SOLVED 4 PROBLEMS!

Granted, they were homework problems from the beginning of the course, but I solved them! Myself! In only 2 hours!

Today has been very productive. :)

Today:

7:30-7:40 -- listened to Dan explain a solution that I had copied

7:40-7:50 -- found lost homework files

Once upon a time, I did math research. I loved it. Part of why it was so enjoyable is inextricably connected to my adviser, Leah W. Berman, who gave me a healthy perspective on productivity. It's been two years since I've done any truly creative math, which really puts a damper on my perseverance...

Nevertheless, there is hope! In the spirit of the work begun two years ago, I re-plunge myself into this blog for the sake of accountability and thus much-hoped-for productivity!

10:30 -- woke up

12:30 -- left for campus

1-1:30-- course calender posted and algebra email list made

1:30-1:45--blogged.

Things left to do:

MAC and teaching 3-8:30. :(

Lesson plan??

... which clearly leaves no time for prelim studying. Or prelim memorizing. You know. $%^#.

BLOGGER IS WORKING IN UST!

While I was sitting in the middle of a conference given in Russian, I was wondering what would happen if you merged two 2-cfgs together. I'm pretty sure it gets you a 4-cfg. The question is whether or not it gets you a set of superimposable ones... and I think I need more of my notes before I determine that. I forget the proof... that is what I'll have to look up. Who knows! Maybe I'll figure out what I couldn't finish!

Wk 6, Tu

Brussel sprouts and lima beans! There are less than 3 weeks left!

Progress, progress... I proved that given a configuration m#(a,b,c,d,b,a,d,c), it has complements m#(b,c,a,d,c,b,d,a) and m#(c,a,b,d,a,c,d,b). Now I'm trying to generalize this.

After a questionable proof and a feeble attempt at coding, I discovered a pair of nontrivial configurations that trivially satisfied the conditions for being superimposable/having extra intersections!

Furthermore, while my code yields non-realizable configurations (i.e., useless), it still (I think) shows where it's not possible, and I think I'm finding a correlation:

1. According to my code (and my analysis of the results it yielded), there are NO superimposable configurations for m = 8, 9, 10, 11, 13, 14, 16, 17, 19, or 20 (15 is questionable: it was too extensive to check according to one coding [this code was yielding pattern 2 cfgs, which was way bad], and the second coding of restrictions said there were none. so, it's very, very unlikely that there are any superimposable cfgs for m=15).

2. Conjecture: a configuration is superimposable if it is half-trivial, i.e., two elements of S are the same as two elements of T.
*I'm having a little bit of trouble determining the direction of this implication, so:

3. According to Angela's data, there are no half-trivial configurations for m = 8, 9, 10, 11, 13, 14, 15, 16, 17, or 19, (haven't checked 20 yet) but there IS one in 12, which just so happens to yield the two pairs of superimposables.

Wk 5, M

8:30-11
2-3
10-11

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

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