Answered step by step
Verified Expert Solution
Question
00
1 Approved Answer
* * * * * * * * * * * * * * * * * * * * * * * * *
You must use all the given methods without changing their signature. However you may add new methods using the given ones if necessary. public class PegSolitaireBoardProfVersion protected char pegBoard; private static final int UP ; private static final int DOWN ; private static final int LEFT ; private static final int RIGHT ; Initializes the board at its starting position XXX XXX XXXXXXX XXXOXXX XXXXXXX XXX XXX public PegSolitaireBoardProfVersion this.pegBoard new char; for int i ; i ; i for int j ; j ; j if i && j this.pegBoardij; else if i && j this.pegBoardij; else if i && j this.pegBoardij; else if i && j this.pegBoardij; else if i && j this.pegBoardijO; else this.pegBoardijX; moves the Xs and Os to the corresponding positions after the jump happens private boolean moveForwardint x int y int direction Implement moveForward here The error will go away once you implement and returns a boolean value return false; moves the Xs and Os back to their initial position before the jump occurred private void moveBackwardint x int y int direction Implement moveBackward here private void addPegint x int y pegBoardxyX; private void removePegint x int y pegBoardxyO; check if the jump is valid checks if the result of the jump will be out of bounds checks that the X performing the jump has one more X in between itself and the O private boolean checkMoveint x int y int newX, int newY Implement checkMove here The error will go away once you implement and returns a boolean value return false; getNewX and getNewY get the x and y coordinates of the position where the X will occupy after the jump private int getNewXint x int direction int newX x; switch direction case RIGHT: newX newX ; break; case LEFT: newX newX ; return newX; private int getNewYint y int direction int newY y; switch direction case UP: newY newY ; break; case DOWN: newY newY ; return newY; private void print Implement print here recursive and backtracking algorithm to compute all the possible moves loops through the rows and columns of the matrix and tests if a jump in each of the directions is a valid move if it is a valid move it will perform the move and update the board and store it into the array at the corresponding move count base case occurs if the move count is equal to AND the centre position is still filled with an X otherwise it calls the recursive method and moves on to the next move if the base case is not reached, than the backtracking method is called and tries a new jump direction public boolean findSolutionint move Implement findSolution here The error will go away once you implement and returns a boolean value return false; considering this code, help me implement the peg solitaire so the solution can look like this: X
You must use all the given methods without changing their signature.
However you may add new methods using the given ones if necessary.
public class PegSolitaireBoardProfVersion
protected char pegBoard;
private static final int UP ;
private static final int DOWN ;
private static final int LEFT ;
private static final int RIGHT ;
Initializes the board at its starting position
XXX
XXX
XXXXXXX
XXXOXXX
XXXXXXX
XXX
XXX
public PegSolitaireBoardProfVersion
this.pegBoard new char;
for int i ; i ; i
for int j ; j ; j
if i && j
this.pegBoardij;
else if i && j
this.pegBoardij;
else if i && j
this.pegBoardij;
else if i && j
this.pegBoardij;
else if i && j
this.pegBoardijO;
else
this.pegBoardijX;
moves the Xs and Os to the corresponding positions after the jump happens
private boolean moveForwardint x int y int direction
Implement moveForward here
The error will go away once you implement and returns a boolean
value
return false;
moves the Xs and Os back to their initial position before the jump
occurred
private void moveBackwardint x int y int direction
Implement moveBackward here
private void addPegint x int y
pegBoardxyX;
private void removePegint x int y
pegBoardxyO;
check if the jump is valid
checks if the result of the jump will be out of bounds
checks that the X performing the jump has one more X in between itself
and the O
private boolean checkMoveint x int y int newX, int newY
Implement checkMove here
The error will go away once you implement and returns a boolean
value
return false;
getNewX and getNewY get the x and y coordinates of the position where the X
will occupy after the jump
private int getNewXint x int direction
int newX x;
switch direction
case RIGHT:
newX newX ;
break;
case LEFT:
newX newX ;
return newX;
private int getNewYint y int direction
int newY y;
switch direction
case UP:
newY newY ;
break;
case DOWN:
newY newY ;
return newY;
private void print
Implement print here
recursive and backtracking algorithm to compute all the possible moves
loops through the rows and columns of the matrix and tests if a jump in each
of the directions is a valid move
if it is a valid move it will perform the move and update the board and store
it into the array at the corresponding move count
base case occurs if the move count is equal to AND the centre position is
still filled with an X otherwise it calls the recursive method and moves
on to the next move
if the base case is not reached, than the backtracking method is called and
tries a new jump direction
public boolean findSolutionint move
Implement findSolution here
The error will go away once you implement and returns a boolean
value
return false;
considering this code, help me implement the peg solitaire so the solution can look like this:
X
Step by Step Solution
There are 3 Steps involved in it
Step: 1
Get Instant Access with AI-Powered Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started