How to calculate corners based on min and max

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Mats
Mats on 2 Nov 2017
Commented: Guillaume on 2 Nov 2017
If i have an n-dimensional space contained by straight walls (distinct max and min in each dimension). I need to find the position of the corner points.
For example in 3D I have the min and max values in each dimension in a 2x3 matrix. Is there any good way from that to calculate the cornerpoints of the shape that encloses that volume?
  2 Comments
Mats
Mats on 2 Nov 2017
Hi Adam,
Yes, but a good way to combine the coordinates to get each cornerpoint on one row with (x,y,z)

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Answers (1)

Jos (10584)
Jos (10584) on 2 Nov 2017
For the 3D case, try this:
mm = [0 2 ; 1 3 ; 4 5] % column 1 is minimum, column 2 is maximum, rows are [x y z]
[px,py,pz] = ndgrid(mm(1,:),mm(2,:),mm(3,:))
P = [px(:) py(:) pz(:)] % the 8 corners of a 3D cube
Flexible ND case:
mm = [0 2 ; 1 3 ; 4 5 ; -1 1 ; 10 20] ; % 5D
nd = size(mm,1) ; % number of dimensions
mmC = mat2cell(mm,ones(1,nd), 2) ;
[p{1:nd}] = ndgrid(mmC{:}) ;
P = reshape(cat(nd,p{:}),[],nd) % the 32 corners of a 5D "cube"
  1 Comment
Guillaume
Guillaume on 2 Nov 2017
To avoid problems if p already exists and is not a cell array (or a larger cell array than required) I'd use:
p = cell(1, nd);
[p{:}] = ndgrid(mmC{:});

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