Online Appendix to Strategy-proof tie-breaking in matching with priorities

Online Appendix to Strategy-proof tie-breaking in matching with priorities Lars Ehlers Alexander Westkamp December 12, 2017 Section 1 contains the omitted proofs of Lemma 5, Lemma 6 and Lemma 7 Subsection 11 shows that neither (A) nor (B) of Assumption 1 can be dispensed of in Theorem 1 Subsection 12 shows that all priority structures having at most a two-way tie at the top (and being otherwise strict) is solvable 1 Omitted proofs Lemma 5 Fix a weak priority structure (1) Let i, j, k I be three distinct agents and o, p O be two distinct objects such that i o j o k and i p k p j Let R be a preference profile such that R i R j R k o o p, and such that for all z I \ {i, j, k} and all õ {o, p} for which z õ ĩ for some ĩ {i, j, k}, zp z õ If f is constrained efficient and strategy-proof, then f i (R) = o Département de Sciences Économiues and CIREQ, Université de Montréal, Montréal, Québec H3C 3J7, Canada; e-mail: larsehlers@umontrealca Department of Management, Economics and Social Sciences, University of Cologne; e-mail: westkamp@wisouni-koelnde 1

(2) Let i, j, k, l I be four distinct agents and o, p, O be three distinct objects such that i o j o k o l, {i, j} p k p l, and i l j Let R be a preference profile such that R i R j R k R l o o p, and such that for all z I \ {i, j, k, l} and all õ {o, p, } for which z õ ĩ for some ĩ {i, j, k, l}, zp z õ If f is constrained efficient and strategy-proof, then f i (R) = o (3) Let i, j, k, l I be four distinct agents and o, p, O be three distinct objects such that i o j o k, i p l p k, and k l j Let R be a preference profile such that R i R j R k R l o o p, and such that for all z I \ {i, j, k, l} and all õ {o, p, } for which z õ ĩ for some ĩ {i, j, k, l}, zp z õ If f is constrained efficient and strategy-proof, then f i (R) = o Proof (1) By definition of R and stability, object o is assigned to agent i or agent j at R Assume to the contrary that f i (R) o and f j (R) = o The following diagram shows how we derive a contradiction: R R i R j R k o o p R 1 Ri 1 R j R k o o p p R 3 Ri 1 R j Rk 2 o o o p p R 2 Ri 1 R j Rk 1 o o p p o In moving from R to R 1, we have used strategy-proofness for agent i to infer f i (R 1 ) o Since i p k, stability then implies f k (R 1 ) = k In moving from R 1 to R 2, we have used strategy-proofness for agent k to infer f k (R 2 ) p This is compatible with stability only when f i (R 2 ) = p But then constrained efficiency reuires that f k (R 2 ) o, since i and j would otherwise form a stable improvement cycle of f(r 2 ) at R 2 Finally, in moving from R 2 to R 3 we have used strategy-proofness for k one more time to infer f k (R 3 ) = k This is compatible with constrained efficiency only when f i (R 3 ) = p and f j (R 3 ) = o However, this assignment at R 3 contradicts Lemma 4 in Ehlers and Westkamp (2017) given that {i, k} p j 2

so that i p j is an (i, j; o, p)-path which is compatible with the (k, j; o, p)-path k p j and i o j o k Hence, f j (R) = o is impossible for any constrained efficient and strategy-proof mechanism f (2) By definition of R and stability, object o is assigned at R to agent i or agent j Assume to the contrary that f i (R) o and f j (R) = o If i l j, we obtain an immediate contradiction to the first part of Lemma 5 Hence, we must have i l j Now consider the profile R 1 that is obtained from R by letting i add as her second most preferred object: R 1 R 1 i R j R k R l o o p By strategy-proofness, f i (R 1 ) o By individual rationality, we are left to consider two possible cases Case 1: f i (R 1 ) = The following diagram shows how to derive a contradiction to the assumed properties of f: R 1 R 1 i R j R k R l R 2 R 1 i R j R k R 1 l R 3 R 1 i R j R k R 2 l o o p o o p o o p o o In moving from R 1 to R 2, we have used strategy-proofness for agent l to infer f l (R 2 ) By definition of R and R 2, this is compatible with stability only when f i (R 2 ) = But then, we must have f l (R 2 ) o, as otherwise i and l would form a stable improvement cycle of f(r 2 ) at R 2 In moving from R 2 to R 3, we have again used strategy-proofness for agent l to infer f l (R 3 ) = l This is compatible with constrained efficiency only when f i (R 3 ) = and f j (R 3 ) = o But the latter is a contradiction to Lemma 4 in Ehlers and Westkamp (2017) given that {i, l} j so that i j is a (i, j; o, )-path which is compatible with the (l, j; o, )-path l j Hence, f i (R 1 ) = is impossible for any constrained efficient and strategy-proof mechanism f Case 2: f i (R 1 ) = i 3

By the definition of R 1 and stability, f i (R 1 ) = i implies f l (R 1 ) = The following diagram shows how to derive a contradiction: R 1 R 1 i R j R k R l R 4 R 2 i R j R k R l R 5 R 2 i R j R k R 2 l o o p o p o p o In moving from R 1 to R 4, we have used strategy-proofness for agent i to infer f i (R 4 ) = i This is compatible with stability only when f l (R 4 ) = By strategy-proofness for agent l, f l (R 4 ) = implies f l (R 5 ) {o, } Suppose first that f l (R 5 ) = In this case, strategy-proofness for agent i would imply that, for R defined by R R i R j R k R 2 l o p o, o we must have f i ( R) Furthermore, if f i ( R) = o, stability would reuire that f l ( R) = But then i and l would form a stable improvement cycle of f( R) at R, contradicting constrained efficiency of f Since f i ( R) / {, o}, we must have f i ( R) = i But then strategy-proofness for agent i implies that, for ˆR defined by ˆR R 1 i R j R k R 2 l o o p o, we must have f i ( ˆR) = i This is compatible with constrained efficiency only if f l ( ˆR) = and f j ( ˆR) = o But f j ( ˆR) = o is a contradiction to Lemma 4 in Ehlers and Westkamp (2017) given that {i, l} j so that i j is an (i, j; o, )-path which is compatible with the (l, j; o, )-path l j (and i o j o l) Since f l (R 5 ) = necessarily leads to a contradiction, we must have f l (R 5 ) = o Strategy-proofness for agent l then reuires that, for R 6 defined by R 6 R 2 i R j R k R 3 l o p o, 4

