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1 EXACT PROCEDURES FOR ENVY-FREE CHORE DIVISION ELISHA PETERSON AND FRANCIS EDWARD SU draft version October 22, 1998 Abstract. We develop the rst explicit procedures for exact envy-free chore division for 4 or more players. We also give a simple 3-person procedure. Our 4- person moving-knife scheme and n-person discrete algorithm are accomplished by exploiting a notion of \irrevocable advantage" for chores, and are motivated by similar solutions for cake-cutting by Brams, Taylor, and Zwicker. We discuss the dierences between cake-cutting and chore division and additional problems encountered in chore division. 1. Introduction In this paper we explore the problem of chore division, which is closely related to a classical question, due to Steinhaus [9], of how to cut a cake fairly. We focus on constructive solutions, i.e., those obtained via a well-dened procedure or algorithm. Among the many notions of fairness is envy-freeness: an envy-free cake division is a set of cuts and an allocation of the pieces that gives each person what she feels is the largest piece. Much progress has been made on nding constructive algorithms for achieiving envy-free cake divisions; most recently, Brams and Taylor [3] have given a general n-person procedure. In contrast to cakes, which are desirable, the dual problem of chore division is concerned with dividing an object deemed undesirable. Here, each player would like to receive what he considers to be the smallest piece, of say, a set of chores. This problem appears to have been rst introduced by Martin Gardner in [6]. Oskui (see [7]) referred to it as the dirty work problem and gave the rst discrete and moving-knife solutions for exact envy-free chore division among 3 people. Su [11] and Zeng [12] have recently given -approximate algorithms that give envy-free chore divisions up to a pre-specied error. Much less work has been done to develop algorithms for chore division than for cake-cutting. In particular, there appear to be no explicit formulations of exact envy-free chore-division algorithms for more than three players in the literature. The purpose of this paper is to ll that gap. The main reason for the gap is that chore division has often been considered a straightforward (though perhaps tedious) extension of cake-cutting methods. For instance, Brams and Taylor suggest in [4] how cake-cutting methods might be adapted to chore division without working out the details. Nevertheless, we believe that it is worthwhile and instructive to explicitly construct such procedures, since the added complications oer some new ideas Mathematics Subject Classication. Primary 90D06; Secondary 90A06, 00A69. Key words and phrases. fair division, chore division, envy-free, irrevocable advantage. Both authors gratefully acknowledge the support of an HMC Research Grant. 1

2 2 ELISHA PETERSON AND FRANCIS EDWARD SU Moreover, as the allocation of undesirables is nearly as common as the problem of fairly dividing desirables, it is helpful to have explicit procedures available. In this paper, we develop several exact chore-division schemes. In Section 2, we give a new 3-person moving-knife scheme which is less complicated and more symmetrical than the procedure of Oskui. Section 3 develops the notion of irrevocable advantage for chores and develops a 4-person envy-free chore-division scheme, requiring 16 cuts. In the nal section, we give an exact n-person procedure for envyfree chore division, adapted from Brams and Taylor's n-person envy-free procedure for cake-cutting [4]. We assume throughout this paper that chores are innitely divisible. This is not unreasonable as a nite set of chores can be partitioned by dividing up each chore (e.g., a lawn to be mowed could be divided just as if it were a cake), or dividing the time spent on them. For ease of expression, we shall call the set to be divided a cake, rather than a set of chores. Implicit in this is the assumption that the players desire the smallest, rather than largest, piece of cake. All algorithms we consider will be envy-free, unless otherwise stated. 