The Complexity of Partial Function Extension for Coverage Functions

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as well as results on approximate extension [9]. Pitt and Valiant show a direct relation between the complexity of partial function extension problem and proper PAC-learning. Informally, a class of (boolean) functions on 2[m] is said to be properly PAC-learnable F [m] if for any distribution µ on 2 and any small enough ǫ > 0, any function f ∗ can be ∈ F learned by a polynomial-time algorithm that returns a function f with a polynomial ∈ F number of samples that differs from f ∗ with probability at most ǫ. Pitt and Valiant show that if partial function extension for a class of functions is NP-hard, then the class F F cannot be PAC-learned unless NP = RP [22].1 They show computational lower bounds for various classes of boolean functions, thereby obtaining lower bounds on the complexity for learning these classes. In this paper, we show lower bounds on partial function extension for coverage functions, which by this relation give lower bounds on proper PAC learning as well. In separate work, we present results on the computational complexity of partial function extension for submodular, subadditive, and convex functions, and show further connections with learning and property testing [5]. Characterizing partial functions extendible to convex functions is widely studied in convex analysis. Here a partial function is given defined on a non-convex set of points, and is required to be extended to a convex function on the convex hull or some other convex domain. Characterizations for extendible partial functions are given in various papers, such as [12, 29]. This finds many applications, including mechanism design [14], decision making under risk [21], and quantum computation [27]. Another example of the ubiquity of partial function extension is in property testing. Given oracle access to a function f, the goal of property testing is to determine by querying the oracle if the function f lies in some class of functions of interest, or is far from it, i.e., F differs from any function in at a large number of points. Partial function extension is a F natural step in property testing, since at any time the query algorithm has a partial function consisting of the points queried and the values at those points. If at any time the partial function thus obtained is not extendible to a function in , the algorithm should reject, and F should accept otherwise. Partial function extension is used to give both upper and lower bounds for property testing [5, 24]. Partial function extension is thus a basic problem that finds application in a wide variety of different fields.

Our Contribution

Our input is a partial function H = (T1,f1),..., (Tn,fn) with Ti [m] and fi 0, and the { } [m] ⊆ ≥ goal is to determine if there exists a coverage function f :2 R 0 such that f(Ti)= fi → ≥ for all i [n]. This is the Coverage Extension problem. Throughout the paper we use [m] ∈ for the ground set, n for the number of defined sets in the partial function, and for the D set of defined sets T1,...,Tn . We also use d = maxi [n] Ti to denote the maximum size { } ∈ | | of sets in , and F := f . D i [n] i Our first result shows that∈ Coverage Extension is NP-hard. Interestingly, we show if there P exists a coverage function extending the given partial function then there is an extension by a coverage function for which the size of the universe U is at most n. This shows that | | Coverage Extension is in NP. In contrast, it is known that minimal certificates for nonextendibility may be of exponential size [10]. Also, unlike property testing, this shows that Coverage Extension does not become easier when restricted to succinct coverage functions.

1 Randomized Polynomial (RP) is the class of problems for which a randomized algorithm runs in polynomial time, always answers correctly if the input is a ‘no’ instance, and answers correctly with probability at least 1/2 if the input is a ‘yes’ instance. U. Bhaskar and G. Kumar 3

◮ Theorem 1. Coverage Extension is NP-complete.

For the hardness, we show a reduction from fractional graph colouring, a problem studied in fractional graph theory. Our hardness for extension also shows the following result for proper learning of succinct coverage functions.

◮ Theorem 2. Unless RP = NP, the class of succinct coverage functions cannot be properly PAC-learned (i.e., cannot be PMAC-learned with approximation factor α =1).

These are the ﬁrst hardness results for learning coverage functions based on standard complexity assumptions. Earlier results showed a reduction from learning disjoint DNF formulas to learning coverage functions [13], however as far as we are aware, there are no known lower bounds for learning disjoint DNF formulas. The following theorem is shown in the appendix. Given the hardness result for Coverage Extension, we study approximation algorithms for two natural optimization versions of the extension problem. In both of these problems, the goal is to determine the distance between the given partial function and the class of coverage functions. Based on the notion of the distance, we study the following two problems. In Coverage Approximate Extension, the goal is to determine minimum value of α 1 [m] ≥ such that there exists a coverage function f :2 R 0 satisfying fi f(Ti) αfi for all → ≥ ≤ ≤ i [n]. ∈ In Coverage Norm Extension, the goal is to determine the minimum L1 distance from a coverage function, i.e., minimize ǫ where ǫ = f(T ) f for all i [n] for some i [n] | i| i i − i ∈ coverage function f. ∈ P The two notions of approximation we study thus roughly correspond to the two prevalent notions of learning real-valued functions. Coverage Approximate Extension corresponds to PMAC learning, where we look for point-wise multiplicative approximations. Coverage Norm Extension corresponds to minimizing the L1 distance in PAC learning. Throughout this paper, the minimum value of α in Coverage Approximate Extension will be denoted by α∗ and minimum value of ǫ in Norm Extension will be denoted i [n] | i| by OPT . As both of these problems are generalisations∈ of Coverage Extension, they are P NP-hard. We give upper and lower bounds for approximation for both of these problem.

◮ Theorem 3. There is a min d, m2/3 log d -approximation algorithm for Coverage Ap- { } proximate Extension. If d is a constant then there is a d-approximation algorithm. In Coverage Norm Extension, OPT = 0 iff the partial function is extendible and hence no multiplicative approximation is possible for OPT unless P = NP (because of Theorem 1). We hence consider additive approximations for Coverage Norm Extension. An algorithm for Coverage Norm Extension is called an α-approximation algorithm if for all instances (partial functions), the value β returned by the algorithm satisfies OPT β OPT + α. We show ≤ ≤ nearly tight upper and lower bounds on the hardness of approximation. As defined before

F = i [n] fi. Note that an F -approximation algorithm is trivial, since the function f =0 ∈ is coverage and satisfies i [n] f(Ti) fi F . P ∈ | − |≤ ◮ Theorem 4. There is aP(1 1/d)F -approximation algorithm for Coverage Norm Extension. − Moreover, a coverage function f can be efficiently computed such that i [n] f(Ti) fi ∈ | − |≤ OPT + (1 1/d)F . − P ◮ Theorem 5. It is NP-hard to approximate Coverage Norm Extension by a factor α = 2poly(n,m)F δ for any fixed 0 δ < 1. This holds even when d =2. ≤ 4 The Complexity of Partial Function Extension for Coverage Functions

Our lower bound is roughly based on the equivalence of validity and membership, where given a convex, compact set K, the validity problem is to determine the optimal value of cT x given a vector c over all x K, while the membership problem seeks to determine if a given ∈ point x is in K or not. The equivalence of optimization and separation is a widely used tool. The reduction from optimization to separation is particularly useful for, e.g., solving linear programs with exponential constraints. Our work is unusual in both the use of validity and membership rather than optimization and separation, and because of the direction — we use the equivalence to show hardness of the validity problem. We hope that our techniques may be useful in future work as well.

Related Work We focus here on work related to partial function extension and coverage functions. In a separate paper, we study partial function extension to submodular, subadditive, and convex functions, showing results on the complexity as well as applications to learning and property testing [5]. Previously, Seshadri and Vondrak [24] introduce the problem of extending partial functions to a submodular function, and note its usefulness in analyzing property testing algorithms. For submodular functions, partial function extension is also useful in optimization [26]. The problem of extending a partial function to a convex function is also studied in convex analysis [29, 12]. As mentioned earlier, both characterizing extendible partial functions, and the complexity of partial function extension has been studied for large classes of Boolean functions [9, 22]. Chakrabarty and Huang study property testing for coverage functions [10]. Here, the goal is to determine whether the input function (given by an oracle) is coverage or far from coverage by querying an oracle, where distance is measured by the number of points at which the function must be changed for it to be coverage. They show that succinct coverage functions can be reconstructed with a polynomial number of queries and hence can be efficiently tested. However, they conjecture that testing general coverage functions requires 2Ω(m) queries, and prove this lower bound under a different notion of distance. They present a particular characterization of coverage functions in terms of the W -transform that we use as well. There has also been interest in sketching and learning coverage functions. Badanidiyuru et al. [1] showed that coverage functions admit a (1 + ǫ)-sketch, i.e., given any coverage function, there exists a succinct coverage function (of size polynomial in m and 1/ǫ) that approximates the original function within (1 + ǫ) factor with high probability. Feldman and Kothari [13] gave a fully polynomial time algorithm for learning succinct coverage functions in the PMAC model if the distribution is uniform. However, if the distribution is unknown, they show learning coverage functions is as hard as learning polynomial size DNF formulas for which no efficient algorithm is known. Balkanski et al [3] study whether coverage functions can be optimized from samples. They consider a scenario where random samples (S ,f(S )) of an unknown coverage func- { i i } tion f are provided and ask if it is possible to optimize f under a cardinality constraint, i.e.,

solve maxS: S k f(S). They prove a negative result: no algorithm can achieve approxima- | |≤ | tion ratio better than 2Ω(√log m) with a polynomial number of sampled points.

