In this section we consider a problem that generalizes Multi-Partial-SC and CIPs while being a special case of Multi-Submod-Cover. We call this problem CCF (Covering Coverage Functions). Bera et al. (2014) already considered this version in the restricted context of Vertex-Cover. Formally the input is a weighted set system \((\mathcal {U}, \mathcal {S})\) and a set of inequalities of the form \(Az \ge b\) where \(A \in [0,1]^{h \times n}\) matrix and \(b \in \mathbb {R}_+^h\) is a positive vector. The goal is to optimize the integer program CCF-IP shown in Fig. 3a. Multi-Partial-SC is a special case of CCF when the matrix A contains only \(\{0,1\}\) entries. On the other hand CIP is a special case when the set system is very restricted and each set \(S_i\) consists of a single element. We say that an instance is r-sparse if each set \(S_i\) “influences” at most r rows of A; in other words the elements of \(S_i\) have non-zero coefficients in at most r rows of A. This notion of sparsity coincides in the case of CIPs with column sparsity and in the case of Multi-Submod-Cover with the sparsity that we saw in Sect. 3. It is useful to explicitly see why CCF is a special case of Multi-Submod-Cover. The ground set \(N = [m]\) corresponds to the sets \(S_1,\ldots ,S_m\) in the given set system \((\mathcal {U},\mathcal {S})\). Consider the row k of the covering constraint matrix \(Az \ge b\). We can model it as a constraint \(f_k(S) \ge b_k\) where the submodular set function \(f_k:2^N \rightarrow \mathbb {R}_+\) is defined as follows: for a set \(X \subseteq N\) we let \(f_k(X) = \sum _{e_j \in \cup _{i \in X} S_i} A_{k,j}\) which is simply a weighted coverage function with the weights coming from the coefficients of the matrix A. Note that when formulating via these submodular functions, the auxiliary variables \(z_1,\ldots ,z_n\) that correspond to the elements \(\mathcal {U}\) are unnecessary.
We prove the following theorem.
Theorem 4.1
Consider an instance of r-sparse CCF induced by a set system \((\mathcal {U}, \mathcal {S})\) from a deletion-closed family with a \(\beta \)-approximation for Set Cover via the natural LP. There is a randomized polynomial-time algorithm that outputs a feasible solution of expected cost \(O(\beta + \ln r) \textsc {OPT}\).
The natural LP relaxation for CCF is shown in Fig. 3b. It is well-known that this LP relaxation, even for CIPs with only one constraint, has an unbounded integrality gap (Carr et al. 2000). For CIPs knapsack-cover inequalities are used to strengthen the LP. KC-inequalities in this context were first introduced in the influential work of Carr et al. (2000) and have since become a standard tool in developing stronger LP relaxations. Bera et al. (2014) adapt KC-inequalities to the setting of PartitionVC, and it is straightforward to extend this to CCF (this is implicit in Bera et al. (2014)).
Remark 4.2
Weighted coverage functions are a special case of sums of weighted rank functions of matroids. The natural LP for CCF can be viewed as using a different, and in fact a tighter extension, than the multilinear relaxation (Calinescu et al. 2007). The fact that one can use an LP relaxation here is crucial to the scaling idea that will play a role in the eventual algorithm. The main difficulty, however, is the large integrality gap which arises due to the partial covering constraints.
We set up and the explain the notation to describe the use of KC-inequalities for CCF. It is convenient here to use the reduction of CCF to Multi-Submod-Cover. For row k in \(Ax \ge b\) we will use \(f_k\) to denote the submodular function that we set up earlier. Recall that \(f_k(D)\) captures the coverage to constraint k if set D is chosen. The residual requirement after choosing D is \(b_k - f_k(D)\). The residual requirement must be covered by elements from sets outside D. The maximum contribution that \(i \not \in D\) can provide to this is \(\min \{f_k(D+i) - f_k(D), b_k - f_k(D)\}\). Hence the following constraint is valid for any \(D \subset N\):
$$\begin{aligned} \sum _{i \not \in D} \min \{f_k(D+i) - f_k(D), b_k - f_k(D)\} x_i \ge b_k - f_k(D) \end{aligned}$$
(1)
Writing the preceding inequality for every possible choice of D and for every k we obtained a strengthened LP that we show in Fig. 4.
