On Fair and Efficient Allocations of Indivisible Public Goods^††thanks: Supported by NSF Grant CCF-1942321 (CAREER)

Let $\mathcal{I}=(\mathcal{A},\mathcal{G},k,\{v_{i}\}_{i\in\mathcal{A}})$ be an instance of the $\mathsf{PublicGoods}$ model. For $k=m$, the MNW problem is trivial, since we can select all the $m$ goods. For $n\leq k<m$, we can construct an instance $\mathcal{I}^{\prime}=(\mathcal{A}^{\prime},\mathcal{G}^{\prime},\{\mathcal{G}_{j}\}_{j\in\mathcal{G}^{\prime}}\{v^{\prime}_{i}\}_{i\in\mathcal{A}^{\prime}})$ of $\mathsf{PublicDecisions}$ from $\mathcal{I}$ in polynomial time, such that given an MNW allocation of $\mathcal{I}^{\prime}$, we can compute an MNW allocation of $\mathcal{I}$ in polynomial time. Let $V=\max_{i,j}v_{ij}$. We create $m$ public issues: corresponding to each good $j\in\mathcal{G}$, we create an issue $j$ with two alternatives $(j,1)$ and $(j,2)$. That is, $\mathcal{G}^{\prime}=[m]$, and $\mathcal{G}_{j}=\{(j,1),(j,2)\}$ for $j\in\mathcal{G}^{\prime}$. We create $\mathcal{A}^{\prime}=[n+mT]$, where $T=\lceil 2mn\log mV\rceil$. The first $n$ agents here correspond to the $n$ agents in $\mathcal{I}$. The last $mT$ agents are of two types: $kT$ agents $\{n+1,\dots,n+kT\}$ of type $A$, and $(m-k)T$ agents $\{n+kT+1,\dots,n+mT\}$ of type $B$. The valuations are as follows: each agent $i\in[n]$ values alternative ‘$1$’ of the issue $j\in\mathcal{G}^{\prime}$ at $v_{ij}$, the agents of type $A$ value only alternative ‘$1$’, agents of type $B$ value only alternative ‘$2$’. Formally, for $i\in\mathcal{A}^{\prime}$, and an alternative $(j,c)$ of the issue $j\in\mathcal{G}^{\prime}$, where $c\in\{1,2\}$:

$$v^{\prime}_{i}(j,c)=\begin{cases}v_{ij},\text{ if }c=1\text{ and }i\in[n];\\ 1,\text{ if }n<i\leq n+kT\text{ and }c=1;\\ 1,\text{ if }n+kT<i\leq n+mT\text{ and }c=2;\\ 0,\text{ otherwise. }\end{cases}$$

Let $\mathbf{x}^{\prime}$ be an allocation for the instance $\mathcal{I}^{\prime}$. For $c\in\{1,2\}$, let $S_{c}$ be the set of issues $j$ with decision $c$ in $\mathbf{x}^{\prime}$. That is, $S_{c}=\{j\in[m]:\mathbf{x}^{\prime}_{j}=c\}$. Let $k^{\prime}=|S_{1}|$. Then we have:

$$\mathsf{NW}(\mathbf{x}^{\prime})=\bigg{(}\prod_{i\in[n]}v^{\prime}_{i}(\mathbf{x}^{\prime})\cdot(k^{\prime})^{kT}\cdot(m-k^{\prime})^{(m-k)T}\bigg{)}^{\frac{1}{n+mT}}.$$

We now relate $\mathbf{x}^{\prime}$ to the $\mathsf{PublicGoods}$ instance $\mathcal{I}$. The decision $(j,1)$ corresponds to selecting the public good $j$. Let $\mathbf{x}=S_{1}\subseteq\mathcal{G}$ be the corresponding set of public goods. Then for any $i\in[n]$ we have that $v_{i}(\mathbf{x})=v^{\prime}_{i}(\mathbf{x}^{\prime})$, since $v^{\prime}_{i}(j,2)=0$ for every $j\in[m]$. Thus:

$$\mathsf{NW}(\mathbf{x}^{\prime})=\big{(}\mathsf{NW}(\mathbf{x})^{n}\cdot(k^{\prime})^{kT}\cdot(m-k^{\prime})^{(m-k)T}\big{)}^{\frac{1}{n+mT}}.$$

