I am trying to implement timing on this program and cannot do it. Wh.pdf
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I am trying to implement timing on this program and cannot do it. Where am I wrong at? #include "stdio.h" #include "stdlib.h" #include "time.h" void swap(int *x, int *y) { if (x == y) return; *x ^= *y; *y ^= *x; *x ^= *y; } void cocktailSort(int *a, size_t n) { while (1) { char flag; size_t start[2] = {1, n - 1}, end[2] = {n, 0}, inc[2] = {1, -1}; for (int it = 0; it < 2; it++) { flag = 1; for (int i = start[it]; i != end[it]; i += inc[it]) if (a[i - 1] > a[i]) { swap(a + i - 1, a + i); flag = 0; } if (flag) return; } } } int main() { int array[] = {5, -1, 101, -4, 0, 1, 8, 6, 2, 3}; size_t n = sizeof(array) / sizeof(array[0]); printf("Original Array: \n"); for (int i = 0; i < n; i++) printf("%d%s", array[i], i == n - 1 ? " \n" : " "); printf("\nSorted Array: \n"); cocktailSort(array, n); for (int i = 0; i < n; ++i) printf("%d ", array[i]); return 0; } void timeAllSorts() { printf("All Times in milliseconds.\n"); printf("| %7s | %10s |\n", "Size", "Cocktail"); int size = 8; while (size <= 131072) { double c = timeASingleSort(cocktailSort, size); printf("| %7d | %10.4f |\n", size, c); size = size * 2; } } double timeASingleSort(void (*func)(int *, int), int size) { int *T = malloc(size * sizeof(int)); for (int i = 0; i < size; i++) { T[i] = rand() % (size * 4); } double before = (double)clock(); func(T, size); double after = (double)clock(); free(T); return (after - before) / (CLOCKS_PER_SEC / 1000); }.
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![I am trying to implement timing on this program and cannot do it. Where am I wrong at?
#include "stdio.h"
#include "stdlib.h"
#include "time.h"
void swap(int *x, int *y) {
if (x == y)
return;
*x ^= *y;
*y ^= *x;
*x ^= *y;
}
void cocktailSort(int *a, size_t n) {
while (1) {
char flag;
size_t start[2] = {1, n - 1}, end[2] = {n, 0}, inc[2] = {1, -1};
for (int it = 0; it < 2; it++) {
flag = 1;
for (int i = start[it]; i != end[it]; i += inc[it])
if (a[i - 1] > a[i]) {
swap(a + i - 1, a + i);
flag = 0;
}
if (flag)
return;
}
}
}
int main() {
int array[] = {5, -1, 101, -4, 0, 1, 8, 6, 2, 3};
size_t n = sizeof(array) / sizeof(array[0]);](https://image.slidesharecdn.com/iamtryingtoimplementtimingonthisprogramandcannotdoit-230330055211-591a886a/85/I-am-trying-to-implement-timing-on-this-program-and-cannot-do-it-Wh-pdf-1-320.jpg)
![printf("Original Array: n");
for (int i = 0; i < n; i++)
printf("%d%s", array[i], i == n - 1 ? " n" : " ");
printf("nSorted Array: n");
cocktailSort(array, n);
for (int i = 0; i < n; ++i)
printf("%d ", array[i]);
return 0;
}
void timeAllSorts() {
printf("All Times in milliseconds.n");
printf("| %7s | %10s |n", "Size", "Cocktail");
int size = 8;
while (size <= 131072) {
double c = timeASingleSort(cocktailSort, size);
printf("| %7d | %10.4f |n", size, c);
size = size * 2;
}
}
double timeASingleSort(void (*func)(int *, int), int size) {
int *T = malloc(size * sizeof(int));
for (int i = 0; i < size; i++) {
T[i] = rand() % (size * 4);
}
double before = (double)clock();
func(T, size);
double after = (double)clock();](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Hello, I need some assistance in writing a java program THAT MUST US.pdf
Hello, I need some assistance in writing a java program THAT MUST USE STACKS (LIFO).
Create a Calculator w/ GUI
Write a program that graphically displays a working calculator for simple infix expressions that
consist of: single-digit operands, the operators: +, -, *, and /, and parentheses.
Make the following assumptions:
unary operators (e.g. -2) are illegal
all operations, including division, are integer operations (and results are integers)
the input expression contains no embedded spaces and no illegal characters
the input expression is a syntactically correct infix expression
division by zero will not occur (consider how you can remove this restriction)
Create a GUI application, the calculator has a display and a keypad of 20 keys, which are
arranged as follows:
C
<
Q
/
7
8
9
*
4
5
6
-
1
2
3
+
0
(
)
=
As the user presses keys to enter an infix expression, the corresponding characters appear in the
display. The C (Clear) key erases all input entered so far; the < (Backspace) key erases the last
character entered. When the user presses the = key, the expression is evaluated and the result
appended to the right end of the expression in the display window. The user can then press C and
enter another expression. If the user presses the Q (Quit) key, the calculator ceases operation and
is erased from the screen.
