A polynomial may be represented as a linked list where each node cont.pdf
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A polynomial may be represented as a linked list where each node contains the coefficient and exponent of a term of the polynomial. The polynomial 4 X^3 - 3 X^2 - 5 would be represented as the linked list. Write a program system that reads two polynomials, stores them as linked lists, adds them together, and prints the result as a polynomial. The result should be a third linked list. Travers both polynomials. If a particular exponent value is present in cither of the two polynomials being summed, then it should be present in the answer. If it is present in both polynomials, then its coefficient is the sum of the corresponding coefficient in both polynomials. (If this sum is zero, the term should be deleted). Solution #include #include typedef struct link { int coeff; int pow; struct link * next; } my_poly; void my_create_poly(my_poly **); void my_show_poly(my_poly *); void my_add_poly(my_poly **, my_poly *, my_poly *); int main(void) { int ch; do { my_poly * poly1, * poly2, * poly3; printf(\"\ Create 1st expression\ \"); my_create_poly(&poly1); printf(\"\ Stored the 1st expression\"); my_show_poly(poly1); printf(\"\ Create 2nd expression\ \"); my_create_poly(&poly2); printf(\"\ Stored the 2nd expression\"); my_show_poly(poly2); my_add_poly(&poly3, poly1, poly2); my_show_poly(poly3); printf(\"\ Add two more expressions? (Y = 1/N = 0): \"); scanf(\"%d\", &ch); } while (ch); return 0; } void my_create_poly(my_poly ** node) { int flag; int coeff, pow; my_poly * tmp_node; tmp_node = (my_poly *) malloc(sizeof(my_poly)); *node = tmp_node; do { printf(\"\ Enter Coeff:\"); scanf(\"%d\", &coeff); tmp_node->coeff = coeff; printf(\"\ Enter Pow:\"); scanf(\"%d\", &pow); tmp_node->pow = pow; tmp_node->next = NULL; printf(\"\ Continue adding more terms to the polynomial list?(Y = 1/N = 0): \"); scanf(\"%d\", &flag); //printf(\"\ FLAG: %c\ \", flag); if(flag) { tmp_node->next = (my_poly *) malloc(sizeof(my_poly)); //Grow the list tmp_node = tmp_node->next; tmp_node->next = NULL; } } while (flag); } void my_show_poly(my_poly * node) { printf(\"\ The polynomial expression is:\ \"); while(node != NULL) { printf(\"%dx^%d\", node->coeff, node->pow); node = node->next; if(node != NULL) printf(\" + \"); } } void my_add_poly(my_poly ** result, my_poly * poly1, my_poly * poly2) { my_poly * tmp_node; tmp_node = (my_poly *) malloc(sizeof(my_poly)); tmp_node->next = NULL; *result = tmp_node; while(poly1 && poly2) { if (poly1->pow > poly2->pow) { tmp_node->pow = poly1->pow; tmp_node->coeff = poly1->coeff; poly1 = poly1->next; } else if (poly1->pow < poly2->pow) { tmp_node->pow = poly2->pow; tmp_node->coeff = poly2->coeff; poly2 = poly2->next; } else { tmp_node->pow = poly1->pow; tmp_node->coeff = poly1->coeff + poly2->coeff; poly1 = poly1->next; poly2 = poly2->next; } if(poly1 && poly2) { tmp_node->next = (my_poly *) malloc(sizeof(my_poly)); tmp_node = tmp_node->next; tmp_node->next = NULL; } } while(poly1 || poly2) { tmp_node->next = (my_poly *) malloc(sizeof(my_poly));.
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I have to write a polynomial class linked list program and i do not .pdf
I have to write a polynomial class linked list program and i do not know what is wrong with my
code. Can someone please help. It is about adding, multiplying and subtraction polynomials.
c++. please change it as needed
#ifndef POLY_H
#define POLY_H
#include
#include
#include
using namespace std;
struct TermNode
{
int exp; //exponent
double coef; //coefficient
TermNode * next;
};
class PolyType
{
public:
PolyType();
PolyType(int r, double c);
PolyType(const PolyType &rhs);
~PolyType();
PolyType operator=(const PolyType &); //assignment
PolyType operator+ (const PolyType &rhs) const; //returns sum of parameter + self
PolyType operator* (const PolyType &rhs) const;
PolyType differ();
PolyType integr();
friend ostream & operator<< (ostream & out, const PolyType & rhs); //coefficients printed to a 2
decimal places
private:
TermNode *polyPtr;
};
PolyType::PolyType() {
polyPtr = nullptr;
}
PolyType::~PolyType() {
delete polyPtr;
polyPtr = nullptr;
}
PolyType::PolyType(int r, double c) {
polyPtr = new TermNode();
polyPtr->exp = c;
polyPtr->coef = r;
}
PolyType::PolyType(const PolyType & rhs) {
polyPtr = rhs.polyPtr;
}
PolyType PolyType::operator+(const PolyType & rhs) const {
PolyType temp;
if (this->polyPtr->exp >= rhs.polyPtr->exp) {
temp.polyPtr = this->polyPtr;
temp.polyPtr->next = rhs.polyPtr;
}
else {
temp.polyPtr = rhs.polyPtr;
temp.polyPtr->next = this->polyPtr;
}
return temp;
}
PolyType PolyType::operator=(const PolyType &rhs)
{
delete polyPtr;
TermNode * orig = rhs.polyPtr;
TermNode * newP = nullptr;
polyPtr = nullptr;
while (orig != nullptr)
{
if (newP == nullptr)
{
polyPtr = new TermNode;
polyPtr->exp = orig->exp;
polyPtr->coef = orig->coef;
polyPtr->next = nullptr;
newP = polyPtr;
}
else
{
newP->next = new TermNode;
newP = newP->next;
newP->exp = orig->exp;
newP->coef = orig->coef;
newP->next = nullptr;
}
orig = orig->next;
}
return *this;
}
PolyType PolyType::operator*(const PolyType &rhs) const
{
PolyType result;
TermNode * ptr = polyPtr;
while (ptr != nullptr)
{
result.polyPtr->exp = ptr->exp * rhs.polyPtr->exp;
result.polyPtr->coef = ptr->coef * rhs.polyPtr->coef;
ptr = ptr->next;
}
return result;
}
ostream & operator<<(ostream & out, const PolyType & rhs) {
TermNode *currentPtr = rhs.polyPtr;
out << currentPtr->coef << \"^\" << currentPtr->exp;
currentPtr = currentPtr->next;
while (currentPtr != nullptr) {
out << \" + \" << currentPtr->coef << \"^\" << currentPtr->exp;
currentPtr = currentPtr->next;
}
return out;
}
PolyType PolyType::integr() {
PolyType temp;
TermNode *currentPtr = this->polyPtr;
TermNode *tempPtr = temp.polyPtr;
while (currentPtr != nullptr) {
tempPtr = new TermNode();
tempPtr->coef = currentPtr->coef / (currentPtr->exp + 1);
tempPtr->exp = currentPtr->exp + 1;
currentPtr = currentPtr->next;
tempPtr = tempPtr->next;
}
return temp;
}
PolyType PolyType::differ() {
PolyType temp;
TermNode *currentPtr = this->polyPtr;
TermNode *tempPtr = temp.polyPtr;
while (currentPtr != nullptr) {
tempPtr = new TermNode();
tempPtr->c.
CSC-1106 Homework 09 (Class and static methods) Due Tuesday, Novembe.pdf
CSC-1106 Homework 09 (Class and static methods) Due: Tuesday, November 22, 2016 Write a
class encapsulating the concept of a rational number, a rational number has the following
attributes: an integer representing the numerator of the number. number, and another integer
representing the denominator of the rational The ratio should always be stored in its simplest
form. For example, the rational number 40/12 should be stored as 103. The class has the
following constructors and methods: A default constructor to set the Rational number to 0/1 A
constructor that has parameters for the numerator and denominator, and converts the resulting
ratio to simplified form simplify 0-a private method to convert the Rational number to simplified
form getGCD (x, y) -a private static method to return the largest common factor of the two
positive integers x and y (their greatest common divisor). For example, the greatest common
divisor of 40 and 12 is 4. value 0-returns the Rational number as a double value o-returns the
Rational number as a string in the form a/b toString multiply (Rational -a public method to return
the Rational number after multiplied by Rational number r add (Rational r)-a public method to
return the Rational number after addition with Rational number r floor (Rational r) -a public
method to return the integer of the floor of the Rational number. max (Rational x, Rational )-a
public static method to return the larger Rational number among x and y.
Solution
public class Rational {
/**
* A public constant that defines the rational value of 0.
*/
public static final Rational ZERO = new Rational(0, 1);
/**
* A public constant that defines the rational value of 1.
*/
public static final Rational ONE = new Rational(1, 1);
/**
* A public constant that defines the rational value of 0.5.
*/
public static final Rational HALF = new Rational(1, 2);
/*
* The Lowest and the Highest levels of precision allowed.
*/
private static final double LWST_PREC = 1E-3;
private static final double HIST_PREC = 1E-16;
private static double precision = 1E-10;
private final long num;
private final long den;
private final double result;
/**
* Creates a new {@code Rational} with the given numerator and
* denominator. It has special cases for Positive Infinity
* ({@link java.lang.Double#POSITIVE_INFINITY Double.POSITIVE_INFINITY}),
* Negative Infinity ({@link java.lang.Double#NEGATIVE_INFINITY
* Double.NEGATIVE_INFINITY}) and Not A Number
* ({@link java.lang.Double#NaN Double.NaN}.
*
* @param numerator The numerator.
* @param denominator The denominator.
*/
public Rational(long numerator, long denominator) {
// for dealing with \"infinities\" and \"NaN\":
if (denominator == 0) {
den = 0;
if (numerator > 0) {
num = 1;
result = Double.POSITIVE_INFINITY;
} else if (numerator < 0) {
num = -1;
result = Double.NEGATIVE_INFINITY;
} else {
num = 0;
result = Double.NaN;
}
return;
}
if (denominator < 0) {
numerator = 0L - numerator;
denominator = 0L - denominator;
}
num = numerator;.
C pointers and references
Computers use their memory for storing instructions of the programs and the values of the variables. Memory is a sequential collection of storage cells. Each cell has an address associated with it. Whenever we declare a variable, the system allocates, somewhere in the memory, a memory location and a unique address is assigned to this location.c pointers lecture
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED
HELP FIXING THIS CODE?
THIS IS THE PROBLEM
Write a program that allows the user to enter two positive integers and outputs their sum. Sounds
simple? Here\'s the catch: The numbers may be any size! They may be way too large to store in a
single variable. You may not assume any particular maximum number of digits.
