Diverse Fair Allocations: Complexity and Algorithms

Abstract

In this work, we initiate the study of diversity of solutions in the context of fair division of indivisible goods. In particular, we explore the notions of disjoint, distinct and symmetric allocations and study their complexity in terms of the fairness notions of envy-freeness and equitability upto one item. We show that for binary valuations, the above problems are polynomial time solvable. In contrast we show NP-hardness of disjoint and symmetric case, when the valuations are additive.

7 Appendix

Proof (of Lemma 3)

The proof idea is similar to that of Lemma 2. We start by an arbitrary EQ1 allocation obtained by the following procedure. The least happy agent comes first and picks an object she values the most. This procedure is repeated until all the objects are allocated. It is easy to verify that the final allocation $\varPhi$ is indeed complete and EQ1. Now any permutation of $\varPhi$ is also EQ1 by the similar argument as in Lemma 2 and this proves the result.    $\square$

Proof (of Theorem 1)

We argue here the equivalence of the reduction.

Forward Direction. Suppose $\mathcal {I}$ is a yes instance. Let $S = \{S_1, S_2, \ldots S_p\}$ be the exact cover. We construct a disjoint allocation $\varPhi ^\prime$ as follows.

$$ \varPhi ^\prime (a) = {\left\{ \begin{array}{ll} S_i \cup \{g_1^i, g_2^i, \ldots g_t^i\} \setminus \{a_i^i\} ~\text {if}~ a=a_i~\text {such that}~i \le p \\ q_{i-p} \cup \{g_1^i, g_2^i, \ldots g_t^i\} \setminus \{g_i^i\} ~\text {if}~ a=a_i~\text {such that}~i > p\\ \{g_i^i, g_i^{t+1} ~\forall ~ i \in [t]\} ~\text {if}~ a=s \\ \end{array}\right. } $$

Notice that $\varPhi ^\prime$ is disjoint from $\varPhi$. To see $\varPhi ^\prime$ is indeed $\textsc {EF1}$, notice that

• The set agent $\{a_i : i \le p\}$ does not envy any other set agent $\{a_j : j \le p\}$, as $u_{a_i}(\varPhi ^\prime (a_i))= 2 = u_{a_i}(\varPhi ^\prime (a_j))$

• The set agent $\{a_i : i \le p\}$ envies $\{a_j : j > p\}$ upto one good, as $u_{a_i}(\varPhi ^\prime (a_i))= 2$ but $u_{a_i}(\varPhi ^\prime (a_j)) = 4$

• The special agent s does not envy any $a_i$, as $u_{s}(\varPhi ^\prime (s))= 0 = u_{s}(\varPhi ^\prime (a_i)$

• $a_i$ envies s up to one good, as $u_{a_i}(\varPhi ^\prime (a_i))= 2 = u_{a_i}(\varPhi ^\prime (s) \setminus \{g_i^{t+1}\}) \le u_{a_i}(\varPhi ^\prime (s))= 4$

Reverse Direction. Suppose $\varPhi ^\prime$ is an $\textsc {EF1}$ allocation, disjoint from $\varPhi$. Consider any set agent $a_i$. Under $\varPhi ^\prime$, $a_i$ can not get any of its $t+1$ identity items. This implies that these $t+1$ items must be allocated among the remaining t agents. By pigeonholing, one of the remaining agents, say x, must get at least two of the identity items of $a_i$. Then $u_{a_i}(\varPhi ^\prime (x)) \ge 4$. As $\varPhi ^\prime$ is $\textsc {EF1}$, it must happen that $u_{a_i}(\varPhi ^\prime (a_i)) \ge 2$. The only way this can happen is either $a_i$ gets a pacifier item $q_i$ or gets all the items in its corresponding set $S_i$. Since this is true for every set agent $\{a_1, a_2, \ldots a_t\}$, and there are only $t-p$ of the pacifier items, therefore, p of the set agents, say $\{a_{i_1}, a_{i_2}, \ldots a_{i_p}\}$, must get their corresponding sets $\{S_{i_1}, S_{i_2}, \ldots S_{i_p}\}$. Since $|U| = 3p$, therefore $\{S_{i_1}, S_{i_2}, \ldots S_{i_p}\}$ must form an exact cover of U. Hence, $\mathcal {I}$ is a yes instance.

