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Envy-free item allocation

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Envy-free item allocation

Since the items are indivisible, an EF assignment may not exist. The simplest case is when there is a single item and at least two agents: if the item is assigned to one agent, the other will envy.

One way to attain fairness is to use monetary transfers. When monetary transfers are not allowed or not desired, there are allocation algorithms providing various kinds of relaxations.

Finding an envy-free allocation whenever it exists ### Preference-orderings on bundles: envy-freeness The undercut-procedure finds a complete EF allocation for two agents, if-and-only-if such allocation exists. It requires the agents to rank bundles of items, but it does not require cardinal utility information. It works whenever the agents' preference relations are strictly monotone, but does not need to assume that they are responsive preferences. In the worst case, the agents may have to rank all possible bundles, so the run-time might be exponential in the number of items.

Preference-orderings on items: necessary/possible envy-freeness It is usually easier for people to rank individual items than to rank bundles. Assuming all agents have responsive preferences, it is possible to lift the item-ranking to a partial bundle-ranking. For example, if the item-ranking is w>x>y>z, then responsiveness implies that {w,x}>{y,z} and {w,y}>{x,z}, but does not imply anything about the relation between {w,z} and {x,y}, between {x} and {y,z}, etc. Given an item-ranking: An allocation is necessarily envy-free (NEF) if it is envy-free according to all* responsive bundle-rankings consistent with the item-ranking; An allocation is possibly envy-free (PEF) if for each agent i, there is at least one responsive bundle-ranking consistent with i's item-ranking, by which i* does not envy; An allocation is weakly possibly envy-free (WPEF) if for each pair of agents i,j, there is at least one responsive bundle-ranking consistent with i's item-ranking, by which i does not envy j*; An allocation is necessarily Pareto-optimal (NPE) if it is Pareto-optimal according to all* responsive bundle-rankings consistent with the item-ranking (see Ordinal Pareto efficiency); An allocation is possibly Pareto-optimal (PPE) if it is Pareto-optimal according to at least one* responsive bundle-ranking consistent with the item-ranking.

The following results are known:

Cardinal utilities The empty allocation is always EF. But if we want some efficiency in addition to EF, then the decision problem becomes computationally hard: * Deciding whether a fair allocation exists requires exponential communication (in the number of goods) when there are more than two agents. When there are two agents, the communication complexity depends on specific combinations of parameters. Deciding whether an EF and Pareto efficient* allocation exists is above NP: it is <math>\Sigma_{2}^{\textrm{P}}</math>-complete even with dichotomous utilities and even with additive utilities. (<math>\Sigma_{2}^{\textrm{P}}</math> is the class of problems that can be solved in nondeterministic time given an oracle that can solve any problem in NP). The decision problem may become tractable when some parameters of the problem are considered fixed small constants:

EFx - envy-free up to at most any item An allocation is called EFx if for every two agents A and B, if we remove any item from the bundle of B, then A does not envy B. EFx is strictly stronger than EF1: EF1 lets us eliminate envy by removing the item most valuable (for A) from B's bundle; EFx requires that we eliminate envy by removing the item least valuable (for A). An EFx allocation is known to exist in some special cases:

Some approximations are known:

It is an open question whether an EFx allocation exists in general. The smallest open case is 4 agents with additive valuations.

In contrast to EF1, which requires a number of queries logarithmic in the number of items, computing an EFx allocation may require a linear number of queries even when there are two agents with identical additive valuations.

EFm - approximate envy-free for a mixture of divisible and indivisible items Some division scenarios involve both divisible and indivisible items, such as divisible lands and indivisible houses. An allocation is called EFm if for every two agents A and B:

An EFm allocation exists for any number of agents. However, finding it requires an oracle for exact division of a cake. Without this oracle, an EFm allocation can be computed in polynomial time in two special cases: two agents with general additive valuations, or any number of agents with piecewise-linear valuations.

In contrast to EF1, which is compatible with Pareto-optimality, EFm may be incompatible with it.

Minimizing the envy Rather than using a worst-case bound on the amount of envy, one can try to minimize the amount of envy in each particular instance. See envy-minimization for details and references.

Finding a partial EF allocation The al-procedure finds an EF allocation for two agents. It may discard some of the items, but, the final allocation is Pareto efficient in the following sense: no other EF allocation is better for one and weakly better for the other. The AL procedure only requires the agents to rank individual items. It assumes that the agents have responsive preferences and returns an allocation that is necessarily envy-free (NEF).

The adjusted-winner-procedure returns a complete and efficient EF allocation for two agents, but it might have to cut a single item (alternatively, one item remains in shared ownership). It requires the agents to report a numeric value for each item, and assumes that they have additive utilities.

When each agent may get at most a single item, and the valuations are binary (each agent either likes or dislikes each item), there is a polynomial-time algorithm that finds an envy-free-matching of maximum cardinality.

Existence of EF allocations with random valuations If the agents have additive utility functions that are drawn from probability distributions satisfying some independence criteria, and the number of items is sufficiently large relative to the number of agents, then an EF allocation exists with high probability. Particularly: * If the number of items is sufficiently large: <math>m = \Omega(n \log n/\log\log n)</math>, then w.h.p. an EF allocation exists (the probability goes to 1 as m goes to infinity) and can be found by the round-robin protocol. * If <math>m = \Omega(n \log n)</math>, then w.h.p. an EF allocation exists and can be found by maximizing the social welfare. This bound is also tight due to connections to the coupon collector's problem. * If <math>m \geq 2n</math> and m is divisible by n, then w.h.p. an EF allocation exists and can be found by a matching-based algorithm.

On the other hand, if the number of items is not sufficiently large, then, with high probability, an EF allocation does not exist.

Envy-freeness vs. other fairness criteria Every EF allocation is min-max-fair*. This follows directly from the ordinal definitions and does not depend on additivity. If all agents have additive utility functions, then an EF allocation is also proportional and max-min-fair*. Otherwise, an EF allocation may be not proportional and even not max-min-fair. Every allocation of a competitive-equilibrium from equal incomes* is also envy-free. This is true regardless of additivity. * Every Nash-optimal allocation (allocation that maximizes the product of utilities) is EF1. **Group-envy-freeness* is a strengthening of envy-freeness, which is applicable to both divisible and indivisible objects.

See Fair item allocation for details and references.

Summary table Below, the following shorthands are used: * <math>n</math> = the number of agents participating in the division; * <math>m</math> = the number of items to divide; * EF = envy-free, EF1 = envy-free except-1-item (weaker than EF), MEF1 = marginal-envy-free except-1-item (weaker than EF1). * PE = Pareto-efficient.

References