CSC373 Fun Asides Fair Division [Image and Illustration Credit: - - PowerPoint PPT Presentation

CSC373 Fun Asides Fair Division

[Image and Illustration Credit: Ariel Procaccia]

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Cake-Cutting

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• A heterogeneous, divisible good

➢ Heterogeneous: it may be valued

differently by different individuals

➢ Divisible: we can share/divide

it between individuals

• Represented as [0,1]

➢ Almost without loss of generality

• Set of players 𝑂 = {1, … , 𝑜}
• Piece of cake 𝑌 ⊆ [0,1]

➢ A finite union of disjoint intervals

Agent Valuations

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• Each player 𝑗 has a valuation 𝑊

𝑗 that

is very much like a probability distribution over [0,1]

• Additive: For 𝑌 ∩ 𝑍 = ∅,

𝑊

𝑗 𝑌 + 𝑊 𝑗 𝑍 = 𝑊 𝑗 𝑌 ∪ 𝑍

• Normalized: 𝑊

𝑗

0,1 = 1

• Divisible: ∀𝜇 ∈ [0,1] and 𝑌,

∃𝑍 ⊆ 𝑌 s.t. 𝑊

𝑗 𝑍 = 𝜇𝑊 𝑗(𝑌)

𝛽 𝜇𝛽 𝛽 β β

𝛽 + 𝛾

Fairness Goals

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• Allocation: disjoint partition 𝐵 = (𝐵1, … , 𝐵𝑜)

➢ 𝐵𝑗 = piece of the cake given to player 𝑗

• Desired fairness properties:

➢ Proportionality (Prop):

∀𝑗 ∈ 𝑂: 𝑊

𝑗 𝐵𝑗 ≥ 1

𝑜

➢ Envy-Freeness (EF):

∀𝑗, 𝑘 ∈ 𝑂: 𝑊

𝑗 𝐵𝑗 ≥ 𝑊 𝑗(𝐵𝑘)

Fairness Goals

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• Prop: ∀𝑗 ∈ 𝑂: 𝑊

𝑗 𝐵𝑗 ≥

Τ 1 𝑜

• EF: ∀𝑗, 𝑘 ∈ 𝑂: 𝑊

𝑗 𝐵𝑗 ≥ 𝑊 𝑗 𝐵𝑘

• Question: What is the relation between

proportionality and EF?

1.

Prop ⇒ EF

2.

EF ⇒ Prop

3.

Equivalent

4.

Incomparable

CUT-AND-CHOOSE

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• Algorithm for 𝑜 = 2 players
• Player 1 divides the cake into two pieces 𝑌, 𝑍 s.t.

𝑊

1 𝑌 = 𝑊 1 𝑍 =

Τ 1 2

• Player 2 chooses the piece she prefers.
• This is envy-free and therefore proportional.

➢ Why?

Input Model

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• How do we measure the “time complexity” of a

cake-cutting algorithm for 𝑜 players?

• Typically, time complexity is a function of the

length of input encoded as binary.

• Our input consists of functions 𝑊

𝑗, which require

infinite bits to encode.

• We want running time as a function of 𝑜.

Robertson-Webb Model

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• We restrict access to valuation 𝑊

𝑗 through two

types of queries:

➢ Eval𝑗(𝑦, 𝑧) returns 𝛽 = 𝑊

𝑗

𝑦, 𝑧

➢ Cut𝑗(𝑦, 𝛽) returns any 𝑧 such that 𝑊

𝑗

𝑦, 𝑧 = 𝛽

• If 𝑊

𝑗

𝑦, 1 < 𝛽, return 1.

𝑦 𝑧

𝛽

eval output cut output

Robertson-Webb Model

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• Two types of queries:

➢ Eval𝑗 𝑦, 𝑧 = 𝑊

𝑗

𝑦, 𝑧

➢ Cut𝑗 𝑦, 𝛽 = 𝑧 s.t. 𝑊

𝑗

𝑦, 𝑧 = 𝛽

• Question: How many queries are needed to find an

EF allocation when 𝑜 = 2?

• Answer: 2

DUBINS-SPANIER

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• Protocol for finding a proportional allocation for 𝑜

players

• Referee starts at 0, and moves a knife to the right.
• Repeat: When the piece to the left of the knife is

worth 1/𝑜 to some player, the player shouts “stop”, gets that piece, and exits.

• The last player gets the remaining piece.

DUBINS-SPANIER

11

1/3 1/3 ≥ 1/3

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DUBINS-SPANIER

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• Robertson-Webb model? Cut-Eval queries?

➢ Moving knife is not really needed.

• At each stage, we want to find the remaining player

that has value 1/𝑜 from the smallest next piece.

➢ Ask each remaining player a cut query to mark a point

where her value is 1/𝑜 from the current point.

➢ Directly move the knife to the leftmost mark, and give

that piece to that player.

