Wk 2, M

More evidence of the BG, RM pairings:
Their orbits are always cyclically adjacent! It even applies when you include the missing 2PIs.

Wk 1, F

9-10 meeting
10-12:30 weekly update
2:30 wireless broke
3-4:30 pattern 2 experiments

Weekly Reflections 1

A General Documentation of our Conjectures:

1. Fixing a,b,c,d as the spans of lines (i.e., m#(a, _; b, _; c, _; d, _)) there are only two ways to fill in the blanks so as to have a coherent/realizable/valid configuration: m#(a, c; b, d; c, a; d, b ) and m#(a, d; b, a; c, b; d, c). The former we will call Pattern 1; the latter, Pattern 2.
2. These two patterns have different properties:
   
3. Pattern 1 configurations seem to always have two extra orbits of 4-fold intersections, (abbreviated here as 2-Pair-intersection, or 2PIs) where 2 of the lines are from one symmetry class of lines, and the other 2 are from another. Furthermore, these 2PIs are different from the ones constructed by the symbol, resulting in every pair existing: 4 symmetry classes of lines, 2 classes through each of these intersections; 4 choose 2 = 6 possible combinations.
   
4. Pattern 2 configurations do not always have two extra 2PIs. In some cases, there is only 1 extra; other cases, there are none. There are also 3-fold intersections, usually of three different symmetry classes. There is a possibility that the properties have connections to the parity of m; however, there are too many possibilities to make any conjectures at this point. Further analysis is needed.
5. You can always check algorithmically to see if a 2PI exists. There are only floor[m/2] distinct possibilities to check. Furthermore, if all the intersections of one color occur inside the innermost intersection of another color, you know certainly that a 2PI of those colors does not exist.
6. Pseudo Conjecture: Based on Branko Grunbaum's work, there are 6 (isomorphically) distinct configurations given a,b,c,d,m, 3 of Pattern 1 and 3 of Pattern 2 (they are polars of each other).
7. Conjecture: The 3 configurations of Pattern 1 given a,b,c,d,m all have the same line structure. What this means is that you could place the configurations right on top of each other and you would get a [6,4] configuration.
8. Conjecture: The 2PIs that are missing from each of the Pattern 1 configurations are always the same: the innermost (orbit 1) and outermost (orbit 6) symmetry classes, orbits 2 & 5, or orbits 3&4.

What i Actually know so Far

there are 4 distinct configurations, but there are also 2 more. lamez. or great!

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.
  

Wk 1, W

9am -- meeting
10-1:15pm --worked on presentation
2:15 -- messed with cfgs and compositions

Wk 1, Tu

9am -- met with berman
10am -- cried to rachel
11am -- listed out possible configuration symbols and found their equivalences
noon -- lunch
noon:30 -- gave tech support my computer
2pm -- retrieved computer; mathematica: drew (all) 12 configurations for m = 10
3pm -- started messing with GS
5:30 -- started logging progress by means of email
6pm -- dinner