Structure and complexity of
ex post efficient random assignments

The following problem is NP-complete: SerialDictatorshipFeasibility — check whether there exists a permutation of agents for which serial dictatorship gives a particular object to an agent [22]. We present a reduction from SerialDictatorshipFeasibility to the problem of checking whether there exists a Pareto optimal deterministic assignment consistent with the given random assignment Consider a random assignment $p$ in which agent $i$ gets $o$ with probability one and all other agents gets each object in $O\setminus\{o\}$ with non-zero probability. We argue that SerialDictatorshipFeasibility has a yes instance if and only if $p$ admits a Pareto optimal deterministic assignment consistent with it.

If there exists a Pareto optimal deterministic assignment consistent with $p$, then this implies that there exists a Pareto optimal assignment in which $i$ gets $o$. This means that there exists some permutation $\pi$ such that $Prio(N,O,\succsim,\pi)(i)(o)=1$.

Assume that there exists some permutation $\pi$ such that $Prio(N,O,\succsim,\pi)(i)(o)=1$. Then the deterministic assignment corresponding to $Prio(N,O,\succsim,\pi)$ is a Pareto optimal permutation matrix that is consistent with $p$. ∎

We first show that testing ex post efficiency of a random assignment is in NP. It is sufficient to show that an ex post efficient random assignment admits a polynomial-sized Pareto optimal decomposition. Consider the $n^{2}$ dimensional Euclidean space of all $n\times n$ matrices. Let $P$ denote the minimal polytope containing all determinisitc Pareto optimal allocations. By definition $P$ is the set of all ex post efficient allocations. Carathéodory’s theorem implies that any ex post efficient allocations must be a convex combination of no more than $n^{2}+1$ deterministic Pareto optimal allocations (because all vertices in $P$ are Pareto optimal allocations).²²2The polytope actually lives in a $n^{2}-2n+1$ dimensional subspace of $R^{n^{2}}$ because its feasible points has to satisfy $2n$ equality constraints and one of which is redundant.

We prove the NP-hardness via a reduction from 3-SAT. Given a 3-SAT instance $F=(X,\mathscr{C})$, where $X=\{x_{1},\ldots,x_{k}\}$ is the set of binary variables and $\mathscr{C}=\{C_{1},\ldots,C_{t}\}$ is the set of clauses. Let each clause $C_{j}=l_{j,1}\vee l_{j,2}\vee l_{j,3}$, where for $s=1,2,3$, $l_{j,s}$ is either $x_{j_{s}}$ or $\neg x_{j_{s}}$ with $j_{1}<j_{2}<j_{3}$. The assumption that $j_{1}<j_{2}<j_{3}$ will be crucial in the proof.

Given a 3-SAT instance $F$, we build an assignment problem $(N,O,\succ)$ as follows.

Let $N_{1}=\cup_{i=1}^{k}\{x_{i},x_{i}^{1},\ldots,x_{i}^{t}\}\cup\{c,c_{1},\ldots,c_{t}\}$. For each agent $x\in N_{1}$, let $d_{x}$ denote the corresponding dummy agent. Let $N_{2}=\{d_{x}:x\in N_{1}\}$ and $N=N_{1}\cup N_{2}$. That is, $N=\cup_{i=1}^{k}\{x_{i},x_{i}^{1},\ldots,x_{i}^{t},d_{x_{i}},d_{x_{i}^{1}},\ldots,d_{x_{i}^{t}}\}\cup\{c,c_{1},\ldots,c_{t},d_{c},d_{c_{1}},\ldots,d_{c_{t}}\}$. We will show that in the decompositions $c_{j}$’s are “copies” of $c$ and $x_{i}^{j}$’s are “copies” of $x_{i}$.

$O=\{+x,-x:\forall x\in N_{1}\}$. For each $x\in N_{1}$, $p(x,+x)=p(x,-x)=p(d_{x},+x)=p(d_{x},-x)=1/2$.

To define the preferences of the agents, we first introduce the following notation. For any literal $l_{j,s}$ in clause $j$, we let $V(l_{j,s})$ denote the item that corresponds to the value of $x_{j_{s}}$ that fails $l_{j,s}$. More precisely,

For each $j\leq t$ and $C_{j}=l_{j,1}\vee l_{j,2}\vee l_{j,3}$, where $l_{j,s}$ is a literal of variable $x_{j_{s}}$, we let $S_{j_{1}}^{j}=\{V(l_{j,2})\}$, $S_{j_{2}}^{j}=\{V(l_{j,3})\}$, and $S_{j_{3}}^{j}=\{+{c_{j}}\}$. For any $i\leq k$ and $j\leq t$, if $S_{i}^{j}$ is not defined above then $S_{i}^{j}=\emptyset$. Moreover, we let $S=\{V(l_{j,1}):\forall j\leq t\}$.

