Question 1: To determine the size of the problem (n) for which algorithm A is faster than algorithm B, I need to compare the number of steps each algorithm takes. Algorithm A takes \(1000n^3\) steps, while algorithm B takes \(2^n\) steps. To find the size of the problem where algorithm A is faster than B, I equate the number of steps for each algorithm: [1000n^3=2^n\] To solve for \(n\), I take the logarithm of both sides to isolate \(n\): \[n^3 = \frac{1} {1000} \times 2^n\] [3\log(n) = \og\left(\frac {1} {1000} \times 2^n\right)\] \[3\log(n) = \log(1) + \log(2^n) - \log(1000)\] [3\log(n) = n - \og(1000)\] [log(n^3) = n - \log(1000)\] [n^3 = 1000 \times 10^n\] [n= \sqrt[3]{1000 \times 10^n}\]
By observation, for large values of n, the exponential function \(2^n\) grows much faster than the cubic function \(1000n^3\). Hence, there's no value of n where algorithm A is faster than algorithm B (Shaffer, 2020). Question 2: The provided code fragment consists of nested loops where the outer loop runs n times, and for each iteration of the outer loop, the inner loop runs i times, where i ranges from 0 to n-1. The total number of iterations of the inner loop can be calculated using the arithmetic sum formula: \[0+1+2+ ... + (n-1) = \frac{n(n-1)} {2}\] Thus, the upper bound (big O notation) for the given code fragment is \(O(n^2)\), as the dominant term is \(n^2\). Regarding the lower bound, it's unlikely to be the same as the upper bound. The lower bound represents the best-case scenario, whereas the upper bound focuses on the worst-case scenario. Since the lower bound often represents the best-case scenario, it's unlikely to be the same as the upper bound, which represents the worst-case scenario (Shaffer, 2020, p. xx). Word count: 302
References: Shaffer, C. A. (2020). A Practical Introduction to Data Structures and Algorithm Analysis (3rd ed.).