Comp Online Algorithms

Transcription

1 Comp Online Algorithms Assignment 1: Introduction, Searching & List Update Shahin Kamalli University of Manitoa - Fall 2018 Due: Thursday, Octoer 4th at 11:59 pm Octoer 29, 2018 The real prolem is not whether machines think ut whether men do. B.F.Skinner Please pay attention to the followings when preparing/sumitting your assignment: All prolems are written prolems. There are six prolems with a total of 74 marks plus 108 onus marks! If you feel the assignment is too hard (or too simple), do not panic! To some extent, this assignment serves to indicate how easy/hard the future assignments/exam should e. More importantly, the assignment is designed to raise a group discussion. I would love to see a discussion on Piazza and when you are confused drop a hint aout a prolem. Think of the assignment as a project that you need to work on together on Piazza. If you are not active on Piazza, you will miss the hints that I plan to drop. Do not approach the onus prolems efore finishing the work on the mandatory parts. Bonus prolems are much harder and you are not expected to know them for your final exam. Indeed one of the onus prolems has 100 marks and is a great topic for a research project. If you are interested, come and talk to me aout it (if you solve it, you make me impressed and very happy). Let me e redundant: if you have any question related to the assignment, you are encouraged to post it on Piazza. Note that you can sumit anonymously. Also, instead of ing me, you can always write a private note in Piazza. It is likely that I drop hints when a question is posted pulicly on Piazza (ecause all students can enefit from it). It is not the case when you ask questions in s or during office hours. Do not post any answer or hint on Piazza. You are welcome to discuss the prolems with your friends (or enemies). But you should write your answers individually. You might e interviewed aout your answers. Be careful not to accidentally copy. Sumit your answers electronically using UMLearn. You should sumit a single pdf file. You are encouraged to prepare your assignment using L A TEX. The LaTex file for this assignment is posted.

2 Prolem 1 Ski-rental & Randomization [4+4 = 8 marks] I) Consider the following randomized algorithm for the ski-rental prolem: with a chance of 1, uy the equipment at the eginning, and with a chance of 1 always rent and never uy. Recall that the cost of uying is and the cost of renting is 1 per day; let x denote the numer of days that the the player goes skiing. a) What is the expected cost of the algorithm in terms of and x? ) Is the algorithm competitive? You need to either prove a constant ound for the competitive ratio OR show via adversarial input that the competitive ratio is not constant. a) The cost of the algorithm is with a chance of 1 and x with a chance of 1 1. On expectation, it is + 1 x = 1+x/. ) Despite using randomization, the algorithm is not competitive. Consider the following adversarial input: the adversary selects a very large x, for example x ω(2 ). The expected cost of the algorithm grows with x while the cost of Opt is (it knows x is pretty large and uys at the eginning). The competitive ratio will e at least ω( x 2 ) which is unounded for large values of. II) consider another randomized algorithm that works as follows. It rents equipment for the first 1 days. For each day that follows, it uys the equipment with a chance of 1/2 and rents it otherwise (note that if the algorithm uys the equipment at some day, it oviously does not make a decision in the days that follows). What is the competitive ratio of the algorithm? To answer this question, put yourself in the shoes of an adversary who wants to make a worst-case scenario for the algorithm. For sufficiently large values of x, the algorithms rents for the first 1 days, and, on average, it rents for another two days ( = 2). So, the adversary chooses x = + 3. The cost of Opt will e and the expected cost of the algorithm will e + 1 (for renting) plus (for uying) which gives The competitive ratio will e 2+1. Note that a smaller value of x decreases the expected cost of algorithm and larger x does not change its cost. So, this particular choice of x reflects the worst-case scenario. 2

