Answer:
The java program is given below.
import java.util.*;
class GradeBook
{
//variables to hold all the given values
static int m11, m12, m13;
static int m21, m22, m23;
static String name1, name2;
//variables to hold the computed values
static double avg1, avg2;
//constructor initializing the variables with the given values
GradeBook()
{
name1 = "Jack";
m11 = 33;
m12 = 55;
m13 = 77;
name2 = "Jill";
m21 = 99;
m22 = 66;
m23 = 33;
}
//method to compute and display the student having highest average grade
static void computeAvg()
{
avg1=(m11+m12+m13)/3;
avg2=(m21+m22+m23)/3;
if(avg1>avg2)
System.out.println("Student with highest average is "+ name1);
else
System.out.println("Student with highest average is "+ name2);
}
}
//contains only main() method
public class Teacher{
public static void main(String[] args)
{
//object created
GradeBook ob = new GradeBook();
//object used to call the method
ob.computeAvg();
}
}
OUTPUT
Student with highest average is Jill
Explanation:
1. The class GradeBook declares all the required variables as static with appropriate data types, integer for grades and double for average grade.
2. Inside the constructor, the given values are assigned to the variables.
3. The method, computeAvg(), computes the average grade of both the students. The student having highest average grade is displayed using System.out.println() method.
4. Inside the class, Teacher, the main() method only has 2 lines of code.
5. The object of the class GradeBook is created.
6. The object is used to call the method, computeAvg(), which displays the message on the screen.
7. All the methods are also declared static except the constructor. The constructor cannot have any return type nor any access specifier.
8. The program does not takes any user input.
9. The program can be executed with different values of grades and names of the students.
Write a Java class to perform the following: 1. Write a method to search the following array using a linear search, ( target elements: 11, 55, 17.). (count the number of comparisons needed). {06, 02, 04, 07, 11, 09, 50, 62, 43, 32, 13, 75, 01, 46, 88, 17} 2. Write a method to sort the array using Selection Sort. (count the number of comparisons needed) 3, Write a method to sort the array using Bubble Sort. (count the number of comparisons needed) 4, Search he sorted array using a binary search (recursive) for the same set of target elements. (count the number of comparisons needed)
Answer:
Check the explanation
Explanation:
Linear search in JAVA:-
import java.util.Scanner;
class linearsearch
{
public static void main(String args[])
{
int count, number, item, arr[];
Scanner console = new Scanner(System.in);
System.out.println("Enter numbers:");
number = console.nextInt();
arr = new int[number];
System.out.println("Enter " + number + " ");
for (count = 0; count < number; count++)
arr[count] = console.nextInt();
System.out.println("Enter search value:");
item = console.nextInt();
for (count = 0; count < number; count++)
{
if (arr[count] == item)
{
System.out.println(item+" present at "+(count+1));
break;
}
}
if (count == number)
System.out.println(item + " doesn't found in array.");
}
}
Kindly check the first attached image below for the code output.
Selection Sort in JAVA:-
public class selectionsort {
public static void selectionsort(int[] array){
for (int i = 0; i < array.length - 1; i++)
{
int ind = i;
for (int j = i + 1; j < array.length; j++){
if (array[j] < array[ind]){
ind = j;
}
}
int smaller_number = array[ind];
array[ind] = array[i];
array[i] = smaller_number;
}
}
public static void main(String a[]){
int[] arr = {9,94,4,2,43,18,32,12};
System.out.println("Before Selection Sort");
for(int i:arr){
System.out.print(i+" ");
}
System.out.println();
selectionsort(arr);
System.out.println("After Selection Sort");
for(int i:arr){
System.out.print(i+" ");
}
}
}
Kindly check the second attached image below for the code output.
Bubble Sort in JAVA:-
public class bubblesort {
static void bubblesort(int[] array) {
int num = array.length;
int temp = 0;
for(int i=0; i < num; i++){
for(int j=1; j < (num-i); j++){
if(array[j-1] > array[j]){
temp = array[j-1];
array[j-1] = array[j];
array[j] = temp;
}
}
}
}
public static void main(String[] args) {
int arr1[] ={3333,60,25,32,55,620,85};
System.out.println("Before Bubble Sort");
for(int i=0; i < arr1.length; i++){
System.out.print(arr1[i] + " ");
}
System.out.println();
bubblesort(arr1);
System.out.println("After Bubble Sort");
for(int i=0; i < arr1.length; i++){
System.out.print(arr1[i] + " ");
}
}
}
Kindly check the third attached image below for the code output.
Binary search in JAVA:-
public class binarysearch {
public int binarySearch(int[] array, int x) {
return binarySearch(array, x, 0, array.length - 1);
}
private int binarySearch(int[ ] arr, int x,
int lw, int hg) {
if (lw > hg) return -1;
int middle = (lw + hg)/2;
if (arr[middle] == x) return middle;
else if (arr[middle] < x)
return binarySearch(arr, x, middle+1, hg);
else
return binarySearch(arr, x, lw, middle-1);
}
public static void main(String[] args) {
binarysearch obj = new binarysearch();
int[] ar =
{ 22, 18,12,14,36,59,74,98,41,23,
34,50,45,49,31,53,74,56,57,80,
61,68,37,12,58,79,904,56,99};
for (int i = 0; i < ar.length; i++)
System.out.print(obj.binarySearch(ar,
ar[i]) + " ");
System.out.println();
System.out.print(obj.binarySearch(ar,19) +" ");
System.out.print(obj.binarySearch(ar,25)+" ");
System.out.print(obj.binarySearch(ar,82)+" ");
System.out.print(obj.binarySearch(ar,19)+" ");
System.out.println();
}
}
Kindly check the fourth attached image below for the code output
Which type of styles can be applied to a word, phrase, or sentence?
Format
Object
Paragraph
Character
Resume
Answer:
I think the answer is going to be paragraph
The types of styles that can be applied to a word, phrase, or sentence are the character style and paragraph style. Thus, the correct options for this question are C and D.
What do you mean by Character style?Character style may be defined as a type of style that can be used for the word, phrase, or sentence in design. This type of style considerably remembers formatting for single characters, words, or phrases.
While Paragraph Style will remember the formatting applied to a whole paragraph. This type of style is most frequently applied to a section of text separated from other text by line breaks with the intention to fragment the literary work into short paragraphs.
Therefore, the character and paragraph styles are the types of styles that can be applied to a word, phrase, or sentence.
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Define a function drawCircle.
This function should expect a Turtle object, the coordinates of the circle's center point, and the circle's radius as arguments.The function should draw the specified circle. The pen color should be changed to yellow before drawing a circle and the width of the pen to 5 pixels. The algorithm should draw the circle's circumference by turning 3 degrees and moving a given distance 120 times. Calculate the distance moved with the formula 2.0*n*radius/120.0.