(3)

We now have to prove that $\mathbf{x}$ satisfies $|\mathbf{x}|\leq k$. Let $W_{\ell}$ be the Nash product of any MNW allocation for the $\mathsf{PublicGoods}$ instance $\mathcal{I}_{\ell}=(\mathcal{A},\mathcal{G},\ell,\{v_{i}\}_{i\in\mathcal{A}})$, $0\leq\ell\leq m$. Clearly, $0=W_{0}\leq W_{1}\leq\dots W_{m}\leq(mV)^{n}$. As $k\geq n$, $W_{k}\geq 1$, since we assume every agent has at least one good that she values positively. Define $g:[m]\rightarrow\mathbb{Z}$, as $g(a)=a^{k}(m-a)^{m-k}$. Then if $\mathbf{x}^{\prime}$ is an MNW allocation for $\mathcal{I}^{\prime}$, (3) becomes:

$$\mathsf{NW}(\mathbf{x}^{\prime})=(W_{k^{\prime}}\cdot g(k^{\prime})^{T})^{1/(n+mT)}.$$

(4)

Hence, the quantity $W_{k^{\prime}}\cdot g(k^{\prime})^{T}$ is maximized when $k^{\prime}=k$. Recalling (4), we conclude that for the MNW allocation $\mathbf{x}^{\prime}$ of $\mathcal{I}^{\prime}$, the corresponding set $\mathbf{x}$ has cardinality exactly $k$. Further $\mathbf{x}$ also maximizes the NW among all allocations of the instance $\mathcal{I}$ satisfying this cardinality constraint. Thus, $\mathbf{x}$ in fact is an MNW allocation for $\mathcal{I}$. Finally, it is clear that this is a polynomial time reduction. ∎

Let $\mathcal{I}=(\mathcal{A}=[n],\mathcal{G}=[m],V)$ be a $\mathsf{PrivateGoods}$ instance, using which we create a $\mathsf{PublicGoods}$ instance $I^{\prime}$ as follows. We create $n+2m$ agents, i.e. $\mathcal{A}^{\prime}=[n+2m]$. The first $n$ agents correspond to the $n$ agents in $\mathcal{I}$. The last $2m$ are dummy agents. We create $n\cdot m$ public goods: for each good $j\in[m]$, we create a set of $n$ copies $S_{j}=\{j_{1},j_{2},\dots,j_{n}\}$, $\mathcal{G}^{\prime}=\bigcup_{j\in\mathcal{G}}S_{j}$. We set $k=m$. The valuations for $i\in\mathcal{A}^{\prime}$, $j_{\ell}\in\mathcal{G}^{\prime}$ are:

$$v^{\prime}_{i}(j_{\ell})=\begin{cases}v_{ij},\text{ if }i=\ell\text{ and }i\in[n];\\ 1,\text{ if }i\in\{n+2j-1,n+2j\};\\ 0,\text{ otherwise, }\end{cases}$$

i.e. each agent $i\in[n]$ values exactly one copy, $j_{i}$ for each $j\in\mathcal{G}$ at $v_{ij}$, and for each good $j\in\mathcal{G}$, there are exactly two dummy agents who value all copies of $j$.

We now state and use the following claim, and prove it immediately after the proof of Theorem 8.

Consider any MNW allocation $\mathbf{x}^{\prime}$ of $\mathcal{I}^{\prime}$. We construct a partition, $\mathbf{x}$ of goods for $\mathcal{I}$ from this in the following way. For $i\in[n]$, $j\in[m]$, define $x_{ij}=1$ if $j_{i}\in\mathbf{x}^{\prime}$, and 0 otherwise. Let $\mathbf{x}_{i}=\{j\in\mathcal{G}:x_{ij}=1\}$. Thus, the value that agent $i$ gets in $\mathbf{x}$ is

