Tuesday, July 7, 2009

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.