# How to caluclate centroid of a voronoi cell of a voronoi diagram.

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Pallov Anand on 3 Jan 2023
Commented: Pallov Anand on 4 Jan 2023
Suppose I have generated a voronoi diagram by following code:
X=[1 2 1.1 1.3 1.4 1.5 1.3 1.2 1.8 2.1 2.2];
Y=[1.5 1.3 1.5 1.8 1.4 1.6 2.5 2.3 2.4 1.1 1.8];
voronoi(X,Y)
Now, once the voronoi diagram is generated, how to find the centroid of each voronoi cell. I know there are lot of algorithms. One of them is Lloyd's Algorithm, the code of which is given here
but after running the script , I am getting this error:
'poly2cw' requires Mapping Toolbox.
Error in lloydsAlgorithm>VoronoiBounded (line 178)
[X2, Y2] = poly2cw(V(C{ij},1),V(C{ij},2));
Error in lloydsAlgorithm (line 89)
[v,c]=VoronoiBounded(Px,Py, crs);
Can anyone help me in this.
Bruno Luong on 3 Jan 2023
Edited: Bruno Luong on 3 Jan 2023
The Vorinoi cells that contain a seed on the hull are unbounded, there is no centroid for such cells.
MATLAB voronoi just cuts them with an empiric bounding box. So of you use those outer vertexes the result is randomly cut.

Matt J on 3 Jan 2023
Edited: Matt J on 3 Jan 2023
The voronoi cells are always convex, so assuming it is bounded, you can just take the mean of all the vertices of each cell.
X=[1 2 1.1 1.3 1.4 1.5 1.3 1.2 1.8 2.1 2.2];
Y=[1.5 1.3 1.5 1.8 1.4 1.6 2.5 2.3 2.4 1.1 1.8];
[V,C]=voronoin([X;Y]');
centroids = cell2mat( cellfun(@(c) mean(V(c,:),1)' , C','uni', 0) )
centroids = 2×11
Inf 1.9180 1.1707 1.2800 1.1995 1.6029 Inf Inf Inf Inf Inf Inf 1.3544 1.4421 1.8802 0.5374 1.6873 Inf Inf Inf Inf Inf
Pallov Anand on 4 Jan 2023

This is not a trivial problem especially because the voronoi algorithm does not give you a series of closed polygons, i.e. change the axis of your current problem
X=[1 2 1.1 1.3 1.4 1.5 1.3 1.2 1.8 2.1 2.2];
Y=[1.5 1.3 1.5 1.8 1.4 1.6 2.5 2.3 2.4 1.1 1.8];
voronoi(X,Y)
axis([-2 4 -3 4])
You will see that the voronoi is finding lines that divide the points that you have provided, but it is not essentially generating polygons. Only some of these would be closed and then a centroid makes sense, but not for all of them
Pallov Anand on 3 Jan 2023
Edited: Pallov Anand on 3 Jan 2023
Ya, right. Lets take one region with vertices:
(1.7038, 1.3731) ; (1.8661, 1.6435); (1.7780, 1.9520); (1.6864, 1.9864); (1.29, 1.59); (1.2929, 1.5786)
For this region, I think centroid can be calculated and it should lie inside the region.

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