulation class that accommodates quadrilaterals.

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Johannes Korsawe am 14 Okt. 2022
allCyclesWithLengthThree = allcycles(g, 'MaxCycleLength', 3);
connectivityList = cell2mat(allCyclesWithLengthThree);
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