Fair-by-design matching

Abstract

Matching algorithms are used routinely to match donors to recipients for solid organs transplantation, for the assignment of medical residents to hospitals, record linkage in databases, scheduling jobs on machines, network switching, online advertising, and image recognition, among others. Although many optimal solutions may exist to a given matching problem, when the elements that shall or not be included in a solution correspond to individuals, it becomes of paramount importance that the solution is selected fairly. In this paper we study individual fairness in matching problems. Given that many maximum matchings may exist, each one satisfying a different set of individuals, the only way to guarantee fairness is through randomization. Hence we introduce the distributional maxmin fairness framework which provides, for any given input instance, the strongest guarantee possible simultaneously for all individuals in terms of satisfaction probability (the probability of being matched in the solution). Specifically, a probability distribution over feasible solutions is maxmin-fair if it is not possible to improve the satisfaction probability of any individual without decreasing it for some other individual which is no better off. Our main contribution is a polynomial-time algorithm building on techniques from minimum cuts, and edge-coloring algorithms for regular bipartite graphs, and transversal theory. In the special case of bipartite matching, our algorithm runs in $O((|V|^2 + |E| |V|^{2/3})\cdot (\log |V|)^2)$ expected time. An experimental evaluation of our fair-matching algorithm shows its ability to scale to graphs with tens of millions of vertices and hundreds of millions of edges, taking only a few minutes on a simple architecture. To the best of our knowledge, this yields the first large-scale implementation of the egalitarian mechanism of Bogomolnaia and Moulin (Econometrica 72(1):257–279, 2004). Our analysis confirms that our method provides stronger satisfaction probability guarantees than non-trivial baselines.

Appendices

Appendix A: Proofs for Sect. 3: fairness and social inequality

Preliminaries In order to prove Theorem 2, we need to recall some additonal facts about matroids [refer to Lawler (1976) for details]. The rank function $\rho :2^L \rightarrow \mathbb {N}$ of a matroid is monotone submodular, meaning that for all $S, T\subseteq L$, it holds that $0 \le \rho (S \cup T) - \rho (S) \le \rho (T) - \rho (S \cap T).$ The dual matroid of M is the matroid with ground set L given by $M^* = \{ L {\setminus } S \mid S \in M \}$; clearly the dual of $M^*$ is M itself. The rank function of $M^*$ is given by $\rho ^*(S) = |S| - (\rho (L) - \rho (L{\setminus } S))$. The contraction of M to the set $L {\setminus } S$ is the matroid M/S with ground set $L {\setminus } S$ and rank function $\rho _{M/S}(X) = \rho (S \cup X) - \rho (S).$ The restriction of M to the set S is the matroid with ground set S and independent sets .

Proof of Lemma 1

Assume that the set

$$\begin{aligned} A = \{u \in {{\mathcal {U}}}\mid {F}[{u}] \ne {D}[{u}]\} \end{aligned}$$

is non-empty. Let

$$\begin{aligned} u = {{\,\mathrm{arg\,min}\,}}\{\, \min ({F}[{u}], {D}[{u}]) \mid u\in A\, \}, \end{aligned}$$

where ties are broken arbitrarily. Then ${D}[{u}] \ne {F}[{u}]$; suppose that ${D}[{u}] > {F}[{u}]$. Then for any $v \in A$, our choice of u implies that ${D}[{v}] \ge \min ({D}[{v}], {F}[{v}]) \ge \min ({D}[{u}], {F}[{u}]) = {F}[{u}]$; and for any $v\notin A$, we have ${D}[{v}] = {F}[{v}]$ by definition. In either case one of the inequalities required by condition (1) fails, so F is not maxmin-fair. Put differently, we have shown the following implication:

$$\begin{aligned} F \text { is maxmin-fair } \implies {D}[{u}] < {F}[{u}]. \end{aligned}$$

Similarly,

$$\begin{aligned} D \text { is maxmin-fair } \implies {F}[{u}] < {D}[{u}]. \end{aligned}$$

But then F and D cannot both be maxmin-fair. The only way out of this contradiction is to conclude that A is empty.