HERE IS THE CODE
#include
#include
struct node /* Linked list node*/
{
int data;
struct node* next;
};
struct node *newNode(int data) /* Function to create a new node with giving the data*/
{
struct node *new_node = (struct node *) malloc(sizeof(struct node));
new_node->data = data;
new_node->next = NULL;
return new_node;
}
void push(struct node** head_ref, int new_data) /* Function to insert a node at the beginning of
the Doubly Linked List*/
{
struct node* new_node = newNode(new_data); /*allocate a node*/
new_node->next = (*head_ref); /* link the old list off to the new node*/
(*head_ref) = new_node; /*move the head to point to the new node*/
}
/* Adding the contents of two linked lists and return the head node of resultant list*/
struct node* addTwoLists (struct node* first, struct node* second)
{
struct node* res = NULL; /* res is head node of the resultant list*/
struct node *temp, *prev = NULL;
int carry = 0, sum;
while (first != NULL || second != NULL) /*while both the lists exists*/
{
/* Calculate value of next digit in resultant list.
// The next digit is sum of following things
// (i) Carry
// (ii) Next digit of first list (if there is a next digit)
// (ii) Next digit of second list (if there is a next digit)*/
sum = carry + (first? first->data: 0) + (second? second->data: 0);
carry = (sum >= 10)? 1 : 0; /* update carry for next calulation*/
sum = sum % 10; /*update sum if it is greater than 10*/
temp = newNode(sum); /*Create a new node with sum as data*/
if(res == NULL) /*if this is the first node then set it as head of the resultant list*/
res = temp;
else /* If this is not the first node then connect it to the rest.*/
prev->next = temp;
prev = temp; /* Set prev for next insertion*/
if (first) first = first->next; /* Move first pointers to next node*/
if (second) second = second->next; /*Move second pointer to the next node*/
}
if (carry > 0)
temp->next = newNode(carry);
return res; /* return head of the resultant list*/
}
void printList(struct node *node) /* function to print a linked list*/
{
while(node != NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
printf(\"\ \");
}
int main(void) /* program to test above function*/
{
struct node* res = NULL;
struct node* first = NULL;
struct node* second = NULL;
printf(\"Enter first positive integer: \");
scanf(\"%d\",&first);
printList(first);
printf(\"Enter second positive integer: \");
scanf(\"%d\",&second);
printList(second);
/*Add the two lists and see result*/
res = addTwoLists(first, second);
printf(\"Sum is: %d\");
printList(res);
return 0;
}
Solution
#include
#include
typedef struct Node
{
int da.
Recommended
I have to write a polynomial class linked list program and i do not .pdf
I have to write a polynomial class linked list program and i do not know what is wrong with my
code. Can someone please help. It is about adding, multiplying and subtraction polynomials.
c++. please change it as needed
#ifndef POLY_H
#define POLY_H
#include
#include
#include
using namespace std;
struct TermNode
{
int exp; //exponent
double coef; //coefficient
TermNode * next;
};
class PolyType
{
public:
PolyType();
PolyType(int r, double c);
PolyType(const PolyType &rhs);
~PolyType();
PolyType operator=(const PolyType &); //assignment
PolyType operator+ (const PolyType &rhs) const; //returns sum of parameter + self
PolyType operator* (const PolyType &rhs) const;
PolyType differ();
PolyType integr();
friend ostream & operator<< (ostream & out, const PolyType & rhs); //coefficients printed to a 2
decimal places
private:
TermNode *polyPtr;
};
PolyType::PolyType() {
polyPtr = nullptr;
}
PolyType::~PolyType() {
delete polyPtr;
polyPtr = nullptr;
}
PolyType::PolyType(int r, double c) {
polyPtr = new TermNode();
polyPtr->exp = c;
polyPtr->coef = r;
}
PolyType::PolyType(const PolyType & rhs) {
polyPtr = rhs.polyPtr;
}
PolyType PolyType::operator+(const PolyType & rhs) const {
PolyType temp;
if (this->polyPtr->exp >= rhs.polyPtr->exp) {
temp.polyPtr = this->polyPtr;
temp.polyPtr->next = rhs.polyPtr;
}
else {
temp.polyPtr = rhs.polyPtr;
temp.polyPtr->next = this->polyPtr;
}
return temp;
}
PolyType PolyType::operator=(const PolyType &rhs)
{
delete polyPtr;
TermNode * orig = rhs.polyPtr;
TermNode * newP = nullptr;
polyPtr = nullptr;
while (orig != nullptr)
{
if (newP == nullptr)
{
polyPtr = new TermNode;
polyPtr->exp = orig->exp;
polyPtr->coef = orig->coef;
polyPtr->next = nullptr;
newP = polyPtr;
}
else
{
newP->next = new TermNode;
newP = newP->next;
newP->exp = orig->exp;
newP->coef = orig->coef;
newP->next = nullptr;
}
orig = orig->next;
}
return *this;
}
PolyType PolyType::operator*(const PolyType &rhs) const
{
PolyType result;
TermNode * ptr = polyPtr;
while (ptr != nullptr)
{
result.polyPtr->exp = ptr->exp * rhs.polyPtr->exp;
result.polyPtr->coef = ptr->coef * rhs.polyPtr->coef;
ptr = ptr->next;
}
return result;
}
ostream & operator<<(ostream & out, const PolyType & rhs) {
TermNode *currentPtr = rhs.polyPtr;
out << currentPtr->coef << \"^\" << currentPtr->exp;
currentPtr = currentPtr->next;
while (currentPtr != nullptr) {
out << \" + \" << currentPtr->coef << \"^\" << currentPtr->exp;
currentPtr = currentPtr->next;
}
return out;
}
PolyType PolyType::integr() {
PolyType temp;
TermNode *currentPtr = this->polyPtr;
TermNode *tempPtr = temp.polyPtr;
while (currentPtr != nullptr) {
tempPtr = new TermNode();
tempPtr->coef = currentPtr->coef / (currentPtr->exp + 1);
tempPtr->exp = currentPtr->exp + 1;
currentPtr = currentPtr->next;
tempPtr = tempPtr->next;
}
return temp;
}
PolyType PolyType::differ() {
PolyType temp;
TermNode *currentPtr = this->polyPtr;
TermNode *tempPtr = temp.polyPtr;
while (currentPtr != nullptr) {
tempPtr = new TermNode();
tempPtr->c.
CSC-1106 Homework 09 (Class and static methods) Due Tuesday, Novembe.pdf
CSC-1106 Homework 09 (Class and static methods) Due: Tuesday, November 22, 2016 Write a
class encapsulating the concept of a rational number, a rational number has the following
attributes: an integer representing the numerator of the number. number, and another integer
representing the denominator of the rational The ratio should always be stored in its simplest
form. For example, the rational number 40/12 should be stored as 103. The class has the
following constructors and methods: A default constructor to set the Rational number to 0/1 A
constructor that has parameters for the numerator and denominator, and converts the resulting
ratio to simplified form simplify 0-a private method to convert the Rational number to simplified
form getGCD (x, y) -a private static method to return the largest common factor of the two
positive integers x and y (their greatest common divisor). For example, the greatest common
divisor of 40 and 12 is 4. value 0-returns the Rational number as a double value o-returns the
Rational number as a string in the form a/b toString multiply (Rational -a public method to return
the Rational number after multiplied by Rational number r add (Rational r)-a public method to
return the Rational number after addition with Rational number r floor (Rational r) -a public
method to return the integer of the floor of the Rational number. max (Rational x, Rational )-a
public static method to return the larger Rational number among x and y.
Solution
public class Rational {
/**
* A public constant that defines the rational value of 0.
*/
public static final Rational ZERO = new Rational(0, 1);
/**
* A public constant that defines the rational value of 1.
*/
public static final Rational ONE = new Rational(1, 1);
/**
* A public constant that defines the rational value of 0.5.
*/
public static final Rational HALF = new Rational(1, 2);
/*
* The Lowest and the Highest levels of precision allowed.
*/
private static final double LWST_PREC = 1E-3;
private static final double HIST_PREC = 1E-16;
private static double precision = 1E-10;
private final long num;
private final long den;
private final double result;
/**
* Creates a new {@code Rational} with the given numerator and
* denominator. It has special cases for Positive Infinity
* ({@link java.lang.Double#POSITIVE_INFINITY Double.POSITIVE_INFINITY}),
* Negative Infinity ({@link java.lang.Double#NEGATIVE_INFINITY
* Double.NEGATIVE_INFINITY}) and Not A Number
* ({@link java.lang.Double#NaN Double.NaN}.
*
* @param numerator The numerator.
* @param denominator The denominator.
*/
public Rational(long numerator, long denominator) {
// for dealing with \"infinities\" and \"NaN\":
if (denominator == 0) {
den = 0;
if (numerator > 0) {
num = 1;
result = Double.POSITIVE_INFINITY;
} else if (numerator < 0) {
num = -1;
result = Double.NEGATIVE_INFINITY;
} else {
num = 0;
result = Double.NaN;
}
return;
}
if (denominator < 0) {
numerator = 0L - numerator;
denominator = 0L - denominator;
}
num = numerator;.
C pointers and references
Computers use their memory for storing instructions of the programs and the values of the variables. Memory is a sequential collection of storage cells. Each cell has an address associated with it. Whenever we declare a variable, the system allocates, somewhere in the memory, a memory location and a unique address is assigned to this location.c pointers lecture
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf
THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED
HELP FIXING THIS CODE?
THIS IS THE PROBLEM
Write a program that allows the user to enter two positive integers and outputs their sum. Sounds
simple? Here\'s the catch: The numbers may be any size! They may be way too large to store in a
single variable. You may not assume any particular maximum number of digits.
HERE IS THE CODE
#include
#include
struct node /* Linked list node*/
{
int data;
struct node* next;
};
struct node *newNode(int data) /* Function to create a new node with giving the data*/
{
struct node *new_node = (struct node *) malloc(sizeof(struct node));
new_node->data = data;
new_node->next = NULL;
return new_node;
}
void push(struct node** head_ref, int new_data) /* Function to insert a node at the beginning of
the Doubly Linked List*/
{
struct node* new_node = newNode(new_data); /*allocate a node*/
new_node->next = (*head_ref); /* link the old list off to the new node*/
(*head_ref) = new_node; /*move the head to point to the new node*/
}
/* Adding the contents of two linked lists and return the head node of resultant list*/
struct node* addTwoLists (struct node* first, struct node* second)
{
struct node* res = NULL; /* res is head node of the resultant list*/
struct node *temp, *prev = NULL;
int carry = 0, sum;
while (first != NULL || second != NULL) /*while both the lists exists*/
{
/* Calculate value of next digit in resultant list.
// The next digit is sum of following things
// (i) Carry
// (ii) Next digit of first list (if there is a next digit)
// (ii) Next digit of second list (if there is a next digit)*/
sum = carry + (first? first->data: 0) + (second? second->data: 0);
carry = (sum >= 10)? 1 : 0; /* update carry for next calulation*/
sum = sum % 10; /*update sum if it is greater than 10*/
temp = newNode(sum); /*Create a new node with sum as data*/
if(res == NULL) /*if this is the first node then set it as head of the resultant list*/
res = temp;
else /* If this is not the first node then connect it to the rest.*/
prev->next = temp;
prev = temp; /* Set prev for next insertion*/
if (first) first = first->next; /* Move first pointers to next node*/
if (second) second = second->next; /*Move second pointer to the next node*/
}
if (carry > 0)
temp->next = newNode(carry);
return res; /* return head of the resultant list*/
}
void printList(struct node *node) /* function to print a linked list*/
{
while(node != NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
printf(\"\ \");
}
int main(void) /* program to test above function*/
{
struct node* res = NULL;
struct node* first = NULL;
struct node* second = NULL;
printf(\"Enter first positive integer: \");
scanf(\"%d\",&first);
printList(first);
printf(\"Enter second positive integer: \");
scanf(\"%d\",&second);
printList(second);
/*Add the two lists and see result*/
res = addTwoLists(first, second);
printf(\"Sum is: %d\");
printList(res);
return 0;
}
Solution
#include
#include
typedef struct Node
{
int da.
