Maximizing Nash product social welfare in allocating indivisible goods

European Journal of Operational Research 247 (2015) 548–559

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European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor

Decision Support

Maximizing Nash product social welfare in allocating indivisible goods Andreas Darmann a,∗, Joachim Schauer b a b

University of Graz, Institute of Public Economics, Universitaetsstr. 15, A-8010 Graz, Austria University of Graz, Department of Statistics and Operations Research, Universitaetsstr. 15, A-8010 Graz, Austria

a r t i c l e

i n f o

Article history: Received 16 June 2014 Accepted 26 May 2015 Available online 3 June 2015 Keywords: Fair division Social welfare Nash product Computational complexity Scoring rules

a b s t r a c t We consider the problem of allocating indivisible goods to agents who have preferences over the goods. In such a setting, a central task is to maximize social welfare. In this paper, we assume the preferences to be additive and measure social welfare by means of the Nash product. We focus on the computational complexity involved in maximizing Nash product social welfare when scores inherent in classical voting procedures such as approval or Borda voting are used to associate utilities with the agents’ preferences. In particular, we show that the maximum Nash product social welfare can be computed efficiently when approval scores are used, while for Borda and lexicographic scores the corresponding decision problem becomes NP-complete. © 2015 Elsevier B.V. and Association of European Operational Research Societies (EURO) within the International Federation of Operational Research Societies (IFORS). All rights reserved.

1. Introduction The allocation of goods (items, resources) to agents who have preferences over these goods (multiagent resource allocation) is a fundamental problem of economics, and, in particular, social choice theory. This problem has been tackled in various scenarios (see, e.g., Chevaleyre et al., 2006 for a survey), where, e.g., we distinguish between divisible and indivisible goods, and centralized and decentralized approaches. Here, we consider the case of m indivisible and nonshareable goods to be distributed among agents who report their preferences to a central authority. Typically, individual utilities of (bundles of) items are associated with the preferences over the items. In this work, this is done via numerical scores used in voting rules. Now, a major task is to find an allocation which maximizes the social welfare achieved. Different notions of social welfare have been introduced, the most important being utilitarian, egalitarian, and Nash product social welfare (cf. Brandt, Conitzer, & Endriss, 2013). Utilitarian social welfare of an allocation is given by the sum of the agents’ utilities resulting from the allocation. A more fine-grained approach is egalitarian social welfare, where the lowest of the agents’ individual utilities in a given allocation is considered. In a certain sense, the Nash product social welfare links these two approaches: by measuring the product of the agents’ utilities in an allocation, maximizing the Nash product social welfare targets at a “balanced” allocation (see also Nguyen, Nguyen, Roos, & Rothe, 2014).

∗

Corresponding author. Tel.: +43 3163807139; fax: +43 3163809530. E-mail addresses: [email protected] (A. Darmann), joachim.schauer@ uni-graz.at (J. Schauer).

The Nash product as a measure for social welfare satisfies several desirable properties (see Moulin, 2003 for an explicit treatment). For example, it satisfies the basic fairness criterion that it increases when inequality among two agents is reduced (given the respective change is mean-preserving; see also Ramezani & Endriss, 2010). Clearly, the Nash product also satisfies the monotonicity property that an increase of an agent’s utility yields an increase in the Nash product. In addition, it is independent of both common utility scale and individual utility scale: the social welfare ordering, i.e., the ordering of the allocations according to their Nash product, remains unchanged both if (i) all agents rescale their utilities with the same factor, and (ii) each agent rescales her utility using a different factor. A central question in maximizing social welfare is the computational complexity involved. We assume that the agents have additive preferences, i.e., for each agent, the utility of a set of goods is the sum of the utilities of the single goods it contains. Clearly, maximizing utilitarian social welfare is an easy task – simply allocate each item to an agent who it yields the highest utility for (see also Brandt et al., 2013). In contrast, it is known that maximizing egalitarian social welfare and Nash product social welfare are NP-complete for additive utilities and general scoring functions (Roos & Rothe, 2010). Recently, Baumeister et al. (2013) have shown that maximizing egalitarian social welfare remains NP-complete for a number of prototypical scoring functions from voting theory: quasiindifference, Borda, and lexicographic scoring. On the positive side, it is known that the maximum egalitarian social welfare can be computed in polynomial time for approval scores (Golovin, 2005). To the best of our knowledge, the computational complexity of maximizing Nash product social welfare under scoring functions such as approval, Borda, or lexicographic scoring has not been considered yet. In

http://dx.doi.org/10.1016/j.ejor.2015.05.071 0377-2217/© 2015 Elsevier B.V. and Association of European Operational Research Societies (EURO) within the International Federation of Operational Research Societies (IFORS). All rights reserved.

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

this paper, we investigate the computational complexity involved in maximizing Nash product social welfare under these classical scoring functions. Related work and our contribution. In the Santa Claus problem (Bansal & Sviridenko, 2006), the goal is to maximize egalitarian social welfare when indivisible items have to be allocated among agents. There, the agents associate an arbitrary numerical value with each item, i.e., the input is numerical, and the utility of an agent is the sum of the values of the items assigned to her. In contrast, as done in our work, in the setting presented in Brams, Edelman, and Fishburn (2003) the agents’ preferences are expressed ordinally. Brams et al. (2003) introduce the notion of a Borda-optimal allocation, which is defined as an allocation maximizing egalitarian social welfare where an agent’s utility is the sum of the Borda scores of the items the agent receives. We translate this concept to maximizing Nash product social welfare, and expand the perspective from Borda scores to approval (and, in particular, k-approval) and lexicographic scores. Informally speaking, in k-approval scoring a distinction between “good” (approved) and “bad” (disapproved) items is made: the top k items in the ranking of an agent receive a score of 1, while the remaining items get score 0. In Borda scoring, an agent’s most preferred item gets a score of m, her second-ranked item a score of m − 1, and so on; her least preferred item has a score of 1. In lexicographic scoring, the position of the item in the ranking is even more crucial: any item r yields a higher score than the total of all items ranked below r. Our goal is to analyze the computational complexity involved in maximizing Nash product social welfare with respect to these types of scores. In the context of maximizing social welfare in multiagent resource allocation, complexity results have been achieved with respect to different types of utility representation: the bundle form, k-additive form, or straight-line programs. For the bundle form representation, NP-completeness results for utilitarian (Chevaleyre, Endriss, Estivie, & Maudet, 2008), egalitarian (Roos & Rothe, 2010), and Nash product social welfare (Ramezani & Endriss, 2010; Roos & Rothe, 2010) are known. For straight-line programs, Dunne, Wooldridge, and Laurence (2005) show that maximizing utilitarian social welfare is NP-complete, while Nguyen et al. (2014) show that maximizing social welfare is NP-complete both for the egalitarian and Nash product approach. Both maximizing egalitarian social welfare and maximizing Nash product social welfare turn out to be NP-complete for 1-additive, i.e., additive utilities already (Lipton, Markakis, Mossel, & Saberi, 2004; Roos & Rothe, 2010). In these works, however, reductions from Partition are given, which do not imply the NP-completeness for any of the scoring functions considered in our work. Given additive utilities, Baumeister et al. (2013), besides many other results, have proven that maximizing egalitarian social welfare is NP-complete for Borda, lexicographic, and quasiindifference scoring. In this paper, we show that maximizing Nash product social welfare is NP-complete for Borda and lexicographic scores, whereas it is polynomially solvable for approval scores. The computational complexity involved when quasi-indifference scores are used is still open.

549

at least close to the agents’ actual utilities. However, starting with numerical inputs instead would have several drawbacks (see also Baumeister et al., 2013); e.g., often it is easier for agents to rank items instead of associating numerical values with each single item, especially in contexts where money is not a key factor. Next, as also pointed out in Baumeister et al. (2013), the use of numerical inputs has the severe disadvantage that it insinuates comparability of interpersonal preferences. Finally, note that our approach is very common in voting theory, as in fact it resembles the way that positional scoring rules proceed.1 In particular, we assume that agents have preferences over the single resources. The preferences are expressed by means of strict orders ai over R, which are summarized by the n-tuple π = (a1 , a2 , . . . , an ) called profile. We denote by rankai (r ) the rank of resource r in the ranking of agent ai . We adopt scores used in voting procedures to evaluate these preferences by means of utility functions ua : R → Q, a ∈ A. We assume that the utility functions are additive, i.e., for any subset R ⊆R we have  ua (R ) = r∈R ua (R ). For the sake of readability, we may write ua (P) instead of ua (P(a)). Given a profile π , we consider the following types of scores (where r ∈ R): •

k-approval scores: For each agent a ∈ A,



ua ( r ) = • •

1

if ranka (r ) ≤ k

0

otherwise

Borda scores: For each agent a ∈ A, ua (r ) = m + 1 − ranka (r ). Lexicographic scores: For each agent a ∈ A, ua (r ) = 2m−ranka (r ) .

Given k-approval scores, for each a ∈ A, ua partitions the set R into a set Sa := {r ∈ R : ua (r ) = 1} (the set of resources agent a approves of) and a set Sac := {r ∈ R : ua (r ) = 0} (the set of resources agent a disapproves of). Conversely, specifying the set Sa (of size k) for each agent a uniquely determines the corresponding k-approval scores. More generally (and slightly abusing notation), given a set S(a)⊆R for each a ∈ A, approval scores are given by ua (r ) = 1 for r ∈ S(a) and ua (r ) = 0 for r ∈ RS(a). Given an allocation P, the Nash product social welfare for P is given  by swN (P ) = 1≤i≤n uai (P ). 2.2. Problem definitions In this paper, we consider the problem of maximizing the Nash product social welfare with respect to the above scores, i.e., utility functions. The corresponding decision problems are defined as follows.

