A market design approach to job rotation

Journal Pre-proof A market design approach to job rotation

Jingsheng Yu, Jun Zhang

PII:

S0899-8256(20)30003-8

DOI:

https://doi.org/10.1016/j.geb.2020.01.002

Reference:

YGAME 3096

To appear in:

Games and Economic Behavior

Received date:

19 November 2018

Please cite this article as: Yu, J., Zhang, J. A market design approach to job rotation. Games Econ. Behav. (2020), doi: https://doi.org/10.1016/j.geb.2020.01.002.

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A Market Design Approach to Job Rotation Jingsheng Yu

Jun Zhang

Yu: School of Economics & China Center for Behavioral Economics and Finance, Southwestern University of Finance and Economics, 555 Liutai Road, Chengdu, 611130, China. Email: [email protected] Zhang (corresponding author): Institute for Social and Economic Research, Nanjing Audit University, 86 Yushan West Road, Nanjing, 211815, China. Email: [email protected]

January 14, 2020

Abstract Organizations often rotate employees’ jobs. This paper proposes a market design approach to organize job rotation. In our model each employee has occupied a position, and if any employee wants to move to another position, the current occupier must leave the position. This requirement is described by a priority structure in which each employee has the lowest priority for his position and others have the equal priority. It is the “opposite” to the famous housing market priority structure in which every owner has the highest priority for his endowment and others have the equal priority. We adapt Top Trading Cycle to solve our model. Our mechanism is novel in that employees are not allowed to point to their positions during the mechanism and cycles are cleared by backward induction after all of them are generated. Our mechanism is stable, constrained efficient and weakly group strategy-proof.

Keywords: job rotation; coarse priority; Backward-induction Top Trading Cycle; constrained efficiency; strategy-proofness JEL Classification: C71, C78, D78

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1

Introduction

Job rotation refers to the phenomenon that an employer rotates employees’ assigned positions from period to period. It happens in both private sectors and public sectors. Japanese firms are well-known examples in private sectors. In the book addressing the success of Japanese firms, Ouchi (1981) wrote “An electrical engineer may go from circuit design to fabrication to assembly; a technician may work on a different machine or in a different division every few years; and all managers will rotate through all areas of business”. Osterman (1994) found that job rotation became increasingly popular among companies in the United States and other OECD countries, and bigger firms were more likely to implement job rotation. In public sectors, Japan is again a good example. In Japanese public schools, teachers are rotated from one school to another within each prefecture on various schedules. It contributes to equalization of faculty resources among schools. Rotation is also a tool to achieve fairness in repeated assignment problems. For example, seat allocations in classrooms are often believed to have an impact on student performance. Many schools in China implement seat rotation so that every student has a chance to sit in the front of classrooms. Ostrom (1990) reported that rotation schemes were used in irrigation systems in Spain and Philippines. Berkes (1992) reported that fishermen in Turkey used rotation schemes to allocate fishing spots. Rotation is different from randomization in static assignment problems that are intensively studied in the literature (Hylland and Zeckhauser, 1979; Abdulkadiroglu and Sonmez, 1998). This paper proposes a market design approach to solve job rotation and the rotation in repeated assignment problems. We consider a model in which every agent (employee) has occupied a distinct position and the employer wants to implement a rotation by eliciting agents’ preferences. To encourage rotation, the employer requires that if an agent prefers another agent’s position, the latter agent should leave the position. We use a priority structure to describe this requirement and call it job rotation priority: for every position the occupier has the unique lowest priority while the others have equal highest priority. Meeting the rotation requirement is equivalent to respecting the priority structure, or in other words, satisfying stability. It is interesting to contrast the job rotation priority with the classical housing market model of Shapley and Scarf (1974) in which for every object the owner has the unique highest priority while the others have equal lowest priority. 2

From this perspective, the job rotation model is the “opposite” to the housing market model. By assuming that agents have strict preferences, we design a mechanism that always finds a stable matching for the job rotation priority and the matching is Pareto undominated among all stable matchings; in this sense we call the mechanism constrained efficient. The mechanism is weakly group strategy-proof, which means that no group of agents can jointly misreport preferences to ensure that all of them become strictly better off. When there are at least four agents, we prove that there do exist stable, Pareto efficient and strategy-proof mechanisms and constrained efficient and strongly group strategy-proof mechanisms. So given stability, our mechanism has achieved the strongest combination of efficiency and (group) incentive compatibility. Our mechanism is an adaption of the famous Top Trading Cycle (TTC) mechanism, which was proposed by Shapley and Scarf (1974) to solve the housing market model. By regarding each agent as the owner of his position, our mechanism has a similar procedure as TTC. But it has two important differences. First, in every step of our mechanism, every agent is not allowed to point to his position except that he is the only remaining agent. Second, when cycles are generated in each step, they are removed but not cleared immediately. After all agents are involved in cycles, cycles are cleared by backward induction. A cycle is cleared either by letting the agents in the cycle exchange their positions or by letting them keep their positions. It depends on the preferences of the agents in the cycle and the preferences of the agents who are involved in cycles in later steps. If all agents in the cycle prefer their positions to those they point to, and all agents in cycles of later steps prefer their assignments to all positions in the cycle, the agents in the cycle keep their positions; otherwise, they exchange their positions. Our mechanism partially answers Ehlers and Westkamp’s (2018) question on the solvability of a special coarse priority structure. Priority-based allocation problems widely exist in real life. When priorities are strict, Deferred Acceptance (Gale and Shapley, 1962; Abdulkadiro˘glu and S¨onmez, 2003) and TTC perform well. However, priorities are coarse in many applications; school choice is a well-known example. Erdil and Ergin (2008) show that coarse priorities may not admit a strategy-proof and constrained efficient mechanism. Ehlers and Westkamp (2018) aim to characterize coarse priorities that admit such a desirable mechanism and call such priorities solvable. Among a large class

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of priority structures,1 when agents may regard some objects as unacceptable, they prove that at most two types are solvable: the house allocation with existing tenants (HET) type, and the task allocation with unqualified agents (TAU) type. HET is widely known to be solvable by TTC and its generalization (Abdulkadiro˘glu and S¨onmez, 1999). TAU is a type in which, for each object, either all agents have equal priority, or only one agent has the lowest priority and the others have equal highest priority. Ehlers and Westkamp discuss the difficulty of solving TAU and strongly suspect it to be unsolvable. When the number of agents is equal to the number of objects, TAU coincides with the job rotation priority.2 So what we prove is that, when agents regard all objects as acceptable, an interesting subclass of TAU is solvable. We consider several extensions of the job rotation priority. The first extension includes vacant positions and new employees. In practice, vacant positions appear if employees retire or resign and new employees can be hired. To address the possibility that employers may not want the occupiers of some important positions to change frequently, we allow such occupiers to be “existing tenants” of their positions. So the corresponding priority structure is a mixture of the job rotation priority and HET. As long as positions are no more than agents, our mechanism can be easily adapted and remains to be constrained efficient and weakly group strategy-proof. In particular, when positions are fewer than agents, our mechanism is strongly group strategy-proof. We also discuss extensions to which our mechanism cannot be easily adapted. Actually, we do not know whether these extensions are solvable. The second extension adds vacant positions to the job rotation priority so that positions are more than agents. Through an example (Example 1) we show that if vacant positions point to agents so that some agent is pointed by several positions during our mechanism, the mechanism fails constrained efficiency or strategy-proofness. A special case of this extension is that agents may regard some positions as unacceptable (i.e., there exist outside options). The third extension allows an agent to have the lowest priority for several positions, and then 1

They assume that there are at least six agents, at least four agents form a priority tie for some

object (four-way tie), and for any two agents i, j, if i has higher priority than j for some object, j has higher priority than i for another object (reversibility). 2 To be precise, TAU allows an agent to have the lowest priority for several objects. But when there are equal numbers of agents and objects, under the reversibility assumption (see footnote 1), every agent has the lowest priority for only one object, so TAU coincides with the job rotation priority.

