1. While a command is executing, the shell waits for the command to finish before executing the next command. If you do not want to wait for a command to finish before running another command, you can run the command in the background by appending an ampersand (&) at the end of the command. This allows you to continue using the shell while the command is executing.
2. Using sort as a filter, the sequence of commands can be rewritten as follows:
```
$ sort list > temp &
$ lpr temp
$ rm temp
```
In this sequence, the `sort` command is run in the background by appending `&` at the end. This allows the shell to execute the next command (`lpr`) without waiting for `sort` to finish. Once `lpr` is executed, the `rm` command removes the temporary file `temp`.
3. PID stands for Process IDentifier. PID numbers are unique numerical identifiers assigned to each running process on a computer system. These numbers are useful when running processes in the background because they allow you to identify and manage individual processes. The `ps` utility (or `ps -e` command) displays the PID numbers of the commands you are running, along with other process information.
4. Using wildcards, the commands to achieve the given tasks are:
a. `ls section*`
b. `ls section[1-3]`
c. `ls intro`
d. `ls section[13] ref[13]`
5. a. `grep -ci 'a'` (displays the count of lines containing the word 'a' or 'A')
b. `ls -d *.[cC]`
c. `ls -r`
d. `ls -S | lpr`
6. a. `sort numbers > phone_list`
b. `tr '[{]' '(' < permdemos.c | tr '[}]' ')' > modified_file`
c. `cat part1 part2 > book`
7. a. `cat`, `awk`
b. `grep`
c. `sort`
d. `cat file.txt | grep 'pattern' | wc -l`
8. a. `grep 'pattern' < input.txt > output.txt`
b. `grep 'pattern' < input.txt`
c. `grep 'pattern' > output.txt`
d. `cat file.txt | grep 'pattern'`
grep is used as a filter in cases (a), (b), and (d).
9. The error message "rm: cannot remove 'abc': No such file or directory" indicates that the file "abc" does not exist in the current directory. A subsequent `ls` command would display the following filenames: "abd", "abe", "abf", "abg", "abh".
10. To demonstrate that the shell creates the output file immediately, you can use the following command:
```
$ echo "Hello, world!" > output.txt
$ ls output.txt
```
Running the `ls` command immediately after the first command will display the "output.txt" file, indicating that it has been created before the command was executed.
11. If Max accidentally deletes his PATH variable, he may encounter problems when trying to execute commands that are not located in the current directory or specified with an absolute path. Without the PATH variable, the shell will not know where to find these commands. To easily return PATH to its original value, Max can open a new shell or terminal session, as it will inherit the default PATH variable from the system's configuration.
12. a. To empty a file without invoking an editor, you can use the following command:
```
$
> file.txt
```
This command uses the shell's output redirection to truncate the file and make it empty.
b. You might have permission to modify a file that you cannot delete if the file's permissions allow write access but do not allow the delete (unlink) operation. In such cases, you can modify the file's content, but you cannot remove the file itself.
13. If you accidentally create a filename that contains a nonprinting character, such as a control character, you can remove the file by specifying the filename using the appropriate escape sequence. For example, if the filename contains a control character represented by '^G', you can remove the file using the following command:
```
$ rm $'filename^G'
```
The `$'...'` syntax allows you to use escape sequences in the filename.
14. The noclobber variable in the shell, when enabled, prevents existing files from being overwritten by redirection operators (`>` or `>>`). However, the `cp` and `mv` commands do not respect the noclobber variable because they are designed to explicitly modify or move files, and not to redirect output. Therefore, the noclobber variable does not protect against overwriting existing files when using `cp` or `mv`.
15. Command names and filenames usually do not have embedded spaces because spaces are used as delimiters by the shell. If you want to create a filename containing a space, you can enclose the filename in quotes or use escape characters. For example, to create a filename "my file.txt", you can do either of the following:
```
$ touch "my file.txt"
$ touch my\ file.txt
```
To remove a filename containing a space, you can use the same quoting or escape character techniques. For example:
```
$ rm "my file.txt"
$ rm my\ file.txt
```
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Listen To increase access to a file a soft link or shortcut can be created for the file. What would happened to the soft link if the original file is deleted? OA) The file system will deallocate the space for the original file and the link will remain broken. B) The file system will deallocate the space for the original file and the space for the link. Both files will bedeleted. OC) The file system will keep the original file until the link is deleted. OD) The file system will delete the link but will keep the original file in the original path.
The correct option is A) The file system will deallocate the space for the original file and the link will remain broken.
A soft link is a special type of file that points to another file. Soft links, also known as symbolic links or symlinks, are pointers to other files or directories in a filesystem. Soft links are just like aliases, shortcuts, or references to other files. Symbolic links, like hard links, do not have any physical attributes or contents of their own. They're just references that point to a particular file's physical location. If the original file is deleted, the symbolic link will be broken or invalid. When you delete the initial file to which the soft link was pointing, the symbolic link still exists, but it is no longer functional because it is no longer linked to a valid file. The operating system does not delete the symbolic link, but instead replaces the file's information with null data. Therefore, the file system will deallocate the space for the original file, and the link will remain broken.
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You must create your own data for this excel project. Create a workbook to contain your worksheets related to this project. Your workbook and worksheets should look professional in terms of formatting and titles, etc. Date the workbook. Name the workbook Excel Project and your first name. Name each worksheet according to the task you are performing (such as subtotals). Put your name on each worksheet. Include the following in your.worksheets: Use a separate worksheet to show results of each task. Directly on the worksheet explain each numbered item and worksheet specifically so that I can follow your logic. For example, the worksheet showing functions - what five functions did you use and what is the purpose for each? Explain the data you are using. 1. Use a minimum of five functions in your first worksheet (such as SUM, MIN, etc.) 2. Create a Chart to help visualize your data. 3. Use the sart command on more than one column. Create conditional formatting along with this sort. 4. Use AutoFilter to display a group of records with particular meaning. 5. Use subtotals to highlight subtotals for particular_sategories. 6. Develop a Pivot Table and Pivot Chart to visualize data in a more meaningful way. 7. Use the If function to return a particular value. 8. Use the Goal Seek command. 9. Submit your workbook on Blackboard so that I can evaluate the cells. Use a text box to explain.
Here are the steps to create your Excel project:
Step 1: Create a new Excel workbook and name it "Excel Project - [First Name]."
Step 2: To keep your worksheets organized, create a separate worksheet for each task you perform and name each worksheet appropriately, such as "Subtotals," "Functions," etc.
Step 3: Begin by entering data into your worksheets for each task. Make sure to include an explanation of each numbered item and worksheet specifically on the worksheet so that it is easy to follow your logic. For instance, you can create a separate worksheet to show the results of each task.
Step 4: In your first worksheet, use a minimum of five functions such as SUM, MIN, MAX, etc. to showcase your skills.
Step 5: In your second worksheet, create a chart to help visualize your data.
Step 6: Utilize the sort command on more than one column and create conditional formatting along with this sort in your third worksheet.
Step 7: Use the AutoFilter feature in your fourth worksheet to display a group of records with specific meanings.
Step 8: Use subtotals to highlight subtotals for particular categories in your fifth worksheet.
Step 9: Create a Pivot Table and Pivot Chart to visualize your data more meaningfully in your sixth worksheet.
Step 10: Use the If function to return a particular value in your seventh worksheet.
Step 11: Use the Goal Seek command in your eighth worksheet.
Step 12: Finally, submit your workbook on Blackboard so that your teacher can evaluate the cells.
Use a text box to explain your work to your teacher.
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Write a JavaScript code that use a while loop to print the following prompt This is a while loop Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycles End of the loop
The prompt consists of the text "This is a while loop" followed by the word "Cycle" and a number indicating the cycle count.
The task is to write a JavaScript code that uses a while loop to print a specific prompt. The loop should continue printing the prompt until it reaches the fourth cycle, and then print "Cycles End of the loop".
To accomplish this task, you can use a while loop in JavaScript along with a counter variable to track the cycle count. Here's an example code snippet that demonstrates the desired behavior:
javascript
let cycleCount = 1;
while (cycleCount <= 4) {
console.log("This is a while loop Cycle " + cycleCount);
cycleCount++;
}
console.log("Cycles End of the loop");
In this code, we initialize the cycleCount variable to 1 before entering the while loop. The loop condition checks if the cycleCount is less than or equal to 4. Inside the loop, we print the prompt using console.log() and concatenate the current cycle count. After each iteration, we increment the cycleCount by 1.
Once the loop finishes executing for the fourth cycle, the loop condition becomes false, and the program proceeds to the next line, which prints "Cycles End of the loop" using console.log().
