Lecture 4: Random-order model for the k-secretary problem

Transcription

1 Algoritmos e Incerteza PUC-Rio INF2979, Lecture 4: Random-order model for the k-secretary problem Lecturer: Marco Molinaro April 3 rd Scribe: Joaquim Dias Garcia In this lecture we continue with the subject of random-order, models initially developed during Lecture 3. We shall go deeper in the models of this class by extending the Secretary problem to the k-secretary problem. The basic idea of Learning a Classifier exposed in the previous lecture will be applied to devise an algorithm. Finally, we will analyze the proposed algorithm using the probability tools of Lecture 3. Now let s go back a little The secretary problem: a brief review Going straight to the point and omitting some story telling presented in Lecture 3, the secretary problem goes like this: A problem instance is given by the set: {v 1, v 2,..., v n } where n is given and all v i are positive real numbers. These values will be presented to us in random order: v I1, v I2, v I3,... At each time step and we either pick the value presented or abandon it. If we pick the value v i that s our solution value and we cannot choose any other v j ; if we discard it we simply wait for the next v k. After n steps the process ends and the solution is the value we have chosen. The off-line optimal is clearly given by: OP T = max i v i, since the whole instance is known in advance. In Lecture 3, we developed the following algorithm A1: during steps 1 to n 2 compute M the maximum values of the numbers that show up; starting from step n + 1 simply select the first value that is larger than M 2 We are assuming n to be even just for simplicity of notation and modeling. Also from Lecture 3 we had: Lemma 0.1. E [A1] = 1 n! σ Σ n A1σ 1OP T, which means that A1 is 1-approximate 4 4 Here, Σ n is the set of permutations of the numbers 1 to n. This notation follows from the observation that every possible input that can be obtained from the instance {v 1, v 2,..., v n } is some permutation of a sequence of these values. 0.2 A digression... Besides the expected value of the algorithm, some other interesting questions could be raised, for instance: can we give a probability bound on the algorithm worst cases performance? 1

2 Example: Can we bound P vala1 1OP T? That is what is the probability upper bound 8 that A1 leads to bad solutions? We start by re-writing the probability of that event to take place as P vala1 18 OP T = P OP T vala1 78 OP T then note, from Lemma 0.1, that E [OP T vala1] 3 OP T. Now we use Markov s inequality: 4 P X a E [X] a, where X = OP T vala1 and a = 7 OP T. Clearly X is always positive since A1 is never 8 strictly better than OP T. Hence, P OP T vala1 78 OP T The final solution Theorem 0.2. There is an algorithm that is 1 -approximated and this is the best one can do even e with randomized algorithms. The main idea is to sample data up to step n and be clever in bounding. e For more on the Secretary problem the reader is referred to [4] for historical review and to [5] for many extensions. The secretary problem also appear in the fun math book [9], in which the problem is also named the marriage problem. 1 Learning a Classifier In this course we will study two main techniques to tackle problems in the random order model class: Learn a Classifier for example [3, 1, 8] and many others The Primal method for example [6] The rest of this lecture will be focused in the first method. 1.1 The k-scretary Problem This problem follows the same set up of the Secretary Problem, but it differs in the following: instead of choosing 1 item we have to choose k of them in order to maximize our objective function, which is the sum of the k items chosen [7, 2]. 2

3 It is worth emphasizing that we keep the rule of selecting an item and never be able to discard it or discarding it and we never have a chance to select it again. Also, the algorithm will do nothing if k items have already been chosen. Once again, the off-line optimal solution is trivial: just select the k most valuable items, the value, OP T, is their sum. The following assumption will useful to simply the analysis: Assumption avoid ties. From now on, all the items in the problem instance will have different values This 1.2 The off-line optimal For the sake of clarity, we describe the off-line optimal algorithm because it will present important insight for the on-line solution. First, let s plot the input data in a line where the largest values are on the left and smaller values in the right: Larger Smaller It is very clear that in order to optimize our objective function we must: Find the k th largest value: λ which is one of the samples v i ; Choose the all items i with v i λ Solution λ Larger Smaller Figure 1: Optimal solution k best values in red, optimal classifier λ is one of the values v i The following definition will be very important to define optimal solution sets and other solution candidates. Definition 1.1. pickλ = {i : v i λ}, i.e., the set of all item with value larger than λ. 3

