By Parshin Shafarevich
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5. Fair division of a necklace with two cuts. 7 (Splitting necklaces). Suppose a pearl necklace has kp white pearls and kq black pearls, for some integers k, p and q. Then one can cut the necklace in at most 2(k − 1) places so that the remaining pieces can be rearranged into k groups, with p white and q black pearls in each group. In fact, we show that each of the k pirates gets either one or two continuous pieces of the necklace. Note also that the bound on the number of cuts in the theorem is tight: if black pearls are separated from white pearls, at least k − 1 cuts are needed to divide pearls of each color.
5) to obtain x, y ∈ [0, k n] such that y − x = n and f (y) − f (x) = 0. Since the number of pearls between x and y must be integral, we can round x, y down and obtain a fair k1 portion of the necklace for the first pirate. 1 Repeat the procedure by cutting out a fair k−1 portion of the remaining necklace, etc. At the end, we have k − 1 pirates who make two cuts each, giving the total of 2(k − 1) cuts. 5. Exercises. 1. [1-] Let f : R → R be a periodic continuous piecewise linear function. Prove that f has inscribed chords of any length.
13In other words, there is a way to place a table of every size on a hill, thus the “table theorem”. is the “chair theorem”. 15Thus, two people can coordinate their movements to move along a path while carrying a ladder. 16Warning: it takes an eﬀort to realize that both parts of the problem are in fact not obvious. 8. Final remarks. 5) is considerably simpler than the general case [Kel]. Both our proofs follow the proof idea in [GPY], stated there in a more general context. 4 follows [Nie]. 9 was ﬁrst proved in [Emch] and generalized in [HLM].
Algebraic Geometry 3 Curves Jaobians by Parshin Shafarevich