On allocations that maximize fairness

Uriel Feige

Microsoft Research and Weizmann Institute

Fair allocation of indivisible goods

m players, n items.

value of item j to player i.

value of bundle S to player i.

Partition all items into m disjoint bundles, and allocate to each player a bundle that she values as at least t.

opt is the maximum feasible value of t.

Three versions

Uniform. Items have intrinsic value.

Restricted assignment case. As above, but some players do not value some of the items.

General case. Arbitrary

Estimation versus Approximation

Computing opt is NP-hard, even in the uniform case.

An approximation algorithm produces an allocation for a value of t that is close to opt (up to the approximation ratio).

An estimation algorithm only estimates opt (up to the estimation factor), without necessarily exhibiting an allocation.

Some known results

Uniform case has a PTAS [Woeginger].

In the restricted assignment case, it is NP-hard to estimate opt within a ratio better than 2 [Bezakova and Dani].

opt can be approximated within an additive term of [BD].

Useless when

Integer linear program

Maximize t subject to:

• Value

• Player

• Item

• 0/1

Feasible allocation = feasible solution to LP.

NP-hard.

The configuration LP [Bansal and Sviridenko]

Maximize t subject to:

• Value

• Player

• Item

• relax

t opt

LP relaxation solvable in polynomial time.

Known results

Gap between LP and opt is sometimes as large as [BS].

Approximation algorithm with ratio

[Asadpour and Saberi].

Restricted assignment case, approximation algorithm with ratio

[BS].

Our result

Theorem: in the restricted assignment case, the configuration LP estimates opt within a constant ratio.

Our proof is nonconstructive. It does not provide a polynomial time approximation algorithm.

Builds heavily on [BS].Repeated applications of the Lovasz local

lemma [Leighton, Maggs, Rao].

A representative special case

Configuration LP value

Player i:

For desirable large items j,

For desirable cheap bundle S,

All constraints satisfied with t=k.

If our theorem is true, then there must exist a feasible allocation with

Previous results [BS]

This special case is hardest for the restricted assignment version, up to constant factors.

Will imply constant estimation for every restricted assignment instance.

[BS] give a Ptime algorithm for

This gives their approximation for the restricted assignment version.

Some intuition

Clearly, in each group g-1 players will receive large items.

1. Need to choose one player in each group to receive her bundle of cheap items.

2. Need to resolve conflicts for cheap items that are in multiple selected bundles.

Given the first task, the second task can be performed optimally.

Hall’s theorem

Given m/g selected bundles (one for each group), if every b selected bundles contain at least bt distinct items, then the items can be distributed such that every player has value at least t.

Moreover, such an allocation can be found in polynomial time (though this is irrelevant to our results).

Consequences

Lemma constant: there is always an allocation of value t max[1, k/g].

Implies the theorem for constant k or constant g.

Theorem proof plan: through a sequence of transformations, reduce either k or g to a constant (preserving the structure of the problem).

Reducing g to 1

Select at random one player in each group (the others receive large items).

For every selected player, for every item in her cheap bundle, in expectation it conflicts with one more selected bundle.

Indicates that we may be able to achieve t = k/2, if events do not deviate from expectation by much.

Challenges

Problem: too many groups. In a small fraction of them something may go wrong.

Observe: each group depends only on other groups. The local lemma would apply if probability of large deviation is smaller than .

Problem: sets overlap, events are correlated. Constant probability of large deviation.

Local Lemma: reducing g

Select g/2 random players in each group (one player from each adjacent pair).

Desired property: each cheap item is demanded by (1 + o(1))g/2 players.

For each cheap item, the probability of large failure is super-polynomially small in g.

Each item depends on 2gk other items.Local lemma applies when g > k.What if k > g?

Local Lemma: reduce k

Select every cheap item independently with probability 1/2.

Desired property: every player demands at most (1 + o(1))k/2 selected cheap items.

For each player, probability of failure is super-polynomially small in k.

Each player depends on gk other players.

Local lemma applies when k > g.

Lifting the solution

After reducing k to k/2, a solution of value t is expected to be reduced to a solution of value at least t’ = (1-o(1))t/2.

In the reduced problem, change t to t’. Need to show:• t in original implies t’ in reduced.• t’ in reduced implies (1 – o(1))t in original.Use general version of local lemma (tricky

definition of dependency graph [BS]).

Proof structure (sketch)

Alternate between:

Lemma reduce g (when g > k).

Lemma reduce k (when k > g). Reduces also t.

When g reaches a constant, finish off using Lemma constant.

For players chosen at the end, lift solution to original value of t.

Some comments

Additional care is needed to make sure that sum of multiple o(1) terms remains bounded.

Lemma reduce k (using the general local lemma to allow lifting of solution) is the obstacle to making the proof constructive. It involves exponentially many events, and many of them involve a linear number of items.

Conclusions

Special case of restricted assignment version has solution of value

Implies (by [BS]) a constant factor estimation for every restricted assignment instance.

No constant factor approximation algorithm known for this problem.