Proof of Theorem 1

$\implies$ Let F be maxmin-fair and consider any other distribution D. We need to show that ${F\uparrow \,} \succeq {D\uparrow \,}$ (that is, ${F\uparrow \,}$ is lexicographically largest). Define

$$\begin{aligned} A = \{u \in {{\mathcal {U}}}\mid {F}[{u}] \ne {D}[{u}]\}. \end{aligned}$$

If A is empty, the claim is trivial; otherwise let

$$\begin{aligned} u = {{\,\mathrm{arg\,min}\,}}\{ {F}[{u}] \mid u\in A \} \; \text { and } \; B = \{v \in {{\mathcal {U}}}\mid {F}[{v}] < {F}[{u}]\}. \end{aligned}$$

Note that $u \in A\subseteq {\overline{B}}$ by our choice of u. If ${D}[{u}] > {F}[{u}]$, from the maxmin-fairness of F we infer the existence of $v\in A\subseteq {\overline{B}}$ such that ${D}[{v}] < {F}[{u}]$. This also holds if ${D}[{u}] < {F}[{u}]$ (then we can take $v = u$). In any case we have

$$\begin{aligned} \min \{ {D}[{v}] \mid v \notin B \}< {F}[{u}] = \min \{ {F}[{v}] \mid v \notin B \} \\ \text { and } \qquad {D}[{v}] = {F}[{v}] < {F}[{u}] \; \forall v \in B. \end{aligned}$$

It is readily verified that this implies ${F\uparrow \,} \succ {D\uparrow \,}$.

$\Longleftarrow$ Let F be a distribution which is not maxmin-fair. We show that F is not lexicographically largest either. Since (1) does not hold for F, there exists another distribution D and a user $u \in {{\mathcal {U}}}$ such that

$$\begin{aligned} {D}[{u}]> {F}[{u}] \text { and } ({D}[{v}] < {F}[{v}] \implies {F}[{v}] > {F}[{u}]) \; \forall v. \end{aligned}$$

(6)

For any $\varepsilon \in (0, 1)$, let $X_\varepsilon$ denote the distribution picking F with probability $1-\varepsilon$ and D with probability $\varepsilon$, so that

$$\begin{aligned} {X_\varepsilon }[{v}] = {F}[{v}] + \varepsilon ({D}[{v}] - {F}[{v}])\quad \forall v. \end{aligned}$$

Choose $\varepsilon > 0$ small enough so as to guarantee that

$$\begin{aligned} ({F}[{v}]< {F}[{u}] \implies {X_{\varepsilon }}[{v}] < {X_{\varepsilon }}[{u}]) \quad \forall v \end{aligned}$$

(7)

and

$$\begin{aligned} ({F}[{v}]> {F}[{u}] \implies {X_{\varepsilon }}[{v}] > {X_{\varepsilon }}[{u}]) \quad \forall v. \end{aligned}$$

(8)

For instance, any

$$\begin{aligned} \varepsilon < \min \left\{ \frac{|{F}[{u}] - {F}[{v}]|}{|{D}[{v}] - {F}[{v}]| + |{D}[{u}] - {F}[{u}]|} \Big \vert {F}[{v}] \ne {F}[{u}] \right\} \end{aligned}$$

will do. We have, by (6),

$$\begin{aligned} ({F}[{v}] \le {F}[{u}] \implies {D}[{v}] \ge {F}[{v}] \implies {X_{\varepsilon }}[{v}] \ge {F}[{v}]) \quad \forall v \end{aligned}$$

(9)

$$\begin{aligned} \text {and} \qquad {D}[{u}] > {F}[{u}]. \end{aligned}$$

(10)

But (7), (8), (9) and (10) say that ${X_\varepsilon \uparrow \,}$ is strictly larger than ${F\uparrow \,}$ in lexicographical order, as we wished to show.

The following two analogues of Lemma 1 and Theorem 1 are also needed for the proof of Theorem 2:

Lemma 7

If F and D are both minmax-Pareto, then ${F}[{u}] = {D}[{u}]$ for all $u \in {{\mathcal {U}}}$.

Proof

Assume that the set

$$\begin{aligned} A = \{u \in {{\mathcal {U}}}\mid {F}[{u}] \ne {D}[{u}]\} \end{aligned}$$

is non-empty. Let

$$\begin{aligned} u = {{\,\mathrm{arg\,max}\,}}\{\, \max ({F}[{u}], {D}[{u}]) \mid u\in A\, \}, \end{aligned}$$

where ties are broken arbitrarily. Then ${D}[{u}] \ne {F}[{u}]$; suppose that ${D}[{u}] < {F}[{u}]$. Then for any $v \in A$, our choice of u implies that ${D}[{v}] \le \max ({D}[{v}], {F}[{v}]) \le \max ({D}[{u}], {F}[{u}]) = {F}[{u}]$; and for any $v\notin A$, we have ${D}[{v}] = {F}[{v}]$ by definition. In either case one of the inequalities required by the definition of minmax-Pareto efficiency fails. Put differently, we have shown the following implication:

$$\begin{aligned} F \text { is minmax-Pareto } \implies {D}[{u}] > {F}[{u}]. \end{aligned}$$

