Java Question: Consider a maze made up of a rectangular array of squares. The maze will
contain a character (either +, -, or |) to represent a blocked square, and to form the walls of the
maze. Mazes will have only one entrance at the Coordinate (0, 1), with only one exit in the lower
right hand corner of the maze. Beginning at the entrance to the maze, find a path to the exit at the
bottom right of the maze. You may only move up, down, left, and right. Each square in the maze
can be in one of four states: clear (space), blocked (X), path (.), or visited (*). Initially, after the
maze has been read in from the file, each square will be either clear or blocked. If a square lies
on a successful path, mark it with a period. If you visit a square but it does not lead to a
successful path, mark it as visited with an asterisk.
Solution
import java.io.*;
class Maze
{
// Any MazeSquare field is true if the corresponding
// wall exists, and false otherwise.
class MazeSquare
{
public boolean hasTopWall;
public boolean hasRightWall;
public boolean hasLeftWall;
public boolean hasBottomWall;
public MazeSquare()
{
hasTopWall = hasRightWall = hasLeftWall = hasBottomWall = false;
}
}
// This Exception class will help pinpoint format errors in the data file.
class MazeFormatException extends Exception
{
public final static int noError = 0;
public final static int badRowAndColumnCounts = 1;
public final static int wrongNumberOfEntriesInRow = 2;
public final static int badEntry = 3;
public int lineNumber;
public int problem;
public MazeFormatException( int line, int prob )
{
lineNumber = line;
problem = prob;
}
}
// Number of rows and columns in the maze.
private int nRows;
private int nColumns;
// The array of maze squares themselves.
MazeSquare[][] square;
public static void main( String[] args )
{
if( args.length != 1 )
{
System.err.println( \"Usage: java Maze mazeFileName\" );
System.exit( 1 );
}
Maze knosos = null;
try
{
knosos = new Maze( args[0] );
}
catch( FileNotFoundException e )
{
System.err.println( \"Can\'t open \" + args[0] + \". Check your spelling.\" );
System.exit( 1 );
}
catch( IOException e )
{
System.err.println( \"Severe input error. I give up.\" );
System.exit( 1 );
}
catch( MazeFormatException e )
{
switch( e.problem )
{
case MazeFormatException.badRowAndColumnCounts:
System.err.println( args[0] + \", line \" + e.lineNumber + \": row and column counts
expected\" );
break;
case MazeFormatException.wrongNumberOfEntriesInRow:
System.err.println( args[0] + \", line \" + e.lineNumber + \": wrong number of entries\"
);
break;
case MazeFormatException.badEntry:
System.err.println( args[0] + \", line \" + e.lineNumber + \": non-hexadecimal digit
detected\" );
break;
default:
System.err.println( \"This should never get printed.\" );
break;
}
System.exit( 1 );
}
knosos.print( System.out );
}
public Maze()
{
square = null;
nRows = nColumns = 0;
}
public Maze( String fileName ) throws FileNotFoundException, IOException,
MazeFormatException
{
Buffered.
Java Question help needed In the program Fill the Add statements.pdfkamdinrossihoungma74
Item 8 Singly ionized (one electron removed) atoms are accelerated and then passed through a
velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 158
V/m and the magnetic field is 3.13 Times 10^2 T. The ions next enter a uniform magnetic field
of magnitude 1.73 Times 10^-2 T that is oriented perpendicular to their velocity. How fast are
the ions moving when they emerge from the velocity selector? If the radius of the path of the
ions in the second magnetic field is 17.4 cm, what is their mass?
Solution
A)
In Velocity selector Electric field & Magnetic field forces on ions are equal & opposite ,this is
because it allows ions with a particular velocity only passes undeflected.
so we can say,
Bqv = Eq
B:magnetic field density
E:Electric field
v:velocity at which the ions remains undeflected.
v=E/B
v = 158V/m / 3.13*10-2 T
=> v = 4.78*103 m/s
B)
as it enteres next in uniform magnetic field of value B=1.73*10-2 T
Work done in uniform magnetic field is zero.It only changes the direction.So B provides
centripetal force for the ions while changing direction in circular path of radius r=17.4cm.
Bqv = mv²/r
m = Bqr / v = (1.73*10-2T)(1.6*10-19C)(0.174m) / (4.78*103m/s)
=>m = 1*10-25 kg.
hello the code given is mostly complete but I need help with the Sol.pdfalmonardfans
hello the code given is mostly complete but I need help with the Solver portion please
Consider a maze made up of rectangular array of squares, such as the following one:
X X X X X X X X X X X X X
X X X X X X
X X X X X
X X X X X X X X
X X X X X
X X X X X X X
X X X X X X X X X X X X X
Figure 1--- A sample maze
The X blocks represent a blocked square and form the walls of the maze. Let us consider mazes
that
have only one entrance and one exit, as in our example. Beginning at the entrance at the top left
side of
the maze, find a path to the exit at the bottom right side. You can move only up, down, left, or
right.
Square is either clear or blocked by an X character.
Let a two-dimensional array represent the maze. Use a stack data structure to find a path through
the
maze. Some mazes might have more than one successful path, while others might have no path.
Hints:
The primary consideration is that if you reach a dead end, you need to be able to backtrack
along the path to the point where you can try a different path. The stack data structure makes it
possible to store the current path and backtrack if necessary. Every clear position visited in the
current path is pushed to the stack and if a dead end is reached, the previous position is popped
from the stack to backtrack. You need to have a loop that begins with the start position and
pushes it into the stack then moves to a neighboring cell until either the goal position is reached,
or the stack becomes empty.
Make sure to mark each clear array cell you move to as visited to avoid checking it again and
getting stuck in an infinite loop. For example, each array cell can have three different values: X
for blocked, V for visited, and for clear.
A dead end is an array cell whose all neighbors are either blocked or already visited.
There are four files you should pay attention to:
Position.java: a class to store the position of an array cell. It has two attributes: row and
column along with their accessor and mutator methods. It also has an equals method that
checks for equality of two Position objects.
Maze.java: a class to store a maze. It has 3 attributes: a two-dimensional character array which
represents the maze, a Position object to represent the start location, and another Position
object to represent the stop location.
MazeSolver: This class contains one static method, solve. It accepts a maze and returns an
array of Positions that is used to traverse the maze if the maze is solvable. If the maze is not
solvable then an empty array is returned.
o Keep in mind that stacks store objects in reverse order so be sure to return the position
array in the correct order.
Tester: a sample driver class which calls the traverse method on a maze in either the
SolvableMaze, UnsolvableMaze, or HugeSolvableMaze files. These should be at the root of the
project folder
SOLVER
import java.util.Stack;
public class MazeSolver {
public static Position[] solve(Maze maze) {
return null;
}
}
POSITION
public class P.
