CSC373 Fun Asides Fair Division [Image and Illustration Credit: Ariel Procaccia] CSC373 - Nisarg Shah 1
Cake-Cutting • 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 CSC373 - Nisarg Shah 2
Agent Valuations • Each player 𝑗 has a valuation 𝑊 𝑗 that is very much like a probability distribution over [0,1] 𝛽 β • Additive: For 𝑌 ∩ 𝑍 = ∅ , 𝑊 𝑗 𝑌 + 𝑊 𝑗 𝑍 = 𝑊 𝑗 𝑌 ∪ 𝑍 β 𝛽 + 𝛾 • Normalized: 𝑊 0,1 = 1 𝑗 𝛽 • Divisible: ∀𝜇 ∈ [0,1] and 𝑌 , ∃𝑍 ⊆ 𝑌 s.t. 𝑊 𝑗 𝑍 = 𝜇𝑊 𝑗 (𝑌) 𝜇𝛽 CSC373 - Nisarg Shah 3
Fairness Goals • Allocation: disjoint partition 𝐵 = (𝐵 1 , … , 𝐵 𝑜 ) ➢ 𝐵 𝑗 = piece of the cake given to player 𝑗 • Desired fairness properties: ➢ Proportionality (Prop): 𝑗 𝐵 𝑗 ≥ 1 ∀𝑗 ∈ 𝑂: 𝑊 𝑜 ➢ Envy-Freeness (EF): ∀𝑗, 𝑘 ∈ 𝑂: 𝑊 𝑗 𝐵 𝑗 ≥ 𝑊 𝑗 (𝐵 𝑘 ) CSC373 - Nisarg Shah 4
Fairness Goals • Prop: ∀𝑗 ∈ 𝑂: 𝑊 Τ 𝑗 𝐵 𝑗 ≥ 1 𝑜 • EF: ∀𝑗, 𝑘 ∈ 𝑂: 𝑊 𝑗 𝐵 𝑗 ≥ 𝑊 𝑗 𝐵 𝑘 • Question: What is the relation between proportionality and EF? Prop ⇒ EF 1. EF ⇒ Prop 2. Equivalent 3. Incomparable 4. CSC373 - Nisarg Shah 5
C UT - AND -C HOOSE • Algorithm for 𝑜 = 2 players • Player 1 divides the cake into two pieces 𝑌, 𝑍 s.t. Τ 𝑊 1 𝑌 = 𝑊 1 𝑍 = 1 2 • Player 2 chooses the piece she prefers. • This is envy-free and therefore proportional. ➢ Why? CSC373 - Nisarg Shah 6
Input Model • 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 . CSC373 - Nisarg Shah 7
Robertson-Webb Model • We restrict access to valuation 𝑊 𝑗 through two types of queries: ➢ Eval 𝑗 (𝑦, 𝑧) returns 𝛽 = 𝑊 𝑦, 𝑧 𝑗 ➢ Cut 𝑗 (𝑦, 𝛽) returns any 𝑧 such that 𝑊 𝑦, 𝑧 = 𝛽 𝑗 o If 𝑊 𝑦, 1 < 𝛽 , return 1 . 𝑗 𝛽 eval output 𝑦 𝑧 cut output CSC373 - Nisarg Shah 8
Robertson-Webb Model • Two types of queries: ➢ Eval 𝑗 𝑦, 𝑧 = 𝑊 𝑦, 𝑧 𝑗 ➢ Cut 𝑗 𝑦, 𝛽 = 𝑧 s.t. 𝑊 𝑦, 𝑧 = 𝛽 𝑗 • Question: How many queries are needed to find an EF allocation when 𝑜 = 2 ? • Answer: 2 CSC373 - Nisarg Shah 9
D UBINS -S PANIER • 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. CSC373 - Nisarg Shah 10
D UBINS -S PANIER 1/3 1/3 ≥ 1/3 CSC373 - Nisarg Shah 11
D UBINS -S PANIER • 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. CSC373 - Nisarg Shah 12
V ISUAL P ROOF OF P ROPORTIONALITY CSC373 - Nisarg Shah 13
V ISUAL P ROOF OF P ROPORTIONALITY 1 3 Τ CSC373 - Nisarg Shah 14
V ISUAL P ROOF OF P ROPORTIONALITY 1 3 Τ 1 3 Τ CSC373 - Nisarg Shah 15
V ISUAL P ROOF OF P ROPORTIONALITY 1 3 Τ 1 3 Τ ≥ 1 3 Τ CSC373 - Nisarg Shah 16
D UBINS -S PANIER • Question: What is the complexity of the Dubins- Spanier protocol in the Robertson-Webb model? Θ 𝑜 1. Θ 𝑜 log 𝑜 2. Θ 𝑜 2 3. Θ 𝑜 2 log 𝑜 4. CSC373 - Nisarg Shah 17
E VEN -P AZ (R ECURSIVE ) • Input: Interval [𝑦, 𝑧] , number of players 𝑜 ➢ For simplicity, assume 𝑜 = 2 𝑙 for some 𝑙 • If 𝑜 = 1 , give [𝑦, 𝑧] to the single player. • Otherwise, let each player 𝑗 mark 𝑨 𝑗 s.t. = 1 𝑊 𝑦, 𝑨 𝑗 2 𝑊 𝑦, 𝑧 𝑗 𝑗 • 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. CSC373 - Nisarg Shah 18
E VEN -P AZ CSC373 - Nisarg Shah 19
