When the order of increase of an algorithm's running time is N log N, the doubling test leads to the hypothesis that the running time is a N for a constant a. Isn't that an issue?
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- The doubling test will result in the hypothesis that the running time is a N for a constant a when an algorithm's order of development is N log N. Is that an issue, then?Al and Bob are arguing about their algorithms. Al claims his O(nlogn)-time method is always faster than Bob's O(n²)-time method. To settle the issue, they perform a set of experiments. To Al's dismay, they find that if n 100 is the O(nlogn)-time one better. Explain how this is possible.The Computer Science club is sponsoring a jigsaw puzzle contest. Jigsaw puzzles are assembled by fitting 2 pieces together to form a small block, adding a single piece to a block to form a bigger block, or fitting 2 blocks together. Each of these moves is considered a step in the solution. Use the second principle of induction to prove that the number of steps required to assemble an n-piece jigsaw puzzle is n − 1.
- A certain computer algorithm executes twice as many operations when it is run with an input of size k as when it is run with an input of size k – 1 (where k is an integer that is greater than 1). When the algorithm is run with an input of size 1, it executes seven operations. How many operations does it execute when it is run with an input of size 26? For each integer n 2 1, let s, -1 be the number of operations the algorithm executes when it is run with an input of size n. Then s, = and s = for each integer k 2 1. Therefore, So, S1, S21 -Select--- with constant Select--- |, which is So, for every integer n 2 0, s, It follows that for an input of size 26, the number of ... is operations executed by the algorithm is s which equals ---Select--- vA certain computer algorithm executes twice as many operations when it is run with an input of size k as when it is run with an input of size k - 1 (where k is an integer that is greater than 1). When the algorithm is run with an input of size 1, it executes seven operations. How many operations does it execute when it is run with an input of size 24? For each integernz 1, let s,-1 be the number of operations the algorithm executes when it is run with an input of size n. Then for each integer 2 1. Therefore, So, S3. Sz. is -Select- and s,= with constant Select- ,which is . So, for every integer n 2 0, s, = It follows that for an input of size 24, the number of operations executed by the algorithm is s -Select-v which equals Need Heln? DesdWhen the order of growth of the running time of an algorithm is N log N, the doubling test will lead to the hypothesis that the running time is ~ a N for a constant a. Isn’tthat a problem?
- 2. Algorithm A has a running time described by the recurrence T(n) = 7T(n/2) + n². A competing algorithm B has a running time described by the recurrence T(n) = aT (n/4) + n². What is the largest integer value for a such that B is asymptotically faster than A? Explain your answer.designed an iterative algorithm for separating n VLSIchips into those that are good and those that are bad by testing two chips at a time andlearning either that they are the same or that they are different. To help, at least half ofthe chips are promised to be good. Now design (much easier) a randomized algorithmfor this problem. Here are some hints. Randomly select one of the chips. What is the probability that the chip is good? How can you learn whether or not the selected chip is good? If it is good, how can you easily partition the chips into good and bad chips? If the chip is not good, what should your algorithm do? When should the algorithm stop? What is the expected running time of this algorithm?Computer science. Correct answer will be upvoted else downvoted. Think about a n by n chessboard. Its columns are numbered from 1 to n from the top to the base. Its sections are numbered from 1 to n from the passed on to one side. A cell on a convergence of x-th line and y-th section is indicated (x,y). The fundamental corner to corner of the chessboard is cells (x,x) for all 1≤x≤n. A stage of {1,2,3,… ,n} is composed on the fundamental slanting of the chessboard. There is actually one number composed on every one of the cells. The issue is to segment the cells under and on the principle askew (there are by and large 1+2+… +n such cells) into n associated areas fulfilling the accompanying imperatives: Each district ought to be associated. That implies that we can move from any cell of a locale to some other cell of a similar area visiting just cells of a similar district and moving from a cell to a neighboring cell. The x-th area ought to contain cell on the fundamental…
- suppose that n is not 2i for any integer i. How would we change the algorithm so that it handles the case when n is odd? I have two solutions: one that modifies the recursive algorithm directly, and one that combines the iterative algorithm and the recursive algorithm. You only need to do one of the two (as long as it works and does not increase the BigOh of the running time.)Prove that the following algorithmfor computing Fibonacci numbers is correct. function fib(n) comment Return (Fn−1, Fn) 1. if n is odd then 2. (a, b) := even(n − 1) 3. return(b, a + b) 4. else return(even(n)) function even(n) comment Return (Fn−1, Fn) when n is even 1. if n = 0 then return(1, 0) 2. else if n = 2 then return(1, 1) 3. else if n = 4 then return(2, 3) 4. (a, b) := fib(n/2 − 1) 5. c := a + b; d := b + c 6. return(b · d + a · c, c · (d + b))Q2. The following algorithm returns the product of two numbers, a and b. The parameters x and y are natural numbers. First, prove the correctness of the algorithm. Then, analyze the time complexity of the algorithm in the worst case scenario. function mult (a, b) if b = 0: return 0 else if b is odd: return (mult (2a, b/2 ) +a) else: return (mult (2a, b/2 ) )