Are Gross Substitutes a Substitute for Submodular Valuations?
AAre Gross Substitutes a Substitute for Submodular Valuations?
Shahar Dobzinski ∗ Uriel Feige † Michal Feldman ‡ March 1, 2021
Abstract
The class of gross substitutes (GS) set functions plays a central role in Economics andComputer Science. GS belongs to the hierarchy of complement free valuations introduced byLehmann, Lehmann and Nisan, along with other prominent classes: GS (cid:40) Submodular (cid:40)
XOS (cid:40)
Subadditive . The GS class has always been more enigmatic than its counterpartclasses, both in its definition and in its relation to the other classes. For example, while it is wellunderstood how closely the Submodular, XOS and Subadditive classes (point-wise) approximateone another, approximability of these classes by GS remained wide open. In particular, thelargest gap known between Submodular and GS valuations was some constant ratio smallerthan 2.Our main result is the existence of a submodular valuation (one that is also budget additive)that cannot be approximated by GS within a ratio better than Ω( log m log log m ), where m is thenumber of items. En route, we uncover a new symmetrization operation that preserves GS,which may be of independent interest.We show that our main result is tight with respect to budget additive valuations. However,whether GS approximates general submodular valuations within a poly-logarithmic factor re-mains open, even in the special case of concave of GS valuations (a subclass of Submodularcontaining budget additive). For concave of Rado valuations (Rado is a significant subclass ofGS, containing, e.g., weighted matroid rank functions and OXS), we show approximability byGS within an O (log m ) factor. A valuation function over a set M of m items is a function f : 2 M → R that assigns a real value f ( S )to every set S ⊆ M , which is additionally monotone ( f ( T ) ≤ f ( S ) for every T ⊆ S ) and normalized ( f ( ∅ ) = 0). Valuation functions are commonly used to describe combinatorial preferences overitems. In many settings it is natural to assume that the valuations additionally belong to a specificclass, e.g., they are submodular or subadditive. Indeed, algorithms and impossibilities for differentclasses of valuation functions were developed for numerous problems, e.g., welfare maximization ∗ Weizmann Institute and Microsoft Research Israel; [email protected] † Weizmann Institite;
[email protected] ‡ Tel-Aviv University and Microsoft Research Israel; [email protected] a r X i v : . [ c s . G T ] F e b
9, 11, 29], truthful mechanisms [2, 8], price of anarchy [7, 12], and learning [5, 4, 13], to name afew examples.In Algorithmic Game Theory, the work that has set the tone for the study of valuations functionsis that of Lehmann, Lehmann and Nisan [19]. They presented a hierarchy of “complement-free”valuation functions, whose five most prominent classes (listed from the least general to the mostgeneral) are OXS, Gross Substitutes (GS), Submodular, XOS, and Subadditive. A typical line ofwork attempts to determine the performance of algorithms, for example, in terms of their approx-imation ratios, in each of the classes of the hierarchy.Among the five classes of the hierarchy, two of them were defined semantically, as the set ofall valuations that satisfy a natural property like submodularity and subadditivity. Two of them(OXS and XOS) were defined syntactically, as the set of all valuations that can be constructed byapplying certain OR- and XOR-like operations. The remaining class of GS valuations stands out inthe sense that its definition is neither syntactic nor semantic but – to some extent – coincidental.It was essentially defined by Kelso and Crawford [18] as a condition on the valuation functions thatis needed for a certain auction to end in an equilibrium.Thus, it may not come as a total surprise that the class of gross substitutes valuations remainedas perhaps the least understood class among the classes considered by [19]. The class is lackingproperties that “natural” classes of valuations tend to possess, e.g., it is not closed under additionand applying a concave function on a GS valuation does not necessarily result in a GS valuation.Despite that, the GS class plays a central role in many settings (for example, it is, in some formalsense, the largest class for which a Walrasian equilibrium exists [16], it guarantees exact welfaremaximization in polynomial time [23], VCG outcomes are guaranteed to be in the core [3]). Oneconsequence to our partial understanding of the class is a lack of “complicated” GS valuationsor a lack of good techniques to construct them, although such valuations are crucial for provinghardness of various tasks . For example, easy-to-construct XOS valuations often serve as the hardestinstances even for subadditive valuations (e.g., [9, 4]). Recent attempts [24, 6] to constructivelyproduce the set of all GS valuations had only partial success.We suggest another direction toward deciphering the enigmatic character of the GS class. In-stead of focusing on exact characterizations of the class, we suggest to study its “closeness” to otherclasses of valuations. We rely on the notion of approximation to explore the proximity of the classof GS valuations to other classes. Definition 1.1.
A valuation function g approximates a valuation function f within a ratio ρ ≥ if for every set S of items we have that g ( S ) ≤ f ( S ) ≤ ρ · g ( S ) . A class C of valuation functionsapproximates a class C of valuation functions within a ratio ρ ≥ if for every function f ∈ C there is a function g ∈ C such that g approximates f within the ratio ρ . The fact that a class C approximates C by some small factor may not have immediate conse-quences in all settings of interest. However, more often than not, a small approximation factor does One exception that proves the rule is the matroid rank function construction of [5] that proves the non-existenceof good sketches for GS valuations. C will work well on valuations from C , perhaps with some modifications,and that impossibility results for C also apply to C .For example, consider the complement-free hierarchy (see also Figure 1): GS (cid:40) Submodular (cid:40)
XOS (cid:40)
Subadditive . It is known that XOS approximates Subadditive within a ratio of O (log m )and no better [8, 7] (recall that m denotes the number of items). Moreover, the approximabilitygap of Submodular and XOS is Ω( √ m ) [4] and this is tight (up to poly-log factors) [4, 15]. Indeed,the large gap between Submodular and XOS is evident when considering, e.g., that value queriesare very effective for various optimization tasks with submodular valuations (maximization subjectto cardinality constraint [22], welfare maximization [29], minimization [17]) but provide only poorapproximation ratios under XOS valuations. Similarly, the relatively small gap between XOSand Subadditive may explain why in many settings the best algorithms for subadditive valuationsachieve comparable results to the best algorithms for XOS valuations [11, 1].As evident from Figure 1, the approximation relationship of GS and the other classes wasunknown before this paper. The first question is to determine the exact approximation ratio ofSubadditive by GS. Previous work implies that it is between √ m (since Submodular functionscannot approximate XOS to within a better factor) and m (every subadditive function v can betrivially approximated within a factor of m by the additive function in which the value of item j is m v ( { j } )). The second question is to determine the approximation factor of Submodular by GS The first question turns out to be much easier than the second one. Previous results [4] alreadytell us that subadditive functions can be approximated to within a factor of ˜ O ( √ m ) by applyinga concave function on an additive valuation. We prove – as part of a more general result – thatany concave function of an additive valuation can be approximated by a GS function to within alogarithmic factor. This establishes that GS can approximate subadditive valuations to within afactor of ˜ O ( √ m ).Answering the second question is more challenging and serves as the main focus of this paper. Theorem 1.1 (Main Result) . The class of Gross Substitutes (GS) valuation functions does notapproximate the class of Submodular valuation functions within a ratio better than Ω( log m log log m ) . To prove this theorem we present a simple-to-describe family of submodular valuations (parametrizedby integers m and d ) that we refer to as BA ( k, d ), and show that members of this family cannotbe approximated well by any GS function. BA ( k, d ) even belongs to the class of budget additive The literature contains examples of submodular functions that are not GS. These examples use a constant numberof items and imply that there is a constant approximation gap between GS and Submodular (see a 3-item examplein [19]). .The proof of Theorem 1.1 makes extensive use of symmetries. We say that items i and j aresymmetric in a valuation f if for every bundle S ⊂ M \ { i, j } it holds that f ( S ∪ { i } ) = f ( S ∪ { j } ).We show that symmetries have far reaching implications on GS functions. Specifically, if f is GS,then every two items that are symmetric are also weak substitutes of each other, in the sense thatthe marginal value of one item given the other cannot be larger than the marginal value of the itemgiven any third item. We refer to functions with this property as symmetric weak substitutes , SWS.(See Definition 3.3 and Proposition 3.4.) Using the SWS property it is trivial to exhibit submodularfunctions that are not GS. For example, consider the budget additive function f defined on threeitems a, b, c , with f ( a ) = f ( b ) = , f ( c ) = 1, and f ( S ) = min[1 , (cid:80) i ∈ S f ( i )]. Items a and b aresymmetric, but they are not weak substitutes, as f ( a | c ) < f ( a | b ). Hence f is not GS. The functions BA ( k, k ) (for k ≥
2) used in the proof of Theorem 1.1 are generalizations of this function f .To derive inapproximability ratios using the SWS framework, we partition the proof into twoparts. One is to show that the approximation ratio of any GS function that preserves the symmetriesof BA ( k, d ) is Ω( log m log log m ) (for appropriate choices of k and d ). For this we extend the SWS propertyto groups of items (see Lemma 3.5). With this extension, and the careful design of the functions BA ( k, d ), proving this inapproximability result is relatively straightforward.The second part of the proof shows that for every function f (and thus also BA ( k, d )), amongGS functions, the one that approximates f best can be assumed to preserve the symmetries of f . Typically, such statements are proved by using symmetrization operations. Symmetrizationis usually relatively easy to obtain in “natural” classes like submodular and subadditive classes,where a symmetrized version of a valuation can be obtained by appropriately permuting the itemsand averaging. However, here we face the following difficulty: the average of two GS functions isnot necessarily GS. Hence, we introduce a different symmetrization operation that preserves theapproximation ratio. Interestingly, applying our new symmetrization operation to a submodularfunction does not guarantee that the symmetrized function is submodular, but if the originalfunction is GS then the symmetrized function is GS as well.Our proof that GS functions are closed under the new symmetrization operation requires adetailed analysis of the possible relations of the values of different bundles. We are able to ana-lytically carry out most parts of the proof. The remaining parts were reduced to a finite set ofclasses of potential counterexamples. Each class of counterexamples takes the form of a subclassof GS valuations over five items, and a counterexample is a member of that subclass whose sym-metrized version is not GS. For every subclass by itself, checking whether such a member existscan be done by solving a simple linear program. To verify that no counterexample exists, we useda computer program that searched over all subclasses, and for each subclass solved the underlyinglinear program. (See more details in Section C.1.)We do not know whether the log m log log m factor is tight, that is, whether submodular valuations canbe approximated by GS to within a factor of O ( log m log log m ). However, we do show that every budget A valuation is budget additive if there exists some B such that for every bundle S , v ( S ) = min(Σ j ∈ S v ( { j } ) , B ). Approximating Concave of GS by GS.
Observe that every budget additive valuation f is acomposition f = g ◦ f (cid:48) of a concave monotone function g : R + → R + (where g ( x ) = x for x ≤ B ,and g = B for x ≥ B ) with an additive valuation function f (cid:48) . Note that the corresponding f (cid:48) ,being additive, belongs to GS.In general, we can consider functions of the form f = g ◦ f (cid:48) , where g is an arbitrary concavemonotone function, and f (cid:48) is an arbitrary GS valuation function. Note that such a function f isnecessarily submodular, as f (cid:48) is submodular, and submodularity (unlike the property of being GS)is preserved under composition with concave functions. We show that GS-ness is approximatelypreserved under composition of concave functions, at least for some GS valuations.Rado valuations can be viewed as weighted matroid matching valuations — a strict superclassof OXS valuations (corresponding to weighted matching in bipartite graphs, see definition in Sec-tion 2). Rado valuations are known to be GS, and until very recently it has been open whetherthey contain the entire GS class; it was recently shown to be false [14]. Theorem 1.2.
Let g : R + → R + be an arbitrary concave monotone function, and let f (cid:48) be anarbitrary Rado valuation function (over m items). The function f = g ◦ f (cid:48) can be approximatedwithin a ratio of O (log m ) by a gross substitutes valuation function. As a special case (withstronger results), if f (cid:48) is additive, or more generally, a weighted Matroid Rank Function, then theapproximation ratio is O (log m ) . Open Questions.