Please copy and paste the code and explain why it won't work- It is su.docxcgraciela1
Please copy and paste the code and explain why it won't work. It is supposed to take a maze and return the exits and number of exits.
import java.util.Scanner;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Arrays;
public class MazeSolver
{
int numRows;
int numCols;
int startRow;
int startCol;
int rows;
int col;
public static void loadMaze(){
int numRows;
int numCols;
int startRow;
int startCol;
int rows;
int col;
try{
Scanner input = new Scanner(System.in);
System.out.println("Please enter file name:");
String filename = input.nextLine();
File file = new File(filename);
Scanner fileInput = new Scanner(file);
numRows = fileInput.nextInt();
numCols = fileInput.nextInt();
ArrayList column= new ArrayList<String>();
ArrayList row = new ArrayList<String>();
char[][] maze= new char[numRows][numCols];
//for the row, get array start row and column index
for (int i = 0; i < numRows; i++) {
String line = fileInput.next();
for (int j = 0; j < numCols; j++) {
char empt = ' ';
maze[i][j] =
if (maze[i][j] == empt){
startRow = i;
startCol = j;
System.out.println(i);
//solveMaze(maze[i][j]);
}
else{
return;
}
}
}
fileInput.close();
}
catch(FileNotFoundException e) {
System.out.println("File not found.");
loadMaze();
}
}
// Method to solve the maze
public void solveMaze(char[][] maze) {
ArrayList<Integer> exitRows = new ArrayList<Integer>();
ArrayList<Integer> exitCols = new ArrayList<Integer>();
boolean[][] visited = new boolean[numRows][numCols];
// Call recursive method to explore the maze
exploreMaze(startRow, startCol, visited, maze, exitRows, exitCols);
// Print number of exits found
System.out.println("Number of exits found: " + exitRows.size());
// Print positions of exits
for (int i = 0; i < exitRows.size(); i++) {
System.out.println("Exit found at: (" + exitRows.get(i) + ", " + exitCols.get(i) + ")");
}
}
public void exploreMaze(int row, int col, boolean[][] visited, char[][] maze, ArrayList<Integer> exitRows, ArrayList<Integer> exitCols) { // Recursive method to explore the maze
if (row < 0 || col < 0 || row >= numRows || col >= numCols || visited[row][col]) { // Base case: if current position is out of bounds or already visited
return;
}
// Base case: if current position is an exit
if (maze[row][col] == 'E'){
exitRows.add(row);
exitCols.add(col);
return;
}
// Mark current position as visited
visited[row][col] = true;
exploreMaze(row-1, col, visited, maze,exitRows, exitCols); // up
exploreMaze(row, col+1, visited, maze,exitRows, exitCols); // right
exploreMaze(row+1, col, visited, maze,exitRows, exitCols); // down
exploreMaze(row, col-1, visited, maze,exitRows, exitCols); // left
// Recursively explore neighboring positions
}
}
.
In this chapter we are going to get familiar with recursion and its applications. Recursion represents a powerful programming technique in which a method makes a call to itself from within its own method body. By means of recursion we can solve complicated combinatorial problems, in which we can easily exhaust different combinatorial configurations, e.g. generating permutations and variations and simulating nested loops. We are going to demonstrate many examples of correct and incorrect usage of recursion and convince you how useful it can be.
Topic Use stacks to solve Maze. DO NOT USE RECURSION. And do not us.pdfambikacollection1084
Topic: Use stacks to solve Maze. DO NOT USE RECURSION. And do not use hashmaps
Position.java
public class Position {
/***************************
* Attributes
****************************/
private int row;
private int column;
/***************************
* Constructor
*
* @param row
* @param column
***************************/
public Position(int row, int column) {
this.row = row;
this.column = column;
}
/**************************
* Checks two positions for equality. Two positions are equal if the have the
* same row and column numbers.
*************************/
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (!(obj instanceof Position)) {
return false;
}
Position otherPosition = (Position) obj;
return (otherPosition.row == row && otherPosition.column == column);
}
public String toString() {
return "row:" + row + " column:" + column;
}
/**************************
* Getter and Setter methods
*
* @return
*/
public int getRow() {
return row;
}
public int getColumn() {
return column;
}
}
Maze.java
public class Maze {
/**
* Two dimensional array to represent a maze
*/
private final char[][] maze;
private final Position start, end;
/**
* Constructor initializing the maze array
*
* @param maze
*/
public Maze(char[][] maze, Position start, Position end) {
this.maze = maze;
if (validPosition(start)) {
this.start = start;
} else {
throw new PositionNotValidException("The start position is valid for the maze: " + start);
}
if (validPosition(end)) {
this.end = end;
} else {
throw new PositionNotValidException("The stop position is valid for the maze: " + end);
}
}
/**
* Returns the start position object.
*
* @return The start position.
*/
public Position getStart() {
return start;
}
/**
* Returns the stop position object.
*
* @return
*/
public Position getEnd() {
return end;
}
/**
* Returns the number of rows the maze has.
*
* @return The number of rows.
*/
public int numRows() {
return maze.length;
}
/**
* Returns the number of columns the row of the maze has.
*
* @param row The row to get the length of.
* @return The length of the row in the maze.
*/
public int numCols(int row) {
return maze[row].length;
}
/**
* Tests if row and column are valid within the two dimensional array. Will
* return true iff row is between 0 and maze.length-1 and column is between 0
* and maze[row].length - 1.
*
* @param row Represents the array to retrieve from maze.
* @param col Represents the character to retrieve from the array at maze[row].
* @return True iff the row and column are valid indices of maze otherwise
* false.
*/
public boolean validPosition(int row, int col) {
if (row >= 0 && row < maze.length) {
if (col >= 0 && col < maze[row].length) {
return true;
}
}
return false;
}
/**
* Tests if the row and column stored in pos are valid. Drops to the
* validPosition(int row, int col) method.
*
* @param pos The position to test.
* @return True iff pos is valid inside the maze otherwise false.
*/
public boolean validPosition(Position pos) {
r.
Requirements Use JavaDo Not Use RecursionDo Not Use Array Lis.pdfalphaagenciesindia
Requirements :
Use Java
Do Not Use Recursion
Do Not Use Array List Use The Methods Provided
Use What Is Provided Only Make And Use The Helper Methods The Instructions Say You Can
Only Provide The Answer Code
Thank You ! I'm Having The Hardest Time Understanding This
Maze Solving with Stacks
Consider a maze made up of rectangular array of squares, such as the following one:
X X X X X X X X X X X X X
X X X X X X
X X X X X
X X X X X X X X
X X X X X
X X X X X X X
X X X X X X X X X X X X X
Figure 1--- A sample maze
The X blocks represent a blocked square and form the walls of the maze. Let us consider mazes
that
have only one entrance and one exit, as in our example. Beginning at the entrance at the top left
side of
the maze, find a path to the exit at the bottom right side. You can move only up, down, left, or
right.
Square is either clear or blocked by an X character.
Let a two-dimensional array represent the maze. Use a stack data structure to find a path through
the
maze. Some mazes might have more than one successful path, while others might have no path.
Hints:
The primary consideration is that if you reach a dead end, you need to be able to backtrack
along the path to the point where you can try a different path. The stack data structure makes it
possible to store the current path and backtrack if necessary. Every clear position visited in the
current path is pushed to the stack and if a dead end is reached, the previous position is popped
from the stack to backtrack. You need to have a loop that begins with the start position and
pushes it into the stack then moves to a neighboring cell until either the goal position is reached,
or the stack becomes empty.
Make sure to mark each clear array cell you move to as visited to avoid checking it again and
getting stuck in an infinite loop. For example, each array cell can have three different values: X
for blocked, V for visited, and for clear.
A dead end is an array cell whose all neighbors are either blocked or already visited.
Maze Solving with Stacks
CSC 385
What you need to do:
There are four files you should pay attention to:
Position.java: a class to store the position of an array cell. It has two attributes: row and
column along with their accessor and mutator methods. It also has an equals method that
checks for equality of two Position objects.
Maze.java: a class to store a maze. It has 3 attributes: a two-dimensional character array which
represents the maze, a Position object to represent the start location, and another Position
object to represent the stop location.
MazeSolver: This class contains one static method, solve. It accepts a maze and returns an
array of Positions that is used to traverse the maze if the maze is solvable. If the maze is not
solvable then an empty array is returned.
o Keep in mind that stacks store objects in reverse order so be sure to return the position
array in the correct order.