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You are to create a web site that will serve the purpose of a fictitious pizza parlor / store that contains the following: 1. a home page to describe the store, its location, its purpose, and its menu offerings in the form of a navigation bar. The navigation bar must be arranged by the type of food (pizza, pasta, roast beef sandwiches, etc.) 2. The store will bear your full name as a banner on the home page. 3. The Banner must include a bakground image, 4. a "Take Out" menu will be provided on a separate page. 5. an on-line order form that will permit the customer to order any product that appears on your menu. The form must be formated in a neatly organized format. Randomly placed input objects is not acceptable. The form input fields must have validation. 6. you must have appropriate, relevant images on each page. 7. background images (if used) cannot be tiled but must fill the entire background. W3Schools background-size. 8. you must use HTML5 semantic sections (not all of them) 9. your stylesheets must be imported or linked. No embedded styles or in-line styles may be used 10. you must submit all content as complete site structure. You may model your page after any of the local pizza stores but you may not copy their banners, images, or menus. You may use appropriate royalty-free images, banners, backgrounds that you can locate with search engines. You must include either a vertical or horizontal navigation menu to allow the user to select the other pages without leaving the site. Your site must be themed, meaning that all pages will have the same basic color scheme or background image. All content on a page must be able to move independently of any background image. Your goal is to present a professional well designed site. You will be graded both on content, functionality, and appearance. I do not expect your site to be as thorough or as detailed as your model sites. You will not have enough time to compose what they have composed over a period of months. But, I do expect a product that is significantly beyond a 'bare minimum'. I also expect that you will not entertain help from any other person with the design and construction of your site. It is yours to own. You must incorporate HTML5 semantic tags in your site. You must incorporate external CSS3 in your site. Thoroughness matters. The better designed and more complete your site is the better the better your grade will be. This is your chance to show your capabilities.
Designing a website for a fictitious pizza parlor/store can be an exciting project to demonstrate your web development skills.
Here's a general outline of the steps you can follow:
Start with the home page: Begin by creating an appealing home page that includes a navigation bar arranged by food categories such as pizza, pasta, sandwiches, etc. Incorporate your full name as a banner and use a background image that suits the theme of the website.
Menu offerings: Within each category on the navigation bar, create separate pages that describe the menu offerings. Include relevant images and descriptions for each item to entice visitors.
Take Out menu: Design a separate page that displays the takeout menu. This page should be easily accessible from the navigation bar and provide clear information about available items, prices, and any special offers.
Online order form: Create a well-formatted order form that allows customers to select items from the menu and input their preferences, such as toppings or customization options. Implement form validation to ensure the input is correct and provide clear instructions for placing the order.
Incorporate relevant images: Use appropriate and high-quality images throughout the website to enhance visual appeal and showcase the food offerings. Ensure the images are relevant and align with the overall theme of the site.
Background images: If you decide to use background images, ensure they fill the entire background without tiling. Utilize CSS3 properties like background-size to achieve this effect.
HTML5 semantic sections: Employ HTML5 semantic tags such as <header>, <nav>, <main>, <section>, and <footer> to structure your web pages in a meaningful way. This will enhance the accessibility and search engine optimization (SEO) of your site.
External CSS3: Create an external CSS file and link it to each HTML page. Use CSS3 properties and selectors to style the elements consistently throughout the website. Avoid inline styles or embedded stylesheets.
Consistent color scheme or background: Maintain a cohesive visual theme by ensuring all pages have the same basic color scheme or background image. This will create a unified experience for the visitors.
Navigation menu: Implement a navigation menu, either vertical or horizontal, to allow users to easily navigate between different pages within the website without leaving the site.
Remember to pay attention to details, such as responsive design for mobile devices, accessibility considerations, and ensuring a professional and polished appearance. Take pride in showcasing your skills and creating a website that goes beyond the "bare minimum" to impress your audience. Good luck with your project!
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Assume that a main memory has 32-bit byte address. A 256 KB cache consists of 4-word blocks. If the cache uses "fully associative", what is the ratio between bits used for management and bits used for storing? O A. 0.23 OB. 0.82 O C.-4.41 O D. All other answers are wrong O E. 1.23
The ratio between bits used for management and bits used for storing is 0.219. The correct answer is not provided in the given options.
To calculate the ratio between bits used for management and bits used for storing in a fully associative cache, we need to determine the number of bits used for management and the number of bits used for storing data.
In a fully associative cache, each block in the cache can hold any data from the main memory. Therefore, the cache needs to store the actual data as well as some additional information for management purposes.
Given:
Main memory address size: 32 bits
Cache block size: 4 words (1 word = 4 bytes)
Cache size: 256 KB
To find the number of bits used for storing data, we can calculate the total number of blocks in the cache and multiply it by the block size (in bytes). Since each block consists of 4 words, the block size in bytes is 4 * 4 = 16 bytes.
Number of blocks in the cache = Cache size / Block size
Number of blocks = 256 KB / 16 bytes = 16,384 blocks
Number of bits used for storing data = Number of blocks * Block size (in bits)
Number of bits used for storing data = 16,384 blocks * 16 bytes * 8 bits/byte = 2,097,152 bits
Next, we need to calculate the number of bits used for management. In a fully associative cache, each block needs to store the data as well as additional information such as tags and flags to manage the cache.
Since each block can hold any data from the main memory, we need to store the full main memory address (32 bits) as the tag for each block.
Number of bits used for management = Number of blocks * Tag size
Tag size = Main memory address size - Offset size (block size)
Offset size = log2(Block size)
Offset size = log2(16 bytes) = 4 bits
Tag size = 32 bits - 4 bits = 28 bits
Number of bits used for management = 16,384 blocks * 28 bits = 458,752 bits
Finally, we can calculate the ratio between the bits used for management and the bits used for storing data:
Ratio = (Number of bits used for management) / (Number of bits used for storing data)
Ratio = 458,752 bits / 2,097,152 bits ≈ 0.219 (rounded to three decimal places)
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QUESTION 38 Let L = {A, b} and M = {ab a), what is ML? O A. {ab, a, bab, ba} O B.{ab, a, abb, ab} O c.{aa, aab, aba} OD. {b, bb)
All the strings in this set are possible as A is there in L and a is there in M. So, the answer is Option C.Option D: {b, bb}This is not possible as there is no A in M. So, this option is also not correct.Hence, the correct option is C.{aa, aab, aba}.
Given that L = {A, b} and M = {ab a). Now we need to find the concatenation of these two sets. That is we need to find ML.Now, the concatenation of two languages L and M is defined as:{xy|x∈L,y∈M}.Therefore, we have to take one string from L and one string from M and concatenate them. Then we can form the set ML.So, we have L = {A, b} and M = {ab, a}Now take one string from L and one string from M:Option A: {ab, a, bab, ba}bab is not possible as there is no A in M.
Also, ba is not possible as there is no b in M.So, Option A is not correct.Option B: {ab, a, abb, ab}abb is not possible as there is no b in M. So, Option B is also incorrect.Option C: {aa, aab, aba}All the strings in this set are possible as A is there in L and a is there in M. So, the answer is Option C.Option D: {b, bb}This is not possible as there is no A in M. So, this option is also not correct.Hence, the correct option is C.{aa, aab, aba}.
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5:02 © * Moda * O Assignment3B 2... a CSIT114 Assignment 3B Assume that you are developing a retailing management system for a store. The following narrative describes the business processes that you learned from a store manager. Your task is to use the Noun Technique to develop a Domain Model Class Diagram. "When someone checkouts with items to buy, a cashier uses the retailing management system to record each item. The system presents a running total and items for the purchase. For the payment of the purchase can be a cash or credit card payment. For credit card payment, system requires the card information card number, name, etc.) for validation purposes. For cash payment, the system needs to record the payment amount in order to return change. The system produces a receipt upon request." (1) Provide a list of all nouns that you identify in the above narrative and indicate which of the following five categories that they belong to: (i) domain class, (ii) attribute, (ii) input/output, (iv) other things that are NOT needed to remember, and (v) further research needed. (2) Develop a Domain Model Class Diagram for the system. Multiplicities must be provided for the associations. Your model must be built with the provided information and use the UML notations in this subject. However, you should make reasonable assumptions to complete your solution. Deliverable: Include your solutions in one PDF document, which is named " .pdf". Submit it to the correct submission dropbox on Moodle before the deadline. E
List of nouns and categories:
Checkout: domain class
Item: domain class
Cashier: domain class
Retailing management system: domain class
Running total: attribute
Purchase: attribute
Payment: domain class
Cash: input/output
Credit card: input/output
Card information: attribute
Validation: input/output
Payment amount: attribute
Change: output
Receipt: output
Domain Model Class Diagram:
+------------------+ +--------------+
| Checkout | | Item |
+------------------+ +--------------+
| | <------> | |
| - purchase | | - name |
| - payment | | - price |
| | | |
+------------------+ +--------------+
^ ^
| |
+----------------+ +-------------------------+
| Retailing | | Payment |
| management | +-------------------------+
| system | | - paymentMethod: String |
| | <-----> | - cardNumber: int |
| | | - cardName: string |
| | | - amount: double |
+----------------+ +-------------------------+
^
|
+-----------------+
| Cashier |
+-----------------+
| |
| - checkout() |
| - recordItem() |
| - makePayment() |
| - printReceipt()|
+-----------------+
In this diagram, there is a many-to-many relationship between Checkout and Item, indicating that one checkout can have multiple items and one item can appear in multiple checkouts. The Retailing management system class has associations with both Payment and Cashier, indicating that it interacts with both of these classes. The Payment class has attributes for payment method, card number, card name, and amount. The Cashier class has methods for checkout, recording items, making payments, and printing receipts.
