# A function smart enough to sort coordinates of block into sub-blocks?

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Xiaohan Du il 12 Nov 2017
Commentato: David Goodmanson il 13 Nov 2017
Hi all,
Imagine there is a 2D shape made of randomly distributed square blocks, with Cartesian coordinates like this:
I can read the coordinates of the nodes, for this case it is:
>> coords
coords =
1 -1 -1
2 -1 1
3 1 -1
4 1 1
5 -1 0
6 0 -1
7 0 0
8 0 1
9 1 0
10 2 0
11 2 1
12 0 2
13 -1 2
14 -2 0
15 -2 -1
I'd like to have a function which is able to read this coordinates data, and sort it into 4-points square blocks, with coordinates in counter-clockwise, so the output would be:
>> otpt{:}
ans =
1 -1 -1
5 -1 0
6 0 -1
7 0 0
ans =
2 -1 1
5 -1 0
7 0 0
8 0 1
ans =
3 1 -1
6 0 -1
7 0 0
9 1 0
ans =
4 1 1
7 0 0
8 0 1
9 1 0
and so on...... (7 cell blocks in this case)
Moreover, for 3D blocks, such as this case:
If I'm able to read the nodal coordinates, can the function also suit for sorting the sub-blocks in 3D? Results would be 9 cell blocks with 8 x-y-z coordinates.
Matt J was very kind to provide me this solution here,
but it only works for uniform shapes. Thanks for any help!
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### Risposte (1)

David Goodmanson il 13 Nov 2017
Modificato: David Goodmanson il 13 Nov 2017
Hi Xiaohan,
Here is some code that creates a 4x3xn 3d matrix sqquares (misspelled on purpose) where each level of sqquares in the third dimension, i.e. (:,:,k) is one of the solutions. I did not want to deal with a connectivity matrix so this assumes that every pair of vertices that can be connected by a line is connected by a line. For example suppose you had two more vertices at (1,2) and (2,2) and they are connected with the vertices below them to make a square. You could also have a line from (0,2) to (1,2) to make another square, or not. This code presumes that that line is there.
I am not going for cubes, life is too short.
coords = ...
[1 -1 -1
2 -1 1
3 1 -1
4 1 1
5 -1 0
6 0 -1
7 0 0
8 0 1
9 1 0
10 2 0
11 2 1
12 0 2
13 -1 2
14 -2 0
15 -2 -1];
xtaxi = -coords(:,2)+coords(:,2)'; % taxi displacements
ytaxi = -coords(:,3)+coords(:,3)';
r = xtaxi==1&ytaxi==0; % point on right or not, etc.
rd = xtaxi==1&ytaxi==-1;
d = xtaxi==0&ytaxi==-1;
ulef = find(any(r,2)&any(rd,2)&any(d,2)); % upper left corner of squares
n = length(ulef);
sqquares = zeros(4,3,n);
for k=1:n
ind = ulef(k);
vtx = [ind find(r(ind,:)) find(rd(ind,:)) find(d(ind,:))]; % cw order
sqquares(:,:,k) = coords(vtx,:)
end
if you don't have one of the later versions of Matlab with implicit expansion you can do
[temp1 temp2] = meshgrid(coords(:,2),coords(:,2));
xtaxi = temp1-temp2;
and similarly for ytaxi with coords(:,3)
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Xiaohan Du il 13 Nov 2017
Hi David,
First thanks a lot for the solution, I'll have a look and give you some feedback.
What is taxi displacements and why do we need it?
David Goodmanson il 13 Nov 2017
Hi Xiaohan,
These are just the usual displacements in x and y from the point in question, to find nearby corners. Taxi is a reference to the the path that a taxi has to take on a grid of city streets, no cutting across the diagonal. See for example Taxicab Geometry in Wikipedia.

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