Thursday, June 4, 2009

What I know so Far

  • You can always check algorithmically to see if a 4-folder exists. Furthermore, if all the points of intersection of one color are inside the innermost intersection of another color, there can be no 4-folder of those two colors.
  • Given a,b,c,d,m, there are 4 distinct configurations.
  • All the 9 configurations have projective intersections of 1 of each of the colors
  • For m=9 (and maybe for all values of m??), Type 1 (abcdbadc) intersections give you the two remaining 4-folders; type 2 sometimes gives you 1, sometimes none. Further analysis needed.
  • 12#(2,4,3,5,4,2,5,3) has a LOT of extra intersections. So does 24#(11,8,3,8,3,7,2,6), even when it's no longer trivial.

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