Fill in the circle with blue color. After drawing the circle, hide the turtle.
import turtle
import math
def drawCircle(centerpoint, radius):
degree = 3
count = 0
centerpoint = (2.0 * math.pi * radius / 120)
t.home()
t.setheading(degree)
while count <= 120:
t.down()
t.forward(2.0 * math.pi * radius / 120)
t.up()
degree += 3
t.setheading(degree)
count += 1
drawCircle(centerpoint, radius)
Final answer:
To draw a circle using a Turtle object in Python, you can define a function called drawCircle that takes in the Turtle object, the center point coordinates, and the radius as arguments. This code uses the turtle module in Python to draw a circle.
Explanation:
Draw a Circle using a Turtle in Python
To draw a circle using a Turtle object in Python, you can define a function called drawCircle that takes in the Turtle object, the center point coordinates, and the radius as arguments. Here's an example of how the function can be implemented:
import turtleThis code uses the turtle module in Python to draw a circle. It sets the pen size to 5 pixels and the color to yellow. The circle is drawn by turning 3 degrees and moving a distance of 2.0 * math.pi * radius / 120.0. After drawing the circumference, the function fills the circle with a blue color and hides the turtle.
Final answer:
The drawCircle function is designed for the Python Turtle library to draw a colored circle based on specified parameters. The Turtle's pen is set to yellow and the pen width to 5 pixels before the circle is drawn and filled with blue. After drawing, the Turtle is hidden.
Explanation:
The function drawCircle is intended to work with the Python Turtle graphics library to draw a circle on the screen. The function will expect a Turtle object, the coordinates of the circle's center point, and the circle's radius as arguments. Below is a corrected version of the function that follows the stated requirements:
import turtleMake sure to create a Turtle object and pass it to the drawCircle function along with the center point's coordinates and the radius of the circle you wish to draw.
Use semaphore(s) to solve the following problem. There are three processes: P1, P2, and P3. Each process Pi has a segment of codes Ci, i=1, 2, 3. These three processes are executed only once, i.e., no repeat or loop at all, and their executions can start at any time. Your goal is to ensure that the execution of C1, C2, and C3 must satisfy the following conditions:
a. If C1 is executed ahead of C2 and C3, C2 must be executed ahead of C3.
b. Otherwise, C1 must be executed after both C2 and C3 are executed. In this case, the order of C2 and C3's execution doesn't matter. One of the possible execution orders is demonstrated below. Obviously, two other possible sequences in time are C2, C3, C1 and C3, C2, C1.
Please write your algorithm level code for semaphore initialization and usage in each code segment. [Hint: no if statements should ever be used. All you need is some semaphore function calls surrounding C1, C2, and C3 and their initial values.]
Answer:
See explaination
Explanation:
Here we will use two semaphore variables to satisfy our goal
We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way
Code:-
s1,s2=1
P1 P2 P3
P(s1)
P(s2)
C1
V(s2) .
P(s2). .
. C2
V(s1) .
P(s1)
. . C3
V(s2)
Explanation:-
The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.
The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process
The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process
Now in the above code:
1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute
So our first condition gets satisfied
2)If C1 is not executed earlier means:-
a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .
b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked
So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.
Create a Visual Logic flow chart with four methods. Main method will create an array of 5 elements, then it will call a read method, a sort method and a print method passing the array to each. The read method will prompt the user to enter 5 numbers that will be stored in the array. The sort method will sort the array in ascending order (smallest to largest). The print method will print out the array.
Answer:
See explaination
Explanation:
import java.util.Scanner;
public class SortArray {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
System.out.println("Enter Size Of Array");
int size = sc.nextInt();
int[] arr = new int[size]; // creating array of size
read(arr); // calling read method
sort(arr); // calling sort method
print(arr); // calling print method
}
// method for read array
private static void read(int[] arr) {
Scanner sc = new Scanner(System.in);
for (int i = 0; i < arr.length; i++) {
System.out.println("Enter " + i + "th Position Element");
// read one by one element from console and store in array
arr[i] = sc.nextInt();
}
}
// method for sort array
private static void sort(int[] arr) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length; j++) {
if (arr[i] < arr[j]) {
// Comparing one element with other if first element is greater than second then
// swap then each other place
int temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
}
}
}
// method for display array
private static void print(int[] arr) {
System.out.print("Your Array are: ");
// display element one by one
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + ",");
}
}
}
See attachment
Which term describes measurable results expected in a given time frame?
Answer: I think a deadline?
The term that describes the results expected in a given time frame is the goal. The correct option is A.
What are goals?The desired states that people want to achieve, maintain, or avoid are referred to as life goals. When we make goals, we commit to imagining, planning for, and obtaining these desired outcomes. The satisfaction with our careers is the satisfaction with our lives because we spend more than half of our waking hours at work.
The most significant professional goal we can set for ourselves is to identify our areas of love and make them into lifelong careers. An objective is something you hope to accomplish. It is the desired outcome that you or a group of people plan for and firmly resolve to attain. The phrase "goal" refers to the outcomes anticipated in a specific time range.
Therefore, the correct option is A, Goals.
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The question is incomplete. Your most probably complete question is given below:
Goals
Mind maps
Purposes
Ideas
The Phonebook class must also have the following functions: Constructor with 1 integer parameter for size (numberOfContacts is initially set to 0; allocate memory for myContacts). Copy constructor. Assignment operator. Destructor. int getNumberOfContacts() // Getter int getSize() // Getter contact getContact(int i) //returns contact at index i (you can assume 0 < i < numberOfContacts) bool addContact(string name, string phonenumber, string email) // add a new contact at the end of the phonebook using the received parameters. Returns true if succesful, false if the contact couldn't be added because the phonebook is already full.