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a similar difficulty in the second extension appears. Related literature This paper contributes to the literature on priority-based allocation problems. When agents may regard some objects as unacceptable, Ehlers and Erdil (2010) characterize priority structures under which constrained efficient matchings are Pareto efficient and priority structures under which the constrained efficient rule is singlevalued and strongly group strategy-proof. The job rotation priority does not belong to such priority structures. On the same preference domain, Han (2018) characterizes priority structures that admit a stable and Pareto efficient mechanism and priority structures that admit a stable, Pareto efficient and strategy-proof mechanism. The job rotation priority belongs to his first characterization, but does not belong to his second when there are at least four agents.3 Our Proposition 1 proves that when there are at least four agents, any stable and Pareto efficient mechanism under the job rotation priority is not strategy-proof, even though agents regard all objects as acceptable. Besides the result mentioned above, Ehlers and Westkamp (2018) also prove that when there are at least four agents, among the class of priority structures that satisfy reversibility (see footnote 1), HET is the only type that admits a constrained efficient and strongly group strategy-proof mechanism. The job rotation priority satisfies reversibility, but because we consider a smaller preference domain, Ehlers and Westkamp’s result does not apply. But we prove that the job rotation priority does not admit a constrained efficient and strongly group strategy-proof mechanism when there are at least four agents. Our mechanism is related to the various extensions of TTC in the literature. P´apai (2000) propose the class of hierarchical exchange mechanisms that augment TTC with a ¨ structure of ownerships. Pycia and Unver (2017) further extend P´apai’s mechanisms by introducing a new control right called brokerage. During their mechanisms, if an agent 3

Han (2018) shows that only three types of priority structures satisfy his first characterization: (1)

the house allocation priority in which all agents have equal priority for all objects; (2) the housing market priority in which, for every agent, there is a nonempty set of objects for which he has the unique highest priority and the remaining agents have equal priority, and the remaining objects give all agents equal priority; (3) the indifference at the top priority in which, for every agent, there is a nonempty set of objects for which he has the unique lowest priority and the remaining agents have equal priority, and the remaining objects give all agents equal priority. Han also shows that the first two types satisfy his second characterization, but the third type satisfies his second characterization only when there are no more than three agents. Our job rotation priority is a special case of his third type.

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brokers an object, he cannot point to his brokered object. So their usage of brokerage bears some similarity to our treatment of occupation. But there are important differences. Their purpose of introducing brokerage is to construct the full class of Pareto efficient and strongly group strategy-proof mechanisms. They require that at most one broker can exist in each step of their mechanisms, and a broker never obtains his brokered object. By contrast, in our mechanism every agent behaves as brokering his position (except that he is the only remaining agent in the last step), and this is to fulfill the rotation requirement. To maintain constrained efficiency, we clear cycles by backward induction and agents are possible to obtain their occupied positions. The rest of the paper is organized as follows. We present the job rotation model in Section 2, and define our mechanism in Section 3. We prove constrained efficiency and weak group strategy-proofness of our mechanism in Section 4. We present extensions in Section 5, and make concluding remarks in Section 6. Appendix includes omitted proofs.

2

Job Rotation Model

A job rotation (JR) problem (I, O, R) consists of • a set of n agents I = {1, 2, . . . , n} with n ≥ 3; • a set of n positions O = {o1 , o2 , . . . , on } in which each oi ∈ O is occupied by i ∈ I; • a preference profile R = (Ri )i∈I in which each Ri is the strict preference relation of i ∈ I on O. For all i ∈ I and all o, o ∈ O, we denote by oPi o that i strictly prefers o to o , and by oRi o that either oPi o or o = o . Let P be the set of all strict preference relations on O. The employer encourages agents to rotate positions. She requires that as long as an agent prefers another agent’s position, the second agent should leave his position. This requirement is represented by a priority structure = (o )o∈O in which, for all oi ∈ O and all j, k ∈ I\{i}, j ∼oi k oi i. In words, i has the lowest priority for oi and the other agents have equal highest priority. We call  job rotation priority (see Figure 1). A matching is a one-to-one mapping μ : I → O. Given a preference profile R ∈ P |I| , μ is stable in R if it respects the rotation priority: for all i ∈ I with μ(i) = oi , μ(j)Pj oi for all j ∈ I\{i}. In words, μ is stable in R if for any agent i who keeps his position 6

 o1

 o2

2, 3, . . . , n

1, 3, . . . , n

1

2

...

 on

. . . 1, 2, . . . , n − 1 ...

n

Figure 1: Job rotation priority oi , any other agent j does not want to move to oi . μ is Pareto dominated by a different matching ν if ν(i)Ri μ(i) for all i ∈ I and ν(j)Pj μ(j) for some j ∈ I. Then μ is Pareto efficient in R if it is not Pareto dominated by a different matching. μ is constrained efficient in R if μ is stable and not Pareto dominated by any other stable matching. Let M be the set of all matchings. A mechanism is a function f : P |I| → M that chooses a matching f (R) ∈ M for each R ∈ P |I| . f is stable, Pareto efficient, or constrained efficient respectively if, for all R ∈ P |I| , f (R) is stable, Pareto efficient, or constrained efficient respectively. For simplicity, we write fi (R) for the assignment of i in f (R). For all R ∈ P |I| and all Iˆ ⊆ I, let RIˆ be the preference profile of Iˆ and R−Iˆ be the preference profile of the other agents. An agent i manipulates a mechanism f in R by reporting Ri ∈ P if fi (R−i , Ri ) Pi fi (R). A group of agents Iˆ ⊆ I manipulates a mechˆ ˆ R = Ri and fi (R ˆ, R ) Ri fi (R), anism f in R by reporting RIˆ ∈ P |I| if, for all i ∈ I, i −I Iˆ

ˆ fj (R ˆ, R ) Pj fj (R). If fi (R ˆ, R ) Pi fi (R) for all i ∈ I, ˆ Iˆ strongly and for some j ∈ I, −I −I Iˆ Iˆ manipulates f . Then we say a mechanism f is • strategy-proof if it is never manipulated by a single agent; • strongly group strategy-proof if it is never manipulated by a group of agents; • weakly group strategy-proof if it is never strongly manipulated by a group of agents.