we must have f l (R 6 ) = o But since j o k o l and j p k p l, f l (R 6 ) = o is a contradiction to the first part of Lemma 5 This completes the proof (3) By the definition of R and non-wastefulness, object o is assigned to agent i or agent j at the preference profile R Assume that, contrary to what we want to show, f i (R) o and f j (R) = o Given that i p l p k and k l, it is easy to see that strategy-proofness and constrained efficiency imply R 1 R 1 i R j R k R l o o p p R 2 Ri 1 R j Rk 1 R l o o p p o R 3 Ri 1 R j Rk 1 Rl 1 o o p p o p Next, consider R 4 R 1 i R j R 1 k R 2 l o o p p p o Strategy-proofness for agent l implies f l (R 4 ) = l This is compatible with constrained efficiency only if f i (R 4 ) = p, f j (R 4 ) = o, and f k (R 4 ) = Strategy-proofness for k implies that R 5 Ri 1 R j Rk 2 Rl 2 o o o p p p By constrained efficiency, f i (R 5 ) = o or f j (R 5 ) = o We argue first that f j (R 5 ) = o is impossible Suppose the contrary Note that i p l j is an (i, j; o, )-path which is compatible with the (k, j; o, )-path k j Furthermore, note that the fact that k ranks p as her second most preferred object is irrelevant since l ranks p first and has strictly higher priority for it than k Given these observations, it is straightforward to modify the arguments in the proof of Lemma 4 in Ehlers and Westkamp (2017) to show that f j (R 5 ) = o implies that f cannot be strategy-proof and constrained efficient Hence, f j (R 5 ) o as was claimed above 5

Since f j (R 5 ) o, we must have f i (R 5 ) = o By constrained efficiency, this implies f l (R 5 ) = p Then R 5 R 1 i R j R 2 k R 2 l R 6 R 1 i R j R 2 k R 1 l R 7 R 1 i R j R 2 k R l o o o p p p o o o p p p o o o p p In moving from R 5 to R 6 we have used strategy-proofness for l to infer f l (R 6 ) l If f l (R 6 ) = p, then stability implies f i (R 6 ) = o and f k (R 6 ) = But then l and k would form a stable improvement cycle, thus contradicting constrained efficiency Hence, we must have f l (R 6 ) = By strategy-proofness, f l (R 7 ) = Since k l, f l (R 7 ) = is compatible with stability only if f k (R 7 ) {o, p} But given that f k (R 2 ) = and k can unilaterally deviate from R 2 to obtain R 7, f k (R 7 ) {o, p} implies that f cannot be strategy-proof Hence, is unsolvable and this completes the proof Lemma 6 Fix a weak priority structure (1) Let i, j I be two distinct agents and o O be an object such that i o j If there is an (i, j; o)-path i p 1 i 1 p M i M o j which is compatible with a (j, i; o)-path j 1 j 1 N j N o i, then is unsolvable (2) Let i, j, k, l I be four distinct agents and o O be an object such that i o j o k o l If there exist two objects p, O such that i p k p j and j l i, then is unsolvable Proof (1) Suppose to the contrary that there exists a constrained efficient and strategy-proof mechanism f Consider the following preference profile: R R i R j R i m R j n o o p m n p 1 1 p m+1 n+1 Lemma 3 in Ehlers and Westkamp (2017) implies f i (R) = p 1 and f j (R) = 1 Now assume that i deviates to R i : o By strategy-proofness, we must have f i (R i, R i ) = i We claim f j (R i, R i ) = o Otherwise, the construction of R would imply that, for all 6

n {0,, N}, f j n(r i, R i ) = n+1 But then j = j 0, j 1,, j N would form a stable improvement cycle of f(r i, R i ) at (R i, R i ) Hence, we must have f j (R i, R i ) = o Next, assume that, starting from R, j deviates to R j : o A completely symmetric argument to the one used to establish that f j (R i, R i ) = o shows that we must have f i (R j, R j ) = o Finally, consider R (R i, R j, R i,j ) Coming from the profile (R j, R j ), strategyproofness for i implies f i (R ) = o Coming from (R i, R i ), strategy-proofness for j implies f j (R ) = o Since o cannot be allocated to more than one agent and i j, we obtain a contradiction Hence, there cannot be a constrained efficient and strategy-proof mechanism (2) This follows immediately from the first part of Lemma 5, since it implies that at the preference profile R R i R j R k R l o o p, a constrained efficient and strategy-proof mechanism would have to (A) assign o to i given that i o j o k and i p k p j, and (B) assign o to j given that i o j o l and j l i Since there is only one copy of o and i j, (A) and (B) imply that there is no constrained efficient and strategy-proof mechanism Lemma 7 Let i 1, i 2, i 3, i 4, i 5 be five distinct agents and o 1, o 2, o 3, o 4, o 5 be five distinct objects Each of the following priority structures is unsolvable: i 1 o1 i 2 o1 i 3 o1 i 4 {i 1, i 2 } o2 i 3 o2 i 4 {i 1, i 2 } o3 i 3 (1) i 2 o4 i 4 o4 i 1 i 1 o5 i 4 o5 i 2 i 1 o1 i 2 o1 i 3 o1 i 4 {i 1, i 2 } o2 i 3 o2 i 4 i 4 o3 {i 2, i 5 } (2) i 2 o4 i 5 o4 i 1 7