2 Preliminaries

As earlier, for m Z , define [m] := 1, 2,...,m . A set function f over a ground set [m] is ∈ + { } a coverage function if there exists a universe U of elements with non-negative weights and U. Bhaskar and G. Kumar 5 m sets A , ..., A U, such that for all S [m],f(S) is the total weight of elements in 1 m ⊆ ⊆ j S Aj . A coverage function is succinct if U is at most a fixed polynomial in m. ∪ ∈ | | Chakrabarty and Huang [10] characterize coverage functions in terms of their W -transform, [m] [m] which we use as well. For a set function f :2 R 0, the W -transform w :2 R → ≥ \ ∅ → is defined as

[m] S T +1 S 2 , w(S)= ( 1)| ∩ | f(T ) . (1) ∀ ∈ \∅ − T :S T =[m] ∪X The set w(S) S 2[m] is called the set of W -coeﬃcients of f. We can also recover the { | ∈ \∅} function f from its W -coeﬃcients.

T [m], f(T )= w(S) . (2) ∀ ⊆ S [m]:S T = ⊆ X∩ 6 ∅

If f is a coverage function induced by the universe U and sets A1,...,Am, then the W -transform w(S) is precisely the weight of the set ( i S Ai) j S Aj , and is hence { ∩ ∈ \ ∪ 6∈ } non-negative. The converse is also true. The set S w(S) > 0 is the called the support of { | } the coverage function, and the elements are exactly the elements of the universe U.

[m] ◮ Theorem 6. [10] A set function f : 2 R 0 is a coverage function iff all of its → ≥ W -coefficients are non-negative. From Theorem 6, given a partial function H, there exists a coverage function f satisfying f(T )= f for all i [n] iff the following linear program is feasible, where the variables are i i ∈ the W -coefficients w(S) for all S 2[m] . ∈ \∅

Extension-P: w(S)= f i [n] , w(S) 0 S 2[m] . i ∀ ∈ ≥ ∀ ∈ \∅ S:S T = X∩ i6 ∅ All missing proofs are in the appendix.

3 Coverage Extension and PAC-Learning

Our first observation is that there is a polynomial-sized certificate of extendibility to a coverage function. This is obtained by observing that at a vertex of the feasible set in Extension-P, at most n of the variables are non-zero. It is interesting to compare this with Chakrabarty and Huang [10], who give an example to show that minimal certificates of nonextendibility may be of exponential size. ◮ Proposition 7. If a partial function is extendible to a coverage function, then it is also extendible to a coverage function with support size n. Hence, Coverage Extension is in ≤ NP. We show the NP-hardness of Coverage Extension by reduction from fractional chromatic number, defined as follows. Given a graph G = (V, E), a set I V is called an independent ⊆ set if no two vertices in I are adjacent. Let be the set of all independent sets. The I fractional chromatic number χ∗(G) of a graph G is the optimal value of the following linear program.

χ∗(G) := min x : x 1 v V (G), 0 x 1 I I I ≥ ∀ ∈ ≤ I ≤ ∀ ∈ I ( I I :v I ) X∈I ∈IX∈ 6 The Complexity of Partial Function Extension for Coverage Functions

Note that if x 0, 1 then the optimal value is just the chromatic number of the graph.2 I ∈{ }

◮ Theorem 8. [15] For graph G = (V, E), there exist nonnegative weights xI I on { } ∈I independent sets such that χ∗(G)= I xI and I :v I xI =1 v V . ∈I ∈I ∈ ∀ ∈

◮ Corollary 9. For graph G = (V, EP) and for anyP value of t such that χ∗(G) t V , ≤ ≤ | | there exist nonnegative weights zI I on independent sets such that I zI = t and { } ∈I ∈I I :v I zI =1 v V . ∈I ∈ ∀ ∈ P ◮PTheorem 10. [18] Given graph G = (V, E) and 1 k V , it is NP-hard to determine ≤ ≤ | | if χ∗(G) k. ≤ We now show the NP-hardness of Coverage Extension.

Proof of Theorem 1. Since membership in NP was shown earlier, we give the reduction from fractional chromatic number. The input is a graph G = (V, E) and 1 k V . ≤ ≤ | | We identify [n′] with the set of vertices V , and therefore E(G) i, j i, j [n′] , and ⊆ {{ }| ∈ } any set S [n′] can be viewed as a set of vertices. The partial function construction is as ⊆ follows. The ground set is [n′] and therefore m = n′. The partial function is deﬁned at all vertices, all edges, and the set consisting of all vertices. Hence = i i [n′] E(G) D {{ }| ∈ } ∪ ∪ [n′] and = n′ + E(G) + 1. The value of the partial function h at these deﬁned sets { } |D| | | is given by

1 if S = i , i [n′] , { } ∈ h(S)= 2 if S E(G) ,  ∈ k if S = [n′] . { }  We claim that χ∗(G) k iff the above partial function is extendible. Suppose χ∗(G) k. ≤ ≤ Therefore by Corollary 9, there exist nonnegative weights xI I such that xI = k { } ∈I I and x = 1 v V (G). For all S 2[m] , define the function w(S∈I) as x if I :v I I ∀ ∈ ∈ \∅ P S S and∈I 0∈ otherwise. Since w(S) 0, this defines the W -transform for a coverage function ∈ IP ≥ g. We have, for any i [n′], ∈

g( i )= w(S)= x =1 , { } I S:S i = I :i I X∩{ }6 ∅ ∈IX∈ for any i, j E(G), { } ∈

g( i, j )= w(S)= x + x =2 { } I I S:S i,j = I :i I I :j I ∩{X }6 ∅ ∈IX∈ ∈IX∈

as no independent set I can contain both i and j; and ﬁnally g( [n′] )= w(S)= { } S:S [n′] = x = k. Therefore g is an extension of the above partial function h. ∩{ }6 ∅ I I P ∈INow suppose there is an extension, i.e., there exists w(S) 0 for all S 2[m] such P ≥ ∈ \∅ that for any i [n′], S:S i = w(S) = 1; for any i, j E(G), S:S i,j = w(S) = 2; ∈ ∩{ }6 ∅ { } ∈ ∩{ }6 ∅ and ﬁnally S:S [n ] = w(S)= k. For any i, j E(G), we have ∩{ ′ }6P∅ { } ∈ P P

2 The chromatic number of a graph is the minimum number of colours required to colour the vertices so that no two adjacent vertices get the same colour. U. Bhaskar and G. Kumar 7

w(S)= w(S)+ w(S) w(S) . − S:S i,j = S:S i = S:S j = S:S i,j ∩{X }6 ∅ X∩{ }6 ∅ X∩{ }6 ∅ X⊇{ }

Therefore, S:S i,j w(S) = 0, i.e., if w(S) > 0 then S must be an independent set. It now ⊇{ } ◭ follows that χ∗(G) S:S [n ] = w(S)= k. P ≤ ∩{ ′ }6 ∅ P Proper PAC-learning of Coverage functions

We now prove Theorem 2. We ﬁrst recall the deﬁnition of PAC-learning.

◮ Deﬁnition 11 ([2]). An algorithm properly PAC-learns a family of functions , if for [m] A F any distribution µ (on 2 ) and any target function f ∗ , and for any suﬃciently small ∈F ǫ,δ > 0:

1. takes the sequence (Si,f ∗(Si)) 1 i l as input where l is poly(m, 1/δ, 1/ǫ) and the A { } ≤ ≤ sequence Si 1 i l is drawn i.i.d. from the distribution µ, { } ≤ ≤ 2. runs in poly(m, 1/δ, 1/ǫ) time, and A 3. returns a function f :2[m] R such that A → ∈F

P rS ,...,S µ P rS µ[f(S)= f ∗(S)] 1 ǫ 1 δ 1 l∼ ∼ ≥ − ≥ − We use the reconstruction algorithm for coverage functions given by Chakrabarty and Huang [10] in our proof. Given a coverage function f as an input, this reconstruction algorithm terminates in O(ms) steps where s is the support size of f, i.e., the number of non-zero W -coefficients of f, and returns these non-zero W -coefficients. Recall the reduction from fractional chromatic number to Coverage Extension (Theorem 1). Given an instance of fractional chromatic number (graph G = (V, E) and rational k′ with V = n′), the instance of Coverage Extension is a set of defined points = i i | | D {{ }| ∈ [n′] E(G) [n′] and a function h on . Let k = = V + E +1. From Theorem 1 } ∪ ∪{ } D |D| | | | | and Proposition 7, χ∗(G) k′ iff h is extendible to a coverage function with support size at ≤ most k. Let be a family of coverage functions with support size at most k. Let ǫ =1/k3 (and F hence ǫ < 1/ ) and µ be a uniform distribution over (S,h(S)) S . Now suppose a |D| { | ∈ D} (randomized) algorithm A properly PAC-learns . We will show that in this case, we can F determine efficiently if the partial function is extendible to a coverage function, and hence RP = NP. Suppose the algorithm A returns a function g. If the partial function is extendible then there exists a function in that has the same value on samples seen by A. Therefore, if F the partial function is extendible then g(S) must be equal to h(S) for all S (since ∈ D ǫ < 1/ and A must satisfyP rS D [f(S) = f ∗(S)] 1 ǫ). We run the reconstruction |D| ∼ ∗ ≥ − algorithm on input g. If the partial function is extendible then g must be in and hence F the reconstruction algorithm must terminate in O(mk) steps. Further, if w(S) S is the { } ∈S output of the algorithm then (i) w(S) > 0 for all S , (ii) k (iii) the coverage function ∈ S |S| ≤ f ′ given by the W -coefficients w′(S)= w(S) if S and 0 otherwise is an extension of the ∈ S partial function h. Condition (iii) should hold because f ′ must be the same as g which we have shown earlier is an extension of h. The converse is also true — if the reconstruction algorithm terminates and (i), (ii), (iii) hold then clearly h is extendible (by f ′). Since all the steps require polynomial time to check, we can efficiently determine if the partial function is extendible. 8 The Complexity of Partial Function Extension for Coverage Functions

4 Coverage Approximate Extension

We now build the framework for Theorem 3. We start with the following lemma.