Tester: a sample driver class which calls the traverse method on a maze in .
Java Question help needed In the program Fill the Add statements.pdfkamdinrossihoungma74
Item 8 Singly ionized (one electron removed) atoms are accelerated and then passed through a
velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 158
V/m and the magnetic field is 3.13 Times 10^2 T. The ions next enter a uniform magnetic field
of magnitude 1.73 Times 10^-2 T that is oriented perpendicular to their velocity. How fast are
the ions moving when they emerge from the velocity selector? If the radius of the path of the
ions in the second magnetic field is 17.4 cm, what is their mass?
Solution
A)
In Velocity selector Electric field & Magnetic field forces on ions are equal & opposite ,this is
because it allows ions with a particular velocity only passes undeflected.
so we can say,
Bqv = Eq
B:magnetic field density
E:Electric field
v:velocity at which the ions remains undeflected.
v=E/B
v = 158V/m / 3.13*10-2 T
=> v = 4.78*103 m/s
B)
as it enteres next in uniform magnetic field of value B=1.73*10-2 T
Work done in uniform magnetic field is zero.It only changes the direction.So B provides
centripetal force for the ions while changing direction in circular path of radius r=17.4cm.
Bqv = mv²/r
m = Bqr / v = (1.73*10-2T)(1.6*10-19C)(0.174m) / (4.78*103m/s)
=>m = 1*10-25 kg.
hello the code given is mostly complete but I need help with the Sol.pdfalmonardfans
hello the code given is mostly complete but I need help with the Solver portion please
Consider a maze made up of rectangular array of squares, such as the following one:
X X X X X X X X X X X X X
X X X X X X
X X X X X
X X X X X X X X
X X X X X
X X X X X X X
X X X X X X X X X X X X X
Figure 1--- A sample maze
The X blocks represent a blocked square and form the walls of the maze. Let us consider mazes
that
have only one entrance and one exit, as in our example. Beginning at the entrance at the top left
side of
the maze, find a path to the exit at the bottom right side. You can move only up, down, left, or
right.
Square is either clear or blocked by an X character.
Let a two-dimensional array represent the maze. Use a stack data structure to find a path through
the
maze. Some mazes might have more than one successful path, while others might have no path.
Hints:
The primary consideration is that if you reach a dead end, you need to be able to backtrack
along the path to the point where you can try a different path. The stack data structure makes it
possible to store the current path and backtrack if necessary. Every clear position visited in the
current path is pushed to the stack and if a dead end is reached, the previous position is popped
from the stack to backtrack. You need to have a loop that begins with the start position and
pushes it into the stack then moves to a neighboring cell until either the goal position is reached,
or the stack becomes empty.
Make sure to mark each clear array cell you move to as visited to avoid checking it again and
getting stuck in an infinite loop. For example, each array cell can have three different values: X
for blocked, V for visited, and for clear.
A dead end is an array cell whose all neighbors are either blocked or already visited.
There are four files you should pay attention to:
Position.java: a class to store the position of an array cell. It has two attributes: row and
column along with their accessor and mutator methods. It also has an equals method that
checks for equality of two Position objects.
Maze.java: a class to store a maze. It has 3 attributes: a two-dimensional character array which
represents the maze, a Position object to represent the start location, and another Position
object to represent the stop location.
MazeSolver: This class contains one static method, solve. It accepts a maze and returns an
array of Positions that is used to traverse the maze if the maze is solvable. If the maze is not
solvable then an empty array is returned.
o Keep in mind that stacks store objects in reverse order so be sure to return the position
array in the correct order.
Tester: a sample driver class which calls the traverse method on a maze in either the
SolvableMaze, UnsolvableMaze, or HugeSolvableMaze files. These should be at the root of the
project folder
SOLVER
import java.util.Stack;
public class MazeSolver {
public static Position[] solve(Maze maze) {
return null;
}
}
POSITION
public class P.
Please copy and paste the code and explain why it won't work- It is su.docxcgraciela1
Please copy and paste the code and explain why it won't work. It is supposed to take a maze and return the exits and number of exits.
import java.util.Scanner;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Arrays;
public class MazeSolver
{
int numRows;
int numCols;
int startRow;
int startCol;
int rows;
int col;
public static void loadMaze(){
int numRows;
int numCols;
int startRow;
int startCol;
int rows;
int col;
try{
Scanner input = new Scanner(System.in);
System.out.println("Please enter file name:");
String filename = input.nextLine();
File file = new File(filename);
Scanner fileInput = new Scanner(file);
numRows = fileInput.nextInt();
numCols = fileInput.nextInt();
ArrayList column= new ArrayList<String>();
ArrayList row = new ArrayList<String>();
char[][] maze= new char[numRows][numCols];
//for the row, get array start row and column index
for (int i = 0; i < numRows; i++) {
String line = fileInput.next();
for (int j = 0; j < numCols; j++) {
char empt = ' ';
maze[i][j] =
if (maze[i][j] == empt){
startRow = i;
startCol = j;
System.out.println(i);
//solveMaze(maze[i][j]);
}
else{
return;
}
}
}
fileInput.close();
}
catch(FileNotFoundException e) {
System.out.println("File not found.");
loadMaze();
}
}
// Method to solve the maze
public void solveMaze(char[][] maze) {
ArrayList<Integer> exitRows = new ArrayList<Integer>();
ArrayList<Integer> exitCols = new ArrayList<Integer>();
boolean[][] visited = new boolean[numRows][numCols];
// Call recursive method to explore the maze
exploreMaze(startRow, startCol, visited, maze, exitRows, exitCols);
// Print number of exits found
System.out.println("Number of exits found: " + exitRows.size());
// Print positions of exits
for (int i = 0; i < exitRows.size(); i++) {
System.out.println("Exit found at: (" + exitRows.get(i) + ", " + exitCols.get(i) + ")");
}
}
public void exploreMaze(int row, int col, boolean[][] visited, char[][] maze, ArrayList<Integer> exitRows, ArrayList<Integer> exitCols) { // Recursive method to explore the maze
if (row < 0 || col < 0 || row >= numRows || col >= numCols || visited[row][col]) { // Base case: if current position is out of bounds or already visited
return;
}
// Base case: if current position is an exit
if (maze[row][col] == 'E'){
exitRows.add(row);
exitCols.add(col);
return;
}
// Mark current position as visited
visited[row][col] = true;
exploreMaze(row-1, col, visited, maze,exitRows, exitCols); // up
exploreMaze(row, col+1, visited, maze,exitRows, exitCols); // right
exploreMaze(row+1, col, visited, maze,exitRows, exitCols); // down
exploreMaze(row, col-1, visited, maze,exitRows, exitCols); // left
// Recursively explore neighboring positions
}
}
.
In this chapter we are going to get familiar with recursion and its applications. Recursion represents a powerful programming technique in which a method makes a call to itself from within its own method body. By means of recursion we can solve complicated combinatorial problems, in which we can easily exhaust different combinatorial configurations, e.g. generating permutations and variations and simulating nested loops. We are going to demonstrate many examples of correct and incorrect usage of recursion and convince you how useful it can be.
Topic Use stacks to solve Maze. DO NOT USE RECURSION. And do not us.pdfambikacollection1084
Topic: Use stacks to solve Maze. DO NOT USE RECURSION. And do not use hashmaps
Position.java
public class Position {
/***************************
* Attributes
****************************/
private int row;
private int column;
/***************************
* Constructor
*
* @param row
* @param column
***************************/
public Position(int row, int column) {
this.row = row;
this.column = column;
}
/**************************
* Checks two positions for equality. Two positions are equal if the have the
* same row and column numbers.