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Main issues with the following code-Im not sure if its reading the fil.pdf
Main issues with the following code:Im not sure if its reading the file correctly because its
adding all terms together. In other words the the first line of terms is adding the second line of
terms but thats not my intention. Each line of term should have their own polynomial expression
as shown below
Basic specifications :
The code should add the polynomial numbers using a linked list. Java . Keep in mind the
polynomials will be gotten from a file.
Things to keep in mind the list should be in decending order and traverse the list after
all coefficient/exponent pair Term nodes from one input line have been
inserted.
Here are the specifications
//The Polynomial class would likely have a single private data member (reference to
class Term) called head.
//The Term class would likely have three private data
members (coefficient and exponent of type int) and next of type reference to
Term.
//The latter class would also have the following public methods:
constructor(s), gets and sets for each data member, and perhaps a formatting
method with the special output requirements.
An example of input is 7 1 2 4 6 2 -3 2 -1 -1 3
Negative negative
pair signals the termination
***NO doubly linked list , it should be a single linked list **
Input file contains the following :
7 1 2 4 6 2 -3 2 -1 -1 3 ----> 2x^4 - 3x^2 +6x^2 + 7x
5 1 2 3 6 0 -3 2 -1 -1 3 ----> 2*X^3 + (-3)*X^2 + 5*X + 6 // thats the ouput for one line of the
the file , each line of the file should have one polynomial output
Code
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Polynomial {
private Term head;
// Constructor for creating an empty polynomial object
public Polynomial() {
head = null;
}
// Method for adding a term to the polynomial, this is inside the polynomial term
public void addTerm(int coefficient, int exponent) {
// Create a new term object
Term newTerm = new Term(coefficient, exponent);
// If the polynomial is empty, set the new term as the head
if (head == null) {
head = newTerm;
} else {
// Traverse the polynomial until we find the correct position for the new term
Term current = head;
Term previous = null;
while (current != null && current.getExponent() > exponent) {
previous = current;
current = current.getNext();
}
// If a term with the same exponent already exists, add the coefficients
if (current != null && current.getExponent() == exponent) {
current.setCoefficient(current.getCoefficient() + coefficient);
} else {
// Otherwise, insert the new term into the polynomial
if (previous == null) {
head = newTerm;
} else {
previous.setNext(newTerm);
}
newTerm.setNext(current);
}
}
// Method for formatting the polynomial as a string
public String toString() {
StringBuilder sb = new StringBuilder();
if (head != null) {
sb.append(head.toString());
Term current = head.getNext();
while (current != null) {
if (current.getCoefficient() > 0) {
sb.append(" + ");
} else {
sb.append(" - ");
}
sb.append(current.toString());
current = cur.
Java AssignmentUsing the ListNode.java file below Write method.pdf
Java Assignment
Using the ListNode.java file below
Write methods called min and max that return the smallest and largest values in the linked list.
These methods will be added to your ListNode class. For example if a variable called list stores
{11, -7, 3, 42, 0, 14], the call of list.min() should return -7 and the call of list.max() should return
42. If the list is empty, return -1. Print the returned value.
Write a method called insertNode that inserts a new node anywhere in your linked list. Display
your linked list with new value.
Write a program to test these methods to ensure accuracy and successful implementation. Upload
your modified version of the ListNode java file only.
Note: Original code should still be included. You will make additions and slight modifications
where necessary. Throw exceptions where you think is necessary. // Listlode is a class for storing
a single node of a linked // list. This node class is for a list of integer values. public class Listlode
\{ public int data; //data stored in this node public Listliode front;//points to head node public
Listliode next; /f link to next node in the list // post: constructs a node with data and null link
public Listlode() \{ 3. // post: constructs a node with given data and null link: public Listlode(int
data) \{ this(data, null); 3 // post: constructs a node with given data and given link. public
Listlode(int data, Listliode next) \{ this.data = data; this. next = next; 3 public String toString() \{
if (front = null) : \} else \{ String result = " ["+ front data; Listliode current = front. next; while
(current != null) \{ result +=m,m+ current data, current = current. next; 3 result += " ]n - return
result; 3. }// post: appends the given value to the end of the list public void add(int value) \&
puolic void aod if (front = nult) \{. front = new Listlode (value); 3 else \{ Listlode current =
front; whistiode current = front; while (current next I= null) \{ current = current. next; 3 current.
next = new Listliode (value ); 3. 3 J// post : returns the position of the first occurrence of the
given /l value (-1 if not found ) public void remove (int index) \{ if ( index =){ \} else \{ Listiode
current = nodeAt ( index - 1); current. next = current. next. next, 3 3 3// pre : <=i< size() // post:
returns a reference to the node at the given index public Listlode nodeAt (int index) : List Node
current = front; for (int i=0;i< index; i++) \{ current = current. next. \}. return current; 3.
Please fill in the code to run the program based on the following in.pdf
Please fill in the code to run the program based on the following instructions(Please make sure it
compiles and runs and gives output in the given format):
(Please note that I don't need an explanation for the code, but the code itself. Thanks in advance)
Input: Take a filed named topo.txt as an input, the contents of the file is below
0 10 5 2
10 0 1 1
5 1 0 3
2 1 3 0
Your simulator should read a topology file (say, topo.txt) with a matrix {Di,j }, i, j {0.., N 1}. N
is the number of nodes (routers) and Di,j is the link cost from node i to node j. If nodes i and j are
the same, Di,j = 0; If nodes i and j are directly connected (adjacent), Di,j = e, where e 0.
Otherwise, nodes i and j are not directly connected and you should assign link cost Di,j = 1.
Implementation:
You need to first implement DV initialization per node via rtinit(struct distance table *dt, int
node id, int *link costs, int num nodes)
For every node, rtinit() will be called only ONCE at the beginning of the simulation (k=0). For
example, to initialize node 0, you need to use rtinit(&dt, 0, link cost, 4) (Please check the sample
code for more information). Node 0 should firstly initialize its distance table by setting all
distances to 1. After that, node 0s distance vector in the distance table is changed to (0,10,5,2),
representing the minimum distance from node 0 to all nodes in the topology. Lastly, node 0
needs to send its distance vector in a routing packet to its connected neighbors using
send2neighbor(rtpkt packet) (you can find this function in the sample code) based on the input
topology.
You then need to implement DV update for each received routing packets. Please assume that
every DV update happens in a synchronous manner while it is not true for DV in reality.
Please use this method to update DV for each node- rtupdate(struct distance table *dt, struct rtpkt
recv pkt)
This method will be called for each received routing packet at each simulation time slot k 1. It
parses the input, which is a routing packet recv pkt to update the distance table based on the DV
algorithm. If its own minimum distance to any of the other nodes is updated, the node informs its
directly connected neighbors of this change in minimum cost by sending them a routing packet
(send2neighbor(rtpkt packet)). Therefore, in the given example, nodes 0 and 1 will communicate
with each other, but nodes 1 and 3 will not.
Convergence: Please make sure that your code can run successfully and produce correct and
converged output under any topology with any N 10.
Test cases and the expected outputs:
Assume the simulation starts at slot k = 0 for initialization and k 1 when updating DVs at each
simulation slot. You need to print out the current slot k followed by DVs of all the nodes in the
ascending order at the end of the simulation slots. Please print them out in the first five slots (k =
0, 1, 2, 3, 4) and then every 10 slots (k = 10, 20, 30, ) until they converge. In this test case, the
expected out.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.next;
}
}
return false;
}
public void printList(){
Node l = first;
while(l.next!=null){
System.out.print(l.number+\" \");
l = l.next;
}
System.out.println();
}
}
Node.java (implements a single node in the linked list; stores number and pointers)
public class Node{
String number;
Node previous;
Node next;
public Node(String num){
number = num;
previous = null;
next = null;
}
public Node(String num, Node p, Node n){
number = num;
previous = p;
next = n;
}
}
Main.java (contains the main method and the helper methods to solve the questions given)
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
static boolean checkHappiness(String num){
Node current = new Node(num, null, null);
NumberList numberList = new NumberList(current);
int len = num.length();
int resultant = 0;
for (int i=0; i happyNumbersfrom1to10000(){
ArrayList numbers = new ArrayList();
for (int j=0; j<10000; j++){
if (checkHappiness(String.valueOf(j+1))){
numbers.add(String.valueOf(j+1));
}
}
return numbers;
}
static void happyNumbersfrom9001to10000(){
for (int j=9001; j<=10000; j++){
//System.out.println(j);
if (checkHappiness(String.valueOf(j))){
System.out.println(String.valueOf(j));
}
}
}
static String getLargeHappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
static String getLargeUnhappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (!checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
public static void main(String[] args){
happyNumbersfrom9001to10000();
System.out.println(happyNumbersfrom1to10000());
}
}
Solution
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.
Python programming workshop
Basic Python Programming สรุปสั้นๆ อ่านง่าย เข้าใจง่าย โดย ผศ ดร ปราโมทย์ กั่วเจริญ เหมาะกับผู้เริ่มต้น
Java Code The traditional way to deal with these in Parsers is the .pdf
Java Code: The traditional way to deal with these in Parsers is the Expression-Term-Factor
pattern:
Expression: TERM {+|- TERM}
Term: FACTOR {*|/ FACTOR}
Factor: number | ( EXPRESSION )
A couple of notation comments:
| means "or" - either + or -, either * or /
{} means an optional repeat. That's what allows this to parse 1+2+3+4
Think of these as functions. Every CAPTIALIZED word is a function call.
Consider our meme expression above: 6/2*(1+2)
We will start by looking at expression. Expression starts with a TERM. We can't do anything
until we resolve TERM, so let's go there.
A term starts with FACTOR. Again, we can't do anything until we deal with that.
A factor is a number or a (EXPRESSION). Now we can look at our token (hint:
MatchAndRemove). We see a number. OK - our factor is a number. We "return" that.
Remember that we got to factor from term. Let's substitute our number in:
TERM: FACTOR(6) {*|/ FACTOR}
Now we deal with our optional pattern. Is the next character * or /? Yes! Now is the next thing a
factor?
It turns out that it is. Let's substitute that in:
TERM: FACTOR(6) / FACTOR(2)
But remember that our pattern is a REPEATING pattern (hint: loop):
TERM: FACTOR(6) / FACTOR(2) {*|/ FACTOR}
We see the * and call factor. But this time, the factor is not a number but a parenthetical
expression.
Factor: number | ( EXPRESSION )
Factor calls expression.