◮ Lemma 12. Given a partial function H and α 1, there is no coverage function f ≥ satisfying f f(T ) αf for all i [n] iﬀ the following program, with variables l for all i ≤ i ≤ i ∈ i i [n] is feasible: ∈ α f l < f l (3) − i i i i i:Xli<0 i:Xli>0

l 0 S [m] (4) i ≤ ∀ ⊆ i:S T = X∩ i6 ∅

Thus the optimal approximation ratio α∗ is the minimum value of α for which (3) and (4) are not feasible together. A natural representation of the partial function H = (T ,f ),..., (T ,f ) is as a { 1 1 n n } weighted bipartite graph H = (A [m], E) with A = n, and an edge between a A and ∪ | | i ∈ j [m] if the set T contains element j [m]. Each vertex a A also has weight f . Then ∈ i ∈ i ∈ i d = max T is the maximum degree of any vertex in A. For the remainder of this section, i | i| we will use this representation of partial functions. We use the following notation given a bipartite graph H = (A [m], E). For any S [m], ∪ ⊆ let N(S) = v A : (v, j) E for some j S be the set of neighbours of set S. { ∈ ∈ ∈ } Similarly for set R A, N(R) = j [m] : (v, j) E for some v R be the set of ⊆ { ∈ ∈ ∈ } neighbours of set R. For any vertex v in H, we use N(v) for N( v ). In this bipartite graph { } representation, the inequality (4) is equivalent to l 0 for all S [m]. i N(S) i ≤ ⊆ We now deﬁne a parameter κ called the replacement∈ ratio for a partial function H. P ◮ Deﬁnition 13. Let H = (A [m], E) be a bipartite graph with weights f on each v A. ∪ v ∈ For v A, let = R A v N(R) N(v) be the set of all subsets of A v that ∈ Fv { ⊆ \{ } | ⊇ } \{ } cover all the neighbours of v. We call each R v a replacement for v. The replacement fw ∈ F w R ratio κ is then the minimum of ∈ over all vertices v A and replacements R v. fv ∈ ∈F P The proof of the upper bound in Theorem 3 will follow from the bounds on α∗ shown in Lemma 14, 16 and 18.

1 ◮ Lemma 14. For any partial function H, α∗ . ≥ κ Proof. By definition of κ, there exists a vertex v A and a replacement R for v such that ∈ w R fw = cfv. Note that setting lw = 1 w R, lv = 1 and all other lw’s to be ∈ − ∀ ∈ [m] zero results in feasibility of the inequalities w N(S) lw 0 for all S 2 . From the P ∈ ≤ ∈ \∅ ◭ definition of α∗ and Lemma 12, α∗ w R fw fv, and hence α∗ 1/κ. ∈ P≥ ≥ min d,m2/3 P Let β = { κ } . Given values lv v A on the vertices in A such that v N(S) lv 0 { } ∈ ∈ ≤ for all S [m], we will show that β fvlv fvlv and hence α∗ β. If lv =0 ⊆ v:lv <0 ≥ v:lv >0 P≤ for any vertex, we simply ignore such a vertex, since it does not affect either (4) or (3). P P By scaling, we can assume that l Z for all v A. At some point, we will use Hall’s v ∈ ∈ theorem to show a perfect matching. To simplify exposition, we replace each v A with l ∈ | v| identical copies, each of which is adjacent to the same vertices as v. Each such new vertex v′ has l = 1 if l > 0 and l = 1 if l < 0, and f = f . Let the new bipartite graph be v′ v v′ − v v′ v H′ = (A′ [m], E′). It is easy to check that in the new bipartite graph, the degree of vertices ∪ U. Bhaskar and G. Kumar 9

in A′ and the values κ, f l , f l and l remain unchanged v A′:lv >0 v v v A′:lv<0 v v v N(S) v for all S [m]. ∈ ∈ ∈ ⊆ P P P Let = v A′ l = 1 and = v A′ l =1 , and let E− be the set of edges with N { ∈ | v − } P { ∈ | v } one end-point in , while E+ are the edges with one end-point in . For any S [m], let N + + P ⊆ N −(S)= N(S) and N (S)= N(S) (so N(S)= N (S) N −(S)). Finally, deﬁne + ∩ N ∩P ∪ E (S) (E−(S)) as the set of edges with one end-point in S and the other end-point in + + P ( ). If S = j , we abuse notation slightly and use N −(j), N (j), E−(j) and E (j). Note P { } + that N −(S) N (S) for all S [m] in H′, since in H, v N(S) lv 0 for all S [m]. | | ≥ | | ⊆ ∈ ≤ ⊆ Our goal is to show β v fv v fv. ∈N ≥ ∈P P + ◮ Lemma 15. SupposeP for some βP′ 1, β′ N −(S) N (j) for all S [m]. Then ≥ | |≥ j S | | ⊆ for each vertex v , there exists a replacement F ∈ such that each vertex in is ∈ P v P⊆ N N contained in Fv for at most β′ vertices v . Hence, β′ v fv κ v fv and so β′ ∈ P ∈N ≥ ∈P α∗ . ≤ κ P P

Proof. By Hall’s theorem, there exists a set of edges M E− such that (i) the degree in ⊆ M of each vertex j [m] is at least N +(j) , and (ii) the degree in M of each vertex v ∈ | | ∈+ N is at most β′. Because of (i), for each j [m] there is an injection hj from edges in E (j) ∈ + to edges in E−(j) M, i.e., each edge in E (j) maps to a distinct edge in E−(j) M. ∩ ∩ Now for a vertex v , consider a neighbouring vertex j N(v). Each such edge (v, j) + ∈ P ∈ is in E (j), and is hence mapped by h to an edge in E−(j) M. Let F be the end- j ∩ v points in of these mapped edges. That is, w F iﬀ there exists j N(v) such that N ∈ v ∈ (w, j) = hj (v, j). Then Fv is a replacement for v, and hence, fw κfv. Further, w Fv ≥ because of (ii), and since each h is an injection, each vertex in appears∈ in F for at most j NP v β vertices v . Then summing the inequality f κf over all v , we get ′ w Fv w v ∈ P ∈ ≥ ∈ P ◭ that β′ v fv κ v fv as required. ∈N ≥ ∈P P d ◮ LemmaP 16. For anyP partial function H, α∗ . ≤ κ + + Proof. Fix S [m]. Since N −(j) N (j) for all j [m], j S N −(j) j S N (j) , ⊆ | | ≥ | | ∈ ∈ | |≥ ∈ | | and since d is the maximum degree of any vertex in A′, d N −(S) N −(j) . The |P | ≥ j S P| | proof follows from Lemma 15. ∈ ◭ P 2/3 + m2/3 If we can show m N −(S) j S N (j) for all S [m] then by Lemma 15, α∗ κ . | |≥ ∈ | | ⊆ ≤ Unfortunately this may not be true. Let = v1 , = v2 , E− = (v1, j) j [m] and + P N + { } P { } { | ∈ } E = (v , j) j [m] . Note that N (j) = m whereas N −([m]) = 1. Notice that { 2 | ∈ } j [m] | | | | in this bad example, the bipartite graph∈ contains a 4-cycle v , j , v , j , v where v P 1 1 2 2 1 1 ∈ N and v . We now deﬁne a subgraph called a diamond which generalises such a 4-cycle. 2 ∈ P A diamond (v , v , J) of size k is a subgraph of H′ where v , v , J [m] ( J = k) p n p ∈P n ∈ N ⊆ | | such that for all j J, both (v , j) and (v , j) are contained in E′. Note that a 4-cycle is ∈ p n a diamond of size two (and the bad example considered above is a diamond of size m). Let a k = m (0 a 1) be the maximum size of any diamond in H′. max ≤ ≤ 1+a + a ◮ Lemma 17. For all S [m], m 2 N −(S) N (j) , where m is the size of the ⊆ | |≥ j S | | largest diamond in H . ∈ ′ P + Proof. Recall that for all j [m], N (j) N −(j) , hence there is an injection hj from + ∈ | | ≤ | + | N (j) to N −(j), i.e, hj maps each vertex in N (j) to a unique vertex in N −(j). Fix S [m] and vertex v , and let Sv := N(v) S be the neighbourhood of v in S. We ⊆ + ∈ P ∩ will consider N (Sv) and N −(Sv), the negative and positive neighbourhoods of Sv. Note that since all vertices in S are adjacent to v , a vertex in N −(S ) is adjacent to at v ∈ P v 10 The Complexity of Partial Function Extension for Coverage Functions

