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On Approximately Fair Allocations of Indivisible Goods

Elchanan Mossel Amin Saberi

Richard Lipton Vangelis Markakis Georgia Tech AUEB

U. C. Berkeley Stanford

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Cake-cutting problems

Divide the cake among a set of people in a fair manner

Fairness measure: Envy [Foley ’67]

Infinitely divisible cakes: Envy-free partitions exist

Cake-cutting procedures: minimize # cuts, achieve additional fairness criteria [Brams, Taylor ’96, Robertson, Webb ‘98]

Mathematical approaches:

[Steinhaus, Banach, Knaster ’48]

Empirically: since Pharaoh times (land division)

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Set of agents N = {1, 2, …, n}

Set of indivisible goods M = {1, 2, …, m}

Discrete version

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Model

For agent p: utility function :

Special cases:

Additive utilities (e.g. probability measures)

Same utility for every agent.

(monotone)

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What is fair?

– Proportionality [Steinhaus - Banach - Knaster ’48]

– Envy-freeness [Foley ’67, Varian ‘74]

– Max-min fairness [Dubins - Spanier ’61]

– Equitability

– …..

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Fairness Concept

Given an allocation A = (A1,…,An):

Envy of p for q:

Envy of A:

Envy-free allocations may not exist

Goal: Algorithms with upper bounds on the envy

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Outline

• Existence of allocations with bounded envy

• Optimization problems: positive and negative results

• Incentive Compatibility

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Outline

• Existence of allocations with bounded envy

• Optimization problems: positive and negative results

• Incentive Compatibility

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Additive Utilities

Theorem [Dall’Aglio - Hill ’03]: There exists an allocation A with e(A) ≤ (2n)3/2.

Proof:

probability measure on [0,1],

Tools: convexity arguments, envy seen as the distance between a certain space and its convex hull.

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A Tight Bound

Theorem: We can compute in time O(mn3) an allocation A, such that e(A) ≤ .

[Dall’Aglio - Hill ’03]: e(A) ≤ (2n)3/2

1 good, 2 players e(A)

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Proof

G(A) = (V, E) : envy graph of A V = {agents} pq E iff p envies q in A.

A: allocation of a subset of the goods S M.

●

●

● ●

●

●

● ●

●

A1

A5

A4

A3A2

A = (A1, A2,…,A5,…)

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●

●

● ●

●

●

● ●

●

# of edges decreases

Envy does not increase

A1

A5

A4

A3A2 ●

●

● ●

●

● ●

A2

A1

A5

A4A3

●

●

Claim: For any allocation A, there exists an allocation B s.t.:

e(B) ≤ e(A).

envy-graph of B is acyclic ( i with in-degree = 0).

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Algorithm

At step i: Eliminate all the cycles from the envy graph. Give good i to an agent that no-one envies (any

node with in-degree = 0).

□

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Remarks

• Bound is tight

• Nonadditive utilities

maximum marginal utility

• Cyclic swaps: used in finding theater sponsors in ancient Greece, (2-cycles)!

€

ντιδοσις

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Outline

• Existence of allocations with bounded envy

• Optimization problems: positive and negative results

• Incentive Compatibility

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Optimization

Problem 1 [envy]: Find an allocation A that minimizes the envy:

Polynomial time algorithms?

Problem 2 [envy-ratio]: Find an allocation A that minimizes the ratio:

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Hardness Results

Both problems are NP-hard.

Proof: Partition; even if n = 2 and both players have the same utility function.

Envy: Also hard to approximate; even for the above case.

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Additive Utilities

Assume agents have the same utility function

Value of good

Envy-ratio(A) =

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Relations with Job Scheduling

People ProcessorsGoods Jobs

[Coffman-Langston ’84]: Graham’s algorithm achieves an approximation factor of 1.4 for the envy-ratio problem.

[Graham ’69]:

Order the goods in decreasing value.

Give next good to the person with the minimum current bundle.

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Polynomial Time Approximation Schemes

PTAS: > 0, algorithm A with cost (1 + )OPT in time poly(| I |), instance I

PTAS’s in job scheduling:

[Hochbaum, Shmoys ’87]: Makespan

[Woeginger ’97]: Maximize min. completion time

[Alon, Azar, Woeginger, Yadid ’98]: Generalizations

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A PTAS for the envy-ratio problem

Theorem: The envy-ratio problem admits a Polynomial Time Approximation Scheme.