C
<
Q
/
7
8
9
*
4
5
6
-
1
2
3
+
0
(
)
=
Solution
import java.awt.*;
import javax.swing.*;
import java.awt.event.*;
public class Calculator extends JFrame implements ActionListener {
JPanel[] row = new JPanel[5];
JButton[] button = new JButton[19];
String[] buttonString = {\"7\", \"8\", \"9\", \"+\",
\"4\", \"5\", \"6\", \"-\",
\"1\", \"2\", \"3\", \"*\",
\".\", \"/\", \"C\", \"\",
\"+/-\", \"=\", \"0\"};
int[] dimW = {300,45,100,90};
int[] dimH = {35, 40};
Dimension displayDimension = new Dimension(dimW[0], dimH[0]);
Dimension regularDimension = new Dimension(dimW[1], dimH[1]);
Dimension rColumnDimension = new Dimension(dimW[2], dimH[1]);
Dimension zeroButDimension = new Dimension(dimW[3], dimH[1]);
boolean[] function = new boolean[4];
double[] temporary = {0, 0};
JTextArea display = new JTextArea(1,20);
Font font = new Font(\"Times new Roman\", Font.BOLD, 14);
Calculator() {
super(\"Calculator\");
setDesign();
setSize(380, 250);
setResizable(false);
setDefaultCloseOperation(EXIT_ON_CLOSE);
GridLayout grid = new GridLayout(5,5);
setLayout(grid);
for(int i = 0; i < 4; i++)
function[i] = false;
FlowLayout f1 = new FlowLayout(FlowLayout.CENTER);
FlowLayout f2 = new FlowLayout(FlowLayout.CENTER,1,1);
for(int i = 0; i < 5; i++)
row[i] = new JPanel();
row[0].setLayout(f1);
for(int i = 1; i < 5; i++)
row[i].setLayout(f2);
for(int i = 0; i < 19; i++) {
button[i] = new JButton();
button[i].setText(buttonString[i]);
button[i].setFont(font);
button[i].addActionListener(this);
}
display.setFont(font);
display.setEditable(false);
display.setComponentOrientation(ComponentOrientation.RIGHT_TO_LEFT);
display.setPreferredSize(displayDimension);
for(int.
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Hello, I need some assistance in writing a java program THAT MUST US.pdf
Hello, I need some assistance in writing a java program THAT MUST USE STACKS (LIFO).
Create a Calculator w/ GUI
Write a program that graphically displays a working calculator for simple infix expressions that
consist of: single-digit operands, the operators: +, -, *, and /, and parentheses.
Make the following assumptions:
unary operators (e.g. -2) are illegal
all operations, including division, are integer operations (and results are integers)
the input expression contains no embedded spaces and no illegal characters
the input expression is a syntactically correct infix expression
division by zero will not occur (consider how you can remove this restriction)
Create a GUI application, the calculator has a display and a keypad of 20 keys, which are
arranged as follows:
C
<
Q
/
7
8
9
*
4
5
6
-
1
2
3
+
0
(
)
=
As the user presses keys to enter an infix expression, the corresponding characters appear in the
display. The C (Clear) key erases all input entered so far; the < (Backspace) key erases the last
character entered. When the user presses the = key, the expression is evaluated and the result
appended to the right end of the expression in the display window. The user can then press C and
enter another expression. If the user presses the Q (Quit) key, the calculator ceases operation and
is erased from the screen.
C
<
Q
/
7
8
9
*
4
5
6
-
1
2
3
+
0
(
)
=
Solution
import java.awt.*;
import javax.swing.*;
import java.awt.event.*;
public class Calculator extends JFrame implements ActionListener {
JPanel[] row = new JPanel[5];
JButton[] button = new JButton[19];
String[] buttonString = {\"7\", \"8\", \"9\", \"+\",
\"4\", \"5\", \"6\", \"-\",
\"1\", \"2\", \"3\", \"*\",
\".\", \"/\", \"C\", \"\",
\"+/-\", \"=\", \"0\"};
int[] dimW = {300,45,100,90};
int[] dimH = {35, 40};
Dimension displayDimension = new Dimension(dimW[0], dimH[0]);
Dimension regularDimension = new Dimension(dimW[1], dimH[1]);
Dimension rColumnDimension = new Dimension(dimW[2], dimH[1]);
Dimension zeroButDimension = new Dimension(dimW[3], dimH[1]);
boolean[] function = new boolean[4];
double[] temporary = {0, 0};
JTextArea display = new JTextArea(1,20);
Font font = new Font(\"Times new Roman\", Font.BOLD, 14);
Calculator() {
super(\"Calculator\");
setDesign();
setSize(380, 250);
setResizable(false);
setDefaultCloseOperation(EXIT_ON_CLOSE);
GridLayout grid = new GridLayout(5,5);
setLayout(grid);
for(int i = 0; i < 4; i++)
function[i] = false;
FlowLayout f1 = new FlowLayout(FlowLayout.CENTER);
FlowLayout f2 = new FlowLayout(FlowLayout.CENTER,1,1);
for(int i = 0; i < 5; i++)
row[i] = new JPanel();
row[0].setLayout(f1);
for(int i = 1; i < 5; i++)
row[i].setLayout(f2);
for(int i = 0; i < 19; i++) {
button[i] = new JButton();
button[i].setText(buttonString[i]);
button[i].setFont(font);
button[i].addActionListener(this);
}
display.setFont(font);
display.setEditable(false);
display.setComponentOrientation(ComponentOrientation.RIGHT_TO_LEFT);
display.setPreferredSize(displayDimension);
for(int.
how can i build this as problem 2 of my code #include iostream.pdf
how can i build this as problem 2 of my code?
