Cakes, Pies, and Fair Division

Walter StromquistSwarthmore College

mail@walterstromquist.com

Rutgers Experimental Mathematics SeminarOctober 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.

It’s 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 rightAnyone 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 > 0, and ask only that

everyone get ((1/n) - ). Now the number of marks can be linear in n.

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Some definitionsCakes 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, whereSi = i-th player’s piece.

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

Always assume measures are nonatomic and absolutely continuous.(vi (S) > 0 S has positive length)

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The value matrixConsider 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.10/4/2007 8

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 2’s piece is better than his own.

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?

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Conway, Guy, and SelfridgeA 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 doesn’t pick reduced glass, B must]

But there’s 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) Sperner’s 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 doesn’t work.

Contrast:5 cuts, 3 players: finite procedure (Conway-Guy-Selfridge)2 cuts, 3 players: no finite procedureWhere’s the boundary?

Wanted: Nice finite procedure for 3 cuts, 3 players --- or k cuts, n players.

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Undominated allocationsA division {Si} = S1, S2, …, Sn is dominated by a division

{Ti} = T1, T2, …, Tn if

vi(Ti) vi(Si) for every iwith 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”

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Envy-free implies undominatedIs 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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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. // 10/4/2007 18

PiesPies 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

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Pies1. Are there envy-free divisions for pies?

YES

2. Does Gale’s proof work? NO

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 continuity”NO for the analogous problem with unequal claims

(Brams, Jones, Klamler, 2006-2007)

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Examples emerge from failed proofsFailed 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.10/4/2007 21

The failed proof1. For every proportions vector in the

simplex, there is an undominated division.

2. In every undominated division, there is at least one player that isn’t envious. (cf Vangelis Markakis)

3. Around each vertex, there’s a set of proportions vectors for which that vertex’s player isn’t envious.

4. Don’t those sets have to overlap? Doesn’t that mean there’s an allocation that satisfies everybody?(Like Weller’s Proof)

B

C

A

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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 exampleWe’ll 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.

We’ll specify the value of each sector to each player. Each player’s measure is uniform over each sector.

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The example

Measures for three players: 0 3 6 9 12 15 18

vA 130 100 69 131 70 100 100 70 131 69 100 130 60 100 100 100 100 60

vB 100 100 115 5 100 100 100 100 100 100 100 100 100 100 100 100 100 100

vC 100 100 100 100 100 100 100 100 115 5 100 100 100 100 100 100 100 100

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SummaryCakes 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.Can’t 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.Can’t do it with a finite procedure.…but can, maybe, if extra cuts are allowed. How many?Envy-free undominated

Pies, envy-freeCan find envy-free divisions for any n, any measures.Can find envy-free, undominated divisions when n=2.…but not always when n 3. So envy-free, undominated can conflict.

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CookiesThis 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?

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