*************************/
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (!(obj instanceof Position)) {
return false;
}
Position otherPosition = (Position) obj;
return (otherPosition.row == row && otherPosition.column == column);
}
public String toString() {
return "row:" + row + " column:" + column;
}
/**************************
* Getter and Setter methods
*
* @return
*/
public int getRow() {
return row;
}
public int getColumn() {
return column;
}
}
Maze.java
public class Maze {
/**
* Two dimensional array to represent a maze
*/
private final char[][] maze;
private final Position start, end;
/**
* Constructor initializing the maze array
*
* @param maze
*/
public Maze(char[][] maze, Position start, Position end) {
this.maze = maze;
if (validPosition(start)) {
this.start = start;
} else {
throw new PositionNotValidException("The start position is valid for the maze: " + start);
}
if (validPosition(end)) {
this.end = end;
} else {
throw new PositionNotValidException("The stop position is valid for the maze: " + end);
}
}
/**
* Returns the start position object.
*
* @return The start position.
*/
public Position getStart() {
return start;
}
/**
* Returns the stop position object.
*
* @return
*/
public Position getEnd() {
return end;
}
/**
* Returns the number of rows the maze has.
*
* @return The number of rows.
*/
public int numRows() {
return maze.length;
}
/**
* Returns the number of columns the row of the maze has.
*
* @param row The row to get the length of.
* @return The length of the row in the maze.
*/
public int numCols(int row) {
return maze[row].length;
}
/**
* Tests if row and column are valid within the two dimensional array. Will
* return true iff row is between 0 and maze.length-1 and column is between 0
* and maze[row].length - 1.
*
* @param row Represents the array to retrieve from maze.
* @param col Represents the character to retrieve from the array at maze[row].
* @return True iff the row and column are valid indices of maze otherwise
* false.
*/
public boolean validPosition(int row, int col) {
if (row >= 0 && row < maze.length) {
if (col >= 0 && col < maze[row].length) {
return true;
}
}
return false;
}
/**
* Tests if the row and column stored in pos are valid. Drops to the
* validPosition(int row, int col) method.
*
* @param pos The position to test.
* @return True iff pos is valid inside the maze otherwise false.
*/
public boolean validPosition(Position pos) {
r.
Requirements Use JavaDo Not Use RecursionDo Not Use Array Lis.pdfalphaagenciesindia
Requirements :
Use Java
Do Not Use Recursion
Do Not Use Array List Use The Methods Provided
Use What Is Provided Only Make And Use The Helper Methods The Instructions Say You Can
Only Provide The Answer Code
Thank You ! I'm Having The Hardest Time Understanding This
Maze Solving with Stacks
Consider a maze made up of rectangular array of squares, such as the following one:
X X X X X X X X X X X X X
X X X X X X
X X X X X
X X X X X X X X
X X X X X
X X X X X X X
X X X X X X X X X X X X X
Figure 1--- A sample maze
The X blocks represent a blocked square and form the walls of the maze. Let us consider mazes
that
have only one entrance and one exit, as in our example. Beginning at the entrance at the top left
side of
the maze, find a path to the exit at the bottom right side. You can move only up, down, left, or
right.
Square is either clear or blocked by an X character.
Let a two-dimensional array represent the maze. Use a stack data structure to find a path through
the
maze. Some mazes might have more than one successful path, while others might have no path.
Hints:
The primary consideration is that if you reach a dead end, you need to be able to backtrack
along the path to the point where you can try a different path. The stack data structure makes it
possible to store the current path and backtrack if necessary. Every clear position visited in the
current path is pushed to the stack and if a dead end is reached, the previous position is popped
from the stack to backtrack. You need to have a loop that begins with the start position and
pushes it into the stack then moves to a neighboring cell until either the goal position is reached,
or the stack becomes empty.
Make sure to mark each clear array cell you move to as visited to avoid checking it again and
getting stuck in an infinite loop. For example, each array cell can have three different values: X
for blocked, V for visited, and for clear.
A dead end is an array cell whose all neighbors are either blocked or already visited.
Maze Solving with Stacks
CSC 385
What you need to do:
There are four files you should pay attention to:
Position.java: a class to store the position of an array cell. It has two attributes: row and
column along with their accessor and mutator methods. It also has an equals method that
checks for equality of two Position objects.
Maze.java: a class to store a maze. It has 3 attributes: a two-dimensional character array which
represents the maze, a Position object to represent the start location, and another Position
object to represent the stop location.
MazeSolver: This class contains one static method, solve. It accepts a maze and returns an
array of Positions that is used to traverse the maze if the maze is solvable. If the maze is not
solvable then an empty array is returned.
o Keep in mind that stacks store objects in reverse order so be sure to return the position
array in the correct order.
Tester: a sample driver class which calls the traverse method on a maze in .
Java Cheat Sheet includes the following contents:
- Data Types
- Statements
- String, ArrayList and HashMap Methods
- Conversion
- Operators
- Exception Handling
I gave this presentation on January 14, 2010 to the Atlanta Scala user group. It covers Scala's implementation of operator overloading, as well as touching on implicit conversions.
This presentation provides an overview of key topics in Java class design; also covers best practices/tips and quiz questions. Based on our OCP 8 book.
The talk is about Kotlin features, which are decompiled to Java code. You’ll understand how all of this ‘magic’ works in depths, which benefits it’ll give to you and what you may loose. Some benchmarks are included as a bonus.
Some languages, like SML, Haskell, and Scala, have built-in support for pattern matching, which is a generic way of branching based on the structure of data.
While not without its drawbacks, pattern matching can help eliminate a lot of boilerplate, and it's often cited as a reason why functional programming languages are so concise.
In this talk, John A. De Goes talks about the differences between built-in patterns, and so-called first-class patterns (which are "do-it-yourself" patterns implemented using other language features).
Unlike built-in patterns, first-class patterns aren't magical, so you can store them in variables and combine them in lots of interesting ways that aren't always possible with built-in patterns. In addition, almost every programming language can support first-class patterns (albeit with differing levels of effort and type-safety).
During the talk, you'll watch as a mini-pattern matching library is developed, and have the opportunity to follow along and build your own pattern matching library in the language of your choice.
Explain what differential cell affinity is, how this process is acco.pdfwasemanivytreenrco51
Explain what differential cell affinity is, how this process is accomplished and why it is
important for morphogenesis.
Solution
Morphogenesis is a process through which the organism develops its shape. The spatial
distribution of organs and cells that belong to a specific group is controlled by this process.
During the stage of embryonic development, the cells are restricted to particular layers
(differential affinity) because all of them are joined by the same adhesion molecules. This
homotypic cell adhesion keeps all the cells of similar type intact and form boundaries with the
cells with different type of adhesion molecules, thus maintaining the organ’s or tissue’s shape..
Explain in detail how OFDM helps mitigates multipath fading effects..pdfwasemanivytreenrco51
Explain in detail how OFDM helps mitigates multipath fading effects.
Thank you! As detailed as possible!
Solution
Multipath fading is a feature that needs to be taken into account when designing or developing a
radio communications system. In any terrestrial radio communications system, the signal will
reach the receiver not only via the direct path, but also as a result of reflections from objects such
as buildings, hills, ground, water, etc that are adjacent to the main path.The overall signal at the
radio receiver is a summation of the variety of signals being received. As they all have different
path lengths, the signals will add and subtract from the total dependent upon their relative
phases.
At times there will be changes in the relative path lengths. This could result from either the radio
transmitter or receiver moving, or any of the objects that provides a reflective surface moving.