#include
#include
#include
#include
#include "./freeglut-3.4.0/include/GL/freeglut.h"
using namespace std;
bool leftDown = false, rightDown = false;
int lastPos[2];
float cameraPos[4] = { 0,1,4,1 };
int windowWidth = 640, windowHeight = 480;
double xRot = 0;
double yRot = 0;
int curProblem = 1; // TODO: change this number to try different examples
float specular[] = { 1.0, 1.0, 1.0, 1.0 };
float shininess[] = { 50.0 };
void problem1() {
// TODO
}
void problem2() {
//FIX HERE
}
void problem3() {
//TO DO
}
void problem4() {
//TO DO
}
void display()
{
glClearColor(0, 0, 0, 0);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glDisable(GL_LIGHTING);
glEnable(GL_DEPTH_TEST);
glBegin(GL_LINES);
glColor3f(1, 0, 0); glVertex3f(0, 0, 0); glVertex3f(1, 0, 0); // x axis
glColor3f(0, 1, 0); glVertex3f(0, 0, 0); glVertex3f(0, 1, 0); // y axis
glColor3f(0, 0, 1); glVertex3f(0, 0, 0); glVertex3f(0, 0, 1); // z axis
glEnd(/*GL_LINES*/);
glEnable(GL_LIGHTING);
glShadeModel(GL_SMOOTH);
glMaterialfv(GL_FRONT, GL_SPECULAR, specular);
glMaterialfv(GL_FRONT, GL_SHININESS, shininess);
glEnable(GL_LIGHT0);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glViewport(0, 0, windowWidth, windowHeight);
float ratio = (float)windowWidth / (float)windowHeight;
gluPerspective(50, ratio, 1, 1000);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt(cameraPos[0], cameraPos[1], cameraPos[2], 0, 0, 0, 0, 1, 0);
glLightfv(GL_LIGHT0, GL_POSITION, cameraPos);
glRotatef(xRot, 1, 0, 0);
glRotatef(yRot, 0, 1, 0);
if (curProblem == 1) problem1();
if (curProblem == 2) problem2();
if (curProblem == 3) problem3();
if (curProblem == 4) problem4();
glutSwapBuffers();
}
void mouse(int button, int state, int x, int y)
{
if (button == GLUT_LEFT_BUTTON) leftDown = (state == GLUT_DOWN);
else if (button == GLUT_RIGHT_BUTTON) rightDown = (state == GLUT_DOWN);
lastPos[0] = x;
lastPos[1] = y;
}
void mouseMoved(int x, int y)
{
if (leftDown) {
xRot += (y - lastPos[1]) * .1;
yRot += (x - lastPos[0]) * .1;
}
if (rightDown) {
for (int i = 0; i < 3; i++)
cameraPos[i] *= pow(1.1, (y - lastPos[1]) * .1);
}
lastPos[0] = x;
lastPos[1] = y;
glutPostRedisplay();
}
void keyboard(unsigned char key, int x, int y)
{
curProblem = key - '0';
if (key == 'q' || key == 'Q' || key == 27) {
exit(0);
}
glutPostRedisplay();
}
void reshape(int width, int height)
{
windowWidth = width;
windowHeight = height;
glutPostRedisplay();
}
int main(int argc, char** argv)
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGB | GLUT_DEPTH);
glutInitWindowSize(windowWidth, windowHeight);
glutCreateWindow("HW2");
glutDisplayFunc(display);
glutMotionFunc(mouseMoved);
glutMouseFunc(mouse);
glutReshapeFunc(reshape);
glutKeyboardFunc(keyboard);
glutMainLoop();
return 0;
}.
#include stdio.h #include stdlib.h #include unistd.h .pdf
#include
#include
#include
#include
#include
#include \"msecond.h\"
#include \"random_int.h\"
float total_time_spent_waiting = 0.0;
int total_number_of_meals = 0;
#define semaphore_create(s,v) sem_init( &s, 0, v )
#define semaphore_wait(s) sem_wait( &s )
#define semaphore_signal(s) sem_post( &s )
#define semaphore_release(s) sem_destroy( &s )
typedef sem_t semaphore;
semaphore chopstick[NUM_PHILOSOPHERS];
semaphore screen;
semaphore mutex;
int screen_row[NUM_PHILOSOPHERS] = { 6, 2, 2, 6, 10 };
int screen_col[NUM_PHILOSOPHERS] = { 31, 36, 44, 49, 40 };
int chopstick_row[5] = { 9, 4, 3, 4, 9 };
int chopstick_col[5] = { 35, 33, 40, 47, 45 };
char chopstick_sym[5] = { \'/\', \'\\\\\', \'|\', \'/\', \'\\\\\' };
for ( i = 0; i < NUM_PHILOSOPHERS; i++ )
{
position( screen_row[i], screen_col[i] );
printf( \"%d\", i );
position( screen_row[i] + 1, screen_col[i] - 1 );
printf( \"(%d)\", 0 );
position( chopstick_row[i], chopstick_col[i] );
printf( \"%c\", chopstick_sym[i] );
}
position_flush( 13, 1 );
}
void draw_thinking( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_hungry( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_eating( int n, int eat_count )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position( screen_row[n] + 1, screen_col[n] - 1 );
printf( \"(%d)\", eat_count );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_up( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
printf( \"%c\", \' \' );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_down( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_done( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
printf( \"D\" );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void chopsticks( int n )
{
if ( n % 2 == 0 ) {
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
} else {
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
}
}
void release_chopsticks( int n )
{
draw_chopstick_down( n );
semaphore_signal( chopstick[n] );
draw_chopstick_down( (n+1) % NUM_PHILOSOPHERS );
semaphore_signal( chopstick[(n+1) % NUM_PHILOSOPHERS] );
}
void philosopher( int *philosopher_data )
{
int start_time;
int eat_count = 0;
int total_hungry_time = 0;
int became_hungry_time;
int n = philosopher_data[0];
int duration = philosopher_data[1];
start_time = msecond();
while( msecond() - start_time < duration * 1000 )
{
became_hungry_time = msecond();
draw_hungry( n );
chopsticks( n );
t.
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Please help with this JAVA Assignment and show output if you can ple.pdf
Please help with this JAVA Assignment and show output if you can please
Complete required scripts based on eclipse and troubleshoot the scripts until the tasks are done.