2. A 3-Person Exact Moving-Knife Procedure Our 3-person chore-division procedure relies on Austin's procedure [1] for dividing a cake into two pieces so that each of two players believes it is a split. For completeness, we review it here. Let one player hold two knives over the cake, with one at the left edge, such that the portion of cake between them is what he believes to be exactly half. If the second player agrees that it is exactly half, we are done. Otherwise, let the rst player move the knives across the cake from left to right, keeping the portion between them exactly half (in his estimation), until the second player agrees it is exactly half. (There must be such a point because when the rightmost knife reaches the right edge, the leftmost knife must be where the rightmost knife began, hence the second player must by that point have changed preferences.) At this point cuts are made and the cake outside the knives are lumped together, yielding two pieces which both players agree are exactly equal. Now we describe our algorithm for 3-person chore division, which will divide the cake into six pieces and assign each player two of the pieces that he feels are smaller than each pair of pieces the other players receive. Step 1. Divide the chores into three portions using any 3-person envy-free cake-division procedure, such as the Stromquist moving-knife procedure (see [10]). Now label each portion by the name of the player who thinks it is largest. Step 2. Let player i divide portion i into 2 pieces (which she feels is exactly half) and assign those pieces to the other two players such they each feel they have gotten no more than half of portion i. (This can be achieved via Austin's procedure: letting player i and one other player, say j, agree on a split, let the remaining player choose the half she thinks is smallest, and give the other half to j.) Step 3. Repeat Step 2 for each player. This ensures that each player is assigned two out of six total pieces such that each feels her share is smallest. We now verify there is no envy. Call the players i, j, and k. Player i will not envy player j because one piece of each of their pairs came from the portion labelled k,

3 EXACT PROCEDURES FOR ENVY-FREE CHORE DIVISION 3 and i feels her half of that portion was no larger than j's. As for her other piece, player i feels it was no more than half of the portion it came from, and therefore cannot be as large as player j's other piece, which i felt was exactly half of the largest portion. The same argument holds for any permutation of i, j, and k. See Figure 1. i j k to j to i to i to k to k to j Figure 1. An envy-free assignment of six pieces (of chores) to three people. This procedure requires at most 8 cuts, and is less complicated that the discrete procedure of Oskui [7]. There are 3-person moving-knife schemes that require fewer cuts [7, 8], but distinguishing features of our approach are that it is symmetrical with respect to the players and it is initially based on a cake-cutting procedure. The former property simplies the verication of envy-freeness, while the latter property may help in generalizing the scheme. The concept of an oracle is dened in [5] to be any cake-cutting scheme whose existence is postulated but not known. Given an oracle that achieves an envy-free chore division for n players such that an additional player thought all n pieces were exactly the same size, we can use the above ideas to construct a procedure (via an (n + 1)-person cake-division scheme) to give an envy-free chore-division method for n + 1 players. In the nal section, we shall give (by a dierent method) an explicit procedure to acheive envy-free chore division for an arbitrary number of players. 