4 We shall study the set pickλ to better understand the problem. Given λ, the k th largest value, its trivial to note that pickλ is the optimal solution, i.e. the one with largest value: Moreover we can easily note that: OP T = valpickλ If λ < λ then pickλ > pickλ = k, thus λ gives an infeasible solution; If λ > λ then valpickλ < valpickλ where X is the cardinality of the set X. Therefore we must choose λ carefully since the first case leads to infeasible solutions and the second one leads to sub-optimal objective values. Now, suppose that we choose λ such that pickλ = k, what is the value of this feasible 2 solution? We can certainly bound the value of such solution. Firstly, the number of elements in OP T not picked by λ, namely pickλ pickλ, is exactly the size of the picked elements pickλ. Moreover, the non-picked elements pickλ pickλ are all smaller or equal to the minimum value picked by λ, namely minpickλ. Together, these imply that valpickλ 1 2 valpickλ = 1 OP T. 2 The same argument can be used to analyze the case where pickλ = βk for any β [0, 1]. Lemma 1.2. If pickλ = βk, with β [0, 1], then valpickλ βop T we use the approximation symbol to emphasize that we are not considering rounding. 1.3 An online algorithm In order to construct an online algorithm via the Learn a classifier technique, we combine what we have learned in the previous section: Let s look for some value of λ such that λ λ and λ > λ so that we have a near optimal and feasible solution; with the core idea of our online algorithm for the secretary problem: use the first values of the input sequence to estimate a classifier or a cutoff value. To be generic, we can use the first αn values of the sequence to estimate ˆλ such that pickˆλ k in the whole data. A ˆλ with such property will, on average, pick αk items among the first αn in the instance: pickˆλ S α αk, where S α = {v I1,, v Iαn } is the set of the first αn elements that appear in the instance. We have the following Algorithm A2: 1. Observe the item from step 1 to αn; 2. Compute the largest value ˆλ that would make us pick 1 ɛαk items in the sample; ˆλ will actually be one of the v i ; 3. From step αn + 1 to n, choose all items with value larger than ˆλ. 4

5 In the second step, the multiplier 1 ɛ is important for some reason we shall see soon. We have the following theorem related to this algorithm: Theorem 1.3. With the appropriate choice of ɛ and α, Algorithm A2 is an 1 approximation. logn k Analyzing Algorithm A2 The main component of the analysis is the following. Theorem 1.4. In Algorithm A2, P { pickˆλ > k} { pickˆλ < 1 2ɛk} < ɛ for an appropriate choice of ɛ, α. The proof of this theorem follows from the Chernoff + union bound approach. So the first piece we need is the following concentration inequality. Lemma 1.5. Given a wrong classifier v i, namely it satisfies either pickv i > k or pickv i < 1 2ɛk. Then P pickv i S α = 1 ɛαk exp{ ɛ 2 αk}. In particular, the probability that such wrong classifier is even considered by Algorithm A2 in its Step 2 is at most exp{ ɛ 2 αk}. Proof sketch. To get the idea, consider a very wrong v i is wrong that picks pickv i 2k items. The number of elements that v i picks in the sample on average is 2αk; picking only 1 ɛαk on the sample is vary far from this average, so by Chernoff/Bernstein s inequality it will happen with tiny probability. We will ignore the multiplier 1 ɛ in this example because it will be inconsequential and makes the expressions cleaner. In a bit more detail, we want to find the probability that P pickv i S α = αk, which is equivalent to P { pickv i S α αk} { pickv i S α αk}, which is at most P pickv i S α αk. We are going to use the Bernstein inequality: P Z E [Z] + t exp { min { t 2, t}}, so 4V [Z] we use the fact that E pickv S α = 2αk and re-write the probability we are trying to bound: P pickv S α 2αk αk. To see we can use Bernstein s inequality, we define the R.V. X t which is one if v picks the element that shows up in step t and zero otherwise. Hence, pickv S α = X X αn If we had sampled the values v i s with replacement the X i s would all be independent and we could use Bernstein s inequality. This is not the case: the random-order model means we are sampling the v i s without replacement. But as we mentioned last lecture, we know that Bernstein do works fine in this sampling without replacement. So we can bound: { { P X X αn 2αk αk exp min 5 2αk 2 4V [X X αn ], αk }},

6 where V [X] is the variance of the random variable X. We also saw in Lecture 3 that given the independent random variables x i [0, 1], V [X X αk ] E [X X αk ] = 2αk, and this also holds in the sampling without replacement case. So, we can re-write the bound we had in the following form { { }} αk 2 exp min V [X X αn ], αk { { }} αk 2 exp min 2αk, αk e αk. This gives the bound of Lemma 1.5 for this case. The same argument can be done for any wrong classifier satisfying the hypothesis of the lemma, in particular for the case pickv i < 1 ɛk, that is, the argument is symmetrical. This concludes the proof sketch of the lemma. Now we go back to prove Theorem 1.4, since we know how to bound a single bad choice of the classifier we will use union bound to bound the probability of learning any bad classifier. Proof of Theorem 1.4. In order to be protected of all bad classifiers we use Union Bound. Firstly, define BAD = {v i : { pick ˆv i > k} { pick ˆv i < 1 2ɛk}}, hence, we can write our ultimate goal P { pickˆλ > k} { pickˆλ < 1 2ɛk} By union bound, we have P ˆλ = v = = P ˆλ = v. P ˆλ = v by definition of v BAD. By construction of the algorithm Step 2 P ˆλ = v = P v picks αk1 ɛ elements in the sample. Now, we apply Lemma 1.5 and we have P v picks αk1 ɛ elements in the sample By choosing appropriate values for α and ɛ we get n exp{ ɛ 2 αk} ɛ. exp{ ɛ 2 αk} n exp{ ɛ 2 αk}. It is possible to convert the probability bound of theorem 1.4 into the following: 6

