Cutting a Pie is Not a Piece of Cake презентация

World Congress of the Game Theory Society
Evanston, IL
July 13, 2008

Steven J. Brams, Walter Stromquist

     Mathematicians enjoy cakes for their own sake and as a metaphor for more general fair division problems.

A cake is cut by parallel planes into n pieces, one for each of n players whose preferences are defined by separate measures. It is known that there is always an envy-free division, and that such a division is always Pareto optimal. So for cakes, equity and efficiency are compatible.

   A pie is cut along radii into wedges. We show that envy-free divisions are not necessarily Pareto optimal --- in fact, for some measures, there may be no division that is both envy-free and Pareto optimal. So for pies, we may have to choose between equity and efficiency.

Julius B. Barbanel (Union College)












Steven J. Brams (New York University)

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 player’s piece.

Player’s preferences are defined by measures v1, v2, …, vn
vi (Sj ) = Player i’s valuation of piece Sj.

These are probability measures.

We always assume that they are non-atomic (single points always have value zero).

respect to each other.

In effect, this assumption means that pieces with positive length also have positive value to every player.

thinks the left piece has reached 1/n says “STOP” …and gets the left piece.
Proceed by induction. (Banach - Knaster ca. 1940)

player’s piece is better than his own:

vi (Si) ≥ vi (Sj) for every i and j.

Can we always find 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) Sperner’s Lemma

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)

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 doesn’t make any player worse off.

{Si} is undominated if it isn’t dominated by any {Ti} .

“undominated” = “Pareto optimal” = “efficient”

(Gale, 1993): Every envy-free division of a cake into n intervals for n players is undominated (assuming absolute continuity).




So for cakes: EQUITY ⇒ EFFICIENCY.

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) < v1(S3) ≤ v1(S1)

so {Ti} doesn’t dominate {Si} after all. //

third of the cake. The first player likes only the left half of the leftmost third.



All other players’ preferences are uniform.



The only envy-free divisions involve cutting the pie in thirds.
None of these divisions is undominated.

Without absolute continuity: We may have to choose between envy-free and undominated.

division is also undominated.
There is always a division that is both envy-free and undominated.

Without absolute continuity:

There is always an envy-free division.
For some measures, there is NO division that is both envy-free
and undominated. We may have to choose!

Unless n = 2, when there is always an envy-free, undominated division, whatever the measures.

pieces for n players.





A cake is an interval.
A pie is an interval with its endpoints identified.

NO
Envy-free does NOT imply undominated

3. Are there pie divisions that are both envy-free and undominated? (“Gale’s question,” 1993)

YES for two players
NO if we don’t assume absolute continuity
NO for the analogous problem with unequal claims
(Brams, Jones, Klamler – next talk!)

NOT exist an envy-free, undominated allocation.

These measures may be chosen to be absolutely continuous.

So, Gale’s question is answered in the negative.

uniform, except…
Each player dislikes certain sectors (grayed out).
Each player perceives positive or negative bonuses (C)
or mini-bonuses (ε) in certain sectors.

The measures for three players:

preferred by Player 2.

That allocation is both envy-free and undominated.

division.
For some measures, there is NO division that is both envy-free
and undominated. We may have to choose!





Unless n = 2, when there is always an envy-free, undominated division, whatever the measures.

can go anywhere. Is there always an envy-free division of the cookie? Envy-free and undominated?