Expression calls term, term calls factor, factor returns the number 1. Term doesn't see a * or / so
it passes the 1 up. Expression sees the + and calls term. Term calls factor which returns the 2.
Expression doesn't see a +|- value so ends the loop and returns 1+2.
So, remember that we were here:
TERM: FACTOR(6) / FACTOR(2) * FACTOR
Our factor is (1+2). That can't be broken down. That math HAS to be done before can multiply
or divide. That's what enforces order of operations.
In code, this looks like something like this (pseudo-code):
Node Factor()
num = matchAndRemove(NUMBER)
if (num.isPresent) return num
if (matchAndRemove(LPAREN).isPresent)
exp = Expression()
if (exp == null) throw new Exception()
if (matchAndRemove(RPAREN).isEmpty)
throw new Exception()
Node Term()
left = Factor()
do
op = MatchAndRemove(TIMES)
if (op.isEmpty) op=MatchAndRemove(DIVIDE)
if (op.isEmpty) return left
right = Factor()
left = MathOpNode(left, op, right)
while (true)
Node Expression()
left = Term()
do
op = MatchAndRemove(PLUS)
if (op.isEmpty) op=MatchAndRemove(MINUS)
if (op.isEmpty) return left
right = Term()
left = MathOpNode(left, op, right)
while (true)
What is this "MathOpNode"? It's "just" a new node type that holds two other nodes (left and
right) and an operation type: *, /, +, -. Notice that the loops use the result of one operation as the
left side of the next operation. This is called left associative - the left most part is done first.
Right associativity is the opposite - the rightmost part is done first, then we work our way left.
Generally, notice the pattern here - we call a function that is the lo.
Unit-I Pointer Data structure.pptx
Cyber security refers to every aspect of protecting an organization and its employees and assets against cyber threats. As cyberattacks become more common and sophisticated and corporate networks grow more complex, a variety of cyber security solutions are required to mitigate corporate cyber risk.
PROVIDE COMMENTS TO FELLOW STUDENTS ANSWERS AND PLEASE DON’T SAY G.docx
PROVIDE COMMENTS TO FELLOW STUDENTS ANSWERS AND PLEASE DON’T SAY GOOD WORK NICE FORMULA OR SOMETHING LIKE THAT, BUT ACTULLY HE CAN USE. THANK YOU.
Hartleys Function Code
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Actions for Hartleys Function Code
Chad Hartley posted Nov 5, 2015 5:10 PM
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This program will add an integer number and a decimal number up to 2 decimal places. I have included notes in the code to explain what each thing does. I hope I did this right. It compiles successfully.
PseudoCode
Start
Declare int O1; Stands for Output1
O1=sum; Sum is the functions name
Int sum()
Declare variables
Int num1;
Float num2;
Write “Enter a number.”
Scanf num1
Write”Enter a decimal number.”
Scanf num2
Return num1+num2
end
C Code
#include <stdio.h>
int sum();//prototype
int main()//calling program
{
//Declare a varaiable
int O1;
O1=sum();//main is calling sum one time.
//if I listed this twice it would run the function 'sum' twice.
// Example: if I add a new int (int O1, O2) and declare O2 to
//be O2=sum then the function would run twice.
}
int sum ()//function 'sum'
{
int num1;// Declare intergers/variables
float num2;
printf("Enter a number.\n");
scanf("%d",&num1);// Take first input and assign it to num1
printf("Enter a decimal number.\n");
scanf("%.2f",&num2);
//Can use the printf statement but when you are calling an integer you can use the return.
//printf("The sum of %d, %d, is %d", num1,num2,num1+num2);
return num1+num2;
}
ADD COMMENT HERE
Chaotic Function
Contains unread posts
Actions for Chaotic Function
Joshua Ray posted Nov 5, 2015 2:33 PM
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float tmp
int i
function float chaos(float num)
{
for i < 20
num = 3.9*num*(1-num)
print num
}
main
print "Program description"
print "Request input btw 0 and 1"
tmp = input
chaos(tmp)
/*
* File: main.c
* Author: JaiEllRei
*
* Created on November 5, 2015, 2:04 PM
*/
#include <stdio.h>
#include <stdlib.h>
float chaos(float num);
int main(void)
{
float tmp;
printf("This program illustrates a choatic function. \n");
printf("Input a number between 0 and 1: ");
scanf("%f", &tmp);
chaos(tmp);
}
float chaos(float num)
{
for (int i=0; i<20; i++){
/*Chaotic Formula*/
num = 3.9 * num * (1-num);
printf("%.3f \n", num);
}
}
This program illustrates a choatic function.
Input a number between 0 and 1: .2
0.624
0.915
0.303
0.824
0.566
0.958
0.156
0.514
0.974
0.098
0.345
0.881
0.409
0.943
0.210
0.647
0.891
0.379
0.918
0.293
ADD COMMENT HERE
//MPH to KPH Conversion Function
Function KPHConv(value) as float
Set KPHConv = value*1.609344
End Function
Pseudocode for simple conversion program calling function
//Declare function
// MPH to KPH Conversion Function
Functio ...
Discuss the benefits of in-built coding software within the Health I.pdf
Discuss the benefits of in-built coding software within the Health Information Managements
Systems?
Note: Mention your references
words limit: 500 words no plagiarism
Solution
Health information management (HIM) is information management applied to health and health
care. It is the practice of acquiring, analyzing and protecting digital and traditional medical
information vital to providing quality patient care. With the widespread computerization of
health records, traditional (paper-based) records are being replaced with electronic health records
(EHRs). The tools of health informatics and health information technology are continually
improving to bring greater efficiency to information management in the health care sector. Both
hospital information systems and health human resources information systems (HRHIS) are
common implementations of HIM.
Paper-based records have been in existence for centuries* and their gradual replacement by
computer-based records has been slowly underway for over twenty years in western healthcare
systems. Computerised information systems have not achieved the same degree of penetration in
healthcare as that seen in other sectors such as finance, transport and the manufacturing and retail
industries. Further, deployment has varied greatly from country to country and from speciality to
specialty and in may cases has revolved around local systems designed for local use.
In record time and at an unprecedented pace, the electronic health record (EHR) has spurred a
revolutionary change in HIM that\'s modifying the profession as each day passes.EHRs and
accompanying technology like health information exchange, computer-assisted coding, voice
recognition software, and patient portals-along with revamped government quality and EHR
incentive programs-are modifying many aspects of HIM.
An information system consists of four interrelated components—data, information technology,
process, and users.14 HIM professionals’ traditional job roles make them the experts in
managing data and processes in an information system. With the digitizing of information
systems in healthcare organizations, the roles of HIM professionals have expanded into
information technology (IT) and user support, which usually are the functions of IT supporting
services. HIM professionals’ training and experience in the intersection of clinical and
management sciences as well as their knowledge about data quality equip them with the
capability to maintain the integrity and accessibility of health information, although they may not
necessarily have the particular skills to support technical operations of a health information
system.
Electronic medical record systems lie at the center of any computerised health information
system. Without them other modern technologies such as decision support systems cannot be
effectively integrated into routine clinical workflow. The paperless, interoperable, multi-
provider, multi-specialty, multi-discipline computerised.
Determine if each system is1. memoryless2. Invertible3. Causal.pdf
Determine if each system is
1. memoryless
2. Invertible
3. Causal
4. Stable
5. Time Invariant
6. Linear y[n] = x[n]2 VIn y(t)= x(2t) y[n-x(-n] y[n] nx| n + 3 y(t)= sin(2*t x(T) dT y( t) -X(T)
dT dx(t) y(t)=-dt -5 VIn 5k tt [ nt_ot, 11 ntnnttttn
Solution
y(t) = R 2 x( )d
Taken t=2(any number)
Causality: The system is NOT causal because for any t < 2, the output depends on a future input.
Invertibility: The system is NOT invertible because only the integral and not the input function
itself determines the output, e.g. the output corresponding to input x1(t) = u(t)u(t1) is the same as
that corresponding to input x2(t) = 2[u(t) u(t 1/2)].
Memorylessness: The system is NOT memoryless because the output at time t depends on input
values at times other than t. Time Invariance: The system is NOT time invariant. To see this, we
let x(t) = u(t) u(t 1) so that y(t) = Z 2 x( )d (18) = Z 1 0 d (19) = 1, (20) i.e. constant for all t, so
in particular, y(t 3) = 1 for any value of t 2 . However, if we let x3(t) = x(t 3) = u(t 3) u(t 4) .
Then y3(t) = Z 2 x3( )d (21) = Z 2 3 x( )d (22) = 0 (23) 6= y(t 3)
Linearity: Assume t=2..any arbitary value The system is linear because if y1(t) = Z 2 x1( )d, and
(12) y2(t) = Z 2 x2( )d, and x(t) = x1(t) + x2(t), then the output y(t) corresponding to the input
x(t) is y(t) = Z 2 (x1 + x2)( )d = Z 2 x1( )d + Z 2 x2( )d = y1(t) + y2(t)
y(t) = (dx/dt)(t) ( causal , invertible, linear , memoryless , time invariant ) Causality: The system
is memoryless, hence causal. Invertibility: The system is NOT invertible because any two inputs
that differ by a constant yield the same output. Linearity: The system is linear because if y1(t) =
(dx1/dt)(t), and (25) y2(t) = (dx2/dt)(t) and x(t) = x1(t) + x2(t), then the output y(t)
corresponding to the input x(t) is y(t) = (d(x1 + x2)/dt)(t) = (dx1/dt)(t) + (dx2/dt)(t) = y1(t) +
y2(t). Memorylessness: The system is memoryless because the output at time t depends on input
values at only time t. Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa(t) = x(t a). Then the output ya(t)
corresponding to the input signal xa(t) is ya(t) = (dxa/dt)(t) = (d(x)/dt)(t a) = y(t a).
(a) y(t) = R t x( )d Invertible: z(t) = d dty(t) = d dt Z t x( ) d = x(t)
y[n] = x[n] (1) y[1] depends on x[1] so the system has memory.
Let x1[n] = x[n N], then y1[n] = x1[n] = x[n N] y[n N] = x[(n N)] = x[n + N] Hence the
system is time-varying. (3) Assume x1[n] y1[n] and x2[n] y2[n] and x3[n] = ax1[n] + bx2[n]
y3[n] = x3[n] = ax1[n] + bx2[n] = ay1[n] + by2[n] Hence the system is linear. (4) y[1] depends
on x[1] so the system is noncausal. (5) If |x[n]| Mx for all n, then |y[n]| = |x[n]| Mx for all n.
Therefore, the system is BIBO stable..
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Similar to A polynomial may be represented as a linked list where each node cont.pdf
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Main issues with the following code-Im not sure if its reading the fil.pdf
Main issues with the following code:Im not sure if its reading the file correctly because its
adding all terms together. In other words the the first line of terms is adding the second line of
terms but thats not my intention. Each line of term should have their own polynomial expression
as shown below
Basic specifications :
The code should add the polynomial numbers using a linked list. Java . Keep in mind the
polynomials will be gotten from a file.
Things to keep in mind the list should be in decending order and traverse the list after
all coefficient/exponent pair Term nodes from one input line have been
inserted.