3. A 4-Person Exact Moving-Knife Procedure Our 4-person chore-division procedure is also a moving-knife procedure, and requires at most 16 cuts. It draws ideas from both the Brams-Taylor-Zwicker 4-person envy-free moving-knife scheme for cakes [5] and the Oskui 3-person envyfree discrete chore-division scheme. We also show how the notion of irrevocable advantage, important in cake-cutting [4], can be applied in chore division. Suppose the players are named Steve, Nathan, Akiko, and Jenni. Assume the cake is a rectangular block which may be divided by vertical cuts. Let Steve and Nathan divide the cake into four pieces they both agree are all equal, by performing three applications of Austin's procedure (using at most 6 cuts). Call the pieces X 1, X 2, X 3, and X 4. Note that if Akiko and Jenni disagree on which piece is the smallest, we can immediately allocate the pieces. Thus we may assume they agree that one piece is strictly smaller than the others, say X 4. Then

4 4 ELISHA PETERSON AND FRANCIS EDWARD SU each person thinks the following: Steve : X 1 = X 2 = X 3 = X 4 Nathan : X 1 = X 2 = X 3 = X 4 Akiko : X 4 < X 1 ; X 2 ; X 3 Jenni : X 4 < X 1 ; X 2 ; X 3 : Now, for each of X 1, X 2, and X 3, let Akiko and Jenni mark how they would trim them to make them the same size as X 4. As each piece is rectangular, assume the trimmings are marked from the top edge, so that a person receives the piece below her mark. See Figure 2. Hence, we can speak of one mark as being \above" Figure 2. This gure shows a possible set of markings made in the rst step of the procedure. In this case, two of Akiko's marks are above Jenni's, so we would follow case 2. another. Now, either two or three of Akiko's marks are at or above Jenni's marks, or vice versa. In the latter case, we can reverse the roles of Akiko and Jenni in what follows. This produces two cases, the rst in which Akiko has three higher marks and the second in which Akiko has two. Each of the following cases follows the same basic pattern: 1. Trim the pieces at the higher marks (3 more cuts), 2. Let Steve \add back" to one piece (1 cut), 3. Let players choose from these pieces (the order Nathan-Steve-Jenni-Akiko will always work, subject to provisos), and then 4. Divide the trimmings (6 cuts), exploiting an irrevocable advantage. Case 1: One player has three higher marks. Assume that Akiko's marks are all at or above Jenni's marks. Let Akiko trim X 1, X 2, and X 3 to obtain a fourway tie for the smallest piece. Call the trimmed pieces X 0 1, X 0 2, and X 0 3, and the trimmings T 1, T 2, and T 3, which are set aside for later. At this point, each person thinks: Steve : X 0 1; X 0 2; X 0 3 < X 4 Nathan : X 0 1; X 0 2; X 0 3 < X 4 Akiko : X 0 1 = X 0 2 = X 0 3 = X 4 Jenni : X 4 X 0 1 ; X 0 2 ; X 0 3 : Jenni must believe X 4 is the smallest, or tied for the smallest, because her marks were all below Akiko's (meaning she believes more should be trimmed to make them

5 EXACT PROCEDURES FOR ENVY-FREE CHORE DIVISION 5 equal to X 4 ). We can immediately assign X 4 to Jenni because Steve and Nathan denitely do not want it, and Akiko is indierent among the pieces. Of the remaining pieces X 0 1 ; X 0 2 ; X 0 3, let Steve return some of the trimmings in one of them, say X 0 3, to create a two-way tie for the smallest piece. (We still call this piece X 0 3, and its trimmings T 3.) This changes Akiko's valuation to Akiko : X 0 1 = X 0 2 = X 4 < X 0 3: Let Nathan choose rst, then Steve, then Akiko, with Steve required to take X 0 3 if it was not chosen by Nathan. Now, Nathan will envy no one because he did not want Jenni's piece, and he chose before Steve and Akiko. Steve will envy no one because he had one of two smallest pieces to choose from, and Akiko will envy no one because she had two smallest pieces to choose from, one of which must still be available. Jenni believed all these pieces were larger than hers, so will envy no one. Dividing the Trimmings. Suppose without loss of generality that Nathan chose X 0 1, Steve chose X 0 3, and Akiko X 0 