Similarly,

$$\begin{aligned} D \text { is minmax-Pareto } \implies {F}[{u}] > {D}[{u}]. \end{aligned}$$

But then F and D cannot both be minmax-Pareto. The only way out of this contradiction is to conclude that A is empty. $\square$

Theorem 14

For matroid problems, a distribution F is minmax-Pareto if and only if F is Pareto-efficient and ${F\downarrow \,} \preceq {D\downarrow \,}$ for all Pareto-efficient distributions D.

Proof

First observe that, for matroids, a distribution is Pareto-efficient if and only if it is supported over bases. For any distribution D of bases over a matroid M with ground set L, consider the distribution $D^*$ of $(L {\setminus } X \mid X \sim D)$ of bases over the dual matroid $M^*$. Then we have ${D}[{u}] + {D^*}[{u}] = 1$ for all $u \in L$, so clearly F is minmax-Pareto if and only if $F^*$ is maxmin-fair, which (by Theorem 1) occurs if and only if $F^*$ is lexicographically largest for $M^*$, which in turn is equivalent to F being lexicographically smallest among distributions of bases of M, as we wished to show. $\square$

Our next result asserts that the only obstruction to achieving high satisfaction probability for every user is the existence of a set of users with small rank-to-size ratio. Finding these obstruction sets will enable us to devise a divide and conquer strategy to obtain fair distributions. For instance, in Example 1 the obstruction set is given by the set of users $\{a_1, a_2, a_3\}$, which force the maximum satisfaction probability to be no larger than $\frac{2}{3}$.

Theorem 15

Let M be a matroid with ground set L and rank function $\rho :2^L \rightarrow \mathbb {N}$. The minimum satisfaction probability in a minmax-fair distribution over M is

$$\begin{aligned} \pi (M) = \min \left\{ \frac{\rho (X)}{|X|} \mid \emptyset \ne X \subseteq L \right\} . \end{aligned}$$

Proof

Any maxmin-fair distribution is supported on the collection $\mathcal {B}$ of bases of M, since extending an independent set to a base containing it never decreases any satisfaction probability. Optimizing the smallest satisfaction probability $\lambda _1$ amounts to finding a suitable distribution over $\mathcal {B}$; let us denote the corresponding probabilities by $\{p_B\}_{B \in \mathcal {B}}$. Since the probability of $v \in {{\mathcal {U}}}$ being included is $\sum _{v \in B} p_B$, maximizing the minimum such probability is modeled by Program (11) below. It may be written as a linear program by introducing an additional variable $\lambda$ to be maximized, and introducing the constraints $\sum _{B \ni v} p_B \ge \lambda$. Its dual is equivalent to (12).

$$\begin{aligned} \begin{array}{rrclcl} \displaystyle \max &{} \displaystyle \min _{v\in {{\mathcal {U}}}} \sum _{B \ni v} p_B \\ \\ \text {s.t.} &{} \displaystyle \sum _{B \in \mathcal {B}} p_B &{} = &{} 1 &{} \\ &{} p_B &{}\ge &{} 0 \\ \end{array} \end{aligned}$$

(11)

$$\begin{aligned} \begin{array}{rrclcl} \displaystyle \min &{} \displaystyle \max _{B\in \mathcal {B}} \displaystyle \sum _{v \in B} z_v \\ \\ \text {s.t.} &{} \displaystyle \sum _{v \in {{\mathcal {U}}}} z_v &{} = &{} 1 &{} \\ &{} z_v &{}\ge &{} 0. \\ \end{array} \end{aligned}$$

(12)

Observe that $\max _{B\in \mathcal {B}} \sum _{v\in \mathcal {B}} z_v$ is the value of a maximum-weight base of M, with weights given by $\{z_v\}_{v \in {{\mathcal {U}}}}$. Thus LP (12) encodes the task of finding an assignment of weights to elements of ${{\mathcal {U}}}$ minimizing the maximum weight of a base. We will turn this min–max problem into a pure minimization problem.