Answer:
See explaination
Explanation:
#include <iostream>
using namespace std;
struct contact
{
string name;
string phoneNumber;
string email;
bool available;
};
class Phonebook
{
private :
int numberOfContacts;
int size;
contact *myContacts;
public :
Phonebook(int size)
{
this->size=size;
this->numberOfContacts=0;
myContacts=new contact[size];
}
~Phonebook()
{
delete[] myContacts;
}
int getNumberOfContacts()
{
return numberOfContacts;
}
int getSize()
{
return size;
}
void addContact(string name,string phonenumber,string email)
{
contact c;
c.name=name;
c.phoneNumber=phonenumber;
c.email=email;
c.available=true;
myContacts[numberOfContacts]=c;
numberOfContacts++;
}
void removeContact(string name)
{
for(int i=0;i<numberOfContacts;i++)
{
if(myContacts[i].name.compare(name)==0)
{
this->myContacts[i].available=false;
contact temp=myContacts[numberOfContacts-1];
myContacts[numberOfContacts-1]=myContacts[i];
myContacts[i]=temp;
numberOfContacts--;
}
}
}
Phonebook(const Phonebook& pb)
{
this->myContacts=pb.myContacts;
this->size=pb.size;
this->numberOfContacts=numberOfContacts;
}
Phonebook& operator=(const Phonebook & pb)
{
myContacts=new contact[size];
this->size=pb.size;
this->numberOfContacts=pb.numberOfContacts;
for(int i=0;i<numberOfContacts;i++)
{
this->myContacts[i]=pb.myContacts[i];
}
return *this;
}
void print()
{
for(int i=0;i<numberOfContacts;i++)
{
if(myContacts[i].available==true)
{
cout<<"Name :"<<myContacts[i].name<<endl;
cout<<"Phone Number :"<<myContacts[i].phoneNumber<<endl;
cout<<"Email :"<<myContacts[i].email<<endl;
cout<<endl;
}
}
}
};
int main(){
Phonebook pb1(10);
pb1.addContact("Williams","9845566778","williamatmail.com");
pb1.addContact("James","9856655445","jamesatmail.com");
cout<<"_____ Displaying Phonebook#1 contacts _____"<<endl;
pb1.print();
pb1.removeContact("Williams");
cout<<"_____ After removing a contact Displaying Phonebook#1 _____"<<endl;
pb1.print();
Phonebook pb2(5);
pb2.addContact("Billy","9845554444","billyatmail.com");
pb2.addContact("Jimmy","9834444444","jimmyatmail.com");
cout<<"_____ Displaying Phonebook#2 contacts _____"<<endl;
pb2.print();
pb2=pb1;
cout<<"____ After Assignment Operator Displaying Phonebook#2 contacts____"<<endl;
pb2.print();
return 0;
}
Nb: Replace the at with at symbol.
Implement the logic function ( , , ) (0,4,5) f a b c m =∑ in 4 different ways. You have available 3to-8 decoders with active high (AH) or active low (AL) outputs and OR, AND, NOR and NAND gates with as many inputs as needed. In every case clearly indicate which is the Most Significant bit (MSb) and which is the Least Significant bit (LSb) of the decoder input.
Answer:
See explaination
Explanation:
Taking a look at the The Logic function, which states that an output action will become TRUE if either one “OR” more events are TRUE, but the order at which they occur is unimportant as it does not affect the final result. For example, A + B = B + A.
Alternatively the Most significant bit which is also known as the alt bit, high bit, meta bit, or senior bit, the most significant bit is the highest bit in binary.
See the attached file for those detailed logic functions designed with relation to the questions asked.
You are working with a MySQL installation in Windows. You plan to use the mysql client utility in batch mode to run a query that has been saved to a file in the C:\mysql_files directory. The name of the file is users.sql, and it includes a command to use the mysql database, which the query targets. You want to save the results of the query to a file named users.txt, which should also be saved to the C:\mysql_files directory. What command should you use to execute the query
Answer:
mysql -t < c:\mysql_files\users.sql > c:\mysql_files\users.txt
Explanation:
MySQL is a system software that is written in programming language such as c++ and c, the software was initially released on the 23rd day of the month of May, in the year 1995 by Oracle corporation. It is a popular software in companies that are commerce related or orientated since it deals with things related to web database.
So, to answer the question we are given that the file that will results of the query is users.txt and the directory is C:\mysql_files, therefore, the command that should you use to execute the query is; mysql -t < c:\mysql_files\users.sql > c:\mysql_files\users.txt
Consider the problem of solving two 8-puzzles.
a. Give a complete problem formulation.
b. How large is the reachable state space? Give an exact numerical expression.
c. Suppose we make the problem adversarial as follows: the two players take turns moving; a coin is flipped to determine the puzzle on which to make a move in that turn; and the winner is the first to solve one puzzle. Which algorithm can be used to choose a move in this setting?
d. Give an informal proof that someone will eventually win if both play perfectly.
Answer:
See Explaination
See Explaination
Explanation:
d. If both players play perfectly, then too each player will play not let the other win. Let there be a state of the game in which the goal state can be achieved in just 2 more moves. If player 1 makes one move then the next move made player 1 will win or player 2 do the same.
Either way whatever the rule states, there must be a winner emerging.
Check attachment for option a to c.
The problem of solving two 8-puzzles involves having a specific goal state, making moves on a puzzle, and using an algorithm like Minimax for an adversarial setting. The reachable state space size of an 8-puzzle is 181,440. Given perfect play, an eventual win is guaranteed due to the finite number of puzzle configurations.
Explanation:a. The problem of solving an 8-puzzle can be formulated as follows: The goal state is a specific configuration of tiles, with an empty space in a particular location. The actions are to move the empty space up, down, left, or right, swapping it with the adjacent tile. The transition model returns the new configuration of tiles resulting from the action. A cost function could return 1 for each move.
b. The size of reachable state space for an 8-puzzle is 9!/2 = 181,440.
c. In the adversarial setting described, a Minimax algorithm could be used to determine the best move.
d. An eventual win in this setting can be informally proven by noting that with each move, the current state of the puzzle moves closer to the goal state. Since there is a limited number of configurations (as determined in part b), the puzzle cannot be played indefinitely and eventually one player will reach the goal state.
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Complexities of communication are organized into successive layers of protocols: lower-level layers are more specific to medium, higher-level layers are more specific to application. Match the layer name to its functionality: Transport layer Network layer Data link layer Physical layer A. controls transmission of the raw bit stream over the medium B. establishes, maintains and terminates network connections C. ensures the reliability of link D. provides functions to guarantee reliable network link
Answer:
Transport layer
establishes, maintains and terminates network connections
Network layer
ensures the reliability of link
Data link layer
provides functions to guarantee reliable network link
Physical layer
controls transmission of the raw bit stream over the medium
Explanation:
see Answer
Write a program that reads students’ names followed by their test scores. The program should output each student’s name followed by the test scores and the relevant grade. It should also find and print the highest test score and the name of the students having the highest test score. Student data should be stored in a struct variable of type studentType, which has four components: studentFName and studentLName of type string, testScore of type int (testScore is between 0 and 100), and grade of type char. Suppose that the class has 20 students. Use an array of 20 components of type studentType. Your program must contain at least the following functions: A function to read the students’ data into the array. A function to assign the relevant grade to each student. A function to find the highest test score. A function to print the names of the students having the highest test score. Your program must output each student’s name in this form: last name followed by a comma, followed by a space, followed by the first name; the name must be left justified. Moreover, other than declaring the variables and opening the input and output files, the function main should only be a collection of function calls.
The program will manage students' test scores using a struct that records each student's name, score, and grade. It involves functions to input data, calculate grades, and identify the highest scorer. Output is formatted as 'LastName, FirstName: Grade'.