The most obvious – and important – question that the paper leaves open isto determine the exact ratio that GS valuations can approximate submodular valuations. Anotherimmediate question is to determine whether the class of all GS valuations is approximately closedunder concave functions; i.e., whether the GS class approximates the class of concave of GS withina poly-logarithmic factor.An interesting subclasses of submodular valuations that is incomparable to GS is that of matroidrank sum (MRS, see Definition A.3). How well can MRS approximate GS valuations, and moregenerally, how well can MRS approximate submodular valuations? In the other direction, how wellcan GS valuations approximate MRS? (Some preliminary results are presented in Appendix G.)
In what follows, we define some valuation functions relevant to the current paper. Additionalvaluations, including matroid rank functions (MRF), weighted matroid rank function (WMRF),matroid rank sum (MRS), and OXS are defined in Appendix A.1. For all functions f below, f : 2 M → R + . We denote by f ( j | X ) the marginal value of j given X ; i.e., f ( j | X ) = f ( X ∪{ j } ) − f ( X ). 5 f is additive is there exist v , . . . , v n ∈ R + such that for every S ⊆ M , f ( S ) = (cid:80) j ∈ S v j . • f is budget additive (BA) if there exist v , . . . , v n , B ∈ R such that for every S ⊆ M , f ( S ) =min { B, (cid:80) j ∈ S v j } . • f is coverage if there exists a finite set Ω, where every element j ∈ Ω is associated with a weight w j ∈ R + , and a function g : M → Ω such that for every set S ⊆ M , f ( S ) = (cid:80) j ∈ (cid:83) i ∈ S g ( i ) w j . • f is Rado if there is a bipartite graph G ( M, V ; E ) with non-negative weights on the edges,and a matroid M = ( V, I ), such that for every S ⊆ M , f ( S ) is the total weight of the maxweight matching on the subgraph induced by S and a subset of V that belongs to I . • f is submodular if for every sets S, T ⊆ M such that S ⊆ T , and item j , f ( j | T ) ≤ f ( j | S ). • f is XOS (also known as fractionally subadditive) if it is a maximum over additive functions;i.e., there exist additive functions f , . . . , f k for some integer k , such that for every S ⊆ M , f ( S ) = max i ∈ [ k ] { f i ( S ) } .We next turn to the gross substitutes (GS) class, whose definition uses the notions of prices,utility and demand. Suppose every item j ∈ M is associated with a price p j . Then, the utility derived from a set S of items, given valuation function f and item prices p = ( p , . . . , p m ), is thenet benefit from S ; i.e., u ( S, p ) = f ( S ) − (cid:80) j ∈ S p j . Consequently, given a valuation function f anditem prices p , the demand is the collection of sets of items that maximize utility. Definition 2.1.
Given a valuation function f , and a price vector p = ( p , . . . , p m ) ∈ R m , the demand of f under p is D ( p ) = argmax S ⊆ M { f ( S ) − (cid:80) j ∈ S p j } . We are now ready to define gross substitutes valuations.
Definition 2.2.
A valuation function f is gross substitutes (GS) if for every pair of price vectors p , q such that p ≤ q , for every set S ∈ D ( p ) , there exists a set T ∈ D ( q ) such that T containsevery item j ∈ S such that p j = q j . Conceptually, this definition says that if S is the demand under a given price vector, and weincrease the prices of some items, then those items whose prices have not increased are still indemand.The following containment relations are known: OXS, W M RF (cid:40)
Rado (cid:40) GS (cid:40) Submodular (cid:40)
XOS (cid:40)
Subadditive.
In addition, it is known that
BA, Coverage, M RS (cid:40) submodular , and every pair among BA,Coverage, GS and MRS is non-comparable, except for
Coverage (cid:40)
M RS . GS characterization
The gross substitute (GS) class has several alternative characterizations.In what follows we give two characterizations that are of particular interest in this paper.A demand query gets a price vector p as input and returns a set S ∈ D ( p ). The first character-ization uses the notion of marginal utility . Given a set S ⊆ M and an item j , the marginal utilityof j with respect to S given prices p is u ( S ∪ { j } , p ) − u ( S, p ). Lemma 2.1. (The greedy characterization of GS)[10, 20] A valuation f is GS if and only if everydemand query can be answered by the following greedy algorithm: at every step pick an item j of ighest marginal utility, until no item has positive marginal utility. The next characterization does not use the notion of prices.
Lemma 2.2. (The triplet characterization of GS)[26] A function f is GS if and only if it issubmodular, and for every set S , and items i, j, k (cid:54)∈ S , it holds that max { f ( i, k | S ) + f ( j | S ) , f ( j, k | S ) + f ( i | S ) } ≥ f ( i, j | S ) + f ( k | S ) . (1) We refer to Equation (1) as the triplet condition . The following lemma (whose proof is deferred to Appendix A) shows that the triplet character-ization of GS holds in an approximate sense for all submodular functions.
Lemma 2.3.
Let f be a submodular function. Then, for every set S , and items i, j, k (cid:54)∈ S , max { f ( i, k | S ) + f ( j | S ) , f ( j, k | S ) + f ( i | S ) } ≥
34 ( f ( i, j | S ) + f ( k | S )) . The ratio of is best possible. Thus, the triplet condition of GS is a local condition that approximately (within a constantfactor) holds for all submodular valuations. In contrast, our main result limits the extent to whichapproximability holds globally, as it shows the existence of submodular valuation functions thatcannot be approximated by a GS function within any constant factor. Throughout this section we denote the set X ∪ { j } by Xj for simplicity of presentation. Definition 3.1.
Let f be an arbitrary set function. For two items i, j ∈ M , we say that f is ( i, j ) -symmetric, denoted by i = f j , if for every set S ⊂ M \ { i, j } it holds that f ( Si ) = f ( Sj ) . Proposition 3.1.
The relation = f is an equivalence relation.Proof. By definition, = f is reflexive ( i = f i ) and symmetric ( i = f j iff j = f i ). It remains toshow that it is transitive, namely, that i = f j and j = f k imply i = f k . There are two cases toconsider. In one j (cid:54)∈ S , and then f ( Si ) = f ( Sk ) follows from f ( Si ) = f ( Sj ) = f ( Sk ). In the other j ∈ S , and then S can be written as S (cid:48) j for S (cid:48) = S \ { j } . Then f ( S (cid:48) ji ) = f ( S (cid:48) jk ) follows from f ( S (cid:48) ji ) = f ( S (cid:48) ij ) = f ( S (cid:48) ik ) = f ( S (cid:48) ki ) = f ( S (cid:48) kj ) = f ( S (cid:48) jk ).If follows that = f partitions M into equivalence classes, where every two items are symmetricwith respect to f if and only if they are in the same equivalence class. We refer to these equivalenceclasses as symmetry classes .Given M , we say that a partition P is a coarsening of a partition P (and P is a refinement of P ), if every class of P is a union of (one or more) classes of P . Equivalently, every class of P Let us remark that if f is XOS, then we might have that f ( i, j | S ) + f ( k | S ) = 1 but max { f ( i, k | S ) + f ( j | S ) , f ( j, k | S ) + f ( i | S ) } = 0. For example, if S contains a single item s , and f ( T ) = max {| T ∩ { s }| , | T ∩ { i, j }|} forevery set T .
7s contained in (or equal to) a class of P . (A partition is both a coarsening and a refinement ofitself.)The following theorem plays a key role in establishing large gaps between various functions andthe class of GS functions. Theorem 3.2.
Let f be a valuation function and, for some ρ ≥ , let g be a GS function thatapproximates f within a ratio of ρ . Then there is a GS function g sym that approximates f withina ratio not worse than ρ , and moreover, the symmetry classes of g sym form a coarsening of thesymmetry classes of f . The standard approach towards proving a theorem such as Theorem 3.2 is to construct addi-tional versions of g that differ from the original g only in the sense that item names are permuted(using permutations that respect the symmetries of f ), and then g sym is taken to be the averageof all versions of g . This gives a function that respects the symmetries of f , and moreover, pre-serves the approximation ratio of g . However, this symmetrization technique does not apply to GSfunctions, because the average of two GS functions is not necessarily a GS function. For example,consider a BA function with budget 2 and item values 2,2,0. This function is GS. Applying theabove symmetrization (with respect to the last two items) implies averaging it with a BA functionwith budget 2 and item values 2,0,2. The average of the two functions gives a BA function withbudget 2 and values 2,1,1, and this function is not GS.Despite the above, our proof of Theorem 3.2 is based on repeatedly applying to g two-item sym-metrization operations until we reach the desired function g sym . Given a function g and two items i and j such that g is not ( i, j )-symmetric, an ( i, j )-symmetrization produces a new function g thatis ( i, j )-symmetric. Necessarily, our two-item symmetrization procedure is based on an operationthat is different from averaging. We refer to our operation as two-item max-symmetrization. Definition 3.2.
Given a set function g and two items i, j ∈ M , two-item max-symmetrization with respect to i and j produces the function g that for every set S ⊂ M \ { i, j } satisfies: • g ( S ) = g ( S ) . • g ( Sij ) = g ( Sij ) . • g ( Si ) = g ( Sj ) = max[ g ( Si ) , g ( Sj )] . Lemma 3.3.
If two-item max-symmetrization is applied to a GS function, the resulting functionis also GS.
Observe the following subtlety in Lemma 3.3. Recall that GS functions are submodular, andhence one aspect of showing that two-item max-symmetrization is GS-preserving is to show thatthe resulting function g is submodular. However, the fact that g is submodular does not implythat g is submodular. A counterexample is presented in Remark C.4. To prove submodularity of g , we need to use the assumption that g is GS, and not just the assumption that g is submodular.Proving that g is GS (and not just submodular) involves also showing that the triplet condition (ofLemma 2.2) holds. Our proof is based on verifying that there are no counterexamples to this fact.8ost of the proof could be done analytically, but the last few hundred potential counterexampleswere ruled out with the aid of a computer program. More details appear in Section C.Given Lemma 3.3, we can prove Theorem 3.2. Proof.
Suppose that valuation function f is approximated within a ratio of ρ ≥ g . Denote g by g , and construct a sequence of functions g t (for t ≥
0) as follows. For agiven t ≥
0, if there is a pair of items i, j ∈ M such that f is ( i, j )-symmetric and g t is not, applythe two-item max-symmetrization procedure (where i and j are the two items) to g t , obtaining g t +1 .By Lemma 3.3, all functions g t are GS.We now consider the approximation ratio. Initially, for every set S it holds that g ( S ) ≤ f ( S ) ≤ ρg ( S ). Values of sets cannot decrease by two-item max-symmetrization. Consequently,we also have f ( S ) ≤ ρg t ( S ) for all t . Moreover, the inequality g t ( S ) ≤ f ( S ) also holds, be-cause when a set increases its value (say, the set Si ), we have the chain of inequalities g t ( Si ) =max[ g t − ( Si ) , g t − ( Sj )] ≤ max[ f ( Si ) , f ( Sj )] = f ( Si ), where the left equality is by definition of two-item max-symmetrization, the middle inequality is by induction, and the last equality is because f is ( i, j )-symmetric.We next show that the sequence { g t } is finite. Consider the potential function ψ that givena function g sums up all its values. That is, ψ ( g ) = (cid:80) S ⊂ M g ( S ). Observe that two-item max-symmetrization increases the value of at least one set and does not decrease the value of any set,and hence the ψ values of the functions in the sequence { g t } form a strictly increasing sequence.Moreover, starting with g = g , the ψ values are supported on only finitely many possible values.This is because for every t ≥ S ⊂ M , there is some T ⊂ M such that g t ( S ) = g t − ( T ),and by induction, there also is some T ⊂ M such that g t ( S ) = g ( T ). Hence every ψ value is a sumof 2 m − M ), each taken from a pool of at most2 m − g ). Consequently, there are less than 2 m possible ψ values. Hence the sequence { g t } has length at most 2 m .The last function in the sequence { g t } can serve as g sym . For every i, j ∈ M for which f is( i, j )-symmetric, so is g sym (as otherwise g sym would not be the last function in the sequence). Forthis g sym , the partition of M into symmetry classes is a coarsening of the symmetry classes of f .Moreover, by being a member of the sequence { g t } , the function g sym is GS and approximates f within a ratio of ρ .Having established Theorem 3.2, we now prove structural properties that must hold for sym-metric GS functions.Two items a and b are considered to be substitutes to each other if the marginal value of anyone of them, given the other, is 0. We introduce a relaxation of the notion of substitutes, that werefer to as weak substitutes . Informally, a and b being weak substitutes means that b is the itemwhose inclusion leads to the most severe decrease in the marginal value of a (compared to including9ny other item), though unlike the case of substitutes, this decrease does not necessarily reduce themarginal value of a all the way down to 0. Definition 3.3.