Edmonds (1971) showed that for any fixed assignment of non-negative weights to the elements of M, a maximum-weight base may be found via the greedy algorithm that examines each element in order of decreasing weight and adds it to the current set if its addition does not violate independence. Let $\varPi$ denote the set of permutations of ${{\mathcal {U}}}= \{1, 2, \ldots , n\}$. Write

$$\begin{aligned} \varDelta = \left\{ z \in \mathbb {R}^{{\mathcal {U}}}\big \vert \sum _{v \in {{\mathcal {U}}}} z_v = 1, \quad z_v \ge 0 \; \forall v \in {{\mathcal {U}}}\right\} \end{aligned}$$

for the probability simplex on ${{\mathcal {U}}}$ and let $G(\pi ) = \{ z \in \varDelta \mid z_{\pi (1)} \ge z_{\pi (2)} \ge \cdots \ge z_{\pi (n)} \}$ denote the elements of $\varDelta$ which become sorted after applying permutation $\pi \in \varPi$.

Note that if $z, z' \in G(\pi )$, then the two bases obtained via the greedy algorithm with vertex weights $\{z_v\}$ and $\{z'_v\}$ are the same. For each $\pi \in \varPi$, let $B(\pi )$ denote the base obtained via the greedy algorithm; Edmond’s result may then be written as

$$\begin{aligned} \max _{B \in \mathcal {B}} \sum _{v \in B} z_v = \sum _{v \in B(\pi )} z_v \quad \text { if } z \in G(\pi ). \end{aligned}$$

By LP (12), the fairness parameter $\pi (M)$ is

$$\begin{aligned} \min _{z \in \varDelta }\; \max _{B \in \mathcal {B}} \sum _{v \in B} z_v = \min _{\pi \in \varPi }\, \min _{z \in G(\pi )}\, \max _{B \in \mathcal {B}} \sum _{v \in B} z_v = \min _{\pi \in \varPi }\, \min _{z \in G(\pi )}\, \sum _{v \in B(\pi )} z_v. \end{aligned}$$

(13)

We claim that, for each $\pi \in \varPi$ and each non-empty $X \subseteq {{\mathcal {U}}}$,

$$\begin{aligned} \min _{z \in G(\pi )} \sum _{v \in X} z_v = \min _{i \in [n]} \frac{|X \cap \pi ([i])|}{i}, \end{aligned}$$

(14)

where $\pi ([i]) = \{ \pi (1), \ldots , \pi (i) \}$.

This means that, if we are given advice on the permutation $\pi$ which sorts an optimal solution z to LP (12), then we can find another solution ${\hat{z}}$ with the same value and whose non-zero weights are evenly distributed among the top t elements of z in sorted order, for some $t \in [n]$. For some optimal t, each of the t non-zero values of ${\hat{z}}_i$ is either 0 or 1/t. To see this assuming 14, notice that we can construct such ${\hat{z}}$ by setting ${\hat{z}}_{\pi (i)} = \frac{1}{t}$ for $1 \le i \le t$ and ${\hat{z}}_j = 0$ for $j \notin \pi ([t])$.

To see why (14) holds, define $d_{n} = z_{\pi (n)}$ and $d_i = z_{\pi (i)} - z_{\pi (i+1)} \ge 0$ for $0< i < n$. Then $z_{\pi (i)} = \sum _{j \ge i} d_i,$ hence

$$\begin{aligned} \sum _{v \in X} z_v = \sum _{i\in [n]} \left( \mathbb {1}[{\pi (i) \in X}] \cdot \sum _{j \ge i} d_i \right) = \sum _{j\in [n]} d_j \cdot |X \cap \pi ([j])|. \end{aligned}$$

The conditions $\sum _{v \in {{\mathcal {U}}}} z_v = 1$ and $z \in G(\pi )$ then become $\sum _i i \cdot d_i = 1$ and $d_i \ge 0$. Therefore

$$\begin{aligned} \min _{z \in G(\pi )} \sum _{v \in X} z_v = \min \left\{ \sum _{i\in [n]} d_i \cdot |X \cap \pi ([i])| \;\Big \vert \; \sum _{i\in [n]} i \cdot d_i = 1, d_i \ge 0 \right\} . \end{aligned}$$