2. Formal framework

Definition 2.1. (Nash Product Social Welfare MaximizationApproval) GIVEN: Quadruple (R, A, S, k): R is a set of resources, A a set of agents, a collection S = {Sa1 , Sa2 , . . . , San } of subsets Sai ⊆ R, and k ∈ N. QUESTION: Is there an allocation P such that swN (P) ≥ k, where uai (r ) = 1 if r ∈ Sai and uai (r ) = 0 otherwise?

2.1. Preliminaries

Analogously, we Maximization-Borda.

Let R = {r1 , r2 , . . . , rm } be a set of m indivisible resources (items) and let A = {a1 , . . . , an } be a set of n agents. An allocation is a mapping that assigns to each agent a subset of resources such that each resource is handed to exactly one agent. Formally, an allocation P is a  mapping P: A → 2R with a∈A P (a ) = R and P (ai ) ∩ P (a j ) = ∅ whenever i = j. Now, in our model, we start with ordinal inputs, i.e., the agents rank resources, and map these ranks to numerical scores then. Note that we do not claim that these numerical scores are equivalent or

defineNash

Product

Social

Welfare

Definition 2.2. (Nash Product Social Welfare MaximizationBorda) GIVEN: Quadruple (R, A, π , k): R is a set of resources, A a set of agents, π is a profile, and k ∈ N. QUESTION: Is there an allocation P such that swN (P) ≥ k for Borda scores? 1 Obviously, with the clear difference that we are finally interested in allocations instead of winners of elections.

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A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

It is straightforward to define Nash Product Social Welfare Maximization-Lexicographic for lexicographic scores. In what follows, we use the shortcut NPSW for Nash Product Social Welfare Maximization.

•

•

3. Complexity of NPSW 3.1. The easy case: NPSW-Approval First, we show that NPSW is in P for approval scores. This is done by a transformation to the polynomially solvable Min Cost Flow problem (cf. Ahuja, Magnanti, & Orlin, 1993). In particular, it follows that in the case of approval scores, an actual allocation that maximizes Nash product social welfare can be found in polynomial time. We begin with some basic definitions and two known properties of a min cost flow (i.e., an optimal solution of the Min Cost Flow problem). Definition 3.1. In an instance M = (G, c, , p, b) of Min Cost Flow, we are given a directed graph G = (V, E ). With each edge e ∈ E, two rational numbers are associated: a cost c(e) and an upper bound p(e) on the capacity of e. For each v ∈ V, we are given the rational-valued vertex demand b(v). The Min Cost Flow problem can be stated as follows:



min

c (u, v ) f (u, v )

(u,v )∈E

s.t.

 v:(u,v )∈E

f (u, v ) −



f (u, v ) = b(u ) for all u ∈ V

v:(v,u )∈E

0 ≤ f (u, v ) ≤ p(u, v )

for all(u, v ) ∈ E (1)

A function f : E → Q is called flow, if f satisfies the conditions stated  in (1). The cost of a flow f is defined by c ( f ) = (u,v )∈E c (u, v ) f (u, v ). In an instance M = (G, c, , p, b) of Min Cost Flow, the capacity constraints on the edges are written by means of [0, p(e)]. The cost of a directed cycle defined as the sum of the costs of the edges in the cycle. In M, we associate a residual network Gf with a flow f. Gf is constructed from G as follows. Each edge (i, j) ∈ E is replaced by the edges (i, j) and (j, i). In Gf , the arc (i, j) has cost c(i, j) and residual capacity [0, p(i, j ) − f (i, j )]; the arc (j, i) has cost c ( j, i ) = −c (i, j ) and residual capacity [0, f(i, j)]. Finally, Gf consists of edges with positive residual capacity only. Theorem 3.1 (Negative cycle optimality condition; cf. Ahuja et al., 1993). A flow f is an optimal solution of Min Cost Flow, if and only if Gf does not contain a negative cost directed cycle. Theorem 3.2 (Integrality property; cf. Ahuja et al., 1993). If all arc capacities and all node demands are integer, then there is an integer min cost flow. Theorem 3.3. NPSW-Approval is in P. Proof. Let I = (R, A, π , k ) be an instance of NPSW-Approval. We assume that each item is approved of by at least one agent (otherwise, items which are not approved by any agent are removed in a preprocessing step). We argue that I can be decided by solving an instance M of Min Cost Flow. M is defined as follows. In the graph G = (V, E ), certain vertices are identified with items/agents of the same label. In particular, V = {s, t } ∪ A ∪ R ∪ {ti, j |i ∈ A, j ∈ R}. The vertex demands are b(s ) = m, b(t ) = −m and b(v ) = 0 for each v ∈ V{s, t}. In order to construct the edge set E, •

For each r ∈ R we introduce edge (s, r) with capacity [0, 1] and zero cost.

For each ai ∈ A and for each r ∈ R with uai (r ) = 1 we introduce the edge (r, ai ) with capacity [0, 1] and zero cost. For each ai ∈ A and 1 ≤ j ≤ m, we introduce • the edge (a , t ) with capacity [0, 1] and cost c (a , t ) = n j i i, j i i, j • the edge (t , t) with capacity [0, 1] and zero cost. i, j

By the integrality property, there is an integer min cost flow f in M. That is, for each e ∈ E, f either does not send flow along e or f sends exactly 1 unit of flow along e. Clearly, due to the choice of the vertex demands and the edge capacities [0, 1] of the edges (s, r), for each r ∈ R there is exactly one unit of flow sent through vertex r. Due to the capacities of the edges (r, a) this means that for each r ∈ R, there is exactly one a ∈ A such that f sends (one unit of) flow along (r, a). Thus, the mapping Pf : A → R defined by r ∈ Pf (a) iff f (r, a ) = 1 is an allocation in I. On the other hand, it is not hard to see that an allocation P induces an integer flow f P in M by • •

•

Sending one unit of flow along (s, r) for each r ∈ R For each ai ∈ A and for each r ∈ R, sending one unit of flow along (r, ai ) iff r ∈ P(ai ) Sending one unit of flow along (ai , ti, h ) and (ti, h , t) for each 1 ≤ h ≤ uai (P )

The proof proceeds in three steps. Step 1: Let f be an integer flow in instance M, where fi denotes the amount of flow sent through vertex ai . We show that the following holds: f is a min cost flow if and only if the two properties 1. For each ai ∈ A, f sends flow along the arcs (ai , ti, h ), for all 1 ≤ h ≤ fi , and 2. There is no sequence (ai1 , r j1 , ai2 , r j2 , . . . , r j−1 , ai ) with fi1 − fi ≥ 2, such that for all 1 ≤ h ≤  − 1 we have (i) (r jh , aih+1 ) ∈ E and (ii) f sends flow along the arc (r jh , aih ) are satisfied. Note that the second property reflects the idea that a more “balanced” and thus cheaper flow cannot be immediately derived from f. In particular, we show that the above conditions are equivalent to the negative cycle optimality condition. First, note that due to the fact that for each edge (s, rj ) demand and upper bound equal to 1, the residual capacity of the edge is 0. That is, the edge is not contained in the residual network H. Thus, H does not contain any edge emanating from s. Hence, s cannot be part of any cycle in R. Step 1a: Assume that one of the two conditions above is not satisfied. Case I considers the case that the first condition is violated. Case II considers the situation that the first condition holds, but the second is violated. Case I: For some ai ∈ A, there is an 1 ≤ h ≤ fi such that f does not send flow along the edges (ai , ti, h ). Then, f must send along an edge (ai , ti,  ) for some  > fi . But then it is easy to see that in the residual network H the cycle γ = (t, ti, , ai , ti,h , t ) is a cycle of cost c (γ ) = −n + nh < 0 due to  > h. Case II: For all ai ∈ A, f sends one unit of flow along the edges (ai , ti, h ), 1 ≤ h ≤ fi . Assume there is a pair (ai , aj ) with fi − f j ≥ 2 such that (i) (r, aj ) ∈ E and (ii) f sends flow along the arc (r, ai ). Then, in the residual network H the cycle γ = (t, ti, fi , ai , r, a j , t j, f j +1 , t ) has negative cost: c (γ ) = −n fi + n f j +1 < 0, because fi − f j ≥ 2 by assumption. Hence, if one of the two conditions is violated, there is a negative cost cycle. Step 1b: On the other hand, assume H contains a negative cost cycle γ . We show that this implies that at least one of the two conditions is violated. Clearly, s cannot be contained in γ . In addition, γ cannot be made up of vertex t and vertices ti, j only, since each edge (ti, j , t) is of zero cost. Assume γ does not contain a vertex rj , 1 ≤ j ≤ m. Then, for some i, x, y, γ = (t, ti,x , ai , ti,y , t ) holds. Note that c (γ ) = −c (ti,x , ai ) +

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

c (ai , ti,y ) = −nx + ny . Thus, c(γ ) < 0 implies

551

With (4) and fk > gπ (k) , we get

x>y

(2)

n 



fi 



gπ ( i ) h

n −



n

h

π (i ) n g 

>−



nh > −ngπ (k) +1 ≥ −n fk

Assume that f sends flow along (ai , ti, h ) for all 1 ≤ h ≤ fi . Then the residual network H must contain (i) the edges (ti, h , ai ) for 1 ≤ h ≤ fi as only edges with head ai . Thus, x ≤ fi follows. In addition, since f is integer and in the original network G the upper bound of the capacity of each of the edges (ai , ti, h ) equals 1, it follows that f sends exactly one unit of flow along each of the arcs (ai , ti, h ), for 1 ≤ h ≤ fi . Hence, H cannot contain any of these edges. Thus, y > fi must hold. Putting things together, we get x ≤ fi < y, in contradiction with (2). Hence, the first condition is violated. Thus, γ contains a vertex r ∈ R. Assume the first condition is not violated (otherwise there is nothing to show). Then, there is a sequence (ai1 , r j1 , ai2 , r j2 , . . . , r j−1 , ai ) such that

Together with (3) we get c ( f ) − c (g) > 0, in contradiction with the fact that f is a min cost flow. Analogously, fk < gπ (k) leads to a contradiction with the fact that g is an integer min cost flow (because of c ( f ) = c (g)). Thus, fi = gπ (i )   holds for all 1 ≤ i ≤ k. Hence, a ∈A fi = a ∈A gi follows. i i Since g is an integer flow of minimum total cost (Step 2), from Step   3 a ∈A fi = a ∈A gi follows for any integer min cost flow f. Hence, i i in order to maximize the Nash product social welfare, it is sufficient to find an integer min cost flow in instance M. This can be done in polynomial time (see, e.g., Ahuja et al., 1993). 

