Problem Overview

You are given two algorithms $A$ and $B$, with their time complexities:

• $T_A(n) = 0.1n^2 \log_{10}(n)$ microseconds
• $T_B(n) = 2.5n^2$ microseconds

We need to:

1. Compare both algorithms in Big-O notation to determine which one performs better asymptotically.
2. Find a threshold size $n_0$ where one algorithm begins to outperform the other.
3. Determine which algorithm to use for a very large input size $n > 10^9$.

Step 1: Big-O Comparison

• Algorithm A: $T_A(n) = 0.1n^2 \log_{10}(n)$
  Growth: $O(n^2 \log(n))$

• Algorithm B: $T_B(n) = 2.5n^2$
  Growth: $O(n^2)$

Since $\log(n)$ grows very slowly compared to $n^2$, Algorithm B (with $O(n^2)$) has a smaller growth rate asymptotically than Algorithm A (with $O(n^2 \log(n))$).

Thus, Algorithm B is better in the Big-O sense for large inputs.

Step 2: Finding the Threshold $n_0$

We need to solve for $n_0$ where: $T_A(n) = T_B(n) \implies 0.1n^2 \log_{10}(n) = 2.5n^2$

Dividing both sides by $n^2$: $0.1 \log_{10}(n) = 2.5$

Solving for $n$: $\log_{10}(n) = \frac{2.5}{0.1} = 25$ $n = 10^{25}$

Thus, $n_0 = 10^{25}$. For any $n > 10^{25}$, Algorithm B will outperform Algorithm A.

Step 3: Recommendation for $n > 10^9$

Since $10^9 < 10^{25}$, we are in the range where Algorithm A performs better. For inputs $n \leq 10^{25}$, Algorithm A is more efficient due to its slower constant factor.

Recommendation:
For inputs greater than $10^9$, but still far below $10^{25}$, Algorithm A is recommended.

Summary

• Big-O Winner: Algorithm B (for very large $n$).
• Threshold $n_0$: $n_0 = 10^{25}$.
• For $n = 10^9$: Use Algorithm A since it is faster in this range.

Would you like a more detailed step-by-step solution or further clarifications?

5 Related Questions:

1. How does the logarithmic function grow compared to polynomial functions?
2. What does "Big-O sense" mean in terms of asymptotic behavior?
3. How can constant factors impact performance in smaller input sizes?
4. What techniques are used to compare two algorithms’ performance empirically?
5. Can there be practical cases where an $O(n^2 \log n)$ algorithm is better than an $O(n^2)$ algorithm?

Tip:

When comparing algorithms, always consider both asymptotic behavior (Big-O) and constant factors—the latter can make a big difference for small to medium input sizes!