3

Backward-induction Top Trading Cycle

We come up with a mechanism called Backward-induction TTC (BTTC) to solve the job rotation problem. We regard each agent as the owner of his occupied position and run the TTC procedure. But as mentioned before, it has two important differences than 7

TTC. First, in every step an agent is not allowed to point to his occupied position except that he is the only remaining agent. Second, when cycles are generated in every step, we remove them but do not clear them immediately. Only after all agents are involved in cycles, we clear cycles by backward induction. The agents in any cycle of the last step keep their occupied positions if all of them prefer their occupied positions to the positions they point to; otherwise, they exchange positions as indicated by the cycle. The agents in any cycle of the second to last step keep their occupied positions if all of them prefer their occupied positions to the positions they point to, and all agents in cycles of the last step prefer their assignments to all positions in the cycle; otherwise, the agents in the cycle exchange positions as indicated by the cycle. We repeat this operation for all cycles of all steps. For simplicity, we denote the mechanism by f B . Backward-induction Top Trading Cycle Stage 1 (generating cycles): • At the beginning of step 1, all agents and positions are present. • In each step d ≥ 1, let each remaining position oi point to agent i. If there are more than one remaining agents, let each remaining agent i point to his most preferred position among the remaining ones different than oi ; otherwise, let the only remaining agent point to his position. Remove the agents and positions involved in cycles. Go to the next step. Stage 1 stops when all agents are involved in cycles. • We define some notations that are useful to describe Stage 2. Let d¯ be ¯ let C(d) be the set of the last step of Stage 1. For each 1 ≤ d ≤ d, ¯ cycles generated in step d, and let C = dd=1 C(d). For each c ∈ C, let I(c) and O(c) be the set of agents and the set of positions involved   in c, respectively. Let I(d) = c∈C(d) I(c) and O(d) = c∈C(d) O(c) be the set of agents and the set of positions involved in cycles of step d, respectively. For all c ∈ C and all i ∈ I(c), let pi ∈ O(c) be the position pointed by i. Stage 2 (clearing cycles): Cycles are cleared as follows.

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¯ if all i ∈ I(c) prefers oi to pi , then let all i ∈ I(c) • For each c ∈ C(d), obtain oi ; otherwise, let all i ∈ I(c) obtain pi .4 ¯ cycles in step d are traded after cycles in later steps • For any 1 ≤ d < d, have been traded. For each c ∈ C(d), if all i ∈ I(c) prefers oi to pi and ¯ all j ∈ dd =d+1 I(d ) prefers his assignment to all o ∈ O(c), then let all i ∈ I(c) obtain oi . Otherwise, let all i ∈ I(c) obtain pi . Our main theorem proves that BTTC is desirable. Theorem 1. BTTC is constrained efficient and weakly group strategy-proof. We prove Theorem 1 in the next section. The job rotation priority satisfies Han’s (2018) first characterization (see footnote 3), so there always exists a stable and Pareto efficient mechanism. When there are only three agents, DA with a preference-based tiebreaking rule proposed by Ehlers (2006) (see also Han (2018)) is stable, Pareto efficient and strongly group strategy-proof.5 But when there are at least four agents, we prove that there do not exist (1) a stable, Pareto efficient and strategy-proof mechanism and (2) a constrained efficient and strongly group strategy-proof mechanism. Proposition 1. When there are at least four agents, there do not exist 1. a stable, Pareto efficient and strategy-proof mechanism; 2. a constrained efficient and strongly group strategy-proof mechanism. Proof. To prove the first statement, let us consider the three preference profiles R, R , R in which four agents 1, 2, 3, 4 have the preferences shown in Table 1 over their occupied positions o1 , o2 , o3 , o4 , and prefer o1 , o2 , o3 , o4 to the other positions, whereas each i > 4 most prefers oi . In each of the three preference profiles, every Pareto efficient matching must assign {o1 , o2 , o3 , o4 } to {1, 2, 3, 4}, and assign each oi with i > 4 to i. 4 5

¯ then pi = oi and i obtains oi . If i is the only remaining agent in step d, In every step of the mechanism, let the three agents apply to their most preferred positions that

have not rejected them. If a matching is not found, at least two agents apply to the same position. Then, (1) if the three agents apply to the same position, let the agent of the lowest priority be rejected; (2) if two agents apply to the same position and one of them has the lowest priority for the object, let the agent of the lowest priority be rejected; (3) if two agents apply to the same position and they have equal priority for the position, the third agent must apply to a position for which one of the former two agents has the lowest priority. Let the agent of the lowest priority be rejected.

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R1

R2

R3

R4

R1

R2

R3

R4

R1

R2

R3

R4

o1

o2

o3

o3

o2

o2

o3

o3

o2

o2

o3

o4

o2

o3

o2

o4

o1

o3

o2

o4

o1

o3

o2

o3

o3

o1

o4

o2

o3

o1

o4

o2

o3

o1

o4

o2

o4

o4

o1

o1

o4

o4

o1

o1

o4

o4

o1

o1

(a) R

(b) R

(c) R

Table 1: Three preference profiles for the proof of Proposition 1 We first prove that in R there is a unique stable and Pareto efficient matching μ in which μ(1, 2, 3, 4) = (o4 , o1 , o2 , o3 ). To see it, note that because 4 most prefers o3 , stability requires that 3 cannot obtain o3 . Then because o2 is 3’s most preferred position among those other than o3 , stability requires that 2 cannot obtain o2 . Because 1 most prefers o1 but both 3 and 4 least prefers o1 (among o1 , o2 , o3 , o4 ), Pareto efficiency requires that 3 and 4 cannot obtain o1 . So suppose 4 obtains o2 in a stable and Pareto efficient matching, then 3 must obtain o4 . But this matching is not Pareto efficient since 3 and 4 can be better off by exchanging their assignments. Suppose 4 obtains o4 in a stable and Pareto efficient matching, then 3 must obtain o2 . Then if 1 obtains o1 and 2 obtains o3 , then the matching is not Pareto efficient because 2 and 3 can be better off by exchanging their assignments. But if 1 obtains o3 and 2 obtains o1 , the matching is still not Pareto efficient because 1 and 2 can be better off by exchanging their assignments. So in any stable and Pareto efficient matching 4 must obtain o3 . Given this fact, suppose 3 obtains o4 , then 2 must obtain o1 and 1 must obtain o2 . But again the matching is not Pareto efficient because 1 and 2 can be better off by exchanging their assignments. So in any stable and Pareto efficient matching 4 must obtain o3 and 3 must obtain o2 . Then 1 must obtain o4 and 2 must obtain o1 because the matching in which 1 obtains o1 and 2 obtains o4 violates stability. By very similar arguments we prove that in R there is a unique stable and Pareto efficient matching μ in which μ (1, 2, 3, 4) = (o2 , o3 , o4 , o1 ). Now suppose f is a stable and Pareto efficient mechanism. Then f (R)(1, 2, 3, 4) = (o4 , o1 , o2 , o3 ), f (R )(1, 2, 3, 4) = (o2 , o3 , o4 , o1 ). 10

Since o4 is the least preferred object (among o1 , o2 , o3 , o4 ) in both R1 and R1 and o1 is the least preferred object (among o1 , o2 , o3 , o4 ) in both R4 and R4 , if f is strategy-proof, it must be that f1 (R ) = f1 (R) = o4 and f4 (R ) = f4 (R ) = o1 . However, it means that f (R ) is not Pareto efficient in R because 1 and 4 can be better off by exchanging their assignments. So f is not strategy-proof. To prove the second statement, we prove that a constrained efficient and strongly group strategy-proof mechanism must be Pareto efficient. Then by the first statement, constrained efficient and strongly group strategy-proof mechanisms do not exist. Claim. A constrained efficient and strongly group strategy-proof mechanism f is Pareto efficient. Proof of Claim. Suppose towards a contradiction that in some preference profile R, f (R) is Pareto dominated by a matching μ. Consider a preference profile R in which all i ∈ I reports μ(i) as most preferred. Obviously, μ is stable and Pareto efficient in R since every agent obtains his most preferred position. Actually, μ is the unique constrained efficient matching in R . So f (R ) = μ. But this means that f is not strongly group strategy-proof: the group of all agents can manipulate f in R by reporting R .