CCF-KC-LP has an exponential number of constraints and the separation problem involves submodular functions. A priori it is not clear that there is even an approximate separation oracle. However, as noted in Bera et al. (2014), one can combine the rounding procedure with the Ellipsoid method to obtain the desired guarantees even though we do not obtain a fractional solution that satisfies all the KC inequalities. This observation holds for rounding as well. In particular, it suffices to obtain a fractional solution that satisfies constraints (2), (3), (5), and (6) (note that there are polynomially many such constraints), and constraint 4 for a particular \(D \subseteq \mathcal {S}\) satisfying certain properties. Then, it can be shown that the ellipsoid algorithm returns such a solution in polynomially many iterations. In the following, we discuss the rounding and the analysis under the assumption that the LP can be solved exactly, and return to the issue of KC inequalities at the end of the section.
Rounding and analysis assuming LP can be solved exactly: Let (x, z) be an optimum solution to CCF-KC-LP. We can assume without loss of generality that for each element \(e_j \in \mathcal {U}\) we have \(z_j = \min \{1, \sum _{i: e_j \in S_i} x_i\}\). As in Sect. 6 we split the elements in \(\mathcal {U}\) into heavily covered elements and shallow elements. For some fixed threshold \(\tau \) that we will specify later, let \(\mathcal {U}_{\text {he}} = \{ e_j \mid z_j \ge \tau \}\), and \(\mathcal {U}_{\text {sh}} = \mathcal {U}\setminus \mathcal {U}_{\text {he}}\). We will also choose another threshold. The rounding algorithm is the following.
-
1.
Solve a Set Cover problem via the natural LP to cover all elements in \(\mathcal {U}_{\text {he}}\). Let \(Y_1\) be the sets chosen in this step.
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2.
Let \(Y_2 = \{ S_i \mid x_i \ge \tau \}\) be the heavy sets.
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3.
Repeat for \(\ell = \varTheta (\ln r)\) rounds: independently pick each set \(S_i\) in \(\mathcal {S}\setminus (Y_1 \cup Y_2)\) with probability \(\frac{1}{\tau }x_i\). Let \(Y_3\) be the sets chosen in this randomized rounding step.
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4.
For \(k \in [h]\) do
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(a)
Let \(b_k - f_k(Y_1 \cup Y_2 \cup Y_3)\) be the residual requirement of k’th constraint.
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(b)
Run the modified Greedy algorithm to satisfy the residual requirement. Let \(F_k\) be the sets chosen to fix the constraint (could be empty).
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5.
Output \(Y_1 \cup Y_2 \cup Y_3 \cup (\cup _{k=1}^h F_k)\).
The algorithm at a high level is similar to that in Bera et al. (2014). There are two main differences. First, we explicitly fix the constraints after the randomized rounding phase using a slight variant of the Greedy algorithm. This ensures that the output of the algorithm is always a feasible solution; this makes it easier to analyze the r-sparse case while a straight forward union bound will not work. Second, the analysis relies on a probabilistic inequality that is simpler in Vertex-Cover case while it requires a more sophisticated approach here. We now describe the modified Greedy algorithm to fix a constraint. For an unsatisfied constraint k we consider the collection of sets that influence the residual requirement for k, and partition them it into \(H_{k}\) and \(L_k\). \(H_{k}\) is the collection of all sets such that choosing any of them completely satisfies the residual requirement for k, and \(L_k\) are the remaining sets. The modified Greedy algorithm for fixing constraint k picks the better of two solutions: (i) the first solution is the cheapest set in \(H_k\) (this makes sense only if \(H_k \ne \emptyset \)) and (ii) the second solution is obtained by running Greedy on sets in \(L_k\) until the constraint is satisfied.