$\displaystyle v_{i}(\mathbf{x}_{i})=\sum_{j\in\mathcal{G}}v_{ij}x_{ij}$

$\displaystyle=\sum_{j\in\mathcal{G}}v_{ij}\mathbf{1}(j_{i}\in\mathbf{x}^{\prime})=\sum_{j\in\mathcal{G}}v^{\prime}_{i}(j_{i})\mathbf{1}(j_{i}\in\mathbf{x}^{\prime})=v^{\prime}_{i}(\mathbf{x}^{\prime}).$

Thus, if $m\geq n$, $\mathsf{NW}(\mathbf{x})={\mathsf{NW}(\mathbf{x}^{\prime})}^{(n+2m)/n}$ and the partition corresponding to $\mathbf{x}^{\prime}$ as defined above gives an MNW solution for $\mathcal{I}$. On the other hand, if $m<n$, then $\mathbf{x}^{\prime}$ already gives non-zero value to all dummy agents by Claim 2. Thus, to maximize the total number of agents who get non-zero value, it maximizes the number of agents in $[n]$ who get non-zero value. Call this set $S^{*}$. Thus partition $\mathbf{x}$ has maximum number of agents getting a non-zero value. Finally, it maximizes the Nash product over $S^{*}\cup\{n+1,\ldots,n+2m\}$. Claim 2 also implies that all dummy agents get value $1$. Thus, $\prod_{i\in S^{*}}v_{i}(\mathbf{x}_{i})=\prod_{i\in S^{*}}v_{i}(\mathbf{x}^{\prime})$. Thus even in this case the allocation $\mathbf{x}$ corresponds to an MNW allocation in $\mathcal{I}$. ∎

Consider first $m\geq n$. Suppose $\exists j\in[m]$ for which two goods $j_{i},j_{i^{\prime}}\in\mathbf{x}^{\prime},i\neq i^{\prime}$. Since exactly $m$ goods are picked in $\mathbf{x}^{\prime}$, there is some $j^{\prime}\in[m]$, for which no good $j^{\prime}_{i}$ is picked in $\mathbf{x}^{\prime}$ for any $i\in[n]$. This implies that the agents $2j^{\prime}+n-1,2j^{\prime}+n$ get zero value in $\mathbf{x}^{\prime}$, making $\mathsf{NW}(\mathbf{x}^{\prime})=0$. However, choosing a good from each $j\in[m]$ gives non-zero value to all dummy agents. At the same time, since $m\geq n$, these goods can be chosen so that they give non-zero value to distinct agents in $[n]$. This makes $\mathsf{NW}(\mathbf{x}^{\prime})\neq 0$ contradicting Nash optimality of $\mathbf{x}^{\prime}$.

Now, if $m<n$ Nash welfare of all allocations in $\mathcal{I}$ is $0$. Thus, the MNW allocation is the one that maximizes the number of agents who get non zero value and then maximizes the product of values for these agents. Consider any allocation $\bar{\mathbf{x}}$, suppose $\exists j\in[m]$ for which two goods $j_{i},j_{i^{\prime}}\in\bar{\mathbf{x}},i\neq i^{\prime}.$ then again for some $j^{\prime}$, agents $n+2j^{\prime}-1$ and $n+2j^{\prime}$ get value 0 making $\mathsf{NW}(\bar{\mathbf{x}})=0$. At the same time, even if $\bar{\mathbf{x}}$ has goods from all different $S_{j}$, since $m<n$, and each one item from $S_{j}$ gives value only to one agent $i\in[n]$, the $\mathsf{NW}(\bar{\mathbf{x}})=0$ even in this case. Thus, if $m<n$, all allocations have Nash welfare $0$ in $\mathcal{I}^{\prime}$ also. Suppose the MNW allocation, $\mathbf{x}^{\prime}$ had two goods from same $S_{j}$ for some $j\in[m]$. Then, there exists a $j^{\prime}\in[m]$ such that no good is selected from $S_{j^{\prime}}$. The two goods from $S_{j}$ give value to exactly four agents - the two dummy agents $2j+n-1,2j+n$ and two agents who receive their copy of good $j$. Instead, if we exchange one of these goods to a good from $S_{j^{\prime}}$, we give non-zero value to at least five agents - dummy agents $2j+n-1,2j+n,2j^{\prime}+n-1,2j^{\prime}+n$ and at least one of the agents in $[n]$. We did not change the value of any other agents in this process. Thus, we increase the number of agents who get non-zero value, contradicting the maximality of $\mathbf{x}^{\prime}$. Thus, in both cases, all $m$ goods are picked from different $S_{j},j\in[m]$. ∎