a most m vertices in Sv, by deﬁnition of a. Thus for a vertex v′ N −(Sv), there are at a ∈ + most m diﬀerent vertices j S for which h maps a vertex in N (j) to v′, and hence ∈ v j ma N (S ) N +(j) . − v j Sv | |≥ ∈ | | 1 a − + + Now if there is a vertex v such that m 2 N (j) N (j) then we P ∈P j Sv | | ≥ j S | | are done, since ∈ ∈ P P + + j S N (j) j S dj ∈ v | | ∈ N −(S) N −(Sv) a 1+a . | | ≥ | |≥ P m ≥ Pm 2 N +(j) + j S | | So assume that for all v , N (j) ∈ 1 a . In this case, note that by j Sv ∈ P | | ≤ m −2 ∈ P reversing the order of summation,P + + 2 + + + j S N (j) N (j) = N (j) = N (j) N (S) ∈ |1 a | . −2 | | + | | + | | ≤ | | m j S j S v N (j) v N (S) j Sv P X∈ X∈ ∈X ∈X X∈ Therefore, using the above inequality for N +(S) , | | 2 1 a + + 2 − + 1 a N (j) 2 j S N (j) N (j) + − j S m j S N −(S) N (S) m 2 ∈ | | ∈ | ∈ | | + + 1+a | | ≥ | |≥ j S N (j) ≥ S Pj S N (j) ≥ m 2 P ∈ | | | | ∈ | | P as required by the lemma. TheP third inequality follows fromP Cauchy-Schwarz. ◭

m2/3 From Lemmas 15 and 17, if a 1/3 then α∗ . Next we show this is true in ≤ ≤ κ general.

m2/3 ◮ Lemma 18. For any partial function H, α∗ . ≤ κ 1/3 m2/3 1/3 Proof. If k m then by Lemma 17 and 15, α∗ . So we assume k >m . max ≤ ≤ κ max In this case, we pick a diamond (v , v , J) of size > m1/3. We remove, for all j J, the p n ∈ edges (vp, j) and (vn, j). We repeat the above procedure (in the new graph) until we are left with a bipartite graph where all diamonds are of size at most m1/3. Note that if a diamond (vp, vn, J) of size k is removed then the degree of vn decreases by k. Hence, for a 2/3 ﬁxed vertex vn, number of removed diamonds is at most m (as at any step we remove 1/3 + diamonds of size at least m ). It is easy to see that after every step, N −(S) N (S) [m] | | ≥ | | (for all S 2 ) still holds in the bipartite graph. Let H∗ be the bipartite graph at the ∈ \∅ end (all diamonds of size at most m1/3). Note that we do not remove any vertex in the above procedure. Fix vertex v . By ∈ P Lemmas 17 and 15 with a = 1/3, there exists Fv such that Fv covers all neighbours ⊆ N 2/3 of v in H∗ and each vertex in appears in Fv for at most m vertices v . Since we N ∈1 P s have removed edges, F may not cover all the neighbours of v in H′. Let v ,...,v v ∈ N be the set of all vertices such that for each i [s], a diamond (v, vi, J i) was removed in a 1 s ∈ removal step. Clearly v ,...,v Fv cover all the neighbour of v in H′. Therefore, we have s { }∪ i 2/3 f i + f κf . Since any v (1 i s) is a part of at most m removed i=1 v w Fv w v ∈ ≥ ≤ ≤ 2/3 diamonds and each vertex in appears in Fv for at most m vertices v , summing P P N 2/3 ∈ P ◭ the above inequality for each v , we get m v fv κ v fv as required. ∈P ∈N ≥ ∈P P P min d,m2/3 It follows from Lemmas 14, 16 and 18 that an algorithm that returns { κ } is a min d, m2/3 -approximation algorithm. However, computing κ corresponds to solving a { } general set cover instance, and is NP-hard. This connection however allows us to show the following result. U. Bhaskar and G. Kumar 11

◮ Lemma 19. Given a partial function, the replacement ratio κ can be eﬃciently approx- imated by κ′ such that κ κ′ κ log d. If d is a constant, the replacement ratio κ can be ≤ ≤ determined eﬃciently. This completes the proof of the upper bound in Theorem 3. We now show that the analysis of our algorithm cannot be substantially improved.

◮ Theorem 20. There is a partial function with α∗ = 1/κ and a partial function with √m d = √m and α∗ = Ω( κ log m ). Proof. Fix r 1, and consider the partial function H = ( 1 , r), ( 2 , r), ( 1, 2 , 1) over ≥ { { } { } { } } ground set 1, 2 . Clearly κ =1/r, and since any coverage function must be monotone (2), { } α∗ = r. √m Now we show existence of a partial function with d = √m,κ = 1 and α∗ Ω( ). ≥ log m Let the ground set be [m] and deﬁned sets be = . Let = 1,..., √m , √m + D P ∪ N P {{ } { 1,..., 2√m ,..., m √m +1,m . Thus = √m and each set in has size √m. We } { − }} |P| P set = k√m log m for some large constant k. Each set in is constructed by randomly |N | N picking exactly one element from the sets 1,..., √m , √m +1,..., 2√m ,..., m √m + { } { } { − 1,m . Thus the size of each set in is also √m. The value of the partial function at each } N set in is set to 1, and at each set in is √m. N P Recall the bipartite graph description of partial functions. In the appendix, we show that P r [ N(S) N(S) S [m]] > 0 . (5) | ∩N|≥| ∩P| ∀ ⊆ + Thus N −(S) N (S) for all S [m]. It can be seen that κ = 1, since for any v | | ≥ | | ⊆ ∈ P every vertex in N(v) has a set in containing it, while for v , all sets in are required N ∈ N P fw w R to cover it. Thus, the minimum value of ∈ over all R v is 1 for v , while fv ∈ F ∈ P P √m for any v , the minimum value is m. Now we show α∗ . Let l = 1 for all ∈ N ≥ k log m v v and lv = 1 for all v . So v N(S) lv 0 for all S. Since l >0 fvlv = m and ∈ P − ∈ N ∈ ≤ v √m fvlv = k√m log m, we have α∗ . ◭ lv<0 − P≥ k log m P P◮ Remark 21. In the above lemma, the bipartite graph shown to exist does not have 4-cycle. For such partial functions, we have α∗ √m/κ (Lemma 15 and 17). Therefore, Theorem 20 min d,√≤m shows that the inequality α∗ { } cannot be improved by more than a log m factor ≤ κ (for such partial functions). It is an interesting open problem to close the gap between our upper bound of m2/3/κ and lower bound of √m/κ log m for general partial functions.

5 Coverage Norm Extension

From Theorem 6, the Norm Extension problem can be stated as the convex program Norm-P. It can be equivalently transformed to a linear program whose dual is Norm-D. n Norm-P: min n ǫ Norm-D: max f y i=1 | i| i i i=1 P X w(S)= fi + ǫi i [n] ∀ ∈ [m] S:S T = yi 0 S 2 (6) X∩ i6 ∅ ≤ ∀ ∈ \∅ i:S T = X∩ i6 ∅ w(S) 0 S 2[m] ≥ ∀ ∈ \∅ 1 yi 1 i [n] (7) Both Norm-P and Norm-D are clearly feasible.− We≤ use≤OPT∀for∈ the optimal value of Norm-P (and Norm-D). As stated earlier, no multiplicative approximation is possible for OPT unless P = NP. Therefore, we consider additive approximations for Norm Extension. 12 The Complexity of Partial Function Extension for Coverage Functions

An algorithm for Norm Extension is called an α-approximation algorithm if for all instances (partial functions), the value β returned by the algorithm satisfies OPT β ≤ ≤ OPT + α. First we prove our upper bound in Theorem 4. Recall that d = maxi [n] Ti ∈ | | and F = i [n] fi. As noted earlier, the function f( ) = 0 is trivially an F -approximation ∈ · algorithm for Norm Extension, since i [n] f(Ti) fi = F . P ∈ | − | Proof of Theorem 4. Consider the linearP programs obtained by restricting Norm-P to variables w(S) for S [m], and similarly restricting the constraints (6) in Norm-D to sets ∈ S [m] only. They are clearly the primal and dual of each other. The optimal values of ∈ these modified problems (say OPT R, wR and yR) can be computed in polynomial time. We will show that OPT OPT R OPT + (1 1/d)F for the proof of the theorem. The first ≤ ≤ − inequality is obvious, since OPT R is the optimal solution to a relaxed (dual) linear program. For the second inequality, define yA = (yA,...,yA) as the vector such that for all i [n], 1 n ∈ yA = yR if yR 0 and yR/d otherwise. Then note that i i i ≤ i

OPT R = f yR = f yA + (1 1/d) f yR f yA + (1 1/d)F , (8) i i i i − i i ≤ i i − i [n] i [n] i:yR 0 i [n] X∈ X∈ Xi ≥ X∈

R A where the last inequality is because each yi 1. We now show that y is a feasible solution A ≤ for Norm-D, and hence i [n] fiyi OPT . Together with (8) this completes the proof. ∈ ≤ Clearly yA satisfies the constraints (7). We will show that yA also satisfies the constraints P (6) for all S 2[m] . Consider any S 2[m] . Let P = i [n] S T = and yR > 0 ∈ \∅ ∈ \∅ { ∈ | ∩ i 6 ∅ i } and N = i [n] S T = and yR 0 . Thus P N are all sets in that have { ∈ | ∩ i 6 ∅ i ≤ } ∪ D nonempty intersection with S. We have for any j S that yR 0. Summing these i:j Ti i ∈ R ∈ ≤R R inequalities over j S, we obtain i P N Ti S yi 0. Thus i P yi + d i N yi 0. ∈ A ∈ ∪ | A∩ | ≤ P ∈ ∈ ≤◭ From the definition of yi , we get i:S T = yi 0, as required. P ∩ i6 ∅ ≤ P P We now prove the lower boundP in Theorem 4. We start with an outline of the proof. In a nutshell, the proof shows the following reductions (for brevity, WM stands for Weak Membership and WV for Weak Validity):