Here are the specifications
//The Polynomial class would likely have a single private data member (reference to
class Term) called head.
//The Term class would likely have three private data
members (coefficient and exponent of type int) and next of type reference to
Term.
//The latter class would also have the following public methods:
constructor(s), gets and sets for each data member, and perhaps a formatting
method with the special output requirements.
An example of input is 7 1 2 4 6 2 -3 2 -1 -1 3
Negative negative
pair signals the termination
***NO doubly linked list , it should be a single linked list **
Input file contains the following :
7 1 2 4 6 2 -3 2 -1 -1 3 ----> 2x^4 - 3x^2 +6x^2 + 7x
5 1 2 3 6 0 -3 2 -1 -1 3 ----> 2*X^3 + (-3)*X^2 + 5*X + 6 // thats the ouput for one line of the
the file , each line of the file should have one polynomial output
Code
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Polynomial {
private Term head;
// Constructor for creating an empty polynomial object
public Polynomial() {
head = null;
}
// Method for adding a term to the polynomial, this is inside the polynomial term
public void addTerm(int coefficient, int exponent) {
// Create a new term object
Term newTerm = new Term(coefficient, exponent);
// If the polynomial is empty, set the new term as the head
if (head == null) {
head = newTerm;
} else {
// Traverse the polynomial until we find the correct position for the new term
Term current = head;
Term previous = null;
while (current != null && current.getExponent() > exponent) {
previous = current;
current = current.getNext();
}
// If a term with the same exponent already exists, add the coefficients
if (current != null && current.getExponent() == exponent) {
current.setCoefficient(current.getCoefficient() + coefficient);
} else {
// Otherwise, insert the new term into the polynomial
if (previous == null) {
head = newTerm;
} else {
previous.setNext(newTerm);
}
newTerm.setNext(current);
}
}
// Method for formatting the polynomial as a string
public String toString() {
StringBuilder sb = new StringBuilder();
if (head != null) {
sb.append(head.toString());
Term current = head.getNext();
while (current != null) {
if (current.getCoefficient() > 0) {
sb.append(" + ");
} else {
sb.append(" - ");
}
sb.append(current.toString());
current = cur.
Java AssignmentUsing the ListNode.java file below Write method.pdf
Java Assignment
Using the ListNode.java file below
Write methods called min and max that return the smallest and largest values in the linked list.
These methods will be added to your ListNode class. For example if a variable called list stores
{11, -7, 3, 42, 0, 14], the call of list.min() should return -7 and the call of list.max() should return
42. If the list is empty, return -1. Print the returned value.
Write a method called insertNode that inserts a new node anywhere in your linked list. Display
your linked list with new value.
Write a program to test these methods to ensure accuracy and successful implementation. Upload
your modified version of the ListNode java file only.
Note: Original code should still be included. You will make additions and slight modifications
where necessary. Throw exceptions where you think is necessary. // Listlode is a class for storing
a single node of a linked // list. This node class is for a list of integer values. public class Listlode
\{ public int data; //data stored in this node public Listliode front;//points to head node public
Listliode next; /f link to next node in the list // post: constructs a node with data and null link
public Listlode() \{ 3. // post: constructs a node with given data and null link: public Listlode(int
data) \{ this(data, null); 3 // post: constructs a node with given data and given link. public
Listlode(int data, Listliode next) \{ this.data = data; this. next = next; 3 public String toString() \{
if (front = null) : \} else \{ String result = " ["+ front data; Listliode current = front. next; while
(current != null) \{ result +=m,m+ current data, current = current. next; 3 result += " ]n - return
result; 3. }// post: appends the given value to the end of the list public void add(int value) \&
puolic void aod if (front = nult) \{. front = new Listlode (value); 3 else \{ Listlode current =
front; whistiode current = front; while (current next I= null) \{ current = current. next; 3 current.
next = new Listliode (value ); 3. 3 J// post : returns the position of the first occurrence of the
given /l value (-1 if not found ) public void remove (int index) \{ if ( index =){ \} else \{ Listiode
current = nodeAt ( index - 1); current. next = current. next. next, 3 3 3// pre : <=i< size() // post:
returns a reference to the node at the given index public Listlode nodeAt (int index) : List Node
current = front; for (int i=0;i< index; i++) \{ current = current. next. \}. return current; 3.
Please fill in the code to run the program based on the following in.pdf
Please fill in the code to run the program based on the following instructions(Please make sure it
compiles and runs and gives output in the given format):
(Please note that I don't need an explanation for the code, but the code itself. Thanks in advance)
Input: Take a filed named topo.txt as an input, the contents of the file is below
0 10 5 2
10 0 1 1
5 1 0 3
2 1 3 0
Your simulator should read a topology file (say, topo.txt) with a matrix {Di,j }, i, j {0.., N 1}. N
is the number of nodes (routers) and Di,j is the link cost from node i to node j. If nodes i and j are
the same, Di,j = 0; If nodes i and j are directly connected (adjacent), Di,j = e, where e 0.
Otherwise, nodes i and j are not directly connected and you should assign link cost Di,j = 1.
Implementation:
You need to first implement DV initialization per node via rtinit(struct distance table *dt, int
node id, int *link costs, int num nodes)
For every node, rtinit() will be called only ONCE at the beginning of the simulation (k=0). For
example, to initialize node 0, you need to use rtinit(&dt, 0, link cost, 4) (Please check the sample
code for more information). Node 0 should firstly initialize its distance table by setting all
distances to 1. After that, node 0s distance vector in the distance table is changed to (0,10,5,2),
representing the minimum distance from node 0 to all nodes in the topology. Lastly, node 0
needs to send its distance vector in a routing packet to its connected neighbors using
send2neighbor(rtpkt packet) (you can find this function in the sample code) based on the input
topology.
You then need to implement DV update for each received routing packets. Please assume that
every DV update happens in a synchronous manner while it is not true for DV in reality.
Please use this method to update DV for each node- rtupdate(struct distance table *dt, struct rtpkt
recv pkt)
This method will be called for each received routing packet at each simulation time slot k 1. It
parses the input, which is a routing packet recv pkt to update the distance table based on the DV
algorithm. If its own minimum distance to any of the other nodes is updated, the node informs its
directly connected neighbors of this change in minimum cost by sending them a routing packet
(send2neighbor(rtpkt packet)). Therefore, in the given example, nodes 0 and 1 will communicate
with each other, but nodes 1 and 3 will not.
Convergence: Please make sure that your code can run successfully and produce correct and
converged output under any topology with any N 10.
Test cases and the expected outputs:
Assume the simulation starts at slot k = 0 for initialization and k 1 when updating DVs at each
simulation slot. You need to print out the current slot k followed by DVs of all the nodes in the
ascending order at the end of the simulation slots. Please print them out in the first five slots (k =
0, 1, 2, 3, 4) and then every 10 slots (k = 10, 20, 30, ) until they converge. In this test case, the
expected out.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.next;
}
}
return false;
}
public void printList(){
Node l = first;
while(l.next!=null){
System.out.print(l.number+\" \");
l = l.next;
}
System.out.println();
}
}
Node.java (implements a single node in the linked list; stores number and pointers)
public class Node{
String number;
Node previous;
Node next;
public Node(String num){
number = num;
previous = null;
next = null;
}
public Node(String num, Node p, Node n){
number = num;
previous = p;
next = n;
}
}
Main.java (contains the main method and the helper methods to solve the questions given)
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
static boolean checkHappiness(String num){
Node current = new Node(num, null, null);
NumberList numberList = new NumberList(current);
int len = num.length();
int resultant = 0;
for (int i=0; i happyNumbersfrom1to10000(){
ArrayList numbers = new ArrayList();
for (int j=0; j<10000; j++){
if (checkHappiness(String.valueOf(j+1))){
numbers.add(String.valueOf(j+1));
}
}
return numbers;
}
static void happyNumbersfrom9001to10000(){
for (int j=9001; j<=10000; j++){
//System.out.println(j);
if (checkHappiness(String.valueOf(j))){
System.out.println(String.valueOf(j));
}
}
}
static String getLargeHappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
static String getLargeUnhappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (!checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
public static void main(String[] args){
happyNumbersfrom9001to10000();
System.out.println(happyNumbersfrom1to10000());
}
}
Solution
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.
Python programming workshop
Basic Python Programming สรุปสั้นๆ อ่านง่าย เข้าใจง่าย โดย ผศ ดร ปราโมทย์ กั่วเจริญ เหมาะกับผู้เริ่มต้น
Java Code The traditional way to deal with these in Parsers is the .pdf
Java Code: The traditional way to deal with these in Parsers is the Expression-Term-Factor
pattern:
Expression: TERM {+|- TERM}
Term: FACTOR {*|/ FACTOR}
Factor: number | ( EXPRESSION )
A couple of notation comments:
| means "or" - either + or -, either * or /
{} means an optional repeat. That's what allows this to parse 1+2+3+4
Think of these as functions. Every CAPTIALIZED word is a function call.
Consider our meme expression above: 6/2*(1+2)
We will start by looking at expression. Expression starts with a TERM. We can't do anything
until we resolve TERM, so let's go there.
A term starts with FACTOR. Again, we can't do anything until we deal with that.
A factor is a number or a (EXPRESSION). Now we can look at our token (hint:
MatchAndRemove). We see a number. OK - our factor is a number. We "return" that.
Remember that we got to factor from term. Let's substitute our number in:
TERM: FACTOR(6) {*|/ FACTOR}
Now we deal with our optional pattern. Is the next character * or /? Yes! Now is the next thing a
factor?
It turns out that it is. Let's substitute that in:
TERM: FACTOR(6) / FACTOR(2)
But remember that our pattern is a REPEATING pattern (hint: loop):
TERM: FACTOR(6) / FACTOR(2) {*|/ FACTOR}
We see the * and call factor. But this time, the factor is not a number but a parenthetical
expression.
Factor: number | ( EXPRESSION )
Factor calls expression.
Expression calls term, term calls factor, factor returns the number 1. Term doesn't see a * or / so
it passes the 1 up. Expression sees the + and calls term. Term calls factor which returns the 2.
Expression doesn't see a +|- value so ends the loop and returns 1+2.
So, remember that we were here:
TERM: FACTOR(6) / FACTOR(2) * FACTOR
Our factor is (1+2). That can't be broken down. That math HAS to be done before can multiply
or divide. That's what enforces order of operations.
In code, this looks like something like this (pseudo-code):
Node Factor()
num = matchAndRemove(NUMBER)
if (num.isPresent) return num
if (matchAndRemove(LPAREN).isPresent)
exp = Expression()
if (exp == null) throw new Exception()
if (matchAndRemove(RPAREN).isEmpty)
throw new Exception()
Node Term()
left = Factor()
do
op = MatchAndRemove(TIMES)
if (op.isEmpty) op=MatchAndRemove(DIVIDE)
if (op.isEmpty) return left
right = Factor()
left = MathOpNode(left, op, right)
while (true)
Node Expression()
left = Term()
do
op = MatchAndRemove(PLUS)
if (op.isEmpty) op=MatchAndRemove(MINUS)
if (op.isEmpty) return left
right = Term()
left = MathOpNode(left, op, right)
while (true)
What is this "MathOpNode"? It's "just" a new node type that holds two other nodes (left and
right) and an operation type: *, /, +, -. Notice that the loops use the result of one operation as the
left side of the next operation. This is called left associative - the left most part is done first.