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Trace the execution of QuickSort on the following list: 81, 42, 22, 15, 28, 60, 10, 75. Your solution should show how the list is split up and how it is merged back together at each step. You may select pivot elements using any strategy you like (as long as it is consistent).
The QuickSort algorithm is applied to the list [81, 42, 22, 15, 28, 60, 10, 75] using the Lomuto partition scheme and selecting the rightmost element as the pivot. At each step, the list is split into sublists and merged back together after sorting. The process continues recursively until the entire list is sorted.
Execution Steps:
Original List: [81, 42, 22, 15, 28, 60, 10, 75]
Pivot: 75
Partitioned Lists: [42, 22, 15, 28, 60, 10] | [81]
Sorted List: [42, 22, 15, 28, 60, 10, 75, 81]
Original List: [42, 22, 15, 28, 60, 10]
Pivot: 10
Partitioned Lists: [10] | [42, 22, 15, 28, 60]
Sorted List: [10, 22, 15, 28, 60, 42]
Original List: [42, 22, 15, 28, 60]
Pivot: 60
Partitioned Lists: [42, 22, 15, 28] | [60]
Sorted List: [42, 22, 15, 28, 60]
Original List: [42, 22, 15, 28]
Pivot: 28
Partitioned Lists: [22, 15] | [28, 42]
Sorted List: [22, 15, 28, 42]
Original List: [22, 15]
Pivot: 15
Partitioned Lists: [15] | [22]
Sorted List: [15, 22]
The sorted sublists are merged back together:
[10, 15, 22, 28, 42, 60, 75, 81]
The final sorted list is [10, 15, 22, 28, 42, 60, 75, 81].
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Write a C++ program that maintains a collection of employees.
Private data that is associated with an Employee class:
Emp_ID
Age (between 25-50)
Salary_per_month
Total_deductions
Annual_salary_with_deductions
Public functions that can be performed on Employee object:
Default Constructor: (constructor without arguments). This should set 0 initially for the data members Emp_ID, Age, Salary_per_month, Total_deductions and Annual_salary_with_deductions for the object created without passing any values.
Constructor: (constructor with arguments). This should set the values for the data members Emp_ID, Age and Salary_per_month as per user’s choice for the object created by using three values. But, the Total_deductions and Annual_salary_with_deductions data members alone should be set to 0 initially in this three parametrized constructor. While setting value to the member data "Age" validation should be performed to check whether the given age is between 25 and 50 only. If so, user input can be set as value to "Age". Otherwise, print "Invalid" and exit the program.
Annual_Salary: This function should compute annual salary of the employee based on the "this" pointer after the following deductions from the annual salary.
Deduction details:-
Income tax: 10% from the annual salary.
Educational chess: 5% from the annual salary.
Professional tax: 3% from the annual salary.
Then set Total_deductions (sum of income tax, educational chess and professional tax) and Annual_salary_with_deductions (Salary_per_month * 12 – Total_deductions) of the employee by using "this" pointer and print the annual salary with the complete deduction details of the employees.
Compare_emp: This function should compare two employees’ annual salaries after all deductions and find which employee is paid high. This function should carry two objects as arguments and after comparison, it should return an object from the function to the main function which is having the highly paid employee details.
Print_emp: This function should displays all data members of the object passed as the argument to it.
Note:-
Write a main function that tests Employee class as follows:
Create two objects of type Employee using argument constructor to initialize it member data as per user’s choice to Emp_ID, Age and Salary_per_month. But Total_deductions and Annual_salary should be set to 0 initially.
Compute the annual salary for these employee objects and print the same with complete deduction details.
Compare both the employees to find the highly paid employee.
Display the details of the highly paid employee.
You may decide the type of the member data as per the requirements.
Output is case sensitive. Therefore, it should be produced as per the sample test case representations.
Age should be between 25 and 50 only. Otherwise, print "Invalid".
In samples test cases in order to understand the inputs and outputs better the comments are given inside a particular notation (…….). When you are inputting get only appropriate values to the corresponding attributes and ignore the comments (…….) section. In the similar way, while printing output please print the appropriate values of the corresponding attributes and ignore the comments (…….) section.
Sample Test cases:-
case=one
input=123 (Emp_ID)
21 (Age)
50000 (Salary)
124 (Emp_ID)
23 (Age)
60000 (Salary)
output=Employee 1
Income tax=5000
Educational chess=2500
Professional tax=1500
Total deductions=9000
Annual salary without deductions=600000
Annual salary with deductions=591000
Employee 2
Income tax=6000
Educational chess=3000
Professional tax=1800
Total deductions=10800
Annual salary without deductions=720000
Annual salary with deductions=709200
Highly paid is Employee 2
Emp_ID=124
Age=23
Salary per month=60000
Total deductions=10800
Annual salary with deductions=709200
grade reduction=15%
case=two
input=123 (Emp_ID)
21 (Age)
50000 (Salary)
124 (Emp_ID)
53 (Age)
60000 (Salary)
output=Invalid
grade reduction=15%
case=three
input=123 (Emp_ID)
51 (Age)
50000 (Salary)
124 (Emp_ID)
23 (Age)
60000 (Salary)
output= Invalid
grade reduction=15%
case=four
input=100 (Emp_ID)
21 (Age)
80000 (Salary)
124 (Emp_ID)
23 (Age)
70000 (Salary)
output= Employee 1
Income tax=8000
Educational chess=4000
Professional tax=2400
Total deductions=14400
Annual salary without deductions=960000
Annual salary with deductions=945600
Employee 2
Income tax=7000
Educational chess=3500
Professional tax=2100
Total deductions=12600
Annual salary without deductions=840000
Annual salary with deductions=827400
Highly paid is Employee 1
Emp_ID=100
Age=21
Salary per month=80000
Total deductions=14400
Annual salary with deductions=945600
grade reduction=15%
The C++ code that implements the Employee class and performs the required operations:
```cpp
#include <iostream>
class Employee {
private:
int Emp_ID;
int Age;
double Salary_per_month;
double Total_deductions;
double Annual_salary_with_deductions;
public:
Employee() {
Emp_ID = 0;
Age = 0;
Salary_per_month = 0;
Total_deductions = 0;
Annual_salary_with_deductions = 0;
}
Employee(int id, int age, double salary) {
Emp_ID = id;
if (age >= 25 && age <= 50) {
Age = age;
} else {
std::cout << "Invalid" << std::endl;
exit(0);
}
Salary_per_month = salary;
Total_deductions = 0;
Annual_salary_with_deductions = 0;
}
void Annual_Salary() {
double income_tax = Annual_salary_without_deductions() * 0.10;
double educational_chess = Annual_salary_without_deductions() * 0.05;
double professional_tax = Annual_salary_without_deductions() * 0.03;
Total_deductions = income_tax + educational_chess + professional_tax;
Annual_salary_with_deductions = Annual_salary_without_deductions() - Total_deductions;
}
double Annual_salary_without_deductions() const {
return Salary_per_month * 12;
}
static Employee Compare_emp(const Employee& emp1, const Employee& emp2) {
if (emp1.Annual_salary_with_deductions > emp2.Annual_salary_with_deductions) {
return emp1;
} else {
return emp2;
}
}
void Print_emp() const {
std::cout << "Emp_ID=" << Emp_ID << std::endl;
std::cout << "Age=" << Age << std::endl;
std::cout << "Salary per month=" << Salary_per_month << std::endl;
std::cout << "Total deductions=" << Total_deductions << std::endl;
std::cout << "Annual salary with deductions=" << Annual_salary_with_deductions << std::endl;
std::cout << "grade reduction=15%" << std::endl;
}
};
int main() {
int id1, age1, id2, age2;
double salary1, salary2;
// Input Employee 1 details
std::cin >> id1 >> age1 >> salary1;
Employee emp1(id1, age1, salary1);
emp1.Annual_Salary();
// Input Employee 2 details
std::cin >> id2 >> age2 >> salary2;
Employee emp2(id2, age2, salary2);
emp2.Annual_Salary();
// Print details and compare employees
std::cout << "Employee 1" << std::endl;
emp1.Print_emp();
std::cout << "Employee 2" << std::endl;
emp2.Print_emp();
Employee highly_paid = Employee::Compare_emp(emp1, emp2);
std::cout << "Highly paid is Employee " << highly_paid.Emp_ID << std::endl;
highly_paid.Print_emp();
return 0;
}
```
This code defines the `Employee` class with the required private data members and public member functions. It includes constructors, the `Annual_Salary` function to calculate deductions and annual salary, the `Compare_emp` function to compare two employees,
and the `Print_emp` function to display the employee details. In the `main` function, two `Employee` objects are created, their details are taken as input, the annual salaries are computed, and the details of the highly paid employee are printed.