γ = (t, ti1 ,x , ai1 , r j1 , ai2 , r j2 , . . . , r j−1 , ai , ti ,y , t )

3.2. The hard cases: NPSW-Borda and NPSW-Lexicographic

for some  ≥ 2 and some x, y, with i1 = i . Since by assumption the first condition is satisfied, x ≤ fi1 and y ≥ fi + 1 hold. c (γ ) = −c (ti1 ,x , ai1 ) + c (ai , ti ,y ) = −nx + ny ≥

3.2.1. NPSW-Lexicographic is NP-complete

f +1

f +1

−nx + n i . Now, c(γ ) < 0 implies −nx + n i < 0, i.e., x > fi + 1. Hence, fi1 ≥ x > fi + 1 holds. Thus, fi1 ≥ fi + 2 holds, since all flow values are integer. That is, the second condition is violated. As a consequence, the negative cycle condition is in fact equivalent to the two above stated conditions. Step 2: Let P be an allocation that maximizes Nash product social welfare. Throughout this proof, let g be the integer flow induced by allocation P . We show that g is a min cost flow. For each ai ∈ A, g sends gi = uai (P ) units of flow through vertex ai and one unit of flow through each of the arcs (ai , ti, h ) for 1 ≤ h ≤ gi . Assume there is a pair (ai , aj ) with gi − g j ≥ 2 such that (i) (r, aj ) ∈ E and (ii) g sends flow along the arc (r, ai ). Then, both ai , aj approve of item r. Consider the assignment P defined by P (a ) = P (a ) for a ∈ A{ai , aj }, P (ai ) = P (ai ) \ {r} and P (a j ) = P (a j ) ∪ {r}. Then,



uai (P )

(uai (P ) − 1 )(ua j (P ) + 1 ) uai (P )(ua j (P ) ai ∈A uai ( ) uai (P )ua j (P ) + uai (P ) − ua j (P ) − 1 = >1 uai (P )(ua j (P )

a ∈A

i

P

=

c ( f ) − c (g ) =

(

gπ ( i )

nh −



nh )

= ( n + n2 + · · · + n f k ) − ( n + n 2 + · · · + n gπ ( k ) ) +

n 



i=k+1

=n

gπ (k) +1

fi 

n −



n

gπ (k) +2

fk

··· + n +

n  i=k+1



fi 



gπ ( i ) h

n −

h=1



π (i ) n g 

i=k+1 h =1



nh ≤ ( n − k )

gπ ( k )



h =1

h i=1

nh =

nh+1 −1 n−1

nh < ( n − k )

Proof. We provide a reduction from the NP-complete problem Cubic Monotone 1-in-3 Sat (cf. Moore & Robson, 2001). An instance I = (X, C ) of that problem consists of a set of variables X and a set C of clauses over X, such that each clause is made up of exactly three variables of X and each variable occurs in exactly three clauses. In Cubic Monotone 1-in-3 Sat we ask if there is a truth assignment for X such that exactly one variable is true in each clause of C. Note that there are no negated literals contained in any clause of C. In addition, observe that |X | = |C | holds. Further note that φ can be a satisfying truth assignment in instance I only if it the number of variables set true under φ is exactly |X3 | . Thus, |X| is a multiple of 3. Given an instance I = (X, C ) of Cubic Monotone 1-in-3 Sat we construct an instance L = (R, A, π , k ) of NPSW-Lexicographic as follows. Abusing notation, within this paper we will identify (the set of) variables and clauses respectively with (sets of) items and agents respectively. Let n = |X | = |C |, and  = 12n. R consists of  + 6n + 3n items: •

•

The items d1 , d2 , . . . , d The item sets X = {x1 , x2 , . . . , xn }, and Y = {x1,1 , x1,2 , x1,3 , x2,1 , x2,2 , x2,3 , . . . , xn,1 , xn,2 , xn,3 } The item sets B = {b1,1 , b1,2 , b2,1 , b2,2 , . . . , bn,1 , bn,2 } and H = {h1 , h2 , . . . , h n } 3

Over the sets B, H, X resp. Y we define the rankings τ B , τ H , τ X , and τ Y as follows: • •

τX = x1  x2  · · ·  xn , τH = h1  h2  · · ·  h n , 3 τB = b1,1  b1,2  b2,1  · · ·  bn,2 , and τY = x1,1  x1,2  x1,3  x2,1  · · ·  xn,3

n

h =1

n ≤ ngπ (k) +1 n−1

For 1 ≤ i ≤ n, the ranking of agent Hi is given by

τH  xi  d1  d2  · · ·  d5n−1  τB  τY  τX \{xi }  d5n  · · ·  d •

(4)

For each 1 ≤ i ≤ , the ranking of agent Di is given by

d1  d2  · · ·  d  τB  τY  τX  τH •

− 1 holds. Thus,

gπ (k) +1

•

h

(3) Note that for any fixed h ∈ N,

i=k+1 h =1

xi3 ,k3 }. A consists of  + 5n agents:

h

h =1

h=1

+n



gπ ( i ) h

h =1

Let S ∈ {B, H, X, Y}. For any subset Z of S, by τ Z we denote the ranking τ S restricted to the subset Z. Within this proof, we represent the clause Ci = (xi1 ,k1 ∨ xi2 ,k2 ∨ xi3 ,k3 ), where kj ∈ {1, 2, 3} denotes the kj -th occurrence of variable xi j in C, by the set Ci = {xi1 ,k1 ∨ xi2 ,k2 ∨



h =1

i=1 h=1

h=1

Theorem 3.4. NPSW-Lexicographic is NP-complete.

•

where the last inequality follows from uai (P ) − ua j (P ) − 1 = gi − g j − 1 ≥ 1. This contradicts with the fact that P maximizes Nash social welfare. With Step 1, if follows that g is a min cost flow.  Step 3: Let f be an integer min cost flow. We show that a ∈A fi = i  ai ∈A gi holds. W.l.o.g., we assume f1 ≥ f2 ≥  ≥ fn . Clearly, there is a permutation π : A → A such that gπ (1) ≥ gπ (2) ≥  ≥ gπ (n) holds. We show that fi = gπ (i ) for each 1 ≤ i ≤ n. Assume the opposite, i.e., there is an index k ≥ 1 such that fi = gπ (i ) for i < k and fk = gπ (k ) . If fk > gπ (k) , then f (i ) n  

i=k+1

For 1 ≤ i ≤ n, the ranking of agent Xi is given by

xi  d1  · · ·  dn  xi,1  xi,2  xi,3  dn+1  · · ·  d5n−6   τB  τY \{xi,1 ,xi,2 ,xi,3 }  τX \{xi }  d5n−5  · · ·  d

552 •

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

For 1 ≤ i ≤ n, the rankings of the agents α i , β i , γ i are as follows. Let

ταi = xi1 ,k1  d2n+5  d2n+6  · · ·  d3n+4  xi2 ,k2  d3n+5  d3n+6  · · ·  d4n+4  xi3 ,k3

ua1 (Q ) · ua2 (Q ) − ua1 (P ) · ua2 (P )

 d3n+6  · · ·  d4n+4  xi1 ,k1

≥ (ua1 (P ) + 2λ+( j+1) − 2λ )(ua2 (P ) + 2μ− j − 2μ ) − ua1 (P ) · ua2 (P ) = ua1 (P )(2μ− j − 2μ ) + (2λ+( j+1) − 2λ )ua2 (P ) + 2λ+μ+1

ταi = xi3 ,k3  d2n+5  d2n+6  · · ·  d3n+4  xi1 ,k1  d3n+5

−2λ+( j+1)+μ − 2λ+μ− j + 2λ+μ

 d3n+6  · · ·  d4n+4  xi2 ,k2

= [ua1 (P )2μ− j − 2λ+μ− j ] + [2λ+μ+1 − 2μ ua1 (P )]

The ranking of α i is given by

+[(2λ+( j+1) − 2λ )ua2 (P ) − 2λ+( j+1)+μ + 2λ+μ ]

d1  · · ·  dn+4  bi,1  bi,2  dn+5  · · · d2n+4  ταi

>[(2λ+( j+1) − 2λ )ua2 (P ) − 2λ+( j+1)+μ + 2λ+μ ]

 d4n+5  · · ·  d5n−1  τB\{bi,1 ,bi,2 }  τY \Ci  τX  τH

where the last inequality follows from (6). Since ua2 (P ) ≥ 2μ , we hence get

 d5n  · · ·  d The ranking of β i (resp. γ i ) results from παi by replacing ταi with τβi (resp. τγi ). Let M := |R| − 1, and

κ = [2

+2

M−(n+2 )

+2

M−(n+3 )

]

k=

−1

 2M−i

n 3

·2 ·2

 n3 −1 i=0

(M−i )

· 2(

M− n3

−2λ+( j+1)+μ + 2λ+μ ] = 0

)

2n 3

·κ

In instance L we ask if there is an allocation P with swN (P) ≥ k.

•

We begin with two simple lemmata.

