DRC abutting edges.

Comments

• Matthias

  February 6

  Hi @kareemfarid,

  Here is one solution that is based on the separation of edges into horizontal and vertical ones:

  report("discussion 2669")
  
  l1 = input(1, 0)  # reddish
  l2 = input(2, 0)  # blueish
  
  l1e = l1.edges
  l2e = l2.edges
  
  # split l2e into horizontal and vertical edges
  l2e_h = l2e.with_angle(0)
  l2e_v = l2e.without_angle(0)
  
  # split l1e into horizontal and vertical edges
  l1e_h = l1e.with_angle(0)
  l1e_v = l1e.without_angle(0)
  
  # selects full edges of L2 that touch L1
  l2e_on_l1e_h = l2e_h.interacting(l1e_h)
  l2e_on_l1e_v = l2e_v.interacting(l1e_v)
  
  # l2 two-side butting l1 horizontally, but not vertically
  l2butt_h = l2.interacting(l2e_on_l1e_h, 2).not_interacting(l2e_on_l1e_v)
  
  # l2 two-side butting l1 vertically, but not horizontally
  l2butt_v = l2.interacting(l2e_on_l1e_v, 2).not_interacting(l2e_on_l1e_h)
  
  # combines both versions
  l2butt = l2butt_h + l2butt_v
  
  l2butt.output("L2 butting L1 opposite")
  

  It renders this result:

  

  If you want the L1 rectangle selected, add:

  l2rect = l2.interacting(l2butt)
  

  The following, somewhat more esoteric solution, renders the same result in a more efficient way:

  report("discussion 2669")
  
  l1 = input(1, 0)  # reddish
  l2 = input(2, 0)  # blueish
  
  # select l2 shapes where exactly two l1 edges are touching
  l2_with_two_l1_edges = l2.interacting(l1.edges, 2, 3)
  
  # from this, use a zero-distance (<1 DBU) separation check 
  # with opposite-only filter to select the l2 edges in opposite
  # configuration
  l2_with_l1_opposite = l2_with_two_l1_edges.sep(l1, 1.dbu, whole_edges, only_opposite).first_edges
  
  l2_with_l1_opposite.output("L2 butting L1 opposite (2)")
  

  Matthias
• kareemfarid

  February 9 edited February 9

  Thanks a lot.

  I got another question. In the case of, this implementation:

  report("discussion 2669")
  
  l1 = input(1, 0)  # reddish
  l2 = input(2, 0)  # blueish
  
  # select l2 shapes where exactly two l1 edges are touching
  l2_with_two_l1_edges = l2.interacting(l1.edges, 2, 3)
  
  # from this, use a zero-distance (<1 DBU) separation check 
  # with opposite-only filter to select the l2 edges in opposite
  # configuration
  l2_with_l1_opposite = l2_with_two_l1_edges.sep(l1, 1.dbu, whole_edges, only_opposite).first_edges
  
  l2_with_l1_opposite.output("L2 butting L1 opposite (2)")
  

  How to compute the distance between l2_with_l1_opposite edges. I thought space will not work since the edges belong to same polygon shape (not sure if this is a right assumption). But then width didn't work either.

  Thank you
• Matthias

  February 9

  Hi @kareemfarid,

  First I should correct the above (second) solution. It has this be this:

  # select l2 shapes where exactly two l1 edges are touching
  # (NOTE: the boolean AND between L1 and L2 edges selects
  # only L1 edges that are parallel to L2 edges)
  l2_with_two_l1_edges = l2.interacting(l1.edges & l2.edges, 2, 3)
  
  # from this, use a zero-distance (<1 DBU) separation check 
  # with opposite-only filter to select the l2 edges in opposite
  # configuration
  l2_with_l1_opposite = l2_with_two_l1_edges.sep(l1, 1.dbu, whole_edges, only_opposite).first_edges
  
  l2_with_l1_opposite.output("L2 butting L1 opposite (2)")
  

  Reason: the interacting test will otherwise fail in the general case as l1.edges that are perpendicular to l2.edges are counted as well.

  However, testing the resulting edges with "width" works for me:

  # Width check
  l2_with_l1_opposite.width(2.um).output("w<2µm")
  

  This gives me this:

  

  Matthias
• kareemfarid

  February 9

  Yes width appears to be working. Perhaps I missed something while testing. Thanks a lot