. Two algorithms takes n 2 days and 2 n seconds respectively, to solve an instance of size n. What is the size of the smallest i
nstance on which the former algorithm outperforms the latter algorithm? Approximately how long does such an instance take to solve?
1 answer:
Answer:
- n = 1
- 1 day
Step-by-step explanation:
n^2 is less than 2^n for n < 2 and for n > 4. The smallest size of n that is of interest is n=1. For that, n^2 = 1^1 = 1.
The n^2 algorithm will outperform the 2^n algorithm for n = 1. That problem size will take 1 day to solve.
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Please note that there are no algebraic methods for solving an inequality of the form x^2 < 2^x. We have solved it using a graphing calculator.
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