Elsevier

Discrete Applied Mathematics

Volume 291, 11 March 2021, Pages 15-35
Discrete Applied Mathematics

Fair multi-cake cutting

https://doi.org/10.1016/j.dam.2020.10.011Get rights and content

Abstract

In the classic problem of fair cake-cutting, a single interval (“cake”) has to be divided among n agents with different value measures, giving each agent a single sub-interval with a value of at least 1n of the total. This paper studies a generalization in which the cake is made of m disjoint intervals, and each agent should get at most k sub-intervals. The paper presents a polynomial-time algorithm that guarantees to each agent at least min(1n,k(m+n1)) of the total value, and shows that this is the largest fraction that can be guaranteed. The algorithm simultaneously guarantees to each agent at least 1n of the value of his or her k most valuable islands. The main technical tool is envy-free matching in a bipartite graph. Some of the results remain valid even with additional fairness constraints such as envy-freeness.

Besides the natural application of the algorithm to simultaneous division of multiple land-estates, the paper shows an application to a geometric problem — fair division of a two-dimensional land estate shaped as a rectilinear polygon, where each agent should receive a rectangular piece.

Introduction

Consider n people who inherit m distant land-estates (henceforth “islands”) and want to divide the property fairly among them. Classic algorithms for proportional cake-cutting [25], [51] can be used to divide each island into n pieces such that the value of piece i, according to the personal value-measure of person i, is at least 1n of the total island value. However, this scheme requires each person to manage properties on m distant locations, which may be quite inconvenient. An alternative scheme is to consider the union of all m islands as a single cake (henceforth “multicake”), and partition it into n pieces using the above-mentioned algorithms. However, this too might give some agents a share that overlaps many distinct islands. For example, with m=5 islands and n=3 agents, a typical partition might look like this:

where Alice’s share contains 4 disconnected pieces and Carl’s share contains 2 disconnected pieces.1 This may require Alice to waste a lot of time flying between her different estates.

Can we find a more convenient division? For example, can we divide the multicake fairly such that each agent receives at most 3 disconnected pieces? In general, what is the smallest number k (as a function of m and n) such that there always exists a fair division of the multicake in which each agent receives at most k disconnected pieces?

This paper studies the following more general question. We are given an integer parameter k1, and it is required to give each agent a share made of at most k disconnected pieces. What is the largest fraction r(m,k,n) such that there always exists an allocation giving each agent at least a fraction r(m,k,n) of his/her total property value?

An obvious upper bound is r(m,k,n)1n. It is attainable when km, since then we can just divide each island separately using classic cake-cutting algorithms. The question is more challenging when k<m. The main result of this paper is:

Theorem 1

For every m1,k1,n1: r(m,k,n)=min1n,km+n1.

In words: it is always possible to guarantee to each agent at least a fraction min1n,km+n1 of the total multicake value, and in some cases it is impossible to guarantee a higher fraction.

The upper bound r(m,k,n)min1n,km+n1 is proved by a simple example (Section 3). The matching lower bound is proved constructively by a polynomial-time algorithm, which is the main technical contribution of the paper (Section 4). The case k=1 is simple and can be handled using classic cake-cutting techniques (Section 4.2). However, for k2 these techniques yield a sub-optimal result, and to match the upper bound we need a new technique using an envy-free matching in a bipartite graph (Section 4.3).

Whenever k1+m1n, Theorem 1 implies that r(m,k,n)=1n. Therefore,

Corollary 1

If k1+m1n (each agent is willing to accept at least 1+m1n disconnected pieces), then a proportional multicake allocation exists and can be computed efficiently.

In particular, in the opening example, it is indeed possible to find a fair allocation in which each person receives at most 3 disconnected pieces.

Theorem 1 provides an absolute value guarantee — a fraction of the total multicake value. One may also be interested in a relative value guarantee — a fraction of the value that an agent could attain without partners, which is the value of his/her k most valuable islands. What is the largest fraction rR(m,k,n) such that there always exists an allocation giving each agent at least a fraction rR(m,k,n) of his/her k best islands? The next result shows that this fraction is 1n.

Theorem 2

For every m1,k1,n1: rR(m,k,n)=1n.