1. Implement linear-sort, merge-sort and quick-sort (write the code for them).
Solution
Linear or Insertion Sort
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.Scanner;
class LinearSort
{
public static void sort( int values[] )
{
int N = values.length;
int i, j, temp;
for (i = 1; i< N; i++)
{
j = i;
temp = values[i];
while (j > 0 && temp < values[j-1])
{
values[j] = values[j-1];
j = j-1;
}
values[j] = temp;
}
}
public static void main(String[] args)
{
Scanner scan = new Scanner( System.in );
int n, i;
System.out.println(\"Enter number of elements\");
n = scan.nextInt();
int values[] = new int[n];
System.out.println(\"\ Enter elements\");
for (i = 0; i < n; i++)
values[i] = scan.nextInt();
sort(values);
System.out.println(\"\ Sorted Elements\");
for (i = 0; i < n; i++)
System.out.print(values[i]+\" \");
System.out.println();
}
}
Output:
Enter number of elements
6
Enter elements
34 56 12 8 9 23
Sorted Elements
8 9 12 23 34 56
Merge Sort
import java.util.Scanner;
public class Mergesort
{
public static void mergesort(int[] a, int low, int high)
{
int N = high - low;
if (N <= 1)
return;
int mid = low + N/2;
mergesort(a, low, mid);
mergesort(a, mid, high);
int[] temp = new int[N];
int i = low, j = mid;
for (int k = 0; k < N; k++)
{
if (i == mid)
temp[k] = a[j++];
else if (j == high)
temp[k] = a[i++];
else if (a[j] pivot)
j--;
if (i <= j)
{
temp = values[i];
values[i] = values[j];
values[j] = temp;
i++;
j--;
}
}
if (low < j)
quickquicksort(values, low, j);
if (i < high)
quickquicksort(values, i, high);
}
public static void main(String[] args)
{
Scanner scan = new Scanner( System.in );
int n, i;
System.out.println(\"Enter number of elements\");
n = scan.nextInt();
int values[] = new int[ n ];
System.out.println(\"\ Enter elements\");
for (i = 0; i < n; i++)
values[i] = scan.nextInt();
quicksort(values);
System.out.println(\"\ qSorted Elements \");
for (i = 0; i < n; i++)
System.out.print(values[i]+\" \");
System.out.println();
}
}
Output:
Enter number of elements
8
Enter elements
89 23 90 56 13 71 46 13
Sorted Elements
13 13 23 46 56 71 89 90.
I need to find run time analysis and description of the algo.pdf
I need to find run time analysis and description of the algorithm complexity (i.e., BigO)
Thank you!
///////////////////////////////////////////////////////////////////////////////////////////////
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Arrays;
import java.util.Scanner;
public class Sorting {
private static int n;
public static void main(String[] args) {
int[] arr = readFile("Data.txt");
// Bubble sort
long startTime = System.nanoTime();
bubbleSort(arr);
long endTime = System.nanoTime();
System.out.println("Bubble sort time: " + (endTime - startTime) + " nanoseconds");
// Selection sort
startTime = System.nanoTime();
selectionSort(arr);
endTime = System.nanoTime();
System.out.println("Selection sort time: " + (endTime - startTime) + " nanoseconds");
// Insertion sort
startTime = System.nanoTime();
insertionSort(arr);
endTime = System.nanoTime();
System.out.println("Insertion sort time: " + (endTime - startTime) + " nanoseconds");
// Merge sort
startTime = System.nanoTime();
mergeSort(arr);
endTime = System.nanoTime();
System.out.println("Merge sort time: " + (endTime - startTime) + " nanoseconds");
// Quick sort
startTime = System.nanoTime();
quickSortTime = quickSort(arr, 0, arr.length-1);
endTime = System.nanoTime();
System.out.println("Quick sort time: " + (endTime - startTime) + " nanoseconds");
}
private static int[] readFile(String filename) {
int[] arr = null;
try {
File file = new File(filename);
Scanner scanner = new Scanner(file);
n = scanner.nextInt();
arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = scanner.nextInt();
}
scanner.close();
} catch (FileNotFoundException e) {
System.out.println("File not found: " + filename);
System.exit(0);
}
return arr;
}
private static void bubbleSort(int[] arr) {
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
int temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
}
}
private static void selectionSort(int[] arr) {
for (int i = 0; i < n - 1; i++) {
int minIndex = i;
for (int j = i + 1; j < n; j++) {
if (arr[j] < arr[minIndex]) {
minIndex = j;
}
}
int temp = arr[i];
arr[i] = arr[minIndex];
arr[minIndex] = temp;
}
}
private static void insertionSort(int[] arr) {
for (int i = 1; i < n; i++) {
int key = arr[i];
int j = i - 1;
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j--;
}
arr[j + 1] = key;
}
}
private static void mergeSortHelper(int[] arr, int left, int right) {
if (left < right) {
int mid = (left + right) / 2;
mergeSortHelper(arr, left, mid);
mergeSortHelper(arr, mid+1, right);
merge(arr, left, mid, right);
}
}
private static void merge(int[] arr, int left, int mid, int right) {
int[] temp = new int[right-left+1];
int i = left, j = mid+1, k = 0;
while (i <= mid && j <= right) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
} else {
temp[k++] = arr[j++];
}
}
while (i <= mid) {
temp[k++] = arr[i++];
}
while (j <= right) {
temp[k++] = arr[j++];
}
for (i = left, k = 0; i <= right; i++, k++).