   $\square$

Proof (of Lemma 8)

Let $a_1$ and $a_2$ be two agents. Let $a_1's$ valuation for the four goods $\{g_1, g_2, g_3, g_4\}$ be given by the vector [1, 0, 0, 0] and $a_2's$ valuation be [1, 1, 1, 1]. Notice that under any EQ1 allocation, $a_2$ can not receive all the four goods. Also, if $a_1$ gets all the four goods under an allocation $\varPhi$, then the allocation $\varPhi ^\star$ obtained by swapping the bundles is not EQ1, therefore, $\varPhi$ can not be symmetric. So both the agents must get at least one good under any symmetric EQ1 allocation. Now consider the following cases:

• Suppose under the EQ1 allocation $\varPhi$, $a_1$ gets $g_1$. Then she must also get one of the three remaining goods else, $u(a_2)=3$ which will violate EQ1. Say, $a_1$ gets $\{g_1, g_2\}$, WLOG, and $a_2$ gets $\{g_3, g_4\}$. But notice that $a_1$ value $\{g_3, g_4\}$ at 0 and hence will violate EQ1 if the bundles are swapped. Therefore $\varPhi$ is not symmetric. Also, if $a_1$ happened to receive $\{g_1, g_2, g_3\}$ under $\varPhi$, then although, $\varPhi$ remains EQ1 as both the agents value their respective bundles at one, but if the bundles are swapped, then $a_1$ derives 0 value, while $a_2$ derives the value of 3, therefore violating EQ1.

• Suppose under the EQ1 allocation $\varPhi$, $a_1$ does not get $g_1$. Then she must get all the remaining 3 goods in order for $\varPhi$ to be EQ1. But then under the allocation $\varPhi ^\star$ obtained by swapping the bundles, $u_{\varPhi ^\star }(a_1) = 1$ and $u_{\varPhi ^\star }(a_2) = 3$, violating EQ1. Therefore, $\varPhi$ is not symmetric.

This concludes the argument.    $\square$

Proof (of Lemma 9)

Consider two agents $a_1$ and $a_2$ and a set of goods $\{g_1. g_2, \ldots g_m\}$. Let X denote the set of items valued at 1 by $a_1$ and 0 by $a_2$. Let Y denote the set of items valued at 0 by $a_1$ and 1 by $a_2$. Let Z be the set of items valued at 1 by both $a_1$ and $a_2$. We argue the correctness of the claim by a case analysis depending on the cardinalities of the sets X, Y and Z. Notice that without loss of generality we can assume that $|X|\ge |Y|$. (If that is not the case, then we can interchange the agents $a_1$ and $a_2$ so that we now have $|X|> |Y|$.)

Also let $z^\prime =\frac{|Z|}{2}$. Let $Z_1$ denote a subset of Z of size $\lfloor z^\prime \rfloor$ and $Z_2$ denote $Z\setminus Z_1$. Note that the size of $Z_2$ will be $\lceil z^\prime \rceil$.

• $|X| \le |Y|+1$. We claim that in this case , a symmetric EQ1 allocation always exist, and can be computed as follows: Consider the following allocation $\varPhi$:

  $$ \varPhi (a)= {\left\{ \begin{array}{ll} X \cup Z_1 &{} a = a_1,\\ Y \cup Z_2 &{} a = a_2, \\ \end{array}\right. } $$