The quantity in the right-hand side equals the smallest ratio (among all i) between the coefficient of $d_i$ in the objective function ($|X \cap \pi ([i])|$) and in the only equality constraint, proving (14). From (13) and (14) the theorem follows, because the greedy algorithm satisfies $|B(\pi ) \cap \pi ([i])| = \rho (\pi [i])$ for all i, so if $\displaystyle S^*= {{\,\mathrm{arg\,min}\,}}_{S\subseteq {{\mathcal {U}}}}\frac{ \rho (S) }{ |S|}$, then for any $\pi \in \varPi$ we have

$$\begin{aligned} \min _{z \in G(\pi )} \sum _{v \in B(\pi )} z_v =\min _{i \in [n]} \frac{|B(\pi ) \cap \pi ([i])|}{i} = \min _{i \in [n]} \frac{\rho (\pi ([i]))}{|\pi ([i])|} \ge \frac{\rho (S^*)}{|S^*|}, \end{aligned}$$

and equality holds for any permutation where the elements of $S^*$ precede those of ${{\mathcal {U}}}{\setminus } S^*$. $\square$

Theorem 16

Let M be a matroid with ground set L and rank function $\rho :2^L \rightarrow \mathbb {N}$. The maximum satisfaction probability in a minmax-Pareto distribution over M is

$$\begin{aligned} \varPi (M) = \max \left\{ \frac{\rho (L) - \rho (X)}{|L {\setminus } X|} \mid {\emptyset \subseteq X \subsetneq L} \right\} . \end{aligned}$$

Proof

Apply Theorem 15 to the dual matroid of M. $\square$

An extension of Theorem 15 allows us to compute the satisfaction probability of every element of L.

Lemma 8

Define a sequence of sets $B_1, B_2, \ldots , B_k$ iteratively by:

$$\begin{aligned}&B_i \text { is a maximal set } X \subseteq L{\setminus } S_{i-1} \text { minimizing } \frac{ \rho (X \cup S_{i-1}) - \rho (S_{i-1}) }{|X|} ,\nonumber \\&\quad \text { where } S_i = \bigcup _{j=1}^i B_j. \end{aligned}$$

(15)

We stop when $S_i = L$ (which will eventually occur as the sequence $\{S_i\}$ is strictly increasing). Then for every $i, u \in B_i$, the satisfaction probability of u in a maxmin-fair distribution F is $\lambda _i = \frac{ \rho (B_i) }{ |B_i| }$.

(Maximality of each $B_i$ is not required for the conclusion to hold, but its inclusion guarantees uniqueness of the sets thus defined, owing to the submodularity of $\rho$.)

Proof

We reason by induction on the number k of sets. First, observe that Theorem 15 implies $F[u] \ge \lambda _1$ for all $u \in L$. As the expected number of satisfied elements within $B_1$, which obviously cannot exceed $\rho (B_1) = \lambda _1 |B_1|$, is equal to

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}_{A \sim F} [|A \cap B_1|] = \sum _{u\in B_1} F[u] \ge \lambda _1 |B_1| \end{aligned}$$

(16)

by linearity of expectation, the equality $F[u] = \lambda _1$ must hold for all $u \in B_1$. If $k = 1$, this shows the result.

If $k > 1$, let $D_1$ be a maxmin-fair distribution for the restriction of M to $B_1$ and let $D_2$ be a maxmin-fair distribution for the contraction $M/(L{\setminus } B_1)$ of M to the remaining elements $L {\setminus } B_1$. Since restriction does not change the rank function within $B_1$, $D_1$ satisfies $D_1[u] = F[u] = \lambda _1$ for all $u \in B_1$. The rank function of the contraction $M/(L{\setminus } B_1)$ is $\rho _{M/B_1}(X) = \rho (X \cup B_1) - \rho (B_1)$, so by applying rule (15) iteratively we obtain the same sequence of sets $B_2, \ldots , B_k$ (excluding $B_1$), and by the induction hypothesis $D_2$ satisfies $D_2[u] = \lambda _i$ for all $i\ge 2, u \in B_i$. It remains to be shown that $D_2[u] = F[u]$ for all $u \notin B_1$.

Denote by $[D_1 \cup D_2]$ the distribution of $(A \cup B \mid A \sim D_1, B \sim D_2)$. This is a distribution over independent sets of M by the following property of matroid contractions [see Lawler (1976)]:

(17)

On the other hand, for any set A in the support of a maxmin-fair distribution F, the set $A \cap B_1$ must be a base of (or else Eq. (16) would fail). Let $F_2$ denote the distribution $(A {\setminus } B_1 \mid A \sim F)$; by (17), $F_2$ is a distribution over elements of the contraction $M/B_1$.

