Mathematical Foundations of AI Lecture 9 – Fair Allocation (cake cutting)

Content Proportional and Envy-free cutting Dividing a cake to 2 diners Dividing a cake to n diners Other issues

Fair division

Fair division: definitions The cake is represented by the unit interval [0,1] The preferences of the i’th Player are represented by a function Vi : [0,1]  R+ Vi has the following properties: Additivity: For 2 distinct portions A,B [0,1] Vi(A U B) = Vi(A) + Vi(B) The value of the portion [x,x] (size 0) is 0 Vi([0, 1]) = 1

Divisions A division S={S1,S2,…,SN} is a partition of the cake + allocation to the N agents Each Si is the union of a finite number of intervals The union of all N parts is [0,1] For i≠j, Si,Sj are distinct

Fairness We use two different concepts of “fairness” A division is proportional, if Vi(Si) ≥ 1/n ,i.e. every agent thinks he gets at least 1/n A division is envy-free, if Vi(Si) ≥ Vi(Sj) for all j≠i, i.e. no agent wants to trade portions with anyone else

Fairness (example) The valuation functions: 2/3 1/3 1/3 1/3 1/3 1/3 [0… ⅓] [⅓…⅔] [⅔ …1] 2/3 1/3 1/3 1/3 1/3 1/3 2/3

Fairness (example) Non-proportional division 2/3 1/3 1/3 1/3 1/3 1/3 [0… ⅓] [⅓…⅔] [⅔ …1] 2/3 1/3 1/3 1/3 1/3 1/3 2/3

Fairness (example) Division is proportional, but not envy-free ≥ ≤ 2/3 [0… ⅓] [⅓…⅔] [⅔ …1] ≥ 2/3 1/3 1/3 1/3 1/3 ≤ 1/3 2/3

Fairness (example) Division is proportional, and envy-free 2/3 1/3 1/3 [0… ⅓] [⅓…⅔] [⅔ …1] 2/3 1/3 1/3 1/3 1/3 1/3 2/3

Fairness (example) Division is proportional, and envy-free 7/12 1/4 [0…¼] [¼…½] [½ …1] 7/12 1/4 1/6 1/4 1/4 1/2 1/4 5/12 1/3

Proportional vs. Envy-free Lemma 1: If a division is Envy free, it is also proportional Proof: For every agent i, there is at least one portion j, for which Vi(Sj) ≥ 1/n, Thus Vi(Si) ≥ Vi(Sj) ≥ 1/n Lemma 2: for n=2, a proportional division is also envy-free

Division protocols Example: N=2 The Cut-and-choose protocol: Agent 1 divides the cake to 2 equal parts, i.e. V1(A)=V1(B) Agent 2 selects the larger portion Claim: The protocol guarantees an envy-free division (and also proportional)

Our protocol is perfect! (but only for 2 agents…) Division Protocols (2) Other properties of a protocol: How many cuts do we need? In Cut-and-choose: just one cut  optimal Are portions contiguous? Yes Do we need any other tools/assumptions? No Our protocol is perfect! (but only for 2 agents…)

N=3 Protocol for N = 3 (Steinhaus, 1943) Agent 1 cuts the cake to three even pieces If there are two parts that worth at least ⅓ to agent 2, then: Agents select pieces in the order 3,2,1 Otherwise, agent 2 marks the two small pieces as “bad” 1/3 1/3 1/3 3/12 2/12 7/12

n=3 (cont.) (similarly) If there are two pieces that worth at least ⅓ to agent 3, then Agents select pieces in the order 2,3,1 Otherwise agent 3 marks two small pieces as “bad” (if players 2+3 marked bad pieces) There is one piece marked twice Agent 1 takes that piece Other two pieces are merged to one smaller cake, and agents 2,3 split it using cut-and-choose.

Protocol properties Cut and choose 2 1 V Steinhaus 3 X name N #cuts Proportional Envy-free contiguous Cut and choose 2 1 V Steinhaus 3 X

N=3 + envy free Envy-Free Protocol for N = 3 (Selfridge-Conway, 1960) A pre stage: (a) agent 1 divides the cake to three equal parts (b) agent 2 trims the largest part, so that now the two largest parts are equal We now have two cakes: A and B

N=3 + envy free 1/3 1/3 1/3 2/10 3/10 3/10 5/10 2/10 A B

Stage One (Dividing cake A) NT T Agent 3 picks first, then 2 If 3 picked the trimmed part, 2 picks the other part of same size. Otherwise 2 picks the trimmed part Agent 1 takes the piece that is left The player that got the trimmed piece is denoted by T. The other one (from {2,3}), is denoted by NT

Stage two (dividing cake B) NT Agent NT divides the cake to three equal parts Agents select pieces in this order: T 1 NT Claim: the protocol is envy free

Protocol properties Cut and choose 2 1 V Steinhaus 3 X name N #cuts Proportional Envy-free contiguous Cut and choose 2 1 V Steinhaus 3 X Selfridge-Conway 5

What if N>3 ! ! ! ! Protocol for arbitrary N (Dubins-Spanier, 1961) Each agent stops the knife when it passes 1/N of its value function ! ! V3(S3)=1/5 V5(S5)=1/5 V2(S2) =1/5 V4(S4) =1/5 ! ! Claim: the protocol is proportional

Protocol properties Cut and choose 2 1 V Steinhaus 3 X name N #cuts Proportional Envy-free contiguous other Cut and choose 2 1 V Steinhaus 3 X Selfridge-Conway 5 Dubins-Spanier (Banach-Knaster) N-1 Moving knife

Envy free for N>3 Is there an envy free allocation for general N? Yes Is there a protocol that guarantees such a partition? Yes, but with an unbounded number of cuts ( Brams &Taylor ‘95) Open question: can we do it with a bounded number of cuts?

More… Optimality Dishonesty Allocation of indivisible goods (next class) Other topologies Independent / correlated value functions