Develop Reference

Algorithm for 2 player game subtracting perfect squares from a given number

I need some help with a problem. I recently got this in an interview and I was trying to solve it but having no luck. Here's the question:
Code an algorithm for a game consisting of two players. The input is a positive integer x. Each round, a player deducts a perfect square from the number. The player can choose any perfect square as long as it is less than or equal to the current number and greater than 0. The current player wins if the number becomes 0 after his deduction.
I suppose this should be done using dynamic programming/greedy approach which I am not very well verse with. Or we might need to come up with a sequence of winning/losing numbers and if a player ends up in a winning sequence, he will win no matter what. But how do we come up with such a sequence?
Can anyone help me solve this question possibly in Java?
UPDATE:
Solution 1 as suggested by DAle:
public int subtractPerfectSquares(int n)
{
if (n <= 0)
return 1;
boolean[] isWinningCase = new boolean[n + 1];
for (int i = 1; i <= n; i++)
{
for (int num = 1; num * num <= i; num++)
{
if (!isWinningCase[i - (num * num)])
{
isWinningCase[i] = true;
break;
}
}
}
return isWinningCase[n] ? 1 : 2;
}
Solution2 modified for better understanding:
public int subtractPerfectSquares2(int n)
{
if (n <= 0)
return 1;
boolean[] isWinningCase = new boolean[n + 1];
// if we reach 0, we win
isWinningCase[0] = true;
// 1 is a win
isWinningCase[1] = true;
// 2 is a losing condition. We must define this as this state dictates losing scenarios for further states
isWinningCase[2] = false;
// we start from 3
for (int i = 3; i <= n; i++)
{
for (int num = 1; num * num <= i; num++)
{
int prefectSquare = num * num;
// if we get to 0 from current number or if we get to a losing scenario (to be faced by next player) from current number, then the current state is a winning position
if (i - prefectSquare == 0 || !isWinningCase[i - prefectSquare])
{
isWinningCase[i] = true;
break;
}
}
}
// return player 1 if given number is a winning state else player 2 wins
return isWinningCase[n] ? 1 : 2;
}

The only difference between players in this game is that player 1 goes first. This type of games is called impartial game and the perfect strategies of both players are identical. Furthermore, we can divide all states (integer x) of the game into two types: winning position or losing position using the following rules:
x=0 is a losing position
position x is a winning position if there is at least one possible move that leads to a losing position.
position x is a losing position if every possible move leads to a winning position.
Then we need to determine the type of all positions from 1 to x. Possible implementation:
boolean[] isWinning = new boolean[x+1];
for (int state = 0; state <= x; ++state) {
isWinning[state] = false;
for (int i = 1; i*i <= state; ++i) {
int perfectSquare = i*i;
if (!isWinning[state - perfectSquare]) {
isWinning[state] = true;
break;
}
}
}
If the current player is in the winning position (isWinning[x] == true), you should choose such perfect square that isWinning[x - perfectSquare] == false. This will bring the game (and another player) to a losing position. If the player is in the losing position, nothing could save him, every possible perfect square is equally bad.

Below program will help you to implement the logic. I've tried implementing the code as per your requirement.Make sure to comment if you need improvements or to clarify any misunderstandings of the logic.
public class Game {
public static void main(String[] args) {
int number = 0;
int count = 1;
int player = 1;
System.out.println("Please Enter a positive integer");
try (Scanner sc = new Scanner(System.in);) {
number = sc.nextInt();
while (number > 0) {
int numberArray[] = generatePerfectSquare(1, number);
System.out.println("===================");
System.out.println("perfect square numbers");
for (int i : numberArray) {
System.out.print(i);
System.out.print(" ");
}
System.out.println("");
System.out.println("===================");
player = ((count % 2 == 0) ? 2 : 1);
System.out.println("Round : " + count + " Player : " + player);
System.out.println("Please Enter your prefered perfect square number");
number = number - sc.nextInt();
if (number <= 0) {
System.out.println("****************");
System.out.println("You won the game");
System.out.println("===================");
} else {
System.out.println("===================");
System.out.println("You should try more");
System.out.println("===================");
}
count++;
}
} catch (Exception e) {
System.out.println(e);
}
}
private static int[] generatePerfectSquare(int start, int end) {
if (start > end || start < 0) {
throw new IllegalArgumentException();
}
int[] perfectSquares = new int[end - start];
int n = 0;
int candidate = (int) Math.ceil(Math.sqrt(start));
int square;
while ((square = candidate * candidate) < end) {
perfectSquares[n++] = square;
candidate++;
}
return Arrays.copyOf(perfectSquares, n);
}
Test outcome

