Minutes for Thursday, March 14, 2002 (Jason Burns, with some input from David Speyer) Jim: There's a formula for the number of perfect matchings of an axbxc semiregular hexagon of smaller hexagons. * * * * (diagram here: a=4, b=3, c=2 * * * * * ; the *'s are hexagons, and * * * * * adjacent ones share an edge) * * * * a b c ---- ---- ---- / \ It's the product | | | | | | | i+j+k-1 | . | | | | | | | ------- | i= 1 j= 1 k= 1 \ i+j+k-2 / (Setting c=1 becomes a binomial coefficient, for example.) If we generalize to four numbers, the quadruple product of (i+j+k+l-1)/(i+j+k+l-3) doesn't even give integers. So this is it: the end of the line. Q: Can we find a unified way of looking at this hexagon result, and all the others we've seen? (Domino tilings of Aztec diamonds, lattice paths, etc.) (Digression: pie-eating techniques.) More to be said about Somos-4: can we get nice formulas like the above? -- The elevator trick for getting to the fifth floor, discussed. (Why would you want to get there? Because here are no locks on the 5th floor, and you can use the Math Dept. staircase to go down to the 4th floor.) Problem: 5th floor button disabled after hours. Trick: One elevator rests on 5th floor. We get in that elevator. Step 1: Take elevator to 4th floor. Call elevators with up button. Step 2: Send non-5th floor elevators to lower floors. Step 3: Get in 5th floor elevator, close door, wait for elevator to return to 5th floor. Step 4: Press "Door Open" button to get out. Business matters: getting in old minutes (Stathis?), Web pages, money (send in your work logs!) David: I'm working on the octahedron recurrence with different IC's can get Aztec diamonds, etc. -- working on general theory of what matchings get counted. (Jim: Get Federico's input.) -- David explains his theory: (Note: this is all conjectural at this point.) Take a diagram as follows, illustrating which initial conditions we take. This is a view from above; the numbers represent the heights. For example, the diagram 1 1 1 0 0 1 1 1 0 0 1 1 1 is (part of) a picture of the initial conditions for an Aztec diamond. Specifically, we're specifying all our initial conditions on the height-0 plane and the height-1 plane. The picture for this grid becomes: 1 >< 1 >< 1 We've drawn a little 'x' between each pair of 1's... v 0 v 0 v ^ ^ ^ 1 >< 1 >< 1 v 0 v 0 v ^ ^ ^ 1 >< 1 >< 1 and... 1 >< 1 >< 1 if we connect the half-edges coming out of those 'x's, / \ / \ we get an infinite square lattice, like we should for / 0 \/ 0 \ the Aztec diamond matchings. \ /\ / \ / \ / 1 >< 1 >< 1 / \ / \ Here's another example: 1 \ / 1 Draw between 1 and -1 0 | 0 0 0 a dash with lines coming out. / \ 1 -1 1 Here we get the diagram 1 >-< -1 >-< 1 0 0 \ / 1 0 | 0 / \ 1 Draw between 1 and 1 a little 'X', and between 1 and -1 a >-< ... connect the ends to get squares, hexagons, and octagons. Diagonal edges get weights, horizontal/vertical ones don't. We know how to label the faces (we label them with the initial conditions), but how do we label the edges? Here's how: 0. Only diagonal edges get weights; horiz/vert edges get weight 1. (We saw this before, in the Somos-3 pictures.) 1. Take the diagonal edge between faces (n1, i1, j1) and (n2, i2, j2). (We can assume that |i1-i2|=|j1-j2|=2, and |n1-n2|=1. (*)) 2. Let i := (i1+i2)/2, j := (j1+j2)/2, and n := n1+n2+1. 3. The edge weight is (i +/- n, j +/- n). (Here, we choose i+1 if the smaller i-value is on the left, i-1 if the right; and we choose j-1 if the smaller j-value is on the bottom, j+1 if on the top.) (*) We assume that |n1-n2|=1, that is, that the face layer doesn't change by more than one at a time, so the recurrence will be well defined -- that is, nothing "slips through the holes" in the surface of initial conditions. Conversely, weighting the edges for any given matching is straightforward (exponent 1 if included, exponent 0 if not), but weighting the faces needs explanation. Suppose the face has k edges included in the matching. If the face is a square, we raise it to the (1-k) power; if the face is a hexagon, to the (2-k) power; an octagon, the (3-k) power. -- A possibly unhelpful addition to our mental picture of the surface of ICs: We can think of the edges between height 1 and 0, say, as at "height 1/2". Kuo condensation approach: changing the point where we stand; urban renewal: changing the surface. Can we get two complementary proofs? Jim again: f(n, , )f(n-4, , )=f(n-1, , )f(n-3, , ) + f(n-1, , )f(n-3, , ) Can we fill in the i's and j's in an essentially symmetric way to get an octahedron recurrence? (This is the other Somos-4, which is less interesting: the numbers are just powers of 2.) Specifically: f(n,i,j)f(n-4,i,j)=f(n-1,i-1,j)f(n-3,i+1,j)+f(n-1,i,j-1)*f(n-3,i,j-1) with initial conditions on four parallel planes. This is essentially the "only" way to fill in the numbers so as to get a centrally symmetric octahedron. -- Progress reports. David: I should be working on my thesis, but... proving this stuff. Roberto: Ehrhart reciprocity extensions of perfect matchings of certain graphs -- I'm not getting the same numbers Jim is, so we're asking why. Also "k-pedes" and cluster algebras. Dan: between projects, at the moment. Gregg: "recurrence graphs": they contain squares, hexagons, octagons... and pentagons. (Trevor: I got something with pentagons too! We should talk.) Bo-Yin: condensation on other graphs... Stathis: Somos-3 and Somos-4, looking through code. Trevor: Q: Can we take these graphs of David's -- if we alternate + and - charges on ions, there's a change in a certain map that's very well described in physics ... Noncommutative Somos sequences, or just adding a coupla indices. Lionel: Constructing a hypergraph matching for arbitrary polynomials. (Meeting adjourned.)