# The Online Set Cover Problem

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3 In the following we analyze the performance of the algorithm and explain which sets to add to the cover C in the weight augmentation step. Lemma. The total number of iterations in which a weight augmentation step is performed is at most C OP T log m + 2. Proof. For each subset S, w S 2 always holds, since the algorithm maintains in all iterations that w j 2 for all elements j. Consider an iteration in which the adversary gives element j. A weight augmentation is performed in this iteration if and only if w j <. When doing a weight augmentation, the weight of at least one set belonging to C OP T is multiplied by a factor greater than or equal to two. Since the weight of each set is initially /2m and at the end at most 2, it follows that each set can participate in at most log4m iterations in which a weight augmentation is performed. Hence, the desired result follows. Lemma 2. Consider an iteration in which a weight augmentation is performed. Let Φ s and Φ e be the values of the potential function Φ before and after the iteration, respectively. Then, there exist at most 4 log n sets that can be added to C during the iteration such that Φ e Φ s. Proof. The proof is probabilistic. Suppose that the adversary gives element j in the iteration. For each set S S j, let w S and w S + δ S denote the weight of S before and after the iteration, respectively. Define δ j = j δ S. The algorithm maintains that w j + δ j = j w S + δ S 2. We now explain which sets from S j are added to C. Repeat 4 log n times: choose at most one set from S j such that each set S S j is chosen with probability δ S/2. This can be implemented by choosing a number uniformly at random in [0, ], since δ j/2. Consider an element j X such that j / C. Let the weight of j before the iteration be w j and let the weight after the iteration be w j +δ j. Element j contributes before the iteration to the potential function the value n 2w j. In each random choice, the probability that we do not choose a set containing element j is δ j. The probability that 2 this happens in all the 4 log n random choices is therefore δ j 2 4 log n n 2δ j. Therefore, the expected contribution of element j to the potential function after the iteration is at most n 2δ j n 2w j +δ j = n 2w j. By linearity of expectation it follows that Exp[Φ e] Φ s. Hence, there exists a choice of at most 4 log n sets such that Φ e Φ s. Theorem 3. At the end of the algorithm, C is a feasible cover of X and C is O C OP T log m log n. Proof. Initially, the value of the potential function Φ is less than n n = n 2. It follows from Lemma 2 that Φ is non-increasing throughout the iterations. Therefore, if w j for an element j during the algorithm, then j C, otherwise Φ n 2w j n 2. Hence, C is a feasible cover. It follows from Lemma that the number of iterations is at most C OP T log m + 2. By Lemma 2, in each iteration we choose at most 4 log n sets to C. Therefore, the total number of sets chosen by the algorithm is as claimed. Remark: If every element appears in at most d sets, then the algorithm can be modified by starting with the weights w S = /2d for each S S, and the competitive factor can be improved in this case to Olog d log n. 3. THE WEIGHTED CASE We describe in this section an Olog m log n competitive algorithm for the weighted case. For each set S S, a positive cost c S is associated with the set. The cost of the optimal solution, cc OP T, is denoted by α. Note, first, that we may assume, by doubling, that the value of α is known up to a factor of 2. Indeed, we can start guessing α = min c S, and run the algorithm with this value of the optimal solution. If it turns out that the value of the optimal solution is already at least twice our current guess for it, that is, the cost of C exceeds Θα log m log n, then we forget about all sets chosen so far to C, update the value of α by doubling it, and continue on. We note that the cost of the sets that we have forgotten about can increase the cost of our solution by at most a factor of 2, since the value of α was doubled in each step. We thus assume that α is known. Hence, we can ignore all sets of cost exceeding α, and also choose all sets of cost at most α/m to C. Thus, we assume that all costs are between α/m and α, and further normalize the costs so that the minimum cost is and hence the maximum cost is at most m. We now describe an online algorithm with competitive factor 6 + o log m log n, assuming that α is known, where the o term tends to zero as n and m tend to infinity. It is worth noting that the constant 6+o can be improved to 2+o by being somewhat more careful, but we prefer to describe the algorithm with the inferior constant, to simplify the computation. From now on, all o terms denote terms that tend to zero as n and m tend to infinity. All logarithms are in the natural basis e. As in the unweighed case, the algorithm maintains a weight w S > 0 for each S S. The weights can only increase during the run of the algorithm. Initially w S = /m 2 for each S