Let $\mathcal{I}=(\mathcal{A},\mathcal{G},k,\{v_{i}\}_{i\in\mathcal{A}})$ be an instance of the $\mathsf{PublicGoods}$ model. For $k=m$, the leximin problem is trivial, since we can select all the $m$ goods. When $k\geq n$, we can construct an instance $\mathcal{I}^{\prime}=(\mathcal{A}^{\prime},\mathcal{G}^{\prime},\{\mathcal{G}_{j}\}_{j\in\mathcal{G}^{\prime}}\{v^{\prime}_{i}\}_{i\in\mathcal{A}^{\prime}})$ of the $\mathsf{PublicDecisions}$ model from $\mathcal{I}$ in polynomial time, such that given a leximin allocation of $\mathcal{I}^{\prime}$, we can compute a leximin allocation of $\mathcal{I}$ in polynomial time. To construct $\mathcal{I}^{\prime}$, we first create a set $\mathcal{A}^{\prime}$ of $n+2$ agents. The first $n$ agents here correspond to the $n$ agents in $\mathcal{I}$. The last $2$ agents are used in the construction, and ensure that exactly $k$ goods are selected in $I$.

We next create $m$ public issues: for each good $j\in\mathcal{G}$, we create an issue $j$ with two alternatives $(j,1)$ and $(j,2)$. That is, $\mathcal{G}^{\prime}=[m]$, and $\mathcal{G}_{j}=\{(j,1),(j,2)\}$ for $j\in\mathcal{G}^{\prime}$.

The valuations are as follows: for an agent $i\in\mathcal{A}^{\prime}$, and an alternative $(j,c)$ of the issue $j\in\mathcal{G}^{\prime}$, where $c\in\{1,2\}$:

$$v^{\prime}_{i}(j,c)=\begin{cases}v_{ij},\text{ if }c=1\text{ and }i\in[n];\\ \alpha\cdot(m-k),\text{ if }i=n+1\text{ and }c=1;\\ \alpha\cdot(k),\text{ if }i=n+2\text{ and }c=2;\\ 0,\text{ otherwise. }\end{cases}$$

where $\alpha<1/m^{2}$ is a sufficiently small constant. Essentially, each agent $i\in[n]$ values the ‘$1$’ decisions of the issue $j\in\mathcal{G}^{\prime}$ at $v_{ij}$, the agent $n+1$ values only the ‘$1$’ decisions, and agent $n+2$ values only the ‘$2$’ decisions.

Let $\mathbf{x}^{\prime}$ be a leximin allocation for the instance $\mathcal{I}^{\prime}$. Clearly $v^{\prime}_{i}(\mathbf{x}^{\prime})>0$ for all agents, since there is some allocation that gives positive utility to all agents, and the minimum utility only improves in the leximin solution. In particular $v^{\prime}_{i}(\mathbf{x}^{\prime})_{i}\geq 1$ for all $i\in[n]$. For $c\in\{1,2\}$, let $S_{c}$ be the set of issues $j$ with decision $c$ in $\mathbf{x}^{\prime}$. That is, $S_{c}=\{j\in[m]:\mathbf{x}^{\prime}_{j}=c\}$. Let $k^{\prime}=|S_{1}|$. we note that $v^{\prime}_{n+1}(\mathbf{x}^{\prime})=\alpha(m-k)k^{\prime}$, and $v^{\prime}_{n+2}(\mathbf{x}^{\prime})=\alpha k(m-k^{\prime})$. Since $\alpha<1/m^{2}$, for each $i\in[n],b\in[2]$, we have $v^{\prime}_{n+b}(\mathbf{x}^{\prime})<v^{\prime}_{i}(\mathbf{x}^{\prime})$. Suppose $k^{\prime}\neq k$. Then any allocation $\mathbf{x}^{\prime\prime}$ with $|\{j:\mathbf{x}^{\prime\prime}_{j}=1\}|=k$ gives $v^{\prime}_{n+1}(\mathbf{x}^{\prime\prime})=v^{\prime}_{n+2}(\mathbf{x}^{\prime\prime})=\alpha k(m-k)$, which is a leximin improvement over $\mathbf{x}^{\prime}$, since $\min(v^{\prime}_{n+1}(\mathbf{x}^{\prime\prime}),v^{\prime}_{n+2}(\mathbf{x}^{\prime\prime}))>\min(v^{\prime}_{n+1}(\mathbf{x}^{\prime}),v^{\prime}_{n+2}(\mathbf{x}^{\prime}))$. Hence $k^{\prime}=k$.