7 [ ] Corollary 1.6. With probability greater than 1 ɛ we have E valpickˆλ 1 2ɛOP T Proof. Let E be the event that the bound from Theorem 1.4 holds. Then [ ] E valpickˆλ = E valpickˆλ E PrE + E valpickˆλ Ē PrĒ [ ] E valpickˆλ E PrE E valpickˆλ E 1 ɛ, where the first inequality follows from the fact that in every scenario valpickˆλ is non-negative. But in every scenario under the event E, we have the number of picked items is pickˆλ [1[ 2ɛk, k], which] together with Lemma 1.2 gives valpickˆλ 1 2ɛOP T. In particular, E valpickˆλ E 1 2ɛOP T. This concludes the proof. However, our Algorithm A2 used the first αn values in the sample to estimate ˆλ, therefore none of these values could be chosen. Thus we have: [ ] E [vala2] = E valpickˆλ E valpickˆλ S α E valpickˆλ E [ OP T 1 ɛα S α ], where OP T 1 ɛα S α is the value of picking the best 1 ɛαk items in the sample S α notice by definition pickˆλ = 1 ɛαk. Finally, we can bound the expected value of this optimal solution in the sample. Lemma 1.7. E [ OP T 1 ɛα S α ] αop T. Proof. The proof is actually a duality-based argument, which actually gives a cleaner and more intuitive interpretation see for example [1]; we have streamlined the exposition so as to not require knowledge of duality. We use [x] + = max{x, 0} to denote the positive part of x. Recalling that OP T is the set of items of value at least λ there are k of them, we break up the value of these items as v i = λ + [v i λ] + and obtain OP T = kλ + i [v i λ ] +. 1 Notice this sum can range over all items because the ones not included in OP T will have [v i λ ] + = 0. Now let U be the set of the top k = 1 ɛαk in the sample S α ; we want to show EvalU αop T. Since for all items we have the upper bound v i λ + [v i λ ] +, we obtain EvalU U λ + E v U[v λ ] + U λ + E v S α [v λ ]

8 The first term in the RHS is exactly 1 ɛαλ, and the second term is exactly 1 ɛα i [v i λ ] +, since each item has probability α of being in the sample S α. Together these imply that EvalU α λ + [v i λ ] +. i Putting this together with inequality 1 we obtain the desired bound. Thus, we finally conclude that E [vala2] 1 2ɛ αop T. Choosing ɛ = α = logn/ɛ 1/3 should recover Theorem 1.3. k An important observation is that the argument can be improved by using some technique other than union bound, which in this case is weak and give really loose bounds, to obtain an optimal 1 1 k -approximation, recovering the result of [7]. References [1] S. Agrawal, Z. Wang, and Y. Ye. A dynamic near-optimal algorithm for online linear programming. Operations Research, 624: , [2] M. Babaioff, N. Immorlica, D. Kempe, and R. Kleinberg. A knapsack secretary problem with applications. In Approximation, randomization, and combinatorial optimization. Algorithms and techniques, pages Springer, [3] N. R. Devenur and T. P. Hayes. The adwords problem: online keyword matching with budgeted bidders under random permutations. In ACM Conference on Electronic Commerce, pages 71 78, [4] T. S. Ferguson. Who solved the secretary problem? Statistical science, pages , [5] P. Freeman. The secretary problem and its extensions: A review. International Statistical Review/Revue Internationale de Statistique, pages , [6] T. Kesselheim, A. Tönnis, K. Radke, and B. Vöcking. Primal beats dual on online packing lps in the random-order model. In Proceedings of the 46th Annual ACM Symposium on Theory of Computing, pages ACM, [7] R. Kleinberg. A multiple-choice secretary algorithm with applications to online auctions. In Proceedings of the sixteenth annual ACM-SIAM symposium on Discrete algorithms, pages Society for Industrial and Applied Mathematics, [8] M. Molinaro and R. Ravi. The geometry of online packing linear programs. Mathematics of Operations Research, 391:46 59, [9] M. Parker. Things to make and do in the fourth dimension. Penguin UK,