2. Then, Steve thinks: Steve : T 1 ; T 2 T 3 Note that because Steve believed X 3 = X 4, he could receive all of T 3 and still not envy Jenni. Therefore, we say that Steve has an irrevocable advantage over Jenni with respect to T 3. In fact, by the above inequality he could receive 3 1(T 1+T 2 +T 3 ) T 3 and still not envy her. So, lump all the trimmings together (say, T = T 1 + T 2 + T 3 ), and let Akiko and Nathan use Austin's procedure to divide T into four pieces that they both agree are all equal. Then let the players choose in the order Jenni, Steve, and then (in any order) Akiko and Nathan. With respect to the trimmings, Jenni will envy no one because she chooses rst. Steve, choosing the smallest of the remaining three pieces, will have a piece that he believes is at most 3 1 T and therefore will not envy Jenni. Akiko and Nathan will not envy Steve or Jenni because they think all four pieces are equal. Case 2: One player has two higher marks. Assume now that Akiko has two marks at or above Jenni's marks. Without loss of generality suppose that Jenni has a higher mark on X 3 than Akiko. Let cuts be made at all three highest marks. Then: Steve : X 0 1 ; X 0 2 ; X 0 3 < X 4 Nathan : X 0 1 ; X 0 2 ; X 0 3 < X 4 Akiko : X 0 1 = X 0 2 = X 4 < X 0 3 Jenni : X 0 3 = X 4 X 0 1; X 0 2 Now, let Steve create a two-way tie for the smallest piece (as before) by returning to the smallest piece some of the corresponding trimmings. He may add either to X 0 1, X 0 2 or X 0 3. The X 0 1 and X 0 2 cases are equivalent, so we have two subcases. Suppose Steve adds to X 0 3 until it is as large as, say, X 0 2. Then: Steve : X 0 2 = X 0 3 X 0 1 < X 4 Nathan : X 0 1; X 0 2; X 0 3 < X 4 Akiko : X 0 1 = X 0 2 = X 4 X 0 3 Jenni : X 4 X 0 1; X 0 2; X 0 3

6 6 ELISHA PETERSON AND FRANCIS EDWARD SU As in the previous case, we may immediately assign X 4 to Jenni, and choose in the order Nathan-Steve-Akiko, with Steve required to take X 0 3 if Nathan didn't. Akiko receives X 0 1 or X 0 2, Steve receives X 0 2 or X 0 3, Nathan chooses rst, and Jenni believes X 4 is the smallest, so there is no envy for these pieces. The trimmings can be handled just as before, since Steve has an irrevocable advantage over Jenni with respect to each of the trimmings. Otherwise, suppose Steve adds to X 0 1 until it is as large as either X 0 2 or X 0 3. Then: Steve : X 0 1 = X 0 2 X 0 3 < X 4 (or X 0 1 = X 0 3 X 0 2 < X 4 ) Nathan : X 0 1; X 0 2; X 0 3 < X 4 Akiko : X 0 2 = X 4 X 0 1 ; X 0 3 Jenni : X 0 3 = X 4 X 0 1 ; X 0 2 Let Nathan choose rst. Then assign Steve X 0 1 if it is available, otherwise give him the other piece he thinks is tied for smallest. Next, assign Jenni X 0 3 if it is available, otherwise give her X 4 (which Nathan and Steve denitely did not choose). Because X 0 1 and X 0 3 are allocated by this point, we know that Akiko will receive either X 0 2 or X 4. Hence there is no envy for these pieces. For the trimmings, note that Steve has an irrevocable advantage with respect to each of the trimmings over whoever receives the X 4 piece (either Jenni or Akiko). Hence the trimmings can be divided using the method discussed earlier. This completes the 4-person moving-knife envy-free chore-division procedure. It is similar to the Brams-Taylor-Zwicker 4-person scheme for cakes, but complications arise from the increased diculty in chore division of constructing an irrevocable advantage. Chore division must also allow for adding back parts of pieces to create \ties" (an idea due to Oskui), rather than the trimming that occurs in cake-division procedures. This complexity has some cost: while Brams-Taylor-Zwicker's cakecutting solution takes at most 11 cuts, our chore-division procedure takes at most 16 cuts (which can be reduced to 15 cuts with a modication much like Brams- Taylor-Zwicker's 5-cut modication [5] to Austin's procedure). 