For example, for $k=5$, $t=2$, $C_{1}=x_{2}\vee\neg x_{4}\vee x_{5}$, and $C_{2}=\neg x_{2}\vee x_{3}\vee x_{4}$, we have

Agents’ preferences are defined in two tables: preferences for $x$’s are in Table 2 and preferences for all $c$’s are in Table 3.

Because for any $x\in N_{1}$, $+x$ and $-x$ are assigned to agents $x$ and $d_{x}$ in $p$ with probability $1$, if $p$ is ex post efficient, then for any deterministic Pareto optimal assignment in the decomposition of $p$, $+x$ and $-x$ must be assigned to $x$ and $d_{x}$. Therefore, for any such assignment $M$, we say that the sign of $x$ (respectively, $d_{x}$) is positive, if $+x$ is allocated to $x$ (respectively, $d_{x}$); otherwise the sign is negative.

In light of Claim 1 in the remainder of this proof, we sometimes only use signs to represent the items, which will be clear from the context.

Suppose $p$ is ex post efficient. We now show that there exists a solution to the 3-SAT instance. Let $M$ be any deterministic Pareto optimal assignment in $p$’s decomposition where the sign of $c$ is positive. In the SAT instance, we let $x_{i}=+$ if and only if the sign of $x_{i}$ is positive in $M$ (or equivalently, $x_{i}$ is assigned item $+x_{i}$). Suppose for the sake of contradiction a clause $C_{j}=l_{j,1}\vee l_{j,2}\vee l_{j,3}$ is not satisfied, where $l_{j,s}$ corresponds to variable $x_{j_{s}}$. By part 2 of Claim 1, $V(l_{j,s})$ is allocated to $x_{j_{s}}$. Then, in $M$ there exists a trading cycle illustrated in Figure 1, which is a contradiction. In Figure 1 $a_{1}(o_{1})\rightarrow a_{2}(o_{2})$ means that currently $o_{1}$ (respectively, $o_{2}$) is allocated to $a_{1}$ (respectively, $a_{2}$), and $a_{1}$ prefers $o_{2}$ to $o_{1}$. Specifically, for $s=1,2,3$, if $l_{j,s}=x_{j_{s}}$, then $V(l_{j,s})=-x_{j_{s}}^{j}$, and $x_{j_{s}}$ prefers $S_{j_{s}}^{j}$ to $-x_{j_{s}}^{j}$ (Table 2); similarly if $l_{j,s}=\neg x_{j_{s}}$, then $V(l_{j,s})=+x_{j_{s}}^{j}$, and $x_{j_{s}}$ prefers $S_{j_{s}}^{j}$ to $+x_{j_{s}}^{j}$ (Table 2).

Therefore, the 3-SAT instance is satisfiable.

Suppose there exists a valuation $v$ that satisfies $F$. We now construct a decomposition of $p$ to two deterministic Pareto optimal allocations $M_{1}$ and $M_{2}$. $M_{1}$ is illustrated in Table 4.

$M_{2}$ is obtained from $M_{1}$ by taking the negation of all signs. More precisely, $M_{2}$ is illustrated in Table 5.

It is easy to check that $p=\frac{1}{2}M_{1}+\frac{1}{2}M_{2}$. The demand graph of $x$’s is illustrated in Figure 2, where all outgoing edges of $x$’s are shown (some incoming edges are not shown). We recall that an edge from agent $a_{1}$ to agent $a_{2}$ means that $a_{1}$ demand the item allocated to $a_{2}$. (a) represents the case for $x_{i}=+$ and (b) represents the case for $x_{i}=-$. Dashed lines in (b) means that it is valid if and only if $x_{i}$ is a literal in $C_{j}$.

We establish that both $M_{1}$ and $M_{2}$ are Pareto optimal in the following two claims.

∎

Since $\mathscr{P}$ is a convex combination of Pareto optimal determistic assignments, it contains all the ex post efficient points. ∎

Since testing membership in $\mathscr{P}$, the statements follows from the equivalence between optimizing over a poltope and implementing a separation oracle over over a polytope [16]. ∎

Consider the following assignment problem.

The following assignment $p$ is a result of RSD and hence ex post efficient:

Now consider the following assignment:

Note that $q$ is a random assignment such that $q(i)(o)>0$ if and only if $p(i)(o)>0$. Then in each Pareto optimal matrix that can be used to decompose the assignment, due to Pareto optimality, agents $1$ and $2$ get an object each from the following sets of objects $\{o_{1},o_{2}\}$, $\{o_{1},o_{3}\}$, and $\{o_{3},o_{4}\}$. Agent 1 and 2 cannot get the set $\{o_{2},o_{4}\}$ in a Pareto optimal assignment because if it were the case then there exists a serial dictatorship in which agent 3 or 4 take $o_{1}$ before $o_{2}$ is allocated. In order for agent $1$ to get $5/12$ of $o_{4}$, we need to use the $5/12$ time the Pareto optimal assignment in which $o_{4}$ gts $5/12$. But this means that agent $2$ gets $o_{3}$ $5/12$ of the time. But this is not possible since agent $2$ gets $o_{3}$ $1/12$ of time. ∎