Densest-Cut Cut WM Span WM Coverage WM Coverage WV Norm Extension . ≤p ≤p ≡ ≤p ≤p Given a graph G = (V, E) and a positive rational M, the Densest-Cut problem asks if δ(S) there is a cut S V such that | | >M. The Densest-Cut problem is known to be NP- ⊂ S V S hard [7], and ultimately we reduce| || the\ Densest-Cut| problem to the problem of approximating the optimal value for Norm-P. We formally define the other problems later. However, to show this reduction, we need to utilize the equivalence of optimization (or validity) over a polytope and membership in the polytope. Typically optimization algorithms use the equivalence of optimization and separation to show upper bounds, e.g., that a linear program with an exponential number of constraints can be optimized. Our work is unique in that we use the less-utilized equivalence of validity and membership; and secondly, we use it to show hardness. In fact, since we are looking for hardness of approximation algorithms, our work is complicated further by the need to use weak versions of this equivalence. Given a convex and compact set K and a vector c, the Strong Validity problem, given a vector c, is to find the maximum value of cT x such that x K (the x which obtains this ∈ maximum is not required). In the Strong Membership problem, the goal is to determine if a given vector y is in K or not. The Weak Validity and Weak Membership problems U. Bhaskar and G. Kumar 13 are weaker versions of the Strong Validity and Strong Membership problems respectively, formally defined later. Then Theorem 4.4.4 in [16] says that for a convex and compact body K, there is an oracle polynomial time reduction from the Weak Membership problem for K to the Weak Validity problem for K. To formally state Theorem 4.4.4 from [16], which will form the basis of our reduction, we need the following notations and definitions. We use . for the Euclidean norm. Let K Rn′ be a convex and compact set. A ball || || ⊆ of radius ǫ> 0 around K is defined as

S(K,ǫ) := x Rn′ x y ǫ for some y in K . { ∈ | || − || ≤ } Thus, for x Rn′ , S(x, ǫ) is the ball of radius ǫ around x. The interior ǫ-ball of K is deﬁned ∈ as

S(K, ǫ) := x K S(x, ǫ) K − { ∈ | ⊆ } Thus S(K, ǫ) can be seen as points deep inside K. − ◮ Deﬁnition 22 ([16]). Given a vector c Qn′ , a rational number γ and a rational number ∈ ǫ> 0, the Weak Validity problem is to assert either (1) cT x γ + ǫ for all x S(K, ǫ), or ≤ ∈ − (2) cT x γ ǫ for some x S(K,ǫ). Note that the vector x satisfying the second inequality ≥ − ∈ is not required.

◮ Deﬁnition 23 ([16]). Given a vector y Rn′ and δ > 0, the Weak Membership problem ∈ is to assert either (1) y S(K,δ), or (2) y S(K, δ). ∈ 6∈ − Intuitively, in the Weak Membership problem, it is required to distinguish between the cases when the given point y is far from the polyhedron K (in which case, the algorithm should return y S(K, δ)) and y is deep inside K (which case the algorithm should return 6∈ − y S(K,δ)). If y is near the boundary of K, then either output can be returned. Our ∈ reduction crucially uses the following result. The notation , <δ > denotes the number of bits used to represent K and δ.

◮ Theorem 24 (Theorem 4.4.4 of [16]). Given a weak validity oracle for K Rn′ that runs ⊆ in time polynomial in (n′, K , c ) and a positive R such that K S(0, R) then the Weak h i h i ⊆ Membership problem for the polyhedron K can be solved in time polynomial in input size (n′, K , δ ). h i h i For our problem K is the polytope of linear program Norm-D.

n K := y R : yi 0 S [m], y 1 . (9)  ∈ ≤ ∀ ⊆ || ||∞ ≤  i:S T =  X∩ i6 ∅ 

Coverage WM Coverage WV Coverage Norm Extension. ≤p ≤p Coverage Weak Membership is the Weak Membership problem for polytope K (9). Given a set = T ,...,T (where T [m]) with weightsy ˆ (ˆy R) associated with T for all D { 1 n} i ⊆ i i ∈ i i [n] and a δ > 0, the goal in this problem is to assert either (ˆy ,..., yˆ ) S(K,δ) or ∈ 1 n ∈ (ˆy ,..., yˆ ) S(K, δ). 1 n 6∈ − Note that Coverage Norm Extension is the Strong Validity problem for K with ci = fi. We show the following lemma (Coverage WV Coverage Norm Extension). ≤p 14 The Complexity of Partial Function Extension for Coverage Functions

◮ Lemma 25. If there is an α = 2poly(n,m)F δ efficient approximation algorithm (for any fixed 0 δ < 1) for Coverage Norm Extension then there is an efficient algorithm for Weak ≤ Validity problem for K. Theorem 24 immediately gives Coverage WM Coverage WV. ≤p Span WM Coverage WM. ≡ In fact, we show that Coverage Weak Membership is NP-hard even for the case when T =2 | i| for all i [n].3 The restriction T = 2 gives us a graphical representation of the membership ∈ | i| problems. We first introduce some notations, which will be used in the remainder. Given a weighted graph G = (V, E) and a set S V , the span E+(S) and cut δ (S) of set S ⊆ G G are the set of edges with at least one endpoint and exactly one endpoint in S respectively. + We use w(EG (S)), w(δG(S)) and w(EG(S)) for the sum of weight of edges with at least one endpoint, exactly one endpoint and both endpoints in S respectively. If the set S is a single vertex v then we use v instead of v . If the graph G is understood from the context we { } drop the subscript G. Given a set = T ,...,T (T [m]) with the property that T = 2 for all i [n], D { 1 n} i ⊆ | i| ∈ we construct a weighted graph G = (V, E) as follows: vertex set V = [m] and i, j E { } ∈ (i, j [m]) iff there exists a T such that T = i, j . The weighty ˆ associated with ∈ k ∈ D k { } k Tk = i, j is now associated to the edge i, j . Now the constraint i:T S= yi 0 (in { } { } i∩ 6 ∅ ≤ the polyhedron K) translates to + y 0 for all S V . Thus Coverage-Weak- e E (S) e ≤ ⊆ P Membership for T = 2 case is equivalent∈ to following problem, which we call Span Weak | i| P Membership. Given a weighted graph G = (V, E) with weightsy ê on the edges and δ > 0, assert either yˆ = (ˆye)e E is in S(Ks,δ)ory ˆ is not in S(Ks, δ), where ∈ −

Ks = ye 0 S V, y 1 . (10)  ≤ ∀ ⊆ || ||∞ ≤  e E+(S)  ∈X  Densest-Cut Cut WM Span WM.  ≤p ≤p We now show that the Span Weak Membership is NP-Hard thereby showing Coverage Weak Membership is also NP-Hard for the restricted setting with T = 2 for all i [n]. We first | i| ∈ define Cut Weak Membership. Given a weighted graph G = (V, E) with weightsy ê on the edges and δ > 0, the goal in Cut Weak Membership is to assert eithery ˆ = (ˆye)e E is in S(Kc,δ)ory ˆ is not in S(Kc, δ) ∈ − where

V Kc = ye 0 S 2 , y 1 . (11)  ≤ ∀ ∈ \∅ || ||∞ ≤  e δ(S)  ∈X 

Note that in the Cut Weak membership problem, we have constraints e δ(S) ye 0 ∈ ≤ instead of e E+(S) ye 0 for all S. ∈ ≤ P ◮ LemmaP 26. There is a reduction from Densest-Cut to Cut Weak Membership and from Cut Weak Membership to Span Weak Membership. Therefore, Coverage Weak Membership is NP-hard even when d =2.

3 There is a relatively easier proof for unrestricted d by reduction from Set Cover, which we show in the appendix. U. Bhaskar and G. Kumar 15

c c

7 7 − − b b 3 1 1

d d 1 s 5 5 3 3 1 − − − − a a 24 t G = (V, E) G′ = (V ′, E′)

Figure 1 Reduction from Cut Strong Membership to Span Strong Membership. The number ′ shown on the edges in E is the weight ye, while on edges in E is the product of L = 24 and weight ′ ye.

We can now complete the proof of Theorem 5.

Proof of Theorem 5. Suppose there is an efficient α-approximation algorithm for Coverage Norm Extension. Then by Lemma 25 there is an efficient algorithm for Weak Validity problem for polytope K (9) and then by Theorem 24 we have an efficient algorithm for Coverage Weak Membership. But by Lemma 26, this is not possible unless P = NP . ◭

We here prove Lemma 27, which is a weaker statement than Lemma 26 to convey the main ideas. Recall that in Strong Membership problem, the goal is to decide if given vector y is in polyhedron K. Following our nomenclature, we define the following Strong Membership problems. An instance of Span Strong Membership and Cut Strong Membership is given by a weighted graph G = (V, E) with weightsy ê on the edges, and the goal is to decide if vector y = (ye)e E is in Ks and Kc respectively, with Ks and Kc as defined in (10), (11). ∈ ◮ Lemma 27. There is a reduction from Densest-Cut to Cut Strong Membership, and from Cut Strong Membership to Span Strong Membership.