S. The weight of each element j X is defined as w j = j w S, where S j denotes the collection of sets containing element j. Initially, the algorithm starts with the empty cover C =. Define C to be the set of all elements covered by the members of C. The following potential function is used throughout the algorithm: Φ = n 2w j + n exp c Sχ CS 3w Sc S log n. j C The function χ C above is the characteristic function of C, that is, χ CS = if S C, and χ CS = 0 otherwise. We now give a high level description of a single iteration of the algorithm in which the adversary gives an element j and the algorithm chooses sets that cover it.. If w j, then do not add any new set to C and do not update the weights of the sets. 2. Else, w j <, perform a sequence of weight augmentation steps as long as w j < : a For each S S j, w S w S + n c S

4 b Choose from S j sets to C so that the value of the potential function Φ does not exceed its value before the weight augmentation. In the following we analyze the performance of the algorithm and explain which sets to add to the cover C in the weight augmentation step. Lemma 4. The total number of weight augmentation steps performed during the algorithm is at most m onα log m. n S C OP T n c S + log Proof. Obviously, for each subset S, w S + n c S always holds. Consider an iteration in which the adversary gives element j. A weight augmentation is performed in this iteration as long as w j <. When doing a weight augmentation, the weight of at least one set S C OP T is multiplied by a factor of + n c S. Since the weight of each set is initially /m 2 and at the end at most + /n, it follows that each set S participates in at most n c S + log m 2 + n steps in which a weight augmentation is performed. Hence, the desired result follows. Lemma 5. The following is maintained throughout the algorithm: w Sc S 2 + oα log m. Proof. Consider an iteration in which the adversary gives element j. We start with weights satisfying j w S, and increase the weight of each set S in S j by w S/n c S in each step. Thus, the total increase of the quantity wscs in a step does not exceed j w S c S = w S nc S n n. j Initially, wscs m m =, and the result thus m follows from Lemma 4 that bounds 2 the number of weight augmentation steps. Lemma 6. Consider a step in which a weight augmentation is performed. Let Φ s and Φ e be the values of the potential function Φ before and after the step, respectively. Then, there exist sets that can be added to C during the step such that Φ e Φ s. Proof. The proof is probabilistic. Suppose that the adversary gives element j in the iteration. For each set S S j, let w S and w S + δ S denote the weight of S before and after the step, respectively. Define δ j = j δ S. We now explain which sets from S j are added to C. Independently, for each S S j, set S is added to C with probabilty n 2δ S. This is roughly the same as choosing S with probability δ S/2 and repeating this 4 log n times. Let C denote the cover obtained from C by adding to it the randomly chosen sets. We first bound the expected value of the first term of the potential function. This is similar to the unweighted case. Consider an element j X such that j / C. Let the weight of j before the step be w j and let the weight after the step be w j + δ j. Element j contributes before the step to the first term of the potential function the value n 2w j. The probability that we do not choose a set containing element j is n 2δ j. Therefore, the expected contribution of element j to the potential function after the step is at most n 2δ j n 2w j +δ j = n 2w j. By linearity of expectation it follows that the expected value of j C n2w j after the step is precisely its value before the step. It remains to bound the expected value of the second term of the potential function. Let T = n exp c Sχ CS 3w Sc S log n denote the value of the second term of the potential function before the step, and let T denote the same term with respect to the cover C. Since the choices of different sets are independent, and the random variable T can be viewed as a product of independent random variables, its expected value is the product of the corresponding expected values. Therefore, Exp[T ] = n exp 3w S + δ Sc S log n Exp [ exp ] csχ C S Fix an S S. If the weight of S was not changed during the step, or if it was changed even though S C i.e., j is covered, but w j <, then the expected value of exp c Sχ C S is precisely its value before the step. Therefore, if we let S denote the family of all sets S S \ C whose weights were changed during the weight augmentation step, then the expected value in is precisely Exp[T ] = T exp 3δ Sc S log n [ Exp exp csχ C S ]. 2 Consider, now, an S S. S did not belong to C before the step, and after the step, the probability that χ C S = is n 2δ S. Thus, [ exp ] csχ C S Exp = n 2δ S + n 2δ S cs exp cs 2δ S log n + 2δ S log n exp cs = + 2δ S log n exp δ S log n 3cS 7 4α 3δSc S log n exp. 8 Here, 5 follows since for all x 0 and z, e x + e x z x + x z, 7 follows since e y 3y/2 for all 0 y /2, and 8 follows since + x e x for all x 0.