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Q.2.1 Consider the snippet of code below, then answer the questions that follow: if customerAge>18 then if employment = "Permanent" then if income > 2000 then output "You can apply for a personal loan" endif endif Q.2.1.1 If a customer is 19 years old, permanently employed and earns a salary of R6000, what will be the outcome if the snippet of code is executed? Motivate your answer. Q.2.2 Using pseudocode, plan the logic for an application that will prompt the user for two values. These values should be added together. After exiting the loop, the total of the two numbers should be displayed. endif (2)
Q.2.1.1: The outcome of executing the snippet of code for a 19-year-old customer who is permanently employed and earns a salary of R6000 will be "You can apply for a personal loan."Q.2.2:The pseudocode logic for the application prompts the user for two values, adds them together, and displays the total after exiting the loop.
Q.2.1.1:The code snippet consists of nested if statements that evaluate specific conditions. In this case, the customer's age of 19 satisfies the condition of being greater than 18. Additionally, their employment status is "Permanent" and their income of R6000 exceeds the threshold of 2000. Therefore, all the nested if statements evaluate to true, resulting in the execution of the output statement "You can apply for a personal loan."
Q.2.2:The pseudocode outlines the step-by-step process of the application. It begins by initializing a variable called "total" to 0. Then, it prompts the user for the first value and stores it in "value1." Next, it prompts for the second value and stores it in "value2." The values of "value1" and "value2" are then added together and stored in the "total" variable. Finally, the application displays the value of "total" and exits the loop, completing the logic for adding the two input values.
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Q4. Consider an array having elements: 10 2 66 71 12 8 52 34 Sort the elements of the array in an ascending order using insertion sort algorithm.
In the case of the given array [10, 2, 66, 71, 12, 8, 52, 34], applying the insertion sort algorithm results in the array being sorted in ascending order as [2, 8, 10, 12, 34, 52, 66, 71].
To sort the given array [10, 2, 66, 71, 12, 8, 52, 34] in ascending order using the insertion sort algorithm, the following steps can be followed:
Start with the second element (index 1) and iterate through the array.For each element, compare it with the previous elements in the sorted portion of the array.If the current element is smaller than any of the previous elements, shift those elements one position to the right.Insert the current element in its correct position within the sorted portion of the array.Repeat steps 2 to 4 until all elements are in their sorted positions.Using the insertion sort algorithm, the sorted array in ascending order would be: [2, 8, 10, 12, 34, 52, 66, 71].
Starting with the array [10, 2, 66, 71, 12, 8, 52, 34], we iterate through the array starting from the second element (2) and compare it with the previous element (10). Since 2 is smaller than 10, we shift 10 one position to the right and insert 2 in the first position.
Next, we move to the next element (66) and compare it with the previous elements (10 and 2). Since 66 is greater than both, we leave it in its position.
We continue this process for each element, comparing it with the previous elements in the sorted portion of the array and inserting it in its correct position.
After completing all iterations, the array will be sorted in ascending order: [2, 8, 10, 12, 34, 52, 66, 71].
The insertion sort algorithm is an efficient method for sorting small or partially sorted arrays. It works by iteratively inserting each element into its correct position within the sorted portion of the array. In the case of the given array [10, 2, 66, 71, 12, 8, 52, 34], applying the insertion sort algorithm results in the array being sorted in ascending order as [2, 8, 10, 12, 34, 52, 66, 71].
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Can someone fix this code? I'm trying to run it in PYTHON and it won't work.
Thanks!
import random
import time
# Initial Steps to invite in the game:
name = input("Enter your name: ")
print("Hello " + name + "! Best of Luck!")
time.sleep(2)
print("The game is about to start!\n Let's play Hangman!")
time.sleep(3)
# The parameters we require to execute the game:
def main():
global count
global display
global word
global already_guessed
global length
global play_game
words_to_guess = ["january","border","image","film","promise","kids","lungs","doll","rhyme","damage"
,"plants"]
word = random.choice(words_to_guess)
length = len(word)
count = 0
display = '_' * length
already_guessed = []
play_game = ""
# A loop to re-execute the game when the first round ends:
def play_loop():
global play_game
play_game = input("Do You want to play again? y = yes, n = no \n")
while play_game not in ["y", "n","Y","N"]:
play_game = input("Do You want to play again? y = yes, n = no \n")
if play_game == "y":
main()
elif play_game == "n":
print("Thanks For Playing! We expect you back again!")
exit()
The given code is a Hangman game. It prompts the player to enter their name, then starts the game by choosing a random word from a list of words. The player is then asked if they want to play again.
The code provided is a basic implementation of a Hangman game using Python. Let's break down the code and understand how it works.
First, the code imports the random and time modules, which are used for choosing a random word and introducing delays in the game, respectively.
Next, the code asks the player to enter their name and greets them with a welcome message. It uses the input() function to read the player's name and concatenates it with a string for the greeting.
After a brief delay of 2 seconds using time.sleep(2), the code prints a message indicating that the game is about to start and prompts the player to play Hangman. Another delay of 3 seconds is introduced using time.sleep(3) to create a pause before starting the game.
The code then defines a function named main(), which holds the core logic of the game. Within this function, several global variables are declared to store important game parameters such as count (number of incorrect guesses), display (current state of the word being guessed, with underscores representing unknown letters), word (the random word chosen from the list), already_guessed (a list to store the letters already guessed by the player), length (length of the word), and play_game (variable to control if the player wants to play again).
The words_to_guess list contains the words that the game will randomly choose from. The random.choice() function is used to select a word from this list and assign it to the word variable. The len() function is then used to determine the length of the chosen word and store it in the length variable.
The count variable is initialized to 0, representing the number of incorrect guesses made by the player so far. The display variable is set to a string of underscores (_) with a length equal to the chosen word, indicating the unknown letters. The already_guessed list is initialized as an empty list to keep track of the letters already guessed by the player.
Finally, the play_game variable is set to an empty string, and the play_loop() function is defined. This function prompts the player if they want to play again and checks their input. If the player enters 'y' or 'Y', indicating they want to play again, the main() function is called to start a new game. If they enter 'n' or 'N', the game ends. If they enter any other input, they are prompted again until they provide a valid choice.
In summary, the given code sets up the initial steps for the Hangman game, including greeting the player, choosing a random word, and providing an option to play again. The core logic of the game is contained within the main() function, and the play_loop() function handles the player's choice to play again.
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KNN questions
Question. Notice the structured patterns in the distance matrix, where some rows or columns are visible brighter. (Note that with the default color scheme black indicates low distances while white indicates high distances.)
What in the data is the cause behind the distinctly bright rows?
What causes the columns?
Your Answer:
The distinctly bright rows in the distance matrix of KNN signify isolated or dissimilar data points, potentially including outliers. The bright columns, on the other hand, indicate instances that are consistently far away from many other data points and may represent clusters or subgroups within the dataset.
1. The distinctly bright rows in the distance matrix of KNN (K-Nearest Neighbors) can be attributed to instances that have high distances from most other data points in the dataset. This indicates that these particular rows represent data points that are relatively isolated or dissimilar compared to the rest of the data. On the other hand, the bright columns in the distance matrix correspond to instances that are consistently far away from many other data points. This suggests the presence of outliers or extreme values that significantly deviate from the overall patterns observed in the dataset.
2. In KNN, the distance matrix represents the distances between each pair of data points in the dataset. Bright rows in the distance matrix indicate instances that have relatively high distances from the majority of other data points. This could occur due to several reasons. One possibility is that these rows represent outliers or anomalies in the data, which are significantly different from the majority of the instances. Outliers can have a substantial impact on the KNN algorithm's performance by influencing the neighborhood of nearby points. Consequently, they may lead to erroneous classifications or predictions.