i 1 o1 i 2 o1 i 3 i 1 o2 i 4 o2 i 5 o2 i 2 o2 i 3 i 4 o3 i 5 o3 i 1 (3) {i 2, i 3 } o4 i 4 i 1 o1 i 2 o1 i 3 {i 2, i 3 } o2 i 4 i 4 o3 i 1 o3 {i 2, i 3 } (4) i 1 o1 i 2 o1 i 3 i 1 o2 i 4 i 2 o3 i 5 (5) {i 2, i 4 } o4 i 3 o4 i 1 {i 1, i 5 } o5 i 3 o5 i 2 i 1 o1 i 2 o1 i 3 {i 1, i 2 } o2 i 5 o2 i 3 i 2 o3 i 4 (6) i 3 o4 i 5 o4 i 1 {i 3, i 4 } o5 i 5 o5 i 2 Proof For each of the six priority structures defined in Lemma 7, we will use a proof by contradiction Throughout the proof, let f be an arbitrary constrained efficient and strategy-proof mechanism 1 The priority structure in E (1) is unsolvable The proof revolves around the following preference profile: R R i1 R i2 R i3 R i4 o 1 o 1 o 3 o 2 Claim 1: f i2 (R) = o 1 We start by considering the following preference profile 8

R R i1 R i2 Ri3 Ri4 o 1 o 1 o 2 o 4 Since {i 1, i 2 } o2 i 3 o2 i 4 and i 2 o4 i 4 o4 i 1, the second part of Lemma 5 implies that we must have f i2 ( R) = o 1 Next, note that strategy-proofness for agent i 2 implies that i 2 must still obtain object o 1 at the following profile: 1 R 1 R i1 R 1 i 2 Ri3 Ri4 o 1 o 1 o 2 o 4 o 4 Since i 2 obtains o 1 at R 1, constrained efficiency reuires f i4 (R 1 ) = o 4 Next, consider the preference profile R 2 R i1 R 1 i 2 Ri3 R 1 i 4 o 1 o 1 o 2 o 2 o 4 o 4 By strategy-proofness for agent i 4, we must have f i4 (R 2 ) {o 2, o 4 } Since i 3 ranks o 2 first and i 3 o2 i 4, stability implies f i4 (R 2 ) o 2 Hence, we must have f i4 (R 2 ) = o 4 We will now show that f i4 (R 2 ) = o 4 is only possible when f i2 (R 2 ) = o 1 If i 2 o4 i 4, then f i2 (R 2 ) = o 1 follows immediately from stability of f(r 2 ) So suppose that i 2 o4 i 4 and that, contrary to what we want to show, f i1 (R 2 ) = o 1 Since f i4 (R 2 ) = o 4 and f i1 (R 2 ) = o 1, we must have f i2 (R 2 ) = i 2 We derive a contradiction to the assumed properties of f using the following seuence of preference profiles: 1 Recall that boxes indicate assigned objects 9

R 2,1 R i1 R 2 i 2 Ri3 R 1 i 4 o 1 o 4 o 2 o 2 R 2,2 R i1 R 2 i 2 Ri3 R 2 i 4 o 1 o 4 o 2 o 1 o 4 o 4 R 2,4 R i1 Ri 1 2 Ri3 Ri 2 4 o 1 o 1 o 2 o 1 o 4 o 4 R 2,3 R i1 R 3 i 2 Ri3 R 2 i 4 o 1 o 4 o 2 o 1 o 1 o 4 Given that f i2 (R 2 ) = i 2, the indicated assignment at R 2,1 follows immediately from strategy-proofness and stability Strategy-proofness implies that f i4 (R 2,2 ) {o 1, o 4 } since f i4 (R 2,1 ) = o 4 If f i4 (R 2,2 ) = o 1, then i 4 would still have to obtain o 1 when she unilaterally deviates to R 3 i 4 = o 1 However, since i 1 o2 i 3 o2 i 4 and i 1 o1 i 3 o1 i 4, f i4 (R 3 i 4, R 2,2 i 4 ) = o 1 is incompatible with the tie-breaking rule in the first part of Lemma 5 Hence, we must have f i4 (R 2,2 ) = o 4 and f i1 (R 2,2 ) = o 1 Strategy-proofness and nonwastefulness imply f i2 (R 2,3 ) o 4 and f i4 (R 2,3 ) = o 4 If f i2 (R 2,3 ) = o 1, then i 2 and i 4 would form a stable improvement cycle of f(r 2,3 ) at R 2,3, a contradiction Hence, we must have f i2 (R 2,3 ) = i 2 By strategy-proofness, we must also have f i2 (R 2,4 ) = i 2 Nonwastefulness then implies that f i4 (R 2,4 ) = o 4 and f i1 (R 2,4 ) = o 1 Given that {i 2, i 4 } o4 i 1, i 2 o4 i 1 is an (i 2, i 1 ; o 1, o 4 )-path which is compatible with the (i 4, i 1 ; o 1, o 4 )-path i 4 o4 i 1 Since i 2 o1 i 4 o1 i 1, f i1 (R 2,4 ) = o 1 contradicts the tie-breaking rule in Lemma 4 in Ehlers and Westkamp (2017) Hence, f i1 (R 2 ) = o 1 also leads to a contradiction when i 2 o4 i 4 and we must have f i2 (R 2 ) = o 1 The following diagram summarizes the remainder of the proof of Claim 1: R 2 R i1 Ri 1 2 Ri3 Ri 1 4 o 1 o 1 o 2 o 2 o 4 o 4 R 7 R 1 i 1 R 4 i 2 R i3 R i4 o 1 o 1 o 3 o 2 o 2 o 3 R 3 Ri 1 1 Ri 1 2 Ri3 Ri 1 4 o 1 o 1 o 2 o 2 o 2 o 4 o 4 R 6 Ri 1 1 Ri 4 2 R i3 Ri 1 4 o 1 o 1 o 3 o 2 o 2 o 3 o 4 R 4 Ri 1 1 Ri 1 2 Ri 1 3 Ri 1 4 o 1 o 1 o 2 o 2 o 2 o 4 o 3 o 4 R 5 Ri 1 1 Ri 4 2 Ri 1 3 Ri 1 4 o 1 o 1 o 2 o 2 o 2 o 3 o 3 o 4 By strategy-proofness, we must have f i1 (R 3 ) o 1 given that f i1 (R 2 ) = i 1 Since 10

