Cakes, Pies, and Fair Division

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Cakes, Pies, and Fair Division

• Walter Stromquist
• Swarthmore College
• mail_at_walterstromquist.com
• Rutgers Experimental Mathematics Seminar
• October 4, 2007

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Cakes, Pies, and Fair Division

• Abstract           Mathematicians enjoy cakes
  for their own sake and as a metaphor for more
  general fair division problems. We describe the
  state of the art of cake cutting, including some
  new results on computational complexity.
             Suppose that a cake is to be divided
  by parallel planes into n pieces, one for each of
  n players whose preferences are defined by
  separate measures. We show that there is always
  an "envy-free" division, meaning that no player
  prefers another player's piece, and that such a
  division is always Pareto optimal. Alas, such a
  division cannot always be found by a finite
  procedure. Even assuring each player 1/n of the
  cake seems to require n log n steps.           
  Pies, of course, have their own attractions. We
  cut them radially into wedges. It turns out that
  pie cutters, unlike cake cutters, may be forced
  to choose between envy-freeness and Pareto
  optimality.

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Why cakes?

• Vote for one
• Cake cutting is a laboratory for studying the
  great issues of mankind, in which we address the
  compatibility of equity and efficiency in a
  mathematically tractable environment.
• Its fun.

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I cut, you choose
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Everybody gets 1/n

• n players
• Referee slides knife from left to right
• Anyone who thinks the left piece has reached 1/n
  says STOP and gets the left piece.
• Proceed by induction. (Banach - Knaster ca.
  1940)

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Finite algorithm for 1/n ?

• Can we guarantee everyone 1/n by a finite
  procedure?
• ( moving knife ? finite )
• Marks, cuts, and queries. How many marks ?
• Divide and conquer (Even Paz)
• Is (n lg n) marks the best we can do?
• Woeginger and Sgall, 2007 Need (n lg n) marks
  and queries.
• Woeginger and Sgall, 2007 Fix ? gt 0, and ask
  only that everyone get ((1/n) - ?). Now the
  number of marks can be linear in n.

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Some definitions

• Cakes are cut by parallel planes.
• The cake is an interval C 0, m .
• Points in interval possible cuts.
• Subsets of interval possible pieces.
• We want to partition the interval into S1, S2,
  , Sn, where
• Si i-th players piece.
• Players preferences are defined by measures
  v1, v2, , vn
• vi (Sj ) Player is valuation of piece
  Sj.
• Always assume measures are nonatomic and
  absolutely continuous.
• (vi (S) gt 0 ? S has positive length)

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The value matrix

• Consider the matrix
• v1(S1) v1(S2) v1(Sn)
• v2(S1) v2(S2) vn(Sn)
• vn(S1) vn(S2) vn(Sn)
• We could think of this as a giant vector in Rn2 .
• Fix the measures, and let (S1,,Sn) range over
  all partitions.
• Amazing fact the value matrices form a convex
  set in Rn2.
• (Lyapounov Dvoretsky-Wald-Wolfovitz.)
• But these are partitions into arbitrary
  measurable sets.

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Consequence of Lyapounov

• If we allow arbitrary measurable sets as pieces,
  then there is always a division in which
• vi(Sj) 1/n for every i, j.
• That is, every player considers the cake to be
  evenly divided.
• (Dubins and Spanier, 1961)

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Envy-free divisions

• In the 1/n procedures, player 1 might think
  player 2s piece is better than his own.
• A division is envy-free if no player thinks any
  other players piece is better than his own
• vi (Si) ? vi (Sj) for every i and j.
• Can we always find an envy-free division?

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Conway, Guy, and Selfridge

• A pours wine into three equal glasses (so he
  says)
• B identifies first choice and second choice, and
  pours enough from first choice back into bottle
  to make them equal (so he says)
• C picks, then B, then A If C doesnt pick
  reduced glass, B must
• But theres still wine in the bottle!
• Suppose B got reduced glass. Then C pours.
• B picks, then A, then C.
• Applied to cakes, there are 5 cuts, and each
  player gets two intervals.

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Two moving knives the squeeze

• A cuts the cake into thirds (by his measure).
• Suppose B and C both choose the center piece.
• A moves both knives in such a way as to keep end
  pieces equal (according to A)
• B or C says STOP when one of the ends becomes
  tied with the middle. (Barbanel and Brams, 2004)

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There is always an envy-free division.

• Theorem (1980) For n players, there is always
  an envy-free division in which each player
  receives a single interval.
• Proofs
• (WRS) The division simplex
• (Francis Edward Su) Sperners Lemma

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Is there a finite procedure for envy-freeness?

