Cake-Cutting with Different Entitlements: How Many Cuts are Needed?

03/14/2018 ∙ by Erel Segal-Halevi, et al. ∙ 0

A cake has to be divided fairly among n agents. When all agents have equal entitlements, it is known that such a division can be implemented with n-1 cuts. When agents may have different entitlements, we show that at least 2 n -2 cuts may be necessary, and 2 n·Θ((n)) cuts are always sufficient.

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1 Introduction

The fair cake-cutting problem involves a continuous resource dubbed “cake” that has to be divided among several agents with different valuations for subsets of the cake. The problem was introduced by Steinhaus (1948) and inspired hundreds of research papers and several books (Brams and Taylor, 1996; Robertson and Webb, 1998; Moulin, 2004; Brams, 2007). See Procaccia (2015); Brânzei (2015); Brams and Klamler (2017); Segal-Halevi (2017) for recent surveys.

An important requirement in dividing a cake is to minimize the number of cuts used to implement the division. When a cake is divided among people, we would ideally like to make cuts, such that each person receives a single connected piece. Making too many cuts might result in each person receiving a bunch of crumbs. This requirement is even more important when the “cake” represents a resource such as land or time. When a land-estate is divided among heirs, a plot made of many disconnected pieces cannot be efficiently built upon or cultivated. Similarly, when the usage-time of a vacation-house is divided among its owners, having many disconnected time-intervals is practically useless.

In the simplest variant of the cake-cutting problem, all agents have equal entitlements. The common fairness criterion in this case is that each agent should receive a piece worth at least of the total cake value (a requirement called proportionality or fair-share guarantee). Steinhaus (1948) already proved that there always exists a proportional cake division with cuts, which is obviously the smallest number of cuts needed for making pieces.

However, in many division problems, agents have different entitlements. As an example, consider a land-estate owned by a partnership in which Alice’s share is 2/5 and George’s share is 3/5. If the partnership is dissolved, then Alice is entitled to 2/5 and George is entitled to 3/5 of the land-estate. As another example, consider a vacation-house bought by several friends, each of whom invested a different sum of money towards the purchase. The time in which the house is used should be divided among the friends in proportion to their investment. The situation of agents with different entitlements has been studied both for cake-cutting and for allocation of indivisible items (Babaioff et al., 2017; Farhadi et al., 2017; Segal-Halevi, 2018); here we focus on cake-cutting.

In general, in the cake-cutting with entitlements setting, each agent has an entitlement such that . A division is called -proportional if each agent values his/her share as at least a fraction of the total cake value. Proportionality is the special case of -proportionality in which for every agent . The present paper studies the following question:

How many cuts are required to divide a cake -proportionally among agents?

A first solution that comes to mind is agent cloning. In the example above, create 2 clones of Alice and 3 clones of George and find a proportional division among the clones. Since each clone receives a value of at least 1/5, each agent receives at least his/her entitlement. However, this solution might result in a large number of cuts. In particular, if the common denominator of all agents’ entitlements is then agent-cloning might require up to cuts. If the entitlements are irrational then the number of cuts is unbounded.

Various authors have tried to improve on the agent-cloning method by decreasing its computational complexity — the number of evaluation queries and intermediate cut-marks required to find a proportional division. See McAvaney et al. (1992), Barbanel (1995), Robertson and Webb (1997), Robertson and Webb (1998, pages 36-–44), Brams et al. (2011). The most recent advancement in this front is by Cseh and Fleiner (2017). After summarizing the computational complexity of previous algorithms, they present a faster one that requires queries, where is the smallest common denominator of the entitlements. They prove that this number is asymptotically optimal.

In contrast, the present paper focuses on the existential question of how many cuts are required in the final division, regardless of how this division is computed. To the best of our knowledge, the only previous answer to this question is by Brams et al. (2011), who prove that cuts may be insufficient.

The present paper provides a lower and an upper bound for this question.

  • Lower bound: cuts might be necessary (Section 3).

  • Upper bound: cuts are always sufficient, where rounded up to the nearest power of two (Section 4).

The bounds coincide when at 2 cuts, but they diverge when , where the lower bound is and the upper bound is .

2 Model

There is a cake , represented by the interval . There is a set of agents, represented by nonatomic value-measures on the cake.

There is a vector

of length where each element represents the entitlement of agent . The sum of entitlements is one: .

The goal is to find a partition of the cake among the agents, , where the are pairwise-disjoint and . The partition should be -proportional, i.e, for every agent :

3 Lower Bound

Theorem 1.

For each , there exists a set of value measures and an entitlement-vector such that at least cuts are required for a -proportional allocation.

Proof.

Suppose that the entitlement of agent 1 is and the entitlement of each of the the other agents is . The cake is an interval consisting of subintervals and the agent valuations in these subintervals are:

Agent 1 1 0 1 0 1 0 1 1 0 1
Agent 2 0 1 0 0 0 0 0 0 0 0
Agent 3 0 0 0 1 0 0 0 0 0 0
Agent 4 0 0 0 0 0 1 0 0 0 0
Agent 0 0 0 0 0 0 0 0 1 0

Agent 1 must receive more than of his cake value, so he must receive a positive slice of each of his positive subintervals. But, he cannot receive a single interval that touches two of his positive subintervals, since such an interval will leave one of the other agents with zero value. Therefore, agent 1 must receive at least different intervals. Each of the other agents must receive at least one interval, so the total number of intervals is at least . Therefore at least cuts are needed. ∎

4 Upper Bound

Our upper bound shows the existence of a partition that satisfies a requirement much stronger than -proportionality. A partition is called -consensus if, for every agents :

Clearly, a -consensus division is also -proportional. The existence of -consensus divisions was already shown by Dubins and Spanier (1961), but they did not consider the number of required cuts.

We will use the Stromquist-Woodall theorem (Stromquist and Woodall, 1985). Their cake model is circular — they assume the left and right endpoints of the cake are identified. In other words, they divide a 1-dimensional “pie”.

Theorem (Stromquist and Woodall (1985)).

Consider a 1-dimensional pie , a set of agents with nonatomic measures on , and a number . There always exists a subset , which is a union of at most intervals, such that for every :

The number is tight when the denominator in the reduced fraction of is at least (in particular, it is tight when is irrational).

We now present our upper bound. Denote rounded up to the nearest power of two .

Theorem 2.

For every , set of value measures, and an entitlement-vector , there exists a -consensus allocation with at most cuts.

Proof.

Let . Partition the group of agents into two sub-groups: with agents and with agents. Define:

By the Stromquist-Woodall theorem, there exists a piece that all agents value as exactly . Define and note that all agents value as exactly . If we identify the two endpoints of , then is a union of at most intervals. Hence, we can separate from using at most cuts — two cuts per interval. The same cuts can be used when the endpoints are not identified (even though in this case might be a union of intervals).

Now, we have to divide to the agents in and divide to the agents in . In the first subdivision, cuts are needed, and in the second subdivision, cuts are needed; all in all, cuts are needed.

We proceed recursively until there is a single agent in each sub-group. Then, each agent holds a piece that is worth exactly for all agents.

In step of the recursion, cuts are required. The number of steps is . Summing up the required cuts gives as claimed. ∎

Since the upper bound is for -consensus division, which is much stronger than -proportional division, we conjecture that the actual number of required cuts is as in the lower bound.

Conjecture.

For every , set of value measures, and an entitlement-vector , there exists a -consensus allocation with at most cuts.

5 Acknowledgments

This paper started as a question in MathOverflow (https://mathoverflow.net/q/242112/34461). I am grateful to Sam Zbarsky for the discussion.

References

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