MATH4999 Capstone Projects in Mathematics and Economics. 1.1 Criteria for fair divisions Proportionality, envy-freeness, equitability and efficiency

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1 MATH4999 Capstone Projects in Mathematics and Economics Topic One: Fair allocations and matching schemes 1.1 Criteria for fair divisions Proportionality, envy-freeness, equitability and efficiency 1.2 Procedures for two-player and multi-player cake-cutting Discrete cut-and-choose procedures Continuous moving-knife procedures 1.3 Adjusted winner for two-party allocation of discrete goods Point allocation procedures Efficiency 1.4 Matching schemes Marriage, college admission and roommates problems Matching algorithms and stable solutions 1

2 1.1 Criteria for fair divisions Fair division is the problem of dividing a set of goods or resources between several people who have an entitlement to them, such that each person receives his due share. This problem arises in cake-cutting, divorce settlements, allocation of scarce resources, etc. Theory of fair division procedures Provide explicit criteria for various different types of fairness. Provide efficient procedures (algorithms) to achieve a fair division; desirable to require the least number of steps (minimum cuts). Study the properties of such divisions both in theory and in real life. Understand the impossibility of achieving fairness based on certain criteria and/or within a given allowable set of procedures. 2

3 There is a set X and a group of n players. A division is a partition of X to n disjoint subsets: X = X 1 X 2... X n, one subset per player. The set X can be of several types: X may be a finite set of indivisible items, for example: X = {piano, car, apartment}, such that each item should be given entirely to a single person. X may be an infinite set representing a divisible resource, for example: money, or a cake. Mathematically, a divisible resource is often modeled as a subset of a real space. For example, the section [0, 1] may represent a long narrow cake, that has to be cut into parallel pieces. The unit disk may represent an apple pie. 3

4 The set to be divided may be homogeneous - such as money, where only the amount matters. heterogeneous - such as a cake that may have different ingredients, different icings, etc. In the general case, different parts may be valued differently by different people. The items to be divided may be desirable - such as a car or a cake. undesirable - such as house works (cleaning floor, washing dishes). 4

5 Desirability, divisibility and homogeneity properties of items When dividing inheritance, or dividing household property during divorce, it is common to have desirable divisible heterogeneous property such as land, and desirable divisible homogeneous property such as money. In the housemates problem, several friends rent a house together, and they have to both allocate the rooms in the apartment (a set of indivisible, heterogeneous, desirable goods), and divide the rent to pay (divisible, homogeneous, undesirable good). This problem is also called the room assignment-rent division. 5

6 Subjective fairness According to the subjective theory of value, there cannot be an objective measure of the value of each item. Therefore, objective fairness is not possible, as different people may assign different values to each item. The most current research on fairness focuses on concepts of subjective fairness. The presence of different measures opens a vast potential for many challenging questions. The i th person in the group of n persons is assumed to have a personal subjective utility function or value function, V i, which assigns a numerical value to each subset of X. Usually the functions are assumed to be normalized, so that every person values the empty set as 0 [V i ( ) = 0 for all i], and the entire set of items as 1 [V i (X) = 1 for all i] if the items are desirable, and 1 if the items are undesirable. 6

7 Examples 1. For the set of indivisible items {piano, car, apartment}, Alice may assign a value of 3 1 to each item, which means that each item is important to her just the same as any other item. Bob may assign the value of 1 to the set X = {car, apartment}, and the value 0 to all other sets except X. This means that he wants to get only the car and the apartment together. The car alone or the apartment alone, or each of them together with the piano, is worthless to him. 2. If X is a long narrow cake (modeled as the interval [0, 1]), then Alice may assign each subset a value proportional to its length, which means that she wants as much cake as possible, regardless of the icings. Bob may assign value only to subsets of [0.4, 0.8] since this part of the cake contains cherries and Bob only cares about cherries. 7

8 Notions of fair divisions 1. A proportional division, also called simple fair division, means that every person gets at least his due share according to his own value function. For instance, if three people divide up a cake, each gets at least a third by their own valuation. That is, each of the n people gets a subset of X which he values as at least n 1 : V i(x i ) n 1 for all i. It is said to be strong fair division if V i (X i ) > n 1 for all i. Obviously, strong fair division is not possible if all measures are equal. 2. An envy-free division guarantees that no-one will want somebody else s share more than their own. That is, every person gets a share that he values at least as much as all other shares: V i (X i ) V i (X j ) for all i and j. In simple language, envy-free means each player receives a piece he or she would not swap for that received by any other players. Dirty work envy-free if V i (X i ) V i (X j ) for all i and j (each gets an envy-free task). 8

9 Super envy-free division means V i (X j ) < 1 wherever i j, 1 n i, j n (each feels everyone else receives less than 1/n), so V i (X i ) > 1 n. 3. An equitable division means each person s subjective valuation of the piece that he receives is the same as the other person s subjective valuation. V i (X i ) = V j (X j ) for all i and j. Equitability may not imply envyfreeness since the subjective valuations of the players may differ. An allocation where each agent assigns value 0 to its own piece and value 1 to another piece is equitable but not proportional (hence not envy free). Most envy free allocations (hence proportional) would not satisfy the stringent equality constraint that equitability requires. 9

10 Proposition Suppose X = (X 1, X 2,..., X n ) is a complete allocation. If an allocation is envy-free, then it must also be proportional. In other words, envy-freeness is the stronger notion of fairness. envy-freeness proportional division We prove by contradiction. Suppose that V i (X i ) < n 1 for some i. Since the allocation is complete, we deduce that V i (X X i ) > n 1 n. Pigeon hole principle: If n items are put into m containers, with n > m, then at least one container must contain more than one item. By the pigeon hole principle, this implies that V i (X j ) > n 1 for some j i. This would give V i (X i ) < V i (X j ), contradicting envy-freeness. Hence, V i (X i ) n 1 for all i. 10

11 For two agents, proportionality and envy-freeness are equivalent. Suppose V i (X i ) 1 2, i = 1, 2, then V 1(X 2 ) 1 2 and V 2(X 1 ) 1 2, so it is envy-free. Note that there is no such equivalence when there are three or more players. It is still possible that player i may think player j receives X j where V i (X j ) > V i (X i ) while V i (X i ) 1 n, n 3. 11

12 Efficient allocation (Pareto optimal) An allocation is efficient if there is no other allocation that is strictly better for at least one player and as good for all the others. A division where one player gets the whole set is optimal by this definition. An efficient allocation needs not be proportional, envyfree or equitable. Suppose player A places no value at all on a portion to which some other player B attaches some value. Taking away that portion from A would keep the same valuation value for the new allocation for A but the new piece received by the other player B gives a higher value for B. The allocation is seen to be Pareto nonoptimal. 12