Analysis: We now analyze the expected cost of the solution output by the algorithm. The lemma below bounds the cost of \(Y_1\).
Lemma 4.3
The cost of \(Y_1\), \(w(Y_1)\) is at most \(\beta \frac{1}{\tau }\sum _i w_i x_i\).
Proof
Recall that \(z^*_j \ge \tau \) for each \(e_j \in \mathcal {U}_{\text {he}}\). Consider \(x'_i = \min \{1,\frac{1}{\tau }x_i\}\). It is easy to see that \(x'\) is a feasible fractional solution for SC-LP to cover \(\mathcal {U}_{\text {he}}\) using sets in \(\mathcal {S}\). Since the set family is deletion-closed, and the integrality gap of the SC-LP is at most \(\beta \) for all instances in the family, there is an integral solution covering \(\mathcal {U}_{he}\) of cost at most \(\beta \sum _i w_i x'_i \le \frac{1}{\tau }\beta \sum _i w_i x_i\). \(\square \)
The expected cost of randomized rounding in the second step is easy to bound.
Lemma 4.4
The expected cost of \(Y_3\) is at most \(\frac{\ell }{\tau } \sum _{i} w_i x_i\).
An analog of the following lemma for PartitionVC was proved by Bera et al. (2014). However, in PartitonVC, each element is contained in at most two sets, which is crucially used in their proof. Consequently, their proof does not readily generalize to set systems with unbounded frequency. We rely on tools from submodularity to prove this lemma even in the general case of CCF.
Lemma 4.5
Fix a constraint k. If \(\tau \) is a sufficiently small but fixed constant, the probability that constraint k is satisfied after one round of randomized rounding is at least a fixed constant \(c_\tau \).
Before we give a proof of this lemma, we finish the rest of the analysis first. Let \(I_k =\{ i \mid S_i \text{ influences } \text{ constraint } k\}\). Note that for any \(i \in N\), \(|\{k \in [h]: S_i \in I_k \}| \le r\) by our sparsity assumption.
Lemma 4.6
Let \(\rho _k\) be the cost of fixing constraint k if it is not satisfied after randomized rounding. Then \(\rho _k \le c'_\tau \sum _{i \in I_k} w_i x_i\) for some constant \(c'_\tau \).
Proof
We will assume that \(\tau < (1-1/e)/2\). Let \(D = Y_1 \cup Y_2\) and let \(b'_k = b_k - f_k(D)\) be residual requirement of constraint k after choosing \(Y_1\) and \(Y_2\). Let \(\mathcal {U}' = \mathcal {U}\setminus \mathcal {U}_D\) be elements in the residual instance; all these are shallow elements. Consider the scaled solution \(x'\) where \(x'_i = 1\) if \(S_i \in D\) and \(x'_i = \frac{1}{\tau } x_i\) for other sets. For any shallow element \(e_j\) let \(z'_j = \min \{1,\sum _{i: j \in S_i} x'_i\}\); since \(e_j\) is shallow we have \(z'_j = \frac{1}{\tau }z_j = \sum _{i: j \in S_i, i \not \in D} x'_i\).