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Proof of Theorem 1

4.1

Constrained efficiency

For any preference profile R, let μ be the matching found by BTTC. Then, for all i ∈ I, either μ(i) = pi or μ(i) = oi . By the rule of clearing cycles in BTTC, μ(i) = oi only if pi = oi or i prefers oi to pi . So μ(i)Ri pi . ¯ suppose there exists i ∈ I(d) such that μ(i) = oi . (Stability) For any step 1 ≤ d ≤ d, We want to prove that for all j ∈ I\{i}, μ(j)Pj oi . For all j ∈ I(d )\{i} with d ≤ d, since oi remains available in step d , all j ∈ I(d ) must point to a position weakly better than oi . Thus, μ(j)Rj pj Rj oi . Since μ(j) = oi , μ(j)Pj oi . For all j ∈ I(d ) with d > d, Stage 2 of BTTC ensures that μ(j)Pj oi . (Constrained efficiency) Suppose there is a stable matching ν such that ν(i)Ri μ(i) for all i ∈ I. We prove by induction that ν(i) = μ(i) for all i ∈ I.

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¯ Since every j ∈ I\I(d) ¯ is involved in a cycle • Base case. Consider the last step d. ¯ it must be that μ(j)Rj pj Pj o for all o ∈ O(d). ¯ Thus, ν(j)Rj μ(j)Pj o before step d, ¯ Consequently, for all i ∈ I(d), ¯ it must be that ν(i) ∈ O(d). ¯ for all o ∈ O(d). ¯ such that ν(i)Pi μ(i). Suppose towards a contradiction that there exists i ∈ I(d) ¯ otherwise, ν(i) = oi = μ(i). Then i cannot be the only remaining agent in step d; ¯ Thus, pi is i’s most preferred position among O(d)\{o i }. Since ν(i)Pi μ(i)Ri pi and ¯ it must be that ν(i) = oi and μ(i) = pi . Let k be the agent such that ν(i) ∈ O(d), μ(k) = oi . Then ν(k) = μ(k), which means that ν(k)Pk μ(k). By a similar argument we have [ν(k) = ok ]Pk [μ(k) = pk = oi ]. Repeating this argument, we conclude that for all agent  belonging to the same cycle as i, [ν() = o ]P [μ() = p ]. But this means that the agents in the cycle should obtain their occupied positions in BTTC, ¯ we must have ν(i) = μ(i). which is a contradiction. So for all i ∈ I(d), ¯ Suppose for all j ∈  + I(d+ ) we • Induction step. Consider any step 1 ≤ d < d. d >d have proved that ν(j) = μ(j), we then prove that for all i ∈ I(d), ν(i) = μ(i).  Similarly as above, every j ∈ d− d I(d+ ) do not obtain positions from O(d). So for all i ∈ I(d), it must be that ν(i) ∈ O(d). Suppose towards a contradiction that there exists i ∈ I(d) such that ν(i)Pi μ(i). Since d is a middle step, i is not the only remaining agent in step d. Thus, pi is i’s most preferred position among O(d)\{oi }. Since ν(i)Pi μ(i)Ri pi , it must be that  [ν(i) = oi ]Pi [μ(i) = pi ]. Since ν is stable, for all j ∈ d+ >d I(d+ ), [ν(j) = μ(j)]Pj oi . Let k be the agent such that μ(k) = oi . Then, by a similar argument we have  [ν(k) = ok ]Pk [μ(k) = pk = oi ], and for all j ∈ d+ >d I(d+ ), [ν(j) = μ(j)]Pj ok . Repeating the argument, we conclude that for all agent  belonging to the same  cycle as i, [ν() = o ]P [μ() = p ] and for all j ∈ d+ >d I(d+ ), [ν(j) = μ(j)]Pj o . But this means that the agents in the cycle should obtain their occupied positions ¯ we must have ν(i) = μ(i). in BTTC, which is a contradiction. So for all i ∈ I(d),

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4.2

Weak group strategy-proofness

Our proof strategy is as follows. We first prove that BTTC is strategy-proof, then prove that it satisfies the conditions of Barber`a et al. (2016). In a general framework, Barber`a et al. prove that if each agent’s preference domain is rich, a strategy-proof mechanism satisfying their conditions is weakly group strategy-proof. Our model belongs to their framework and our preference domain satisfies their richness. So BTTC is weakly group strategy-proof. Before presenting our proof, it is useful to compare BTTC and TTC to understand why proving strategy-proofness for BTTC is different from the proof for TTC. In TTC, suppose an agent i is involved in a cycle in step d. Then the owners of the objects removed before step d must form cycles among themselves no matter how i misreports preferences. So the best object i can obtain in TTC is the best object among the remaining ones in step d. This object is exactly the one pointed and obtained by i in step d. In BTTC, it is also true that if i is involved in a cycle in step d, i has no way to obtain any position removed before step d by misreporting preferences, since the occupiers of such positions form cycles among themselves. However, in BTTC the position obtained by i may not be i’s most preferred position among the remaining ones in step d. There are two cases. First, if the position obtained by i is his most preferred position among the remaining ones in step d, then as in TTC, i has no way to obtain a better position by misreporting preferences. Second, if the position obtained by i is not his most preferred position among the remaining ones in step d, then he must obtain pi and prefer oi to pi . Then we need to check whether i can obtain oi by misreporting preferences and changing the set of cycles generated in step d and later steps. We prove that there is no chance for such manipulation. Formally, to understand how a group of agents Iˆ ⊆ I can change the outcome of BTTC by misreporting preferences, we introduce an alternative algorithm that is equivalent to BTTC. In the algorithm, we first run the generating cycles stage of BTTC but let the agents in Iˆ not point to any position. After all agents not in Iˆ are either involved in cycles or linked to agents in Iˆ through directed paths, we remove all cycles.6 These cycles are generated no matter how Iˆ misreport preferences. The positions in these cycles are those 6

A directed path that links j to i is a sequence of agents {1 , . . . , m } and their positions such that

j → o1 → 1 → o2 → · · · → m → oi → i.

13

that Iˆ have no chances to obtain by misreporting preferences. We then run the generating cycles stage of BTTC again by letting Iˆ reveal their preferences. Lastly, we clear cycles by backward induction as in BTTC. Alternative Procedure (APIˆ) for given Iˆ ⊆ I Stage 1 (generating cycles): • Step 1 to s: Run Stage 1 of BTTC but let all i ∈ Iˆ not point to any position. Let s be the step after which every j ∈ I\Iˆ is either involved in a cycle, or is linked to some i ∈ Iˆ through a directed path. Remove the agents and positions involved in cycles. Let AIˆ and OIˆ be the set of remaining agents and the set of remaining positions respectively. Note that Iˆ ⊆ AIˆ. • Step s + 1 to s + t¯: Run Stage 1 of BTTC when the set of agents is AIˆ and the set of objects is OIˆ. In particular, we let Iˆ point to positions as in BTTC. Let s + t¯ be the last step after which all agents in AIˆ are involved in cycles. Stage 2 (clearing cycles): Run Stage 2 of BTTC (using the same notations defined for BTTC). In words, I\AIˆ is the set of agents who form cycles among themselves irrespective of ˆ and A ˆ is the set of agents who are affected by the preferences of I. ˆ the preferences of I, I For any nonempty Iˆ ⊆ I, we denote by fIˆA the mechanism induced by the above algorithm APIˆ. Lemma 1 below proves that BTTC and APIˆ are equivalent. So we can use APIˆ to analyze the outcome of BTTC. The proof of the lemma is presented in Appendix. Lemma 1. For all R ∈ P |I| and all nonempty Iˆ ⊆ I, f B (R) = fIˆA (R). Lemma 2 proves that Iˆ obtain their occupied positions in APIˆ if and only if every agent in AIˆ most prefers his occupied position among OIˆ. Lemma 2. For all R ∈ P |I| and all nonempty Iˆ ⊆ I, fiB (R) = oi for all i ∈ Iˆ if and only if 14