13 1.2 Procedures for two-player and multi-player cake-cutting A fair division procedure lists the actions to be performed by the players based on available data and their valuations. Where an action depends on a player s valuation, the procedure is describing the strategy that a rational player will follow. A valid procedure is one that guarantees a fair division for every player who acts rationally according to their valuations. Procedures can be divided into finite and continuous procedures. A finite procedure would only involve one person at a time cutting a cake. Continuous procedures may involve one player moving a knife and the other saying stop. 13

14 Operational properties Does the procedure guarantee that each agent receives a single continguous slice (rather than the union of several subintervals)? We prefer continguous procedures, which also minimize the number of cuts to be made. Note that a procedure for n players will require at least n 1 cuts. If the number of cuts is not minimal, can we provide an upper bound on the number of cuts? Does the procedure require an active referee, or can all actions be performed by the players themselves? 14

15 Cake-cutting problems Cake-cutting is the problem of fair division of a single divisible and heterogeneous good between n players. The cake is represented by the unit interval [0, 1]: Each agent i has a valuation function V i defined for each subinterval of [0, 1]. 15

16 Properties of a valuation function Non-negativity: V i (B) 0 for all B [0, 1] Normalization: V i ( ) = 0 and V i ([0, 1]) = 1 Additivity: V i (B B ) = V i (B) + V i (B ) for disjoint B, B [0, 1] V i is continuous: The Intermediate-Value Theorem applies and single points do not have any value. The Intermediate-Value Theorem states that if a continuous function f with an interval [a, b] as its domain takes values f(a) and f(b) at each end of the interval, then it also takes any value between f(a) and f(b) at some point within the interval. Application: Suppose player i assigns a subinterval S with a value less than 1/n. Recall V i (S) < 1 n and V i([0, 1]) = 1. By the Intermediate-Value Theorem, there exists a subinterval that is enlarged continuously from S whose value to player i is exactly 1/n. 16

17 Discrete cut-and-choose procedures The classical approach for dividing a (heterogeneous) cake between two agents: One agent (chosen at random) cuts the cake in two pieces (she considers to be of equal value based on her valuation), and the other chooses one of them (the piece he prefers). The chooser always takes the piece with higher or at least equal valuation. The cutter is indifferent to the two pieces. Therefore, the procedure is not equitable. However, it satisfies Proportionality : Each agent is guaranteed at least one half according to her own valuation. Envy-freeness: No agent will envy the other. Even if the role of the cutter is determined by the flip of a coin, which is presumably a fair procedure not favoring either player, the cutter may believe that the chooser has a definite advantage. This is a consequence of the failure of equitability. 17

18 Extension to 3 agents: Steinhaus procedure 1. Agent 1 (chosen at random) cuts the cake into three pieces (which she values equally). 2. Agent 2 passes (if she thinks at least two of the pieces are 1 3 ) or labels two of them as bad. If agent 2 passed, then agents 3, 2, 1 each choose a piece (in that order). The procedure is then completed. 3. If agent 2 did not pass, then agent 3 can also choose between passing and labelling. If agent 3 passed, then agents 2, 3, 1 each chooses a piece (in that order) and we are done. In other words, the one who passes would not choose first. 4. If neither agent 2 nor agent 3 passed, then agent 1 has to take (one of) the piece(s) labelled as bad by both 2 and 3. The rest is reassembled, then 2 and 3 play cut-and-choose. 18

19 All valuations of the divisions are personal, which may differ among the 3 players. 19

20 There are two mutually exclusive cases that can arise (since it is impossible for either Agent 2 or Agent 3 to think that all 3 pieces cut by Agent 1 are unacceptable). 1. One of Agent 2 or Agent 3 passed (that is, two or more of the pieces are acceptable). Suppose Agent 2 passed, then A- gent 3 takes any of the pieces that he considers acceptable. Agent 2 has available at least one of the pieces she considered acceptable. Lastly, Agent 1 takes the remaining piece that is acceptable since he made the cut initially. The same argument can be applied when Agent 3 passed. 2. Both Agents 2 and 3 did not pass, and this indicates that at most one piece is acceptable. In this case, there is at least one piece that both agents agree that it is not acceptable. This non-acceptable piece is given to Agent 1. Now, both Agents 2 and 3 would think they are resembling more than 2/3 of the cake to divide between them. 20

21 Properties of the Steinhaus procedure It guarantees a proportional division of the cake, where V i (X i ) 1, i = 1, 2, 3. 3 It is not envy-free. Suppose Agent 2 passed, it may be possible that Agent 2 may envy Agent 3 if Agent 3 may choose the larger of the two pieces that Agent 2 considered acceptable. In another case, the cut-and-choose played by Agent 2 and 3 may not be 50 50% in Agent 1 s own valuation. In this case, there exists another piece received by Agent 2 or Agent 3 that has player 1 s valuation higher than 1/3. This violates envy-free division. The resulting pieces might not be contiguous. If both Agents 2 and 3 label the middle piece as bad and Agent 1 takes it. One additional cut is required if the cut-and-choose cut is different from Agent 1 s original cut. 21

22 Proportional procedure with trimming for arbitrary n players (Banach-Knaster) Step 1. Step 2. Step 3. Aside. Step 4. Player 1 cuts a piece P 1 (of size 1/n) from the cake. Player 2 is given the choice of either passing (which he does if he thinks P 1 is of size less than 1/n), or trimming a piece from P 1 to create a smaller piece (that he thinks is of size exactly 1/n). The piece P 1, now perhaps trimmed, is renamed P 2. The trimmings are set aside. For 3 i n, Player i takes the piece P i 1 and proceeds exactly as Player 2 did in Step 2, with the resulting piece now called P i. For 1 i n, Player i thinks that P i is of size less than or equal to 1/n. We also have that P 1 P n. Thus, every player thinks P n is of size at most 1/n. The last player to trim the piece, or Player 1 if no one trimmed it, is given P n. 22

23 Aside. Step 5. Aside. Step 6. The player receiving P n thinks it is of size exactly 1/n. The trimmings are reassembled, and Steps 1-4 are repeated for the remainder of the cake, and with the remaining n 1 players in place of the original n players. The player who gets a piece at this second stage is getting exactly 1/(n 1) of the remainder of the cake. He, and everyone else, thinks this remainder is of size at least (n 1)/n. Hence, he thinks his piece is of size at least 1/n. Step 5 is iterated until there are only 2 players left. The last 2 players use the cut-and-choose procedure. 23