Given a set function f , two items a and b are weak substitutes if for every set S ⊂ M \ { a, b } and item c (cid:54)∈ S , both f ( a | Sc ) ≥ f ( a | Sb ) and f ( b | Sc ) ≥ f ( b | Sa ) hold. We saythat a function f is symmetric weak substitutes (SWS) if every two items that belong to the samesymmetry class of f are weak substitutes. Proposition 3.4.
Every gross substitutes (GS) function is also symmetric weak substitute (SWS).Proof.
Let g be a GS function that is ( a, b )-symmetric. It suffices to prove that g ( a | Sc ) ≥ g ( a | Sb )holds for every set S ⊂ M \ { a, b } and item c (cid:54)∈ S . (The proof that g ( b | Sc ) ≥ g ( b | Sa ) is identical.)The triplet condition (Lemma 2.2) with the triplet ( a, b, c ) relative to the set S implies that: g ( c | S ) + g ( ab | S ) ≤ max[ g ( a | S ) + g ( bc | S ) , g ( b | S ) + g ( ac | S )]As g is ( a, b )-symmetric, we have that g ( a | S ) + g ( bc | S ) = g ( b | S ) + g ( ac | S ). It follows that g ( c | S )+ g ( ab | S ) ≤ g ( b | S )+ g ( ac | S ). Using g ( ab | S ) = g ( b | S )+ g ( a | Sb ) and g ( ac | S ) = g ( c | S )+ g ( a | Sc )we have: g ( c | S ) + g ( b | S ) + g ( a | Sb ) ≤ g ( b | S ) + g ( c | S ) + g ( a | Sc )Cancelling identical terms from both sides we get the desired g ( a | Sb ) ≤ g ( a | Sc ).We have established that every GS function g is also SWS. This implies that pairs of items thatare symmetric under g are also weak substitutes. The following lemma extends this property frompairs of items to sets of items, of arbitrary size. Lemma 3.5.
Let g be a SWS function (or a GS function, as a special case), and let B ⊂ M be a set of items that all belong to the same symmetry class in f . Partition B into B and B arbitrarily. Let C be an arbitrary set of items, disjoint from B , and with | C | = | B | . Then g ( B | C ) ≥ g ( B | B ) .Proof. We may assume without loss of generality that B contains only a single item b . (Forexample, if B contains also an additional item b , then g ( B | C ) = g ( b | C ) + g ( b | C b ). We firstadd b , and then apply the lemma again by taking B (cid:48) = { b } , C (cid:48) = C b , and B (cid:48) = B b .)Assume for the sake of contradiction that g ( b | C ) < g ( b | B ). For 0 ≤ k ≤ | B | , consider thehybrid sets H k that contain the first k items of C , and | B | − k items of B . Then H = B and H | B | = C , and by a hybrid argument, there is a value of k for which g ( b | H k +1 ) < g ( b | H k ). Let c be the item in H k +1 \ H k , let b the item in H k \ H k +1 , and let H = H k ∩ H k +1 . We get that g ( b | Hc ) < g ( b | Hb ), contradicting Proposition 3.4.We remark that SWS is not closed under neither average nor convolution; see Appendix D.10 .1 Gap of Ω( log m log log m ) between GS and Budget-Additive via Symmetrization In this section we prove Theorem 1.1 using the symmetrization technique. By Theorem 3.2, toshow that no GS function approximates a given function f within a ratio better than ρ , it sufficesto show that no GS function that respects the symmetries of f approximates f within a ratio betterthan ρ .Consider the following class of budget-additive (BA) functions, parameterized by k and d . Definition 3.4 ( BA ( k, d )) . For integers k ≥ and d ≥ the budget-additive function denoted by BA ( k, d ) is defined as follows. Its items are partitioned into d + 1 levels. Level h , for ≤ h ≤ d ,has k h items, each of value k − h . The budget is 1. In Theorem 3.7 we establish a lower bound on the ratio by which BA ( k, d ) can be approximatedby a GS function. This lower bound is an expression that depends on the parameters k and d . Thedesired lower bound of (1 − o (1)) log m log log m is then obtained for k = d log d . For the special case of BA ( k,
1) (i.e., d = 1) we obtain a lower bound of kk +1 , which is tight (by Proposition B.1). Thiscase is presented next as a warmup. Recall that BA ( k,
1) has one item a of value 1, and k items b , . . . , b k , each of value 1 /k , and a budget of 1. Theorem 3.6. (warmup) BA ( k, cannot be approximated by GS within a ratio better than kk +1 .Proof. Let f = BA ( k, f approx-imates f within a ratio better than kk +1 . Let g be a GS function that respects the symmetries of f . Let B denote the set of items b , . . . , b k . For simplicity of presentation, we sometimes omit theparenthesis of sets when referring to singletons.If g ( a ) < k +12 k , then f ( a ) g ( a ) > kk +1 and we are done. Hence we may assume that g ( a ) ≥ k +12 k .Since all sets have value at most 1, g ( B | a ) ≤ k − k . Let B = B \ { b } . Observe that g ( B | a ) ≤ g ( B | a ) ≤ k − k . By Lemma 3.5, g ( B | b ) ≤ g ( B | a ). In addition, g ( b ) ≤ k . We get g ( B ) = g ( b ) + g ( B | b ) ≤ k + k − k = k +12 k , whereas f ( B ) = 1.We now extend this result to BA ( k, d ). Theorem 3.7.
Let ρ ( k, d ) > be the smallest ratio by which BA ( k, d ) can be approximated by aGS function. Then: ρ ( k, d ) ≤ d + 1 (cid:32) d (cid:88) h =1 hk h − k d (cid:33) < d + 1 + 1 k − In particular, for every (cid:15) > , for sufficiently large d and k = d log d , the function BA ( k, d ) cannotbe approximated by GS within a ratio better than (1 − (cid:15) ) log m log log m .Proof. Let f = BA ( k, d ). For convenience, we denote ρ ( k,d ) by ρ . Hence 0 < ρ < ρ ( k, d ) = ρ is the smallest value) such that there is a GS function g thatapproximates f within a ratio of ρ . Namely, for every set S we have that ρf ( S ) ≤ g ( S ) ≤ f ( S ).By Theorem 3.2, we may assume that g respects the symmetries of f . We shall derive linear11onstraints that can be used in order to upper bound the value of ρ . For this purpose, we introducesome notation. • L h , for 0 ≤ h ≤ d , denotes the set of items in level h . Note that | L h | = k h . • S h , for 0 ≤ h ≤ d , denotes the set of items up to (and including) level h . Hence S h = (cid:83) hi =0 L h ,and | S h | = (cid:80) hi =0 k i . • L ph ( p stands for prefix) denotes an arbitrary subset of (cid:80) h − i =0 k i = | S h − | items from L h . Bysymmetry, it will not matter for us which items of L h are in the set L ph . • Given a set L ph , we denote L h \ L ph by L sh ( s stands for suffix). Observe that L sh contains k h − (cid:80) h − i =0 k i items from L h .For every 1 ≤ h ≤ d , Lemma 3.5 implies that g ( L sh | L ph ) ≤ g ( L sh | S h − ). This is because | L ph | = | S h − | . Hence using the notation of Lemma 3.5, L sh can serve as B , L ph can serve as B , and S h − can serve as C .Using the facts that L h = L ph ∪ L sh and that g ( L ph ) ≤ f ( L ph ) ≤ k h | L ph | we then have that g ( L h ) = g ( L ph ) + g ( L sh | L ph ) ≤ (cid:80) h − i =0 k i k h + g ( L sh | S h − ) ≤ (cid:80) h − i =0 k i k h + g ( L h | S h − ).Using the above we have that: (cid:88) ≤ h ≤ d g ( L h ) ≤ g ( L ) + d (cid:88) h =1 (cid:32) (cid:80) h − i =0 k i k h + g ( L h | S h − ) (cid:33) = d (cid:88) h =1 (cid:80) h − i =0 k i k h + g ( L ) + d (cid:88) h =1 g ( L h | S h − ) . Observe that g ( S d ) = g ( M ) ≤ f ( M ) ≤
1. Consequently, g ( L ) + (cid:80) dh =1 g ( L h | S h − ) ≤
1. We get (cid:88) ≤ h ≤ d g ( L h ) ≤ d (cid:88) h =1 (cid:80) h − i =0 k i k h + 1 . Consequently, ρ ≤ min ≤ h ≤ d g ( L h ) ≤ d + 1 (cid:32) d (cid:88) h =1 (cid:80) h − i =0 k i k h (cid:33) = 1 d + 1 (cid:32) d (cid:88) h =1 hk h − k d (cid:33) The above upper bound on ρ can be replaced by a simpler upper bound. Observe that (cid:80) dh =1 hk h − ≤ d (cid:80) dh =1 k h − < d k d k − . Hence ρ ≤ d + 1 (cid:32) d (cid:88) h =1 hk h − k d (cid:33) < d + 1 + 1 k − BA ( k, d ) with sufficiently large d and k = d log d , we have that d = (1 + o (1)) log m log log m , where m = (1 + o (1)) k d is the number of items. (Here o (1) denotes terms that tend to 0 as d grows.)Hence in that case ρ < d +1 + d log d − < (1 + o (1)) log log m log m , as desired.12 emark: In Appendix B we show that the same lower bound of kk +1 with respect to BA ( k, BA ( k, d )with d > √ m between GS and XOS via Symmetrization In this section we show that the known gap of Ω( √ m ) between GS and XOS valuations can beproved quite easily using the symmetrization techniques, and moreover, with an improved leadingconstant. Proposition 3.8. If m is a perfect square, then XOS cannot be approximated by GS within a ratiobetter than √ m .Proof. Items are partitioned into √ m groups M , . . . , M √ m , each of size √ m . The XOS functionis f = max i { f i } , where f i is additive (with item values 1) over the i th group. We show that no GSfunction that respects the symmetries of f approximates f within a ratio better than √ m .Let g be a GS function that respects the symmetries of f . Let T be a set with one item fromeach group (all such sets T have the same value by symmetry), and observe that g ( T ) ≤ f ( T ) = 1.Consider a random item x ∈ T and a random permutation π over the items of T . The expected(expectation taken both over choice of x and choice of π ) marginal value of x (marginal valuecompared to the prefix of π that precedes x ) is exactly g ( T ) | T | ≤ √ m . Consequently, there exists anitem x ∈ T , for which the above expectation (now taken only over choice of π ) is at most √ m .Consider the group M i that x belongs to. By Proposition 3.4, the marginal contribution of the j thitem to g ( M i ) is not larger than the marginal contribution of x to g ( T ) when x is in the j th location.As there are √ m locations and the expected contribution at a random location (corresponding tothe random permutation π ) is at most √ m , we have that g ( M i ) ≤
1. But f ( M i ) = √ m , showingthat the approximation ratio cannot be better than √ m . Remark:
It is known that submodular functions do not approximate f of Proposition 3.8 withina ratio better than Ω( √ m ), but they do achieve a ratio somewhat better than √ m . For example,when m = 4 the submodular function that is on sets of size 1, is on sets of size 3 or more, andeither 1 or on sets of size 2 (the latter holding if both items are from the same group), gives a < In this section we establish approximability results for concave functions of some GS functions, byGS functions. Section 4.1 establishes an upper bound of (1 + o (1)) log m log log m with respect to budget-additive functions. This bound is tight due to Theorem 3.7. Budget additive functions are concave13unctions of additive functions. In Section 4.2 we present a unified approach for establishing upperbounds for concave functions of a more general class of GS valuations. The generality of thisapproach may come at some loss in the approximation factor (e.g., it gives log m approximation forBA functions, compared with the log m/ log log m given in Section 4.1). Finally, in Section 4.3 weuse the techniques developed in Section 4.2 to prove Theorem 1.2, establishing that every concavefunction of a Rado function can be approximated by a GS function within a ratio of O (log m ). O ( log m log log m ) Theorem 4.1.