3. On the other hand, the bright columns in the distance matrix represent instances that are consistently far away from many other data points. This suggests the presence of patterns or characteristics that make these instances distinct from the rest of the data. Such columns may indicate the existence of specific clusters or groups within the dataset, where the instances in the column share similar attributes or properties. These clusters might represent meaningful subgroups or subclasses in the data, reflecting inherent structures or patterns that are of interest for further analysis or interpretation.
4. Analyzing these patterns can provide valuable insights into the characteristics and structure of the data, aiding in the interpretation and refinement of the KNN algorithm's outcomes.
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Vehicles are increasingly connected to different types of networks, making them targets for
potential attacks. Consider a smart vehicle prototype that works as follows:
- Multiple types of sensors, including cameras, lidar sensors, and infrared sensors, are used to detect
road conditions to provide varying degrees of autonomous driving support;
- All data from sensors are transmitted to the on-board computer for decision making. An on-board
backup server stores all data in the backend;
- The user can interact with the on-board computer via a touchscreen;
- When the driver is not in the vehicle, the vehicle sets up an alarm mode. Drivers get alarms
through their smartphones. Optionally, alarms can also be sent to the police;
- The software on-board can be updated remotely by the vehicle manufacturer, with the permission
of the driver.
- The operation of the vehicle will be simplified as follows: the on-board computer processes the
sensor readings and makes decisions such as speed maintenance and braking operation. The
driver’s input will override the computer decisions and will take priority. Once the driver exits the
vehicle, the doors should be automatically locked and the vehicle enters alarm mode.
Based on this description, plot a level 0 and level 1 DFD diagram with the following external
entities: vehicle manufacturer, driver, and police. You may assume that there is only one on-board
computer for decision making, and one on-board backup server for storage of all data. You may
add additional details and assumptions as you see necessary. In the level 0, all entities including
sensors, the on-board computer and the on-board server should be plotted. In level 1, you should
focus on the operations of the vehicle and plot the basic functions and processes including speed
maintenance, braking, and alarm mode.
Level 0 DFD Diagram: It is a high-level data flow diagram that represents the overall view of a system, and it displays external entities, processes, and data flows that enter and exit the system.
The DFD is developed to present a view of the system at a high level of abstraction, with minimal information on the process. The DFD diagram for the given system is given below: As shown in the above diagram, there are three external entities, vehicle manufacturer, driver, and police, and the three processes involved in the system are the onboard computer for decision making, onboard backup server for data storage, and sensors for data collection.
Level 1 DFD Diagram: It represents a low-level data flow diagram that provides an in-depth view of a system. It contains all information on the process required to construct the system and breaks down the process into smaller, more explicit sub-processes. The DFD diagram for the given system is given below: As shown in the above diagram, the driver can interact with the on-board computer via a touchscreen. When the driver is not in the vehicle, the vehicle sets up an alarm mode, which sends alarms to the driver's smartphones. The software on-board can be updated remotely by the vehicle manufacturer, with the permission of the driver. Once the driver exits the vehicle, the doors are automatically locked, and the vehicle enters alarm mode. The on-board computer processes the sensor readings and makes decisions such as speed maintenance and braking operation. The driver's input will override the computer decisions and will take priority.
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Which of the following function calls would successfully call this function? (Select all that apply) void swapShellsFirstInArray(int basket) { int temp basket [0]; basket [0] basket [1] basket [1]; temp; a. int collection [3] - [3, 2, 1); swapShellsFirst InArray(collection); b. int collection []; swapShellsFirstInArray(collection); c. int collection [5] (3, 2, 1, 4, 6}; swapShellsFirstInArray(collection); d. int collection] =(3, 2); swapShellsFirstInArray(collection); - e. int collection [10] (3, 2, 1, 4, 6); swapShellsFirstInArray(collection); > 3
The following function calls would successfully call the swapShellsFirstInArray() function:
int collection[3] = {3, 2, 1}; swapShellsFirstInArray(collection);
int collection[5] = {3, 2, 1, 4, 6}; swapShellsFirstInArray(collection);
int collection[10] = {3, 2, 1, 4, 6}; swapShellsFirstInArray(collection);
The swapShellsFirstInArray() function takes an array of integers as its input. The function then swaps the first two elements of the array. The function returns nothing.
The three function calls listed above all pass an array of integers to the swapShellsFirstInArray() function. Therefore, the function calls will succeed.
The function call int collection = {}; swapShellsFirstInArray(collection); will not succeed because the collection variable is not an array. The collection variable is an empty object. Therefore, the swapShellsFirstInArray() function will not be able to access the elements of the collection variable.
The function call int collection = (3, 2); swapShellsFirstInArray(collection); will not succeed because the collection variable is not an array. The collection variable is a tuple. A tuple is a data structure that can store a fixed number of elements. The elements of a tuple are accessed by their index. The swapShellsFirstInArray() function expects an array as its input. Therefore, the function will not be able to access the elements of the collection variable.
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Answer only in R coding language, please. Thank you
Q1. Write a function margin_index_power() that takes as input a matrix A and an argument rows, TRUE/FALSE with a default value of TRUE. margin_index_power(A, rows = TRUE) outputs the matrix A with the elements in the ith row of A taken to the ith power. If rows = FALSE, then do the same but with the columns instead of rows of A.
Please test your function on the following test inputs in your submission:
# test case 1: A = matrix(6:1, 3, 2)
# test case 2: A = matrix(2:7, 3, 2), rows = FALSE
# test case 3: A = matrix(2:5, 3, 4)
Q2.
Write a function is_anti_diagonal() that takes as input a matrix A and outputs TRUE if it is anti-diagonal and FALSE otherwise. While you can assume A is a matrix, you cannot assume that it is square.
Q3.
Write a function called set_border_NA() that takes as input a matrix A and outputs A with its borders set to NA. If A has exactly one row or exactly one column, throw an error of your choosing.
1. The function `margin_index_power()` in R takes a matrix `A` as input along with an argument `rows` (default value: TRUE). It computes the element-wise power of the elements in each row or column of `A`, depending on the value of `rows`. If `rows = TRUE`, it raises each element in the ith row to the power of i. If `rows = FALSE`, it performs the same operation on the columns of `A`. The function returns the modified matrix `A`.
2. The function `is_anti_diagonal()` in R determines whether a given matrix `A` is anti-diagonal. It checks if all the elements on the main diagonal are zero and all the elements outside the main diagonal are non-zero. The function returns TRUE if `A` is anti-diagonal and FALSE otherwise. It handles non-square matrices as well.
3. The function `set_border_NA()` in R takes a matrix `A` as input and sets the border elements of `A` to NA. It first checks if `A` has exactly one row or one column. If so, it throws an error. Otherwise, it identifies the border elements of `A` and replaces them with NA. The modified matrix `A` is then returned as the output.
1. Function `margin_index_power()`: The function takes a matrix `A` and an argument `rows` indicating whether to operate on rows (default) or columns. It uses the `apply()` function to iterate over the rows or columns of `A`. Within each iteration, it raises the elements in the current row or column to the power of the corresponding index. The modified matrix `A` is returned.
2. Function `is_anti_diagonal()`: The function checks if a matrix `A` is anti-diagonal by comparing the elements on the main diagonal with zero and the elements outside the main diagonal with non-zero values. It uses a combination of indexing and logical operators to perform this check. The function returns TRUE if `A` is anti-diagonal and FALSE otherwise, even for non-square matrices.
3. Function `set_border_NA()`: The function first checks if `A` has exactly one row or one column. If it does, it throws an error indicating that the function cannot handle matrices with only one row or column. Otherwise, it identifies the border elements of `A` using indexing and replaces them with NA values. The modified matrix `A` with NA values at the borders is returned as the output.
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Two hosts simultaneously send data through the network with a capacity of 1 Mpbs. Host A uses UDP and transmits 100 bytes packet every 1 msec. Host B generates data with a rate of 600 kpbs and uses TCP. Which host will obtain higher throughput and why? Explain your answer.
Host A using UDP will obtain higher throughput compared to Host B using TCP. UDP (User Datagram Protocol) is a connectionless protocol that does not guarantee reliable delivery of data.
It has lower overhead compared to TCP and does not require acknowledgment of packets. This allows Host A to send data more frequently, with smaller packet sizes, resulting in higher throughput. Host A sends 100-byte packets every 1 millisecond, which translates to a data rate of 100 kilobits per second (kbps). Since the network capacity is 1 Mbps (1,000 kbps), Host A's data rate of 100 kbps is well below the network capacity, allowing it to achieve higher throughput.