i 1 o2 {i 3, i 4 }, stability reuires f i1 (R 3 ) = o 2 and f i3 (R 3 ) = i 3 Strategy-proofness and stability then imply f i3 (R 4 ) = o 3 and f i1 (R 4 ) = o 2 Hence, f i2 (R 4 ) = o 1 and another application of strategy-proofness yields f i2 (R 5 ) = o 1 Since f i3 (R 5 ) = o 3 if f i2 (R 5 ) = o 1, strategy-proofness reuires that f i3 (R 6 ) = o 3 as well Given that i 2 o3 i 3, f i3 (R 6 ) = o 3 implies f i2 (R 6 ) = o 1 Finally, f i2 (R 6 ) = o 1 and i 1 o2 i 4 imply f i1 (R 6 ) = o 2 and f i4 (R 6 ) o 2 Strategy-proofness allows us to infer f i4 (R 7 ) = i 4, which is compatible with constrained efficiency only if f i1 (R 7 ) = o 2 and f i2 (R 7 ) = o 1 It is now straightforward to see that two more applications of strategy-proofness from R 7 yield the desired statement that f i2 (R) = o 1 This completes the proof of Claim 1 Claim 2: f i1 (R) = o 1 Consider the following preference profile ˆR R i1 R i2 R i3 R i 4 o 1 o 1 o 2 o 5 Since {i 1, i 2 } o2 i 3 o2 i 4 and i 1 o5 i 4 o5 i 2, the second part of Lemma 5 implies that we must have f i1 ( ˆR) = o 1 A completely analogous argument to that used in the proof of Claim 1 shows that f i1 ( ˆR) = o 1 implies f i1 (R) = o 1 2 Combining Claims 1 and 2, we find that a constrained efficient and strategy-proof mechanism for has to satisfy f i1 (R) = f i2 (R) = o 1 Since there is only one copy of o 1, this is a contradiction Hence, there exists no constrained efficient and strategy-proof mechanism for priority structure in E (1) 2 The priority structure in E (2) is unsolvable As usual, arrows indicate how we move between profiles and boxes indicate object assignments 2 In the arguments so far, we have used that i 2 o4 i 4 o4 i 1, i 1 o2 i 3 o2 i 4, and i 2 o3 i 3 Since i 1 o5 i 4 o5 i 2, i 2 o2 i 3 o2 i 4, and i 1 o3 i 3, one just has to switch the roles of i 1 and i 2 and the roles of o 4 and o 5 in the proof of Claim 1 to establish that f i1 ( ˆR) = o 1 implies f i1 (R) = o 1 We omit the details 11

R R i1 R i2 R i3 R i4 R i5 o 4 o 1 o 2 o 1 o 3 R 1 R i1 R i2 R i3 R 1 i 4 R i5 o 4 o 1 o 2 o 1 o 3 o 3 R 3 Ri 1 1 R i2 R i3 Ri 1 4 Ri 1 5 o 4 o 1 o 2 o 1 o 3 o 1 o 3 o 4 R 4 Ri 1 1 Ri 1 2 R i3 Ri 1 4 Ri 1 5 o 4 o 1 o 2 o 1 o 3 o 1 o 4 o 3 o 4 R 2 R i1 R i2 R i3 Ri 1 4 Ri 1 5 o 4 o 1 o 2 o 1 o 3 o 3 o 4 R 5 Ri 2 1 Ri 1 2 R i3 Ri 1 4 Ri 1 5 o 1 o 1 o 2 o 1 o 3 o 4 o 4 o 3 o 4 Note that f i2 (R) = o 1 follows from the first part of Lemma 5 since i 2 o2 i 3 o2 i 4 and i 2 o1 i 3 o1 i 4 By strategy-proofness for i 4, we must have f i4 (R 1 ) o 1 Since i 4 o3 i 5, stability then implies that f i4 (R 1 ) = o 3 By strategy-proofness for i 5, we must have f i5 (R 2 ) o 3 Since i 5 o4 i 1, stability then implies that f i5 (R 2 ) = o 4 By strategy-proofness for i 1, we must have f i1 (R 3 ) o 4 If f i1 (R 3 ) = o 1, then nonwastefulness would imply f i4 (R 3 ) = o 3 and f i5 (R 3 ) = o 4 But then i 1, i 4, and i 5 would form a stable improvement cycle at R 3, thus contradicting constrained efficiency of f(r 3 ) Hence, f i1 (R 3 ) = i 1 (and the indicated assignments at R 3 then follow from constrained efficiency) Strategy-proofness for i 2 implies f i2 (R 4 ) = o 1 The indicated assignments then follow immediately from stability given that i 4 o3 i 5 and i 5 o4 i 1 Strategy-proofness for i 1 implies that f i1 (R 5 ) = i 1 We will now argue that we must have f i2 (R 5 ) = o 1 : Otherwise, strategy-proofness for i 2 would imply that for the preference profile, R 5,1 Ri 2 1 R i2 R i3 Ri 1 4 Ri 2 5 o 1 o 1 o 2 o 1 o 3 o 4 o 3 o 4 we must have f i2 (R 5,1 ) = i 2 Since i 1 can obtain R 5,1 from R 3 by a unilateral deviation (from R 1 i 1 to R 2 i 1 ) and since f i1 (R 3 ) = i 1, strategy-proofness for i 1 would then imply f i1 (R 5,1 ) = i 1 Now note that f i2 (R 5,1 ) = i 2 and f i1 (R 5,1 ) = i 1 could be compatible with non-wastefulness only if f i4 (R 5,1 ) = o 1 and f i5 (R 5,1 ) = o 4 But then o 3 would remain unassigned even though o 3 P i5 (R 5,1 )o 4 Hence, f(r 5,1 ) cannot be stable if f i2 (R 5 ) o 1 12

