Uneven Splitting of Ham Sandwiches

Abstract

Let m_1,...,m_n be continuous probability measures on R^n and a_1,...,a_n in [0,1]. When does there exist an oriented hyperplane H such that the positive half-space H^+ has m_i(H^+)=a_i for all i in [n]? It is well known that such a hyperplane does not exist in general. The famous ham sandwich theorem states that if a_i=1/2 for all i, then such a hyperplane always exists. In this paper we give sufficient criteria for the existence of H for general a_i in [0,1]. Let f_1,...,f_n:S^{n-1}->R^n denote auxiliary functions with the property that for all i the unique hyperplane H_i with normal v that contains the point f_i(v) has m_i(H_i^+)=a_i. Our main result is that if Im(f_1),...,Im(f_n) are bounded and can be separated by hyperplanes, then there exists a hyperplane H with m_i(H^+)=a_i for all i. This gives rise to several corollaries, for instance if the supports of m_1,...,m_n are bounded and can be separated by hyperplanes, then H exists for any choice of a_1,...,a_n in [0,1]. We also obtain results that can be applied if the supports of m_1,...,m_n overlap.Comment: 18 pages, 2 figure

Strategyproof Linear Regression in High Dimensions

“…, S d+1 , there exists a unique hyperplane H which contains a given number of points from each set in its negative closed half-space. This discrete result builds upon previous continuous variants [1,4]. We first define a condition they require, which also plays a key role in our result.…”

“…Definition 7 (Well Separable Sets [26] Well separable sets are sometimes called affinely independent sets [4]. Well separability is equivalent to various other conditions [4,37]. In what follows, Conv(·) denotes the convex hull.…”

“… Moulin [33] shows that for the single dimensional setting, strategyproofness is equivalent to group strategyproofness 4. A mechanism is anonymous if permuting the reports of the agents does not change the output of the mechanism.…”

The hamburger theorem

“…For the case of five measures on R 3 , our approach from the proof of the hamburger theorem fails for the following reason. If µ i (R 3 ) = 1/5 for each i ∈ [5], then the target polytope, now in R 5 , intersected with the boundary of the box B, does not contain a closed curve symmetric with respect to the center of B. If such a curve existed, we could apply a generalization of the Borsuk-Ulam theorem saying that if f : S k → S k+l and g : S l → S k+l are antipodal maps, then their images intersect [14, Exercise 3.…”

“…For these more general measures, the condition µ i (H) = µ i (R d )/2 must be replaced by the inequality µ i (H) ≤ µ i (R d )/2. Breuer [5] gave sufficient conditions for the existence of more general splitting ratios. In particular, he showed that for absolutely continuous measures whose supports can be separated by hyperplanes, there is a hyperplane splitting the measures in any prescribed ratio.…”