On the other hand, Host B using TCP (Transmission Control Protocol) is a connection-oriented protocol that ensures reliable delivery of data. TCP establishes a connection between the sender and receiver, performs flow control, and handles packet loss and retransmission. This additional overhead reduces the available bandwidth for data transmission. Host B generates data at a rate of 600 kbps, which is closer to the network capacity of 1 Mbps. The TCP protocol's mechanisms for reliability and congestion control may cause Host B to experience lower throughput compared to Host A.
In summary, the higher throughput is achieved by Host A using UDP because of its lower overhead and ability to transmit data more frequently. Host B using TCP has additional protocols and mechanisms that reduce the available bandwidth for data transmission, resulting in potentially lower throughput.
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Write a C++ program that reads the user's name and his/her body temperature for the last three hours. A temperature value should be within 36 G and 42.0 Celsus. The program calculates and displays the maximum body temperature for the last three hours and it he/she is normal or might have COVID19 The program must include the following functions: 1. Max Temp() function: takes three temperature values as input parameters and returris the maximum temperature value 2. COVID190) function takes the maximum temperature value and the last temperature value as input parameters, and displays in the user might have COVID19 or not according to the following instructions: If the last temperature value is more than or equal to 37.0, then display "You might have COVID19, visit hospital immediately Else If the maximum temperature value is more than or equal to 37.0 and the last temperature value is less than 37.0, then display "You are recovering! Keep monitoring your temperature!" Otherwise, display "You are good! Keep Social Distancing and Sanitizer 3. main() function Prompts the user to enter the name. Prompts the user to enter a temperature value from 36.0-42.0 for each hour separately (3hrs). If the temperature value is not within the range, it prompts the user to enter the temperature value again. Calls the Max Temp() function, then displays the user name and the maximum temperature value Calls the COVID19() function Sample Run 2 Sample Run 3. Please enter your name: Arwa Please enter your name Saed Enter temperature for 3 hours ago (36.0-42.0) 36.8 Enter temperature for 2 hours ago (36 0-42.0) 36.5 Enter temperature for last hour (36.0-42.0) 37.1 Enter temperature for 3 hours ago (36.0-42.0) 38.5 Enter temperature for 2 hours ago (36.0-42.01: 37.6 Enter temperature for last nour (36.0-42.0) 36.0 Arwa, your max body temperature in the last 3 hours was (37.1. Saed your max body temperature in the last 3 hours was 38.5 You are recovering! Keep monitoring your temperature You might have COVID19, visit hospital immediately! Sample Run 1: Please enter your name: Ahmed Enter temperature for 3 hours ago (36.0-42.0): 36.5 Enter temperature for 2 hours ago (36.0-42.0) 46.4 Enter temperature for 2 hours ago (36.0-42.0) 32.1 Enter temperature for 2 hours ago (36.0-42.0): 36.9 Enter temperature for last hour (36.0-42.0) 36.5 Ahmed, your max body temperature in the last 3 hours was 36.9. You are good! Keep Social Distancing and Sanitize!
it displays the maximum temperature and the corresponding message based on the temperature readings.
Certainly! Here's a C++ program that reads the user's name and their body temperature for the last three hours. It calculates the maximum body temperature and determines whether the user might have COVID-19 or not, based on the temperature readings:
```cpp
#include <iostream>
#include <string>
// Function to calculate the maximum temperature among three values
double maxTemp(double temp1, double temp2, double temp3) {
double max = temp1;
if (temp2 > max) {
max = temp2;
}
if (temp3 > max) {
max = temp3;
}
return max;
}
// Function to check if the user might have COVID-19 based on temperature readings
void COVID19(double maxTemp, double lastTemp) {
if (lastTemp >= 37.0) {
std::cout << "You might have COVID-19. Visit the hospital immediately." << std::endl;
} else if (maxTemp >= 37.0 && lastTemp < 37.0) {
std::cout << "You are recovering! Keep monitoring your temperature!" << std::endl;
} else {
std::cout << "You are good! Keep social distancing and sanitize!" << std::endl;
}
}
int main() {
std::string name;
double temp1, temp2, temp3;
std::cout << "Please enter your name: ";
getline(std::cin, name);
do {
std::cout << "Enter temperature for 3 hours ago (36.0-42.0): ";
std::cin >> temp1;
} while (temp1 < 36.0 || temp1 > 42.0);
do {
std::cout << "Enter temperature for 2 hours ago (36.0-42.0): ";
std::cin >> temp2;
} while (temp2 < 36.0 || temp2 > 42.0);
do {
std::cout << "Enter temperature for last hour (36.0-42.0): ";
std::cin >> temp3;
} while (temp3 < 36.0 || temp3 > 42.0);
double maxTemperature = maxTemp(temp1, temp2, temp3);
std::cout << name << ", your max body temperature in the last 3 hours was " << maxTemperature << "." << std::endl;
COVID19(maxTemperature, temp3);
return 0;
}
```
This program prompts the user to enter their name and their body temperature for the last three hours, ensuring that the temperature values are within the range of 36.0-42.0. It calculates the maximum temperature using the `maxTemp()` function and then determines if the user might have COVID-19 or not using the `COVID19()` function. Finally, it displays the maximum temperature and the corresponding message based on the temperature readings.
Please note that in the `COVID19()` function, the logic is based on the assumption that a temperature of 37.0 or higher indicates a potential COVID-19 case. You can modify this logic according to the specific guidelines or requirements of your application.
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- All answers (either Microsoft Word or answer on text pad) must be converted to a PDF file (one file only) and upload through Spectrum within stipulated times. - The lecturer has the right not to accept the submission of plagiarized work either from internet or amongst peers. . 1. Bin the age variable using the bins for below 28, 28-65, and over 65. Create a bar chart and normalized bar chart of the binned age variable with response overlay. Work with bank_marketing_training data set for this question.
2. For the following questions, work with the cereals data set. Here example to load data csv in Spyder: cereals = pd.read_csv("C:/.../cereals.csv") cereals = pd.read_csv("C:/Users/Soon SV/Desktop/DSPR_Data_Sets/cereals.csv") a) Create a bar graph of the Manuf variable with Type overlay. b) Create a contingency table of Manuf and Type. c) Create normalized histogram of Calories with Manuf overlay. d) Bin the Calories variable using bins for 0-90, 90-110, and over 110 calories. Create a normalized bar chart of the binned calories variable with Manuf overlay.
In the first question, the age variable from the bank_marketing_training dataset is binned into three categories and a bar chart, as well as a normalized bar chart, is created with response overlay. In the second question, using the cereals dataset, a bar graph of the Manuf variable with Type overlay is created
For the first question, the age variable from the bank_marketing_training dataset is categorized into three bins: below 28, 28-65, and over 65. A bar chart is created to visualize the distribution of the binned age variable, and a normalized bar chart is generated to compare the distribution with the response variable overlay.
Moving on to the second question, the cereals dataset is used. In part (a), a bar graph is created to display the distribution of the Manuf variable, and the Type variable is overlaid to show the distribution of cereal types by manufacturer. In part (b), a contingency table is generated to analyze the relationship between the Manuf and Type variables.
In part (c), a normalized histogram is created to visualize the distribution of Calories, and the Manuf variable is overlaid to observe how different manufacturers contribute to the calorie distribution. Finally, in part (d), the Calories variable is binned into three categories, and a normalized bar chart is generated to compare the binned calorie distribution with the manufacturer overlay.
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Write a python program: that writes how often an ETF rebalances?
An ETF rebalance is the process of bringing an ETF back to its original target asset allocation. The purpose of a rebalance is to maintain the desired asset allocation and maintain diversification. To determine how often an ETF rebalances, we must look at the fund's prospectus or research its holdings and look at its portfolio turnover ratio. A portfolio turnover ratio is the percentage of a fund's assets that have been bought and sold over a specific time period. It is a measure of how often an ETF rebalances its portfolio. Here is the Python program that writes how often an ETF rebalances:
```
import pandas as pd
# Read ETF holdings data
holdings = pd.read_csv("ETF_holdings.csv")
# Calculate the portfolio turnover ratio
portfolio_turnover_ratio = len(holdings) / holdings["Ticker"].nunique()
# Print the portfolio turnover ratio
print("The ETF rebalances approximately", round(portfolio_turnover_ratio, 2), "times per year.")```The program reads the ETF holdings data from a CSV file and calculates the portfolio turnover ratio. Then, it prints out the number of times per year the ETF rebalances.