Proof. For the second reduction, the instance of Cut Strong Membership is weighted graph G = (V, E) with weights ye on the edges. We assume y 1 as otherwise clearly y Kc. || ||∞ ≤ 6∈ Let L =2 E + V E . We construct an instance of Span Strong Membership (see Figure | | | || | 1), i.e., graph G′ = (V ′, E′) and weights ye′ as follows:

ye L if e E(G) 1 ∈ V ′ = V s,t , E′ = E s,t v,s v V,y′ = w(δ (v)) if e = v,s , v = t ∪{ } ∪{ }∪{ } ∀ ∈ e  − 2L G { } 6 1 if e = s,t .  − { }

Then y′ 1.  || ||∞ ≤ Assume y K , i.e., there exists S V s.t. w(δ (S)) > 0. We need to show there 6∈ c ⊆ G exists S′ V ′ s.t. e E+(S ) ye′ > 0. For S′ = S, L e E+(S ) ye′ = w(EG(S))+ w(δG(S))+ ⊆ ∈ ′ ∈ ′ 1 1 w(δG(S)) v S 2 w(δG(v)) = w(EG(S)) + w(δG(S)) 2 (2w(EG(S)) + w(δG(S))) = 2 > 0. ∈ − · P − · P Now assume y Kc, i.e., S V, w(δG(S)) 0. We need to show S′ V ′, e E+(S ) ye′ P ∈ ∀ ⊆ ≤ ∀ ⊆ ∈ ′ ≤ 0. Since y′ s,t = 1 (and L is suﬃciently large), we need to consider only those S′ which do { } − P w(δG(S′)) not contain either s or t. But we have shown that for such S′, + y′ = 0. e E (S′) e 2L ≤ Now we ﬁnish the proof by giving a reduction from Densest-Cut∈ to Cut Strong Member- P ship. Given an undirected graph G = (V, E) and rational M, we want to know if there exists 16 The Complexity of Partial Function Extension for Coverage Functions

δG(S) S V s.t. S V S >M. Consider the complete graph G′ = (V, E′) where the weight of an ⊂ 1 M| || \ | M edge is − if it existed in E, and is otherwise (note that edges may now have positive, L − L negative, or zero weight). Let L′ = 2max M, 1 M be a suﬃciently large quantity so { | − |} that yˆ < 1 . It is easy to see that L w(δG′ (S)) = δG(S) M S V S . Therefore, || ||∞ δ (S) | |− | || \ | | G | ◭ S V s.t. w(δG′ (S)) > 0 S V s.t. S V S >M. ∃ ⊂ ⇔ ∃ ⊂ | || \ |

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18 Subhash Khot. Improved inapproximability results for maxclique, chromatic number and approximate graph coloring. In Proceedings 2001 IEEE International Conference on Cluster Computing, pages 600–609. IEEE, 2001. 19 Andreas Krause, H Brendan McMahan, Carlos Guestrin, and Anupam Gupta. Robust submodular observation selection. Journal of Machine Learning Research, 9(Dec):2761–2801, 2008. 20 Benny Lehmann, Daniel Lehmann, and Noam Nisan. Combinatorial auctions with decreasing marginal utilities. Games and Economic Behavior, 55(2):270–296, 2006. 21 Hans JM Peters and Peter P Wakker. Convex functions on non-convex domains. Economics letters, 22(2-3):251–255, 1986. 22 Leonard Pitt and Leslie G. Valiant. Computational limitations on learning from examples. J. ACM, 35(4):965–984, 1988. 23 Lior Seeman and Yaron Singer. Adaptive seeding in social networks. In 2013 IEEE 54th Annual Symposium on Foundations of Computer Science, pages 459–468. IEEE, 2013. 24 C. Seshadhri and Jan Vondrák. Is submodularity testable? Algorithmica, 69(1):1–25, 2014. 25 Comandur Seshadhri and Jan Vondrák. Is submodularity testable? Algorithmica, 69(1):1–25, 2014. 26 Donald M Topkis. Minimizing a submodular function on a lattice. Operations research, 26(2):305–321, 1978. 27 Armin Uhlmann. Roofs and convexity. Entropy, 12(7):1799–1832, 2010. 28 Min Yan. Extension of convex function. arXiv preprint arXiv:1207.0944, 2012. To appear in the Journal of Convex Analysis. 29 Min Yan. Extension of convex function. arXiv preprint arXiv:1207.0944, 2012. To appear in the Journal of Convex Analysis.

A Appendix

Proof of Proposition 7

Consider the polyhedron Extension-P. If the partial function is extendible, then Extension-P is nonempty. Since the variables are non-negative, the polyhedron must have a vertex [4], and in particular there is a vertex in which at most n variables w(S) are non-zero. This is because the dimension of the problem is 2m, hence at a vertex at least 2m constraints must be tight. But then at least 2m n of constraints w(S) 0 must be tight. − ≥ Proof of Corollary 9

Consider the polytope P = I :v I xI = 1 v V (G), 0 xI 1 I . By { ∈I ∈ ∀ ∈ ≤ ≤ ∀ ∈ I} the Theorem 8, there exists x = xI I in P such that χ∗(G) = I xI . Consider P { } ∈I ∈I y = yI I given by y v = 1 for all v V (G) and 0 otherwise. Therefore, y P and { } ∈I { } ∈ V (G) t P ∈ V (G) = I yI . Consider z = λx + (1 λ)y where λ = V (|G) χ|−(G) . Therefore, z P | | ∈I − | |− ∗ ∈ and I zI = λ I xI + (1 λ) I yI = t. ∈IP ∈I − ∈I P P P Proof of Lemma 12 From Theorem 6, given a partial function H and α 1, there exists a coverage function f ≥ satisfying f f(T ) αf for all i [n] iﬀ the following linear program is feasible, where i ≤ i ≤ i ∈ the variables are the W -coeﬃcients w(S) for all S 2[m] : ∈ \∅ f w(S) αf i [n] i ≤ ≤ i ∀ ∈ S:S T = X∩ i6 ∅ 18 The Complexity of Partial Function Extension for Coverage Functions

w(S) 0 S 2[m] . ≥ ∀ ∈ \∅ By Farkas’ Lemma, it follows that the above linear program is feasible iﬀ the following linear program is infeasible, with variables y and z for all i [n]: i i ∈ n n α fiyi < fizi (12) i=1 i=1 X X y z S 2[m] (13) i ≥ i ∀ ∈ \∅ i:S T = i:S T = X∩ i6 ∅ X∩ i6 ∅ y ,z 0 i i ≥ .

Now we proceed towards proving the claim. Suppose li’s satisfy (4) and (3). Set yi and z as follows: If l 0 then let y = l and z = 0. Else if l > 0 then let y = 0 and i i ≤ i − i i i i zi = li. It is easy to see that yi,zi 0 and li = zi yi and hence (13) is satisﬁed by yi’s and n ≥ − zi’s. Further, α i=1 fiyi = α( i:l 0 fiyi + i:l >0 fiyi) = α i:l 0 fili and similarly n i≤ i − i≤ fizi = fili. Thus (12) is also satisﬁed by yi’s and zi’s. i=1 i:lPi>0 P P P For the other direction observe that if the vector y = (y , .., y ),z = (z , ..., z ) 0 P P 1 n 1 n ≥ satisfy (12) and (13) then wlog we can assume for any i, the minimum of yi and zi is 0 (otherwise we can decrease both y and z by the minimum of y and z , and α 1 allows i i i i ≥ (12) to remain true). Note that i fiyi = i:y z fiyi + i:y >z fiyi = i:y >z fiyi, since i≤ i i i i i min yi,zi = 0 by the previous observation. Now suppose y,z 0 satisfy (12) and (13). { } n nP P P ≥ P We thus have α fiyi < fizi α fiyi < fizi. Now let li = zi yi. i=1 i=1 ⇔ yi>zi zi>yi − This makes both (4) and (3) true. P P P P

Proof of Lemma 19

Suppose we are given a weighted bipartite graph G = (A [m], E) with weight f on each ∪ v fw w R v A. Recall that κ is the minimum of ∈ over vertices v A and R v where ∈ fv ∈ ∈ F = R A v N(R) N(v) is the setP of all R A v that covers all the neighbours Fv { ⊆ \{ }| ⊇ } ⊆ \{ } of v. We will use f(R) (R A) to denote the summation f . If d is a constant then ⊆ v R v for each v A, we can ﬁnd minimum of f(R) over all R ∈in O(nd) time where n = A . ∈ ⊆FP v | | Therefore, by taking the minimum of the above minimum value over all vertices v A, we ∈ get the value of κ. For general d, we use an approximation algorithm for Set-Cover to ﬁnd, for each vertex v A, a set R′ such that f(R′ ) f(R )log d where R is the optimal ∈ v ∈Fv v ≤ v v f(Rv′ ) set. It can be seen that κ′ = minv A has the property κ′ κ log d. ∈ fv ≤