Next, consider R 6 R 2 i 1 R 1 i 2 R i3 R 1 i 4 R 2 i 5 o 1 o 1 o 2 o 1 o 4 o 4 o 4 o 3 Strategy-proofness and f i5 (R 5 ) = o 4 imply f i5 (R 6 ) = o 4 Since i 2 o4 implies f i2 (R 6 ) = o 1 Finally, consider the profile i 5, stability R 7 Ri 2 1 Ri 2 2 R i3 Ri 1 4 Ri 2 5 o 1 o 3 o 2 o 1 o 4 o 4 o 1 o 3 By strategy-proofness for i 2, we must have f i2 (R 7 ) {o 1, o 3 } If f i2 (R 7 ) = o 1, then non-wastefulness would reuire f i4 (R 7 ) = o 3 But then i 2 and i 4 would form a stable improvement cycle, thus contradicting the constrained efficiency of f(r 7 ) Hence, we must have f i2 (R 7 ) = o 3 Since i 4 o3 i 2, this reuires f i4 (R 7 ) = o 1 It is straightforward to show that strategy-proofness implies that i 4 must still obtain o 1 when, starting from R 7, i 2 first deletes o 1 from her preferences (again since i 4 o3 i 2 ), i 1 then deletes o 4 from her preferences and, finally, i 4 deletes o 3 from her preferences Since i 1 o2 i 3 o2 i 4 and i 1 o1 i 3 o1 i 4, we obtain a contradiction to the first part of Lemma 5 3 The priority structure in E (3) is unsolvable Consider the following preference profile R R i1 R i2 R i3 R i4 R i5 o 1 o 1 o 1 o 4 o 3 o 2 o 4 o 4 o 2 Since, for k {2, 3} and l {2, 3} \ {k}, i 1 o2 i k and i l o4 i 4 o2 i k are two compatible paths, Lemma 4 in Ehlers and Westkamp (2017) and constrained efficiency immediately imply f i1 (R) = o 1 We will now complete the proof by showing that f i1 (R) = o 1 is impossible as well The proof relies on the following seuence of profiles: 13

R 1 R 1 i 1 R i2 R i3 R i4 R i5 o 2 o 1 o 1 o 4 o 3 o 1 o 4 o 4 o 2 R 3 R 1 i 1 R 1 i 2 R i3 R i4 R i5 o 2 o 1 o 1 o 4 o 3 o 1 o 4 o 2 R 2 Ri 1 1 R i2 Ri 1 3 R i4 R i5 o 2 o 1 o 1 o 4 o 3 o 1 o 4 o 2 R 4 Ri 1 1 Ri 1 2 Ri 2 3 R i4 R i5 o 2 o 1 o 1 o 4 o 3 o 1 o 2 By strategy-proofness for i 1, we must have f i1 (R 1 ) {o 1, o 2 } Assume first that f i1 (R 1 ) = o 2 Given that {i 2, i 3 } o4 i 4, stability would then imply f i4 (R 1 ) = i 4 It is easy to see that two applications of strategy-proofness (for i 4 and then for i 1 ) yield R 1 Ri 2 1 R i2 R i3 Ri 2 4 R i5 o 2 o 1 o 1 o 2 o 3 o 4 o 4 Since i 4 o3 i 5 o3 i 1 and i 1 o2 i 4 o2 i 5, we obtain a contradiction to the first part of Lemma 5 Hence, we must have f i1 (R 1 ) = o 1 By strategy-proofness for i 3 and constrained efficiency, f i1 (R 1 ) = o 1 implies f i2 (R 2 ) = o 1 Similarly, f i3 (R 3 ) = o 1 But then strategy-proofness for i 2 and i 3 would imply that f i2 (R 4 ) = f i3 (R 4 ) = o 1, which is impossible Since every possible allocation of o 1 at R necessarily leads to a contradiction of either strategy-proofness or constrained efficiency, the priority structure in E (3) is unsolvable 4 The priority structure in E (4) is unsolvable Consider the following preference profile R R i1 R i2 R i3 R i4 o 1 o 1 o 1 o 2 o 3 o 2 o 2 o 3 Note that i 2 o2 i 4 o3 i 1 is an (i 2, i 1 ; o 1, o 3 )-path which is compatible with the (i 3, i 1 ; o 1, o 3 )-path i 3 o2 i 4 o3 i 1, and Lemma 4 in Ehlers and Westkamp (2017) implies f i1 (R) o 1 Similarly, i 1 o3 i 2 and i 3 o2 i 4 o3 i 2 are compatible paths, 14