Supplemental knowledge conversion processes have been added to the E.pdf
Supplemental knowledge conversion processes have been added to the Extended Nonaka (1991)
organizational learning model by Ray (2014) to form a new Enhanced Organizational Learning
model. Match the supplement knowledge conversion process titles to the correct definitions:
Acquisition Assimilation Absorption Utilization Exploitation A. the process of scanning,
identifying, and selecting data to support propositions that become new knowledge. B. the
process of analyzing, processing and interpreting info so it can be understood. C. the process of
translating explicit group knowledge in transferrable states into organizational remembered tacit
knowledge within specific contexts of interest. D. represents applying organizationally
remembered knowledge to new experiences to extend and leverage existing competencies. E. is
the process of creating new competencies by incorporating new transformed knowledge into
operations..
Supongamos que Musashi, un economista de un programa de radio AM, y .pdf
Supongamos que Musashi, un economista de un programa de radio AM, y Rina, una economista
de una universidad de Massachusetts, estn discutiendo sobre el seguro mdico. El siguiente dilogo
muestra un extracto de su debate:
Rina: Un tema popular de debate entre polticos y economistas es la idea de brindar asistencia
gubernamental para beneficios de salud.
Musashi: Creo que es opresivo que el gobierno cobre impuestos a las personas que se cuidan a s
mismas para pagar el seguro mdico de las personas obesas.
Rina: No estoy de acuerdo. Creo que la financiacin gubernamental del seguro de salud es til para
garantizar la equidad bsica.
El desacuerdo entre estos economistas probablemente se deba a:
A: Diferencias en los juicios cientficos
B: Diferencias en valores
C: Diferencias entre percepcin VS. realidad
A pesar de sus diferencias, con qu proposicin es ms probable que estn de acuerdo dos
economistas elegidos al azar?
A.Los abogados constituyen un porcentaje excesivo de los funcionarios electos.
B. Las leyes de salario mnimo hacen ms para daar a los trabajadores poco calificados que para
ayudarlos.
C. Los aranceles y las cuotas de importacin generalmente reducen el bienestar econmico..
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Similar to I am trying to implement timing on this program and cannot do it. Wh.pdf
how can i build this as problem 2 of my code #include iostream.pdf
how can i build this as problem 2 of my code?
#include
#include
#include
#include
#include "./freeglut-3.4.0/include/GL/freeglut.h"
using namespace std;
bool leftDown = false, rightDown = false;
int lastPos[2];
float cameraPos[4] = { 0,1,4,1 };
int windowWidth = 640, windowHeight = 480;
double xRot = 0;
double yRot = 0;
int curProblem = 1; // TODO: change this number to try different examples
float specular[] = { 1.0, 1.0, 1.0, 1.0 };
float shininess[] = { 50.0 };
void problem1() {
// TODO
}
void problem2() {
//FIX HERE
}
void problem3() {
//TO DO
}
void problem4() {
//TO DO
}
void display()
{
glClearColor(0, 0, 0, 0);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glDisable(GL_LIGHTING);
glEnable(GL_DEPTH_TEST);
glBegin(GL_LINES);
glColor3f(1, 0, 0); glVertex3f(0, 0, 0); glVertex3f(1, 0, 0); // x axis
glColor3f(0, 1, 0); glVertex3f(0, 0, 0); glVertex3f(0, 1, 0); // y axis
glColor3f(0, 0, 1); glVertex3f(0, 0, 0); glVertex3f(0, 0, 1); // z axis
glEnd(/*GL_LINES*/);
glEnable(GL_LIGHTING);
glShadeModel(GL_SMOOTH);
glMaterialfv(GL_FRONT, GL_SPECULAR, specular);
glMaterialfv(GL_FRONT, GL_SHININESS, shininess);
glEnable(GL_LIGHT0);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glViewport(0, 0, windowWidth, windowHeight);
float ratio = (float)windowWidth / (float)windowHeight;
gluPerspective(50, ratio, 1, 1000);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt(cameraPos[0], cameraPos[1], cameraPos[2], 0, 0, 0, 0, 1, 0);
glLightfv(GL_LIGHT0, GL_POSITION, cameraPos);
glRotatef(xRot, 1, 0, 0);
glRotatef(yRot, 0, 1, 0);
if (curProblem == 1) problem1();
if (curProblem == 2) problem2();
if (curProblem == 3) problem3();
if (curProblem == 4) problem4();
glutSwapBuffers();
}
void mouse(int button, int state, int x, int y)
{
if (button == GLUT_LEFT_BUTTON) leftDown = (state == GLUT_DOWN);
else if (button == GLUT_RIGHT_BUTTON) rightDown = (state == GLUT_DOWN);
lastPos[0] = x;
lastPos[1] = y;
}
void mouseMoved(int x, int y)
{
if (leftDown) {
xRot += (y - lastPos[1]) * .1;
yRot += (x - lastPos[0]) * .1;
}
if (rightDown) {
for (int i = 0; i < 3; i++)
cameraPos[i] *= pow(1.1, (y - lastPos[1]) * .1);
}
lastPos[0] = x;
lastPos[1] = y;
glutPostRedisplay();
}
void keyboard(unsigned char key, int x, int y)
{
curProblem = key - '0';
if (key == 'q' || key == 'Q' || key == 27) {
exit(0);
}
glutPostRedisplay();
}
void reshape(int width, int height)
{
windowWidth = width;
windowHeight = height;
glutPostRedisplay();
}
int main(int argc, char** argv)
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGB | GLUT_DEPTH);
glutInitWindowSize(windowWidth, windowHeight);
glutCreateWindow("HW2");
glutDisplayFunc(display);
glutMotionFunc(mouseMoved);
glutMouseFunc(mouse);
glutReshapeFunc(reshape);
glutKeyboardFunc(keyboard);
glutMainLoop();
return 0;
}.
