# Nov13 - TODAY Short recurrence for counting lattice paths...

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Unformatted text preview: TODAY: Short recurrence for counting lattice paths by q-weight Number partitions Short recurrence for counting lattice paths by q-weight Warm-up: combinatorial interpretation of (a choose b) b = a (a-1 choose b-1) Consequence: (a choose b) = (a/b) (a-1 choose b-1) 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(m-1,n;1) This is a special case of a q-version you proved in HW #10: P(m,n) = ((1-q^(m+n))/(1-q^m)) P(m-1,n) = ((1+q+...+q^(m+n-1))/(1+q+...+q^(m-1))) P(m-1,n) (just send q → 1) Definition: [n] = 1+q+...+q^(n-1). We call this the q-analogue of n. (Some authors prefer to define [n] as 1-q^n.) [1], [2], [3], ... are the q-integers. (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^(m-1)) P(m,n) = (1+q+...+q^(m+n-1)) P(m-1,n): Motivation: non-q version: m P(m,n) = (m+n-1) P(m-1,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 x-coordinate of the left endpoint of the marked edge. RHS: Counts lattice paths from (0,0) to (m-1,n) with a marked vertex, where the area associated with such a marked path is defined as its ordinary area plus the x-coordinate of the vertex plus the y-coordinate of the vertex. Weight-preserving bijection from LHS-objects to RHS-objects: ... Replace each marked dot by a marked horizontal edge, shifting everything after it to the right. Do example with (m-1,n) = (1,1), (m,n) = (2,1). Discuss why it’s a bijection. Analysis: Consider a lattice-path from (0,0) to (m,n) with weight x^A. When you mark a vertex (i,j) on a lattice-path from (0,0) to (m-1,n), you get a vertex-marked 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 edge-marked path of weight x^(A+i+j). path of weight x^(A+i+j)....
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Nov13 - TODAY Short recurrence for counting lattice paths...

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