We now explain how we can relate $\mathbf{x}^{\prime}$ of $\mathcal{I}^{\prime}$ to the public goods instance $\mathcal{I}$. Intuitively, the decision $(j,1)$ corresponds to selecting the public good $j$, and $(j,2)$ corresponds to not selecting $j$. Let $\mathbf{x}=S_{1}\subseteq\mathcal{G}$ be a set of public goods of cardinality $k^{\prime}$. Then for any $i\in[n]$ we have that $v_{i}(\mathbf{x})=v^{\prime}_{i}(\mathbf{x}^{\prime})$, since $v^{\prime}_{i}(j,2)=0$ for every $j\in[m]$. Further since $k^{\prime}=k$, $|\mathbf{x}|=k$. Hence $\mathbf{x}$ is a feasible solution for $I$. Since for all $i\in[n]$, $v_{i}(\mathbf{x})=v^{\prime}_{i}(\mathbf{x}^{\prime})$, $\mathbf{x}$ is a leximin allocation for $I$.

Since the number of agents and goods created in the reduction are polynomially many in the size of the instance $\mathcal{I}$, and all other computations can also be carried out in polynomial time, this is a polynomial time leximin-preserving reduction. ∎

The proof follows from essentially the same reduction used to show Theorem 5. ∎

Fix any agent $i$. Let $\mathbf{x}=\{h_{1},h_{2},\ldots,h_{k}\}$ be any allocation that satisfies $\mathsf{RRS}$. Let $\mathbf{x}_{k}^{*}=\{g_{1},g_{2},\ldots,g_{k}\}$ denote the top $k$ goods for agent $i$. We assume that the goods both in $\mathbf{x}$ and $\mathbf{x}_{k}^{*}$ are ordered in decreasing order of valuations according to agent $i$. Now, suppose that top $\ell$ goods of $\mathbf{x}$ match with top $\ell$ goods of $\mathbf{x}_{k}^{*}$, i.e. $v_{i}(h_{j})=v_{i}(g_{j}),\forall j\leq\ell$ and $v_{i}(h_{\ell+1}<v_{i}(g_{\ell+1}))$. Note that since $\mathbf{x}_{k}^{*}$ is the top $k$ goods of agent $i$, we cannot have that $v_{i}(h_{j})>v_{(}g_{j})$ for any $j\leq\ell$. We want to prove that $\mathsf{RRS}$ implies $\mathsf{Prop}1$. If $\mathbf{x}$ was already satisfying proportionality, it is obvious that $\mathbf{x}$ is $\mathsf{Prop}1$. If $\ell\geq d$, it is again easy to see that $\mathbf{x}$ is $\mathsf{Prop}1$. This is because, if $k=d$ then we already have top $k$ goods, giving a proportional allocation. If $k>d$, then we can remove any good from $h_{d+1},\ldots,h_{k}$ and exchange it with $g_{d+1}$ to ensure proportionality, making the original allocation $\mathsf{Prop}1$. Finally, if $n$ divides $k$ then we have proportionality implied by $\mathsf{RRS}$ from Lemma 10.

