On the solvability of three-agent task allocation with unqualified agents priority structures

Abstract

In this paper, we study the problem of solvability for task allocation with unqualified agents (TAU) priority structures proposed by Ehlers and Westkamp (Theor Econ 13:1009-1041, 2018). In the TAU priority structure, at any position, either all agents have equal priority, or there exists exactly one agent who has the lowest priority and all others have equal highest priority. A priority structure is solvable if it admits a constrained efficient and strategy-proof mechanism, where a constrained efficient mechanism always produces a stable matching which can not be Pareto dominated by any other stable matching. We show that TAU priority structures with three agents are solvable via a top trading cycles mechanism with endogenous tie-breaking rules.

Appendix A

A.1. Preliminaries

According to the cycles clearing rules in the eTTC mechanism, any single-agent cycle generated in intermediate steps may not be cleared, and it will go to the next step. Therefore, a single-agent cycle which is cleared in the terminal step may be generated in an earlier step. We sort all cleared cycles according to the ordering of their generating steps. For a cycle cleared in the eTTC algorithm, it is called a first cleared cycle if it is generated in the first step among all steps in which all cleared cycles are generated.⁸ Similarly, we can define second or third cleared cycles. Note that there may be multiple first or second cleared cycles in the eTTC algorithm. Moreover, after all first cleared cycles are cleared, any second cleared cycle in the original problem becomes a first cleared cycle in the remaining subproblem.

In the eTTC algorithm, due to the pointing modification rule and the pointing rules of vacant positions, some agent in a first cleared cycle may not point to her favorite position. The following result characterizes the situation when an agent in a first cleared cycle of the eTTC mechanism points to her third favorite position.

Lemma 1

Given an allocation problem $(I,O,\succeq ,P)$ with $|I|=3$ and an exogenous ordering $\sigma$, if an agent in a first cleared cycle of the eTTC mechanism points to her third favorite position, then her favorite and second favorite positions are both occupied by herself.

Proof

Denote by $c_1$ a first cleared cycle in the eTTC algorithm, and in which agent i is pointing to her third favorite position. Let $o_a$ and $o_b$ be the favorite and second favorite positions of i respectively. According to the mechanism, the pointing of i to $o_a$ and $o_b$ has been modified or rejected before pointing to her third favorite position. We prove this result by showing that the following two cases are both impossible: (i) only one position in $\{o_a, o_b\}$ is occupied by i; (ii) neither $o_a$ nor $o_b$ is occupied by i.

Suppose Case (i) happens. Without loss of generality, assume that $o_a$ is occupied by i, but $o_b$ is not. By the pointing modification rule of the eTTC algorithm, there exists an agent, say j, pointing to $o_a$ which triggers the pointing modification of i. Then, agent i next points to $o_b$ which is also pointed by the third agent, say k, and i is rejected. In such a situation, it is only possible that $o_b$ is a vacant position and k and $o_b$ form a unique first cleared cycle; this contradicts with the fact that i forms a first cleared cycle with her third favorite position. Suppose Case (ii) happens. Assume that $o_a,o_b$ are eventually assigned to the other two agents j, k respectively. Agents j, k may be involved in a two-agent cycle or two single-agent cycles, in either situation, the cycle $c_1$ involving agent i and her third favorite position is not possible to be a first cleared cycle, a contradiction. $\square$

A.2. Proof of Theorem 1

Let $I = \{i, j,k\}$. Take as given any TAU priority structure. In the eTTC mechanism, according to the pointing modification rule and cycles clearing rule, the stability of this mechanism is naturally satisfied. We will next prove that the eTTC mechanism is constrained efficient and strategy-proof respectively.

Proof of constrained efficiency

Take as given any preference profile P. Denote by $\mu$ the resulted eTTC outcome. We will show that in this problem, for any stable matching $\nu$ which weakly Pareto dominates $\mu$, then $\nu =\mu$.