#include stdio.h #include stdlib.h #include unistd.h .pdf
#include
#include
#include
#include
#include
#include \"msecond.h\"
#include \"random_int.h\"
float total_time_spent_waiting = 0.0;
int total_number_of_meals = 0;
#define semaphore_create(s,v) sem_init( &s, 0, v )
#define semaphore_wait(s) sem_wait( &s )
#define semaphore_signal(s) sem_post( &s )
#define semaphore_release(s) sem_destroy( &s )
typedef sem_t semaphore;
semaphore chopstick[NUM_PHILOSOPHERS];
semaphore screen;
semaphore mutex;
int screen_row[NUM_PHILOSOPHERS] = { 6, 2, 2, 6, 10 };
int screen_col[NUM_PHILOSOPHERS] = { 31, 36, 44, 49, 40 };
int chopstick_row[5] = { 9, 4, 3, 4, 9 };
int chopstick_col[5] = { 35, 33, 40, 47, 45 };
char chopstick_sym[5] = { \'/\', \'\\\\\', \'|\', \'/\', \'\\\\\' };
for ( i = 0; i < NUM_PHILOSOPHERS; i++ )
{
position( screen_row[i], screen_col[i] );
printf( \"%d\", i );
position( screen_row[i] + 1, screen_col[i] - 1 );
printf( \"(%d)\", 0 );
position( chopstick_row[i], chopstick_col[i] );
printf( \"%c\", chopstick_sym[i] );
}
position_flush( 13, 1 );
}
void draw_thinking( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_hungry( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_eating( int n, int eat_count )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position( screen_row[n] + 1, screen_col[n] - 1 );
printf( \"(%d)\", eat_count );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_up( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
printf( \"%c\", \' \' );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_down( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_done( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
printf( \"D\" );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void chopsticks( int n )
{
if ( n % 2 == 0 ) {
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
} else {
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
}
}
void release_chopsticks( int n )
{
draw_chopstick_down( n );
semaphore_signal( chopstick[n] );
draw_chopstick_down( (n+1) % NUM_PHILOSOPHERS );
semaphore_signal( chopstick[(n+1) % NUM_PHILOSOPHERS] );
}
void philosopher( int *philosopher_data )
{
int start_time;
int eat_count = 0;
int total_hungry_time = 0;
int became_hungry_time;
int n = philosopher_data[0];
int duration = philosopher_data[1];
start_time = msecond();
while( msecond() - start_time < duration * 1000 )
{
became_hungry_time = msecond();
draw_hungry( n );
chopsticks( n );
t.
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Please help with this JAVA Assignment and show output if you can ple.pdf
Please help with this JAVA Assignment and show output if you can please
Complete required scripts based on eclipse and troubleshoot the scripts until the tasks are done.
1. Implement linear-sort, merge-sort and quick-sort (write the code for them).
Solution
Linear or Insertion Sort
import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.Scanner;
class LinearSort
{
public static void sort( int values[] )
{
int N = values.length;
int i, j, temp;
for (i = 1; i< N; i++)
{
j = i;
temp = values[i];
while (j > 0 && temp < values[j-1])
{
values[j] = values[j-1];
j = j-1;
}
values[j] = temp;
}
}
public static void main(String[] args)
{
Scanner scan = new Scanner( System.in );
int n, i;
System.out.println(\"Enter number of elements\");
n = scan.nextInt();
int values[] = new int[n];
System.out.println(\"\ Enter elements\");
for (i = 0; i < n; i++)
values[i] = scan.nextInt();
sort(values);
System.out.println(\"\ Sorted Elements\");
for (i = 0; i < n; i++)
System.out.print(values[i]+\" \");
System.out.println();
}
}
Output:
Enter number of elements
6
Enter elements
34 56 12 8 9 23
Sorted Elements
8 9 12 23 34 56
Merge Sort
import java.util.Scanner;
public class Mergesort
{
public static void mergesort(int[] a, int low, int high)
{
int N = high - low;
if (N <= 1)
return;
int mid = low + N/2;
mergesort(a, low, mid);
mergesort(a, mid, high);
int[] temp = new int[N];
int i = low, j = mid;
for (int k = 0; k < N; k++)
{
if (i == mid)
temp[k] = a[j++];
else if (j == high)
temp[k] = a[i++];
else if (a[j] pivot)
j--;
if (i <= j)
{
temp = values[i];
values[i] = values[j];
values[j] = temp;
i++;
j--;
}
}
if (low < j)
quickquicksort(values, low, j);
if (i < high)
quickquicksort(values, i, high);
}
public static void main(String[] args)
{
Scanner scan = new Scanner( System.in );
int n, i;
System.out.println(\"Enter number of elements\");
n = scan.nextInt();
int values[] = new int[ n ];
System.out.println(\"\ Enter elements\");
for (i = 0; i < n; i++)
values[i] = scan.nextInt();
quicksort(values);
System.out.println(\"\ qSorted Elements \");
for (i = 0; i < n; i++)
System.out.print(values[i]+\" \");
System.out.println();
}
}
Output:
Enter number of elements
8
Enter elements
89 23 90 56 13 71 46 13
Sorted Elements
13 13 23 46 56 71 89 90.
I need to find run time analysis and description of the algo.pdf
I need to find run time analysis and description of the algorithm complexity (i.e., BigO)
Thank you!
