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For each of the problems (a)-(d) below: (i) Write an algorithm in pseudocode to solve the problem (be sure your algorithm works correctly if \(m=0\) or \(n=0\); it should not make any assignments to clements of the array), and (ii) Calculate how many assignment statements and how many comparisons the algorithm causes to be executed as a function of \(m\) and \(n\). In this case, count assignments to and comparisons of index variables, as well as assignments to and comparisons of positions in the array. Simplify your answers. (a) Initialize all the elements of an \(m \times n\) array to \(0 .\) (b) Initialize all the elements of an \(m \times n\) array that lie on or above the diagonal to \(0 .\) (Here, by "diagonal" we mean positions \([r, c]\) where \(r=c .)\) (c) Initialize all the elements of an \(m \times n\) array that lic above the diagonal to \(0 .\) (d) Initialize all the elements on the diagonal of an \(m \times n\) array on the diagonal to \(0 .\)

Short Answer

Expert verified
For problem (a): Total assignments = \(m \times n\), comparisons = \(m(n+1)\). For (b): Assignments = \(\frac{m(2n-m+1)}{2}\), comparisons = same. For (c): Assignments = \(\frac{(m-1)m}{2}\), comparisons \(mn-\frac{m(m+1)}{2}\). For (d): Both = \(\min(m,n)\).

Step by step solution

01

Problem (a) Pseudocode

The goal is to initialize all elements of an \(m \times n\) array to 0. Here is the pseudocode:1. **For** each row \(i\) from 0 to \(m-1\): - **For** each column \(j\) from 0 to \(n-1\): - Set \(array[i][j] = 0\).This algorithm initializes every element in the matrix to 0, whether \(m\) or \(n\) is 0 or not.
02

Problem (a) Analysis

For each element \(array[i][j]\), one assignment occurs, and there are \(m \times n\) elements. Hence, there are \(m \times n\) assignments.The total number of comparisons is the sum of the iterations of the two loops:- The outer loop runs \(m\) times, performing \(m\) comparisons.- The inner loop runs \(n\) times per each outer loop iteration, so it runs a total of \(m \times n\) times.Thus, there are \(m + m \times n = m(n + 1)\) comparisons.
03

Problem (b) Pseudocode

The task is to initialize elements on or above the diagonal to 0.1. **For** each row \(i\) from 0 to \(m-1\): - **For** each column \(j\) from \(i\) to \(n-1\): - Set \(array[i][j] = 0\).This approach ensures only elements \([i, j]\) where \(j \geq i\) are set to 0.
04

Problem (b) Analysis

The assignments correspond to entries \([i, j]\) for \(j \geq i\). Thus, for each row, \((n-i)\) assignments are performed.Summed over all rows, the total number of assignments is:\[ \sum_{i=0}^{m-1} (n-i) = mn - \frac{(m-1)m}{2} \]Comparisons include:- The outer loop (\(m\) comparisons).- The inner loop within each row contributing \((m\cdot (m-1)/2 + m)\) comparisons for the range \(j = i\) to \(j = n-1\).Thus, the total number of comparisons is \(m + \sum_{i=0}^{m-1} (n-i) = mn - \frac{m(m-1)}{2} + m\).
05

Problem (c) Pseudocode

To initialize elements above the diagonal to 0:1. **For** each row \(i\) from 0 to \(m-1\): - **For** each column \(j\) from \(i+1\) to \(n-1\): - Set \(array[i][j] = 0\).This ensures that only elements above the diagonal, \([i, j]\) for \(j > i\), are modified.
06

Problem (c) Analysis

Assignments occur for elements where \(j > i\), counting for each row as:\[ \sum_{i=0}^{m-1} (n-i-1) = \begin{align*}(n-1) + (n-2) + \cdots + 1 \= \frac{(m-1)m}{2} - 0. \end{align*} \]Comparisons include:- The outer loop: \(m\) comparisons.- The inner loop (each row contributing individually):\[ \sum_{i=0}^{m-1} (n-i-1) = \frac{m(m-1)}{2} - m. \]This results in a total comparison count of approximately \(\frac{m(m-1)}{2}\) added with the initial outer loop, leaving: \(mn - \frac{m(m+1)}{2}\), considering every element where \(j > i\).
07