We first prove that for any agent i in a first cleared cycle, say $c_1$, it must be that $\nu (i)=\mu (i)$, where $\mu (i)$ is the position that agent i points to in $c_1$. Note that $\mu (i)$ could be either the favorite, second or third favorite of i. If $\mu (i)$ is the favorite position of i, because $\nu$ weakly Pareto dominates $\mu$, it is clear that $\nu (i)=\mu (i)$. We next discuss the other two possible cases respectively.

Assume that $\mu (i)$ is the second favorite position of agent i, and let $o_a$ be her favorite position. Because $\nu$ weakly Pareto dominates $\mu$, $\nu (i)$ can be either $o_a$ or $\mu (i)$. We only need to show that $\nu (i)\ne o_a$. There are two possible situations for $o_a$ that $o_a$ is either occupied by i, or not. In the former situation that $o_a$ is occupied by i, due to the pointing modification rule, there exists an agent, say j, pointing to $o_a$ which triggers the pointing modification of i. Similarly, $o_a$ is either the favorite or second favorite of j. If $o_a$ is the favorite of j, then it must be the case that $\nu (i)\ne o_a$; otherwise, $\nu (j)$ is strictly worse than $o_a$ for agent j, a contradiction with the stability of $\nu$. If $o_a$ is the second favorite of j, by the eTTC algorithm, the third agent, say k, points to the favorite position of j, say $o_b$, and the favorite position of k is also $o_b$.⁹ Because $\nu$ weakly Pareto dominates $\mu$ and $\nu$ is stable, it implies that $\nu (k)=\mu (k)$ thus $\nu (j)=\mu (j)=o_a$. As a result, we have $\nu (i) \ne o_a$. In the other situation that $o_a$ is not occupied by i, according to the eTTC algorithm, $o_a$ will be eventually assigned to another agent, say j, and it is also the favorite of j. Because $\nu$ weakly Pareto dominates $\mu$, it must be the case that $\nu (j)=\mu (j) = o_a$. As a result, $\nu (i)\ne o_a$.

If $\mu (i)$ is the third favorite of i, by Lemma 1, i’s favorite position, say $o_a$, and second favorite position, say $o_b$, are both occupied by herself. According to the pointing modification rule in the eTTC algorithm, the other two agents j, k have pointed to $o_a$ and $o_b$ respectively which trigger the pointing modification of i twice; moreover, either j or k is pointing to her favorite position. Similar arguments in the previous paragraph can be applied to show that in the matching $\nu$, agents j and k are assigned the same positions in {$o_a,o_b$} as in $\mu$. As a result, $\nu (i)$ is the third favorite position of i, and $\nu (i) = \mu (i)$.

For any agent, say j, in a second cleared cycle, say $c_2$, if any, we next show that $\nu (j)=\mu (j)$. After all first cleared cycles are cleared, any second cleared cycle in the original problem becomes a first cleared cycle in the remaining subproblem. Note that in the remaining subproblem (with at most two agents), the assignment of j in this mechanism is either her favorite or second favorite. Therefore, similar arguments as above can be applied to show that $\nu (j)=\mu (j)$. Furthermore, after clearing all first and second cleared cycles, if there exists a third cleared cycle, say $c_3$, the agent in $c_3$ must point to her favorite position among the remaining ones. It is clear that $\nu (k)=\mu (k)$. $\square$

Proof of Strategy-proofness

For simplicity, let f be the eTTC mechanism. Take as given any preference profile P, we will prove the property of strategy-proofness of f by discussing the incentives of agents in all cleared cycles.

Claim 1

No agent in a first cleared cycle can obtain a strictly better position in f by unilaterally misreporting her preference.

For any agent i in a first cleared cycle, say $c_1$, we will show that $f_i(P) R_i f_i(P'_i,P_{-i})$ where $P'_i\in \mathcal {P}$. Note that $f_i(P)$ is either the favorite, second or third favorite of agent i in $P_i$. If $f_i(P)$ is the favorite of i, it is clear that i has no incentive to misreport preference. If $f_i(P)$ is the third favorite of i, by Lemma 1, i’s favorite and second favorite positions are both occupied by herself, denote by $o_a$ and $o_b$ respectively. By the pointing modification rule in the eTTC algorithm, the other two agents j, k have also pointed to $o_a$ and $o_b$ respectively which trigger the pointing modification of i twice. Given that j, k truthfully report their preferences, it is clear that i can not obtain $o_a$ or $o_b$ by misreporting her preference in this mechanism.

