TicTacToe -- Check winner?

[Shazer2]

I designed a TicTacToe console game, and I'm trying to figure out how to check if somebody has one. My current code for the game is below.

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package tictactoe; import java.util.Scanner; public class TicTacToe {    private String[][] board = new String[3][3];         private String turn = "O";    private int turns = 0;        public void gameLoop() {       for (int y = 0; y < 3; y++) {          board[y][0] = " ";          board[y][1] = " ";          board[y][2] = " ";          for (int x = 0; x < 3; x++) {             board[0][x] = " ";             board[1][x] = " ";             board[2][x] = " ";          }       }             while (true) {          try {             Scanner scan = new Scanner(System.in);                         System.out.println("      |      |      ");             System.out.println("   " + board[0][0] + "  |   " + board[0][1] + "  |   " + board[0][2] + "  ");             System.out.println("______|______|______");             System.out.println("      |      |      ");             System.out.println("   " + board[1][0] + "  |   " + board[1][1] + "  |   " + board[1][2] + "  ");             System.out.println("______|______|______");             System.out.println("      |      |      ");             System.out.println("   " + board[2][0] + "  |   " + board[2][1] + "  |   " + board[2][2] + "  ");             System.out.println("      |      |      ");                         if (turns != 9) {                if (scan.hasNextInt()) {                   int input = scan.nextInt();                                     if (!board[input/3][input%3].equals(" ")) {                      System.out.println("Place already taken.");                   } else {                      board[input/3][input%3] = turn;                      turn();                   }                }             } else if (turns == 9) {                System.out.println(checkWin());                System.exit(0);             }          } catch (Exception ex) {             ex.printStackTrace();          }       }    }        public String turn() {       if (turn == "O") {          setTurns(getTurns() + 1);          return turn = "X";       }             setTurns(getTurns() + 1);       return turn = "O";    }    public String checkWin() {    }        public int getTurns() {       return turns;    }    public void setTurns(int turns) {       this.turns = turns;    } }

I'm not sure how to check for the win, all I know is that I will need to loop through the board array to check if there is 3 in a row horizontally, vertically or diagonally.

~Shazer2

[Riven]

On a sidenote: this is extremely dangerous:

1	if (turn == "O") {

as for example

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"O" != "OO".substring(0,1) "O" != ""+"O" "O" != "o".toUpperCase()

because they are equal in contents but different objects.

to reliably check strings for equality, you must use

1	if(turn.equals("O"))

[Shazer2]

Ah, I'll fix that up. Thanks Riven. Any idea on checking the winner?

~Shazer2

Games published by our own members! Check 'em out!

[Riven]

there are 3+3+2 ways to win: 3x horizontal, 3x vertical, 2x diagonal

so first you need to check whether all strings in the 3 horizontal/vertical lines are equal:
using: board[ y ][ x ]

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lineH0 = board[0][0].equals(board[0][1]) && board[0][0].equals(board[0][2]); lineH1 = board[1][0].equals(board[1][1]) && board[1][0].equals(board[1][2]); lineH2 = board[2][0].equals(board[2][1]) && board[2][0].equals(board[2][2]); lineV0 = board[0][0].equals(board[0][0]) && board[0][0].equals(board[2][0]); lineV1 = board[0][1].equals(board[0][1]) && board[0][1].equals(board[2][1]); lineV2 = board[0][2].equals(board[0][2]) && board[0][2].equals(board[2][2]);

you can see the repetition here... so let's write a loop:

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for(int i=0; i<3; i++) {    boolean h = board[i][0].equals(board[i][1]) && board[i][0].equals(board[i][2]);    boolean v = board[0][i].equals(board[1][i]) && board[0][i].equals(board[2][i]);    // if H or V is true, we have a row/column with the same chars - that means either a winner, or they are all empty }

etc etc etc

Adding a check for diagonal is simple, just like adding the check that it's not an empty line.

[Riven]

As you can see, you mixed 'visuals' and 'logic' and it is going to be extremely annoying to code with all these strings.