To complete the proof, observe that $F\uparrow \succeq [D_1 \cup D_2]\uparrow$ by Theorem 1 because F is maxmin-fair. As ${F}[{u}] = {D_1}[{u}] < {F}[{v}], {D}[{v}]$ for $u \in B_1, v \in L{\setminus } B_1$, the fact that $F\uparrow \succeq [D_1 \cup D_2]\uparrow$ implies $F_2\uparrow \succeq D_2\uparrow$, and the maxmin-fairness of $D_2$ allows us to deduce that $F_2\uparrow = D_2\uparrow$. Hence, by Lemma 1, for all $v \notin B_1$ we have ${F}[{v}] = {F_2}[{v}] = {D_2}[{v}]$.

$\square$

Similarly, we have the following for minmax-fairness.

Lemma 9

Define a sequence of sets $B'_1, \ldots , B'_{k'}$ iteratively by:

$$\begin{aligned}&B'_i \text { is a maximal set } X\subseteq S'_i \text { maximizing } \frac{ \rho (S'_{i}) - \rho (S'_{i}{\setminus } X) }{|X|},\nonumber \\&\quad \text { where}~ {S'_i = L{\setminus } \bigcup _{j=1}^{i-1} B'_j} \end{aligned}$$

(18)

We stop when $S'_k = \emptyset$. Then for every $i, u \in B'_i$, the satisfaction probability of u in a minmax-Pareto distribution F is $\lambda '_i = \frac{ \rho (B'_i) }{ |B'_i| }$.

Proof

We argue by induction on k. First, observe that Theorem 16 implies $F[u] \le \lambda '_1$ for all $u \in L$. The expected number of satisfied elements within $B_1$ cannot be below $\lambda '_1 |B_1|$ for any Pareto-efficient distribution F, otherwise we would have the contradiction

$$\begin{aligned} \rho (L) = {{\,\mathrm{\mathbb {E}}\,}}_{A \sim F} |A| = {{\,\mathrm{\mathbb {E}}\,}}_{A \sim F} [|A \cap B_1|] + {{\,\mathrm{\mathbb {E}}\,}}_{A \sim F} [|A {\setminus } B_1|] < \lambda '_1 |B_1| + \rho (L{\setminus } B_1) = \rho (L). \end{aligned}$$

On the other hand,

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}_{A \sim F} [|A \cap B_1|] = \sum _{u\in B_1} F[u] \le \lambda '_1 |B_1| = \rho (B_1) \end{aligned}$$

(19)

by linearity of expectation, so the equality $F[u] = \lambda '_1$ must hold for all $u \in B_1$. If $k = 1$, this shows the result. The rest of the proof is completely analogous to that of Lemma 8, except that we use Theorem 14 and Lemma 7 in place of Theorem 1 and Lemma 1. $\square$

Proof of Theorem 2

It suffices to prove the equivalence (1) $\Leftrightarrow$ (2). Inded, if it holds, then a maxmin-fair distribution simultaneously maximizes the minimum satisfaction probability and minimizes the maximum satisfaction probability (among Pareto-efficient distributions), hence it also minimizes the largest difference between two satisfaction probabilities. An easy inductive argument (omitted) shows that the equivalence (1) $\Leftrightarrow$ (3) then follows.

To show that (1) $\Leftrightarrow$ (2), consider the sequence $B_1, \ldots , B_k$ from Lemma 8 and the sequence $B'_1, \ldots , B'_{k'}$ from Lemma 9. It suffices to show that they are the same sequence in reverse: $k=k'$ and $B_i=B_{k'+1-i}$ for all i. We proceed from top to bottom, showing by induction that for each value of i from 1 to k, $B'_i = B_{k+1-i}$. Consider any $Z \in S_{i+1} = S_i \cup B_{i+1}$, which may be split into $Z = X \cup Y$ where $X \subseteq S_i = \bigcup _{j \le i} B_j$ and $Y\in B_{i+1}$. Then we have

$$\begin{aligned} \rho (Z) - \rho (S_i) \ge \rho (S_i \cup Y) - \rho (S_i) \ge \lambda _{i+1} |Y|, \end{aligned}$$

where the first inequality is by submodularity of $\rho$, and the second by construction of $B_{i+1}$ (15). On the other hand,

$$\begin{aligned} \rho (S_{i+1}) - \rho (S_i) = \lambda _{i+1} |B_{i+1}|, \end{aligned}$$

hence

$$\begin{aligned} \frac{ \rho (S_{i+1}) - \rho (Z) }{|S_{i+1} {\setminus } Z|} \le \frac{\lambda _{i+1} |B_{i+1}| + \rho (S_i) - (\lambda _{i+1} |Y| + \rho (S_i)) }{|B_{i+1} {\setminus }Y|} = \lambda _{i+1}. \end{aligned}$$

