Monday, June 29, 2009

Wk 5, M


i have two pages of notes trying to figure out how to generalize my theorem! geeze!

Thursday, June 25, 2009

Wk 4, Th

9-11:30 -- paper revisions; 6-celestial pre-work
1:30-3 -- made a (3,4) cfg
3-4 -- 6-celestial exps

Wednesday, June 24, 2009

Wk 4, W

7:20am-9 -- typed paper
9-10 -- meeting
10-11:30 -- surfed and slept
11:30-12 -- configured
2-3 -- i think i was working on categorizing the properties?

i was up at 6 this morning! how have i only worked 4 hours?!

Friday, June 19, 2009

Wk 3, F

8:30-9:30--"finished" proof, i.e., wrote up angle problem solution
9:30-10:30--meeting w/berman
10:30... added more pictures to paper

Thursday, June 18, 2009

Wk 3, Th

9-12 -- wrote half a page; other half of page: picture
12-3:30 -- lunch, talk, tutored
3:30-5:30 -- papered
6-7-- papered futile-y

i hate papers.

Tuesday, June 16, 2009

Wk 3, Tu

9:30-11 -- tried to fix "101 error(s)" in Latex in my tables (i was missing some dollar signs)
i was wondering about formatting. how do my tables look? (i was also thinking they could be equation lists, like to show the middle step, and forget about having columns.
1-2:30 -- worked on showing (and typing up) that the angles must be the same. got half way there.
i think this also leads to some interesting questions about configurations... i wonder if anyone's done any research on the placement of vertices? (especially with the pairing that occurs with trivial configurations). for instance, when are the symmetry classes of points collinear?
2:30-3:30 -- outlined proof, for organizational purposes; blogged.
11-12 -- reviewed proof for tomorrow

today, like yesterday, was very much a writing day.

Friday, June 12, 2009

Wk 2, F

8:30-9 -- worked on paper
9 -- meeting
9-12:15 -- worked on writing down exactly what i was proving; chose better forms for the complements (b,c,a,d,... instead of a,d,c,b,...); made a table of spans for the three cfgs; hashed out branko's explanation of how to calculate where the next point goes. hopefully, this will be enlightening for why there must be the third (fifth and sixth 2PI rings) set of intersections.

2-2:30 -- blogged
2:30-5 -- conjectured that the innermost radius of the L_i intersections was determined by t_(i-1); further modified to hold only for pattern 1 (first look at pattern 2 yielded a counterexample). looked at more examples. figured out why the s_i weren't mapping to the predicted color: it's s_i and t_(i-1) dependent, not just s_i dependent. Fun.

Thursday, June 11, 2009

Wk 2, Th

9-10 -- meeting
10-11 -- made final decisions about the colors:
Lines: Blue, Red, (Dark)Green, (Dark)Magenta
Points: Cyan, Orange, Green, Magenta, Yellow, Purple
Additionally, mathematica is STILL like, NO, go figure out spans and colors and whatnot yourself. :P see notebook.
11:30-12:30 -- fiddled with
4:30-5:30 -- wasted. like duh. obviously you can't preserve the line labels: take your initial 2PIs, BdG & RdM. They become your new points. Your new 2PIs, however, are still BdG and RdM, which are supposed to be in locations that were points in the initial cfg. But they couldn't be points in the initial cfg because they were 2PIs!

Monday, June 8, 2009

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.

Friday, June 5, 2009

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.

Thursday, June 4, 2009

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.

Wednesday, June 3, 2009

Wk 1, W

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

Tuesday, June 2, 2009

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