///////////////////////////////////////////////////////////////////////////////////////////////
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Arrays;
import java.util.Scanner;
public class Sorting {
private static int n;
public static void main(String[] args) {
int[] arr = readFile("Data.txt");
// Bubble sort
long startTime = System.nanoTime();
bubbleSort(arr);
long endTime = System.nanoTime();
System.out.println("Bubble sort time: " + (endTime - startTime) + " nanoseconds");
// Selection sort
startTime = System.nanoTime();
selectionSort(arr);
endTime = System.nanoTime();
System.out.println("Selection sort time: " + (endTime - startTime) + " nanoseconds");
// Insertion sort
startTime = System.nanoTime();
insertionSort(arr);
endTime = System.nanoTime();
System.out.println("Insertion sort time: " + (endTime - startTime) + " nanoseconds");
// Merge sort
startTime = System.nanoTime();
mergeSort(arr);
endTime = System.nanoTime();
System.out.println("Merge sort time: " + (endTime - startTime) + " nanoseconds");
// Quick sort
startTime = System.nanoTime();
quickSortTime = quickSort(arr, 0, arr.length-1);
endTime = System.nanoTime();
System.out.println("Quick sort time: " + (endTime - startTime) + " nanoseconds");
}
private static int[] readFile(String filename) {
int[] arr = null;
try {
File file = new File(filename);
Scanner scanner = new Scanner(file);
n = scanner.nextInt();
arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = scanner.nextInt();
}
scanner.close();
} catch (FileNotFoundException e) {
System.out.println("File not found: " + filename);
System.exit(0);
}
return arr;
}
private static void bubbleSort(int[] arr) {
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
int temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
}
}
private static void selectionSort(int[] arr) {
for (int i = 0; i < n - 1; i++) {
int minIndex = i;
for (int j = i + 1; j < n; j++) {
if (arr[j] < arr[minIndex]) {
minIndex = j;
}
}
int temp = arr[i];
arr[i] = arr[minIndex];
arr[minIndex] = temp;
}
}
private static void insertionSort(int[] arr) {
for (int i = 1; i < n; i++) {
int key = arr[i];
int j = i - 1;
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j--;
}
arr[j + 1] = key;
}
}
private static void mergeSortHelper(int[] arr, int left, int right) {
if (left < right) {
int mid = (left + right) / 2;
mergeSortHelper(arr, left, mid);
mergeSortHelper(arr, mid+1, right);
merge(arr, left, mid, right);
}
}
private static void merge(int[] arr, int left, int mid, int right) {
int[] temp = new int[right-left+1];
int i = left, j = mid+1, k = 0;
while (i <= mid && j <= right) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
} else {
temp[k++] = arr[j++];
}
}
while (i <= mid) {
temp[k++] = arr[i++];
}
while (j <= right) {
temp[k++] = arr[j++];
}
for (i = left, k = 0; i <= right; i++, k++).
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Supplemental knowledge conversion processes have been added to the E.pdf
Supplemental knowledge conversion processes have been added to the Extended Nonaka (1991)
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process of analyzing, processing and interpreting info so it can be understood. C. the process of
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Rina: Un tema popular de debate entre polticos y economistas es la idea de brindar asistencia
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Musashi: Creo que es opresivo que el gobierno cobre impuestos a las personas que se cuidan a s
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Rina: No estoy de acuerdo. Creo que la financiacin gubernamental del seguro de salud es til para
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El dlar estadounidense se deprecia
Las exportaciones estadounidenses aumentarn
Las importaciones estadounidenses disminuirn
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Debe llenar los espacios en blanco con una "I" (sin la comilla) para aumentar y "D" (sin la
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Suppose that there are two groups of people in the economy. In group.pdf
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H'i = b(1 u)[aHi+(1-a)Hj])
where H'i is the individual is human capital in the future period, 0 < a < 1, Hi is the individuals
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in the economy choose the same u.
A. What happens to incomes in groups 1 and 2 in the long run?
B. What does this say about the role of human capital accumulation in income inequality?.
Suppose that the Fed will increase the money supply. Which of the fo.pdf
Suppose that the Fed will increase the money supply. Which of the following would make the
effect of this increase on aggregate demand larger in the short-run?
A. the interest rate falls and SRAS relatively steep
B. the interest rate rises and SRAS is relatively steep
C. the interest rate falls and SRAS is relatively flat
D. the interest rate rises and SRAS is relatively flat.
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Suppose that the body weights of Roborovski dwarf hamsters are normally distributed with a m
of 22 g and a standard deviation of 1.2 g. A scientist thinks that conditions in the Ordos desert
affecting the hamsters in the region, She collects a random sample of n=25 hamsters from the Or
and finds the sample mean weight to be 21.4 g. Find the probability of observing a sample mean
than or equal to 21.4 if the mean weights of the hamsters in the Ordos desert is 22g as reported..
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Suppose that in a particular country, the TFR fell to zero and remained there. Also suppose that
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population that was working age would change over time. Immediately following the fertility
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Suppose that disposable income, consumption, and saving in some country are $200 billion,
$150 billion, and $50 billion, respectively. Next, assume that disposable income increases by $20
billion, consumption rises by $18 billion and saving goes up by $2 billion. What is the economys
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Suppose that 60% of students at Kansas State University have listened to Funtabi's album "The
Creatures We Are" (available to listen to on Spotify, Apple Music, and other music streaming
platforms). Additionally, suppose that, to gauge interest in the band, a major record label
randomly samples 400 students.
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Suppose Maria and Jamal both face the following individual loss dist.pdf
Suppose Maria and Jamal both face the following individual loss distribution: 30% probability of
losing $60. The individual Standard Deviation (without pooling) is approx:
27
18
22
19
27
18
22
19.
Suppose a random variable X has the following probability distributi.pdf
Suppose a random variable X has the following probability distribution:
a. Find the expected value of X, E(X).
b. Find the variance of X, V ar(X).
c. Find the cumulative distribution function of X, F(x) = P(X x).xf(x)100.20110.40120.30130.10.