What remains is to prove the case when $f_i(P)$ is the second favorite position of i. The favorite position of i is either occupied by i or not. We discuss these two cases respectively.

Case 1-1: The favorite position of i is not occupied by herself, and $f_i(P)$ is her second favorite.

Fig. 3

Illustration of two possible situations in Case 1-1

According to the eTTC algorithm, there exists an agent j who also firstly points to the favorite position of i, at which agent i is rejected. According to the mechanism, the favorite position of i is either occupied by the third agent k, say $o_k$, or a vacant position, say $o_v$; see Fig. 3 for some illustration of these two possible situations. For the situation (1) that i’s favorite is occupied by k, because agent i is involved in a first cleared cycle, the favorite position of agent k may be either another position she occupies, say $o'_k$, or a vacant position, say $o'_v$.¹⁰ After agent i is rejected at $o_k$, she next points to $o'_k$ (or $o'_v$). In such a case, whatever preference $P_i'$ agent i might report, agent j will be eventually assigned $o_k$ in the matching $f(P_i', P_{-i})$. For the situation (2) that i’s favorite is a vacant position $o_v$, because agent i is in a first cleared cycle, $o_v$ must be the favorite of all three agents, and agent i should be rejected at first. Note that this situation happens only when i has the lowest rank in $\sigma$. Take as given any $P'_i$ agent i could misreport. Because $o_v$ is still the favorite position of j, k, either j or k would eventually obtain $o_v$ in the matching $f(P'_i,P_{-i})$. Therefore, in this Case 1-1, i can not obtain her favorite position in $P_i$ by unilaterally misreporting her preference $P_i'$.

Case 1-2: The favorite position of i is occupied by herself, say $o_i$, and $f_i(P)$ is her second favorite.

By the pointing modification rule in the eTTC algorithm, there exists another agent, say j, pointing to $o_i$ to trigger the pointing modification of i. Because $c_1$ is a first cleared cycle, $o_i$ is either the favorite or second favorite position of j; see Fig. 4 for some illustration.¹¹

Fig. 4

Illustration of two possible situations in Case 1-2

Take as given any preference $P'_i$ agent i could misreport. In the eTTC mechanism running with $(P'_i,P_{-i})$, because j, k still point to the same positions as in P, by the pointing modification rule, it is not possible for agent i to be eventually assigned $o_i$ in $f(P'_i,P_{-i})$. The best position that i can obtain is $f_i(P)$ which is her second favorite position. As a result, we have $f_i(P) R_i f_i(P'_i,P_{-i})$.

We here complete the proof of Claim 1. We next consider the incentives for agents in second cleared cycles, if any.

Claim 2

No agent in a second cleared cycle, if any, can obtain a strictly better position by unilaterally misreporting her preference in f.

If there is a second cleared cycle, say $c_2$, and j is an agent in this cycle, we will prove that $f_j(P) R_j f_j(P'_j,P_{-j})$ for any $P'_j\in \mathcal {P}$. We first show that if j prefers some position in a first cleared cycle to her assignment $f_j(P)$, it is not possible for her to obtain this position by unilaterally misreporting her preference. Let $o_a$ be this position such that $o_a P_j f_j(P)$, and it is assigned to agent i in a first cleared cycle $c_1$, i.e., $f_i(P) =o_a$. It is not possible for this $o_a$ to be the third favorite position of i, otherwise, j can not prefer $o_a$ to her assignment $f_j(P)$. We discuss the following two cases respectively.

Case 2-1: $o_a$ is the favorite position of i.

It is clear that $o_a$ is not occupied by i. Neither $o_a$ is occupied by j. Suppose not. The pointing of j at $o_a$ should be modified as i also points to $o_a$. In such a situation, whatever the preference of the third agent is, it is not possible for agent i to be involved in a first cleared cycle with $o_a$; a contradiction. As a result, $o_a$ is either a position occupied by the third agent k, or a vacant position. We then discuss these two possible cases respectively.