Remaining proof of Theorem 20

We will show that

P r[ N(S) N(S) S [m]] > 0 | ∩N|≥| ∩P| ∀ ⊆ + Let = S [m] S P 1 P . Also let N −(S) = N(S) and N (S) = S { ⊆ || ∩ | ≤ ∀ + ∈ P} ∩ N + N(S) . Note that N −(S) N (S) for all S implies N −(S) N (S) for all ∩P | | ≥ | | ∈ S | | ≥ | | S [m]. ⊆ U. Bhaskar and G. Kumar 19

4 Given S of size s and a set M of size s, the probability that N −(S) M is (1 ( s∈)s S ⊆ N + ⊆ − 1/√m) |N|− . Therefore, P r[ N −(S) N (S) S [m]]=1 S M , M = S (1 | | ≥ | | ∀ ⊆ − ∈S ⊆N | | | | − ( s)s √m √m s k√m log m (k√m log m s)s 1/√m) |N|− =1 s=1 s √m s (1 1/√m) P− P √me s− s ke√m log m s − √m ( s ) √m ( s ) √m mk√m log m s > 1 Pks log m >1 (O( k )) which is strictly greater − s=1 e − s=1 m than 0 for suﬃciently large k. P P

Proof of Lemma 25

The instance of weak validity problem are vector c Qn and rational numbers γ and ǫ> 0. ∈ We show that there is a reduction from general Weak Validity problem to Weak Validity problem with instances satisfying c 0 for all i [n]. i ≥ ∈ Let N = i [n] c 0 . Consider a vector c′ such that c′ = 0 for i N and c { ∈ | i ≤ } i ∈ i otherwise and γ′ = γ i N ci . If x is in S(K,ǫ) then clearly x¯ deﬁned as x¯i = 1 if − ∈ | | T − i N and xi otherwise, is also in S(K,ǫ). If for some x in S(K,ǫ), we have (c′) x γ′ ǫ ∈ P T T ≥ − then for x¯ S(K,ǫ), we have c x¯ = i N ci + (c′) x γ ǫ. Also if for all x S(K, ǫ), ∈T T ∈ | | T ≥ − ∈ − we have (c′) x γ′ + ǫ then c x i N ci + (c′) x γ + ǫ. This shows the reduction ≤ ≤P ∈ | | ≤ and hence we assume c 0 in the instance of Weak Validity problem. i ≥ P Let OPT and OPT ′ be the optimal value of Norm-P for (f1,...,fn) = (c1,...,cn) and (f1,...,fn) = (Lc1,...,Lcn) respectively (L will be chosen later). Obviously OPT ′ = L OPT . Let the approximation algorithm for Coverage Norm Extension (and hence Norm- · P) return β for instance (f1,...,fn) = (Lc1,...,Lcn). Let C = i ci. Therefore, OPT ′ β poly(n,m) ≤ δ poly(n,m) δ poly(n,m) δ 2 (C) OPT ′+2 (LC) = L OPT +2 (LC) and hence β/L OPT + 1 δ . P L − ≤ 1/1· δ ≤ 2poly(n,m)(C)δ − 2poly(n,m) (C)δ We set L := so that 1 δ < 2ǫ. Note that the number of bits to 2ǫ L − specify L is polynomial in c , ǫ ,n,m. Thus, OPT β/L OPT +2ǫ. Now if γ +ǫ β/L h i h i T ≤ β ≤ ≤ then for the optimal solution x∗ K, c x∗ = OPT 2ǫ γ ǫ. If γ + ǫ β/L then ∈ ≥ L − ≥ − ≥ for all x in K (and hence S(K, ǫ)), we have cT x OPT β/L γ + ǫ. Since at least one − ≤ ≤ ≤ of these two conditions must hold, the conditions of weak validity problem can be correctly asserted.

In the following proofs, for any vector y, recall that we use y for maxi yi and yˆ y || ||∞ | | || − || for the Euclidean distance between yˆ and y. We will frequently use the fact that the distance T T w x0+b of a point x0 from the hyperplane w x + b =0 is equal to | w | . || ||

Proof of hardness of Coverage Weak Membership for unrestricted d

We give a reduction from Set Cover. An instance of Set Cover consists of a universe of n′ elements = u ,...,u , a family of sets = S ,...,S such that S for each U { 1 n′ } F { 1 m′ } i ⊆ U i [m′], and a positive integer k. We need to determine if there exists ′ and ′ k ∈ F ⊆F |F |≤ such that ′ covers the universe , i.e., S S = . F U ∪ ∈F ′ U The instance of Coverage-Weak-Membership requires a set = T ,...,T where each D { 1 n} T [m], real values yˆ associated with each T and δ > 0. The goal is to assert either i ⊆ i i the point yˆ = (ˆyi)i [n] is in S(K,δ) or is not in S(K, δ). Given an instance of Set Cover, ∈ − we construct an instance of Coverage Weak Membership as follows. We set m = m′, the number of sets in , hence each set T can be thought of as a subset of . The instance has F i F n = m′ +1+ m′ sets in , as follows (the value of L will be chosen later): D

4 The analysis from here on is similar to existence proof of a expander on a bipartite graph. 20 The Complexity of Partial Function Extension for Coverage Functions

1 1. The first m′ sets correspond to sets in : Ti = i with yî = L F 1 { } − k kn′ + 2 2. Set T = 1,...,m′ with yˆ = − m′+1 { } m′+1 L 3. For each of the n′ elements in , there is a set in : For each u , contains the set U D i ∈U D j [m] S contains u with yˆ-value k . { ∈ | j i} L . 1 1 The value of L is set to 2max 1, k kn′ + , k = kn′ k 1/2 (to ensure yˆ = {− − 2 } − − || ||∞ 2 in view of the constraint y 1 of K). We set δ to be 1 . || ||∞ ≤ 4L√n Suppose that this is a YES instance of the Set Cover problem, i.e., there exists a ′ F ⊆F and ′ k such that ′ covers the universe . In our instance, if we take the set of |F | ≤ F U elements S = j [m′] Sj ′ then this intersects with at most k sets of the first kind, { ∈ | ∈ F } 1 k+(k kn′ + 2 )+kn′ 1 set Tm +1, and all sets of the third kind. Hence, yî − − = . ′ i:Ti S= ≥ L 2L So the point yˆ violates the constraint y 0 ∩and6 ∅ is at least 1 distance away i:S Ti= i 2L√n ∩ 6 ∅ P≤ from the corresponding hyperplane y = 0. Therefore, yˆ y 1 for all i:S Ti= i 2L√n P ∩ 6 ∅ || − || ≥ y K. Hence yˆ S(K,δ) as δ < 1 . ∈ 6∈ 2L√Pn Now suppose that this is a NO instance of the Set Cover problem. For any S 2[m′] 1 k 1+(k kn′ + )+kn′ ⊆ \ , if S k +1 then yˆ − − − 2 = 1 and if S k then i:Ti S= i L 2L ∅ | | ≥ ∩ 6 ∅1 ≤ − | | ≤ (n′ 1)k 1+(k kn′ + ) yˆ − − − 2 = 1 , since any set of size strictly less than k +1 i:Ti S= i PL 2L ∩ 6 ∅ ≤ − covers at most n′ 1 elements of universe . Thus the point yˆ satisfies the constraints P − U y 0 for all S and is at least 1 distance away from all the hyperplanes i:Ti S= i 2L√n ∩ 6 ∅ ≤ yi =0. Also, for any y S(ˆy,δ), we have y yˆ y yˆ y yˆ δ. Pi:Ti S= ∈ || ||∞ −|| ||∞ ≤ || − ||∞ ≤ || − || ≤ So y∩ 6 ∅ δ +1/2 1 (as δ < 1 ). Thus yˆ S(K, δ). P || ||∞ ≤ ≤ 2 ∈ − Proof of Lemma 26 Recall the definitions of Span Weak Membership, Cut Weak Membership and Densest Cut:

1. Given a weighted graph G = (V, E) with weights yˆe on the edges and δ > 0, a. The goal in Span Weak Membership is to assert either yˆ = (ˆye)e E is in S(Ks,δ) or ∈ yˆ is not in S(K , δ) where s −

V Ks = ye 0 S 2 , y 1 ,  ≤ ∀ ∈ \∅ || ||∞ ≤  e E+(S)  ∈X  b. The goal in Cut Weak Membership is to assert either yˆ = (ˆye)e E is in S(Kc,δ) or yˆ   ∈ is not in S(K , δ) where c −

V Kc = ye 0 S 2 , y 1 .  ≤ ∀ ∈ \∅ || ||∞ ≤  e δ(S)  ∈X 

Note that in the Cut Weak membership, we have constraints e δ(S) ye 0 instead of ∈ ≤ + y 0 for all S. e E (S) e ≤ P 2. In the∈ Densest-Cut problem, given a graph G = (V, E) and a positive rational M, the P δ(S) goal is to decide if there exist a set S V s.t. S| V S| M. ⊂ | || \ | ≥ δ(S) The Densest-Cut is known to be NP-Hard [7]. Note that S| V S| called the density of | || \ | r cut (S, V S) can take values only from s( V s) 1 r E , 1 s V 1,r,s Z+ . \ | |− | ≤ ≤ | | ≤ ≤ | |− ∈ Thus there are only polynomially many possiblen values of cut densities. We will use thiso fact in our proof. U. Bhaskar and G. Kumar 21

c c

7 7 − − b b 3 1 1

d d 1 s 5 5 3 3 1 − − − − a a 24 t G = (V, E) G′ = (V ′, E′)

Figure 2 Reduction from Cut-Weak-Membership to Span-Weak-Membership. The number ′ ′ shown on the edges in E is the weight ye, while on edges in E is product of L = 48 and weight ye.