so that Lemma 4 in Ehlers and Westkamp (2017) implies f i2 (R) o 1, and i 1 o3 i 3 and i 2 o2 i 4 o3 i 3 are compatible paths, so that Lemma 4 in Ehlers and Westkamp (2017) implies f i3 (R) o 1 Thus, o 1 must remain unallocated at R, contradicting non-wastefulness of f(r) 5 The priority structure in E (5) is unsolvable Consider the following preference profile R R i1 R i2 R i3 R i4 R i5 o 1 o 1 o 4 o 2 o 3 o 2 o 3 o 5 o 4 o 5 We claim that f i2 (R) = o 1 Consider first the preference profile R R i 1 R i 2 R i 3 R i 4 R i 5 o 1 o 1 o 4 o 2 o 3 Since i 2 o4 i 3 o4 i 1, the first part of Lemma 5 implies f i2 (R ) = o 1 Two applications of strategy-proofness, once for i 1 and once for i 2, imply that f i2 (R i1, R i2, R {i 1,i 2 } ) = o 1 Non-wastefulness then implies that f i5 (R i1, R i2, R {i 1,i 2 } ) = o 3 By strategy-proofness for i 5, we must have f i5 (R i1, R i2, R i5, R {i 1,i 2,i 5 } ) = o 3 Since i 2 o3 i 5 and i 1 o2 i 4, stability implies that f i1 (R i1, R i2, R i5, R {i 1,i 2,i 5 } ) = o 2, f i2 (R i1, R i2, R i5, R {i 1,i 2,i 5 } ) = o 1, and f i4 (R i1, R i2, R i5, R {i 1,i 2,i 5 } ) = i 4 By strategy-proofness for i 4, stability, and the assumption that i 4 o4 i 3, we must have f i4 (R i1, R i2, R i 3, R i4, R i5 ) = o 4 and f i3 (R i1, R i2, R i 3, R i4, R i5 ) = i 3 By strategy-proofness for i 3, we must have f i3 (R i1, R i2, R i3, R i4, R i5 ) o 4 Stability is easily seen to imply f i4 (R i1, R i2, R i3, R i4, R i5 ) = o 4, f i1 (R i1, R i2, R i3, R i4, R i5 ) = o 2, and, as we claimed above, f i2 (R i1, R i2, R i3, R i4, R i5 ) = o 1 Next, consider the following preference profile R R i1 R i2 Ri3 R i4 R i5 o 1 o 1 o 5 o 2 o 3 o 2 o 3 o 4 o 5 15

We claim that f i1 ( R) = o 1 The proof of the claim proceeds from the preference profile R R i 1 R i Ri3 2 R i 4 R i 5 o 1 o 1 o 5 o 2 o 3 and is similar to the proof that f i2 (R) = o 1 Given that f i1 ( R) = o 1, i 2 o3 i 5, and i 5 o5 i 3, stability implies f i3 ( R) = i 3 Furthermore, given that f i2 (R) = o 1, i 1 o2 i 4, and i 4 o4 i 3, stability implies f i3 (R) = o 5 But then, since i 3 can obtain R from R by a unilateral deviation from R i3 to R i3, f cannot be strategy-proof 6 The priority structure in E (6) is unsolvable Consider first the following preference profile R 1 R i1 R i2 R i3 R i4 R i5 o 1 o 1 o 2 o 3 o 5 Because i 1 o1 i 2 o1 i 3, i 1 o2 i 5 o2 i 3 and i 3 o5 i 5 o5 i 2, the third part of Lemma 5 implies f i1 (R 1 ) = o 1 It is straightforward to verify that strategy-proofness and constrained efficiency imply R 1 R 1 i 1 R 1 i 2 R 1 i 3 R 1 i 4 R i5 o 1 o 1 o 2 o 3 o 5 o 2 o 3 o 4 o 5 Next, consider first the following preference profile R 2 R i1 R i2 R i3 R i4 Ri 1 5 o 1 o 1 o 2 o 3 o 4 Because i 1 o1 i 2 o1 i 3, i 2 o2 i 5 o2 i 3 and i 3 o4 i 5 o4 i 1, the third part of Lemma 5 implies f i2 (R 2 ) = o 1 It is straightforward to verify that strategy-proofness 16

and constrained efficiency imply R 2 Ri 1 1 Ri 1 2 Ri 1 3 Ri 1 4 Ri 2 5 o 1 o 1 o 2 o 3 o 4 o 2 o 3 o 4 o 5 o 5 Since i 5 can obtain R 2 from R 1 by a unilateral deviation (from R i5 to Ri 2 5 ), we obtain that f cannot be strategy-proof This shows that f cannot be constrained efficient and strategy-proof 11 Necessity of Assumption 1 in Ehlers and Westkamp (2017) In this section, we present two examples: in the first example, the priority structure satisfies Assumption 1 (B) in Ehlers and Westkamp (2017) but violates Assumption 1 (A) in Ehlers and Westkamp (2017); in the second example, the priority structure satisfies Assumption 1 (A) in Ehlers and Westkamp (2017) but violates Assumption 1 (B) in Ehlers and Westkamp (2017) For both examples, we show that a constrained efficient and strategy-proof mechanism exists even though the priority structures are not strict, HET, or TAU This shows that both parts of Assumption 1 in Ehlers and Westkamp (2017) are necessary for Theorem 1 in Ehlers and Westkamp (2017) to hold Example 1 Let I = {1,, 6} and O = {o, p 1, p 2 } Priorities are as follows: o p1 p2 1, 2, 3 6 6 4 5 5 5 4 4 6 3 1 2 2 1 3 This priority structure violates Assumption 1 (A) (because is not strict and does not contain any four-way tie), but satisfies in Assumption 1 (B) in Ehlers and Westkamp 17