The upper bound of 1n is obvious.2 The lower bound is proved using the same algorithm as Theorem 1. Moreover, the algorithm can even allow each agent to choose between the two guarantees. This may be useful, as for each agent, a different guarantee may be preferred. As an example, suppose there are n=10 agents and m=11 islands and k=1. Theorem 1 guarantees to each agent a fraction 120 of the total value. Suppose Alice values all 11 islands uniformly at 111 of the total value, while George thinks that all the multicake value is concentrated at a single island. Theorem 2 guarantees Alice 1110 and George 110 of the total value. Our algorithm can give the better guarantee to each agent, i.e., at least 120 to Alice and at least 110 to George (Section 5).

Besides proportionality, an additional fairness condition is envy-freeness. It requires that every agent values his/her share at least as much as the share of every other agent. In the classic setting where m=k=1, it is known that envy-freeness implies proportionality, and an envy-free allocation always exists [52], [54], [57]. What is the largest fraction rE(m,k,n) such that there always exists an envy-free allocation giving each agent at least a fraction rE(m,k,n) of his/her total property value?

An obvious upper bound is rE(m,k,n)r(m,k,n): adding a constraint on the allocations cannot help attain a higher value-guarantee. A lower bound is presented below.

Theorem 3

For every m1,k1:

(a) When n=2: rE(m,k,n)=min1n,km+n1.

(b) For any n3: rE(m,k,n)min1n,km+nkk.

Part (a) is proved constructively by a polynomial-time algorithm. It exactly matches the upper bound of Theorem 1: when there are two agents, adding an envy-freeness constraint does not decrease the per-agent value guarantee.

Part (b) is proved existentially by a reduction to the classic setting (Section 6). The bound is tight for k=1 but not for k2. The exact value of rE(m,k,n) for n3 and k2 remains an open question.

Besides the natural application of the multicake division algorithm to simultaneous division of multiple land-estates, it can be applied to the problem of dividing a two-dimensional land estate shaped as a rectilinear polygon, when each agent requires a rectangular land-plot. If the land-estate has T reflex vertices (vertices with internal angle 270°), then each agent can be allocated a rectangle worth at least 1(n+T) of the total polygon value, and this is tight (Section 7).

Various ideas for future work are presented at ending subsections of Sections 4, 5, 6, 7.

Fair division problems have inspired hundreds of research papers and several books [12], [14], [37], [40]. See Procaccia [39], Brânzei [16], Lindner and Rothe [34], Brams and Klamler [13], Segal-Halevi [43] for recent surveys.3 Some works that are more closely related to the present paper are surveyed below.

Most works on cake-cutting either require to give each agent a single connected piece, or ignore connectivity altogether. We found two works that consider the natural intermediate case, in which each agent should receive at most a fixed number k of disconnected pieces:

(a) Arunachaleswaran and Gopalakrishnan [6] consider a connected cake, i.e, a single island. They quantify the gain, as a function of k, in the optimal social welfare, defined as the sum of values of all agents or the minimum value per agent. Based on their results, they conjecture that this gain grows linearly with min(n,k).

(b) Bei and Suksompong [11] consider a cake modeled as a connected graph, where the agents prefer to receive a connected subset of edges. This is a generalization of the classic model of an interval cake. They prove that, for every connected graph, it is possible to give each agent a connected piece worth at least 12n1 of his/her total value. For a star with m<2n1 edges, the fraction improves to 1n+m21. When there are n=2 agents, and the agents may get disconnected pieces whose total count (for both agents together) is at most k+1, the fraction improves to 12(113k).

The present paper is more general in that it considers a cake that may itself be disconnected. It is less general in that each connected component must be an interval (i.e., a single edge). A potential future work avenue is to generalize both papers by letting the cake be an arbitrary (possibly disconnected) graph.

Some works aim to minimize the number of cuts required to attain various fairness and efficiency goals [2], [8], [9], [20], [42], [45], [49], [55]. These usually assume that the cake itself is connected. Moreover, their goal is to minimize the global number of cuts, while the goal in the present paper is to ensure that every individual agent is not given too many disconnected crumbs. Nevertheless, our algorithm makes at most n1 cuts, which in some cases is the smallest possible amount (see Remark 4.11).

Some works consider a different multi-cake division problem, in which each agent must get a part of every sub-cake [19], [32], [38]. This is opposite to the present paper, in which each agent wants to overlap as few cakes (islands) as possible.

In the multi-layered cake-cutting model [29], there are several overlapping cakes, representing time-intervals in which different facilities are available. An agent can get pieces in two different cakes (representing time-intervals in two different facilities) only if these pieces do not overlap.

