Question
This question asks you to perform competitive analysis of transpose (TR) and frequency count (FC).
(i) [5 marks] Suppose that you are maintaining a list of n elements under access operation only. The cost of access to the i-th element in the list is i. Let S be a
request sequence of m access operations over this list. For any sufficiently large m, construct a request sequence S such that for this request sequence, the total cost of TR divided by the total cost of MTF is ω(1).
(ii) [5 marks] Use the result of (i) to argue that TR is not competitive.
Hint: You just have to use the definition of competitiveness. Therefore, You can
solve this subquestion without having to solve (i) first.
(iii) [5 marks] Further prove that FC is not competitive, either.
Asked By StormySkies20 at
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Jay
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Step 1/1
Step 1 The question is related to computer science, specifically in the area of algorithm analysis and competitive analysis of caching algorithms. This question asks you to perform a competitive analysis of transpose (TR) and frequency count (FC). (i) Suppose that you are maintaining a list of n elements under access operation only. The cost of access to the i-th element in the list is i. Let S be a request sequence of m access operations over this list. For any sufficiently large m, construct a request sequence S such that for this request sequence, the total cost of TR divided by the total cost of MTF is ω(1). (ii) Use the result of (i) to argue that TR is not competitive. Hint: You just have to use the definition of competitiveness. Therefore, You can solve this subquestion without having to solve (i) first. (iii) Further prove that FC is not competitive, either. Step 2 (i) To construct a request sequence S such that the total cost of TR divided by the total cost of MTF is ω(1), we can design S as follows: Access the elements in reverse order, so the cost of accessing the i-th element is n - i + 1. Repeat the access sequence S multiple times. By accessing the elements in reverse order and repeating the sequence multiple times, the total cost of TR will be O(n^2) because each access will require swapping n-i+1 elements. The total cost of MTF, on the other hand, will be O(nm) because each access may require moving up to m elements to the front of the list. Therefore, the total cost of TR divided by the total cost of MTF is ω(1). (ii) To show that TR is not competitive, we can use the definition of competitiveness, which states that a caching algorithm A is c-competitive if, for any request sequence S of length m, the total cost of A on S is at most c times the optimal offline algorithm OPT on S, plus some constant factor. From part (i), we have shown that there exists a request sequence S such that the total cost of TR divided by the total cost of MTF is ω(1). Therefore, TR cannot be c-competitive for any constant c, because for sufficiently large m, the total cost of TR on S will be much larger than the total cost of MTF on S. (iii) To show that FC is not competitive, we can use a similar argument as in part (ii). Suppose we have a request sequence S such that the first m/2 accesses are to the same element x, and the second m/2 accesses are to another element y. Then, the total cost of FC on S will be O(m), because it will need to count the frequency of each element before moving x and y to the front of the list. However, the optimal offline algorithm OPT can simply move x to the front of the list for the first m/2 accesses, and then move y to the front for the remaining accesses, resulting in a total cost of O(m/2). Therefore, the total cost of FC on S divided by the total cost of OPT on S is ω(1), and FC is not c-competitive for any constant c.
Final Answer
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