The Communication Complexity of Fair Division Simina Brânzei and Noam Nisan Hebrew University Noam Nisan
Cake Cutting n Metaphor for fair division: q Cake = [0,1] q Each player i has a non-atomic probability measure v i on [0,1] q Challenge: partition the cake to the n players so that each feels that he has been treated “fairly” . q Many applications, from land to divorces to cloud computing… q Long research history (Steinhaus 1948 … Aziz Mackenzie 17) Noam Nisan
Some Fairness Criteria n Proportional: v i (S i ) ≥ 1/n . n Envy Free: v i (S i ) ≥ v i (S j ) . n Equitable: v i (S i ) = v j (S j ) . n Perfect: v i (S j ) = 1/n . Theorem: For any tuple of atomless player valuations, each of these exists. Proof: Usually some fixed-point theorem Noam Nisan
Example: Perfect, n=2 players Austin v 1 ([x,f(x)])=1/2 x f(x) Noam Nisan
Example: Perfect, n=2 players v 1 ([x,f(x)])=1/2 x f(x) v 2 ([x,f(x)]) >? 1/2 Noam Nisan
Example: Perfect, n=2 players v 1 ([x*,f(x*)])=1/2 x* f(x*) v 2 ([x*,f(x*)]) = 1/2 Noam Nisan
Models for Cake Cutting Algorithms n Atom-less Input valuations n Infinite-precision output n Infinite-precision Operations/Queries: q Robertson-Webb model: n Value Query: v i ([a,b]) ? n Cut Query: Find x such that v i ([0,x])=c. q Moving Knife Procedures: n Various examples where players continuously move knives as a function of their valuation n No single accepted formal model Noam Nisan
Discrete Models For Cake Cutting n Discrete Queries n Output to satisfy fairness to within e : q Proportional: v i (S i ) ≥ 1/n - e . q Envy Free: v i (S i ) ≥ v i (S j ) - e . q Equitable: | v i (S i ) - v j (S j )| ≤ e . q Perfect: | v i (S j ) - 1/n| ≤ e . n Input valuations have bounded density q Otherwise nothing is possible n Complexity is a function of e (and n ) Noam Nisan
Communication Complexity Which discrete query model should we choose? q v i ([a,b]) > c ? q Any query that depends on a single valuation But “charge” for the length of the answer in bits n Communication Complexity model (Yao 79) q Each player “knows” his valuation q They exchange messages according to a fixed protocol q Equivalent to “any query on a single valuation” model n A message from i corresponds to a query on v i n Number of bits corresponds to answer length Noam Nisan
Simulating infinite-precision queries Proposition: For any notion of fairness that can be found using O(1) Robertson-Webb queries, an e -fair cutting can be found with O(1) rounds of communication of O(log e -1 ) bits. Proof: “Round” the valuations (carefully). Comment: Simulation only works for “notions of fairness” Noam Nisan
Simulating Moving Knife Procedures Definition : Moving Knife Procedure…. (proofs that use intermediate value theorem…) Theorem: A moving knife procedure can be simulated with O(log e -1 ) rounds of communication each using O(log e -1 ) bits. Noam Nisan
Proof of Theorem (by example) v 1 ([x,f(x)])=1/2 x f(x) v 2 ([x,f(x)]) >? 1/2 Binary search over knife positions. Requires O(log e -1 ) “queries” each requiring O(log e -1 ) bits of communication. Noam Nisan
Deduced CC of popular fairness notions Easy (RW protocols): Q (log e -1 ) q Contiguous Proportional (any fixed n) Even-Paz q Envy Free (any fixed n) Aziz-Mackenzie Medium (Moving Knife Protocol): O(log 2 e -1 ) q Contiguous Envy Free ( n=3 ) Stromquist q Contiguous Equitable (any fixed n ) q Perfect, 2 cuts (n=2) Austin Hard : O( e -1 log e -1 ) q Everything else Noam Nisan
New Upper bounds Easy (RW protocols): Q (log e -1 ) q Contiguous Proportional (any fixed n) q Envy Free (any fixed n) Medium (Moving Knife Protocol): q Contiguous Envy Free ( n=3 ) Q (log e -1 ) q Contiguous Equitable ( n=2 ) O(log e -1 loglog e -1 ) q Perfect, 2 cuts (n=2) randomized Hard : O( e -1 log e -1 ) q Everything else Noam Nisan
New Lower Bounds Easy (RW protocols): Q (log e -1 ) (with O(1) rounds) q Contiguous Proportional (any fixed n) q Envy Free (any fixed n) Medium (Moving Knife Protocol): q Contiguous Envy Free ( n=3 ) q Contiguous Equitable ( n=2 ) Require many rounds q Perfect, 2 cuts (n=2) Hard : O( e -1 log e -1 ) Main Open Problem q Everything else Noam Nisan