Every BA function f can be approximated by a GS function within a ratio of (1 + o (1)) log m log log m , where here o (1) denotes a term that tends to 0 as m grows. Before proving Theorem 4.1, we introduce a new family of GS functions. We then show that amember of this family serves to prove Theorem 4.1.Let g : M → R ≥ be a function that assigns a real value to every item, and let T : M → R ≥ bea monotone non-increasing threshold function, that assigns a real value to every position 1 , . . . , m .Given an item j , and a position r , we define the marginal value of j with respect to position r as µ ( j, r ) = min { g ( j ) , T ( r ) } . (2)The function µ satisfies the following property. Claim 4.2.
Let j, k be two items such that g ( j ) ≥ g ( k ) . Then, for every position r , µ ( j, r ) + µ ( k, r + 1) ≥ µ ( k, r ) + µ ( j, r + 1) . Proof.
We show that for every a, b, α, β such that a ≥ b, α ≥ β ,min { a, α } + min { b, β } ≥ min { b, α } + min { a, β } . (3)Indeed, if b ≤ β , then (3) holds iff min { a, α } + b ≥ b + min { a, β } , which holds by α ≥ β . Otherwise, β < b , and (3) holds iff min { a, α } + β ≥ min { b, α } + β , which holds by a ≥ b .Given a set of items S , sort them in a non-increasing order according to g (breaking tiesarbitrarily), and let S r be the item in position r . Consider the function h : 2 M → R ≥ defined as h ( S ) = | S | (cid:88) r =1 µ ( S r , r ) . (4)I.e., h sums up the marginal values of its items w.r.t. their corresponding positions. The followingproposition is proved in Appendix F. Proposition 4.3.
For every function g : M → R ≥ and non-increasing threshold function T :[ m ] → R ≥ , the function h defined in Equation (4) is monotone GS. emark: The proof of Proposition 4.3 does not use the specific expression of µ in Equation 2,but only the fact that it satisfies the property specified in Claim 4.2. Thus, for any µ functionsatisfying this property, the function h defined in Equation 4 is monotone GS.We are now ready to prove Theorem 4.1. Proof.
In this proof, it will be convenient for us to use natural logarithms, denoted by ln. Observethat even though log m and ln m differ by a constant multiplicative factor, the values ln m ln ln m and log m log log m are the same, up to additive terms that become negligible as m grows. Hence in thestatement of Theorem 4.1 we may replace log m log log m by ln m ln ln m .Consider an arbitrary BA function f . By scaling, we may assume without loss of generality thatthe budget of f is 1 + ln ln m ln m . (Choosing this value, which is 1 + o (1) instead of the more naturalvalue of 1, simplifies the definition of T .) Consider the threshold function T : [ m ] → R , where T (1) = ln ln m ln m , and for every i > T ( i ) = i ln m . Observe that (cid:80) j ≤ m T ( j ) ≤ ln ln m ln m .Given the BA function f , let g : M → R ≥ be the induced function on the singletons, and let h be the function defined in Eq. (4) with respect to g and T . h is GS by Proposition 4.3.Consider an arbitrary set S of items. By the definitions of h and g we have h ( S ) ≤ g ( S ). Wealso have that g ( S ) = f ( S ), unless g ( S ) > ln ln m ln m , in which case f ( S ) = 1 + ln ln m ln m . However, h ( S ) ≤ ln ln m ln m for every set S , because T ( S ) ≤ ln ln m ln m . This shows that h ( S ) ≤ f ( S ). Itremains to show that h ( S ) ≥ f ( S ) · (1 − o (1)) ln ln m ln m .Recall that S denotes the highest ranked item in S . We may assume that g ( S ) < ln ln m ln m , asotherwise we have h ( S ) ≥ h ( S ) ≥ ln ln m ln m ≥ (1 − o (1)) ln ln m ln m f ( S ), as desired.Let i be the smallest position for which T ( i ) < ln ln m ln m g ( S i ). (We may assume that such a positionexists, as otherwise h ( S ) ≥ ln ln m ln m g ( S ) ≥ ln ln m ln m f ( S ).) As g ( S i ) ≤ g ( S ) < ln ln m ln m , the inequality i ln m = T ( i ) < ln ln m ln m g ( S i ) implies that i > ln m (ln ln m ) . Let k = (ln ln m ) ln m i . Observe that k >
1, that T ( k ) ≤ g ( S i ) ≤ g ( S k ), and that ln ik = (1 − o (1)) ln ln m . It follows that h ( S ) ≥ (cid:88) k ≤ j ≤ i T ( i ) = 1ln m (cid:88) k ≤ j ≤ i j (cid:39) m ln ik (cid:39) ln ln m ln m ≥ (1 − o (1)) ln ln m ln m f ( S ) . To prove that a given function f can be approximated by a GS function within a polylogarithmicratio, we follow an approach that is fairly standard for approximating a function by a function from adifferent class. However, implementing this approach in the context of GS functions involves varioussubtleties that are not commonly encountered when other classes are concerned.Let f be a set function that is not GS, and we wish to find a GS function h that approximates f well. Our approach starts off as follows. For a parameter T that is at most polylogarithmic in m ,we find GS functions h , . . . , h T with the following sandwich property holding for every set S ⊂ M :15ax t ≤ T h t ( S ) ≤ f ( S ) ≤ (cid:88) t ≤ T h t ( S )We remark that sometimes we use an extension of the sandwich property in which either theright hand side or the left hand side is multiplied by some constant, but this does not significantlyaffect the discussion below.The sandwich property naturally suggests considering a lower bound function h L defined as h L ( S ) = max t ≤ T h t ( S ) and an upper bound function h U defined as h U ( S ) = (cid:80) t ≤ T h t ( S ). We havethat h L ( S ) ≤ f ( S ) ≤ h U ( S ), and for every S we have h U ( S ) ≤ T · h L ( S ). Consequently, both h L and T h U approximate f from below within a factor of T . The problem is that neither h L nor h U need to be GS, as GS is not preserved neither under the max operation (in fact, using the maxoperation one gets from GS the whole class XOS), nor under summation.To overcome this problem, we use one of two approaches. The first approach is to select thefunctions h t in such a way that each function is supported over a subset of the items (items notin the support have value 0 under h t ), with no intersections between the subsets that support twodifferent functions. Under this condition, the function h U is indeed GS, by the following observation(that can easily be verified using the triplet condition). Observation 4.4.
Let h , . . . , h T be GS functions over T pairwise disjoint sets of items M , . . . , M T .Then, the function h U = (cid:80) Tt =1 h t is GS over the set of items (cid:85) Ti =1 M i . In many interesting cases, we do not know how to implement the first approach (selecting thefunctions h t in such a way that each function is supported over a disjoint subset of the items) whilekeeping T polylogarithmic in m . Consequently, we need to deal with the situation in which h U is notGS (and neither is h L ). Our second approach is to find a GS function that is sandwiched between h L and h U . Luckily, the existence of such a function is guaranteed by the following argument.View each function h t as a valuation function of a bidder in a combinatorial auction, andconsider the maximum welfare W ( S ) function for the auction with items M and the above bidders.For every set S of items, W ( S ) is the maximum welfare that can be obtained by allocating S to thebidders. (This function W is also referred to as the convolution of the collection of functions h t ,see [20].) As each h t is GS, so is the welfare function W (see [19, 20, 21]). Moreover, max t ≤ T h t ( S ) ≤ W ( S ) ≤ (cid:80) t ≤ T h t ( S ). As both f and H enjoy the same sandwich property, it follows that T W approximates f from below by a ratio no worse than T . (Note that here the approximation ratiois T , whereas in the first approach it was T . The difference stems from the following fact. In bothapproaches we identify a GS function that is within a factor of T of f , but in the first approachthis function h U satisfies T h U ( S ) ≤ f ( S ) ≤ h U ( S ) for every S , whereas in the second approachthis function W satisfies T W ( S ) ≤ f ( S ) ≤ T · W ( S ) for every S .)The above discussion is summarised (and slightly generalized to an extended sandwich prop-erty) in the following lemma (whose proof can easily be completed by the reader, given the abovediscussion). Lemma 4.5.
Suppose that for function f , there are α, β > , and a collection of GS functions , . . . , h T satisfying the following extended sandwich property for every set S ⊂ M : α · max t ≤ T h t ( S ) ≤ f ( S ) ≤ β · (cid:88) t ≤ T h t ( S ) Let h U be the function defined as h U ( S ) = (cid:80) t ∈ T h t ( S ) , and let W be the welfare function (theconvolution of the functions h t ) as defined above. Then:1. W is a GS function, and the GS function αT W approximates f from below within a ratio of βα T .2. If every item is in the support of at most one of the functions h t , then h U is a GS function,and the GS function αT h U approximates f from below within a ratio of βα T . As an illustration of the use of Lemma 4.5, we prove Proposition 4.6 (which proves a weakerbound than that proved in Theorem 4.1, but does so via a simpler proof).
Proposition 4.6.
Every BA function f can be approximated by a GS function within a ratio of O (log m ) .Proof. Suppose without loss of generality that the budget limitation of f is 1, that there is no itemof value larger than 1, and that m is a power of 2. Round the value of each item down to thenearest power of 2 (namely, to 1, to , and so on), thus obtaining f (cid:48) . For every set S we havethat f (cid:48) ( S ) ≤ f ( S ) ≤ f (cid:48) ( S ). Partition items into T = 1 + log m classes M , . . . M log m by theirvalue, where all items of value 2 − t are in class M t , and class M log m contains also all items of valuesmaller than m . Let h t be the function f (cid:48) restricted to the items of M t . Observe that h log m isan additive function (as the sum of item values cannot reach the budget), and hence GS. Each of h t for t < log m remains BA, but is also GS. This is because all items in M t for t < log m havethe same value, and hence the triplet condition holds even with the budget limit. We have theextended sandwich property: max ≤ t ≤ log m h t ( S ) ≤ f ( S ) ≤ (cid:88) ≤ t ≤ log m h t ( S )Let h U be the function satisfying h U ( S ) = (cid:80) ≤ t ≤ log m h t ( S ). By item 2 of Lemma 4.5, thefunction T h U ( S ) is GS and approximates f (from below) within a ratio of 2(1 + log m ).Lemma 4.5 does not suffice for the proof of Theorem 1.2 (approximating any concave function g ( f ) of a Rado function f by GS within a ratio of O (log m )). One problem is that the range ofvalues of f can span values that differ from each other by more than a polynomial factor in m .(The function g ( f ) that we wish to approximate may still have a range of values that is polynomialin m , if composing the concave function g with f shrinks the range of values.) Handling suchsituations requires a version of Lemma 4.5 in which T is polynomial in m rather than logarithmicin m . Lemma 4.7, whose proof is considerably more complicated than the proof of Lemma 4.5,provides such a version. 17 emma 4.7. Let H = { h , . . . , h T } be a collection of submodular functions, where for every func-tion h t , all marginals are either 0 or t . (This condition is known to imply that h t is an MRF,scaled by t .) Suppose that for a valuation function f , there are α, β > satisfying the followingextended sandwich property for every set S ⊂ M : α · max t ≤ T h t ( S ) ≤ f ( S ) ≤ β · (cid:88) t ≤ T h t ( S ) Let g be an arbitrary normalized monotone concave function. Then:1. The function g ( f ) can be approximated by a GS function within a ratio of O ( βα (log m ) ) .2. If every item is in the support of at most one of the functions h t , then the function g ( f ) canbe approximated by a GS function within a ratio of O ( βα log m ) . The proof of Lemma 4.7 is deferred to Appendix E. O (log m ) In this section we prove Theorem 1.2, concerning approximating a concave function of a Radofunction by a GS function within a ratio of O (log m ).Before doing so, we consider several natural subclasses of Rado functions, namely matroid rankfunctions, additive functions, and weighted matroid rank functions. For additive and weightedmatroid rank functions we provide a better approximation ratio of O (log m ). For matroid rankfunctions, it is a known fact that applying a concave function to such functions results in a GSfunction; we provide a proof for completeness. Proposition 4.8.