Problem (d) Pseudocode

For this problem, initialize all elements on the diagonal to 0:1. **For** each row \(i\) from 0 to \(\min(m, n) - 1\): - Set \(array[i][i] = 0\).This sets the diagonal elements \(array[i][i]\) to 0.
08

Problem (d) Analysis

There is one assignment for each element on the diagonal, hence \(\min(m,n)\) assignments.For comparisons:- The loop runs \(\min(m,n)\) times, each requiring one comparison.Therefore, there are \(\min(m,n)\) comparisons across the process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algorithm analysis
Algorithm analysis involves examining the efficiency of an algorithm in terms of time and space usage. It provides a mathematical means to predict the resources required by different algorithms to compare them and select the best one for a given problem.

For instance, in the exercise on matrix initialization, each problem involves calculating the number of assignments and comparisons made by the algorithm as a function of the matrix dimensions, \(m\) and \(n\).

- **Time Complexity**: This refers to the time an algorithm takes to complete its task. For example, initializing all elements of an \(m \times n\) matrix to zero involves \(m \times n\) operations, indicating a time complexity of \(O(mn)\).
- **Space Complexity**: Although not a primary concern in this scenario, space complexity would consider additional memory usage apart from the space taken by the input data, which is mostly constant since we’re operating in-place on the given matrix.

Understanding these aspects helps in efficient algorithm design, ensuring tasks complete fast, even with large data inputs.
Diagonal matrix
A diagonal matrix is a special type of square matrix where all elements outside the principal diagonal are zero. The principal diagonal itself can contain zeroes or any other values. In simpler terms, a matrix \(A\) is diagonal if all its elements \(A[i][j]\) are zero for \(i eq j\).

In one of the problems, you are tasked with initializing only the diagonal elements to zero in an \(m \times n\) matrix. Only the elements \([i, i]\) for \(0 \leq i < \min(m, n)\) need to be accessed and modified.

This operation is efficient as it takes advantage of the matrix's structural properties, simplifying computation with fewer assignments and comparisons.
Matrix initialization
Matrix initialization is the process of setting every element in a matrix to a specific value. This procedure is a common step in algorithms where a matrix is used to store computed values or prepare the space for subsequent operations.

- **Full Initialization**: In the given exercise, this entails setting all elements of an \(m \times n\) matrix to zero. This requires nested loops, running from 0 to \(m-1\) for rows and 0 to \(n-1\) for columns.
- **Partial Initialization**: Sometimes, only specific parts of the matrix, like elements above a diagonal, need initialization. Such tasks are less computationally intensive since fewer elements are altered.

Efficient initialization ensures that the matrix starts in a predictable state, which is essential for the correctness of algorithms relying on these structures.
Assignment statements
Assignment statements are commands that allocate a specific value to a variable. They are fundamental to algorithms as they modify data within programs.

In initializing matrices, assignments play a crucial role. For example, setting each matrix element \(array[i][j] = 0\) is an assignment.

- **Counting Assignments**: In the provided problems, analyzing assignment complexity means counting how many times values are set within nested loops. This count gives us insight into the operations required, directly relating to the execution, speed, and efficiency of the algorithm.

Analyzing assignment statements allows us to understand operations' depth and potential bottlenecks within an algorithm.
Comparison operations
Comparison operations are crucial elements of control structures in algorithms where conditions have to be checked, such as loops and if-statements. They help in determining when to continue or halt a loop, making them indispensable for iterative tasks, like matrix manipulation scenarios.

- **Loop Comparisons**: In each loop, comparisons determine whether another iteration should occur, such as "is \(i < m\)?" or "is \(j < n\)?" These checks are performed on every loop iteration, contributing to the total computational cost.
- **Counting Comparisons**: In algorithm analysis, summing these comparisons gives a sense of the decision-making workload at runtime. For instance, initializing a full matrix entails \(m(n + 1)\) such checks.

Efficiently managing comparison operations can significantly optimize algorithms, especially when processing large datasets.

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