4. (a) State clearly the lower bound proved for the worst-case number of key- comparisons for any comparison-based sorting algorithm. 4 (b) For n = 4, use the above result to determine the lower bound (exact number) for the worst-case number of key-comparisons. 4 (c) Show how Mergesort is used to sort n = 4 elements. Show the merge-tree and compute the worst-case number of key-comparisons. How does your result compare with the above lower bound?
4. (a) State clearly the lower bound proved for the worst-case number of key- comparisons for any comparison-based sorting algorithm. 4 (b) For n = 4, use the above result to determine the lower bound (exact number) for the worst-case number of key-comparisons. 4 (c) Show how Mergesort is used to sort n = 4 elements. Show the merge-tree and compute the worst-case number of key-comparisons. How does your result compare with the above lower bound?
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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can you solve 2020 image one I also provided refrences how MY prof solved so based on that pleease solve this image 2020 one step by step formet please use from refrnecs
![4. Consider sorting \( n = 6 \) real-valued keys. Compute the worst-case number of key comparisons (exact number, not order) for each of the following cases. Show the work.
(a) **Insertion Sort**
- Sequence: \( O\ O\ O\ O\ O\ O \)
- Comparisons: \( 1\ 2\ 3\ 4\ 5 \)
- Total: \( 1 + 2 + 3 + 4 + 5 = 15 \)
(b) **Mergesort** (Assume recursive implementation of mergesort.) Show the merge tree. Show the number of comparisons for each merge and compute the total.
- Merge Tree:
- First level: Two pairs of \( O\ O \)
- Second level: Merges of each pair resulting in two nodes
- Final merge to complete the sort
- Comparisons: \( 1 + 1 + 2 + 2 + 5 \)
- Total: \( 11 \)
(c) **Theoretical Lower Bound**
- What is the theoretical lower bound on the worst-case number of comparisons? (Give the exact worst-case number.)
- Calculation:
\[
\lceil \log_2(6!) \rceil = \lceil \log_2(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2) \rceil
\]
\[
= \lceil \log_2(720) \rceil = 10
\]
- Explanation:
\[
2^9 < 720 < 2^{10} \Rightarrow 2^9 = 512,\, 2^{10} = 1024
\]
- Comparison with upper bounds indicates:
- Insertion Sort: 15
- Mergesort: 11
**Conclusion:**
- The theoretical lower bound of 10 comparisons is less than both the insertion sort and mergesort worst-case scenarios.
- Mergesort is closer to the lower bound than insertion sort.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bad4e48-dad8-4710-a64b-b24b80d1efcf%2Fe88569d1-bf11-4f32-a1f6-e03aa07dfa01%2F0st4yd_processed.png&w=3840&q=75)
Transcribed Image Text:4. Consider sorting \( n = 6 \) real-valued keys. Compute the worst-case number of key comparisons (exact number, not order) for each of the following cases. Show the work.
(a) **Insertion Sort**
- Sequence: \( O\ O\ O\ O\ O\ O \)
- Comparisons: \( 1\ 2\ 3\ 4\ 5 \)
- Total: \( 1 + 2 + 3 + 4 + 5 = 15 \)
(b) **Mergesort** (Assume recursive implementation of mergesort.) Show the merge tree. Show the number of comparisons for each merge and compute the total.
- Merge Tree:
- First level: Two pairs of \( O\ O \)
- Second level: Merges of each pair resulting in two nodes
- Final merge to complete the sort
- Comparisons: \( 1 + 1 + 2 + 2 + 5 \)
- Total: \( 11 \)
(c) **Theoretical Lower Bound**
- What is the theoretical lower bound on the worst-case number of comparisons? (Give the exact worst-case number.)
- Calculation:
\[
\lceil \log_2(6!) \rceil = \lceil \log_2(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2) \rceil
\]
\[
= \lceil \log_2(720) \rceil = 10
\]
- Explanation:
\[
2^9 < 720 < 2^{10} \Rightarrow 2^9 = 512,\, 2^{10} = 1024
\]
- Comparison with upper bounds indicates:
- Insertion Sort: 15
- Mergesort: 11
**Conclusion:**
- The theoretical lower bound of 10 comparisons is less than both the insertion sort and mergesort worst-case scenarios.
- Mergesort is closer to the lower bound than insertion sort.
Transcribed Image Text:### Educational Content on Comparison-Based Sorting Algorithms
#### Problem 4: Understanding Lower Bounds and Mergesort
**4 (a) Lower Bound for Sorting Algorithms**
State clearly the lower bound proved for the worst-case number of key comparisons for any comparison-based sorting algorithm.
**4 (b) Specific Case for \( n = 4 \)**
For \( n = 4 \), use the above result to determine the lower bound (exact number) for the worst-case number of key comparisons.
**4 (c) Mergesort Example for \( n = 4 \)**
Show how Mergesort is used to sort \( n = 4 \) elements. Illustrate the merge-tree and compute the worst-case number of key comparisons. How does your result compare with the above lower bound?
---
### Detailed Explanation
#### Lower Bound Concept
- **Definition:** In comparison-based sorting algorithms, the lower bound is the minimum number of comparisons necessary to sort the input in the worst-case scenario.
- **Theoretical Lower Bound:** The proven lower bound for these algorithms is \( \Omega(n \log n) \).
#### Applying the Lower Bound for \( n = 4 \)
- **Calculation:** For four elements (\( n = 4 \)), calculate the actual lower bound number of key comparisons.
#### Mergesort with \( n = 4 \)
1. **Step-by-Step Mergesort Process:**
- Split the list into smaller lists until each one contains a single element.
- Recursively merge lists by comparing the smallest elements of each.
2. **Merge-Tree Visualization:**
- Draw and explain a tree diagram that visually represents the merging process at each step.
3. **Comparison Calculation:**
- Compute the exact number of key comparisons performed during the sorting process.
- Compare the calculation with the theoretical lower bound to assess efficiency.
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