•

Lemma 3.5. Let P be an allocation. Let Q result from P by handing an item p ∈ P(a2 ) to agent a1 such that the following properties are satisfied: •

∃q ∈ P(a2 ) such that a2 ranks q higher than p a1 ranks p higher than the highest-ranked item of P(a1 )   Then, a ∈ A ua (P) < a ∈ A ua (Q).

(5)

With (5), we can conclude that



a ∈ A ua (Q).



Lemma 3.6. Let P be an allocation. Let Q result from P by handing an item p ∈ P(a2 ) to agent a1 and q ∈ P(a1 ) to agent a2 such that for some j ∈ N the following properties are satisfied: • •

a2 ranks q at most j positions lower than p a1 ranks p at least ( j + 1 ) positions higher than the highest-ranked item of P(a1 )   Then, a ∈ A ua (P) < a ∈ A ua (Q).

Proof. Let λ denote the rank of the highest-ranked item of P(a1 ) in the ranking of a1 . Then,

2λ ≤ ua1 (P ) < 2λ+1

n

n

(2M ) 3 · (2M−(3n+6) · 2M−(n+4) · 2M−(n+5) ) 3 · (2M · 2M−1 · · · 2M− 3 +1 ) n

For each xi ∈φ , let P (Hi ) = xi and P (Xi ) = {xi,1 , xi,2 , xi,3 }. The total product of the utilities of these agents is 2n

 Thus, a∈A ua (P ) = k, implying that L is a “yes”-instance.  “If”-part: Let P be an allocation with a ∈ A ua (P) ≥ k. This implies that the maximum Nash product social welfare achieved exceeds the threshold k. W.l.o.g. we assume that P is an allocation of maximum Nash product social welfare. We show that P must satisfy several properties:

> ua1 (P ) · ua2 (P ) + 2t+s+1 − 2s 2t − 2s+t <



P (Di ) = di for each 1 ≤ i ≤ . Thus, the total product of the utilities  M−i . of these agents is −1 i=0 2 For each xi ∈ φ , let P (Xi ) = xi , – for the q-th clause Cj that contains xi , q ∈ {1, 2, 3}, assign xi, q to the one among α j , β j , γ j that ranks xi, q highest (i.e., directly below d2n+4 , in position 3n + 7); for the two remaining agents among α j , β j , γ j , allocate bi, 1 to one and bi, 2 to the other agent. – allocate exactly one of {h1 , . . . , h n } to agent Hi .

2n

= ua1 (P ) · ua2 (P ) + 2t ua2 (P ) − 2s ua1 (P ) − 2s+t

a ∈ A ua (P)

a ∈ A ua (Q).

(2M ) 3 · (2M−(n+1) + 2M−(n+2) + 2M−(n+3) ) 3

ua1 (Q ) · ua2 (Q ) = [ua1 (P ) + 2t )] · [ua2 (P ) − 2s ]





3

•

ua2 (P ) > 2s+1 and ua1 (P ) < 2t

holds. Therefore,

<

Hence, the total product of the utilities of these agents is

Proof. Let ua2 ( p) = 2s and ua1 ( p) = 2t for some s, t ∈ N. Clearly, the stated properties imply

= ua1 (P ) · ua2 (P )

a ∈ A ua (P)

Proof of Claim. “Only-if”-part: Let φ be a truth assignment that sets true exactly one variable in each clause. Abusing notation, we identify φ with the set of variables set true under φ . Recall that |φ| = n3 . We define the allocation P as follows.

i=0

•



Claim. I is a “yes”-instance of Cubic Monotone 1-in-3 Sat if and only if L is a “yes”-instance of NPSW-Lexicographic.

n



ua1 (Q )·ua (Q )−ua (P )·ua (P ) > [(2λ+( j+1) − 2λ )2μ 2 1 2 As a consequence,

2n 3

·[2M−(n+4) · 2M−(n+5) · 2M−(3n+6) ] 3

Set

ua1 (Q ) · ua2 (Q ) − ua1 (P ) · ua2 (P ) since for the remaining the agents the utilities of P and P coincide. Now,

τβi = xi2 ,k2  d2n+5  d2n+6  · · ·  d3n+4  xi3 ,k3  d3n+5

M−(n+1 )

holds. Let μ denote the rank of item p in the ranking of a2 ; 2μ ≤ ua2 (P ) holds. Comparing the Nash product social welfare achieved by the allocations, it is enough to consider

(6)

1. di is allocated to Di for each 1 ≤ i ≤ n: Assume there is an agent Di such that P (Di ) ∩ {d1 , . . . , d } = ∅. Let r ∈ P(Di ) denote the item which Di ranks highest among the items in P(Di ). We distinguish the following cases. (a) There is a Dj who gets allocated at least two elements of {d1 , . . . , d }. Let dmin be the lowest ranked of these items in the ranking of Dj . Consider the allocation P which results  from P by handing dmin to Di . With Lemma 3.5, a ∈ A ua (P) <  a ∈ A ua (P ) holds which contradicts with the choice of P. (b) There is an agent a = D j who gets allocated at least one of {d1 , . . . , d }. Take an arbitrary such d ∈ P(a). Consider the allocation P which results from P by handing r to a and d to Di . If

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

a ranks r above d, trivially ua (P ) > ua (P) and uDi (P ) > uDi (P ) follow, since Di by construction ranks d above r. Assume a ranks r below d. We can observe that in the ranking of Di , r is among the last (6n + 3n ) positions. For any other agent, r is ranked higher by construction. Thus, a ranks r higher than Di does; also by construction, Di ranks d at least as high as a does. Hence, the number μ of items between d and r in the ranking of Di exceeds the number of items between d and r in the ranking of a by at least one item. In other words, Lemma 3.6 can be applied, again leading to a contradiction with the choice of P. As a consequence, each agent Di gets at least (and thus exactly) one item of {d1 , . . . d }. Since the rankings of the agents Di , 1 ≤ i ≤ , coincide, w.l.o.g. we assume that di is allocated to Di . 2. hi is allocated to one of {H1 , . . . , Hn }, for each 1 ≤ i ≤ n3 . Assume hi is allocated to an agent a ∈ / {H1 , . . . , Hn }. Then, take an arbitrary H j ∈ {H1 , . . . , Hn } who is not allocated any item of {h1 , . . . , h n }. 3

Obviously, such an agent Hj exists. It is easy to see that for the allocation P¯ which results from P by handing hi to Hj , and, in turn, any item of P(Hj ) to a satisfies uH j (P¯ ) > uH j (P ) and ua (P¯ ) > ua (P ),   i.e., a∈A ua (P ) < a∈A ua (P¯ ).

3. xi is allocated to one of {Hi , Xi }, for each 1 ≤ i ≤ n. Assume xi is assigned to an agent a∈{Hi , Xi }. Consider the allocation P˜ which results from P by handing xi to Xi , and, in turn, the item r of P(Xi ) which Xi ranks highest to agent a. Recall that r ∈ / {d1 , . . . , d , h1 , . . . h n }. If r ∈ / {xi+1 , . . . , xn }, then obvi3  ously uXi (P˜ ) > uXi (P ) and ua (P˜ ) > ua (P ) hold, i.e., a∈A ua (P ) <  ˜ a∈A ua (P ). Let r ∈ {xi+1 , . . . , xn }. Then, Xi ranks xi more than r .

r

Note that any agent a∈{Hi , Xi } ranks at 3n positions above most (n − 1 ) positions below xi . Thus, the conditions stated in Lemma 3.6 are satisfied, and again we get a contradiction with the choice of P. 4. Hj is allocated exactly one of {h1 , . . . , h n } ∪ {x j }, for each 1 ≤ j 3

≤ n. Note that with Step 3 this means thatHj is allocated exactly one of {h1 , . . . , h n } ∪ {x1 , . . . , xn }. This step is split in three 3 parts: (a) Hj is allocated at most one of {h1 , . . . , h n }, for each 1 ≤ j ≤ n. 3

Assume there is an agent Hj who is allocated at least two items of {h1 , . . . , h n }. Let hg be the lower-ranked of the two items in 3

the ranking of Hj . For the allocation Pˆ which results from P by handing hg to an agent H j who P does not allocate an item  of {h1 , . . . , h n } to, we get with Lemma 3.5 that a∈A ua (P ) < 3  ˆ a∈A ua (P ) holds. (b) If Hj is allocated one of{h1 , . . . , h n } , then Hj is not allocated xj . 3

Let h ∈ {h1 , . . . , h n } be allocated to Hj . Assume the opposite. 3

Consider the allocation Q which results from P by handing xj to Xj . Again, with Lemma 3.5 uH j (Q ) · uXi (Q ) > uH j (P ) · uXi (P ) follows. (c) If Hj is allocated none of {h1 , . . . , h n }, then Hj is allocated xj . 3  Assume the opposite. Since by assumption a ∈ A ua (P) > 0, P must allocate an item r to agent Hj . Again, let r be the item highest-ranked by Hj that Hj receives under P. From Steps 1–3, we can conclude that r ∈ {b1,1, . . . , bn,2 } ∪ {x1,1 , . . . , xn,3 } holds. Consider the allocation Q which results from P by handing xj to Hj , and, in turn, item r to Xj . By construction, Xi ranks r more than n3 positions higher than Hj does. On the other hand, Hj ranks xj exactly n3 positions lower than Xi does. Therefore, the number μ of items between xj and r in the ranking of Hj exceeds the number of items between xj and r in the ranking of Xi by at least one item. As a consequence, Lemma 3.6 yields a contradiction with the choice of P.