Recall from the description of the modified Greedy algorithm that a set \(S_i\) is in \(H_k \subseteq I_k\) iff adding \(S_i\) to D satisfies constraint k. In other words \(i \in H_k\) iff \(f_k(D+i) - f_k(D) \ge b'_k\). Suppose \(\sum _{i \in H_k} x'_i \ge 1/2\). Then it is not hard to see that the cheapest set from \(H_k\) will cover the residual requirement and has cost at most \(2 \sum _{i \in H_k} w_i x'_i\) and we are done. We now consider the case when \(\sum _{i \in H_k} x'_i < 1/2\). Let \(L_k = I_k \setminus H_k\). For each \(j \in \mathcal {U}'\) let \(z''_j = \sum _{i: j \in S_i, i \in L_k} x'_i\). We claim that \(\sum _{j \in \mathcal {U}'}A_{k,j}z''_j \ge \frac{1}{2 \tau } b'_k\). Since \(\tau \le (1-1/e)/2\) this implies \(\sum _{j \in \mathcal {U}'}A_{k,j}z''_j \ge \frac{1}{(1-1/e)} b'_k\). Assuming the claim, if we run Greedy on \(L_k\) to cover at least \(b'_k\) elements then the total cost, by Lemma 2.1, is at most \((1+e)\sum _{i \in L_k} w_i x'_i\); note that we use the fact that no set in \(L_k\) has coverage more than \(b'_k\) and hence \(c=1\) in applying Lemma 2.1.
We now prove the claim. Since the x, z satisfy KC inequalities:
$$\begin{aligned} \sum _{i \not \in D, i \in I_k} \min \{f_k(D+i) - f_k(D), b'_k\} x_i \ge b'_k. \end{aligned}$$
We split the LHS into two terms based on sets in \(H_k\) and \(L_k\). Note that if \(i \in H_k\) then \(f_k(D+i) -f_k(D) \ge b'_k\) and if \(i \in L_k\) then \(f_k(D+i) -f_k(D) < b'_k\). Furthermore, \(f_k(D+i) -f_k(D) \le \sum _{e_j \in S_i} A_{k,j}\). We thus have
$$\begin{aligned} \sum _{i \not \in D, i \in I_k} \min \{f_k(D+i) - f_k(D), b'_k\} x_i\le & {} \sum _{i \in H_k} b'_k x_i + \sum _{i \in L_k} x_i \sum _{e_j \in S_i} A_{k,j} \\\le & {} b'_k \sum _{i \in H_k} x_i + \sum _{i \in L_k} x_i \sum _{e_j \in S_i} A_{k,j} \end{aligned}$$
Putting together the preceding two inequalities and the condition that \(\sum _{i \in H_k} x'_i < 1/2\) (recall that \(x'_i = x_i/\tau \) for each \(i \in I_k\)),
$$\begin{aligned} \sum _{i \in L_k} x'_i \sum _{e_j \in S_i} A_{k,j} \ge \frac{1}{2\tau } b'_k. \end{aligned}$$
We have, by swapping the order of summation,
$$\begin{aligned} \sum _{i \in L_k} x'_i \sum _{e_j \in S_i} A_{k,j} \!=\! \sum _{e_j \in \cup _{i \in L_k}S_i} A_{k,j} \sum _{i \in L_k: e_j \in S_i} x'_i \!\le \! \sum _{j \in \mathcal {U}'} A_{k,j} \sum _{i \in L_k: e_j \in S_i} x'_i \!=\! \sum _{j \in \mathcal {U}'} A_{k,j} z''_j. \end{aligned}$$
The preceding two inequalities prove the claim. \(\square \)
With the preceding lemmas we can finish the analysis of the total expected cost of the sets output by the algorithm. From Lemma 4.5 the probability that any fixed constraint k is not satisfied after the randomized rounding step is at most \(c^{\ell }\), for some constant \(c < 1\). By choosing \(\ell \ge 1+ \log _{1/c} r\) we can reduce this probability to at most 1/r. Thus, as in the preceding section, the expected fixing cost is \(\sum _{k} \frac{1}{r} w(F_k)\). From Lemma 4.6,
$$\begin{aligned} \sum _k w(F_k) \le c' \sum _k \sum _{i \in I_k} w_i x_i \le c' r \sum _i w_i x_i \end{aligned}$$
since the given instance is r-sparse. Thus the expected fixing cost is at most \(c' \sum _i w_i x_i\). The cost of \(Y_1\) is \(O(\beta ) \sum _i w_i x_i\), the cost of \(Y_2\) is \(O(1) \sum _i w_ix_i\), and the expected cost of \(Y_3\) is \(O(\log r) \sum _i w_i x_i\). Putting together, the total expected cost is at most \(O(\beta + \log r) \sum _i w_i x_i\) where the constants depend on \(\tau \). We need to choose \(\tau \) to be sufficiently small to ensure that Lemma 4.5 holds. We do not attempt to optimize the constants or specify them here.