3 Prolem 2 Path-cow Prolem [6+6+8 = 20 marks] I) Consider the following algorithm for path-cow prolem. The cow starts at the origin, moves x = 1 unit to the right. If the target is not found, the cow comes ack to the origin and goes x = 1 unit to the left. If the target is not found, the cow comes ack to the origin and repeats this procedure with x = 2, 4,..., 2 i,... until the target is found. a) Assume the cow finds the hole at distance u from the origin on its left. Assume the largest power of 2 which is smaller than u is 2 k. What is the total distance moved y the cow? ) Where does the adversary place the hole in order to harm the algorithm? c) What is the competitive ratio of this algorithm? a) The total moved distance would e: 4( k ) k+1 + u = 6 2 k+1 + u 4. ) Adversary places the hole at distance u = 2 k + ɛ on the left. The cost of Opt will e u = 2 k + ɛ. c) The competitive ratio would e: 6 2k+1 +u 4 u small ɛ and large u). = k k +ɛ k+1 2 k = 13 (to get this ratio, the adversary chooses II) In part I we assumed the algorithm is deterministic and the first move is to the right. Consider the same algorithm in which the first move is randomly selected to e to the right or left (each with a chance of 1/2). What the competitive ratio of this randomized algorithm is? As efore, let 2 k e the largest power of 2 smaller than u. We consider the algorithm Alg1 whose first move is to the right and Alg2 whose first move is to the left. The total cost of the two algorithms is Alg1 + Alg2 = 4( k ) k+1 + u + 4( k ) + u = 8 2 k u k+1 So, the expected cost of the randomized algorithm is 4 2 k k + u 4 = 2 k k + u 4. As efore, the adversary places the target at distance 2 k + ɛ (this time it does not matter on which side). The competitive ratio will e: 2 k k+1 + u 4 u = 1 + 2k k k 11 III) Assume instead of a path, we have a inary tree with each edge having a length of 1. Originally, the cow is located at the root. First, she moves to the left child; if the target is not found, she goes ack to the root and then moves to the right child. If the target is still not found, she returns to the root and goes to visit all nodes of depth 2 on the left (i.e., at distance 2 from the root on the left). After checking these nodes, the cow returns to the root and repeats the same for nodes of depth 2 in the right side of the root. This procedure is repeated until at some point the target is found. Assume the cow uses depth first strategy to check nodes of depth i. In a nutshell, the algorithm works y working in rounds, where at round i it visits all vertices of depth 2 i. What is the competitive ratio of this algorithm? To answer, assume the target is at depth k and write the competitive ratio in terms of k. As efore, you have to indicate where the adversary places the target and deduce the competitive ratio accordingly. At step i, nodes of depth i are visited. The induced inary sutree is formed y 2 i leaves and 2 i+1 2 edges. Each of these edges are visited 2 times in a round. So, the cost of the algorithm at round i is 2 i+2 4. In the worst case, the target is located at depth k and is the last vertex checked y the algorithm at round k. In this case, the distance moved y round k is 2 k+2 4 k (the last deduction is ecause the algorithm does not return to the root after finding the target). On the other hand, the cost of Opt is k. The competitive ratio will e ( ) + ( ) (2 (k 1)+2 2) + (2 k+2 2 k) = (2 k+3 1) ( ) 2k k = 2 k+3 3k 8 The competitive ratio will e 2k+3 3k 8 k. 3

4 Prolem 3 Online Bidding and Advice [4 + 6 = 10 marks] We saw in the class that a simple douling approach gives the est competitive ratio that a deterministic online idding algorithm can achieve. That ratio was 4. In this prolem, we examine the power of advice and randomization for this prolem; asically we want to show that advice is stronger that randomization. Consider two determnistic algorithms Alg1 and Alg2, where Alg1 guesses are 1, 4, 16,... 4 i and Alg2 guesses are 2, 8, 32,..., 2 4 i. a) Consider an algorithm that flips a fair coin at the eginning and randomly chooses etween Alg1 and Alg2, and uses the guesses of the selected algorithm. What is the competitive ratio of this randomized algorithm? The adversary chooses u = 2 k + ɛ. The sum of the costs of the two algorithms is k + 2 k k+2 = 2 k+3 1. The expected cost is therefore 2 k+2 1/2 and the competitive ratio is 2k Note that this no etter 2 k +ɛ than the deterministic algorithm. ) Assume an algorithm that receives 1 it of advice. For each instance of the prolem the advice it indicates the algorithm which has smaller cost etween Alg1 and Alg2. What is the competitive ratio of the algorithm with 1 it of advice? As efore, assume the adversary chooses u = 2 k + ɛ. If k is even, Alg2 has cost k k+1 = 2( (k 2)/2 + 4 k/2 ) = 2 4k/ /3 2 k. The ratio etween the cost of Alg2 and Opt will converge to 8/3 for large values of k. If k is odd, Alg1 has cost k k+1 = (k 1)/2 + 4 (k+1)/2 = 4(k+1)/ k. Like efore the ratio etween the cost of the algorithm and Opt approaches to 8/3 for large values of k. Note that only one it of advice resulted in an algorithm with etter competitive ratio than any purely-online algorithm. 4