E VEN -P AZ • Theorem: E VEN -P AZ returns a Prop allocation. • Inductive Proof: ➢ Hypothesis: With 𝑜 players, E VEN -P AZ ensures that for Τ each player 𝑗 , 𝑊 𝑗 𝐵 𝑗 ≥ 1 𝑜 ⋅ 𝑊 𝑦, 𝑧 𝑗 o Prop follows because initially 𝑊 𝑦, 𝑧 = 𝑊 0,1 = 1 𝑗 𝑗 ➢ Base case: 𝑜 = 1 is trivial. ➢ Suppose it holds for 𝑜 = 2 𝑙−1 . We prove for 𝑜 = 2 𝑙 . ➢ Take the 2 𝑙−1 left players. 𝑦, 𝑨 ∗ o Every left player 𝑗 has 𝑊 ≥ 1 2 𝑊 Τ 𝑦, 𝑧 𝑗 𝑗 1 1 𝑦, 𝑨 ∗ o If it gets 𝐵 𝑗 , by induction, 𝑊 𝑗 𝐵 𝑗 ≥ 2 𝑙−1 𝑊 ≥ 2 𝑙 𝑊 𝑦, 𝑧 𝑗 𝑗 CSC373 - Nisarg Shah 20
E VEN -P AZ • Theorem: E VEN -P AZ 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! CSC373 - Nisarg Shah 21
Complexity of Proportionality • Theorem [Edmonds and Pruhs, 2006]: Any proportional protocol needs Ω(𝑜 log 𝑜) operations in the Robertson-Webb model. • Thus, the E VEN -P AZ protocol is (asymptotically) provably optimal! CSC373 - Nisarg Shah 22
Envy-Freeness? • “ 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 𝑜 2 lower bound for EF. ➢ Last year, the long-standing major open question of “ bounded EF protocol ” was resolved! ➢ [Aziz and Mackenzie, 2016]: 𝑃(𝑜 𝑜 𝑜𝑜𝑜𝑜 ) protocol! o Yes, it ’ s not a typo! CSC373 - Nisarg Shah 23
Pareto Optimality • Pareto Optimality ➢ We say that 𝐵 is Pareto optimal if for any other allocation 𝐶 , it cannot be that 𝑊 𝑗 𝐶 𝑗 ≥ 𝑊 𝑗 𝐵 𝑗 for all 𝑗 and 𝑊 𝑗 𝐶 𝑗 > 𝑊 𝑗 (𝐵 𝑗 ) 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 ” . CSC373 - Nisarg Shah 24
PO + EF • 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: argmax 𝐵 ς 𝑗∈𝑂 𝑊 𝑗 𝐵 𝑗 ➢ Obviously, this is PO. The fact that it is EF is non-trivial. ➢ This is named after John Nash. o Nash social welfare = product of utilities o Different from utilitarian social welfare = sum of utilities CSC373 - Nisarg Shah 25
Nash-Optimal Allocation 2 3 ൗ 0 1 • Example: 2 3 ➢ Green player has value 1 distributed evenly over 0, Τ ➢ Blue player has value 1 distributed evenly over [0,1] ➢ Without loss of generality (why?) suppose: o Green player gets [0, 𝑦] for 𝑦 ≤ Τ 2 3 2 3 ∪ 2 3 , 1 = [𝑦, 1] o Blue player gets 𝑦, Τ Τ 𝑦 2 3 , blue ’ s utility = 1 − 𝑦 ➢ Green ’ s utility = Τ 3 1 2 ➢ Maximize: 2 𝑦 ⋅ (1 − 𝑦) ⇒ 𝑦 = Τ 1 2 Green has utility 3 ൗ 4 0 1 Allocation Blue has utility 1 2 CSC373 - Nisarg Shah 26
Indivisible Goods • Goods cannot be shared / divided among players ➢ E.g., house, painting, car, jewelry, … • Problem: Envy-free allocations may not exist! CSC373 - Nisarg Shah 27
Indivisible Goods: Setting 8 7 20 5 9 11 12 8 9 10 18 3 Given such a matrix of numbers, assign each good to a player. We assume additive values. So, e.g., 𝑊 , = 8 + 7 = 15 CSC373 - Nisarg Shah 28
Indivisible Goods 8 7 20 5 9 11 12 8 9 10 18 3 CSC373 - Nisarg Shah 29
Indivisible Goods 8 7 20 5 9 11 12 8 9 10 18 3 CSC373 - Nisarg Shah 30
Indivisible Goods 8 7 20 5 9 11 12 8 9 10 18 3 CSC373 - Nisarg Shah 31
Indivisible Goods 8 7 20 5 9 11 12 8 9 10 18 3 CSC373 - Nisarg Shah 32
Indivisible Goods • Envy-freeness up to one good (EF1): ∀𝑗, 𝑘 ∈ 𝑂, ∃ ∈ 𝐵 𝑘 ∶ 𝑊 𝑗 𝐵 𝑗 ≥ 𝑊 𝑗 𝐵 𝑘 \{} ➢ Technically, ∃ ∈ 𝐵 𝑘 only applied if 𝐵 𝑘 ≠ ∅ . ➢ “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? CSC373 - Nisarg Shah 33
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