(2-1a) i’s favorite position $o_a$ is occupied by the third agent k.

Because i and $o_a$ are involved in a first cleared cycle $c_1$, $o_a$ is occupied by the third agent k, and j is in a second cleared cycle, it must be the case that both agents i and k are involved in the first cleared cycle $c_1$ in which k points to an occupied position of i, say $o_i$. As a result, the preference of k may be $P_k: o_i, \cdots$, or $P_k: o_a, o_i, \cdots ,$ or in $P_k$, k prefers her another occupied position $o_k$ to $o_i$; see Fig. 5 for some illustration of these situations.

Fig. 5

Illustration of Case 2-1a

In situation (1), the two-agent cycle $c_1$ is formed in the starting step of the mechanism. In this situation, whatever $P_j'$ agent j may misreport, this first cleared cycle $c_1$ will be formed and cleared in the problem $(P_j', P_{-j})$. As a result, $f_j(P_j', P_{-j})$ is not a position in this cycle $c_1$. In situation (2), after the pointing of j forces the pointing modification of k, agent k then points to $o_i$ occupied by i, and forms a two-agent cycle $c_1$ with i. Therefore, whatever preference $P_j'$ agent j might misreport, agent i would obtain the same $o_a$ in both matchings $f(P_j', P_{-j})$ and f(P). As a result, $f_j(P_j', P_{-j}) \ne o_a$.

(2-1b) The favorite position $o_a$ of agent i is vacant.

By the pointing rules, at some step, both i and j have pointed to $o_a$, but the pointing of j at $o_a$ is rejected, due to the pointing of the third agent k. At this time, the pointing of k may be the following situations: (1) a position occupied by j, say $o_j$, or (2) a position occupied by k, say $o_k$, or another vacant position, say $o_v$, or (3) the same vacant position $o_a$; see Fig. 6 for illustration.

Fig. 6

Illustration of some situations in Case 2-1b

In the situations (1) and (2), whatever preference $P'_j$ that j might unilaterally misreport, i is still involved in a first cleared cycle with $o_a$ in the eTTC mechanism running with $(P'_j,P_{-j})$. As a result, $f_j (P'_j,P_{-j}) \ne o_a$. In the situation (3), according to the pointing rules of vacant positions, it must be the case that j is ranked lowest in $\sigma$. Whatever preference $P'_j$ is, because i, k still most prefer $o_a$, in the algorithm running with $(P'_j,P_{-j})$, either agent i or k will eventually obtain $o_a$. As a result, it is still the case that $f_j (P'_j,P_{-j}) \ne o_a$. We therefore complete the proof of Case 2-1b.

We next consider the following case.

Case 2-2: $o_a$ is the second favorite position of agent i.

In such a case, according to the pointing rules of positions, the favorite position of agent i is either occupied by j, say $o_j$, or occupied by i, say $o_i$; see Fig. 7 for illustration of the two situations.

Fig. 7

Illustration of two situations in Case 2-2

For the situation (1), it must be the case that j first forms a non-cleared single-agent cycle with $o_a$ (a vacant position), and $o_j$ is the favorite of both i and k. After $o_j$ becomes tentatively vacant, i is rejected and next points to $o_a$, then j is rejected at $o_a$ because k is pointing to $o_j$. In this situation, no matter what preference $P_j'$ is, it is clear that $f_j(P_j', P_{-j}) \ne o_a$. For the situation (2), it must be the case that $o_i$ is pointed by the third agent k, $o_a$ is a position occupied by k, and $c_1$ is a two-agent cycle containing i and k. In this situation, no matter which preference $P_j'$ is, the position $o_a$ will be assigned to either i or k in the matching $f(P_j', P_{-j})$, as a result, $f_j(P_j', P_{-j}) \ne o_a$ still holds.

We next argue that when all first cleared cycles are cleared, the best possible position that j obtains is her assignment $f_j(P)$. Among the remaining positions, $f_j(P)$ is the favorite or second favorite of j. Similar arguments in the proof of Claim 1 can be applied to show that in this mechanism, agent j can not obtain a strictly better position than $f_j(P)$ by unilaterally misreporting her preference. Therefore, we complete the proof of Claim 2.