Explanation:
Program Structure for Student Grade Records
To create a program for managing student test scores and grades, we would define a struct named studentType with components studentFName, studentLName, testScore, and grade. We'd use an array of studentType of size 20 to store each student's details. The program would include functions to read student data, assign grades, find the highest test score, and print students with the highest score. Grades would be assigned based on the test scores in a typical A-F scale.
The main function should be clean and comprise primarily of function calls to handle various operations such as reading data and processing grades.
The output format for each student's name and grade should be "LastName, FirstName: Grade" with the last name and first name left justified, following the requirements specified for the assignment.
Where can we buy a cryptocurrency? from below options
A) Through a private transaction
B) All the options
C) Cryptocurrency Exchanges
D) In a smart contract
it would be C
hope this helps!
what is the molarity of a solution prepared by dissolving 15.0g of sodium hydroxide in enough water to make a total of 225 ml of solution
Answer:
1.6666 g/mol = 1 [tex]\frac{2}{3}[/tex] g/mol
Explanation:
Molar mass of NaOH= 23+16+1 =40g/mol
Mols in 15g = 15/40 mol
If this was dissolved in 225ml of water molarity of the solution is
[tex]\frac{15}{40}[/tex] ÷ 225 x 1000 = 1.6666 g/mol = 1 [tex]\frac{2}{3}[/tex] g/mol
Final answer:
The molarity of the sodium hydroxide solution is calculated by dividing the number of moles of NaOH (0.375 moles) by the volume of the solution in liters (0.225 L), resulting in 1.667 M.
Explanation:
The molarity of a solution is determined by the number of moles of solute per liter of solution. To find the molarity of the solution, we need to know the molar mass of sodium hydroxide (NaOH), which is approximately 40 g/mol. First, we convert the mass of NaOH to moles:
Mass of NaOH = 15.0 g
Molar mass of NaOH = 40 g/mol
Moles of NaOH = Mass / Molar mass = 15.0 g / 40 g/mol = 0.375 moles
Second, we convert the volume of the solution from milliliters to liters:
Volume of solution = 225 mL
1 liter = 1000 mL
Volume in liters = 225 mL / 1000 = 0.225 liters
Finally, we calculate the molarity of the solution using the formula:
Molarity (M) = Moles of solute / Volume of solution in liters
Molarity (M) = 0.375 moles / 0.225 liters
Molarity (M) = 1.667 M
Therefore, the molarity of the sodium hydroxide solution is 1.667 M.
Suppose you are currently in the /home/hnewman/os/fall/2013 directory and would like to navigate to /home/hnewman/discreteStructures/spring/2011 directory and would like to navigate using absolute path names. What command will you type in the shell? A. cd /home/hnewman/discreteStructures/spring/2011 B. cd discreteStructures/spring/2009 C. cd ../../..discreteStructures/spring/2009 D. cd ../../discreteStructures/spring/2009
Answer:
A. cd /home/hnewman/discreteStructures/spring/2011
Explanation:
Nothing with 2009 in the pathname will get you where you want to go. The only reasonable answer choice is the first one:
cd /home/hnewman/discreteStructures/spring/2011
A lamp outside a front door comes on automatically when it is dark, and when someone stands on the doormat outside the front door. A pressure sensor under the mat changes from OFF (0) to ON (1) when someone stands on the doormat. The light sensor is ON (1) when it is light and OFF (0) when it is dark. Design a program to show what would happen to the lamp. Account for all possible scenarios by determining whether the pressure sensor and light sensor are ON or OFF. (HINT: Ask the user "Is it dark?" and "Is someone standing on the doormat outside the front door?")
Answer:
See explaination
Explanation:
#include <iostream>
#include <string>
using namespace std;
int main()
{
string raptor_prompt_variable_zzyz;
?? standing;
?? dark;
raptor_prompt_variable_zzyz ="Is it dark?";
cout << raptor_prompt_variable_zzyz << endl;
cin >> DARK;
if (DARK=="Yes")
{
raptor_prompt_variable_zzyz ="Is someone standing on the doormat outside the front door?";
cout << raptor_prompt_variable_zzyz << endl;
cin >> STANDING;
if (STANDING=="Yes")
{
cout << "LAMP IS ON" << endl; }
else
{
cout << "LAMP IS OFF" << endl; }
}
else
{
cout << "LAMP IS OFF" << endl; }
return 0;
}
Here's a simple program in pseudo-code to demonstrate the behavior of the lamp based on the inputs from the light sensor and pressure sensor:
Ask user: "Is it dark?" (yes or no)
Read user's response and store it in variable 'is_dark'
Ask user: "Is someone standing on the doormat outside the front door?" (yes or no)
Read user's response and store it in variable 'is_someone_on_doormat'
if is_dark equals yes and is_someone_on_doormat equals yes:
Turn on the lamp
Display message: "Lamp turned on because it is dark and someone is standing on the doormat."
else if is_dark equals yes and is_someone_on_doormat equals no:
Keep the lamp off
Display message: "Lamp remains off because it is dark but no one is standing on the doormat."
else if is_dark equals no and is_someone_on_doormat equals yes:
Turn on the lamp
Display message: "Lamp turned on because someone is standing on the doormat."
else:
Keep the lamp off
Display message: "Lamp remains off because it is not dark and no one is standing on the doormat."
This program covers all possible scenarios based on the inputs from the light sensor and pressure sensor and determines whether the lamp should be turned on or off accordingly.
3.27 LAB: Exact change (FOR PYTHON PLEASE)
Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line. The coin types are Dollars, Quarters, Dimes, Nickels, and Pennies. Use singular and plural coin names as appropriate, like 1 Penny vs. 2 Pennies.
Ex: If the input is:
0
(or less than 0), the output is:
No change
Ex: If the input is:
45
the output is:
1 Quarter
2 Dimes
Answer:
amount = int(input())
#Check if input is less than 1
if amount<=0:
print("No change")
else: #If otherwise
#Convert amount to various coins
dollar = int(amount/100) #Convert to dollar
amount = amount % 100 #Get remainder after conversion
quarter = int(amount/25) #Convert to quarters
amount = amount % 25 #Get remainder after conversion
dime = int(amount/10) #Convert to dimes
amount = amount % 10 #Get remainder after conversion
nickel = int(amount/5) #Convert to nickel
penny = amount % 5 #Get remainder after conversion
#Print results
if dollar >= 1:
if dollar == 1:
print(str(dollar)+" Dollar")
else:
print(str(dollar)+" Dollars")
if quarter >= 1:
if quarter == 1:
print(str(quarter)+" Quarter")
else:
print(str(quarter)+" Quarters")
if dime >= 1:
if dime == 1:
print(str(dime)+" Dime")
else:
print(str(dime)+" Dimes")
if nickel >= 1:
if nickel == 1:
print(str(nickel)+" Nickel")
else:
print(str(nickel)+" Nickels")
if penny >= 1:
if penny == 1:
print(str(penny)+" Penny")
else:
print(str(penny)+" Pennies")
Explanation:
The first answer is almost right, but this should give you the complete answer with the proper starting point and correct capitalizations.