24 Summary 1. In the first round, we observe P n P n 1 P 1, and P n = 1 n to the one who receives the final piece. 2. The same procedure is repeated with the remainder part n 1 n to every one staying behind in the division game. 3. Repeating the procedure until down to 2 players, which is then finally settled by the cut-and-choose procedure. Proportionality is guaranteed since every player receives a piece that he thinks to be of size at least 1/n. There is no guarantee on envy free, and equitability is not under consideration in this procedure. 24

25 Envy-free discrete cuts procedure for 3-player division (Selfridge- Conway) This is a combination of the trimming idea of Banach-Knaster and the basic framework of Steinhaus. Initialization: 1. Agent 1 divides the cake into three equally-valued pieces X 1, X 2, X 3 : V 1 (X 1 ) = V 1 (X 2 ) = V 1 (X 3 ) = 1/3. 2. Agent 2 trims the most valuable piece according to V 2 to create a tie for most valuable. For example, if V 2 (X 1 ) > V 2 (X 2 ) V 2 (X 3 ), agent 2 removes X X 1 such that V 2 (X 1 \X ) = V 2 (X 2 ). We call the three pieces one of which is trimmed cake 1 (X 1 \X, X 2, X 3 in the example), and we call the trimmings cake 2 (X in the example). 25

26 Exact division for Agent 1 V 1 (X 1 ) = V 1 (X 2 ) = V 1 (X 3 ) = 1 3 Cake 1 V 2 (X 1 ) > V 2 (X 2 ) V 2 (X 3 ) Trim X from X 1 so that V 2 (X 1 \X ) = V 2 (X 2 ) 26

27 Division of cake 1: Agent 3 chooses first from one of the three pieces of cake 1. If Agent 3 chose X 1 \X, then Agent 2 chooses between the two other pieces of cake 1. Otherwise, Agent 2 receives X 1 \X. Between Agents 2 and 3, we call them T and T according to Agent T takes X 1 \X ; Agent T the other person. Agent 1 receives the remaining piece of cake 1 (always an untrimmed piece). Division of cake 2: Agent T divides cake 2 into three equally-valued pieces. Agents T, 1 and T select a piece of cake 2 each, in that order. 27

28 Proof of envy-freeness The division of cake 1 is clearly envy free: Agent 3 chooses first; agent 2 receives one of the two pieces that it views as tied for largest; and agent 1 definitely receives an untrimmed piece, which he also views as tied for largest. Now consider the division of cake 2. Agent T (who received the trimmed piece) chooses first, and agent T is indifferent between the three pieces. Agent 1 will never envy the combined pieces received by T (even if T received all of cake 2). Combining envy-free divisions of the two disjoint pieces of cake yields an envy-free division of the combined cake. Hence, agents T and T are not envious overall. 28

29 Continuous moving knife procedures Single-knife procedure: proportional but not envy-free Suppose there are three kids who are to split the cake. One strategy is for Mom to place a knife over one corner of the cake and begin to move it slowly across the cake. When any of the kids says stop, that kid (K1 let s say) gets the piece. Presumably K1 thinks she got at least 3 1 and K2 and K3 (who did not speak up) believe that remainder is at least 2 3. Note that K 1 should be refrained from saying stop too late. This is because she runs into the risk of not being able to achieve 1/3 with the remaining portion with valuation less than 2/3 when K 2 or K 3 initiates the first call of stop. Mom keeps on moving the knife, until K2 says stop. Now, K2 is happy, because he thinks he got at least and the same for K3. The division is seen to be proportional where every player envisions to receive at least 1/3. K1 might be envious, say, K1 might think the piece received by K3 has a value more than 1 3. Although each is guaranteed to receive at least 1/3 in his evaluation, there might not be an envy-free division. 29

30 Austin s two-knife procedure: equitable division for two players Player 1 moves the two knives from left to right so that the piece between the knives remains of size 1/2 in his opinion. 30

31 Procedure There is a single knife that moves slowly across the cake from the left edge toward the right edge, until one of the players (say, player 1) calls stop (at the point when the piece so determined is of size exactly 1/2). At this time, a second knife is placed at the left edge of the cake. Player 1 then moves both knives across the cake in parallel fashion (in such a way that the piece between the two knives remains of size exactly 1/2 in player 1 s measure), subject to the requirement that when the knife on the right arrives at the right-hand edge of the cake, the left-hand knife lines up with the position that the first knife was in at the moment when player 1 first called stop. While the two knives are moving, player 2 can call stop at any time (which he does precisely when the measure of the piece between the two knives is of size exactly 1/2 in his measure). What guarantees that there will be a point where player 2 thinks the piece between the knives is of size exactly 1/2? 31

32 Notice that at the instant when the two knives start moving, player 2 thinks the piece between the knives is of size strictly less than 1/2 (since player 2 does not initiate stop ). If the first knife were to reach the right edge of the cake, the piece between the knives would be the complement of what it was when the knives started moving. Hence, player 2 would think the piece between the knives is now of measure strictly greater than 1/2. Thus, with an appropriate continuity assumption, there must have been a point where the measure of the piece between the knives is exactly 1/2. Query Can someone game around the procedure by delaying the call to stop? The piece between the two knives has evaluation higher than 1/2 since the point of evaluation of 1/2 has passed. As soon as the two parallel knives start moving (sooner the better), call stop. If both players adopt this strategy, then the first knife will travel from the left side all the way to the right side of the cake. 32

33 Equitable division for two players and general fractions The two partners can find a single piece of cake that both of them value as exactly 1 k, for any integer k 2. We call this procedure Cut 2 ( 1 k ). Alice makes k 1 parallel marks on the cake such that k pieces so determined have a value of exactly 1 k. If there is a piece that George also values as 1 k, then we are done. Otherwise, there must be a piece that George values as less than 1 k, and an adjacent piece that George values as more than 1 k. It is not possible to have all adjacent pairs to be more than 1 k or less than 1 k. Let Alice place two knives on the two marks of one of these pieces, and move them in parallel, keeping the value between them at exactly 1 k, until they meet the marks of the other piece. There must be a point at which George agrees that the value between the knives is exactly 1 k. 33

34 By recursively applying Cut 2, the two partners can divide the entire cake to k pieces, each of which is worth exactly 1 k for both of them. Use Cut 2 ( 1 k ) to cut a piece which is worth exactly 1 k partners. for both Now the remaining cake is worth exactly k 1 k for both partners; use Cut 2 ( k 1 1 ) to cut another piece worth exactly 1 k of the original cake for both partners. Continue the procedure until there are k pieces. 34