Notice that equality holds when $Z=S_i$, so $S_i$ maximizes $\frac{ \rho (S_{i+1}) - \rho (Z) }{|S_{i+1} {\setminus } Z|}$ over all $Z \subseteq S_{i+1}$. Using (18) and the induction hypothesis, this means that $B'_{k'+1-i} = B_i$, as we wished to show.

Appendix B: Proofs for Sect. 4: a polynomial-time algorithm for maxmin-fair matching

Proof of Corollary 3

From Theorem 3 it follows that, when $\rho (L) = |R|$, the rank function of a bipartite matching problem is given by

$$\begin{aligned} \rho (S) = \min _{T \subseteq S} |\varGamma (T)| + |S| - |T|. \end{aligned}$$

(20)

Define

$$\begin{aligned} \alpha = \max _{S\subsetneq L} \frac{|\varGamma (L)| - |\varGamma (S)|}{|L {\setminus } S|}; \qquad \beta = \max _{S\subsetneq L} \frac{\rho (L) - \rho (S)}{|L {\setminus } S|}. \end{aligned}$$

In view of Theorem 16 and the equivalence between maxmin and minmax fairness for matroid problems (Theorem 2), it suffices to show that $\alpha = \beta$. Since $\rho (L) = |\varGamma (L)|$ and $\rho (S) \le |\varGamma (S)|$ for all S, inequality $\alpha \le \beta$ is immediate. To show that $\beta \le \alpha$, it suffices to prove that $\rho (L) - \rho (S) - \alpha |L {\setminus } S| \le 0$ for all $S \subseteq L$. This follows from (20) and the fact that $\alpha \le \beta \le 1$:

$$\begin{aligned} \rho (L) - \rho (S) - \alpha |L {\setminus } S|&= \max _{T \subseteq S}{\; |\varGamma (L)| - |\varGamma (T)| - |S| + |T|- \alpha |L {\setminus } S|}\\&\le \max _{T \subseteq S}{\; \alpha |L {\setminus } T| - |S {\setminus } T|- \alpha |L {\setminus } S|}\\&\le 0. \end{aligned}$$

Lemma 10

Define a sequence of sets $B_1, B_2, \ldots , B_k$ iteratively by:

$$\begin{aligned}&B_i \text { is a maximal set } X \subseteq L{\setminus } S_{i-1} \text { minimizing } \frac{ |\varGamma (X \cup S_{i-1})| - |\varGamma (S_{i})| }{|X|},\nonumber \\&\quad \text { where}~{ S_i = \bigcup \nolimits _{j <i} B_i.} \end{aligned}$$

(21)

Stop when $S_k = L$. Then for every $i, u \in B_i$, the satisfaction probability of u in a maxmin-fair distribution F is $\lambda _i = \frac{ | \varGamma (B_i)| }{ |B_i| }$, and any $w \in \varGamma (B_i){\setminus } \varGamma (S_{i-1})$ is matched to some $u \in B_i$ with probability 1.

Proof

Since the sequence $S_0, S_1, \ldots$ is strictly increasing (with respect to inclusion) and L is finite, there exists some k such that $S_k = L$.

For each $i=0,\ldots , k$, let $H_i$ denote the graph , i.e., the result of removing the vertices in $S_{i-1}$ and all their incident edges. For $i = 1, \ldots , k$, we argue by induction on i that the coverage probabilities of F outside of $S_{i-1}$ coincide with those of a maxmin-fair distribution for $H_i$; and and moreover the probabilities are as prescribed by the statement of the lemma.

