2. [3] A program takes 10 seconds for input size 500 (i.e., n=500). Ignoring the effect of constants, approximately how much time can the same program be expected to take if the input size is increased to 2000 given the following run-time complexities? 1. O(N) 2. O(N log N) 3. O(N³)
2. [3] A program takes 10 seconds for input size 500 (i.e., n=500). Ignoring the effect of constants, approximately how much time can the same program be expected to take if the input size is increased to 2000 given the following run-time complexities? 1. O(N) 2. O(N log N) 3. O(N³)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![**Problem 2: Time Complexity Analysis**
A program takes 10 seconds for an input size of 500 (i.e., n = 500). Ignoring the effect of constants, determine approximately how much time the same program is expected to take if the input size is increased to 2000, given the following run-time complexities:
1. **O(N)**
2. **O(N log N)**
3. **O(N³)**
*Solution Explanation*:
1. **O(N)**: If the time complexity is linear, the time taken is directly proportional to the input size. For n = 500, the time taken is 10 seconds. When n = 2000, the expected time will be:
\[
\text{Time} = 10 \times \frac{2000}{500} = 40 \text{ seconds}
\]
2. **O(N log N)**: This complexity involves a logarithmic multiplication factor. For n = 500, the time taken is 10 seconds. When n = 2000, the expected time will be:
\[
\text{Time} = 10 \times \frac{2000 \cdot \log(2000)}{500 \cdot \log(500)}
\]
Approximating \(\log\) values,
\[
\log(500) \approx 2.698 \quad \text{and} \quad \log(2000) \approx 3.301
\]
The new time is:
\[
\text{Time} = 10 \times \frac{2000 \times 3.301}{500 \times 2.698} \approx 40.97 \text{ seconds}
\]
3. **O(N³)**: With cubic complexity, the increase in input size has a much greater impact. For n = 500, the time was 10 seconds. When n = 2000, the expected time will be:
\[
\text{Time} = 10 \times \frac{2000^3}{500^3}
\]
Simplifying,
\[
\text{Time} = 10 \times \left(\frac{2000}{500}\right)^3 = 10 \times 64 = 640 \text{ seconds}
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4f8485b-d935-478d-91e7-b806502d9702%2Feb1ac1f7-2fb9-4489-8545-abb4729dd328%2Fgp59ufe.png&w=3840&q=75)
Transcribed Image Text:**Problem 2: Time Complexity Analysis**
A program takes 10 seconds for an input size of 500 (i.e., n = 500). Ignoring the effect of constants, determine approximately how much time the same program is expected to take if the input size is increased to 2000, given the following run-time complexities:
1. **O(N)**
2. **O(N log N)**
3. **O(N³)**
*Solution Explanation*:
1. **O(N)**: If the time complexity is linear, the time taken is directly proportional to the input size. For n = 500, the time taken is 10 seconds. When n = 2000, the expected time will be:
\[
\text{Time} = 10 \times \frac{2000}{500} = 40 \text{ seconds}
\]
2. **O(N log N)**: This complexity involves a logarithmic multiplication factor. For n = 500, the time taken is 10 seconds. When n = 2000, the expected time will be:
\[
\text{Time} = 10 \times \frac{2000 \cdot \log(2000)}{500 \cdot \log(500)}
\]
Approximating \(\log\) values,
\[
\log(500) \approx 2.698 \quad \text{and} \quad \log(2000) \approx 3.301
\]
The new time is:
\[
\text{Time} = 10 \times \frac{2000 \times 3.301}{500 \times 2.698} \approx 40.97 \text{ seconds}
\]
3. **O(N³)**: With cubic complexity, the increase in input size has a much greater impact. For n = 500, the time was 10 seconds. When n = 2000, the expected time will be:
\[
\text{Time} = 10 \times \frac{2000^3}{500^3}
\]
Simplifying,
\[
\text{Time} = 10 \times \left(\frac{2000}{500}\right)^3 = 10 \times 64 = 640 \text{ seconds}
\
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