Comp Online Algorithms

Transcription

1 Comp Online Algorithms Assignment 2: Compression, Splay Trees, Caching, and k-server Shahin Kamalli University of Manitoba - Fall 2018 Due: Monday, October 22th at 11:59 pm October 27, 2018 [Computer science] is not really about computers and it s not about computers in the same sense that physics is not really about particle accelerators, and biology is not about microscopes and Petri dishes...and geometry isn t really about using surveying instruments. Now the reason that we think computer science is about computers is pretty much the same reason that the Egyptians thought geometry was about surveying instruments: when some field is just getting started and you don t really understand it very well, it s very easy to confuse the essence of what you re doing with the tools that you use. Hal Abelson (CS professor at MIT) Please pay attention to the followings when preparing/submitting your assignment: All problems are written problems. There are six problems with a total of 101 marks. The assignment is designed to be shorter than Assignment 1, and hopefully you will find it easier. Most questions require smart observations after which the solution is easy to derive and write. If you have any question related to the assignment, you are encouraged to post it on Piazza. Note that you can submit 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 publicly on Piazza (because all students can benefit from it). It is not the case when you ask questions in s or during office hours. You are welcome to discuss the problems with your friends (or enemies). But you should write your answers individually. You might be interviewed about your answers. Be careful not to accidentally copy. Submit your answers electronically using UMLearn. You should submit a single pdf file. You are encouraged to prepare your assignment using L A TEX. The LaTex file for this assignment is posted.

2 Problem 1 Compression [ = 20 marks] a) Apply the Burrows-Wheeler transform on the following string; show your work and the output Assume $ precedes all characters when you sort rotations. BARBAPAPA$ $BARBAPAPA ARBAPAPA$B A$BARBAPAP RBAPAPA$BA APA$BARBAP BAPAPA$BAR = sort APAPA$BARB APAPA$BARB ARBAPAPA$B PAPA$BARBA BAPAPA$BAR APA$BARBAP BARBAPAPA$ PA$BARBAPA PA$BARBAPA A$BARBAPAP PAPA$BARBA $BARBAPAPA RBAPAPA$BA The result will be the last column, i.e., AP P BBR$AAA BARBAP AP A$ b) Assume an initial list $ A B P R, i.e., initially $ is at index 0, A is at index 1, etc. Assume we use Move-To-Front on the above list to encode the outcome of the BWT transform from part (a). Show what numbers will be encoded (you need to show how the list is updated). $ A B P R = A : 1 is encoded A $ B P R = P : 3 is encoded P A $ B R = P : 0 is encoded P A $ B R = B : 3 is encoded B P A $ R = B : 0 is encoded B P A $ R = R : 4 is encoded R B P A $ = $ : 4 is encoded $ R B P A = A : 4 is encoded A $ R B P = A : 0 is encoded A $ R B P = A : 0 is encoded So, the text is encoded as c) Assume an initial list $ A B P R, i.e., initially $ is at index 0, A is at index 1, etc. Assume we use Timestamp on the above list to encode BA$AP BP. Show what numbers will be encoded (you need to show how the list is updated). $ A B P R = BA$ : the first accesses to these items does not change the list, 2, 1, 0 are encoded $ A B P R = A : 1 is encoded A $ B P R = P : 3 is encoded A $ B P R = B : 2 is encoded B A $ P R = P : 3 is encoded P B A $ P R So, the text is encoded as

3 d) Assume an initial list $ A B C D. A compressing scheme that uses Move-To-Front has encoded the following numbers for a text T. Show what the actual text is. The numbers are So, the text is decoded as CC$CA. 3:C is decoded $ A B C D = 0:C is decoded 1:$ is decoded 1:C is decoded $ C A B D = 2:A is decoded A C $ B D Problem 2 Splay Trees [7+7+7 = 21 marks] a) Apply the splay operation on the following splay tree when there is a request to node 30. Show your steps. b) Prove or disprove the following statement: the root of a splay tree has always two children. This is wrong. After the access to the largest/smallest item, that item becomes the root and there will be no node on its right/left. c) Prove or disprove the following statement: after a splay operation, the old root is always at depth 1 or 2 (i.e., it is at distance 1 or 2 of the new root). It is true; the old root s position is changed as a result of the very last rotation; the old root will be at depth 1 in case of the last operation being zig or zig-zag, and at depth 2 in case of the last operatin being zig-zig. Problem 3 Paging & Resource Augmentation [15 marks] Sometimes when we analyse online algorithms, we reduce the power of Opt to be more fair to the online algorithm; it is called resource augmentation. For example, for the paging algorithm, instead of comparing an online algorithm with an optimal algorithm which has the same cache-size, we assume the size of the cache of Opt is smaller than that of algorithm. Clearly, when comparing with a weaker optimal algorithm, the competitive ratio of algorithms is expected to improve. 3

