1. A pie that can’t be cut fairly Walter Stromquist Swarthmore College mail@walterstromquist.com Fair Division Seminar Dagstuhl, Deutschland June 26, 2007 (last slide added 8/21/07)

2. Cake cutting as a metaphor Early results: 1980: There is always an envy-free division. (Best current proof: Follow Francis Su’s argument based on Sperner’s Lemma, 1999) 1993 (Gale): Every envy-free division is undominated ( = efficient = Pareto optimal maximal ) So: In this model, there is always a division that is both envy free and undominated. A cake is cut by n-1 cuts forn players. Players’ preferences are represented by nonatomic measures.

3. Cake-cutting with measurable pieces Pieces need not be intervals; they can be any measurable sets. Julius Barbanel told us about Weller’s Theorem: There is always a division that is both envy free and undominated. ( Where have we seen Kakutani’s fixed point theorem before ? )

4. Pies are an alternative metaphor 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

5. Can pies be cut fairly? Are there envy-free divisions? YES. Many of them. Cut the pie anywhere, and then treat it as a cake. Are envy-free divisions necessarily undominated (like for cakes) ? NO. Must there be even ONE division that is both envy-free and undominated? (Gale, 1993) Today’s answer: NO.

6. 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.

7. 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

8. 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.

9. 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. The example