Connect 4 Minimax Algorithm (with Alpha-Beta Pruning) in Java

I'm trying to write a connect four game with AI using the minimax algorithm and alpha-beta pruning. I know that the heuristic works OK (at the very least it scores 4 in the row very highly), but for some reason my minimax isn't working.
It seems like the problem is that the current score and the minimax score are always equal, when this obviously should not be the case.
The computer frequently makes stupid decisions like going to a random spot when the human is about to get 4 in a row.
Any and all help would be greatly appreciate.
public class AI {
public Move minimax(Board b, int depth, int alpha, int beta, String player, Move move) {
if(depth == 0 || b.boardScore >= 10000) {
return move;
}
else if(player.equals("computer")) {
Move maxMove = new Move(100,100);
maxMove.moveScore = -10000000;
//Do this for every possible column
for(int x = 0; x < 7; x++) {
//If column is filled, move on.
while(b.findYSpot(x) == -1) {
x++;
if(x > 6) {
x = 0;
}
}
Move m = new Move(b.findYSpot(x), x); // Create the move
b.putPiece(player, m.xMove); // Put the piece down
m.moveScore = b.boardScore;//Set the score the move
Move miniMaxMove = minimax(b, depth-1, alpha, beta, "human", m);
// Maximize the score
if (m.moveScore >= miniMaxMove.moveScore) {
if(m.moveScore >= maxMove.moveScore) {
maxMove = m;
}
}
else {
if(miniMaxMove.moveScore >= maxMove.moveScore) {
maxMove = miniMaxMove;
}
}
b.removePiece(m); //Remove the piece
//Adjust for alpha-beta pruning
alpha = Math.max(alpha, maxMove.moveScore);
if (alpha >= beta) {
break;
}
}
return maxMove;
}
else {
Move minMove = new Move(100, 100);
minMove.moveScore = 10000000;
//Do this for every possible column
for(int x = 0; x < 7; x++) {
// If column is filled, move on
while(b.findYSpot(x) == -1) {
x++;
if(x>6) {
x = 0;
}
}
Move m = new Move(b.findYSpot(x), x); // Make the move
b.putPiece(player, m.xMove); // Put the piece down
m.moveScore = b.boardScore; //Set the score the move
Move miniMaxMove = minimax(b, depth-1, alpha, beta, "computer", m);
//Minimize the score
if (m.moveScore <= miniMaxMove.moveScore) {
if(m.moveScore <= minMove.moveScore) {
minMove = m;
}
}
else {
if(miniMaxMove.moveScore <= minMove.moveScore) {
minMove = miniMaxMove;
}
}
b.removePiece(m);
//Adjust for alpha-beta pruning
beta = Math.min(beta, minMove.moveScore);
if(alpha >= beta) {
break;
}
}
return minMove;
}
}
}