The Crossing Problem n Alice gets x 0 , x 1 , …, x m Î {0, 1, … m} n Bob gets y 0 , y 1 , …, y m Î {0, 1, … m} n x 0 ≤ y 0 ; x m ≥ y m n Find i so that x i ≤ y i ; x i+1 ≥ y i+1 Noam Nisan
The Crossing Problem n Alice gets x 0 , x 1 , …, x m Î {0, 1, … m} n Bob gets y 0 , y 1 , …, y m Î {0, 1, … m} n x 0 ≤ y 0 ; x m ≥ y m n Find i so that x i ≤ y i ; x i+1 ≥ y i+1 Monotone Variant: q 0 = x 0 ≤ x 1 ≤ … ≤ x m = m q m = y 0 ≥ y 1 ≥ … ≥ y m = 0 Noam Nisan
Monotone Crossing ≈ Equitable(n=2) v 1 ([0,x*]) = v 2 ([x*,1]) x i =v 1 ([0,i/m]) y i =v 2 ([i/m,1]) Noam Nisan
Crossing ≈ Perfect(n=2, 2 cuts) 1/2 i/m x i x i chosen so that v 1 ([i/m,x i ]) ≈ 1/2 y i chosen so that v 2 ([i/m,y i ]) ≈ 1/2 Noam Nisan
Monotone Crossing problem Lemma: The Monotone Crossing problem has an O(log m) bit communication protocol. Corollary: Contiguous equitable cutting for n=2 can be done with O(log e -1 ) bits. Contiguous Envy Free cutting for n=3 can be done with O(log e -1 ) bits. Proof: Consider crossing with m numbers in range 0…k . n Query x m/2 >k/2? And y m/2 >k/2? n Answers will allow to either cut m by factor of 2 or cut k by factor of 2, giving a total of O(log m + log k) queries. Noam Nisan
General Crossing Problem Lemma: The Crossing problem has a x 0 ≤ y 0 randomized communication protocol that x 1 y 1 uses O(log m loglog m) bits. . . . . Corollary: Perfect cutting with 2-cuts for n=2 x i y i ? can be done with O(log e -1 loglog e -1 ) bits . . (randomized). . . . . Proof: Binary search using “ x i <y i ?” queries. x m y m ≥ Randomized CC of each “x<y?” is only loglogm (Nisan-Safra 1993) . Noam Nisan
Communication Rounds Recall: CC rounds correspond to queries and number of bits corresponds to length of answers. n General simulation of Robertson-Webb protocols gives r=O(1) rounds and t=O(log e -1 ) bits for the “easy problems”. n General simulation of moving-knife protocols, as well as our new upper bounds require r=O(log e -1 ) ) rounds for the “medium problems”. Theorem: The monotone crossing problem requires r= W (log m/log t) rounds of most t bits each. Corollary: For n=2 , Perfect and Equitable division require r= W (log e -1 /loglog e -1 ) rounds of t=polylog e -1 bits Noam Nisan
Round-elimination proofs Main Lemma: r -round protocol for input size m è (r-1)- round protocol for input size W ( m/t). Proof Structure: Step 1: “Encode” k instances of problem of size m as one instance of problem of size km . Step 2: If t<<k then first round of t bits of communication cannot “help” all k instances. Noam Nisan
Round elimination lemma (Miltersen et al 1998, Sen-Venkatesh 2008) n Let f(x,y) be an arbitrary communication problem. n f k is defined as follows: q Alice gets x 1 …x k q Bob gets z Î {1…k} and y q Need to solve f(x z ,y) Lemma: A randomized communication protocol for f k where Alice sends first round of t<<k bits of communication implies a similar protocol for f with the first round removed . Noam Nisan
Encoding multiple instances Notation: MC(m) – the monotone crossing problem on inputs of size m. q 0 = x 0 ≤ x 1 ≤ … ≤ x m = m q m = y 0 ≥ y 1 ≥ … ≥ y m = 0 q Goal: Find i so that x i ≤ y i ; x i+1 ≥ y i+1 Noam Nisan
Encoding multiple instances Notation: MC(m) – the monotone crossing problem on inputs of size m. q 0 = x 0 ≤ x 1 ≤ … ≤ x m = m q m = y 0 ≥ y 1 ≥ … ≥ y m = 0 q Goal: Find i so that x i ≤ y i ; x i+1 ≥ y i+1 Lemma: Protocol for MC(mk) è similar protocol for (MC(m)) k Noam Nisan
Encoding multiple instances Notation: MC(m) – the monotone crossing problem on inputs of size m. q 0 = x 0 ≤ x 1 ≤ … ≤ x m = m q m = y 0 ≥ y 1 ≥ … ≥ y m = 0 q Goal: Find i so that x i ≤ y i ; x i+1 ≥ y i+1 Lemma: Protocol for MC(mk) è similar protocol for (MC(m)) k Proof: … … x 1 x z x k Alice: y z Bob: Noam Nisan
Summary n Study communication requirements of fair division notions n Main open problem: Prove a super-logarithmic CC lower bound for some fairness notion n Other open problems: q Determine the exact randomized/deterministic CC of the general crossing problem. q Lower bound for contiguous envy-free cutting for n=3? Noam Nisan
Thanks Noam Nisan
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