Proof outline:

1. Rounding step ( I IR ).

2. Solve IR optimally: IP with constant # of variables

3. Transform allocation of rounded instance to an

allocation in I.

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Step 1: Rounding (I IR)

Let L be the average utility:

3 types of goods:

1. Large:

2. Medium:

3. Small:

Rounding parameter: integer constant

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Step 1: Rounding (I IR)

1. Large: give to some agent, remove agent

We may assume there are no large goods in I

Claim: There exists an optimal solution in which every large good is assigned to a person with no other goods in her bundle.

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Step 1: Rounding (I IR)

1. Large: WLOG no large goods in I

2. Medium: round to next integer multiple of

(ignore some of the least significant digits)

3. Small: merge together and round:

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Step 1: Rounding (I IR)

1. Large: WLOG no large goods in I

2. Medium: round to next integer multiple of

(ignore some of the least significant digits)

3. Small: merge together and round:

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Step 2: Solve IR optimally

Constant number of distinct values for the goods in IR :

Claim: optimal allocation A in IR s.t.

# distinct bundles with 2λ goods is constant(exp(λ) but still constant)

# goods in

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Step 2: Solve IR optimally

For , solve the decision problem: Is there an allocation A = (A1,…,An) with

?

Integer program, constant number of variables Lenstra’s algorithm

Repeat only for a constant number of pairs (t1, t2).

Pick solution with best envy-ratio.

Integer variable XS: # agents with bundle S, for each S with 2λ goods

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Step 3 (IR I)

Lemma 1: Given an optimal solution of IR, we can find an allocation in I, B = (B1,…,Bn), such that:

OPTR: Optimal solution of the rounded instance.

Lemma 2: OPTR OPT

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Finally…

Which turns out to be:

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Non-additive utilities

Theorem 3: Any deterministic algorithm that computes a finite approximation to minimum envy or minimum envy-ratio needs an exponential number of queries for the players’ utilities.

Proof: Counting argument, similar to [Nisan-Segal ’03]

Note: Not dependent on any complexity theory assumption.

Input: exponential in size

Use only polynomial amount of input? (query model)

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Related Work and Extensions

• Envy-ratio

– Additive non-identical utilities: O(m)-approximation

– Nonadditive (e.g. submodular) ?

• Max-min fairness:

– [Bezakova, Dani ’05, Saberi, Asadpour ’07]: new approximations + hardness results

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Incentive Compatibility

Definition: An algorithm is truthful if being honest is always a dominant strategy for every player.

So far we have assumed that players report their true utilities.

Theorem 4: An algorithm that outputs a minimum envy allocation is not truthful.

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Conclusions There exist allocations, in which the envy is bounded by the

maximum marginal utility.

Minimizing the envy is hard in general.

If all players have the same (additive) utility function the envy ratio can be well approximated.

Any algorithm that computes a minimum envy allocation is not truthful.

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Post-mortem

Economic Theory: models and solution concepts Rationality, fairness, incentive compatibility,… Mathematically rich; however mostly non-

constructive

Discrete math and theory of algorithms: Dealing with indivisibilities Computational complexity

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Post-mortem

Finding efficient algorithms for computing / approximating economic solution concepts:

Fair division (partially here, [DH ’88, BD ’05, AS ’07])

Nash Equilibria [P ’94, LMM ’03, LM ’04, PT ‘04, DMP ’06, BBM ’07]

Market Equilibria [DPS ’02, DPSV ’02, JMS ’03, DV ’03]

Cost Sharing [MS ’97, FPS ’00, JV ‘01]

……

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The End!

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Proof of Theorem 4Proof: Construction of an example in which any such algorithm will fail.

biscuit muffin k eggs

0.45 0.35 0.2/k each

0.35 0.45 0.2/k each

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Proof of Theorem 4Proof: Construction of an example in which any such algorithm will fail.

biscuit muffin k eggs

0.45 - δ 0.35 + δ 0.2/k each

0.35 0.45 0.2/k each

By misreporting Homer will receive the biscuit and more eggs than before.