The program is an illustration of conditional statements.
Conditional statements are statements whose execution depends on its truth value.
The program in Python, where comments are used to explain each line is as follows:
#This gets input for the amount
amount = int(input("Enter Amount: "))
#This checks if the amount is less than 1
if amount<=0:
print("No Change")
else: #If otherwise
#The next lines convert amount to various coins
dollar = int(amount/100) #Convert to dollar
amount = amount % 100 #Get remainder after conversion
quarter = int(amount/25) #Convert to quarters
amount = amount % 25 #Get remainder after conversion
dime = int(amount/10) #Convert to dimes
amount = amount % 10 #Get remainder after conversion
nickel = int(amount/5) #Convert to nickel
penny = amount % 5 #Get remainder after conversion
#The following if statements print the change
#The prints dollars
if dollar >= 1:
if dollar == 1:
print(str(dollar)+" dollar")
else:
print(str(dollar)+" dollars")
#The prints quarters
if quarter >= 1:
if quarter == 1:
print(str(quarter)+" quarter")
else:
print(str(quarter)+" quarters")
#The prints dimes
if dime >= 1:
if dime == 1:
print(str(dime)+" dime")
else:
print(str(dime)+" dimes")
#The prints nickels
if nickel >= 1:
if nickel == 1:
print(str(nickel)+" nickel")
else:
print(str(nickel)+" nickels")
#The prints pennies
if penny >= 1:
if penny == 1:
print(str(penny)+" penny")
else:
print(str(penny)+" pennies")
Read more about similar programs at:
https://brainly.com/question/16839801
Develop a crawler that collects the email addresses in the visited web pages. You can write a function emails() that takes a document (as a string) as input and returns the set of email addresses (i.e., strings) appearing in it. You should use a regular expression to find the email addresses in the document.
Answer:
see explaination
Explanation:
import re
def emails(document):
"""
function to find email in given doc using regex
:param document:
:return: set of emails
"""
# regex for matching emails
pattern = r'[\w\.-]+at[\w\.-]+\.\w+'
# finding all emails
emails_in_doc = re.findall(pattern, document)
# returning set of emails
return set(emails_in_doc)
# Testing the above function
print(emails('random text ertatxyz.com yu\npopatxyz.com random another'))
Implement a Breadth-First Search of the people in the network who are reachable from person id 3980. You can implement BFS with a queue and the pseudocode is given in CLRS. Print out the distance (number of connections) from person 3980 to every person who is reachable from 3980 via edge traversals. Note that not all people in this network may be reachable via edge traversals from user id 3980, so users that are not accessible can be ignored in BFS.
Answer:
Check the explanation
Explanation:
import java.io.File;
import java.io.FileNotFoundException;
import java.util.LinkedList;
import java.util.Scanner;
import static org.junit.Assert.assertEquals;
/**
* CS146 Assignment 3 Node class This class is used for undirected graphs
* represented as adjacency lists The areFriends() method checks if two people
* are friends by checking if an edge exists between the two
*
*/
public class NetworkAdjList {
// Initialize array with max number of vertices taken from SNAP
static int max_num_vertices = 88234;
static LinkedList<Integer>[] adjacencyList = new LinkedList[max_num_vertices];
public static void createAdjacencyList() {
// Initialize array elements
for (int j = 0; j < max_num_vertices; j++) {
adjacencyList[j] = new LinkedList<Integer>();
}
// Get file path of the 3980.edges file
String filePath = "C:\\Users\\shahd\\Documents\\CS146\\Assignment 3\\Question 3\\Question3\\src\\a3\\3980.edges";
File f = new File(filePath);
// Use Scanner to read edges from the file and put it into adjacency list
int a;
int b;
try {
Scanner fileIn = new Scanner(f);
while (fileIn.hasNext()) {
a = fileIn.nextInt();
b = fileIn.nextInt();
adjacencyList[a].add(b);
adjacencyList[b].add(a); // We need to add the edges both ways
}
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public static boolean areFriends(int A, int B) {
// If the adjacency list contains (A, B) edge, then return true, else false
if (adjacencyList[A].contains(B)) {
return true;
} else {
return false;
}
}
private static void bfsHelper(boolean visited[], int currentNode, int dis, int sourceNode) {
dis++;
if(!visited[currentNode]) {
visited[currentNode] = true;
for(int neighbor: adjacencyList[currentNode]) {
System.out.println(neighbor + " is at a distance of " + dis + " from " + sourceNode);
bfsHelper(visited, neighbor, dis++, sourceNode);
}
}
}
public static void BFStraversal(int start) {
boolean visited[] = new boolean[max_num_vertices];
bfsHelper(visited, start, 0, start);
}
public static void main(String[] args) {
/**
* These test cases assume the file 3980.edges was used
*/
createAdjacencyList();
System.out.println("Testing...");
assertEquals(areFriends(4038, 4014), true);
System.out.println("1 of 7");
System.out.println("Testing...");
assertEquals(areFriends(3982, 4037), true);
System.out.println("2 of 7");
System.out.println("Testing...");
assertEquals(areFriends(4030, 4017), true);
System.out.println("3 of 7");
System.out.println("Testing...");
assertEquals(areFriends(4030, 1), false);
System.out.println("4 of 7");
System.out.println("Testing...");
assertEquals(areFriends(1, 4030), false);
System.out.println("5 of 7");
System.out.println("Testing...");
assertEquals(areFriends(4003, 3980), true);
System.out.println("6 of 7");
System.out.println("Testing...");
assertEquals(areFriends(3985, 4038), false);
System.out.println("7 of 7");
System.out.println("Success!");
}
}
**************************************************
2. Use inheritance to create a hierarchy of Exception classes -- EndOfSentenceException, PunctuationException, and CommaException. EndOfSentenceException is the parent class to PunctuationException, which is the parent class to CommaException. Test your classes in a Driver class with a main method that asks the user to input a sentence. If the sentence ends in anything except a period (.), exclamation point (!), or question mark (?), the program should throw a PunctuationException. If the sentence specifically ends in a comma, the program should throw a CommaException. Use a catch block to catch all EndOfSentenceExceptions, causing the program to print a message and terminate. If a general PunctuationException is caught, the program should print "The sentence does not end correctly." If a CommaException is caught, the program should print "You can't end a sentence in a comma." If there are no exceptions, the program should print "The sentence ends correctly." and terminate.