4 Assume the cache of a marking algorithm A has size k and the cache of Opt has size k/2. Prove that the competitive ratio of A is at most 2 under this setting (note that this is a big improvement over the competitive ratio k in the classic setting without resource augmentation). Use a phase partitioning technique as we did for LRU (so that there are k distinct pages in each phase). As discussed in the class, the number of faults by any marking algorithm for each phase is at most k. Opt, however, has a cache of size k/2; each phase contains k distinct pages, and only k/2 of them can be hits by Opt in their first access. So, Opt has to incur a cost of k k/2 = k/2 per phase. This gives a ratio of at most k/(k/2) = 2 for each phase, which can be extended to the whole sequence for an upper bound. Problem 4 Double-Coverage-Algorithm for Cycles [15 marks] Note that the double-coverage algorithm is well-defined for cycles: for each request to a node x, move the two servers on its two sides (in the cycle) with the equal speed towards x. Show that DCA has unbounded competitive ratio for cycles. You might prove this by considering the problem for a cycle of length 5 with k = 3 servers in it. Consider a cycle formed by vertices A, B, C, D, E in the same order. Assume servers are initially at A, C, and E. Assume a sequence starting with requests B, D, E. For these three requests, the algorithm incurs a cost of 6, and its configuration will be as before (i.e., servers at A, C, E). So, repeating these requests n times yields a total cost of 6 for the algorithm, while an optimal algorithm pays a cost of 2 to move servers once to B, D, and E. The competitive ratio of such sequence will be 6n 2 = 3n. Problem 5 Competitive k-server Algorithm for Cactus and Block Graphs [15 marks] From the following two questions, choose one and submit a solution for it. There is no advantage is submitting solutions to both: A block graph is a graph in which every biconnected component (that is, any collection of vertices for which there are at least two paths between any pair of vertices) is a clique (consult Wikipedia to see a picture of a block graph). Prove the k-sever conjecture for block graphs, that is, described that there is a deterministic algorithm that has a competitive ratio of k for these metrics (your solution should include a description on why the algorithm is k-competitive). Just embed the block graph into a tree by replacing each clique with a star in a way that the distances remain the same (the edges of the clique should have length 0.5 OR edges in the original graph have size 2 while the ones in the new stars have length 1). Since the distances are the same (or all scaled similarly), a k-competitive algorithm in the new graph is equivalent to a k-competitive algorithm in the original graph. Apply the Double-Coverage-Algorithm on the resulting tree. A cactus graph is a graph in which every biconnected component (that is, any collection of vertices for which there are at least two paths between any pair of them) is a cycle. In other words, it is a graph in which every edge belongs to at most one cycle (consult Wikipedia to see a picture of a cactus graph). Provide a randomized k-server algorithm with competitive ratio of at most 2k for cactus graphs (your solution should include a description on why the algorithm is at most 2k-competitive). Take a random edge from each cycle and remove it. The result will be a tree on which we can apply Double-Coverage- Algorithm (DCA). We claim the result will be a 2k-competitive randomized algorithm. The analysis is similar to that of Circ algorithm we saw earlier. Assume Opt-Tree be the optimal algorithm for the same sequence but for the resulting tree metric. From DCA analysis, we know cost(alg) k cost(opt-tree) (1) Assume Opt makes moves of lengths d 1,..., d n to serve the input sequence of length n. Consider for the move d i, opt s server moves from cycles c 1, c 2,..., c q. An offline algorithm for the tree metric can mimic these moves except that it pays a penalty of c j when Opt passes a deleted edge at cycle c j (because it might take the other direction). Let d j i be the distance opt server moved on the ith move on cycle c j. The chance of the server passing a deleted edge 4

5 is d j i /c j. So, the penalty incured by the offline algorithm is expected to be at most d j i /c j c j = d j i. Summing up, the penalty of the algorithm is at most d i for the i th move and hence its total cost is at most 2d i. Summing up for all requests, the cost of the offline algorithm, and hence the cost of Opt-Tree is at most twice the cost of Opt, that is cost(opt-tree) 2cost(Opt) (2) From Equations (1), (2), we get the desired results, that is cost(alg) 2kcost(Opt) Problem 6 Balance for Paths [15 marks] We learned in the class that Balance is not a competitive algorithm. The example we saw in the class was based on the metric which was a cycle. A clever student comes to the conclusion that Balance might be competitive when the metric space is a tree or a path. Show that he is wrong. For that, consider a path of length m (assume vertices are labelled from 1 to m) and show a worst-case sequence for the 2-server problem which shows Balance is not competitive. Assume servers s 1 and s 2 are respectively located at vertices 1 and 2, and the request is (m, m 1, 1, 2) n For the first request, s 2 moves a distance of m 1; for the second request, s 1 moves the same distance. The next requests to 1 and 2 are served by respectively by s 1 and s 2 and each move a distance of m 1 to their original location. In total, Balance incurs a cost of 4(m 1), which sums up to 4n(m 1). An optimal algorithm maintains a server on each side of the path and incurs a cost of m + 4n. Hence, the competitive of Balance will be at least 4n(m 1) m+4n, which approaches to m 1 for large values of n. 5