oj Rj o for all j ∈ AIˆ and all o ∈ OIˆ, where AIˆ and OIˆ are defined in APIˆ. Proof. If all j ∈ AIˆ most prefers oj among OIˆ, then it is obvious that all j ∈ AIˆ obtains oj in APIˆ. This proves the “if” part. Below we prove the “only if” part. Suppose towards a contradiction that there exists j ∈ AIˆ whose most preferred position among OIˆ is ok = oj . There are two cases to consider. ˆ After step s of AP ˆ, j must point to ok and there is a directed path • Case 1: j ∈ AIˆ\I. I ˆ If j obtains ok , j → ok → k → ok1 → · · · → km → oi → i that links j to some i ∈ I. k cannot obtain ok . If j does not obtain ok , stability requires that k cannot obtain ok . Thus, we conclude that fkB (R) ∈ OIˆ\{ok }. Since k’s most preferred position among OIˆ\{ok } is ok1 , we can similarly conclude that fkB1 (R) = ok1 . By repeating this argument, we conclude that fiB (R) = oi , which is a contradiction. ˆ Thus, f B (R) = oj = ok . Then, stability requires that f B (R) = ok , • Case 2: j ∈ I. j k ˆ Thus, k ∈ A ˆ\I. ˆ Similarly as above, there is a path which means that k ∈ / I. I ˆ By repeating the above argument, we conclude that that links k to an agent i ∈ I. fiB (R) = oi , which is a contradiction.

Lemma 2 answers why the second case of manipulating BTTC discussed at the beginning of this subsection cannot happen. If i obtains his occupied position by misreporting preferences, all agents in A{i} must most prefer their occupied positions among O{i} . So if i reports true preferences, i must still obtain his occupied position. Now we are ready to prove strategy-proofness. Lemma 3. BTTC is strategy-proof. Proof. Suppose towards a contradiction that there exist R ∈ P |I| , i ∈ I, and Ri ∈ P such that fiB (R−i , Ri ) Pi fiB (R). To ease notations, let f B (R−i , Ri ) = μ and f B (R) = μ. When i reports Ri , let pi be the position pointed by i when he is removed in BTTC. We first prove that pi = pi . We consider the equivalent procedure AP{i} . No matter i reports Ri or Ri , the first s steps of AP{i} do not change. In step s+1 of AP{i} , since every agent in A{i} \{i} is linked 15

to i through a directed path, i must be involved in a cycle no matter what position he points to. Suppose towards a contradiction that pi = pi . That is, by reporting Ri , i still points to pi in step s + 1 of AP{i} . Then there is no change in the generating cycles stage of AP{i} . If Ri and Ri have equal rankings of oi and pi , then it must be that μ(i) = μ (i). Thus, μ (i)Pi μ(i) holds only if Ri and Ri have different rankings of oi and pi . If μ (i) = oi and μ(i) = pi , it means that oi Pi pi and pi Pi oi , which contradicts μ (i) = oi . If μ (i) = pi and μ(i) = oi , it means that pi Pi oi and oi Pi pi , which contradicts μ(i) = oi . So it must be that pi = pi . Since pi = pi , |O{i} | ≥ 2. So pi is i’s most preferred position among O{i} \{oi }, which implies that pi Pi pi . Thus, μ (i)Pi μ(i) holds only if μ (i) = oi , μ(i) = pi and oi Pi pi . Since μ (i) = oi , by Lemma 2, all j ∈ A{i} most prefers oj among O{i} . So if i reports Ri , i must also obtain oi in AP{i} . It means that μ(i) = oi , which contradicts μ(i) = pi . So BTTC is strategy-proof. To prove that BTTC is weakly group strategy-proof, we prove that our preference domain satisfies the richness condition defined by Barber`a et al. (2016) and BTTC satisfies their joint monotonicity and respectfulness conditions. The definitions of the three conditions and the proof of Lemma 4 are presented in Appendix. Lemma 4. The domain P satisfies richness in Barber` a et al. (2016), and BTTC satisfies joint monotonicity and respectfulness. So BTTC is weakly group strategy-proof.

5

Extensions

In this section we discuss some extensions of the job rotation priority. BTTC can be adapted to solve an extension in which vacant positions and new employees are present as long as positions are no more than agents, but cannot solve other extensions in which positions are more than agents or an agent occupies several positions.

5.1

Vacant positions, new employees and “existing tenants”

In our main model, all employees occupy positions and all positions are occupied. But in real-life firms, positions can become vacant because of retirements and new employees can be hired. So we consider an extended model with vacant positions and new em16

ployees. Moreover, we allow some employees to be “existing tenants” of their positions. This addresses the possibility that firms may discourage the rotation of some important positions. We show that BTTC can be easily adapted to solve the extended model when positions are no more than employees. Formally, an extended job rotation (EJR) problem (I, O, , R) consists of • a set of n + m + q agents I = {1, 2, . . . , n + m + q} with n ≥ 3, m ≥ 0 and q ≥ 1; • a set of n + m + q  positions O = {o1 , o2 , . . . , on+m+q } with 0 ≤ q  ≤ q; • = (o )o∈O is a priority structure (see Figure 2) such that – for all 1 ≤ i ≤ n and j, k ∈ I\{i}, j ∼oi k oi i; – for all n + 1 ≤ i ≤ n + m and j, k ∈ I\{i}, i oi j ∼oi k; – for all n + m + 1 ≤ i ≤ n + m + q  and j, k ∈ I, j ∼oi k.  o1

...

on

on+1

...

on+m

on+m+1

...

on+m+q

I\{1}

...

I\{n}

n+1

...

n+m

I

...

I

1

...

n

I\{n + 1}

...

I\{n + m}

Figure 2: Extended job rotation priority • R = (Ri )i∈I is a preference profile in which Ri is a strict preference relation on O. In words, IC = {1, 2, . . . , n} is the set of current employees who respectively occupy {o1 , o2 , . . . , on } and are encouraged to rotate their positions. So each of them has the lowest priority for his position. IE = {n + 1, n + 2, . . . , n + m} is the set of current employees who respectively occupy {on+1 , on+2 , . . . , on+m } and are discouraged to rotate their positions. So each of them has the highest priority for his position and is interpreted as an existing tenant. IN = {n+m+1, n+m+2, . . . , n+m+q} is the set of new employees and OV = {on+m+1 , on+m+2 , . . . , on+m+q } is the set of vacant positions. Note that q  ≤ q means that positions may not be enough. Now a matching μ is a mapping from I to O ∪ {∅} such that for every distinct i, j ∈ I, μ(i) = μ(j) if and only if μ(i) = μ(j) = ∅. μ is stable if (1) for all i ∈ IC such that μ(i) = oi , μ(j)Pj oi for all j ∈ I\{i}, and (2) for all i ∈ IE , μ(i)Ri oi . In particular, if positions are not enough (q  < q), in every stable matching every i ∈ IC must not obtain oi . 17