Let f be a matroid rank function. Let g be a function obtained by composing aconcave function with f . Then g is GS.Proof. Clearly, g is submodular; thus, by the local characterization of GS functions it suffices toprove Inequality (1). If f ( j, k | S ) = 0, then (1) follows by the DC property of matroids. Similarly,if f ( j, k | S ) = 1, then all terms in the RHS of (1) are at least g (1), and the inequality follows. Itremains to prove the inequality for the case where f ( j, k | S ) = 2. By the DC property of matroids, f ( j | S ) = f ( k | S ) = 1. If f ( i | S ) = 1, then by the exchange property of matroids, either f ( i, j | S ) = 2or f ( i, k | S ) = 2. The LHS of (1) is g (1) + g (2), and one of the terms in the RHS of (1) is at least g (1) + g (2) as well, as desired. If f ( i | S ) = 0, then the LHS of (1) is g (0) + g (2), and both terms inthe RHS of (1) are g (1) + g (1). By concavity of g , g (0) + g (2) ≤ g (1) + g (1), as desired.Proposition 4.8 turns out to be very useful in our context, because every submodular function(and hence also every GS function) in which the marginals are either 0 or 1 is an MRF. It willbe used in the proof of Lemma 4.7, where we encounter GS functions that are scaled versions ofMRFs (marginals are 0 or c for some value c ). Consequently, applying a concave function on thesefunctions also results in a GS function. 18e next consider additive and weighted matroid rank functions. Proposition 4.9.
Let g be a normalized monotone concave function and let f be an additive setfunction. The function g ( f ) can be approximated by a GS function within a ratio of O (log m ) .Proof. Let R be the ratio between the maximum value and minimum value item in f . Withoutloss of generality, we assume that the smallest value that an item has is 1.Round the value of each item down to the nearest power of 2, thus obtaining f (cid:48) . For every set S we have that f (cid:48) ( S ) ≤ f ( S ) ≤ f (cid:48) ( S ). By monotonicity and concavity of g , it also holds that g ( f (cid:48) ( S )) ≤ g ( f ( S )) ≤ g ( f (cid:48) ( S )). Consequently, for simplicity of notation and losing only a factorof 2 in the approximation ratio, we assume that in f the value of every item is a power 2.Partition items into T = 1 + log R classes M , . . . M log R by their value, where all items of value2 t are in class M t . For every 0 ≤ t ≤ log R , let f t be the the function defined by f t ( S ) = f ( S ∩ M t ).Then for every set S ⊂ M we have the sandwich propertymax ≤ t ≤ R f t ( S ) ≤ f ( S ) ≤ (cid:88) ≤ t ≤ R f t ( S )The proof of the proposition now follows from item 2 of Lemma 4.7. Proposition 4.10.
Let g be a normalized monotone concave function and let f be a weighted MRFfunction. g ( f ) can be approximated by a GS function within a ratio of O (log m ) .Proof. The proof follows that of Proposition 4.9, but with one change. Recall the classes M t defined in the proof of Proposition 4.9. Every item in the class M t had f value 2 t . In the proof ofProposition 4.9, f restricted to the items of M t was an additive function. In our current context,when f is a WMRF, f restricted to the items of M t is an MRF (scaled by 2 t ). Consequently, item 2of Lemma 4.7 still applies.We now prove Theorem 1.2, establishing that if g is a normalized monotone concave functionand f is a Rado function, then the function g ( f ) can be approximated by a GS function within aratio of O (log m ).Recall the definition of Rado valuations. A set function f : 2 M → R + is a Rado valuationif there is a bipartite graph G ( M, V ; E ) with non-negative weights on the edges, and a matroid M = ( V, I ), such that for every S ⊆ M , f ( S ) is the total weight of the max weight matching onthe subgraph induced by S and a subset of V that belongs to I . Proof. (of Theorem 1.2) Let f be a Rado function. Recall that its representation involves a bipartitegraph G ( U, V ; E ) with non-negative weights on the edges, with m = | U | left side vertices, that weshall refer to as items .Let R be the ratio between the maximum weight and minimum weight edge in G . Without lossof generality, we assume that the smallest weight that an edge has is 1, and that R is a power of 2.Round down the weight of each edge to the nearest power of 2, thus obtaining f (cid:48) . For everyset S we have that f (cid:48) ( S ) ≤ f ( S ) ≤ f (cid:48) ( S ). By monotonicity and concavity of g , it also holds that19 ( f (cid:48) ( S )) ≤ g ( f ( S )) ≤ g ( f (cid:48) ( S )). Consequently, for simplicity of notation and losing only a factorof 2 in the approximation ratio, we assume that in G the weight of every edge is a power 2.Partition the graph into a collection { G t } of uniform weight graphs, where for every t graph G t contains all edges of weight 2 t . The corresponding Rado function for each such graph G t will bereferred to as f t , and it is a matroid rank function (scaled by 2 t , because f t is submodular, and itsmarginals are 0 and 2 t ).We have the sandwich property max t ≤ log R f t ( S ) ≤ f ( S ) ≤ (cid:80) t ≤ log R f t ( S ). Using item 1 ofLemma 4.7 we have that there is a GS function that approximates g ( f ) within a ratio of O ((log m ) ). Acknowledgements
We are grateful to Noam Guterman and Ittay Toledo who each independently wrote code verifyingClaim C.8.
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APPENDIXA Appendix for Section 2
Proof of Lemma 2.3:
We distinguish between two cases.Case (i): f ( i, j | S ) < f ( k | S ). First observe that max { f ( i | S ) , f ( j | S ) } ≥ ( f ( i | S ) + f ( j | S )) ≥ f ( i, j | S ), where the last inequality follows by submodularity. Combining this with monotonic-ity and the assumption of case (i), we get that max { f ( i, k | S ) + f ( j | S ) , f ( j, k | S ) + f ( i | S ) } ≥ max { f ( k | S ) + f ( j | S ) , f ( k | S ) + f ( i | S ) } = f ( k | S ) + max { f ( j | S ) , f ( i | S ) } ≥ f ( k | S ) + f ( i, j | S ) ≥ ( f ( i, j | S ) + f ( k | S )). 22ase (ii): f ( i, j | S ) ≥ f ( k | S ). By submodularity, f ( i, k | S ) + f ( j, k | S ) ≥ f ( i, j, k | S ) + f ( k | S ) andalso f ( i | S ) + f ( j | S ) ≥ f ( i, j | S ). By these inequalities and monotonicity we get that f ( i, k | S ) + f ( j | S ) + f ( j, k | S ) + f ( i | S ) ≥ f ( i, j | S ) + f ( k | S ). Hence max[ f ( i, k | S ) + f ( j | S ) , f ( j, k | S ) + f ( i | S )] ≥ f ( i, j | S ) + f ( k | S ) ≥ ( f ( i, j | S ) + f ( k | S )), where the last inequality follows by the assumption ofcase (ii).To show that is the best possible ratio, consider a budget additive function f with f ( i ) = f ( j ) = 1, f ( k ) = 2, and a budget of 2. For S = ∅ , it holds that max { f ( i, k | S ) + f ( j | S ) , f ( j, k | S ) + f ( i | S ) } = 3, whereas f ( i, j | S ) + f ( k | S )) = 4. A.1 Matroids and Additional Valuation Functions A matroid M is a pair ( M, I ), where M is a finite set of elements, and I is a non-empty collectionof subsets of M (often termed the collection of independent sets ) satisfying the following twoconditions: • Downward-closed (DC): If S ∈ I and S (cid:48) ⊆ S , then S (cid:48) ∈ I . • Exchange property: For any two sets
S, T ∈ I such that | S | < | T | , there exists an element j ∈ T \ S such that S ∪ { j } ∈ I .Given a matroid M = ( M, I ), the rank function of M is a function rank M : 2 M → N , where rank M ( S ) = max T ∈I | S ∩ T | . Definition A.1.
A set function f : 2 M → R is a matroid rank function (MRF) if there exists amatroid M = ( M, I ) such that f ( S ) = rank M ( S ) for every S ⊂ M . A more general class of valuations is weighted matroid rank (WMRF) functions. Given a matroid M = ( M, I ) and a weight function w that associates a non-negative weight w j with every element j ∈ M , the weighted rank function of M with respect to w is a function wrank M ,w : 2 M → R ,where wrank M ( S ) = max T ∈I ,T ⊆ S (cid:80) j ∈ T w j . Claim A.1. [25] A valuation function is MRF if and only if it is a submodular valuation withbinary marginal values.
With a slight abuse of notation, we shall refer to a submodular valuation with marginal valuesin { , c } (for some constant c ) as MRF as well. Definition A.2.
A set function f : 2 M → R is a weighted matroid rank function (WMRF) ifthere exist a matroid M = ( M, I ) , and a weight function w over the elements in M such that f ( S ) = wrank M ,w ( S ) for every S ⊆ M . Finally we define the class of matroid rank sum (MRS), which is the class of sum over matroidrank functions.
Definition A.3.
A set function f : 2 M → R is a matroid rank sum (MRS) if there exists acollection of MRF functions f , . . . f k : 2 M → R , and associated non-negative weights w , . . . , w k ∈ R + such that f ( S ) = (cid:80) kj =1 w j f j ( S ) for every S ⊂ M . Note that coverage valuations are MRS over matroid rank functions of rank 1.23 efinition A.4.
A set function f : 2 M → R is OXS if there exists a bipartite graph G ( M, V ; E ) with non-negative weights on the edges, such that for every S ⊆ M , f ( S ) is the total weight of themaximum weighted matching on the subgraph induced by S and V . B Gap of kk +1 between GS and Submodular via the Greedy Char-acterization In this section we show that the submodular (and in fact, budget additive, BA) function BA ( k, kk +1 . Wedo so without making use of symmetrization (Theorem 3.2). Hereafter, ρ f denotes the best ratioby which any GS function can approximate f . Proposition B.1.
Let f be the budget additive function with budget 1, one item a of value 1, anda set B of k items b , . . . , b k , each of value /k . Then ρ f = kk +1 .Proof. For presentation simplicity, for an item j , we write g ( j ) instead of g ( { j } ) to denote its valueunder valuation g .To see that ρ f ≤ kk +1 , consider the function g , where g ( a ) = k +12 k , g ( b j ) = k for every j ∈ [ k ], andthe marginal value of every additional item from B is k (except for the last item that contributes0). That is, for a set S ⊆ B of size (cid:96) ≥ g ( S ) = (cid:96) +12 k ; for a set S that contains a and (cid:96) items from B , g ( S ) = min[ k +1+ (cid:96) k , g approximates f within a ratio kk +1 .The function g is clearly submodular. To see that it is GS, we show that it satisfies the tripletcondition. Since the items in B are symmetric, it suffices to prove that for every i, j ∈ { , . . . , k } and item set S s.t. a, b i , b j (cid:54)∈ S , g ( a | S ) + g ( b i b j | S ) = g ( b i | S ) + g ( ab j | S ). If S = ∅ , then g ( a | S ) + g ( b i b j | S ) = g ( b i | S ) + g ( ab j | S ) = k +42 k . Else (1 ≤ | S | ≤ k − g ( a | S ) + g ( b i b j | S ) = g ( b i | S ) + g ( ab j | S ) = k +22 k . It follows that g is GS.To prove that ρ f ≥ kk +1 , let g be a GS function that approximates f from below. W.l.o.g., g ( b ) ≥ g ( b j ) for all 2 ≤ j ≤ k . Also, we may assume that g ( a ) > g ( b ), as otherwise theapproximation ratio of g is no better than k ≥ > kk +1 .Consider a vector of prices p with p a = g ( a ) − g ( b ), and p b j = 0 for all j ∈ [ k ]. The greedyalgorithm first selects a (and pays g ( a ) − g ( b )), and then the remaining items. As g is upperbounded by 1, the profit is at most 1 − g ( a ) + g ( b ). Alternatively, one can select the set B andobtain a profit of g ( B ). By Lemma 2.1 (the greedy characterization of GS functions), we have that1 − g ( a ) + g ( b ) ≥ g ( B ), thus 1 + g ( b ) ≥ g ( B ) + g ( a ). Since g ( b ) ≤ f ( b ) = k it follows that k +1 k ≥ g ( B ) + g ( a ). Consequently, min[ g ( a ) , g ( B )] ≤ k +12 k , whereas f ( a ) = f ( B ) = 1. C Proof of Lemma 3.3
In this section we prove Lemma 3.3. 24et h be the two-item max-symmetrization function introduced in Definition 3.2. We showthat if h is applied to a monotone GS function, the resulting function is a monotone GS function.Monotonicity is proved in Lemma C.1, submodularity is proved in Lemma C.2, and the tripletcondition is proved in Lemma C.5. Lemma C.1.