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4. Convert the following grammar to Chomsky normal form. (20 pt) S → ABC A + aC | D B → bB | A | e C → Cc | Ac | e
D → aa
Eliminate e-productions: First, find nullable symbols and then eliminate. Remove chain productions: First, find chain sets of each nonterminal and then do the removals. Remove useless symbols: Explicitly indicate nonproductive and unreachable symbols. Convert to CNF: Follow the two. Do not skip any of the steps. Apply the algorithm as we did in class.
Here is the solution to convert the given grammar to Chomsky Normal Form:(1) Eliminate e-productions:There are three variables A, B, and C, which produce ε. So, we will remove the productions with e.First, eliminate productions with A→ε. We have two productions: A → aC and A → D.
Now, we need to replace A in other productions by the right-hand sides of these productions:A → aC | D | aSecond, eliminate productions with B→ε. We have three productions: B→ bB, B → A, and B → ε. Now, we need to replace B in other productions by the right-hand sides of these productions:B → bB | A | b | εThird, eliminate productions with C→ε. We have three productions: C → Cc, C → Ac, and C → ε. Now, we need to replace C in other productions by the right-hand sides of these productions:C → Cc | Ac | c(2) Remove chain productions:A → aC | D | aB → bB | A | b | εC → Cc | Ac | cD → aa(3) Remove useless symbols:There are no nonproductive or unreachable symbols.(4) Convert to CNF:S → XYX → AB | ADY → BC | AXZ → a | b | cA → a | D | bB → b | aC → C | A | cD → aaTherefore, the grammar is converted into CNF.
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Is the following statement True or False?
It is guaranteed that Dynamic Programming will generate an optimal solution as it generally considers all possible cases and then choose the best. However, in Greedy Method, sometimes there is no such guarantee of getting global optimal solution.
O True
O False
The statement : It is guaranteed that Dynamic Programming will generate an optimal solution as it generally considers all possible cases and then choose the best, is false.
False. The statement is incorrect. While it is true that dynamic programming generally considers all possible cases and chooses the best solution, it does not guarantee an optimal solution in all cases. Dynamic programming is based on the principle of optimality, where the optimal solution to a larger problem can be constructed from optimal solutions to its subproblems. However, this assumption holds true only if the problem exhibits the optimal substructure property. If the problem lacks this property, dynamic programming may not generate an optimal solution.
On the other hand, the statement's claim about the Greedy Method is not entirely accurate either. While it is true that the Greedy Method does not always guarantee a global optimal solution, it can still provide satisfactory solutions in many cases. The Greedy Method makes locally optimal choices at each step, hoping that these choices will lead to a global optimum. However, the lack of a systematic consideration of all possibilities may result in a suboptimal solution. Therefore, while the Greedy Method may not guarantee an optimal solution in all scenarios, it can still be effective in certain situations and provide reasonably good solutions.
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I have a quick question to ask here: It's apparently that (-5 * 0.2 + 1 / 1000) = -0.999 approximately -1. Yet the SQL shows it's 1.0 so I try (1000 / 1000) which is correct. Can somebody explain it here in detail and I will give you a good rating if you can answer it correctly? PostgreSQL 13.4 : TLSv1.2 : max_sql_connection : postgres : SQL Query 1 -- singleton case 1: 2 select ABS( -5* 0.2 + 1 / 1000); 4 * line 2, column 25, location 46 abs 1.0 Ô e PostgreSQL 13.4 : TLSv1.2 : max_sql_connection : postgres : SQL Query 1 singleton case 1: 2 select ABS( 1000 / 1000); * line 2, column 18, location 39 abs 1
The discrepancy in the results between the two SQL queries arises due to the order of operations in arithmetic calculations and the data types used in the calculations. The first query `ABS(-5 * 0.2 + 1 / 1000)` yields 1.0, while the second query `ABS(1000 / 1000)` also yields 1. The difference in the intermediate steps of the calculations leads to different results.
In the first query, `-5 * 0.2` is calculated first, resulting in -1. Then, `1 / 1000` is computed, which evaluates to 0.001. Finally, the sum of these two values (-1 + 0.001) is taken and passed to the `ABS` function, resulting in 1.0.
In the second query, `1000 / 1000` is calculated, which equals 1. The result is then passed to the `ABS` function, which also yields 1.
The difference in results arises from the fact that the first calculation involves floating-point arithmetic with decimal numbers, while the second calculation involves integer division. The order of operations and data types used in the calculations lead to the discrepancy in the final results.
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package p1; public class Parent{ private int x; public int y; protected int z; int w; public Parent() { System.out.println("In Parent"); } public void print() { System.out.print(x + + y); } }// end class = package p2; public class Child extends Parent{ private int a; public Child() { System.out.println("In Child"); } public Child(int a) { this.a = a; System.out.print("In Child with parameter"); } public void print() { // 1 System.out.print(a); // 2 System.out.print(x); // 3 System.out.print(z); // 4 System.out.print (w); // end class In the method print() of the child class. Which statement is illegal ?? O All statements are illegal. O // 2 System.out.print (x); // 4 System.out.print (w); O // 2 System.out.print (x); // 3 System.out.print (z); // 2 System.out.print (x): // 3 System.out.print(z); 77 4 System.out.print (w); // 1 System.out.print(a); // 2 System.out.print (x); // 2 System.out.print (x); O
In the given code, the statement "// 2 System.out.print(x);" is illegal in the method print() of the child class.
The class Child extends the class Parent, which means that it inherits the members (fields and methods) of the parent class. However, there are certain restrictions on accessing these members depending on their access modifiers.
In the code provided:
The statement "// 2 System.out.print(x);" tries to access the private member x of the parent class Parent from the child class Child. Private members are only accessible within the class in which they are declared and are not visible to the child classes. Therefore, accessing x directly in the child class is illegal.
To fix this issue, you can modify the accessibility of the member x in the parent class Parent to be protected or public. For example:
package p1;
public class Parent {
protected int x; // Modified the access modifier to protected
// Rest of the class code...
}
With this modification, the child class Child will be able to access the member x using the statement "// 2 System.out.print(x);".
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Consider the following code which is part of a multi-threaded program and will be executed concurrently. int private_count [MAX_THREADS]; void* count3s_thread (void *arg) { int id= (int) arg; int length_per_thread = length/t; int start = id*length_per_thread; int end = start+length_per_thread; int i; if (end>length) end length; for (i start; i
The given code is a part of a multi-threaded program that counts the number of occurrences of the digit 3 in an array. The array is divided into several sub-arrays, and each thread counts the number of 3s in its assigned sub-array.
The private_count array keeps track of the number of 3s counted by each thread.
The count3s_thread function takes a void pointer argument arg, which is cast to an integer id representing the thread ID. The length of the total array is divided by the number of threads t to determine the length of the sub-array assigned to each thread. The start and end indices of the sub-array are calculated using the thread ID and the sub-array length.
The for loop iterates over the elements of the sub-array from start to end, and checks if each element is equal to 3. If it is, the corresponding element of the private_count array is incremented.
It’s important to note that the private_count array is declared outside of the function, but since each thread will access a different portion of the array (i.e., their respective index), there won't be any data race as each thread is updating its own element only.
Overall, this code implements a parallel approach to counting the number of 3s in an array, which can significantly reduce the running time of the program compared to a sequential implementation. However, it's important to ensure that the threads do not interfere with each other while accessing the shared data (i.e., private_count array) to avoid any race conditions or synchronization errors.
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Which one is not a keyword?
A. double B. if C. return D. Float
The keyword in C programming language has a defined meaning and it can not be used for any other purpose. float, double, and if are keywords of C programming language. So the answer is D.float.
Whereas, Return is not a keyword in C programming language.A keyword is a reserved word that has a special meaning and it cannot be used as a variable name, function name, or any other identifier. In C programming language, there are a total of 32 keywords.Here are the keywords of C programming language:auto double if long else break enum int char extern float case continue default const for goto do while return signed sizeof static struct switch union unsigned void volatile typedefApart from these 32 keywords, there are other identifiers and reserved words. These are the words that have some meaning or definition in C programming language, but they are not keywords. Return is an example of a reserved word. It is used to return a value from a function but it is not a keyword.
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You must use JFLAP to answer this question. 2. Do not hand-draw the required state diagram. 3. Make sure you pick the Turing Machine option on JFLAP. 1. Using the Finite Automaton option on JFLAP is not acceptable. 5. A scanned image of a hand-drawn state will not be acceptable. 3. Name your JFLAP project file as Q3.jff and upload Q3.jff. Use JFLAP to draw the state diagram of a Turing Machine that recognizes the lang 2n ³n | n ≥ 0}. {a b²c3n
The Turing Machine (TM) for the language {a^b^2c^3n | n ≥ 0} has multiple states and transitions to process the input string. It starts in the initial state, reads 'a' symbols and moves to a state where it expects 'b' symbols. After reading two 'b' symbols, it transitions to a state to read 'c' symbols. Once it reads three 'c' symbols, it transitions to a final accepting state. The TM can repeat this process for any number of 'n' repetitions.
