CSC2556 Lecture 7 Fair Division 2: Indivisible Goods Leximin Allocation
CSC2556 - Nisarg Shah 1
Fair Division 2: Indivisible Goods Leximin Allocation CSC2556 - - - PowerPoint PPT Presentation
CSC2556 Lecture 7 Fair Division 2: Indivisible Goods Leximin Allocation CSC2556 - Nisarg Shah 1 Cake-Cutting (contd) Indivisible Goods CSC2556 - Nisarg Shah 2 Pareto Optimality (PO) Definition We say that an allocation =
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β’ We say that an allocation π΅ = (π΅1, β¦ , π΅π) is PO if there is
no alternative allocation πΆ = (πΆ1, β¦ , πΆπ) such that
π πΆπ β₯ π π(π΅π), βπ β π
π πΆπ > π π(π΅π), βπ β π
β’ I.e., an allocation is PO if there is no βbetterβ allocation.
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β’ There always exists an allocation of the cake that is both
envy-free and Pareto optimal.
β’ Nash-optimal allocation: argmaxπ΅ Οπβπ π
π π΅π
β’ Obviously, this is PO. The fact that it is EF is non-trivial. β’ This is named after John Nash.
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β’ Green player has value 1 distributed over 0, Ξ€
2 3
β’ Blue player has value 1 distributed over [0,1] β’ Without loss of generality (why?) suppose:
2 3]
2 3] AND all of [ Ξ€ 2 3 , 1].
β’ Greenβs utility = π¦, blueβs utility = 1 β x β 2 3 + 1 3 = 3β2π¦ 3 β’ Maximize: π¦ β 3β2π¦ 3
β π¦ = Ξ€
3 4 ( Ξ€ 3 4 fraction of Ξ€ 2 3 is Ξ€ 1 2).
1
ΰ΅ 2 3
Allocation 1
ΰ΅ 1 2
Green has utility 3
4
Blue has utility 1
2
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β’ I believe it should require an unbounded number of
queries in the Robertson-Webb model. But I canβt find such a result in the literature.
β’ For piecewise constant valuations, the Nash-optimal
solution can be computed in polynomial time.
1
The density function of a piecewise constant valuation looks like this
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β’ Each good can be divided fractionally between the agents
β’ Homogeneous = agent doesnβt care which βpartβ
β’ Line up the goods on [0,1] β piecewise uniform
valuations
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β’ Polynomial time solvable
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β’ E.g., house, painting, car, jewelry, β¦
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We assume additive values. So, e.g., π , = 8 + 7 = 15 Given such a matrix of numbers, assign each good to a player.
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π π΅π β₯ π π π΅π\{π}
β’ Technically, we need either this or π΅π = β . β’ βIf π envies π, there must be some good in πβs bundle such
that removing it would make π envy-free of π.β
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β’ Agents take turns in cyclic order: 1,2, β¦ , π, 1,2, β¦ , π, β¦ β’ In her turn, an agent picks the good she likes the most
among the goods still not picked by anyone.
β’ Informal proof on the board.
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β’ The allocation argmaxπ΅ Οπβπ π
π π΅π is EF1 + PO.
β’ Note: This maximization is over only βintegralβ allocations
β’ Note: Subtle tie-breaking if all allocations have zero Nash
welfare.
it is possible to give a positive utility to every player in π simultaneously.
π π΅π
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β’ That is, remains NP-hard even if all values in the matrix
are bounded
β’ Not sure. But a recent paper gives a pseudo-polynomial
π,π π π
π .
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β’ Generalization of βcut and chooseβ for π players β’ MMS value of player π =
β’ Let π¬
π π denote the family of partitions of the set of
goods π into π bundles. ππππ = max
πΆ1,β¦,πΆπ βπ¬π π
min
πβ 1,β¦,π π π(πΆπ) .
β’ An allocation is π½-MMS if every player π receives value at
least π½ β ππππ.
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β’ [Procaccia, Wang β14]:
There is an example in which no MMS allocation exists.
β’ [Procaccia, Wang β14]:
2 3 - MMS allocation always exists.
β’ [Ghodsi et al. β17]:
A Ξ€
3 4 - MMS allocation always exists.
β’ [Caragiannis et al. β16]:
The Nash-optimal solution is
2 1+ 4πβ3 βMMS, and this is
the best possible guarantee.
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π π΅π β₯ π π π΅π\{π}
β’ Intuitively, π doesnβt envy π if she gets to remove her most
π π΅π β₯ π π π΅π\{π}
β’ Note: Need to quantify over π such that π
π
π > 0.