35 Multiple partners It is possible to divide a cake to n partners, such that each partner receives a piece worth exactly n 1 for him. Partners #1 and #2 use Cut 2 ( 2 1 ) to give each one of them a piece worth exactly 1 2 for them. Partner #3 uses Cut 2 ( 1 3 ) with partner #1 to get exactly 3 1 of his share and then Cut 2 ( 1 3 ) with partner #2 to get exactly 3 1 of her share. The first piece is worth exactly 1 6 for partner #1 and so partner #1 remains with exactly 3 1 ; the same is true for partner #2. As for partner #3, he gets 3 1 of the two pieces whose sum is one, so he gets exactly 1 3 of the entire cake. 35

36 Envy-free moving-knife procedure for three players 1. Webb s procedure (combined with Austin s procedure) Step 1: Step 2: An unbiased fourth party slowly moves a knife across the cake until someone yells Cut! to indicate that he or she values the piece to be cut off at one-third of the cake. Suppose that Annie is the one who yells cut, and let P 1 represent the piece of cake that is cut off. Annie and Ben now use Austin s procedure to divide the remaining cake into two pieces that they both consider equally valuable. Let P 2 and P 3 denote these two pieces. Step 3: Chris chooses first from the three pieces P 1, P 2, and P 3. Ben chooses next, and Annie chooses last. 36

37 Envy freeness Chris envies no one since he gets to choose first. Since Annie yelled cut the first time, she believes that P 1 is exactly one-third of the cake. She thinks that P 2 and P 3 are equally valuable and together are worth two-thirds of the cake, so she thinks P 2 and P 3 are each exactly one-third of the cake as well. Since she considers each of the three pieces to be equally valuable, she envies no one. Finally, Ben considers P 1 to be less than one-third the cake since he was not the one to yell cut. So he thinks P 2 and P 3 together make up more than two-thirds of the cake. So Ben values P 2 and P 3 equally, and strictly more than P 1. Since Ben chooses second, at least one of P 2 and P 3 will be available, so he envies no one. 37

38 2. Stromquist s procedure with minimum cuts It requires only two cuts, the minimum for three pieces. There is no natural generalization to more than three players which divides the cake without extra cuts. For example, 11 cuts are required for 4-player envy free moving knife procedure. The resulting partition is not necessarily efficient. For example, it cannot produce the efficient allocation of cutting a cake when the vanilla strips are on the two edges, while the chocolate strip and banana strip are in the middle. Suppose Alice only favors chocolate, Ben only favors banana and Chris favors only vanilla. The allocation that allocates the parts according to the sole flavor is the only efficient allocation since players valuations of all other allocations can always be improved by choosing this efficient allocation, where V i (X i ) = 1, i = 1, 2, 3. However, this efficient allocation cannot be achieved by this 3-player moving knife procedure. 38

39 A referee moves a sword from left to right over the cake, hypothetically dividing it into a small left piece and a large right piece. Each player moves a knife over the right piece, always keeping it parallel to the sword. The players must move their knives in a continuous manner without making any jumps. When any player shouts cut, the cake is cut by the sword and by whichever of the players knives happens to be the central one of the three (that is, the second in order from the sword). 39

40 The cake is divided in the following manner: The piece to the left of the sword, which we denote Left, is given to the player who first shouted cut. We call this player the shouter and the other two players the quieters. The piece between the sword and the central knife, which we denote Middle, is given to the remaining player whose knife is closest to the sword. The remaining piece, Right, is given to the third player. 40

41 Strategy Each player can act according to his own measure that guarantees no other player receives more than him based on his personal valuation. Always hold your knife such that it divides the part to the right of the sword to two pieces that are equal in your eyes (hence, your knife initially divides the entire cake to two equal parts and then moves rightwards as the sword moves rightwards). Shout cut when Left becomes equal to the piece you are about to receive if you remain quiet. That is, if your knife is leftmost, shout out if Left = Middle; if your knife is rightmost, shout if Left = Right; if your knife is central, shout out if Left = Middle = Right. If the player does not shout, she may receive a smaller piece with certain probability though delaying shout might yield a larger piece. Based on the risk averse assumption of the player, the players should play honestly. 41

42 Envy-free share First, consider the two quieters. Each of them receives a piece that contains his knife, so they do not envy each other. Additionally, because they remained quiet, the piece they receive is larger in their eyes then Left, so they also do not envy the shouter. The shouter receives Left, which is equal to the piece he could receive by remaining silent and larger than the third piece. Hence, the shouter does not envy any of the quieters. Following this strategy each person gets the largest of one of the largest pieces by their own valuation. Therefore, the division is envy-free. 42

43 1.3 Adjusted winner procedures for two-party allocation of discrete goods The adjusted winner procedure is a method of dispute resolution (division of individual goods) for two parties that guarantee an outcome that is envy-free, equitable and efficient. Suppose that Annie and Ben are getting divorced. Each party has 100 points to distribute over all the items according to which they value most. Annie and Ben s point distribution are below. Annie Item Ben 35 House Investments Piano 25 5 TV Dog 10 5 Car Total

44 Stability of a division procedure Fairness Difficulty of manipulating a procedure that produces a division (providing intrinsic incentive to be truthful about one s evaluations of item values) Two-stage division During the first stage, each item is initially awarded to the person who values it most. So Annie receives the house and the dog, and Ben receives the investment account, baby grand piano, plasma TV, and the car. At this point, Annie has 60 points, and Ben has 75 points. Since Ben has more points, we say that Ben is the initial winner. The next stage is the equitability adjustment. We need to transfer items, or fractions thereof, from Ben to Annie until the point totals of each are equal and the allocation is thus equitable. 44

45 The order of the items to be transferred is important. To determine the order, for each of Ben s items, we consider the ratio of the points assigned by Ben to the item to the points assigned by Annie to the item. Note that each of these ratios will be at least 1, since Ben received the items to which he had assigned more points. The ratios for each of Ben s items are as follows: Investment : Piano : TV : Car : = = = = 2 45

46 The transfer of items starts with the item for which the ratio above is the smallest, then the next smallest, and so on. Intuitively, this is the fairest way to proceed since the cost to Ben per point transferred to Annie is smallest. For example, transferring the TV requires lowering Ben s point total by 3 points for every 1 point transferred to Annie, while transferring the car would only lower Ben s point total by 2 for every 1 point transferred to Annie. This order of transfer is crucial to the proof that the resulting allocation is efficient. 46