The function h is monotone.Proof. Consider sets
S, T ⊆ M such that S ⊆ T . We need to show that h ( T ) ≥ h ( S ). If { a, b } ∈ S or { a, b }∩ S = φ , then h ( S ) = g ( S ), and the monotonicity of h follows because h ( T ) ≥ g ( T ) ≥ g ( S ).It remains to handle the cases in which |{ a, b } ∩ S | = 1.Suppose that S = S (cid:48) a and T = T (cid:48) a such that S (cid:48) ⊆ T (cid:48) , and b (cid:54)∈ T (cid:48) . It holds that h ( S ) = max { g ( S (cid:48) a ) , g ( S (cid:48) b ) } ≤ max { g ( T (cid:48) a ) , g ( T (cid:48) b ) } = h ( T ) , where the inequality follows by the monotonicity of g .For the case where S = S (cid:48) a and T = T (cid:48) ab with S (cid:48) ⊆ T (cid:48) , we get h ( S ) = max { g ( S (cid:48) a ) , g ( S (cid:48) b ) } ≤ max { g ( T (cid:48) a ) , g ( T (cid:48) b ) } ≤ g ( T (cid:48) ab ) = h ( T ) . Lemma C.2.
The function h is submodular.Proof. To prove submodularity, it suffices to prove that for every set S and every two items c, d (cid:54)∈ S ,it holds that h ( c | S ) ≥ h ( c | Sd ). The following four cases can be addressed without using the factthat g is GS.1. If { a, b } ⊂ S then the claim follows from the submodularity of g , and the fact that h = g forthe sets of interest.2. If { a, b } ∩ S = φ there are several subcases to consider:(a) If d (cid:54)∈ { a, b } and c (cid:54)∈ { a, b } , the claim follows from the submodularity of g , and the factthat h = g for the sets of interest.(b) If d (cid:54)∈ { a, b } and c ∈ { a, b } (w.l.o.g., c = a ), the claim follows because h ( a | S ) =max { g ( a | S ) , g ( b | S ) } ≥ max { g ( a | Sd ) , g ( b | Sd ) } = h ( a | Sd ), where the inequality followsby submodularity of g .(c) If d ∈ { a, b } and c (cid:54)∈ { a, b } , suppose without loss of generality that g ( Sac ) ≥ g ( Sbc ).Then the claim follows because h ( c | S ) = g ( c | S ) ≥ g ( Sac ) − g ( Sa ) ≥ g ( Sac ) − max { g ( Sa ) , g ( Sb ) } = h ( c | Sa ), where the last equality follows from g ( Sac ) ≥ g ( Sbc ).(d) If d ∈ { a, b } and c ∈ { a, b } , suppose without loss of generality that g ( Sa ) ≥ g ( Sb ). Thenthe claim follows because for c ∈ { a, b } we have that h ( c | S ) = max { g ( a | S ) , g ( b | S ) } ≥ g ( b | S ) ≥ g ( b | Sa ) ≥ g ( Sab ) − max { g ( Sa ) , g ( Sb ) } = h ( c | Sd ), where the last equalityfollows from h ( Sd ) = max { g ( Sa ) , g ( Sb ) } .25. If |{ a, b } ∩ S | = 1 (w.l.o.g., a ⊂ S , and we let S = S (cid:48) a ) and d = b , then h ( c | Sb ) = g ( c | Sb )(as S = S (cid:48) a ). If g ( S (cid:48) a ) ≥ g ( S (cid:48) b ) the claim follows because then h ( c | S ) ≥ g ( c | S ) ≥ g ( c | Sb ) = h ( c | Sb ). If g ( S (cid:48) b ) ≥ g ( S (cid:48) a ), the claim follows because then h ( c | S ) ≥ g ( S (cid:48) bc ) − g ( S (cid:48) b ) = g ( c | S (cid:48) b ) ≥ g ( c | Sb ) = h ( c | Sb ).4. If |{ a, b }∩ S | = 1 (w.l.o.g., a ⊂ S ) and c = b , let S = S (cid:48) a . Then h ( b | S ) = min { g ( b | S (cid:48) a ) , g ( a | S (cid:48) b ) } whereas h ( b | Sd ) = min { g ( b | S (cid:48) ad ) , g ( a | S (cid:48) bd ) } . The claim follows by submodularity of g .It remains to deal with the case that |{ a, b } ∩ S | = 1 (w.l.o.g., a ⊂ S , and we let S = S (cid:48) a )and b (cid:54)∈ { c, d } . This is the only case in which we will need to use the fact that g is not onlysubmodular, but also GS. The following claim addresses this case. For simplicity of notation (butwithout affecting the proof), its formulation assumes that S (cid:48) = φ . Claim C.3.
The function h satisfies h ( c | a ) ≥ h ( c | da ) .Proof. Based on the definition of h , proving the claim amounts to proving thatmax { g ( ac ) , g ( bc ) } + max { g ( ad ) , g ( bd ) } (cid:124) (cid:123)(cid:122) (cid:125) LHS ≥ max { g ( acd ) , g ( bcd ) } + max { g ( a ) , g ( b ) } (cid:124) (cid:123)(cid:122) (cid:125) RHS . (5)For each term of the form max { g ( aS ) , g ( bS ) } we say that a realizes the max term if g ( aS ) ≥ g ( bS ); otherwise, we say that b realizes the max term. We distinguish between the following cases.Case 1 is addressed without using the fact that g is GS. The remaining cases use this fact.The first cases are ones where either (i) a realizes both max terms in the RHS of (5), or (ii) b realizes both max terms in the RHS of (5). We establish (5) for (i) (case 1 below); (ii) is provedanalogously. Case 1: g ( acd ) ≥ g ( bcd ) and g ( a ) ≥ g ( b ). We get: g ( acd ) + g ( a ) ≤ g ( ac ) + g ( ad ) ≤ max { g ( ac ) , g ( bc ) } + max { g ( ad ) , g ( bd ) } , where the first inequality follows by submodularity.The next cases are ones where a realizes exactly one of the max terms in the LHS of (5), andexactly one of the max-terms in the RHS of (5). We establish (5) for the cases where a realizes thefirst max term of LHS (cases 2,3 below); the other two cases are proved analogously. Case 2: (i) g ( ac ) ≥ g ( bc ), (ii) g ( bd ) > g ( ad ), (iii) g ( a ) ≥ g ( b ) and (iv) g ( bcd ) ≥ g ( acd ). Supposetoward contradiction that Equation (5) is violated. That is, (v) g ( ac )+ g ( bd ) < g ( a )+ g ( bcd ). Then,by submodularity we get g ( ac ) − g ( a ) < g ( cd ) − g ( d ) . (6)We get: g ( bc ) + g ( ad ) < g ( ac ) + g ( bd ) < g ( a ) + g ( bcd ), where the first inequality follows from (i) and(ii), and the second inequality is (v). By submodularity it follows that g ( ad ) + g ( c ) < g ( a ) + g ( cd ).By the triplet condition w.r.t. acd , it follows that g ( d ) + g ( ac ) = g ( a ) + g ( cd ), in contradiction toEquation (6). 26 ase 3: (i) g ( ac ) ≥ g ( bc ), (ii) g ( bd ) > g ( ad ), (iii) g ( b ) ≥ g ( a ) and (iv) g ( acd ) ≥ g ( bcd ). Supposetoward contradiction that Equation (5) is violated. That is, (v) g ( ac )+ g ( bd ) < g ( b )+ g ( acd ). Then,by submodularity we get g ( bd ) − g ( b ) < g ( cd ) − g ( c ) . (7)We get: g ( bc ) + g ( ad ) < g ( ac ) + g ( bd ) < g ( b ) + g ( acd ), where the first inequality follows from (i) and(ii), and the second inequality is (v). By submodularity it follows that g ( bc ) + g ( d ) < g ( b ) + g ( cd ).By the triplet condition w.r.t. bcd , it follows that g ( c ) + g ( bd ) = g ( b ) + g ( cd ), in contradiction toEquation (7).The remaining cases are ones where either (i) a realizes both max terms in the LHS of (5), andexactly one of the max terms in the RHS, or (ii) b realizes both max terms in the LHS of (5), andexactly one of the max terms in the RHS. We establish (5) for the two cases of (i) (cases 4,5 below);the two cases of (ii) are proved analogously. Case 4: (i) g ( ac ) ≥ g ( bc ), (ii) g ( ad ) ≥ g ( bd ), (iii) g ( a ) ≥ g ( b ) and (iv) g ( bcd ) > g ( acd ). Supposetoward contradiction that Equation (5) is violated. That is, (v) g ( ac )+ g ( ad ) < g ( a )+ g ( bcd ). Then,by (i), (v) and submodularity we get g ( ad ) − g ( a ) < g ( cd ) − g ( c )Consequently, by the triplet condition w.r.t. acd we get g ( ad ) + g ( c ) < g ( ac ) + g ( d ) . (8)Similarly, by (ii), (v) and submodularity we get g ( ac ) − g ( a ) < g ( cd ) − g ( d )Consequently, by the triplet condition w.r.t. acd we get g ( ac ) + g ( d ) < g ( ad ) + g ( c ), contradicting(8). Case 5: (i) g ( ac ) ≥ g ( bc ), (ii) g ( ad ) ≥ g ( bd ), (iii) g ( a ) < g ( b ) and (iv) g ( acd ) ≥ g ( bcd ).Suppose toward contradiction that Equation (5) is violated. That is, (v) g ( ac ) + g ( ad ) There exists a monotone submodular function g on items such that the induced h function is not submodular. That is, the triplet condition of g is necessary for submodularityof h . To see this, consider the following function: g ( ∅ ) = 0 , g ( a ) = 3 , g ( b ) = 2 , g ( c ) = 1 , g ( d ) =2 , g ( ab ) = 5 , g ( ac ) = 3 , g ( ad ) = 3 , g ( bc ) = 3 , g ( bd ) = 4 , g ( cd ) = 3 , g ( abc ) = 5 , g ( abd ) = 5 , g ( acd ) =3 , g ( bcd ) = 5 , g ( abcd ) = 5 . One can verify that g is submodular, but h ( c | a ) = 0 while h ( c | da ) = 1 ,violating submodularity. Lemma C.5. The function h satisfies the triplet condition of GS.Proof. The triplet condition involves a base set S , and a triple T of three items that are not in S .There are five cases to consider (see below). Among the five, four are quite simple to address. Incontrast, the fifth case is not easy to establish, and we prove it by a computer assisted code. • { a, b } ∩ ( S ∪ T ) = φ . In this case the triplet condition for h follows from that for g . • |{ a, b } ∩ S | = 2. In this case the triplet condition for h follows from that for g . • |{ a, b } ∩ T | = 2. This case is handled in Claim C.6. • |{ a, b } ∩ T | = 1 and |{ a, b } ∩ S | ∈ { , } . This case is handled in Claim C.7. • |{ a, b } ∩ T | = 0 and |{ a, b } ∩ S | = 1. This case is handled in Claim C.8, via a computerassisted proof. Claim C.6. Let S be a set s.t. S ∩ { a, b } = ∅ . Then, h satisfies the triplet condition for everytriple T with { a, b } ⊂ T .Proof. Observe that h ( a | S ) = h ( b | S ) = max { g ( a | S ) , g ( b | S ) } and that h ( bc | S ) = h ( ac | S ) = max { g ( ac | S ) , g ( bc | S ) } . Consequently h ( a | S ) + h ( bc | S ) = h ( b | S ) + h ( ac | S ). Hence itsuffices to prove that h ( c | S ) + h ( ab | S ) ≤ max { h ( a | S ) + h ( bc | S ) , h ( b | S ) + h ( ac | S ) } . Thelast inequality is verified as follows: h ( c | S ) + h ( ab | S ) = g ( c | S ) + g ( ab | S ) ≤ max { g ( a | S ) + g ( bc | S ) , g ( b | S ) + g ( ac | S ) }≤ max { h ( a | S ) + h ( bc | S ) , h ( b | S ) + h ( ac | S ) } , where the first inequality follows by the triplet condition w.r.t. abc . Claim C.7. Let S, T be sets such that |{ a, b } ∩ T | = 1 and |{ a, b } ∩ S | ∈ { , } . Then, h satisfiesthe triplet condition.Proof. Given a set S and triple T = { x, y, z } , there are the following three pairs : ( Sx, Syz ),( Sy, Sxz ), ( Sz, Sxy ). Suppose that in each of the three pairs, exactly one of the two terms containsexactly one item from { a, b } . This is indeed the case when |{ a, b } ∩ T | = 1, where if |{ a, b } ∩ S | = 0the other term in the pair is disjoint from { a, b } , whereas if |{ a, b } ∩ S | = 1 the other term in the28air contains both a and b . Consider a table with three rows and three columns. Each columncorresponds to one of the pairs. For j ∈ { , , } , the j th entry in the first row gives the sum ofvalues of g applied to the two terms in the j th pair, when the item from { a, b } (in the term thathas one such item) is taken to be a . The entries in the second row are constructed in an analogousway, with a replaced by b . The third row corresponds to values under h , hence each of its entriesis the maximum among the two other entries in its column. Let M denote the maximum value inthis table. M must appear in at least one of the first two rows, and by the triplet property of g ,it must appear at least twice in that row. Consequently, it appears at least twice also in the