553

5. Two of {α i , β i , γ i } are allocated exactly one of {bi, 1 , bi, 2 }, for each 1 ≤ i ≤ n.This step is proven in two parts. (a) bi, 1 (resp. bi, 2 ) is allocated to α i , β i or γ i , for each 1 ≤ i ≤ n. Assume bi, 1 is not allocated to one of these agents. Clearly, at most one of α i , β i , γ i is allocated bi, 2 . W.l.o.g. assume bi, 2 is not allocated to α i . By Step (1), this implies that α i ranks bi, 1 more than 2n positions higher than the highest-ranked among the items in P(α i ). Take an arbitrary p ∈ P(α i ). Note that with Steps 1–3, p ∈ / {d1 , . . . , d , x1 , . . . , xn , h1 , . . . , h n } ∪ {bi,2 } fol3

lows. Consider the allocation Q which results from P by handing p to the agent a with bi, 1 ∈ P(a) and, in turn, bi, 1 to agent α i . By construction (in particular, by the items dn+5 , . . . , d2n+4 in the ranking of α i ), it follows that the conditions of Lemma   3.6 are satisfied. Thus a ∈ A ua (P) < a ∈ A ua (Q) holds, in contradiction with the choice of P. (b) bi, 1 and bi, 2 are not allocated to the same agent, for each 1 ≤ i ≤ n. Assume the opposite. Then, analogously to above, by the use of Lemma 3.6 we can find an allocation with a higher Nash product social welfare than P. 6. For each 1 ≤ i ≤ n and a ∈ {α i , β i , γ i }, the following holds: If bi, 1 or bi, 2 is allocated to a, then a is allocated no further item.It remains to show that no item of Y is allocated to a. We provide a proof for agent α i and bi, 2 ∈ P(α i ) (the other cases follow analogously). Assume at least one of Y is allocated to α i . Let xg, j be allocated to α i . Consider the allocation Q which results from P by handing xg, j to agent Xg . Let uα i (xg, j ) = 2ε for some ε ∈ N. Note that uXg (xg, j ) ≥ 2ε+2n and

uαi (P ) ≥ 2M−(n+4) + 2ε

(7)

hold. We get

uXg (Q ) · uαi (Q ) − uXg (P ) · uαi (P ) ≥ (uXg (P ) + 2ε+2n ) · (uαi (P ) − 2ε ) − uXg (P ) · uαi (P ) ≥ [uXg (P ) · uαi (P )−2ε uXg (P )+2ε+2n 2M−(n+4) ]−uXg (P ) · uαi (P ) > − 2ε+M+1 + 2ε+M+n−4 >0 where the third line follows from (7), the fourth from uXg (P ) < 2M+1 , and the last from n > 5. 7. For each 1 ≤ i ≤ n and a ∈ {α i , β i , γ i }, the following holds: If a is allocated an item of {xi1 ,k1 , xi2 ,k2 , xi3 ,k3 }, then a is allocated exactly one item.This follows analogously to Step 6. 8. None of the items in Y is allocated to an agent a ∈ {D1 , . . . , D , H1 , . . . , Hn }. Assume the opposite. Take an arbitrary item xg, j ∈ P(a)∩Y. Note that ua (xg, j ) = 2ε and uXg (xg, j ) ≥ 2ε+3n for some ε < M − . Thus,

ua (P ) ≥ 2ε + 2M− 3

n

Consider the allocation agent Xg . With (8),

(8) Q

which results from P by handing xg, j to

uXg (Q ) · ua (Q ) − uXg (P ) · ua (P ) ≥ (uXg (P ) + 2ε+3n ) · (ua (P ) − 2ε ) − uXg (P ) · ua (P ) > uXg (P ) · ua (P ) − 2ε uXg (P ) + 2ε+3n 2M− 3 − uXg (P ) · ua (P ) n

> − 2ε+M+1 + 2ε+M+2n >0 and thus a contradiction with the choice of P is implied. As an immediate consequence, we know that (i) each of D1 , . . . , D , H1 , . . . , Hn is allocated exactly one item (follows from Steps 1–5 and Step 8), and (ii) there are at most 2n items available for the agents X1 , . . . , Xn (by the pigeonhole principle), all of which

554

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

belonging to the set Y. From (i), it follows with Steps 1 and 4 that



ua ( P )·

a∈{D1 ,...,D }





ua ( P ) =

a∈{H1 ,...,Hn }

−1

 2

M−i

• •

·2

 n3 −1 i=0

3

(M−i )

·2

(M− n3 ) 2n 3

• •

i=0

(9) From Steps 3 and 4 we know that there are exactly 3n agents among the agents Xi that are allocated xi , 1 ≤ i ≤ n, while the remaining 2n 3 agents among X1 , . . . , Xn are not allocated any item of X. Keeping in mind that the Nash product is maximized for the most balanced allocation, it is not difficult to verify that the following observation holds. Observation. If at most  a∈{X1 ,...,Xn } P (a ), then



2n

items

of

Y

are

contained

ua (P ) ≤ (2M ) 3 · (2M−(n+1) + 2M−(n+2) + 2M−(n+3) ) 3 n

2n

in

(10)

a∈{X1 ,...,Xn }

holds; equality in (10) is achieved if and only if for all 1 ≤ i ≤ n, all the items {xi,1 , xi,2 , xi3 } are allocated to the agent Xi satisfying xi ∈P(Xi ). Now, assume for some i, there is an a ∈ {α i , β i , γ i } which is allocated none of {bi,1 , bi,2 } ∪ {xi1 ,k1 , xi2 ,k2 , xi3 ,k3 }. With Step 1, ua (P ) < 2M−7n follows. With Steps 6 and 7, we get



ua (P ) < (2M−7n · 2M−(n+4) · 2M−(n+5) )

a∈{αi ,βi ,γi |1≤i≤n} n−1

·(2M−(3n+6) · 2M−(n+4) · 2M−(n+5) )

(11)



Combining (9), (10), (11), a ∈ A < k follows. Thus, for each i and a ∈ {α i , β i , γ i }, a is allocated at least – by Steps 6 and 7, that means exactly – one of {bi,1 , bi,2 } ∪ {xi1 ,k1 , xi2 ,k2 , xi3 ,k3 }. With Step 5, we can conclude that exactly one agent of {α i , β i , γ i } is allocated exactly one of {xi1 ,k1 , xi2 ,k2 , xi3 ,k3 }, obviously yielding an utility of 2M−(3n+6 ) . Hence, we get



ua (P ) = (2M−(3n+6) · 2M−(n+4) · 2M−(n+5) )n

(12)

a∈{αi ,βi ,γi |1≤i≤n}

With (9) and (12), the above observation implies that •

•

For each clause Ci , exactly one of {xi1 ,k1 , xi2 ,k2 , xi3 ,k3 } is allocated to one of {α i , β i , γ i } (i.e., one of the variables xi1 , xi2 , xi3 is set “true”), and Either all or none of {xi, 1 , xi, 2 , xi, 3 } are allocated to some agents of the set C = {α j , β j , γ j |1 ≤ i, j ≤ n}, i.e., either {xi,1 , xi,2 , xi,3 } ⊂   ( a∈C P (a )) or {xi,1 , xi,2 , xi,3 } ∩ ( a∈C P (a )) = ∅ holds.

Therewith, the truth assignment φ which sets xi “true” if and only if xi ∈ P(Xi ) (i.e., xi, 1 , xi, 2 , xi, 3 are allocated to some agent {α j , β j , γ j }, 1 ≤ j ≤ n), is a feasible truth assignment that sets “true” exactly one variable of each clause. Hence, I is a “yes”-instance of Cubic Monotone 1-in-3 Sat.  3.2.2. NPSW-Borda is NP-complete Next, we show that maximizing Nash Product Social Welfare remains NP-complete in the case of Borda scores.2 Theorem 3.7. NPSW-Borda is NP-complete. Proof. The proof proceeds again by a reduction from Cubic Mono˜ C˜ ) of Cubic Monotone 1tone 1-in-3 Sat. Given an instance I = (X, in-3 Sat we construct an instance B = (R, A, π , k ) of NPSW-Borda as follows. Let n = |X˜ | = |C˜|, and  = 10n10 . R consists of  + 5n + n3 items: Note that in classical voting theory, the Borda scores range from 0 to m − 1 (see, e.g., Brams & Fishburn, 2002, Chap. 4). We point out that our NP-completeness result also holds for this classical version of Borda scores. 2

The item set D = {d1 , d2 , . . . , d } The item set H = {h1 , h2 , . . . , h n }

The item set E = {e1 , e2 , . . . , en } The item sets X = {x1 , x2 , . . . , xn } x2,1 , x2,2 , x2,3 , . . . , xn,1 , xn,2 , xn,3 }

and Y = {x1,1 , x1,2 , x1,3 ,

A consists of  + 3n agents: • • • •

The agent set L = {L1 , L2 , . . . , L } The agent set H = {H1 , H2 , . . . , Hn } The agent set X = {X1 , X2 , . . . , Xn } The agent set C = {C1 , C2 , . . . , Cn }

The ranking of the agents is given in Table 1, where a d in a column denotes that the next available item di is used (where the items are placed in the increasing order of their index). Note that in any case where we use this abbreviation the index of the item itself is irrelevant, since, as we will show, in an allocation of maximum Nash product social welfare all items di are allocated to agents from L. Again, we represent the clauses of instance I in the form Ci = (xi1 ,k1 ∨ xi2 ,k2 ∨ xi3 ,k3 ), where kj ∈ {1, 2, 3} denotes the kj -th occurrence of variable xi j in C, by the set Ci = {xi1 ,k1 ∨ xi2 ,k2 ∨ xi3 ,k3 }. Thus, the items ci, 1 , ci, 2 and ci, 3 of agent Ci thus correspond to the xi1 ,k1 , xi2 ,k2 , xi3 ,k3 (and in this way to the three variables of the clause Ci of instance I). In particular, the items ci, 1 , ci, 2 and ci, 3 of an agent Ci are elements of Y. In Table 1, we have c1,1 = x1,1 , c1,2 = x2,1 , c1,3 = xn,3 and c2,1 = x1,2 . The main features of the construction of the NPSW-Borda instance B can be briefly described as follows: agents Hi rank all hj and the item xi very high when compared to agents from L. The agents Xi rank item xi in a very high position and also the items xi, j in a high position when compared to agents from L. Finally agents Ci rank item ei high when compared to agents from L and also the three items xk, l corresponding to variables in clause Ci (and represented by the three ci, j ). Let a be an agent, l an item and S a set of items. Recall that now ua (l) denotes the Borda score of item l in the ranking of a. Again, let P be an allocation. Claim. I is a “yes”-instance of Cubic Monotone 1-in-3 Sat if and only if B is a “yes”-instance of NPSW-Borda, where

 n −1

∗

l

k =M ·

3



 2n

n n 3 n 3 (M − i ) · M − · M− 3

i=0

·

1

2

M−

n −2 3

 2n3 5 ·

4

3

(M − 9n10 − 2 )

n

Proof of the Claim. “⇐” We will first show that in any “yes” instance B, there is an allocation of Nash product social welfare at least k∗ which satisfies: 1. 2. 3. 4. 5. 6.