Thus, we now assume that $\ell<d$, $k=nd+r$ with $r\leq n-1$ and that $\mathbf{x}$ is not already a proportional allocation. We know that $v(h_{1},\ldots,h_{\ell})=v(g_{1},\ldots,g_{\ell})$ and $v(h_{1},\ldots,h_{k})<\frac{1}{n}v(g_{1},g_{2},\ldots,g_{k})$. Thus,

$$v(h_{\ell+1},\ldots,h_{k})<\frac{1}{n}v(g_{\ell+1},\ldots,g_{k})$$

(5)

Now, $v(h_{k})\leq\frac{1}{k-\ell}v(h_{\ell+1},\ldots,h_{k})$. Thus,

$$v(h_{k})\leq\frac{1}{n\cdot(k-\ell)}v(g_{\ell+1},\ldots,g_{k})$$

(6)

Now, consider the good $g_{\ell+1}$. It is the good with highest value that is not in $\mathbf{x}$. We prove that removing $h_{k}$ and adding $g_{\ell+1}$ gives us an allocation that is proportional. Since $\ell<d$, $v_{i}(g_{\ell+1})\geq v_{i}(g_{nd+j}),\forall j\leq r$. Combining with the fact that $r<n$,

$$(n-1)\cdot v_{i}(g_{\ell+1})\geq v_{i}(g_{nd+1},\ldots,g_{nd+r}).$$

(7)

Again since the goods are arranged in decreasing order of valuations, we have $v_{i}(g_{1},\ldots,g_{d})\geq v_{i}(g_{jd+1},\ldots,g_{(j+1)d}),\forall 1\leq j\leq(n-1)$. Thus,

$$(n-1)\cdot v_{i}(g_{1},\ldots,g_{d})\geq v_{i}(g_{d+1},\ldots,g_{nd}).$$

(8)

Define, $LHS=(n-1)v_{i}(g_{\ell+1})+(n-1)v_{i}(g_{1},\ldots,g_{d})$. Combining (7) and (8),

	$\displaystyle LHS$	$\displaystyle\geq v_{i}(g_{nd+1},\ldots,g_{nd+r})+v_{i}(g_{d+1},\ldots,g_{nd})$
		$\displaystyle=v_{i}(g_{d+1},\ldots,g_{k})$
		$\displaystyle=v_{i}(g_{\ell+1},\ldots,g_{k})-v_{i}(g_{\ell+1},\ldots,g_{d})$

Thus we get,

$$(n-1)v_{i}(g_{\ell+1})+(n-1)v_{i}(g_{1},\ldots,g_{\ell})\geq v_{i}(g_{\ell+1},\ldots,g_{k})-nv_{i}(g_{\ell+1},\ldots,g_{d})$$

Now adding $v_{i}(g_{\ell+1})$ on both sides and using $v_{i}(g_{\ell+1})\geq\frac{1}{k-\ell}v_{i}(g_{\ell+1},\ldots,g_{k})$ gives:

	$\displaystyle nv_{i}(g_{\ell+1})$	$\displaystyle+(n-1)v_{i}(g_{1},\ldots,g_{\ell})$
		$\displaystyle\geq v_{i}(g_{\ell+1},\ldots,g_{k})-nv_{i}(g_{\ell+1},\ldots,g_{d})+\frac{1}{k-\ell}v_{i}(g_{\ell+1},\ldots,g_{k})$
		$\displaystyle\geq v_{i}(g_{\ell+1},\ldots,g_{k})-nv_{i}(h_{\ell+1},\ldots,h_{k})+nv_{i}(h_{k}),$

where the second inequality follows because $\mathbf{x}$ is $\mathsf{RRS}$ and from (6). Rearranging the above terms and using the fact that $v_{i}(g_{1},\ldots,g_{\ell})=v_{i}(h_{1},\ldots,h_{\ell})$, we get

$\displaystyle nv_{i}(g_{\ell+1})+nv_{i}(h_{1},\ldots,h_{k})-nv_{i}(h_{k})\geq v_{i}(g_{1},\ldots,g_{k})$

which implies that $\mathbf{x}$ is $\mathsf{Prop}1$. ∎