Although big-O notation determines a growth rate of a function, sometimes a worse algorithm may outperform...

 

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Although big-O notation determines a growth rate of a function, sometimes a worse algorithm may outperform a better algorithm for very small input sizes n. Consider each of the following big-O estimates f(n) for algorithm A and g(n) for algorithm B, all of which g(n) = O(f(n)). For each instance of f and g, find the smallest integer input n beyond which, algorithm B will always outperform algorithm A. Show your work. For this question, you may use algebra, a graphing calculator or online resources to justify your answer. = a) g(n) 4x log2(n) = b) g(n) 100 x nx log (n) c) g(n) 6xn5 f(n) = f(n) = n f(n) = n6 Although big-O notation determines a growth rate of a function, sometimes a worse algorithm may outperform a better algorithm for very small input sizes n. Consider each of the following big-O estimates f(n) for algorithm A and g(n) for algorithm B, all of which g(n) = O(f(n)). For each instance of f and g, find the smallest integer input n beyond which, algorithm B will always outperform algorithm A. Show your work. For this question, you may use algebra, a graphing calculator or online resources to justify your answer. = a) g(n) 4x log2(n) = b) g(n) 100 x nx log (n) c) g(n) 6xn5 f(n) = f(n) = n f(n) = n6

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a For better performance gn must be less fn Hence 4log 2 ... View the full answer