Some works consider an endowment made of several discrete objects, and try to minimize the number of items that have to be shared. Brams and Taylor [15] present the Adjusted Winner procedure, which divides items fairly and efficiently among two agents such that at most one item is shared. Wilson [56], in an unpublished manuscript, shows that this can be generalized to n agents such that at most n1 items are shared, and this is the smallest number that can be guaranteed. Segal-Halevi [46] extends these bounded sharing results to other fairness criteria, while Sandomirskiy and Segal-Halevi [41] present an algorithm for finding a fair and efficient division with minimal sharing. The present paper focuses on the convenience of each individual agent, rather than the global number of items (islands) that are shared.

Finally, the present work is a part of an ongoing endeavor of extending cake-cutting algorithms to handle more realistic division scenarios such as allocation of land [23], [26], [30], [33], [50] and other natural resources [18], [21], [31].

Section snippets

Preliminaries

There is a set C (“the multicake”) that contains m1 disjoint subsets (“the islands”). It is convenient to assume that C is a subset of the real line and the islands are pairwise-disjoint intervals of unit length. Given such an island BC and a real number d[0,1], we denote by B[d] the sub-interval of length d at the left-hand side of B.

There are n1 agents. The preferences of each agent i are represented by a non-negative integrable function vi:CR+ called a value-density. The agent’s value

Upper bound

Before presenting an algorithm for dividing a multicake, it is useful to have an upper bound on what such an algorithm can hope to achieve.

Lemma 3.1

For every m1,n1,k1, there is a multicake instance with m islands and n agents in which, in every k-feasible allocation, at least one agent receives at most the following fraction of the total multicake value: min1n,kn+m1

Proof

Consider a multicake with m1 “small” islands and one “big” island. All n agents have the same value-measure: they value every small

Algorithm

Motivated by the upper bound of Lemma 3.1, this section aims to solve the following problem.

Below, Section 4.1 presents the main algorithm: it is a recursive algorithm that, in each iteration, finds a partial allocation of a subset of C to a subset of the agents. The main challenge is to find the appropriate partial allocation in each step. As a warm-up, Sections 4.2 One piece per agent, 4.3 Two pieces per agent show how to find a partial allocation in the cases k=1 and k=2 respectively.

Relative value guarantee

This section aims to prove Theorem 2 by solving Problem 4. As in the previous section, it is convenient to work with the normalized format specified by Problem 5.

Lemma 5.1

Any instance of Problem 4 can be reduced to an instance of Problem 5.

Proof

Given an instance of Problem 4, construct an instance of Problem 5 with the same n,k and with m=max(m,nkn+1); if m<nkn+1, then add dummy islands whose value for all agents is 0, such that the total number of islands becomes nkn+1. Construct value-measures V1,,

Envy-free allocation

The goal in this section is to find a k-feasible allocation which not only guarantees to each agent a value above some threshold, but is also envy-free: each agent should value his/her piece at least as much as the piece of every other agent. Formally, we want that for every two agents i,j[n]: Vi(Zi)Vi(Zj).

Define the utility that an agent i[n] gains from a piece ZC as Uik(Z)Vi(Bestik(Z)).Observe that any general allocation can be converted to a k-feasible allocation by letting each agent i

Dividing a rectilinear polygon

This section shows an application of the multicake division algorithm for fairly dividing a two-dimensional land-estate. A natural requirement in land division settings is to give each agent a rectangular piece [48]. We assume that the cake C is a rectilinear polygon — a polygon whose angles are 90° and 270°. Rectilinearity is a common assumption in practical partition problems [36]. The complexity of a rectilinear polygon is measured by the number of its reflex vertices — vertices with a 270°

CRediT authorship contribution statement

Erel Segal-Halevi: Conceptualization, Formal analysis, Writing - original draft, Writing - review & editing.

Acknowledgments

I am grateful to Elad Aigner-Horev, Chris Culter,10 Zur Luria, Yuval Filmus, Gregory Nisbet and lulu for their helpful ideas, and three anonymous reviewers of Discrete Applied Mathematics for their many constructive comments. The research was partly funded by the Doctoral Fellowships of Excellence Program at Bar-Ilan University, the Mordecai and Monique Katz Graduate Fellowship Program, and the Israel Science Foundation grants 1083/13 and 712/20.

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