Maximizing Coverage While Ensuring Fairness: A Tale of Conflicting Objectives

Abstract

Ensuring fairness in computational problems has emerged as a key topic during recent years, buoyed by considerations for equitable resource distributions and social justice. It is possible to incorporate fairness in computational problems from several perspectives, such as using optimization, game-theoretic or machine learning frameworks. In this paper we address the problem of incorporation of fairness from a combinatorial optimization perspective. We formulate a combinatorial optimization framework, suitable for analysis by researchers in approximation algorithms and related areas, that incorporates fairness in maximum coverage problems as an interplay between two conflicting objectives. Fairness is imposed in coverage by using coloring constraints that minimizes the discrepancies between number of elements of different colors covered by selected sets; this is in contrast to the usual discrepancy minimization problems studied extensively in the literature where (usually two) colors are not given a priori but need to be selected to minimize the maximum color discrepancy of each individual set. Our main results are a set of randomized and deterministic approximation algorithms that attempts to simultaneously approximate both fairness and coverage in this framework.

Appendices

Appendix

Proof of Lemma 1

(a) We describe the proof for $\chi =2$; generalization to $\chi >2$ is obvious. The reduction is from the Exact Cover by 3-sets (X3C) problem which is defined as follows. We are given an universe ${\mathscr {U}}'=\{u_1,\dots ,u_{n'}\}$ of $n'$ elements for some $n'$ that is a multiple of 3, and a collection of $n'$ subsets ${\mathscr {S}}_1,\dots ,{\mathscr {S}}_{n'}$ of ${\mathscr {U}}$ such that $\bigcup _{j=1}^{n'} {\mathscr {S}}_j={\mathscr {U}}$, every element of ${\mathscr {U}}'$ occurs in exactly 3 sets and $|{\mathscr {S}}_j|=3$ for $j=1,\dots ,n'$. The goal is to decide if there exists a collection of ${n'}/{3}$ (disjoint) sets whose union is ${\mathscr {U}}'$. X3C is known to be $\textsf {NP}$-complete [19]. Given an instance $\langle {\mathscr {U}}',{\mathscr {S}}_1,\dots ,{\mathscr {S}}_{n'}\rangle$ of X3C as described, we create the following instance $\langle {\mathscr {U}},{\mathscr {S}}_1,\dots ,{\mathscr {S}}_{n'+1},k\rangle$ of Fmc($2, k$):

(i):

The universe is ${\mathscr {U}}=\{u_1,\dots ,u_{n'}\}\cup \{u_{n'+1},\dots ,u_{2n'}\}$ (and thus $n=2n'$),

(ii):

$w(u_j)=1$ for $j=1,\dots ,2n'$,

(iii):

the sets are ${\mathscr {S}}_1,\dots ,{\mathscr {S}}_{n'}$ and a new set ${\mathscr {S}}_{n'+1}=\{u_{n'+1},\dots ,u_{2n'}\}$,

(iv):

the coloring function is given by ${\mathscr {C}}(u_j) = \left\{ \begin{array}{r l} 1, &{} \text{ if } 1\le j\le n' \\ 2, &{} \text{ otherwise } \end{array} \right.$, and

(v):

$k=\frac{n'}{3}+1 = \frac{n}{6}+1$.

Clearly, every element of ${\mathscr {U}}$ occurs in no more than 3 sets and all but the set ${\mathscr {S}}_{n'+1}$ contains exactly 3 elements. The proof is completed once the following is shown:

($*$) the given instance of X3C has a solution if and only if the transformed instance of Fmc($2, 1+{n}/{6}$) has a solution.

A proof of ($*$) is easy: since the set ${\mathscr {S}}_{n'+1}$ must appear in any valid solution of Fmc, a solution ${\mathscr {S}}_{i_1},\dots ,{\mathscr {S}}_{i_{n'/3}}$ of X3C corresponds to a solution ${\mathscr {S}}_{i_1},\dots ,{\mathscr {S}}_{i_{n'/3}},{\mathscr {S}}_{n'+1}$ of Fmc($2, k$) and vice versa.

