1 Fair Allocations of Indivisible Goods Part I: Minimizing Envy Elchanan Mossel Amin Saberi Richard Lipton Vangelis Markakis Georgia Tech CWI U. C. Berkeley Stanford

2 Cake-cutting problems Divide the cake among a set of people in a fair manner Fairness measure: Envy [Foley ’67, Varian ’74] 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)

3 Set of agents K = {1, 2, …, k} Set of indivisible goods M = {1, 2, …, m} Discrete version

4 Model For agent p: utility function :  Special cases:  Additive utilities (e.g. probability measures)  Same utility for every agent. (monotone)

5 What is fair? –Proportionality [Steinhaus - Banach - Knaster ’ 48] –Envy-freeness [Foley ’ 67, Varian ‘ 74] –Max-min fairness [Dubins - Spanier ’ 61] –Equitability –…..

6 Fairness Concept Given an allocation A = (A 1,…,A k ): Envy of p for q: Envy of A: Envy-free allocations may not exist Goal: Polynomial time algorithms with upper bounds on the envy

7 Outline Existence of allocations with bounded envy Optimization questions: positive and negative results Incentive Compatibility

8 Outline Existence of allocations with bounded envy Optimization questions: positive and negative results Incentive Compatibility

9 Additive Utilities     Theorem [Dall’Aglio - Hill ’03]: There exists an allocation A with e(A) ≤  (2k) 3/2. Proof: probability measure on [0,1], Tools: convexity arguments, envy seen as the distance between a certain space and its convex hull.

10 A Tight Bound Theorem: We can compute in time O(mk 3 ) an allocation A, such that e(A) ≤ . [Dall’Aglio - Hill ’03]: e(A) ≤  (2k) 3/2 1 good, 2 players  e(A)  

11 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. ● ● ●● ● ● ● ● ● A1A1 A5A5 A4A4 A3A3 A2A2 A = (A 1, A 2,…,A 5,…) 

12 ● ● ●● ● ● ● ● ● # of edges decreases Envy does not increase A1A1 A5A5 A4A4 A3A3 A2A2 ● ● ●● ● ● ● A2A2 A1A1 A5A5 A4A4 A3A3 ● ● 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).

13 Algorithm At step i:  Find and eliminate all the directed cycles from the envy graph.  Give good i to an agent that no-one envies (any node with in-degree = 0). □

14 Remarks Bound is tight Nonadditive utilities maximum marginal utility Cyclic swaps: used in finding theater sponsors in ancient Greece, (2-cycles)!

15 Outline Existence of allocations with bounded envy Optimization questions: positive and negative results Incentive Compatibility

16 Optimization Problem 1 [envy]: Find an allocation A that minimizes the envy: Problem 2 [envy-ratio]: Find an allocation A that minimizes the ratio:

17 Hardness Results Both problems are NP-hard. Proof: Partition; even if k = 2 and both players have the same additive utility function. Definition: An algorithm A, for a minimization problem , achieves an approximation factor of  (   1), if for every instance I of , the solution returned by A satisfies: Envy: Also hard to approximate with better than exponential approximation factor; even for the above case. SOL(I)   OPT(I) Approximation algorithms?

18 Envy-ratio: Identical Additive Utilities Assume agents have the same utility function Value of good Envy-ratio(A) =

19 Relations with Job Scheduling People  Processors Goods  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.

20 Polynomial Time Approximation Schemes PTAS: A family of algorithms {A  } s.t.   >0 A  returns a solution with error  (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

21 A PTAS for the envy-ratio problem Theorem: The envy-ratio problem admits a Polynomial Time Approximation Scheme. Proof outline: 1.Rounding step ( I  I R ). 2.Solve I R optimally: Integer Programming with constant # of variables 3.Transform allocation of rounded instance to an allocation in I.

22 Proof Outline Cont ’ d 1.Rounding step ( I  I R ) (with respect to  )  Large goods: give each to some agent and remove these agents from I  Small goods: Merge together and divide into equal pieces  Medium goods: delete some least significant digits and round up 2.Solve I R optimally  New instance has constant number of different bundles an agent can have in an optimal solution  Integer programming formulation with constant number of variables  Lenstra ’ s algorithm 3.Transform allocation of rounded instance to an allocation in I.  Rounding error incurs at most 1 +  loss

23 More General Utilities Theorem: Any deterministic algorithm needs an exponential number of queries to produce any finite approximation. Additive non-identical utilities: O(m)-approximation Non-additive utilities: (assuming access to the utilities via queries) Proof: Counting argument, similar to [Nisan-Segal ’03]. Not dependent on any complexity theory assumption.

24 Summary of Approximability PositiveNegative Identical AdditivePTASNP-hard General AdditiveO(m)NP-hard General Non-additive Hard for any f(k, m) Other (e.g. submodular) ??

25 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: Any algorithm that always outputs a minimum envy allocation cannot be truthful.

26 A Related Problem Problem 3 [max-min fairness]: Find an allocation A that maximizes the utility of the least happy person:

27 Comparisons with envy-ratio Envy-ratioMax-min fairness PositiveNegativePositiveNegative Additive Identical PTASNP-hardPTASNP-hard General Additive O(m)NP-hardO(m) [BD ‘05] O(k) [G ‘05]  k [AS ‘07] 2-hard General Non-additive Hard for any f(k, m) Other (e.g. submodular) ??O(k) [G ‘05, KP ‘07] ?

28 Difficulties with Existing Techniques Integrality Gap: gap between LP optimum and true optimum Linear Programming relaxation for max-min fairness:

29 Why we need better LP Relaxations Consider instances with a good of very high value Fractionally: Everybody can get a piece Integrally: Somebody will be unhappy

30 Conclusions  There exist allocations, in which the envy is bounded by the maximum marginal utility.  Envy and max-min fairness are computationally hard in general.  If all players have the same (additive) utility function both problems can be well approximated.  There can be no truthful algorithm that computes a minimum envy allocation.

31 Thank You!

32 Step 1: Rounding (I  I R ) Let L be the average utility: 3 types of goods: 1.Large: 2.Medium: 3.Small: Rounding parameter: integer constant

33 Step 1: Rounding (I  I R ) 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:         

34 Step 1: Rounding (I  I R ) 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:                  

35 Step 2: Solve I R optimally Constant number of distinct values for the goods in I R : Claim:  optimal allocation A in I R s.t. # distinct bundles with  2 λ goods is constant (exp( λ) but still constant)  # goods in Integer program formulation with constant number of variables  Lenstra ’ s algorithm

36 Step 3 (I R  I) Lemma 1: Given an optimal solution of I R, we can find an allocation in I, B = (B 1,…,B n ), such that: OPT R : Optimal solution of the rounded instance. Lemma 2: OPT R  OPT