It turns out that BTTC can be easily adapted to solve this model. We choose q  new employees as the existing tenants of the q  vacant positions and let the other q − q  new employees occupy ∅. Stage 0 (generating ownership): For each oi ∈ {on+m+1 , . . . , on+m+q }, regard i ∈ IN as the existing tenant of oi . Let IN = {n+m+1, . . . , n+m+q  }. Stage 1 (generating cycles): In each step, let all remaining oi ∈ O point to i, all remaining i ∈ IN ∪ IE point to his most preferred remaining position, and all remaining i ∈ IC point to his most preferred remaining position other than oi if there are at least two remaining agents and otherwise point to oi . In particular, if in some step the position oi of some i ∈ IC becomes the only remaining position and q  < q, let i point to ∅, ∅ point to agent n + m + q  + 1, and n + m + q  + 1 point to oi (that is, the two agents form a cycle). Remove all agents and positions involved in cycles and go to the next step. Stage 1 stops when all positions are involved in cycles. Stage 2 (clearing cycles): If q  = q, clear cycles by backward induction as in BTTC. If q  < q, clear all cycles by letting every agent obtain the position he points to. Agents not involved in any cycle do not obtain positions. When positions are not enough (q  < q), because every i ∈ IC cannot obtain oi in any stable matching, the agents in every cycle exchange their positions as indicated by the cycle. In particular, if the position oi of some i ∈ IC becomes the only remaining position in the last step, we let i obtain ∅ and n + m + q  + 1 obtain oi . So BTTC essentially coincides with TTC except that every i ∈ IC cannot point to oi . This means that BTTC is strongly group strategy-proof. Proposition 2. In the extended job rotation model, BTTC is constrained efficient and weakly group strategy-proof. When positions are not enough, BTTC is strongly group strategy-proof. Proof. By almost the same arguments as before we can prove that BTTC is constrained efficient. In any cycle that involves new employees or existing tenants, the agents in the cycle must obtain the positions they point to. So from the perspective of new employees and existing tenants, BTTC coincides with TTC, which means that BTTC is strategyproof for them. By almost the same arguments as before we can prove that BTTC is 18

strategy-proof for IC . So BTTC is strategy-proof. We can prove similarly as before that BTTC satisfies joint monotonicity and respectfulness. So BTTC is weakly group strategy-proof. When positions are not enough, we prove that BTTC is non-bossy; that is, for any R ∈ P |I| , i ∈ I and Ri ∈ P, if fiB (R) = fiB (Ri , R−i ), we prove that f B (R) = f B (Ri , R−i ). Note that fiB (R) = fiB (Ri , R−i ) means that, i points to the same position in (Ri , R−i ) as in R when he is involved in a cycle. This implies that in (Ri , R−i ) the agents other than i must point to the same positions in every step of BTTC as they do in R. So the cycles generated in (Ri , R−i ) are same as those generated in R, and thus it must be that f B (R) = f B (Ri , R−i ). By P´apai (2000), strategy-proofness and non-bossiness together imply strong group strategy-proofness. Abdulkadiro˘glu and S¨onmez (1999) propose the you request my house - I get your turn (YRMH-IGYT) mechanism to solve HET. When positions are not enough, YRMH-IGYT can be adapted to solve the extended job rotation problem. We first select an ordering of all agents r : I → {1, 2, . . . , n + m + q}. Then we assign the first agent r(1) his favorite position (other than or(1) if r(1) ∈ IC ), the second agent r(2) his favorite position among remaining ones (other than or(2) if r(2) ∈ IC ), and so on, until some i demands the occupied position of an existing tenant j. If at that point j is already assigned a position, then we assign oj to i and proceed. Otherwise, modify the remainder of the ordering by inserting j to the top and proceed. If at any point a cycle forms, it must consist of existing tenants and we let them exchange their positions. It is not hard to show that YRMH-IGYT is also constrained efficient and strongly group strategy-proof. Remark 1. For HET, Abdulkadiro˘glu and S¨ onmez (1999) prove that YRMH-IGYT equipped with an ordering r is equivalent to TTC equipped with r as priority ranking.7 When positions are not enough, there does not exist an ordering r such that BTTC is equivalent to YRMH-IGYT equipped with r. It is because BTTC does not use any ordering to break priority ties.8 7

Each vacant house uses r as its priority ranking and each occupied house gives the highest priority

to its existing tenant and uses r to rank the remaining agents. 8 By contradiction, suppose BTTC is equivalent to YRMH-IGYT equipped with an ordering r. For all preference profiles in which all agents prefer vacant positions to other positions, vacant positions must   in BTTC. So the ordering r must place IN as the first q  agents. But for all preference be assigned to IN

19

5.2

Extensions to which BTTC cannot be adapted

In this subsection we present two extensions to which BTTC cannot be adapted. Actually, we do not know whether these extensions are solvable. Positions are more than agents. We consider an extension that adds vacant positions to our main model (without adding new employees). So there are n ≥ 3 agents I = {1, 2, . . . , n} and more than n positions O = {o1 , o2 , . . . , on , on+1 , . . . , on+m } where m > 0. OV = {on+1 , . . . , on+m } is the set of vacant positions. In HET, if there are more objects than agents, TTC can be easily adapted by generating priority rankings for vacant objects. Suppose in BTTC we similarly let vacant positions point to agents according to some priority rankings. Then some agent is pointed by more than one positions. When some occupied positions become vacant after their occupiers are removed, they should point to remaining agents according to some priority rankings. However, the following example illustrates the difficulty in this idea. Example 1. There are three agents 1, 2, 3 and four positions o1 , o2 , o3 , o4 . o4 is vacant. Agents have the preferences shown in Figure 3a. R1

R2

R3

o1

o4

o2

o2

o3

o3

o3

o2

o1

o4

o1

o4

3

o2

1

o3

2

o4

o1

(b) Step one of BTTC

(a) Preferences

Figure 3: Example 1 Suppose in step one of BTTC, o4 points to 1. Then we obtain Figure 3b. Observe that there is one cycle consisting of 1, 2 and o2 , o4 . Remove this cycle. In step two, 3 becomes the only remaining agent. Since o1 becomes vacant, let it point to 3. Then 3 and o3 form a cycle. By clearing cycles, we obtain the matching μ(1, 2, 3) = (o2 , o4 , o3 ). However, μ is profiles in which all agents prefer the positions of IC to other positions and IC exchange their positions among themselves in BTTC, r must place IC as the first n agents since otherwise some agent not in IC must obtain some position of IC . This is a contradiction.

20

not constrained efficient: since 1 most prefers o1 , if we assign o1 to 1, it does not violate any other agent’s priority. On the other hand, suppose in step one of BTTC, o4 points to 3. Then there is one cycle consisting of 2, 3 and o2 , o4 . In step two, 1 becomes the only remaining agent. Since o3 becomes vacant, let it point to 1. Then 1 and o1 form a cycle. By clearing the two cycles, we obtain the unique constrained efficient matching for this problem: μ (1, 2, 3) = (o1 , o4 , o2 ). The above example seems to suggest that to ensure constrained efficiency, the priority rankings used by vacant positions should depend on agents’ preferences. But preferencebased priority may create the chance of manipulation. We are not aware of any intricate preference-based priority rule to overcome this difficulty. We are neither aware of other mechanisms to solve this model. The only solvable case we know is when there are only three agents (as in the above example). As mentioned before, DA with a preferencebased priority rule provided by Ehlers (2006) is stable, Pareto efficient and weakly group strategy-proof.9 If agents may regard some positions as unacceptable (Ehlers and Westkamp, 2018), it is equivalent to assuming that agents have outside options and so positions are more than agents. Ehlers and Westkamp use the priority structure in Example 1 to explain why the usual methods in the literature do not work and they strongly suspect TAU to be unsolvable. Occupying multiple positions. In firms it is natural that each employee occupies one position. But it is interesting to examine whether BTTC can solve a TAU problem of Ehlers and Westkamp (2018) in which an agent has the lowest priority for several positions (objects). In this extension, an agent will be pointed by several positions in BTTC. Through the following example we show that similar difficulty can appear. Example 2. There are four agents 1, 2, 3, 4 and four positions o1 , o2 , o3 , o4 . The priority structure is shown in Figure 4a. Agent 1 has the lowest priority for both o1 and o2 . Agents have the preferences shown in Figure 4b. 9

The mechanism is described in footnote 5 and its generalization is described in Han (2018). Given

three agents, when there are three positions, the mechanism is strongly group strategy-proof; when there are more than three positions, the mechanism is weakly group strategy-proof.