Right associativity is the opposite - the rightmost part is done first, then we work our way left.
Generally, notice the pattern here - we call a function that is the lo.
Unit-I Pointer Data structure.pptx
Cyber security refers to every aspect of protecting an organization and its employees and assets against cyber threats. As cyberattacks become more common and sophisticated and corporate networks grow more complex, a variety of cyber security solutions are required to mitigate corporate cyber risk.
PROVIDE COMMENTS TO FELLOW STUDENTS ANSWERS AND PLEASE DON’T SAY G.docx
PROVIDE COMMENTS TO FELLOW STUDENTS ANSWERS AND PLEASE DON’T SAY GOOD WORK NICE FORMULA OR SOMETHING LIKE THAT, BUT ACTULLY HE CAN USE. THANK YOU.
Hartleys Function Code
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Chad Hartley posted Nov 5, 2015 5:10 PM
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This program will add an integer number and a decimal number up to 2 decimal places. I have included notes in the code to explain what each thing does. I hope I did this right. It compiles successfully.
PseudoCode
Start
Declare int O1; Stands for Output1
O1=sum; Sum is the functions name
Int sum()
Declare variables
Int num1;
Float num2;
Write “Enter a number.”
Scanf num1
Write”Enter a decimal number.”
Scanf num2
Return num1+num2
end
C Code
#include <stdio.h>
int sum();//prototype
int main()//calling program
{
//Declare a varaiable
int O1;
O1=sum();//main is calling sum one time.
//if I listed this twice it would run the function 'sum' twice.
// Example: if I add a new int (int O1, O2) and declare O2 to
//be O2=sum then the function would run twice.
}
int sum ()//function 'sum'
{
int num1;// Declare intergers/variables
float num2;
printf("Enter a number.\n");
scanf("%d",&num1);// Take first input and assign it to num1
printf("Enter a decimal number.\n");
scanf("%.2f",&num2);
//Can use the printf statement but when you are calling an integer you can use the return.
//printf("The sum of %d, %d, is %d", num1,num2,num1+num2);
return num1+num2;
}
ADD COMMENT HERE
Chaotic Function
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Joshua Ray posted Nov 5, 2015 2:33 PM
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float tmp
int i
function float chaos(float num)
{
for i < 20
num = 3.9*num*(1-num)
print num
}
main
print "Program description"
print "Request input btw 0 and 1"
tmp = input
chaos(tmp)
/*
* File: main.c
* Author: JaiEllRei
*
* Created on November 5, 2015, 2:04 PM
*/
#include <stdio.h>
#include <stdlib.h>
float chaos(float num);
int main(void)
{
float tmp;
printf("This program illustrates a choatic function. \n");
printf("Input a number between 0 and 1: ");
scanf("%f", &tmp);
chaos(tmp);
}
float chaos(float num)
{
for (int i=0; i<20; i++){
/*Chaotic Formula*/
num = 3.9 * num * (1-num);
printf("%.3f \n", num);
}
}
This program illustrates a choatic function.
Input a number between 0 and 1: .2
0.624
0.915
0.303
0.824
0.566
0.958
0.156
0.514
0.974
0.098
0.345
0.881
0.409
0.943
0.210
0.647
0.891
0.379
0.918
0.293
ADD COMMENT HERE
//MPH to KPH Conversion Function
Function KPHConv(value) as float
Set KPHConv = value*1.609344
End Function
Pseudocode for simple conversion program calling function
//Declare function
// MPH to KPH Conversion Function
Functio ...
Similar to A polynomial may be represented as a linked list where each node cont.pdf (20)
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Data Structures and Agorithm: DS 05 Doubly Linked List.pptx
Main issues with the following code-Im not sure if its reading the fil.pdf
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Java AssignmentUsing the ListNode.java file below Write method.pdf
Java AssignmentUsing the ListNode.java file below Write method.pdf
Please fill in the code to run the program based on the following in.pdf
Please fill in the code to run the program based on the following in.pdf
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)public class NumberLis.pdf
Java Code The traditional way to deal with these in Parsers is the .pdf
Java Code The traditional way to deal with these in Parsers is the .pdf
PROVIDE COMMENTS TO FELLOW STUDENTS ANSWERS AND PLEASE DON’T SAY G.docx
PROVIDE COMMENTS TO FELLOW STUDENTS ANSWERS AND PLEASE DON’T SAY G.docx
More from fashionfolionr
Discuss the benefits of in-built coding software within the Health I.pdf
Discuss the benefits of in-built coding software within the Health Information Managements
Systems?
Note: Mention your references
words limit: 500 words no plagiarism
Solution
Health information management (HIM) is information management applied to health and health
care. It is the practice of acquiring, analyzing and protecting digital and traditional medical
information vital to providing quality patient care. With the widespread computerization of
health records, traditional (paper-based) records are being replaced with electronic health records
(EHRs). The tools of health informatics and health information technology are continually
improving to bring greater efficiency to information management in the health care sector. Both
hospital information systems and health human resources information systems (HRHIS) are
common implementations of HIM.
Paper-based records have been in existence for centuries* and their gradual replacement by
computer-based records has been slowly underway for over twenty years in western healthcare
systems. Computerised information systems have not achieved the same degree of penetration in
healthcare as that seen in other sectors such as finance, transport and the manufacturing and retail
industries. Further, deployment has varied greatly from country to country and from speciality to
specialty and in may cases has revolved around local systems designed for local use.
In record time and at an unprecedented pace, the electronic health record (EHR) has spurred a
revolutionary change in HIM that\'s modifying the profession as each day passes.EHRs and
accompanying technology like health information exchange, computer-assisted coding, voice
recognition software, and patient portals-along with revamped government quality and EHR
incentive programs-are modifying many aspects of HIM.
An information system consists of four interrelated components—data, information technology,
process, and users.14 HIM professionals’ traditional job roles make them the experts in
managing data and processes in an information system. With the digitizing of information
systems in healthcare organizations, the roles of HIM professionals have expanded into
information technology (IT) and user support, which usually are the functions of IT supporting
services. HIM professionals’ training and experience in the intersection of clinical and
management sciences as well as their knowledge about data quality equip them with the
capability to maintain the integrity and accessibility of health information, although they may not
necessarily have the particular skills to support technical operations of a health information
system.
Electronic medical record systems lie at the center of any computerised health information
system. Without them other modern technologies such as decision support systems cannot be
effectively integrated into routine clinical workflow. The paperless, interoperable, multi-
provider, multi-specialty, multi-discipline computerised.
Determine if each system is1. memoryless2. Invertible3. Causal.pdf
Determine if each system is
1. memoryless
2. Invertible
3. Causal
4. Stable
5. Time Invariant
6. Linear y[n] = x[n]2 VIn y(t)= x(2t) y[n-x(-n] y[n] nx| n + 3 y(t)= sin(2*t x(T) dT y( t) -X(T)
dT dx(t) y(t)=-dt -5 VIn 5k tt [ nt_ot, 11 ntnnttttn
Solution
y(t) = R 2 x( )d
Taken t=2(any number)
Causality: The system is NOT causal because for any t < 2, the output depends on a future input.
Invertibility: The system is NOT invertible because only the integral and not the input function
itself determines the output, e.g. the output corresponding to input x1(t) = u(t)u(t1) is the same as
that corresponding to input x2(t) = 2[u(t) u(t 1/2)].
Memorylessness: The system is NOT memoryless because the output at time t depends on input
values at times other than t. Time Invariance: The system is NOT time invariant. To see this, we
let x(t) = u(t) u(t 1) so that y(t) = Z 2 x( )d (18) = Z 1 0 d (19) = 1, (20) i.e. constant for all t, so
in particular, y(t 3) = 1 for any value of t 2 . However, if we let x3(t) = x(t 3) = u(t 3) u(t 4) .
Then y3(t) = Z 2 x3( )d (21) = Z 2 3 x( )d (22) = 0 (23) 6= y(t 3)
Linearity: Assume t=2..any arbitary value The system is linear because if y1(t) = Z 2 x1( )d, and
(12) y2(t) = Z 2 x2( )d, and x(t) = x1(t) + x2(t), then the output y(t) corresponding to the input
x(t) is y(t) = Z 2 (x1 + x2)( )d = Z 2 x1( )d + Z 2 x2( )d = y1(t) + y2(t)
y(t) = (dx/dt)(t) ( causal , invertible, linear , memoryless , time invariant ) Causality: The system
is memoryless, hence causal. Invertibility: The system is NOT invertible because any two inputs
that differ by a constant yield the same output. Linearity: The system is linear because if y1(t) =
(dx1/dt)(t), and (25) y2(t) = (dx2/dt)(t) and x(t) = x1(t) + x2(t), then the output y(t)
corresponding to the input x(t) is y(t) = (d(x1 + x2)/dt)(t) = (dx1/dt)(t) + (dx2/dt)(t) = y1(t) +
y2(t). Memorylessness: The system is memoryless because the output at time t depends on input
values at only time t. Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa(t) = x(t a). Then the output ya(t)
corresponding to the input signal xa(t) is ya(t) = (dxa/dt)(t) = (d(x)/dt)(t a) = y(t a).
(a) y(t) = R t x( )d Invertible: z(t) = d dty(t) = d dt Z t x( ) d = x(t)
y[n] = x[n] (1) y[1] depends on x[1] so the system has memory.
Let x1[n] = x[n N], then y1[n] = x1[n] = x[n N] y[n N] = x[(n N)] = x[n + N] Hence the
system is time-varying. (3) Assume x1[n] y1[n] and x2[n] y2[n] and x3[n] = ax1[n] + bx2[n]
y3[n] = x3[n] = ax1[n] + bx2[n] = ay1[n] + by2[n] Hence the system is linear. (4) y[1] depends
on x[1] so the system is noncausal. (5) If |x[n]| Mx for all n, then |y[n]| = |x[n]| Mx for all n.
Therefore, the system is BIBO stable..
According to the Fick equation, what happens to J when X (the distanc.pdf
According to the Fick equation, what happens to J when X (the distance between the two
concentrations) is increased? The diffusion coefficient increases The net rate of diffusion
decreases The concentration increases The net rate of diffusion increases
Solution
The rate of diffusion is directly proportional to a particle\'s distance.
The net rate of diffusion increases..
A 14 foot ladder is leaning against a building. If the bottom of the.pdf
A 14 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the
pavement directly away from the building at 3 ft/sec, how fast is the top of the ladder moving
down when the foot of the ladder is 5 feet from the wall?
answer in ft/sec, rounded to the nearest thousandth
Solution
velocity along the rod will remain same at both ens thus we have
3*5/14=v*13.07/14 v=1.147 ft/sec.
Briefly discuss operating system, firmware, and utility programs..pdf
Briefly discuss operating system, firmware, and utility programs.