47 We start with the ratio for the Investment, since it is the smallest. Notice that if we were to transfer the entire investment portfolio to Annie, then Annie would have more points than Ben. Let x be the fraction of the investments transferred to Annie, so that 1 x is the fraction retained by Ben. After the transfer, Annie will have 60 points (from the house and dog) plus 20x (her portion of the investments), while Ben will have 50 points (from the piano, TV, and car) plus 25(1 x) (his portion of the investments). To guarantee that the resulting point totals are equal, we need to ensure that x = (1 x) = 75 25x. Solving the equation, we obtain x = 3 1. Annie receives the house, the dog, and one-third of the investment portfolio, while Ben keeps the piano, TV, car and two-thirds of the investments. Each person walks away with an impressive total of over half the total value. points, well 47

48 Indivisible goods If we had needed to split the piano, it certainly would not be simple since a third of a piano is not very valuable to anyone! Together, then Annie and Ben might decide to sell the piano and split the profits according to the prescribed proportions. Or they might decide that if Annie receives the larger half, they will sell the piano, but if Ben receives the larger share, he will buy out Annie s share. 48

49 Summary of steps 1. Each item, for which there is no tie in point values, is initially awarded to the party who awarded it more points. Next, in any order and one at a time, the tied items are given to whomever has fewer points at the time. If the point totals are equal, then a tied item can be given to either party. 2. If the point totals of each party are equal at the end of Step 1, then the procedure is done; an equitable allocation has been achieved. Otherwise, the equitability adjustment (Step 3) occurs. 3. Call the party with more points at the end of Step 1 the initial winner. Calculate the ratio, for each item awarded to the initial winner during Step 1, of the points awarded to the item by the initial winner to the points awarded to the item by the other party. In order to ascending ratios, transfer items (or fractions thereof) from the initial winner to the other party until the point totals are equal. 49

50 Since only finitely many items are under dispute, the procedure does in fact end. If after an item is transferred, the initial winner still has the larger point total, then the next item is transferred. If transferring an item results in equal point total, then the procedure is finished. If transferring an item would result in the initial winner having fewer points than the other party, then that item must be split; the procedure is then finished. The procedure can be modified in the case of unequal entitlements, for instance if a prenuptial agreement indicated that the shared property be divided 60%-40%. 50

51 Israeli-Palestinian conflict in the Middle East 1. West Bank: Several areas of the West bank are inhabited by Israelis who have no desire to leave their homes. The Palestinians, however, believe that these settlements are illegal, and that the Israelis should evacuate. 2. East Jerusalem: In 1967, Israel unified control over all the Jerusalem by defeating Jordanian forces in the Six Days War. A majority of the residents of east Jerusalem are Palestinian, however, and both Israelis and Palestinians argue that East Jerusalem is central to their sovereignty. 3. Palestinian Refugees: Israel has refused to recognize that its establishment and expansion in 1948 and 1967 displace Palestinian villages and communities. The Palestinians insist that Israel recognizes the refugees right to return to Israel, and provides compensation for the refugees and to Arab states that have hosted the refugees. 51

52 4. Palestinian Sovereignty : Israel does not recognize Palestine as a sovereign nation. 5. Security : Some Israelis fear that terrorism would flourish under a Palestinian state that lacks the means to effectively fight terrorism. Specific security issues include: border control, control of airspace, security in Jerusalem, and early warning stations in the West Bank and Gaza that would assuage Israeli concerns against surprise attacks but provide insufficient military capability to threaten Palestinian forces. 52

53 Point allocation By examining the expert opinions, interim agreements, and working plans, one may arrive at the following reasonable estimates of possible point allocations by each side. Israel Item Palestine 22 West Bank East Jerusalem Palestinian Refugees Palestinian Sovereignty Security Total

54 In the first stage of the adjusted winner procedure, Israel wins the issues of the West Bank, East Jerusalem and security, while Palestine wins the issues of refugees and sovereignty. After the first stage, Israel has 73 points and Palestine has 42 points. Since Israel is the initial winner, then we look at the ratios of points for the issues won by Israel: West Bank : East Jerusalem : Security : The equitability adjustment begins with the West Bank since < <. Transferring the entire West Bank would give the Palestinians more points than the Israelis. 54

55 To determine the percentage x of the West Bank retained by Israel, we solve for x in the following equation: x = (1 x) = 63 21x 43x = 12 x = The Israelis are left with the issues of East Jerusalem, security, and roughly 2 7 of the issue of the West Bank. The Palestinians are left with the issues of refugees, sovereignty, and roughly 5 7 of the issues of the West Bank. 55

56 Equitability, efficiency and envy-freeness Equitability : The procedure is equitable by design. The procedure ends when the point totals of each party are equal. Efficiency (in Pareto sense): There exist no other allocations that give higher point to one player and at least as good for the other player when compared to the allocations based on the adjusted winner procedure. Envy-freeness: This property follows from the other two when exactly two parties are involved. 56

57 Proof of envy-freeness We prove by contradiction. Suppose that the allocation is equitable and efficient, but not envy-free. Since envy-freeness and proportionality are equivalent for two parties, then it must be the case that at least one of the parties received less than half according to his own valuation. Note that equitability then implies that both parties received less than half. This allocation is not efficient because we can find another division in which both players do better: give each party s share to the other party. If each party originally received x points, where x < 50, then now each receives 100 x > 50 points, so this allocation is strictly better for both parties involved, contradicting efficiency of the original division. 57

58 Efficiency in the adjusted winner procedure In the first stage of adjusted winner, every item is first given to the person who valued it most. Items are then transferred from the initial winner to the other party until both have an equal number of points. The proof of efficiency hinges on the order in which the items are transferred: the transfer begins with the item with the smallest ratio of points given by the initial winner to points given by the other party. In this way, we minimize the effective cost to the initial winner for all points transferred to the other party. Intuitively, the adjusted winner procedure is efficient since the initial stage of the procedure is efficient. It suffices to show that efficiency is not affected during the equitablility adjustment. 58

59 Parties: Annie and Ben. Items: G 1,..., G n to be divided between Annie and Ben. For G i, fraction a i goes for Annie and fraction b i goes for Ben. A i = points allocated by Annie for G i B i = points allocated by Ben for G i Lemma 1 Suppose that we have an allocation of the items in which (i) Annie values item G i at least as much as Ben does (ii) Ben values item G j at least as much as Annie does Suppose that Annie trades her portion of G i for Ben s portion of G j. If this trade is strictly better for one player, then it is strictly worse for the other. 59

60 Proof Since Annie values item G i at least as much as Ben does, then we know that A i B i. Similarly, since Ben values item G j at least as much as Annie does, then we know that B j A j. We can ignore all items except G i and G j since they are not involved in the trade. During the trade, Annie gives away a total of a i A i points, and gains a total of b j A j points. If the trade is strictly better for Annie, then b j A j > a i A i. (1) We compare Ben s points before or after trade, where Ben s points after trade Ben s point before trade = a i B i b j B j a i A i b j A j since B j A j and B i A i < 0 by virtue of (1), so Ben is strictly worse off after the trade. Similarly, if the trade is strictly better for Ben, then it is strictly worse for Annie. 60