21

R1

R2

R3

R4

 o1

 o2

 o3

 o4

o2

o4

o3

o1

2, 3, 4

2, 3, 4

1, 2, 4

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o2

1

1

3

4

o1

o3

o2

o3

o4

o1

o4

o4

(a) Priority structure

(b) Preferences

Figure 4: Example 2 In step one of BTTC, if both o1 and o2 point to 1, then we obtain Figure 5a. In particular, 1 points to his most preferred position o3 other than his occupied positions o1 , o2 . There is one cycle consisting of 1, 3 and o1 , o3 . Remove this cycle.

2

4

o1

1

o4

3

o3

o2

(a) Step one

o2

4

2

o4

(b) Step two

Figure 5: Steps of BTTC In step two, 2 points to o4 and 4 points to o2 , so we obtain Figure 5b. If o2 points to 2, then remaining agents and positions are involved in one cycle. If o2 points to 4, then 4 and o2 are involved a cycle in step two, and 2 and o4 must be in a cycle in step three. Thus, no matter which agent is pointed by o2 in step two, in the final matching 4 must obtain o2 and 2 must obtain o4 . Then we must clear the cycle in step one by letting 1 obtain o3 and 3 obtain o1 . However, this matching is not constrained efficient since it is Pareto dominated by another stable matching in which 1 obtains o2 , 2 obtains o4 , 3 obtains o3 , and 4 obtains o1 .

6

Conclusion

This paper provides a market design approach to solve the job rotation problem. The problem is described by a priority structure in which each employee has the lowest priority for his position and the other employees have equal highest priority. It is an interesting counterpart to the house exchange problem in the sense that the former allows agents 22

to “veto” the status-quo allocation and the latter allows agents to “veto” changes to the status-quo allocation. We adapt the famous Top Trading Cycle mechanism to solve this problem. Every employee is regarded as the owner of his position, and during our mechanism every employee is not allowed to point to his position except that he is the only remaining agent. Different from TTC, only after all employees are involved in cycles, we clear cycles by backward induction. Our mechanism is stable, constrained efficient and weakly group strategy-proof. When there are at least four agents, there do not exist stable, Pareto efficient and strategy-proof mechanisms and constrained efficient and strongly group strategy-proof mechanisms. In this sense our mechanism is the best stable mechanism one can design for the job rotation problem. We discuss several extensions of the model. Our mechanism partially answers an open question of Ehlers and Westkamp (2018) on the solvability of coarse priority structures. In our model, rotation happens when at least one employee prefers to do so. If rotation is mandatory irrespective of employees’ preferences, then the problem becomes easy to solve. We can arbitrarily order employees and then let the first employee choose his most preferred matchings from the set of feasible matchings, the second employee choose his most preferred matchings from the set of matchings chosen by the first employee, and so on. In the end, one matching will be selected. This is an adaption of the serial dictatorship mechanism. To address fairness we can draw the ordering of employees uniformly at random as in Abdulkadiroglu and Sonmez (1998). On the other hand, TTC becomes not appropriate here because if only one employee remains in the last step of TTC, the employee has to keep his position, which violates mandatory rotation. In real life, employers rotate employees’ positions from period to period. Our analysis considers only one period and takes as input employees’ current positions and their preferences for the next period. This approach simplifies the analysis but also ignores the inherent dynamic feature of job rotation. Employees may have intertemporal preferences and the employer may have dynamic objectives. How to model job rotation in a dynamic setup and how to design an efficient and incentive compatible mechanism are interesting questions for future research.

23

Appendix Proof of Lemma 1 For all R ∈ P |I| and all nonempty Iˆ ⊆ I, it is easy to prove as in Section 4.1 that fIˆA (R) is stable. Below we prove that the same set of cycles appear in BTTC and APIˆ, and every cycle is cleared in the same way in the two algorithms. It means that f B (R) = fIˆA (R). Step 1: same cycles appear in the two algorithms. In APIˆ, I\AIˆ form cycles among themselves. In every step d of BTTC, every i ∈ I\AIˆ must point to the same position as he does in step d of APIˆ until i is involved in a cycle. So in BTTC I\AIˆ must form the same cycles among themselves as they do in APIˆ, and the ordering of steps in which these cycles appear does not change. The above fact implies that AIˆ must form cycles among themselves in the two algorithms. We prove by induction that AIˆ must form the same cycles in the two algorithms. In each step s + t (1 ≤ t ≤ t¯) of APIˆ, we denote by CIA ˆ (s + t) the set of cycles generated A in step s + t, and by IIA ˆ (s + t) and OIˆ (s + t) the set of agents and the set of positions

involved in these cycles, respectively. Base case. We first prove that every i ∈ IIA ˆ (s + 1) points to the same position when he is removed in the two algorithms. If |AIˆ| = 1, then i must point to oi in APIˆ. Since AIˆ form cycles among themselves in the two algorithms, i also points to oi in BTTC. If |AIˆ| > 1, in APIˆ, when i ∈ IIA ˆ (s + 1) is removed in step s + 1, there is a cycle i → oi1 → i1 → oi2 → · · · → im → oi → i such that oi1 is i’s most preferred position among OIˆ\{oi }, oi+1 is i ’s most preferred position among OIˆ\{oi } for all 1 ≤  < m, and oi is im ’s most preferred position among OIˆ\{oim }. Let d be the step of BTTC in which i is removed. We have shown that in BTTC, im must point to a position in OIˆ when he is removed. Since oi is his most preferred position among OIˆ\{oim }, im and oim must remain in the algorithm at the beginning of step d of BTTC. Similarly, for all 1 ≤  ≤ m, i and oi must remain in the algorithm at the beginning of step d of BTTC. Since i points to some position in OIˆ when he is removed in BTTC, i must point to oi1 in step d of BTTC. Induction step. Suppose it is true that, for all step τ with 1 ≤ τ < t ≤ t¯, every agent in IIA ˆ (s + τ ) points to the same position when he is removed in the two algorithms. We then prove that every i ∈ IIA ˆ (s + t) points to the same position when he is removed 24