Li gets item di , for each 1 ≤ i ≤ . Ci gets ei , for each 1 ≤ i ≤ n. Each hi is given to some Hj , and each H j ∈ H gets at most one hi . If Hj does not get an item of H, then it gets xj . If Hj gets an item of H, then Xj gets xj . All items from Y are allocated to agents from X ∪ C.

Proof of 1. Li gets item di , for each 1 ≤ i ≤ . Note that all agents Li rank all items l ∈ (H∪E∪X) in the same way. Assume that one agent Lj did not get an item of the set D. Let a be the agent such that dj ∈ P(a). We can re-write the utility achieved by agent a as ua (P ) = ua (d j ) + ua (Ra ), where3 Ra := (P(a)dj ). Hence, a

3 In the following, we frequently use the notation Ra in order to sum up the “rest” of items allocated to an agent a, i.e., a set of items distinguished from some dedicated items allocated to a. In what follows, it should be clear from the context which item set Ra refers to.

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

555

Table 1 Rankings and respective Borda scores in instance B of NPSW-Borda. Borda score

L1

L2



L

H1

H2



Hn

X1

X2



Xn

C1

C2



Cn

M   M − n3   1 (M − 3n ) 6     M − 9n10   1 (M − 9n10 ) 4      5n + n3    4n + n3      n + 3n   n   1

d1 d2                  d x1 x2  xn x1, 1 x1, 2 x1, 3 x2, 1  xn, 3 h1  h n3 e1   en

d2 d3

 

d d1                  d−1 x1 x2  xn x1, 1 x1, 2 x1, 3 x2, 1  xn, 3 h1  h n3 e1   en

h1  h n3 x1 d1              

h1  h n3 x2 d1



h1  h n3 xn d1              

d1

   

x2, 1 x2, 2 x2, 3 d

   

e2 d

 

c2, 1 c2, 2 c2, 3 d

   

d− n3 −1 x1 d− n3

  

d−4 x1 x2

  

xn x1, 1 x1, 2 x1, 3 x2, 1

    

xn x1, 1 d−3 x1, 3 x2, 1

    

xn, 3 d− n3 +1

 

xn, 3 h1

 

d e1

 

h 3n e1 d

  

en



d1 d2          e1 d  c1, 1 c1, 2 c1, 3 d  d−4 x1 x2  xn d−3 x1, 2 x1, 3 d−2  d−1 h1  h n3 d e2  en

 

d n3 x2 d n3 +1

d1  d n3 xn d n3 +1  xn, 1 xn, 2 xn, 3 d          d−4 x1 x2  d−3 x1, 1 x1, 2 x1, 3 x2, 1  d h1  h n3 e1   en

d1 d2

d− n3 −1 d− n3 x2  xn x1, 1 x1, 2 x1, 3 x2, 1  xn, 3 d− n3 +1  d e1   en

d1  d n3 x1 d n3 +1  x1, 1 x1, 2 x1, 3 d          d−4 d−3 x2  xn d−2 d−1 d x2, 1  xn, 3 h1  h n3 e1   en

en



d1 d2          en d  cn, 1 cn, 2 cn, 3 d  d−4 x1 x2  xn x1, 1 x1, 2 x1, 3 x2, 1  xn, 3 h1  h n3 e1 e2  d

d1 x1 x2

  

xn x1, 1 x1, 2 x1, 3 x2, 1

    

xn, 3 h1

 

h n3 e1

 

en



  

and Lj together contribute to the Nash product social welfare:

ua ( P ) · uL j ( P ) = ua ( h j ) · uL j ( I ) + ua ( R ) · uL j ( I ) a

However, if we handle dj over to Lj and all items I of Lj to a we get a total contribution of the two agents of:

(ua (I ) + ua (R )) · M = ua (I ) · M + ua (R ) · M a

a

Note that ua (I ) ≥ uL j (I ), since a does not rank any item l ∈ {H∪E∪X} n 2 3)

< M by construction. lower than Lj . In addition, uL j (I ) < (5n + Hence this exchange does not diminish the total contribution of a and Lj to the objective value and hence there exists an assignment such that any agent Li has one dj allocated. By the pigeonhole principle this means that each agent Lj gets exactly one di . Since Lj ranks dj on top position, in an allocation that maximizes Nash product social welfare clearly Lj gets item dj , 1 ≤ j ≤ .  Proof of 2. Ci gets ei , 1 ≤ i ≤ n. Assume the opposite, i.e., there is an agent Ci such that ei ∈P(Ci ). Case 1. Assume that ei ∈ P(Lj ): the total contribution of Lj to the Nash product achieved by P can be bounded by:

uL j (P ) = uL j (ei ) + uL j (d j ) + uL j (RL j ) < M + n3 In this respect, RL j denotes the set of items allocated to Lj under P with the exception of ei and dj . Here, n3 is a very rough bound for the sum over the profits of all items with weight smaller or equal to 5n + 3n . Let imin denote the item of smallest Borda score allocated to Ci . The contribution of Ci can be bounded as follows:

uCi (P ) = uCi (imin ) + uCi (RCi ) < 3 ·

(M − 9n10 ) 4

+ n3

d− n3 −1 x1 x2  d− n3 x1, 1 x1, 2 x1, 3 x2, 1  xn, 3 d− n3 +1  d e1   en

d−4 x1 d−3

  

xn x1, 1 x1, 2 x1, 3 d−2

    

xn, 3 h1

 

h n3 e1

 

en



The bound comes from taking the three items ci, j and all the items of profit at most 5n + n3 . Exchanging imin with ei , however, yields a contribution of Ci and Lj to the Nash product of at least:

M · (M − 9n10 ) It remains to show that uCi (P ) · uL j (P ) ≤ M · (M − 9n10 ) holds:



uCi (P ) · uL j (P ) <

3·

(M − 9n10 ) 4

+ n3

· ( M + n3 )

3 (M − 9n10 )(M + n3 ) + n3 (M + n3 ) 4 < M · (M − 9n10 ) =

where the last inequality follows from the fact that

1 1 (M − 9n10 )M = n10 · 10n10 4 4 3 > n10 · n3 + n3 (n10 + n3 ) 4 is satisfied. Case 2. Assume that ei ∈ P(Hj ):

uH j (P ) = uH j (ei ) + uH j (RH j ) ≤ n + uH j (RH j ) uCi (P ) = uCi (imin ) + uCi (RCi ) Case 2.1 If RH j contains an item of Borda score larger than n, we handle ei over to Ci and get a new product contribution of Hj and Ci of at least:

uH j (RH j ) · (uCi (imin ) + uCi (RCi ) + (M − 9n10 ))

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Note that



uH j (RH j ) · (M − 9n10 ) > n ·

3·

M − 9n 4

10

If xl ∈ P(Hl ), we get

uXl (P ) <

+ n3

≥ n · (uCi (imin ) + uCi (RCi )) holds; i.e., handling ei to Ci yields a higher Nash product. Case 2.2 If RH j contains items of Borda score at most n only, then exchange ei with imax , where imax denotes the item of P(Ci ) with highest Borda score. If uH j (imax ) < uH j (ei ), then imax ∈ {e2 , . . . , en }. Thus,

uH j (P ) · uCi (P ) ≤ (uH j (RH j ) + uH j (ei )) · (uCi (RCi ) + uCi (P )(imax ))





Xj

If imax ∈ E, we get

uC j (P ) · uCi (P ) ≤ (uC j (ei ) + uC j (RC j )) · n2 ≤ uC j (RC j ) · (M − 9n10 ) In the case RC j = ∅, this proves that the new contribution is higher. If RC j = ∅ we apply the same argument as in Case 3.  Proof of 3. Each hi is given to some Hk and each Hk ∈ H gets at most one hi . First, we show that each hi is given to an agent of H. Assume the opposite, i.e., hi is given to an agent a ∈ (L ∪ X ∪ C ). Let Hk be an agent without any item from H (obviously, such an agent must exist). Handle the item imax ∈ P(Hk ) with the highest Borda score to a, and in turn handle hi to Hk . Note that uHk (imax ) < uHk (hi ) and ua (imax ) > ua (hi ), since imax ∈(D∪E) by 1 and 2. Therefore, this exchange yields an allocation with a higher Nash product social welfare. Thus, hi is allocated to an agent of H. Second, we show that each H ∈ H gets at most one hi . Assume there exists an agent Hk ∈ H with at least two items from H, say hi and hj . By the pigeonhole principle, thus at least one Hl has no item from H. If xl ∈P(Hl ), we have uHl (P ) < n3 . Handle hj over to Hl . Thus,

3





+ 1 M · n3 < M −

n 3

2

M−











n + 1 + uHk (RHk ) · 3





⇔ M−



n n +1 · M− 3 3







> uXl (RXl ) · 2M − M −

< (uC j (imax ) + uC j (R )) · (uCi (ei ) + uCi (R ))

n

M−





Case 4. ei ∈ P(Cj ), for some i = j. Due to Cases 1–3, we can assume that ei is allocated to one of the agents of C. Assume imax ∈E. Note that uCi (ei ) > uCi (imax ) and uC j (ei ) < uC j (imax ) hold. Thus, if we exchange ei with imax of Ci we get:

uHk (P ) · uHl (P ) <



> 2M + uHk (RHk ) · uXl (RXl )

uX j (P ) · uCi (P ) ≤ uX j (ei ) · n2 < n3 < (M − 9n10 )

Ci

< uHk (hi )uHl (h j )

implying that there is an allocation with a higher Nash product social welfare.