4. An n-person Exact, Discrete Procedure Finally, we construct an n-person chore-division scheme. Unlike the 3 and 4- person schemes we have given, it is a completely discrete procedure. It closely mirrors Brams and Taylor's n-person cake-cutting procedure [3], but diers from theirs in the creation of ties by \adding on", rather than trimming. This necessitates the existence of reserves to draw on. Though our 4-person procedure possessed a natural set of reserves due to the initial trimming, for this procedure we need to carefully create enough reserves for use by specic players. The approach of creating reserves is apparent in Oskui's 3-person procedure and suggested by Brams and Taylor [2, 4] for chore-division schemes. The purpose of this section is to show explicitly how it can be made to work for exact chore division among n players. A brief sketch of our procedure runs as follows. Let one player divide the cake and allocate the pieces. As long as there are objections, we shall iterate a procedure that gives an envy-free allocation of part of the cake, and also gives the player who

7 EXACT PROCEDURES FOR ENVY-FREE CHORE DIVISION 7 objected an irrevocable advantage over another player with respect to the part of the cake that has not yet been allocated. With enough iterations, there will be enough players with irrevocable advantages to allow the allocation of the remainder of the cake. For ease of comparison with Brams and Taylor's cake-cutting procedure, we include step numbers in our procedure that correspond their step numbers. (See [3]). The signicant departures occur in Steps 6.1, 7.1, 7.3, 8, and Following their example, we distinguish rules from strategies by placing strategies in parentheses. Again, for ease of expression, we refer to the chore set as a cake, bearing in mind that each person wants the piece he/she thinks is smallest. We shall give the procedure for n = 4 case. The generalization to more players will be discussed subsequently. Step 1. Let Player 2 cut the cake into 4 pieces (that she considers equal), and then assign one piece to each player. Step 2. Player 2 asks the other 3 players if anyone objects to this allocation. (A player objects i he/she envies another player.) Step 3. If no one objects, then each keeps the piece he was given, and we are done. Step 4. If someone objects, say Player 1, let Player 1 choose two pieces (that he thinks are not equal in size) and name them A (for the larger piece) and B (for the smaller piece). Aside. The other pieces are reassembled for allocation later. Note that Player 1 thinks A is larger than B but Player 2 thinks they are the same size. Step 5. Let Player 1 name an r 10 (chosen such that, even if A were divided into r pieces and the 7 smallest pieces of A were removed, he would still prefer B). Aside. This is possible because the union of the 7 smallest pieces is no larger than 7 times the average size of all r pieces. Hence Player 1 can choose r large enough so that 7(A)=r = (A)? (B) where is Player 1's measure. (The same fact was used in Brams and Taylor's algorithm.) Step 6. Player 2 divides each of A and B into r sets (that she considers equal). Step 6.1. From the pieces in B, let Player 3 choose a piece (he considers largest). This will be used as Player 3's reserves. Step 7. From the remaining pieces, Player 1 chooses (what he thinks are the smallest) 3 sets in B, calling these Y 1, Y 2, and Y 3. Step 7.1. The rest of B is set aside as Player 1's reserves. If necessary, Player 1 uses his reserves to add to two of the Y i (to make them all equally sized the reserves will be enough because it originated from at least 5 pieces in B and the smallest two are larger than the largest of the Y i in Player 1's opinion.) Step 7.2. If Player 1 considers the three largest pieces in A all strictly larger than these pieces, the three largest pieces in A are identied as Z 1, Z 2, and Z 3. Otherwise, Player 1 cuts the largest piece in A into three (equal) pieces, calling them Z 1, Z 2, and Z 3 (which are each bigger than each of the Y i, in Player 1's opinion.)