61 Lemma 2 Suppose that we have an allocation of the items in which A j B A i j B. i If Annie trades her portion of G i for Ben s portion of G j, and this trade is strictly better for one player, then the trade is strictly worse for the other. Proof If the trade is better for Annie, then b j A j > a i A i. then A j B i A i B j. Now, consider Since A j B j A i B i, Ben s points after trade Ben s point before trade = a i B i b j B j < B i ( bj A j A i ) = b j ( Bi A j B j A i A i b j B j since b j A j > a i A i ) 0 since A j B i A i B j, so Ben is strictly worse off after the trade. 61

62 If the trade strictly benefits Ben, however, then it follows that a i B i > b j B j. We then have Annie s points after trade Annie s point before trade = b j A j a i A i < b j A j A i ( bj B j B i = b j ( Aj B i A i B j B i ) ) 0 since A j B i A i B j, since a i B i > b j B j so Annie is strictly worse off after the trade. 62

63 Lemma 3 If a given allocation is not efficient, then there exist goods G i and G j and some portions thereof such that if Annie exchanges her fraction a i of G i for Ben s fraction b j of G j, the resulting trade yields an allocation that is at least as good for both players and strictly better for at least one of the players. Proof Since the given allocation is not efficient, there is an alternative allocation that is at least as good for both Annie and Ben and strictly better for a least one of the two, say Annie. It suffices to show that there exist disjoint sets S and T of goods belonging to Annie and Ben, respectively, such that an exchange of S and T makes Annie better off without hurting Ben. We just need to show that S and T can each be taken to be (possibly a fraction of) a single item. 63

64 Assumption of weak additivity of preferences: If A and B are disjoints sets of goods, and Annie values A at least as much as some set X of goods and B at least as much as some set Y of goods, then she must value A B at least as much as X Y. Write S = S 1... S n, where S i s are pairwise disjoint, and each is a fraction of a single item. Ben can now break up T into a disjoint union T = T 1... T n (not necessarily subsets of a single item) such that an exchange of S i for T i yields an allocation that is no worse for him than the current allocation. It suffices for Ben to choose each T i such that the value to Ben of T i is no more than the value to Ben of S i. 64

65 There exists an i such that Annie prefers the allocation obtained by exchanging S i for T i to the existing allocation. If such an i did not exist, then the existing allocation is at least as good for Annie as the one obtained by exchanging S i for T i for all i. By additivity of preferences again, then the existing allocation is at least as good for Annie as the one obtained by exchanging S 1... S n = S for T... T n = T, contrary to the assumption that an exchange of S and T makes Annie better off without hurting Ben. Relabeling if necessary, suppose that Annie prefers the allocation obtained by exchanging S 1 for T 1 to the existing allocation. Now S 1 consists of some portion of a single good, but T 1 may consist of portions of several goods. 65

66 Division of T 1 into a disjoint union Write T 1 as a disjoint union of sets T T 1m and S 1 as a disjoint union of sets S S 1m such that each T 1j is some portion of a single good and Annie is better off with the allocation obtained by exchanging S 1j for T 1j than with the existing allocation. By the same reasoning, there must exist a j such that the allocation obtained by exchanging S 1j for T 1j is at least as good for Ben as the existing allocation. Otherwise, the existing allocation is better for Ben than the one obtained by exchanging S 1j for T 1j for all j. It follows by additivity of preferences that the existing allocation is better for Ben than the one obtained by exchanging S 1 for T 1, which is a contradiction. Thus, we have found subsets S 1j for T 1j each consisting of a portion of a single item for which a trade of S 1j for T 1j yields an allocation that is strictly better for Annie and no worse for Ben than the existing allocation. 66

67 Proof of efficiency We prove by contradiction. Suppose otherwise, by Lemma 3, there exist goods G i and G j and portions thereof such that if Annie exchanges her fraction a i of G i for Ben s fraction b j of G j, the resulting trade yields an allocation that is at least as good for both and strictly better for at least one. Suppose that Annie was the initial winner after the first step of the adjusted winner procedure. Since Annie still has at least a i of item G i after any necessary transfers, then Annie must value item G i at least as much as Ben does, so A i B i. Now if Ben values item G j at least as much as Annie does, then Lemma 1 implies that the trade will not benefit both parties as we are assuming. 67

68 It must be the case that Ben values item G i less than Annie does, that is, B j < A j. But Ben has part of G j, so he must have received that during the transfer stage of the adjusted winner procedure. Since only one item is split among the parties during adjusted winner, then G j must be that item, so Annie must have all of item G i. Since item G i was not involved in the transfer stage, it follows that the ratio of points for item G i is the adjusted winner procedure is at least as big as the ratio of points for item G j. Thus A i B A j i B. j By Lemma 2, this contradicts our assumption that the trade does not hurt neither party. 68

69 Manipulability Determining point totals is itself not an easy task. The situation is still more stressful if the parties involved need to worry about strategies as well, especially in the case of a divorce where each party has in depth knowledge of the other s like and dislike. It is natural to wonder whether this knowledge would enable one party to manipulate the system, and achieve a better outcome by submitting dishonest point allocations. Unless knowledge of the other s party s valuations is strictly onesided, then honesty is the best policy in the adjusted winner procedure. 69

70 Example: Honesty is the best policy Suppose that Annie and Ben are getting a divorce, and currently share the following items: a townhouse in Central Square, season passes to the Red Sox, and a painting by Klee. They value the items as follows: Annie Item Ben 50 Townhouse Red Sox Tickets Klee painting Total

71 Applying the adjusted winner procedure, we see that Annie is initially awarded the townhouse and the painting, while Ben gets the Red Sox tickets. Annie currently has 80 points, while Ben has 50, so Annie is the initial winner. The ratio of points for the townhouse is 5 3, while the ration for the painting is 3 2, so the painting needs to be divided. Solving for x in the following equation gives the fraction of the painting that Annie keeps: x = (1 x) = 70 20x giving x = 2 5. Annie ends up with the townhouse and 2 5 of the painting (Annie and Ben decide that she will buy out his share of the painting), and Ben gets the Red Sox tickets and 3 5 of the painting. Each with a total of 62 points. 71