Answer:
See explaination
Explanation:
EndOfSentenceException.java
//Create a class EndOfSentenceException that
//extends the Exception class.
class EndOfSentenceException extends Exception
{
//Construtor.
EndOfSentenceException(String str)
{
System.out.println(str);
}
}
CommaException.java
//Create a class CommaException that extends the class
//EndOfSentenceException.
class CommaException extends EndOfSentenceException
{
//Define the constructor of CommaException.
public CommaException(String str)
{
super(str);
}
}
PunctuationException.java
//Create a class PunctuationException that extends the class
//EndOfSentenceException.
class PunctuationException extends EndOfSentenceException
{
//Constructor.
public PunctuationException(String str)
{
super(str);
}
}
Driver.java
//Include the header file.
import java.util.Scanner;
//Define the class Driver to check the sentence.
public class Driver {
//Define the function to check the sentence exceptions.
public String checkSentence(String str)
throws EndOfSentenceException
{
//Check the sentence ends with full stop,
//exclamation mark
//and question mark.
if(!(str.endsWith(".")) && !(str.endsWith("!"))
&& !(str.endsWith("?")))
{
//Check the sentence is ending with comma.
if(str.endsWith(","))
{
//Throw the CommaException.
throw new CommaException("You can't "
+ "end a sentence in a comma.");
}
//Otherwise.
else
{
//Throw PunctuationException.
throw new PunctuationException("The sentence "
+ "does not end correctly.");
}
}
//If the above conditions fails then
//return this message to the main function.
return "The sentence ends correctly.";
}
//Define the main function.
public static void main(String[] args)
{
//Create an object of Scanner
Scanner object = new Scanner(System.in);
//Prompt the user to enter the sentence.
System.out.println("Enter the sentence:");
String sentence=object.nextLine();
//Begin the try block.
try {
//Call the Driver's check function.
System.out.println(new
Driver().checkSentence(sentence));
}
//The catch block to catch the exception.
catch (EndOfSentenceException e)
{}
}
}
Write a Java program that prompts the user to enter a sequence of non-negative numbers (0 or more), storing them in an ArrayList, and continues prompting the user for numbers until they enter a negative value. When they have finished entering non-negative numbers, your program should return the mode (most commonly entered) of the values entered.
Answer: provided in explanation segment
Explanation:
the code to carry out this program is thus;
import java.util.ArrayList;
import java.util.Scanner;
public class ArrayListMode {
public static int getMode(ArrayList<Integer>arr) {
int maxValue = 0, maxCount = 0;
for (int i = 0; i < arr.size(); ++i) {
int count = 0;
for (int j = 0; j < arr.size(); ++j) {
if (arr.get(j) == arr.get(i))
++count;
}
if (count > maxCount) {
maxCount = count;
maxValue = arr.get(i);
}
}
return maxValue;
}
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
ArrayList<Integer>list= new ArrayList<Integer>();
int n;
System.out.println("Enter sequence of numbers (negative number to quit): ");
while(true) {
n=sc.nextInt();
if(n<=0)
break;
list.add(n);
}
System.out.println("Mode : "+getMode(list));
}
}
⇒ the code would produce Mode:6
cheers i hope this helps!!!!
Create an application that creates a report from quarterly sales. Console The Sales Report application Region Q1 Q2 Q3 Q4 1 $1,540.00 $2,010.00 $2,450.00 $1,845.00 2 $1,130.00 $1,168.00 $1,847.00 $1,491.00 3 $1,580.00 $2,305.00 $2,710.00 $1,284.00 4 $1,105.00 $4,102.00 $2,391.00 $1,576.00 Sales by region: Region 1: $7,845.00 Region 2: $5,636.00 Region 3: $7,879.00 Region 4: $9,174.00 Sales by quarter: Q1: $5,355.00 Q2: $9,585.00 Q3: $9,398.00 Q4: $6,196.00 Total sales: $30,534.00
Answer:
See explaination
Explanation:
package miscellaneous;
import java.text.NumberFormat;
import java.util.Currency;
import java.util.Locale;
public class sales {
public static void main(String[] args) {
double[][] sales= {
{1540.0,2010.0,2450.0,1845.0},//region1
{1130.0,1168.0,1847.0,1491.0},//region2
{1580.0,2305.0,2710.0,1284.0},//region3
{1105.0,4102.0,2391.0,1576.0}};//region4
//object for NumberFormat class
//needed in $
NumberFormat defaultFormat = NumberFormat.getCurrencyInstance(java.util.Locale.US);
System.out.println("The sales report application: ");
//the j(th) element in i(th) row in sales matrix contains sales value for
//sales in j(th) quarter
System.out.println("Sales by quarter: ");
System.out.println("Region\tQ1\t\tQ2\t\tQ3\t\tQ4");
for(int i=0;i<sales.length;i++) {
System.out.print(i+1+"\t");
for(int j=0;j<sales[i].length;j++) {
System.out.print(defaultFormat.format(sales[i][j])+"\t");
}
System.out.println();
}
//i(th) row in the matrix has sales for i(th) region
System.out.println("Sales by region: \n");
for(int i=0;i<sales.length;i++) {
System.out.print("Region "+(i+1)+":");
double region_sale=0;
for(int j=0;j<sales[i].length;j++) {
region_sale+=sales[i][j];
}
System.out.println(defaultFormat.format(region_sale));
}
System.out.println("\nSales by Quarter: \n");
//we have quarters so for adding up their sales we need 4 variables
double q1=0;
double q2=0;
double q3=0;
double q4=0;
for(int i=0;i<sales.length;i++) {
for(int j=0;j<sales[i].length;j++) {
//j=0 implies the sales data for first quarter
if(j==0) {
q1+=sales[i][j];
}
//j=1 implies the sales data for second quarter
if(j==1) {
q2+=sales[i][j];
}
//j=2 implies the sales data for third quarter
if(j==2) {
q3+=sales[i][j];
}
//j=3 implies the sales data for fourth quarter
if(j==3) {
q4+=sales[i][j];
}
}
}
System.out.println("Q1: "+defaultFormat.format(q1));
System.out.println("Q2: "+defaultFormat.format(q2));
System.out.println("Q3: "+defaultFormat.format(q3));
System.out.println("Q4: "+defaultFormat.format(q4));
//with the help of 2 loops every sales data
//in the matrix can be accessed, which can be added
//to total_sales variable
double total_sales=0;
for(int i=0;i<sales.length;i++) {
for(int j=0;j<sales[i].length;j++) {
total_sales+=sales[i][j];
}
}
System.out.println("\nTotal Sales: "+defaultFormat.format(total_sales));
}
}
Suppose that the first number of a sequence is x, where x is an integer. Define ; if is even; if is odd. Then there exists an integer k such that . Write a program that prompts the user to input the value of x. The program outputs the integer k such that and the numbers . (For example, if , then , and the numbers , respectively, are 75, 226, 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1.) Test your program for the following values of x: 75, 111, 678, 732, 873, 2048, and 65535.
