1. Cakes, Pies, and Fair Division Walter Stromquist Swarthmore College mail@walterstromquist.com Rutgers Experimental Mathematics Seminar October 4, 2007 10/4/20071

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. 10/4/20072

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. 10/4/20073

4. I cut, you choose 10/4/20074

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) 10/4/20075

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. 10/4/20076

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 S 1, S 2, …, S n, where S i = i-th player’s piece. Player’s preferences are defined by measures v 1, v 2, …, v n v i (S j ) = Player i’s valuation of piece S j. Always assume measures are nonatomic and absolutely continuous. (v i (S) > 0  S has positive length) 10/4/20077

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 R n2. Fix the measures, and let (S1,…,Sn) range over all partitions. Amazing fact: the value matrices form a convex set in R n2. (Lyapounov; Dvoretsky-Wald-Wolfovitz.) But these are partitions into arbitrary measurable sets. 10/4/20078

9. Consequence of Lyapounov If we allow arbitrary measurable sets as pieces, then there is always a division in which v i (S j ) = 1/nfor every i, j. That is, every player considers the cake to be evenly divided. (Dubins and Spanier, 1961) 10/4/20079

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: v i (S i )  v i (S j )for every i and j. Can we always find an envy-free division? 10/4/200710

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. 10/4/200711

12. 10/4/200712

13. 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) 10/4/200713

14. 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 10/4/200714

15. 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. 10/4/200715

16. Undominated allocations A division {S i } = S 1, S 2, …, S n is dominated by a division {T i } = T 1, T 2, …, T n if v i (T i )  v i (S i )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. {S i } is undominated if it isn’t dominated by any {T i }. “undominated” = “Pareto optimal” = “efficient” 10/4/200716

17. 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. 10/4/200717

18. Gale’s proof Theorem (Gale, 1993): Every envy-free division of a cake into n intervals for n players is undominated. Proof: Let {S i } be an envy-free division. Let {T i } be some other division that we think might dominate {S i }. S2S3 S1 T3T1 T2 v1(T1) < v1(S3)  v1(S1) so {T i } doesn’t dominate {S i } after all. // 10/4/200718

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 10/4/200719

20. 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) 10/4/200720

21. Examples emerge from failed proofs Failed proof that there IS an envy-free, undominated allocation: Call the players A, B, C. Call their measures v A, v B, v C. Given a division P A, P B, P C, define: The values vector is ( v A (P A ), v B (P B ), v C (P C ) ). (The possible values vectors are the IPS. ) The sum is v A (P A ) + v B (P B ) + v C (P C ). The proportions vector is ( v A (P A )/sum, v B (P B )/sum, v C (P C )/sum). The possible proportions vectors form a simplex. 10/4/200721

22. 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) B C A 10/4/200722

23. 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. 10/4/200723

24. 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. 10/4/200724

25. The example 10/4/200725

26. 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. 10/4/200726

27. 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? 10/4/200727