72 Now Annie has known Ben for ten years, and knows how much his Red Sox tickets mean to him. She is confident that she can estimate Ben s point allocations fairly well, and decides to submit the following false valuations, rather than her true preferences given above Annie s fake valuations Item Ben s sincere valuations 32 Townhouse Red Sox Tickets Klee painting Total

73 Intuitively, Annie might do better under this scenario. By indicating that she values the townhouse only slightly more than Ben, she hopes to win the townhouse but at a lower cost, thereby winning a higher percentage of the painting as well. In the first step of the process, Annie still gets the townhouse and the painting, and Ben gets the Red Sox tickets. Annie has 52 points (according to her false point allocations), and Ben has 50. Solving for x give the fraction of the painting that Annie keeps: x = (1 x) = 70 20x giving x = By lowering the point to the Townhouse as much as possible, Annie has 18 = points of difference. Furthermore, she gains more by lowering the point to painting from 30 to 20. This is achieved by increasing the point to tickets. 73

74 With this kind of knowledge on both sides, it becomes much riskier to submit false preferences. While it may be to someone s advantage to be dishonest (Annie might still get lucky if Ben chooses to submit his true point allocations even with knowledge of Annie s preferences), this strategy can also backfire, resulting in an outcome that is worse than the honest outcome. For example, if Ben thinks that Annie will be honest, he may submit the following point allocations: Annie s sincere valuations Item Ben s fake valuations 50 Townhouse Red Sox Tickets Klee painting Total

75 If Annie were honest, then Ben and Annie would each get 52.5 points, though this would really constitute over 68 points for Ben according to his true valuations. The resulting point is higher than 62 if both are honest. But if Annie and Ben both submit these false preferences, the result is not good for either. In the first step of the process, Annie receives the Red Sox tickets, and Ben gets the townhouse and the painting. The ratio for the painting is 1.5 while the ratio for the townhouse is 45 32, strictly less. The following calculation gives the fraction of the townhouse to be given to Annie: x = (1 x) = 75 45x giving x =

76 So Annie gets just over a third of the townhouse and the Red Sox tickets, while Ben gets just under 2 3 of the townhouse and the painting. Although this appears to be just over 59 points for each with the false point allocations, both Annie and Ben do much worse according to their true preferences. Annie s share give her , roughly 37.5 points and Ben s share gives him , just under 39.5 points. Both Annie and Ben would have fared much better had they been honest! Final remark In addition to guaranteeing an allocation that is envy-free, equitable, and efficient, the adjusted winner procedure also promotes honesty. This is true at least when knowledge of the other party s preference is not strictly one-sided. 76

77 1.4 Matching schemes 2012 Nobel Awards in Economics go to Shapley and Roth on their works on matching schemes We consider matching between two sets of elements given a set of preferences for each element. A matching is a mapping from the elements of one set to the elements of the other set. A matching is stable whenever it is not the case that both: (i) some given element A of the first matched set prefers some given element B of the second matched set over the element to which A is already matched, and (ii) B also prefers A over the element to which B is already matched In other words, a matching is stable when there does not exist any alternative pairing (A, B) in which both A and B are individually better off than they would be with the element to which they are currently matched. 77

78 Marriage markets A set of n men and a set of n women (n = size of marriage market) Each man has strict preferences over the women Each woman has strict preferences over the men A matching is a bijection (one-to-one correspondence) M between m and w. A man (woman) prefers M to M if he (she) prefers the partner he (she) is matched to in M to the one he (she) is matched to in M. A man-woman pair blocks M if they prefer each other to their spouses under M. A matching M is stable if there is no man-woman pair blocking M. For M, a stable matching p M (i) is also called a stable partner of i. 78

79 Example Men s Preferences Women s Preferences The matching {(1, 4), (2, 3), (3, 2), (4, 1)} is stable. Stability may be verified by considering each man in turn as a potential member of a blocking pair. Man 1 could form a blocking pair only with woman 2, but she prefers her partner, man 3 to man 1. Each of men 2 and 3 is matched with his favorite woman, so neither can be in a blocking pair. Finally, man 4 could form a blocking pair only with woman 4, but she would rather stick with her partner, man 1. 79

80 A second example of a stable matching, indeed the only other stable matching in this case, is {(1, 4), (2, 1), (3, 2), (4, 3)}, as may be verified in a similar way. The matching {(1, 1), (2, 3), (3, 2), (4, 4)}, for example, is unstable because of the blocking pair (1, 4); man 1 prefers woman 4 to his partner, woman 1, and woman 4 prefers man 1 to her partner, man 4. Some other unstable matchings may have many more blocking pairs: for example, the matching {(1, 1), (2, 2), (3, 4), (4, 3)} has six. Stable pair and fixed pair A man m and a woman w constitute a stable pair if and only if m and w are partners in some stable matching. In these circumstances, m is a stable partner of w, and vice versa. If some man m and woman w are partners in all stable matchings, then (m, w) is called a fixed pair. 80

81 Gale-Shapley algorithm: Deferred acceptance algorithm (DAA) 1. Each man proposes to his favorite woman. 2. Each woman engages her favorite man among her suitors and rejects the others. 3. Each rejected man proposes to his next favorite woman. 4. Repeat step 2 and 3 until all women have been proposed to. 81

82 Example 82

83 Each boy approaches the first girl on their lists. In the first round, Mary gets three offers, and holds on to Adam, while rejecting the other two. Jane receives one offer, from Bob, so she asks him to wait. Kate receives no offers, so at the end of the day she has nobody. The girls do not commit to anyone. They just say something like hold on while I think about it. 83

84 In the second round, the two boys who are not being held onto approach the second girls on their lists. In this case, Kate receives both proposals, holds onto Don and sends Charlie away. Since nobody proposed to Mary or Jane, they hold onto their boys from the first round. In the third round, Charlie (the only boy not currently held by a girl), asks Jane. Since Jane ranks Charlie ahead of Bob (who she s held since the first round), she releases Bob, and holds onto Charlie. In the fourth round, Bob asks Mary, but is rejected by her since she ranks Adam higher. In the fifth round Bob asks his last choice, Kate, but she rejects him as well, since she ranks Don higher. 84

85 All three girls have boys on hold, and the one unattached boy has been rejected by all three girls. The process is done. Adam and Mary end up together, as do Don and Kate, and Charlie and Jane. Bob ends up alone. Since no boy proposes to any girl after she has rejected him, this algorithm will reach a stable solution in a finite number of steps. In this case it took five rounds. Since each attached boy has been rejected by all the girls who he ranked ahead of the one he ended up with, this match is stable, since no boy could improve his matching by switching to a partner he ranked higher than the one he ended up with. This works with any different numbers of boys and girls. It turns out that if the boys do the proposing, the resulting pairing would be better for the boys; while if the girls do the proposing, it ends up better overall for the girls. 85

