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Unformatted text preview: TODAY: Short recurrence for counting lattice paths by qweight Number partitions Short recurrence for counting lattice paths by qweight Warmup: combinatorial interpretation of (a choose b) b = a (a1 choose b1) Consequence: (a choose b) = (a/b) (a1 choose b1) Put a=m+n, b=m (motivation: counting lattice paths from (0,0) to (m,n)) Let P(m,n;1) = the polynomial P(m,n) from the homework due today, evaluated at q=1; i.e. P(m,n;1) = the number of lattice paths from (0,0) to (m,n). P(m,n;1) = (m+n)/(m) P(m1,n;1) This is a special case of a qversion you proved in HW #10: P(m,n) = ((1q^(m+n))/(1q^m)) P(m1,n) = ((1+q+...+q^(m+n1))/(1+q+...+q^(m1))) P(m1,n) (just send q → 1) Definition: [n] = 1+q+...+q^(n1). We call this the qanalogue of n. (Some authors prefer to define [n] as 1q^n.) [1], [2], [3], ... are the qintegers. (Compare with an earlier definition of [n] as {1,...,n}. I’ll continue to use both, but it should always be clear which one I have in mind.) Digress for a moment: We defined P(m,n) as a sum of weights of lattice paths, where the weight of a lattice path was defined as q to the power of the sum of the heights of the horizontal edges. (Give example: compute P(2,2).) Is there another way to think about the weight? ...: q to the power of the area under the path. Let’s find a direct proof of (1+q+...+ q^(m1)) P(m,n) = (1+q+...+q^(m+n1)) P(m1,n): Motivation: nonq version: m P(m,n) = (m+n1) P(m1,n) LHS: Counts lattice paths from (0,0) to (m,n) with a marked horizontal edge, where the area associated with such a marked path is defined as its ordinary area plus the xcoordinate of the left endpoint of the marked edge. RHS: Counts lattice paths from (0,0) to (m1,n) with a marked vertex, where the area associated with such a marked path is defined as its ordinary area plus the xcoordinate of the vertex plus the ycoordinate of the vertex. Weightpreserving bijection from LHSobjects to RHSobjects: ... Replace each marked dot by a marked horizontal edge, shifting everything after it to the right. Do example with (m1,n) = (1,1), (m,n) = (2,1). Discuss why it’s a bijection. Analysis: Consider a latticepath from (0,0) to (m,n) with weight x^A. When you mark a vertex (i,j) on a latticepath from (0,0) to (m1,n), you get a vertexmarked path of weight x^(A+i+j). On the other hand, if you replace the marked vertex by a marked horizontal edge, you get an edgemarked path of weight x^(A+i+j). path of weight x^(A+i+j)....
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 Fall '08
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 Algebra, Combinatorics, Counting, Recurrence relation, Generating function, lattice paths

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