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TRUTH, JUSTICE, AND CAKE CUTTING Ariel Procaccia (Harvard SEAS) 1.

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Presentation on theme: "TRUTH, JUSTICE, AND CAKE CUTTING Ariel Procaccia (Harvard SEAS) 1."— Presentation transcript:

1 TRUTH, JUSTICE, AND CAKE CUTTING Ariel Procaccia (Harvard SEAS) 1

2 Standing on the shoulders of giants Superman: “I’m here to fight for truth, justice, and the American Way.” Lois Lane: “You’re gonna wind up fighting every elected official in this country!” Superman (1978) 2

3 Truth, justice, and cake cutting  Division of a heterogeneous divisible good  The cake is the interval [0,1]  Set of agents N={1,...,n}  Piece of cake X  [0,1] = finite union of disjoint intervals  Each agent has a valuation function V i over pieces of cake  Integral over a value density function v i   i  N, V i (0,1) = 1  Find an allocation A 1,...,A n 3

4 Truth, justice, and cake cutting  Proportionality:  i  N, V i (A i )  1/n  Envy-freeness:  i,j  N, V i (A i )  V i (A j )  Assuming free disposal the two properties are incomparable  Envy-free but not proportional: throw away cake  Proportional but not envy-free 1/3 1/2 1/6 1 1 1 1 4

5 Some childhood nostalgia  Assume that n=2  The cut and choose algorithm [Procaccia&Procaccia, circa 1987?]:  Player 1 cuts the cake into two pieces X 1,X 2 s.t. V 1 (X 1 )=V 1 (X 2 ) = ½  Player 2 chooses the piece that he prefers  Player 1 gets the other piece  Not a bad algorithm!  Envy-free  proportional  (Contiguous pieces  one cut) 1/2 1/3 2/3 5

6 Cake cutting is not a piece of cake  Very cool envy-free algorithm for n=3 [Selfridge&Conway, circa 1960]  Envy-free algorithm for n  4 [Brams&Taylor, 1995]  May require an unbounded number of steps!  Recent lower bounds in a concrete complexity model  Envy-free unbounded assuming contiguous pieces [Stromquist, 2008]   (n 2 ) lower bound for envy-free cake cutting [Procaccia, 2009] 6

7 Truth, justice, and cake cutting  Previous work considered strategyproof cake cutting [Brams, Jones & Klamler 2006, 2008]  Their notion: agents report the truth if there exist valuations for others s.t. agent does not gain by lying  Prior-free!  Truthful algorithm = truthfulness is a dominant strategy  Cut and choose is “strategyproof” but not truthful 7

8 An inconvenient truth  Goal: design truthful, fair (envy-free and proportional), and tractable cake cutting algorithms  Requires restricting the valuation functions  Valuation V i is piecewise constant if its value density function v i is piecewise constant  Valuation is piecewise uniform if moreover v i is some uniform constant or zero  Agent is uniformly interested in piece of cake U i  Representation: boundaries of these intervals  A natural (?) restriction and also proof of concept 8

9 Restricted valuations illustrated Piecewise constant valuation that is not piecewise uniform Piecewise uniform valuation 00.51 0 1 2 0 1 0 1 2 V i ([0,0.1]  [0.5,0.7]) = 0.4 9

10 The case of two agents: take 1  We first assume n=2 (and piecewise uniform valuations)  A simple algorithm:  For each agent, make a mark at the beginning and end of each of the agent’s desired intervals  For each subinterval between consecutive marks, allocate left half to agent 1and right half to agent 2  Each agent gets value ½  envy-free and proportional ... but not truthful  If U 1 = [0,0.5] and U 2 = [0,1] then A 1 = [0,0.25]  [0.5,0.75] and A 2 = [0.25,0.5]  [0.75,1]  Agent 1 can gain by reporting [0,1]  A 1 = [0,0.5] 10