Why don't you make board an int[][] and make the empty cells 0, and the players: 1 & 2

Suddenly your code would become a lot more readable:

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for(int i=0; i<3; i++) {    boolean h = board[i][0]==board[i][1] && board[i][0]==board[i][2];    boolean v = board[0][i]==board[1][i] && board[0][i]==board[2][i];    // if H or V is true, we have a row/column with the same ints - that means either a winner, or they are all zero }

when you visualize the board, you simply write:

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private String getLabelForPlayer(int p) {    if(p == 1) return "X";    if(p == 2) return "O";    throw new IllegalArgumentException("ehmpf"); } private String getLabelForCell(int x, int y) {    int cell = board[y][x];    if(cell == 0) return " ";    return getLabelForPlayer(cell); } private void drawBoardRow(int y) {             System.out.println("   " + getLabelForCell(0,y) + "  |   " + getLabelForCell(1,y) + "  |   " + getLabelForCell(2,y) + "  "); } private void drawBoard() {             System.out.println("      |      |      ");             drawBoardRow(0);             System.out.println("______|______|______");             drawBoardRow(1);             System.out.println("______|______|______");             drawBoardRow(2);             System.out.println("      |      |      "); }

[Shazer2]

I know how my code works though, but I do need to optimize in future if I want to get anywhere. I'll see if I can take that win check code in. Thanks.

Also, I understand that boolean check if iterating through each cell on the board, but how do we determine who has the pieces there? This is very confusing for me, sorry I might need some extra explanation if you're willing to help.

~Shazer2

[ra4king]

How about you make checkWin() return either 0 for tie, 1 for X and 2 for O?
Then inside the for loop:

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for(int i = 0; i < 3; i++) {     boolean h = ...     boolean v = ...         if(h)         return board[i][0];     if(v)         return board[0][i]; } //Check for diagonals here return 0;

[Shazer2]

So I either need to make board an int, or return a string for 0 instead.

~Shazer2

[ra4king]

If you still want to use Strings, make it return "0".

[Shazer2]

Ah yes. Now, how do I tell who has 3 lined up? As of now it only really returns whether one or the other is lined up vertically or horizontally.

~Shazer2

Games published by our own members! Check 'em out!

[ra4king]

if(checkWin().equals("X"))
    .... x wins
else if(checkWin().equals("O"))
    .... o wins
else
    .... tie

[Shazer2]

Alright, now what about diagonals?

~Shazer2

[ra4king]

You're checking each separately.
They should be easy enough if you think about it

[ReBirth]

Repeat Riven's way for diagonal. Write them in stupid way and see the pattern of... index number for example... then turn into loop. I never try to create board game though.

and I'm here not to stalk you ra4king. geez.

[Shazer2]

Yay, I got it to work.

Many thanks to ra4king and Riven.

~Shazer2

[sproingie]

X is 1, O is 0, the board is a nine bit int.  Write a bitmask for each win condition and test that the bitwise AND of the condition and the board XOR the mask is zero.

That's if I wanted to be clever.  Truth is, I'd just hardwire the logic with some 'if' statements.

[Riven]

X is 1, O is 0, the board is a nine bit int.

Each cell has three possible states

You can only use this approach by setting 1 = current player & 0 = anything else, and unleach your bitmasks on that.



If I were to do it, it would do it totally different, properly, with a Player class, etc.

[ra4king]

Using OO with tic-tac-toe = overkill

[Riven]

Using OO with tic-tac-toe = overkill

Unless you want to get the job done and get a good grade (if applicable).

[ra4king]

Blegh, school assignments....

[counterp]

you don't necessarily win in 9 turns

[Riven]

you don't necessarily win in 9 turns

nobody claimed that

[ra4king]

Good eye!

[Shazer2]

Well 9 squares = 9 moves doesn't it? :/ Unless I'm mistaken.

~Shazer2

[Riven]

That all the cells are filled, doesn't mean there is a winner, it can be a tie. In the same way, you can win when there are less than 9 cells filled, so you have to call you checkWin( ) method after every turn, not after 9 turns.

[loom_weaver]

Using OO with tic-tac-toe = overkill

OOX ... take that!

[Shazer2]

That's true, Riven. I'm having a problem, I dropped in my win check after each turn and now when I do check the win (place the final mark in a row of 3) it checks the win, before it's placed (redraw of board).

So I am checking the win correctly but it is checking the win before the final piece is drawn to the board. The board is drawn BEFORE the last move.

EDIT: I fixed the drawing issue. Now, when checking for a tie or win I have the following check.

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public void actualCheckWin() {    if (checkWin().equals("X")) {       System.out.println("X wins!");       System.exit(0);    } else if (checkWin().equals("O")) {       System.out.println("O wins!");       System.exit(0);    } else {       System.out.println("It's a tie!");       System.exit(0);    } }

The else gets called straight away on the first move. I'm not sure why.

~Shazer2

[ra4king]

Try

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else if(turns == 9) {     // tie! }

[Shazer2]

Ah good point, because I just assumed I didn't need that variable anymore and removed it conpletely. However I shall re-add it when I get home.

[ReBirth]

What variable? you meant "turns"? although if it's not matter on logic, I think it's good to show it for player