Suppose a country had a smaller increase in debt in 2011 than it had.pdf
Suppose a country had a smaller increase in debt in 2011 than it had in 2010. Then other things
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b. lower interest rates and greater investment in 2011 than in 2010.
c. higher interest rates and greater investment in 2011 than in 2010.
d. higher interest rates and lower investment in 2011 than in 2010.
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demonstrate empowerment in practice.
a. give the client limited range of choices of services and activities
b. ask the client to follow their care plan, even if they wish to change it
c. give the client information about different choices available to them and support them to
choose independently
d. support the client to understand their own needs and decide what they need to be able to meet
those needs.
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How would I create this diagramRentTool Viewpoints � Technology .pdf
How would I create this diagram
RentTool Viewpoints
Technology Usage Viewpoint: How applications are supported by hardware and software.
RentTool is a tool/equipment rental company.
They provide their service to construction companies.
Use the description of RentTools application and technology layers in the next slide.
Use an Architecture library.
Draw the Infrastructure Viewpoint of ToolRent.
Draw the Technology Usage Viewpoint of ToolRent.
Application Layer
The Rental company uses two applications:
RMS: Rental Management System
CRM: Customer Relation Management
The following three application services are provided:
Rental Management Service
Financial Service
CRM Data Access Service
Data object defined in the applications:
Customer Data
Application layer services
Three Applications services:
CRM Application Component realizes CRM Data Access (application layer) service.
RMS Application Component realizes Rental Management Service (application layer) and
Financial Service (application layer).
CRM Data Access Service serves RMS application.
CRM Application Component accesses Customer Data (application layer passive structure
element).
Technology Layer (computers and services)
Three computers in total
Two computers run the RMS software application
General-purpose Server (GP Server #1)
Back-up Server Cluster
One computer runs the CRM software application
General purpose server (GP Server #2)
Two technology layer services
Application Hosting Service
Database Management Service
Technology Layer (system software)
Two system software elements
Application Server
Database Management (DBMS)
All server computers run Application Server system software ( is assigned to relationship)
Application Server system software realizes Application Hosting service (Technology layer)
GP Server runs Database Management (DBMS) system software (is assigned to relationship)
Database Management (DBMS) system software realizes Database Management service
(Technology layer)
Technology layer (artifacts)
Related artifacts: RMS system artifact and CRM system artifact
Shows the software files distributions (e.g. zip file)
The assignment relationships with the related computers
Customer Database Tables artifact represents the implementation of the Customer Data object
in the application layer
Technology-Application Alignment
Technology layer provides the following technology layer services:
Application Hosting service serves all application components in the application layer
use a relationship connector
Database Management service serves only CRM Application application component.
Database Management service (technology layer service) accesses Customer Database Table
(an artifact that realizes Customer Data in the application layer).
How well did Thaldorf interact with each member of the DMUOn what.pdf
How well did Thaldorf interact with each member of the DMU?
On what date did Thaldorf effectively lose the sale to Lohmann University? On December 18,
Kurt Thaldorf, a sales engineer for can diagnose cancer of the pancreas in less than two the
German sales subsidiary of Mediquip, S.A., was weeks after it develops. This was not possible
before informed by Lohmann University Hospital in Stuttgart CT scanners." that it had decided
to place an order with Sigma, a Mediquip was a subsidiary of Technologie UniDutch competitor,
for a CT scanner. The hospital's verselle, a French conglomerate. The company's proddecision
came as disappointing news to Thaldorf, who uct line included, in addition to CT scanners, X-
ray, had worked for nearly eight months on the account. ultrasonic, and nuclear diagnostic
equipment. Mediquip The order, if obtained, would have meant a sale of enjoyed a worldwide
reputation for advanced technol(Euro) 2,370,000 for the sales engineer. 1 He was con- ogy and
competent after-sales service. vinced that Mediquip's CTscanner was technologically "Our
competitors are mostly from other European superior to Sigma's and, overall, a better product.
countries," commented Mediquip's Sales Director for Thaldorf began a review of his call reports
in order Europe. "In some markets they have been there longer to better understand the factors
that had led to Lohmann than we have, and they know the decision-makers betUniversity
Hospital's decision. He wanted to apply the ter than we do. But we are learning fast." Sigma, the
lessons from this experience to future sales situations. subsidiary of a diversified Dutch company
under the same name, was the company's most serious competiBACKGROUND tor. Other
major contenders in the CT scanner market were FNC, Eldora, Magna, and Piper. At the time,
the computer tomography (CT) scanner Mediquip executives estimated the European marwas a
relatively recent product in the field of diagnostic ket for CT scanners to be around 200 units per
year. imaging. This medical device, used for diagnostic pur- They pointed out that prices ranged
from 1.5 to 3.0 poses, allowed examination of cross sections of the million per unit. The
company's CT scanner sold in the human body through display of images. CT scanners upper
end of the price range. "Our equipment is at least combined sophisticated X-ray equipment with
a com- two years ahead of our most advanced competition," puter to collect the necessary data
and translate them explained a sales executive. "And our price reflects this into visual images.
When computer tomography was first introduced in Mediquip's sales organization in Europe
included the late 1960s, radiologists had hailed it as a major tech- eight country sales
subsidiaries, each headed by a mannologiqal breakthrough. Commenting on the advantages
aging director. Within each country, sales engineers of CT scanners, a product specialist with
Mediquip said, reported to regional sales managers who, in turn, ".
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Increasing atmospheric CO2 leads to more CO2 becoming dissolved in the ocean, which makes
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Increasing atmospheric CO2 leads to destruction of the ozone layer, which increases UV
radiation and coral bleaching
Increasing atmospheric CO2 increases sedimentation on reefs, which shades out photosynthetic
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