java:implement 8 queen using depth first search

i am try to implement 8 queen using depth search for any initial state it work fine for empty board(no queen on the board) ,but i need it to work for initial state if there is a solution,if there is no solution for this initial state it will print there is no solution
Here is my code:
public class depth {
public static void main(String[] args) {
//we create a board
int[][] board = new int[8][8];
board [0][0]=1;
board [1][1]=1;
board [2][2]=1;
board [3][3]=1;
board [4][4]=1;
board [5][5]=1;
board [6][6]=1;
board [7][7]=1;
eightQueen(8, board, 0, 0, false);
System.out.println("the solution as pair");
for(int i=0;i<board.length;i++){
for(int j=0;j<board.length;j++)
if(board[i][j]!=0)
System.out.println(" ("+i+" ,"+j +")");
}
System.out.println("the number of node stored in memory "+count1);
}
public static int count1=0;
public static void eightQueen(int N, int[][] board, int i, int j, boolean found) {
long startTime = System.nanoTime();//time start
if (!found) {
if (IsValid(board, i, j)) {//check if the position is valid
board[i][j] = 1;
System.out.println("[Queen added at (" + i + "," + j + ")");
count1++;
PrintBoard(board);
if (i == N - 1) {//check if its the last queen
found = true;
PrintBoard(board);
double endTime = System.nanoTime();//end the method time
double duration = (endTime - startTime)*Math.pow(10.0, -9.0);
System.out.print("total Time"+"= "+duration+"\n");
}
//call the next step
eightQueen(N, board, i + 1, 0, found);
} else {
//if the position is not valid & if reach the last row we backtracking
while (j >= N - 1) {
int[] a = Backmethod(board, i, j);
i = a[0];
j = a[1];
System.out.println("back at (" + i + "," + j + ")");
PrintBoard(board);
}
//we do the next call
eightQueen(N, board, i, j + 1, false);
}
}
}
public static int[] Backmethod(int[][] board, int i, int j) {
int[] a = new int[2];
for (int x = i; x >= 0; x--) {
for (int y = j; y >= 0; y--) {
//search for the last queen
if (board[x][y] != 0) {
//deletes the last queen and returns the position
board[x][y] = 0;
a[0] = x;
a[1] = y;
return a;
}
}
}
return a;
}
public static boolean IsValid(int[][] board, int i, int j) {
int x;
//check the queens in column
for (x = 0; x < board.length; x++) {
if (board[i][x] != 0) {
return false;
}
}
//check the queens in row
for (x = 0; x < board.length; x++) {
if (board[x][j] != 0) {
return false;
}
}
//check the queens in the diagonals
if (!SafeDiag(board, i, j)) {
return false;
}
return true;
}
public static boolean SafeDiag(int[][] board, int i, int j) {
int xx = i;
int yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy++;
xx++;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy--;
xx--;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy--;
xx++;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy++;
xx--;
}
return true;
}
public static void PrintBoard(int[][] board) {
System.out.print(" ");
for (int j = 0; j < board.length; j++) {
System.out.print(j);
}
System.out.print("\n");
for (int i = 0; i < board.length; i++) {
System.out.print(i);
for (int j = 0; j < board.length; j++) {
if (board[i][j] == 0) {
System.out.print(" ");
} else {
System.out.print("Q");
}
}
System.out.print("\n");
}
}
}
for example for this initial state it give me the following error:
Exception in thread "main" java.lang.StackOverflowError
i am stuck, i think the error is infinite call for the method how to solve this problem.
any idea will be helpful,thanks in advance.
note:the broad is two dimensional array,when i put (1) it means there queen at this point.
note2:
we i put the initial state as the following it work:
board [0][0]=1;
board [1][1]=1;
board [2][2]=1;
board [3][3]=1;
board [4][4]=1;
board [5][5]=1;
board [6][6]=1;
board [7][1]=1;

[EDIT: Added conditional output tip.]
To add to #StephenC's answer:
This is a heck of a complicated piece of code, especially if you're not experienced in programming Java.
I executed your code, and it outputs this over and over and over and over (and over)
back at (0,0)
01234567
0
1 Q
2 Q
3 Q
4 Q
5 Q
6 Q
7 Q
back at (0,0)
And then crashes with this
Exception in thread "main" java.lang.StackOverflowError
at java.nio.Buffer.<init>(Unknown Source)
...
at java.io.PrintStream.print(Unknown Source)
at java.io.PrintStream.println(Unknown Source)
at Depth.eightQueen(Depth.java:56)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
...
My first instinct is always to add some System.out.println(...)s to figure out where stuff is going wrong, but that won't work here.
The only two options I see are to
Get familiar with a debugger and use it to step through and analyze why it's never stopping the loop
Break it down man! How can you hope to deal with a massive problem like this without breaking it into digestible chunks???
Not to mention that the concept of 8-queens is complicated to begin with.
One further thought:
System.out.println()s are not useful as currently implemented, because there's infinite output. A debugger is the better solution here, but another option is to somehow limit your output. For example, create a counter at the top
private static final int iITERATIONS = 0;
and instead of
System.out.println("[ANUMBERFORTRACING]: ... USEFUL INFORMATION ...")
use
conditionalSDO((iITERATIONS < 5), "[ANUMBERFORTRACING]: ... USEFUL INFORMATION");
Here is the function:
private static final void conditionalSDO(boolean b_condition, String s_message) {
if(b_condition) {
System.out.println(s_message);
}
}
Another alternative is to not limit the output, but to write it to a file.
I hope this information helps you.
(Note: I edited the OP's code to be compilable.)