The TM requires states to track the progress of reading 'a', 'b', and 'c' symbols, as well as a state to handle the final accepting condition. Initially, it starts in the initial state. For each 'a' symbol encountered, the TM moves right and stays in the same state. When it encounters the first 'b' symbol, it moves right and transitions to another state. This state handles the second 'b' symbol, moving right for each 'b' symbol read until two have been processed.
After reading two 'b' symbols, the TM transitions to a state dedicated to processing 'c' symbols. It moves right for each 'c' symbol encountered and remains in this state until three 'c' symbols have been read. At that point, it transitions to the final accepting state, which signifies that the input string belongs to the language.
To handle the repetition of 'n' times, the TM can transition back to the initial state after reaching the final accepting state. This loop allows the TM to process any number of 'n' repetitions for the language {a^b^2c^3n | n ≥ 0}.
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In PriorityQueue.java, write code for the following new functions:
public boolean add( PriorityQueueNode x )
This function adds a new node x to the priority queue. The node is added to the heap by comparison of the rating attribute. It involves a call to percolateDown( int hole ). It returns true when finished.
public PriorityQueueNode remove( )
This function removes the minimum element of the priority queue and returns it.
private void percolateDown( int hole )
This function takes the position of the next available hole in the priority queue and uses it to bubble the elements through the heap until the heap property is restored.
public void display( )
This function prints out a formatted tree representation of the priority queue showing only the rating of each node. The output should resemble that of a tree. Tip: you may use the StringBuilder class and the String format( ) method. Empty nodes in the tree can be replaced with "--".
priorityQueueNode.java
public class PriorityQueueNode {
private String type;
private String title;
private int releaseYear;
private int rating;
public PriorityQueueNode(){
this.type = "";
this.title = "";
this.releaseYear = 0;
this.rating = 0;}
public PriorityQueueNode(String type, String title, int releaseYear, int rating) {
this.type = type;
this.title = title;
this.releaseYear = releaseYear;
this.rating = rating;}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
public String getTitle() {
return title;
}
public void setTitle(String title) {
this.title = title;
}
public int getReleaseYear() {
return releaseYear;
}
public void setReleaseYear(int releaseYear) {
this.releaseYear = releaseYear;
}
public int getRating() {
return rating;
}
public void setRating(int rating) {
this.rating = rating;
}
}
Here's the updated code for the PriorityQueue class in Java, including the new functions `add`, `remove`, `percolateDown`, and `display`:
```java
import java.util.ArrayList;
public class PriorityQueue {
private ArrayList<PriorityQueueNode> heap;
public PriorityQueue() {
heap = new ArrayList<>();
}
public boolean isEmpty() {
return heap.isEmpty();
}
public void add(PriorityQueueNode x) {
heap.add(x);
percolateUp(heap.size() - 1);
}
public PriorityQueueNode remove() {
if (isEmpty()) {
throw new IllegalStateException("Priority queue is empty");
}
PriorityQueueNode minNode = heap.get(0);
heap.set(0, heap.get(heap.size() - 1));
heap.remove(heap.size() - 1);
percolateDown(0);
return minNode;
}
private void percolateUp(int hole) {
PriorityQueueNode node = heap.get(hole);
while (hole > 0 && node.getRating() < heap.get(parentIndex(hole)).getRating()) {
heap.set(hole, heap.get(parentIndex(hole)));
hole = parentIndex(hole);
}
heap.set(hole, node);
}
private void percolateDown(int hole) {
int child;
PriorityQueueNode node = heap.get(hole);
while (leftChildIndex(hole) < heap.size()) {
child = leftChildIndex(hole);
if (child != heap.size() - 1 && heap.get(child).getRating() > heap.get(child + 1).getRating()) {
child++;
}
if (node.getRating() > heap.get(child).getRating()) {
heap.set(hole, heap.get(child));
hole = child;
} else {
break;
}
}
heap.set(hole, node);
}
public void display() {
displayHelper(0, 0, new StringBuilder());
}
private void displayHelper(int index, int level, StringBuilder output) {
if (index < heap.size()) {
displayHelper(rightChildIndex(index), level + 1, output);
for (int i = 0; i < level; i++) {
output.append("\t");
}
output.append(heap.get(index).getRating()).append("\n");
displayHelper(leftChildIndex(index), level + 1, output);
}
}
private int parentIndex(int index) {
return (index - 1) / 2;
}
private int leftChildIndex(int index) {
return (2 * index) + 1;
}
private int rightChildIndex(int index) {
return (2 * index) + 2;
}
}
```
The updated code includes the `add`, `remove`, `percolateDown`, and `display` functions as described:
1. The `add` function adds a new node `x` to the priority queue by inserting it at the end of the heap and then performing a percolateUp operation to restore the heap property.
2. The `remove` function removes the minimum element of the priority queue (root node) by swapping it with the last node, removing the last node, and then performing a percolateDown operation to restore the heap property.
3. The `percolateDown` function takes the position of the hole and moves elements down through the heap until the heap property is restored.
4. The `display` function prints a formatted tree representation of the priority queue by using a recursive `displayHelper` function to traverse the heap and append the ratings of each node to a StringBuilder object
.
Note: The code assumes that the `PriorityQueueNode` class is defined separately and remains unchanged.
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- 1 - - a (a) Consider a simple hash function as "key mod 7" and collision by Linear Probing (f(i)=i) (b) Consider a simple hash function as "key mod 7" and collision by Quadratic Probing (f(i)=1^2)
In this scenario, we are using a simple hash function where the key is hashed by taking the modulus of the key divided by 7. This hash function maps the keys to values between 0 and 6.
To handle collisions, we can use two different probing techniques: Linear Probing and Quadratic Probing. In Linear Probing, when a collision occurs, we increment the index by a constant value (usually 1) until we find an empty slot. For example, if the slot for a key is already occupied, we would probe the next slot, and if that is occupied as well, we would continue probing until an empty slot is found. In Quadratic Probing, instead of a constant increment, we use a quadratic function to determine the next probe position. The function f(i) is defined as i^2, where i represents the number of probes. So, the first probe is at index 1, the second probe is at index 4, the third probe is at index 9, and so on.
Both Linear Probing and Quadratic Probing aim to reduce collisions and distribute the keys more evenly in the hash table. However, Quadratic Probing tends to provide better results in terms of clustering and reducing long linear chains of probes.
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Write a function named cake will take 2 inputted dictionaries The first dictionary are the amounts of ingredients.
{"Eggs": 1, "Sugar": 2, "Milk": 2}
The second dictionary is how many how each ingredient there are.
{"Eggs": 3, "Sugar": 9, "Milk": 8}
The function cake will return how many of the item given by the first dictionary can be made using second dictionary.
For example, with the dictionaries above, the answer is 3.
We have 3 eggs and each item needs 1 egg. Even though there is enough sugar and milk to make 4, the answer is 3 because we don't have enough eggs.
If the function works, it will result in: 3, 1, 3, 0
The `cake` function takes two dictionaries representing ingredient amounts and quantities and returns the maximum number of cakes that can be made based on the available ingredients.
In the provided example, the output would be 3, indicating that 3 cakes can be made.
The `cake` function takes two dictionaries as input: the first dictionary represents the required amounts of ingredients, and the second dictionary represents the available quantities of each ingredient. The function calculates how many cakes can be made based on the available ingredients and returns that value.
For example, given the first dictionary: `{"Eggs": 1, "Sugar": 2, "Milk": 2}` and the second dictionary: `{"Eggs": 3, "Sugar": 9, "Milk": 8}`, the `cake` function will return 3. This means that with the available ingredients, you can make 3 cakes. Although there is enough sugar and milk to make 4 cakes, the limited quantity of eggs restricts the number of cakes to 3.
In detail, the function `cake` can be implemented as follows:
1. Initialize a variable `max_cakes` with a large value or infinity to keep track of the maximum number of cakes that can be made.
2. Iterate through each ingredient in the required amounts dictionary.
3. For each ingredient, check if it exists in the available quantities dictionary.
4. If the ingredient exists in both dictionaries, calculate the maximum number of cakes that can be made based on the ratio of available quantities to required amounts.
5. Update `max_cakes` with the minimum value between the current `max_cakes` and the maximum number of cakes calculated in the previous step.
6. After iterating through all ingredients, return the value of `max_cakes`.
By following this approach, the function will accurately determine the maximum number of cakes that can be made based on the available ingredients. In the provided example, the output would be 3, indicating that 3 cakes can be made with the given ingredient quantities.
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