The case $i = 1$ is trivial, so assume $i > 1$. By Corollary 2, there is a distribution of matchings in $H_i$ with minimum satisfaction probability at least $\lambda _i$; the expected number of covered elements from $B_i$ is then at least $\lambda _i |B_i| = |\varGamma (B_i) {\setminus } \varGamma (S_{i-1})| = |\varGamma _{H_i}(B_i)|$. Hence equality must always hold, and the maxmin-fair distribution $F_i$ for $H_i$ has satisfaction probability precisely $\lambda _i$ for all $u \in B_i$. By the induction hypothesis, $F[u] = F_i[u] = \lambda _i$ for all $u \in B_i$. Now observe that the neighbors of $B_i$ that belong to $S_{i-1}$ are already matched with probability 1. There are only $|\varGamma _{H_i}(B_i)|$ other neighbors, and since the expected number of covered neighbours of $B_i$ in F is equal to $|\varGamma _{H_i}(B_i)|$, it follows that any $w \in \varGamma _{H_i}(B_i)$ is matched to some $v \in B_i$ with probability 1 in F. In particular, in F no element of $\varGamma _{H_i}(B_i)$ is matched to any vertex outside $B_i$ with non-zero probability, so the satisfaction probabilities of F outside of $S_i$ must coincide with those of of a maxmin-fair distribution for $H_{i+1}$. $\square$

Lemma 11

For any two distinct fairly isolated sets X and Y, either $X \subseteq Y$ or $Y \subseteq X$ holds.

Proof

By Corollaries 2 and 3, $X \ne \emptyset$ is fairly isolated if and only if $X = L$ or

For any two sets A, B such that $A \subsetneq B$, define $d(A \mid B) = \frac{ |\varGamma (A \cup B)| - |\varGamma (B)| }{|A {\setminus } B|}.$ Then we can rewrite the definition of fair isolation (including the case $X = L$) as:

$$\begin{aligned} X\text { is fairly isolated} \qquad \Leftrightarrow \qquad d(X \mid S) < d(T \mid X) \qquad \forall \; S\subsetneq X, T\supsetneq X. \end{aligned}$$

Now assume for contradiction X and Y are fairly isolated but $X {\setminus } Y$ and $Y{\setminus } X$ are both non-empty. Then $d(X \mid X \cap Y)$ and $d(Y\mid X \cap Y)$ are well defined; assume without loss of generality that $d(X \mid X \cap Y) \le d(Y\mid X \cap Y)$. Then

$$\begin{aligned} d(X\mid X \cap Y) \le d(Y\mid X \cap Y) < d(X \cup Y\mid Y), \end{aligned}$$

where we used the fair isolation of Y. But this contradicts the submodularity of $\varGamma$. $\square$

Proof of Theorem 5

Let $B'_1, \ldots , B'_k$ be the sequence of sets given by Lemma 10 and define $S'_i = \bigcup _{j \le i} B'_i,$$\lambda '_i = \frac{\varGamma (S'_i \cup B'_i) - \varGamma (S'_i) }{|B'_i|}$ and $\lambda '_0 = 0$. We show that $S'_1, \ldots , S'_{k}$ comprise all fairly isolated sets. Assuming this for the moment, notice that by definition these sets form a chain, and the sets $B'_1, \ldots , B'_k$ satisfy property (a) by Lemma 10. Part (b) follows then by applying the fair decomposition to the dual of the matching matroid, using Corollary 3, and recalling that maxmin-fairness and minmax-Pareto efficiency are equivalent for matroids (Theorem 2).

To see that $S'_1, \ldots , S'_{k-1}$ are fairly isolated, notice that $S'_k = L$ indeed is by definition, whereas for $i < k$ we have This meets the definition of fair separation from Sect. 4.4.

The fact that the fairly isolated sets form a chain is a direct consequence of Lemma 11. Finally, assume for contradiction that some fairly isolated set X exists other than $S'_1, \ldots , S'_k$. Then $S'_i \subsetneq X \subsetneq S'_{i+1}$ for some $i, 0 \le i < k$. Then we have

$$\begin{aligned} \lambda '_{i+1} \le \frac{ |\varGamma (X)| - |\varGamma (S'_i)| }{|X{\setminus } S'_i|} < \frac{ |\varGamma (S'_{i+1})| - |\varGamma (X)| }{|S'_{i+1}{\setminus } X|}, \end{aligned}$$

(22)

where the first inequality is by construction of $B'_{i+1}$ and $S'_{i+1}$, and the second by the fair isolation of X. But then

$$\begin{aligned} \lambda '_{i+1} {|S'_{i+1}{\setminus } X|} + |\varGamma (X)| < |\varGamma (S'_{i+1})| =|\varGamma (S'_{i})| + \lambda '_{i+1} |B'_i|, \end{aligned}$$

i.e.,

$$\begin{aligned} |\varGamma (X)| - |\varGamma (S'_{i})|< \lambda '_{i+1} {|X{\setminus } S'_{i}|}, \end{aligned}$$

contradicting (22).