We next consider the incentive issues of the agent in a third-cleared cycle, if any.

Claim 3

No agent in a third cleared cycle, if any, can obtain a strictly better position in the mechanism f by unilaterally misreporting her preference.

If there exists a third cleared cycle, because there are only three agents, each agent must be involved in a single-agent cycle which is generated in some step. For $n = 1,2,3$, denote by $c_n$ the n-th cleared cycle. Let $c_1$ consist of agent i and position $o_a$, and $c_2$ of j and $o_b$. We will show that if the agent in the third cleared cycle $c_3$, say agent k, prefers $o_a$ or $o_b$ to $f_k(P)$ in $P_k$, she can not obtain them by unilaterally misreporting her preference. Note that the favorite of k may be either $o_a$ or $o_b$.¹² We then consider the following two cases:

Case 3-1: For agent k, $o_a P_k f_k(P)$, and $o_a$ is k’s favorite position in $P_k$.

We first show that $o_a$ is a vacant position. It is clear that $o_a$ is not a position occupied by i because k has also pointed to $o_a$ at some step, and this would force i to modify her pointing there. Moreover, it is neither possible for $o_a$ to be a position occupied by agent j or k. Suppose not, according to the pointing rules of eTTC algorithm, it is not possible for i and $o_a$ to form a cycle which is eventually cleared earlier than the occupant of $o_a$.

Because $o_a$ is a vacant position, by the pointing rules of vacant positions, i and k both point to $o_a$ at some step, and the pointing of $o_a$ to i is determined by the pointing of the third agent j. There are two possible situations for the pointing of j: either (1), j points to a position occupied by j, say $o_j$, or (2), $o_a$ is the favorite of all three agents, k is rejected first due to $\sigma$, then k points to another position such that j is rejected at $o_a$; see Fig. 8 for illustration.

Fig. 8

Illustration of Case 3-1

In situation (1), after k is rejected at $o_a$ and next points to $o_j$, agent j modifies her pointing and forms the second cleared cycle $c_2$ with $o_b$. Moreover, k and $o_j$ form the third cleared cycle $c_3$, i.e., $o_j = f_k(P)$. In this situation, whatever preference $P_k'$ is, in the matching $f(P_k', P_{-k})$, the assignment of agent k can not be strictly better than $f_k(P)$ in the preference $P_k$. In situation (2), $o_a$ is the favorite position of all three agents, and k is rejected first. It happens only when k has the lowest rank in $\sigma$. In this situation, the preference of k is either $P_k: o_a, o_j, \cdots$, or $P_k: o_a, o_b, \cdots$, where $o_b$ is another vacant position. In the latter case, after j is rejected at $o_a$, she next points to $o_b$ with k, and then k is rejected at $o_b$. In both cases, whatever preference $P_k'$ is, either i or j is finally assigned the vacant position $o_a$, in other words, it is impossible for k to obtain $o_a$ in the mechanism f by unilaterally misreporting preference $P'_k$.

We thus complete the proof in Case 3-1. We next consider the other case.

Case 3-2: In the preference $P_k$, $o_b$ is the favorite position of k, in particular, $o_b P_k f_k(P)$.

In this case, $o_b$ is not occupied by j, otherwise, the pointing of j to $o_b$ should have been modified. Similarly, $o_b$ is not occupied by k, otherwise, the cycle with j and $o_b$ should be cleared later than that with k, a contradiction with the hypothesis that j and $o_b$ are in a second cleared cycle. Therefore, there are two possible situations that $o_b$ is either occupied by i, or a vacant position; see Fig. 9 for illustration of these two possible situations.