You asked for ideas on how to solve it (as distinct from solutions!) so, here's a couple of hints:
Hint #1:
If you get a StackOverflowError in a recursive program it can mean one of two things:
your problem is too "deep", OR
you've got a bug in your code that is causing it to recurse infinitely.
In this case, the depth of the problem is small (8), so this must be a recursion bug.
Hint #2:
If you examine the stack trace, you will see the method names and line numbers for each of the calls in the stack. This ... and some thought ... should help you figure out the pattern of recursion in your code (as implemented!).
Hint #3:
Use a debugger Luke ...
Hint #4:
If you want other people to read your code, pay more attention to style. Your indentation is messed up in the most important method, and you have committed the (IMO) unforgivable sin of ignoring the Java style rules for identifiers. A method name MUST start with a lowercase letter, and a class name MUST start with an uppercase letter.
(I stopped reading your code very quickly ... on principle.)

Try to alter your method IsValid in the lines where for (x = 0; x < board.length - 1; x++).
public static boolean IsValid(int[][] board, int i, int j) {
int x;
//check the queens in column
for (x = 0; x < board.length - 1; x++) {
if (board[i][x] != 0) {
return false;
}
}
//check the queens in row
for (x = 0; x < board.length - 1; x++) {
if (board[x][j] != 0) {
return false;
}
}
//check the queens in the diagonals
if (!SafeDiag(board, i, j)) {
return false;
}
return true;
}

8 Non-Attacking Queens Algorithm with Recursion

I'm having trouble coding the 8 queens problem. I've coded a class to help me solve it, but for some reason, I'm doing something wrong. I kind of understand what's supposed to happen.
Also, we have to use recursion to solve it but I have no clue how to use the backtracking I've read about, so I just used it in the methods checking if a position is legitimate.
My board is String [] [] board = { { "O", "O"... etc etc with 8 rows and 8 columns.
If I'm getting anything wrong conceptually or making a grave Java mistake, please say so :D
Thanks!
public void solve () {
int Queens = NUM_Queens - 1;
while (Queens > 0) {
for (int col = 0; col < 8; col++) {
int row = -1;
boolean c = false;
while (c = false && row < 8) {
row ++;
c = checkPos (row, col);
}
if (c == true) {
board[row][col] = "Q";
Queens--;
}
else
System.out.println("Error");
}
}
printBoard ();
}
// printing the board
public void printBoard () {
String ret = "";
for (int i = 0; i < 8; i++) {
for (int a = 0; a < 8; a++)
ret += (board[i][a] + ", ");
ret += ("\n");
}
System.out.print (ret);
}
// checking if a position is a legitimate location to put a Queen
public boolean checkPos (int y, int x) {
boolean r = true, d = true, u = true, co = true;
r = checkPosR (y, 0);
co = checkPosC (0, x);
int col = x;
int row = y;
while (row != 0 && col != 0 ) { //setting up to check diagonally downwards
row--;
col--;
}
d = checkPosDD (row, col);
col = x;
row = y;
while (row != 7 && col != 0 ) { //setting up to check diagonally upwards
row++;
col--;
}
d = checkPosDU (row, col);
if (r = true && d = true && u = true && co = true)
return true;
else
return false;
}
// checking the row
public boolean checkPosR (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && x == 7)
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y, x+1);
}
// checking the column
public boolean checkPosC (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && y == 7)
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y+1, x);
}
// checking the diagonals from top left to bottom right
public boolean checkPosDD (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && (x == 7 || y == 7))
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y+1, x+1);
}
// checking the diagonals from bottom left to up right
public boolean checkPosDU (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && (x == 7 || y == 0))
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y-1, x+1);
}
}

As this is homework, the solution, but not in code.
Try to write a method that only handles what needs to happen on a single column; this is where you are supposed to use recursion. Do backtracking by checking if a solution exists, if not, undo your last change (i.e. change the queen position) and try again. If you only focus on one part of the problem (one column), this is much easier than thinking about all columns at the same time.
And as Quetzalcoatl points out, you are assigning false to your variable in the first loop. You probably do not want to do that. You should always enable all warnings in your compiler (run javac with -Xlint) and fix them.

