By N Vilenkin
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Clearly, since φs is a permutation, so is φs . We summarize the results of this subsection as follows: 28 2. Simple Juggling The Inverse of a Juggling Sequence Let s = {ak }p−1 k=0 be a juggling sequence, and let φs be its associated permutation. Then, its inverse is the juggling sequence s = {aφ−1 }p−1 s (p−1−k) k=0 and φs (k) = p − 1 − φ−1 s (p − 1 − k) for k = 0, 1, . . , p − 1. 3 Pick a Pattern Procedure In [94], page 168, Martin Probert gives a recipe for constructing some of the b-ball juggling sequences of period p that boils down to the following.
In fact, it turns out that this group is the affine Weyl group A˜p−1 . Two juggling sequences of period p correspond to the same element of A˜p−1 if and only if one is the vertical shift of the other; see page 23 for a definition of the term vertical shift. 7 How Many Ways to Juggle? 43 Clearly, A˜0 is the one-element group. For p ≥ 2, it can be shown that A˜p−1 is generated by the simple reflections ⎧ ⎨ i + 1 for i mod p = k mod p, i − 1 for i mod p = (k + 1) mod p, tk : Z → Z : i → ⎩ i for i mod p = k, (k + 1) mod p, for k = 0, 1, .
If the number is a 0, choose a copy of card C0 and place it above the 0. Choose any of the numbers that are not 0, say the 1 (boxed in the diagram). 40 2. Simple Juggling From the circle that indicates the corresponding beat, trace backwards on the incoming arc and inspect the crossings with other arcs. When we cross another arc, we either cross from the inside to the outside or in the other direction. In the first case, we mark the crossing by a gray circle, otherwise by a gray square. Count the number c of gray circles.
De cuántas formas? Combinatoria by N Vilenkin
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