  If |Z| is even, the agent $a_1$ values her bundle at $|X|+\frac{|Z|}{2}$ and the agent $a_2$ values her bundle at $|Y|+\frac{|Z|}{2}$. Since $|X|\le |Y|+1$, the value of the bundle of $a_1$ (according to $a_1$) is at most 1 more than the value of the bundle of $a_2$ (according to $a_2$). Therefore, the value of the bundle of $a_1$ after removing one good is at most as much as the bundle of $a_2$. Thus this allocation is EQ1. For showing that it is symmetric, we need to show that the allocation obtained after swapping the bundles of agents $a_1$ and $a_2$ is also EQ1. In the new allocation, both $a_1$ and $a_2$ value their bundles at $\frac{|Z|}{2}$ each. Thus it is also EQ1. Now consider the case where |Z| is odd. Thus $|Z_1|=|Z_2|-1$. The agent $a_1$ will value her bundle at $|X|+|Z_1|$ and the agent $a_2$ values her bundle at $|Y|+|Z_2|$. If $|X|=|Y|$, then the valuation of agent $a_2$ for her bundle will be one more than the valuation of agent $a_1$ for her bundle. But after removing one item from the bundle of $a_2$, both agents will have equal valuation for their respective bundle (according to themselves). If $|X|=|Y|+1$, then both the agents will have an equal valuation for their bundle. Hence this is an EQ1 allocation. After swapping the bundles, $a_1$ will value her bundle at $|Z_2|$ and $a_2$ will value her bundle at $|Z_1|$. Since we have $|Z_1|=|Z_2|-1$, the value assigned by $a_1$ to her bundle is one more than the value assigned by $a_2$ to her bundle. Thus, after removing one good from the bundle of $a_2$, both agents will have same value for their bundle. Thus it is an EQ1 allocation even after swapping the bundles. So we have shown that $\varPhi$ is a symmetric EQ1 allocation.

• $|X| = |Y| + 2$. We claim that in this case, a symmetric EQ1 allocation always exists, and can be computed as follows: If |Z| is even, consider the following allocation $\varPhi$:

  $$ \varPhi (a)= {\left\{ \begin{array}{ll} \{|Y|+1 ~\text {many goods from}~ X\} \cup Z_1 &{} a = a_1,\\ Y \cup Z_2 \cup \{1~\text {good from X}\} &{} a = a_2, \\ \end{array}\right. } $$

  The value assigned by $a_1$ to her bundle will be $|Y|+1+|Z_1|$ and the value assigned by $a_2$ to her bundle will be $|Y|+|Z_2|$. Since $|Z_1|=|Z_2|$ as |Z| is even, the value assigned by $a_1$ to her bundle is one more than the value assigned by $a_2$ to her bundle. Therefore, after removing one good from the bundle of $a_1$, both the agents assign the same value to their respective bundles. Thus $\varPhi$ is an EQ1 allocation. After swapping the bundles of the two agents, the value assigned by $a_1$ to her bundle is $|Z_2|+1$ and the value assigned by $a_2$ to her bundle is $|Z_1|$. Thus after removing one good from the bundle of $a_1$, both the agents assign the same value to their respective bundles. Thus we get an EQ1 allocation even after swapping the bundles. Hence we have shown that $\varPhi$ is a symmetric EQ1 allocation. If |Z| is odd, consider the following allocation $\varPhi$:

  $$ \varPhi (a)= {\left\{ \begin{array}{ll} X \cup Z_1 &{} a = a_1,\\ Y \cup Z_2 &{} a = a_2, \\ \end{array}\right. } $$

  Since |Z| is odd, we have $|Z_2|=|Z_1|-1$. Now the value assigned by $a_1$ to her bundle is $|X|+|Z_1|$ which is equal to $|Y|+|Z_2|+1$. The value assigned by $a_2$ to her bundle will be $|Y|+|Z_2|$. So, after removing one good from the bundle of $a_1$, both the agents will assign the same value to their respective bundles. Thus the allocation $\varPhi$ is EQ1. After exchanging the bundles, the agent $a_1$ will value her bundle at $|Z_2|$ and the agent $a_2$ will value her bundle at $|Z_1|$. Thus, after removing one good from the bundle of $a_1$, both the agents will assign the same value to their respective bundles. Thus the allocation obtained after the exchange is also EQ1. Hence the allocation $\varPhi$ is a symmetric EQ1 allocations.

• $|X| \ge |Y|+3$. We claim that in this case, a symmetric EQ1 allocation does not exist, as argued below. For the sake of contradiction, assume that there exists a symmetric EQ1 allocation, say $\varPhi$. Let x, y and z denote the number of items from X, Y and Z respectively allocated to $a_1$ under $\varPhi$. Note that under $\varPhi$,

  • $a_1$ values her own bundle at x+z.

  • $a_2$ values her own bundle at $|Y|-y+|Z|-z$.

  Since $\varPhi$ is EQ1, we have $x+z\le \big (|Y|-y+|Z|-z\big ) +1$. That is,

  $$\begin{aligned} x+y+2z-|Z| \le |Y|+1 \end{aligned}$$

  (4)

  Also, under the swapped allocation, say $\varPhi ^{\star }$,

  • $a_1$ values her own bundle at $|X|-x + |Z|-z$.