Solution
Operating system :An operating system is a software which performs all the basic operations of a
user such as file management, process management ,memory management, handling input and
output.
File management:
The operating system performs the following basic tasks in file management.
Process management:
The operating system performs the following basic tasks in file management.
Memory management:
The operating system performs the following basic tasks in file management.
Handling input and output:
Using drivers the o/s can manage input and o/p devices.
Firmware: Firmware is a software of hard ware i.e it is s/w program or set of instructions
programmed on a hardware device. It provides the necessary instructions for how the device
communicates with the other computer hardware.
Utility programs: utility program is a program perform the basic functions of a computer like
playing multimedia files, viewing images finding files and backing up data. Generally this
programs are installed by the user ,it can be uninstalled as per user requirements..
Assume that scores on the LSAT are approximately normally distribute.pdf
Assume that scores on the LSAT are approximately normally distributed in the population with a
mean of 545 and a standard deviation of 110. What proportion of all people who take the exam
score over 700?
A) .9207
B) .0793
C) .9500
D) .2061
E) The question cannot be answered from the information given.
Solution
Z=X-MEAN/SD
Z=(700-545)/110
Z=1.41
We take the value from the table of z
P(Z>1.41)=0.0793.
All answers must be in your own wordsWhat iswas The Great Law Wh.pdf
All answers must be in your own words
What is/was The Great Law? Who wrote it and why?
Solution
Under the Iroquois Confederacy, the Great Law of Peace formed an Oral Constitution that was
written by the Great Peacemaker, Dekanawidah in association with his spokesman, Hiawatha.
With the initial recording and transmission of the Law in wampum symbolic language, it got
gradually translated into English. The Great Law revolves around such noting laws that are
narrative in nature and also describes what types of ceremonies that are to be performed at a
given point of time. This law comprises of more than a hundred of articles which are known as
constitution. Since the language is particular the nation following the same, it has varied from its
member to member nations, in order to make the laws more comprehendible to the members.
Most of the preaching is also adopted by the US Constitution. Since this law gave importance to
the freedom of expression and believed in the concept of servant leadership, it is believed that it
inspired various significant leaders to create their roots of democracy into the system. It was
primarily formed to teach the people of the nations that the Constitution shall not tolerate any
offensive actions or criticism but shall instead believe in peace and prestige and the leaders shall
be formed to act not only as mentor but also to dedicate their lives for service and welfare of the
People. The determination should be coupled with the warmth of the people so that they accept
these leaders who shall deliberate calmly to deal with the issues faced by the nation overall..
4. An upward shift in the Reserve Bank of Australias policy reactio.pdf
4. An upward shift in the Reserve Bank of Australia\'s policy reaction function at a time when
the United States Federal Reserve did not shift its policy reaction function is likely to have which
of the following effects? a. Increase the demand for Australian financial assets and so cause the
Australiarn dollar to depreciate. b. Decrease the demand for Australian financial assets and so
cause the Australian dollar to appreciate. c. Decrease the demand for Australian financial assets
and so cause the Australian dollar to depreciate d. Increase the demand for Australian financial
assets and so cause the Australiarn dollar to appreciate.
Solution
An upward shift in the reaction function will increase the interest rate of Australian financial
assets because the rate of return from these assets will rise.This will result in capital inflows.This
will increase the demand of Australian currency and thus, Australian currency would appreciate.
Answer-d.
Write a function prototype for each of the following functions (Jus.pdf
Write a function prototype for each of the following functions: (Just the line, not the body.)
1.) function area that takes two values of type in as parameters, called side1 and side2 and
returns a value of type int as its results.
2.)function showMenu that does not receive any parameters and does not return a value.
3.)function convert that takes a float parameter, called number, and returns a value of type char
as its results.
Solution
1.) function area that takes two values of type in as parameters, called side1 and side2 and
returns a value of type int as its results
sol:int area(int side1,int side2);
2)function showMenu that does not receive any parameters and does not return a value.
sol:void showMenu();
3)function convert that takes a float parameter, called number, and returns a value of type char as
its results.
sol:char convert(float number);.
Why is white collar crime is so difficult to address with law enforc.pdf
Why is white collar crime is so difficult to address with law enforcement and regulatory efforts?
Solution
The reasons why white collar crime is so difficult to address with law enforcement and
regulatory efforts is mentioned below:
- White collar crimes are seen as non threatening. People who commit the crimes are citizens of
respect in society.
- They are also non-violent crimes with difficult to identify victims. Thus, the law and regulatory
authorities perceives it as a mere bad judgement on part of some individuals.
- White collar crimes do not draw an intense response from the public as well as the authorities.
They are not seen as priority.
- Due to lack of personnel and resources, white collar crimes are just not given the importance
they deserve..
What is Eulers number and who is it named after Can you name some.pdf
What is Euler\'s number and who is it named after? Can you name some uses of Euler\'s number?
Also, list some formulas containing Euler\'s number?
Solution
e is known as Euler\'s number and is named after Leonhard euler .
Formula\'s containing euler\'s number
ei pi+1=0
ei theta=cos theta + i sin theta.
Using C++...Hints- Use binary file readwrite to store employee .pdf
Using C++...
Hints
- Use binary file read/write to store employee structures effeciently
- You can find the size of struct Employee using sizeof(Employee).
- Structs can be written out to a file using write() function.
- write() function takes address to a memory and the number of bytes to be written to the file.
It may be easier to get first and last name using std::string as follows:
- You can easily convert a char[] to a string as follows
- It is possble to get a const char[] from an std::string using c_str() function. Step 1 Create a
struct that can store employee id, wages and first and last names. To keep things simple, we
assume that employee names are stored in charll, and we only store the first 16 characters of the
name. See below. struct Employee char fname [16] char Iname [16] int id double wages Step 2
write a program that adds an employee to the employees at file. add-employee Enter firstname
Captain Enter lastname: America. Enter id: 2 Enter wage: 5.00 This program will append this
record to the employees dat Step 3 write a program that prints the info of the employee with the
passed employee id. s prn-employee 2 America, Captain id wage $5.00
Solution
as C++ is known as powerful lanuage because of classess (superior in comparison with struct) so
class employee is created.
#include
class Employee
{
int id;
char fname[16];
char lname[16];
double wages;
public:
//To display record based on employee id
void diplay(int );
//To append the data of employee
void append();
};
void Empoyee::append()
{
fstream File;
File.open( “EMP.DAT”, ios::binary | ios::in | ios::out);
cin>>id;
cin.getline(fname, 20);
cin.getline(lname, 20);
cin>>wages;
File.write( (char*)this, sizeof(Employee));
File.close();
}
void Employee::Search(int data)
{
fstream File;
File.open( “EMP.DAT”, ios::binary | ios::in);
File.read((char*)this, sizeof(Employee));
cout << id<< ”==> ” << Employee<< endl;
File.close();
}
Still if you need structure , here is the code
#include
#include
using namespace std;
typedef struct employee
{
char fname[16];
char lname[16];
int id;
double wages
}emp;
int main()
{
//values are hard coded
emp apprentice[3];
strcpy(apprentice[0].fname, \"john\");
strcpy(apprentice[0].fname, \"sina\");
apprentice[0].id = 21;
apprentice[0].wages=2000;
strcpy(apprentice[1].fname, \"under\");
strcpy(apprentice[1].fname, \"taker\");
apprentice[1].id = 20;
apprentice[1].wages=3000;
strcpy(apprentice[2].fname, \"under\");
strcpy(apprentice[2].fname, \"taker\");
apprentice[2].id = 20;
apprentice[2].wages=3000;
// Serializing struct to employee.data
ofstream output_file(\"employee.data\", ios::binary);
output_file.write((char*)&apprentice, sizeof(apprentice));
output_file.close();
// Reading from it
ifstream input_file(\"employee.data\", ios::binary);
emp master[3];
input_file.read((char*)&master, sizeof(master));
for (emp e = 0; e < 3; e++)
{
cout << \"Id #\" << e << endl;
cout << \"fname: \" << master[e].fname << endl;
cout << \"lname: \" << master[e].lname <<.
Unless specifically legally permitted, agreements to suppress or eli.pdf
Unless specifically legally permitted, agreements to suppress or eliminate completion are illegal
and unenforceable, however, there are monopolies that exist, and thrive. Some examples may be
a patented medical device, the USPS, MLB, NFL, and public utilities, write a long paragraph on
a “legal monopoly.” All sources should be cited
Solution
A legal monopoly is set up in the beginning as a perceived best option for both government and
its citizens. For example, AT&T operated as a legal monopoly until 1982 because it was deemed
vital to have cheap and reliable service for everyone. Railroads and airlines have also been
operated as legal monopolies at different periods in history.
A legal monopoly is a company that is operating as a monopoly under a government mandate. A
legal monopoly offers a specific product or service at a regulated price and can either be
independently run and government regulated, or government run and regulated.
A legal monopoly is also known as a \"statutory monopoly.
A legal monopoly is set up in the beginning as a perceived best option for both government and
its citizens. For example, AT&T operated as a legal monopoly until 1982 because it was deemed
vital to have cheap and reliable service for everyone. Railroads and airlines have also been
operated as legal monopolies at different periods in history.
The prevailing idea behind instituting legal monopolies is that too many competitors investing in
their own delivery infrastructure would mean that prices across the board would be too high.
While this idea proves true, it\'s not true indefinitely. In most cases, capitalism has eventually
wins out over legal monopolies as technology and the economy become more advanced, which
levels the playing field for the industry, while slashing costs and barriers to entry. When the
doors open to competition, it always proves to be in the best interest of consumers.
Examples of Legal Monopolies
Throughout history, various governments have imposed legal monopolies on a variety of
commodities, including salt, iron and tobacco. The Statute of Monopolies of 1623 was an early
step in an English movement to convert letters patent from a method of rewarding royal favorites
at other than royal expense, to a method of encouraging inventors.
The Dutch East India Company, British East India Company and similar national trading
companies were granted exclusive trade rights by their respective national governments. Private
freelance traders operating outside the scope of those two companies were subject to criminal
penalties, and those companies fought wars in the 17th century to delineate and defend their
monopoly territories.
Legal monopolies on alcohol remain fairly common, both as a source of public revenue and as a
means of control. Meanwhile, monopolies on opium and cocaine, formerly important for
revenue, were converted or reinstituted during the twentieth century to curb the abuse of
controlled substances. For example, Mallinckrodt Inco.
There are tremendous pressures on selected industrialized countries,.pdf
There are tremendous pressures on selected industrialized countries, as well as MNCs and NGOs
to be more socially and environmentally responsive to world problems.
Do governments and companies in developed countries have an ethical responsibility to
contribute to economic growth and social development in developing countries?
Luthans, F. & Doh, J. P. (2018). International Management Culture, strategy and behavior (10th
ed.). Ethics, Social Resposibility and Sustainability
There are tremendous pressures on selected industrialized countries, as well as MNCs and
NGOs to be more socially and environmentally responsive to world problems.
Do governments and companies in developed countries have an ethical responsibility to
contribute to economic growth and social development in developing countries?
