Minutes for 2/21/02: For the first hour, two presentations were given in the computer room. The first was a demonstration of mathematica given by David and Seth; the second was an overview of topics relevant to this semester's research given by Jim Propp. Mathematica Presentation ------------------------ Entry and Syntax 'Enter' key on the number pad ends input; 'Enter' key on the main section of the keyboard moves the input to a new line. If pressed together with 'shift' it will end the input. Mathematica is case sensitive. To avoid collision with pre-defined functions, use names that begin with lower-case letters. ( ) are simple parentheses. [ ] calls a function eg. sin[Pi/3] { } marks a list (not a set). { }[[2]] will output the second member of the list. % calls the last answer. %% calls the last answer but one. Out[n] recalls the nth output. Similarly In[m] recalls the mth input. Defining a function: e.g. f[x_] := (x^2 + 1)/2 defines the function f:x->(x^2 +1)/2 points to note: -- the underscore in f[x_] tells mathematica that x is a dummy variable -- ':=' implies a definition. -- type 'Control+2' instead of '^,2' to get the x^2 to look smooth. -- type 'Control+/' to input your fraction to look like a proper fraction. Once f is defined, the input f[2] gives 5/2 as expected. The input g[x_] := g[x] = 3x^3 - 1 defines the function g and gives you assigned and remembered values. Recursive definition: g[x_] := g[x] = 3g[x-1]^3 - 1 g[1] = 1 the above input defines the recurrence and an initial value. Mathematica will output 1 unless you suppress the output by ending with a semi-colon. then g[4] will output 36500, the fourth term of this sequence. Type '?g' to get mathematica to tell you what it knows about the variable/function g. 'Clear[g]' wipes all knowledge of g (the equivalent command in maple is 'unassign') Last week's homework recurrence: h[n_] := h[n] = ( (h[n-1]^2 +z^n) / h[n-2] ) h[0] = x; h[1] = y; (semicolons repress output) defines the recurrence. h[3] gave an output in a continued fraction style. There are several tools for changing the style of the output: Together; Apart; Simplify; FullSimplify; Expand. Together(h[n]) gave a single fraction with expanded numerator and denominator. Expand(h[n]) and Apart(Together(h[n])) gave the ouptut as Laurent polynomials. - - Numerator(P) gives the numerator of a polynomial P. - - CoefficientList[P , {x,y,z}] gives the coefficients of a polynomial P with variables x, y and z in incomprehensible form. - - Flatten takes out the parentheses in a list. - - DeleteCases[L,0] extracts all the zeros from a list L. (there is an opposite function Select) - - Length[..] gives the number of terms. - - Max[..] gives the maximum of a given list. e.g. we defined the function hh[n] to list the coefficients of the Laurent polynomials of h[n] in comprehensible form: hh[n_] := DeleteCases[Flatten[CoefficientList[Numerator[Together[h[n]], {w,y,z}, 0] you can use the Table function to tabulate results. the MatrixForm function below merely centers the columns here: MatrixForm[Table[hh[k], {k,0,7}]] this input gives a table of values of hh[n] as n ranges from 0 to 7. Odds & Ends: Factor[P] factorises a polynomial P. FactorInteger[n] gives the prime factorisation of the integer n. The 'If' function gives Mathematica programming ability. The help function is reached by typing 'shift+F1' End of Presentation - ----------------------------------------------------------------- Next, Jim Propp discussed the items that may lead to a paper: - - Diamond Pattern / Antifrieze pattern recurrences - - The Number Wall rerurrence - - Dodgson Condensation - - Kuo Condensation (like Dodgson but with 'edge variables') - - Tilted Diamond Patterns have initial conditions in some diagonal region, generate Laurent polynomials with coefficients = 1. The number of terms in numerator of f(n,0) is 2, 3, 11, 26, 97, 153 which lists two thirds of the terms of the sequence 1, 1, 1, 2, 3, 7, 11, 26, 41, 97, ... (which satisfies a linear as well as a quadratic recurrence). To see the missing terms, try f(n,1) or f(n,2) for various values of n. - - S-arrays: these give the first clues to a Somos-4 interpretation, but need to be replace by something 3-dimensional. - - 3-dimensional Somos-4 (face-variables version): gives