5 Prolem 4 Clustering & Advice [5 + 7 = 12 marks] Considering the online clustering prolem. Recall that in this prolem, we need to partition a sequence of online points into k clusters so that the maximum diameter of clusters is minimized. Assume the length of the input sequence (i.e., the numer of points) is n. a) Show that O(n log k) its of advice are sufficient to achieve an optimal clustering. To get the full mark, you need to precisely indicate i) what the advice encodes? ii) how the algorithm works? iii) why the algorithm is optimal? Consider an optimal clustering Opt. Let c(x) indicates the cluster at which x elongs to in the final clustering of Opt. The four steps for an algorithm with advice are as follows: what is the advice? for each point x, the advice indicates c(x). what is the size of advice? the value of c(x) can e encoded in O(log k). This is ecause there are at most k clusters in the final packing of Opt. So, the total advice size is O(n log k). how the algorithm uses advice? the algorithm places each input point x in cluster c(x) (indicated y advice). why the algorithm is optimal? the algorithm creates the same clustering as Opt. ) The competitive ratio of the est deterministic algorithm that we saw in the class was 8. Consider inputs for which the cost of Opt is a large value. Show that for any such input σ, provided with O(log(Opt(σ))) its of advice, we can achieve a competitive ratio of 2. Here, Opt(σ) is the cost of Opt for σ. To get the full mark, you need to precisely indicate i) what the advice encodes? ii) how the algorithm works? iii) why the algorithm is 2-competitive? Let Opt e an optimal algorithm, and d Opt e the maximum diameter of any cluster in the clustering of Opt, i.e., cost(opt) = d Opt. Here are the four steps for an scheme with advice: what is the advice? the value of d Opt. what is the size of advice? the value of d Opt can e encoded in O(log(d Opt )). So, the size of advice is O(log(d Opt ). how the algorithm uses advice? the algorithm works like the last step of the algorithm we saw in the class. It maintains k clusters and for each cluster considers a center as the first point in the cluster. Now, for any input point x, it works as follows: if x is within distance d Opt from any center c i, place x in the same cluster as c i. otherwise, create a new cluster with center x. We claim that the numer of clusters is no more than k. Consider otherwise; since the pairwise distance of all centres is more than d Opt, there will e at least k + 1 centres with pairwise distance more than d Opt. By the pigeon-hole principle, Opt places at least two of these centres in the same cluster. It means the maximum diameter of a cluster in Opt s clustering is more than d Opt, which is a contradiction. why the algorithm is 2-competitive? the distance of any two points x and y that elong to the same cluster is at most d(x, c) + d(c, y) where c is the center of the cluster. By the definition of the algorithm, d(x, c) d Opt. So, the d(x, y) 2d Opt. This implies the diameter of any cluster is at most 2d Opt. 5

6 Prolem 5 List-Update Algorithms [ = marks] Consider the Move-By-Bit algorithm for the list update prolem. Here, each item has a it associated with it. At the eginning, all its are 0. After an access to an item x, Move-By-Bit moves it to the front if the it of x is 1; otherwise it keeps x at its position. In addition, after each access, the it of the accessed item is flipped. a) Use an adversarial argument to show that competitive ratio of Move-By-Bit is at least 2. You should show the worst-case sequence and indicate the cost of Opt and Move-By-Bit. The adversarial sequence for list a 1 a 2... a m 1 a m will repeat the following sequence: a m, a m, a m 1, a m 1,..., a 1, a 1 The cost of the algorithm for a sequence of length n is n m. The cost of Mtf (or the static offline algorithm we saw in the class) will e nm/2, i.e., the cost of Opt is at most nm/2. Hence, the competitive ratio of Move-To-Front- Every-Other-Access algorithm is at least 2. ) Use a potential function argument to show the competitive ratio of Move-By-Bit is at most 3. You should clearly indicate what your potential function is, what the amortized cost of the algorithm for different scenario is, and provide an uuper ound for the ratio etween the amortized cost of the algorithm and Opt. Define inversions as we did for the analysis of MTF. Consider a pair of items which appear as x y in the list of Move-By-Bit and y x in the list of Opt (hence they form an inversion). Define the weight of the inversion etween x and y as 2 if the it of y is 0 and 1 if the it of y is 1. Define the potential as the total weight of all inversions. Assume there is a request to an item a. As in the analysis of MTF, assume a is at index i of the algorithm and index j of Opt, and assume Opt makes k paid exchanges efore serving a. Note that the cost of Opt is j + k. Any paid exchange of Opt creates at most one inversion and increases the potential y 2. So, the increase in potential ecause of Opt s actions is opt Φ = 2k. For the algorithm, we need to consider two events: Assume Move-y-Bit moves a to the front (it means the it of a was 1 efore the access). In this case, at least i j inversions are removed (all having weight 1 since it of a is 1), and at most j inversions are added (each having a weight of at most 2). So, the difference in potential ecause of the actions of the algorithm is Alg Φ (i j)+2j = i+3j. The amortized cost will e actual cost+delta Alg Φ+ opt Φ (i)+( i+3j)+2k = 3j+2k. Comparing with the cost j + k of Opt, we conclude that amortized cost is at most 3 times the cost of opt. Assume Move-y-Bit does not move a to the front (it means the it of a was 0 efore the access). In this case, the numer of inversions does not change; ut the weight of all inversions including a will decrease from 2 to 1. As efore, the numer of such inversions is at least i j. Hence for the increase of potential for actions of the algorithm we have Alg Φ (i j) = i + j. The amortized cost will e actual cost + Delta Alg Φ + opt Φ (i) + ( i + j) + 2k = j + 2k. Comparing with the cost j + k of Opt, we conclude that amortized cost is at most twice the cost of opt in this case. c (onus) Show that, with 1 it of advice, it is possile to achieve a deterministic list-update algorithm with a competitive ratio of 2. Consider two variants of Move-By-Bit: one starts with all its (for all items) eing 0 while these its are 1 in the other algorithm. We can use the same potential function as part (). But we need to find the sum of the cost of the two algorithms and show the total amortized costs of the two algorithms is at most 4 times that of Opt. This will e followed from the same potential argument in part (). The total amortized cost of the two algorithms is (3j + 2k) + (j + 2k) = 4j + 4k, which is 4 times cost j + k of Opt. Hence, the total cost of oth algorithms is 4 times that of Opt. This means that the etter of them has a competitive ratio of 2. 6