◮ Lemma 28. There is a reduction from Cut Weak Membership to Span Weak Membership.

Proof. Our goal in Cut Weak Membership, given a graph a G = (V, E) with weights yˆe on edges and δ > 0, is to assert either yˆ = (ˆye)e E is in S(Kc,δ) or yˆ is not in S(Kc, δ). If ∈ − the point yˆ violates the constraint y 1 of Kc then it can be asserted that yˆ is not in || ||∞ ≤ S(Kc, δ). So we assume yˆ 1. Given this assumption, we have w(δG(v)) E . − || ||∞ ≤ ≤ | | We construct an instance of Span-Weak-Membership (see Figure 2), i.e., graph G′ =

(V ′, E′), yˆe′ and δ′ as follows (the values of B and L will be set later): V ′ = V s,t ∪{ } E′ = E s,t v,s v V ′, v = s,t ∪ye {{ }}∪{{ }} ∀ ∈ 6 { } L if e E 1 w(δ (v)) ∈ yˆ = 2 G e′  L if e = v,s , v = t  − B { } 6  L if e = s,t . − { } B+ E +1/2 V E The value of B is set to 2 E + V E so that e E+ (S) yˆe′ − | | L | || | 0 for all S  | | | || | ∈ G′ ≤ ≤ containing either s or t. Further, L =2B so that yˆ′ =1/2 where yˆ′ = (ˆye′ )e E′ . Finally P || ||∞ ∈ √ E δ E +1/2 V E 1 | | 1 we choose δ′ = 2 min , | | | || | , 2 . 2L√ E′ √ E′ L | | | | ⊲ Claim 29. For all S V , w(E+ (S)) = w(δG(S)) . ⊆ G′ 2L Proof. This is because

+ 1 L w(E (S)) = L w′ = w(EG(S)) + w(δG(S)) + w(δG(v)) , G′ e −2 e E+ (S) v S ∈ XG′ X∈ and since w(δG(v)) counts edges in EG(S) twice and edges in δG(S) once,

+ 1 w(δG(S)) L w(E (S)) = w(EG(S)) + w(δG(S)) (2w(EG(S)) + w(δG(S))) = . G′ − 2 · 2 ◭

Suppose the algorithm for Span Weak Membership asserts that the point yˆ′ is in S(Ks,δ′). V If yˆ′ satisﬁes all the constraints e E+ (S) ye 0 for all S 2 then the point yˆ must ∈ G ≤ ∈ \∅ ′ V satisfy all the constraints ye 0 for all S 2 (because by the Claim 29 e δGP(S) + ∈ ≤ ∈ \∅ w(δG(S))=2L w(E (S))) and hence yˆ Kc . Thus yˆ S(Kc,δ). Now suppose yˆ′ violates · G′ P ∈ ∈ 22 The Complexity of Partial Function Extension for Coverage Functions

V a constraint e E+ (R) ye 0 for some R 2 . Since yˆ′ S(Ks,δ′), it is at most δ′ ∈ G ≤ ∈ \∅ ∈ distance away from′ the hyperplanes corresponding to the violated constraints. Therefore, P+ we have w(E (R)) = + yˆ′ δ′ E . By Claim 29, w(δ (R)) 2Lδ′ E . G′ e E (R) e ′ G ′ ∈ G′ ≤ | | ≤ | | 2Lδ′√ E′ √ E δ Therefore, the point yˆ Pis at most | p| distance from Kc. Since δ′ < | |p, so √ E 2L√ E′ | | | | yˆ S(K ,δ). ∈ c Suppose the algorithm for Span Weak Membership problem asserts that the point yˆ′ is V not in S(Ks, δ′). If yˆ′ violates a constraint e E+ (S) ye 0 for some S 2 then the − ∈ G′ ≤ ∈ \∅ point yˆ also violates e δ (S) ye 0 for S (by Claim 29). Hence, it can be asserted that ∈ G ≤ P yˆ is not in S(Kc, δ). So now assume that yˆ′ satisfies all the constraints e E+ (S) ye 0 − P ∈ G′ ≤ V for all S 2 . Also, as shown earlier, yˆ′ satisfies the other constraints of K . Since ∈ \∅ P s yˆ′ is in K but not in S(K , δ′), some y S(ˆy′,δ′) must have distance < δ′ from some s s − ∈ hyperplane of Ks. The distance of yˆ′ from the hyperplane e E+ (S) ye =0 for S containing ∈ G′ B+ E +1/2 V E′ E +1/2 V E′ s or t is at least |− | | | || || = | | | || | >δ′. AlsoP for any y S(ˆy′,δ′), we have √ E′ L √ E′ L | | | | ∈ y yˆ′ y yˆ y yˆ′ . So y δ +1/2 1 for all y S(ˆy′,δ′). || ||∞ − || ||∞ ≤ || − ||∞ ≤ || − || || ||∞ ≤ ≤ ∈ Therefore, it must be the case that distance of yˆ′ from the hyperplane e E+ (S) ye = 0 ∈ G′ V for some S 2 is <δ′. By Claim 29, the distance of the point yˆ from the hyperplane ∈ \∅ P 2L√ E′ δ′ ye =0 is at most | | <δ. Hence, yˆ is not in S(Kc, δ). ◭ e δG(S) √ E ∈ | | − P Now we finish the proof by giving reduction from Densest-Cut to Cut Weak Membership.

◮ Lemma 30. There is a reduction from Densest-Cut to Cut Weak Membership.

Proof. In the Densest Cut problem, a graph G = (V, E) and a positive rational M are given and the goal is to determine if there exists a set S V s.t. the density of the cut (S, V S) δ (S) p ⊂ \ is at least M, i.e., | G | M. Let M = for positive integer p, q. S V S ≥ q Given the graph| ||G \=| (V, E) and M, the instance of the Cut Weak Membership is a V ( V 1) complete graph G′ = (V, E′) (so E′ = | | |2 |− ), weight yê on each edge e E′ such that 1 M | M| 1 1 t ∈ yê is −L if it existed in E and −L otherwise and δ = 2 min 2 , . Let yˆ = (ˆye)e E′ . √ E′ ∈ { | | } We set L to 2max M, 1 M so that yˆ =1/2 and t to 1 . ∞ qL { | − |} 1|| || It is easy to see that w(δG′ (S)) = L ( δG(S) M S V S ). Therefore, S V s.t. δ (S) | |− | || \ | ∃ ⊂ | G | w(δG′ (S)) 0 S V s.t. S V S M. ≥ ⇔ ∃ ⊂ p | || \ | ≥ q p Since M is equal to q for some p, q Z+, therefore the weight of an edge is either qL− p ∈ − or qL . So if a cut value w(δG′ (S)) is strictly positive for any S then w(δG′ (S)) must be at least 1 = t. Similarly, if w(δ (S)) < 0 then we have w(δ (S)) t. qL G′ G′ ≤− Now suppose an algorithm for Cut-Weak-Membership asserts yˆ is in S(Kc,δ). Thus for t all S, w(δG′ (S)) = e δ (S) yê E′ δ. Since δ < , so it must be the case that for G′ √ E′ ∈ ≤ | | | | δ (S) all S, w(δ (S)) 0P. This impliesp that for all S, the cut density | G | M. G′ ≤ S V S ≤ Now suppose the algorithm for Cut Weak Membership asserts th| ||at\yˆ |is not in S(K , δ). c − If yˆ Kc (and since yˆ 1) then clearly there exists a set S such that w(δG′ (S)) > 0. 6∈ || ||∞ ≤ δ (S) This implies there is a cut (S, V S) with density | G | >M. Now assume yˆ is in K . Now \ S V S c for any y S(ˆy,δ), we have y yˆ y yˆ| || \ | y yˆ δ. So y δ+1/2 < 1. ∈ || ||∞−|| ||∞ ≤ || − ||∞ ≤ || − || ≤ || ||∞ ≤ So there must exist a hyperplane e δ (S) ye =0 for some S with at most δ distance from G′ ∈ t yˆ. Therefore, there exist a set S with 0 w(δG′ (S)) E′ δ. Since δ < , this P √ E′ ≥ ≥ − | | | | means w(δG (S))=0 and hence density of cut (S, V S) is Mp. Thus, if an algorithm for Cut ′ \ U. Bhaskar and G. Kumar 23

Weak Membership asserts that yˆ is not in S(K , δ) then there exists a cut with density at c − least M. Therefore, assuming an eﬃcient algorithm for Cut Weak Membership, it can be determined if there exists a cut with density at least M or all cuts have density at most M. However, the goal in Densest Cut is to determine if there is a cut with density M or all cuts have ≥ density strictly less than M. But since the density can take only polynomial number of r values s( V s) 1 r E , 1 s V 1,r,s Z+ (as noted before), by using at most | |− | ≤ ≤ | | ≤ ≤ | |− ∈ two oraclen calls to the Cut-Weak-Membership problem weo can solve the original problem. ◭