11 The case of two agents: take 2 Initialization phase: 1. Discard [0,1]\U 1  U 2 2. Make a mark at the beginning and end of each desired interval 3. Allocate half of each subinterval between consecutive marks to agent 1 and half to agent 2  Denote:  len(X) = the total length of intervals in X  X 1 = U 1 \U 2, X 2 = U 2 \U 1, X 12 = U 1  U 2  Assume len(U 1 )  len(U 2 )  Another simple algorithm (that works) 11

12 The case of two agents: take 2 Swapping phase: 1. Swap pieces Y,Z of equal length where agent 1 owns Y, agent 2 owns Z, Y  X 2, Z  X 1 2. Swap pieces Y,Z of equal length where agent 1 owns Y, agent 2 owns Z, Y  X 2, Z  X 12 3. If there are still pieces of X 2 owned by agent 1, give them to agent 2 U1U1 U1U1 U2U2 X1X1 X2X2 X 12 X2X2 Initialization phase: 1. Discard [0,1]\U 1  U 2 2. Make a mark at the beginning and end of each desired interval 3. Allocate half of each subinterval between consecutive marks to agent 1 and half to agent 2 12

13 Properties of the algorithm (n=2)  Envy-free and proportional: obvious  There are two cases (given len(U 1 )  len(U 2 )):  len(U 1 )  len(U 1  U 2 )/2: the agents receive a desired piece of length len(U 1  U 2 )/2 (an exact allocation)  len(U 1 )  len(U 1  U 2 )/2: agent 1 gets U 1 and agent 2 gets X 2 13 Swapping phase: 1. Swap pieces Y,Z of equal length where agent 1 owns Y, agent 2 owns Z, Y  X 2, Z  X 1 2. Swap pieces Y,Z of equal length where agent 1 owns Y, agent 2 owns Z, Y  X 2, Z  X 12 3. If there are still pieces of X 2 owned by agent 1, give them to agent 2

14 The algorithm is truthful (n=2)  Assume agent 1 misreports U’ 1  we have X’ 1, X’ 2, X’ 12  Can assume len(U 1 )  len(U 1  U 2 )/2  Originally got len(U 1  U 2 )/2 = (len(X 1 )+len(U 2 ))/2  Now gets  len(U’ 1  U 2 )/2 = (len(X’ 1 )+len(U 2 ))/2  len(X’ 1 ) = len(X 1 )  k  increases length of piece by k/2 but length of k is useless  Crucial: Agent 1 first trades for X 1  len(X’ 1 ) = len(X 1 )  k  decreases length of A 1 by k/2, before all of A 1 was desired 14

15 The general algorithm: setup 15  Let S  N, X is a piece of cake  D(S,X) = (  i  S U i )  X = portions of X desired by at least one agent in S  avg(S,X) = len(D(S,X))/|S|  A 1,...,A n is exact wrt S,X if  i  S, len(A i )=avg(S,X) and A i is desired by agent i  For example, S={1,2} and X=[0,1]  U 1 =U 2 =[0,0.2]  A 1 =[0,0.1], A 2 =[0.1,0.2] is exact  U 1 =[0,0.2], U 2 =[0.3,0.7]  no exact allocation

16 The general algorithm U1U1 Initialization: 1. S  N, X  [0,1] While S  1. S min  argmin avg(S’,X) 2. Let E 1,...,E n be an exact allocation wrt S min,X 3.  i  S min, A i  E i 4. S  S\S min 5. X  X\D(S min,X) 16 S’  S U3U3 U2U2 0.39 00 0.1 0.6

17 The case of two agents revisited  Assume len(U 1 )  len(U 2 )  S min is either {1} or {1,2}  len(U 1 )  len(U 1  U 2 )/2: S min is {1,2}, give exact allocation wrt {1,2},[0,1]  len(U 1 ) < len(U 1  U 2 )/2: S min is {1}, give 1 exact allocation wrt {1},[0,1] (U 1 ), the rest to 2 in next iteration 17 Initialization: 1. S  N, X  [0,1] While S  1. S min  argmin avg(S,’X) 2. Let E 1,...,E n be an exact allocation w.r.t. S min,X 3.  i  S min, A i  E i 4. S  S\S min 5. X  X\D(S min,X) S’  S