(14)

n + 1 + uHk (RHk ) · 3

holds:

If RX j = ∅, we handle ei over to Ci and imax over to Xj and get a new contribution of at least M − 9n10 . Obviously, this exceeds the former contribution, since

Cj

M−



10

uC j (P ) · uCi (P ) = (uC j (ei ) + uC j (RC j )) · (uCi (imax ) + uCi (RCi ))



n + 1 + uHl (RHl ) 3

n 3





+ uHl (RHl ) · (uXl (RXl ))

(15)

M−

n 3

M−

n 3



+ uXl (RXl )



> 2 · M + uHk (RHk ) · uXl (RXl )

uX j (P ) · uCi (P ) ≤ (uX j (R ) + uX j (ei )) · n ≤ uX j (R ) · (M − 9n ) 2

M−

In order to show that the new contribution, i.e., the term in (14), exceeds the contribution under P, it is sufficient to show that (14) exceeds the right-hand term of (15). Clearly, to prove this it is enough to show that



Xj



≤ (2 · M + uHk (RHk )) ·

Xj

9n10 ). Clearly,



M−

uHk (P ) · uHl (P ) · uXl (P )

Thus, we can assume uCi (imax ) ≤ n, i.e., imax ∈ E. If RX j = ∅, handle

ei over to Ci which yields a new contribution of at least uX j (R ) · (M −

(13)

However, we have

Ci

because of uCi (P )(imax ) ≤ uCi (P )(ei ). Case 3. Assume that ei ∈ P(Xj ). If uCi (imax ) > n, then exchange ei with imax . Note that uX j (imax ) > uX j (ei ) and uCi (imax ) < uCi (ei ) hold, which clearly implies that the new contribution exceeds uX j (P ) · uCi (P ).

+ n3

n + 1 + uHk (RHk ) · 3

  n · M− + uXl (RXl ) 3

≤ (uH j (R ) + uH j (imax )) · (uCi (R ) + uCi (P )(ei )) Hj



Handing xl to Xl and hj to Hl , we get a new contribution of Hk , Hl and Xl to the Nash product of at least

uH j (P ) · uCi (P ) < n4 , while the exchange leads to a contribution of at least (M − 9n10 ). If uH j (imax ) ≥ uH j (ei ), then we have



1 n M− 2 3





+ M−

n +1 3







+ uXl (RXl )





n uHk (RHk ) 3





n n n +1 · M− + M− uHk (RHk ) 3 3 3   n > uXl (RXl ) · M + − 1 3

⇔ M−

With uXl (RXl ) < follows from



1 2 (M



− n3 ) + n3 (stated in (13)), the last inequality





 



n n n n 1 +1 · M− > M− + n3 · M + − 1 3 3 2 3 3



 1

   n n n n ⇔ M− M− +1 − M + − 1 >n3 M + − 1 3 3 2 3 3

 1    n 3 n n 3 M− + ⇔ M− · >n M + − 1 3 2 2 2 3

M−

where the last inequality is clearly satisfied by the choice of M.



Proof of 4. If Hj does not get an item of H, then it gets xj . Assume that Hj does not get an item of H. In addition, assume xj ∈P(Hj ). Let a be the agent xj is allocated to under P. Recall that uH j (P ) ≤ n3 holds. Case 1. a ∈ (L∪C). It is easy to verify that handling over xj to Hj improves the total Nash product. Case 2. a = Hk ∈ H: If xj is allocated to some Hk and xk or some hi is also allocated to Hk , it is again easy to check that handling xj over to Hj improves the total contribution. If Hk does neither have xk nor any hi , then uHk (P ) ≤ n3 holds. Hence handling xj over to Hj and any item from Hj to Hk improves the total profit by reaching a value of at least (M − n3 ). Case 3. a ∈ X : Case 3.1. If a = Xj , then ua (P ) ≤ (5n + 3n ) + ua (Ra ) holds. If Ra = ∅ we handle xj over to Hj and clearly augment the total Nash product. If Ra = ∅ the total contribution of a and H j < n3 · (5n + 3n ), hence handling xj over to Hj and any other item from Hj to a improves the contribution as well.

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

Case 3.2. If a = X j we have:

ua ( P ) · uH j ( P ) ≤



 n

M−

since



+ ua ( R ) · n a

3

3

(16)

ua ( P ) · uX j ( P ) ≤

n ) 3

(17)

If Ra contains at least one of the three items xj, l , then (17) is larger than the right hand side of (16), i.e., the contribution of a and Hj to the Nash product after handling xj over to is higher than before. If Ra does not contain one of these items we exchange all the items of a with all the items of Hj and get a contribution that is at least as high as before (the contribution becomes strictly higher if any xj, l was allocated to Hj ).  Proof of 5. If Hj gets an item of H, then Xj gets xj . Assume that Hj gets some hi , but Xj does not get xj . Let a be the agent with xj ∈ P(a). Note that

uX j ( P ) <

1 2

M−

 n

3

+ n3

ua ( P ) · uX j ( P ) ≤ <

M + 5n +



M−

n 3



 n 3



+ ua (Ra ) · uX j (P )



+ ua (Ra ) ·

n 3

M−



uH j ( P ) · uX j ( P ) < M −



+ uX j ( P )

n + 1 ) + ua (Ra ) · 3

M−

 n



+ uX j ( P )

3

≤

holds. Clearly, uH j (P ) · uX j (P ) < (2M + ua ever,

(2M + ua (Ra )) · uX j (P )

 n



X j (P )





holds. How-

With (19), the last inequality follows, since



uX j ( P ) M +

n −1 3



<

1

2



M−

 n

3

1 n2 < M2 − 2 9

+ n3



M+

+ n3 M +

n −1 3

M−

n 3



+ uX j ( P )







5n +



n 3

 1

+ ua (Ra ) ·

M−

n 3

M−

n 3

6 

 n n a n + + 1 + ua ( R ) · M − 3 3

ua ( P ) · uX j ( P ) <





5n +





n 3

 2

+ ua (Ra ) ·

n 2· n+ +1 3



6



+ ua (Ra ) · M −



 n 3

+ n3

+ n3





n 3

 1

n

( M − ) + n3 2 3





 n n a ≤ 3· n+ +1 + ua ( R ) · M − 5n +

+ ua (Ra ) ·

3





If |{x j,1 , x j,2 , x j,3 } ∩ P (X j )| = 3, we get

ua ( P ) · uX j ( P ) <





3

Proof of 6. All items from Y are allocated to agents from X ∪ C. Assume that some xi, j is allocated to an agent a from H ∪ L. We handle xi, j over to Xi and get:

ua (P ) · uXi (P ) ≤



4n +

n 3







+ ua (Ra ) · uXi (P )

≤ ua (Ra ) · uXi (P ) +



1 n M− −2 6 3



The inequality holds since ua (P ) ≥ (M − n3 ) and uXi (P ) ≤ n3 . 

8 6 (M

− 3n ) +

Proof of the Claim (continued). “⇐” Recall that B is a “yes”-instance of NPSW-Borda, with an allocation P such that the Nash product social welfare for P is greater or equal to:



n < M − + 1 + ua ( R ) · M− + uX j ( P ) 3 3



  n n ⇔ 2M · uX j (P ) < M − + 1 M− + uX j ( P ) 3 3

 n a +ua (R ) · M − 3





 n n n ⇔ uX j ( P ) M + − 1 < M − + 1 M − 3 3 3

 n a +ua (R ) · M − 3 a



If |{x j,1 , x j,2 , x j,3 } ∩ P (X j )| = 2, we get

(19)

(Ra )) · u



+ ua (Ra ) · uX j (P )

Thus, also in the case uX j (P ) > n3 , exchanging items yields an allocation with a higher Nash product. 

If a = H j handling xj over also improves the Nash product – this is proven by showing that



ua ( P ) · uX j ( P ) <

(18)





n 3

If uX j (P ) > n3 , P allocates at least one of {xj, 1 , xj, 2 , xj, 3 } to Xj . In this case, we handle them over to a in exchange for xj . If |{x j,1 , x j,2 , x j,3 } ∩ P (X j )| = 1, we get

≤

holds. Case 1. a ∈ (L∪C). In this case, handling over xj to Xj clearly improves the Nash product social welfare. Case 2. a ∈ H. If a ∈ H \ H j , then with (18) is it not hard to check that handling over xj to Xj improves the Nash product social welfare:



5n +

≤ (ua (Ra ) + 1 ) ·

Handling xj over to Hj gives a contribution of at least:

ua (Ra ) · (M −



557

 n −1

k∗ = M ·

3



 2n

n n 3 n 3 (M − i ) · M − · M− 3

i=0

·



n 3

< 50n20 + n13 20 11 n < 100n20 − 3



 n n < M− +1 M− 3 3 Case 3. a ∈ X \ X j . If uX j (P ) ≤ n3 , then a hands xj to Xj , and Xj in turn hands an arbitrary item to a. This augments the Nash product,

1 2

(M −

n − 2) 3

 2n3 5 ·

4

3

(M − 9n10 − 2 )

n

In addition, we can assume that P satisfies the properties 1–6 and is an allocation of maximum Nash product. We will show that the corresponding instance of Cubic Monotone 1-in-3 Sat a “yes”-instance. Due to the six properties, we already know the exact contribution of agents from L and H. Hence, we know that in order to reach a total profit of k the agents from X and C have to contribute to the Nash product a factor of at least



M−

n 3

 n3 1

·

2

M−

n −2 3

 2n3 5 ·

4

(M − 9n10 − 2 )

n

Now, we show that we can assume that P satisfies the following additional properties: 7. Each item xi, l is assigned to an agent a with ua (xi,l ) > 5n + n3 . 8. If Xi gets xi , then Xi does not get an item of {xi, 1 , xi, 2 , xi, 3 }.

