Consider the following algorithm Algorithm Mystery(A[0..n-1, 0..n - 1]) //Input: A matrix A[0..n-1, 0..n - 1] of real numbers for i 0 to n - 2 do for ji+ 1 to n - 1 do if A[i, j] = A[j,i] return true return false a. What does this algorithm compute? b. What is its input size? c. What is its basic operation?
Consider the following algorithm Algorithm Mystery(A[0..n-1, 0..n - 1]) //Input: A matrix A[0..n-1, 0..n - 1] of real numbers for i 0 to n - 2 do for ji+ 1 to n - 1 do if A[i, j] = A[j,i] return true return false a. What does this algorithm compute? b. What is its input size? c. What is its basic operation?
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
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![### Algorithm Analysis
**Algorithm: Mystery**
- **Input:** A matrix \( A[0..n-1, 0..n-1] \) of real numbers.
**Code:**
```plaintext
for i ← 0 to n − 2 do
for j ← i + 1 to n − 1 do
if A[i, j] ≠ A[j, i]
return false
return true
```
### Explanation
This algorithm checks whether a given matrix is symmetric. A matrix is symmetric if it is equal to its transpose, meaning \( A[i, j] = A[j, i] \) for all \( i \) and \( j \).
### Questions
**a. What does this algorithm compute?**
This algorithm checks if the input matrix is symmetric.
**b. What is its input size?**
The input size is \( n \), where the matrix is \( n \times n \).
**c. What is its basic operation?**
The basic operation is comparing two matrix elements, \( A[i, j] \) and \( A[j, i] \).
**d. How many times is the basic operation executed?**
The basic operation is executed \(\frac{n(n-1)}{2}\) times (the sum of the first \( n-1 \) integers).
**e. What is the efficiency class of this algorithm?**
The efficiency class of this algorithm is \( O(n^2) \), as the algorithm performs comparisons for each pair of indices in the upper triangle of the matrix.
**f. Suggest an improvement, or a better algorithm altogether, and indicate its efficiency class. If you cannot do it, try to prove that, in fact, it cannot be done.**
A potential improvement could involve parallelizing the comparison operations, as each comparison is independent of the others. However, in terms of algorithmic time complexity, the operation of checking symmetry inherently requires inspecting all required pairs, thus \( O(n^2) \) remains the lower bound in sequential execution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7757f974-220a-4bcf-b905-f534f24cec51%2F75f1aef2-e109-49b3-9955-7a9fe5b6c8e6%2Fcqby2jg_processed.png&w=3840&q=75)
Transcribed Image Text:### Algorithm Analysis
**Algorithm: Mystery**
- **Input:** A matrix \( A[0..n-1, 0..n-1] \) of real numbers.
**Code:**
```plaintext
for i ← 0 to n − 2 do
for j ← i + 1 to n − 1 do
if A[i, j] ≠ A[j, i]
return false
return true
```
### Explanation
This algorithm checks whether a given matrix is symmetric. A matrix is symmetric if it is equal to its transpose, meaning \( A[i, j] = A[j, i] \) for all \( i \) and \( j \).
### Questions
**a. What does this algorithm compute?**
This algorithm checks if the input matrix is symmetric.
**b. What is its input size?**
The input size is \( n \), where the matrix is \( n \times n \).
**c. What is its basic operation?**
The basic operation is comparing two matrix elements, \( A[i, j] \) and \( A[j, i] \).
**d. How many times is the basic operation executed?**
The basic operation is executed \(\frac{n(n-1)}{2}\) times (the sum of the first \( n-1 \) integers).
**e. What is the efficiency class of this algorithm?**
The efficiency class of this algorithm is \( O(n^2) \), as the algorithm performs comparisons for each pair of indices in the upper triangle of the matrix.
**f. Suggest an improvement, or a better algorithm altogether, and indicate its efficiency class. If you cannot do it, try to prove that, in fact, it cannot be done.**
A potential improvement could involve parallelizing the comparison operations, as each comparison is independent of the others. However, in terms of algorithmic time complexity, the operation of checking symmetry inherently requires inspecting all required pairs, thus \( O(n^2) \) remains the lower bound in sequential execution.
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