VISUAL PROOF OF PROPORTIONALITY

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VISUAL PROOF OF PROPORTIONALITY

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Τ 1 3

VISUAL PROOF OF PROPORTIONALITY

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Τ 1 3 Τ 1 3

VISUAL PROOF OF PROPORTIONALITY

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Τ 1 3 Τ 1 3 ≥ Τ 1 3

DUBINS-SPANIER

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• Question: What is the complexity of the Dubins-

Spanier protocol in the Robertson-Webb model?

1.

Θ 𝑜

2.

Θ 𝑜 log 𝑜

3.

Θ 𝑜2

4.

Θ 𝑜2 log 𝑜

EVEN-PAZ (RECURSIVE)

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• Input: Interval [𝑦, 𝑧], number of players 𝑜

➢ For simplicity, assume 𝑜 = 2𝑙 for some 𝑙

• If 𝑜 = 1, give [𝑦, 𝑧] to the single player.
• Otherwise, let each player 𝑗 mark 𝑨𝑗 s.t.

𝑊

𝑗

𝑦, 𝑨𝑗 = 1 2 𝑊

𝑗

𝑦, 𝑧

• Let 𝑨∗ be mark 𝑜/2 from the left.
• Recurse on [𝑦, 𝑨∗] with the left 𝑜/2 players, and on [𝑨∗, 𝑧]

with the right 𝑜/2 players.

EVEN-PAZ

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EVEN-PAZ

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• Theorem: EVEN-PAZ returns a Prop allocation.
• Inductive Proof:

➢ Hypothesis: With 𝑜 players, EVEN-PAZ ensures that for

each player 𝑗, 𝑊

𝑗 𝐵𝑗 ≥

Τ 1 𝑜 ⋅ 𝑊

𝑗

𝑦, 𝑧

• Prop follows because initially 𝑊

𝑗

𝑦, 𝑧 = 𝑊

𝑗

0,1 = 1

➢ Base case: 𝑜 = 1 is trivial. ➢ Suppose it holds for 𝑜 = 2𝑙−1. We prove for 𝑜 = 2𝑙. ➢ Take the 2𝑙−1 left players.

• Every left player 𝑗 has 𝑊

𝑗

𝑦, 𝑨∗ ≥ Τ 1 2 𝑊

𝑗

𝑦, 𝑧

• If it gets 𝐵𝑗, by induction, 𝑊

𝑗 𝐵𝑗 ≥ 1 2𝑙−1 𝑊 𝑗

𝑦, 𝑨∗ ≥

1 2𝑙 𝑊 𝑗

𝑦, 𝑧

EVEN-PAZ

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• Theorem: EVEN-PAZ uses 𝑃 𝑜 log 𝑜 queries.
• Simple Proof:

➢ Protocol runs for log 𝑜 rounds. ➢ In each round, each player is asked one cut query. ➢ QED!

Complexity of Proportionality

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• Theorem [Edmonds and Pruhs, 2006]: Any

proportional protocol needs Ω(𝑜 log 𝑜) operations in the Robertson-Webb model.

• Thus, the EVEN-PAZ protocol is (asymptotically)

provably optimal!

Envy-Freeness?

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• “I suppose you are also going to give such cute

algorithms for finding envy-free allocations?”

• Bad luck. For 𝑜-player EF cake-cutting:

➢ [Brams and Taylor, 1995] give an unbounded EF protocol. ➢ [Procaccia 2009] shows Ω 𝑜2 lower bound for EF. ➢ Last year, the long-standing major open question of

“bounded EF protocol” was resolved!

➢ [Aziz and Mackenzie, 2016]: 𝑃(𝑜𝑜𝑜𝑜𝑜𝑜

) protocol!

• Yes, it’s not a typo!

Pareto Optimality

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• Pareto Optimality

➢ We say that 𝐵 is Pareto optimal if for any other allocation

𝐶, it cannot be that 𝑊

𝑗 𝐶𝑗 ≥ 𝑊 𝑗 𝐵𝑗 for all 𝑗 and 𝑊 𝑗 𝐶𝑗

> 𝑊

𝑗(𝐵𝑗) for some 𝑗.

• Q: Is it PO to give the entire cake to player 1?
• A: Not necessarily. But yes if player 1 values “every

part of the cake positively”.

PO + EF

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• Theorem [Weller ‘85]:

➢ There always exists an allocation of the cake that is both

envy-free and Pareto optimal.

• One way to achieve PO+EF:

➢ Nash-optimal allocation: argmax𝐵 ς𝑗∈𝑂 𝑊

𝑗 𝐵𝑗

➢ Obviously, this is PO. The fact that it is EF is non-trivial. ➢ This is named after John Nash.

• Nash social welfare = product of utilities
• Different from utilitarian social welfare = sum of utilities

Nash-Optimal Allocation

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• Example:

➢ Green player has value 1 distributed evenly over 0, Τ

2 3

➢ Blue player has value 1 distributed evenly over [0,1] ➢ Without loss of generality (why?) suppose:

• Green player gets [0, 𝑦] for 𝑦 ≤ Τ

2 3

• Blue player gets 𝑦, Τ

2 3 ∪

Τ

2 3 , 1 = [𝑦, 1]

➢ Green’s utility =

𝑦 Τ

2 3, blue’s utility = 1 − 𝑦

➢ Maximize: 3 2 𝑦 ⋅ (1 − 𝑦) ⇒ 𝑦 = Τ 1 2

1

ൗ 2 3

Allocation 1

ൗ 1 2

Green has utility 3

4

Blue has utility 1

2

Indivisible Goods

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• Goods cannot be shared / divided among players

➢ E.g., house, painting, car, jewelry, …

• Problem: Envy-free allocations may not exist!

