where each branch contains all the points generated by dividing each curve, so if you divide into 10 segments, you'll get:
{0;0}(N = 11)
{0;1}(N = 11)
{0;2}(N = 11)
{0;3}(N = 11)
{0;4}(N = 11)
Where the second integer in the curly brackets refers back to the index of the curve in the original list.
Another way to look at this data is to see it as a table. It's got 5 rows (one for each original curve) and 11 columns, where every column contains a specific division point.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Maybe with numbers it would be easier to understand.
If there is 1 insert to the list 5
If there is a 5 insert to the list 6 or 4
So if i start from 1 one possible list would be: {1, 5, 6}
element connectivities in text, when matched with the corresponding coordinates of the nodes from dupPt, it makes all the polygons again (Green lines in the picture below), exactly as the original.
So Im thinking its like reconstructing everything again from the original shape with new relation between coordinates and element connectivities.
So in this case, eg: 0 1 2 3 4 5 makes the hexagon, 21 20 0 5 to makes the trapz.
Well here it misses the '0' in the end to close the hexagon here, the same with trapz. However the 'string join' and cset somehow close it to 0 1 2 3 4 5 0 to makes the hexagon, and 21 20 0 5 21 for trapz.
Im very glad now.
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