18 Exact allocations and network flow  There are two problematic steps in while loop:  Step 1: computing S min ?  Step 2: existence and computation of exact allocation?  Solution: use network flow 18 Initialization: 1. S  N, X  [0,1] While S  1. S min  argmin avg(S,’X) 2. Let E 1,...,E n be an exact allocation w.r.t. S min,X 3.  i  S min, A i  E i 4. S  S\S min 5. X  X\D(S min,X) S’  S

19 Let it flow 19  Define a graph G(S,X)  Mark beginning and end of every interval in U i  X  Nodes: consecutive markings, agents, s and t  For each I, edge (s,I) with capacity len(I)  Each i  N connected to t with capacity avg(S,X)  Edge (I,i) with capacity  if agent i desires interval I s s t t 1 1 2 2 0.25,0.4 0.5,1 0,0.1 0.1,0.25 0.4,0.5 U 1 = [0,0.25]  [0.5,1], U 2 = [0.1,0.4] 0.5 0.15 0.1 0.15 0.45     

20 A lemma 20  Lemma: Let S  N, a piece of cake X. If for all S’  S, avg(S’,X)  avg(S,X) then there is a network flow of size len(D(S,X)) in G(S,X)  Proof:  Max Flow = Min Cut  Disconnect subset T  S from t at cost |T|  avg(S,X)  Need to additionally disconnect len(D(S\T,X)) =|S\T|  avg(S\T,X)  |S\T|  avg(S,X) s s t t 1 1 2 2 0.25,0.4 0.5,1 0,0.1 0.1,0.25 0.4,0.5 0.5 0.15 0.1 0.15 0.45     

21 Properties of the algorithm 21  Lemma: Let S  N, a piece of cake X. If there exists a network flow of size len(D(S,X)) in G(S,X) then there is an exact allocation wrt S,X  If S min minimizes avg(S’,X) then there is an exact flow wrt S min,X, can be computed using network flow algorithms  Computing S min is similar but more involved  Theorem: assume that the agents have piecewise uniform valuations, then the algorithm is truthful, proportional, envy-free, and polynomial-time

22 Randomized algorithms 22  A randomized alg is universally envy-free (resp., universally proportional) if it always returns an envy- free (resp., proportional) allocation  A randomized alg is truthful in expectation if an agent cannot gain in expectation by lying  Looking for universal fairness and truthfulness in expectation  Does it make sense to look for fairness in expectation and universal truthfulness?  Theorem: assume that the agents have piecewise linear valuations, then there is a randomized alg that is truthful in expectation, universally proportional, universally envy-free, and polynomial-time

23 Discussion 23  Conceptual contributions  Truthful cake cutting  Restricted valuations functions and tractable algorithms  Communication model  Many previous discrete algorithms can be simulated using eval and cut queries  Our algorithms are centralized  Future work  Generalize deterministic algorithm  Piecewise uniform valuations with minimum interval length

24 Bibliographic notes 24  Yiling Chen, John K. Lai, David C. Parkes and Ariel D. Procaccia. Truth, Justice, and Cake Cutting. In the proceedings of AAAI 2010  Full version coming soon, will be posted online (rought draft available on request)

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26 Properties of the algorithm  There are two problematic steps in while loop:  Step 1: computing S min ?  Step 2: existence and computation of exact allocation?  Solution: use network flow / max flow min cut  Theorem: assume that the agents have piecewise uniform valuations, then the algorithm is truthful, proportional, envy- free, and polynomial-time 26 Initialization: 1. S  N, X  [0,1] While S  1. S min  argmin avg(S,’X) 2. Let E 1,...,E n be an exact allocation w.r.t. S min,X 3.  i  S min, A i  E i 4. S  S\S min 5. X  X\D(S min,X) S’  S


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