5 Plugging in 2, we conclude that the expected value of the second term after the step and random choices is at most Exp[T ] = T exp 3δ Sc S log n exp 3δScS log n T. By linearity of expectation it now follows that Exp[Φ e] Φ s. Therefore, there exists a choice of sets from S j such that Φ e Φ s. Theorem 7. Throughout the algorithm, the following properties hold. ievery j X of weight w j is covered, that is, j C. ii S C cs 6 + oα log m log n. Proof. Initially, the value of the potential function Φ is at most n n 2/m + n < n 2, and hence it stays smaller than n 2 during the whole algorithm. Therefore, if w j for some j during the process, then j C, since otherwise the contribution of the term n 2w j itself would be at least n 2. This proves part i. To prove part ii, note that by the same argument, throughout the algorithm n exp Therefore, c Sχ CS 3w Sc S log n < n 2. c Sχ CS 3w Sc S log n + log n, and the desired result follows from Lemma DERANDOMIZATION The choices of the various sets S to be added to C after each iteration can be done deterministically and efficiently, by the method of conditional probabilities, c.f., e.g., [], chapter 5. In fact, this can be done here in a very simple way. A close look at the proof in the last section shows that we can simply decide, after each weight augmentation, for each set whose weight has been increased in its turn, if we add it to C or not, making sure that the potential function will not increase after each such choice. The details will appear in the full version of the paper. We note that knowing in advance the set cover instance is crucial for making the online algorithm deterministic. To see this, consider the following example where the sets containing an element j are revealed only when the adversary gives element j to the algorithm. The set cover instance contains elements {, 2,..., n} and sets S,... S n. The adversary gives the algorithm the elements in the order, 2,..., n. Assume inductively that the algorithm has used sets S,... S i for covering elements,..., i. When element i + is given, the sets containing it are S i+,..., S n. Without loss of generality we can assume that the algorithm used set S i+ for covering element i +. Clearly, the adversary can cover all the elements with a single set, S n, yielding a competitive ratio of n. 5. LOWER BOUND In this section we show that for every fixed δ > 0 and every m and n satisfying log n m e n/2 δ, 9 there is a family F of m subsets of X, X = n, so that the competitive ratio of any deterministic online algorithm for the unweighted online set cover problem with X and F is at least Ω log n log m log log m + log log n. 0 Before describing the proof, we note that the assumption 9 is essentially optimal. Let OP T denote the value of the optimum off-line solution to the problem. Note, first, that the problem has a trivial algorithm with competitive ratio m that simply takes all sets after the first element appears, showing that for m < log n ɛ the above lower bound 0 fails. In fact, we may always assume that m log 2 n, since all the elements that lie in the same cell of the Venn diagram of the sets in F can be replaced by a single element, without any change in the problem. It is also easy to see that the problem has a simple algorithm with competitive ratio O n; when the first element arrives, we pick, repeatedly, all sets that cover at least n members of X among those not covered so far. Note that after this process terminates, there are at most OP T n yet uncovered elements that can appear, and hence even if from now on we pick an arbitrarily chosen set for each new element, the algorithm will choose at most n+op T n sets altogether. By being a bit more careful we can actually get an O n/op T -competitive algorithm this way. The details are left to the reader. This discussion shows that for m > e n/2+δ the lower bound 0 also fails. Therefore, both inequalities in the assumption 9 are needed. Proposition 8. Let X = {0,, 2,..., n } be a set of n = 2 k elements. For each i k, let F i be the set of all elements j of X so that the ith bit in the binary representation of j is on, and let F be the family of all k sets F i. Then, the competitive ratio of the best deterministic algorithm for the online set cover problem X, F is F = k = log 2 n. Proof. The adversary starts by giving the number n in which all bits are on. If the algorithm covers it by F i, then the adversary gives the number in which all bits are on besides the i th bit. The algorithm covers it by F i2 and the adversary gives the number in which all bits are on besides the i th and i 2th bits, and so on. Clearly, the algorithm will have to choose this way all k sets, while the optimal solution consists of only one set: the last set chosen by the algorithm. The above proposition and some obvious modification of the family F for bigger values of m implies that the lower bound 0 holds for all m satisfying, say, log 2 n m log 2 n 3. It thus remains to establish the lower bound for pairs n, m satisfying log n 3 m e n/2 δ. This is done in what follows. Let k, r be positive integers. Suppose n 2 k kr 2, and let X, X 2,..., X kr 2 be kr 2 pairwise disjoint blocks of elements in X = {, 2,..., n}, each of size 2 k. For a block X b and a bit location t, with t k, let X b t denote the set

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