From propp@math.wisc.edu Sat Feb 9 10:36:36 2002 Date: Fri, 9 Nov 2001 14:50:40 -0600 (CST) From: propp@math.wisc.edu To: reach@math.harvard.edu Subject: [reach] minutes Minutes for 11/6, written by David Speyer Bureaucracy Prof. Propp would like feedback on how REACH is running Drop it off anonymously (or not) in his mailbox Questions of especial interest are how to encourage collaboration, how coordination with 192 is working, anything else Efstathios will compile a list of what people are interested in working on, to aid in collaboration. He also has sent out a note on how to run Mathematica in idle time. Presentation for the NSF presentations will be at 2 on Friday (11/16), probably in the Math Library should last roughly 5 miunutes afterwards, informal discussion until 3, show up if at all possible Greg, Trevor and David seem to be the people available/interested to do presentations There will be practice presentations on Tues for REACH and Wed (11/17) for 192 Prof Gross would like to talk to the presenters beforehand. It is not clear if this is possible as he will be away and hard to reach until Wednesday. Prof Propp will also be away this weekend, but available by email. A problem that Prof. Propp would liek someone to work on: consider the number of paths from (0,0) to (i,j) by steps of (0,2), (1,1) or (2,0). For i or j fixed, this is a polynomial in the other variable. What happens for i or j negative? Specifically, can we get a good extension to both negative? Prof Propp points out that there is an explicit correspondence between lattice paths from (0,0) to (i,j) and subsets of size i of an i+j element set: just take the indices of the horizontal edges. We have reciprocity for binomial coefficents both via negative sets and by running lattice paths backwards. Do we get the same bijection? Similarly, for domino tilings, there are three equivalent things: matchings, routings, and spanning trees! It turns out that spanning trees of a 2xn rectangle are in bijection with matchings of a 3x(2n-1) rectangle wih a corner removed. Once again, does the bijecion reverse? There was some confusion about the proper indexing, which was resolved. Trevor asked whether one could/should approach domino reciprocity via the product of trig function formula. David pointed out that this would give fractions. Professor Propp answered that in fact, for 2-adic reasons, one should expect fractions. If A(n) is the number of tilings of a 2n x 2n rectangle, we should apparently have A(-n)=2^(-2n+1) A(n-1). The first few values should be A(-1)=1/2, A(-2)=1/4, A(-3)=9/8. Trevor wants a fast algorithm to compute the determinant of a matrix with nonzero diagonal terms. Greg has been studying the recurrence g(n)=(g(n-1)^2+k)/g(n-2). Greg conjectures g(n)=sum k^m (n-1+m choose n-1-m); g(n)=(2+k)g(n-1)-g(n-2). g(n)=sqrt(k)h(n) where h are the polynomials on the first 192 problem set with x=1/sqrt(K). Stav has been computing paths through graphs of the form * - * - * / \ / \ / \ * - * - * - * When he runs them backwards, he gets pictures that aren't paths!