  • $a_2$ values her own bundle at y+z.

  Since $\varPhi ^{\star }$ is EQ1, we have $|X|-x+|Z|-z \le \big (y+z\big )+1$. That is,

  $$\begin{aligned} x+y+2z-|Z|\ge |X|-1 \end{aligned}$$

  (5)

  Using inequalities (1) and (2), we get $|X|\le |Y|+2$, a contradiction.

Proof (of Lemma 10)

Notice that for any pair of items X and Y that do not appear consecutively in $a_1$’s valuation, the relative order of X and Y is the same in $a_1$’s and $a_2$’s valuations. This is true because there’s only one inversion.

$a_1: g_m\ge g_{m-1}\ge g_{m-2}\ge \ldots \ge g_3\ge g_2\ge g_1$

$a_2:$ any of the $(m-1)$ valuations that have 1 inversion

The following is a symmetric EF1 allocation:

$a_1$ gets $g_m, g_{m-3}, g_{m-4}, g_{m-7}, g_{m-8}, \ldots$

$a_2$ gets $g_{m-1}, g_{m-2}, g_{m-5}, g_{m-6}, g_{m-9}, \ldots$

This is because in $a_2$’s (and also $a_1$’s) valuation,

- $g_m\ge g_{m-2}$, $g_{m-3}\ge g_{m-5}$, $g_{m-4}\ge g_{m-6}$, $g_{m-7}\ge g_{m-9},\ldots$ and

- $g_{m-1}\ge g_{m-3}, g_{m-2}\ge g_{m-4}, g_{m-5}\ge g_{m-7}, g_{m-6}\ge g_{m-8},\ldots$ so on.    $\square$

Proof (of Lemma 11)

Notice that for any pair of items X and Y that do not appear consecutively in $a_1$’s valuation, the relative orders of X and Y are different in $a_1$’s and $a_2$’s valuations - this is true because there are ${m \atopwithdelims ()2} -1$ inversions.

$a_1: g_m\ge g_{m-1}\ge g_{m-2}\ge \ldots \ge g_3\ge g_2\ge g_1$

$a_2:$ any of the $(m-1)$ valuations that have $\left( {\begin{array}{c}m\\ 2\end{array}}\right) -1$ inversions

The following is a symmetric EF1 allocation:

$a_1$ gets $g_m, g_{m-3}, g_{m-4}, g_{m-7}, g_{m-8}, \ldots$

$a_2$ gets $g_{m-1}, g_{m-2}, g_{m-5}, g_{m-6}, g_{m-9},\ldots$

This is because

• In $a_1$’s valuation, - $g_m\ge g_{m-2}, g_{m-3}\ge g_{m-5}, g_{m-4}\ge g_{m-6}, g_{m-7}\ge g_{m-9}, \ldots$ and - $g_{m-1}\ge g_{m-3}, g_{m-2}\ge g_{m-4}, g_{m-5}\ge g_{m-7}, g_{m-6}\ge g_{m-8},\ldots$

• In $a_2$’s valuation, - $g_{m-2}\ge g_m$, $g_{m-5}\ge g_{m-3}$, $g_{m-6}\ge g_{m-4}, g_{m-9}\ge g_{m-7},\ldots$and -$g_{m-3}\ge g_{m-1}, g_{m-4}\ge g_{m-2}, g_{m-7}\ge g_{m-5}, g_{m-8}\ge g_{m-6},\ldots$

   $\square$

Proof

(Example for Theorem 2)    $\square$

Fig. 2.

Consider an instance of 2-partition with $S=\{1,2,3,4\}$. We create set goods $g_1,g_2,g_3$ and $g_4$ such that both $a_1$ and $a_2$ value them at 1, 2, 3 and 4 respectively. These are shown by the dark pink circles labelled 1, 2, 3 and 4 respectively. We create two dummy goods $d_1$ and $d_2$ which are valued at 11 each by agent $a_1$ and at 0 each by agent $a_2$. These are denoted by green circles labelled 11. It can be verified that both the allocations shown (the allocation on the right is obtained by swapping the one on the left) are disjoint and EQ1. (Color figure online)