This will result in the phases of the signals arriving at the receiver changing, and in turn this will
result in the signal strength varying as a result of the different way in which the signals will sum
together. It is this that causes the fading that is present on many signals.
Multipath fading can affect radio communications channels in two main ways. This can give the
way in which the effects of the multipath fading are mitigated..
More Related Content
Similar to Java Question Consider a maze made up of a rectangular array of squ.pdf
Java Cheat Sheet includes the following contents:
- Data Types
- Statements
- String, ArrayList and HashMap Methods
- Conversion
- Operators
- Exception Handling
I gave this presentation on January 14, 2010 to the Atlanta Scala user group. It covers Scala's implementation of operator overloading, as well as touching on implicit conversions.
This presentation provides an overview of key topics in Java class design; also covers best practices/tips and quiz questions. Based on our OCP 8 book.
The talk is about Kotlin features, which are decompiled to Java code. You’ll understand how all of this ‘magic’ works in depths, which benefits it’ll give to you and what you may loose. Some benchmarks are included as a bonus.
Some languages, like SML, Haskell, and Scala, have built-in support for pattern matching, which is a generic way of branching based on the structure of data.
While not without its drawbacks, pattern matching can help eliminate a lot of boilerplate, and it's often cited as a reason why functional programming languages are so concise.
In this talk, John A. De Goes talks about the differences between built-in patterns, and so-called first-class patterns (which are "do-it-yourself" patterns implemented using other language features).
Unlike built-in patterns, first-class patterns aren't magical, so you can store them in variables and combine them in lots of interesting ways that aren't always possible with built-in patterns. In addition, almost every programming language can support first-class patterns (albeit with differing levels of effort and type-safety).
During the talk, you'll watch as a mini-pattern matching library is developed, and have the opportunity to follow along and build your own pattern matching library in the language of your choice.
Explain what differential cell affinity is, how this process is acco.pdfwasemanivytreenrco51
Explain what differential cell affinity is, how this process is accomplished and why it is
important for morphogenesis.
Solution
Morphogenesis is a process through which the organism develops its shape. The spatial
distribution of organs and cells that belong to a specific group is controlled by this process.
During the stage of embryonic development, the cells are restricted to particular layers
(differential affinity) because all of them are joined by the same adhesion molecules. This
homotypic cell adhesion keeps all the cells of similar type intact and form boundaries with the
cells with different type of adhesion molecules, thus maintaining the organ’s or tissue’s shape..
Explain in detail how OFDM helps mitigates multipath fading effects..pdfwasemanivytreenrco51
Explain in detail how OFDM helps mitigates multipath fading effects.
Thank you! As detailed as possible!
Solution
Multipath fading is a feature that needs to be taken into account when designing or developing a
radio communications system. In any terrestrial radio communications system, the signal will
reach the receiver not only via the direct path, but also as a result of reflections from objects such
as buildings, hills, ground, water, etc that are adjacent to the main path.The overall signal at the
radio receiver is a summation of the variety of signals being received. As they all have different
path lengths, the signals will add and subtract from the total dependent upon their relative
phases.
At times there will be changes in the relative path lengths. This could result from either the radio
transmitter or receiver moving, or any of the objects that provides a reflective surface moving.
This will result in the phases of the signals arriving at the receiver changing, and in turn this will
result in the signal strength varying as a result of the different way in which the signals will sum
together. It is this that causes the fading that is present on many signals.
Multipath fading can affect radio communications channels in two main ways. This can give the
way in which the effects of the multipath fading are mitigated..
Entamoeba histolytica is an amoeba responsible for the gastrointestin.pdfwasemanivytreenrco51
Entamoeba histolytica is an amoeba responsible for the gastrointestinal disease amebic
dysentery. While E. histolytica is found in the environment as both an active trophozoite and as
an inactive cyst, virtually all cases of amebic dysentery can be traced to infection with the cyst
form of the organism. Speculate on why the active form of the pathogen causes far fewer cases
of disease.
Solution
Mature cysts are more resistant to atmosphere as well as for antibiotics and Immune system than
mature organism, upon ingestion of active form of organism which gets killed while if mature
cyst enters the human body it remains as it is and later it develops to an active organism which
can cause actual disease..
Describe briefly each mouth part of the grasshopper. b. Describe how.pdfwasemanivytreenrco51
Describe briefly each mouth part of the grasshopper. b. Describe how each of the following
mouth parts of the moth, flesh fly. and mosquito have been modified in comparison to the
grasshopper. Vocabulary used in drawings: lbr = labrum lbm = labium md = mandible mx =
maxillae hyp = hypopharyn p = palp: mxp = maxillary palp lbp = labial palp fc = food
channel sc = salivary channel clp = lower part of the head capsule (not a mouth part) Parts of
labium: bk, hst, lbl
Solution
Answer:
2a)Mouthparts of grass hoper:
1)Mandibles:A grasshoper has a pair of mandibles on each side of head. the mandibles are the
largest and most robust mouthparts of a chewing insect, and it uses them to cut, tear, crush, chew
food items. Two sets of muscles move the mandibles in the coronal plane abductor muscles
move insects mandibles apart adductor muscles bring them together . This they do mainly in
opening and closing their jaws in feeding, but also useful the mandibles as tools, or possibly help
in fighting .The mandibles are caudal to the labrum and anterior to the maxillae.
2)LABRUM: A simple plate-like sclerite that serves as a front lip to help contain the food.it is
broad ,roughly rectangular shaped.
3)MAXILLAE:—pair of structures lying outside and behind the mandibles. Each of them consist
of 5-segmented sensory maxillary palp in addition to other parts. The maxillae are used to
manipulate the food before it enters the mouth. paired appendages with the following parts:
Forms the broad median lower lip consisting of several parts in addition to a pair of 3-segmented
labial palps on either side.
The mandibles and the maxillae grind the food by moving it laterally.
5)HYPOPHARYNX: a tongue-like process that helps mix food and saliva.it is found attached to
the labrum.The hypopharynx is a somewhat globular structure, located medially to the mandibles
and the maxillae. it is membranous and associated with salivary glands. It assists in swallowing
the food.....
Contrast the views of Piaget and Bandura on how children develop..pdfwasemanivytreenrco51
Contrast the views of Piaget and Bandura on how children develop.
Solution
Piaget\'s theory of child - cognitive development:
According to Piaget, cognitive reversibility is a tangible operation state in which children
initially habituate and gain knowledge to understand the things how it works and alter they gain
knowledge to return to their original state.
Paiget illustrated children’s cognitive development in four stages
They are 1. Sensory motor stage
2. Pre-operational stage (2 to7 years of old)
3. Concrete operational (children of 7 to 11 years old)
4. Formal operations (11 to 16 years adolescence stage)
Preoperational stage defined as the stage at which children mainly habituate and gain knowledge
to speak at the age of 2 to 7 years. In this stage child develops to use symbols initially and finally
form enduring and manipulating symbols. In this stage, children do not develop logical thinking
or reasoning. In concrete operational stage, children often develop logical thinking and efficient
cognitive abilities.
Preoperational stage includes the following stages of features such as a. animism; b. artificialism
c. transductive reasoning.
According Piaget conservation has illustrated as the process in which children at the age of 2 to
seven are unable to think logically. But their logical thinking starts at the age of seven to eleven
years. The formal operational stage starts after 11 years in which children often learn about
mathematical models and apply them to solve a variety of problems.
According to Bandura, children learning perspective is mainly associated with anticipated
outcomes based on observational learning where environmental cues has specific impact on
them.
Behavioral cognitive theory: According to Bandura, behavior is resultant response due to
complex interplay between environmental cues & inner processes in any species finally attain
empirical validity.