86 Summary Each man proposes to the women on his preference list, in their order of appearance, until he becomes engaged. If ever that engagement is broken (by the woman), then he becomes free again, and he resumes his sequence of proposals, starting with the next woman on his list. The algorithm terminates when everyone is engaged, and we will see that this will happen before any man exhausts his preference list. Furthermore, we will show that, on termination, the engaged couples constitute a stable matching. The Gale-Shapley algorithm involves an element of nondeterminism, since the order in which the free men propose is not specified. However, it turns out, as we will see, that this nondeterminism is of no consequence: the order in which the free men propose is immaterial to the outcome. 86

87 Man oriented version of the Gale-Shapley algorithm assign each person to be free; while some man m is free do begin w := first woman on m s list to whom m has not yet proposed; if w is free then assign m and w to be engaged {to each other} else if w prefers m to her fiancé m then assign m and w to be engaged and m to be free else w rejects m {and m remains free} end; output the stable matching consisting of the n engaged pairs 87

88 The algorithm may be expressed in terms of a sequence of proposals from men to women. At any point during the algorithm s execution, each person is either engaged or free; each man may alternate between being engaged and being free, but once a woman is engaged, she is never again free, though the identity of her fiancé may change. A man who is engaged more than once obtains fiancées who are successively less desirable to him, while each successive engagement brings a woman a more favored partner. When a free woman receives a proposal, she will immediately accept it, becoming engaged to the proposer. When an engaged woman receives a proposal, she compares the proposer with her current fiancé and rejects the less favored of the two men; that is, if she prefers her fiancé, she rejects the new proposal, but if she prefers the proposer, she breaks her current engagement, setting her ex-fiancé free, and becomes engaged to the current proposer. 88

89 Theorem For any given instance of the stable marriage problem, the Gale- Shapley algorithm terminates, and, on termination, the engaged pairs constitute a stable matching. Proof First, we show that no man can be rejected by all the women. A woman can reject only when she is engaged, and once she is engaged she never again becomes free. So the rejection of a man by the last woman on his list would imply that all the women were already engaged. But since there are equal numbers of men and women, and no man has two fiancées, all the men would also be engaged, which is a contradiction. Also, each iteration involves one proposal, and no man ever proposes twice to the same woman, so the total number of iterations cannot exceed n 2 (for an instance involving n men and n women). Termination is therefore established. 89

90 We write p M (m) as the M-partner of m and p M (w) is the M-partner of w. It is clear that, on termination, the engaged pairs specify a matching, which we denote by M. If man m prefers woman w to p M (m), then w must have rejected m at some point during the execution of the algorithm. But this rejection implies that w was, or became, engaged to a man she prefers to m, and any subsequent change of her fiancé brings her a still better partner. So w cannot prefer m to p M (m), and therefore (m, w) cannot block M. It follows that there are no blocking pairs for M, and therefore that M is a stable matching. 90

91 All possible executions of the Gale-Shapley algorithm (with the men as proposers) lead to the same stable matching. Furthermore, this stale matching has the remarkable property that every man achieves in it the best partner that he can possibly have in any stable matching. It is perhaps surprising that all the men, who are essentially in competition with each other for the women, can agree on a stable matching that is simultaneously optimal for all of them. This result is stated formally in the next theorem, which also establishes the insignificance of the nondeterminism in the algorithm. Theorem All possible executions of the Gale-Shapley algorithm (with the men as proposers) yield the same stable matching, and in this stable matching, each man has the best partner that he can have in any stable matching. 91

92 Proof Suppose that an arbitrary execution E of the algorithm yields the stable matching M, and that, in contradiction of the theorem, there is a stable matching M and a man m such that m prefers w = p M (m) to w = p M (m). Then during E, w must have rejected m. Suppose, without loss of generality, that this was the first occasion, during E, that a woman rejected a stable partner, and suppose that this rejection took place because of the engagement of w to m (so that w prefers m to m). Then m can have no stable partner whom he prefers to w (for no woman had previously rejected a stable partner). So m prefers w to his partner in M, and the supposed stable matching M is blocked by (m, w ). Each man m is therefore matched in M with his favorite stable partner W, and since E was an arbitrary execution of the algorithm, it follows that all possible executions of the algorithm leads to this same stable matching. 92

93 This is a remarkable result. It implies that if each man is independently given his best stable partner, then the result is a stable matching. Yet there seems no a prior reason why this should even be a matching. The stable matching generated by the man-oriented version of the Gale-Shapley algorithm is called man-optimal. If the roles of the sexes in the algorithm are interchanged, then the resulting womanoptimal stable matching, obtained by the woman-oriented version of the Gale-Shapley algorithm, is analogously optimal for the women. It may happen that the man and woman optimal stable matchings are identical, but this will not, in general, be the case. The optimality property from the point of view of the members of one sex is gained at the expense of the members of the other sex. Specifically, in the man-optimal stable matching, each woman has the worst partner that she can have in any stable matching, so that, to coin what seems an appropriate term, man-optimal is also woman- pessimal ; likewise, woman-optimal is man-pessimal. 93

94 Theorem In the man-optimal stable matching, each woman has the worst partner that she can have in any stable matching. Proof Suppose not. Let M 0 be the man-optimal stable matching, and suppose there is a stable matching M and a woman w such that w prefers m = p M0 (w) to m = p M (w). But then (m, w) blocks M unless m prefers p M (m) to w = p M0 (m), in contradiction of the fact that m has no stable partner better than his partner in M 0. Remark It can happen that the man-oriented and woman-oriented versions of the algorithm yield the same stable matching, in which case it is immediate, by combining the optimality and pessimality properties, that this is the unique stable matching for that instance. 94

95 College admission problems Characteristics of a stable solution 1. For candidates who are not assigned to any study programmes, they are inferior to all the selected ones in all the programmes they have applied for. 2. For a candidate who is assigned to a study programme which is not his first choice, then in all his more preferred choices, he is inferior to all the candidates who have been accepted. 95

96 In other words, a student cannot find a more preferred curricula which is willing to accept him, and an institution cannot get a more eligible student willing to accept its offer to replace the weakest one already accepted. Stable solutions exist but not unique. However, in all the stable solutions, it is always the same group of applicants that are selected for admission. 96

97 Program optimal approach 97

98 Each admissions officer gives out all the K offers to the top K candidates (represented by representatives). If there are more than one offer is given to the candidate, then they will return the less preferred offers to the admissions officers, who will then give them to the next eligible candidates in the queues. 98

99 Student optimal approach 99