• Theorem (2007) There is no finite protocol for
  finding an envy-free division among 3 or more
  players, if each player is to receive an
  interval.
• Proof If you think you have a finite protocol,
  I can construct a set of measures for which it
  doesnt work.
• Contrast
• 5 cuts, 3 players finite procedure
  (Conway-Guy-Selfridge)
• 2 cuts, 3 players no finite procedure
• Wheres the boundary?
• Wanted Nice finite procedure for 3 cuts, 3
  players
• --- or k cuts, n players.

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Undominated allocations

• A division Si S1, S2, , Sn is dominated by
  a division
• Ti T1, T2, , Tn if
• vi(Ti) ? vi(Si) for every i
• with strict inequality in at least one case.
• That is T makes some player better off, and
  doesnt make any player worse off.
• Si is undominated if it isnt dominated by
  any Ti .
• undominated Pareto optimal efficient

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Envy-free implies undominated

• Is there an envy-free allocation that is also
  undominated?
• Theorem (Gale, 1993) Every envy-free division
  of a cake into n intervals for n players is
  undominated.
• So for cakes EQUITY ? EFFICIENCY.

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Gales proof

• Theorem (Gale, 1993) Every envy-free division
  of a cake into n intervals for n players is
  undominated.
• Proof Let Si be an envy-free division.
• Let Ti be some other division that we think
  might
• dominate Si.
• S2 S3 S1
• T3 T1 T2
• v1(T1) lt v1(S3) ? v1(S1)
• so Ti doesnt dominate Si after all. //

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Pies

• Pies are cut along radii. It takes n cuts to
  make pieces for n players.
• A cake is an interval.
• A pie is an interval with its endpoints
  identified.

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Pies

• 1. Are there envy-free divisions for pies?
• YES
• 2. Does Gales proof work?
• NO
• 3. Are there pie divisions that are both
  envy-free and undominated? (Gales
  question, 1993)
• YES for two players
• NO if we dont assume absolute continuity
• NO for the analogous problem with unequal
  claims
• (Brams, Jones, Klamler, 2006-2007)

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Examples emerge from failed proofs

• Failed proof that there IS an envy-free,
  undominated allocation
• Call the players A, B, C. Call their measures
  vA, vB, vC.
• Given a division PA, PB, PC, define
• The values vector is ( vA(PA), vB(PB), vC(PC)
  ).
• (The possible values vectors are the IPS. )
• The sum is vA(PA) vB(PB) vC(PC).
• The proportions vector is
• ( vA(PA)/sum, vB(PB)/sum, vC(PC)/sum).
• The possible proportions vectors form a simplex.

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The failed proof

• 1. For every proportions vector in the simplex,
  there is an undominated division.
• 2. In every undominated division, there is at
  least one player that isnt envious. (cf
  Vangelis Markakis)
• 3. Around each vertex, theres a set of
  proportions vectors for which that vertexs
  player isnt envious.
• 4. Dont those sets have to overlap? Doesnt
  that mean theres an allocation that satisfies
  everybody?
• (Like Wellers Proof)

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The failed proof

• NO! The sets can be made to overlap. But for
  that proportions vector, there may be TWO
  undominated allocations, each satisfying
  different sets of players.
• Lesson for a counterexample There must be at
  least one instance of a proportions vector with
  two or more (tied) undominated allocations.

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The example

• Well represent the pie as the interval 0, 18
  with the endpoints identified.
• By the sectors we mean the intervals 0, 1,
  1, 2, , 17, 18.
• The players are still A, B, C.
• Well specify the value of each sector to each
  player. Each players measure is uniform over
  each sector.

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The example
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Summary

• Cakes with arbitrary measurable pieces
• Can get all players to agree that everyone has
  1/n.
• Cakes, 1/n fairness
• Can guarantee everyone 1/n with (n lg n) marks
  and queries.
• Cant do better than (n lg n) marks and queries.
• Can guarantee 1/n - ? with cn marks and queries.
• What about 1/n with cn marks ??
• Cakes, envy-free
• Can find envy-free divisions for any n, any
  measures.
• Cant do it with a finite procedure.
• but can, maybe, if extra cuts are allowed. How
  many?
• Envy-free ? undominated
• Pies, envy-free
• Can find envy-free divisions for any n, any
  measures.
• Can find envy-free, undominated divisions when
  n2.
• but not always when n ? 3. So envy-free,
  undominated can conflict.

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Cookies

• This cookie cutter has blades at fixed 120-degree
  angles.
• But the center can go anywhere. Is there always
  an envy-free division of the cookie? Envy-free
  and undominated?