β’ Intuitively, π doesnβt envy π even if she removes her least
positively valued item from πβs bundle.
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β’ Suppose there are two players and three goods with
values as follows.
β’ If you give {A} β P1 and {B,C} β P2, itβs EF1 but not EFx.
β’ Instead, {A,B} β P1 and {C} β P2 would be EFx.
A B C P1 5 1 10 P2 1 10
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β’ Let ππ,π be the cost of player π for bad π.
β’ EF: βπ, π π·π π΅π β€ π·π π΅π β’ PO: There should be no alternative allocation in which no
player has more cost, and some player has less cost.
β’ EF + PO allocation always exists, like for divisible goods.
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β’ EF1: βπ, π βπ β π΅π ππ π΅π\ π
β€ ππ π΅π
β’ EFx: βπ, π βπ β π΅π ππ π΅π\ π
β€ ππ π΅π
β’ Open Question 1:
β’ Open Question 2:
β’ More open questions related to relaxations of
proportionality
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β’ Player 1 requires (2 CPU, 1 RAM) for each copy of task β’ Indifferent between (4,2) and (5,2), but prefers (5,2.5) β’ βfractionalβ copies are allowed
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β’ 0 < πππ β€ 1 for every π , πππ = 1 for some π
β’ Utility to player π βΆ π£π π΅π = min
π βπ π΅ππ /πππ .
β’ Weβll assume a non-wasteful allocation
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β’ Allocate maximal resources while maintaining equal
36
Total
1 2
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β’ Why? [Note: EF no longer implies proportionality.]
β’ Why?
β’ If a group of players manipulate, it canβt be that none of
β’ Weβll skip this proof.
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β’ Choose an allocation π΅ that
π
π£π π΅π
minimum utility.
utility.
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β’ In the previous illustration, we didnβt need tie-breaking
because we assumed πππ > 0 for every π β π, π β π.
β’ In practice, not all the players need all the resources. β’ When πππ = 0 is allowed, we need to continue allocating
even after some agents are saturated.
β’ When πππ = 0 is allowed, the leximin mechanism still
retains all four properties (proportionality, envy-freeness, Pareto optimality, group strategyproofness).
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β’ A dynamic version of the leximin mechanism satisfies
proportionality, Pareto optimality, and strategyproofness along with a relaxed version of envy-freeness when agents arrive one-by-one.
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β’ Designing fair, efficient, and game-theoretic mechanisms
in dynamic environments is a relatively new research area, and we do not know much.
β’ E.g., what if agents can depart, demands can change over
β’ Lots of open questions!
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β’ We assumed ranked preferences, and showed that the
Gale-Shapley algorithm produces a stable matching.
β’ What if agent preferences werenβt ranked?
β’ Each man finds a subset of women βacceptableβ (utility
1), and the rest βunacceptableβ (utility 0).
β’ Same for womenβs preferences over men.
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β’ If a perfect matching exists, itβs awesome. β’ What if there is no perfect matching?
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β’ βSelectβ or βnot selectβ each edge such that the number
β’ βPut a weightβ on each edge such that the total weight of
edges incident on each vertex is at most 1.
β’ Every fractional matching can be βimplementedβ as a
probability distribution over integral matchings.
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β’ Compute the leximin fractional matching, and implement
it as a distribution over integral matchings.
β’ Both steps are doable in polynomial time!
β’ The randomized leximin mechanism satisfies
proportionality, envy-freeness, Pareto optimality, and group-strategyproofness (for both sides).
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β’ βUnused resources in public schools should be fairly
allocated to local charter schools that desire them.β
β’ If the demand is met, the charter school can relocate to
the public school facility.
π unused classrooms.
β’ We assume facilities donβt have preferences over agents.
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Number of unused classrooms
6 3 8 4 11 7
2015/2016 request form: βprovide a description of the district school site and/or general geographic area in which the charter school wishes to locateβ
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β’ The randomized leximin mechanism satisfies
proportionality, envy-freeness, Pareto optimality, and group strategyproofness.
β’ Unlike DRF and matching under dichotomous
preferences.
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β’ Convexity: If two utility vectors are feasible, then so should be their convex
combinations.
β’ Equality: The maximum utility of each agent should be the same.
β’ Shifting Allocations: Swapping allocations of two agents should be allowed. β’ Maximal Utilization: No agent should have a higher utility for agent πβs
allocation than agent π has.
allocation, and many other settings from the literature.