Luthans, F. & Doh, J. P. (2018). International Management Culture, strategy and behavior (10th
ed.).
Solution
Developed countries are those countries which have high level of economic growth and GDP.and
they are independent and developing nations are those which have poor infrastructure who look
towards someone for thier help .therefore it\'s a ethical responsibility of developed nations
companies and goverment to put forward thier hands towards developing nations becuase people
in developin nation really looking for help rhey are one who do not get two time proper meal
where as in develpoed countires people throw thier excess food in dustbin,which is incorrect.
inspite of wasting excess money its better to invest in grwoth and development of developing
countries..
This week we covered Accounting and Payroll liabilities and the diff.pdf
This week we covered Accounting and Payroll liabilities and the different classifications that fall
under them. We also deductions, register, and earning records. State the definition of liabilities.
Describe the difference between accounting and payroll liabilities. Discuss the implications of
not adhering to the regulations of payroll liabilities.
Solution
Liabilites are the financial obligations which arise over the course of buisness proceeding to be
paid in foreseable future by the company.
Accounting liabilities include loans, accounts payable, mortgages, deferred revenues and
accrued expenses which are created because of business transactions
Payroll liabilities include items which are deducted from the payroll of emplpoyees and typically
include taxes withheld from employees, such as federal and state income tax, Social Security tax,
401k retirement fund contributions, , health insurance contributions Medicareunion dues, and
payroll garnishments
Failure to deposit payroll taxes can lead to action agains the company as well the person
responsible for paying those money into the trust of the government for these purpose. Penalty is
also imposed as follows:
Days late Penalty
1 to 5 days 2%
6 to 15 days 5%
16 or more days 10%
Also applies to deposits made within
10 days of the date of the first IRS
notice requesting the tax.
Amounts unpaid more than 10 days after the date of the first IRS notice requesting the tax, or the
day when you received notice and demand for immediate payment, whichever is earlier would be
taxed penalised at 15%.
t Assignment 2 Econ 201 Importance of scarcity. SolutionScarc.pdf
t: Assignment 2 Econ 201: Importance of scarcity.
Solution
Scarcity refers to the problem when the unlimited human wants are not fulfilled with the limited
resources in the world. This leads us to the question of how these resources need to be used. This
leads human beings to choosing to allocate the limited resources for their unlimited wants. This
also brings us closer to the concept of opportunity cost where we choose one alternative over the
other ie. forgo the other option. Scarcity is an important concept because had there been no
scarcity then all the unlimited wants would have been fulfilled by the unlimited resources.
Economics brings the concept of scarcity to understand the purpose of allocation of resources
judiciously and making the most of the available resources. Scarcity basically brings up the
concept of giving up an alternative and choosing the other one. It leads to trade off or
opportunity cost..
Stark Company has five employees. Employees paid by the hour receive.pdf
Stark Company has five employees. Employees paid by the hour receive a $12 per hour pay rate
for the regular 40-hour workweek plus one and one-half times the hourly rate for each overtime
hour beyond the 40 hours per week. Hourly employees are paid every two weeks, but salaried
employees are paid monthly on the last biweekly payday of each month. FICA Social Security
taxes are 6.2% of the first $118,500 paid to each employee, and FICA Medicare taxes are 1.45%
of gross pay. FUTA taxes are 0.6% and SUTA taxes are 5.4% of the first $7,000 paid to each
employee. The company has a benefits plan that includes medical insurance, life insurance, and
retirement funding for employees. Under this plan, employees must contribute 5 percent of their
gross income as a payroll withholding, which the company matches with double the amount.
Following is the partially completed payroll register for the biweekly period ending August 31,
which is the last payday of August.
Solution
Some workers are entitled to acquire extra time pay once they work additional hours. Eligibility
is centered on weekly profits and hours worked. Extra time pay is calculated situated on a forty-
hour workweek and overtime pay for eligible employees is required for any hours labored over
those 40 hours. Time beyond regulation pay rules are governed by way of the reasonable Labor
necessities Act (FLSA) and, in some states, by way of state overtime laws.
In states the place an employee is subject to each state and federal additional time laws,
additional time is paid according to the commonplace to be able to furnish the higher amount of
pay. Verify your State division of Labor website for knowledge on time beyond regulation pay
standards to your vicinity. Employers have got to comply with each federal and state law to stay
compliant.
What is time beyond regulation Pay?
Staff currently earning less than $455 per week, which is $23,660 per yr, are presently
guaranteed federal overtime safety. There are exemptions for tremendously compensated
employees who normally and most often perform any person or more of the exempt tasks or
tasks of an government, administrative, or reputable worker.
In step with the division of Labor, workers blanketed by way of the Act must acquire extra time
pay for hours labored in way over forty in a workweek at a cost now not not up to time and one-
half their common charges of pay.
Double Time Pay
Double time is a rate of pay double the average amount a person receives for natural hours
worked. So, in case your common price of pay was $eleven.00 an hour, double time pay could be
$22.00 per hour. Double time is normally paid for engaged on federal holidays or when hours
work exceeding the common workday.
When Double Time is Paid
There are no federal laws that require an enterprise to pay double time for extra time worked.
The reasonable Labor requirements Act (FLSA) has no requirement for double time pay.
Nevertheless, state legal guidelines may just furnish for .
Silas, Cole, and Addison are on the same Learning Team at UOP. If ea.pdf
Silas, Cole, and Addison are on the same Learning Team at UOP. If each worked on the team
assignments alone, it would take Silas 10 hours, Cole 8 hours, and Addison 12 hours to finish the
assignments. They usually started the same time on Monday night after work. But last week,
Silas started early and he had worked for 2 hours on the assignments when Cole and Addison
began. Later, all three finished the team assignments. How long did they take to finish by
working together?
Solution
In one hour Silas would finish 1/10th work
In two hours Silas would finish 2/10th work
Remaining (1-2/10) = 8/10th work is finished by all of them
If all of them started at same time the would finish the entire work in
1/(1/10+1/8+1/12)
=1/(80+120+96)/960
=960/296days
0.8 work they finish in
960*0.8/296 days
768/296
In one hour Silas would finish 1/10th work
In two hours Silas would finish 2/10th work
Remaining (1-2/10) = 8/10th work is finished by all of them
If all of them started at same time the would finish the entire work in
1/(1/10+1/8+1/12)
=1/(80+120+96)/960
=960/296days
0.8 work they finish in
960*0.8/296 days
768/296.
Q1 Discuss three types of accounting anomalies and define why they .pdf
Q1: Discuss three types of accounting anomalies and define why they are important to a fraud
investigation and why the fraud examiner should be mindful of these anomalies?
Q2: Define circumstantial evidence and describe why this type of evidence is so useful to a
forensic accountant during an investigation.
Solution.
QUESTION 2727. A profit center is the same as a revenue center.T.pdf
QUESTION 27
27. A profit center is the same as a revenue center.
True
False
QUESTION 1
1. Step costs relate to the range of activity for which cost behavior assumptions are reasonably
valid.
True
False
Solution
Revenue center are subunits of a business responsible for generating revenues whereas profit
centers are responsible for managing costs in addition to generaring revenue.
Hence, the answer is false.
Pleass raise a different question for the second question.
Thanks..
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A polynomial may be represented as a linked list where each node cont.pdf
- 1. A polynomial may be represented as a linked list where each node contains the coefficient and exponent of a term of the polynomial. The polynomial 4 X^3 - 3 X^2 - 5 would be represented as the linked list. Write a program system that reads two polynomials, stores them as linked lists, adds them together, and prints the result as a polynomial. The result should be a third linked list. Travers both polynomials. If a particular exponent value is present in cither of the two polynomials being summed, then it should be present in the answer. If it is present in both polynomials, then its coefficient is the sum of the corresponding coefficient in both polynomials. (If this sum is zero, the term should be deleted). Solution #include #include typedef struct link { int coeff; int pow; struct link * next; } my_poly; void my_create_poly(my_poly **); void my_show_poly(my_poly *); void my_add_poly(my_poly **, my_poly *, my_poly *); int main(void) { int ch; do { my_poly * poly1, * poly2, * poly3; printf(" Create 1st expression "); my_create_poly(&poly1); printf(" Stored the 1st expression"); my_show_poly(poly1);
- 2. printf(" Create 2nd expression "); my_create_poly(&poly2); printf(" Stored the 2nd expression"); my_show_poly(poly2); my_add_poly(&poly3, poly1, poly2); my_show_poly(poly3); printf(" Add two more expressions? (Y = 1/N = 0): "); scanf("%d", &ch); } while (ch); return 0; } void my_create_poly(my_poly ** node) { int flag; int coeff, pow; my_poly * tmp_node; tmp_node = (my_poly *) malloc(sizeof(my_poly)); *node = tmp_node; do { printf(" Enter Coeff:"); scanf("%d", &coeff); tmp_node->coeff = coeff; printf(" Enter Pow:"); scanf("%d", &pow); tmp_node->pow = pow; tmp_node->next = NULL;
- 3. printf(" Continue adding more terms to the polynomial list?(Y = 1/N = 0): "); scanf("%d", &flag); //printf(" FLAG: %c ", flag); if(flag) { tmp_node->next = (my_poly *) malloc(sizeof(my_poly)); //Grow the list tmp_node = tmp_node->next; tmp_node->next = NULL; } } while (flag); } void my_show_poly(my_poly * node) { printf(" The polynomial expression is: "); while(node != NULL) { printf("%dx^%d", node->coeff, node->pow); node = node->next; if(node != NULL) printf(" + "); } } void my_add_poly(my_poly ** result, my_poly * poly1, my_poly * poly2) { my_poly * tmp_node; tmp_node = (my_poly *) malloc(sizeof(my_poly)); tmp_node->next = NULL; *result = tmp_node; while(poly1 && poly2) {
- 4. if (poly1->pow > poly2->pow) { tmp_node->pow = poly1->pow; tmp_node->coeff = poly1->coeff; poly1 = poly1->next; } else if (poly1->pow < poly2->pow) { tmp_node->pow = poly2->pow; tmp_node->coeff = poly2->coeff; poly2 = poly2->next; } else { tmp_node->pow = poly1->pow; tmp_node->coeff = poly1->coeff + poly2->coeff; poly1 = poly1->next; poly2 = poly2->next; } if(poly1 && poly2) { tmp_node->next = (my_poly *) malloc(sizeof(my_poly)); tmp_node = tmp_node->next; tmp_node->next = NULL; } } while(poly1 || poly2) { tmp_node->next = (my_poly *) malloc(sizeof(my_poly)); tmp_node = tmp_node->next;
- 5. tmp_node->next = NULL; if(poly1) { tmp_node->pow = poly1->pow; tmp_node->coeff = poly1->coeff; poly1 = poly1->next; } if(poly2) { tmp_node->pow = poly2->pow; tmp_node->coeff = poly2->coeff; poly2 = poly2->next; } } printf(" Addition Complete"); } output: Create 1st expression Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff:
- 6. Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0): Enter Coeff: Enter Pow: Continue adding more terms to the polynomial list?(Y = 1/N = 0):