As per cognitive theories behavioral concern always involves a variety of environmental
stimulus and concerned response, it could be voluntary response or reflexive response. These
responses are associated with cues. This behavior is mainly based on external rewards through
cortico-striatal pathway..
Compute the probability of event E if the odds in favor of E are 31.pdfwasemanivytreenrco51
Compute the probability of event E if the odds in favor of E are 3/19 14/27 6/11 11/7 P(E) =
(Type the probability as a fraction. Simplify your answer)
Solution
(A) Given odds in the favour of event E are 3/19
Therefore, probability = 3/(3+19) = 3/22
(B) Given odds in favour of event E are 14/27. Therefore,
probability = 14/(14+27) = 14/41
(c) Given odds in favour of event E are 6/11. Therefore,
probability = 6/(6+11) = 6/17
(d) Given odds in favour of event E are 11/7. Therefore,
probability =11/(11+7) = 11/18.
Compare and contrast the world population with that of the United St.pdfwasemanivytreenrco51
Compare and contrast the world population with that of the United States. Is the United States
becoming a larger or a smaller part of the world\'s population?
Solution
According to the results of the 2015 Revision, the world population reached 7.3 billion as of
mid-
2015 (table 1), implying that the world has added approximately one billion people in the span
of the last
twelve years. Sixty per cent of the global population lives in Asia (4.4 billion), 16 per cent in
Africa (1.2
billion), 10 per cent in Europe (738 million), 9 per cent in Latin America and the Caribbean (634
million),
and the remaining 5 per cent in Northern America (358 million) and Oceania (39 million). China
(1.4
billion) and India (1.3 billion) remain the two largest countries of the world, both with more
than 1 billion
people, representing 19 and 18 per cent of the world’s population, respectively.
It is becoming larger part of worlds population.
In 2013 its was 4.42% of worlds population..
8. What protocol is are layers 6 and 7 of the OSI model based on.pdfwasemanivytreenrco51
8. What protocol is are layers 6 and 7 of the OSI model based on?
1. TCP/IP
2. IPsec
3. FTP
4. PPTP
9. Which network services provides the mapping and translation of domain names to IP
addresses?
1. HTTP
2. DHCP
3. DNS
4. FTP
10. With a proxy server enabled on a firewall, the firewall stores web pages where?
1. Local usb drive
2. RAM
3. Local hard drive
4. FTP server
11 The ________ provides definitions for seen layers, which each layer provides services that
depend on higher level layers depend on.
2. TCP
3. FTP
4. OSI
5. DNS
12. Which of the ways listed below can firewall security be completely bypassed? (Multiple
answers)
1. HTTP sites are never secure
2. E-mail is never secure
3. A computer not owned by the enterprise should never be connected to the enterprise
4. Files from home computers and files from office computers should never commingle.
Solution
Osi model 6 and 7 layer consist of FTP as the 7 th layer is application layer.This layer is used to
access network services.
It is useful for remote file sharing.
remote printer access etc..
Which statement is true of serous membranesA) They line closed cavi.pdfwasemanivytreenrco51
Which statement is true of serous membranes?A) They line closed cavities such as the pleural,
peritoneal, and pericardial cavities.B) They are dry and form the epithelia of the skin.C) They
cover small areas within the body.D) Their openings are made of stratified squamous epithelial
tissue.E) They secrete mucus.
Solution
A,C and E statements are correct. They are not dry and will not form epithelia of skin. They
secrete mucous like substance. They are not made of compound epithelium.they are made of
simple epithelium..
Which of the following is not included in the calculation of the VIX .pdfwasemanivytreenrco51
Which of the following is not included in the calculation of the VIX cross-asset rank? O
Treasury bond volatility Gold volatility High yield corporate bond volatility EUR volatility
Solution
which of the following is not involed in VIX Cross asset rank
Answer:high yeild coporate bond volatility.
What is an Accountable Care Organizations (ACO) How does an ACOs .pdfwasemanivytreenrco51
What is an Accountable Care Organizations (ACO)? How does an ACO\'s economics work to
manage costs and quality?
Solution
Accountable Care Organizations (ACOs) are groups of doctors, hospitals, and other health care
providers, who come together voluntarily to give coordinated high quality care to their Medicare
patients.
An accountable care organization (ACO) is a healthcare organization characterized by a payment
and care delivery model that seeks to tie provider reimbursements to quality metrics and
reductions in the total cost of care for an assigned population of patients. A group of coordinated
health care providers forms an ACO, which then provides care to a group of patients. The ACO
may use a range of payment models (capitation, fee-for-service with asymmetric or symmetric
shared savings, etc.). The ACO is accountable to the patients and the third-party payer for the
quality, appropriateness and efficiency of the health care provided. According to the Centers for
Medicare and Medicaid Services (CMS), an ACO is \"an organization of health care providers
that agrees to be accountable for the quality, cost, and overall care of Medicare beneficiaries who
are enrolled in the traditional fee-for-service program who are assigned to it.
Design and Structure
There is no single organizational model for developing an ACO. ACOs may be formed and
organized by health systems using employed and contracted physicians, by integrated delivery
systems, by physician groups (either primary care or multispecialty) or through joint ventures or
contractual relationships among providers. Regardless of the organizational structure, an ACO
must be physician-led and physician-driven. Physician leadership is critical because an ACO is
primarily a vehicle for clinical integration, not financial or risk integration. Only physicians are
able to develop, monitor and adjust clinical care protocols that can more efficiently use resources
based on documented effectiveness.
Qualifying ACOs will be assigned a pool of patients whose care the ACO will be responsible for
managing in a cost-effective and clinically appropriate manner. The ACO will need to develop
internal mechanisms for monitoring and managing costs and quality that cut across traditional
reporting lines and result in a higher degree of clinical interdependence than is typical in a less-
integrated medical community.
The PPACA states that any of the following groups of providers of services and suppliers that
have established a mechanism for shared governance are eligible to participate, in accordance
with regulations to be developed by the Secretary of Health and Human Services (HHS):
ACOs Under Health Reform
Section 3022 of PPACA requires HHS to establish a shared savings program under which
qualifying ACOs may be eligible for incentive payments. The criteria in the statute, which will
need to be further defined by regulation, include:
ACOs will be required to measure and report their progress to HHS, includ.
True or false A selective force (such as antibiotics) must be prese.pdfwasemanivytreenrco51
True or false: A selective force (such as antibiotics) must be present for DNA mutations to occur.
Solution
Answer: False
Reason:
Selective forces are not going to exert any impact on the alterations of genome mainly via
causing DNA mutations. For example, antibiotics are going to cause change in the RNA
structure or change in enzyme moieties inside the cells of microbial species result in inhibition of
growth without causing damage or random change (mutation) in the DNA. Natural selection
always trying to eliminate any deleterious alleles in the host cells but do not cause any selective
DNA mutation when taking antibiotics.
The receptors in the feedback loop regulating ADH secretion are osmor.pdfwasemanivytreenrco51
The receptors in the feedback loop regulating ADH secretion are osmoreceptors in the kidneys
True False
Solution
Ans 39: The receptor in feedback loop regulating ADH secretion are osmoreceptors in kidneys-
False because hypothalmus regulate vasopressin (ADH) secretion from the posterior pituitary to
return blood concentration to normal.