(b) The proof is similar to that in (a) but now instead of X3C we reduce the node cover problem for cubic (i.e., 3-regular) graphs ($\hbox {VC}_3$) which is defined as follows: given a cubic graph $G=(V,E)$ of $n'$ nodes and $3n'/2$ edges and an integer $k'$, determine if there is a set of $k'$ nodes that cover all the edges. $VC_3$ is known to be $\textsf {NP}$-complete even if G is planar [19]. For the translation to an instance of Fmc($2, k$), edges of G are colored with color 1, we add a new connected component ${\mathscr {K}}_{(3n'/2)+1}$ to G that is a complete graph of $(3n'/2)+1$ nodes with every edge having color 2, transform this to the set-theoretic version of Fmc using the standard transformation from node cover to set cover and set $k=k'+1$; note that $n=3n'/2 + \left( {\begin{array}{c}(3n'/2)+1\\ 2\end{array}}\right) =\varTheta ((n')^2)$ and $a=3n'/2=O(\sqrt{n}\,)$. To complete the proof, note that any feasible solution for the Fmc($2, k$) instance must contain exactly one node from ${\mathscr {K}}_{(3n'/2)+1}$ covering $3n'/2$ edges and therefore the solution for the edges with color 1 must correspond to a node cover in G (and vice versa).

(c) We given a different reduction from X3C. Given an instance $\langle {\mathscr {U}}',{\mathscr {S}}_1,\dots ,{\mathscr {S}}_{n'}\rangle$ of X3C as in (a), we create the following instance $\langle {\mathscr {U}},{\mathscr {T}}_1,\dots ,{\mathscr {T}}_{n'},k\rangle$ of Fmc($n', k$):

(i):

For every set ${\mathscr {S}}_i=\big \{u_{i_1},u_{i_2},u_{i_3}\big \}$ of X3C we have three elements $u_{i_1}^i,u_{i_2}^i,u_{i_3}^i$ and a set ${\mathscr {T}}_i=\big \{ u_{i_1}^i,u_{i_2}^i,u_{i_3}^i\big \}$ in Fmc (and thus $n=3n'$, $a=3$ and $f=1$),

(ii):

$w(u_{i_j}^i)=1$ and ${\mathscr {C}}(u_{i_j}^i) = i_j$ for $i\in \{1,\dots ,n'\},j\in \{1,2,3\}$ (and thus $\chi =n'={n}/{3}$),

(iii):

$k={n'}/{3} = {n}/{9}$.

The proof is completed by showing the given instance of X3C has a solution if and only if the transformed instance of Fmc(${n}/{3}, {n}/{9}$) has a solution. This can be shown as follows. We include the set ${\mathscr {T}}_i$ in the solution for Fmc if and only if the set ${\mathscr {S}}_i$ is in the solution for X3C. For any valid solution of X3C and every $j\in \{1,\dotsc ,n'\}$ the element $u_j\in {\mathscr {U}}'$ appears in exactly one set, say ${\mathscr {S}}_\ell =\big \{u_{\ell _1},u_{\ell _2},u_{\ell _3}\big \}$, of X3C where one of the elements, say $u_{\ell _1}$, is $u_j$. Then, the solution of Fmc contains exactly one element, namely the element $u_{\ell _1}^\ell$, of color $\ell _1=j$. Conversely, given a feasible solution of Fmc with at most $k\le {n}/{9}$ sets, first note that if $k<{n}/{9}$ then the total number of colors of various elements in the solution is $3k<n'$ and thus the given solution is not valid. Thus, $k={n}/{9}$ and therefore the solution of X3C contains ${n}/{9}={n'}/{3}$ sets. Now, for every color j the solution of Fmc contains a set, say ${\mathscr {T}}_\ell =\big \{u_{\ell _1}^\ell ,u_{\ell _2}^\ell ,u_{\ell _3}^\ell \big \}$, containing an element of color j, say the element $u_{\ell _1}^\ell$. Then $\ell _1=j$ and the element $u_j$ appears in a set in the solution of X3C. To see that remaining claims about the reduction, there is no solution of Fmc that includes at least one element of every color and that is not a solution of X3C.