8 8 ELISHA PETERSON AND FRANCIS EDWARD SU Aside. The proof to show this is possible is almost exactly the same argument as established in [3] for usage in cake-cutting the only dierence being that since Player 3's reserves were chosen before the Y i 's, we can only say that the Y i 's are among the smallest four sets of B in Player 1's opinion. Step 7.3. The remaining pieces of A are set aside for Player 2's reserves. If necessary, Player 2 uses her reserves to add on to two of the Z i (to make them all equally sized the reserves are enough because they originated from pieces in B that were each the same size as all three Z i 's combined). Aside. At this stage, Player 1 believes Y 1 = Y 2 = Y 3 < Z 1 ; Z 2 ; Z 3 (note the strict inequality). Player 2 believes Z 1 = Z 2 = Z 3 Y 1 ; Y 2 ; Y 3. Step 8. Player 3 takes this collection of 6 pieces and, if necessary, adds on to one of the pieces from his reserves (to make a two-way tie for smallest piece). (Player 3 has enough reserves because his strategy from Step 6:1 gave him reserves that he feels are at least as large as all 3 Y i 's, hence at least as large as the second-smallest piece of all 6 pieces.) Step 9. Let the players choose in the order , with Player 3 required to take a piece he augmented if it is available. Player 2 must choose one of the Z i 's, and Player 1 must choose one of the Y i 's. Aside. This yields a partition fx 1 ; X 2 ; X 3 ; X 4 ; L 1 g of the cake such that the X i 's are an envy-free partial allocation, and L 1 is the leftover piece consisting of all cake not yet allocated. Moreover, note that Player 1 thinks his piece is strictly smaller than Player 2's piece, say, by an amount. Step 10. Player 1 names s (such that [ 1 2 (L 1)] s <, where is Player 1's measure). Steps will result in Player 1 thinking that at least half of L 1 is Aside. allocated. Hence s represents the number of times to iterate Steps to make the leftover piece smaller than. Step 11. Player 1 cuts L 1 into 8 (equal) pieces. Player 2 sets aside (the largest) 2 pieces for her reserves, and Player 3 sets aside a piece from those remaining (that he feels is largest) for his reserves. Step 12. Of the remaining 5 pieces, Player 2 returns part of her reserves to 2 of the pieces, if necessary (to create a 3-way tie for the smallest piece). Step 13. Player 3 returns, if necessary, some from his reserves to one of the 5 pieces (to create a 2-way tie for the smallest piece). Step 14. Let the players choose in the order , with Players 2 and 3 required to take a piece they augmented if one is available. Step 15. Repeat steps (s? 1) additional times, with each application of these four steps applied to the leftover piece from the preceding application. Aside. According to player 1, the leftover piece L 2 is now smaller than. Hence, we have an envy-free partial allocation of the cake except for the leftover piece L 2, and moreover, Player 1 feels that his portion together with L 2 would still be smaller than Player 2's portion. Thus, we say that player 1 has an irrevocable advantage

9 EXACT PROCEDURES FOR ENVY-FREE CHORE DIVISION 9 over player 2 with respect to L 2, and we create a subset of f1; 2; 3; 4g f1; 2; 3; 4g called IA, and place (1; 2) 2 IA. Step 16. Player 2 cuts L 2 into 12 pieces. Step 17. Each player who agrees that all pieces are the same size is placed in the set A. Otherwise, players who disagree are placed in the set D. Step 18. If D A IA, we divide the pieces among the players in A, each receiving the same number of pieces, and we are done. Aside. Players in A do not envy each other since they agree each of the 12 pieces were the same size. None of the players in D envy those in A by the denition of IA. Players in D do not envy each other because they have not received any new pieces in this step. Step 19. Otherwise, we choose the lexographically least pair (i; j) from DA that is not in IA, and return to step 4, with Player i in place of Player 1, Player j in place of Player 2, and L 2 in place of the cake. Step 20. Repeat Steps Aside. Since each pass through Step 15 adds a new ordered pair to IA), eventually IA will contain D A, no matter what D A currently is. When this occurs, the algorithm concludes at Step 18 with an envy-free chore division of the entire cake. This ends the procedure. The extension of this procedure to n players is relatively straightforward, but we briey mention the changes that ensue and leave the verication to the reader. Let k = (n? 3). The 7-piece criterion in Step 5 needs to be increased (to k 2 + 4k + 2 pieces) for later use in Step 7. This necessitates an increase in r. In Step 6.1, all players except Players 1, 2, and n will choose, in reverse order, pieces of B to form their reserves. Specically, player i will choose what she thinks are the (n? i) largest pieces of what remains in B. This ensures that these players have enough reserves for Step 8. In Step 7, there need to be more Y i 's and Z i 's chosen (k + 2 of each), so that after the augmentation in Step 8 there is at least one Y i and one Z i untouched. Also, in the case where Player 1 feels the biggest k + 2 pieces of A are not all larger than the Y i 's just chosen, r needs to be large enough (r k 2 + 5k + 4) so that the largest piece of A can be split into k + 2 pieces all larger than the Y i 's. (Some care needs to be exercised here, since the reserves are chosen before the Y i 's, and hence Player 1 may not think the Y i 's are the smallest pieces of B.) The size of r also guarantees that Player 1 and 2 have enough reserves from the remainders of B and A. In Step 8, there are enough reserves because each person feels his reserves are larger than at least half the Y i 's and Z i 's. In Step 9, the players choose pieces in reverse order and players are required to take a piece they augmented if available. Steps can be modied analogously (let Player 1 start by cutting the leftovers into n 2? 3n + 4 equal pieces) to accomodate more people in the iterative part of the procedure. Curiously, though our chore-division procedure is more complicated than its cake-cutting counterpart, it may converge faster and require fewer cuts overall. For instance, for n = 4, each pass through the iterative part of the procedure (Steps 10-15) guarantees at least half the cake is apportioned rather than only one-fth. The authors have noted that faster convergence of chore division is also a feature of the -approximate algorithm of Su [11].