You are trying some kind of brute-force, but, as you already mentioned, you have no recursion.
Your programs tries to put a queen on the first possible position. But at the end no solution is found. It follows that your first assumption (the position of your first queen) is invalid. You have to go back to this state. And have to assume that your checkPos(x,y) is false instead of true.
Now some hints:
As mentioned before by NPE int[N] queens is more suitable representation.
sum(queens) have to be 0+1+2+3+4+5+6+7=28, since a position has to be unique.
Instead of checking only the position of the new queen, you may check a whole situation. It is valid if for all (i,j) \in N^2 with queen(i) = j, there exists no (k,l) != (i,j) with abs(k-i) == abs(l-j)

Recursive Searching in Java

So I've been writing a program for the game boggle. I create a little board for the user to use, but the problem is I don't know how to check if that word is on the board recursively. I want to be able to check if the word the entered is indeed on the board, and is valid. By valid I mean, the letters of the word must be adjacent to each other. For those who have played boggle you'll know what I mean. All I want to do is check if the word is on the board.
This is what I have so far ....
import java.io.*;
public class boggle {
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
private String s = "";
private int [][] lettersNum = new int [5][5];
private char [][] letters = new char [5][5];
private char [] word = new char [45]; // Size at 45, because the longest word in the dictionary is only 45 letters long
private char [] temp;
public void generateNum()
{
for (int row = 0; row < 5; row ++)
{
for (int col = 0; col < 5; col++)
{
lettersNum [row][col] = (int) (Math.random() * 26 + 65);
}
}
}
public void numToChar()
{
for (int row = 0; row < 5; row ++)
{
for (int col = 0; col < 5; col++)
{
letters [row][col] = (char)(lettersNum[row][col]);
}
}
}
public void display()
{
for (int row = 0; row < 5; row ++)
{
for (int col = 0; col < 5; col++)
{
System.out.print(letters[row][col]);
}
System.out.println("");
}
}
public void getInput() throws IOException
{
System.out.println("Please enter a word : ");
s=br.readLine();
s=s.toUpperCase();
word = s.toCharArray();
}
public int search(int row, int col)
{
if((row <0) || (row >= 5) || (col < 0) || (col >= 5))
{
return (0);
}
else
{
temp = word;
return (1+ search(row +1, col) +
search(row -1, col) +
search(row, col + 1) +
search(row, col-1) +
search(row +1, col +1)+
search(row +1, col -1)+
search(row -1, col +1)+
search(row -1, col -1));
}
}
}
The search was my searching algorithm to check if the word is on the board but I don't know if it is correct or if it will work. Furthermore, I don't know how to actually tell the user that the word is valid !
Thanks for all the help :)
SO I tried to use what you suggested below but I dont really understand the int [5][5] thing. So this is what I tried, but I keep getting out of bounds errors ! Here is the soruce ...
public void locate()
{
temp = word[0];
for (int row = 0; row <5; row++)
{
for (int col = 0; col <5; col++)
{
if(temp == letters[row][col])
{
search(row,col);
}
}
}
}
public int search(int row, int col)
{
if(letters[row][col-1]==word[count]) // Checks the letter to the left
{
count++;
letters[row][col-1] = '-'; // Just to make sure the program doesn't go back on itself
return search(row, col-1);
}
else if (letters[row][col+1] == word[count])// Checks the letter to the right
{
count++;
letters[row][col+1] = '-';// Just to make sure the program doesn't go back on itself
return search(row, col +1);
}
else if (letters[row+1][col]== word[count])// Checks the letter below
{
count++;
letters[row+1][col] = '-';// Just to make sure the program doesn't go back on itself
return search(row +1 , col);
}
else if (letters[row-1][col]== word[count])// Checks the letter above
{
count++;
letters[row-1][col] = '-';// Just to make sure the program doesn't go back on itself
return search(row -1 , col);
}
else if (letters[row-1][col-1]== word[count])// Checks the letter to the top left
{
count++;
letters[row-1][col-1] = '-';// Just to make sure the program doesn't go back on itself
return search(row -1 , col-1);
}
else if (letters[row-1][col+1]== word[count])// Checks the letter to the top right
{
count++;
letters[row-1][col+1] = '-';// Just to make sure the program doesn't go back on itself
return search(row -1 , col+1);
}
else if (letters[row+1][col-1]== word[count])// Checks the letter to the bottom left
{
count++;
letters[row+1][col-1] = '-';// Just to make sure the program doesn't go back on itself
return search(row +1 , col-1);
}
else if (letters[row+1][col+1]== word[count])// Checks the letter to the bottom right
{
count++;
letters[row+1][col+1] = '-';// Just to make sure the program doesn't go back on itself
return search(row +1 , col+1);
}
return 0;
}
private int count = 0; (was declared at the top of the class, in case you were wondering where I got the word[count] from