The UV spectrum of the hot B0V star is significantly below the conti.pdfwasemanivytreenrco51
The UV spectrum of the hot B0V star is significantly below the continuum level due to a large
number of absorption lines caused by highly ionized heavy elements. Can you explain why the
B0V star, rather than the cooler stars, has all those lines?
http://i67.tinypic.com/2u9hn2v.jpg
Solution
In B0V stars have these emission and absorption lines due to radiative recombination between
Hydrogen and Helium ions with electrons and other excitd metal ions present withine the star.
Hot and young stars are able to ionize the gas around them which then produces emission
spectra. This region is known as H|| region. This occurs only around hot stars as UV photons are
required to ionize hydrogen atoms..
Q2 For any drosophila population, what is the proportion of live ho.pdfwasemanivytreenrco51
Q2: For any drosophila population, what is the proportion of live homozygous (Cy/Cy) flies?
Explain.
Solution
The proportion of live homozygous (Cy/Cy) flies in a drosophila population will be 0.This is
because Cy acts as a mutation that is recessive and lethal.In homozygous condition,Cy/Cy flies
will die even before they will reach the adult stage..
Prove asymptotic upper and lower hounds for each of the following sp.pdfwasemanivytreenrco51
Prove asymptotic upper and lower hounds for each of the following specified otherwise, assume
that in each case, T(n) = 1 (or any small constant) for small value You may assume that n = c^k
for some constant c that you choose. Make your bounds as tight as (No need to specify the
origin of your guess.) T(n0 = 8T(n/3) + n^1.83838383... T(n) = T(n - 1) = 1/n T(n) = 16T(n/2)
+ (n log n)^4. T(n) = 2T(n/2) + n/lg n. T(n) = T(n - 1) + T(n - 2) + 1 with base case of T(1) = 1
and T(2) = 2
Solution
A statement are often outlined as a declaratory sentence, or a part of a sentence, that\'s capable of
getting a truth-value, like being true or false. So, as an example, the subsequent area unit
statements:
George W. Bush is that the forty third President of the us.
Paris is that the capital of France.
Everyone born on Monday has purple hair.
Sometimes, a press release will contain one or a lot of alternative statements as elements.
contemplate as an example, the subsequent statement:
Either Ganymede may be a moon of Jupiter or Ganymede may be a moon of Saturn.
While the on top of sentence is itself a press release, as a result of it\'s true, the 2 elements,
\"Ganymede may be a moon of Jupiter\" and \"Ganymede may be a moon of Saturn\", area unit
themselves statements, as a result of the primary is true and therefore the second is fake.
The term proposition is typically used synonymously with statement. However, it\'s typically
accustomed name one thing abstract that 2 totally different statements with an equivalent which
means area unit each aforementioned to \"express\". during this usage, nation sentence, \"It is
raining\", and therefore the French sentence \"Il pleut\", would be thought-about to specific an
equivalent proposition; equally, the 2 English sentences, \"Callisto orbits Jupiter\" and \"Jupiter
is orbitted by Callisto\" would even be thought-about to specific an equivalent proposition.
However, the character or existence of propositions as abstract meanings continues to be a matter
of philosophical dispute, and for the needs of this text, the phrases \"statement\" and
\"proposition\" area unit used interchangeably.
Propositional logic, conjointly referred to as linguistic string logic, is that branch of logic that
studies ways that of mixing or neutering statements or propositions to create a lot of difficult
statements or propositions. change of integrity 2 easier propositions with the word \"and\" is one
common approach of mixing statements. once 2 statements area unit joined along side \"and\",
the advanced statement fashioned by them is true if and as long as each the element statements
area unit true. owing to this, associate argument of the subsequent kind is logically valid:
Paris is that the capital of France and Paris contains a population of over 2 million.
Therefore, Paris contains a population of over 2 million.
Propositional logic for the most part involves learning logical connectives like the words \"and\"
and \"or\" and therefo.
pls show details.thx 1. Consider the vectors v1 (a Find the projecti.pdfwasemanivytreenrco51
pls show details.thx 1. Consider the vectors v1 (a Find the projection, proj1(v2), of v2 on v1 and
draw a somewhat accurate picture to show the vectors and the projection. (b) Find the projection,
proju2(v1), of vi on v2 and draw a somewhat accurate picture to show the vectors and the
projection.
Solution
1).
Please create an infographic for medical fraud! Thank you so much.pdfwasemanivytreenrco51
Please create an infographic for medical fraud! Thank you so much!
Please create an infographic for medical fraud! Thank you so much!
Solution
causing unintended injury or illness. Duplicate medical records and overlays can also lead to
reimbursement losses, reductions in medical data integrity, administrative inefficiencies,
resource drains, liability concerns, and perhaps most significantly, compromised care delivery
and threats to patient safety. Adding fuel to the fire, the biggest threat to the creation of
duplicates and overlays in healthcare is when a hospital adopts an electronic health record system
(EHR) and these numbers threaten to grow exponentially as hospitals share more information
through Integrated Delivery Networks (IDNs) and Health Information Exchanges (HIEs).
The continued threat that duplicate medical records and overlays place on patient safety and
hospital liability causes unintended injuries and illnesses from patient data matching errors.
Despite the absence of patient identification industry standards and a lack of national leadership
on mandating unique patient identifiers, the healthcare industry is increasingly deploying
modernized patient identification technologies that are drastically reducing the number of
duplicate medical records and overlays to help secure patient safety, reduce the overall cost of
care delivery, and improve medical data integrity
Technologies such as iris biometrics for patient identification and smart cards are rising stars of
securing patient data matching accuracy and provide healthcare providers with competitive
advantages to help stop duplicate medical records and overlays. This Infographic on the effect
that duplicate medical records and overlays have on the healthcare industry is a stark reminder
that this problem still persists and negatively impacts patient safety, medical data integrity, and
the cost of care delivery but modern patient data matching tools are available to help prevent
their creation and keep patients safe..
Internet Programming. For event-driven architecture, how does pollin.pdfwasemanivytreenrco51
Internet Programming. For event-driven architecture, how does polling work?
Solution
Answer:-
Event-driven Architecture :
The service-oriented architecture (SOA), combining the intelligence and proactiveness of event-
driven architecture with the organizational capabilities found in service offerings.
Before event-driven SOA, the typical SOA platform orchestrated services centrally, through pre-
defined business processes, assuming that what should have already been triggered is defined in
a business process.
This older approach does not account for events that occur across, or outside of, specific business
processes.
Prototypical Service-oriented Architecture :
The prototypical service-oriented architecture paradigm contains four essential elements:
1) multiple low-level system events that, separately do not appear to have any relationship but
through pattern detection by comparing these many events some unusual or less obvious
correlation becomes clear.
2) some amount of data enrichment by infusion of related information to each event to more
clearly illustrate how the many events are related.
3) a trigger condition which when not met, the business-level event is not created, but when the
trigger condition is met, the higher-level business event is created.
4) some human or automated process that is invoked when the trigger event is reached.
Multiple low-level system events:
1) The human intelligence processes and gathers these relationships faster than current artificial
computational capability
2) The engine uses preponderant causal vector rate-of-change propagation to code the
relationship among the events and establishes a partial order in which it validates the causality
perceived between multiple occurrences.
3) Multiple low-level system events are processed by the Causal Vector Engine and compared
against these rules to trigger higher-level Business Events
4) a Service Component Architecture (SCA). Like any SCA component, the program accesses a
mediation flow through exports that it provides, and the mediation flow forwards messages to
other external services via imports.
5) Once the developers have coded the custom binding for both export and import, they can start
to focus on the mediation flow component.
6) In the WebSphere Integration Developer assembly editor, this is done by the JMS Custom
Binding Mediation Component where each operation on the flow component\'s interface is
represented by a request and a response .