10 10 ELISHA PETERSON AND FRANCIS EDWARD SU Note that unlike our 4-person moving-knife procedure (which required at most 16 cuts), this n-person algorithm may take arbitrarily long depending on player preferences. It remains an open question whether a bounded procedure exists for either cake or chore division among n-people. 5. Conclusion In this paper we have shown explicitly how cake-cutting algorithms can be translated, with complications, into exact envy-free chore-division algorithms. There are two important features of this translation. First, the use of reserves is very important, and we showed that the creation of such reserves is generally possible, though it requires either clever trimming or the removal of reserves before the division. Secondly, the notion of an irrevocable advantage for chores, developed in Section 3, allows one to terminate what would otherwise be an innite procedure. The nature of the irrevocable advantage in our 4-person moving-knife procedure was slightly dierent from that in our discrete n-person procedure; namely, in the latter case we could achieve an irrevocable advantage with respect to the entire leftover piece. Given the numerous cuts needed to implement both the Brams-Taylor n-person cake-cutting procedure [4] and our n-person chore-division procedure, one may rightfully question their practicality. There are two possible responses. First, the number of cuts can be reduced if one is willing to accept an -approximate solution, by rotating players through the roles of Steps 10-15, and quitting when satised. Secondly, the construction of an initial solution, however complex, is always the rst step towards nding useful simplications. References [1] A. K. Austin, Sharing a cake, Mathematical Gazette 6, no. 437 (October 1982), [2] S. J. Brams and A. D. Taylor, An envy-free cake division algorithm, Economic Research Reports, C. V. Starr Center for Applied Economics, New York University, [3] S. J. Brams and A. D. Taylor, An envy-free cake division protocol, Amer. Math. Monthly 102 (1995), [4] S. J. Brams and A. D. Taylor, Fair Division: from Cake-Cutting to Dispute Resolution, Cambridge University Press, [5] S. J. Brams, A. D. Taylor, and W. S. Zwicker, A moving-knife solution to the four-person envy-free cake-division problem, Proc. Amer. Math. Soc. 125 (1997), [6] M. Gardner, aha! Insight, New York: W.F. Freeman and Co., [7] J. M. Robertson and W. A. Webb, Cake-Cutting Algorithms: Be Fair If You Can, A. K. Peters Ltd., [8] F. W. Simmons, personal communication, [9] H. Steinhaus, The problem of fair division, Econometrica 16, no. 1 (January 1948), [10] W. Stromquist, How to cut a cake fairly, Amer. Math. Monthly, 87 (1980), [11] F. E. Su, Rental harmony: Sperner's lemma in fair division, preprint, in submission, [12] D. Z. Zeng, -approximate envy-free algorithms, preprint, Department of Mathematics, Harvey Mudd College, Claremont, CA address: epeterso@hmc.edu, su@math.hmc.edu