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A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

9. If Xi does not get xi , then Xi gets all three items of {xi, 1 , xi, 2 , xi, 3 }. Proof of 7. Each item xi, l is allocated to an agent a with ua (xi,l ) > 4n + n 3 . Assume the opposite, i.e., for some indices j, l, xj, l is allocated to an agent a with ua (x j,l ) ≤ 4n + n3 . Case 1. Assume that an agent Xi gets such an item xj, l with uXi (x j,l ) ≤ 4n + n3 . If P(Xi ) contains another item of Borda score at least 2 ), we handle xj, l over to Xj :





uX j (P ) · uXi (P ) ≤ uX j (RX j ) · uXi (RXi ) + 4n +



≤

uX j (RX j ) +



1 n M− −2 6 3

n 3



1 6 (M

−

n 3

−





· uXi (RXi )





uXk (RXk ) + 4n +

< (uXk (RXk ) + 1 ) ·



n 3



· n3

1 n M− −2 6 3



C with ua (x j,l ) ≥ 14 (M − 9n10 − 2 ) (note that by construction such an agent exists). In this way, we get a higher Nash product because







n +1 · ua (Ra ) 3

 1 < uCk (RCk ) · ua (Ra ) + (M − 9n10 − 2 ) 4 uCk (RCk ) + 4n +

7 (M − 9n10 ) · 4 6 ≤ (M − 9n10 ) · 4

6 (M − 9n10 ) · 4 5 ≤ (M − 9n10 ) · 4

uCk (P ) · uXi (P ) ≤

≤

1 4

6

M−

n 3



+ uXi (RXi )

7 (M − 9n10 ) · 4 6 ≤ (M − 9n10 ) · 4



 (M − 9n10 − 2 ) + ua (Ra ) · uXi (RXi )

(20)

Note that (20) holds if ua (Ra ) · 16 (M − n3 ) ≤ 14 (M − 9n10 − 2 ) · uXi (RXi ). The latter inequality, however, is clearly satisfied since

uXi (RXi ) ≥ (M − n3 ) and ua (Ra ) <

3 2 (M





− 9n10 − 2 ) hold. 

Proof of 9. If Xi does not get xi , then Xi gets all three items of {xi, 1 , xi, 2 , xi, 3 }. Assume that Xi does not get xi . In addition, assume that Xi does not have all items of {xi, 1 , xi, 2 , xi, 3 }. With 6 and 7, it follows that there is an agent Ck ∈ C who has at least one such item xi, l . Again from 7 we can conclude that uCk (xi,l ) ≥ 14 (M − 9n10 ) − 2 holds. We handle it over to Xi and have to distinguish a few cases: Case 1. Xi has none of the items {xi, 1 , xi, 2 , xi, 3 }. It follows that Xi  has no items at all, and thus a∈A ua (P ) = 0 holds, in contradiction  with a ∈ A ua (P) ≥ k∗ .



2 M− 6

1 M− 2



n 3  n −3 3

Case 7. Xi has two of the items xi, j and Ck has its item from E, xi, l and exactly two items of {ck, 1 , ck, 2 , ck, 3 }{xi, l }:

uCk (P ) · uXi (P ) ≤

Proof of 8. If Xi gets xi , then Xi does not get an item of {xi, 1 , xi, 2 , xi, 3 }. Assume that Xi gets both xi and an item xi, l for some l ∈ {1, 2, 3}. We handle xi, l to an agent a ∈ C with ua (x j,l ) ≥ 14 (M − 9n10 − 2 ), which again implies that an allocation with a higher Nash product exists, since we will argue that the following inequality holds:

1



n 3  n −3 3

5 2 n (M − 9n10 ) · M− 4 6 3

 1 n ≤ (M − 9n10 ) · M− −3 2 3

uCk (P ) · uXi (P ) ≤



ua (P ) · uXi (P ) ≤ ua (Ra ) ·



1 M− 6

2 M− 6

Case 6. Xi has two of the items xi, j and Ck has its item from E, xi, l and exactly one item of {ck, 1 , ck, 2 , ck, 3 }{xi, l }:

Case 2. Assume that an agent Ck ∈ C gets assigned such an item xj, l with uCk (x j,l ) ≤ 4n + n3 . We handle xj, l over to an arbitrary agent a ∈

uCk (P ) · ua (P ) ≤



n 3  n − 32 3

Case 5. Xi has two of the items xi, j and Ck has only its item from E and xi, l :

Thus, with 6 we can assume that xi, l is allocated to an agent of Xk ∈ X . By construction, uXk (xi,l ) ≤ 4n + 3n . Exchanging xi, l with xj, l , we get a higher Nash product social welfare allocation:

uXk (P ) · uXi (P ) <



1 M− 6

2 M− 6

Case 4. Xi has one of the items xi, j and Ck has its item from E, xi, l and exactly two of the items {ck, 1 , ck, 2 , ck, 3 }{xi, l }:

uCk (P ) · uXi (P ) ≤





Case 3. Xi has one of the items xi, j and Ck has its item from E, xi, l and exactly one item c ∈ {ck, 1 , ck, 2 , ck, 3 }, c = xi,l :

6 (M − 9n10 ) · 4 5 ≤ (M − 9n10 ) · 4

1 (M − 9n10 ) 4



 1

 1 n n ≤ uX j (RX j ) + M− −2 · M − − 2 · ua (Ra ) 6 3 6 3 < uX j (RX j ) · n3 · ua (Ra ) +



5 1 n (M − 9n10 ) · M− 4 6 3

 2 n ≤ (M − 9n10 ) · M− −3 6 3

uCk (P ) · uXi (P ) ≤

uCk (P ) · uXi (P ) ≤

Otherwise, uXi (P ) < n3 and P(Xi ) contains items xj, l with j = i only. Let a be the agent with xi, l ∈ P(a). If a ∈ C, handling xi, l over to Xi and xj, l over to Xj yields a higher Nash product:

uX j (P ) · uXi (P ) · ua (P )

Case 2. Xi has one of the items xi, j and Ck has only its item from E and xi, l :



2 M− 6

1 M− 2



n 3  n −3 3

 But now we also know the exact contribution of the agent set X to the Nash product. By the pigeonhole principle, we know that exactly n of the items of Y are still available for agents from C. The agents of C have to contribute a factor of at least ( 45 (M − 9n10 − 2 ))n in order to  enable that a ∈ A ua (P) ≥ k∗ holds. But this means that every agent Ci ∈ C has to get exactly one of the items {ci, 1 , ci, 2 , ci, 3 }, since otherwise the required profit cannot be contributed. “⇒” This direction is straightforward.  4. Conclusion When indivisible items are to be allocated to agents, a major task is to allocate them in such a way that social welfare is maximized. In this respect three notions of social welfare are of particular interest: utilitarian social welfare, egalitarian social welfare, and Nash product social welfare. For each notion of social welfare, a central question is the computational complexity involved in finding an allocation that maximizes social welfare. Clearly, maximizing social welfare is an easy task for the utilitarian approach. However, the problem is known to be computationally hard for egalitarian social welfare under the use of classical scores such as Borda and lexicographic scores already (Baumeister et al., 2013); on the positive side, an allocation maximizing egalitarian

A. Darmann, J. Schauer / European Journal of Operational Research 247 (2015) 548–559

social welfare can be found in polynomial time for approval scores (Golovin, 2005). In this paper we have analyzed the computational complexity involved in allocating indivisible items to agents in order to maximize Nash product social welfare with respect to classical scores used in voting rules. We have shown that, on the negative side, maximizing Nash product social welfare is NP-complete when Borda or lexicographic scores are used. On the positive side, the problem is solvable in polynomial time for approval scores. Hence, for approval scores an allocation maximizing social welfare can be found in polynomial time for utilitarian, egalitarian, and Nash product social welfare. In addition, for the types of scores considered, i.e., for approval, Borda, and lexicographic scores, maximizing Nash product social welfare shows the same computational complexity behavior as maximizing egalitarian social welfare. However, whether or not the computational complexity statuses of the two problems also coincide when quasi-indifference scores are used is an interesting question for future research: while maximizing egalitarian social welfare is known to be computationally intractable for quasiindifference scores (Baumeister et al., 2013), the computational complexity involved in maximizing Nash product social welfare for quasiindifference scores is still open. In addition, the hardness results provided in this paper hint at another interesting direction for future research: for the problem of finding an allocation of maximum Nash product social welfare under the use of Borda and lexicographic scores respectively, investigate the existence of approximation algorithms that run in polynomial time. Acknowledgments Andreas Darmann was supported by the Austrian Science Fund (FWF) [P 23724-G11] “Fairness and Voting in Discrete Optimization”. Joachim Schauer was supported by the Austrian Science Fund (FWF) [P 23829-N13] “Graph Problems with Knapsack Constraints”.

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