Answer:
The program is written in c++ , go to the explanation part for it, the output can be found in the attached files.
Explanation:
C++ Code:
#include <iostream>
using namespace std;
int main() {
int x;
cout<<"Enter a number: ";
cin>>x;
int largest = x;
int position = 1, count = 0;
while(x != 1)
{
count++;
cout<<x<<" ";
if(x > largest)
{
largest = x;
position = count;
}
if(x%2 == 0)
x = x/2;
else
x = 3*x + 1;
}
cout<<x<<endl;
cout<<"The largest number of the sequence is "<<largest<<endl;
cout<<"The position of the largest number is "<<position<<endl;
return 0;
}
The program prompts the user for an integer ( x ). It calculates ( k ), the number of iterations until ( x ) becomes 1 in a sequence generated by specific rules: if ( x ) is even, it's divided by 2; if odd, it's multiplied by 3 and incremented by 1. .
Here's a Python program that implements the described sequence and finds the integer ( k ) for the given ( x ) value:
```python
def find_k(x):
k = 0
while x != 1:
if x % 2 == 0:
x = x // 2
else:
x = 3 * x + 1
k += 1
return k
def generate_sequence(x):
sequence = [x]
while x != 1:
if x % 2 == 0:
x = x // 2
else:
x = 3 * x + 1
sequence.append(x)
return sequence
def main():
test_values = [75, 111, 678, 732, 873, 2048, 65535]
for x in test_values:
k = find_k(x)
sequence = generate_sequence(x)
print(f"For x = {x}, k = {k}, sequence: {sequence}")
if __name__ == "__main__":
main()
```
This program calculates the integer \( k \) for each given value of \( x \) and generates the sequence according to the rules provided. Then it prints the \( k \) value and the sequence for each test value. You can run this program to verify the results for the provided test values.
(1) The given program outputs a fixed-height triangle using a * character. Modify the given program to output a right triangle that instead uses the user-specified triangle_char character. (1 pt) (2) Modify the program to use a loop to output a right triangle of height triangle_height. The first line will have one user-specified character, such as % or *. Each subsequent line will have one additional user-specified character until the number in the triangle's base reaches triangle_height. Output a space after each user-specified character, including a line's last user-specified character. (2 pts) Example output for triangle_char = % and triangle_height = 5:
Answer:
Following are the code to this question can be described as follows:
c= input('Input triangle_char: ') #defining variable c that input character value
length = int(input('Enter triangle_height: ')) # Enter total height of triangle
for i in range(length): # loop for column values
for j in range(i+1): #loop to print row values
print(c,end=' ') #print value
print()# for new line
Output:
please find the attachment.
Explanation:
In the above python code, two variable "c and length" variables are declared, in variable c is used to input char variable value, in the next line, length variable is defined, that accepts total height from the user.
In the next line, two for loop is declared, it uses as nested looping, in which the outer loop prints column values and inner the loop is used to prints rows. To prints all the value the print method is used, which prints the user input character triangle.The question asks how to modify a program to print a right triangle using a character and height defined by the user. The solution is to use a nested loop, wherein an outer loop specifies the number of rows equal to the triangle's height and the inner loop iterates over each row to print the specified character. The range of the inner loop increases with each outer loop iteration.
Explanation:To modify the given program to output a right triangle using the user-specified character and of a specified height, we would implement a nested loop in the program. An outer loop would control the number of rows, which equals the triangle's height, and an inner loop would handle the printing of the user-specified character per line. The inner loop's range would increase with each iteration of the outer loop. Here's an example in Python:
triangle_char = input('Enter a character: ') triangle_height = int(input('Enter triangle height: ')) for i in range(1, triangle_height + 1):for j in range(i): print(triangle_char, end=' ') print()
The triangle_char variable is the character used to generate the triangle. The triangle_height variable determines the number of rows. The range() function in the loops determines how many times the loop executes.
Learn more about Programming Nested Loops here:https://brainly.com/question/35123300
#SPJ11
Write a function called find_max that takes in a single parameter called random_list, which should be a list. This function will find the maximum (max) value of the input list, and return it. This function will assume the list is composed of only positive numbers. To find the max, within the function, use a variable called list_max that you initialize to value 0. Then use a for loop to loop through random_list. Inside the list, use a conditional to check if the current value is larger than list_max, and if so, re-assign list_max to store this new value. After the loop, return list_max, which should now store the maximum value from within the input list.
Answer:
see explaination
Explanation:
python code
def find_max(random_list):
list_max=0
for num in random_list:
if num > list_max:
list_max=num
return list_max
print(find_max([1,45,12,11,23]))
Implement the A5/1 algorithm. Suppose that, after a particular step, the values of the register are: X = (x0, x1, …, x18) = (1010101010101010101) Y = (y0, y1, …, y21) = (1100110011001100110011) Z = (z0, z1, …, z22) = (11100001111000011110000) List the next 32 keystream bits and give the contents of X, Y, and Z after these 32 bits have been generated.
Answer:
Check the explanation
Explanation:
After a particular step the registers X, Y and Z values are as it is in the first attached image below.
Now calculate the key stream bit, s using the following formula:
key stream bit , s= x0 XOR y0 XOR z0
s= 1 XOR 1 XOR 1
Hence, the 1st key bit stream ,s= 1
Now, for the next step we have to re calculate the contents of registers X, Y and Z as it is in the second attached image below.
For register X:
t= x5 XOR x2 XOR x1 XOR x0
= 0 XOR 1 XOR 0 XOR 1
t=0
For register Y:
t= y1 XOR y0
=1 XOR 1
t=0
For register Z:
t= z15 XOR z2 XOR z1 XOR z0
=1 XOR 1 XOR 1 XOR 1
t=0
Now, the contents of X, Y and X are as it is in the third attached image below.
Key stream bit, s= x0 XOR y0 XOR z0
S= 0 XOR 1 XOR 1
Hence the 2nd key stream bit, s= 0
The A5/1 algorithm generates keystream bits by shifting three LFSRs based on a majority bit mechanism and producing output bits through XOR operations. Following this process, the registers X, Y, and Z are updated, and the keystream is generated. The new register states and keystream give us the final output.
