Combinatorial Necklace Splitting

We give a new, combinatorial proof for the necklace splitting problem for two thieves using only Tucker's lemma (a combinatorial version of the Borsuk-Ulam theorem). We show how this method can be applied to obtain a related recent result of Simonyi and even generalize it. 1 Necklace Splitting This paper was inspired by the combinatorial proof of Matou2ek [7] of the Lovász-Kneser theorem [6]. He used Tucker's lemma [10] that was proved combinatorically by Freund and Todd [4]. A combinatorial proof for a generalization of Tucker's lemma can be found in a paper of Ziegler [11]. We start by stating a version of this lemma and then show how it can be used to give a simple proof for the necklace splitting theorem. We denote the set {1, . . . , n} by [n] and {−1, . . . ,−n} by −[n]. For four subsets of [n], A1, B1, A2, B2 we say that the set-pair (A1, B1) is smaller than the set-pair (A2, B2) if A1 ⊂ A2 and B1 ⊂ B2 and at least one of the inclusions is strict. We denote this by (A1, B1) ⊂ (A2, B2). A family of set-pairs is said to form a chain, if any two members of the family are comparable. Lemma 1.1. (Octahedral Tucker's lemma, in [11] Lemma 4.1) If for any set-pair A,B ⊂ [n], A∩B = ∅, A ∪ B 6= ∅ we have a λ(A,B) ∈ ±[n − 1] color, such that λ(A,B) = −λ(B,A), then there are two set-pairs, (A1, B1) and (A2, B2), such that (A1, B1) ⊂ (A2, B2) and λ(A1, B1) = −λ(A2, B2). Two set-pairs for which (A1, B1) ⊂ (A2, B2) and λ(A1, B1) = −λ(A2, B2) are said to form an opposing inclusion. In the necklace splitting problem, we have an open necklace with k types of beads, ai beads of the ith kind and p thieves wants to split it by using as few cuts as possible, such that each one of them gets bai/pc or dai/pe beads of the ith kind. They are allowed to cut the necklace between any two beads ∗ELTE, Budapest and EPFL, Lausanne. Supported by OTKA NK 67867.