Indivisible Goods: Setting

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8 7 20 5 9 11 12 8 9 10 18 3

We assume additive values. So, e.g., 𝑊 , = 8 + 7 = 15 Given such a matrix of numbers, assign each good to a player.

8 7 20 5 9 11 12 8 9 10 18 3

Indivisible Goods

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8 7 20 5 9 11 12 8 9 10 18 3

Indivisible Goods

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8 7 20 5 9 11 12 8 9 10 18 3

Indivisible Goods

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8 7 20 5 9 11 12 8 9 10 18 3

Indivisible Goods

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Indivisible Goods

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• Envy-freeness up to one good (EF1):

∀𝑗, 𝑘 ∈ 𝑂, ∃𝑕 ∈ 𝐵𝑘 ∶ 𝑊

𝑗 𝐵𝑗 ≥ 𝑊 𝑗 𝐵𝑘\{𝑕}

➢ Technically, ∃𝑕 ∈ 𝐵𝑘 only applied if 𝐵𝑘 ≠ ∅. ➢ “If 𝑗 envies 𝑘, there must be some good in 𝑘’s bundle such

that removing it would make 𝑗 envy-free of 𝑘.”

• Does there always exist an EF1 allocation?

EF1

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• Yes! We can use Round Robin.

➢ Agents take turns in a cyclic order, say

1,2, … , 𝑜, 1,2, … , 𝑜, …

➢ An agent, in her turn, picks the good that she likes the

most among the goods still not picked by anyone.

➢ [Assignment Problem] This yields an EF1 allocation

regardless of how you order the agents.

• Sadly, the allocation returned may not be Pareto
• ptimal.

EF1+PO?

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• Nash welfare to the rescue!
• Theorem [Caragiannis et al. ‘16]:

➢ Maximizing Nash welfare achieves both EF1 and PO. ➢ But what if there are two goods and three players?

• All allocations have zero Nash welfare (product of utilities).
• But we cannot give both goods to a single player.

➢ Algorithm in detail:

• Step 1: Choose a subset of players 𝑇 ⊆ 𝑂 with the largest |𝑇| such

that it is possible to give every player in 𝑇 positive utility simultaneously.

• Step 2: Choose argmax𝐵 ς𝑗∈𝑇 𝑊

𝑗 𝐵𝑗

8 7 20 5 9 11 12 8 9 10 18 3

Integral Nash Allocation

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8 7 20 5 9 11 12 8 9 10 18 3

20 * 8 * (9+10) = 3040

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8 7 20 5 9 11 12 8 9 10 18 3

(8+7) * 8 * 18 = 2160

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8 7 20 5 9 11 12 8 9 10 18 3

8 * (12+8) * 10 = 1600

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8 7 20 5 9 11 12 8 9 10 18 3

20 * (11+8) * 9 = 3420

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Computation

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• For indivisible goods, Nash-optimal solution is

strongly NP-hard to compute

➢ That is, remains NP-hard even if all values are bounded.

• Open Question: Can we find an allocation that is

both EF1 and PO in polynomial time?

➢ A recent paper provides a pseudo-polynomial time

algorithm, i.e., its time is polynomial in 𝑜, 𝑛, and max

𝑗,𝑕 𝑊 𝑗

𝑕 .

Stronger Fairness Guarantees

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• Envy-freeness up to the least valued good (EFx):

➢ ∀𝑗, 𝑘 ∈ 𝑂, ∀𝑕 ∈ 𝐵𝑘 ∶ 𝑊

𝑗 𝐵𝑗 ≥ 𝑊 𝑗 𝐵𝑘\{𝑕}

➢ “If 𝑗 envies 𝑘, then removing any good from 𝑘’s bundle

eliminates the envy.”

➢ Open question: Is there always an EFx allocation?

• Contrast this with EF1:

➢ ∀𝑗, 𝑘 ∈ 𝑂, ∃𝑕 ∈ 𝐵𝑘 ∶ 𝑊

𝑗 𝐵𝑗 ≥ 𝑊 𝑗 𝐵𝑘\{𝑕}

➢ “If 𝑗 envies 𝑘, then removing some good from 𝑘’s bundle

eliminates the envy.”

➢ We know there is always an EF1 allocation that is also PO.

Stronger Fairness

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• Difference between EF1 and EFx:

➢ Suppose there are two players ➢ They are dividing one diamond and two rocks ➢ Giving a diamond and a rock to P1 and only a rock to P2

satisfies EF1, but seems unfair

➢ The only way to get EFx is to give diamond to one player

and both rocks to the other

Diamond Rock 1 Rock 2 P1 100 1 1 P2 100 1 1