T RUTH J USTICE A LGOS Fair Division V: Indivisible Goods Teachers: Ariel Procaccia (this time) and Alex Psomas
INDIVISIBLE GOODS β’ Set π» of π goods π» β’ Each good is indivisible β’ Players π = 1, β¦ , π have valuations π π for bundles of goods β’ Valuations are additive if for all π β π» and π β π π = Ο πβπ» π π , π π π β’ Assume additivity unless noted otherwise β’ An allocation is a partition of the goods, denoted π© = (π΅ 1 , β¦ , π΅ π ) β’ Envy-freeness and proportionality are infeasible!
MAXIMIN SHARE GUARANTEE Total: Total: Total: $50 $30 $20 $3 $2 $50 $30 $5 $5 $5
MAXIMIN SHARE GUARANTEE Total: Total: Total: $50 $30 $20 $3 $2 $50 $30 $5 $5 $5 $3 $2 $5 $40 $10 $20 $20 Total: Total: Total: $30 $30 $40
MAXIMIN SHARE GUARANTEE β’ Maximin share (MMS) guarantee [Budish 2011] of player π : π 1 ,β¦,π π min max π π (π π ) π β’ An MMS allocation is such that π π (π΅ π ) is at least πβs MMS guarantee for all π β π β’ For π = 2 an MMS allocation always exists β’ Theorem [Kurokawa et al. 2018]: βπ β₯ 3 there exist additive valuation functions that do not admit an MMS allocation
COUNTEREXAMPLE FOR π = 3 17 25 12 1 17 25 12 1 17 25 12 1 17 17 25 25 12 12 1 1 2 22 3 28 2 22 3 28 2 22 3 28 2 22 3 28 2 22 3 28 11 0 21 23 11 0 21 23 11 0 21 11 0 21 23 11 0 21 23 23
COUNTEREXAMPLE FOR π = 3 1 1 1 1 17 25 12 1 Γ 10 6 Γ 10 3 + + 1 1 1 1 2 22 3 28 1 1 1 1 11 0 21 23 3 -1 -1 -1 3 -1 0 0 3 0 -1 0 0 0 0 0 -1 0 0 0 0 0 -1 0 0 0 0 0 -1 0 0 0 0 0 0 -1 Player 1 Player 2 Player 3
APPROXIMATE ENVY-FREENESS β’ Assume general monotonic valuations, i.e., for all π β π β π», π π π β€ π π (π) β’ An allocation π΅ 1 , β¦ , π΅ π is envy free up to one good (EF1) if and only if βπ, π β π, βπ β π΅ π s.t. π€ π π΅ π β₯ π€ π π΅ π \{π} β’ Theorem [Lipton et al. 2004]: An EF1 allocation exists and can be found in polynomial time
PROOF OF THEOREM β’ A partial allocation is an allocation of a subset of the goods β’ Given a partial allocation π© , we have an edge (π, π) in its envy graph if π envies π β’ Lemma: An EF1 partial allocation π© can be transformed in polynomial time into an EF1 partial allocation πͺ of the same goods with an acyclic envy graph
PROOF OF LEMMA β’ If π» has a cycle π· , shift allocations along π· to obtain π 4 π 1 π©β² ; clearly EF1 is maintained β’ #edges in envy graph of π©β² decreased: π 3 π 2 β¦ Same edges between π β π· β¦ Edges from π β π· to π· shifted π 4 π 2 β¦ Edges from π· to π β π· can only decrease β¦ Edges inside C decreased π 1 π 3 β’ Iteratively remove cycles β
PROOF OF THEOREM β’ Maintain EF1 and acyclic envy graph β’ In round 1, allocate good π 1 to arbitrary agent β’ π 1 , β¦ , π πβ1 are allocated in acyclic π© β’ Derive πͺ by allocating π π to source π β’ π π πΆ π = π π π΅ π β₯ π π π΅ π = π π πΆ π β π π β’ Use lemma to eliminate cycles β
ROUND ROBIN β’ Let us return to additive valuations β’ Now proving the existence of an EF1 allocation is trivial β’ A round-robin allocation is EF1: Phase 1 Phase 2
IMPLICATIONS FOR CAKE CUTTING β’ In cake cutting, we can define an allocation to be π -envy free if for all π, π β π, π π π΅ π β₯ π π π΅ π β π β’ The foregoing result has interesting implications for cake cutting! Poll 1 ? Complexity of π -EF in the RW model? 1 π β’ π β’ π π 2 π π 2 1 β’ π β’ π π 2 π
MAXIMUM NASH WELFARE β’ An allocation π© is Pareto efficient if there is no allocation π©β² such that β² β₯ π π π π΅ π π π΅ π for all π β π , and β² > π π π π΅ π π π΅ π for some π β π β’ Round Robin is not efficient β’ Is there a rule that guarantees both EF1 and efficiency?
MAXIMUM NASH WELFARE β’ The Nash welfare of an allocation π© is the product of values NW π© = ΰ· π π (π΅ π ) πβπ β’ The maximum Nash welfare (MNW) solution chooses an allocation that maximizes the Nash welfare β’ For ease of exposition we ignore the case of NW π© = 0 for all π© β’ Theorem [Caragiannis et al. 2016]: Assuming additive valuations, the MNW solution is EF1 and efficient
PROOF OF THEOREM β’ Efficiency is obvious, so we focus on EF1 β’ Assume for contradiction that π envies π by more than one good β’ Let π β β argmin πβπ΅ π ,π π π >0 π π (π)/π π (π) β’ Move π β from π to π to obtain π©β² , we will show that NW π© β² > NW(π©) β² ) for all π β π, π , β’ It holds that π π π΅ π = π π (π΅ π β² = π π π β , and π π π΅ π π π΅ π + π β² = π π π β π π π΅ π π π΅ π β π
PROOF OF THEOREM β’ NW π΅ β² π π π β π π π β NW π΅ > 1 β 1 β 1 + > 1 β π π π΅ π π π π΅ π π π π β π π β π π π΅ π + π < π π π΅ π π π π β β’ Due to our choice of π β , Ο πβπ΅ π π π π β π π π π π = π π π΅ π π π β β€ Ο πβπ΅ π π π π π π΅ π β’ Due to EF1 violation, we have π π β < π π π π΅ π + π π π΅ π β’ Multiply the last two inequalities to get the first β
TRACTABILITY OF MNW 30 25 20 Time (s) 15 10 5 0 5 10 15 20 25 30 35 40 45 50 Number of players [Caragiannis et al., 2016]
INTERFACE
AN OPEN PROBLEM β’ An allocation π΅ 1 , β¦ , π΅ π is envy free up to any good (EFX) if and only if βπ, π β π, βπ β π΅ π , π€ π π΅ π β₯ π€ π π΅ π \{π} β’ Strictly stronger than EF1, strictly weaker than EF β’ An EFX allocation exists for two players with monotonic valuations β’ Existence is an open problem for π β₯ 3 players with additive valuations
Recommend
More recommend