(2017) We will show below that is solvable by constructing a variant of the DA that yields a constrained efficient and strategy-proof mechanism 3 Example 2 Let I = {1,, 6} and O = {o, p 1, p 2, p 3, p 4 } Priorities are as follows: o p1 p2 p3 p4 1, 2, 3, 4, 5, 6 6 5 6 6 5 6 4 5 4 4 5 2 3 1 2 4 2 2 1 1 1 3 3 3 Note that this priority structure violates Assumption 1 (B) (since, eg, 6 p1 1 and 6 1 for all O) but satisfies Assumption 1 (A) in Ehlers and Westkamp (2017) (because there is a four-way tie at o ) 4,5 We will now proceed to construct a deferred acceptance algorithm with tie-breaking (DAT) that, as we will show below, is constrained efficient and strategy-proof mechanism for both examples Let R be an arbitrary preference profile for the six agents in one of the above examples Step 1: Exogenous tie-breaking For Example 1, define the weak priority structure by setting 1 o 3 o 2 o 4 o 5 o 6 and p= p for p {p 1, p 2 } For Example 2, define the weak priority structure by setting 6 o 5 o 4 o 2 o 1 o 3 and p= p for p {p 1, p 2, p 3, p 4 } 3 This example can be extended to an arbitrary number of agents 1,, N as follows: Let 1 o 2 o 3 o 4 o o N, N p1 p1 4 p1 3 p1 2 p1 1, and N p2 p2 4 p2 1 p2 2 p2 3 It is easy to see that all arguments below continue to hold for this extended example 4 The priority structure does, however, satisfy the weaker reuirement that the priority structure is connected in the sense that there is no subset of agents J I such that J o I \ J for all o 5 In order to extend this type of example to an arbitrary number of agents 1,, N, let there be N 1 objects o, p 1,, p N 2 such that N o o 1, N p1 p1 1, N 1 p2 N p2 N 2 p2 p2 4 p2 1 p2 2 p2 3, N p3 N 2 p3 N 1 p3 N 3 p3 p3 4 p3 2 p3 1 p3 3,, N pn 2 pn 2 5 pn 2 2 pn 2 4 pn 2 1 pn 2 3 As shown in our earlier working paper, Ehlers and Westkamp (2011), the just described construction can be used to characterize all solvable priority structures within the class of priority structures where ties are restricted to occur only at the bottom of priority rankings 18

Step 2: DA without tie-breaking Run a DA in which rejections are determined by and let µ 1 be the temporary assignment at the end of this algorithm 6 Stop, if µ 1 is a matching and proceed to Step 3 otherwise Note that, given our construction of, the only possibility for the procedure to proceed to Step 3 is that µ 1 (1) = µ 1 (3) = o Step 3: Endogenous tie-breaking If there is an object p such that µ 1 (2) = p and 3 p 2 p 1, let o reject 1 If there is an object O that is acceptable to agent 1 and most preferred among the ones which have not received any proposals in Step 2, match 1 to and stop Otherwise, leave 1 unmatched and stop In any other case, let o reject 3 If there is an object O that is acceptable to agent 3 and most preferred among the ones which have not received any proposals in Step 2, match 3 to and stop Otherwise, leave 3 unmatched and stop Given a preference profile R, let DAT (R) denote the outcome of the above procedure Claim 1: For any R, DAT (R) is constrained efficient Fix a preference profile R and let µ DAT (R) We argue first that µ is stable Note that all rejections in Step 2 respect all strict priority rankings for the priority structures in Example 1 and Example 2 Hence, the only possibility for µ to violate the stability condition is that the algorithm reaches Step 3 and there is an object p O \ {o} and an agent j i st pp i µ(i), µ(j) = p, and i p j for some i {1, 3}, say for i = 1 But by the rules of Step 3, 1 will only be rejected by o if µ(2) = p for some p such that 3 p 2 p 1 For both examples, the last observation implies that j µ 1 (j) 1 for all j I\{1} such that µ 1 (j) j Hence, µ must be stable Next, we will show that µ is also constrained efficient Suppose to the contrary that there is a stable improvement cycle i 1,, i m Assume wlog that i 1 is among the first (in the 6 Formally, in each round, let (A) each agent apply to the most preferred object that has not rejected him yet, (B) each object p reject all but the highest priority agents according to p, and (C) stop, if there were no new proposals in (A) 19

course of above algorithm) agents in the stable improvement cycle to be rejected by the object he is pointing to in the cycle There has to be an agent j who causes the rejection of i 1, ie an agent j who was rejected by all objects he prefers to o 2 when he applied to o 2 in the DA with tie-breaking and for whom j o2 i 1 By our assumption that i 1 was among the first agents to be rejected by the object he is pointing to in the stable improvement cycle, we must have j i 2 and hence µ(j) o 2 Furthermore, since the welfare of agents is weakly decreasing during the course of the above algorithm, we must have o 2 P j µ(j) These arguments imply that j o2 i 1 and, conseuently, that o 2 = o: otherwise, we would have i 2 o2 j o2 i 1 and i 1 could not be among the highest priority agents who desires o 2 = p at µ, thus contradicting the definition of a stable improvement cycle Next, we will show that {i 1, i 2, j} = {1, 2, 3} For Example 1, we can immediately infer the just mentioned statement from the original priority ranking o For Example 2, it is easy to see that if {i 1, i 2, j} = {1, 2, 3}, then the form of o implies i 1 = 4, j = 5, and i 2 = 6 However, by our previous arguments, 6 was temporarily matched to an object p O \ {o} st pr 6 o 3 when 4 was rejected by o Since the only agent who can displace j at o is 6 and the only agent who can displace 6 at p is 5, 6 can t be rejected by o 3 during the DA with tie-breaking Hence, for both examples, we must have {i 1, i 2, j} = {1, 2, 3} Now assume first that i 1 = 2 This is possible only in Example 1 and we can assume wlog that j = 1 and i 2 = 3 Since each object can appear at most once in a stable improvement cycle and since no no agent in {1, 2, 3} can displace an agent in {4, 5, 6} at either p 1 or p 2 in Example 1, we must have {i 1,, i m } {1, 2, 3} This is easily seen to imply that m = 2 and 1 o1 2 o1 3 But then, the rules of Step 3 of the DA with tie-breaking immediately imply that 3 could not have displaced 1 at o 2 = o subseuently to being rejected by o 1 Hence, we must have, wlog, i 1 = 1 In Example 1, the only agent who can displace 1 at o is agent 3 Hence, it would have to be the case that j = 3 But then, 3 could not have subseuently been displaced by i 2 = 2, thus contradicting µ(i 2 ) = o 2 = o We are left to consider the case of i 1 = 1 for Example 2 Here, we must have j = 3 and i 2 = 2 given that 2 o 1 o 3 But by the rules of Step 3, 1 cannot displace 2 subseuently to losing a tie-breaking decision against 3 This completes the proof of constrained efficiency Claim 2: DAT is strategy-proof Suppose that, contrary to what we want to show, there are a preference profile R, an agent i, and a misreport ˆR i such that DAT i ( ˆR i, R i )P i DAT i (R) Let µ DAT (R) and 20

