algorithmic modeling for Rhino
Hello community!
I have a head scratcher that I desperately need help with.
The Setting: I have a set of partially overlapping surfaces.
The Problem: I would like to cull out the parts of the surfaces that overlap one another. (reference images below).
This is part of a much larger definition that i am working on that contains hundreds of sets of overlapping surfaces and so I have to be weary of computational load and would request you factor that in with your suggestions.
I attached a definition with an example set of surfaces.
Any help would be much appreciated as this is the last piece of the puzzle for my definition.
mohammed
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Beautifully simple! my concern is the Solid difference.. the hair on my neck stands straight when im forced to use solid intersections, they are frighteningly heavy! (atleast in my case when i have to apply this to hundreds of surfaces).
Yeah, two duplicate surfaces! I finally noticed that too. Would have had this much earlier if I had noticed that sooner. :(
OH NO, Im soooo Sorry about that.. I removed the duplicates from my original definition and re-uploaded it just incase this discussion evolves into one of those competitions of achieving an end result with the least number of components and more people wanna take a crack at it!
This is fantastic,, im yet to apply it to all the surfaces because i need to modify it so that it works with my existing tree structure (which my mind seems to not allow me to.. it seems that it knows that its Friday! all im hearing, "leave it till monday"!). Once i do so, ill reupload with all surfaces, perhaps someone may find it useful in the furture.
Here is a method that uses two Anemone loops. The outer loop (white group) interates through each branch, the inner loop (yellow group) iterates through each surface and applies 'RDiff (Region Difference)' to subtract all other regions in that branch, except for itself.
WARNING!!! The 'DeDup' cluster removes duplicate surfaces in one branch and is very specific to this set of data. For this problem, duplicate surfaces will cause any algorithm to fail.
P.S. Once again though, I have to hand it to Hyungsoo Kim for a very clever solution that returns both sets of surfaces - those removed and those remaining.
ok ok, im loving the creativity behind this one...
*runs away to the corner with his laptop to study it further!
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