thirdrow, implying the h satisfies the triplet condition. Claim C.8. Let S, T be such that |{ a, b } ∩ T | = 0 and |{ a, b } ∩ S | = 1 . Then, h satisfies the tripletcondition.Proof. We first observe that it is sufficient to prove the claim for the case of 5 items. Indeed, ifthere is a counterexample consisting of a triplet T and a base set S , then the function h (cid:48) defined as h (cid:48) ( X ) = h ( X | S \ { a, b } ) (i.e., the marginal value with respect to S \ { a, b } ), defined over T ∪ { a, b } ,would also serve as a counterexample. (It is known that GS is closed under marginal value; i.e., if h is GS, then h (cid:48) is GS as well; this follows easily by Lemma 2.2.)Thus, without loss of generality we need to prove that h ( c | a ) + h ( de | a ) ≤ max { h ( d | a ) + h ( ce | a ) , h ( e | a ) + h ( cd | a ) } , or equivalently, h ( ca ) + h ( dea ) ≤ max { h ( da ) + h ( cea ) , h ( ea ) + h ( cda ) } . We say that a term of the form h ( aS ) is realized by a if g ( aS ) ≥ g ( bS ); otherwise, we say that itis realizes by b . Observe that if either a realizes both terms in the left hand side or b realizes bothterms, then the inequality follows by the triplet condition for g . Thus, it remains to show that theinequality holds when one of these terms is realizes by a and the other is realized by b . Withoutloss of generality assume that h ( ca ) = g ( ca ) and h ( dea ) = g ( deb ). We need to show that g ( ca ) + g ( deb ) ≤ max { h ( da ) + h ( cea ) , h ( ea ) + h ( cda ) } . (10)Recall that each one of the 4 terms of the form h ( Sa ) in the RHS of (10) is realized by either a or b . A counterexample to Inequality (10) takes the form of an instantiation of g , where an instan-tiation of g is a function that assigns a value g ( B ) to every non-empty subset B ⊆ { a, b, c, d, e } in such a way that g is GS (i.e., satisfies monotonicity, submodularity and the triplet condition),and g has the property that applying two-item max-symmetrization to the items a and b resultsin a function h that violates Inequality (10). A violation of Inequality (10) is encoded by 8 strictlinear inequalities, namely g ( ca ) + g ( deb ) > g ( d ∗ ) + g ( ce ∗ ), and g ( ca ) + g ( deb ) > g ( e ∗ ) + g ( cd ∗ ),for ∗ ∈ { a, b } . We refer to these inequalities as the 8 fixed constraints.29o show that no counterexample exists, it is sufficient to show that there does not exist aninstantiation of g that satisfies the triplet condition, and also the 8 fixed constraints.We actually prove a stronger result, but let us first introduce some additional notation. A combination is a tuple of a triplet from { a, b, c, d, e } and a base set from the remaining two items.Let ( ijk ; S ) denote the combination consisting of the triplet { i, j, k } and the base set S (if S is asingleton { (cid:96) } we write ( ijk ; (cid:96) ) for simplicity). For g to satisfy the triplet condition, it should satisfythe triplet inequality (1) for a total of 40 combinations ( (cid:0) (cid:1) = 10 possible triplets, and 4 possiblebase sets for every triplet). We show that even the more modest requirement to satisfy the tripletcondition with respect to only 6 (particular) combinations leads to a contradiction. In particular,we show the following: Claim C.9. There does not exist an instantiation of g that satisfies the 8 fixed constraints, and thetriplet condition with respect to the following 6 combinations: ( acd ; b ) , ( ace ; b ) , ( ade ; b ) , ( bce ; a ) , ( bde ; a ) , ( cde ; b ) . We prove this claim by a linear programming-based computer assisted code. The code isnow available in GitHub [27]. Claim C.9 implies Inequality (10), which concludes the proof ofClaim C.8. Readers who are interested in more details about the LP-based code are referred toSubsection C.1. C.1 The LP-Based Code In this section we give details regarding the code used to prove Claim C.9.We wish to encode the fact that g satisfies the triplet condition using linear inequalities. Foreach combination, there are 3 ways to choose the minimum out of the 3 terms (recall, the maximumshould not be unique). Every such choice can be encoded by one equality and one weak inequality.For example, consider the combination ( abd ; e ). For this combination, the minimum term iseither (i) g ( a | e ) + g ( bd | e ), (ii) g ( b | e ) + g ( ad | e ), or (iii) g ( d | e ) + g ( ab | e ). To encode that, say, g ( d | e ) + g ( ab | e ) is a minimum, we write: g ( a | e ) + g ( bd | e ) ≥ g ( d | e ) + g ( ab | e ) and g ( b | e ) + g ( ad | e ) = g ( a | e ) + g ( bd | e ) . A naive implementation would run 3 linear programs, each with 31 variables (one for eachnon-empty subset of { a, b, c, d, e } ) and 88 linear constraints (the 8 fixed constraints for the violationof Inequality (10), plus 80 constraints, 2 for each one of the 40 combinations).To overcome the associated computational challenge, our code gradually grows a ternary treeof height 40, where each node corresponds to a combination, the three edges from a given nodecorrespond to the 3 ways to choose the minimum term of that combination, and each combinationappears in all nodes in a given level of the tree. A node v of height h is associated with an LP with The same claim holds with respect to the following 6 combinations as well:( acd ; e ) , ( ace ; d ) , ( ade ; c ) , ( bcd ; e ) , ( bce ; d ) , ( cde ; a ). Recall that these constraints are strict inequalities. In the LP, we encoded a strict inequality of the form x > y by the weak inequality x ≥ y + 1. One can verify that this choice does not affect the feasibility of the LP. 30 + 2 h constraints, corresponding to the 8 fixed constraints and the 2 h constraints correspondingto the path from v to the root. A feasible solution to the LP of a leaf node corresponds to aninstantiation of g that satisfies the triplet condition and violates Inequality (10).Thus, a necessary condition for violating the assertion of Claim C.8 is a leaf node whose cor-responding LP is feasible. Consequently, to prove the claim it suffices to prove that no such leafexists. We grow the tree from the root (using BSF/DSF), and once infeasibility is obtained for anode v , we trim off its subtree and backtrack. The advantage of growing the tree gradually is thatsome branches of the tree may be trimmed off early on, substantially reducing the number of LPsthat we need to solve. Note that the order of the 40 combinations may be crucial in determiningthe runtime of the program. Indeed, we observe high variability in the depth reached by differentorders.We ran two different implementations of the program, written by two different programmers,one in matlab, the other in C++. The full code is available in GitHub [27]. The program terminatedwithout finding any feasible leaf, concluding the proof of Claim C.8. The best order resulted ina search of depth 6 (i.e., 6 combinations). Interestingly, there are exactly two (out of (cid:0) (cid:1) > . { ( acd ; b ) , ( ace ; b ) , ( ade ; b ) , ( bce ; a ) , ( bde ; a ) , ( cde ; b ) } , (2) { ( acd ; e ) , ( ace ; d ) , ( ade ; c ) , ( bcd ; e ) , ( bce ; d ) , ( cde ; a ) } . D (Non)-Closedness of SWS Valuations Claim D.1. The class of SWS functions is not closed under average, neither is it closed underconvolution.Proof. To see that SWS is not closed under average, consider two budget additive functions f , f over 3 items a, b, c , where f gives value 2 to a, b and value 0 to c , and f gives value 2 to a, c andvalue 0 to b . Both functions have budget 2. f and f are SWS, but the average of them (whichgives value 2 to a and value 1 to b, c ) is not SWS, as b and c belong to the same symmetry class,but the marginal value of b with respect to c is greater than its marginal value with respect to a .We next show that SWS is not closed under convolution. The idea is to take a function f over aset of items M that has symmetries but is not SWS, then add an item j that breaks all symmetriesin f . The obtained function, call it f , is trivially SWS. Now let f be a function that gives item j a high value, and all other items 0 ( f is SWS), and consider the convolution over f , f ; callit g . Given a set S , g will assign j to f and all other items to f , so its value on subsets of M coincides with f , which is not SWS. For concreteness, let f be the budget additive function overitems a, b, c with respective values 2 , , f be a function over a, b, c, d , such that f ( S ) = f ( S ) for all S ⊆ { a, b, c } , and the marginal value of d with respect to all subsets is 0, except for its marginal value with respect to b , which is 1 (thus,31he symmetry between b, c breaks). Let f be an additive function with value 100 for d and 0 forall other items. E Proof of Lemma 4.7 Proof. (of Lemma 4.7) Color the sequence { , . . . , T } periodically by log m colors { , . . . , log m − } ,where t belongs to color class C i if t = i modulo log m . Decompose H into log m collections H i (for 0 ≤ i < log m ), where collection H i has those functions h t with t ∈ C i . Let f i be the functionsatisfying f i ( S ) = (cid:80) t ∈ C i h t ( S ). Then we have the following sandwich property for every set S : α · max ≤ i< log m f i ( S ) ≤ f ( S ) ≤ β · (cid:88) ≤ i< log m f i ( S )By monotonicity and concavity of g , we also have: α · max ≤ i< log m g ( f i ( S )) ≤ g ( f ( S )) ≤ β · (cid:88) ≤ i< log m g ( f i ( S ))We shall need the following claim: Claim E.1. For every i , the function g ( f i ) can be approximated by a GS function h i within aconstant factor. Claim E.1 implies that for some constant c and GS functions h i , we have that: α max ≤ i< log m h i ( S ) ≤ g ( f ( S )) ≤ cβ (cid:88) ≤ i< log m h i ( S )Using the above sandwich property, Lemma 4.7 is an immediate corollary of Lemma 4.5.It remains to prove Claim E.1.Consider the function f i = (cid:80) t ∈ C i h t . (The functions h t are as defined in Lemma 4.7, not tobe confused with the functions h i of Claim E.1.) Observe that for every set S ⊂ M and for every t, t (cid:48) ∈ C i , if t < t (cid:48) and h t (cid:48) ( S ) > 0, then h t ( S ) ≤ m t ≤ t (cid:48) ≤ h t (cid:48) ( S ).Let τ : M → ( N ∪⊥ ) be a function that maps every item x ∈ M to the highest value of t ∈ C i forwhich x is in the support of h t , and to ⊥ if there is no such t . Let τ − ( t ) = { x | x ∈ M , τ ( x ) = t } .For every t ∈ C i , let h (cid:48) t be the function satisfying h (cid:48) t ( S ) = h t ( S ∩ τ − ( t )) for every S ⊂ M . Thenevery item x ∈ M is in the support of at most one of the functions h (cid:48) t .Note that max t ∈ C i h t ( S ) = max t ∈ C i h (cid:48) t ( S ), because max t ∈ C i h t ( S ) is attained at a t ∗ that satisfies t ∗ = max x ∈ S τ ( x ), and h (cid:48) t ∗ ( S ) = h t ∗ ( S ). Moreover, (cid:80) t ∈ C i h (cid:48) t ( S ) ≤ f i ( S ) ≤ (cid:80) t ∈ C i h (cid:48) t ( S ) holdsfor every S , where the right inequality holds because the combined marginal values of items of S dropped from all functions (with t < t ∗ ) does not exceed h t ∗ ( S ) = h (cid:48) t ∗ ( S ).For every t we use (cid:96) t to denote g (2 t ) rounded down to the nearest power of 2, and u t todenote g (2 t m ) rounded down to the nearest power of 2. For S ⊂ M , if | S ∩ τ − ( t ) | ≥ 1, then32 t ≤ g ( h (cid:48) t ( S )) ≤ u t . For two consecutive members t, t (cid:48) ∈ C i (hence t (cid:48) ≥ t + log m ), we have that u t ≤ (cid:96) t (cid:48) .We now define a merging operation. For s > 2, we refer to a sequence t < . . . < t s of indexesin color class C i as mergeable if it satisfies (cid:96) t = (cid:96) t s . As long as a mergeable sequence exists, wepick a maximal mergeable sequence (that is not contained in any longer mergeable sequence) andmerge the functions h (cid:48) t , . . . , h (cid:48) t s − into one new function h (cid:48) t ,...,t s − . The items in the support of thisfunction are τ − ( t , . . . , t s − ) = (cid:83) j ≤ s − τ − ( t j ). We define h (cid:48) t ,...,t s − ( S ) = 2 t for sets that containan item from τ − ( t , . . . , t s − ), and 0 otherwise. Observe that for every set S ⊂ M : g ( h (cid:48) t ,...,t s − ( S )) ≤ g ( (cid:88) j ≤ s − h (cid:48) t j ( S )) ≤ g ( h (cid:48) t ,...,t s − ( S ))The right inequality follows because u t s − ≤ (cid:96) t s = (cid:96) t . We consequently have that (cid:96) t ,...,t s − = u t ,...,t s − = g (2 t ).Observe that the function g ( h (cid:48) t ,...,t s − ) is a GS function (as h (cid:48) t ,...,t s − has value either 0 or 2 t ).Likewise, for every unmerged function h (cid:48) t , Lemma 4.8 implies that the corresponding function g ( h (cid:48) t )is a GS function, as h (cid:48) t is an MRF (scaled by 2 t ).Given a color class C i , after performing all applicable merge operations (and without changingthe order of the functions), rename the functions that remain in C i as f i , f i , . . . . Observe thathaving performed the merge operations, for every j ≥ u ij ≤ (cid:96) ij +1 < (cid:96) ij +3 (where (cid:96) ij and u ij are the natural renaming for the notation (cid:96) t and u t ).With each function f ij we associate the function h ij = g ( f ij ). As discussed above, every h ij function is GS. Moreover, we have the following sandwich property for every set S ⊂ M :max j h ij ( S ) ≤ g ( f i ( S )) ≤ (cid:88) j h ij ( S )(The factor 4 is a product of a factor 2 that is paid for switching from h t to h (cid:48) t , and a factor 2 thatis paid for the merging operation.)We claim that for every S it holds that (cid:80) j h ij ( S ) ≤ j h ij ( S ) (the constant 6 can beimproved). Recall that the supports of the function h ij are disjoint (for different j ), and let M j denote the support of function h ij . To prove the claim, let k be the largest index for which | S ∩ M k | ≥ 1. Then max j h ij ( S ) = h ik ( S ) ≥ (cid:96) ik . For every k (cid:48) < k we have that h ik (cid:48) ( S ) ≤ u ik (cid:48) ≤ (cid:96) ik . Moreover, forevery j , u ij ≤ u ij +3 . Hence the u ij values can be partitioned into three geometric series that eachcontributes at most 2 (cid:96) ik to (cid:80) j h ij ( S ), proving the claim.Let h iu denote the function satisfying h iu ( S ) = (cid:80) j h ij ( S ) for every S . Observe that h iu is a GSfunction, as it is a sum over GS functions that have disjoint supports. Following the above claimand using concavity of g , for every S ⊂ M we have that:16 h iu ( S ) ≤ g ( f i ( S )) ≤ h iu ( S )Taking h i = h iu proves Claim E.1. 33 Proof of Proposition 4.3 In this section we prove Proposition 4.3. Proof. We first show that h is monotone, i.e., that h ( a | S ) ≥ S and item a . Byadding a last in the sorted order of S , a non-negative value is added to h . Then, keep swapping a with its neighbor until it reaches its place (according to g ). By Claim 4.2, each such swap cannotdecrease h .We next show that h is submodular. It suffices to prove that for every set S and two items a, b , h ( b | S ) ≥ h ( b | Sa ). Adding an item j to a set T has two effects: (i) the value goes up by themarginal value of j w.r.t. its position within T , denote this position r , (ii) the value goes down bythe sum, over all items positioned r or higher, of the difference in the marginal values due to theshift in position. Let r and q be the respective positions of items a and b when added to S , and let k = | S | . We distinguish between two cases, as follows. Case 1: g ( a ) ≥ g ( b ). In this case b ’s position when added to Sa is q + 1. The respective marginalvalues of b with respect to S and Sa are h ( b | S ) = µ ( b, q ) − k (cid:88) i = q ( µ ( S i , i ) − µ ( S i , i + 1)) h ( b | Sa ) = µ ( b, q + 1) − k (cid:88) i = q ( µ ( S i , i + 1) − µ ( S i , i + 2)) . We need to show that h ( b | S ) − h ( b | Sa ) ≥ h ( b | S ) − h ( b | Sa ) = µ ( b, q ) − µ ( b, q + 1) − ( µ ( S q , q ) − µ ( S q , q + 1))+ k − (cid:88) i = q ( µ ( S i , i + 1) − µ ( S i , i + 2) − ( µ ( S i +1 , i + 1) − µ ( S i +1 , i + 2)))+ µ ( S k , k + 1) − µ ( S k , k + 2) . (11) The first two lines of (11) are non-negative by Claim 4.2, and the last line is non-negative by themonotonicity of T . Case 2: g ( a ) < g ( b ). In this case b ’s position when added to Sa is still q . The respective marginalvalues of b with respect to S and Sa are h ( b | S ) = µ ( b, q ) − k (cid:88) i = q ( µ ( S i , i ) − µ ( S i , i + 1)) h ( b | Sa ) = µ ( b, q ) − r − (cid:88) i = q ( µ ( S i , i ) − µ ( S i , i + 1)) − ( µ ( a, r ) − µ ( a, r + 1)) − k (cid:88) i = r ( µ ( S i , i + 1) − µ ( S i , i + 2)) . We need to show that h ( b | S ) − h ( b | Sa ) ≥ h ( b | S ) − h ( b | Sa ) = µ ( a, r ) − µ ( a, r + 1) − ( µ ( S r , r ) − µ ( S r , r + 1))+ k − (cid:88) i = r ( µ ( S i , i + 1) − µ ( S i , i + 2) − ( µ ( S i +1 , i + 1) − µ ( S i +1 , i + 2)))+ µ ( S k , k + 1) − µ ( S k , k + 2) . (12) T .It remains to show that h satisfies the triplet condition; that is, for every set S , and 3 items a, b, c , h ( a | S ) + h ( bc | S ) ≤ max { h ( b | S ) + h ( ac | S ) , h ( c | S ) + h ( ab | S ) } . Assume without loss of generality that g ( a ) ≥ g ( b ) ≥ g ( c ). We show that h ( a | S ) + h ( bc | S ) = h ( b | S ) + h ( ac | S ) ≥ h ( c | S ) + h ( ab | S ) . Let q (resp., t ) be the position of item b (resp., c ) when added to S , and let k = | S | .We first show that h ( a | S ) + h ( bc | S ) = h ( b | S ) + h ( ac | S ), or equivalently, that h ( c | Sb ) = h ( c | Sa ). Indeed, h ( c | Sb ) = h ( c | Sa ) = µ ( c, t ) + (cid:80) ki = t ( µ ( S i , i + 1) − µ ( S i , i + 2)).We next show that h ( a | S ) + h ( bc | S ) ≥ h ( c | S ) + h ( ab | S ), or equivalently, that h ( b | S ) − h ( b | Sa ) + h ( c | Sb ) − h ( c | S ) ≥ 0. After canceling out terms, we get h ( b | S ) − h ( b | Sa ) + h ( c | Sb ) − h ( c | S ) = µ ( b, q ) − µ ( b, q + 1) − ( µ ( S q , q ) − µ ( S q , q + 1)) + t − (cid:88) i = q ( µ ( S i , i + 1) − µ ( S i , i + 2) − ( µ ( S i +1 , i + 1) − µ ( S i +1 , i + 2))) + µ ( S t − , t ) − µ ( S t − , t + 1) − ( µ ( c, t ) − µ ( c, t + 1)) ≥ , where the last inequality follows since every line in this expression is non-negative by Claim 4.2;this concludes the proof of Proposition 4.3. G Approximation of Budget Additive Valuations by CoverageValuations Theorem G.1. Every BA function can be approximated by a coverage function within a ratio of ρ = ee − .Proof. Let f be a BA function over n items, where item i has value v i , and the budget is B . Wenow create a coverage function g . Let N be sufficiently large. We shall have N elements. Eachitem i covers v i NB elements chosen at random. Every element has weight ρBN . Hence for every item i we have g ( i ) = ρf ( i ). For every set we have g ( S ) ≤ ρf ( S ), because g is subadditive and upperbounded by ρB .It remains to show that g ( S ) ≥ f ( S ) for every S . This holds by definition for sets that containonly one item, and likewise, for every set that contains an item of value at least B . For theremaining sets, suppose first that f ( S ) < B . The expected value of g ( S ) (expectation over therandomness of the construction) is: 35 [ g ( S )] = ρBN N (1 − (cid:89) i ∈ S (1 − v i B )) ≥ ρB (1 − e − (cid:80) i ∈ S viB ) = ρB (1 − e − f ( S ) B )Denoting x = f ( S ) B we need to show that ρ (1 − e − x ) ≥ x in the range 0 ≤ x < 1. This indeedholds for ρ = ee − because we then have equality for x ∈ { , } , and the function 1 − e − x is concave.If f ( S ) ≥ B , the same argument as above works, by first reducing the value of one or more ofthe items in S such that after that f ( S ) = B even without the budget constraint (details omitted).Finally, we note that by taking N sufficiently large, g ( S ) becomes arbitrarily close to E [ g ( S )]simultaneously for all S . As a lower bound derived E [ g ( S )] has some slackness (the inequality inthe derivation is strict because S contains an item of positive value strictly less than B ), there is achoice of g that satisfies g ( S ) ≥ f ( S ) for every S .We complement Theorem G.1 by showing that a ratio better than e +1 e cannot be guaranteed,even by matroid rank sum (MRS) functions [28, 6] — sum over matroid rank functions (see Def-inition A.3) — which is a strict superclass of coverage functions (coverage is the class of matroidrank functions of rank 1). Closing the gap between e +1 e (cid:39) . 368 and ee − (cid:39) . 582 remains open. Proposition G.2. Consider the BA function f with budget B , a set X = { x , . . . , x n } of n items,each of value 1, and an additional item y of value B . For large n and B (e.g., B = √ n ), f cannotbe approximated by MRS within a ratio better than ρ = e +1 e .Proof. Let B (cid:48) ≥ B be the total weight of matroids containing y , and let X (cid:48) be the total weight thatmatroids that do not contain y can contribute (when all is X present). We require X (cid:48) + B (cid:48) ≤ ρB .Given an arbitrary solution, we can symmetrize it (average over all permutations) so that all x i are treated in exactly the same way.Assume that all matroids that contain y have rank 1. (The bounds that follow hold withoutchange even without this assumption, because the access above rank 1 can go into X (cid:48) ). Let w ≤ ρ be the weight of matroids that contain both y and x .Consider now a random set of B items. The fraction of uncovered weight of matroids thatcontain y is roughly B (cid:48) (1 − wB (cid:48) ) B . Hence X (cid:48) ≥ B − B (cid:48) (1 − wB (cid:48) ) B . For this X (cid:48) , we get two lowerbounds on ρ . Considering the grand bundle we have that Bρ ≥ B (cid:48) + X (cid:48) , whereas considering only x we have that ρ ≥ w + X (cid:48) B .If w = 1 and B (cid:48) = B , then X (cid:48) = Be and ρ = 1 + ee