SSL Notes for meeting #6 (9/25) Today's note-taker: Abby (Martin next time) Today's snack-provider: Emilie (Stephen next time) We went around with names and asked everyone to talk about the courses they're taking (mathematical and non-). John: teaching calc, masters class Emilie: French, CS 302, Math 491, Math 521, Genetics Abby: CS 564, Conservation 339, Math 521, Stat 641 Martin: CS 719, Math 491, CS 302 ta-ing Josh: don't go to school - works at library & star liquor, records Jim: teach 491, maybe take sign language Hal: Math 744, Math 751 Carl: Math 741, CS 412, Antropology 104, Greek 103, Psych 311 Melania: Assessor to see if people are ready for classes, Cultural Anthropology training, qualitative research Marta: Milania's helper Sam: Math 541, Math 491, Art History, Italian Lit Stephen: Com. Algebra, Seminar - Yangians, Lie Theory, working on research Paul: Math 541, Math 491, Eng. 467, Psych 202, Genetics 466 Next Thursday there'll be ten-minute meetings with everyone. (Jim and Stephen, plus maybe Melania.) 3:30: Paul 3:45: Emilie 4:00: Carl 4:15: Hal 4:30: Sam 4:45: Martin 5:00: Abby 5:15: Josh We started making lists of the number of perfect matchings (pm's) with two or four of different colored vertices, alternating around the boundary, removed. We got: 29 5 34 3 8 7 2 13 12 10 3 9 2 5 1 1 6 4 4 2 15 1 3 1 1 4 4 10 2 9 1 3 3 1 6 6 4 2 15 1 3 4 1 4 6 4 2 9 1 3 1 1 4 4 where the first row is with all the vertices, the next 4 rows are without 2 vertices, and the last row is without all 4 vertices. each column is a different graph. We also discovered that Carl was working on a different problem where the different colored vertices do not alternate around the boundary. Jim assigned weights to a small graph and everyone saw the relationship: AF = BD + CE where A is with all the vertices, F is without all 4 vertices, B is without vertices (a,b), D is without vertices (c,d), C is without vertices (a,c), and E is without vertices (b,d). Small graph: 3 boxes across and one up, with the bottom right 3 vertices colored and the vertex on the top left of the top box colored. Assigned weights a-g to horizontal edges, and h-m to vertical edges. Got these pm polynomials A: efg + acg + bde = 3 B: ac + ef = 2 C: b + e = 1 D: e + g = 1 E: d + e = 1 F: e = 1 A/B approx. = D/F which leads us to AF approx. = BD and AF - BD = CD which is equivalent to AF = BD + CD, our conjectural formula. Paul and Sam had found this relationship and described their methods: Paul: Looked at a large list of numbers of pm's Sam: Looked at patterns with ones, and at numbers closer together and farther apart (extremes) In general, to find relationships we hypothesize: - big examples are not good in this case - pattern may jump out more with bigger numbers - if you are proving pattern you know about then large numbers help show you the proof - law of small numbers: there are too few small integers for tasks assigned them, so assigning weights can narrow down the possibilities - helpful to assign weights according to various methods (eg. horizontal vs. vertical, attached vs. unattached) ------------------------------------------ Break Conjecture with colored vertices not alternating around the boundary: Hal showed at the board that AF = BD - CE works on a straight three boxes, colored vertices so that one remains at the top left and three remain in the bottom right corner. Next, Hal looked at a straight snake with four boxes. We colored the bottom left two boxes and the top right two boxes. Found: A: (x^5) + 3x(y^4) + 4(x^3)(y^2) B: y(2(y^2)x + (x^3)) C: (y^4) D: same as B E: (x^2)(y^2) F: (y^2)x where vertical edges are x and horizontal edges are y This holds for our conjecture when AF = BD - CE We tried proving the first conjecture. For next time, 1) Make straight snake graphs and find the sum of the weights of the pm's and look for formulas for the coefficients. 2) Make chains of hexagons where at most two hexagons are attached and count the pm's of them and try to find pattern 3) Look at new snake of connected triangles, where each triangle becomes a Y. Get rid of two vertices from the boundaries because otherwise there is a coloring imbalance. Think about these graphs.