Fig. 9

Illustration of Case 3-2

Because $o_b$ is k’s favorite position, and $o_b$ is involved in the second cleared cycle $c_2$ with agent j, it must be the case that in some step, both j and k point to $o_b$, and then k is rejected at $o_b$.¹³ As a result, k has lower rank than j in $\sigma$. After being rejected at $o_b$, agent k may point to the position $f_k(P)$ and eventually form the third cleared cycle $c_3$. That is, $f_k(P)$ is the second favorite of k in $P_k$. In this situation, no matter what $P_k'$ is, the best possible position that k can obtain in the matching $f(P_k', P_{-k})$ is $f_k(P)$. Hence, k can not be strictly better off by unilaterally misreporting $P_k'$ in this mechanism. Finally, it is also possible that after being rejected at $o_b$, k may next point to $o_a$, but is rejected for another time, see the situation (2) in Fig. 9. In such a situation, k also has lower rank than i in $\sigma$, thus she is ranked lowest in $\sigma$. Similar arguments as above can be applied to show that k can not manipulate her preference to obtain a strictly better position.

We therefore complete the proof of strategy-proofness. $\square$

A.3. Proof of Proposition1

Proof

We prove this result by contradiction. Suppose that the weakly group strategy-proofness of the eTTC mechanism is violated. There exist exactly two agents who can obtain strictly better positions by jointly manipulating their preferences. Without loss of generality, assume these two agents are i and j, both of whom do not obtain their favorite positions in this mechanism.

Let $o_a$ be the favorite position of i in her true preference $P_i$, and $o_b$ be the favorite position of j in her true preference $P_j$. There are two cases, either (1) $o_a = o_b = o$, or (2) $o_a \ne o_b$. In the first case, o has been pointed by all three agents, the pointing of both i and j at o is rejected or modified, and o is eventually assigned to the third agent k. In the second case $o_a \ne o_b$, because both i and j do not obtain their favorite positions, there will be two situations that either the third agent k most prefers a position in {$o_a,o_b$}, without loss of generality, assume that it is $o_a$, and k is finally assigned her favorite position $o_a$, or agents i and j eventually obtain $o_b$ and $o_a$ respectively.

In both cases (1) and (2), without loss of generality, suppose that agent i finally obtains her second favorite position in $P_i$. According to the analysis above, i’s favorite position $o_a$ is assigned to either k or j. In the former situation, by the pointing rules of the eTTC algorithm, i, k firstly point to the same position $o_a$, and i is modified or rejected at $o_a$. If i is modified at $o_a$, then $o_a$ is a position occupied by i, by the pointing modification rule, i can not obtain $o_a$ by jointly manipulating her preference with j. Then, $o_a$ is either occupied by j or a vacant position. Suppose that both i and j jointly misreport preferences $P_i'$ and $P_j'$ respectively such that both of them obtain strictly better positions in this mechanism. This implies that i would obtain her favorite position $o_a$ in $P_i$. To make this happen, it is only possible that j firstly points to an occupied position of i if $o_a$ is occupied by j, or an occupied position of k if $o_a$ is a vacant position. However, this way agent j can never obtain a strictly better position in $P_j$ in this mechanism. We then consider the latter situation that $o_a$ is assigned to j, it should be the case that $o_a,o_b$ are occupied by i, j respectively, the second favorite positions of i, j are $o_b,o_a$ respectively, and the pointing of k trigger the pointing modification of i and j to form a two-agent cycle. Given that k truthfully reports her preference $P_k$, by the pointing modification rule, it is clear that i, j can not both obtain strictly better positions, i.e., $o_a$ and $o_b$ respectively, by jointly manipulating their preferences. $\square$

A.4. Proof of Proposition 2

Proof

Consider the situation that $|I| =3$ and each agent occupies exactly one position, and there is no vacant position. In the eTTC algorithm, the part of pointing rules of vacant or tentatively vacant positions does not have bites. It is easy to check that any multiple-agent cycle cleared in the eTTC mechanism also appears in the BTTC algorithm, and this cycle will be eventually cleared; as a result, any agent in such a cycle is finally assigned the position she points to in these two algorithms. Moreover, for any single-agent cycle involving an agent and her occupied position which is cleared in the terminal step of the eTTC algorithm, the agent in such a cycle also obtains her occupied position in the BTTC algorithm. Therefore, the two mechanisms yield the same outcome. $\square$