1 / 27

Cakes, Pies, and Fair Division

Cakes, Pies, and Fair Division. Walter Stromquist Swarthmore College mail@walterstromquist.com Rutgers Experimental Mathematics Seminar October 4, 2007. Cakes, Pies, and Fair Division.

ginata
Download Presentation

Cakes, Pies, and Fair Division

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Cakes, Pies, and Fair Division Walter Stromquist Swarthmore College mail@walterstromquist.com Rutgers Experimental Mathematics Seminar October 4, 2007

  2. 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.

  3. 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.

  4. I cut, you choose

  5. 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)

  6. 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.

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

  8. 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.

  9. 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)

  10. 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?

  11. 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 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.

  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)

  13. 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

  14. 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 procedure Where’s the boundary? Wanted: Nice finite procedure for 3 cuts, 3 players --- or k cuts, n players.

  15. 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”

  16. 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.

  17. Gale’s 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) < v1(S3)  v1(S1) so {Ti} doesn’t dominate {Si} after all. //

  18. Cuts meet at center 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.

  19. Pies 1. 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 players NO if we don’t assume “absolute continuity” NO for the analogous problem with unequal claims (Brams, Jones, Klamler, 2006-2007)

  20. 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.

  21. B C A 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 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)

  22. 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.

  23. The example We’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.

  24. The example

  25. 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. 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-free Can 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.

  26. 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?

More Related