A pie that can

1
A pie that cant be cut fairly

• Walter Stromquist
• Swarthmore College
• mail_at_walterstromquist.com
• Fair Division Seminar
• Dagstuhl, Deutschland
• June 26, 2007 (last slide added 8/21/07)

2
Cake cutting as a metaphor
A cake is cut by n-1 cuts for n players.
Players preferences are represented by nonatomic
measures.

• Early results
• 1980 There is always an envy-free division.
• (Best current proof Follow Francis Sus
• argument based on Sperners Lemma, 1999)
• 1993 (Gale) Every envy-free division is
  undominated
• ( efficient Pareto optimal maximal )
• So In this model, there is always a division
  that is both envy free and undominated.

3
Cake-cutting with measurable pieces

• Pieces need not be intervals they can be any
  measurable sets.
• Julius Barbanel told us about Wellers Theorem
  There is always a division that is both envy free
  and undominated.
• ( Where have we seen Kakutanis fixed point
  theorem before ? )

4
Pies are an alternative metaphor

• 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.

5
Can pies be cut fairly?

• Are there envy-free divisions?
• YES. Many of them. Cut the pie anywhere, and
  then treat it as a cake.
• Are envy-free divisions necessarily undominated
  (like for cakes) ?
• NO.
• Must there be even ONE division that is both
  envy-free and undominated?
• (Gale, 1993)
• Todays answer NO.

6
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.

7
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)

8
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.

9
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.

10
The example