7 Prolem 6 List Update Algorithms [ = marks] In this prolem, we consider a variant of the list update prolem where no free exchange is allowed, i.e., after each access the algorithm can adjust the list only using paid exchanges. Note that algorithms like Move-To-Front are still well-defined ut they need to simulate their free-exchanges with paid-exchanges, for example, Move-To-Front moves a requested item to the front using a sequence of paid-exchanges instead of a single free-exchange. a) Show that the competitive ratio of Move-To-Front is exactly 4. We need to figure out how the competitive ratio of MTF and Opt change in the new model. For MTF, after accessing an item at index i, the algorithm has to make i 1 paid-exchanges to move the item to the front. So, if the cost of MTf was X under standard model, it ecomes 2X n under the new model. On the other hand, the cost of Opt does not change at all (recall that there is an optimal algorithm which does not use free exchanges; such algorithm has the same cost under the new model). For the upper ound, we have cost new model(mt F ) cost new model (Opt) 2 cost standard model(mt F ) cost standard model (Opt) 2cost standard model(mt F ) n cost standard model (Opt) = 2. The last equation comes from the analysis of MTF under standard model. For the lower ound, consider the same adversarial sequence we saw in the class for standard model (keep asking for the last item in the list of MTF). For a list of length m, the access cost for each request will e m and moving to front adds another cost of m 1, summing up to 2m 1. The total cost for n requests will e 2mn n. The cost of opt, is at most equal to the cost of an static offline algorithm, that is, n(m/2). The competitive ratio of MTF will e at least 2mn n nm/2 = 4mn 2n mn = 4 2/m. This ratio approaches to 4 for large values of m. ) Show that any online algorithm A for this model has a competitive ratio of at least 3. For that, consider an offline algorithm OFF which maintains a list that is reverse of that of A. For example, if the list of A is a c d, then the list of OFF is d c a. If A uses k paid exchanges to update its list, OFF can also use k paid exchanges to maintain a list reverse of A. To answer the question, use OFF and optimal static algorithms to find an upper ound for the cost of Opt. Consider an adversarial sequence of length n in which we keep asking for the last item in the list. Let m denote the length of the list. The cost of A is n m + X; here n m indicates the total accessing cost and let X denotes the numer of paid exchanges that X makes. We consider two possiilities for X and show that the competitive ratio of A is at least 3 in oth cases. Assume X nm/2. Hence, we have cost(a) nṁ + nṁ/2 = 3nm/2. Note that an static offline algorithm has a cost of nm/2 + O(m 2 ). So, the cost of Opt is at most nm/2 + O(m 2 ), and the competitive ratio of A will e 3nm/2 at least nm/2+o(m 2 ) = 3 O(m 2 ) nm/2+o(m 2 ), which approaches to 3 when n is sustantially larger than m. Assume X nm/2. Then the offline algorithm that maintains the reverse list has a total access cost of n and paid-exchange cost of X which sums to n + X = nm/2 + n. Hence, the cost of Opt is at most nm/2 + n, and 3nm/2 the competitive ratio of A will e at least nm/2+n = 3 n nm/2+n, which approaches to 3 when m is sufficiently large. c (onus) Provide an algorithm which has a competitive ratio etter than 4 (this is an open research question.) Let me know if you know the answer :) 7