ˆµ DAT ( ˆR i, R i ), and let µ 1 and ˆµ 1 be the temporary assignments at the end of Step 2 of the DAT under R and ˆR ( ˆR i, R i ), respectively Note first that the DA with tie-breaking has to reach Step 3 for R and ˆR, and that the tie-breaking decision in Step 3 has to be different for R and ˆR: if, say, 1 wins the tie-breaking decision at R and ˆR, the DA with tie-breaking would be euivalent to a DA with strict priorities in which 1 has strictly higher priority for o than 3 Since the DA with strict priorities is strategy-proof, there cannot be a an agent who can profitably manipulate at R Next, we will argue that i / {2, 4, 5, 6} Note first that all agents apart from 1 and 3 receive their final allocation in the second step of DAT Now consider first Example 1 Here, the temporary assignment at the end of Step 2 of the DAT under R and ˆR is euivalent to the final outcome of a DA in which the priority-ranking of object o is given by {1, 3} o {1} o {3} o {2} o {4} o {5} o {6} and the priority ranking for all other objects is as in Since these priorities induce substitutable preferences that satisfy the law of aggregate demand, such a DA is strategy-proof (Hatfield and Milgrom, 2005) implies that no agent in {2, 4, 5, 6} can manipulate DAT for Example 1 Next, consider Example 2 By our construction of o for Example 2, we can infer that no agent in {2, 4, 5, 6} could have applied to o under R and ˆR But then the temporary assignment at the end of Step 2 of the DAT is euivalent to a DA in which the priorities of o are given by {1, 3} o {1} o {3} o {2} o {4} o {5} o {6} and the priority ranking for all other objects is as in By the same arguments as above, this implies that no agent in {2, 4, 5, 6} can profitably manipulate DAT This To complete the proof, we will now show that i = 1 is impossible (the arguments in the case of i = 3 are completely symmetric) Assume first that µ(1) = o and ˆµ(1) o For both examples, we must have ˆµ 1 (2) = p 1 by the rules of Step 3 Furthermore, since the tie-breaking decision between 1 and 3 is different at R and ˆR, we must have µ 1 (2) ˆµ 1 (2) It is easy to see that for both examples, we must have µ 1 (2) = p 2 since otherwise, 1 would either fail to win the tie-breaking decision against 3 at R (if µ 1 (2) = p 1 ) or would not be able to affect the pre tie-breaking assignment (if µ 1 (2) O \ {o, p 1, p 2 }) We immediately obtain that op 1 p 2 and p 2 ˆP1 o But then, in order for the deviation to ˆR 1 to be profitable for 1, it would have to be the case that, subseuently to displacing 2 at p 2, 1 is rejected by p 2 and 2 ends up temporarily assigned to p 1 at the end of Step 2 It is straightforward to check that the just mentioned configuration is impossible for both types of examples Next, assume that 21

µ(1) o and ˆµ(1) = o By the rules of Step 3, we must have µ 1 (2) = p 1 and ˆµ 1 (2) p 1 It is easy to see that in order for 1 to be able to influence the pre tie-breaking assignment, we must have ˆµ 1 (2) = p 2 But then, it has to be the case that 1 displaced 2 at p 2 during Step 2 of the DAT for profile R For Example 1, this immediately implies µ(1) = p 2 and we obtain a contradiction to the assumption that 1 and 3 compete for o in Step 3 For Example 2, 1 must have been displaced at some point of Step 2 of the DAT under profile R But this is possible only if µ 1 (2) = p 3 and µ 1 (4) = p 2, thus contradicting our assumption that µ 1 (2) = p 1 This completes the proof 12 Two-way ties at the top We say that is a two-way ties at the top (TWT) priority structure, if, for all o O, there exist i(o), j(o) I such that (A) i(o) o j(o), (B) for all k I \ {i(o), j(o)}, j(o) o k, and (C) o I\{i(o)} is strict We will now argue that if is a TWT priority structure, then the following two-step procedure induces a constrained efficient and strategy-proof mechanism: 1 Let be a strict priority structure that respects all strict priority rankings in, ie assume that i o j whenever i o j 2 For any preference profile R, choose the outcome of the DA-algorithm with respect to R and Note that the mechanism induced by the procedure just described is strategy-proof since the DA mechanism for strict priority structures is strategy-proof and since the same strict priority ranking is used for all preference profiles To see that the outcome of above procedure is always constrained efficient, let R be a preference profile and µ be the matching chosen by the above procedure If µ is not constrained efficient, then µ contains a SIC, say i 1,, i m Note that i l desires µ(i l+1 ) and i l D µ(il+1 )(µ) for all l {1,, m} (where m + 1 := 1) Choose an agent from i 1,, i m who is among the first ones rejected in DA by the object he desires, say i 1 But then i 1 is rejected by µ(i 2 ) because some other agent j I\{i 1, i 2 } applied to µ(i 2 ) Note that µ(j) µ(i 2 ) and µ(i 2 )P j µ(j) If j µ(i2 ) i 1, then µ(i 2 )P j µ(j) implies i 1 / D µ(i2 )(µ), a contradiction Thus, we must have j µ(i2 ) i 1 and i 1 and j are tied at the top of µ(i2 ) But then j is never rejected by µ(i 2 ) and we must have µ(j) = µ(i 2 ), again a contradiction 22

References [1] Ehlers, Lars and Alexander Westkamp (2011), Strategy-proof tie-breaking, CIREQ Discussion Paper 09-2011, Université de Montréal [2] Ehlers, Lars and Alexander Westkamp (2017), Strategy-proof tie-breaking in matching with priorities, Theoretical Economics, forthcoming [3] Hatfield, John and Paul Milgrom (2005), Matching with contracts, Amer Econ Rev 95, 913 935 23