The A5/1 algorithm uses three Linear Feedback Shift Registers (LFSRs) named X, Y, and Z. Here are the initial states of the registers:
X = (1010101010101010101)Y = (1100110011001100110011)Z = (11100001111000011110000)To generate the next 32 keystream bits and update the registers, the following steps are followed:
Identify the majority bit of X<11>, Y<11>, and Z<11>.Only the registers with bits equal to the majority bit are shifted.Shift each register, updating the bits according to their feedback taps (for X: positions 13, 16, 17, 18; for Y: positions 20, 21; for Z: position 7, 20, 21, 22).Compute the output bit as the XOR of bits X<18>, Y<21>, Z<22>.Repeat until 32 bits are produced.After generating 32 keystream bits, the contents of the registers and the keystream are:
Keystream = (Provide the actual calculation here)X = (Updated state)Y = (Updated state)Z = (Updated state)Bob is stationed as a spy in Cyberia for a week and wants to prove that he is alive every day of this week and has not been captured. He has chosen a secret random number, x, which he memorized and told to no one. But he did tell his boss the value y = H(H(H(H(H(H(H(x))))))), where H is a one-way cryptographic hash function. Unfortunately, he knows that the Cyberian Intelligence Agency (CIA) was able to listen in on that message; hence, they also know the value of y. Explain how he can send a single message every day that proves he is still alive and has not been captured. Your solution should not allow anyone to replay any previous message from Bob as a (false) proof he is still alive.
Bob can use a hash chain method with a cryptographic hash function to send a message daily, validating his aliveness without risk of replay attacks. Each day he reveals the previous hash in the chain, which can be verified by hashing to obtain the initially shared value.
In order for Bob to prove he is still alive and has not been captured, without the risk of replay attacks, a strategy called hash chain can be employed using the cryptographic hash function. Bob starts with his secret random number x and computes a hash of it, H(x). He then repeatedly applies the hash function to this result a number of times equal to the number of days he needs to send a message (in this case, 7 times since he is there for a week), resulting in y = H(H(H(H(H(H(H(x))))))).
Each day, Bob unveils the previous hash in the chain. For example, on the first day, he sends H(H(H(H(H(H(x)))))), which anyone can hash once to verify it leads to y, the value his boss already knows. On the second day, he sends H(H(H(H(H(x))))), and so on. This proves that each message can only have come from someone with knowledge of the previous day's secret -- presumably, Bob himself. Because hash functions are one-way, even if someone knows the value H(x), they cannot reverse-engineer to find x.
This method ensures that even if the Cyberian Intelligence Agency (CIA) intercepted a message, they could not fake a future message since they would not be able to produce the next hash in the sequence without knowing the current one.
Suppose you design an algorithm to multiply two n-bit integers x and y. The general multiplication technique takes T(n) = O(n2) time. For a more efficient algorithm, you first split each of x and y into their left and right halves, which are n=2 bits long. For example, if x = 100011012, then xL = 10002 and xR = 11012, and x = 24 xL + xR. Then the product of x and y can be re-written as the following: x y = 2n (xL yL) + 2n=2 (xL yR + xR yL) + (xR yR)
Answer:
See Explaination
Explanation:
a) Assume there are n nits each in x and y.SO we divide them into n/2 and n/2 bits.
x = xL * 2^{n/2} + xR
y = yL * 2^{n/2} + yR
x.y = 2^{n}.(xL.yL) + 2^{n/2}.(xL.yR + xR.yL) +(xR.yR)
If you see there are 4 multiplications of size n/2 and we took all other additions and subtractions as O(n).
So T(n) = 4*T(n/2) + O(n)
Now lets find run time using master theorem.
T(n) = a* T(n/b) + O(n^{d})
a = 4
b = 2
d = 1
if a > b^{d}
T(n) = O(n^v) where v is log a base b
In our case T(n) = O(n^v) v = 2
=> T(n) = O(n^{2})
The splittinng method is not benefecial if we solve by this way as the run time is same even if go by the naive approach
b)
x.y = 2^{n}.(xL.yL) + 2^{n/2}.((xL+xR).(yL+yR)-(xL.yL) - (xR.yR)) +(xR.yR)
Here we are doing only three multipliactions as we changed the term.
So T(n) = 3*T(n/2) + O(n)
a = 3
b = 2
d = 1
if a > b^{d}
T(n) = O(n^v) where v is log a base b
In our case T(n) = O(n^v) v = log 3 base 2
v = 1.584
So T(n) = O(n^{1.584})
As we can see this is better than O(n^{2}).Therefore this algorithm is better than the naive approach.
//Add you starting comment block public class BubbleBubbleStarter //Replace the word Starter with your initials { public static void main (String[] args) { //Task 1: create an input double array list named mylist with some values pSystem.out.println("My list before sorting is: "); //Task 2: print the original list //Use println() to start and then replace with your printList() method after Task 4a is completed. p//Task 3: call the bubblesort method for mylist p//Task 4b: print the sorted list p} //Task 4a: create a method header named printlist to accept a formal parameter of a double array //create a method body to step through each array element println each element p//printList method header p//for loop p//println statement static void bubbleSort(double[] list) { boolean changed = true; do { changed = false; for (int j = 0; j < list.length - 1; j++) if (list[j] > list[j+1]) { //swap list[j] with list[j+1] double temp = list[j]; list[j] = list[j + 1]; list[j + 1] = temp; changed = true; } } while (changed); } }
Answer:
See explaination
Explanation:
import java.util.Scanner;
public class BubbleBubbleStarter {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
double arr[] = new double[10];
System.out.println("Enter 10 GPA values: ");
for (int i = 0; i < 10; i++)
arr[i] = sc.nextDouble();
sc.close();
System.out.println("My list before sorting is: ");
printlist(arr);
bubbleSort(arr);
System.out.println("My list after sorting is: ");
printlist(arr);
}
static void bubbleSort(double[] list) {
boolean changed = true;
do {
changed = false;
for (int j = 0; j < list.length - 1; j++) {
if (list[j] > list[j + 1]) {
double temp = list[j];
list[j] = list[j + 1];
list[j + 1] = temp;
changed = true;
}
}
} while (changed);
}
static void printlist(double list[]) {
for (int j = 0; j < list.length; j++) {
System.out.println(list[j]);
}
}
}
A company has offices in Honolulu, Seattle, Ogden, and Dublin, Ireland. There are transmission links between Honolulu and Seattle, Seattle and Ogden, and Ogden and Dublin. Seattle needs to communicate at 1 Gbps with each other site. Seattle and Dublin only need to communicate with each other at 1 Mbps. Ogden and Dublin need to communicate at 2 Gbps, and Ogden and Seattle need to communicate with each other at 10 Mbps. How much traffic will each transmission link have to carry? Show your work.
Answer:
See explaination
Explanation:
Looking at telecommunications network, a link is a communication channel that connects two or more devices for the purpose of data transmission. The link is likely to be a dedicated physical link or a virtual circuit that uses one or more physical links or shares a physical link with other telecommunications links.
Please check attachment for further solution.