Submodularity and proof of Lemma 4.5: We follow some notation that we used in the proof of Lemma 4.6. Let \(D = Y_1 \cup Y_2\) and consider the residual instance obtained by removing the elements covered by D and reducing the coverage requirement of each constraint. The lemma is essentially only about the residual instance. Fix a constraint k and recall that \(b'_k\) is the residual coverage requirement and that each set in \(H_k\) fully satisfies the requirement by itself. Recall that \(x'_i = \frac{1}{\tau }x_i \le 1\) for each set \(i \not \in D\) and \(z'_j = \frac{1}{\tau } z_j = \sum _{i: e_j \in S_i} x'_i\) for each residual element \(e_j\). As in the proof of Lemma 4.6 we consider two cases. If \(\sum _{i \in H_k} x'_i \ge 1/2\) then with probability \((1-1/\sqrt{e})\) at least one set from \(H_k\) is picked and will satisfy the requirement by itself. Thus the interesting case is when \(\sum _{i \in H_k} x'_i < 1/2\). Let \(\mathcal {U}'' = \cup _{i \in L_k} S_i\). As we saw earlier, in this case
$$\begin{aligned} \sum _{j \in \mathcal {U}''} A_{k,j} \min \left\{ 1,\sum _{i: j \in S_i} x'_i\right\} \ge \frac{1}{2\tau }b'_k. \end{aligned}$$
For ease of notation we let \(N = L_k\) be a ground set. Consider the weighted coverage function \(g: 2^N \rightarrow \mathbb {R}_+\) where g(T) for \(T \subseteq L_k\) is given by \(\sum _{j \in \cup _{i \in T} S_i} A_{k,j}\). Then for a vector \(y \in [0,1]^N\) the quantity \(\sum _{j \in \mathcal {U}''} A_{k,j} \min \{1,\sum _{i: j \in S_i} y_i\}\) is the continuous extension \(\tilde{g}(y)\) discussed in Sect. 2. Thus we have \(\tilde{g}(x') \ge \frac{1}{2\tau } b'_k\). From Lemma 2.2, we have \(G(x') \ge (1-1/e)\frac{1}{2\tau } b'_k\) where G is the multilinear extension of g. If we choose \(\tau \le (1-1/e)/4\) then \(G(x') \ge 2b'_k\). Let Z be the random variable denoting the value of g(R) where \(R \simeq x'\). Independent random rounding of \(x'\) preserves \(G(x')\) in expectation by the definition of the multilinear extension, therefore \(\mathrm {E}[Z] = G(x') \ge 2b'_k\). Moreover, by Lemma 2.4, Z is concentrated around its expectation since \(G(i) \le b'_k\) for each \(i \in L_k\). An easy calculation shows that \(\Pr [Z< b'_k] \le e^{-1/4} < 0.78\). Thus with constant probability \(g(R) \ge b'_k\).
Solving the LP with KC inequalities: The proof of the performance guarantee of the algorithm relies on the fractional solution satisfying KC inequalities with respect to the set \(D = Y_1 \cup Y_2\). Thus, given a fractional solution (x, z) for the LP we can check the easy constraints in polynomial time and implement the first two steps of the algorithm. Once \(Y_1,Y_2\) are determined we have D and one can check if (x, z) satisfies KC inequalities with respect to D (for each row of A). If it does then the rest of the proof goes through and performance guarantee holds with respect to the cost of (x, z) which is a lower bound on \(\textsc {OPT}\). If some constraint does not satisfy the KC inequality with respect to D we can use this as a separation oracle in the Ellipsoid method.