Laurent polynomials with all coefficients equal to 1 (we think). - - 3-dimensional Somos-4 (edge-variable version): gives polynomials with all coefficients equal to 1 (we think). ------------------------------------------End of part 1------------------ There were two hand outs: (1) 'Alternating-Sign Matrices and Domino Tilings' Recommended reading parts 1,2,3,6,7,8 (2) REACH Projects list Professor Propp spent the rest of the meeting outlining connections between different subjects that we are interested in and possible routes to the solutions we seek. I will attempt to outline at least the main areas. Julian West is looking at matchings of the Aztec Diamond graph and attempting to give Somos-4 some interpretation via this route. He is looking for cliques in the compatability graph, moving forward on the assumption that a graph of the perfect matchings we have exists. While Julian does this, REACH will look at the "Somos-3" and "Somos-2" sequences: "Somos-3": a(n)a(n-3) = a(n-1)a(n-2) + a(n-2)a(n-1) Of course this equation is a(n) = 2a(n-1)a(n-2) / a(n-3), but we write it in the above fashion to show how it relates to Somos-4 and higher. "Somos-2": a(n)a(n-2) = a(n-1)a(n-1) + a(n-1)a(n-1) Again we can write this a(n) = a(n-1)^2 / a(n-2), and we see that it is the Aztec Diamond recurrence. So we understand "Somos-2", but we want to extend this understanding to "Somos-3". The recurrence used in Dodgson Condensation can be drawn as below: f / |\ / | | \ / | | \ b X-------------X c / | | / / | | / d X-------------X e \ | / \ | / \ | / a where the middle square is all considered to be on one level. If we take this middle square to be on two seperate levels, with b and c on the higher plane, and d and e on the lower plane, the same recurrence rules give the "Somos-3" recurrence. Our first succinctly stated problem was then stated: - If you define f := proc(n,i,j) option remember; if n<0 then undefined elif n<4 then x(n,i,j); else simplify( ( f(n-1,i-2,j+2)*f(n-3,i+2,j-2) + f(n-2,i+2,j+2)*f(n-2,i-2,j-2)) / f(n-4,i,j) ); fi; end; and evaluate f(n,i,j) (for n,i,j integers with n positive): (i) do you get a Laurent polynomial? (answer: yes, proved by Fomin and Zelevinsky) (ii) are all coefficients > zero? (iii) are all coefficients = 1? (iv) are all exponents of the Laurent Polynomials bounded by some value that is independent of n? The answer to parts (ii), (iii) and (iv) seems to be yes, but as yet there are no proofs. Propp believes that the best way to answer these questions is all in one go, via a direct combinatorial interpretation. We saw an example of such an interpretation, where an algebraic relation could be interpreted as perfect matchings of hexagons. Once this was known, integrality/positivity etc. follows immediately. * S-Arrays: x a / The pattern propagates upwards and b x the recurrence is given by: \ x c ae = bd + c^2 \ x d / e x b, c, d, e are known and everything above them can be determined in terms of Laurent polynomials. However, these polynomials do not have every coefficient equal to 1. The coefficients tend to be composite, and this hints at a combinatorial link. * The Cube Recurrence currently, Federico is working on this. We assume a relationship between the values of the vertices of the cube. Each vertex has a value (i,j,k) where i = m or m+1 j = n or n+1 k = o or o+1 and the top corner is (m+1, n+1, o+1). If the corners of the cube are labeled a, b, c, d, e, f, g, h with a (top corner) unknown, the cube recurrence is ah = bg + cf + de and has properties (i) - (iv) stated earlier when you start with initial values in the planes x+y+z = +1 and -1. The terms are powers of 3 - is this some higher dimensional brother to the Aztec Diamond? At this point, Propp stated a _Fundamental_Principle_; that if you have a polynomial in which every coefficient is 1, puting all variables = 1 is equivalent to counting the terms of the polynomial. OUR GOAL (or one of them, anyway) is a uniform combinatorial interpretation of the terms of sequences a(1), a(2), a(3), ... satisfying a recurrence relation of the form a(n)a(n-k) = a(n-i)a(n-k+i) + a(n-j)a(n-k+j) as introduced by Gale and Robinson. And finally, with regards to T-shirt design, if we could see it now, it should show us the way to the solution we seek. END.