Upload
paul-houston
View
216
Download
1
Embed Size (px)
Citation preview
1
Cutting a Pie is Not a Piece of Cake
Walter StromquistSwarthmore College
Third World Congress of the Game Theory SocietyEvanston, ILJuly 13, 2008
1
2
Cutting a Pie is Not a Piece of CakeJulius B. Barbanel, 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.
2
6
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 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). 6
7
“Absolutely continuous”
Sometimes we assume that the measures are absolutely continuous with respect to each other.
In effect, this assumption means that pieces with positive length also have positive value to every player.
10
n players:Everybody gets 1/n
Referee slides knife from left to rightAnyone who thinks the left piece has reached 1/n says
“STOP” …and gets the left piece.
Proceed by induction. (Banach - Knaster ca.
1940) 10
11
Envy-free divisions
A division is envy-free if no player thinks any other 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
11
12
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)
12
13
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 doesn’t make any player worse off.
{Si} is undominated if it isn’t dominated by any {Ti} .
“undominated” = “Pareto optimal” = “efficient”
13
14
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 (assuming absolute continuity).
So for cakes: EQUITY EFFICIENCY.
14
15
Gale’s proofTheorem (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) < v1(S3) v1(S1)
so {Ti} doesn’t dominate {Si} after all. // 15
16
Cakes without absolute continuity
First player’s preference: Uniform, EXCEPT on the leftmost 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.
17
Summary for cakes
With absolute continuity:
There is always an envy-free division.Every envy-free 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.
19
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.
Cuts meet at center
19
21
Pies
1. Are there envy-free divisions for pies?YES
2. Does Gale’s proof work? 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 playersNO if we don’t assume absolute continuityNO for the analogous problem with unequal claims
(Brams, Jones, Klamler – next talk!)
21
22
Pies
For n 3, there are measures for which there does 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.
24
The example
24
Partition the pie into 18 tiny sectors.
Each player’s preference is uniform, except…Each player dislikes certain sectors (grayed out).Each player perceives positive or negative bonuses (C)
or mini-bonuses () in certain sectors.
The measures for three players:
+C–
–C +C –C+
–C +
+C –C +C–
+C
+C
25
Pies for two players
Of all envy-free allocations, pick the one most preferred by Player 2.
That allocation is both envy-free and undominated.
26
Summary for pies
With or 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.
28
Summary:
When must there be an envy-free,undominated allocation?
CAKE PIE
2 players YES YES
3 players
YES, assuming absolute
continuity(otherwise NO)
NO