7) Service Data Objects (SDO) framework provides a unified framework for data application
development. With SDO, developers do not need to be familiar with any specific API in order to
access and utilize data.
Business-level trigger condition :
A business-level trigger condition enables the service-oriented architecture, architecture to
establish real-time customer intelligence, marketing automation and customer loyalty solutions,
among other features..
PCAOB Please respond to the followingGo to the PCAOB Website..pdfwasemanivytreenrco51
\"PCAOB\" Please respond to the following:
Go to the PCAOB Website. Next, analyze the role of the PCAOB to protect the public. Based on
your analysis, give your opinion as to whether you believe they possess enough of the requisite
power to be effective in accomplishing their mission. Provide support for your opinion.
Solution
The Public Company Accounting Oversight Board is a nonprofit corporation set up to protect
investors and the public interest by providing informative, accurate, and independent audit
reports and to oversee the audits of public companies and broker-dealers.
ROLE OF PCAOB
1.It explore ways to improve its audit requirements and their implementation by audit firms,
while preserving the intended benefits.
2. It helps to inform a current standards setting project dealing with firms’ systems of quality
controls.
3. helps in setting standard for the project on the supervision of other auditors that
strengthen the standards in that area.
4.looks after auditing requirements that will aid users and investors of financial statements to
make informed decisions and assist in restoring the confidence
5.The ultimate mission of PCAOB is to protect public interest and to protect investors by
overseeing whether set out norms are being adhered to or not.
OPINION
There has been a period of extreme economic stress, triggered and then compounded by a
meltdown of financial system.PCAOB shall succeed in meeting its goal by Improving the
transparency of the financial condition of registered accounting firms .Another way to enhance
transparency is to improve the reporting model for financial statements which would contribute
to the investor’s understanding of the financial statements and the audit process..
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Java Question Consider a maze made up of a rectangular array of squ.pdf
1. Java Question: Consider a maze made up of a rectangular array of squares. The maze will
contain a character (either +, -, or |) to represent a blocked square, and to form the walls of the
maze. Mazes will have only one entrance at the Coordinate (0, 1), with only one exit in the lower
right hand corner of the maze. Beginning at the entrance to the maze, find a path to the exit at the
bottom right of the maze. You may only move up, down, left, and right. Each square in the maze
can be in one of four states: clear (space), blocked (X), path (.), or visited (*). Initially, after the
maze has been read in from the file, each square will be either clear or blocked. If a square lies
on a successful path, mark it with a period. If you visit a square but it does not lead to a
successful path, mark it as visited with an asterisk.
Solution
import java.io.*;
class Maze
{
// Any MazeSquare field is true if the corresponding
// wall exists, and false otherwise.
class MazeSquare
{
public boolean hasTopWall;
public boolean hasRightWall;
public boolean hasLeftWall;
public boolean hasBottomWall;
public MazeSquare()
{
hasTopWall = hasRightWall = hasLeftWall = hasBottomWall = false;
}
}
// This Exception class will help pinpoint format errors in the data file.
class MazeFormatException extends Exception
{
public final static int noError = 0;
public final static int badRowAndColumnCounts = 1;
public final static int wrongNumberOfEntriesInRow = 2;
public final static int badEntry = 3;
2. public int lineNumber;
public int problem;
public MazeFormatException( int line, int prob )
{
lineNumber = line;
problem = prob;
}
}
// Number of rows and columns in the maze.
private int nRows;
private int nColumns;
// The array of maze squares themselves.
MazeSquare[][] square;
public static void main( String[] args )
{
if( args.length != 1 )
{
System.err.println( "Usage: java Maze mazeFileName" );
System.exit( 1 );
}
Maze knosos = null;
try
{
knosos = new Maze( args[0] );
}
catch( FileNotFoundException e )
{
System.err.println( "Can't open " + args[0] + ". Check your spelling." );
System.exit( 1 );
}
catch( IOException e )
{
System.err.println( "Severe input error. I give up." );
System.exit( 1 );
}
catch( MazeFormatException e )
3. {
switch( e.problem )
{
case MazeFormatException.badRowAndColumnCounts:
System.err.println( args[0] + ", line " + e.lineNumber + ": row and column counts
expected" );
break;
case MazeFormatException.wrongNumberOfEntriesInRow:
System.err.println( args[0] + ", line " + e.lineNumber + ": wrong number of entries"
);
break;
case MazeFormatException.badEntry:
System.err.println( args[0] + ", line " + e.lineNumber + ": non-hexadecimal digit
detected" );
break;
default:
System.err.println( "This should never get printed." );
break;
}
System.exit( 1 );
}
knosos.print( System.out );
}
public Maze()
{
square = null;
nRows = nColumns = 0;
}
public Maze( String fileName ) throws FileNotFoundException, IOException,
MazeFormatException
{
BufferedReader in = null;
in = new BufferedReader( new FileReader( fileName ) );
load( in );
in.close();
4. }
public void load( BufferedReader in ) throws MazeFormatException
{
String line;
String[] tokens;
int lineNumber = 0;
// Get the number of rows and columns. Protect
// against out-of-range values.
try
{
line = in.readLine();
lineNumber++;
tokens = line.split( "s+" );
nRows = Integer.parseInt( tokens[0] );
nColumns = Integer.parseInt( tokens[1] );
if( nRows <= 0 || nColumns <= 0 )
throw new Exception();
}
catch( Exception e )
{
throw new MazeFormatException( lineNumber,
MazeFormatException.badRowAndColumnCounts );
}
// Allocate the 2D array of MazeSquares.
square = new MazeSquare[nRows][nColumns];
for( int i=0; i < nRows; i++ )
for( int j=0; j < nColumns; j++ )
square[i][j] = new MazeSquare();
// Read the square values from input into the array of MazeSquares.
for( int i=0; i < nRows; i++ )
{
try
{
line = in.readLine();
lineNumber++;
5. tokens = line.split( "s+" );
if( tokens.length != nColumns )
throw new Exception();
}
catch( Exception e )
{
throw new MazeFormatException( lineNumber,
MazeFormatException.wrongNumberOfEntriesInRow );
}
for( int j=0; j < nColumns; j++ )
{
int squareValue;
try
{
squareValue = Integer.parseInt( tokens[j], 16 );
}
catch( NumberFormatException e )
{
throw new MazeFormatException( lineNumber, MazeFormatException.badEntry );
}
// These are "bitwise AND" operations. We'll discuss them in class.
square[i][j].hasTopWall = ((squareValue & 1) != 0);
square[i][j].hasRightWall = ((squareValue & 2) != 0);
square[i][j].hasBottomWall = ((squareValue & 4) != 0);
square[i][j].hasLeftWall = ((squareValue & 8) != 0);
}
}
}
public void print( PrintStream out )
{
int i, j;
for( i=0; i < nRows; i++ )
{
// Draw the top walls of this row of squares.
out.print( '+' );
![public int lineNumber;
public int problem;
public MazeFormatException( int line, int prob )
{
lineNumber = line;
problem = prob;
}
}
// Number of rows and columns in the maze.
private int nRows;
private int nColumns;
// The array of maze squares themselves.
MazeSquare[][] square;
public static void main( String[] args )
{
if( args.length != 1 )
{
System.err.println( "Usage: java Maze mazeFileName" );
System.exit( 1 );
}
Maze knosos = null;
try
{
knosos = new Maze( args[0] );
}
catch( FileNotFoundException e )
{
System.err.println( "Can't open " + args[0] + ". Check your spelling." );
System.exit( 1 );
}
catch( IOException e )
{
System.err.println( "Severe input error. I give up." );
System.exit( 1 );
}
catch( MazeFormatException e )](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)