Your current search function doesn't actually do anything. I'm assuming this is homework so, no free lunch ;)
The simplest approach would be to have two recursive functions:
public boolean findStart(String word, int x, int y)
This will do a linear search of the board looking for the first character in word. If your current location doesn't match, you call yourself with the next set of coords. When it finds a match, it calls your second recursive function using word, the current location, and a new, empty 4x4 matrix:
public boolean findWord(String word, int x, int y, int[][] visited)
This function first checks to see if the current location matches the first letter in word. If it does, it marks the current location in visited and loops through all the adjoining squares except ones marked in visited by calling itself with word.substring(1) and those coords. If you run out of letters in word, you've found it and return true. Note that if you're returning false, you need to remove the current location from visited.
You can do this with one function, but by splitting out the logic I personally think it becomes easier to manage in your head. The one downside is that it does do an extra comparison for each first letter in a word. To use a single function you would need to keep track of what "mode" you were in either with a boolean or a depth counter.
Edit: Your longest word should only be 16. Boggle uses a 4x4 board and a word can't use the same location twice. Not that this really matters, but it might for the assignment. Also note that I just did this in my head and don't know that I got it 100% right - comments appreciated.
Edit in response to comments:
Here's what your iterative locate would look like using the method I outline above:
public boolean locate(String word)
{
for (int row = 0; row < 4; row++)
{
for (int col =0; col < 4; col++)
{
if (word.charAt(0) == letters[row][col])
{
boolean found = findWord(word, row, col, new boolean[4][4]);
if (found)
return true;
}
}
}
return false;
}
The same thing recursively looks like the following, which should help:
public boolean findStart(String word, int x, int y)
{
boolean found = false;
if (word.charAt(0) == letters[x][y])
{
found = findWord(word, x, y, new boolean[4][4]);
}
if (found)
return true;
else
{
y++;
if (y > 3)
{
y = 0;
x++;
}
if (x > 3)
return false;
}
return findStart(word, x, y);
}
So here's findWord() and a helper method getAdjoining() to show you how this all works. Note that I changed the visited array to boolean just because it made sense:
public boolean findWord(String word, int x, int y, boolean[][] visited)
{
if (letters[x][y] == word.charAt(0))
{
if (word.length() == 1) // this is the last character in the word
return true;
else
{
visited[x][y] = true;
List<Point> adjoining = getAdjoining(x,y,visited);
for (Point p : adjoining)
{
boolean found = findWord(word.substring(1), p.x, p.y, visited);
if (found)
return true;
}
visited[x][y] = false;
}
}
return false;
}
public List<Point> getAdjoining(int x, int y, boolean[][] visited)
{
List<Point> adjoining = new LinkedList<Point>();
for (int x2 = x-1; x2 <= x+1; x2++)
{
for (int y2 = y-1; y2 <= y+1; y2++)
{
if (x2 < 0 || x2 > 3 ||
y2 < 0 || y2 > 3 ||
(x2 == x && y2 == y) ||
visited[x2][y2])
{
continue;
}
adjoining.add(new Point(x2,y2));
}
}
return adjoining;
}
So now, after you get input from the user as a String (word), you would just call:
boolean isOnBoard = findStart(word,0,0);
I did this in my head originally, then just went down that path to try and show you how it works. If I were to actually implement this I would do some things differently (mainly eliminating the double comparison of the first letter in the word, probably by combining the two into one method though you can do it by rearranging the logic in the current methods), but the above code does function and should help you better understand recursive searching.