in the two algorithms. Define I  = AIˆ\



A τ
+ τ ). The induction assumption

implies that I  must form cycles among themselves in the two algorithms. Define O =  OIˆ\ τ 1, in APIˆ, when i ∈ IIA ˆ (s + t) is removed in step s + t, there is a cycle i → oi1 → i1 → oi2 → · · · → im → oi → i such that oi1 is i’s most preferred position among O \{oi }, oi+1 is i ’s most preferred position among O \{oi } for all 1 ≤  < m, and oi is im ’s most preferred position among O \{oim }. By a similar argument as above, i must point to oi1 when he is removed in BTTC. Step 2: every cycle is cleared in the same way in the two algorithms. Define E = {i ∈ I : fIˆA (R)(i) = oi = f B (R)(i) = pi } and F = {i ∈ I : f B (R)(i) = oi = fIˆA (R)(i) = pi }. If f B (R) = fIˆA (R), then either E = ∅ or F = ∅. If E = ∅, define d∗ = max{d : i ∈ I(d) ∩ E}, and let i be an agent such that i ∈ I(d∗ ) ∩ E. That is, i is an agent in E who is removed in BTTC no earlier than any other agent in E. Let c be the cycle in BTTC that involves i. Since fIˆA (R) is stable, fIˆA (R)(i) = oi implies that every j ∈ I \ {i} prefers fIˆA (R)(j) to oi , and every k ∈ I(c) prefers ok to pk . Since f B (R)(i) = pi , it must be that d < d¯ and there exists an agent

j ∈ I(d ) with d > d who prefers oi to f B (R)(j); otherwise, we should have f B (R)(i) = oi . By stability of fIˆA (R), we must have [fIˆA (R)(j) = oj ]Pj oi Pj [f B (R)(j) = pj ]. Thus, j ∈ E. However, the fact that j ∈ I(d ) ∩ E and d > d contradicts the selection of i. If F = ∅, we obtain a contradiction by using a similar argument. So f B (R) = fIˆA (R).

Proof of Lemma 4 For any Ri ∈ P and any o ∈ O, let U (Ri , o) = {o ∈ O : o Pi o} be the strict upper contour set of Ri at o, and L(Ri , o) = {o ∈ O : oPi o } be the strict lower contour set of Ri at o. In our model, a set of preferences P  ⊂ P satisfies richness if for any Ri , Ri ∈ P  and any o, o ∈ O such that oPi o , there exists Ri ∈ P  such that U (Ri , o) ⊂ U (Ri , o) ∩ U (Ri , o), L(Ri , o) ⊂ L(Ri , o), U (Ri , o ) = U (Ri , o ), and L(Ri , o ) = L(Ri , o ). In our model, a mechanism f satisfies • joint monotonicity if, for all R ∈ P |I| , Iˆ ⊆ I, and RIˆ ∈ P |I| (Ri = Ri for all

ˆ such that L(Ri , fi (R)) ⊆ L(R , fi (R)) and U (Ri , fi (R)) ⊇ U (R , fi (R)) for i ∈ I) i i 25

ˆ we have all i ∈ I, ˆ fi (RIˆ, RI\Iˆ)Ri fi (R) for all i ∈ I. • respectfulness if, for all R ∈ P |I| , i ∈ I, and Ri ∈ P such that U (Ri , fi (R)) = U (Ri , fi (R)) and L(Ri , fi (R)) = L(Ri , fi (R)), we have fi (R) = fi (Ri , R−i ) implies fj (R) = fj (Ri , R−i ) for all j ∈ I\{i}. Proof of Lemma 4. (Richness) Because P consists of all strict preference relations, for any Ri , Ri ∈ P and any o, o ∈ O such that oPi o , move o to the top of Ri and denote the resulting preference relation by Ri . It is clear that Ri satisfies the requirement in the definition of richness. ˆ (Joint monotonicity) For all R ∈ P |I| , Iˆ ⊆ I, and RIˆ ∈ P |I| (Ri = Ri for all i ∈ I)

ˆ such that L(Ri , fiB (R)) ⊆ L(Ri , fiB (R)) and U (Ri , fiB (R)) ⊇ U (Ri , fiB (R)) for all i ∈ I, there are two cases to consider. ˆ • Case 1: fiB (R) = pi = oi for all i ∈ I. ˆ by reporting R , all i ∈ Iˆ must Since U (Ri , fiB (R)) ⊇ U (Ri , fiB (R)) for all i ∈ I, Iˆ

ˆ be involved in the same cycles as when they report RIˆ. Then for all i ∈ I, fiB (RIˆ, RI\Iˆ) Ri pi = fiB (R). So f B satisfies joint monotonicity. ˆ • Case 2: fiB (R) = oi for some i ∈ I.

Define I˜ = {i ∈ Iˆ : fiB (R) = oi }. We consider the equivalent procedure API˜. By Lemma 2, all i ∈ AI˜ most prefers oi among OI˜. Thus, all i ∈ AI˜ must obtain oi in ˜ API˜, or equivalently, fiB (R) = oi . Since L(Ri , fiB (R)) ⊆ L(Ri , fiB (R)) for all i ∈ I, all i ∈ I˜ still most prefers oi among OI˜ in Ri . So by reporting RIˆ, all i ∈ I˜ must ˜ still obtain oi in API˜. Thus, for all i ∈ I,

fiB (RIˆ, RI\Iˆ) = oi = fiB (R). ˆ I˜ is nonempty, for all j ∈ I\ ˆ I, ˜ f B (R) = pj = oj . So I\ ˆ I˜ ⊆ I\A ˜, and thus I\ ˆ I˜ If I\ j I are involved in cycles before step s + 1 of API˜. Since U (Ri , fiB (R)) ⊇ U (Ri , fiB (R)) 26

ˆ when Iˆ report R , every j ∈ I\ ˆ I˜ must be involved in the same cycle for all i ∈ I, Iˆ ˆ I, ˜ as when Iˆ report RIˆ. Thus, for all j ∈ I\

fjB (RIˆ, RI\Iˆ) Rj pj = fjB (R). So f B satisfies joint monotonicity. (Respectfulness) For all R ∈ P |I| , i ∈ I, and Ri ∈ P such that U (Ri , fiB (R)) = U (Ri , fiB (R)) and L(Ri , fiB (R)) = L(Ri , fiB (R)), there are two cases to consider. • Case 1: fiB (R) = pi = oi . U (Ri , fiB (R)) = U (Ri , fiB (R)) and L(Ri , fiB (R)) = L(Ri , fiB (R)) ensure that by reporting Ri , i must be involved in the same cycle in the same step of BTTC as when he reports Ri , and moreover, the outcome of BTTC does not change. So f B (R) = f B (Ri , R−i ). • Case 2: fiB (R) = oi . We consider the equivalent procedure AP{i} . By Lemma 2, all j ∈ A{i} most prefers oj among O{i} , which implies that fjB (R) = fiP (R)(j) = oj . Since L(Ri , fiB (R)) = L(Ri , fiB (R)), i still most prefers oi among O{i} in Ri . Thus, for all j ∈ A{i} , we have fjB (Ri , R−i ) = fiP (Ri , R−i )(j) = oj = fjB (R). I\A{i} form same cycles in AP{i} no matter i report Ri or Ri . U (Ri , fiB (R)) = U (Ri , fiB (R)) and L(Ri , fiB (R)) = L(Ri , fiB (R)) ensure that when i reports Ri , these cycles must be cleared in the same way as when i reports Ri . So for all  ∈ I\A{i} , fB (Ri , R−i ) = fB (R). Thus, f B satisfies respectfulness. By Theorem 1 of Barber`a et al. (2016), f B is weakly group strategy-proof.

Acknowledgment We are grateful to two anonymous referees and an Advisory Editor for useful comments. We also thank Ning Sun and Ning Yu for suggesting the example of teacher rotation in Japanese education system. We got useful feedbacks from the third East Asia Game 27

Theory (EAGT) International Conference and the 2019 Chengdu Economic Theory Workshop. All remaining errors are ours. Jun Zhang is supported by National Natural Science Foundation of China (Grant No. 71903093). Jingsheng Yu is supported by the Fundamental Research Funds for the Central Universities (Grant No. JBK2001072).

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